Worked Example Of Cable Calculation
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Worked example of cable calculation (see Fig. G65) The installation is supplied through a 630 kVA transformer. The process requires a high degree of supply continuity and part of the installation can be supplied by a 250 kVA standby generator. The global earthing system is TN-S, except for the most critical loads supplied by an isolation transformer with a downstream IT configuration. The single-line diagram is shown in Figure G65 below. The results of a computer study for the circuit from transformer T1 down to the cable C7 is reproduced on Figure G66. This study was carried out with Ecodial (a Schneider Electric software). This is followed by the same calculations carried out by the simplified method described in this guide.
Fig. G65: Example of single-line diagram
Calculation using software Ecodial General network characteristics
Number of poles and protected poles
4P4d
Earthing system
TN-S
Tripping unit
Micrologic 2.3
Neutral distributed
No
Overload trip Ir (A)
510
Voltage (V)
400
Short-delay trip Im / Isd (A)
5100
Frequency (Hz)
50
Cable C3
Upstream fault level (MVA)
500
Length
20
Resistance of MV network (mΩ)
0.0351
Maximum load current (A)
509
Reactance of MV network (mΩ)
0.351
Type of insulation
PVC
Ambient temperature (°C)
30
Transformer T1
Rating (kVA)
630
Conductor material
Copper
Short-circuit impedance voltage (%)
4
Single-core or multi-core cable
Single
Transformer resistance RT (mΩ)
3.472
Installation method
F
Transformer reactance XT (mΩ)
10.64
Phase conductor selected csa (mm2)
2 x 95
3-phase short-circuit current Ik3 (kA)
21.54
Neutral conductor selected
2 x 95
csa (mm2)
Cable C1
PE conductor selected csa (mm2)
1 x 95
Length (m)
5
Cable voltage drop ΔU (%)
0.53
Maximum load current (A)
860
Total voltage drop ΔU (%)
0.65
Type of insulation
PVC
3-phase short-circuit current Ik3 (kA)
19.1
Ambient temperature (°C)
30
1-phase-to-earth fault current Id (kA)
11.5
Conductor material
Copper
Switchboard B6
Single-core or multi-core cable
Single
Reference
Linergy 800
Installation method
F
Rated current (A)
750
Number of layers
1
Circuit-breaker Q7
Phase conductor selected csa (mm2)
2 x 240
Load current (A)
255
Neutral conductor selected csa (mm2)
2 x 240
Type
Compact
PE conductor selected csa (mm2)
1 x 120
Reference
NSX400F
Voltage drop ΔU (%)
0.122
Rated current (A)
400
3-phase short-circuit current
21.5
Number of poles and
3P3d
Ik3 (kA)
Courant de défaut phase-terre Id (kA)
protected poles
15.9
Circuit-breaker Q1
Tripping unit
Micrologic 2.3
Overload trip Ir (A)
258
2576
Load current (A)
860
Short-delay trip Im / Isd (A)
Type
Compact
Cable C7
Reference
NS1000N
Length
5
Rated current (A)
1000
Maximum load current (A)
255
Number of poles and protected poles
4P4d
Type of insulation
PVC
Tripping unit
Micrologic 5.0
Ambient temperature (°C)
30
Overload trip Ir (A)
900
Conductor material
Copper
Short-delay trip Im / Isd (A)
9000
Single-core or multi-core cable
Single
Tripping time tm (ms)
50
Installation method
F
Phase conductor selected csa (mm2)
1 x 95
Switchboard B2
Reference
Linergy 1250
Neutral conductor selected csa (mm2)
-
Rated current (A)
1050
PE conductor selected csa
1 x 50
(mm2)
Circuit breaker Q3
Cable voltage drop ΔU (%)
0.14
Load current (A)
509
Total voltage drop ΔU (%)
0.79
Type
Compact
3-phase short-circuit current Ik3 (kA)
18.0
Reference
NSX630F
1-phase-to-earth fault current Id (kA)
10.0
Rated current (A)
630
Fig. G66: Partial results of calculation carried out with Ecodial software (Schneider Electric). The calculation is performed according to Cenelec TR50480
The same calculation using the simplified method recommended in this guide Dimensioning circuit C1 The MV/LV 630 kVA transformer has a rated no-load voltage of 420 V. Circuit C1 must be suitable for a current of:
per phase Two single-core PVC-insulated copper cables in parallel will be used for each phase.These cables will be laid on cable trays according to method F. Each conductor will therefore carry 433A. Figure G21a indicates that for 3 loaded conductors with PVC isolation, the required c.s.a. is 240mm2. The resistance and the inductive reactance, for the two conductors in parallel, and for a length of 5 metres, are:
(cable resistance: 23.7 mΩ.mm2/m) (cable reactance: 0.08 mΩ/m)
Dimensioning circuit C3 Circuit C3 supplies two 150kW loads with cos φ = 0.85, so the total load current is:
Two single-core PVC-insulated copper cables in parallel will be used for each phase. These cables will be laid on cable trays according to method F. Each conductor will therefore carry 255A. Figure G21a indicates that for 3 loaded conductors with PVC isolation, the required c.s.a. is 95mm2. The resistance and the inductive reactance, for the two conductors in parallel, and for a length of 20 metres, are:
(cable resistance: 23.7 mΩ.mm2/m) (cable reactance: 0.08 mΩ/m) Dimensioning circuit C7 Circuit C7 supplies one 150kW load with cos φ = 0.85, so the total load current is:
One single-core PVC-insulated copper cable will be used for each phase. The cables will be laid on cable trays according to method F. Each conductor will therefore carry 255A. Figure G21a indicates that for 3 loaded conductors with PVC isolation, the required c.s.a. is 95mm2. The resistance and the inductive reactance for a length of 20 metres is:
(cable resistance: 23.7 mΩ.mm2/m) (cable reactance: 0.08 mΩ/m) Calculation of short-circuit currents for the selection of circuit-breakers Q1, Q3, Q7 (see Fig. G67) Circuit components
R (mΩ)
X (mΩ)
Z (mΩ)
Ikmax (kA)
Upstream MV network, 500MVA fault level (see Fig. G34)
0,035
0,351
Transformer 630kVA, 4% (see Fig. G35)
2.9
10.8
Cable C1
0.23
0.4
Sub-total
3.16
11.55
Cable C3
2.37
1.6
Sub-total
5.53
13.15
Cable C7
1.18
0.4
Sub-total
6.71
13.55
11.97
20.2
14.26
17
15.12
16
Fig. G67: Example of short-circuit current evaluation
The protective conductor When using the adiabatic method, the minimum c.s.a. for the protective earth conductor (PE) can be calculated by the formula given in Figure G58:
For circuit C1, I = 20.2kA and k = 143. t is the maximum operating time of the MV protection, e.g. 0.5s This gives:
A single 120 mm2 conductor is therefore largely sufficient, provided that it also satisfies the requirements for indirect contact protection (i.e. that its impedance is sufficiently low).
Generally, for circuits with phase conductor c.s.a. Sph ≥ 50 mm2, the PE conductor minimum c.s.a. will be Sph / 2. Then, for circuit C3, the PE conductor will be 95mm2, and for circuit C7, the PE conductor will be 50mm2. Protection against indirect-contact hazards For circuit C3 of Figure G65, Figures F41 and F40, or the formula given TN system - Protection against indirect contact may be used for a 3-phase 4-wire circuit. The maximum permitted length of the circuit is given by:
(The value in the denominator 630 x 11 is the maximum current level at which the instantaneous short-circuit magnetic trip of the 630 A circuit-breaker operates). The length of 20 metres is therefore fully protected by “instantaneous” over-current devices. Voltage drop The voltage drop is calculated using the data given in Figure G28, for balanced three-phase circuits, motor power normal service (cos φ = 0.8). The results are summarized on Figure G68 The total voltage drop at the end of cable C7 is then: 0.77%. C1
C3
C7
2 x 240mm2
2 x 95mm2
1 x 95mm2
0.22
0.43
0.43
Load current (A)
866
509
255
Length (m)
5
20
5
Voltage drop (V)
0.48
2.19
0.55
Voltage drop (%)
0.12
0.55
0.14
c.s.a.
∆U per conductor(V/A/km) see Fig. G28
Fig. G68: Voltage drop introduced by the different cables.
3-phase short-circuit current (Isc) at any point within a LV installation. In a 3-phase installation Isc at any point is given by:
where U20 = phase-to-phase voltage of the open circuited secondary windings of the power supply transformer(s). ZT = total impedance per phase of the installation upstream of the fault location (in Ω)
Method of calculating ZT Each component of an installation (MV network, transformer, cable, busbar, and so on...) is characterized by its impedance Z, comprising an element of resistance (R) and an inductive reactance (X). It may be noted that capacitive reactances are not important in short-circuit current calculations. The parameters R, X and Z are expressed in ohms, and are related by the sides of a right angled triangle, as shown in the impedance diagram of Figure G33.
Fig. G33: Impedance diagram The method consists in dividing the network into convenient sections, and to calculate the R and X values for each. Where sections are connected in series in the network, all the resistive elements in the section are added arithmetically; likewise for the reactances, to give RT and XT.
The impedance (ZT) for the combined sections concerned is then calculated from Any two sections of the network which are connected in parallel, can, if predominantly both resistive (or both inductive) be combined to give a single equivalent resistance (or reactance) as follows: Let R1 and R2 be the two resistances connected in parallel, then the equivalent resistance R3 will be given by:
or for reactances It should be noted that the calculation of X3 concerns only separated circuit without mutual inductance. If the circuits in parallel are close togother the value of X3 will be notably higher.
Determination of the impedance of each component Network upstream of the MV/LV transformer (see Fig. G34) The 3-phase short-circuit fault level PSC, in kA or in MVA[1] is given by the power supply authority concerned, from which an equivalent impedance can be deduced. Psc
Uo (V)
Ra (mΩ)
Xa (mΩ)
250 MVA
420
0.07
0.7
500 MVA
420
0.035
0.351
Fig. G34: The impedance of the MV network referred to the LV side of the MV/LV transformer
A formula which makes this deduction and at the same time converts the impedance to an equivalent value at LV is given, as follows:
where Zs = impedance of the MV voltage network, expressed in milli-ohms Uo = phase-to-phase no-load LV voltage, expressed in volts Psc = MV 3-phase short-circuit fault level, expressed in kVA
The upstream (MV) resistance Ra is generally found to be negligible compared with the corresponding Xa, the latter then being taken as the ohmic value for Za. If more accurate calculations are necessary, Xa may be taken to be equal to 0.995 Za and Ra equal to 0.1 Xa. Figure G36 gives values for Ra and Xa corresponding to the most common MV[2] short-circuit levels in utility power-supply networks, namely, 250 MVA and 500 MVA. Transformers (see Fig. G35) The impedance Ztr of a transformer, viewed from the LV terminals, is given by the formula:
where: U20 = open-circuit secondary phase-to-phase voltage expressed in volts Sn = rating of the transformer (in VA) Usc = the short-circuit impedance voltage of the transformer expressed in % The transformer windings resistance Rtr can be derived from the total load-losses as follows:
so that
in milli-ohms
where Pcu = total load-losses in watts In = nominal full-load current in amps Rtr = resistance of one phase of the transformer in milli-ohms (the LV and corresponding MV winding for one LV phase are included in this resistance value).
Note: for an approximate calculation, in the absence of more precise information on transformer characteristics, Cenelec 50480 suggests to use the following guidelines:
if U20 is not known, it may be assumed to be 1.05 Un
in the absence of more precise information, the following values may be used: Rtr = 0.31 Ztr and Xtr = 0.95 Ztr Example: for a transformer of 630kVA with Usc=4% / Un = 400V, approximate calculation gives:
U20 = 400 x 1.05 = 420V
Ztr = 4202 / 630000 x 4% = 11 mΩ
Rtr = 0.31 x Ztr = 3.5 mΩ and Xtr = 0.95 x Ztr = 10.6 mΩ Rated Power kVA)
Oil-immersed
Cast-resin
Usc (%)
Rtr (mΩ)
Xtr (mΩ)
Ztr (mΩ)
Usc (%)
Rtr (mΩ)
Xtr (mΩ)
Ztr (mΩ)
100
4
37.9
59.5
70.6
6
37.0
99.1
105.8
160
4
16.2
41.0
44.1
6
18.6
63.5
66.2
200
4
11.9
33.2
35.3
6
14.1
51.0
52.9
250
4
9.2
26.7
28.2
6
10.7
41.0
42.3
315
4
6.2
21.5
22.4
6
8.0
32.6
33.6
400
4
5.1
16.9
17.6
6
6.1
25.8
26.5
500
4
3.8
13.6
14.1
6
4.6
20.7
21.2
630
4
2.9
10.8
11.2
6
3.5
16.4
16.8
800
6
2.9
12.9
13.2
6
2.6
13.0
13.2
1,000
6
2.3
10.3
10.6
6
1.9
10.4
10.6
1,250
6
1.8
8.3
8.5
6
1.5
8.3
8.5
1,600
6
1.4
6.5
6.6
6
1.1
6.5
6.6
2,000
6
1.1
5.2
5.3
6
0.9
5.2
5.3
Fig. G35: Resistance, reactance and impedance values for typical distribution 400 V transformers (no-load voltage = 420 V) with MV windings ≤ 20 kV
Busbars The resistance of busbars is generally negligible, so that the impedance is practically all reactive, and amounts to approximately 0.15 mΩ/metre[3] length for LV busbars (doubling the spacing between the bars increases the reactance by about 10% only). In practice, it's almost never possible to estimate the busbar length concerned by a short-circuit downstream a switchboard. Circuit conductors The resistance of a conductor is given by the formula:
where ρ = the resistivity of the conductor material at the normal operating temperature ρ has to be considered:
at cold state (20°C) to determine maximum short-circuit current,
at steady state (normal operating temperature) to determine minimum short-circuit current. L = length of the conductor in m S = c.s.a. of conductor in mm2 20 °C
PR/XLPE 90 °C
PVC 70 °C
Copper
18.51
23.69
22.21
Alu
29.41
37.65
35.29
Fig. G35b: Values of ρ as a function of the temperature, cable insulation and cable core material, according to IEC60909-0 and Cenelec TR 50480 (in mΩ mm2/m).
Cable reactance values can be obtained from the manufacturers. For c.s.a. of less than 50 mm2 reactance may be ignored. In the absence of other information, a value of 0.08 mΩ/metre may be used (for 50 Hz systems) or 0.096 mΩ/metre (for 60 Hz systems). For busways (busbar trunking systems) and similar pre-wired ducting systems, the manufacturer should be consulted.
Motors At the instant of short-circuit, a running motor will act (for a brief period) as a generator, and feed current into the fault. In general, this fault-current contribution may be ignored. However, if the total power of motors running simultaneously is higher than 25% of the total power of transformers, the influence of motors must be taken into account. Their total contribution can be estimated from the formula: Iscm = 3.5 In from each motor i.e. 3.5m In for m similar motors operating concurrently. The motors concerned will be the 3-phase motors only; single-phase-motor contribution being insignificant. Fault-arc resistance Short-circuit faults generally form an arc which has the properties of a resistance. The resistance is not stable and its average value is low, but at low voltage this resistance is sufficient to reduce the fault-current to some extent. Experience has shown that a reduction of the order of 20% may be expected. This phenomenon will effectively ease the current-breaking duty of a CB, but affords no relief for its fault-current making duty. Recapitulation table (see Fig. G36) Parts of power-supply system
R (mΩ)
X (mΩ) Xa = 0.995 Za
Supply network Figure G34 Transformer where
Figure G35
Rtr is often negligible compared to Xtr for transformers > 100 kVA Circuit-breaker
Not considered in practice
Busbars
Negligible for S > 200 mm2 in the
XB = 0.15 mΩ/m
[a]
formula: Circuit conductors[b]
Cables: Xc = 0.08 mΩ/m [a]
Motors
See 3-phase short-circuit current (Isc) at any point within a LV installation Motors (often negligible at LV)
Three-phase maximum circuit current in kA [a] ρ = resistivity at 20°C [b] If there are several conductors in parallel per phase, then divide the resistance of one conductor by the number of conductors. The reactance remains practically unchanged. U20: Phase-to-phase no-load secondary voltage of MV/LV transformer (in volts). Psc: 3-phase short-circuit power at MV terminals of the MV/LV transformers (in kVA). Pcu: 3-phase total losses of the MV/LV transformer (in watts). Sn: Rating of the MV/LV transformer (in kVA). Usc: Short-circuit impedance voltage of the MV/LV transfomer (in %). RT : Total resistance. XT: Total reactanc Fig. G36: Recapitulation table of impedances for different parts of a power-supply system
Example of short-circuit calculations (see Fig. G37) LV installation
R (mΩ)
X (mΩ)
0.035
0.351
2.35
8.5
MV network Psc = 500 MVA Transformer 20 kV / 420 V Pn = 1000 kVA Usc = 5% Pcu = 13.3 x
RT (mΩ)
XT (mΩ)
103 watts Single-core cables 5 m copper 4 x 240 mm2/phase
Xc = 0.08 x 5 = 0.40
2.48
9.25
Isc1 = 25 kA
Three-core cable 100 m 95 mm2copper
Xc = 100 x 0.08 = 8
22
17.3
Isc3 = 8.7 kA
Three-core cable 20 m 10 mm2copper final circuits
Xc = 20 x 0.08 = 1.6
59
18.9
Isc4 = 3.9 kA
Main circuitbreaker
Not considered in practice
Busbars 10 m
Not considered in practice
RT : Total resistance. XT: Total reactance. Isc : 3-phase maximum short-circuit current Calculations made a G36 Fig. G37: Example of maximum short-circuit current calculations for a LV installation supplied at 400 V (nominal) from a 1000 kVA MV/LV transformer
Notes 1. ^ Short-circuit MVA:
EL Isc where:
EL = phase-to-phase nominal system voltage expressed in kV (r.m.s.)
Isc = 3-phase short-circuit current expressed in kA (r.m.s.)
2. ^ up to 36 kV 3. ^ For 50 Hz systems, but 0.18 mΩ/m length at 60 Hz.
Location of protective devices. General rule (see Fig. G7a) A protective device is necessary at the origin of each circuit where a reduction of permissible maximum current level occurs.
Possible alternative locations in certain circumstances (see Fig. G7b) The protective device may be placed part way along the circuit:
If AB is not in proximity to combustible material, and
If no socket-outlets or branch connections are taken from AB Three cases may be useful in practice: Consider case (1) in the diagram
AB ≤ 3 metres, and
AB has been installed to reduce to a practical minimum the risk of a short-circuit (wires in heavy steel conduit for example) Consider case (2)
The upstream device P1 protects the length AB against short-circuits in accordance
Consider case (3)
The overload device (S) is located adjacent to the load. This arrangement is
convenient for motor circuits. The device (S) constitutes the control (start/stop) and overload protection of the motor while (SC) is: either a circuit-breaker (designed for motor protection) or fuses type aM The short-circuit protection (SC) located at the origin of the circuit conforms with the
principles of Calculation of minimum levels of short-circuit current .
Circuits with no protection (see Fig. G7c) Either The protective device P1 is calibrated to protect the cable S2 against overloads and short-
circuits Or Where the breaking of a circuit constitutes a risk, e.g.
Excitation circuits of rotating machines
circuits of large lifting electromagnets
the secondary circuits of current transformers
No circuit interruption can be tolerated, and the protection of the cabling is of secondary importance.
a.
b.
c.
Short-circuit current at the secondary terminals of a MV/LV distribution transformer. The case of one transformer In a simplified approach, the impedance of the MV system is assumed to be negligibly small,
so that:
where
and:
S = kVA rating of the transformer U20 = phase-to-phase secondary volts on open circuit In = nominal current in amps Isc = short-circuit fault current in amps Usc = short-circuit impedance voltage of the transformer in %. Typical values of Usc for distribution transformers are given in Figure G31 Transformer rating (kVA)
Usc in %
Oil-immersed
Cast-resin dry type
50 to 750
4
6
800 to 3,200
6
6
Fig. G31: Typical values of Usc for different kVA ratings of transformers with MV windings ≤ 20 kV
Example 400 kVA transformer, 420 V at no load Usc = 4%
The case of several transformers in parallel feeding a busbar The value of fault current on an outgoing circuit immediately downstream of the busbars (see Fig. G32) can be estimated as the sum of the Isc from each transformer calculated separately. It is assumed that all transformers are supplied from the same MV network, in which case the values obtained from Figure G31when added together will give a slightly higher fault-level value than would actually occur. Other factors which have not been taken into account are the impedance of the busbars and of the cable between transformers and circuit breakers.
The conservative fault-current value obtained however, is sufficiently accurate for basic installation design purposes. The choice of circuit breakers and incorporated protective devices against shortcircuit and fault currents is described in Selection of a circuit-breaker .
Fig. G32: Case of several transformers in parallel.
Conductor sizing: methodology and definition. Methodology (see Figure G1) Following a preliminary analysis of the power requirements of the installation, as described in The consumer substation with LV metering, a study of cabling[1] and its electrical protection is undertaken, starting at the origin of the installation, through the intermediate stages to the final circuits. The cabling and its protection at each level must satisfy several conditions at the same time, in order to ensure a safe and reliable installation, e.g. it must:
Carry the permanent full load current, and normal short-time overcurrents
Not cause voltage drops likely to result in an inferior performance of certain loads, for example: an excessively long acceleration period when starting a motor, etc. Moreover, the protective devices (circuit-breakers or fuses) must:
Protect the cabling and busbars for all levels of overcurrent, up to and including short-circuit currents
Ensure protection of persons against indirect contact hazards, particularly in TN- and ITearthed systems, where the length of circuits may limit the magnitude of short-circuit currents,
thereby delaying automatic disconnection (it may be remembered that TT- earthed installations are necessarily protected at the origin by a RCD, generally rated at 300 mA). The cross-sectional areas of conductors are determined by the general method described in Practical method for determining the smallest allowable cross-sectional area of circuit conductors of this Chapter. Apart from this method some national standards may prescribe a minimum cross-sectional area to be observed for reasons of mechanical endurance. Particular loads (as noted in Chapter Characteristics of particular sources and loads) require that the cable supplying them be oversized, and that the protection of the circuit be likewise modified.
Fig. G1: Flow-chart for the selection of cable size and protective device rating for a given circuit
Definitions Maximum load current: IB
At the final circuits level, this design current (according to IEV "International Electrotechnical Vocabulary" ref 826-11-10) corresponds to the rated kVA of the load. In the case of motor-starting, or other loads which take a high in-rush current, particularly where frequent starting is concerned (e.g. lift motors, resistance-type spot welding, and so on) the cumulative thermal effects of the overcurrents must be taken into account. Both cables and thermal type relays are affected.
At all upstream circuit levels this current corresponds to the kVA to be supplied, which takes account of the diversity and utilization factors, ks and ku respectively, as shown in Figure G2.
Fig. G2: Calculation of maximum load current IB Maximum permissible current: Iz Current carrying capacity Iz is the maximum permissible that the cabling for the circuit can carry indefinitely, without reducing its normal life expectancy. The current depends, for a given cross sectional area of conductors, on several parameters:
Constitution of the cable and cable-way (Cu or Alu conductors; PVC or EPR etc. insulation; number of active conductors)
Ambient temperature
Method of installation
Influence of neighbouring circuits Overcurrents An overcurrent occurs each time the value of current exceeds the maximum load current IB for the load concerned. This current must be cut off with a rapidity that depends upon its magnitude, if permanent damage to the cabling (and appliance if the overcurrent is due to a defective load component) is to be avoided. Overcurrents of relatively short duration can however, occur in normal operation; two types of overcurrent are distinguished:
Overloads These overcurrents can occur in healthy electric circuits, for example, due to a number of small short-duration loads which occasionally occur co-incidentally: motor starting loads, and so on. If either of these conditions persists however beyond a given period (depending on protective-relay settings or fuse ratings) the circuit will be automatically cut off.
Short-circuit currents These currents result from the failure of insulation between live conductors or/and between live conductors and earth (on systems having low-impedance-earthed neutrals) in any combination, viz: 3 phases short-circuited (and to neutral and/or earth, or not) 2 phases short-circuited (and to neutral and/or earth, or not) 1 phase short-circuited to neutral (and/or to earth).
Overcurrent protection principles. A protective device is provided at the origin of the circuit concerned (see Fig. G3 and Fig. G4 ).
Acting to cut-off the current in a time shorter than that given by the I2t characteristic of the circuit cabling
But allowing the maximum load current IB to flow indefinitely The characteristics of insulated conductors when carrying short-circuit currents can, for periods up to 5 seconds following short-circuit initiation, be determined approximately by the formula: I2t = k2 S2
which shows that the allowable heat generated is proportional to the squared cross-sectional-area of the condutor. where t = Duration of short-circuit current (seconds) S = Cross sectional area of insulated conductor (mm2) I = Short-circuit current (A r.m.s.) k = Insulated conductor constant (values of k2 are given in Figure G52) For a given insulated conductor, the maximum permissible current varies according to the environment. For instance, for a high ambient temperature (θa1 > θa2), Iz1 is less than Iz2 (see Fig. G5). θ means “temperature”. Note: ISC: 3-phase short-circuit current ISCB: rated 3-ph. short-circuit breaking current of the circuit-breaker Ir (or Irth)[1]: regulated “nominal” current level; e.g. a 50 A nominal circuit-breaker can be regulated to have a protective range, i.e. a conventional overcurrent tripping level (see Fig. G6) similar to that of a 30 A circuit-breaker.
Fig. G3: Circuit protection by circuit breaker
Fig. G4: Circuit protection by fuses
Fig. G5: I2t characteristic of an insulated conductor at two different ambient temperatures.
Practical values for a protective scheme. General rules A protective device (circuit-breaker or fuse) functions correctly if:
Its nominal current or its setting current In is greater than the maximum load current IB but less than the maximum permissible current Iz for the circuit, i.e. IB ≤ In ≤ Iz corresponding to zone “a” in Figure G6
Its tripping current I2 “conventional” setting is less than 1.45 Iz which corresponds to zone “b” in Figure G6 The “conventional” setting tripping time may be 1 hour or 2 hours according to local standards and the actual value selected for I2. For fuses, I2 is the current (denoted If) which will operate the fuse in the conventional time.
Its 3-phase short-circuit fault-current breaking rating is greater than the 3-phase short-circuit current existing at its point of installation. This corresponds to zone “c” in Figure G6.
Fig. G6: Current levels for determining circuir breaker or fuse characteristics IB ≤ In ≤ Iz zone a I2 ≤ 1.45 Iz zone b ISCB ≥ ISC zone c
Applications
Protection by circuit-breaker Criteria for circuit-breakers: IB ≤ In ≤ Iz and ISCB ≥ ISC. By virtue of its high level of precision the current I2 is always less than 1.45 In (or 1.45 Ir) so that the condition I2 ≤ 1.45 Iz (as noted in the “general rules” above) will always be respected.
Particular case If the circuit-breaker itself does not protect against overloads, it is necessary to ensure that, at a time of lowest value of short-circuit current, the overcurrent device protecting the circuit will operate correctly. This particular case is examined in Calculation of minimum levels of short-circuit current. Protection by fuses Criteria for fuses: IB ≤ In ≤ Iz/k3 and ISCF ≥ ISC. The condition I2 ≤ 1.45 Iz must be taken into account, where I2 is the fusing (melting level) current, equal to k2 x In (k2 ranges from 1.6 to 1.9) depending on the particular fuse concerned.
A further factor k3 has been introduced
such that I2 ≤ 1.45 Iz
will be valid if In ≤ Iz/k3. For fuses type gG: In < 16 A → k3 = 1.31 In ≥ 16 A → k3 = 1.10 Moreover, the short-circuit current breaking capacity of the fuse ISCF must exceed the level of 3phase short-circuit current at the point of installation of the fuse(s). Association of different protective devices The use of protective devices which have fault-current ratings lower than the fault level existing at their point of installation are permitted by IEC and many national standards in the following conditions:
There exists upstream, another protective device which has the necessary short-circuit rating, and
The amount of energy allowed to pass through the upstream device is less than that which
can be withstood without damage by the downstream device and all associated cabling and appliances. In pratice this arrangement is generally exploited in:
The association of circuit-breakers/fuses
The technique known as “cascading” or “series rating” in which the strong current-limiting performance of certain circuit-breakers effectively reduces the severity of downstream shortcircuits Possible combinations which have been tested in laboratories are indicated in certain manufacturers catalogues.
General method for cable sizing. Possible methods of installation for different types of conductors or cables The different admissible methods of installation are listed in Figure G8, in conjonction with the different types of conductors and cables.
Conductors cables
Method of installation
Without fixings
Bare conductors
[b]
Clipped direct
-
Condu it syste ms
-
Cable trunking systems (including skirting trunking,
Cable ducting systems
Cable ladder,
tray, Cable
trunking)
rackets
[a]
Sup port wire
Cable
flush floor
-
On insulat ors
-
-
+
-
+
-
+
-
Insulated conductors
-
-
+
+
Sheathed cables
Multicore
+
+
+
+
+
+
0
+
(including
Single -core
0
+
+
+
+
+
0
+
armoured and mineral insulated)
+ : Permitted. – : Not Permitted. 0 : Not applicable, or not normally used in practice. [a] Insulated conductors are admitted if the cable trunking systems provide at least he degree of protection IP4X or IPXXD and if the cover can only be removed by means of a tool or a deliberate action. [b] Insulated conductors which are used as protective conductors or protective bonding conductors may use any appropriate method of installation and need not be laid in conduits, trunking or ducting systems. Fig. G8: Selection of wiring systems (table A.52.1 of IEC 60364-5-52)
Possible methods of installation for different situations: Different methods of installation can be implemented in different situations. The possible combinations are presented in Figure G9. The number given in this table refer to the different wiring systems considered. Situations
Method of installation
Without fixings
Clipped direct
Conduit System s
Cable trunking (including skirting trunking, flush floor trunking)
Cable ducting systems
Cable ladder, cable tray, cable brackets
On insulators
Support wire
Accessible
40
33
41, 42
6, 7, 8, 9,12
43, 44
30, 31, 32, 33, 34
-
0
Not accessible
40
0
41, 42
0
43
0
0
0
Cable channel
56
56
54, 55
0
30, 31, 32, 34
-
-
Buried in ground
72, 73
0
70, 71
-
70, 71
0
-
-
Embedded in structure
57, 58
3
1, 2, 59, 60
50, 51, 52, 53
46, 45
0
-
-
Building voids
Surface mounted
-
20, 21, 22, 23, 33
4, 5
6, 7, 8, 9, 12
6, 7, 8, 9
30, 31, 32, 34
36
-
Overhead/free in air
-
33
0
10, 11
10, 11
30, 31, 32, 34
36
35
Window frames
16
0
16
0
0
0
-
-
Architrave
15
0
15
0
0
0
-
-
Immersed 1
+
+
+
-
+
0
-
-
– : Not permitted. 0 : Not applicable or not normally used in practice. + : Follow manufacturer’s instructions. Note: The number in each box, e.g. 40, 46, refers to the number of the method of installation in Table A.52.3. Fig. G9: Erection of wiring systems (table A.52.2 of IEC 60364-5-52)
Examples of wiring systems and reference methods of installations An illustration of some of the many different wiring systems and methods of installation is provided in Figure G10. Several reference methods are defined (with code letters A to G), grouping installation methods having the same characteristics relative to the current-carrying capacities of the wiring systems. Item No.
Methods of installation
1
Room
Description
Reference method of installation to be used to obtain current-carrying capacity
Insulated conductors or single-core cables in conduit in a thermally insulated wall
A1
2
Multi-core cables in conduit in a thermally insulated wall
A2
4
Insulated conductors or single-core cables in conduit on a wooden, or masonry wall or spaced less than 0,3 x conduit diameter from it
B1
5
Multi-core cable in conduit on a wooden, or mansonry wall or spaced less than 0,3 x conduit diameter from it
B2
20
Single-core or multi-core cables: fixed on, or sapced less than 0.3 x cable diameter from a wooden wall
C
30
Single-core or multi-core cables:
Room
C
On unperforated tray run horizontally or vertically
31
Single-core or multi-core cables:
E or F
On perforated tray run horizontally or vertically
36
Bare or insulated conductors on insulators
G
70
Multi-core cables in conduit or in cable ducting in the ground
D1
71
Single-core cable in conduit or in cable ducting in the ground
D1
Fig. G10: Examples of methods of installation (part of table A.52.3 of IEC 60364-5-52)
Maximum operating temperature: The current-carrying capacities given in the subsequent tables have been determined so that the maximum insulation temperature is not exceeded for sustained periods of time. For different type of insulation material, the maximum admissible temperature is given in Figure G11. Type of insulation
Temperature limit °C
Polyvinyl-chloride (PVC)
70 at the conductor
Cross-linked polyethylene (XLPE) and ethylene propylene rubber
90 at the conductor
(EPR)
Mineral (PVC covered or bare exposed to touch)
70 at the sheath
Mineral (bare not exposed to touch and not in contact with combustible material)
105 at the seath
Fig. G11: Maximum operating temperatures for types of insulation (table 52.1 of IEC 60364-5-52)
Correction factors In order to take environment or special conditions of installation into account, correction factors have been introduced. The cross sectional area of cables is determined using the rated load current IB divided by different correction factors, k1, k2, ...:
I’B is the corrected load current, to be compared to the current-carrying capacity of the considered cable. Ambient temperature The current-carrying capacities of cables in the air are based on an average air temperature equal to 30 °C. For other temperatures, the correction factor is given in FigureG12 for PVC, EPR and XLPE insulation material. The related correction factor is here noted k1. Ambient temperature °C
Insulation
PVC
XLPE and EPR
10
1.22
1.15
15
1.17
1.12
20
1.12
1.08
25
1.06
1.04
30
1
1
35
0.94
0.96
40
0.87
0.91
45
0.79
0.87
50
0.71
0.82
55
0.61
0.76
60
0.50
0.71
65
-
0.65
70
-
0.58
75
-
0.50
80
-
0.41
Fig. G12: Correction factors for ambient air temperatures other than 30 °C to be applied to the current-carrying capacities for cables in the air (from table B.52.14 of IEC 60364-5-52)
The current-carrying capacities of cables in the ground are based on an average ground temperature equal to 20 °C. For other temperatures, the correction factor is given in Figure G13 for PVC, EPR and XLPE insulation material. The related correction factor is here noted k2.
Ground temperature °C
Insulation
PVC
XLPE and EPR
10
1.10
1.07
15
1.05
1.04
20
1
1
25
0.95
0.96
30
0.89
0.93
35
0.84
0.89
40
0.77
0.85
45
0.71
0.80
50
0.63
0.76
55
0.55
0.71
60
0.45
0.65
65
-
0.60
70
-
0.53
75
-
0.46
80
-
0.38
Fig. G13: Correction factors for ambient ground temperatures other than 20 °C to be applied to the current-carrying capacities for cables in ducts in the ground (from table B.52.15 of IEC 60364-5-52)
Soil thermal resistivity The current-carrying capacities of cables in the ground are based on a ground resistivity equal to 2.5 K•m/W. For other values, the correction factor is given in Figure G14. The related correction factor is here noted k3.
Thermal resistivity, K•m/W
0.5
0.7
1
1.5
2
2.5
3
Correction factor for cables in buried ducts
1.28
1.20
1.18
1.1
1.05
1
0.96
Correction factor for direct buried cables
1.88
1.62
1.5
1.28
1.12
1
0.90
Note 1: The correction factors given have been averaged over the range of conductor sizes and types of installation included in Tables B.52.2 to B.52.5. The overall accuracy of correction factors is within ±5 %. Note 2: The correction factors are applicable to cables drawn into buried ducts; for cables laid direct in the ground the correction factors for thermal resistivities less than 2.5 K•m/W will be higher. Where more precise values are required they may be calculated by methods given in the IEC 60287 series. Note 3: The correction factors are applicable to ducts buried at depths of up to 0.8 m. Note 4: It is assumed that the soil properties are uniform. No allowance had been made for the possibility of moisture migration which can lead to a region of high thermal resistivity around the cable. If partial drying out of the soil is foreseen, the permissible current rating should be derived by the methods specified in the IEC 60287 series. Fig. G14: Correction factors for cables in buried ducts for soil thermal resistivities other than 2.5 K.m/W to be applied to the current-carrying capacities for reference method D (table B.52.16 of IEC 60364-5-52)
Based on experience, a relationship exist between the soil nature and resistivity. Then, empiric values of correction factors k3 are proposed in Figure G15, depending on the nature of soil. Nature of soil
k3
Very wet soil (saturated)
1.21
Wet soil
1.13
Damp soil
1.05
Dry soil
1.00
Very dry soil (sunbaked)
0.86
Fig. G15: Correction factor k3 depending on the nature of soil
Grouping of conductors or cables The current-carrying capacities given in the subsequent tables relate to single circuits consisting of the following numbers of loaded conductors: Two insulated conductors or two single-core cables, or one twin-core cable (applicable to single-phase circuits); Three insulated conductors or three single-core cables, or one three-core cable (applicable to three-phase circuits). Where more insulated conductors or cables are installed in the same group, a group reduction factor (here noted k4) shall be applied. Examples are given in Figures G16 to G18 for different configurations (installation methods, in free air or in the ground). Figure G16 gives the values of correction factor k4 for different configurations of unburied cables or conductors, grouping of more than one circuit or multi-core cables. Arrangement (cables touching)
Number of circuits or multi-core cables
1
2
3
4
5
6
Reference methods
7
8
9
12
16
20
Bunched in air, on a surface, embedded orenclosed
1.00
0.8 0
0.7 0
0.6 5
0.6 0
0.5 7
0.54
0.5 2
0.5 0
0.4 5
0.41
0.38
Single layer on wall, floor or unperforated tray
1.00
0.8 5
0.7 9
0.7 5
0.7 3
0.7 2
0.72
0.7 1
0.7 0
No further reduction factor for more than nine circuits or multi-core cables
Single layer fixed directly under a wooden ceiling
0.95
0.8 1
0.7 2
0.6 8
0.6 6
0.6 4
0.63
0.6 2
0.6 1
Single layer on a perforated horizontal or vertical tray
1.00
0.8 8
0.8 2
0.7 7
0.7 5
0.7 3
0.73
0.7 2
0.7 2
Single layer on ladder support or cleats etc.
1.00
0.8 7
0.8 2
0.8 0
0.8 0
0.7 9
0.79
0.7 8
0.7 8
Methods A to F
Method C
Methods E and F
Fig. G16: Reduction factors for groups of more than one circuit or of more than one multi-core cable (table B.52.17 of IEC 60364-5-52)
Figure G17 gives the values of correction factor k4 for different configurations of unburied cables or conductors, for groups of more than one circuit of single-core cables in free air. Method of installation
Perforated trays
Number of tray
31
Number of three-phase circuits
1
2
3
1
0.98
0.91
0.87
2
0.96
0.87
0.81
3
0.95
0.85
0.78
Use as a multiplier to rating for
Three cables in horizontal formation
Vertical perforated trays
Ladder supports, cleats, etc... formation
Perforated trays
Vertical perforated trays
Ladder supports, cleats, etc...
31
1
0.96
0.86
2
0.95
0.84
32
1
1.00
0.97
0.96
33
2
0.98
0.93
0.89
34
3
0.97
0.90
0.86
31
1
1.00
0.98
0.96
2
0.97
0.93
0.89
3
0.96
0.92
0.86
1
1.00
0.91
0.89
2
1.00
0.90
0.86
32
1
1.00
1.00
1.00
33
2
0.97
0.95
0.93
34
3
0.96
0.94
0.90
31
Three cables in vertical formation
Three cables in horizontal formation
Three cables in trefoil formation
Fig. G17: Reduction factors for groups of more than one circuit of single-core cables to be applied to reference rating for one circuit of single-core cables in free air - Method of installation F. (table B.52.21 of IEC 60364-5-52)
Figure G18 gives the values of correction factor k4 for different configurations of cables or conductors laid directly in the ground. Number of circuits
Cable to cable clearance(a)
Nil (cables touching)
One cable diameter
0.125 m
0.25 m
0.5 m
2
0.75
0.80
0.85
0.90
0.90
3
0.65
0.70
0.75
0.80
0.85
4
0.60
0.60
0.70
0.75
0.80
5
0.55
0.55
0.65
0.70
0.80
6
0.50
0.55
0.60
0.70
0.80
7
0.45
0.51
0.59
0.67
0.76
8
0.43
0.48
0.57
0.65
0.75
9
0.41
0.46
0.55
0.63
0.74
12
0.36
0.42
0.51
0.59
0.71
16
0.32
0.38
0.47
0.56
0.38
20
0.29
0.35
0.44
0.53
0.66
(a) for Multicore cables
(a) for Singlecore cables
Fig. G18: Reduction factors for more than one circuit, single-core or multi-core cables laid directly in the ground. Installation method D. (table B.52.18 of IEC 60364-5-52)
Harmonic current The current-carrying capacity of three-phase, 4-core or 5-core cables is based on the assumption that only 3 conductors are fully loaded. However, when harmonic currents are circulating, the neutral current can be significant, and even higher than the phase currents. This is due to the fact that the 3rd harmonic currents of the three phases do not cancel each other, and sum up in the neutral conductor. This of course affects the current-carrying capacity of the cable, and a correction factor noted here k5 shall be applied. In addition, if the 3rd harmonic percentage h3 is greater than 33%, the neutral current is greater than the phase current and the cable size selection is based on the neutral current. The heating effect of harmonic currents in the phase conductors has also to be taken into account. The values of k5 depending on the 3rd harmonic content are given in Figure G19. Third harmonic content of phase current %
Correction factor
Size selection is based on phase current
0 - 15
1.0
15 - 33
0.86
Size selection is based on neutral current
33 - 45
0.86
> 45
1.0[a]
[a] If the neutral current is more than 135 % of the phase current and the cable size is selected on the basis of the neutral current then the three phase conductors will not be fully loaded. The reduction in heat generated by the phase conductors offsets the heat generated by the neutral conductor to the extent that it is not necessary to apply any reduction factor to the current carrying capacity for three loaded conductors. Fig. G19: Correction factors for harmonic currents in four-core and five-core cables (table E.52.1 of IEC 60364-5-52)
Admissible current as a function of nominal cross-sectional area of conductors IEC standard 60364-5-52 proposes extensive information in the form of tables giving the admissible currents as a function of cross-sectional area of cables. Many parameters are taken into account, such as the method of installation, type of insulation material, type of conductor material, number of loaded conductors. As an example, Figure G20 gives the current-carrying capacities for different methods of installation of PVC insulation, three loaded copper or almunium conductors, free air or in ground. Nominal cross-sectional area of conductors(mm2)
1
Installation methods of Table B.52.1
A1
A2
B1
B2
C
D1
D2
2
3
4
5
6
7
8
13.5
13
15.5
15
17.5
18
19
Copper
1.5
2.5
18
17.5
21
20
24
24
24
4
24
23
28
27
32
30
33
6
31
29
36
34
41
38
41
10
42
39
50
46
57
50
54
16
56
52
68
62
76
64
70
25
73
68
89
80
96
82
92
35
89
83
110
99
119
98
110
50
108
99
134
118
144
116
130
70
136
125
171
149
184
143
162
95
164
150
207
179
223
169
193
120
188
172
239
206
259
192
220
150
216
196
262
225
299
217
246
185
245
223
296
255
341
243
278
240
286
261
346
297
403
280
320
300
328
298
394
339
464
316
359
Aluminium
2.5
14
13.5
16.5
15.5
18.5
18.5
4
18.5
17.5
22
21
25
24
6
24
23
28
27
32
30
10
32
31
39
36
44
39
16
43
41
53
48
59
50
53
25
57
53
70
62
73
64
69
35
70
65
86
77
90
77
83
50
84
78
104
92
110
91
99
70
107
98
133
116
140
112
122
95
129
118
161
139
170
132
148
120
149
135
186
160
197
150
169
150
170
155
204
176
227
169
189
185
194
176
230
199
259
190
214
240
227
207
269
232
305
218
250
300
261
237
306
265
351
247
Note: In columns 3, 5, 6, 7 and 8, circular conductors are assumed for sizes up to and including 16 mm2. Values for larger sizes relate to shaped conductors and may safely be applied to circular conductors. Fig. G20: Current-carrying capacities in amperes for different methods of installation, PVC insulation, three loaded conductors, copper or aluminium, conductor temperature: 70 °C, ambient temperature: 30 °C in air, 20 °C in ground (table B.52.4 of IEC 60364-5-52).
3-phase short-circuit current (Isc) at any point within a LV installation. In a 3-phase installation Isc at any point is given by:
where U20 = phase-to-phase voltage of the open circuited secondary windings of the power supply transformer(s). ZT = total impedance per phase of the installation upstream of the fault location (in Ω)
Method of calculating ZT Each component of an installation (MV network, transformer, cable, busbar, and so on...) is characterized by its impedance Z, comprising an element of resistance (R) and an inductive reactance (X). It may be noted that capacitive reactances are not important in short-circuit current calculations. The parameters R, X and Z are expressed in ohms, and are related by the sides of a right angled triangle, as shown in the impedance diagram of Figure G33.
282
Fig. G33: Impedance diagram The method consists in dividing the network into convenient sections, and to calculate the R and X values for each. Where sections are connected in series in the network, all the resistive elements in the section are added arithmetically; likewise for the reactances, to give RT and XT. The impedance (ZT) for the combined sections concerned is then calculated from Any two sections of the network which are connected in parallel, can, if predominantly both resistive (or both inductive) be combined to give a single equivalent resistance (or reactance) as follows: Let R1 and R2 be the two resistances connected in parallel, then the equivalent resistance R3 will be given by:
or for reactances It should be noted that the calculation of X3 concerns only separated circuit without mutual inductance. If the circuits in parallel are close togother the value of X3 will be notably higher.
Determination of the impedance of each component Network upstream of the MV/LV transformer (see Fig. G34) The 3-phase short-circuit fault level PSC, in kA or in MVA[1] is given by the power supply authority concerned, from which an equivalent impedance can be deduced. Psc
Uo (V)
Ra (mΩ)
Xa (mΩ)
250 MVA
420
0.07
0.7
500 MVA
420
0.035
0.351
Fig. G34: The impedance of the MV network referred to the LV side of the MV/LV transformer
A formula which makes this deduction and at the same time converts the impedance to an equivalent value at LV is given, as follows:
where Zs = impedance of the MV voltage network, expressed in milli-ohms Uo = phase-to-phase no-load LV voltage, expressed in volts Psc = MV 3-phase short-circuit fault level, expressed in kVA The upstream (MV) resistance Ra is generally found to be negligible compared with the corresponding Xa, the latter then being taken as the ohmic value for Za. If more accurate calculations are necessary, Xa may be taken to be equal to 0.995 Za and Ra equal to 0.1 Xa. Figure G36 gives values for Ra and Xa corresponding to the most common MV[2] short-circuit levels in utility power-supply networks, namely, 250 MVA and 500 MVA. Transformers (see Fig. G35) The impedance Ztr of a transformer, viewed from the LV terminals, is given by the formula:
where: U20 = open-circuit secondary phase-to-phase voltage expressed in volts Sn = rating of the transformer (in VA) Usc = the short-circuit impedance voltage of the transformer expressed in % The transformer windings resistance Rtr can be derived from the total load-losses as follows:
so that
in milli-ohms
where Pcu = total load-losses in watts In = nominal full-load current in amps Rtr = resistance of one phase of the transformer in milli-ohms (the LV and corresponding MV winding for one LV phase are included in this resistance value).
Note: for an approximate calculation, in the absence of more precise information on transformer characteristics, Cenelec 50480 suggests to use the following guidelines:
if U20 is not known, it may be assumed to be 1.05 Un
in the absence of more precise information, the following values may be used: Rtr = 0.31 Ztr and Xtr = 0.95 Ztr Example: for a transformer of 630kVA with Usc=4% / Un = 400V, approximate calculation gives:
U20 = 400 x 1.05 = 420V
Ztr = 4202 / 630000 x 4% = 11 mΩ
Rtr = 0.31 x Ztr = 3.5 mΩ and Xtr = 0.95 x Ztr = 10.6 mΩ Rated Power kVA)
Oil-immersed
Cast-resin
Usc (%)
Rtr (mΩ)
Xtr (mΩ)
Ztr (mΩ)
Usc (%)
Rtr (mΩ)
Xtr (mΩ)
Ztr (mΩ)
100
4
37.9
59.5
70.6
6
37.0
99.1
105.8
160
4
16.2
41.0
44.1
6
18.6
63.5
66.2
200
4
11.9
33.2
35.3
6
14.1
51.0
52.9
250
4
9.2
26.7
28.2
6
10.7
41.0
42.3
315
4
6.2
21.5
22.4
6
8.0
32.6
33.6
400
4
5.1
16.9
17.6
6
6.1
25.8
26.5
500
4
3.8
13.6
14.1
6
4.6
20.7
21.2
630
4
2.9
10.8
11.2
6
3.5
16.4
16.8
800
6
2.9
12.9
13.2
6
2.6
13.0
13.2
1,000
6
2.3
10.3
10.6
6
1.9
10.4
10.6
1,250
6
1.8
8.3
8.5
6
1.5
8.3
8.5
1,600
6
1.4
6.5
6.6
6
1.1
6.5
6.6
2,000
6
1.1
5.2
5.3
6
0.9
5.2
5.3
Fig. G35: Resistance, reactance and impedance values for typical distribution 400 V transformers (no-load voltage = 420 V) with MV windings ≤ 20 kV
Busbars The resistance of busbars is generally negligible, so that the impedance is practically all reactive, and amounts to approximately 0.15 mΩ/metre[3] length for LV busbars (doubling the spacing between the bars increases the reactance by about 10% only). In practice, it's almost never possible to estimate the busbar length concerned by a short-circuit downstream a switchboard. Circuit conductors The resistance of a conductor is given by the formula:
where ρ = the resistivity of the conductor material at the normal operating temperature ρ has to be considered:
at cold state (20°C) to determine maximum short-circuit current,
at steady state (normal operating temperature) to determine minimum short-circuit current. L = length of the conductor in m S = c.s.a. of conductor in mm2 20 °C
PR/XLPE 90 °C
PVC 70 °C
Copper
18.51
23.69
22.21
Alu
29.41
37.65
35.29
Fig. G35b: Values of ρ as a function of the temperature, cable insulation and cable core material, according to IEC60909-0 and Cenelec TR 50480 (in mΩ mm2/m).
Cable reactance values can be obtained from the manufacturers. For c.s.a. of less than 50 mm2 reactance may be ignored. In the absence of other information, a value of 0.08 mΩ/metre may be used (for 50 Hz systems) or 0.096 mΩ/metre (for 60 Hz systems). For busways (busbar trunking systems) and similar pre-wired ducting systems, the manufacturer should be consulted. Motors At the instant of short-circuit, a running motor will act (for a brief period) as a generator, and feed current into the fault. In general, this fault-current contribution may be ignored. However, if the total power of motors running simultaneously is higher than 25% of the total power of transformers, the influence of motors must be taken into account. Their total contribution can be estimated from the formula: Iscm = 3.5 In from each motor i.e. 3.5m In for m similar motors operating concurrently. The motors concerned will be the 3-phase motors only; single-phase-motor contribution being insignificant. Fault-arc resistance Short-circuit faults generally form an arc which has the properties of a resistance. The resistance is not stable and its average value is low, but at low voltage this resistance is sufficient to reduce the fault-current to some extent. Experience has shown that a reduction of the order of 20% may be expected. This phenomenon will effectively ease the current-breaking duty of a CB, but affords no relief for its fault-current making duty. Recapitulation table (see Fig. G36) Parts of power-supply system
R (mΩ)
X (mΩ) Xa = 0.995 Za
Supply network Figure G34 Transformer with Figure G35
where
Rtr is often negligible compared to Xtr for transformers > 100 kVA Circuit-breaker
Not considered in practice
Busbars
Negligible for S > 200 mm2 in the
XB = 0.15 mΩ/m
[a]
formula: Circuit conductors[b]
Cables: Xc = 0.08 mΩ/m [a]
Motors
See 3-phase short-circuit current (Isc) at any point within a LV installation Motors (often negligible at LV)
Three-phase maximum circuit current in kA [a] ρ = resistivity at 20°C [b] If there are several conductors in parallel per phase, then divide the resistance of one conductor by the number of conductors. The reactance remains practically unchanged. U20: Phase-to-phase no-load secondary voltage of MV/LV transformer (in volts). Psc: 3-phase short-circuit power at MV terminals of the MV/LV transformers (in kVA). Pcu: 3-phase total losses of the MV/LV transformer (in watts). Sn: Rating of the MV/LV transformer (in kVA). Usc: Short-circuit impedance voltage of the MV/LV transfomer (in %). RT : Total resistance. XT: Total reactanc Fig. G36: Recapitulation table of impedances for different parts of a power-supply system
Example of short-circuit calculations (see Fig. G37) LV installation
R (mΩ)
X (mΩ)
RT (mΩ)
XT (mΩ)
0.035
0.351
2.35
8.5
MV network Psc = 500 MVA Transformer 20 kV / 420 V Pn = 1000 kVA Usc = 5% Pcu = 13.3 x 103 watts Single-core cables 5 m copper 4 x 240 mm2/phase
Xc = 0.08 x 5 = 0.40
Main circuitbreaker
Not considered in practice
Busbars 10 m
Not considered in practice
2.48
9.25
Isc1 = 25 kA
Three-core cable 100 m 95 mm2copper
Xc = 100 x 0.08 = 8
22
17.3
Isc3 = 8.7 kA
Three-core cable 20 m 10 mm2copper final circuits
Xc = 20 x 0.08 = 1.6
59
18.9
Isc4 = 3.9 kA
RT : Total resistance. XT: Total reactance. Isc : 3-phase maximum short-circuit current Calculations made as described in figure G36 Fig. G37: Example of maximum short-circuit current calculations for a LV installation supplied at 400 V (nominal) from a 1000 kVA MV/LV transformer
Notes 1. ^ Short-circuit MVA:
EL Isc where:
EL = phase-to-phase nominal system voltage expressed in kV (r.m.s.)
Isc = 3-phase short-circuit current expressed in kA (r.m.s.)
2. ^ up to 36 kV 3. ^ For 50 Hz systems, but 0.18 mΩ/m length at 60 Hz.
Isc at the receiving end of a feeder as a function of the Isc at its sending end. The network shown in Figure G38 typifies a case for the application of Figure G39 , derived by the «method of composition» (mentioned in chapter Protection against electric shocks and electric fires ). These tables give a rapid and sufficiently accurate value of short-circuit current at a point in a network, knowing:
The value of short-circuit current upstream of the point considered
The length and composition of the circuit between the point at which the short-circuit current level is known, and the point at which the level is to be determined It is then sufficient to select a circuit-breaker with an appropriate short-circuit fault rating immediately above that indicated in the tables. If more precise values are required, it is possible to make a detailed calculation or to use a software package, such as Ecodial. In such a case, moreover, the possibility of using the cascading technique should be considered, in which the use of a current limiting circuit-breaker at the upstream position would allow all circuit-breakers downstream of the limiter to have a short-circuit current rating much lower than would otherwise be necessary (See chapter LV switchgear: functions and selection ).
Method Select the c.s.a. of the conductor in the column for copper conductors (in this example the c.s.a. is 47.5 mm2). Search along the row corresponding to 47.5 mm2 for the length of conductor equal to that of the circuit concerned (or the nearest possible on the low side). Descend vertically the column in which the length is located, and stop at a row in the middle section (of the 3 sections of the Figure) corresponding to the known fault-current level (or the nearest to it on the high side). In this case 30 kA is the nearest to 28 kA on the high side. The value of short-circuit current at the downstream end of the 20 metre circuit is given at the intersection of the vertical column in which the length is located, and the horizontal row corresponding to the upstream Isc (or nearest to it on the high side). This value in the example is seen to be 14.7 kA. The procedure for aluminium conductors is similar, but the vertical column must be ascended into the middle section of the table.
In consequence, a DIN-rail-mounted circuit-breaker rated at 63 A and Isc of 25 kA (such as a NG 125N unit) can be used for the 55 A circuit in Figure G38. A Compact rated at 160 A with an Isc capacity of 25 kA (such as a NS160 unit) can be used to protect the 160 A circuit.
Fig. G38: Determination of downstream short-circuit current level Isc using Figure G39 Copper 230 V / 400 V
c.s.a.of phase conductor s (mm2)
Length of circuit (in metres)
1.5
2.5
4
1.2
1.3
1.8
2.6
3. 6
5.2
7.3
1 0 3
1.1
1.5
2.1
3.0
4.3
6. 1
8.6
12. 1
1 7 2
1.7
2.4
3.4
4.9
6.9
9. 7
13. 7
19. 4
2 7
6
10
16
1.8
2.6
3.6
5.2
7.3
10. 3
1 4. 6
21
29
4 1
2. 2
3.0
4.3
6.1
8.6
12.2
17. 2
2 4
34
49
6 9
1. 7
2 . 4
3. 4
4.9
6.9
9.7
13. 8
19.4
27
3 9
55
78
1 1 0
25
1 . 3
1. 9
2. 7
3 . 8
5. 4
7.6
10. 8
15. 2
21
30
43
6 1
86
121
1 7 2
35
1 . 9
2. 7
3. 8
5 . 3
7. 5
10. 6
15. 1
21
30
43
60
8 5
120
170
2 4 0
3 2 6
47.5
1 . 8
2 . 6
3. 6
5. 1
7 . 2
10 .2
14. 4
20
29
41
58
82
1 1 5
163
231
70
2 . 7
3 . 8
5. 3
7. 5
1 0 . 7
15 .1
21
30
43
60
85
120
1 7 0
240
340
2. 6
3 . 6
5 . 1
7. 2
10 .2
1 4 . 5
20
29
41
58
82
115
163
2 3 1
326
461
95
120
150
1.2
1. 6
2. 3
3. 2
4 . 6
6 . 5
9. 1
12 .9
1 8 . 3
26
37
52
73
103
146
206
2 9 1
412
1. 8
2. 5
3. 5
5 . 0
7 . 0
9. 9
14 .0
1 9 . 8
28
40
56
79
112
159
224
3 1 7
448
185
1.5
2. 1
2. 9
4. 2
5 . 9
8 . 3
11 .7
16 .6
2 3
33
47
66
94
133
187
265
3 7 4
529
240
1.8
2. 6
3. 7
5. 2
7 . 3
1 0 . 3
4. 6
21
2 9
41
58
83
117
165
233
330
4 6 6
659
300
2.2
3. 1
4. 4
6. 2
8 . 8
1 2 . 4
17 .6
25
3 5
50
70
99
140
198
280
396
5 6 1
2x120
2.3
3. 2
4. 6
6. 5
9 . 1
1 2 . 9
18 .3
26
3 7
52
73
103
146
206
292
412
5 8 3
2x150
2.5
3. 5
5. 0
7. 0
9 . 9
1 4 . 0
20
28
4 0
56
79
112
159
224
317
448
6 3 4
2x185
2.9
4. 2
5. 9
8. 3
1 1 . 7
1 6 . 6
23
33
4 7
66
94
133
187
265
375
530
7 4 9
3x120
3.4
4. 9
6. 9
9. 7
1 3 . 7
1 9 . 4
27
39
5 5
77
110
155
219
309
438
619
3x150
3.7
5. 3
7. 5
10 .5
1 4 . 9
2 1
30
42
6 0
84
119
168
238
336
476
672
3x185
4.4
6. 2
8. 8
12 .5
1 7 . 6
2 5
35
50
7 0
10 0
141
199
281
398
562
Isc upstream (in kA)
Isc downstream (in kA)
100
93
9 0
87
82
7 7
7 0
62
54
4 5
37
29
22
17. 0
12. 6
9.3
6.7
4. 9
3.5
2.5
1 8
90
84
8 2
79
75
7 1
6 5
58
51
4 3
35
28
22
16. 7
12. 5
9.2
6.7
4. 8
3.5.
2.5
1 8
80
75
7 4
71
68
6 4
5 9
54
47
4 0
34
27
21
16. 3
12. 2
9.1
6.6
4. 8
3.5
2.5
1 8
70
66
6 5
63
61
5 8
5 4
49
44
3 8
32
26
20
15. 8
12. 0
8.9
6.6
4. 8
3.4
2.5
1 8
60
57
5 6
55
53
5 1
4 8
44
39
3 5
29
24
20
15. 2
11. 6
8.7
6.5
4. 7
3.4
2.5
1 8
50
48
4 7
46
45
4 3
4 1
38
35
3 1
27
22
18. 3
14. 5
11. 2
8.5
6.3
4. 6
3.4
2.4
1 7
40
39
3 8
38
37
3 6
3 4
32
30
2 7
24
20
16. 8
13. 5
10. 6
8.1
6.1
4. 5
3.3
2.4
1 7
35
34
3 4
33
33
3 2
3 0
29
27
2 4
22
18. 8
15. 8
12. 9
10. 2
7.9
6.0
4. 5
3.3
2.4
1 7
30
29
2 9
29
28
2 7
2 7
25
24
2 2
20
17. 3
14. 7
12. 2
9.8
7.6
5.8
4. 4
3.2
2.4
1 7
25
25
2 4
24
24
2 3
2 3
22
21
1 9 . 1
17 .4
15. 5
13. 4
11. 2
9.2
7.3
5.6
4. 2
3.2
2.3
1 7
20
20
2 0
19 .4
19 .2
1 8 . 8
1 8 . 4
17 .8
17 .0
1 6 . 1
14 .9
13. 4
11. 8
10. 1
8.4
6.8
5.3
4. 1
3.1
2.3
1 7
15
14. 8
4 1. 8
14 .7
14 .5
1 4 . 3
1 4 . 1
13 .7
3. 3
1 2 . 7
11 .9
11. 0
9.9
8.7
7.4
6.1
4.9
3. 8
2.9
2.2
1 6
10
9.9
9. 9
9. 8
9. 8
9 . 7
9 . 6
9. 4
9. 2
8 . 9
8. 5
8.0
7.4
6.7
5.9
5.1
4.2
3. 4
2.7
2.0
1 5
7
7.0
6. 9
6. 9
6. 9
6 . 9
6 . 8
6. 7
6. 6
6 . 4
6. 2
6.0
5.6
5.2
4.7
4.2
3.6
3. 0
2.4
1.9
1 4
5
5.0
5. 0
5. 0
4. 9
4 . 9
4 . 9
4. 9
4. 8
4 . 7
4. 6
4.5
4.3
4.0
3.7
3.4
3.0
2. 5
2.1
1.7
1 3
4
4.0
4. 0
4. 0
4. 0
4 . 0
3 . 9
3. 9
3. 9
3 . 8
3. 7
3.6
3.5
3.3
3.1
2.9
2.6
2. 2
1.9
1.6
1 2
3
3.0
3. 0
3. 0
3. 0
3 . 0
3 . 0
2. 9
2. 9
2 . 9
2. 9
2.8
2.7
2.6
2.5
2.3
2.1
1. 9
1.6
1.4
1 1
2
2.0
2. 0
2. 0
2. 0
2 . 0
2 . 0
2. 0
2. 0
1 2 . 0
1. 9
1.9
1.9
1.8
1.8
1.7
1.6
1. 4
1.3
1.1
1 0
1
1.0
1. 0
1. 0
1. 0
1 . 0
1 . 0
1. 0
1. 0
1 . 0
1. 0
1.0
1.0
1.0
0.9
0.9
0.9
0. 8
0.8
0.7
0 6
Note: for a 3-phase system having 230 V between phases, divide the above lengths by Fig. G39: Isc at a point downstream, as a function of a known upstream fault-current value and the length and c.s.a. of the intervening conductors, in a 230/400 V 3-phase system.
Calculation of minimum levels of short-circuit current. If a protective device in a circuit is intended only to protect against short-circuit faults, it is essential that it will operate with certainty at the lowest possible level of short-circuit current that can occur on the circuit In general, on LV circuits, a single protective device protects against all levels of current, from the overload threshold through the maximum rated short-circuit current breaking capability of the device. The protection device should be able to operate in a maximum time to ensure people and circuit safety, for all short-circuit current or fault current that may occur. To check that behavior, calculation of minimal short-circuit current or fault current is mandatory. In addition, in certain cases overload protective devices and separate short-circuit protective devices are used.
Examples of such arrangements Figures G40 to G42 show some common arrangements where overload and short-circuit protections are achieved by separate devices.
Fig. G40: Circuit protected by aM fuses
Fig. G41: Circuit protected by circuit breaker without thermal overload relay
Fig. G42a: Circuit breaker D provides protection against short-circuit faults as far as and including the load As shown in Figures G40 and G41, the most common circuits using separate devices control and protect motors. Figure G42a constitutes a derogation in the basic protection rules, and is generally used on circuits of prefabricated bustrunking, lighting rails, etc. Variable speed drive Figure G42b shows the functions provided by the variable speed drive, and if necessary some additional functions provided by devices such as circuit-breaker, thermal relay, RCD.
Protection to be provided
Protection generally provided by the variable speed drive
Additional protection if not provided by the variable speed drive
Cable overload
Yes
CB / Thermal relay
Motor overload
Yes
CB / Thermal relay
Downstream shortcircuit
Yes
Variable speed drive overload
Yes
Overvoltage
Yes
Undervoltage
Yes
Loss of phase
Yes
Upstream short-circuit
Circuit-breaker (short-circuit tripping)
Internal fault
Circuit-breaker (short-circuit and overload tripping)
Downstream earth fault (indirect contact)
(self protection)
Direct contact fault
RCD ≥ 300 mA or CB in TN earthing system RCD ≤ 30 mA
Fig. G42b: Protection to be provided for variable speeed drive applications
Conditions to be fulfilled The protective device must fulfill:
instantaneous trip setting Im < Iscmin for a circuit-breaker
fusion current Ia < Iscmin for a fuse
The protective device must therefore satisfy the two following conditions: Its breaking capacity must be greater than Isc, the 3-phase short-circuit current at its point of
installation Elimination of the minimum short-circuit current possible in the circuit, in a time tc compatible
with the thermal constraints of the circuit conductors, where:
(valid for tc < 5 seconds) where S is the cross section area of the cable, k is a factor depending of the cable conductor material, the insulation material and initial temperature. Example: for copper XLPE, initial temperature 90 °C, k = 143 (see IEC60364-4-43 §434.3.2 table 43A). Comparison of the tripping or fusing performance curve of protective devices, with the limit curves of thermal constraint for a conductor shows that this condition is satisfied if:
Isc (min) > Im (instantaneous or short timedelay circuit-breaker trip setting current level), (see Fig. G43 )
Isc (min) > Ia for protection by fuses. The value of the current Ia corresponds to the crossing point of the fuse curve and the cable thermal withstand curve (see Fig. G44 and Fig. G45)
Fig. G43: Protection by circuit breaker
Fig. G44: Protection by aM-type fuses
Fig. G45: Protection by gG-type fuses
Practical method of calculating Lmax In practice this means that the length of circuit downstream of the protective device must not exceed a calculated maximum length: The limiting effect of the impedance of long circuit conductors on the value of short-circuit currents must be checked and the length of a circuit must be restricted accordingly. The method of calculating the maximum permitted length has already been demonstrated in TN- and IT- earthed schemes for single and double earth faults, respectively. Two cases are considered below: 1 - Calculation of Lmax for a 3-phase 3-wire circuit
The minimum short-circuit current will occur when two phase wires are short-circuited at the remote end of the circuit (see Fig. G46).
Fig. G46: Definition of L for a 3-phase 3-wire circuit Using the “conventional method”, the voltage at the point of protection P is assumed to be 80% of the nominal voltage during a short-circuit fault, so that 0.8 U = Isc Zd, where: Zd = impedance of the fault loop Isc = short-circuit current (ph/ph) U = phase-to-phase nominal voltage
For cables ≤ 120 mm2, reactance may be neglected, so that
[1]
where: ρ = resistivity of conductor material at the average temperature during a short-circuit, Sph = c.s.a. of a phase conductor in mm2 L = length in metres The condition for the cable protection is Im ≤ Isc with Im = magnetic trip current setting of the CB.
This leads to
which gives
with U = 400 V ρ = 0.023 Ω.mm2/m[2] (Cu) therefore
with Lmax = maximum circuit length in metres In general, the value of Im is given with +/- 20% tolerance, so Lmax should be calculated for Im+20% (worst case).
k factor values are provided in the following table, taking into account these 20%, and as a function of cross-section for Sph > 120 mm2[1] Cross-section (mm2)
≤ 120
150
185
240
300
k (for 400 V)
5800
5040
4830
4640
4460
2 - Calculation of Lmax for a 3-phase 4-wire 230/400 V circuit The minimum Isc will occur when the short-circuit is between a phase conductor and the neutral at the end of the circuit. A calculation similar to that of example 1 above is required, but for a single-phase fault (230V).
If Sn (neutral cross-section) = Sph Lmax = k Sph / Im with k calculated for 230V, as shown in the table below
Cross-section (mm2)
≤ 120
150
185
240
300
k (for 400 V)
3333
2898
2777
2668
2565
If Sn (neutral cross-section) < Sph, then (for cable cross-section ≤ 120mm2)
Tabulated values for Lmax Figure G47 below gives maximum circuit lengths (Lmax) in metres, for:
3-phase 4-wire 400 V circuits (i.e. with neutral) and
1-phase 2-wire 230 V circuits protected by general-purpose circuit-breakers. In other cases, apply correction factors (given in Figure G51) to the lengths obtained. In general, the value of Im is given with +/- 20% tolerance.
Lmax values below are therefore calculated for Im+20% (worst case). For the 50 mm2 c.s.a., calculation are based on a 47.5 mm2 real c.s.a. Operating current level Im of the instantaneous magnetic tripping element (in A)
c.s.a. (nominal cross-sectional-area) of conductors (in mm2)
1.5
2.5
4
6
10
16
25
35
50
100
167
26 7
400
63
79
133
21 2
317
80
63
104
16 7
250
41 7
100
50
83
13 3
200
33 3
125
40
67
10 7
160
26 7
427
160
31
52
83
125
20 8
333
200
25
42
67
100
16 7
267
417
250
20
33
53
80
13 3
213
333
46 7
320
16
26
42
63
10 4
167
260
36 5
50
495
70
95
1
400
13
21
33
50
83
133
208
29 2
396
500
10
17
27
40
67
107
167
23 3
317
560
9
15
24
36
60
95
149
20 8
283
41 7
630
8
13
21
32
53
85
132
18 5
251
37 0
700
7
12
19
29
48
76
119
16 7
226
33 3
452
800
6
10
17
25
42
67
104
14 6
198
29 2
396
875
6
10
15
23
38
61
95
13 3
181
26 7
362
4
1000
5
8
13
20
33
53
83
11 7
158
23 3
317
4
1120
4
7
12
18
30
48
74
10 4
141
20 8
283
3
1250
4
7
11
16
27
43
67
93
127
18 7
253
3
1600
5
8
13
21
33
52
73
99
14 6
198
2
2000
4
7
10
17
27
42
58
79
11 7
158
2
5
8
13
21
33
47
63
93
127
1
2500
3200
4
6
10
17
26
36
49
73
99
1
4000
5
8
13
21
29
40
58
79
1
5000
4
7
11
17
23
32
47
63
8
6300
5
8
13
19
25
37
50
6
8000
4
7
10
15
20
29
40
5
10000
5
8
12
16
23
32
4
12500
4
7
9
13
19
25
3
Fig. G47: Maximum circuit lengths in metres for copper conductors (for aluminium, the lengths must be multiplied by 0.62)
Figures G48 to G50 give maximum circuit length (Lmax) in metres for:
3-phase 4-wire 400 V circuits (i.e. with neutral) and
1-phase 2-wire 230 V circuits protected in both cases by domestic-type circuit-breakers or with circuit-breakers having similar tripping/current characteristics. In other cases, apply correction factors to the lengths indicated. These factors are given in Figure G51. Circuit-breaker rating (A)
c.s.a. (nominal cross-sectional-area) of conductors (in mm2)
1.5
2.5
4
6
10
16
25
6
200
333
533
800
10
120
200
320
480
800
16
75
125
200
300
500
800
20
60
100
160
240
400
640
25
48
80
128
192
320
512
800
32
37
62
100
150
250
400
625
40
30
50
80
120
200
320
500
50
24
40
64
96
160
256
400
63
19
32
51
76
127
203
317
80
15
25
40
60
100
160
250
100
12
20
32
48
80
128
200
125
10
16
26
38
64
102
160
Fig. G48: Maximum length of copper-conductor circuits in metres protected by B-type circuit-breakers
Circuit-breaker rating (A)
6
c.s.a. (nominal cross-sectional-area) of conductors (in mm2)
1.5
2.5
4
6
10
100
167
267
400
667
16
25
10
60
100
160
240
400
640
16
37
62
100
150
250
400
625
20
30
50
80
120
200
320
500
25
24
40
64
96
160
256
400
32
18.0
31
50
75
125
200
313
40
15.0
25
40
60
100
160
250
50
12.0
20
32
48
80
128
200
63
9.5
16.0
26
38
64
102
159
80
7.5
12.5
20
30
50
80
125
100
6.0
10.0
16.0
24
40
64
100
125
5.0
8.0
13.0
19.0
32
51
80
Fig. G49: Maximum length of copper-conductor circuits in metres protected by C-type circuit-breakers
Circuit-breaker rating (A)
c.s.a. (nominal cross-sectional-area) of conductors (in mm2)
1.5
2.5
1
429
714
2
214
357
4
6
571
857
10
16
25
3
143
238
381
571
952
4
107
179
286
429
714
6
71
119
190
286
476
762
10
43
71
114
171
286
457
714
16
27
45
71
107
179
286
446
20
21
36
57
86
143
229
357
25
17.0
29
46
69
114
183
286
32
13.0
22
36
54
89
143
223
40
11.0
18.0
29
43
71
114
179
50
9.0
14.0
23
34
57
91
143
63
7.0
11.0
18.0
27
45
73
113
80
5.0
9.0
14.0
21
36
57
89
100
4.0
7.0
11.0
17.0
29
46
71
125
3.0
6.0
9.0
14.0
23
37
57
Fig. G50: Maximum length of copper-conductor circuits in metres protected by D-type circuit-breakers
Circuit detail
3-phase 3-wire 400 V circuit or 1-phase 2-wire 400 V circuit (no neutral)
1-phase 2-wire (phase and neutral) 230 V circuit
3-phase 4-wire 230/400 V circuit or 2-phase 3-wire 230/400 V circuit (i.e with neutral)
Sph / S neutra
Sph / S neutra
Fig. G51: Correction factor to apply to lengths obtained from Figures G47 to G50
Note: IEC 60898 accepts an upper short-circuit-current tripping range of 10-50 In for type D circuitbreakers. European standards, and Figure G50 however, are based on a range of 10-20 In, a range which covers the vast majority of domestic and similar installations.
Examples Example 1 In a 3-phase 3-wire 400 V installation the protection is provided by a 50 A circuit-breaker type NS80HMA, the instantaneous short-circuit current trip, is set at 500 A (accuracy of ± 20%), i.e. in the worst case would require 500 x 1,2 = 600 A to trip. The cable c.s.a. = 10 mm2 and the conductor material is copper. In Figure G47, the row Im = 500 A crosses the column c.s.a. = 10 mm2 at the value for Lmax of 67 m. The circuit-breaker protects the cable against short-circuit faults, therefore, provided that its length does not exceed 67 metres. Example 2 In a 3-phase 3-wire 400 V circuit (without neutral), the protection is provided by a 220 A circuitbreaker type NSX250N with an instantaneous short-circuit current trip unit type MA set at 2,000 A (± 20%), i.e. a worst case of 2,400 A to be certain of tripping. The cable c.s.a. = 120 mm2 and the conductor material is copper. In Figure G47 the row Im = 2,000 A crosses the column c.s.a. = 120 mm2 at the value for Lmax of 200 m. Being a 3-phase 3-wire 400 V circuit (without neutral), a correction factor from Figure G51 must be applied. This factor is seen to be 1.73. The circuit-breaker will therefore protect the cable against short-circuit current, provided that its length does not exceed 200 x 1.73 = 346 metres.
Notes 1. ^ a b For larger c.s.a.’s, the resistance calculated for the conductors must be increased to account for the non-uniform current density in the conductor (due to “skin” and “proximity” effects Suitable values are as follows:
150 mm2: R + 15 %
185 mm2: R + 20 %
240 mm2: R + 25 %
300 mm2: R + 30 %
2. ^ Resistivity for copper EPR/XLPE cables when passing short-circuit current, eg for the max temperature they can withstand = 90°C (cf Figure G35b).
Verification of the withstand capabilities of cables under short-circuit conditions. Thermal constraints When the duration of short-circuit current is brief (several tenths of a second up to five seconds maximum) all of the heat produced is assumed to remain in the conductor, causing its temperature to rise. The heating process is said to be adiabatic, an assumption that simplifies the calculation and gives a pessimistic result, i.e. a higher conductor temperature than that which would actually occur, since in practice, some heat would leave the conductor and pass into the insulation. For a period of 5 seconds or less, the relationship I2t = k2S2 characterizes the time in seconds during which a conductor of c.s.a. S (in mm 2) can be allowed to carry a current I, before its temperature reaches a level which would damage the surrounding insulation. The factor k is given in Figure G52 below. Conductor insulation
PVC ≤ 300 mm2
PVC > 300 mm2
EPR XLPE
Rubber 60 °C
Initial temperature °C
70
70
90
60
Final temperature °C
160
140
250
200
Material of conductor
Copper
115
103
143
141
Aluminium
76
68
94
93
Fig. G52: Value of the constant k according to table 43A of IEC 60364-4-43
The method of verification consists in checking that the thermal energy I2t per ohm of conductor material, allowed to pass by the protecting circuit-breaker (from manufacturers catalogues) is less than that permitted for the particular conductor (as given in Figure G53 below). S (mm2)
PVC
XLPE
Copper
Aluminium
Copper
Aluminium
1.5
0.0297
0.0130
0.0460
0.0199
2.5
0.0826
0.0361
0.1278
0.0552
4
0.2116
0.0924
0.3272
0.1414
6
0.4761
0.2079
0.7362
0.3181
10
1.3225
0.5776
2.0450
0.8836
16
3.3856
1.4786
5.2350
2.2620
25
8.2656
3.6100
12.7806
5.5225
35
16.2006
7.0756
25.0500
10.8241
50[a]
29.839
13.032
46.133
19.936
[a] For 50mm2 cable, the values are calculated for the actual cross-section of 47.5mm2 Fig. G53: Maximum allowable thermal stress for cables I2t (expressed in ampere2 x second x 106)
Example Is a copper-cored XLPE cable of 4 mm2 c.s.a. adequately protected by a iC60N circuit-breaker? (see Fig. G53b) Figure G53 shows that the I2t value for the cable is 0.3272 x 106, while the maximum “let-through” value by the circuit-breaker, as given in the manufacturer’s catalogue, is considerably less ( < 0.1.106 A2s). The cable is therefore adequately protected by the circuit-breaker up to its full rated breaking capability.
Electrodynamic constraints For all type of circuit (conductors or bus-trunking), it is necessary to take electrodynamic effects into account. To withstand the electrodynamic constraints, the conductors must be solidly fixed and the connection must be strongly tightened. For bus-trunking, rails, etc. it is also necessary to verify that the electrodynamic withstand performance is satisfactory when carrying short-circuit currents. The peak value of current, limited by the circuit-breaker or fuse, must be less than the busbar system rating. Tables of coordination ensuring adequate protection of their products are generally published by the manufacturers and provide a major advantage of such systems.
Fig. G53b: Example of energy limitation of a MCB for different ratings.
Connection and choice for protective earthing conductor. Protective (PE) conductors provide the bonding connection between all exposed and extraneous conductive parts of an installation, to create the main equipotential bonding system. These conductors conduct fault current due to insulation failure (between a phase conductor and an exposed conductive part) to the earthed neutral of the source. PE conductors are connected to the main earthing terminal of the installation. The main earthing terminal is connected to the earthing electrode (see Chapter E) by the earthing conductor (grounding electrode conductor in the USA). PE conductors must be:
Insulated and coloured yellow and green (stripes)
Protected against mechanical and chemical damage In IT and TN-earthed schemes it is strongly recommended that PE conductors should be installed in close proximity (i.e. in the same conduits, on the same cable tray, etc.) as the live cables of the
related circuit. This arrangement ensures the minimum possible inductive reactance in the earth-fault current carrying circuits. It should be noted that this arrangement is originally provided by bus-trunking.
Connection PE conductors must:
Not include any means of breaking the continuity of the circuit (such as a switch, removable links, etc.)
Connect exposed conductive parts individually to the main PE conductor, i.e. in parallel, not in series, as shown in Figure G54
Have an individual terminal on common earthing bars in distribution boards.
Fig. G54: A poor connection in a series arrangement will leave all downstream appliances unprotected TT scheme The PE conductor need not necessarily be installed in close proximity to the live conductors of the corresponding circuit, since high values of earth-fault current are not needed to operate the RCDtype of protection used in TT installations. IT and TN schemes
The PE or PEN conductor, as previously noted, must be installed as close as possible to the corresponding live conductors of the circuit and no ferro-magnetic material must be interposed between them. A PEN conductor must always be connected directly to the earth terminal of an appliance, with a looped connection from the earth terminal to the neutral terminal of the appliance (see Fig. G55).
TN-C scheme (the neutral and PE conductor are one and the same, referred to as a PEN conductor) The protective function of a PEN conductor has priority, so that all rules governing PE conductors apply strictly to PEN conductors
TN-C to TN-S transition The PE conductor for the installation is connected to the PEN terminal or bar (see Figure G56) generally at the origin of the installation. Downstream of the point of separation, no PE conductor can be connected to the neutral conductor.
Fig. G55: Direct connection of the PEN conductor to the earth terminal of an appliance
Fig. G56: The TN-C-S scheme
Types of materials Materials of the kinds mentioned below in Figure G57 can be used for PE conductors, provided that the conditions mentioned in the last column are satisfied. Type of protective earthing conductor(PE)
IT scheme
TN scheme
TT scheme
Conditions to be respected
Supplementary conductor
In the same cable as the phases, or in the same cable run
Strongly ecommended
Strongly recommended
Correct
Independent of the phase conductors
Possible[a]
Possible[a][b]
Correct
The PE conductor must be insulated to the same level as the phases
The PE conductor may
be bare or insulated[b] The electrical
[c]
Metallic housing of bus-trunking or of other prefabricated prewired ducting[e]
Possible
External sheath of extruded, mineralinsulated conductors (e.g. «pyrotenax» type systems)
Possible[c]
Certain extraneous conductive elements[f] such as:
Possible[d]
[c]
PE possible PEN possible[h]
Correct
PE possible[c] PEN not recommended[b][c]
Possible
PE possible[d]
Possible
Steel building structure
Machine frames
Water pipes[g] Metallic cable ways, such as, conduits[i]ducts, trunking, trays, ladders, and so on…
by protection against deterioration by mechanical, chemical and electrochemical hazards
PEN forbidden
continuity must be assured
Their conductance
must be adequate
Possible[d]
PE possible[d] PEN not recommended[b][d]
Possible
Forbidden for use as PE conductors, are: metal conduits[i], gas pipes, hot-water pipes, cable-armouring tapes[i]or wires[i] [a] In TN and IT schemes, fault clearance is generally achieved by overcurrent devices (fuses or circuit-breakers) so that the impedance of the fault-current loop must be sufficiently low to assure positive protective device operation. The surest means of achieving a low loop impedance is to use a supplementary core in the same cable as the circuit conductors (or taking the same route as the circuit conductors). This solution minimizes the inductive reactance and therefore the impedance of the loop. [b] The PEN conductor is a neutral conductor that is also used as a protective earth conductor. This means that a current may be flowing through it at any time (in the absence of an earth fault). For this reason an insulated conductor is recommended for PEN operation. [c] The manufacturer provides the necessary values of R and X components of the impedances (phase/PE, phase/PEN) to include in the calculation of the earth-fault loop impedance. [d] Possible, but not recomended, since the impedance of the earth-fault loop cannot be known at the design stage. Measurements on the completed installation are the only practical means of assuring adequate protection for persons. [e] It must allow the connection of other PE conductors. Note: these elements must carry an indivual green/yellow striped visual indication, 15 to 100 mm long (or the letters PE at less than 15 cm from each extremity). [f] These elements must be demountable only if other means have been provided to ensure uninterrupted continuity
of protection. [g] With the agreement of the appropriate water authorities. [h] In the prefabricated pre-wired trunking and similar elements, the metallic housing may be used as a PEN conductor, in parallel with the corresponding bar, or other PE conductor in the housing. [i] Forbidden in some countries only. Universally allowed to be used for supplementary equipotential conductors. Fig. G57: Choice of protective conductors (PE).
Choice of earthing method – implementation. After consulting applicable regulations, Figure E16 and Figure E17 can be used as an aid in deciding on divisions and possible galvanic isolation of appropriate sections of a proposed installation.
Division of source This technique concerns the use of several transformers instead of employing one high-rated unit. In this way, a load that is a source of network disturbances (large motors, furnaces, etc.) can be supplied by its own transformer. The quality and continuity of supply to the whole installation are thereby improved. The cost of switchgear is reduced (short-circuit current level is lower). The cost-effectiveness of separate transformers must be determined on a case by case basis.
Network islands The creation of galvanically-separated “islands” by means of LV/LV transformers makes it possible to optimise the choice of earthing methods to meet specific requirements (see Fig. E18 and Fig. E19).
Fig. E18: TN-S island within an IT system
Fig. E19: IT islands within a TN-S system
Conclusion The optimisation of the performance of the whole installation governs the choice of earthing system. Including:
Initial investments, and
Future operational expenditures, hard to assess, that can arise from insufficient reliability, quality of equipment, safety, continuity of service, etc. An ideal structure would comprise normal power supply sources, local reserve power supply sources (see Selection criteria for the TT, TN and IT systems ) and the appropriate earthing arrangements.
Installation and measurements of earth electrodes. A very effective method of obtaining a low-resistance earth connection is to bury a conductor in the form of a closed loop in the soil at the bottom of the excavation for building foundations.
The resistance R of such an electrode (in homogeneous soil) is given (approximately) in ohms by: where L = length of the buried conductor in metres ρ = soil resistivity in ohm-metres The quality of an earth electrode (resistance as low as possible) depends essentially on two factors:
Installation method
Type of soil
Installation methods Three common types of installation will be discussed: Buried ring (see Fig. E20) This solution is strongly recommended, particularly in the case of a new building. The electrode should be buried around the perimeter of the excavation made for the foundations. It is important that the bare conductor be in intimate contact with the soil (and not placed in the gravel or aggregate hard-core, often forming a base for concrete). At least four (widely-spaced) vertically arranged conductors from the electrode should be provided for the installation connections and, where possible, any reinforcing rods in concrete work should be connected to the electrode. The conductor forming the earth electrode, particularly when it is laid in an excavation for foundations, must be in the earth, at least 50 cm below the hard-core or aggregate base for the concrete foundation. Neither the electrode nor the vertical rising conductors to the ground floor, should ever be in contact with the foundation concrete. For existing buildings, the electrode conductor should be buried around the outside wall of the premises to a depth of at least 1 metre. As a general rule, all vertical connections from an electrode to above-ground level should be insulated for the nominal LV voltage (600-1,000 V). The conductors may be:
Copper: Bare cable ( ≥ 25 mm2) or multiple-strip (≥ 25 mm2) and (≥ 2 mm thick)
Aluminium with lead jacket: Cable ( ≥ 35 mm2)
Galvanised-steel cable: Bare cable ( ≥ 95 mm2) or multiple-strip ( ≥ 100 mm2 and ≥ 3 mm thick) The approximate resistance R of the electrode in ohms: where L = length of conductor in metres ρ = resistivity of the soil in ohm-metres (see Influence of the type of soil )
Fig. E20: Conductor buried below the level of the foundations, i.e. not in the concrete Earthing rods (see Fig. E21)
For n rods: Vertically driven earthing rods are often used for existing buildings, and for improving (i.e. reducing the resistance of) existing earth electrodes. The rods may be: Copper or (more commonly) copper-clad steel. The latter are generally 1 or 2 metres long
and provided with screwed ends and sockets in order to reach considerable depths, if necessary (for instance, the water-table level in areas of high soil resistivity) Galvanised[1] steel pipe ≥ 25 mm diameter or rod ≥ 15 mm diameter, ≥ 2 metres long in each
case.
Fig. E21: Earthing rods connected in parallel
It is often necessary to use more than one rod, in which case the spacing between them should exceed the depth to which they are driven, by a factor of 2 to 3. The total resistance (in homogeneous soil) is then equal to the resistance of one rod, divided by the number of rods in question.
The approximate resistance R obtained is:
if the distance separating the rods > 4L
where L = the length of the rod in metres ρ = resistivity of the soil in ohm-metres (see Influence of the type of soil) n = the number of rods Vertical plates (see Fig. E22)
For a vertical plate electrode: Rectangular plates, each side of which must be ≥ 0.5 metres, are commonly used as earth electrodes, being buried in a vertical plane such that the centre of the plate is at least 1 metre below the surface of the soil. The plates may be:
Copper of 2 mm thickness
Galvanised[1] steel of 3 mm thickness The resistance R in ohms is given (approximately), by: where L = the perimeter of the plate in metres ρ = resistivity of the soil in ohm-metres (see Influence of the type of soil)
Fig. E22: Vertical plate - 2 mm thickness (Cu)
Influence of the type of soil Measurements on earth electrodes in similar soils are useful to determine the resistivity value to be applied for the design of an earth-electrode system Type of soil
Mean value of resistivity in Ωm
Swampy soil, bogs
1 - 30
Silt alluvium
20 - 100
Humus, leaf mould
10 - 150
Peat, turf
5 - 100
Soft clay
50
Marl and compacted clay
100 - 200
Jurassic marl
30 - 40
Clayey sand
50 - 500
Siliceous sand
200 - 300
Stoney ground
1,500 - 3,000
Grass-covered-stoney sub-soil
300 - 500
Chalky soil
100 - 300
Limestone
1,000 - 5,000
Fissured limestone
500 - 1,000
Schist, shale
50 - 300
Mica schist
800
Granite and sandstone
1,500 - 10,000
Modified granite and sandstone
100 - 600
Fig. E23: Resistivity (Ωm) for different types of soil
Type of soil
Average value of resistivity in Ωm
Fertile soil, compacted damp fill
50
Arid soil, gravel, uncompacted non-uniform fill
500
Stoney soil, bare, dry sand, fissured rocks
3000
Fig. E24: Average resistivity (Ωm) values for approximate earth-elect
Measurement and constancy of the resistance between an earth electrode and the earth The resistance of the electrode/earth interface rarely remains constant Among the principal factors affecting this resistance are the following: Humidity of the soil
The seasonal changes in the moisture content of the soil can be significant at depths of up to 2 meters. At a depth of 1 metre the resistivity and therefore the resistance can vary by a ratio of 1 to 3 between a wet winter and a dry summer in temperate regions Frost
Frozen earth can increase the resistivity of the soil by several orders of magnitude. This is one reason for recommending the installation of deep electrodes, in particular in cold climates Ageing
The materials used for electrodes will generally deteriorate to some extent for various reasons, for example: Chemical reactions (in acidic or alkaline soils) Galvanic: due to stray DC currents in the earth, for example from electric railways, etc. or due to dissimilar metals forming primary cells. Different soils acting on sections of the same conductor can also form cathodic and anodic areas with consequent loss of surface metal from the latter areas. Unfortunately, the most favourable conditions for low earthelectrode resistance (i.e. low soil resistivity) are also those in which galvanic currents can most easily flow. Oxidation
Brazed and welded joints and connections are the points most sensitive to oxidation. Thorough cleaning of a newly made joint or connection and wrapping with a suitable greased-tape binding is a commonly used preventive measure.
Measurement of the earth-electrode resistance There must always be one or more removable links to isolate an earth electrode so that it can be tested.
There must always be removable links which allow the earth electrode to be isolated from the installation, so that periodic tests of the earthing resistance can be carried out. To make such tests, two auxiliary electrodes are required, each consisting of a vertically driven rod. Ammeter method (see Fig. E25)
Fig. E25: Measurement of the resistance to earth of the earth electrode of an installation by means of an ammeter
When the source voltage U is constant (adjusted to be the same value for each test) then:
In order to avoid errors due to stray earth currents (galvanic -DC- or leakage currents from power and communication networks and so on) the test current should be AC, but at a different frequency to that of the power system or any of its harmonics. Instruments using hand-driven generators to make these measurements usually produce an AC voltage at a frequency of between 85 Hz and 135 Hz. The distances between the electrodes are not critical and may be in different directions from the electrode being tested, according to site conditions. A number of tests at different spacings and directions are generally made to cross-check the test results.
Use of a direct-reading earthing-resistance ohmmeter These instruments use a hand-driven or electronic-type AC generator, together with two auxiliary electrodes, the spacing of which must be such that the zone of influence of the electrode being tested should not overlap that of the test electrode (C). The test electrode (C) furthest from the electrode (X) under test, passes a current through the earth and the electrode under test, while the second test electrode (P) picks up a voltage. This voltage, measured between (X) and (P), is due to the test current and is a measure of the contact resistance (of the electrode under test) with earth. It is clear that the distance (X) to (P) must be carefully chosen to give accurate results. If the distance (X) to (C) is increased, however, the zones of resistance of electrodes (X) and (C) become more remote, one from the other, and the curve of potential (voltage) becomes more nearly horizontal about the point (O). In practical tests, therefore, the distance (X) to (C) is increased until readings taken with electrode (P) at three different points, i.e. at (P) and at approximately 5 metres on either side of (P), give similar values. The distance (X) to (P) is generally about 0.68 of the distance (X) to (C).
[a] the principle of measurement is based on assumed homogeneous soil conditions. Where the zones of influence of electrodes C and X overlap, the location of test electrode P is difficult to determine for satisfactory results.
[b] showing the effect on the potential gradient when (X) and (C) are widely spaced. The location of test electrode P is not critical and can be easily determined.
Fig. E26: Measurement of the resistance to the mass of earth of electrode (X) using an earth-electrode-testing ohmmeter
Notes 1. ^ a b Where galvanised conducting materials are used for earth electrodes, sacrificial cathodic protection anodes may be necessary to avoid rapid corrosion of the electrodes where the soil is aggressive. Specially prepared magnesium anodes (in a porous sack filled with a suitable “soil”) are available for direct connection to the electrodes. In such circumstances, a specialist should be consulted.
Selection criteria for the TT, TN and IT systems. Selection does not depend on safety criteria. The three systems are equivalent in terms of protection of persons if all installation and operating rules are correctly followed. The selection criteria for the best system(s) depend on the regulatory requirements, the required continuity of service, operating conditions and the types of network and loads In terms of the protection of persons, the three system earthing arrangements (SEA) are equivalent if all installation and operating rules are correctly followed. Consequently, selection does not depend on safety criteria. It is by combining all requirements in terms of regulations, continuity of service, operating conditions and the types of network and loads that it is possible to determine the best system(s) (see Fig. E16). Selection is determined by the following factors:
Above all, the applicable regulations which in some cases impose certain types of SEA
Secondly, the decision of the owner if supply is via a private MV/LV transformer (MV subscription) or the owner has a private energy source (or a separate-winding transformer) If the owner effectively has a choice, the decision on the SEA is taken following discussions with the network designer (design office, contractor). The discussions must cover:
First of all, the operating requirements (the required level of continuity of service) and the operating conditions (maintenance ensured by electrical personnel or not, in-house personnel or outsourced, etc.)
Secondly, the particular characteristics of the network and the loads (see Fig. E17 ).
TNS
TNC
IT1[a]
IT2[b]
Comments
Fault current
-
--
--
+
--
Fault voltage
-
-
-
+
-
Touch voltage
+/-
-
-
+
-
Protection of persons against indirect contact
+
+
+
+
+
Protection of persons with emergency generating sets
+
-
-
+
-
TT
Electrical characteristics
Protection
Protection against fire (with an RCD)
+
+
Not allowed
+
+
Continuous overvoltage
+
+
+
-
+
Transient overvoltage
+
-
-
+
-
Overvoltage if transformer breakdown(primary/secondary)
-
+
+
+
+
-
+
+
+
+
Overvoltages
Electromagnetic compatibility Immunity to nearby lightning strikes
Immunity to lightning strikes on MV lines
-
-
-
-
-
Continuous emission of an electromagnetic field
+
+
-
+
+
Transient non-equipotentiality of the PE
+
-
-
+
-
Interruption for first fault
-
-
-
+
+
Voltage dip during insulation fault
+
-
-
+
-
Continuity of service
Installation
Special devices
-
+
+
-
-
Number of earth electrodes
-
+
+
-/+
-/+
Number of cables
-
-
+
-
-
Cost of repairs
-
--
--
-
--
Installation damage
+
-
-
++
-
Maintenance
[a] IT-net when a first fault occurs. [b] IT-net when a second fault occurs. Fig. E16: Comparison of system earthing arrangements
Type of network
Advised
Very large network with high-quality earth electrodes for exposed conductive parts (10 Ω max.)
Very large network with low-quality earth electrodes for exposed conductive parts (> 30 Ω)
TN
Disturbed area (storms) (e.g. television or radio transmitter)
TN
Network with high leakage currents (> 500 mA)
TN[d]
Network with outdoor overhead lines
TT[e]
Emergency standby generator set
IT
Type of loads
Loads sensitive to high fault currents (motors, etc.)
IT
Loads with a low insulation level (electric furnaces,welding machines, heating elements, immersion heaters, equipment in large kitchens)
TN[i]
Numerous phase-neutral single-phase loads (mobile, semi-fixed, portable)
TT[j] TN-S
Loads with sizeable risks (hoists, conveyers, etc.)
TN[k]
Numerous auxiliaries (machine tools)
TN-S
Miscellaneous
Supply via star-star connected power transformer[n]
TT
Premises with risk of fire
IT[o]
Increase in power level of LV utility subscription, requiring a private substation
TT[p]
Installation with frequent modifications
TT[q]
Installation where the continuity of earth circuits is uncertain (work sites, old installations)
TT[s]
Electronic equipment (computers, PLCs)
TN-S
Machine control-monitoring network, PLC sensors and actuators
IT[t]
[a] When the SEA is not imposed by regulations, it is selected according to the level of operating characteristics (continuity of service that is mandatory for safety reasons or desired to enhance productivity, etc.). Whatever the SEA, the probability of an insulation failure increases with the length of the network. It may be a good idea to break up the network, which facilitates fault location and makes it possible to implement the system advised above for each type of application. [b] The risk of flashover on the surge limiter turns the isolated neutral into an earthed neutral. These risks are high for regions with frequent thunder storms or installations supplied by overhead lines. If the IT system is selected to ensure a higher level of continuity of service, the system designer must precisely calculate the tripping conditions for a second fault. [c] Risk of RCD nuisance tripping.
[d] Whatever the SEA, the ideal solution is to isolate the disturbing section if it can be easily identified. [e] Risks of phase-to-earth faults affecting equipotentiality. [f] Insulation is uncertain due to humidity and conducting dust. [g] The TN system is not advised due to the risk of damage to the generator in the case of an internal fault. What is more, when generator sets supply safety equipment, the system must not trip for the first fault. [h] The phase-to-earth current may be several times higher than In, with the risk of damaging or accelerating the ageing of motor windings, or of destroying magnetic circuits. [i] To combine continuity of service and safety, it is necessary and highly advised, whatever the SEA, to separate these loads from the rest of the installation (transformers with local neutral connection). [j] When load equipment quality is not a design priority, there is a risk that the insulation resistance will fall rapidly. The TT system with RCDs is the best means to avoid problems. [k] The mobility of this type of load causes frequent faults (sliding contact for bonding of exposed conductive parts) that must be countered. Whatever the SEA, it is advised to supply these circuits using transformers with a local neutral connection. [l] Requires the use of transformers with a local TN system to avoid operating risks and nuisance tripping at the first fault (TT) or a double fault (IT). [l2] With a double break in the control circuit. [m] Excessive limitation of the phase-to-neutral current due to the high value of the zero-phase impedance (at least 4 to 5 times the direct impedance). This system must be replaced by a star-delta arrangement. [n] The high fault currents make the TN system dangerous. The TN-C system is forbidden. [o] Whatever the system, the RCD must be set to Δn ≤ 500 mA. [p] An installation supplied with LV energy must use the TT system. Maintaining this SEA means the least amount of modifications on the existing network (no cables to be run, no protection devices to be modified). [q] Possible without highly competent maintenance personnel. [r] This type of installation requires particular attention in maintaining safety. The absence of preventive measures in the TN system means highly qualified personnel are required to ensure safety over time. [s] The risks of breaks in conductors (supply, protection) may cause the loss of equipotentiality for exposed conductive parts. A TT system or a TN-S system with 30 mA RCDs is advised and is often mandatory. The IT system may be used in very specific cases. [t] This solution avoids nuisance tripping for unexpected earth leakage. Fig. E17: Influence of networks and loads on the selection of system earthing arrangements.
Characteristics of TT, TN and IT systems. TT system
(see Fig. E12) The TT system:
Technique for the protection of persons: the exposed conductive parts are earthed and residual current devices (RCDs) are used
Operating technique: interruption for the first insulation fault
Fig. E12: TT system Note: If the exposed conductive parts are earthed at a number of points, an RCD must be installed for each set of circuits connected to a given earth electrode. Main characteristics
Simplest solution to design and install. Used in installations supplied directly by the public LV distribution network.
Does not require continuous monitoring during operation (a periodic check on the RCDs may be necessary).
Protection is ensured by special devices, the residual current devices (RCD), which also prevent the risk of fire when they are set to ≤ 500 mA.
Each insulation fault results in an interruption in the supply of power, however the outage is limited to the faulty circuit by installing the RCDs in series (selective RCDs) or in parallel (circuit selection).
Loads or parts of the installation which, during normal operation, cause high leakage currents, require special measures to avoid nuisance tripping, i.e. supply the loads with a separation transformer or use specific RCDs (see TT system - Protective measures).
TN system
see (Fig. E13 and Fig. E14) The TN system: Technique for the protection of persons:
Interconnection and earthing of exposed conductive parts and the neutral are mandatory
Interruption for the first fault using overcurrent protection (circuit-breakers or fuses) Operating technique: interruption for the first insulation fault
Fig. E13: TN-C system
Fig. E14: TN-S system Main characteristics Generally speaking, the TN system:
Requires the installation of earth electrodes at regular intervals throughout the
installation
Requires that the initial check on effective tripping for the first insulation fault be carried out by calculations during the design stage, followed by mandatory measurements to confirm tripping during commissioning
Requires that any modification or extension be designed and carried out by a qualified electrician
May result, in the case of insulation faults, in greater damage to the windings of
rotating machines May, on premises with a risk of fire, represent a greater danger due to the higher
fault currents In addition, the TN-C system:
At first glance, would appear to be less expensive (elimination of a device pole and of
a conductor)
Requires the use of fixed and rigid conductors
Is forbidden in certain cases: Premises with a risk of fire For computer equipment (presence of harmonic currents in the neutral) In addition, the TN-S system:
May be used even with flexible conductors and small conduits
Due to the separation of the neutral and the protection conductor, provides a clean PE (computer systems and premises with special risks)
IT system (see Fig. E15) IT system: Protection technique:
Interconnection and earthing of exposed conductive parts
Indication of the first fault by an insulation monitoring device (IMD)
Interruption for the second fault using overcurrent protection (circuit-breakers or fuses) Operating technique:
Monitoring of the first insulation fault
Mandatory location and clearing of the fault
Interruption for two simultaneous insulation faults
Fig. E15: IT system Main characteristics
Solution offering the best continuity of service during operation
Indication of the first insulation fault, followed by mandatory location and clearing, ensures systematic prevention of supply outages
Generally used in installations supplied by a private MV/LV or LV/LV transformer
Requires maintenance personnel for monitoring and operation
Requires a high level of insulation in the network (implies breaking up the network if it is very large and the use of circuit-separation transformers to supply loads with high leakage currents)
The check on effective tripping for two simultaneous faults must be carried out by calculations during the design stage, followed by mandatory measurements during commissioning on each group of interconnected exposed conductive parts
Protection of the neutral conductor must be ensured as indicated in Protection of the neutral conductor inside chapter Sizing and protection of conductors.
Definition of standardised earthing schemes. he different earthing schemes (often referred to as the type of power system or system earthing arrangements) described characterise the method of earthing the installation downstream of the secondary winding of a MV/LV transformer and the means used for earthing the exposed conductive-parts of the LV installation supplied from it The choice of these methods governs the measures necessary for protection against indirect-contact hazards. The earthing system qualifies three originally independent choices made by the designer of an electrical distribution system or installation:
The type of connection of the electrical system (that is generally of the neutral conductor) and of the exposed parts to earth electrod (s)
A separate protective conductor or protective conductor and neutral conductor being a single conductor
The use of earth fault protection of overcurrent protective switchgear which clear only relatively high fault currents or the use of additional relays able to detect and clear small insulation fault currents to earth In practice, these choices have been grouped and standardised as explained below. Each of these choices provides standardised earthing systems with three advantages and drawbacks:
Connection of the exposed conductive parts of the equipment and of the neutral conductor to the PE conductor results in equipotentiality and lower overvoltages but increases earth fault currents
A separate protective conductor is costly even if it has a small cross-sectional area but it is much more unlikely to be polluted by voltage drops and harmonics, etc. than a neutral conductor is. Leakage currents are also avoided in extraneous conductive parts
Installation of residual current protective relays or insulation monitoring devices are much more sensitive and permits in many circumstances to clear faults before heavy damage occurs (motors, fires, electrocution). The protection offered is in addition independent with respect to changes in an existing installation
TT system (earthed neutral) (see Fig. E3) One point at the supply source is connected directly to earth. All exposed- and extraneousconductive-parts are connected to a separate earth electrode at the installation. This electrode may or may not be electrically independent of the source electrode. The two zones of influence may overlap without affecting the operation of protective devices.
Fig. E3: TT System
TN systems (exposed conductive parts connected to the neutral) The source is earthed as for the TT system (above). In the installation, all exposed- and extraneousconductive-parts are connected to the neutral conductor. The several versions of TN systems are shown below. TN-C system (see Fig. E4) The neutral conductor is also used as a protective conductor and is referred to as a PEN (Protective Earth and Neutral) conductor. This system is not permitted for conductors of less than 10 mm2 or for portable equipment. The TN-C system requires an effective equipotential environment within the installation with dispersed earth electrodes spaced as regularly as possible since the PEN conductor is both the neutral conductor and at the same time carries phase unbalance currents as well as 3rd order harmonic currents (and their multiples). The PEN conductor must therefore be connected to a number of earth electrodes in the installation. Caution: In the TN-C system, the “protective conductor” function has priority over the “neutral function”. In particular, a PEN conductor must always be connected to the earthing terminal of a load and a jumper is used to connect this terminal to the neutral terminal.
Fig. E4: TN-C system TN-S system (see Fig. E5) The TN-S system (5 wires) is obligatory for circuits with cross-sectional areas less than 10 mm2 for portable equipment. The protective conductor and the neutral conductor are separate. On underground cable systems where lead-sheathed cables exist, the protective conductor is generally the lead sheath. The use of separate PE and N conductors (5 wires) is obligatory for circuits with cross-sectional areas less than 10 mm2 for portable equipment.
Fig. E5: TN-S system TN-C-S system (see Fig. E6 and Fig. E7) The TN-C and TN-S systems can be used in the same installation. In the TN-C-S system, the TN-C (4 wires) system must never be used downstream of the TN-S (5 wires) system, since any accidental interruption in the neutral on the upstream part would lead to an interruption in the protective conductor in the downstream part and therefore a danger.
Fig. E6: TN-C-S system
Fig. E7: Connection of the PEN conductor in the TN-C system
IT system (isolated or impedance-earthed neutral) IT system (isolated neutral) No intentional connection is made between the neutral point of the supply source and earth (see Fig. E8).
Fig. E8: IT system (isolated neutral) Exposed- and extraneous-conductive-parts of the installation are connected to an earth electrode. In practice all circuits have a leakage impedance to earth, since no insulation is perfect. In parallel with this (distributed) resistive leakage path, there is the distributed capacitive current path, the two paths together constituting the normal leakage impedance to earth (see Fig. E9).
Fig. E9: IT system (isolated neutral) Example (see Fig. E10) In a LV 3-phase 3-wire system, 1 km of cable will have a leakage impedance due to C1, C2, C3 and R1, R2 and R3 equivalent to a neutral earth impedance Zct of 3,000 to 4,000 Ω, without counting the filtering capacitances of electronic devices.
Fig. E10: Impedance equivalent to leakage impedances in an IT system IT system (impedance-earthed neutral) An impedance Zs (in the order of 1,000 to 2,000 Ω) is connected permanently between the neutral point of the transformer LV winding and earth (see Fig. E11). All exposed- and extraneousconductive-parts are connected to an earth electrode. The reasons for this form of power-source earthing are to fix the potential of a small network with respect to earth (Zs is small compared to the leakage impedance) and to reduce the level of overvoltages, such as transmitted surges from the MV windings, static charges, etc. with respect to earth. It has, however, the effect of slightly increasing the first-fault current level.
Fig. E11: IT system (impedance-earthed neutral).
Earthing connections. In a building, the connection of all metal parts of the building and all exposed conductive parts of electrical equipment to an earth electrode prevents the appearance of dangerously high voltages between any two simultaneously accessible metal parts
Definitions National and international standards (IEC 60364) clearly define the various elements of earthing connections. The following terms are commonly used in industry and in the literature. Bracketed numbers refer to Figure E1
Fig. E1: An example of a block of flats in which the main earthing terminal (6) provides the main equipotential connection; the removable link (7) allows an earth-electrode-resistance check
Earth electrode (1): A conductor or group of conductors in intimate contact with, and providing an electrical connection with Earth (cf details in section 1.6 of Chapter E.)
Earth: The conductive mass of the Earth, whose electric potential at any point is conventionally taken as zero
Electrically independent earth electrodes: Earth electrodes located at such a distance from
one another that the maximum current likely to flow through one of them does not significantly affect the potential of the other(s)
Earth electrode resistance: The contact resistance of an earth electrode with the Earth
Earthing conductor (2): A protective conductor connecting the main earthing terminal (6) of an installation to an earth electrode (1) or to other means of earthing (e.g. TN systems); Exposed-conductive-part: A conductive part of equipment which can be touched and which
is not a live part, but which may become live under fault conditions Protective conductor (3): A conductor used for some measures of protection against electric
shock and intended for connecting together any of the following parts:
Exposed-conductive-parts
Extraneous-conductive-parts
The main earthing terminal
Earth electrode(s)
The earthed point of the source or an artificial neutral Extraneous-conductive-part: A conductive part liable to introduce a potential, generally earth potential, and not forming part of the electrical installation (4). For example: Non-insulated floors or walls, metal framework of buildings Metal conduits and pipework (not part of the electrical installation) for water, gas, heating, compressed-air, etc. and metal materials associated with them
Bonding conductor (5): A protective conductor providing equipotential bonding
Main earthing terminal (6): The terminal or bar provided for the connection of protective conductors, including equipotential bonding conductors, and conductors for functional earthing, if any, to the means of earthing.
Connections The main equipotential bonding system The bonding is carried out by protective conductors and the aim is to ensure that, in the event of an incoming extraneous conductor (such as a gas pipe, etc.) being raised to some potential due to a fault external to the building, no difference of potential can occur between extraneous-conductiveparts within the installation.
The bonding must be effected as close as possible to the point(s) of entry into the building, and be connected to the main earthing terminal (6). However, connections to earth of metallic sheaths of communications cables require the authorisation of the owners of the cables. Supplementary equipotential connections These connections are intended to connect all exposed-conductive-parts and all extraneousconductive-parts simultaneously accessible, when correct conditions for protection have not been met, i.e. the original bonding conductors present an unacceptably high resistance. Connection of exposed-conductive-parts to the earth electrode(s) The connection is made by protective conductors with the object of providing a low-resistance path for fault currents flowing to earth.
Components (see Fig. E2) Effective connection of all accessible metal fixtures and all exposed-conductive-parts of electrical appliances and equipment, is essential for effective protection against electric shocks. Component parts to consider:
as exposed-conductive-parts
as extraneous-conductive-parts Elements used in building construction
Cableways Metal or reinforced concrete (RC):
Conduits
Impregnated-paper-insulated lead-covered cable, armoured or unarmoured Mineral insulated metal-sheathed cable (pyrotenax,
Reinforcement rods
Prefabricated RC panels Surface finishes: Floors and walls in reinforced concrete without further
cradle of withdrawable switchgear Appliances Exposed metal parts of class 1 insulated appliances
Switchgear
Steel-framed structure
etc.)
surface treatment
Tiled surface Metallic covering:
Metallic wall covering Building services elements other than electrical
Non-electrical elements metallic fittings associated with cableways (cable
trays, cable ladders, etc.) Metal objects:
Close to aerial conductors or to busbars
In contact with electrical equipment.
Metal pipes, conduits, trunking, etc. for gas,water
and heating systems, etc. Related metal components (furnaces,
tanks,reservoirs, radiators) Metallic fittings in wash rooms, bathrooms, toilets,
etc.
Metallised papers
Component parts not to be considered: as exposed-conductive-parts Diverse service channels, ducts, etc.
Conduits made of insulating material
Mouldings in wood or other insulating material
Conductors and cables without metallic sheaths Switchgear
Wooden-block floors
Rubber-covered or linoleum-covered floors
Dry plaster-block partition
Brick walls
Carpets and wall-to-wall carpeting
Enclosures made of insulating material Appliances
as extraneous-conductive-parts
All appliances having class II insulation regardless of the type of exterior envelope
Fig. E2: List of exposed-conductive-parts and extraneous-conductive-parts.
Cables and busways. Distribution and installation methods (see Fig. E36) Distribution takes place via cableways that carry single insulated conductors or cables and include a fixing system and mechanical protection.
Fig. E36: Radial distribution using cables in a hotel
Busbar trunking (busways) Busways, also referred to as busbar trunking systems, stand out for their ease of installation, flexibility and number of possible connection points Busbar trunking is intended to distribute power (from 20 A to 5000 A) and lighting (in this application, the busbar trunking may play a dual role of supplying electrical power and physically holding the lights).
Busbar trunking system components A busbar trunking system comprises a set of conductors protected by an enclosure (see Fig. E37). Used for the transmission and distribution of electrical power, busbar trunking systems have all the necessary features for fitting: connectors, straights, angles, fixings, etc. The tap-off points placed at regular intervals make power available at every point in the installation.
Fig. E37: Busbar trunking system design for distribution of currents from 25 to 4000 A
The various types of busbar trunking: Busbar trunking systems are present at every level in electrical distribution: from the link between the transformer and the low voltage switch switchboard (MLVS) to the distribution of power sockets and lighting to offices, or power distribution to workshops.
Fig. E38: Radial distribution using busways We talk about a distributed network architecture. There are essentially three categories of busways.
Transformer to MLVS busbar trunking Installation of the busway may be considered as permanent and will most likely never be modified. There are no tap-off points. Frequently used for short runs, it is almost always used for ratings above 1,600 /2,000 A, i.e. when the use of parallel cables makes installation impossible. Busways are also used between the MLVS and downstream distribution switchboards. The characteristics of main-distribution busways authorize operational currents from1,000 to 5,000 A and short-circuit withstands up to 150 kA.
Sub-distribution busbar trunking with low or high tap-off densities Downstream of main-distribution busbar trunking , two types of applications must be supplied: Mid-sized premises (industrial workshops with injection presses and metalwork machines or large supermarkets with heavy loads). The short-circuit and current levels can be fairly high (respectively 20 to 70 kA and 100 to 1,000 A) Small sites (workshops with machine-tools, textile factories with small machines,supermarkets with small loads). The short-circuit and current levels are lower (respectively 10 to 40 kA and 40 to 400 A) Sub-distribution using busbar trunking meets user needs in terms of: Modifications and upgrades given the high number of tap-off points Dependability and continuity of service because tap-off units can be connected under energized conditions in complete safety The sub-distribution concept is also valid for vertical distribution in the form of 100 to 5,000 A risers in tall buildings.
Lighting distribution busbar trunking Lighting circuits can be distributed using two types of busbar trunking according to whether the lighting fixtures are suspended from the busbar trunking or not.
busbar trunking designed for the suspension of lighting fixtures These busways supply and support light fixtures (industrial reflectors, discharge lamps, etc.). They are used in industrial buildings, supermarkets, department stores and warehouses. The busbar trunkings are very rigid and are designed for one or two 25 A or 40 A circuits. They have tap-off outlets every 0.5 to 1 m. busbar trunking not designed for the suspension of lighting fixtures Similar to prefabricated cable systems, these busways are used to supply all types of lighting fixtures secured to the building structure. They are used in commercial buildings (offices, shops, restaurants, hotels, etc.), especially in false ceilings. The busbar trunking is flexible and designed for one 20 A circuit. It has tap-off outlets every 1.2 m to 3 m. Busbar trunking systems are suited to the requirements of a large number of buildings.
Industrial buildings: garages, workshops, farm buildings, logistic centers, etc.
Commercial areas: stores, shopping malls, supermarkets, hotels, etc.
Tertiary buildings: offices, schools, hospitals, sports rooms, cruise liners, etc.
Standards Busbar trunking systems must meet all rules stated in IEC 61439-6. This defines the manufacturing arrangements to be complied with in the design of busbar trunking systems (e.g.: temperature rise characteristics, short-circuit withstand, mechanical strength, etc.) as well as test methods to check them. The new standard IEC61439-6 describes in particular the design verifications and routine verifications required to ensure compliance. By assembling the system components on the site according to the assembly instructions, the contractor benefits from conformity with the standard.
The advantages of busbar trunking systems Flexibility
Easy to change configuration (on-site modification to change production line configuration or extend production areas).
Reusing components (components are kept intact): when an installation is subject to major modifications, the busbar trunking is easy to dismantle and reuse.
Power availability throughout the installation (possibility of having a tap-off point every meter).
Wide choice of tap-off units. Simplicity
Design can be carried out independently from the distribution and layout of current consumers.
Performances are independent of implementation: the use of cables requires a lot of derating coefficients.
Clear distribution layout
Reduction of fitting time: the trunking system allows fitting times to be reduced by up to 50% compared with a traditional cable installation.
Manufacturer’s guarantee.
Controlled execution times: the trunking system concept guarantees that there are no unexpected surprises when fitting. The fitting time is clearly known in advance and a quick solution can be provided to any problems on site with this adaptable and scalable equipment.
Easy to implement: modular components that are easy to handle, simple and quick to connect. Dependability
Reliability guaranteed by being factory-built
Fool-proof units
Sequential assembly of straight components and tap-off units making it impossible to make any mistakes Continuity of service
The large number of tap-off points makes it easy to supply power to any new current consumer. Connecting and disconnecting is quick and can be carried out in complete safety even when energized. These two operations (adding or modifying) take place without having to stop operations.
Quick and easy fault location since current consumers are near to the line
Maintenance is non existent or greatly reduced Major contribution to sustainable development
Busbar trunking systems allow circuits to be combined. Compared with a traditional cable
distribution system, consumption of raw materials for insulators is divided by 4 due to the busbar trunking distributed network concept (see Fig. E39).
Reusable device and all of its components are fully recyclable.
Does not contain PVC and does not generate toxic gases or waste.
Reduction of risks due to exposure to electromagnetic fields.
Fig. E39: Example of a set of 14 x 25A loads distributed along 34 meters (for busway, Canalis KS 250A)
New functional features for Canalis Busbar trunking systems are getting even better. Among the new features we can mention:
Increased performance with a IP55 protection index and new ratings of 160 A through to 1000 A (Ks).
New lighting offers with pre-cabled lights and new light ducts.
New fixing accessories. Quick fixing system, cable ducts, shared support with “VDI” (voice, data, images) circuits. Busbar trunking systems are perfectly integrated with the environment:
white color to enhance the working environment, naturally integrated in a range of electrical distribution products.
conformity with European regulations on reducing hazardous materials (RoHS).
Examples of Canalis busbar trunking systems
Fig. E41: Rigid busbar trunking able to support light fittings: Canalis KBA or KBB (25 and 40 A)
Fig. E43: A busway for medium power distribution: Canalis KN (40 up to 160 A)
Fig. E44: A busway for medium power distribution: Canalis KS (100 up to 1000 A)
Fig. E45: A busway for high power distribution: Canalis KT (800 up to 5000 A).
The architecture design. The architecture design considered in this document starts at the preliminary design stage (see Fig. D3 step1). It generally covers the levels of MV/LV main distribution, LV power distribution, and exceptionally the terminal distribution level. (see Fig. D2). In buildings all consumers are connected in low voltage. It means that MV distribution consists in:
connection to utility,
distribution to MV/LV substation(s),
MV/LV substation(s) itself.
Fig. D2: Example of single-line diagram The design of an electrical distribution architecture can be described by a 3-stage process, with iterative possibilities. This process is based on taking account of the installation characteristics and criteria to be satisfied.
Internal MV circuits. Internal MV circuits are dedicated to the supply of the secondary MV/LV substations dispersed in the installation. They are three typical principles commonly used for this purpose Fig. D11:
Single feeder
Dual feeder
Open ring.
Comparison of these three typical principles of internal distribution is given Fig. D12. MV circuit configuration
Characteristic to consider
Single feeder
Open ring
Dual feeder
Site topology
Any
Single or several buildings
Single or several buildings
Power demand
Any
> 1250kVA
> 2500kVA
Disturbance sensitivity
Long interruption acceptable
Short interruption acceptable
Short interruption not acceptable
Fig. D12: Comparison of the typical internal circuits.
Number of MV/LV transformers. For every MV/LV substation, the definition of the number of MV/LV transformers takes into account the following criteria:
Total power supplied by the substation
Standardization of the rated power to reduce the number of spare transformers
Limit of the rated power. It is recommended to set this limit at 1250 kVA in order to facilitate the handling and the replacement of the transformers
Scalability of the installation
Need to separate the loads having a high level of sensitivity to the electrical perturbations
Need to dedicate a transformer to the load generating a high level of perturbation such as voltage dips, harmonics, flicker
Need for partial or total redundancy. When required, two transformers each sized for the full load and equipped with an automatic change-over are installed
Loads requiring a dedicated neutral system. IT for example to ensure the continuity of operation in case of phase to earth fault.
LV distribution - centralized or distributed layout. Layout Position of the main MV and LV equipment on the site or in the building. This layout choice is applied to the results of stage 1. Selection guide As recommended in IEC60364-8-1 §6.3, MV/LV substation location can be determined by using the barycenter method:
taking into account service conditions: in dedicated premises if the layout in the workshop is too restrictive (temperature, vibrations, dust, etc.)
Placing heavy equipment (transformers, generators, etc.) close to walls or to main exits for ease of maintenance. A layout example is given in the following diagram (Fig. D13):
Fig. D13: The position of the global load barycentre guides the positioning of power sources
Centralized or distributed layout of LV distribution In centralized layout, each load is connected directly to the power source. (Fig. D14):
Fig. D14: Example of centralized layout with point to point links In distributed layout, loads are connected to sources via a busway. This type of distribution is well adapted to supply many loads that are spread out, where easy change is requested or future new connection (need of flexibility) (Fig. D15):
Fig. D15: Example of distributed layout, with busway Factors in favour of centralized layout (see summary table in Fig. D16):
Installation flexibility: no,
Load distribution: localized loads (high unit power loads). Factors in favor of distributed layout:
Installation flexibility: "Implementation" flexibility (moving of workstations, etc…),
Load distribution: uniform distribution of low or medium unit power loads
Load distribution
Flexibility (see Installation flexibilityfor definition of the flexibility levels)
Localized loads
No flexibility
Centralized
Intermediate distribution loads
Uniformly distributed loads
Decentralized
Flexibility of design
Implementation flexibility
Centralized
Decentralized
Operation flexibility
Fig. D16: Recommendations for centralized or distributed layout
Centralized distribution gives greater independence of circuits, reducing the consequences of a failure from power availability point of view. The use of decentralized distribution with busway is a way to merge all the circuits in one: it makes it possible to take into account the diversity factor (ks), which means cost savings on conductor sizing (See Fig. D17). The choice between centralized and decentralized solutions, according to the diversity factor, allows to find an economic optimum between investment costs, installation costs and operating costs. These two distribution modes are often combined.
Fig. D17: Example of a set of 14 x 25A loads distributed along 34 meters (for busway, Canalis KS 250A).
Presence of LV back-up generators. LV backup-up generator is the association of an alternator mechanically powered by a thermal engine. No electrical power can be delivered until the generator has reached its rated speed. This type of device is therefore not suitable for an uninterrupted power supply. Depending, if the generator is sized to supply power to all or only part of the installation, there is either total or partial redundancy. A back-up generator runs generally disconnected from the network. A source changeover and an interlocking system is therefore required (see Fig. D18). The generator back-up time depends on the quantity of available fuel.
Fig. D18: Connection of a back-up generator The main characteristics to consider for implementing LV back-up generator:
Sensitivity of loads to power interruption (see Voltage Interruption Sensitivity for definition),
Availability of the public distribution network (see Service reliability for the definition),
Other constraints (e.g.: generators compulsory in hospitals or high buildings) In addition the presence of generators can be decided to reduce the energy bill or due to the opportunity for co-generation. These two aspects are not taken into account in this guide. The presence of a back-up generator is essential if the loads cannot be shed (only short interruption acceptable) or if the utility network availability is low. Determining the number of back-up generator units is in line with the same criteria as determining the number of transformers, as well as taking account of economic and availability considerations (redundancy, start-up reliability, maintenance facility). Determining the generator apparent power, depends on:
installation power demand of loads to be supplied,
transient constraints that can occur by motors inrush current for example.
Configuration of LV circuits. Single feeder configuration
Fig. D20 This is the reference configuration and the most simple. A load is connected to one single source. This configuration provides a minimum level of availability, since there is no redundancy in case of power source failure.
Fig. D20: Single feeder configuration
Parallel transformers configuration Fig. D21 The power supply is provided by more than 1 transformer generally connected in parallel to the same main LV switchboard.
Fig. D21: Parallel transformers configuration
Variant: Normally open coupled transformers Fig. D22 In order to increase the availability it is possible to split the main LV switchboard into 2 parts, with a normally open bus-coupler (NO). This configuration may require an Automatic Transfer Switch between the coupler and transformer incomers. These 2 configurations are more often used when power demand is greater than 1 MVA.
Fig. D22: Normally open coupled transformers
Main LV switchboard interconnected by a busway Fig. D23 Transformers are physically distant, and operated in parallel. They are connected by a busway, the load can always be supplied in the case of failure of one of the sources. The redundancy can be:
Total: each transformer being able to supply all of the installation,
Partial: each transformer only being able to supply part of the installation. In this case, part of the loads must be disconnected (load-shedding) in the case of one of transformer failure.
Fig. D23: Main LV switchboard interconnected by a busway
LV ring configuration Fig. D24 This configuration can be considered as an extension of the previous configuration with interconnection between switchboards. Typically, 4 transformers connected in parallel to the same MV line, supply a ring using busway. A given load is then supplied by several transformers. This configuration is well suited to large sites, with high load density (in kVA/m2). If all of the loads can be
supplied by 3 transformers, there is total redundancy in the case of failure of one of the transformers. In fact, each busbar can be fed by one or other of its ends. Otherwise, downgraded operation must be considered (with partial load shedding). This configuration requires special design of the protection plan in order to ensure discrimination in all of the fault circumstances. As the previous configuration this type of installation is commonly used in automotive industry or large site manufacturing industry.
Fig. D24: Ring configuration
Double-ended power supply Fig. D25 This configuration is implemented in cases where maximum availability is required. The principle involves having 2 independent power sources, e.g.:
2 transformers supplied by different MV lines,
1 transformer and 1 generator,
1 transformer and 1 UPS.
An automatic transfer switch (ATS) is used to avoid the sources being parallel connected. This configuration allows preventive and curative maintenance to be carried out on all of the electrical distribution system upstream without interrupting the power supply.
Fig. D25: Double-ended configuration with automatic transfer switch
Configuration combinations Fig. D26 An installation can be made up of several sub-asssemblies with different configurations, according to requirements for the availability of the different types of load. E.g.: generator unit and UPS, choice by sectors (some sectors supplied by cables and others by busways).
Fig. D26: Example of a configuration combination 1: Single feeder, 2: Main LV switchboard interconnected by a busway, 3: Double-ended For the different possible configurations, the most probable and usual set of characteristics is given in the following table:
Characteristic to be considered
Configuration
Single feeder (fig. D20)
Parallel transformer or transformers connected via a coupler (fig. D21-D22)
Main LV switchboard interconnected by a busway (fig D24)
LV ring
Site topology
Any
Any
1 level 5000 to 25000 m2
1 level 5000 to m2
Power demand
< 2500kVA
Any
≥ 2500kVA
> 2500kVA
Location latitude
Any
Any
Medium or high
Medium or high
Load distribution
Localized loads
Localized loads
Intermediate or uniform load distribution
Intermediate or uniform load distribution
Maintainability
Minimal
Standard
Standard
Standard
Disturbances sensitivity
Low sensitivity
High sensitivity
High sensitivity
High sensitivity
Fig. D27: Recommendations for the configuration of LV circuits.
Example: electrical installation in a printworks. Brief description Printing of personalized mailshots intended for mail order sales.
Installation characteristics Characteristic
Category
Activity
Mechanical
Site topology
single storey building 10000m2 (8000m2 dedicated to the process,
2000m2 for ancillary areas)
Layout latitude
High
Service reliability
Standard
Maintainability
Standard
Installation flexibility
No flexibility planned:
HVAC
Process utilities
Office power supply Possible flexibility:
finishing, putting in envelopes
special machines, installed at a later date
rotary machines (uncertainty at the draft design stage)
Power demand
3500kVA
Load distribution
Intermediate distribution
Power interruptions sensitivity
Sheddable circuits:
offices (apart from PC power sockets)
air conditioning, office heating
social premises
maintenance premises long interruptions acceptable:
printing machines
workshop HVAC (hygrometric control
Finishing, envelope filling
Process utilities (compressor, recycling of cooled water) No interruptions acceptable:
servers, office PCs
Disturbance sensitivity
Average sensitivity:
motors, lighting
High sensitivity:
IT
No special precaution to be taken due to the connection to the EdF network (low level of disturbance) Disturbance capability Other constraints
Non disturbing Building with lightning classification: lightning surge
arresters installed Power supply by overhead single feeder line
Technological characteristics Criteria
Service conditions
Category
IP: standard (no dust, no water protection)
IK: standard (use of technical pits, dedicated premises) °C: standard (temperature regulation)
Required service index
211
Offer availability by country
No problem (project carried out in Europe)
Other criteria
Not applicable
Architecture assessment criteria Criteria
Category
On-site work time
Standard
Environmental impact
Minimal: compliance with European standard regulations
Preventive maintenance costs
Standard
Power supply availability
Tier 1
Step 1: Architecture fundamentals Choice
Main criteria
Solution
Connection to upstream network
Isolated site, 3500 kVA
MV single-line service
MV Circuits
Layout + criticality
single feeder
Number of transformers
Power > 2500kVA
2 x 2000kVA
Number and distribution of substations
Surface area and power distribution
2 possible solutions: 1 substation or 2 substations
if 1 substations: Normaly open bus-coupler
between MLVS if 2 substations: Main LV switchboard
interconnected by a busway MV Generator
Site activity
No
Fig. D31: Two possible single-line diagrams Step 2: Architecture details Choice
Main criteria
Solution
Layout
Service conditions
Dedicated premises
LV circuit configuration
2 transformers, requested by the power demand
Solution from fig.D22 or D23 are possible
Centralized or distributed layout
Uniform loads, distributed power, scalability possibilities Non-uniform loads, direct link from MLVS
Decentralized with busbar trunking: finishing sector, envelope filling Centralized with cables:
special machines, rotary machines, HVAC, process utilities, offices (2 switchboards), office air conditioning, social premises, maintenance
Presence of backup generator
Criticality ≤ low
No back-up generator
Network availability: standard Presence of UPS
Criticality
UPS unit for IT devices and office workstations
Fig. D32: Detailed single-line diagram (1 substation based on fig.D22)
Fig. D33: Detailed single-line diagram (2 substation based on fig.D24)
Choice of technological solutions Choice
Main criteria
Solution
MV/LV substation
Service conditions
indoor (dedicated premises)
MV switchboard
Offer availability by country
SM6 (installation in Europe)
Transformers
Service conditions
cast resin transfo (avoids constraints related to oil)
LV switchboard
Service conditions, service index for LV switchboards
MLVS: Prisma P Sub-distribution: Prisma
Busway
Load distribution
Canalis KS (fig.D32 or D33) Canalis KT for main distribution (fig D33)
UPS units
Installed power to be supplied, back-up
Galaxy PW
time Power factor correction
Reactive power to provide for the minimum up to the full load without harmonic (see chapter Power Factor Correction for more information), presence of harmonics.
MCB’s. Instantaneous Trip Setting. 6, 8, 10, 13, 16, 20, 25, 32, 40, 50, 63, 80, 100 and 125 A. Instantaneous tripping – less than 0.1 sec. for tripping. B.
3 – 5 of the rated continuous currents.
C.
5 – 10 of the rated continuous currents.
D.
10 – 20 of the rated continuous currents.
Suitable for resistive loads, heaters, lighting etc. Suitable for the reactive loads – motors, general lighting etc. Suitable for high reactive loads – transformers, motors etc.
Permissible I2t (let-through energy) in A2s for circuit-breakers type B with rated current up to and including 63A.
Permissible I2t (let-through energy) in A2s for circuit-breakers type C with rated current up to and including 63A.
Cables withstanding capacities.
The application of fault current protective devices to cable protection is detailed in BS 7671 and is given by:
I2t k2S2 Where:
I2t is the energy let-through value of the protective device k2S2 is the energy withstand of the cable.
The code allows the breaking capacity of a circuit-breaker to be less than its associated prospective fault current when back-up protection is employed in cascading through a suitable upstream protective devices. Back-up protection consists of an upstream short-circuit protective device (SCPD) that helps a downstream circuit-breaker to break fault currents greater than its maximum breaking current. However, where an MCCB, MCB or fuse is the upstream SCPD, and the downstream SCPD is an MCB,
coordination tests can be used to validate that the I2t of the specific combination will not exceed the I2t value of the downstream MCB at its maximum breaking capacity. The I 2t of the upstream SCPD “A” and downstream MCB “B” operating together at 20 kA, will be equal to or less than the I 2t of MCB “B” at 10 kA. The I 2t to be used in the conductor fault current assessment would be that of MCB “B” at 10 kA.
MCCB’s. (MCCBs) may have fixed or adjustable protection settings, normally a three position toggle operating handle giving on-off-tripped indication plus reset function, and a performance level relative to the incoming supply such that they can be installed at a point close to the supply transformer. Ratings: 16A to 1600A (may be upto 3200A), with the short circuit withstanding capacities upto 100kA in selections.
Rated Short-Time Withstand Current. Circuit-breakers of Selectivity Category B have a short-time delay (STD) allowing timegraded selectivity between circuit-breakers in series.
Icw is the fault current the circuit-breaker will withstand for the maximum short-time delay time. Preferred times are: 0.05, 0.1, 0.25, 0.5 and 1.0 second.
In order to provide protection against electric shock in accordance with the Wiring, it is necessary to determine the maximum value of Zs that will give the required disconnection time. This will give the max. length of the circuits downstream of the device for achieving the protections at around 80% of the phase – to – N voltages. No tolerance may be required for the IT – trippings, but a 20% may be considered in the case of instantaneous types.
Calculation Methodology
This calculation is based on IEEE Std 80 (2000), "Guide for safety in AC substation grounding". There are two main parts to this calculation:
Earthing grid conductor sizing
Touch and step potential calculations
IEEE Std 80 is quite descriptive, detailed and easy to follow, so only an overview will be presented here and IEEE Std 80 should be consulted for further details (although references will be given herein).
Prerequisites The following information is required / desirable before starting the calculation:
A layout of the site
Maximum earth fault current into the earthing grid
Maximum fault clearing time
Ambient (or soil) temperature at the site
Soil resistivity measurements at the site (for touch and step only)
Resistivity of any surface layers intended to be laid (for touch and step only)
Earthing Grid Conductor Sizing Determining the minimum size of the earthing grid conductors is necessary to ensure that the earthing grid will be able to withstand the maximum earth fault current. Like a normal power cable under fault, the earthing grid conductors experience an adiabatic short circuit temperature rise. However unlike a fault on a normal cable, where the limiting temperature is that which would cause permanent damage to the cable's insulation, the temperature limit for earthing grid conductors is the melting point of the conductor. In other words, during the worst case earth fault, we don't want the earthing grid conductors to start melting! The minimum conductor size capable of withstanding the adiabatic temperature rise associated with an earth fault is given by re-arranging IEEE Std 80 Equation 37:
Where (mm2)
is the minimum cross-sectional area of the earthing grid conductor
is the energy of the maximum earth fault (A2s) is the maximum allowable (fusing) temperature (ºC) is the ambient temperature (ºC) is the thermal coefficient of resistivity (ºC - 1) is the resistivity of the earthing conductor (μΩ.cm)
is is the thermal capacity of the conductor per unit volume(Jcm - 3ºC - 1) The material constants Tm, αr, ρr and TCAP for common conductor materials can be found in IEEE Std 80 Table 1. For example. commercial hard-drawn copper has material constants:
Tm = 1084 ºC
αr = 0.00381 ºC - 1
ρr = 1.78 μΩ.cm
TCAP = 3.42 Jcm - 3ºC - 1.
As described in IEEE Std 80 Section 11.3.1.1, there are alternative methods to formulate this equation, all of which can also be derived from first principles). There are also additional factors that should be considered (e.g. taking into account future growth in fault levels), as discussed in IEEE Std 80 Section 11.3.3.
Touch and Step Potential Calculations When electricity is generated remotely and there are no return paths for earth faults other than the earth itself, then there is a risk that earth faults can cause dangerous voltage gradients in the earth around the site of the fault (called ground potential rises). This means that someone standing near the fault can receive a dangerous electrical shock due to:
Touch voltages - there is a dangerous potential difference between the earth and a metallic object that a person is touching
Step voltages - there is a dangerous voltage gradient between the feet of a person standing on earth
The earthing grid can be used to dissipate fault currents to remote earth and reduce the voltage gradients in the earth. The touch and step potential calculations are performed in
order to assess whether the earthing grid can dissipate the fault currents so that dangerous touch and step voltages cannot exist.
Step 1: Soil Resistivity The resistivity properties of the soil where the earthing grid will be laid is an important factor in determining the earthing grid's resistance with respect to remote earth. Soils with lower resistivity lead to lower overall grid resistances and potentially smaller earthing grid configurations can be designed (i.e. that comply with safe step and touch potentials). It is good practice to perform soil resistivity tests on the site. There are a few standard methods for measuring soil resistivity (e.g. Wenner four-pin method). A good discussion on the interpretation of soil resistivity test measurements is found in IEEE Std 80 Section 13.4. Sometimes it isn't possible to conduct soil resistivity tests and an estimate must suffice. When estimating soil resistivity, it goes without saying that one should err on the side of caution and select a higher resistivity. IEEE Std 80 Table 8 gives some guidance on range of soil resistivities based on the general characteristics of the soil (i.e. wet organic soil = 10 Ω.m, moist soil = 100 Ω.m, dry soil = 1,000 Ω.m and bedrock = 10,000 Ω.m).
Step 2: Surface Layer Materials Applying a thin layer (0.08m - 0.15m) of high resistivity material (such as gravel, blue metal, crushed rock, etc) over the surface of the ground is commonly used to help protect against dangerous touch and step voltages. This is because the surface layer material increases the contact resistance between the soil (i.e. earth) and the feet of a person standing on it, thereby lowering the current flowing through the person in the event of a fault. IEEE Std 80 Table 7 gives typical values for surface layer material resistivity in dry and wet conditions (e.g. 40mm crushed granite = 4,000 Ω.m (dry) and 1,200 Ω.m (wet)). The effective resistance of a person's feet (with respect to earth) when standing on a surface layer is not the same as the surface layer resistance because the layer is not thick enough to have uniform resistivity in all directions. A surface layer derating factor needs to be applied in order to compute the effective foot resistance (with respect to earth) in the presence of a finite thickness of surface layer material. This derating factor can be approximated by an empirical formula as per IEEE Std 80 Equation 27:
Where
is the surface layer
derating factor is the soil resistivity (Ω.m) is the resistivity of the surface layer material (Ω.m) is the thickness of the surface layer (m) This derating factor will be used later in Step 5 when calculating the maximum allowable touch and step voltages.
Step 3: Earthing Grid Resistance A good earthing grid has low resistance (with respect to remote earth) to minimise ground potential rise (GPR) and consequently avoid dangerous touch and step voltages. Calculating the earthing grid resistance usually goes hand in hand with earthing grid design - that is, you design the earthing grid to minimise grid resistance. The earthing grid resistance mainly depends on the area taken up by the earthing grid, the total length of buried earthing conductors and the number of earthing rods / electrodes. IEEE Std 80 offers two alternative options for calculating the earthing grid resistance (with respect to remote earth) - 1) the simplified method (Section 14.2) and 2) the Schwarz equations (Section 14.3), both of which are outlined briefly below. IEEE Std 80 also includes methods for reducing soil resistivity (in Section 14.5) and a treatment for concrete-encased earthing electrodes (in Section 14.6). Simplified Method IEEE Std 80 Equation 52 gives the simplified method as modified by Sverak to include the effect of earthing grid depth:
Where
is the earthing grid resistance with respect to remote earth (Ω)
is the soil resistivitiy (Ω.m) is the total length of buried conductors (m) is the total area occupied by the earthing grid (m2) is the depth of the earthing grid (m)
Schwarz
Equations
The Schwarz equations are a series of equations that are more accurate in modelling the effect of earthing rods / electrodes. The equations are found in IEEE Std 80 Equations 53, 54, 55(footnote) and 56, as follows:
Where
is the earthing grid resistance with respect to remote earth (Ω)
is the earth resistance of the grid conductors (Ω) is the earth resistance of the earthing electrodes (Ω) is the mutual earth resistance between the grid conductors and earthing electrodes (Ω) And the grid, earthing electrode and mutual earth resistances are:
Where
is the soil resistivity (Ω.m)
is the total length of buried grid conductors (m) is radius
for conductors buried at depth metres, or simply
metres and with cross-sectional
for grid conductors on the surface
is the total area covered by the grid conductors (m2) is the length of each earthing electrode (m) is number of earthing electrodes in area is the cross-sectional radius of an earthing electrode (m) and
are constant coefficients depending on the geometry of the grid
The coefficient
(1) For depth
(2) For depth
can be approximated by the following: :
:
(3) For depth
The coefficient
(1) For depth
(2) For depth
(3) For depth
: can be approximated by the following: :
:
: Where in both cases,
is
the length-to-
width ratio of th earthing grid.
Step 4: Maximum Gri Current The maximum
grid current is t
worst case eart
fault current tha would flow via
the earthing gri back to remote earth. To calculate the maximum grid current, you firstly need to calculate the worst case symmetrical earth fault current at the
facility that wou
have a return path through
remote earth (c this
). This
can be found
from the power
systems studies
or from manual calculation. Generally speaking, the
highest relevan
earth fault level will be on the primary side of the largest distribution
transformer (i.e either the
terminals or the
delta windings).
Current Division Factor Not all of the earth fault
current will flow back through
remote earth. A portion of the earth fault
current may ha
local return pat (e.g. local generation) or there could be
alternative retu
paths other tha remote earth (e.g. overhead earth return cables, buried
pipes and cable etc). Therefore
current division factor
must
be applied to account for the
proportion of th fault current flowing back
through remote earth. Computing the
current division factor is a task
that is specific t
each project an
the fault locatio and it may
incorporate som
subjectivity (i.e "engineeing
judgement"). In any case, IEEE Std 80 Section
15.9 has a good discussion on calculating the
current division factor. In the
most conservative case, a current division factor of
can
be applied, meaning that 100% of earth fault current flows back
through remote earth.
The symmetrica grid current calculated by:
Decrement Fact
The symmetrica
current is not th
maximum grid c
because of asym
in short circuits
namely a dc cur
offset. This is ca
by the decreme
factor, which ca
calculated from
Std 80 Equation
Where
is t
decrement facto is the duration of the fault (s)
is the dc time offset constant (see below)
The dc time offs
derived from IE Equation 74:
Where
is th
fault location is the system frequency (Hz)
The maximum g calculated by:
Step 5: Touch
One of the goal
protect people a the event of an
ac electric curre
human body ca
range of 60 to 1
fibrillation and h
duration of an e
the risk of mort
faults are cleare
need to prescrib touch and step lethal shocks.
The maximum t
touch scenarios
from IEEE Std S
50kg and 70kg:
Touch voltage li
difference betw
the potential of
during a fault (d
50kg person:
70kg person:
Step voltage lim
surface potentia distance of 1m
earthed object:
50kg person:
70kg person: Where is the step voltage limit (V) is the surface layer derating factor (as calculated in Step 2) is the soil resistivity (Ω.m) is the maximum fault clearing time (s)
The choice of bo
expected weigh
women are exp
to choose 50kg.
Step 6: Groun
Normally, the p around the site they are at the
(where the faul flow of current
gradients in and
difference betw
ground potentia
a maximum po
potentials aroun fault.
The maximum G
Where is the maximum grid current found earlier in Step 4 (A) is the earthing grid resistance found earlier in Step 3 (Ω)
Step 7: Earth
Now we just ne
and step potent
exceed either o the grid design
However if it do
further analysis
of the maximum 16.5.
Mesh Voltage C
The mesh volta
earthing grid an
Where ::
is
is the maximum grid current found earlier in Step 4 (A) is the geometric spacing factor (see below) is the irregularity factor (see below) is the effective buried length of the grid (see below)
Geometric Sp
The geometric s
Where
is th
is the depth of buried grid conductors (m) is the cross-sectional diameter of a grid conductor (m) is a weighting factor for depth of burial = is a weighting factor for earth electrodes /rods on the corner mesh for grids with earth electrodes along the grid perimeter or corners
for grids with no earth electrodes on the corners or on the perimeter
is a geometric factor (see below)
Geometric Fa
The geometric f
With
for square grids, or otherwise
for square and rectangular grids, or otherwise
for square, rectangular and L-shaped grids, or otherwise Where
is th
is the length of grid conductors on the perimeter (m) is the total area of the grid (m2) and
are the maximum length of the grids in the x and y directions (m)
is the maximum distance between any two points on the grid (m)
Irregularity Fa
The irregularity
Where
is the
Effective Buri
The effective bu
For grids wi
Where
is th
is the total length of earthing electrodes / rods (m)
For grids wi
Where
is th
is the total length of earthing electrodes / rods (m) is the length of each earthing electrode / rod (m) and
are the maximum length of the grids in the x and y directions (m)
Step Voltage Ca
The maximum a
Where ::
is
is the maximum grid current found earlier in Step 4 (A) is the geometric spacing factor (see below) is the irregularity factor (as derived above in the mesh voltage calculation) is the effective buried length of the grid (see below)
Geometric Sp
The geometric s
Where is the depth of buried grid conductors (m) is a geometric factor (as derived above in the mesh voltage calculation)
is th
Effective Buri
The effective bu
Where
is th
is the total length of earthing electrodes / rods (m) What Now?
Now that the m
, and
then the earthin
If not, however
Redesign the earthing grid to lower the grid resistance (e.g. more grid conductors, more earthing electrodes, increasing cross-sectional area of conductors, etc). Once this is done, re-compute the earthing grid resistance (see Step 3) and re-do the touch and step potential calculations.
Limit the total earth fault current or create alternative earth fault return paths
Consider soil treatments to lower the resistivity of the soil
Greater use of high resistivity surface layer materials
Worked Ex
In this example
line and a delta
Step 1: Soi
The soil resistiv
Step 2: Sur
A thin 100mm l
Step 3: Ear
Figure 1. Propose
A rectangular e
Length of 90m and a width of 50m
6 parallel rows and 7 parallel columns
Grid conductors will be 120 mm2 and buried at a depth of 600mm
22 earthing rods will be installed on the corners and perimeter of the grid
Each earthing rod will be 3m long
Using the simpl
Step 4: Ma
Suppose that th
The X/R ratio at
The decrement
Fianlly, the max
kA
Step 5: Tou
Based on the av
The maximum a
V
The maximum a
V
Step 6: Gro
The maximum g
V
The GPR far exc
Step 7: Ear
Mesh Voltage
The component
and
is the number of parallel rows and columns respectively (e.g. 6 and 7)
m
V
m
V.
ntroduction Number of Earthing Electrode and Earthing Resistance depends on the resistivity of soil and time for fault current to pass through (1 sec or 3 sec). If we divide the area for earthing required by the area of one earth plate gives the number of earth pits required. There is no general rule to calculate the exact number of earth pits and size of earthing strip, but discharging of leakage current is certainly dependent on the cross section area of the material so for any equipment the earth strip size is calculated on the current to be carried by that strip. First the leakage current to be carried is calculated and then size of the strip is determined. For most of the electrical equipment like transformer, diesel generator set etc., the general concept is to have 4 number of earth pits. 2 no’s for body earthing with 2 separate strips with the pits shorted and 2 nos for Neutral with 2 separate strips with the pits shorted. The Size of Neutral Earthing Strip should be capable to carry neutral current of that equipment. The Size of Body Earthing should be capable to carry half of neutral Current.
For example for 100kVA transformer, the full load current is around 140A. The strip connected should be capable to carry at least 70A (neutral current) which means a strip of GI 25x3mm should be enough to carry the current and for body a strip of 25×3 will do the needful. Normally we consider the strip size that is generally used as standards. However a strip with lesser size which can carry a current of 35A can be used for body earthing. The reason for using 2 earth pits for each body and neutral and then shorting them is to serve as back up. If one strip gets corroded and cuts the continuity is broken and the other leakage current flows through the other run thery by completing the circuit.
Similarly for panels the no of pits should be 2 nos. The size can be decided on the main incomer circuit breaker. For example if main incomer to breaker is 400A, then body earthing for panel can have a strip size of 25×6 mm which can easily carry 100A.
Number of earth pits is decided by considering the total fault current to be dissipated to the ground in case of fault and the current that can be dissipated by each earth pit. Normally the density of current for GI strip can be roughly 200 amps per square cam. Based on the length and dia of the pipe used the number of earthing pits can be finalized.
1. Calculate numbers of pipe earthing A. Earthing resistance and number of rods for isolated earth pit (without buried earthing strip) The earth resistance of single rod or pipe electrode is calculated as per BS 7430:
R=ρ/2×3.14xL (loge (8xL/d)-1) Where: ρ = Resistivity of soil (Ω meter), L = Length of electrode (meter), D = Diameter of electrode (meter)
Example: Calculate isolated earthing rod resistance. The earthing rod is 4 meter long and having 12.2mm diameter, soil resistivity 500 Ω meter. R=500/ (2×3.14×4) x (Loge (8×4/0.0125)-1) =156.19 Ω. The earth resistance of single rod or pipe electrode is calculated as per IS 3040:
R=100xρ/2×3.14xL (loge(4xL/d)) Where: ρ = Resistivity of soil (Ω meter), L = Length of electrode (cm), D = Diameter of electrode (cm)
Example: Calculate number of CI earthing pipe of 100mm diameter, 3 meter length. System has fault current 50KA for 1 sec and soil resistivity is 72.44 Ω-Meters. Current Density At The Surface of Earth Electrode (As per IS 3043):
Max. allowable current density I = 7.57×1000/(√ρxt) A/m2 Max. allowable current density = 7.57×1000/(√72.44X1) = 889.419 A/m2 Surface area of one 100mm dia. 3 meter Pipe = 2 x 3.14 x r x L = 2 x 3.14 x 0.05 x3 = 0.942 m2 Max. current dissipated by one Earthing Pipe = Current Density x Surface area of electrode Max. current dissipated by one earthing pipe = 889.419x 0.942 = 837.83 A say 838 Amps Number of earthing pipe required = Fault Current / Max.current dissipated by one earthing pipe. Number of earthing pipe required = 50000/838 = 59.66 Say 60 No’s. Total number of earthing pipe required = 60 No’s. Resistance of earthing pipe (isolated) R = 100xρ/2×3.14xLx(loge (4XL/d)) Resistance of earthing pipe (isolated) R = 100×72.44 /2×3.14x300x(loge (4X300/10)) = 7.99 Ω/Pipe Overall resistance of 60 no of earthing pipe = 7.99/60 = 0.133 Ω. Top
B. Earthing resistance and number of rods for isolated earth pit (with buried earthing strip) Resistance of earth strip (R) As per IS 3043:
R=ρ/2×3.14xLx (loge (2xLxL/wt)) Example: Calculate GI strip having width of 12mm , length of 2200 meter buried in ground at depth of 200mm, soil resistivity is 72.44 Ω-meter.
Resistance of earth strip(Re) = 72.44/2×3.14x2200x(loge (2x2200x2200/.2x.012)) = 0.050 Ω From above calculation overall resistance of 60 no of earthing pipes (Rp) = 0.133 Ω. And it connected to bury earthing strip. Here net earthing resistance = (RpxRe)/(Rp+Re) Net eatrthing resistance = (0.133×0.05)/(0.133+0.05) = 0.036 Ω.
C. Total earthing resistance and number of electrode for group (parallel) In cases where a single electrode is not sufficient to provide the desired earth resistance, more than one electrode shall be used. The separation of the electrodes shall be about 4 m. The combined resistance of parallel electrodes is a complex function of several factors, such as the number and configuration of electrode the array. The total resistance of group of electrodes in different configurations as per BS 7430:
Ra=R (1+λa/n) where a=ρ/2X3.14xRxS Where: S = Distance between adjustment rod (meter), λ = Factor given in table below, n = Number of electrodes, ρ = Resistivity of soil (Ω meter), R = Resistance of single rod in isolation (Ω) Factors for parallel electrodes in line (BS 7430) Number of electrodes (n)
Factor (λ)
2
1.0
3
1.66
4
2.15
5
2.54
6
2.87
7
3.15
8
3.39
9
3.61
10
3.8
For electrodes equally spaced around a hollow square, e.g. around the perimeter of a building, the equations given above are used with a value of λ taken from following table.
For three rods placed in an equilateral triangle, or in an L formation, a value of λ = 1.66 may be assumed. Factors for electrodes in a hollow square (BS 7430) Number of electrodes (n)
Factor (λ)
2
2.71
3
4.51
4
5.48
5
6.13
6
6.63
7
7.03
8
7.36
9
7.65
10
7.9
12
8.3
14
8.6
16
8.9
18
9.2
20
9.4
For Hollow square total number of electrodes (N) = (4n-1). The rule of thumb is that rods in parallel should be spaced at least twice their length to utilize the full benefit of the additional rods. If the separation of the electrodes is much larger than their lengths and only a few electrodes are in parallel, then the resultant earth resistance can be calculated using the ordinary equation for resistances in parallel. In practice, the effective earth resistance will usually be higher than calculation.
Typically, a 4 spike array may provide an improvement 2.5 to 3 times. An 8 spike array will typically give an improvement of maybe 5 to 6 times. The Resistance of Original Earthing Rod will be lowered by Total of 40% for Second Rod, 60% for third Rod,66% for forth rod.
Example: Calculate Total Earthing Rod Resistance of 200 Number arranges in Parallel having 4 Meter Space of each and if it connects in Hollow Square arrangement. The Earthing Rod is 4 Meter Long and having 12.2mm Diameter, Soil Resistivity 500 Ω. First Calculate Single Earthing Rod Resistance:
R = 500/ (2×3.14×4) x (Loge (8×4/0.0125)-1) =136.23 Ω.
Now calculate total resistance of earthing rod of 200 number in parallel condition:
a = 500/(2×3.14x136x4) =0.146 Ra (Parallel in Line) =136.23x (1+10×0.146/200) = 1.67 Ω.
If earthing rod is connected in Hollow square than rod in each side of square is 200 = (4n-1) so n = 49 No. Ra (in hollow square) =136.23x (1+9.4×0.146/200) = 1.61 Ω.
(1) Demand factor (in IEC, Max.Utilization factor (Ku)):
The word “demand” itself says the meaning of Demand Factor. The ratio of the maximum coincident demand of a system, or part of a system, to the total connected load of the system.
Demand Factor = Maximum demand / Total connected load
For example, an over sized motor 20 Kw drives a constant 15 Kw load whenever it is ON. The motor demand factor is then 15/20 =0.75= 75 %. Demand Factor is express as a percentage (%) or in a ratio (less than 1).
Demand factor is always < =1. Demand Factor is always change with the time to time or hours to hours of use and it will not constant. The connected load is always known so it will be easy to calculate the maximum demand if the demand factor for a certain supply is known at different time intervals and seasons. The lower the demand factor, the less system capacity required to serve the connected load.
Calculation: (1) A Residence Consumer has 10 No’s Lamp of 400 W but at the same time It is possible that only 9 No’s of Bulbs are used at the same time. Here Total Connected load is 10×40=400 W. Consumer maximum demand is 9×40=360 W. Demand Facto of this Load = 360/400 =0.9 or 90%. (2) One Consumer have 10 lights at 60 Kw each in Kitchen, the load is 60 Kw x 10 = 600 KW. This will be true only if All lights are Turns ON the same time (Demand factor=100% or 1) For this Consumer it is observed that only half of the lights being turned ON at a time so we can say that the demand factor is 0.5 (50%). The estimated load = 600 Kw X 0.5 = 300 Kw.
Use of demand factors: Feeder conductors should have sufficient Ampere Capacity to carry the load. The Ampere Capacity does not always be equal to the total of all loads on connected branch-circuits. This factor must be applied to each individual load, with particular attention to electric motors, which are very rarely operated at full load. As per National Electrical Code (NEC) demand factor may be applied to the total load. The demand factor permits a feeder ampearcity to be less than 100 percent of all the branch-circuit loads connected to it. Demand factor can be applied to calculate the size of the sub-main which is feeding a Sub panel or a fixed load like a motor etc. If the panel have total load of 250 kVA , considering a Demand factor of 0.8, we can size the feeder cable for 250 x 0.8= 200 kVA. Demand factors for buildings typically range between 50 and 80 % of the connected load. In an industrial installation this factor may be estimated on an average at 0.75 for motors.
For incandescent-lighting loads, the factor always equals 1.
Demand Factor For Industrial Load Text Book of Design of Elect. Installation- Jain Electrical Load
Demand Factor
1 No of Motor
1
Up to 10 No’s of Motor
0.75
Up to 20 No’s of Motor
0.65
Up to 30 No’s of Motor
0.6
Up to 40 No’s of Motor
0.5
Up to 50 No’s of Motor
0.4
Demand Factor Text Book of Design of Elect. Installation- Jain Utility
Demand Factor
Office ,School
0.4
Hospital
0.5
Air Port, Bank, Shops,
0.6
Restaurant, Factory,
0.7
Work Shop, Factory (24Hr Shift)
0.8
Arc Furnace
0.9
Compressor
0.5
Hand tools
0.4
Inductance Furnace
0.8
Demand Factor Saudi Electricity Company Distribution Standard Utility
Demand Factor
Residential
0.6
Commercial
0.7
Flats
0.7
Hotel
0.75
Mall
0.7
Restaurant
0.7
Office
0.7
School
0.8
Common Area in building
0.8
Public Facility
0.75
Street Light
0.9
Indoor Parking
0.8
Outdoor Parking
0.9
Park / Garden
0.8
Hospital
0.8
Workshops
0.6
Ware House
0.7
Farms
0.9
Fuel Station
0.7
Factories
0.9
Demand Factor Text Book of Principal of Power System-V.K.Mehta Utility
Demand Factor
Residence Load (<0.25 KW)
1
Residence Load (<0.5 KW)
0.6
Residence Load (>0.1 KW)
0.5
Restaurant
0.7
Theatre
0.6
Hotel
0.5
School
0.55
Small Industry
0.6
Store
0.7
Motor Load (up to 10HP)
0.75
Motor Load (10HP to 20HP)
0.65
Motor Load (20HP to 100HP)
0.55
Motor Load (Above 100HP)
0.50
(2) Diversity factor:
Diversity Factor is ratio of the sum of the individual maximum demands of the various sub circuit of a system to the maximum demand of the whole system.
Diversity Factor = Sum of Individual Maximum Demands / Maximum Demand of the System.
Diversity Factor = Installed load / Running load.
The diversity factor is always >= 1. Diversity Factor is always >1 because sum of individual max. Demands >Max. Demand. In other terms, Diversity Factor (0 to 100%) is a fraction of Total Load that is particular item contributed to peak demand. 70% diversity means that the device operates at its nominal or maximum load level 70% of the time that it is connected and turned ON.
It is expressed as a percentage (%) or a ratio more than 1. If we use diversity value in % than it should be multiply with Load and if we use in numerical value (>1) than it should be divided with Load. Diversity occurs in an operating system because all loads connected to the System are not operating simultaneously or are not simultaneously operating at their maximum rating. The diversity factor shows that the whole electrical load does not equal the sum of its parts due to this time Interdependence (i.e. diverseness). In general terms we can say that diversity factor refers to the percent of time available that a machine. 70% diversity means that the device operates at its nominal or maximum load level 70% of the time that it is connected and turned ON. Consider two Feeders with the same maximum demand but that occur at different intervals of time. When supplied by the same feeder, the demand on such is less the sum of the two demands. In electrical design, this condition is known as diversity. Diversity factor is an extended version of demand factor. It deals with maximum demand of different units at a time/Maximum demand of the entire system. Greater the diversity factor, lesser is the cost of generation of power. Many designers prefer to use unity as the diversity factor in calculations for planning conservatism because of plant load growth uncertainties. Local experience can justify using a diversity factor larger than unity, and smaller service entrance conductors and transformer requirements chosen accordingly. The diversity factor for all other installations will be different, and would be based upon a local evaluation of the loads to be applied at different moments in time. Assuming it to be 1.0 may, on some occasions, result in a supply feeder and equipment rating that is rather larger than the local installation warrants, and an overinvestment in cable and equipment to handle the rated load current. It is better to evaluate the pattern of usage of the loads and calculate an acceptable diversity factor for each particular case.
Calculation: One Main Feeder have two Sub feeder (Sub Feeder A and Sub Feeder B), Sub Feeder-A have demand at a time is 35 KW and Sub Feeder-B have demands at a time is 42 KW, but the maximum demand of Main Feeder is 70 KW. Total individual Maximum Demand =35+42=77 KW. Maximum Demand of whole System=70 KW So Diversity factor of The System= 77/70 =1.1 Diversity factor can shoot up above 1.
Use of diversity factor: The Diversity Factor is applied to each group of loads (e.g. being supplied from a distribution or subdistribution board). Diversity factor is commonly used for a complete a coordination study for a system. This diversity factor is used to estimate the load of a particular node in the system. Diversity factor can be used to estimate the total load required for a facility or to size the Transformer Diversity factors have been developed for main feeders supplying a number of feeders, and typically 1.2 to 1.3 for Residence Consumer and 1.1 to 1.2 for Commercial Load. 1.50 to 2.00 for power and lighting loads. Note: Reciprocal of the above ratio (will be more than 1) also is used in some other countries. Diversity factor is mostly used for distribution feeder size and transformer as well as to determine the maximum peak load and diversity factor is always based on knowing the process. You have to understand what will be on or off at a given time for different buildings and this will size the feeder. Note for typical buildings diversity factor is always one. You have to estimate or have a data records to create 24 hours load graph and you can determine the maximum demand load for node then you can easily determine the feeder and transformer size. The diversity factor of a feeder would be the sum of the maximum demands of the individual consumers divided by the maximum demand of the feeder. In the same manner, it is possible to compute the diversity factor on a substation, a transmission line or a whole utility system. The residential load has the highest diversity factor. Industrial loads have low diversity factors usually of 1.4, street light practically unity and other loads vary between these limits.
Diversity Factor in distribution Network (Standard Handbook for Electrical Engineers” by Fink and Beaty) Elements of System
Residential
Commercial
General Power
Large Industrial
Between individual users
2.00
1.46
1.45
Between transformers
1.30
1.30
1.35
1.05
Between feeders
1.15
1.15
1.15
1.05
Between substations
1.10
1.10
1.10
1.10
From users to transformers
2.00
1.46
1.44
From users to feeder
2.60
1.90
1.95
1.15
From users to substation
3.00
2.18
2.24
1.32
From users to generating station
3.29
2.40
2.46
1.45
Diversity Factor for Distribution Switchboards Number of circuits
Diversity Factor in % (ks)
Assemblies entirely tested 2 and 3
90%
4 and 5
80%
6 to 9
70%
10 and more
60%
Assemblies partially tested in every case choose
100%
Diversity Factor as per IEC 60439 Circuits Function
Diversity Factor in % (ks)
Lighting
90%
Heating and air conditioning
80%
Socket-outlets
70%
Lifts and catering hoist
For the most powerful motor
100%
For the second most powerful motor
75%
For all motors
80%
Diversity Factor for Apartment block Apartment
Diversity Factor in % (ks)
2 To 4
1
5To 19
0.78
10To 14
0.63
15To 19
0.53
20To 24
0.49
25To 29
0.46
30 To 34
0.44
35 To 39
0.42
40To 40
0.41
50 To Above
0.40
Diversity Factor Text Book of Principal of Power System-V.K.Mehta Area
Residence Ltg
Commercial Ltg
Ind. Ltg
Between Consumer
3
1.5
1.5
Between Transformer
1.3
1.3
1.3
Between Feeder
1.2
1.2
1.2
Between S.S
1.1
1.1
1.1
(3) Load factor:
The ratio of the Actual Load of equipment to Full load of equipment.
Load Factor=Actual Load / Full Load
It is the ratio of actual kilowatt-Hours used in a given period, divided by the total possible kilowatt -hours that could have been used in the same period at the peak KW level. Load Factor = ( energy (kWh per month) ) / ( peak demand (kW) x hours/month ) In other terms Load factor is defined as the ratio of Average load to maximum demand during a given period. Load Factor= Average Load / Maximum Demand during given Time Period
The Load factor is always <=1. Load Factor is always less than 1 because maximum demand is always more than average demand. Load Factor can be calculated for a single day, for a month or for a year. Load factor in other terms of efficiency. It is used for determining the overall cost per unit generated. Higher the load factor is GOOD and it will more Output of Plan, lesser the cost per unit which means an electricity generator can sell more electricity at a higher spark spread, Fixed costs are spread over more kWh of output. A power plant may be highly efficient at High load factors. Low load factor is a BED. A low load factor will use electricity inefficiently relative to what we could be if we were controlling our peak demand. A power plant may be less efficient at low load factors. For almost constant loads, the load factor is close to unity. For Varying Load Factor is closed Zero. Load Factor is a measure of the effective utilization of the load and distribution equipment, i.e. higher load factor means better utilization of the transformer, line or cable. A high load factor means power usage is relatively constant. Low load factor shows that occasionally a high demand is set. To service that peak, capacity is sitting idle for long periods, thereby imposing higher costs on the system. Electrical rates are designed so that customers with high load factor are charged less overall per kWh. Sometimes utility companies will encourage industrial customers to improve their load factors. Load factor is term that does not appear on your utility bill, but does affect electricity costs. Load factor indicates how efficiently the customer is using peak demand. Calculation: Motor of 20 hp drives a constant 15 hp load whenever it is on. The motor load factor is then 15/20 = 75%.
Demand Factor & Load Factor
Introduction to Power Requirement for Building – J. Paul Guyer, Utility
Demand Factor (%)
Load Factor (%)
Communications – buildings
60-65
70-75
Telephone exchange building
55-70
20-25
Air passenger terminal building
65-80
28-32
Aircraft fire and rescue station
25-35
13-17
Aircraft line operations building
65-80
24-28
Academic instruction building
40-60
22-26
Applied instruction building
35-65
24-28
Chemistry and Toxicology Laboratory
70-80
22-28
Materials Laboratory
30-35
27-32
Physics Laboratory
70-80
22-28
Electrical and electronics laboratory
20-30
3-7
Cold storage warehouse
70-75
20-25
General warehouse
75-80
23-28
Controlled humidity warehouse
60-65
33-38
Hazardous/flammable storehouse
75-80
20-25
Disposal, salvage, scrap building
35-40
25-20
Hospital
38-42
45-50
Laboratory
32-37
20-25
K-6 schools
75-80
10-15
7-12 schools
65-70
12-17
Churches
65-70
5-25
Post Office
75-80
20-25
Retail store
65-70
25-32
Bank
75-80
20-25
Supermarket
55-60
25-30
Restaurant
45-75
15-25
Auto repair shop
40-60
15-20
Hobby shop, art/crafts
30-40
25-30
Bowling alley
70-75
10-15
Gymnasium
70-75
20-45
Skating rink
70-75
10-15
Indoor swimming pool
55-60
25-50
Theatres
45-55
8-13
Library
75-80
30-35
Golf clubhouse
75-80
15-20
Museum
75-80
30-35
(4) Coincidence factor (in IEC, Factor of simultaneity (ks)):
The reciprocal of diversity factor is coincidence factor The coincidence factor is the ratio of the maximum demand of a system, or part under consideration, to the sum of the individual maximum demands of the subdivisions
Coincidence factor = Maximum demand / Sum of individual maximum demands
Expressed as a percentage (%) or a ratio less than 1.
The Confidence Factor is always <=1. Usually Confidence Factor will decrease as the number of connected customer’s increases. The factor ks is applied to each group of loads (e.g. distribution or sub-distribution board). The determination of these factors is the responsibility of the designer, since it requires a detailed knowledge of the installation and the conditions in which the individual circuits are to be exploited. For this reason, it is not possible to give precise values for general application.
(5) Maximum demand:
The maximum demand of an installation is the maximum rate of consumption expressed in amperes, kW or kVA. It is generally taken as the average rate of consumption over a period of time. Example the 15-minute maximum kW demand for the week was 150 kW. Maximum demand does not include motor starting currents or other transient effects. Fault currents and overload currents are also excluded. Maximum demand in KW is relevant only for metering/tariff purposes. Maximum demand (often referred to as MD) is the largest current normally carried by circuits, switches and protective devices. It does not include the levels of current flowing under overload or short circuit conditions. Maximum Demand is a greatest of all demands that occur during a specific time The major disadvantage of allocating load using the diversity factors is that most utilities will not have a table of diversity factors and sometime it is not viable to determine accurate Diversity Factor. In this situation Maximum Demand is very helpful to calculate size of Feeder or TC. The kVA rating of all distribution transformers is always known for a feeder. The metered readings can be taken to each transformer based upon the transformer rating. An “allocation factor” (AF) can be calculate. Allocation Factor= Metered Demand (KVA) / Total KVA. Equipment Demand= AF x Total KVA of Equipments Calculation: Actual Loading or Size of TC-1 and TC-2.
Total Load on TC-1 =10+11+12+08= 41 KW. Maximum Diversity Demand of TC-1= 41 / 1.1 = 37.3 KW. Total Load on TC-2 =4+3+12+02= 21 KW. Maximum Diversity Demand of TC-2= 21 / 1.2 = 17.5 KW. Total Load= 37.3 + 17.5 =54.8 KW. Allocating Factor (AF)= M.D / Total Load Allocating Factor (AF)= 0.27.
Actual Load on TC-1=0.27×37.3 = 1.20 KW. Actual Load on TC-2=0.27×17.5 = 4.8 KW. Assessment of maximum demand is very easy for Resistive Load , For example, the maximum demand of a 240 V single-phase 8 kW shower heater can be calculated by dividing the power (8 kW) by the voltage (240 V) to give a current of 33.3 A. This calculation assumes a power factor of unity, which is a reasonable assumption for such a purely resistive load. Lighting circuits pose a special problem when determining MD. Discharge lamps are particularly difficult to assess, and current cannot be calculated simply by dividing lamp power by supply voltage. The reasons for this are Control gear losses result in additional current, the power factor is usually less than unity so current is greater, and Chokes and other control gear usually distort the waveform of the current so that it contains harmonics which are additional to the fundamental supply current. So long as the power factor of a discharge lighting circuit is not less than 0.85, the current demand for the circuit can be calculated from: current (A) = (lamp power x 1.8) / supply voltage (V) For example, the steady state current demand of a 240 V circuit supplying ten 65 W fluorescent lamps would be: I = 10X65X1.8A / 240 = 4.88A Switches for circuits feeding discharge lamps must be rated at twice the current they are required to carry, unless they have been specially constructed to withstand the severe arcing resulting from the switching of such inductive and capacitive loads.
Where to use Demand and Diversity factor:
There is generally confusion between Demand factor and Diversity factor. Demand factors should be ideally applied to individual loads and diversity factor to a group of loads. When you talk about ‘diversity’, there are naturally more than one or many loads involved. Demand factor can be applied to calculate the size of the sub-main, which is feeding a Sub panel or a fixed load like a motor etc, individual Load. Demand factors are more conservative and are used by NEC for service and feeder sizing. If the Sub panel have total load is 250 kVA , considering a Demand factor of 0.8, we can size the feeder cable for 250 x 0.8= 200 kVA. The Diversity Factor is applied to each group of loads (e.g. being supplied from a distribution or subdistribution board), size the Transformer. Demand factors and diversity factors are used in design. For example, the sum of the connected loads supplied by a feeder is multiplied by the demand factor to determine the load for which the feeder must be sized. This load is termed the maximum demand of the feeder. The sum of the maximum demand loads for a number of sub feeders divided by the diversity factor for the sub feeders will give the maximum demand load to be supplied by the feeder from which the sub feeders are derived.
Calculate Size of Electrical Switchgear by Demand & Diversity Factor:
The estimated electrical demand for all feeders served directly from the service entrance is calculated by multiplying the total connected loads by their demand factors and then adding all of these together. This sum is divided by the diversity factor (frequently assumed to be unity) to calculate the service entrance demand which is used to determine ampacity requirements for the service entrance conductors. When used Diversity and Demand Factor in an electrical design it should be applied as follows, the sum of the connected loads supplied by a feeder-circuit can be multiplied by the demand factor to determine the load used to size the components of the system. The sum of the maximum demand loads for two or more feeders is divided by the diversity factor for the feeders to derive the maximum demand load. Example-1: Calculate Size of Transformer having following details:
Feeder Breaker-1 Demand Load= Feeder Breaker-1xDemand Factor. Feeder Breaker-1 Demand Load=2000×0.7=1400 KVA Feeder Breaker-2 Demand Load= Feeder Breaker-2xDemand Factor. Feeder Breaker-2 Demand Load=1500×0.6=900 KVA Feeder Breaker-3 Demand Load= Feeder Breaker-3xDemand Factor. Feeder Breaker-2 Demand Load=1000×0.5=500 KVA Total Feeder Breaker Demand=1400+900+500=2800KVA Transformer Demand Load= Total Feeder Breaker Demand / Diversity Factor. Transformer Demand Load=2800/1.1 =2545 KVA If we Calculated Total Load on Transformer without any Demand & Diversity=2000+1500+1000=4500KVA. But after Calculating Demand & Diversity Factor Total Load on Transformer =2545 KVA Example-2: Calculate Size of Main Feeder of Main Transformer having following Details:
Sum of Maximum Demand of Customer on TC-1 =10 KWx0.65 =6.5 KW Sum of Maximum Demand of Customer on TC-2 =20 KWx0.75 =15 KW Sum of Maximum Demand of Customer on TC-3 =30 KWx0.65 =19.5 KW As Diversity of Consumer Connected on TC-1 is 1.5 so, Maximum Demand on TC-1 =6.5 KW/1.5 = 4 KW. As Diversity of Consumer Connected on TC-2 is 1.1 so, Maximum Demand on TC-2 =15 KW/1.1 = 14 KW As Diversity of Consumer Connected on TC-3 is 1.5 so, Maximum Demand on TC-3 =19.5 KW/1.5 = 13 KW.
Individual Maximum Demand on Main Transformer =04+14+13= 31 KW. Maximum Demand on Main Feeder =04+14+13 / 1.3 =24 KW
Significance of Load Factor and Diversity Factor
Load factor and diversity factor play an important part in the cost of the supply of electrical energy. Higher the values of load factor and diversity factors, lower will be the overall cost per unit generated. The capital cost of the power station depends upon the capacity of the power station. Lower the maximum demand of the power station, the lower is the capacity required and therefore lower is the capital cost of the plant. With a given number of consumers the higher the diversity factor of their loads, the smaller will be the capacity of the plant required and consequently the fixed charges due to capital investment will be much reduced. Similarly higher load factor means more average load or more number of units generated for a given maximum demand and therefore overall cost per unit of electrical energy generated is reduced due to distribution of standing charges which are proportional to maximum demand and independent of number of units generated. Thus the suppliers should always try to improve the load factor as well as diversity factor by inducing the consumers to use the electrical energy during off peak hours and they may be charged at lower rates for such schemes.
Main-Tie-Main.
Load Configuration. Both Bus#1 & Bus#2 are supplying normal loads that mean interruption for PT1 or PT2 is accepted for fault located between M2-PT2 and main supply. No critical load (instantaneous interruption is not allowed) connected on Bus #1 and Bus #2. It shall be supplied from UPS. Load on Bus #1 has a standby load on Bus#2 or vice versa, so if the bus #2 fail, load on bus #1 is operated. Basic Operation. This diagram may be useful for our discussion. The basic M1-T-M2 configuration is shown. During normal operation M1 & M2 breakers are closed and tie breaker T is opened. Supply coming from PT-1 and PT-2. This drawing indicate when M2 open, T and M1 CBs are closed (abnormal condition). This condition is done for maintenance purpose for equipment located between M2 to upstream (main source). Load transfer from bus #2 to bus #1 can be carried out without interruption done by ATS scheme. If fault located on bus #2 to tie breaker T or bus #1 to tie breaker T, load transfer is prohibited by ATS scheme. But for fault located from M2 to upstream load transfer is allowed with deenergizing bus #2 loads first, then tie-breaker T closed by ATS scheme. Loads may be in service after this tranfer, if the loads is set in auto position. We cannot maintain the supply on fault bus (e.g. bus #2 or bus #1) before correction is made. A redundant bus tie or switch isolator acting as maintenance bypass to ATS operation. Based on discussion above, I do not know, where we have to install those equipment to maintain supply for fault on respectively bus. Normally Closed Tie Breaker Operation. It is possible to operate tie breaker in closed position, but we have to consider a short-circuit level on that bus. By calculation (Short-circuit study), a fault on bus #1 or bus #2 the magnitude become double. So, we
have to ensure the equipment s.c. rating (buse, breakers, feeder loads, feeder breakers, and etc) meet the requirements for tie-in in closed position. Note: Temporary closing three breakers for maintenance purpose is allowed within 3 cycles to 1 (one) second is accepted. Relay application. 1). Bus differential for bus#1 and bus #2 may be applied (we apply on 4.16 kV systems). 2). Directional relay may be applied on incoming breaker M1 & M2 if the NC for tie breaker T is applied. 3). Restrictive earth fault is applied for transformer with low resistance grounding. 4). Please consider to provide better coordination for instantaneous relay between incoming breaker and load breakers as well as ground fault protection. 5). Syncheck relay is required for synchronising bus # & bus #2 before closing tie breaker T. We provide permissive closed for ATS schecme. ATS can only be operated if the upstream system is in synchronising condition (Generating buses are in remote but located closed to each other). Conclusions. 1). We cannot maintain load on bus faulted before repairing is made. 2). I do not know the location for instaling redudant bus tie breaker or isolator to prevent faulted bus total failure. 3). Comprehensive study shall be caried out to operate tie breaker in NC. Especially in selecting electrical equipment and relay coordination. 4). Pay more attention on safety aspect in establishing the ATS scheme. http://www.sayedsaad.com/substation/index_SF6circuitbreaker.html.
Basic electrical design of a PLC panel (Wiring diagrams).
Building the PLC panel It is uncommon for engineers to build their own PLC panel designs (but not impossible of course). For example, once the electrical designs are complete, they must be built by an electrician. Therefore, it is your responsibility to effectively communicate your design intentions to the electricians through drawings. In some factories, the electricians also enter the ladder logic and do debugging. This article discusses the design issues in implementation that must be considered by the designer.
Electrical wiring diagrams of a PLC panel In an industrial setting a PLC is not simply “plugged into a wall socket”. The electrical design for each machine must include at least the following components. 1. 2.
Transformers – to step down AC supply voltages to lower levels Power contacts – to manually enable/disable power to the machine with e-stop buttons
3. Terminals – to connect devices 4. Fuses or circuit breakers – will cause power to fail if too much current is drawn 5. Grounding – to provide a path for current to flow when there is an electrical fault 6. Enclosure – to protect the equipment, and users from accidental contact A control system of a PLC panel will normally use AC and DC power at different voltage levels. Control cabinets are often supplied with single phase AC at 220/440/550V, or two phase AC at 220/440V AC, or three phase AC at 330/550V.
This power must be dropped down to a lower voltage level for the controls and DC power supplies. 110Vac is common in North America, and 220 V AC Is common in Europe and the Commonwealth countries. It is also common for a control cabinet to supply a higher voltage to other equipment, such as motors.
Motor controller example An example of a wiring diagram for a motor controller is shown in Figure 1. Note that symbols are discussed in detail later). Dashed lines indicate a single purchased component. This system uses 3 phase AC power (L1, L2 and L3) connected to the terminals. The three phases are then connected to a power interrupter. Next, all three phases are supplied to a motor starter that contains three contacts, M, and three thermal overload relays (breakers).
Figure 1 – A Motor Controller Schematic
The contacts, M, will be controlled by the coil, M. The output of the motor starter goes to a three phase AC motor. Power is supplied by connecting a step down transformer to the control electronics by connecting to phases L2 and L3. The lower voltage is then used to supply power to the left and right rails of the ladder below. The neutral rail is also grounded. The logic consists of two push buttons:
Start push button is normally open, so that if something fails the motor cannot be started. Stop push button is normally closed, so that if a wire or connection fails the system halts safely.
The system controls the motor starter coil M, and uses a spare contact on the starter, M, to seal in the motor starter. The diagram also shows numbering for the wires in the device. This is essential for industrial control systems that may contain hundreds or thousands of wires. These numbering schemes are often particular to each facility, but there are tools to help make wire labels that will appear in the final controls cabinet.
Figure 2 – A Physical Layout for the Control Cabinet Once the electrical design is complete, a layout for the controls cabinet is developed, as shown in Figure 2. The physical dimensions of the devices must be considered, and adequate space is needed to run wires between components. In the cabinet the AC power would enter at the terminal block, and be connected to the main breaker.
It would then be connected to the contactors and overload relays that constitute the motor starter. Two of the phases are also connected to the transformer to power the logic. The start and stop buttons are at the left of the box (note: normally these are mounted elsewhere, and a separate layout drawing would be needed). The final layout in the cabinet might look like the one shown in Figure 1.
Figure 3 – Final PLC Panel Wiring
When being built the system will follow certain standards that may be company policy, or legal requirements. This often includes items such as;
Hold downs – the will secure the wire so they don’t move Labels – wire labels help troubleshooting Strain reliefs – these will hold the wire so that it will not be pulled out of screw terminals Grounding – grounding wires may be needed on each metal piece for safety
A photograph of an industrial controls cabinet is shown in Figure 4:
Figure 4 – An industrial control cabinet with wire runs, terminal strip, buttons on PLC panel front, etc.
When including a PLC in the ladder diagram still remains. But, it does tend to become more complex. Figure 5 below shows a schematic diagram for a PLC based motor control system, similar to the previous motor control example. This figure shows the E-stop wired to cutoff power to all of the devices in the circuit, including the PLC. All critical safety functions should be hardwired this way.
Figure 5 – An Electrical Schematic with a PLC.
Measurement of N–PE loop resistance in TN, TT and IT systems.
N–PE loop resistance Up-to-date test instruments, with built in modern electronics, can measure loop resistance even between the neutral N and protection PE conductors in spite of
possible high currents in the neutral conductor. The current, which is driven by phase voltages through different linear and non-linear loads, causes voltage drops of extremely irregular (non sine wave) shape. The voltage drops interfere with the test voltage and thereby disturb the measurement. Internal test voltage (approx. 40V, AC, <15 mA) is used, as there is no mains voltage between neutral and protection conductors. Important advantage of this measurement against Fault Loop test (L–PE) is, that the RCD does definitly not trip during the measurement, this is due to the low test current.
Used test instrument Eurotest 61557 uses special (patented) measurement principle to filter the test signal and therefore assures correct measurement results.
What can we deduce from the measurement? The following conclusions can be reached on the basis of the measurement result: 1. 2. 3.
Type of used protection conductor connection (TN, TT or IT-system) Earth Resistance value for TT-system In case of TT or TN-system, the result is quite similar to the Fault Loop
Resistance value, this is why the test instrument can also calculate the Fault Loop Prospective Short circuit current.
Generally about the measurement principle As there is no mains voltage between the N and PE terminals which could be used as a test voltage the instrument must generate an internal one. This voltage may be either DC or AC. Used instrument uses AC test voltage, measurement is done following the U-I method according to the figure below.
Fi gure 1 – N–PE loop resistance measurement principle
Result = Ut / It = RN-PE Where:
Ut – Test voltage measured by the V-meter. It – Test current measured by the A-meter. RN-PE – Resistance of N-PE loop.
1. Measurement of N–PE loop resistance in TNsystem The test instrument measures the resistance of the neutral and the protection conductors from the power transformer to the measurement site (the loop is marked with a bold line on upper figure). The test result in this case is quite low (maximum a couple of ohms), showing that a TN-system is involved.
Figure 2 – Resistance measurement between neutral and protection conductor in TN-system
Result 1 = RN + RPE Result 2 = Ipsc = 230V × 1,06 / (RN + RPE) Where:
RN – Resistance of neutral conductor (marked with bold line) RPE – Resistance of protection conductor (marked with bold dotted line) Ipsc – Prospective short circuit current of fault loop
2. Measurement of N–PE loop resistance in TTsystem The test instrument measures the resistance in the following loop – Neutral conductor from power transformer to measurement site (mains outlet), protection conductor from the mains outlet to earth electrode and then back to the power transformer via soil and the transformer’s earthing system (the loop is marked with a bold line on the figure 3 below). The test result in this case is quite high (in excess of ten ohms), showing that a TTsystem is involved.
Figure 3 – Resistance measurement between the neutral and the protection conductor in a TT-system
Result 1 = RN + RPE + RE + RO Result 2 = Ipsc = 230V × 1,06 / (RN + RPE + RE + RO) As it could be presumed, that resistance R E is much higher than the sum of all other resistances, the following can be noted:
Result 1 ≈ RE Result 2 = Ipsc = 230V × 1,06 / RE Where:
RN – Resistance of neutral conductor from power transformer to measurement site (mains outlet) RPE – Resistance of protection conductor from the mains outlet to earthing electrode RE – Earth resistance of protection earth electrode RO – Earth resistance of transformer’s earthing system Ipsc – Prospective short circuit fault loop current
3. Measurement of N–PE loop resistance in ITsystem As can be seen from the Figure 4, there is no hard wired connection between the neutral and protection conductor in an IT-system. The test result is therefore very high (it can even be out of display range), showing that an IT-system is involved.
Figure 4 – Resistance measurement between neutral and protection conductor in IT-system Attention! – A high test result in itself is not sufficient evidence that an IT-system is involved (it could be just an interrupted protection conductor in a TN or TT-system).
Detailed calculation of currents and power according to the type of load.
Calculation of currents & power Currents and power analysis are key factors in any design or redesign of an installation they will enable the source(s) to be sized according to the purpose of the installation, the intended use of the circuits and the receivers to be supplied. The current consumed Ia corresponds to the nominal current consumed by a receiver independently of the utilisation factor and the coincidence factor, but taking into account the aspects of efficiency (η factor), displacement factor or phase shift (cos φ) for motors or other inductive or capacitive loads. For non-linear (or distorting) loads, the quadratic sum of the fundamental current and the harmonic currents must be calculated in order to obtain the actual rms current.
Let’s break the calculation of the power into few parts, so we can easily follow:
1. Purely resistive load The current consumed Ia of a purely resistive load is calculated by simply applying the formulas. For single phase:
and for the three-phase:
But beware, very few loads are totally resistive. Incandescent lighting is losing ground to solutions that offer higher performance levels, but which are on the other hand less “pure” from an electrical viewpoint.
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2. Non-distorting load that is not purely resistive The nominal power (Pn) of a motor corresponds to the mechanical power available on its shaft. The actual power consumed (Pa) corresponds to the active power carried by the line. This is dependent on the efficiency of the motor:
The current consumed (Ia) is given by the following formulae. For single-phase:
and for the three-phase:
Where:
Ia – rms current consumed (in A) Pn – nominal power (in W; this is the useful power)
U – voltage between phases in three-phase, and between phase and neutral in singlephase (in V) η – efficiency cosφ – displacement factor.
3. Calculation of the current consumed by several receivers The example described below shows that the current and power calculations must be carried out in accordance with precise mathematical rules in order to clearly distinguish the different components.
Example of asynchronous motors A group of circuits consists of two three-phase asynchronous motors M1 and M2 connected to the same panel (mains supply: 400V AC – 50 hz). The nominal power of the motors are respectively: Pn1 = 22 kW and Pn2 = 37 kW. The displacement factors are cosφ1 = 0.92 for M1 and cosφ2 = 0.72 for M2 the efficiencies are η1 = 0.91 and η2 = 0.93 respectively.
Calculation of the power consumed:
The reactive power can in this case be calculated by determining the value of tanφ from cosφ. the relationship with the tangent is given by the formula:
Calculation of the reactive power:
Calculation of the apparent power:
Calculation of the total current consumption for M1, M2, M1 + M2 and the corresponding power factor:
The active power (in W) and the reactive power (in VAr) can be added together algebraically, while the apparent power and currents can only be added together geometrically.
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Presentation of the results All power analyses must show, as in the table below, at least for each group of:
Active power circuits which corresponds (to the nearest efficiency) to the energy supplied, Reactive power so that the compensation devices (capacitors) can be sized, Apparent power so that the power of the source can be determined and Current consumed so that the trunking and protection devices can be calculated. M1 M2 M1 + M2 (Total t)
Active power: P [kW]
Pa1 = 24.18
Pa2 = 39.78
Pt = 63.96
Reactive power: Q [kVAR]
Q1 = 10.30
Q2 = 38.35
Qt = 48.65
Apparent power: S [kVA]
S1 = 26.28
S2 = 55.26
St = 80.36
Current consumed: Ia [A]
Ia1 = 38
Ia2 = 80
Iat = 116
0.92
0.72
0.80
cosφ
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4. Overloads on conductors according to the total harmonic distortion The current circulating in each phase is equal to the quadratic sum of the fundamental current (referred to as 1st harmonic order) and all the harmonic currents (of the following orders):
The THDi (Total Harmonic Distortion) expresses the ratio between the share of all the harmonic currents and the total current as a percentage.
I1 being the rms value of the fundamental and in In the rms value of the nth order harmonic. The principle is to apply a current reduction factor that can be calculated based on the THDi. For a permissible THDi value of 33%, the current must thus in theory be reduced in each phase by a factor K:
If the factor is not applied, the current will then be increased by:
This remains acceptable and explains why the standard does not recommend any derating or oversizing of cross-sections up to 33% THDi. Above 33%. the standard recommends an increase in the current IB which results in necessary oversizing of the neutral conductor.
Reduction of the current or oversizing of multi-core cables may also be necessary for the phase conductors. It should be noted that the standard recommends a reduction factor of 0.84. which in fact corresponds to a pessimistic THDi of 65%.
Related to the neutral conductor, it is considered that if all the harmonics are 3rd order and its multiples, they will be added together and the current due to the harmonics in the neutral will then be IN = 3 × Iph, which can be expressed using an equivalent notation, THDn = 3 THDi. Devices whose load is said to be non-linear do not consume a current that is a reflection of the voltage applied. This leads to unnecessary power consumption: the distorting power that generates an additional current, the consequences of which must not be overlooked. But this current is never expressed directly because it involves a fairly complex mathematical calculation, the fourier transform, to ascertain its relative overall part (THDi: total harmonic distortion) or the value order by order: ih2, ih3, ih4, ih5,..ihn.
With no precise measurements, it is difficult to know exactly the current level that corresponds to each harmonic order. It is therefore preferable to simply increase the cross-section of the neutral conductor as a precaution, since it is known that the main 3rd order harmonics and their multiples are added together in the neutral. and to adapt the protection of this conductor. Standard IEC 60364 indicates the increasing factors to be applied to the cross-section of the neutral conductor according the percentage of 3rd order harmonics.
In principle, the neutral must be the same cross-section as the phase conductor in all single-phase circuits. In three-phase circuits with a cross-section greater than 16 mm2 [25 mm2 aluminium]. The cross-section of the neutral can be reduced to crosssection/2. However this reduction is not permitted if:
The loads are not virtually balanced The total 3rd order harmonic currents are greater than 15%
If this total is greater than 33%, the cross-section of the active conductors of multi-core cables is chosen by increasing the current In by a fixed multiplication factor of 1.65. For single-core cables, only the cross-section of the neutral is increased. In practice, the increase of the current Ia in the neutral is compensated by an increase of its cross-section. When the neutral is loaded, a reduction factor of 0.86 is applied to the permissible current of cables with 3 or 1 conductors.
The current reduction factor KN or rather its inverse which will be used to oversize the neutral conductor will then be:
With a total 3rd order harmonic distortion of 65%, the current of the phase conductors must be increased by 119% and that in the neutral conductor by 163%. If the THDi were to reach 100%, 1/KN would theoretically reach 2.12. This value would be impossible to reach as it would mean that the harmonic had totally replaced the fundamental. The theoretical overcurrent limit for the neutral in relation to the phases is:
These calculations demonstrate that the harmonic currents above all must not be ignored both in terms of “hidden” power consumption and in terms of sizing the conductors which may be overloaded. The relative complexity of the calculations leads to the use of generic derating values which normally cover most cases, just as software is used elsewhere.
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Example of following the standards for defining a protection device with neutral overloaded by harmonics For a 3P+N circuit, intended for 170 A, with TNS system, with total 3rd order harmonic distortion of more than 33%. When sizing the phase cables, the reduction factor of 0.84 (loaded neutral, see above) must be included. This requires a minimum cross-section of 70 mm2 per phase. The neutral conductor must be sized to withstand a current of 1.45 × 170 A = 247 A, i.e. a cross-section of 95 mm2. A circuit breaker must therefore be chosen that is capable of withstanding the current that may cross the neutral:
In device ≥ IB neutral ⇒ In = 250 A But the device must be set according to the current that may flow in the phases:
Ir ≥ IB phases ⇒ Ir ≥ 170 A (and < 206 A, limit of the cable) A 250 A unprotected interrupted neutral circuit breaker, set to 0.7 is therefore suitable for this application.
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5. Distorting load that is not purely resistive The current consumed (Ia) is given by the following formulae:
where:
Ia – rms current consumed (in A) Pn – nominal power (in W; this is the useful power) U – voltage between phases in three-phase, and between phase and neutral in singlephase (in V) η – efficiency PF – power factor
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Example of a fluorescent luminaire and electronic ballast The nominal active power consumed by the luminaire is 9 W, and the measured apparent power is 16 VA. The measured displacement factor is cosφ = 0.845 and the power factor PF = 0.56. The measured current consumed Ia is 0.07 A. As cosφ and the power factor are different, it is not possible to calculate the value of the tanφ or that of the reactive power Q (VAR) for the receiver in question. The measured cosφ and power Q which would be calculated can only be calculated for the reactive power part connected with the sinusoidal component of the signal, in fact the current of the fundamental at 50Hz: 0.045 A measured in this case.
The powers relative to this linear and sinusoidal part of the load can be calculated as follows
S = 230 × 0.045 = 10.3 VA P = S × cosφ = 10.3 × 0.85 = 8.7 W Q = 5.5 VAR which is confirmed by the calculation of the power triangles Q2 = P2 – S2 or by the tanφ: Q = P × tanφ = 8.7 × 0.63 = 5.5 VAR Therefore not all the apparent power consumed is linear as there is a significant difference between the measured total apparent power S (16 VA) and the calculated theoretical sinusoidal power (10.3 VA).
It can also be seen that the sinusoidal active power of the device 8.7 W is very similar to the measured total active power 9 W. It can therefore be deduced that a large part of power S (16 – 10.3 = 5.7 VA) is consumed without producing any active power. The fluorescent luminaire and electronic ballast in the example consumes unproductive power in the form of harmonic currents. The total harmonic distortion is easy to calculate and represents expressed as a rate.
The spectral decomposition of the signal carried out on this luminaire shows that the main harmonic is 3rd order (34 mA) but that all the following odd-order harmonics are present and decaying. The main purpose of the above example is to demonstrate that active power information (in W) only for a non-linear receiver is very inadequate. The cosφ has no real relevance or meaning as it is only applicable to the fundamental signal. Only the apparent power and power factor (PF or ?\.,) information can really quantify and qualify the power that must be supplied by the source.
In the example given, it can be seen that an active power of approximately 9 W corresponds to a consumed power of 16 VA. Many modern devices (light bulbs, computer equipment, domestic appliances and electronic equipment) have this particular feature of consuming non-linear currents. For domestic use, where only the power in W is billed (sic), the power savings shown for these products is attractive. In practice, the currents consumed are higher than it seems and the energy distributor is supplying wasted energy. In large commercial or industrial installations the situation is different. A poor power factor results in consumption of reactive power that is billed. Compensation of non-linear loads thus becomes meaningful and useful here, but also at the design stage when it prevents oversizing of the energy sources, which it must be remembered supply VA (volt-amperes) and not W (watts). Important: Unlike linear loads (page 29), for non-linear loads the active powers (in W) can be added together algebraically, the apparent powers must only be added together geometrically, and likewise the currents which must be the same order. The reactive powers Q must not be added together except to as certain the relative part of the power associated with the sinusoidal fundamental signal and the part connected with the harmonic signals. Go back to currents and power calculations ↑
Reference // Power balance and choice of power supply solutions by Legrand. Pick Up Current | Current Setting | Plug Setting Multiplier and Time Setting Multiplier of Relay. Pick Up Current of Relay In all electrical relays, the moving contacts are not free to move. All the contacts remain in their respective normal position by some force applied on them continuously. This force is called controlling force of the relay. This controlling force may be gravitational force, may be spring force, may be magnetic force. The force applied on the relay’s moving parts for changing the normal position of the contacts, is called deflecting force. This deflecting force is always in opposition of controlling force and presents always in the relay. Although the deflecting force always presents in the relay directly connected to live line, but as the magnitude of this force is less than controlling force in normal condition, the relay does not operate. If the actuating current in the relay coil increases gradually, the deflecting force in electro mechanical relay, is also increased. Once, the deflecting force crosses the controlling force, the moving parts of the relay initiate to move to change the position of the contacts in the relay. The current for which the relay initiates it operation is called pick up current of relay. Current Setting of Relay The minimum pick up value of the deflecting force of an electrical relay is constant. Again the deflecting force of the coil is proportional to its number of turns and current flowing through the coil. Now, if we can change the number of active turns of any coil, the required current to reach at minimum pick value of the deflecting force, in the coil also changes. That means if active turns of the relay coil is reduced, then proportionately more current is required to produce desired relay actuating force. Similarly if active turns of the relay coil is increased, then proportionately reduced current is required to produce same desired deflecting force. Practically same model relays may be used in different systems. As per these systems requirement the pick up current of relay is adjusted. This is known as current setting of relay. This is achieved by providing required number of tapping in the coil. These taps are brought out to a plug bridge. The number of active turns in the coil can be changed by inserting plug in different points in the bridge. The current setting of relay is expressed in percentage ratio of relay pick up current to rated secondary current of CT.
That means, For example, suppose, you want that, an over current relay should operate when the system current just crosses 125% of rated current. If the relay is rated with 1 A, the normal pick up current of the relay is 1 A and it should be equal to secondary rated current of current transformer connected to the relay. Then, the relay will be operated when the current of CT secondary becomes more than or equal 1.25 A. As per definition, The current setting is sometimes referred as current plug setting. The current setting of over current relay is generally ranged from 50 % to 200 %, in steps of 25 %. For earth fault relay it is from 10% to 70% in steps of 10%. Plug Setting Multiplier of Relay Plug setting multiplier of relay is referred as ratio of fault current in the relay to its pick up current.
Suppose we have connected on protection CT of ratio 200/1 A and current setting is 150%. Hence, pick up current of the relay is, 1 × 150 % = 1.5 A Now, suppose fault current in the CT primary is 1000 A. Hence, fault current in the CT secondary i.e. in the relay coil is, 1000 × 1/200 = 5A Therefore PSM of the relay is, 5 / 1.5 =3.33 Time Setting Multiplier of Relay The operating time of an electrical relay mainly depends upon two factors : 1.
How long distance to be traveled by the moving parts of the relay for closing relay contacts and 2. How fast the moving parts of the relay cover this distance. So far adjusting relay operating time, both of the factors to be adjusted. The adjustment of travelling distance of an electromechanical relay is commonly known as time setting. This adjustment is commonly known as time setting multiplier of relay. The time setting dial is calibrated from 0 to 1 in steps 0.05 sec. But by adjusting only time setting multiplier, we can not set the actual time of operation of an electrical relay. As we already said, the time of operation also depends upon the speed of operation. The speed of moving
parts of relay depends upon the force due to current in the relay coil. Hence, it is clear that, speed of operation of an electrical relay depends upon the level of fault current. In other words, time of operation of relay depends upon plug setting multiplier. The relation between time of operation and plug setting multiplier is plotted on a graph paper and this is known as time / PSM graph. From this graph one can determine, the total time taken by the moving parts of an electromechanical relay, to complete its total travelling distance for different PSM. In time setting multiplier, this total travelling distance is divided and calibrated from 0 to 1 in steps of 0.05. So when time setting is 0.1, the moving parts of the relay has to travel only 0.1 times of the total travelling distance, to close the contact of the relay. So, if we get total operating time of the relay for a particular PSM from time / PSM graph and if we multiply that time with the time setting multiplier, we will get, actual time of operation of relay for said PSM and TSM. For getting clear idea, let us have a practical example. Say a relay has time setting 0.1 and you have to calculate actual time of operation for PSM 10. From time / PSM graph of the relay as shown below, we can see the total operating time of the relay is 3 seconds. That means, the moving parts of the relay take total 3 seconds to travel 100 % travelling distance. As the time setting multiplier is 0.1 here, actually the moving parts of the relay have to travel only 0.1 × 100% or 10% of the total travel distance, to close the relay contacts. Hence, actual operating time of the relay is 3 × 0.1 = 0.3 sec. i.e. 10% of 3 sec. Time vs PSM Curve of Relay This is relation curve between operating time and plug setting multiplier of an electrical relay. The x-axis or horizontal axis of the Time / PSM graph represents, PSM and Y-axis or vertical axis represents time of operation of the relay. The time of operation represents in this graph is that, which required to operate the relay when time setting multiplier set at 1. From the Time / PSM curve of a typical relay shown below, it is seen that, if PSM is 10, the time of operation of relay is 3 sec. That means, the relay will take 3 seconds to complete its operation, with time setting 1. It is also seen from the curve that, for lower value of plug setting multiplier, i.e. for lower value of fault current, the time of operation of the relay is inversely proportional to the fault current. But when PSM becomes more than 20, the operating time of relay becomes almost constant. This feature is necessary in order to ensure discrimination on very heavy fault current flowing through sound feeders. Calculation of Relay Operation Time For calculating actual relay operating time, we need to know these following operation.
1. Current setting. 2. Fault current level. 3. Ratio of current transformer. 4. Time / PSM curve. 5. Time setting. Step-1 From CT ratio, we first see the rated secondary current of CT. Say the CT ratio is 100 / 1 A, i.e. secondary current of CT is 1 A. Step-2 From current setting we calculate the trick current of the relay. Say current setting of the relay is 150 % therefore pick up current of the relay is 1 × 150% = 1.5 A. Step-3 Now we have to calculate PSM for the specified faulty current level. For that, we have to first divide primary faulty current by CT ratio to get relay faulty current. Say the faulty current level is 1500 A, in the CT primary, hence secondary equivalent of faulty current is 1500/(100/1) = 15
A Step-4 Now, after calculating PSM, we have to find out the total time of operation of the relay from Time / PSM curve. From the curve, say we found the time of operation of relay is 3 second for PSM = 10. Step-5 Finally that operating time of relay would be multiplied with time setting multiplier, in order to get actual time of operation of relay. Hence say time setting of the relay is 0.1. Therefore actual time of operation of the relay for PSM 10, is 3 × 0.1 = 0.3 sec or 300 ms. Working Principle of Over Current Relay In an over current relay, there would be essentially a current coil. When normal current flows through this coil, the magnetic effect generated by the coil is not sufficient to move the moving element of the relay, as in this condition the restraining force is greater than deflecting force. But when the current through the coil increased, the magnetic effect increases, and after certain level of current, the deflecting force generated by the magnetic effect of the coil, crosses the restraining force, as a result, the moving element starts moving to change the contact position in the relay. Types of Over Current Relay Depending upon time of operation, there are various types of Over Current relays, such as, 1. Instantaneous over current relay. 2. Definite time over current relay. 3. Inverse time over current relay.
Inverse time over current relay or simply inverse OC relay is again subdivided as inverse definite minimum time (IDMT), very inverse time, extremely inverse time over current relay or OC relay. Instantaneous Over Current Relay Construction and working principle of instantaneous over current relay quite simple. Here generally a magnetic core is wound by current coil. A piece of iron is so fitted by hinge support and restraining spring in the relay, that when there is not sufficient current in the coil, the NO contacts remain open. When current in the coil crosses a present value, the attractive force becomes sufficient to pull the iron piece towards the magnetic core and consequently the no contacts are closed. The preset value of current in the relay coil is referred as pick up setting current. This relay is referred as instantaneous over current relay, as ideally, the relay operates as soon as the current in the coil gets higher than pick up setting current. There is no intentional time delay applied. But there is always an inherent time delay which can not be avoided practically. In practice the operating time of an instantaneous relay is of the order of a few milliseconds. Fig. Definite Time Over Current Relay This relay is created by applying intentional time delay after crossing pick up value of the current. A definite time over current relay can be adjusted to issue a trip output at definite amount of time after it picks up. Thus, it has a time setting adjustment and pick up adjustment. Inverse Time Over Current Relay Inverse time is a natural character of any induction type rotating device. This means the speed of rotation of rotating art of the device is faster if input current is increased. In other words, time of operation inversely varies with input current. This natural characteristic of electromechanical induction disc relay in very suitable for over current protection. This is because, in this relay, if fault is more severe, it would be cleared more faster. Although time inverse characteristic is inherent to electromechanical induction disc relay, but the same characteristic can be achieved in microprocessor based relay also by proper programming. Inverse Definite Minimum Time Over Current Relay or IDMT O/C Relay Ideal inverse time characteristics can not be achieved, in an over current relay. As the current in the system increases, the secondary current of the current transformer is increased proportionally. The secondary current is
fed to the relay current coil. But when the CT becomes saturated, there would not be further proportional increase of CT secondary current with increased system current. From this phenomenon it is clear that from trick value to certain range of faulty level, an inverse time relay shows exact inverse characteristic. But after this level of fault, the CT becomes saturated and relay current does not increase further with increasing faulty level of the system. As the relay current is not increased further, there would not be any further reduction in time of operation in the relay. This time is referred as minimum time of operation. Hence, the characteristic is inverse in the initial part, which tends to a definite minimum operating time as the current becomes very high. That is why the relay is referred as inverse definite minimum time over current relay or simply IDMT relay. Motor Protection Relay for High Voltage Induction Motor. Above 90% of motors used in an industry are induction motors, because they are cheap, robust & easy to maintain. For higher HP (>250HP) motors we prefer high voltage, because it will reduce operating current and the size of the motor. Why we require protection of motors? To understand this we need to know the cost associated with the failure of motor, i.e. Loss of production (Cost of production) Replacement of motor (Replacement cost) Cost of repair Cost of man hours due to this emergency The basic function of a protective relay is to identify the fault and isolate it from the healthy part of the system. This will improve the reliability of power system. For protection of motor, we have to identify the various causes of failure and to address the same. The various causes of failure are as below.
Thermal stress on winding Single phasing Earth fault Short circuit Locked rotor Number of hot starts Bearing failure Thermal Stress on Winding – If a motor runs continuously more than its rated capacity then this will over heat the winding and insulation. Subsequently deteriorate the winding insulation resulting in failure of
motor. If the voltage is less than the designed value then also it will over heat the winding at rated load and fail the motor. Single Phasing – Loss of one phase supplied to the motor (in case of 3-phase motor) leads to single phasing. If we start the motor on load, then the motor will fail due to unbalance. Earth Fault – If any part of winding comes in contact with the ground then we can say the motor is earthed. If we start the motor then it will lead to failure of motor. Short Circuit – If there is a contact between two phases of a three phase winding or between the turns of a phase, then this will termed as short circuit. Locked Rotor – If the driven equipment is in jammed condition or the motor shaft is jam, then this is known as locked rotor. If we start the motor then it will fail. Number of Hot Start – Each motor is designed to withstand a certain number of hot starts. Consider a motor is in running condition, if we stop the motor & immediately start the same, then this is called as hot start. Depending upon the thermal curve of a motor we have to give certain time to bring down the temperature of winding. Bearing Failure – If bearing fails then rubbing of rotor on stator will occur, resulting physical damage of insulation and winding. The bearing failure can be avoided by monitoring the bearing temperature. Bearing temperature detector (BTD) is used for monitoring and tripping of motor in case of abnormality. All motor protection relays operate on the basis of current taken by the motor. Motor protection relay is used for high voltage area having the following features Thermal overload protection Short circuit protection Single phasing protection Earth fault protection Locked rotor protection Number of start protection For setting of the relay we require the CT ratio & full load current of the motor. The setting of different element is listed below
Thermal over Load Element - To set this element we have to identify the % of Full load current on which the motor is running continuously. Thermal setting = (Full load current x % of Full load) / CT ratio. Short circuit Element - The range available for this element is 1 to 5 times of starting current. Time delay is also available. We normally set at 2 times of I starting with a time delay of 0.1 second.
Single Phasing Element - This element will operate, if there is an unbalance in current of three phases. It is also called as unbalance protection. The element is set for 1/3rd of starting current. If it tripped during starting, then the parameter will changed to 1/2 of starting current. Earth Fault Protection - This element measures the neutral current of star connected CT secondary. The range available for this element is 0.02 to 2 times of CT primary current. Time delay is also available. We normally set at 0.1 times of CT primary current with a time delay of 0.2 seconds. If tripped during starting of motor, then the time setting can be raised to 0.5 sec. Locked rotor protection - The range available for this element is 1 to 5 times of full load current. Time delay is also available. We normally set at 2 times of FLC. The time delay will be more than the starting time of the motor. "Starting time means the time require by the motor to reach its full speed." Number of hot start protection - Here we will provide the number of start allowed in specified time duration. By this we will limit the number of hot start given to the motor. The schematic diagram to connect a motor protection relay is as below
Modern digital motor protection relays are having some extra features, i.e. protection against no load running of a motor and thermal protection. In case of no load running, the relay senses the motor current. If it is less than the specified value then it will trip the motor. We can also connect the temperature probe to the relay, which will monitor the bearing and winding temperature and trip the motor if it exceeds the specified value. Feeder Protection Relays. Distance Protection Relay One of the important protections in Power System Protection is Feeder Protection. Different types of relays were used for feeder protection like electromagnetic type relays, static type relays etc. But now a day we are using Numerical relays for all protections. The benefits of Numerical relays are, 1. Accurate tripping, 2. Less tolerance, 3. Fault events and counter storage 4. Display of fault parameters on screen (Fault parameters means current, voltage, resistance and reactance values during fault and fault distance, Numerical relays can store thousands of tripping events). The main inputs needed for distance protection is Voltage and Current from corresponding feeder PT and CT. According to site condition we will set certain impedance values in relay settings (i.e., R and X values) for fault detection. Relay will monitor Current & Voltage in the feeder line (PT and CT secondary), and from these values, the relay will calculate Impedance value Z. i.e., Z=V/I. In normal load condition the impedance values on line will be high. But when fault comes on the feeder line, the impedance will decrease and becomes less than the impedance setting in the relay, then the distance relay will trip with in 40 ms in zone 1 (different zones are there and that will be explained later) and isolate power equipments from fault. I.e., during fault, relay will trip and shows the fault parameters like fault current, voltage, reactance, resistance and fault distance on relay screen. Suppose if the fault is on 25km, then relay will show Fault Distance (FD) = 25km, and thus it becomes easy to identify the location where there is fault. For Distance protection now a day Quadrilateral characteristics is used. We already discussed that for fault identification we have to set different parameters in relay. i.e., 1. Forward and Backward Resistance (RF, RB), 2. Forward and backward Reactance (XF, XB), 3. RCA (Relay Characteristics Angle) and 4. Line Impedance per km.
These parameters are used for making Quadrilateral characteristics. Suppose if RCA=70° and we are using parallelogram characteristics (Quadrilateral), we can plot the graph by setting Forward Resistance (RF) value in positive X axis, Backward Resistance (RB) in negative X axis, Forward Reactance (XF) value in positive Y axis, and Backward Reactance (XB) in negative Y axis. And plot parallelogram with a slope of RCA angle. Thus we will get a parallelogram graph and the protection zone is inside the parallelogram. Means during fault the impedance will reach inside the parallelogram then the relay will trip. In graph there are 4 quadrants of operation 1. First quadrant (R and X values + ve) If the load is inductive and the fault is in forward direction from Relay, then the relay will trip in this quadrant values. 2. Second quadrant (R – ve and X + ve) If the load is capacitive and the fault is in reverse direction from Relay, then the relay will trip in this quadrant values. 3. Third quadrant (R – ve and X - ve) If the load is inductive and the fault is in reverse direction from Relay, then the relay will trip in this quadrant values. 4. Fourth quadrant (R + ve and X - ve) If the load is capacitive and the fault is in forward direction from Relay, then the relay will trip in this quadrant values. Different Zones of operation, fault distance calculation and other feeder protection relays etc. will be explained in next article... Remark: A model of
quadrilateral characteristics is shown below….
Types of Electrical Protection Relays or Protective Relays. Definition of Protective Relay A relay is automatic device which senses an abnormal condition of electrical circuit and closes its contacts. These contacts in turns close and complete the circuit breaker trip coil circuit hence make the circuit breaker tripped for disconnecting the faulty portion of the electrical circuit from rest of the healthy circuit. Now let’s have a discussion on some terms related to protective relay. Pickup Level of Actuating Signal: The value of actuating quantity (voltage or current) which is on threshold above which the relay initiates to be operated. If the value of actuating quantity is increased, the electromagnetic effect of the relay coil is increased and above a certain level of actuating quantity the moving mechanism of the relay just starts to move. Reset Level: The value of current or voltage below which a relay opens its contacts and comes in original position. Operating Time of Relay: Just after exceeding pickup level of actuating quantity the moving mechanism (for example rotating disc) of relay starts moving and it ultimately close the relay contacts at the end of its journey. The time which elapses between the instant when actuating quantity exceeds the pickup value to the instant when the relay contacts close. Reset Time of Relay: The time which elapses between the instant when the actuating quantity becomes less than
the reset value to the instant when the relay contacts returns to its normal position. Reach of Relay: A distance relay operates whenever the distance seen by the relay is less than the pre-specified impedance. The actuating impedance in the relay is the function of distance in a distance protection relay. This impedance or corresponding distance is called reach of the relay. Power system protection relays can be categorized into different types of relays. Types of Relays Types of protection relays are mainly based on their characteristic, logic, on actuating parameter and operation mechanism. Based on operation mechanism protection relay can be categorized as electromagnetic relay, static relay and mechanical relay. Actually relay is nothing but a combination of one or more open or closed contacts. These all or some specific contacts the relay change their state when actuating parameters are applied to the relay. That means open contacts become closed and closed contacts become open. In electromagnetic relay these closing and opening of relay contacts are done by electromagnetic action of a solenoid. In mechanical relay these closing and opening of relay contacts are done by mechanical displacement of different gear level system. In static relay it is mainly done by semiconductor switches like thyristor. In digital relay on and off state can be referred as 1 and 0 state. Based on Characteristic the protection relay can be categorized as1. Definite time relays 2. Inverse time relays with definite minimum time(IDMT) 3. Instantaneous relays. 4. IDMT with inst. 5. Stepped characteristic. 6. Programmed switches. 7. Voltage restraint over current relay. Based on of logic the protection relay can be categorized as1. Differential. 2. Unbalance. 3. Neutral displacement. 4. Directional. 5. Restricted earth fault. 6. Over fluxing. 7. Distance schemes. 8. Bus bar protection. 9. Reverse power relays. 10. Loss of excitation. 11. Negative phase sequence relays etc. Based on actuating parameter the protection relay can be categorized as-
1. Current relays. 2. Voltage relays. 3. Frequency relays. 4. Power relays etc. Based on application the protection relay can be categorized as1. Primary relay. 2. Backup relay. Primary relay or primary protection relay is the first line of power system protection whereas backup relay is operated only when primary relay fails to be operated during fault. Hence backup relay is slower in action than primary relay. Any relay may fail to be operated due to any of the following reasons, 1. 2. 3. 4.
The protective relay itself is defective. DC Trip voltage supply to the relay is unavailable. Trip lead from relay panel to circuit breaker is disconnected. Trip coil in the circuit breaker is disconnected or defective. 5. Current or voltage signals from CT or PT respectively is unavailable. As because backup relay operates only when primary relay fails, backup protection relay should not have anything common with primary protection relay. Some examples of Mechanical Relay are1. o o o 2. o o o o 3. 4. 5.
Thermal OT trip (Oil Temperature Trip) WT trip (Winding Temperature Trip) Bearing temp trip etc. Float type Buchholz OSR PRV Water level Controls etc. Pressure switches. Mechanical interlocks. Pole discrepancy relay.
List Different Protective Relays are used for Different Power System Equipment Protection Now let’s have a look on which different protective relays are used in different power system equipment protection schemes. Relays for Transmission & Distribution Lines Protection SL
Lines to be
Relays to be used
protected 400 KV Transmission Line
Main-I: Non switched or Numerical Distance Scheme Main-II: Non switched or Numerical Distance Scheme
220 KV Transmission Line
Main-I : Non switched distance scheme (Fed from Bus PTs) Main-II: Switched distance scheme (Fed from line CVTs) With a changeover facility from bus PT to line CVT and vice-versa.
3
132 KV Transmission Line
Main Protection : Switched distance scheme (fed from bus PT). Backup Protection: 3 Nos. directional IDMT O/L Relays and 1 No. Directional IDMT E/L relay.
4
33 KV lines
Non-directional IDMT 3 O/L and 1 E/L relays.
5
11 KV lines
Non-directional IDMT 2 O/L and 1 E/L relays.
1
2
Relays for Transformer Protection Voltage Ratio and SL Capacity of Transformer 1
11/132 KV Generator Transformer
Relays on HV Side
Relays on LV Side
3 nos Non-Directional O/L Relay 1 no Non-Directional E/L Relay and/or standby E/F + REF Relay
--
Common Relays
Differential Relay or Overall differential Relay Overflux Relay Buchholz Relay OLTC Buchholz Relay PRV Relay
OT Trip Relay WT Trip Relay
2
3
4
13.8/220 KV 15.75/220 KV 18/400 KV 21/400 KV Generator Transformer
3 nos Non-Directional O/L Relay 1 no Non-Directional E/L Relay and/or standby E/F + REF Relay
220 /6.6KV Station Transformer
3 nos Non-Directional O/L Relay 1 no Non-Directional E/L Relay and/or standby E/F + REF Relay
Gen-volt/6.6KV UAT
3 nos Non-Directional O/L Relay
--
Differential Relay or Overall differential Relay Overflux Relay Buchholz Relay OLTC Buchholz Relay PRV Relay OT Trip Relay WT Trip Relay
3 nos NonDirectional O/L Relay
Differential Relay Overflux Relay Buchholz Relay OLTC Buchholz Relay PRV Relay OT Trip Relay WT Trip Relay
3 nos NonDirectional O/L Relay
Differential Relay Overflux Relay Buchholz Relay OLTC Buchholz
Relay PRV Relay OT Trip Relay WT Trip Relay
5
6
7
8
132/33/11KV upto 8 MVA
132/33/11KV above 8 MVA & below 31.5 MVA
3 nos O/L Relay 1 no E/L Relay
3 nos O/L Relay 1 no Directional E/L Relay
2 nos O/L Relays 1 no E/L Relay
Buchholz Relay OLTC Buchholz Relay PRV Relay OT Trip Relay WT Trip Relay
3 nos O/L Relay 1 no E/L Relay
Differential Relay Buchholz Relay OLTC Buchholz Relay PRV Relay OT Trip Relay WT Trip Relay Differential Relay Overflux Relay Buchholz Relay OLTC Buchholz Relay PRV Relay OT Trip Relay WT Trip Relay
132/33KV, 31.5 MVA & above
3 nos O/L Relay 1 no Directional E/L Relay
3 nos O/L Relay 1 no E/L Relay
220/33 KV, 31.5MVA &
3 nos O/L Relay 1 no Directional E/L
3 nos O/L Relay Differential 1 no Directional E/L Relay
50MVA 220/132KV, 100 Relay MVA
9
400/220KV 315MVA
3 nos Directional O/L Relay (with dir.highset) 1 no Directional E/L relay. Restricted E/F relay 3 nos Directional O/L Relay for action
Relay
Overflux Relay Buchholz Relay OLTC Buchholz Relay PRV Relay OT Trip Relay WT Trip Relay
Differential Relay Overflux Relay Buchholz 3 nos Directional Relay O/L Relay OLTC (with dir.highset) Buchholz 1 no Directional E/L Relay relay. PRV Relay Restricted E/F relay OT Trip Relay WT Trip Relay Over Load (Alarm) Relay
Points to be remembered in respect of protection of transformers 1.
2. 3.
4.
5. 6. 7. 8.
No Buchholz relay for transformers below 500 KVA capacity. Transformers up to 1500 KVA shall have only Horn gap protection. Transformers above 1500 KVA and upto 8000 KVA of 33/11KV ratio shall have one group control breaker on HV side and individual LV breakers if there is more than one transformer. Transformers above 8000 KVA shall have individual HV and LV circuit breakers. The relays indicate above shall be provided on HV and LV. LAs to be provided on HV and LV for transformers of all capacities and voltage class. OLTC out of step protection is to be provided where master follower scheme is in operation. Fans failure and pumps failure alarms to be connected.
9.
Alarms for O.T., W.T., Buchholz (Main tank AND OLTC) should be connected.
Equivalent Circuit of Transformer referred to Primary and Secondary. Equivalent Circuit of Transformer Equivalent impedance of transformer is essential to be calculated because the electrical power transformer is an electrical power system equipment for estimating different parameters of electrical power system which may be required to calculate total internal impedance of an electrical power transformer, viewing from primary side or secondary side as per requirement. This calculation requires equivalent circuit of transformer referred to primary or equivalent circuit of transformer referred to secondary sides respectively. Percentage impedance is also very essential parameter of transformer. Special attention is to be given to this parameter during installing a transformer in an existing electrical power system. Percentage impedance of different power transformers should be properly matched during parallel operation of power transformers. The percentage impedance can be derived from equivalent impedance of transformer so, it can be said that equivalent circuit of transformer is also required during calculation of % impedance. Equivalent Circuit of Transformer Referred to Primary For drawing equivalent circuit of transformer referred to primary, first we have to establish general equivalent circuit of transformer then, we will modify it for referring from primary side. For doing this, first we need to recall the complete vector diagram of a transformer which is shown in the
figure below. consider the transformation ratio be,
Let us
In the figure above, the applied voltage to the primary is V1 and voltage across the primary winding is E1. Total current supplied to primary is I1. So the voltage V1 applied to the primary is partly dropped by I1Z1 or I1R1 + j.I1X1 before it appears across primary winding. The voltage appeared across winding is countered by primary induced emf E1. So voltage equation of this portion of the transformer can be written as,
The equivalent circuit for that equation can be drawn as below,
From the vector diagram above, it is found that the total primary current I1 has two components, one is no load component Io and the other is load component I2′. As this primary current has two components or branches, so there must be a parallel path with primary winding of transformer. This parallel path of current is known as excitation branch of equivalent circuit of transformer. The resistive and reactive branches of the excitation circuit can be represented as
The load component I2′ flows through the primary winding of transformer and induced voltage across the winding is E1 as shown in the figure right. This induced voltage E1transforms to secondary and it is E2 and load component of primary current I2′ is transformed to secondary as secondary current I2. Current of secondary is I2. So the voltage E2 across secondary winding is partly dropped by I2Z2 or I2R2 + j.I2X2 before it appears across load. The load voltage is V2.
The
complete
equivalent
circuit
of
transformer
is
shown
below.
Now if we see the voltage drop in secondary from primary side, then it would be ′K′ times greater and would be written as K.Z2.I2. Again I2′.N1 = I2.N2
Therefore,
From above equation, secondary impedance of transformer referred to primary is,
So, the complete equivalent circuit of transformer referred to primary is shown in the figure below,
Approximate Equivalent Circuit of Transformer Since Io is very small compared to I1, it is less than 5% of full load primary current, Iochanges the voltage drop insignificantly. Hence, it is good
approximation to ignore the excitation circuit in approximate equivalent circuit of transformer. The winding resistance and reactance being in series can now be combined into equivalent resistance and reactance of transformer, referred to any particular side. In this case it is side 1 or primary side.
Equivalent Circuit of Transformer Referred to Secondary In similar way, approximate equivalent circuit of transformer referred to secondary can be drawn. Where equivalent impedance of transformer referred to secondary, can be derived as.
Knee Point Voltage of Current Transformer PS Class. Current Transformer PS Class Before understanding Knee Point Voltage of Current Transformer and current transformer PS class we should recall the terms instrument security factor of CT and accuracy limit factor. Instrument Security Factor or ISF of Current Transformer Instrument security factor is the ratio of instrument limit primary current to the rated primary current. Instrument limit current of a metering current transformer is the maximum value of primary current beyond which current transformer core becomes saturated. Instrument security factor of CT is the significant factor for choosing the metering instruments which to be connected to the secondary of the CT. Security or Safety of the measuring unit is better, if ISF is low. If we go through the example below it would be clear to us. Suppose one current transformer has rating 100/1 A and ISF is 1.5 and another current transformer has same rating with ISF 2. That means, in first CT, the metering core would be saturated at 1.5 × 100 or 150 A, whereas is second CT, core will be saturated at 2 × 100 or 200 A. That means whatever may be the primary current of both CTs, secondary current will not increase further after 150 and 200 A of primary current of the CTs respectively. Hence maximum secondary current of the CTs would be 1.5 and 2.0 A. As the maximum current can flow through the instrument connected to the first CT is 1.5 A which is less than the maximum value of current can flow through the instrument connected to the second CT i.e. 2 A. Hence security or safety of the instruments of first CT is better than later. Another significance of ISF is during huge electrical fault, the short circuit current, flows through primary of the CT does not affect destructively, the measuring instrument attached to it as because, the secondary current of the CT will not rise above the value of rated secondary current multiplied by ISF. Accuracy Limit Factor or ALF of Current Transformer
For protection current transformer, the ratio of accuracy limit primary current to the rated primary current. First we will explain, what is rated accuracy limit primary current?. Broadly, this is the maximum value of primary current, beyond which core of the protection CT or simply protection core of of a CT starts saturated. The value of rated accuracy limit primary current is always many times more than the value of instrument limit primary current. Actually CT transforms the fault current of the electrical power system for operation of the protection relays connected to the secondary of that CT. If the core of the CT becomes saturated at lower value of primary current, as in the case of metering CT, the system fault will not reflect properly to the secondary, which may cause, the relays remain inoperative even the fault level of the system is large enough. That is why the core of the protection CT is made such a way that saturation level of that core must be high enough. But still there is a limit as because, it is impossible to make one magnetic core with infinitely high saturation level and secondly most important reason is that although the protection care should have high saturation level but that must be limited up to certain level otherwise total transformation of primary current during huge fault may badly damage the protection relays. So it is clear from above explanation, rated accuracy limit primary current, should not be so less, that it will not at all help the relays to be operated on the other hand this value must not be so high that it can damage the relays. So, accuracy limit factor or ALF should not have the value nearer to unit and at the same time it should not be as high as 100. The standard values of ALF as per IS-2705 are 5, 10, 15, 20 and 30. Knee Point Voltage of Current Transformer This is the significance of saturation level of a CT core mainly used for protection purposes. The sinusoidal voltage of rated frequency applied to the secondary terminals of current transformer, with other winding being open circuited, which when increased by 10% cause the exiting current to increase 50%. The CT core is made of CRGO steel. It has its won saturation level. The EMF induced in the CT secondary windings is E2 = 4.44φfT2 Where, f is the system frequency, φ is the maximum magnetic flux in Wb. T2 is the number of turns of the secondary winding. The flux in the core, is produced by excitation current Ie. We have a non-liner relationship between excitation current and magnetizing flux. After certain value of excitation current, flux will not further increase so rapidly with increase in excitation current. This non-liner relation curve is also called B - H curve. Again from the equation above, it is found that, secondary voltage of acurrent
transformer is directly proportional to flux φ. Hence one typical curve can be drawn from this relation between secondary voltage and excitation current as shown below. It is clear from the curve that, linear relation between V and Ie is maintained from point A and K. The point ′A′ is known as ′ankle point′ and point ′K′ is known as ′Knee Point′.
In differential and restricted earth fault (REF) protection scheme, accuracy class and ALF of the CT may not ensure the reliability of the operation. It is desired that, differential and REF relays should not be operated when fault occurs outside the protected transformer. When any fault occurs outside the differential protection zone, the faulty current flows through the CTs of both sides of electrical power transformer. The both LV and HV CTs have magnetizing characteristics. Beyond the knee point, for slight increase in secondary emf a large increasing in excitation current is required. So after this knee point excitation current of both current transformers will be extremely high, which may cause mismatch between secondary current of LV & HV current transformers. This phenomena may cause unexpected tripping of power transformer. So the magnetizing characteristics of both LV & HV sides CTs, should be same that means they have same knee point voltage Vk as well as same excitation current Ieat Vk/2. It can be again said that, if both knee point voltage of current transformer and magnetizing characteristic of CTs of both sides of power transformer differ, there must be a mismatch in high excitation currents of the CTs during fault which ultimately causes the unbalancing between secondary current of both groups of CTs and transformer trips. So for choosing CT for differential protection of transformer, one should consider current transformer PS class rather its convectional protection
class. PS stands for protection special which is defined by knee point voltage of current transformer Vk and excitation current Ie at Vk/2. Why CT Secondary Should Not Be Kept Open? The electrical power system load current always flows through current transformer primary; irrespective of whether the current transformer is open circuited or connected to burden at its secondary.
If CT secondary is open circuited, all the primary current will behave as excitation current, which ultimately produce huge voltage. Every current transformer has its won non-linear magnetizing curve, because of which secondary open circuit voltage should be limited by saturation of the core. If one can measure the rms voltage across the secondary terminals, he or she will get the value which may not appear to be dangerous. As the CT primary current is sinusoidal in nature, it zero 100 times per second.(As frequency of the current is 50 Hz). The rate of change of flux at every current zero is not limited by saturation and is high indeed. This develops extremely high peaks or pulses of voltage. This high peaks of voltage may not be measured by conventional voltmeter. But these high peaks of induced voltage may breakdown the CT insulation, and may case accident to personnel. The actual open-circuit voltage peak is difficult to measure accurately because of its very short peaks. That is why CT secondary should not be kept open. Theory of Transformer We have discussed about the theory of ideal transformer for better understanding of actual elementary theory of transformer. Now we will go through the practical aspects one by one of an electrical power transformer and try to draw vector diagram of transformer in every step.
As we said that, in an ideal transformer; there are no core losses in transformer i.e. loss free core of transformer. But in practical transformer, there are hysteresis and eddy current losses in transformer core. Theory of Transformer on No-Load Theory of Transformer On No-load, and Having No Winding Resistance and No Leakage Reactance of Transformer Let us consider one electrical transformer with only core losses, which means, it has only core losses but no copper loss and no leakage reactance of transformer. When an alternating source is applied in the primary, the source will supply the current for magnetizing the core of transformer. But this current is not the actual magnetizing current, it is little bit greater than actual magnetizing current. Actually, total current supplied from the source has two components, one is magnetizing current which is merely utilized for magnetizing the core and other component of the source current is consumed for compensating the core losses in transformer. Because of this core loss component, the source current in transformer on noload condition supplied from the source as source current is not exactly at 90° lags of supply voltage, but it lags behind an angle θ is less than 90°. If total current supplied from source is I o, it will have one component in phase with supply voltage V1 and this component of the current I w is core loss component. This component is taken in phase with source voltage, because it is associated with active or working losses in transformer. Other component of the source current is denoted as I μ. This component produces the alternating magnetic flux in the core, so it is watt-less; means it is reactive part of the transformer source current. Hence I μ will be in quadrature with V1 and in phase with alternating flux Φ. Hence, total primary current in transformer on no-load condition can be represented as
Now you have seen how simple is to explain the theory of transformer in no-load.
Theory of Transformer on Load Theory of Transformer On Load But Having No Winding Resistance and Leakage Reactance Now we will examine the behavior of above said transformer on load, that means load is connected to the secondary terminals. Consider, transformer having core loss but no copper loss and leakage reactance. Whenever load is connected to the secondary winding, load current will start to flow through the load as well as secondary winding. This load current solely depends upon the characteristics of the load and also upon secondary voltage of the transformer. This current is called secondary current or load current, here it is denoted as I2. As I2 is flowing through the secondary, a self mmf in secondary winding will be produced. Here it is N2I2, where, N2 is the number of turns of the secondary winding of transformer.
This mmf or magneto motive force in the secondary winding produces flux φ2. This φ2 will oppose the main magnetizing flux and momentarily weakens the main flux and tries to reduce primary self induced emf E1. If E1 falls down below the primary source voltage V1, there will be an extra current flowing from source to primary winding. This extra primary current I2′ produces extra flux φ′ in the core which will neutralize the secondary counter flux φ2. Hence the main magnetizing flux of core, Φ remains unchanged irrespective of load. So total current, this transformer draws from source can be divided into two components, first one is utilized for magnetizing the core and compensating the core loss i.e. Io. It is no-load component of the primary current. Second one is utilized for compensating the counter flux of the secondary winding. It is known as load component of the primary current. Hence total no load primary current I1 of a electrical power transformer having no winding resistance and leakage reactance can be represented as follows Where, θ2 is the angle between Secondary Voltage and Secondary Current of transformer. Now we will proceed one further step toward more practical aspect of a transformer.
Theory of Transformer On Load, With Resistive Winding, But No Leakage Reactance Now, consider the winding resistance of transformer but no leakage reactance. So far we have discussed about the transformer which has ideal windings, means winding with no resistance and leakage reactance, but now we will consider one transformer which has internal resistance in the winding but no leakage reactance. As the windings are resistive, there would be a voltage drop in the windings.
We have proved earlier that, total primary current from the source on load is I1. The voltage drop in the primary winding with resistance, R1 is R1I1. Obviously, induced emf across primary winding E1, is not exactly equal to source voltage V1. E1 is less than V1 by voltage drop I1R1.
Again in the case of secondary, the voltage induced across the secondary winding, E2 does not totally appear across the load since it also drops by an amount I2R2, where R2 is the secondary winding resistance and I2 is secondary current or load current. Similarly, voltage equation of the secondary side of the transformer will be
Theory of Transformer On Load, With Resistance As Well As Leakage Reactance in Transformer Windings Now we will consider the condition, when there is leakage reactance of transformer as well as winding resistance of transformer.
Let leakage reactances of primary and secondary windings of the transformer are X1 and X2respectively. Hence total impedance of primary and secondary winding of transformer with resistance R1 and R2 respectively, can be represented as,
We have already established the voltage equation of a transformer on load, with only resistances in the windings, where voltage drops in the windings occur only due to resistive voltage drop. But when we consider leakage reactances of transformer windings, voltage drop occurs in the winding not only because of resistance, it is because of impedance of transformer windings. Hence, actual voltage equation of a transformer can
easily be determined by just replacing resistances R1 & R2 in the previously established voltage equations by Z1 and Z2. Therefore, the voltage equations are,
Resistance drops are in the direction of current vector but, reactive drop will be perpendicular to the current vector as shown in the above vector diagram of transformer. Small and Large Motor Protection Scheme. The abnormalities in motor or motor faults may appear due to mainly two reasons 1. Conditions imposed by the external power supply network, 2. Internal faults, either in the motor or in the driven plant. Unbalanced supply voltages, under-voltage, reversed phase sequence and loss of synchronism (in the case of synchronous motor) come under former category. The later category includes bearing failures, stator winding faults, motor earth faults and overload etc. The degree of motor protection system depends on the costs and applications of the electrical motor. Small Motor Protection Scheme Generally motors up to 30 hp are considered in small category. The small motor protection in this case is arranged by HRC fuse, bimetallic relay and under voltage relay - all assembled into the motor contractor - starter itself. Most common cause of motor burn outs on LV fuse protected system is due to single phasing. This single phasing may remain undetected even if the motors are protected by conventional bimetallic relay. It can not be detected by a set of voltage relays connected across the lines. Since, even when one phase is dead, the motor maintains substantial back emf on its faulty phase terminal and hence voltage across the voltage relay is prevented from dropping - off. The difficulties of detecting single phasing can be overcome by employing a set of three current operated relays as shown in the small
motor protection circuit given below. The current operated relays are very simple instantaneous relays. There are mainly two parts in this relay one is a current coil and other is one or more normally open contacts (NO Contacts). The NO contacts are operated by the mmf of the current coil. This relay is connected in series with each phase of the supply and backup by HRC fuse. When the electrical motor starts and runs the supply current passes through the current coil of the protective relay. The mmf of the current coil makes the NO contacts closed. If suddenly a single phasing occurs the corresponding current through the current coil will falls and the contacts of the corresponding relay will become to its normal open position. The NO contacts of the all three relays are connected in series to hold - in the motor contractor. So if any one relay contact opens, results to release of motor contractor and hence motor will stop running.
Large Motor Protection Large motor specially induction motors require protection against1. 2. 3. 4.
Motor bearing failure, Motor over heating, Motor winding failure, Reverse motor rotation.
Motor Bearing Failure
Ball and roller bearings are used for the motor up to 500 hp and beyond this size sleeve bearings are used. failure of ball or roller bearing usually causes the motor to a standstill very quickly. Due to sudden mechanical jamming in motor bearing, the input current of the motor becomes very high. Current operated protection, attached to the input of the motor can not serve satisfactorily. Since this motor protection system has to be set to override the high motor starting current. The difficulty can be over come by providing thermal over load relay. As the starting current of the motor is high but exists only during starting so for that current the there will be no over heating effect. But over current due mechanical jamming exists for longer time hence there will be a over heating effect. So stalling motor protection can be offered by the thermal overload relay. Stalling protection can also be provided by separate definite time over current relay which is operated only after a certain predefined time if over current persists beyond that period. In the case of sleeve bearing, a temperature sensing device embedded in the bearing itself. This scheme of motor protection is more reliable and sensitive to motor bearing failure since the thermal withstand limit of the motor is quite higher than that of bearing. If we allow the bearing over heating and wait for motor thermal relay to trip, the bearing may be permanently damaged. The temperature sensing device embedded in the bearing stops the motor if the bearing temperature rises beyond its predefined limit. Motor Over Heating The main reason of motor over heating that means over heating of motor winding is due to either of mechanical over loading, reduced supply voltage, unbalanced supply voltage and single phasing. The over heating may cause deterioration of insulation life of motor hence it must be avoided by providing proper motor protection scheme. To avoid over heating, the motor should be isolated in 40 to 50 minutes even in the event of small overloads of the order of 10 %. The protective relay should take into account the detrimental heating effects on the motor rotor due to negative sequence currents in the stator arising out of unbalance in supply voltage. The motor should also be protected by instantaneous motor protection relay against single phasing such as a stall on loss of one phase when running at full load or attempting to start with only two of three phases alive. Motor Winding Failure The motor protection relay should should have instantaneous trip elements to detect motor winding failure such as phase to phase and phase to earth faults. Preferably phase to phase fault unit should be energized from positive phase sequence component of the motor current and
another instantaneous unit connected in the residual circuit of the current transformers be used for earth faults protection. Reverse Motor Rotation Specially in the case of conveyor belt, the reverse motor rotation must be avoided. The reverse rotation during starting can be caused due to inadvertent reversing of supply phases. A comprehensive motor protection relay with an instantaneous negative sequence unit will satisfy this requirement. If such relay has not been provided, a watt-meter type relay can be employed. NB: However, we have to provide some additional motor protection system for synchronous motor which is discussed in details in synchronous motor protection topic. Motor Thermal Overload Protection. For understanding motor thermal overload protection in induction motor we can discuss the operating principle of three phase induction motor. There is one cylindrical stator and a three phase winding is symmetrically distributed in the inner periphery of the stator. Due to such symmetrical distribution, when three phase power supply is applied to the stator winding, a rotating magnetic field is produced. This field rotates at synchronous speed. The rotor is created in induction motor mainly by numbers solid copper bars which are shorted at both ends in such a manner that they form a cylinder cage like structure. This is why this motor is also referred as squirrel cage induction motor. Anyway let's come to the basic point of three phase induction motor - which will help us to understand clearly about motor thermal overload protection.As the rotating magnetic flux cuts each of the bar conductor of rotor, there will be an induced circulating current flowing through the bar conductors. At starting the rotor is stand still and stator field is rotating at synchronous speed, the relative motion between rotating field and rotor is maximum. Hence, the rate of cuts of flux with rotor bars is maximum, the induced current is maximum at this condition. But as the cause of induced current is, this relative speed, the rotor will try to reduce this relative speed and hence it will start rotating in the direction of rotating magnetic field to catch the synchronous speed. As soon as the rotor will come to the synchronous speed this relative speed between rotor and rotating magnetic field becomes zero, hence there will not be any further flux cutting and consequently there will not be any induced current in the rotor bars. As the induced current becomes zero, there will not be any further need of maintaining zero relative speed between rotor and rotating magnetic field hence rotor speed falls. As soon as the rotor speed falls the relative speed between rotor and rotating magnetic field again acquires a non zero value which again causes induced
current in the rotor bars then rotor will again try to achieve the synchronous speed and this will continue till the motor is switch on. Due to this phenomenon the rotor will never achieve the synchronous speed as well as it will never stop running during normal operation. The difference between the synchronous speed with rotor speed in respect of synchronous speed, is termed as slip of induction motor. The slip in a normally running induction motor typically varies from 1 % to 3 % depending upon the loading condition of the motor. Now we will try to draw speed current characteristics of induction motor – let’s have an
example of large boiler fan. In the characteristic Y axis is taken as time in second, X axis is taken as % of stator current. When rotor is stand still that is at starting condition, the slip is maximum hence the induced current in the rotor is maximum and due to transformation action, stator will also draw a heavy current from the supply and it would be around 600 % of the rated full load stator current. As the rotor is being accelerated the slip is reduced, consequently the rotor current hence stator current falls to around 500 % of the full load rated current within 12 seconds when the rotor speed attains 80 % of synchronous speed. After that the stator current falls rapidly to the rated value as the rotor reaches its normal speed. Now we will discuss about thermal over loading of electrical motor or over heating problem of electric motor and the necessity of motor thermal overload protection. Whenever we think about the overheating of a motor, the first thing strikes in our mind is over loading. Due to mechanical over loading of the motor draws higher current from the supply which leads to excessive over heating of the motor. The motor can also be excessively over heated if the rotor is mechanically locked i.e. becomes stationary by any external mechanical force. In this situation the motor will draw excessively high current from the supply which also leads to thermal over loading of
electrical motor or excessive over heating problem. Another cause of overheating is low supply voltage. As the power id drawn by the motor from the supply depends upon the loading condition of the motor, for lower supply voltage, motor will draw higher current from mains to maintain required torque. Single phasing also causes thermal over loading of motor. When one phase of the supply is out of service, the remaining two phases draw higher current to maintain required load torque and this leads to overheating of the motor. Unbalance condition between three phases of supply also causes over heating of the motor winding, as because unbalance system results to negative sequence current in the stator winding. Again, due to sudden loss and reestablish of supply voltage may cause excessive heating of the motor. Since due to sudden loss of supply voltage, the motor is deaccelerated and due to sudden reestablishment of voltage the motor is accelerated to achieve its rated speed and hence for that motor draws higher current form the supply. As the thermal over loading or over heating of the motor may lead to insulation failure and damage of winding, hence for proper motor thermal overload protection, the motor should be protected against the following conditions 1. Mechanical over loading, 2. Stalling of motor shaft, 3. Low supply voltage, 4. Single phasing of supply mains, 5. Unbalancing of supply mains, 6. Sudden Loss and rebuilding of supply voltage. The most basic protection scheme of the motor is thermal over load protection which primarily covers the protection of all the above mentioned condition. To understand the basic principle of thermal over load protection let’s examine the schematic diagram of basic motor control scheme.
In the figure above, when START push is closed, the starter coil is energized through the transformer. As the starter coil is energized, normally open (NO) contacts 5 are closed hence motor gets supply voltage at its terminal and it starts rotating. This start coil also closes contact 4 which makes the starter coil energized even the START push button contact is released from its close position. To stop the motor there are several normally closed (NC) contacts in series with the starter coil as shown in the figure. One of them is STOP push button contact. If the STOP push button is pressed, this button contact opens and breaks the continuity of the starter coil circuit consequently makes the starter coil de-energized. Hence the contact 5 and 4 come back to their normally open position. Then, in absence of voltage at motor terminals it will ultimately stop running. Similarly any of the other NC contacts (1, 2 & 3) connected in series with starter coil if open; it will also stop the motor. These NC contacts are electrically coupled with various protection relays to stop operation of the motor in different abnormal conditions. Let’s look at the thermal over load relay and its function in motor thermal overload protection. The secondary of the CTs in series with motor supply circuit, are connected with a bimetallic strip of the thermal over load relay (49). As shown in the figure below, when current through the secondary of any of the CTs, crosses it’s predetermined values for a predetermined time, the bimetallic strip is over heated and it deforms which ultimately causes to operate the relay 49. As soon as the relay 49 is operated, the NC contacts 1 and 2 are opened which de-energizes the starter coil and hence stop the
motor. An other thing we have to remember during providing motor thermal overload protection. Actually every motor does have some predetermined overload tolerance value. That means every motor may run beyond its rated load for a specific allowable period depending on its loading condition. How long a motor can run safely for a particular load is specified by the manufacturer. The relation between different loads on motor and corresponding allowable periods for running the same in safe condition is referred as thermal limit curve of the motor. Let’s look at the curve of a particular motor, given
below. Here Y axis or vertical axis represents the allowable time in seconds and X axis or horizontal axis represents percentage of overload. Here it is clear from the curve that, motor can run safely without any damage due to overheating for prolonged period at 100 % of the rated load. It can run safely 1000 seconds at 200 % of normal rated load. It can run safely 100 seconds at 300 % of normal rated load. It can run safely 15 seconds at 600 % of normal rated load. The upper portion of the curve represents the normal running condition of the rotor and the lower most portion represents the mechanical locked condition of the rotor. Now the operating time Vs actuating current curve of the chosen thermal over load relay should be situated below the thermal limit curve of the motor for satisfactory and safe operation. Let’s have a discussion on more details-
Remember the characteristics of starting current of the motor – During start up of the induction motor, the stator current goes beyond 600 % of normal rated current but it stays up to 10 to 12 seconds after that stator current suddenly falls to normal rated value. So if the thermal overload relay is operated before that 10 to 12 seconds for the current 600 % of normal rated then the motor cannot be started. Hence, it can be concluded that the operating time Vs actuating current curve of the chosen thermal over load relay should be situated below the thermal limit curve of the motor but above the starting current characteristics curve of the motor. Probable position of the thermal current relay characteristics is bounded by these two said curves as shown in the graph by highlighted area. Another thing has to be remembered during choosing of thermal overload relay. This relay is not an instantaneous relay. It has a minimum delay in operation as the bimetallic strip required a minimum time to be heated up and deformed for maximum value of operating current. From the graph it is found that the thermal relay will be operated after 25 to 30 seconds if either the rotor is suddenly mechanically blocked or motor is fail to start. At this situation the motor will draw a huge current from the supply. If the motor is not isolated sooner, severer damage may occur.
This problem is overcome by providing time over current relay with high pickup. The time current characteristics of these over current relays are so chosen that for lower value of over load, the relay will not operate since thermal overload relay will be actuated before it. But for higher value of overload and for blocked rotor condition time over load relay will be operated instead of thermal relay because former will actuate much before the latter. Hence both the bimetallic over load relay and time over current relay are provided for complete motor thermal overload protection. There is one main disadvantage of bimetallic thermal over load relay, as the rate of heating and cooling of bi-metal is affected by ambient temperature, the performance of the relay may differ for different ambient temperatures. This problem can be overcome by using RTD or resistance temperature detector. The bigger and more sophisticated motors are protected against thermal over load more accurately by using RTD. In stator slots, RTDs are placed along with stator winding. Resistance of the RTD changes with changing temperature and this changed resistive value is sensed by a Wheatstone bridge circuit. This motor thermal overload protection scheme is very simple. RTD of stator is used as one arm of balanced Wheatstone bridge. The amount of current through the relay 49 depends upon the degree of unbalancing of the bridge. As the temperature of the stator winding is increased, the electrical resistance of the detector increases which disturbs the balanced condition of the bridge. As a result current start flowing through the relay 49 and the relay will be
actuated after a predetermined value of this unbalanced current and ultimately starter contact will open to stop the supply to the motor.
Electrical Fault Calculation | Positive Negative Zero Sequence Impedance. Before applying proper electrical protection system, it is necessary to have through knowledge of the conditions of electrical power system during faults. The knowledge of electrical fault condition is required to deploy proper different protective relays in different locations of electrical power system. Information regarding values of maximum and minimum fault currents, voltages under those faults in magnitude and phase relation with respect to the currents at different parts of power system, to be gathered for proper application of protection relay system in those different parts of the electrical power system. Collecting the information from different parameters of the system is generally known as electrical fault calculation. Fault calculation broadly means calculation of fault current in any electrical power system. There are mainly three steps for calculating faults in a system. 1. Choice of impedance rotations. 2. Reduction of complicated electrical power system network to single equivalent impedance. 3. Electrical fault currents and voltages calculation by using symmetrical component theory. Impedance Notation of Electrical Power System If we look at any electrical power system, we will find, these are several voltage levels. For example, suppose a typical power system where electrical power is generated at 6.6 kV then that 132 kV power is transmitted to terminal substation where it is stepped down to 33 kV and 11 kV levels and this 11 kV level may further step down to 0.4kv. Hence from this example it is clear that a same power system network may have different voltage levels. So calculation of fault at any location of the said system becomes much difficult and complicated it try to calculate impedance of different parts
of the system according to their voltage level. This difficulty can be avoided if we calculate impedance of different part of the system in reference to a single base value. This technique is called impedance notation of power system. In other wards, before electrical fault calculation, the system parameters, must be referred to base quantities and represented as uniform system of impedance in either ohmic, percentage, or per unit values. Electrical power and voltage are generally taken as base quantities. In three phase system, three phase power in MVA or KVA is taken as base power and line to line voltage in KV is taken as base voltage. The base impedance of the system can be calculated from these base power and base voltage, as
follows, Per unit is an impedance value of any system is nothing but the radio of actual impedance of the system to the base
impedance value.
Percentage impedance value can be
calculated by multiplying 100 with per unit value. Again it is sometimes required to convert per unit values referred to new base values for simplifying different electrical fault calculations. In that case,
The choice of impedance notation depends upon the complicity of the system. Generally base voltage of a system is so chosen that it requires minimum number of transfers. Suppose, one system as a large number of 132 KV over head lines, few numbers of 33 KV lines and very few number of 11 KV lines. The base voltage of the system can be chosen either as 132 KV or 33 KV or 11 KV, but here the best base voltages 132 KV, because it requires minimum number of transfer during fault calculation. Network Reduction After choosing the correct impedance notation, the next step is to reduce network to a single impedance. For this first we have to convert the impedance of all generators, lines, cables, transformer to a common base value. Then we prepare a schematic diagram of electrical power system showing the impedance referred to same base value of all those generators, lines, cables and transformers. The network then reduced to a common
equivalent single impedance by using star/delta transformations. Separate impedance diagrams should be prepared for positive, negative and zero sequence networks. There phase faults are unique since they are balanced i.e. symmetrical in three phase, and can be calculated from the single phase positive sequence impedance diagram. Therefore three phase fault current
is obtained by, Where, I f is the total three phase fault current, v is the phase to neutral voltage z 1 is the total positive sequence impedance of the system; assuming that in the calculation, impedance are represented in ohms on a voltage base. Symmetrical Component Analysis The above fault calculation is made on assumption of three phase balanced system. The calculation is made for one phase only as the current and voltage conditions are same in all three phases. When actual faults occur in electrical power system, such as phase to earth fault, phase to phase fault and double phase to earth fault, the system becomes unbalanced means, the conditions of voltages and currents in all phases are no longer symmetrical. Such faults are solved by symmetrical component analysis. Generally three phase vector diagram may be replaced by three sets of balanced vectors. One has opposite or negative phase rotation, second has positive phase rotation and last one is co-phasal. That means these vectors sets are described as negative, positive and zero sequence, respectively.
The equation between phase and sequence quantities are,
Therefore,
Where all quantities are referred to the reference phase r. Similarly a set of equations can be written for sequence currents also. From, voltage and current equations, one can easily determine the sequence impedance of the system. The development of symmetrical component analysis depends upon the fact that in balanced system of impedance, sequence currents can give rise only to voltage drops of the same sequence. Once the sequence networks are available, these can be converted to single equivalent impedance. Let us consider Z 1, Z2 and Z0 are the impedance of the system to the flow of positive, negative and zero
sequence current respectively. For earth fault
Phase to phase faults
Double phase to earth
faults Three phase faults If fault current in any particular branch of the network is required, the same can be calculated after combining the sequence components flowing in that branch. This involves the distribution of sequence components currents as determined by solving the above equations, in their respective network according to their relative impedance. Voltages it any point of the network can also be determine once the sequence component currents and sequence impedance of each branch are known. Sequence Impedance Positive Sequence Impedance The impedance offered by the system to the flow of positive sequence current is called positive sequence impedance. Negative Sequence Impedance The impedance offered by the system to the flow of negative sequence current is called negative sequence impedance. Zero Sequence Impedance The impedance offered by the system to the flow of zero sequence current is known as zero sequence impedance. In previous fault calculation, Z1, Z2 and Z0 are positive, negative and zero sequence impedance respectively. The sequence impedance varies with the type of power system components under consideration:1.
In static and balanced power system components like transformer and lines, the sequence impedance offered by the system are the same for positive and negative sequence currents. In other words, the positive sequence impedance and negative sequence impedance are same for transformers and power lines.
2.
But in case of rotating machines the positive and negative sequence impedance are different. 3. The assignment of zero sequence impedance values is a more complex one. This is because the three zero sequence current at any point in a electrical power system, being in phase, do not sum to zero but must return through the neutral and /or earth. In three phase transformer and machine fluxes due to zero sequence components do not sum to zero in the yoke or field system. The impedance very widely depending upon the physical arrangement of the magnetic circuits and winding. 1. The reactance of transmission lines of zero sequence currents can be about 3 to 5 times the positive sequence current, the lighter value being for lines without earth wires. This is because the spacing between the go and return(i.e. neutral and/or earth) is so much greater than for positive and negative sequence currents which return (balance) within the three phase conductor groups. 2. The zero sequence reactance of a machine is compounded of leakage and winding reactance, and a small component due to winding balance (depends on winding tritch). 3. The zero sequence reactance of transformers depends both on winding connections and upon construction of core. External and Internal Faults in Transformer. External Faults in Power Transformer External Short - Circuit of Power Transformer The short - circuit may occurs in two or three phases of electrical power system. The level of fault current is always high enough. It depends upon the voltage which has been short - circuited and upon the impedance of the circuit up to the fault point. The copper loss of the fault feeding transformer is abruptly increased. This increasing copper loss causes internal heating in the transformer. Large fault current also produces severe mechanical stresses in the transformer. The maximum mechanical stresses occurs during first cycle of symmetrical fault current. High Voltage Disturbance in Power Transformer High voltage disturbance in power transformer are of two kinds, 1. 2.
Transient Surge Voltage Power Frequency Over Voltage
Transient Surge Voltage High voltage and high frequency surge may arise in the power system due to any of the following causes,
Arcing ground if neutral point is isolated. Switching operation of different electrical equipment. Atmospheric Lightening Impulse. Whatever may be the causes of surge voltage, it is after all a traveling wave having high and steep wave form and also having high frequency. This wave travels in the electrical power system network, upon reaching in the power transformer, it causes breakdown the insulation between turns adjacent to line terminal, which may create short circuit between turns. Power Frequency Over Voltage There may be always a chance of system over voltage due to sudden disconnection of large load. Although the amplitude of this voltage is higher than its normal level but frequency is same as it was in normal condition. Over voltage in the system causes an increase in stress on the insulation of transformer. As we know that, voltage V = 4.44Φ.f.T ⇒ V ∝ Φ, increased voltage causes proportionate increase in the working flux. This therefore causes, increased in iron loss and dis - proportionately large increase in magnetizing current. The increase flux is diverted from the transformer core to other steel structural parts of the transformer. Core bolts which normally carry little flux, may be subjected to a large component of flux diverted from saturated region of the core alongside. Under such condition, the bolt may be rapidly heated up and destroys their own insulation as well as winding insulation. Under Frequency Effect in Power Transformer As, voltage V = 4.44Φ.f.T ⇒ V ∝ Φ.f as the number of turns in the winding is fixed. Therefore, Φ ∝ V/f From, this equation it is clear that if frequency reduces in a system, the flux in the core increases, the effect are more or less similar to that of the over voltage. Internal Faults in Power Transformer The principle faults which occurs inside a power transformer are categorized as, 1. 2. 3. 4.
Insulation breakdown between winding and earth Insulation breakdown in between different phases Insulation breakdown in between adjacent turns i.e. inter - turn fault Transformer core fault
Internal Earth Faults in Power Transformer
Internal Earth Faults in a Star Connected Winding with Neutral Point Earthed through an Impedance In this case the fault current is dependent on the value of earthing impedance and is also proportional to the distance of the fault point from neutral point as the voltage at the point depends upon, the number of winding turns come under across neutral and fault point. If the distance between fault point and neutral point is more, the number of turns come under this distance is also more, hence voltage across the neutral point and fault point is high which causes higher fault current. So, in few words it can be said that, the value of fault current depends on the value of earthing impedance as well as the distance between the faulty point and neutral point. The fault current also depends up on leakage reactance of the portion of the winding across the fault point and neutral. But compared to the earthing impedance,it is very low and it is obviously ignored as it comes in series with comparatively much higher earthing impedance. Internal Earth Faults in a Star Connected Winding with Neutral Point Solidly Earthed In this case, earthing impedance is ideally zero. The fault current is dependent up on leakage reactance of the portion of winding comes across faulty point and neutral point of transformer. The fault current is also dependent on the distance between neutral point and fault point in the transformer. As said in previous case the voltage across these two points depends upon the number of winding turn comes across faulty point and neutral point. So in star connected winding with neutral point solidly earthed, the fault current depends upon two main factors, first the leakage reactance of the winding comes across faulty point and neutral point and secondly the distance between faulty point and neutral point. But the leakage reactance of the winding varies in complex manner with position of the fault in the winding. It is seen that the reactance decreases very rapidly for fault point approaching the neutral and hence the fault current is highest for the fault near the neutral end. So at this point, the voltage available for fault current is low and at the same time the reactance opposes the fault current is also low, hence the value of fault current is high enough. Again at fault point away from the neutral point, the voltage available for fault current is high but at the same time reactance offered by the winding portion between fault point and neutral point is high. It can be noticed that the fault current stays a very high level throughout the winding. In other word, the fault current maintain a very high magnitude irrelevant to the position of the fault on winding. Internal Phase to Phase Faults in Power Transformer
Phase to phase fault in the transformer are rare. If such a fault does occur, it will give rise to substantial current to operate instantaneous over current relay on the primary side as well as the differential relay. Inter Turns Fault in Power Transformer Power Transformer connected with electrical extra high voltage transmission system, is very likely to be subjected to high magnitude, steep fronted and high frequency impulse voltage due to lightening surge on the transmission line. The voltage stresses between winding turns become so large, it can not sustain the stress and causing insulation failure between inter - turns in some points. Also LV winding is stressed because of the transferred surge voltage. Very large number of Power Transformer failure arise from fault between turns. Inter turn fault may also be occurred due to mechanical forces between turns originated by external short circuit. Core Fault in Power Transformer In any portion of the core lamination is damaged, or lamination of the core is bridged by any conducting material causes sufficient eddy current to flow, hence, this part of the core becomes over heated. Some times, insulation of bolts (Used for tightening the core lamination together) fails which also permits sufficient eddy current to flow through the bolt and causing over heating. These insulation failure in lamination and core bolts causes severe local heating. Although these local heating, causes additional core loss but can not create any noticeable change in input and output current in the transformer, hence these faults can not be detected by normal electrical protection scheme. This is desirable to detect the local over heating condition of the transformer core before any major fault occurs. Excessive over heating leads to breakdown of transformer insulating oil with evolution of gases. These gases are accumulated in Buchholz relay and actuating Buchholz Alarm. Backup Protection of Transformer | Over Current and Earth Fault. Over Current and Earth Fault Protection of Transformer Backup protection of electrical transformer is simple Over Current and Earth Fault protection applied against external short circuit and excessive over loads. These over current and earth Fault relays may be of Inverse Definite Minimum Time (IDMT) or Definite Time type relays. Generally IDMT relays are connected to the in-feed side of the transformer. The over current relays can not distinguish between external short circuit, over load and internal faults of the transformer. For any of the above fault, backup protection i.e. over current and earth fault protection connected to in-feed side of the transformer will operate.
Backup protection is although generally installed at in feed side of the transformer, but it should trip both the primary and secondary circuit
breakers of the transformer. Over Current and Earth Fault protection relays may be also provided in load side of the transformer too, but it should not inter trip the primary side circuit breaker like the case of backup protection at in-feed side. The operation is governed primarily by current and time settings and the characteristic curve of the relay. To permit use of over load capacity of the transformer and coordination with other similar relays at about 125 to 150 % of full load current of the transformer but below the minimum short circuit current. Backup protection of transformer has four elements, three over current relays connected each in each phase and one earth fault relay connected to the common point of three over current relays as shown in the figure. The normal range of current settings available on IDMT over current relays is 50 % to 200 % and on earth fault relay 20 to 80 %.
Another range of setting on earth fault relay is also available and may be selected where the earth fault current is restricted due to insertion of impedance in the neutral grounding. In the case of transformer winding with neutral earthed, unrestricted earth fault protection is obtained by connecting an ordinary earth fault relay across a neutral current transformer. The unrestricted over current and earth fault relays should have proper time lag to co-ordinate with the protective relays of other circuit to avoid indiscriminate tripping.
Transformer Protection and Transformer Fault. There are different kinds of transformers such as two winding or three winding electrical power transformers, auto transformer, regulating transformers, earthing transformers, rectifier transformers etc. Different transformers demand different schemes of transformer protection depending upon their importance, winding connections, earthing methods and mode of operation etc.It is common practice to provide Buchholz relay protection to all 0.5 MVA and above transformers. While for all small size distribution transformers, only high voltage fuses are used as main protective device. For all larger rated and important distribution transformers, over current protection along with restricted earth fault protectionis applied. Differential protection should be provided in the transformers rated above 5 MVA. Depending upon the normal service condition, nature of transformer faults, degree of sustained over load, scheme of tap changing, and many other factors, the suitable transformer protection schemes are chosen. Nature of Transformer Faults Although an electrical power transformer is a static device, but internal stresses arising from abnormal system conditions, must be taken into
consideration. A transformer generally suffers from following types of transformer fault1. Over current due to overloads and external short circuits, 2. Terminal faults, 3. Winding faults, 4. Incipient faults. All the above mentioned transformer faults cause mechanical and thermal stresses inside the transformer winding and its connecting terminals. Thermal stresses lead to overheating which ultimately affect the insulation system of transformer. Deterioration of insulation leads to winding faults. Some time failure of transformer cooling system, leads to overheating of transformer. So the transformer protection schemes are very much required. The short circuit current of an electrical transformer is normally limited by its reactance and for low reactance, the value of short circuit current may be excessively high. The duration of external short circuits which a transformer can sustain without damage as given in BSS 171:1936. Transformer % reactance
Permitted fault duration in seconds
4%
2
5%
3
6%
4
7 % and over
5
The general winding faults in transformer are either earth faults or interturns faults. Phase to phase winding faults in a transformer is rare. The phase faults in an electrical transformer may be occurred due to bushing flash over and faults in tap changer equipment. Whatever may be the faults, the transformer must be isolated instantly during fault otherwise major breakdown may occur in the electrical power system. Incipient faults are internal faults which constitute no immediate hazard. But it these faults are over looked and not taken care of, these may lead to major faults. The faults in this group are mainly inter-lamination short circuit due to insulation failure between core lamination, lowering the oil level due to oil leakage, blockage of oil flow paths. All these faults lead to overheating. So transformer protection scheme is required for incipient transformer faults also. The earth fault, very nearer to neutral point of transformer star winding may also be considered as an incipient fault. Influence of winding connections and
earthing on earth fault current magnitude. There are mainly two conditions for earth fault current to flow during winding to earth faults, 1. A current exists for the current to flow into and out of the winding. 2. Ampere-turns balance is maintained between the windings. The value of winding earth fault current depends upon position of the fault on the winding, method of winding connection and method of earthing. The star point of the windings may be earthed either solidly or via a resistor. On delta side of the transformer the system is earthed through an earthing transformer. Grounding or earthing transformer provides low impedance path to the zero sequence current and high impedance to the positive and negative sequence currents. Star Winding with Neutral Resistance Earthed In this case the neutral point of the transformer is earthed via a resistor and the value of impedance of it, is much higher than that of winding impedance of the transformer. That means the value of transformer winding impedance is negligible compared to impedance of earthing resistor. The value of earth current is, therefore, proportional to the position of the fault in the winding. As the fault current in the primary winding of the transformer is proportional to the ratio of the short circuited secondary turns to the total turns on the primary winding, the primary fault current will be proportional to the square of the percentage of winding short circuited. The variation of fault current both in the primary and secondary winding is shown below. Star Winding with Neutral Solidly Earthed In this case the earth fault current magnitude is limited solely by the winding impedance and the fault is no longer proportional to the position of the fault. The reason for this non linearity is unbalanced flux linkage. Protection of Lines or Feeder. As the length of electrical power transmission line is generally long enough and it runs through open atmosphere, the probability of occurring fault in electrical power transmission line is much higher than that of electrical power transformers and alternators. That is why a transmission line requires much more protective schemes than a transformer and an alternator.Protection of line should have some special features, such asDuring fault, the only circuit breaker closest to the fault point should be tripped. 2. If the circuit breaker closest the faulty point, fails to trip the circuit breaker just next to this breaker will trip as back up. 1.
3.
The operating time of relay associated with protection of line should be as minimum as possible in order to prevent unnecessary tripping of circuit breakers associated with other healthy parts of power system.
These above mentioned requirements cause protection of transmission line much different from protection of transformer and other equipment of power systems. The main three methods of transmission line protection are 1. Time graded over current protection. 2. Differential protection. 3. Distance protection. Time Graded Over Current Protection This may also be referred simply as over-current protection of electrical power transmission line. Let' discuss different schemes of time graded over current protection. Protection of Radial Feeder In radial feeder, the power flows in one direction only, that is from source to load. This type of feeders can easily protected by using either definite time relays or inverse time relays. Line Protection by Definite Time Relay This protection scheme is very simple. Here total line is divided into different sections and each section is provided with definite time relay. The relay nearest to the end of the line has minimum time setting while time setting of other relays successively increased, towards the source. For example, suppose there is a source at point A, in the figure below
At point D the circuit breaker CB-3 is installed with definite time of relay operation 0.5 sec. Successively, at point C an other circuit breaker CB-2 is installed with definite time of relay operation 1 sec. The next circuit breaker CB-1 is installed at point B which is nearest of the point A. At point B, the relay is set at time of operation 1.5 sec. Now, assume a fault occurs at point F. Due to this fault, the faulty current flow through all the current transformers or CTs connected in the line. But as the time of operation of relay at point D is minimum the CB-3, associated with this relay will trip first to isolate the faulty zone from rest part of the line. In case due to any reason, CB-3 fails to trip, then next higher timed relay will operate the associated CB to trip. In this case, CB-2 will trip. If CB2 also fails to trip, then next circuit breaker i.e. CB-1 will trip to isolate major portion of the line. Advantages of Definite Time Line Protection The main advantage of this scheme is simplicity. The second major advantage is, during fault, only nearest CB towards the source from fault point will operate to isolate the specific position of the line. Disadvantage of Definite Time Line Protection If the number of sections in the line is quite large, the time setting of relay nearest to the source, would be very long. So during any fault nearer to the source will take much time to be isolated. This may cause severe destructive effect on the system.
Over Current Line Protection by Inverse Relay The drawback as we discussed just in definite time over current protection of transmission line, can easily be overcome by using inverse time relays. In inverse relay the time of operation is inversely proportional to fault current.
In the above figure, overall time setting of relay at point D is minimum and successively this time setting is increased for the relays associated with the points towards the point A. In case of any fault at point F will obviously trip CB-3 at point D. In failure of opening CB-3, CB-2 will be operated as overall time setting is higher in relay at point C. Although, the time setting of relay nearest to the source is maximum but still it will trip in shorter period, if major fault occurs near the source, as the time of operation of relay is inversely proportional to faulty current. Over Current Protection of Parallel Feeders For maintaining stability of the system it is required to feed a load from source by two or more than two feeders in parallel. If fault occurs in any of the feeders, only that faulty feeder should be isolated from the system in order to maintain continuity of supply from source to load. This requirement makes the protection of parallel feeders little bit more complex than simple non direction over current protection of line as in the case of radial feeders. The protection of parallel feeder requires to use directional relays and to grade the time setting of relay for selective tripping.
There are two feeders connected in parallel from source to load. Both of the feeders have non-directional over current relay at source end. These relays should be inverse time relay. Also both of the feeders have directional relay or reverse power relay at their load end. The reverse power relays used here should be instantaneous type. That means these relays should be operated as soon as flow of power in the feeder is reversed. The normal direction of power from source to load. Now, suppose a fault occurs at point F, say the fault current is If. This fault will get two parallel paths from source, one through circuit breaker A only and other via CB-B, feeder-2, CB-Q, load bus and CB-P. This is clearly shown in figure below, where IA and IB are current of fault shared by feeder-1 and feeder-2 respectively.
As per Kirchoff's current law, IA + IB = If. Now, IA is flowing through CB-A, IB is flowing through CB-P. As the direction of flow of CB-P is reversed it will trip instantly. But CB-Q will not trip as flow of current (power) in this circuit breaker is not reversed. As soon as CB-P is tripped, the fault current IB stops flowing through feeder and hence there is no question of further operating of inverse time over current relay. I A still continues to flow even CB-P is tripped. Then because of over current I A, CBA will trip. In this way the faulty feeder is isolated from system. Differential Pilot Wire Protection This is simply a differential protection scheme applied to feeders. Several differential schemes are applied for protection of line but Mess Price Voltage balance system and Translay Scheme are most popularly used. Merz Price Balance System The working principle of Merz Price Balance system is quite simple. In this scheme of line protection, identical CT is connected to each of the both ends of the line. The polarity of the CTs are same. The secondary of these current transformer and operating coil of two instantaneous relays are formed a closed loop as shown in the figure below. In the loop pilot wire is used to connect both CT secondary and both relay coil as shown.
Now, from the figure it is quite clear that when the system is under normal condition, there would not be any current flowing through the loop. As the secondary current of one CT will cancel out secondary current of other CT. Now, if any fault occurs in the portion of the line between these two CTs, the secondary current of one CT will no longer equal and opposite of secondary current of other CT. Hence there would be a resultant circulating current in the loop. Due this circulating current, the coil of both relays will close the trip circuit of associate circuit breaker. Hence, the faulty line will be isolated from both ends. Busbar Protection | Busbar Differential Protection Scheme. In early days only conventional over current relays were used for busbar protection. But it is desired that fault in any feeder or transformer connected to the busbar should not disturb busbar system. In viewing of this time setting of busbar protection relays are made lengthy. So when faults occurs on busbar itself, it takes much time to isolate the bus from source which may came much damage in the bus system.In recent days, the second zone distance protection relays on incoming feeder, with operating time of 0.3 to 0.5 seconds have been applied for busbar protection. But this scheme has also a main disadvantage. This scheme of protection can not discriminate the faulty section of the busbar. Now days, electrical power system deals with huge amount of power. Hence any interruption in total bus system causes big loss to the company. So it becomes essential to isolate only faulty section of busbar during bus fault.
Another drawback of second zone distance protection scheme is that, sometime the clearing time is not short enough to ensure the system stability. To overcome the above mentioned difficulties, differential busbar protection scheme with an operating time less than 0.1 sec., is commonly applied to many SHT bus systems. Differential Busbar Protection Current Differential Protection The scheme of busbar protection, involves, Kirchoff’s current law, which states that, total current entering an electrical node is exactly equal to total current leaving the node. Hence, total current entering into a bus section is equal to total current leaving the bus section. The principle of differential busbar protection is very simple. Here, secondaries of CTs are connected parallel. That means, S1 terminals of all CTs connected together and forms a bus wire. Similarly S 2 terminals of all CTs connected together to form another bus wire. A tripping relay is connected across these two bus wires.
Here, in the figure above we assume that at normal condition feed, A, B, C, D, E and F carries current IA, IB, IC, ID, IE and IF. Now, according to Kirchoff’s current law, Essentially all the CTs used for differential busbar protection are of same current ratio. Hence, the summation of all secondary currents must also be equal to zero. Now, say current through the relay connected in parallel with
all CT secondaries, is iR, and iA, iB, iC, iD, iE and iF are secondary currents. Now, let us apply KCL at node X. As per KCL at node X,
So, it is clear that under normal condition there is no current flows through the busbar protection tripping relay. This relay is generally referred as Relay 87. Now, say fault is occurred at any of the feeders, outside the protected zone. In that case, the faulty current will pass through primary of the CT of that feeder. This fault current is contributed by all other feeders connected to the bus. So, contributed part of fault current flows through the corresponding CT of respective feeder. Hence at that faulty condition, if we apply KCL at node K, we will still get, iR = 0.
That means, at external faulty condition, there is no current flows through relay 87. Now consider a situation when fault is occurred on the bus itself.
At this condition, also the faulty current is contributed by all feeders connected to the bus. Hence, at this condition, sum of all contributed fault current is equal to total faulty current. Now, at faulty path there is no CT. (in external fault, both fault current and contributed current to the fault by different feeder get CT in their path of flowing).
The sum of all secondary currents is no longer zero. It is equal to secondary equivalent of faulty current. Now, if we apply KCL at the nodes, we will get a non zero value of iR. So at this condition current starts flowing through 87 relay and it makes trip the circuit breakercorresponding to all the feeders connected to this section of the busbar. As all the incoming and outgoing feeders, connected to this section of bus are tripped, the bus becomes dead. This differential busbar protection scheme is also referred as current differential protection of busbar. Differential Protection of Sectionalized Bus During explaining working principle of current differential protection of busbar, we have shown a simple non sectionalized busbar. But in moderate high voltage system electrical bus sectionalized in than one sections to increase stability of the system. It is done because, fault in one section of bus should not disturb other section of the system. Hence during bus fault, total bus would be interrupted.
Let us draw and discuss about protection of busbar with two sections.
Here, bus section A or zone A is bounded by CT 1, CT2 and CT3 where CT1 and CT2 are feeder CTs and CT3 is bus CT. Similarly bus section B or zone B is bounded by CT4, CT5 and CT6where CT4 is bus CT, CT5 and CT6 are feeder CT. Therefore, zone A and B are overlapped to ensure that, there is no zone left behind this busbar protection scheme. ASI terminals of CT1, 2 and 3 are connected together to form secondary bus ASI BSI terminals of CT 4, 5 and 6 are connected together to form secondary bus BSI. S 2 terminals of all CTs are connected together to form a common bus S 2. Now, busbar protection relay 87A for zone A is connected across bus ASI and S 2. Relay 87B for zone B is connected across bus BSI and S 2. This section busbar differential protection scheme operates in some manner simple current differential protection of busbar. That is, any fault in zone A, with trip only CB 1, CB2and bus CB. Any fault in zone B, will trip only CB 5, CB6 and bus CB. Hence, fault in any section of bus will isolate only that portion from live system. In current differential protection of busbar, if CT secondary circuits, or bus wires is open the relay may be operated to isolate the bus from live system. But this is not desirable. DC Circuit of Differential Busbar Protection
A typical DC circuit for busbar differential protection scheme is given
below. Here, CSSA and CSSB are two selector switch which are used to put into service, the busbar protection system for zone A and zone B respectively. If CSSA is in “IN” position, protection scheme for zone A is in service. If CSSB is in “IN” position, protection for zone B is in service. Generally both of the switches are in “IN’ position in normal operating condition. Here, relay coil of 96A and 96B are in series with differential busbar protection relay contact 87A-1 and 87B-1 respectively. 96A relay is multi contacts relay. Each circuit breaker in
zone A is connected with individual contact of 96A. Similarly, 96B is multi contacts relay and each circuit breaker in zone-B is connected with individual contacts of 96B. Although here we use only one tripping relay per protected zone, but this is better to use one individual tripping relay per feeder. In this scheme one protective relay is provided per feeder circuit breaker, whereas two tripping relays one for zone A and other for zone B are provided to bus section or bus coupler circuit breaker. On an interval fault in zone A or bus section A, the respective bus protection relay 87A, be energized whereas during internal fault in zone B, the respective relay 87B will be energized. As soon as relay coil of 87A or 87B is energized respective no. contact 87A-1 or 87B-1 is closed.Hence, the tripping relay 96 will trip the breakers connected to the faulty zone. To indicate whether zone A or B busbar protection operated, relay 30 is used. For example, if relay 87A is operated, corresponding “No” contact 87A-2 is closed which energized relay 30A. Then the No contact 30A-1 of relay 30A is closed to energized alarm relay 74. Supervision relay 95 of respective zone is also energized during internal fault, but it has a time delay of 3 second. So, it reset as soon as the fault is cleared and therefore does not pick up zone bus wire shorting relay 95x which in turn shorts out the bus wires. An alarm contact is also given to this auxiliary 95x relay to indicate which CT is open circuited. No volt relay 80 is provided in both trip and non-trip section of the DC circuit of differential busbar protection system to indicate any discontinuity of D. C. supply. Voltage Differential Protection of Busbar The current differential scheme is sensitive only when the CTs do not get saturated and maintain same current ratio, phase angle error under maximum faulty condition. This is usually not 80, particularly, in the case of an external fault on one of the feeders. The CT on the faulty feeder may be saturated by total current and consequently it will have very large errors. Due to this large error, the summation of secondary current of all CTs in a particular zone may not be zero. So there may be a high chance of tripping of all circuit breakers associated with this protection zone even in the case of an external large fault. To prevent this maloperation of current differential busbar protection, the 87 relays are provided with high pick up current and enough time delay. The greatest troublesome cause of current transformer saturation is the transient dc component of the short circuit current. This difficulties can be overcome by using air core CTs. This current transformer is also called linear coupler. As the core of the CT does not use iron the secondary characteristic of these CTs, is straight line. In voltage differential busbar protection the CTs of all incoming and outgoing feeders are connected in series instead of connecting them in parallel.
The secondaries of all CTs and differential relay form a closed loop. If polarity of all CTs are properly matched, the sum of voltage across all CT secondaries is zero. Hence there would be no resultant voltage appears across the differential relay. When a buss fault occurs, sum of the all CT secondary voltage is no longer zero. Hence, there would be current circulate in the loop due to the resultant voltage. As this loop current also flows through the differential relay, the relay is operated to trip all the circuit beaker associated with protected bus zone. Except when ground fault current is severally limited by neutral impedance there is usually no selectivity problem when such a problem exists, it is solved by use of an additional more sensitive relaying equipment including a supervising protective relay. Protection of Capacitor Bank. Like other electrical equipments, shunt capacitor may also be subjected to internal and external electrical faults. Hence this equipment is also to be protected from internal and external faults. There are numbers of schemes available for protection of capacitor bank, but during applying any of the schemes, we should remember the initial investment on that capacitor for economical point of view. We should compare the initial investment in the capacitor and the cost of the protection applying on it. There are mainly 3 types of protection arrangement are applied to a capacitor bank. 1. 2.
3.
Element Fuse. Unit. Fuse. Bank Protection.
Element Fuses Manufacturers of capacitor unit commonly provide inbuilt fuse in each element of the unit. In this case, if any fault occurs in any element itself, it is automatically disconnected from rest of the unit. In this case, the unit still serves its purpose, but with smaller output. In smaller rated capacitor bank only these inbuilt protection scheme is applied to avoid the expenditure of other special protective equipments. Unit Fuse The unit fuse protection is generally provided to limit the duration of arc inside a faulty capacitor unit. As the arc duration is limited, there is less chance of major mechanical deformation and huge production of gas in the faulty unit, and hence the neighborhood units of the bank are saved. If each unit of a capacitor bank is individually protected against fuse, then in case of failure of one unit, the capacitor bank can still be running without interruption before removing and replacing the faulty unit. Another major advantage of providing fuse protection to each unit of the bank is that, it indicates the exact location of the faulty unit. But during choosing the size of the fuse for this purpose, it should be taken into consideration that the fuse element must withstand the excessive loading due to harmonics in the system. In the view of that the current rating of the fuse element for this purpose is taken as 65 % above the full load current. Whenever the individual unit of capacitor bank is protected by fuse, it is necessary to provide discharge resistance in each of the units. Bank Protection Although in general fuse protection is provided with each of the capacitor units, but when a capacitor unit is under fault and the associated fuse element is blown out, the voltage stress increases to the other capacitor units connected in series in same row. Generally, each capacitor unit is designed for withstanding 110% of its normal rated voltage. If any other capacitor unit further becomes out of service, in the same row where previously one unit is damaged, the voltage stress upon other healthy units of that row will increase further and easily crosses the limit of maximum allowable, voltage of these units. Hence it is always desirable to replace damaged capacitor unit from the bank as soon as possible to avoid excess voltage stress on the other healthy units. Hence, there must be some indicating arrangement to identify the exact faulty unit. As soon as the faulty unit is identified in a bank, the bank should be removed from the service for replacing the faulty unit. There are several methods of sensing unbalance voltage caused by failure of capacitor unit. The figure below is showing the
most common arrangement of capacitor bank protection. Here, the capacitor bank is connected in star formation. Primary of a potential transformer is connected across each phase. The secondary of all three potential transformers are connected in series to form an open delta and a voltage sensitive relay is connected across this open delta. In exact balanced condition there must not be any voltage appears across the voltage sensitive relay because summation of balanced 3 phase voltages is zero. But when there would be any voltage unbalancing due to failure of capacitor unit, the resultant voltage will appear across the relay and the relay will be actuated for providing an alarm and trip signals. The voltage sensitive relay can be so adjusted that up to a certain voltage unbalancing only alarm contacts would be closed and for certain higher voltage level the trip contacts along with alarm contacts would be closed. The potential transformer connected across the capacitors of each phase also serves for discharging of the bank after being switched off.
In another scheme, the
capacitors in each phase are divided into two equal parts connected in series. Discharge coil is connected across each of the parts as shown in the figure. In between the secondary of discharge coil and the sensitive voltage unbalance the relay an auxiliary transformer is connected which serves to regulate the voltage difference between secondary voltages of discharge coil under normal conditions.
Here the capacitor bank is connected in star and the neutral point is connected to the ground through a potential transformer. A voltage sensitive relay is connected across the secondary of the potential transformer. As soon as there is any unbalance between the phases, the resultant voltage will appear across the potential transformer and hence the voltage sensitive relay will be actuated
beyond a preset value. Here, the capacitor bank of each phase is divided into two equal parts connected in parallel and the star points of both parts are interconnected through a current transformer. The secondary of the current transformer are connected across a current sensitive relay. In case any misbalancing occurs between the two parts of the bank, there would be a unbalance current flowing through the current transformer and hence the current sensitive relay will actuate. In this scheme for discharging the bank after switching off, discharge coil may be connected across the capacitors in each phase.
In another scheme of protection of capacitor bank, the star point of a three phase capacitor bank is connected to the ground through a current transformer and a current sensitive relay is connected across the secondary of the current transformer. As soon as there is any unbalancing between the phases of capacitor bank, there must be a current flowing to the ground through the current transformer and hence the current sensitive relay will be actuated to trip the circuit breaker associated with the capacitor bank.
Basic electrical design of a PLC panel (Wiring diagrams).
Building the PLC panel It is uncommon for engineers to build their own PLC panel designs (but not impossible of course). For example, once the electrical designs are complete, they must be built by an electrician. Therefore, it is your responsibility to effectively communicate your design intentions to the electricians through drawings. In some factories, the electricians also enter the ladder logic and do debugging. This article discusses the design issues in implementation that must be considered by the designer.
Electrical wiring diagrams of a PLC panel In an industrial setting a PLC is not simply “plugged into a wall socket”. The electrical design for each machine must include at least the following components. 1. 2.
Transformers – to step down AC supply voltages to lower levels Power contacts – to manually enable/disable power to the machine with e-stop buttons 3. Terminals – to connect devices
4. Fuses or circuit breakers – will cause power to fail if too much current is drawn 5. Grounding – to provide a path for current to flow when there is an electrical fault 6. Enclosure – to protect the equipment, and users from accidental contact A control system of a PLC panel will normally use AC and DC power at different voltage levels. Control cabinets are often supplied with single phase AC at 220/440/550V, or two phase AC at 220/440V AC, or three phase AC at 330/550V.
This power must be dropped down to a lower voltage level for the controls and DC power supplies. 110Vac is common in North America, and 220 V AC Is common in Europe and the Commonwealth countries. It is also common for a control cabinet to supply a higher voltage to other equipment, such as motors.
Motor controller example An example of a wiring diagram for a motor controller is shown in Figure 1. Note that symbols are discussed in detail later). Dashed lines indicate a single purchased component. This system uses 3 phase AC power (L1, L2 and L3) connected to the terminals. The three phases are then connected to a power interrupter. Next, all three phases are supplied to a motor starter that contains three contacts, M, and three thermal overload relays (breakers).
Figure 1 – A Motor Controller Schematic
The contacts, M, will be controlled by the coil, M. The output of the motor starter goes to a three phase AC motor. Power is supplied by connecting a step down transformer to the control electronics by connecting to phases L2 and L3. The lower voltage is then used to supply power to the left and right rails of the ladder below. The neutral rail is also grounded. The logic consists of two push buttons:
Start push button is normally open, so that if something fails the motor cannot be started. Stop push button is normally closed, so that if a wire or connection fails the system halts safely.
The system controls the motor starter coil M, and uses a spare contact on the starter, M, to seal in the motor starter. The diagram also shows numbering for the wires in the device. This is essential for industrial control systems that may contain hundreds or thousands of wires. These numbering schemes are often particular to each facility, but there are tools to help make wire labels that will appear in the final controls cabinet.
Figure 2 – A Physical Layout for the Control Cabinet Once the electrical design is complete, a layout for the controls cabinet is developed, as shown in Figure 2. The physical dimensions of the devices must be considered, and adequate space is needed to run wires between components. In the cabinet the AC power would enter at the terminal block, and be connected to the main breaker.
It would then be connected to the contactors and overload relays that constitute the motor starter. Two of the phases are also connected to the transformer to power the logic. The start and stop buttons are at the left of the box (note: normally these are mounted elsewhere, and a separate layout drawing would be needed). The final layout in the cabinet might look like the one shown in Figure 1.
Figure 3 – Final PLC Panel Wiring
When being built the system will follow certain standards that may be company policy, or legal requirements. This often includes items such as;
Hold downs – the will secure the wire so they don’t move Labels – wire labels help troubleshooting Strain reliefs – these will hold the wire so that it will not be pulled out of screw terminals Grounding – grounding wires may be needed on each metal piece for safety
A photograph of an industrial controls cabinet is shown in Figure 4:
Figure 4 – An industrial control cabinet with wire runs, terminal strip, buttons on PLC panel front, etc.
When including a PLC in the ladder diagram still remains. But, it does tend to become more complex. Figure 5 below shows a schematic diagram for a PLC based motor control system, similar to the previous motor control example. This figure shows the E-stop wired to cutoff power to all of the devices in the circuit, including the PLC. All critical safety functions should be hardwired this way.
Figure 5 – An Electrical Schematic with a PLC.
Electrical Distribution Architecture In Water Treatment Plants.
Water treatment plants For both drinking water and wastewater treatment, 4 different sizes of plants have been distinguished. The size of plants can be expressed in quantity of treated water per day, or in corresponding number of inhabitants. Four different types of (waste) water treatment plants have been distinguished, depending on destination and size:
T1 – Autonomous water treatment plant // See single-line diagram T2 – Small water or wastewater treatment plant // See single line diagram T3 – Medium sized water or wastewater treatment plant See single line diagrams // – Double-radial architecture and – Open medium voltage loop T4 – Large water or wastewater treatment plant See single line diagrams // – Radial-double feed – Open medium voltage loop Characteristics T1 T2 T3
T4
m3/day (drinking water or waste water)
1K-5K
5K-50K
50K-200K
200K-1000K
Inhabitants
1K-10K
10K-100K
100K-500K
500K-1000K
25-125 kVA
125-1250 kVA
1.25-5 MVA
5-25 MVA
Power demand
Electrical Distribution Guidance is given for the selection of Electrical Distribution architecture in water treatment plants. This includes the selection between different possible configurations of MV and LV circuits and the implementation of back-up power sources. The most relevant characteristics of the electrical installation are taken into account, such as typology, power demand, sensitivity to power interruptions, …
Connection to the utility network MV circuit configuration Configurations of LV circuits Backup generators Presence of uninterruptible power supply (UPS).
Commissioning tests of protection relays at site (before set to work).
Installation of protection relays Installation of protection relays at site creates a number of possibilities for errors in the implementation of the scheme to occur. Even if the scheme has been thoroughly tested in the factory, wiring to the CTs and VTs on site may be incorrectly carried out, or the CTs/VTs may have been incorrectly installed. The impact of such errors may range from simply being a nuisance (tripping occurs repeatedly on energisation, requiring investigation to locate and correct the errors) through to failure to trip under fault conditions, leading to major equipment damage, disruption to supplies and potential hazards to personnel.
The strategies available to remove these risks are many, but all involve some kind of testing at site. Commissioning tests at site are therefore invariably performed before protection equipment is set to work. The aims of commissioning tests are: 1. 2. 3.
To ensure that the equipment has not been damaged during transit or installation To ensure that the installation work has been carried out correctly To prove the correct functioning of the protection scheme as a whole
The tests carried out will normally vary according to the protection scheme involved, the relay technology used, and the policy of the client. In many cases, the tests actually conducted are determined at the time of commissioning by mutual agreement between the client’s representative and the commissioning team. The following tests are invariably carried out, since the protection scheme will not function correctly if faults exist.
Wiring diagram check, using circuit diagrams showing all the reference numbers of the interconnecting wiring General inspection of the equipment, checking all connections, wires on relays terminals, labels on terminal boards, etc. Insulation resistance measurement of all circuits [details] Perform relay self-test procedure and external communications checks on digital/numerical relays [details] Test main current transformers Polarity check Magnetisation Curve Test main voltage transformers Polarity check Ratio check
Phasing check Check that protection relay alarm/trip settings have been entered correctly [details] Tripping and alarm circuit checks to prove correct functioning In addition, the following checks may be carried out, depending on the factors noted above (not covered in this technical article):
Secondary injection test on each relay to prove operation at one or more setting values Primary injection tests on each relay to prove stability for external faults and to determine the effective current setting for internal faults (essential for some types of electromechanical relays) Testing of protection scheme logic
Insulation resistance tests All the deliberate earth connections on the wiring to be tested should first be removed, for example earthing links on current transformers, voltage transformers and DC supplies. Some insulation testers generate impulses with peak voltages exceeding 5kV. In these instances any electronic equipment should be disconnected while the external wiring insulation is checked. The insulation resistance should be measured to earth and between electrically separate circuits. The readings are recorded and compared with subsequent routine tests to check for any deterioration of the insulation. The insulation resistance measured depends on the amount of wiring involved, its grade, and the site humidity. Generally, if the test is restricted to one cubicle, a reading of several hundred megohms should be obtained. If long lengths of site wiring are involved, the reading could be only a few megohms. Protection relay self-test procedure Digital and numerical relays will have a self-test procedure that is detailed in the appropriate relay manual. These tests should be followed to determine if the relay is operating correctly.
This will normally involve checking of the relay watchdog circuit, exercising all digital inputs and outputs and checking that the relay analogue inputs are within calibration by applying a test current or voltage. For these tests, the relay outputs are normally disconnected from the remainder of the protection scheme, as it is a test carried out to prove correct relay, rather than scheme, operation.
To shorten testing and commissioning times of SIPROTEC relays, extensive test and diagnostic functions are available to the user in DIGSI 5
Unit protection schemes involve relays that need to communicate with each other. This leads to additional testing requirements. The communications path between the relays is tested using suitable equipment to ensure that the path is complete and that the received signal strength is within specification. Numerical relays may be fitted with loopback test facilities that enable either part of or the entire communications link to be tested from one end.
After completion of these tests, it is usual to enter the relay settings required. This can be done manually via the relay front panel controls, or using a portable PC and suitable software. Whichever, method is used, a check by a second person that the correct settings have been used is desirable, and the settings recorded. Programmable scheme logic that is required is also entered at this stage.
SIPROTE C relay wiring test editor for monitoring and testing of binary inputs, binary outputs and LED (click to expand)
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Current transformer tests
The following tests are normally carried out prior to energisation of the main circuits: checking of polarity and current transformer magnetisation curve.
Polarity checks. Each current transformer should be individually tested to verify that the primary and secondary polarity markings are correct (see Figure 1). The ammeter connected to the secondary of the current transformer should be a robust moving coil, permanent magnet, centre-zero type. A low voltage battery is used, via a single-pole push-button switch, to energise the primary winding. On closing the push-button, the DC ammeter, A, should give a positive flick and on opening, a negative flick.
Figure 1 – Current transformer polarity check
Go back to Commissioning tests ↑
Checking of magnetisation curve Several points should be checked on each current transformer magnetisation curve. This can be done by energising the secondary winding from the local mains supply through a variable auto-transformer while the primary circuit remains open. See Figure 2.
The characteristic is measured at suitable intervals of applied voltage, until the magnetising current is seen to rise very rapidly for a small increase in voltage. This indicates the approximate knee-point or saturation flux level of the current transformer.
The magnetising current should then be recorded at similar voltage intervals as it is reduced to zero.
Figure 2 – Testing current transformer magnetising curve
Care must be taken that the test equipment is suitably rated. The short-time current rating must be in excess of the CT secondary current rating, to allow for measurement of the saturation current. This will be in excess of the CT secondary current rating. As the magnetising current will not be sinusoidal, a moving iron or dynamometer type ammeter should be used. It is often found that current transformers with secondary ratings of 1A or less have a kneepoint voltage higher than the local mains supply. In these cases, a step-up interposing transformer must be used to obtain the necessary voltage to check the magnetisation curve.
Go back to Commissioning tests ↑
Voltage transformer tests Voltage transformers require testing for polarity, ratio and phasing.
Polarity check of voltage transformer The voltage transformer polarity can be checked using the method for CT polarity tests. Care must be taken to connect the battery supply to the primary winding, with the polarity ammeter connected to the secondary winding. If the voltage transformer is of the capacitor type, then the polarity of the transformer at the bottom of the capacitor stack should be checked. Go back to Commissioning tests ↑
Ratio check of VT This check can be carried out when the main circuit is first made live. The voltage transformer secondary voltage is compared with the secondary voltage shown on the nameplate.
Nam plate of a single phase voltage transformer (photo credit: emadrlc.blogspot.com)
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Phasing check of VT The secondary connections for a three-phase voltage transformer or a bank of three single-phase voltage transformers must be carefully checked for phasing. With the main circuit alive, the phase rotation is checked using a phase rotation meter connected across the three phases, as shown in Figure 3 below. Provided an existing proven VT is available on the same primary system, and that secondary earthing is employed, all that is now necessary to prove correct phasing is a voltage check between, say, both ‘A’ phase secondary outputs. There should be nominally little or no voltage if the phasing is correct. However, this test does not detect if the phase sequence is correct, but the phases are displaced by 120o from their correct position, i.e. phase A occupies the position of phase C or phase B in Figure 3.
This can be checked by removing the fuses from phases B and C (say) and measuring the phase-earth voltages on the secondary of the VT. If the phasing is correct, only phase A should be healthy, phases B and C should have only a small residual voltage.
Figure 3 – Voltage transformer phasing check
Correct phasing should be further substantiated when carrying out ‘on load’ tests on any phase-angle sensitive relays, at the relay terminals. Load current in a known phase CT secondary should be compared with the associated phase to neutral VT secondary voltage.
The phase angle between them should be measured, and should relate to the power factor of the system load. If the three-phase voltage transformer has a broken-delta tertiary winding, then a check should be made of the voltage across the two connections from the broken delta VN and VL, as shown in Figure 3 above. With the rated balanced three- phase supply voltage applied to the voltage transformer primary windings, the broken-delta voltage should be below 5V with the rated burden connected. Go back to Commissioning tests ↑
Protection relay setting checks (alarm and trip settings) At some point during commissioning, the alarm and trip settings of the relay elements involved will require to be entered and/or checked. Where the complete scheme is engineered and supplied by a single contractor, the settings may already have been entered prior to despatch from the factory, and hence this need not be repeated. The method of entering settings varies according to the relay technology used. For electromechanical and static relays, manual entry of the settings for each relay element is required. This method can also be used for digital/numerical relays. However, the amount of data to be entered is much greater, and therefore it is usual to use appropriate software, normally supplied by the manufacturer, for this purpose. The software also makes the essential task of making a record of the data entered much easier.
The maximum allowable step potential is 5,664V, which exceeds the step voltage calculated above and the earthing system passes the step potential criteria. Having passed both touch and step potential criteria, we can conclude that the earthing system is safe. Once the data has been entered, it should be checked for compliance with the recommended settings as calculated from the protection setting study. Where appropriate software is used for data entry, the checks can be considered complete if the data is checked prior to download of the settings to the relay. Otherwise, a check may required subsequent to data entry by inspection and recording of the relay settings, or it may be considered adequate to do this at the time of data entry. The recorded settings form an essential part of the commissioning documentation provided to the client.
FAULT CURRENT COMPUTATION USING SEQUENCE NETWORKS In this section we shall demonstrate the use of sequence networks in the calculation of fault currents using sequence network through some examples.
Example 8.4 Consider the network shown in Fig. 8.10. The system parameters are given below Generator G : 50 MVA, 20 kV, X" = X1 = X2 = 20%, X0 = 7.5%
Motor M : 40 MVA, 20 kV, X" = X1 = X2 = 20%, X0 = 10%, Xn = 5%
Transformer T1 : 50 MVA, 20 kV Δ /110 kVY, X = 10%
Transformer T2 : 50 MVA, 20 kV Δ /110 kVY, X = 10%
Transmission line: X1 = X2 = 24.2 Ω , X0 = 60.5 Ω
We shall find the fault current for when a (a) 1LG, (b) LL and (c) 2LG fault occurs at bus-2.
Fig. 8.10 Radial power system of Example 8.4. Let us choose a base in the circuit of the generator. Then the per unit impedances of the generator are:
The per unit impedances of the two transformers are
The MVA base of the motor is 40, while the base MVA of the total circuit is 50. Therefore the per unit impedances of the motor are
For the transmission line
Therefore
Let us neglect the phase shift associated with the Y/ Δ transformers. Then the positive, negative and zero sequence networks are as shown in Figs. 8.11-8.13.
Fig. 8.11 Positive sequence network of the power system of Fig. 8.10.
Fig. 8.12 Negative sequence network of the power system of Fig. 8.10.
Fig. 8.13 Zero sequence network of the power system of Fig. 8.10. From Figs. 8.11 and 8.12 we get the following Ybus matrix for both positive and negative sequences
Inverting the above matrix we get the following Zbus matrix
Again from Fig. 8.13 we get the following Ybus matrix for the zero sequence
Inverting the above matrix we get
Hence for a fault in bus-2, we have the following Thevenin impedances
Alternatively we find from Figs. 8.11 and 8.12 that
(a) Single-Line-to-Ground Fault : Let a bolted 1LG fault occurs at bus-2 when the system is unloaded with bus voltages being 1.0 per unit. Then from (8.7) we get
per unit
Also from (8.4) we get
per unit Also I fb = I fc = 0. From (8.5) we get the sequence components of the voltages as
Therefore the voltages at the faulted bus are
(b) Line-to-Line Fault : For a bolted LL fault, we can write from (8.16)
per unit
Then the fault currents are
Finally the sequence components of bus-2 voltages are
Hence faulted bus voltages are
(c) Double-Line-to-Ground Fault : Let us assumes that a bolted 2LG fault occurs at bus-2. Then
Hence from (8.24) we get the positive sequence current as
per unit The zero and negative sequence currents are then computed from (8.25) and (8.26) as
per unit
per unit Therefore the fault currents flowing in the line are
Furthermore the sequence components of bus-2 voltages are
Therefore voltages at the faulted bus are
Example 8.5 Let us now assume that a 2LG fault has occurred in bus-4 instead of the one in bus-2. Therefore
Also we have
Hence
per unit Also
per unit
per unit Therefore the fault currents flowing in the line are
We shall now compute the currents contributed by the generator and the motor to the fault. Let us denote the current flowing to the fault from the generator side by Ig , while that flowing from the motor by Im . Then from Fig. 8.11 using the current divider principle, the positive sequence currents contributed by the two buses are
per unit
per unit
Similarly from Fig. 8.12, the negative sequence currents are given as
per unit
per unit Finally notice from Fig. 8.13 that the zero sequence current flowing from the generator to the fault is 0. Then we have
per unit Therefore the fault currents flowing from the generator side are
and those flowing from the motor are
It can be easily verified that adding Ig and Im we get If given above. In the above two examples we have neglected the phase shifts of the Y/ Δ transformers. However according to the American standard, the positive sequence components of the high tension side lead
those of the low tension side by 30° , while the negative sequence behavior is reverse of the positive sequence behavior. Usually the high tension side of a Y/ Δ transformer is Y-connected. Therefore as we have seen in Fig. 7.16, the positive sequence component of Y side leads the positive sequence component of the Δ side by 30° while the negative sequence component of Y side lags that of the Δ side by 30° . We shall now use this principle to compute the fault current for an unsymmetrical fault. Let us do some more examples.
Example 8.6 Let us consider the same system as given in Example 8.5. Since the phase shift does not alter the zero sequence, the circuit of Fig. 8.13 remains unchanged. The positive and the negative sequence circuits must however include the respective phase shifts. These circuits are redrawn as shown in Figs. 8.14 and 8.15. Note from Figs. 8.14 and 8.15 that we have dropped the √3 α vis-à-vis that of Fig. 7.16. This is because the per unit impedances remain unchanged when referred to the either high tension or low tension side of an ideal transformer. Therefore the per unit impedances will also not be altered.
Fig. 8.14 Positive sequence network of the power system of Fig. 8.10 including transformer phase shift.
Fig. 8.15 Negative sequence network of the power system of Fig. 8.10 including transformer phase shift. Since the zero sequence remains unaltered, these currents will not change from those computed in Example 8.6. Thus
and
per unit
Now the positive sequence fault current from the generator Iga1 , being on the Y-side of the Y/ Δ transformer will lead I ma1 by 30° . Therefore per unit
per unit Finally the negative sequence current I ga2 will lag I ma2 by 30° . Hence we have per unit per unit
Therefore
Also the fault currents flowing from the motor remain unaltered. Also note that the currents flowing into the fault remain unchanged. This implies that the phase shift of the Y/ Δ transformers does not affect the fault currents.
Example 8.7 Let us consider the same power system as given in Example 1.2, the sequence diagrams of which are given in Figs. 7.18 to 7.20. With respect to Fig. 7.17, let us define the system parameters as: Generator G1 : 200 MVA, 20 kV, X" = 20%, X0 = 10% Generator G2 : 300 MVA, 18 kV, X" = 20%, X0 = 10% Generator G3 : 300 MVA, 20 kV, X " = 25%, X0 = 15% Transformer T1 : 300 MVA, 220Y/22 kV, X = 10% Transformer T2 : Three single-phase units each rated 100 MVA, 130Y/25 kV, X = 10% Transformer T3 : 300 MVA, 220/22 kV, X = 10% Line B-C : X1 = X2 = 75 Ω , X0 = 100 Ω
Line C-D : X1 = X2 = 75 Ω , X0 = 100 Ω Line C-F : X1 = X2 = 50 Ω , X0 = 75 Ω Let us choose the circuit of Generator 3 as the base, the base MVA for the circuit is 300. The base voltages are then same as those shown in Fig. 1.23. Per unit reactances are then computed as shown below.
Generator G1 :
, X0 = 0.15
Generator G2 : , X0 = 0.0656 Generator G3 :
, X0 = 0.15
Transformer T1 :
Transformer T2 :
Transformer T3 :
Line B-C :
,
Line C-D :
,
Line C-F :
,
Neglecting the phase shifts of Y/ Δ connected transformers and assuming that the system is unloaded, we shall find the fault current for a 1LG fault at bus-1 (point C of Fig. 7.17). From Figs. 7.18 and 7.19, we can obtain the positive and negative sequence Thevenin impedance at point C as (verify)
X1 = X2 = j 0.2723 per unit Similarly from Fig. 7.20, the Thevenin equivalent of the zero sequence impedance is
X0 = j 0.4369 per unit Therefore from (8.7) we get
per unit
Then the fault current is Ifa = 3 Ifa0 = 3.0565 per unit.
Double- Line -to Ground Fault The faulted segment for a 2LG fault is shown in Fig. 8.7 where it is assumed that the fault has occurred at node k of the network. In this the phases b and c got shorted through the impedance Zf to the ground. Since the system is unloaded before the occurrence of the fault we have the same condition as (8.8) for
(8.17)
the phase-a current. Therefore
Fig. 8.7 Representation of 2LG fault. Also voltages of phases b and c are given by
(8.18)
(8.19)
Therefore We thus get the following two equations from (8.19)
(8.20)
(8.21)
Substituting (8.18) and (8.20) in (8.21) and rearranging we get Also since I fa = 0 we have
(8.22)
(8.23)
(8.24)
The Thevenin equivalent circuit for 2LG fault is shown in Fig. 8.8. From this figure we get
(8.25)
The zero and negative sequence currents can be obtained using the current divider principle as
(8.26)
Fig. 8.8 Thevenin equivalent of a 2LG fault.
Example 8.3
Let us consider the same generator as given in Examples 8.1 and 8.2. Let us assume that the generator is operating without any load when a bolted 2LG fault occurs in phases b and c. The equivalent circuit for this fault is shown in Fig. 8.9. From this figure we can write
Fig. 8.9 Equivalent circuit of the generator in Fig. 8.4 for a 2LG fault in phases b and c. Combining the above three equations we can write the following vector-matrix form
Solving the above equation we get
Hence
We can also obtain the above values using (8.24)-(8.26). Note from Example 8.1 that
Then
Now the sequence components of the voltages are
Also note from Fig. 8.9 that
and Vb = Vc = 0. Therefore
which are the same as obtained before.
Line-to-Line Fault The faulted segment for an L-L fault is shown in Fig. 8.5 where it is assumed that the fault has occurred at node k of the network. In this the phases b and c got shorted through the impedance Zf . Since the system is unloaded before the occurrence of the fault we have
(8.8)
Fig. 8.5 Representation of L-L fault. Also since phases b and c are shorted we have
(8.9)
(8.10)
Therefore from (8.8) and (8.9) we have We can then summarize from (8.10)
(8.11)
Therefore no zero sequence current is injected into the network at bus k and hence the zero sequence remains a dead network for an L-L fault. The positive and negative sequence currents are negative of each other. Now from Fig. 8.5 we get the following expression for the voltage at the faulted point Again
(8.12)
(8.13)
Moreover since I fa0 = I fb0 = 0 and I fa1 = - I fb2 , we can write
(8.14)
Therefore combining (8.12) - (8.14) we get
(8.15)
Equations (8.12) and (8.15) indicate that the positive and negative sequence networks are in parallel. The sequence network is then as shown in Fig. 8.6. From this network we get
(8.16)
Fig. 8.6 Thevenin equivalent of an LL fault.
Example 8.2 Let us consider the same generator as given in Example 8.1. Assume that the generator is unloaded when a bolted ( Zf = 0) short circuit occurs between phases b and c. Then we get from (8.9) I fb = - I fc . Also since the generator is unloaded, we have I fa = 0. Therefore from (7.34) we get
Also since V bn = V cn , we can combine the above two equations to get
Then
We can also obtain the above equation from (8.16) as
Also since the neutral current I n is zero, we can write V a = 1.0 and
Hence the sequence components of the line voltages are
Also note that
which are the same as obtained before.
Single-Line-to-Ground Fault Let a 1LG fault has occurred at node k of a network. The faulted segment is then as shown in Fig. 8.2 where it is assumed that phase-a has touched the ground through an impedance Zf . Since the system is unloaded before the occurrence of the fault we have
(8.1)
Fig. 8.2 Representation of 1LG fault. Also the phase-a voltage at the fault point is given by
(8.2)
(8.3)
From (8.1) we can write Solving (8.3) we get
(8.4)
This implies that the three sequence currents are in series for the 1LG fault. Let us denote the zero, positive and negative sequence Thevenin impedance at the faulted point as Z kk0 , Z kk1 and Z kk2respectively. Also since the Thevenin voltage at the faulted phase is Vf we get
(8.5)
three sequence circuits that are similar to the ones shown in Fig. 7.7. We can then write
Then from (8.4) and (8.5) we can write
(8.6)
Again since
We get from (8.6)
(8.7)
The Thevenin equivalent of the sequence network is shown in Fig. 8.3.
Fig. 8.3 Thevenin equivalent of a 1LG fault.
Example 8.1 A three-phase Y-connected synchronous generator is running unloaded with rated voltage when a 1LG fault occurs at its terminals. The generator is rated 20 kV, 220 MVA, with subsynchronous reactance of 0.2 per unit. Assume that the subtransient mutual reactance between the windings is 0.025 per unit. The neutral of the generator is grounded through a 0.05 per unit reactance. The equivalent circuit of the generator is shown in Fig. 8.4. We have to find out the negative and zero sequence reactances.
Fig. 8.4 Unloaded generator of Example 8.1. Since the generator is unloaded the internal emfs are
Since no current flows in phases b and c, once the fault occurs, we have from Fig. 8.4
Then we also have
From Fig. 8.4 and (7.34) we get
Therefore
From (7.38) we can write Z1 = j ω ( Ls + Ms ) = j 0.225. Then from Fig. 7.7 we have
Also note from (8.4) that
Therefore from Fig. 7.7 we get
Comparing the above two values with (7.37) and (7.39) we find that Z 0 indeed is equal to j ω ( Ls - 2 Ms ) and Z2 is equal to j ω ( Ls + Ms ). Note that we can also calculate the fault current from (8.7) as
Introduction The sequence circuits and the sequence networks developed in the previous chapter will now be used for finding out fault current during unsymmetrical faults.
Three Types of Faults
Calculation of fault currents
Let us make the following assumptions:
The power system is balanced before the fault occurs such that of the three sequence networks only the positive sequence network is active. Also as the fault occurs, the sequence networks are connected only through the fault location. The fault current is negligible such that the pre-fault positive sequence voltages are same at all nodes and at the fault location. All the network resistances and line charging capacitances are negligible. All loads are passive except the rotating loads which are represented by synchronous machines.
Based on the assumptions stated above, the faulted network will be as shown in Fig. 8.1 where the voltage at the faulted point will be denoted by Vf and current in the three faulted phases are Ifa , I fb and I fc . We shall now discuss how the three sequence networks are connected when the three types of faults discussed above occur.
Fig. 8.1 Representation of a faulted segment. Calculation of the cables capacitances & inductance values.
Capacitance of Cables, Charging Current, and Charging Reactive Power.
.
. The reactances of the insulated cables.
.
.
.
.
Three branches in an electrical network can be connected in numbers of forms but most common among them is either star or delta form. In delta connection, three branches are so connected, that they form a closed loop. As these three branches are connected nose to tail, they form a triangular closed loop, this configuration is referred as delta connection. On the other hand, when either terminal of three branches is connected to a common point to form a Y like pattern is known as star connection. But these star and delta connections can be transformed from one form to another. For simplifying complex network, delta to star or star to delta transformation is often required.
Delta - Star Transformation
The replacement of delta or mesh by equivalent star connection is known as delta - star transformation. The two connections are equivalent or identical to each other if the impedance is measured between any pair of lines. That means, the value of impedance will be the same if it is measured between any pair of lines irrespective of whether the delta is connected between the lines or its equivalent star is connected between that lines.
Consi der a delta system that's three corner points are A, B and C as shown in the figure. Electrical resistance of the branch between points A and B, B and C and C and A are R1, R2and R3 respectively. The resistance between
the
points
A
and
B
will
be,
Now, one star system is connected to these points A, B, and C as shown in the figure. Three arms R A, RB and RC of the star system are connected with A, B and C respectively. Now if we measure the resistance value between points A and B, we will get, Since the two systems are identical, resistance measured between terminals A and B in both systems must be equal.
Similarly, resistance between points B and C being equal in the two systems, points C
and
A
being
equal
And resistance between in the two systems,
Adding equations (I), (II) and
(III) we get, Subtracting equations (I), (II) and (III) from equation (IV) we get,
The relation of delta - star transformation can be expressed as follows. The equivalent star resistance connected to a given terminal, is equal to the product of the two delta resistances connected to the same terminal divided by the sum of the delta connected resistances. If the delta connected system has same resistance R at its three sides then equivalent star resistance r will be,
Star - Delta Transformation For star - delta transformation we just multiply equations (v), (VI) and (VI), (VII) and (VII), (V) that is by doing (v) × (VI) + (VI) × (VII) + (VII) × (V) we get,
Now dividing equation (VIII) by equations (V), (VI) and equations (VII)
separately we get,
.
Concept of Subtransient, Transient & Steady State. The concept of Subtransient, Transient and Steady State arises in case of fault in an Alternator. Let us assume a sudden short circuit in three phase of alternator. The fault current will flow in all the three phases of alternator and its waveform will be as shown in figure below.
.
When the alternator is short-circuited, the currents in all the three-phases rise rapidly to a high value of about 10 to 18 times of full load current, during the first quarter cycle. The flux crossing the air gap is large during a first couple of cycles. The reactance during these first two or three cycle is least and the short circuit current is high. This reactance is called subtransient reactance and is denoted by X”. The first few cycles come under subtransient state. After a first few cycles, the decrement in the r.m.s. value of short circuit current is less rapid than the decrements during the first few cycles. This state is called the Transient State and the reactance in this state is called transient reactance X’. The circuit breaker contacts separate in the transient state. Finally the transient dies out and the current reaches a steady sinusoidal state called the Steady State. The reactance in this state is called steady state reactance Xd. Since the short circuit current of the alternator lags behind the voltage by 90 degree, the reactance involved are direct axis reactance. As clear from the figure above, the d.c. components in the three phases are different; hence the waveforms of the three phases are not identical. If voltage of phase, say, Y, is maximum at the instant of short circuit, the DC component of short circuit current is zero. Hence the waveform is symmetrical as shown in figure below.
.
The currents and reactance are given by the following expressions,.
.
Where I = Steady state current, r.m.s. value I’ = Transient current, r.m.s. value I” = Sub-transient current, r.m.s. value Ea = Induced e.m.f. per phase Xd = Direct axis synchronous reactance Xd’ = Direct axis transient reactance Xd” = Direct axis sub-transient reactance As the short circuit occurs, the short-circuit current attains high value. The circuit breakercontact starts separating after the operation of the protective relay. The contacts of thecircuit breaker separate during ‘transient state.’ The r.m.s. value of the current at the instant of the contact separation is called the breaking current of the circuit-breaker and is expressed in kA. If a circuit-breaker closes on existing fault, the current would increase to a high value during the first, half cycle. The highest peak value of the current is reached during the peak of the first current loop. This peak value is called making current of the circuit breaker and is expressed in kA. This is the reason making current of Circuit Breaker is higher than the Breaking Current.
Now we will go into the discussion back while taking an example of Electrical circuit. Consider the circuit given below.
If the battery is ON at t=0 and kept ON. Inductor has no energy stored initially. So it has no voltage across it.
Inductor voltage = 0 Rate of change of current = 0 Initially current in inductor = 0 Energy stored in inductor = 0 As we know that Inductor does not allow sudden change in currents. So just after the moment when switch is ON, the current in inductor is zero. That means at t=0+ inductor acts as open circuit. But as the time passes inductor allows current. This happens until coil voltage drop is equal to applied voltage.
But after some time, i.e. when coil voltage drop is equal to applied voltage, then current flowing through the circuit is constant as inductor voltage is zero which means rate of change of current is zero. This state is called Steady State.
So if we solve the differential equation of the circuit, we’re actually finding the response as a function of time, which includes both transient and steady state response. Basic Principle of Relay Operation. Relay is a switch which senses fault in a system and once fault is sensed by the Relay, it issues trip command to the Circuit Breaker, CB to isolate the faulty section of the network from the healthy section.
The Relay detects the abnormal condition by continuously monitoring electrical quantities which are different for healthy and faulty condition. The electrical quantities which may change during fault condition are voltage, current, frequency and phase angle. If one or more of the above electrical quantities change, that signals the presence, type and location of the fault to the Relay. After detecting the fault condition, Relay pick-up, its contact will change from NO to NC or vice versa. So we can wire up a particular kind of Relay contact to Breaker tripping circuit. So whenever, the Relay picks up, the tripping of Breaker will take place. You
may
like
to
Read, Why
CT
Secondary
Shall
Never
be
Kept
Open?
A simplified Relay circuit is shown in figure below. Figure below shows one of the three phase system for simplicity.
As shown in the figure above, Current Transformer CT secondary winding is directly connected to the Relay coil. Under normal condition, the current through the Relay coil is not sufficient enough to pull the plunger and close the circuit of Breaker Tripping Coil. Notice here that Breaker Tripping coil is solely responsible for the tripping of Circuit Breaker. If trip coil of breaker fails, then tripping of Breaker will not take place. This is the reason, two trip coils are normally provided in Circuit Breaker to get reliable operation of Breaker. Not only two Trip Coils are provided in CB rather a Trip Coil monitoring Relay is also used. If case of fault i.e. if it happens to be any open circuit in Trip Coil, then the Trip Coil Supervision Relay will be flagged to attract the attention of the operator.
In case of fault, the current through the CT secondary will go up which will cause increased current through the Relay coil. If it happens that the current through the Relay coil exceeds the setting value or pick-up value then the coil will get produce sufficient magnetic pull to the plunger and thus plunger will complete the CB trip circuit. As soon as the CB trip circuit is complete, current will start flowing in the Trip Coil which in turn will pull a lever to trip the Circuit Breaker CB. In the above figure, it is shown that Relay coil is directly pulling the plunger to complete the Breaker Trip Coil circuit but in actual practice, Relay coil when picked up will change its contact status. Let us say Relay Normally Open (NO) contact is wired to the Breaker Trip Coil Circuit. Therefore when the Relay coil is in de-energized state, the circuit of Trip Coil of CB is not complete and hence no tripping of the CB. During fault condition as the current through the Relay coil exceeds the pick-up value, the Relay coil will get actuated which in turn will force its
contact to change over i.e. NO contact will change to Normally Close (NC) thereby closing the Trip Coil circuit of the Breaker.
Since Trip Coil circuit of Breaker is complete, current will flow through the Trip Coil causing CB to trip.
Introduction and Architecture of Numerical Relay. Most of us are aware of Electromagnetic Relays and Static Relays but most us may not be well acquainted with Numerical Relay. If I define a Numerical Relay, honestly speaking it will seem to be quite tough but in reality they are very user friendly and easy to implement different types of protection scheme. However I am going to define Numerical Relays.
Numerical Relay is a device in which measured electrical quantities are sequentially sampled and then converted into numerical data which are mathematically or logically processed to take decision for issuing trip command.
Numerical Relay is basically Digital Relay for which manufacturers have developed specified hardware which can be used in conjunction with suitable Software o meet different protection needs. A Digital Relay comprises both Hardware and Software. The Hardware part is briefly described below. CPU: CPU stand for Central Processing Unit which is responsible for the processing of protection algorithms and digital filtering. Memory:Memory is of two types. One is RAM (Random Access Memory) and ROM (Read Only Memory). RAM serves for the purpose of retaining the input data to the Relay and processing the data during the compilation of algorithm. ROM is used to store Software needed for the working of Relay. ROM is also needed for storing Event and Disturbance data. Event and Disturbance Recording is a must feature for a digital relay because these data are used for troubleshooting any event. A typical Numerical Relay can store as much as 520 Events and 50 Disturbances. The most attractive feature of such relay is that it works on FIFO (First In, First Out). Suppose if it happens to be the number of disturbances exceeds 50 then the Relay will delete the last Disturbance and will register new disturbance. Input Module: The analog single from the Power System is stepped down using Current Transformer and Potential Transformer and then fed to the Numerical Relay using low pass filter. Low pass filter is incorporated in the input module to eliminate any noise single induced in the line due to corona or induction effect of nearby high voltage line. The output from the Filter is then fed to Sample and Hold (S/H)circuit.
A Sample and Hold (S/H) circuit is used to keep the rapidly changing instantaneous value constant during the period of conversion for processing. In addition to the analog input, Numerical Relay is designed to accept digital input too. Separate terminals are provided for the analog and digital inputs. Multiplexer and Analog to Digital Converter: The CPU accepts the input in digital form but the input from Current Transformer CT and Potential Transformer PT are analog in nature. Therefore and A/D converter is used to convert the analog signal to digital signal. In case more than one analog quantity is to be converted into digital form, Multiplexer is used for selecting any analog input at a time to convert into digital form.
Output Module: Output module provided in Numerical Relay is digital contacts which are actuated when a trip decision is taken by the CPU. These output digital contacts are a pulse which is generated as a response signal. The timing of pulse can be changed by the user. Digital Input / Communication Module: Numerical Relay is provided with serial and parallel ports for the interconnection with control and communication system of the substation. Digital output contacts of Numerical Relay is used for wiring with the Auxiliary Relays to extend tripping command to the Circuit Breaker.
Software: Numerical Relay is equipped with software to communicate with external device to program to Relay or one can program by navigating through the Relay Menu. Hardware for Metering: In principle, the hardware setup discussed above can be used for both measurement and protection function. However, considering the order of difference between current magnitudes in case of fault and load, there can be loss of accuracy during metering applications. Consider a hypothetical case where in maximum load current is 100 A and maximum fault current is 20 times this load current i.e. 2000 A. Let a 12 bit unipolar A/D converter be used for sampling current signal. This implies that resolution of A/D converter is 2000/(212-1)=0.488 A. This resolution may be inadequate for metering purposes. One solution is to increase resolution i.e. the number of bits in A/D converter. For example, one may use 16 bit A/D converter in place of 12 bit A/D converter. However, increasing the number of bits of A/D converter also affects the selection of processor. A good design guideline is to choose a processor with double the number of bits of A/D converter. This ensures that truncation and numerical precision problems associated with finite precision arithmetic do not cause significant loss of accuracy. For example, with 16 bit A/D converter, 32 bit processor is the natural choice. Alternatively, a variable gain amplifier can be used along with the A/D converter. At low currents, high gain setting is used and at high currents low gain setting is preferred. However, during the change from one setting to another, loss of information can take place. Therefore, a simple solution would be to keep metering and protection functionality separate. In the next post we will be discussing about some interesting features of Numerical Relay. So be there and follow ELECTRICAL CONCEPTS.
PLC Step By Step: Your Complete Guide. This course will help know TIMERS, Counters, Relays, Coils and Ladder Logic and build you FIRST PLC Program From Scratch. Click below to register. Why CT Secondary Shall Never be Kept Open?. The most important precaution which shall be taken care while working with a CT is that its secondary should never be kept open. In this post we will discuss this aspect of CT. As we know that Current Transformers (CTs) are always used with secondary winding connected with Ammeter, Relays or Wattmeter Coils. A precaution which shall always be taken is that Never open the secondary winding circuit of a Current Transformer while its Primary Winding is energized. If the secondary winding circuit of a CT is kept open then it will lead to severe consequence to the personnel opening the CT secondary and to the CT itself. The question arises why?
To understand this, first we should know the basic difference between a Power Transformer and a CT. The basic difference between a Power Transformer and a CT is that, in Power Transformer the primary current is the reflection of the secondary current by N1I1 = N2I2while in CT the primary current is dependent on the load current or line current as Ct is connected in series with the line. So primary current of CT (assuming constant line current) is constant irrespective of whether the secondary of CT is connected with burden or not. During normal operation of CT, the primary and secondary winding produces mmf which by lenze’s law opposes each other. As the secondary mmf is slightly less than the primary mmf, the net mmf is small. This net mmf is the working / magnetizing mmf of the core of CT. Now, in case secondary winding is kept open then secondary current will be zero while the primary current of CT will remain same. Therefore the opposing mmf of secondary will no longer exist. Hence the net mmf is due to primary current only i.e. N1I1 which is very large. This large mmf will produce large flux in the core and will saturate the core. Again, due to large flux in the core the flux linkage of secondary winding will be large which in turn will produce a large voltage across the secondary terminals of the CT. This large voltage across the secondary terminals will be very dangerous and will lead to the insulation failure and there is a good chance that the person who is opening the CT secondary while primary is energized will die due to shock. Also, because of excessive core flux, the hysteresis and eddy current loss will be very high and the CT will get overheated. As CT is oil filled, because of overheating, the oil of CT will get boil and start to vaporize. Because of vaporization of CT oil, the CT housing will get pressurized and blast. This blasting will lead to fire and smoke. Again, it is not the end here but due to smoke, the nearby lines will trip due to earth fault which may trip the Power Generating Station.
Difference between Isolators and Sectionalizers in Power System. An isolator is a switch used in power lines, to completely open (physically) a specific section of line, when maintenance or modification work is to be done. The isolator can be operated manually, or in some cases, opened by energizing a motor. With a threephase system, the isolator opens all three lines. A picture of Isolator is shown below to visualize.
A sectionalizer is an automatic switch, installed downstream of a recloser. The recloser is a feature provided in a circuit breaker, which opens when a short circuit occurs on the line such as a falling tree branch. Ordinarily, such short circuits clear themselves, for example by the branch falling to the ground. Thus the recloser opens the circuit when the fault occurs, and after a short interval, re-closes itself and restores power to the system. A picture of sectionalizer is shown below to visualize.
However, not all short-circuit faults clear themselves. For example, a fallen branch may lie across the lines and not fall to the ground. A sectionalizer counts the successive openings and closings of the recloser, and after a pre-set number, it opens. Thus it isolates a particular section of line where a continuing fault exists, leaving most of the larger area protected by the recloser still with power.
So both of these are devices that isolate a portion of a power distribution system. But the first is a safety device, to protect people working on the lines. The second is an automatic device, to cut off power to a section where a continuing fault has occurred. Circuit Breaker and Arc Phenomenon. What is Circuit Breaker?
Circuit Breaker is switch capable of making or breaking the circuit under no-load as well as onload condition. It can make or break circuit either manually or by remote control. A Circuit Breaker in conjunction with Relay can break the circuit under fault condition.
You may also like to read, Basic Principle of Relay Operation
Operating Principle of Circuit Breaker:
A Circuit Breaker CB consists of two contacts which are called electrodes, one of which remain fixed, called fixed contact and another moving contact. Under normal operating condition, this contact will remain closed to supply power but as soon as fault is sensed by the Relay, trip coil of Circuit Breaker energizes and the moving contact of CB is pulled apart by some mechanism to open the CB.
When contacts of CB are separated under fault condition, an arc is stuck between the fixed and moving contacts. The current is thus able to continue till the arc persists. The production of arc not only delays the current interruption but it also produces huge amount of heat which if exceeds a limit may damage the system or CB itself. Therefore, the design of CB is done in such a way to minimize the arcing period so that
1) Heat produced during arcing may not exceeds the dangerous value.
2) To have fast fault clearing.
It is worth here to mention that a typical Breaker opening and closing time remain around 30-35 ms and 60-70 ms respectively. Notice that CB opening time is less than the closing time to ensure fast fault clearing.
Arcing Phenomenon in Circuit Breaker:
When a short circuit occurs, heavy current flows through the contacts of circuit breaker before they are opened by the protective system. At the instant when the contacts begin to open after getting trip command from the Relay, the contact area decreases rapidly and large fault current causes increased current density and hence rise in temperature. The heat produced in the medium in between the contacts is sufficient enough to ionize the medium. This ionized medium acts as a conductor and arc is stuck in between the contacts of the circuit breaker. It shall be noted here that the potential difference between the fixed and moving contacts is quite small and just enough to maintain the arc. This arc provides a low resistance path to the current and thus due to arcing the current in the circuit remain uninterrupted as long as arcing persists.
During the arcing period the current flowing through the contacts of circuit breaker deepens upon the arc resistance. The greater the arc resistance the smaller will be the current flowing through the contacts of CB. The arcing resistance depends upon the following factors:
Degree of Ionization:
The more the ionization of medium between the contacts, the less will be the arcing resistance.
Length of Arc:
The arc resistance increases as the length of arc increases i.e. as the separation between the contacts of Breaker increases the arcing resistance also increases.
Cross Section of the Arc:
The arcing resistance increases with decrease in the cross sectional area of the arc.
Principle of Arc Extinction:
As we discussed earlier in this post, ionization of medium in between the contacts and potential difference across the contacts are responsible for the production and maintenance of arc. Thus for arc extinction, we can increase the separation between the contacts to such an extent that potential difference across the contacts is not sufficient enough to maintain the arc. But this philosophy is impractical as in EHV (Like 220 kV, 400 kV, 765 kV etc.) system; the separation between the contacts to extinguish the arc will be many meters which is not practically achievable.
Another way for extinction of arc is to demonize the medium in between the contacts. If the arc path is demonized the arc extinction will definitely be facilitated. This may be achieved by cooling the arc or by quickly removing the ionized particles from the space in between the contacts. This principle of arc extinction is used in all modern Circuit Breakers.
SF6 Circuit Breaker – Construction, Working Principle and Types. In Sulpher Hexafluoride or SF6 Circuit Breaker, sulpher hexafluoride (SF6) gas is used as an insulating and arc quenching medium. SF6 gas has many superior properties which makes it perfect for arc quenching.Sulpher Hexafluoride or SF6 Circuit Breaker is most popular and widely used breaker. This type of breaker is mostly used for EHV systems like 220 kV, 400 kV and 765 kV. It is suggested to read “Why SF6 gas used in HV/EHV Circuit Breakers“.
Construction of SF6 Circuit Breaker Like other circuit breaker viz. Vacuum Circuit Breaker, Air Blast Circuit Breaker etc., SF6Circuit Breaker has fixed contact as well as moving contact. Theses fixed and moving contacts are known as MAIN CONTACT. There exists one another contact which is known as ARCING CONTACT. Arcing Contact is part of fixed contact. Basically, Arcing contacts are only designed to withstand arcing. It is not designed for carrying load current. In spite, main contacts are designed to carry load current and not the arcing. Therefore, it can be said that, while closing of SF6 circuit breaker, first Arcing Contact will close. Thereafter arcing contact will close. Similarly while opening, first main contact will open and then arcing contact will open. Notice here that, during opening operation of SF6Circuit Breaker, the order of opening main and arcing contact is revered as that in closing operation. This is because, while opening if the main contact opens first, there will not be any arcing as the current is getting path through the arcing contact. But if the arcing contact open first then during opening of main contact there will be arcing and as discussed main contacts are not meant to withstand arcing. Apart from Fixed contact and moving contacts, SF6 Circuit Breaker has following main components:
Interrupter Insulating Nozzle SF6 Gas Chamber
An interrupt is a chamber which encloses the breaker contacts, insulating nozzle, SF6 gas chamber. Interrupter is made of porcelain. Figure below shows the basic parts of SF6 circuit breaker.
Figure-1 Carefully observe the figure and notice the different parts, though some parts like SF6 gas chamber, nozzle, valve etc are not shown in the above figure but they will be shown while discussing the working principle. So, please be patient till then and read further. Working principle of SF6 Circuit Breaker The contacts of SF6 Circuit Breaker are surrounded in an environment of SF6 gas at some pressure. Actually, the dielectric strength of SF6 gas is directly proportional to its
pressure. In 220 kV, 400 kV and 765 kV applications, the gas pressure is maintained at 6.5 bar. Let’s consider breaker opening operation for better understanding of operating mechanism. First have a look at the contacts when the breaker is in fully close position as shown in Figure-1. Now we will open the breaker and will observe its mechanism step by step. Step-1: Main Contact Open
As discussed earlier in the post, main contact will open first. This is shown in the figure above. Observe in figure that, though main contacts are open, arcing contacts are still close. As main contacts open, the piston in the cylinder moves causes the SF6 gas to compress due to reduction of volume Vp. Step-2: Arcing Contacts Open
As soon as arcing contacts separates from contact 1, an arc is strikes. Due to this arcing, heat is produced. This heat of arc further increases the pressure of SF 6 in the chamber Vt. Mind that, the pressure of arc extinguisher i.e. SF 6 is increased by the heat of arc. This is the reason; such breaker is called self compensating type. Here self compensating means that, the capacity of breaker to interrupt the fault is proportional to fault current. Step-3: Arcing Contact separates from Nozzle
When arcing contact separates from the insulating Nozzle, the pressurized SF6 gas in volume Vt is released in the arc. This causes the arc to extinguish at the moment the current passes though the natural zero. Thus, the pressurized SF6 gas extinguishes the arc and hence circuit is interrupted. In case of small current like in unloaded transformer or reactor, the thermal energy of arc is not enough to pressurize the SF6 gas. In such case the pressure developed in the SF6 gas chamber Vp in Step-1 is extinguishes the arc. Types of SF6 Circuit Breaker As discussed, in 220 kV, 400 kV and 765 kV applications, the SF6 gas pressure is maintained at 6.5 bar. You will be amazed that, even though voltage level is increasing, same pressure of SF6 i.e. 6.5 bar is used for 220, 400 and 765 kV applications. Actually as we go up at higher voltage level, the number of contacts increases in SF6 circuit Breaker. Based on this philosophy, SF6 circuit breaker can be classified into following types: Single
Single Breaker Circuit Breaker Double Break Circuit Breaker Multi Break Circuit Breaker Break SF6 Circuit Breaker
In Single Break Circuit Breaker, only one moving and fixed contacts are present. This means that, there will only be one interrupter unit in such breaker. Single break SF6 circuit breaker is used for 220 kV applications. Double Break SF6 Circuit Breaker
In such type of breaker, there are two set of moving and fixed contacts connected in series. Therefore, to enclose two set of contacts, there must be two interrupt unit in series. This type of breaker is used in 400 kV applications. In double break circuit breaker, grading capacitorsare used to equalize the voltage distribution across each contact. Thus for 400 kV application, the voltage across each contact will be 200 kV. Therefore it is logical to use SF6 gas at a pressure same as used in 200 kV application. Ha ha..got it? Multi Break SF6 Circuit Breaker
In multi break circuit break, more than two set of fixed and moving contacts are used. Such type of breaker is used in 765 kV applications.
Vacuum Circuit Breaker- Construction and Working. Vacuum Circuit Breaker Vacuum Circuit Breaker (VCB) is one where vacuum of the order of 10 -6 to 10-10 torr is used as an arc quenching medium. 1 torr is equivalent to a pressure represented by a barometric head of 1 mm mercury. Vacuum Circuit Breaker is used for low and medium voltage applications. Construction of Vacuum Circuit Breaker
Vacuum Circuit Breaker consists of Enclosure, Contacts, Vapor Condensing Shield, Metallic Bellows and Seal. Enclosure. The enclosure is made of impermeable insulating material like glass. The enclosure must not be porous and should retain high vacuum of the order of 10-7 torr. Contacts. There are two types of contacts, moving and fixed. The moving contact is connected with large stem connected to operating mechanism of breaker. Contacts of Vacuum Circuit Breaker have generally disc shaped faces. The disc is provided with symmetrical grooves in such a way that the segments of the two contacts are not in the same line. The magnetic field set-up by the components of currents with such geometry causes the plasma of the arc to move rapidly over the contacts instead of remaining stable at one point. The concentration of the arc is thus prevented and the arc remains in diffused state. The sintered material used for contact tip are generally copperchromium or copper bismuth alloy. Vapor Condensing Shield. These metallic shields are supported on insulating housing such that they cover the contact region. The metal vapor released from the contact surface during arcing is condensed on these shields and is prevented from condensing on the insulting enclosure. Metallic Bellows. One end of the bellows is welded to the enclosure. The other end is welded to the moving contact. The bellows permit the sealed construction of the interrupter and yet permit movement of the contact. Stainless steel bellows are generally used in vacuum interrupters. Carefully observe every component of Vacuum Circuit Breaker as shown in figure below.
Arc Extinction in Vacuum Circuit Breaker (VCB)
The arc interruption process in Vacuum Circuit Breaker interrupter is quite different from that in other types of circuit breakers. The vacuum as such is a dielectric medium and arc cannot persist in ideal vacuum. However, the separation of current carrying contacts causes the vapor to be released from the contacts. Thus, as the contacts separate, the contact space is filled with vapor of positive ions liberated from the contact material. The vapor density depends on the current in the arc. During the decreasing mode of the current wave the rate of release of the vapor reduces and after the current zero, the medium regains the dielectric strength provided vapor density around contacts has substantially reduced. While interrupting a current of the order of a few hundred amperes by separating flat contacts under high vacuum, the arc generally has several parallel paths. Thus the total current is divided in several parallel arcs. The parallel arcs repel each other so that the arc tends to spread over the contact surface. Such an arc is called diffused arc. The diffused arc can get interrupted easily. At higher values of currents of the order of a few thousand amperes, the arc gets concentrated on a small region and becomes self-sustained arc. The concentrated arc around a small area causes rapid vaporization of the contact surface. The transition from diffused arc to the concentrated arc depends upon the material and shape of contact, the magnitude of current and the condition of electrodes. The interruption of arc is possible when the vapor density varies in phase with the current and the arc remains in the diffused state. The arc does not strike again if the metal vapor is quickly removed from the contact zone. Thus the arc extinction process in vacuum circuit breaker is related to a great extent to the material and shape of the contacts and the technique adopted in condensing the metal vapor. The contact geometry is so designed that the root of the arc keeps on moving so that the temperature at one point on the contact does not reach a very high value. The rapid building up of dielectric strength after final arc extinction is a unique advantage of vacuum circuit breaker. They are ideally suitable for capacitor switching as they can give restrike free performance. Degree of Vacuum in VCB Interrupters
The breakdown voltage of contact gap varies with the absolute pressure in the vacuum circuit breaker interrupter unit. As the absolute pressure is reduced from 10-1 Torr to 103 Torr, the dielectric strength (kV/mm) increases but above 10-4 Torr, the breakdown strength and pressure characteristic becomes almost flat as shown in figure below.
The dielectric strength in this region is above 12 kV /mm. In vacuum interrupters vacuum level of the order of 10-6 to 10-10 Torr is used. This is called high vacuum range. During the passage of time and after arc interruptions, the vacuum level goes on reducing. However it remains in the range of 10– 5 Torr and 10-8 Torr. Vacuum in the range of 10-3 is sufficient for interruption. Gas Insulated Switchgear or GIS. Gas Insulated Switchgear (GIS) is a compact metal encapsulated switchgear consisting of high voltage components such as circuit breakers, disconnectors, Current Transformer, Earth Switch, Bus bar etc. which can be safely operated in confined spaces. GIS is used where space is limited, for example, extensions, in city buildings, on roofs, on offshore platforms etc.
In Gas Insulated Switchgear SF6 gas is used as an insulating medium and as an arc quenching medium in Circuit Breaker. But there are some GIS where Clean Air is used an insulating medium and Vacuum as an interrupting medium. For example, in Siemens 8VN1GIS (up to 145 kV), vacuum is used as an interrupting medium while clean air is used as insulating medium. Due to advancement of technology, today Gas Insulated Switchgear or GIS are available for voltage ranging from 12 kV to 1200 kV. GIS is basically modular switchgear as shown in figure below.
Siemens press picture
The various modules are factory assembled and are filled with SF 6 gas. Thereafter, they are taken to site for final assembly. Such sub-stations are compact and can be installed conveniently on any floor of a multi-storied building or in an underground sub-station. As the units are factory assembled, the installation time is substantially reduced. Such installations are preferred in cosmopolitan cities, industrial townships, hydro-stations where land is very costly and higher cost of SF6 insulated switchgear is justified by saving due to reduction in floor-area requirement. SF6 insulated switchgear or GIS is also preferred in heavily polluted areas where dust chemical fumes and salt layers can cause frequent flashovers in conventional outdoor substations. Types of Gas Insulated Switchgear
Based on the constructional feature, GIS can be classified into following types: Isolated Phase GIS Basically in such kind of GIS, the modules are assembled phase wise i.e. Breaker, CT, Isolators of individual phases are assembled separately. Modules for each phase of a bay are separate. That is why such GIS are called Isolated Phase GIS. It is quite obvious that the space requirement for Isolated Phase GIS is more. Integrated Three Phase GIS As the name suggests, all three phases’ equipment like Breaker, CT, Isolators etc. are kept in a single module filled with gas. The space requirement for such GIS is 1/3rd of space requirement in Isolated Phase GIS. Hybrid GIS Hybrid GIS is an optimal combination of Isolated Phase and Integrated Three Phase GIS. Highly Integrated Switchgear (HIS) In HIS type of Gas Insulated Switchgear, total substation equipments are encapsulated together in single enclosure. With the HIS, the circuit breakers, disconnectors, earthing switches and instrument transformers are accommodated in compressed gas tight enclosures, which make the switchgear extremely compact. HIS, the gas insulated switchgear can be used for indoor and outdoor purpose as HIS requires less than half the space of a comparable air insulated switchgear. Figure below shows the HIS.
Siemens press picture
Notice that, the GIS in above figure is used for outdoor purpose but the same could be used for indoor purpose. I used this picture to just show you that GIS does not mean
that every equipment is kept in a room filled with SF6 gas rather GIS mean equipment kept in a module filled with SF6 gas. Advantages of SF6 Gas Insulated Switchgear The space occupied by Gas Insulated Switchgear installation is only about 10% of that of conventional outdoor sub-station. Thus the high cost of GIS is partly compensated by saving in cost of space. Protection from pollution. The moisture, pollution, dust etc., have little influence on Gas Insulated Switchgear. However, to facilitate installation and maintenance, such substations are generally housed inside a small building. The construction of the building need not be very strong like convention power houses. Reduced Switching overvoltages. The overvoltages while closing and opening line, cables, motors, capacitors etc. are low. Reduced Installation Time. The principle of modular construction reduces the installation time to a few weeks. Conventional sub-stations require a few months for installation. The gas pressure (4 kg/cm2) is relatively low and does not pose serious leakage problems. Increased Safety. As the enclosures are at earth potential, there is no possibility of accidental contact by service personnel to live parts. Disadvantage of SF6 Gas Insulated Switchgear High cost compared to conventional outdoor sub-station. Excessive damage in case of internal fault. Long outage periods as repair of damaged part at site may be difficult. Requirement of cleanliness are very stringent. Dust or moisture can cause internal flashovers. Procurement of gas and supply of gas to site is problematic. Adequate stock of gas must be maintained.
Directional Earth Fault Protection. Generally earth faults are Single Line to Ground (SLG) and Line-Line to Ground (LLG) faults. Earth faults are characterized by the presence of Zero Sequence Curren I 0. Since, except for unbalance, normal system operation is not having Zero Sequence Current I 0, much more sensitive pick-up is possible for earth fault by using zero sequence current component I 0 = (Ia + Ib + Ic) / 3 and declaring a fault if I0 exceeds a threshold.
We know that I0 = (Ia + Ib + Ic) / 3
However, in a system with multiple sources or parallel paths, we require earth fault relays to be directional as discussed in earlier post How to Incorporate Directional Featurein a Relay.
As we discussed in earlier post How to Incorporate Directional Feature in a Relay, that for making a Relay directional we need Reference Phasor. The reference phasor is called asPolarizing Quantity. For ground fault relaying both Voltage and Current Polarization can be used.
We will consider each Voltage and Current Polarization separately for Earth Fault Protection.
Voltage Polarization:
Let the system be initially unloaded and a ground fault occur on phase A. Therefore Ib = Ic = 0 and Ia = 3I0. For Single Line to Ground fault there is a drop in voltage of phase A while phase B and C voltages remain unchanged. Phasor diagram for Voltage and current for SLG fault can be drawn as below.
Voltage and Current Phasor under Single Line to Ground Fault:
In the phasor diagram only 3I0 is shown as Ib = Ic =0 and Ia = 3I0 for Single Line to Ground fault.
Now we will find the Zero Sequence Voltage under the fault.
As V0 = (Vag+Vbg+Vcg)/3, phasor sum of Va,Vb and Vc is to be taken. 3V0 = Vag+Vbg+Vcg, phasor sum
From phasor diagram it is clear that Zero Sequence Voltage 3V 0 is in phase opposition with Vag (Phase Voltage of A). Therefore it is appropriate to take -3V0 as a reference phasor. In normal power system V0is not present but available only during the fault. Let the maximum torque be drawn at 60 degrees lag with respect to -3V0 phasor as shown in figure below.
As we know that Zero Torque Line is perpendicular to the Maximum Torque Line, therefore we draw Zero Torque Line as shown in figure above.
It is then clear that zero Torque Line which separates the plane into Operate and Do Not Operate zone leads -3V0by 30 degrees. Thus, for fault in the correct region, 3I0 lags -3V0 hence falls in operate region. If fault is behind the relay, 3I 0 will lead -3V0 by about 45 to 60 degrees and hence will lie in do not operate region. Hence, earth fault directional unit will not pick-up.
Current Polarization:
For providing direction feature in earth fault relay we can also use current as refrenec phasor which is called current polarization. It is an alternative for voltage polarization. It does not require an additional Potential Transformer (PT).
For balanced system, Ia+Ib+Ic = 0, phasor sum is taken here.
Therefore, I0 = (Ia+Ib+Ic)/3 = 0 which means absence of Zero Sequence Current in balanced system.
During ground fault say at phase A, 3I0 flows from ground to neutral of a Wye connection of Transformer. If we assume for simplicity that Ib = Ic = 0, then 3I0 and Ia are in phase. This indicates that directional unit for ground relay should pick-up as Ia is in phase with 3I 0. Thus we place maximum torque line at zero degrees with respect to I0 phasor. The correspondingOperate and Do Not Operate zones are marked in figure below.
If fault is behind the relay, then the Ia will fall in Do Not Operate region and hence relay will not pickup as Zero sequence Current through the neutral of Wye connection and Relay will be in phase opposition.
Sensitive Earth Fault Protection. Sensitive Earth fault Protection scheme is used for the detection of earth fault. Protection against phase to ground faults can be a difficult problem since ground fault currents vary within a large range, becoming almost negligible in some situations. The ground fault current magnitude depends on the power system grounding which can vary from solidly grounded to ungrounded neutral. The sensitive earth fault protection is usually used in alternators and
transformers with high resistance grounding. High resistance grounding restricts the earth fault current to a very less value and permits the operation of equipment. Also Read, Concept of Neutral Grounding As we know that, the earth fault current magnitude is given by, IF = 3VLN/ (Z0 + Z1 + Z2 + 3ZE) Where Z0 = Zero sequence impedance Z1 = Positive sequence impedance Z2 = Negative sequence impedance ZE = Impedance of earth fault ZE represents the impedance of the ground return circuit including the fault arc, the grounding circuit, and the intentional neutral impedance, if present. If we consider a solidly grounded system, then ZE = 0 and if Z0 = Z1= Z2 then earth fault current will be given as shown below. IF = VLN/ Z1
The Sensitive Earth Fault protection scheme works by measuring the residual current across the three phases in a system. Measurement of three phase residual current is done either by using Core Balance Current Transformer (CBCT) or three CTs connected in parallel. In the ideal condition, the residual current will be zero as all the currents flow through the three phases. Here Residual current means current flowing through neutral or zero sequence current.
As we know that IN = 3I0= IR + IY +IB
But during normal operation, IR + IY+IB = 0 therefore, no residual current will flow. The most important part in SEF or Sensitive Earth Fault Protection is to make proper setting of the Relay. The protection setting should take into consideration that the three CTs do not have identical characteristics and will perform differently for heavy phase-to-phase faults or for initial asymmetrical motor starting currents. This can produce false residual currents. The setting should also be above the line maximum unbalance current. The above conditions must be satisfied to avoid nuisance tripping. In addition, the ground fault protection must be sensitive to minimum ground fault current at the end of the line. Sensitive earth Fault protection scheme is very sensitive to detection of earth fault in the sense that its setting can be as low as 0.2%.
Concept of Neutral Grounding. The concept of system grounding is extremely important, as it affects the susceptibility of the system to voltage transients, determines the types of loads the system can accommodate, and helps to determine the system protection requirements. The system grounding arrangement is determined by the grounding of the power source. For commercial and industrial systems, the types of power sources generally fall into four broad categories:
Utility Service – The system grounding is usually determined by the secondary winding configuration of the upstream utility substation transformer. Generator – The system grounding is determined by the stator winding configuration. Transformer– The system grounding on the system fed by the transformer is determined by the transformer secondary winding configuration. Neutral grounding is generally of three types:
Solid Grounding Resistance Grounding Reactance Grounding
Each of the grounding method serves a specific purpose and based on the suitability of our need, we use any one of the grounding method.
Solidly Grounded Systems: The solidly grounded system is the most common system arrangement, and one of the most used. The most commonly used configuration is the solidly grounded star, because it support single-phase phase to neutral loads. In this type of grounding method, the star point is directly connected to the ground. The figure below, shows the relationship between the phase and line voltage for Solidly Grounded System.
It can be seen from the above figure that the system voltage with respect to ground is fixed by the phase-to-neutral winding voltage. It means that the line-to-ground insulation level of equipment need only be as large as the phase-to-neutral voltage, which is 57.7% (100/1.732 = 57.7 %) of the phase-to-phase voltage. It also means that the system is less susceptible to phase-to-ground voltage transients. This is very important benefit of Solidly Ground System. A common characteristic of solidly-grounded system is that a short circuit to ground will cause a large amount of short circuit current to flow. This condition is known as a ground fault. As can be seen from figure below the voltage on the faulted phase is depressed, and large current flows in the faulted phase since the phase and fault impedance are small.
The voltage and current on the other two phases are not affected. Thus a solidly grounded system supports a large ground fault current. Statistically, 90-95% of all system short-circuits are ground faults. The occurrence of a ground fault on a solidly grounded system necessitates the removal of the fault as quickly as possible. This is the major disadvantage of the solidly-grounded system as compared to other types of system grounding.
A solidly-grounded system is very effective at reducing the possibility of line-to-ground voltage transients. However, to do this the system must be effectively grounded. One measure of the effectiveness of the system grounding is the ratio of the available ground-fault current to the available three-phase fault current. For effectively grounded systems this ratio is usually at least 60%. To summarize, The solidly grounded system is the most popular, is required where single-phase phaseto-neutral loads must be supplied, and has the most stable phase-to-ground voltage characteristics. However, the large ground fault current is a disadvantage and can be hindrance to system reliability.
Resistance Grounded Systems: In Resistance Grounding method, the neutral point is connected to the ground by using a Resistor. The resistor is sized to allow 1-10 A to flow continuously if a ground fault occurs.
The resistor is sized to be less than or equal to the magnitude of the system charging capacitance to ground. If the resistor is thus sized, the high-resistance grounded system is usually not susceptible to the large transient overvoltages that an ungrounded system can experience. If no ground fault current is present, the phasor diagram for the system is the same as for a solidly grounded system. However, if a ground fault occurs on one phase the system response is as shown in figure below. As can be seen from figure below, the ground fault current is limited by the grounding resistor.
If the approximation is made that ZA (impedance of winding) and ZF (Fault impedance) are very small compared to the ground resistor resistance value R, then the ground fault current is approximately equal to the phase-to-neutral voltage of the faulted phase divided by R. The faulted phase voltage to ground in that case would be zero and the unfaulted phase voltages to ground would be 173% of their values without a ground fault present. The ground fault current is not large enough to force its removal by taking the system off-line. Therefore, the high resistance grounded system has the same operational advantage in this respect as the ungrounded system.
Reactance Grounding: A Reactance Grounded system is one in which the neutral point is grounded through an impedance which is highly inductive. Reactance Grounding lies between the effective grounding and Resonant Grounding (will be discussed in next post). Reactance is provided to keep the fault current within safe limit. This method of grounding is used where the charging current is high like in capacitor bank, line reactors used for voltage control of transmission line etc. Restricted Earth Fault Protection of Transformer. Restricted Earth Fault (REF) protection is basically a Differential Protection. The only difference in between the Differential Protection and REF Protection is that, latter protection is more sensitive as compared to the former protection scheme. In earlier posts we have already discussed Differential Protection of Transformer and various characteristics of Differential Protection. In this post we will focus on Restricted earth Fault protection. Also Read,
Transformer Differential Protection Percentage Differential Protection – Slope in Differential Protection Harmonic Restraining in Differential Protection For the sake of understanding REF Protection, we take a Transformer of configuration DYn i.e. HV side of Transformer is Delta connected while the LV side is Start connected and neutral is grounded solidly.
As shown in figure above, there are a total of four Current Transformers (CTs), three CTs connected in each phase i.e. R, Y and B and one CT connected in neutral. The secondary of these four CTs are connected in parallel. The parallel connected CT secondary are then connected to REF Relay Coil. Basically REF protection Relay element is an over current element. Under balanced condition i.e. under normal operation the sum of currents through the secondary of CTs will be zero and current in neutral CT will also be zero. But as soon as a fault takes place in the secondary winding of Transformer, the current in R, Y and B phase will no longer be balanced. Also under earth fault a current will flow through the neutral CT. Because of this unbalance, the summation of current will not be zero but it will have some finite value and hence the relay will pick up. It shall be noted that for a fault outside the Transformer i.e. for through fault Restricted Earth Fault Protection will not operate as in this case of through fault, the vector sum of currents in CT secondary will be zero. This is the reason; such kind of protection scheme is for restricted zone and hence called Restricted Earth Fault Protection.
Now, it is normal to ask that Differential Protection is also a zone protection and it shall operate for any internal fault in Transformer, then why do we need extra Restricted Earth Fault Protection? This is really a very smart question. See, what happens is, the setting of differential protection is normally kept at 20%. So, differential relay shall pick if the differential current exceeds 0.2 A. Now let us consider a case where earth fault occurs just near the neutral point as shown in figure below.
Since the location of fault is very near to the neutral point, the voltage driving the fault current will be very less and hence the reflection of such a low current in primary side of transformer will also will be low. Thus in such case, Transformer differential protection may not operate as its setting is quite high at 20%. Therefore for protection of Transformer from such a fault we need more sensitive protection scheme which is implemented by using Restricted Earth Fault Protection. The sensitivity of REF protection is superior as compared to Differential Protection. Normally the setting of REF protection is kept as low as 5%. Basically the sensitivity of REF protection increases as we are using CT in neutral of transformer and whenever an earth fault take place it is damn sure that current will complete its path through the neutral and hence increasing the sensitivity of REF protection.
Difference between Earthing, Grounding and Neutral. Earthing means connecting the dead part, it means the part which does not carries current under normal condition to the earth. For example electrical equipment’s frames, enclosures, supports etc.
While Grounding means connecting the live part (it means the part which carries current under normal condition) to the earth. For example neutral of power transformer. The purpose of Earthing is to minimize risk of receiving an electric shock if touching metal parts when a fault is present. While the purpose of Grounding is the protections of power system equipment and to provide an effective return path from the machine to the power source. For example grounding of neutral point of a star connected transformer. Ground is a source for unwanted currents and also as a return path for main current some times. While Earthing is done not for return path but only for protection of delicate equipment. Harmonic Restraining in Differential Protection. In this post we will discuss about the feature of Harmonic Restraining feature of Differential Protection. First we should know why Harmonic Restraining is needed?
Suppose we are going to energize the Transformer, obviously the Transformer will have an Inrush current which is around 6 times of full load current. Refer “Transformer Inrush Current” for detail on Transformer Inrush Current. Figure given below shows waveform of Inrush Current of Transformer.
Therefore, the Differential protection will operate. Thus we won’t ever be able to energize the Transformer or we need to bypass the Transformer Differential Protection when we are going to energize. Is it a good practice to Bypass Transformer Protection? You will say NO.So what we need to do for preventing the Differential Protection operation due to Inrush Current?
This requirement calls for Harmonic Restraining in Differential Protection of Transformer. The inrush current of a Transformer, if analyzed, is rich in 2 nd harmonic component. So we can use this fact to prevent the operation of Differential Protection. Therefore a 2nd harmonic Restraining is provided in Transformer Differential Protection. Normally the setting of 2ndHarmonic Restraining is kept at 20% which means that if the 2nd harmonic component in the differential current of Relay is more than 20% of differential current Idthen Differential Protection Relay will not operate as it will think that it is because of Transformer Inrush Current but if the 2nd harmonic component in the differential current of Relay is less than 20% of differential current Id then Differential Protection Relay will operate.
Similarly, during overfluxing of a Transformer the Transformer current is rich in 5th harmonic current. But it may happen so that Transformer is overfluxed for some short time period say around 2 sec. So for this short period of overfluxing should not cause the operation of Differential Protection. Note that during overluxing, the differential current Id will increase which will operate the Differential protection Relay.
Therefore to prevent the operation of Differential protection Relay due to overfluxing, 5thharmonic Restraining is provided in such a manner that if the 5th harmonic component in the differential current of Relay is more than 25% (say) of differential current Id then Differential
Protection Relay will not operate but if 5th harmonic component in the differential current of Relay is less than 25% of differential current Id then Differential Protection Relay will operate.
Percentage Differential Protection – Slope in Differential Protection. As discussed in the post “Transformer Differential Protection”, differential protection is supposed to operate for the internal faults or for the zone of protection it is intended for. Differential protection is not supposed to operate for a through fault. Through fault means a fault outside the zone of protection. Thus as discussed in earlier post, for a through fault the differential current through the overcurrent element of the Differential Protection Relay is zero while there is some definite value of Differential Current for internal faults. But actually there are many limitations due to which a differential current flows the Differential Relay in normal operation also.
A practical transformers and CTs pose some challenge to Differential Protection. They are as follows:
The primary of transformer will carry no load current even when the secondary is open circuited. This will lead to differential current on which the protection scheme should not operate. It is not possible to exactly match the CT ratio as per equation. This would also lead to differential currents under healthy conditions. If the transformer is used with an off nominal tap, then differential currents will arise as the CT ratio calculated for a particular Tap (Nominal Tap) will be different for different Tap, even under healthy conditions.
Thus we see that because of the above reasons a differential current will flow through the Differential Protection Relay. So Differential Protection will operate which is not expected to operate for the above said reasons. So
what to do to prevent tripping because of the differential current caused by the above mentioned reasons?
To prevent the Differential Protection scheme from picking up under such conditions, aPercentage Differential Protectionscheme is used. It improves security at the cost of sensitivity.
In Percentage Differential Protection, we provide a slope feature to the Differential Protection Relay. In modern Numerical Differential protection Relay two slopes are provided. A typical Slope characteristic is shown in figure below.
Notice an offset of to account for the no load current of Transformer. If we don’t provide this offset then the Differential protection will operate during no load of Transformer and will trip the Transformer Primary side Breaker which is not desired.
The current on the X-axis is the average current of primary and secondary winding referred to primary. It indicates the restraining current called the Biasing Current, I bwhile the corresponding difference on Y-axis represents the differential current. The Differential Protection Relay will pick up if magnitude of differential current is more than a fixed percentage of the restraining current.
Let for differential Protection to operate, Id should be greater than the x% of Ib. Therefore,
Id/Ib> 0.0x
But Id/Ib = Slope of the curve
Thus Differential Protection will operate if the Slope is greater than some fixed value which is set in the Differential Protection Relay. Carefully observe the operating zone in the Slope characteristic of Differential Protection Relay. Consider the figure below.
Suppose, the current in the secondary of CT is 1A at normal operating condition. Therefore the Biasing Current Ib = (1+1)/2 = 1A
While the Differential Current Id = (1-1) = 0A
Thus as discussed above, the restraining current is more than the differential current, Differential Protection Relay will not operate.
Now assume a through fault, so the primary side CT current will be 2 A (say) and secondary side CT gets saturated so current in secondary side CT = 0 A.
Thus, Differential current Id = (2-0) = 2A
and Biasing Current Ib = 2/2 = 1A
Thus we see that Differential protection will operate but it is not expected to operate as the fault is through fault. Thus to prevent tripping on through fault we provide a slope so that Differential current increases as the Biasing current increases and Differential protection will operate if the slope exceeds a particular value (which can be set in the Relay).
Calculation of Stabilizing Resistor in High Impedance Differential Protection. Before going into the calculation part of Stabilizing Resistor, I will first explain the purpose of Stabilizing Resistor in High Impedance Differential Protection.
Stabilizing Resistor in High Impedance Differential Protection is used to prevent the operation of Relay in case of through fault. Through fault is a fault outside the zone of protection. Lets us assume that High Impedance Differential protection is used to protect a Bus bar as shown in figure.
It shall be noted here that, in High Impedance Differential Protection, all the CTs are connected in parallel and then the four wires i.e. R, Y, B and N are connected with the Relay as shown in figure above. If there is any fault in the bus, the according to Kirchhoff’s current law, the summation of current will not be zero and a net current will flow through the Relay coil to operate it. In normal condition, the summation of current will be zero and hence no current will flow through the Relay coil and hence the Relay will be stable. Mathematically under normal condition, I1+ I2 + I3 = 0 As Relay sees only summation of current hence we normally employ an overcurrent element in High Impedance Differential Protection. This is the main difference between a high impedance and low impedance differential protection.
Let us consider a through fault i.e. fault outside the zone of protection. To be more specific, let a fault take place after the CT of any feeder. If all the CT’s maintain the same nominal ratio for all external faults the assumed scheme is perfectly valid since no current can flow in the relay coil. However, when the instantaneous overcurrent relay is set low enough to give useful sensitivity to internal faults the Relay may in practice operate falsely on external faults due to a reduction of the nominal ratio of the fault CT resulting from fault CT core saturation. This reduction of the fault CT nominal ratio results in a “false” differential relay current that may operate the instantaneous overcurrent relay. The wort condition will be when a CT gets completely saturated. Thus we need to make Relay insensitive for through fault. To do this we use Stabilizing Resistor. How Stabilizing Resistor makes Relay Insensitive to through Fault? Well, the main cause for the flow of current through the Relay coil in High Impedance Differential Protection is the Voltage across the terminals of CT. We consider the worst case here when a CT gets completely saturated for through fault. When a CT gets completely saturated, it will no longer will be a source of current rather it will behave purely as a Resistor having a value equal to the CT secondary winding. Thus the fault current will not go toward the relay rather it will go circulate through the saturated CT secondary only as current always chooses a path having least resistance. Let the fault current be IF and the resistance of CT secondary be RCT. Therefore the voltage developed across the saturated CT will be, Vs = IFRCTwhen looping of CT secondaries are done at CT Junction Box only. Or, Vs = IF(RCT+ 2RL) when looping of CT is done at Panel or near Relay end. Here looping of CT secondaries means parallel connection of CT secondaries. It may happen so that we are doing the paralleling at the CT Junction Box (JB) or at Panel (Relay end). If paralleling is done at Relay end then lead resistance of saturated CT up to panel shall be considered for the calculation of driving voltage across the common point of CTs but if paralleling is done at CT JB only then lead resistance of saturated CT from CT core to CT JB shall only be considered which is very less and can be ignored. Now, let us assume that the setting of High Impedance Differential Relay for internal fault be Is. So to make Relay insensitive for through fault, the voltage developed shall not drive a current Is through the Relay, hence we put a Stabilizing Resistor R stbin series with the Relay Coil and the value of Stabilizing Resistor Rstbis given as Rstb= Vs / Is
So, Rstb = IFRCT / Is when paralleling is done at CT JB
Or Rstb= Vs = IF(RCT + 2RL) / Iswhen looping of CT is done at Panel or near Relay end. Thus during through fault, for the worst condition of CT saturation, the current through the Relay coil will not be enough to cross setting value of Is and thus will not operate. Rated Normal, Short Circuit Making & Breaking, Short Time Current Rating & Operating Duty Cycle of Circuit Breaker. Rated Normal Current:
The rated normal current of a circuit-breaker is the r.m.s. value of the current which the circuit breaker can carry continuously and with temperature rise of the various parts within specified limits.
The design of contacts and other current carrying parts in the interrupter of the circuit breaker are generally based on the limits of temperature rise. For a given cross section of the conductor and a certain value of current, the temperature rise depends upon the conductivity of the material. Hence, high conductivity material is preferred for current carrying parts. The crosssection of the conductors should be increased for materials with lower conductivity.
The use of magnetic materials in close circuits should be avoided to prevent heating due to hysteresis loss and eddy currents. The rated current of a circuit-breaker is verified by conducting temperature-rise tests.
Rated Short Circuit Breaking Current:
The rated short-circuit-breaking-current of a circuit-breaker is the highest value of short circuit current which a circuit-breaker is capable of breaking under specified conditions of transient recovery voltage and power frequency voltage. It is expressed in kA r.m.s. at contact separation.
Rated Short Circuit Making Current:
It may so happen that circuit-breaker may close on an existing fault. In such cases the current increase to the maximum value at the peak of first current loop. The circuit breaker should be able to close without hesitation as contacts touch. The circuit breaker should be able to withstand the high mechanical forces during such a closure. These capabilities are proved by carrying out making current test. The rated short-circuit making current of a circuit-breaker is the peak value of first current loop of short-circuit current (I pk) which the circuit-breaker is capable of making at its rated voltage.
Rated making current = 1.8 x √2 x Rated short circuit breaking
= 2.5 x Rated short circuit breaking current.
In the above equation, the factor √2 converts the r.m.s. value to peak value. Factor 1.8 takes into account the doubling effect of short-circuit-current with considerations to slight drop in current during the first quarter cycle.
You may like to read, Why Making Current of Circuit Breaker is more than Breaking Current?
Rated Short Time Current:
The short time current of a circuit-breaker is the r.m.s. value of current that the circuit breaker can carry in a fully closed position during a specified time under prescribed conditions of use and behavior. It is normally expressed in terms of kA for a period of one second. Adjacent poles experience mechanical force during the test.
The rated duration of short circuit is generally 1 second and the circuit breaker should be able to carry short-circuit current equal to its rated breaking-current for one second. During the shorttime current test, the contacts should not get damaged or welded.
The current carrying parts and insulation should not get deteriorated. Generally, the crosssection of conductors based on normal current rating requirements is quite adequate for carrying the rated short-circuit current for the duration of 1 second.
Operating Duty Cycle:
The operating sequence denotes the sequence of opening and closing operations which the circuit-breaker can perform under specified conditions. The operating mechanism experiences severe mechanical stresses during the auto reclosure duty. As per IEC, the circuit-breaker should be able to perform the operating sequence as per one of the following two alternatives:
(i) O-t-CO-T-CO
where, O=opening operation
C=closing opeartion
CO=closing followed by opening
t=3 mintes for circuit-breaker not to be used for rapid auto reclosure
t=0.3 second for circuti-breaker to be used for rapid auto reclosure.
T=3 minutes.
(ii) CO-t’-CO
where t’-15 second for circuit-breaker not to be used for rapid auto reclosure.
You may like to read, AutoreclosingPhilosophy in Distance Protection.
Why Transformer Open Circuit Test Conducted on LV side and Short Circuit on HV Side?. As we know that Transformer Open Circuit Test is conducted on LV side and that Short Circuit Test on HV side. To understand the reason behind this, we will consider a Single Phase Transformer of rating 3300 / 220 Volt, 33 kVA. Thus the voltage at the LV side of this Transformer is 220 Volt. Therefore, for Open Circuit Test on LV side the range of Voltmeter will be 220 V.
Now, Full Load Current = (33×1000) / 220 = 150 A
So, excitation current = 4% of Full Load Current
= 0.06×150 = 6 A
Here note that Excitation Current is taken as 4% of the Full Load Current as the range of excitation current is 2-6%.
Also, the range of Wattmeter will be 220 V and 6A.
We see that the rating of instruments required for the testing are standard and easily available. Furthermore, using standard instruments, more accurate results can be obtained. If the Open Circuit Test is conducted on HV side then a source of 3300 Volt may not be readily available. At the same time the instrument ranges required will be 3300 V, 0.4 A [4% of(33×1000) / 3300 = 4% of 10 = 0.4 A] and 3300 V & 0.4A which are not within the range of ordinary instruments and hence result obtained may not be accurate. Also, it is not safe to work on HV side from safety point of view.
Now, coming to Short CircuitTest. For a Short Circuit Test conducted on HV side the range of voltmeter required will be,
Range of Voltmeter = 5% of Rated Voltage
= 0.05×3300
= 165 V
Note that voltage required to circulate rated current is around 2-12% of the rated voltage, that is why we have considered 5% for selection of range of voltmeter.
Range of Ammeter = Rated Current
= 33×1000 / 3300
= 10 A
Range of Wattmeter = 165 V , 10A
Thus we observe that the range of instruments required to perform the test fall within the range of standard instruments which are easily available and accurate.
At the same time, if we conduct the Short Circuit Test on LV side the
Range of Voltmeter = 5% of Rated Voltage
= 0.05×220
= 11 V
Range of Ammeter = Rated Current
= 33×1000 / 220
= 150 A
Range of Wattmeter = 11 V , 150A
Instruments of such range and Auto Transformer capable of handling 150 A may not be readily available and at the same time result may not be much accurate.
It is for these reasons, Open Circuit Test is conducted on LV side and Short Circuit Test on HV side.
Short Circuit Test of Transformer. Open Circuit Test and Short Circuit Test are two important tests which are carried out on a Transformer to determine its equivalent circuit parameters, Voltage Regulation and Efficiency.
Short Circuit Test Short Circuit Test of Transformer is performed on HV side and the supply voltage is so adjusted that rated current flows through the shorted secondary. As rated current is flowing through the shorted secondary that means that rated current will also flow in the primary because of Transformer action.
The supply voltage required for the flow of rated current in the shorted secondary is around 212% of rated voltage. Thus the supply voltage i.e. primary voltage is very less which in turn means that core losses during short circuit test will be negligible (as core loss is directly proportional to the square of primary voltage.). Thus short circuit test gives us information of Ohmic loss of Transformer and the power measured by the Wattmeter is Ohmic loss. An equivalent circuit of the Short Circuit Test referred to secondary side is shown in figure below. Mind that shunt branch in the equivalent circuit is not shown as the core loss taking place during Short Circuit Test is negligible.
Let us assume that, Vsc, Iscand Psc be the Voltage, Current and Power measured by the Voltmeter, Ammeter and Wattmeter respectively. Therefore from the equivalent circuit diagram, Equivalent leakage impedance referred to HV side, ZeH = (R1+R2) + j(X1+X2) = Vsc / Isc
But Equivalent Resistance referred to HV side, ReH = R1+R2
= Psc / Isc2
Hence Equivalent Leakage Reactance referred to HV side, XeH = (X1+X2)
But in case leakage impedance parameters for both primary and secondary side is required then usually, R1 = R2 = ReH / 2 and X1 = X2 = XeH / 2 referred to same side is taken.
Related Posts: Why Transformer Open Circuit Test Conducted on LV side and Short Circuit on HV Side? 2. Open Circuit Test of Transformer 3. Transformer Load Test or Back to Back Test or Sumpner’s Test 4. Rated Normal, Short Circuit Making & Breaking, Short Time Current Rating & Operating Duty Cycle of Circuit Breaker.
1.
Open Circuit Test of Transformer. Open Circuit Test and Short Circuit Tests are two important tests which are carried out on a Transformer to determine its equivalent circuit parameters, Voltage Regulation and Efficiency. Though theses parameters can also be found using the physical dimension of Core and Winding detail but using Open Circuit and Short Circuit Tests are quite easy and simple.
The circuit diagram for Open Circuit Test is shown below. As clear from the figure below, Voltmeter, Ammeter and Wattmeter are connected in LV side of the Transformer and HV side is left open. Rated LV voltage is applied to the LV side of the Transformer and the reading of Voltmeter, Ammeter & Wattmeter is noted for further analysis.
As the Transformer Secondary i.e. HV side is kept open therefore Transformer will only take excitation current to set up magnetic flux in the core. Therefore Ammeter A will
read Excitation Current Ie which is around 2-6% of the full load current. As Ie is very less therefore, Primary leakage impedance drop is negligible and we can say that applied voltage V1 to LV side is equal to the voltage induced in the Primary winding i.e. E1. Therefore, the equivalent circuit when referred to Primary side reduces to as shown in figure below.
In input power as metered by Wattmeter consists of two components, one is Core Loss and another Ohmic Loss. The exciting current being 2 to 6% of the full load current, the ohmic loss in the primary (Ie2r1) varies from 0.04 to 0.36% of full load Primary ohmic loss. In view of this, the ohmic loss in Primary is negligible when compared to the normal core loss which is being directly proportional to square of applied Voltage. Therefore wattmeter reading can directly be taken as the core loss in the Transformer. Thus we see that Open Circuit Test gives us Core Loss of Transformer. The phasor diagram for Open Circuit Test of Transformer is shown below.
Let, V1 = Applied Rated Voltage on Primary Ie = Excitation Current Pc = Core Loss The Core Loss, Pc= V1IeCosƟ So, no load power factor CosƟ = Pc / (V1Ie) Also from phasor diagram, Ic = IeCosƟ Im =IeSinƟ But Pc = V1Ic
So, Ic = Pc/ V1
Therefore, Core Loss Resistance Rc= V1 / Ic
= V1/ IeCosƟ = V12/ V1IeCosƟ
= V12/ Pc Now, Magnetizing Reactance Xm = V1 / Im
= V1 / IeSinƟ Here Rc and Xmare values referred to LV side. Sometimes a voltmeter is placed in the open HV side of the Transformer to measure the Secondary voltage to get turn ratio. Thus, Open Circuit Test gives us the following information: Transformer Core Loss at rated voltage and frequency. Transformer Shunt Branch Parameters like Rc and Xm. Turn Ratio of the Transformer. Transformer Load Test or Back to Back Test or Sumpner’s Test. Transformer Load Test is basically carried out for determining the maximum temperature rise of the Transformer. However is not viable to conduct this test by connecting the secondary of the Transformer to rated load. In case of small Transformer, rated load can be connected to the secondary of Transformer but for large Transformer, rated load capable of consuming rated power is not easily available and also this will lead to wastage of energy. Therefore the best and smart way to load a Transformer is to conduct Back to Back Test or Sumpner’s Test. It shall be noted here that, Back to Back Test or Sumpner’s test can also be conducted for calculating the efficiency of a Transformer but it is better to calculate the efficiency of Transformer using Open Circuit Test and Short Circuit Test of Transformer as this will give more accurate result. In this post we will focus on Back to Back Test or Sumpner’s Test of single phase Transformer. The Back to back Test on single phase transformer requires two identical Transformers. In this Test, the primaries of two Transformers are connected in parallel and are energized at rated voltage and frequency as shown in figure below.
With Secondary open, the wattmeter W1 records the total core loss occurring in both the Transformers. Now, secondary of both the Transformers are connected in series with their polarity in phase opposition so that the voltage across the terminals ab is zero as indicated by the voltmeter V2. Here it shall be noted that, the basis for selection of range of voltmeter V2 shall be the double of rated secondary voltage of the Transformers. In case, the reading of voltmeter V2 is not zero, this means that the series connection of secondary of Transformer are not in phase opposition and therefore we need to change the connection by just connecting the terminals ad together and check that voltage across bc is zero as indicated by voltmeter V2. Now, if we short the secondary of Transformers, no current will flow as Vab = 0 and therefore the reading of wattmeter W1 will not change. Now we will inject voltage in the secondary circuit by means of voltage regulator, fed from the source connected to the primary or from a separate source. The injected voltage is regulated till rated current flow in the secondary circuit. By Transformer Action, rated current now will start flowing in primary circuit too. Note here that, the full load or rated current in primary completes its path as shown by the dotted line in figure above. As this full load current in the primary is not flowing through the current coil of the wattmeter, therefore the reading of wattmeter W1will remain unaltered. The reading of voltmeter V2will now show the voltage drop due to leakage impedance of both the Transformers. As full load current is flowing in the secondary circuit, the wattmeter W2 will read the total full load ohmic loss of Transformers. Let us suppose that, Pc = Core Loss in individual Transformer
Poh = Ohmic Loss in individual Transformer then, Reading of wattmeter W1=2Pc Reading of wattmeter W2= 2Poh
Therefore, from the above two reading efficiency of each Transformer can be determined. From the above discussion, it is quite clear that, even though Transformer is not connected to any load in Sumpner’s Test, rated current is flowing in the primary as well as in the secondary of Transformers and hence full load ohmic and core loss is taking place. If temperature rise of the two Transformers is to be measured, then two Transformers are kept under rated loss condition for several hours till maximum stable temperature is reached.
Types of Relay Output Contact. We know that Relay is a switch which senses fault in a system and once fault is sensed by the Relay, it issues trip command to the Circuit Breaker to isolate the faulty section of the network from the healthy section. The Relay detects the abnormal condition by continuously monitoring electrical quantities which are different for healthy and faulty condition. The electrical quantities which may change during fault condition are voltage, current, frequency and phase angle. If one or more of the above electrical quantities change, that signals the presence, type and location of the fault to the Relay. After detecting the fault condition, Relay pick-up, its contact will change from NO to NC or vice versa. So we can wire up a particular kind of Relay contact to Breaker tripping circuit. So whenever, the Relay picks up, the tripping of Breaker will take place. Relay coil when picked up will change its contact status. Let us say Relay Normally Open (NO) contact is wired to the Breaker Trip Coil Circuit. Therefore when the Relay coil is in de-energized state, the circuit of Trip Coil of CB is not complete and hence no tripping of the CB. During fault condition as the current through the Relay coil exceeds the pick-up value, the Relay coil will get actuated which in turn will force its contact to change over i.e. NO contact will change to Normally Close (NC) thereby closing the Trip Coil circuit of the Breaker. Relays may be fitted with a variety of contact systems for providing electrical outputs for tripping and remote indication purposes. The most common types of relay Output contacts are as follows: a) Self-reset Type. The contacts remain in the operated condition only while the controlling quantity is applied, returning to their original condition when it is removed
b) Hand or Electrical Reset. These contacts remain in the operated condition after the controlling quantity is removed. They can be reset either by hand or by an auxiliary electromagnetic element. The majority of protection relay elements have self-reset contact systems, which, if so desired, can be modified to provide hand reset output contacts by the use of auxiliary elements. Hand or electrically reset relays are used when it is necessary to maintain a signal or lockout condition. Contacts are generally shown on engineering diagrams in the position corresponding to the un-operated or de-energized condition, regardless of the continuous service condition of the equipment. For example, an under-voltage relay, which is continually energized in normal circumstances, would still be shown in the deenergized condition. A ‘make’ contact is one that closes when the relay picks up, whereas a ‘break’ contact is one that is closed when the relay is de-energized and opens when the relay picks up. Examples of these conventions and variations are shown in figure below.
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How does Trip Circuit Supervision Work?. Trip Circuit Supervision Circuit senses any fault either in the trip coil of breaker or trip circuit. On sensing a fault, Trip Circuit Supervision Relay changes its contact status to window annunciation on the panel. Here fault in the circuit means any break or open circuit. In this post we will be discussing the working principle of trip circuit supervision.
Basically, a breaker is equipped with two trip coils. Both the trip coil energizes if relay issues a tripping command. On energization of trip coil, breaker mechanism opens the circuit breaker. Therefore, it is very important to monitor the trip coil healthiness otherwise during the requirement the breaker may not open to clear the fault. Figure below, shows the simplified diagram of trip circuit supervision.
Trip Circuit Supervision Relay is provided to monitor the healthiness of trip coil. As there are two trip coils therefore, there will be two Trip Circuit Supervision Relays and hence two different circuits. This Relay has three coils namely RLA, RLB and main coil (TCS) of the Relay as shown in the figure above. Now, breaker may either be open or close, therefore we need to monitor the healthiness in both the state of the breaker. Thus, trip circuit supervision is divided as pre-close and post close supervision. As shown in the figure, contacts A and B are breaker auxiliary contacts. Mind that theses contacts are connected to the breaker mechanism and therefore their status depends on the breaker position. If the breaker is open, contact A will be OPEN and contact B will be CLOSE. Similarly if breaker is close, contact B will be OPEN and contact A will be CLOSE. Now we will consider pre-close and post close supervision separately. Pre-close Trip Circuit Supervision:
Pre-close means that breaker is open. Thus the status of contact A will be OPEN and that of B will be CLOSE. If you observe the circuit carefully, you will notice that, in this case current will be flowing through both the coils RLA and RLB. As coils RLA and RLB are energized, their contact will be close.
Now carefully observe the circuit of TCS Relay. Since RLA and RLB are close, DC supply will be extended and hence TCS relay will be energized. Therefore, its output contact 1-2 will be open and hence no window will appear. This means that trip circuit is healthy. Assume there is any open circuit, in such case as no current could flow through the coils RLA and RLB, hence relay TCS will not be energized. Because of this its output contact 1-2 will be close for window annunciation . This means to the operator that either DC supply has failed or there is some problem in the trip circuit. Post Close Trip Circuit Supervision:
Post close means that breaker is close. Thus the status of contact A will be CLOSE and that of B will be OPEN. If you observe the circuit carefully, you will notice that, in this case current will be flowing through both the coils RLA.
Now carefully observe the circuit of TCS Relay. Since output contact of RLA is close, DC supply will be extended to TCS relay and hence TCS relay will be energized. Therefore, itsoutput contact 1-2 will be open and hence no window will appear. This
means that trip circuit is healthy. Assume there is any open circuit, in such case as no current could flow through the coils RLA, hence relay TCS will not be energized. Because of this its output contact 1-2 will be close to annunciate window. This means to the operator that either DC supply has failed or there is some problem in the trip circuit. It shall be noticed that, as the resistance of relay coil is very less therefore a high resistanceshall be connected in series so that less current flows through the circuit to operate the trip coil of the breaker. It shall also be noticed that, in case of protection trip, separately positive DC voltage is extended to the trip coil of circuit breaker so that full current flows through the trip coil to operate the breaker as clear from the figure. Basic Criteria for Selection of 33 kV/ 11 kV Or 66kV/11kV Substation. Basically 33 kV / 11 kV is used for distribution of power at the substation and 66 kV / 11 kV is also used for the distribution. But whether to use 66/11 kV or 33/11 kV depends on many factors. Following are some of the factors which are taken care for the selection of voltage of substation.
a) The substation rating is defined as per the power handling capacity, location and purpose of substation. b) Thumb rule for the economical voltage rating has been categorized for different power range to be received as below:
Load up to 150 MVA – voltage rating of 132 kV. Load up to 80 MVA – voltage rating of 66 kV. Load up to 5 MVA – Voltage rating of 33 kV.
Thus, when a substation is of rating 66/11 kV, means substation has been designed to receive 80 MVA on 66 kV and it will distribute the received power on 11 kV. Again, substation of rating 33/11kV means, the substation has been designed to receive 5 MVA of power at 33 kV and it will distribute the same on 11 kV.
Advantages & disadvantages of Harmonics in Power System. Harmonic voltages and currents in a Power System are a result of non-linear electric loads. In a normal AC Power System, the current varies sinusoidally at a specific frequency, usually 50 or 60 Hertz. When a linear electrical load is connected to the system, it draws a sinusoidal current at the same frequency as the voltage, though usually not in phase with the voltage.
Current harmonics are caused by non-linear loads. When a non-linear load, such as a rectifier / inverter, is connected to the system, it draws a current that is not necessarily sinusoidal. The current waveform can become quite complex, depending on the type of load and its interaction with other components of the system. Regardless of how complex the current waveform becomes, as described through Fourier Series analysis, it is possible to decompose it into a series of simple sinusoids, which start at the power system fundamental frequency and occur at integer multiples of the fundamental frequency. Further examples of non-linear loads include common office equipment such as computers and printers, Fluorescent lighting, battery chargers, electronic ballasts, variable frequency drives, and switching mode power supplies. Total Harmonic Distortion or THD is a common measurement of the level of harmonic distortion present in power systems. THD is defined as the ratio of total harmonics to the value at fundamental frequency.
where Vn is the RMS voltage of nth harmonic and n = 1 is the fundamental frequency.
Effects of Harmonics: One of the major effects of power system harmonics is to increase the current in the system. This is particularly the case for the third harmonic, which causes a sharp increase in the zero sequence current, and therefore increases the current in the neutral conductor. Electric motors experience losses due to hysteresis and losses due to eddy currents set up in the iron core of the motor. These are proportional to the frequency of the current. Since the harmonics are at higher frequencies, they produce higher core losses in a motor than the power frequency would. This results in increased heating of the motor core, which (if excessive) can shorten the life of the motor. The 5th harmonic causes a Counter Electromotive Force in large motors which acts in the opposite direction of rotation. The Counter Electromotive Force is not large enough to counteract the rotation; however it does play a small role in the resulting rotating speed of the motor. How Harmonics Eliminated from Alternator Generated Voltage?. In this post we will discuss about how to eliminate or suppress the harmonics from the emf waveform of an Alternator. In an Alternator, the primary source of harmonics in the emf waveform is due to non sinusoidal field waveform. If the field waveform would have been sinusoidal then there have been no harmonics in generated emf of an Alternator. Therefore, first attempt is made to make the field waveform sinusoidal as far as possible and them means of reducing or suppressing the harmonics is adopted. Field waveform of an Alternator can be made sinusoidal by the following methods:
Small air gap at the pole centre and large air gap at the pole end in an Salient Pole Synchronous machine tens to make the field flux sinusoidal. Skew pole faces if possible.
In Turbo-Alternator or cylindrical pole synchronous machine, the air gap is uniform and hence field winding is distributed in slots in such a manner to make the field waveform sinusoidal. Figure below shows the Rotor of an Alternator.
Having adopted all the above mentioned methods, the filed waveform along the air gap periphery is still not purely sinusoidal but it is flat topped. As a result, harmonic emf is always generated in the Alternator. These harmonics can however be eliminated / suppressed by the following methods:
The distribution of armature winding along the air gap periphery tends to make the generated emf waveform sinusoidal. With chorded coil, harmonics can be eliminated. If the ξ be the chording angle for fundamental flux wave, then for nth space harmonics the chording angle becomes nξ electrical. Therefore, pitch factor for nth harmonics,
Kp= Cos(nξ/2) If we want to eliminate 5th harmonics then, ξ = 36° as Kp = Cos(180/2) = 0 Therefore, generated emf E = Where Kp = Pitch factor Kd = Winding Distribution Factor f =Frequency Nph = Number of turns per phase Ø =Flux
KpKdπfNphØ
A chording angle of 30° is most useful in an Alternator as it gives the following pitch factors, 0.966 for fundamental, 0.707 for 3rd harmonic, 0.259 for 5th and 7th harmonics and 0.7.7 for 9th harmonics. Mind that triplen harmonics is eliminated from the generated line emf by Star connection, though 5thand 7th harmonics of reduced magnitude are present in the line emf. By Skewing the Armature Slot tooth / slot harmonics are eliminated. Thus, using the above mentioned ways, harmonics in the generated emf of a synchronous generator is reduced.
Why Output Voltage of Alternator can’t have Even Harmonics?. Output voltage of a Synchronous Generator cannot contain even harmonics. For understanding the reason, carefully observe the figure below.
As shown in the figure, one full pitched coil a and a’ are placed. The fundamental component of field flux wave induce maximum emf in coil sides a and a’ as these are cutting the maximum filed flux Øm1. If the rms value of emf in each coil be E1 then the resultant emf of fundamental frequency across the coil ends A and B = 2E1. Now, the second harmonics component of field flux wave also induces emf in coil sides a and a’. The induced emf because of second harmonics in coil sides a and a’ are maximum as because these are cutting the maximum field flux Øm2. But here it shall be noted that in both the coil sides the emf induced will oppose each other as both coil sides are cutting maximum positive filed flux wave of second harmonic component. Therefore the net / resultant emf because of second harmonic component across the coil sides A and B = 0 This shows that second / even though even / second harmonic is present in the field flux wave, second harmonic cannot be there in the output voltage of an Alternator. It shall however be noted that field flux waveform of an Alternator is symmetrical and hence it do not have any even harmonics and hence there will not be any even harmonics in the generated voltage of an Alternator. Suppression of Harmonics in Transformer. In this post we will focus on various methods adopted to suppress harmonics in a Transformer. Harmonics in a Transformer can be suppressed by flowing methods: Lower Flux Density: Harmonics in Transformer can be suppressed by having lower flux density in core but it has one disadvantage that, this method of reducing flux density in core will lead to higher cross sectional area of the core and more number of turn for maintain voltage ratio. Because V = 1.414πfNØ So if B is reduced then for having a constant value of V, we need to increase cross sectional area A to maintain flux Ø (as Ø = BA) and number of turns N. Thus this method of reducing harmonics is uneconomical and is not used.
Type of Connection: Triplen harmonics are those which are having frequency multiple of 3. Triplen harmonic current and voltage in the line are suppressed by using Star or Delta connection of the winding in the Transformer. Primary or Secondary Winding in Delta: As we know that harmonic voltages have more nuisance effect than harmonic current. Thus every effort is made to suppress the third harmonic voltages in the Transformer. When any of the Primary or Secondary Winding is connected in Delta, triplen harmonic voltages are suppressed considerably. In view of this, one of the Transformers winding either Primary or Secondary must be connected in Delta to suppress triplen harmonics. Use of Tertiary Winding: As seen in previous paragraph, we need to connect either Primary or Secondary must be connected in Delta to suppress triplen harmonics but in case it is not possible to connect either of the Primary or Secondary in Delta then three phase Transformer must be designed to have an extra i.e. tertiary winding connected in Delta. This closed Delta formed by the tertiary winding provides a path for the third harmonic currents and therefore the EMF and flux becomes approximately sinusoidal in nature. It shall be noted that tertiary winding has no any effect on the fundamental frequency component voltages because the phasor sum of three EMFs mutually displaced by an angle of 120° is zero in the closed Delta connected tertiary winding. Star Delta Earthing Transformer: The third harmonic voltages in the two winding Star-Star Transformer can also be suppressed by using Star-Delta Earthing Transformer as shown in figure below.
The closed Delta provides a path for the third harmonic currents and hence the EMF and flux remains sinusoidal. For Star-Star Transformer, the third harmonic voltages can also be suppressed by using 4 wire supply or earthing Transformer Neutral point if Alternator / Generator neutral is grounded. The neutral wire in this case provide a path for the flow of third harmonic current and hence the flux & EMF remain sinusoidal. Transformer Overflux Protection. Transformer Overflux Protection is provided to protect the Transformer core from overfluxing. A Transformer is designed to operate at a particular flux level. In case the flux in the core of Transformer exceeds a certain level, the core loss increases which may lead to overheating of components which in turn may result into internal fault. Therefore, overflux protection is provided.
A transformer is designed to operate at or below a maximum magnetic flux density in the transformer core. Above this design limit the eddy currents in the core and nearby conductive components cause overheating which within a very short time may cause severe damage. The magnetic flux in the core is proportional to the voltage applied to the winding divided by the impedance of the winding. The flux in the core increases with either increasing voltage or decreasing frequency. During start-up or shutdown of generator-connected transformers, or following a load rejection, the transformer may experience an excessive ratio of Volts to Hertz (V/f), that is, become overexcited. When a transformer core is overexcited, the core is operating in a non-linear magnetic region, and creates
harmonic components in the exciting current. A significant amount of current at the 5th harmonic is characteristic of overexcitation.
Assuming Number of turns constant, Flux is directly proportional to V/f. Here V is supply voltage and f is frequency of supply. In case of any Transformer, signal for supply voltage V is taken from PT. Let us assume that Transformer Primary is connected with 220 kV. Thus normal voltage of primary of Transformer will be 220 kV at a frequency of 50 Hz. Also assume that the PT ratio is 220 kV/110 V. Therefore, V/f ratio = 110/50 = 2.2 Thus at a V/f ratio of 2.2 the Transformer will operate satisfactorily. So the question arises which V/f ratio may cause the overfluxing. For answering this we need to have a look at the Hysteresis curve of the core material and from the curve we can judge at which flux level Transformer can be subject for a particular time safely. Normally the setting of overfluxing is kept 110% of nominal value or 1.1 pu. This means at a flux level of 1.1×2.2 = 2.42 the Transformer will operate safely but above 2.42 the Transformer core will be subjected to overflux. Does this mean that at a V/f ratio of 2.5 Transformer shall be tripped instantaneously? No it doesn’t mean so. Because Transformer core may tolerate such an overflux for some short time duration and hence instantaneous tripping is not required. Therefore, wise decision is to give anINVERSEcharacteristics to the tripping which mean more the ratio of V/f less will be time of tripping. Now we consider two cases: Case1: Transformer Primary voltage rises to 247 kV while frequency is 50.1 Hz As primary of Transformer rises to 247 kV at a frequency f = 50.1 Hz The PT secondary Voltage = 247×110/220 = 123.5 V Hence, V / f = 123.5/50.1 = 2.465 Thus the Relay will pick-up and as the characteristics is inverse, the relay will trip after some time because we have kept the setting 2.42. If the Primary Voltage is maintaining at 247 kV , then we can do nothing and the Relay will definitely trip. Case2: Transformer is provide with Tap Changer Suppose the Transformer is provided with Tap Changer. As the Transformer is provided with Tap Changer in the primary side, we can increase the Tap position from the nominal value which will result in increase in the value of N1 (Primary number of turns) and hence,
But this is not going to help us as we have taken the voltage signal from the PT which is connected to the Primary side i.e. and primary side voltage is maintained at 247 kV, hence V/f will be same. Thus we observe that, even though we have Tap Changer, in the present scenario we can do nothing to prevent tripping of Transformer on overfluxing though the Transformer is not actually in overflux condition (as we have increased the number of turns in the primary side.) Therefore, to take advantage of Tap Changer, we can make a provision of taking voltage signal from the secondary side PT of Transformer Relay. In such case, if the primary turn of Transformer is
increased then its reflection on secondary side PT will be observed proportionally and tripping on Overflux protection can be prevented. In case of no load operation of Transformer, we can give voltage signal to the Relay from the Primary side PT. In this way the purpose will be served without compromising the overflux protection. Thus we see, how important is tap changer in preventing tripping of Transformer from Overfluxing.
Concept of Tap-Changers in Transformer. The modern loads are designed to operate at satisfactorily at one voltage level. It therefore of great importance to keep the consumers terminal voltage within prescribed limits. The Transformers output voltage and hence the terminal voltage the consumer can be controlled by using tap either on primary or secondary side Transformer.
is a of of
As we know that V1 / V2= N1 / N2 Where V1 = Primary Voltage V2 = Secondary Voltage N1 = Primary Number of Turns N2 = Secondary Number of Turns Thus V2 = V1(N2 / N1) Thus,
If we increase the Primary Number of Turns N , output voltage of Transformer V decreases. If we decrease the Primary Number of Turns N , output voltage of Transformer V increases. If we increase the Secondary Number of Turns N , output voltage of Transformer V increases. If we decrease the Secondary Number of Turns N , output voltage of Transformer V decreases. 1
2
1
2
2
2
2
2
Now, whether will we change Primary Number of Turns N1 or Secondary Number of Turns N2depends whether we have provided Tap on Primary side or Secondary side. The choice of providing tap on Primary side or Secondary side is based on maintaining voltage per turn constant as far as possible. The flux in the core of Transformer depends on voltage applied in the primary, V1 = 1.414πfN1Ø So, Ø = V1 / 1.414 πfN1 Thus the flux in the core shall be maintained constant. If the Primary voltage per turn i.e. Flux decrease, which means poor utilization of core while in case the Primary voltage per turn increases that means overflux which may cause heating and saturation of the core. Let us take an example of Generating Transformer. As the primary of the Generating Transformer is connected with the Generator output terminals therefore variation in the Primary voltage will be very less. Therefore, the flux in the core of Transformer will be constant and hence the wise decision will be to put the Taps on the secondary side. Other factors which shall also be taken care while deciding upon the side Taps should be provided are:
Transformer Taps are provided on HV side as in this case Tag changing Gera will handle low current and chance of sparking will be less. If we see the construction of Transformer, we will observe that LV winding are placed just after the core to limit the insulation requirement to be provided and HV winding are placed on the LV winding. Thus it is quite difficult to provide the Taps on the LV winding of the Transformer.
Now after deciding the side where Tap is to be provided in Transformer, next question is that whether Tap shall be provided in the center of the winding or at the end of the winding? A general sense says that Tap shall be provided in the middle of the winding because in this case the forces on the winding will be less. Since the current flowing in the Primary and Secondary coils are in opposite direction, these currents interact with the leakage flux in between the two windings and produce a radial force repelling each other as shown in the figure below.
Now, suppose the winding is tapped at one end. When some of the winding is cut out by tap changer, axial force in addition to radial force is also developed as shown in figure below.
Under short circuit condition, the axial force tending to compress the winding against the core is very large which may damage the winding insulation. In order to eliminate this, physical position of the Tapped winding should be in the middle of the Transformer winding so that no axial force arises after some of the turns are cut out.
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Advantage & Disadvantage of Auto Transformer over Two Winding Transformer. The weight of conductor for any winding depends upon two parameters:
Current carried by the winding Number of turns required in the winding.
Thus we can say that weight of conductor in a winding is directly proportional to its Ampere Turns i.e. MMF. Now, we will focus our attention on the weight of conductor required for Auto Transformer and a Two Winding Transformer.
As discussed in earlier post Basic principle of Auto Transformer, the current carried by winding AC i.e. IAC= I1 and that of winding BC i.e. IBC = (I2-I1). Also the number of turns for winding AC is (N1-N2) and that for winding BC is N2. Therefore, The weight of conductor required for Auto Transformer, WCAT ∝Ampere turns of winding AC + Ampere turns of winding BC ∝ (N1-N2)xI1+ N2x(I2-I1) ∝ N1I1+N2I2-2N2I1 But N2 / N1= I1 / I2 So, N2I2= N1I1 Therefore, The weight of conductor required for Auto Transformer, WCAT
∝ N1I1+N2I2-2N2I1 ∝ 2N1I1 – 2N2I1 ∝ 2(N1 – N2)xI1 …………………….(1) If we want to compare the weight of conductor required for Auto Transformer and Two Winding Transformer, then both the Transformers must have same voltage ratio (V2 / V1), current ratio (I2 / I1), input VA i.e. V1I1and output VA i.e. V2I2. Assuming all the above parameters to be same for Two Winding and Auto Transformer, Weight of conductor in Two Winding Transformer WCTWT, ∝Ampere Turns of Primary + Ampere Turns of Secondary ∝ N1I1+ N2I2 But N1I1= N2I2 So, Weight of conductor in Two Winding Transformer WCTWT, ∝ 2N1I1 ……………………………………….(2) Therefore from equation (1) and (2), WCAT / WCTWT= [2(N2 – N1)xI1] / 2N1I1 = 1 – N2/N1 = (1 – k)
As for step down Auto Transformer k<1, hence the weight of conductor required for Auto Transformer is less than that required for Two Winding Transformer. Saving in Conductor = 1 – WCAT/ WCTWT = 1 – (1 – k) = k Hence there is net saving of conductor in case of Auto Transformer. Let us assume that k =0.1, thus the saving in conductor for Auto Transformer will be only 10% but if k = 0.9 then saving in conductor will be 90% which is quite lucrative. Thus we can conclude that Auto Transformer is more economical when the voltage ratio k is more near to unity. Another important aspect is core. As the conductor required for Auto Transformer is less than that required for Two Winding Transformer that means for Auto Transformer lower window dimension will be required. Thus by using Auto Transformer, there is net saving in Core material as well as conductor material, the saving will be more and more as we tend toward voltage ratio unity. Owing to reduction in conductor and core material, the Ohmic loss in conductor and core loss is reduced considerably. Therefore an Auto Transformer has higher efficiency than Two Winding Transformer of same output. Reduction in conductor material means lower value of winding resistance. Since a part of winding is common to both the Primary and Secondary circuit in Auto Transformer, leakage reactance will be less. Because of lower value of leakage reactance, a superior voltage regulation is achieved with Auto Transformer. Disadvantage of Auto Transformer: As seen earlier in this post that saving of conductor in Auto Transformer is k so saving in conductor decreases as the value of voltage ratio k decreases. Another biggest disadvantage of Auto Transformer is direct electrical connection between the Primary and Secondary circuit. If Primary is supplied with high voltage then any open circuit in common winding will lead to dangerously high voltage on LV side which may damage load as well as this dangerously high voltage will be very
harmful of working personnel. Thus special protection must be provided to prevent such an occurrence. Auto-Transformer Starting of an Induction Motor. It can easily be seen from the slip torque characteristics of an induction motor that, there is some finite torque when the slip s=1 i.e. speed is zero. This simply means that Induction Motor is a self-starting motor and begins to rotate on its own when connected to a 3 phase supply. At the instant of starting, a three phase Induction Motor behaves like a Transformer with its secondary winding shorted. Therefore, Induction Motor during starting takes a high current from the supply mains. To limit this high starting current of Induction Motor, different starting methods are used. In this post we will have a look at the AutoTransformer Starting Method of Induction Motor. The main philosophy of starting any Induction Motor is to start it at a reduced voltage and as soon as the motor reaches its rated speed, full supply voltage is applied to the terminals of Induction Motor. A schematic diagram for AutoTransformer Starting of an Induction Motor is given below.
It shall be observed that, using Auto-Transformer we are only applying a reduced voltage xV1 to the Stator terminal of Induction Motor. Here x is less than 1. As soon as the Induction Motor reaches its rated speed, full supply voltage is applied to the terminals of stator. Therefore, per phase starting current of Motor = xV1/Zsc = xIsc Here Isc is the current through the stator during direct switching of motor.
Thus we observe that starting current of Motor has reduced and is x times that of current during DOL (Direct Online) starting. Again, Input VA of Auto-Transformer = Output VA of the Auto-Transformer Ist.V1= xV1(xIsc) Therefore, Per phase starting current from the Supply Mains Ist = x2Isc Thus per phase starting current from Supply Mains has now became x2 times that of DOL current. Mind that it has reduced as x is less than 1 so x2 will be much less than 1. Thus the main advantage of using Auto-Transformer is that it reduces the starting current from the Supply Mains by x2 times. Note that starting current is the motor winding is x times while the starting current from the supply mains has became x2 time of DOL starting current. Now we will have a look at Torque of Induction Motor. As the torque of an Induction Motor is directly proportional of square of applied voltage at the stator terminals, therefore Starting Torque with Auto-Transformer Test = K(xV1)2 where K is constant of proportionality. Starting Torque with DOL Test = KV12 where K is constant of proportionality. Therefore, Test with Auto-Transformer/ Test with DOL = x2 Thus starting torque with Auto Transformer is less than the starting torque with DOL starting by a factor of x2.
Why Star Delta Starter Preferred in Induction Motor?. The main purpose of any starter is to reduce the requirement of high starting current. Normally the starting current of an induction motor is 6 to 7 times of the full load current. If one has an induction motor with a DOL starter, drawing a high current from the line, which is higher than the current for which this line is designed. This will cause a drop in the line voltage, all along the line, both for the consumers between the substation and this consumer, and those, who are in the line after this consumer. This is the reason for which a starter is to be used. In a squirrel cage induction motor, the starter is used only to decrease the input voltage to the motor so as to decrease the starting current.
It is T.P.D.T switch used to first start the motor with the winding connected in star and then switch for delta connection in running position. TPDT stands for Triple Pole, Double Throw.
Why Star Delta Starter Preferred in Induction Motor? If the winding of Induction Motor is connected in star, the voltage per phase supplied to each winding is reduced by 0.577.In general the voltage per phase in delta connection is Vs, the phase current in each stator winding is (Vs/Zs), where Zs represents the impedance per phase of the motor at standstill or start.
The line current or the input current to the motor is Ist (starting current) = (1.732*Vs)/Zs which is the current when it has to be started by DOL starter. Now, if the stator winding is connected as star, the phase or line current drawn from supply at start (standstill) = (Vs/Zs)/(1.732) which is 0.333= (0.577*0.577) of the starting current, if DOL starter is used. The voltage per phase in each stator winding is now Vs/1.732. So the starting current is reduced by 33.3%. Because of the reduction in starting current, starting torque reduces. Therefore we can conclude that by using Star Delta starter, the starting current is reduced to approximately two-thirds. Since starting current is reduced, the voltage drops during the starting of motor in systems are reduced. Why Induction Motor Star Point not Grounded in Industries?. In any electrical system, we do the neutral grounding at the power source e.g. the starpoints of generators or transformers. By keeping the grounded neutrals at the power source, earth fault current will have a return path from the point of short-circuit at downstream to the source. In this way the direction of earth fault current flow can be easily identified and the earth fault protection relays in the distribution system can easily be coordinated.
As clear from the figure, there will not be any earth fault current in the case where there is no neutral grounding of the source because of absence of return path. But as we can
see, if the source is having neutral grounding then earth fault current will have a return path and earth fault current will flow from the point of fault to the source. After reading the above paragraph, I am sure that a question will strike your smart brain, why do we do neutral grounding as there won’t be any earth fault current in absence of neutral grounding? Yes, it is correct that there will not be any earth fault current in absence of neutral grounding but neutral grounding has many advantages, they are as follows:
Voltage of the phases is limited to phase to Ground Voltage The high voltage due to arching ground or transient line to ground fault are eliminated. Sensitive protective Relays for earth fault protection can be used. Over voltage due to lightening are discharged to ground otherwise there would have been a positive reflection at the isolated neutral of the system.
Hope your doubt is clear now. This is the reason Neutral grounding is done in a system. Now coming to grounding of Star Point of Induction Motor, Grounding a motor star point will create an earth path for earth fault current to flow through that motor’s star point. If there are 10 motors in a process plant and their star points are all grounded then obviously there are 10 additional paths for earth fault currents to flow through. If all the motors’ star points are grounded in this way the earth fault current detections by the protection relays will be complicated and it is most likely that Relay will trip at the incorrect locations because earth fault currents are flowing in many directions toward multiple grounded neutral points. Therefore the electrical consumers i.e. the load, including the capacitor banks, even if they are star connected are not to be grounded. Motor is a balanced 3-phase load. However when the system supply voltage is unbalanced caused by unbalanced loads somewhere else or due to network conductors problem, the motor operating under unbalance voltage will result in unbalance current in the 3 windings. The same is true for the generator windings under that condition. The
design engineer may then decide that individual machines should be fixed with negative phase sequence current protection. Even if there is a neutral voltage shift in the induction motor, we should not ground the motor’s neutral point. If we ground the induction Motor, it may create nuisance trip on earth fault protection relays. Why LT Motors are Delta connected but HT Motors are Star?. In industries Low Tension i.e. low voltage Motors like Motors supplied by three phase 415 V are normally have Stator connected in Delta while High Tension i.e. Motors supplied by high voltage like 6.6 kV have Stator connected in Star configuration. The reason behind this is technical while making the Motor economical. Following are the main reasons due to which high voltage Motors stator are connected in Star:
As the Stator winding of Motor is to be connected with high voltage, it is better to configure the Stator in STAR as in this configuration, the phase current remains the same as the line current but the phase voltage reduces by to Vph = Vline/1.732 which means that insulation requirement from phase winding will be less. The second most important reason is that, starting current for Motors is 6 to 7 times of full load current. So start-up power will be large if HT motors are delta connected. It may cause instability i.e. voltage dip in case of small Power System. In STAR connected HT Motors starting current will be less compared to delta connected motor as voltage is V and current is line current. So starting power and starting torque are reduced. As current is less in STAR configuration, copper (Cu) required for winding will be less. ph
Following are the main reasons due to which low voltage Motors stator are connected in Delta:
In Delta connection, the insulation requirement will not be problem as voltage level is less in LT Motors. Starting current will not be problem as starting power in all will be less. So no problem of voltage dips. Starting torque should be large, as motors are of small capacity and hence Stator should be connected in Delta to have more current and hence more starting torque.
Difference between an Induction Motor and a Synchronous Motor. The basic difference is that an induction motor is an asynchronous machine whereas the other one, as the name suggests is a synchronous machine. Following are some important differences between a Induction Motor and a Synchronous Motor:
Synchronous motors operate at synchronous speed (RPM=120f/p) while induction motors operate at less than synchronous speed (RPM=120f/p – slip). Slip is nearly zero at zero load torque and increases as load torque increases. Synchronous motors require a DC power source for the rotor excitation. Synchronous motors require slip rings and brushes to supply rotor excitation. Induction motors don’t require slip rings, but some induction motors have them for soft starting or speed control. Synchronous motors require rotor windings while induction motors are most often constructed with conduction bars in the rotor that are shorted together at the ends to form a “squirrel cage.”
Synchronous motors require DC excitation to be supplied to the rotor windings; induction motors don’t.
Synchronous motors require a starting mechanism in addition to the mode of operation that is in effect once they reach synchronous speed. Three phase induction motors can start by simply applying power, but single phase motors require an additional starting circuit. The power factor of a synchronous motor can be adjusted to be lagging, unity or leading while induction motors must always operate with a lagging power factor. Synchronous motors are generally more efficient that induction motors. Synchronous motors can be constructed with permanent magnets in the rotor eliminating the slip rings, rotor windings, DC excitation system and power factor adjustability. Synchronous motors are usually built only is sizes larger than about 1000 Hp (750 kW) because of their cost and complexity. However, permanent magnet synchronous motors and electronically controlled permanent synchronous motors called brushless DC motors are available in smaller sizes.
Difference between Squirrel Cage and Slip Ring Induction Motor. Please have a look at the below figures of Squirrel Cage Induction and Wound Rotor / Slip Ring Induction Motor. Squirrel Cage Induction Motor:
Slip Ring / Wound Rotor Induction Motor:
Sr. No Squirrel Cage Induction Motor . 1) 1. In Squirrel cage induction motors the rotor is simplest and most rugged in construction.
Slip Ring / Wound Rotor Induction Motor In slip ring induction motors the rotor is wound type. In the motor the slip rings, brushes are provided. Compared to squirrel cage rotor the rotor construction is not simple.
2)
Cylindrical laminated core rotor with Cylindrical laminated core rotor is heavy bars or copper or Aluminum wound like winding on the stator. or alloys are used for conductors.
3)
Rotor conductors or rotor bars are At starting the 3 phase windings are short circuited with end rings. connected to a star connected rheostat and during running condition, the windings are short circuited at the slip rings.
4)
Rotor bars are permanently short It is possible to insert additional circuited and hence it is not possible resistance in the rotor circuit. to connect external resistance in the Therefore it is possible to increase
circuit in series with the rotor the torque; the additional series conductors. resistance is used for starting purposes. 5)
Cheaper cost.
Cost is slightly higher.
6)
No moving contacts in the rotor.
Carbon brushes, slip rings etc are provided in the rotor circuit.
7)
Higher efficiency.
Comparatively less efficiency.
8)
Low starting torque. It has 1.5 time High starting torque. It can be full load torque. obtained by adding external resistance in the rotor circuit.
9)
Speed control by rotor resistance is Speed control by rotor resistance is not possible. possible.
10)
Starting current is 5 to 7 times the Less starting current compared to full load current. squirrel cage Induction Motor.
.
Difference between Synchronous Motors with Damper Winding and Induction Motor. A Squirrel Cage Induction Motor (SCIM) only has a cage rotor winding, whereas a synchronous machine has both a cage (damper) and a wound DC field winding as shown in figure below. Synchronous Motor with Damper Winding:
Squirrel Cage Induction Motor:
In both cases, and when using only the cage winding, both motors will run close to synchronous speed just like a normal SCIM. However, when the synchronous motor field winding is supplied with DC, it will lock into synchronism and its cage or damper winding will carry zero current as the relative speed between the damper winding and rotating flux is zero. Under load disturbances the damper winding will have current induced in it, develop additional torque, and thus help to damp out transient disturbances. In smaller synchronous motors the damper winding is used to start the motor just like a normal SCIM and when close to synchronous speed the DC field is applied and the rotor locks into and runs at synchronous speed. Capacitor Split Phase Motors or Starting of Single Phase Induction Motor by Capacitor. We have already discussed about staring methods of Single Phase Induction Motor. If you miss that please read here, 1) Revolving Field Theory of Single Phase Induction Motors 2) Starting Methods of Single Phase Induction Motors The schematic diagram of Capacitor Split Phase Motor is shown in figure below.
Like in resistor split phase motor, there are two windings, Main and Auxiliary winding but the basic difference between the two method is that in Capacitor Split Phase Motors a capacitor of suitable value is connected in series with the auxiliary winding. Capacitor is connected in series with the auxiliary winding to obtain the desired time phase displacement between the auxiliary winding current Ia and main winding current Im. A centrifugal Switch is also provided the cut out auxiliary winding when the speed of Single Phase Induction Motor reaches 70 to 80% of synchronous speed. You may like to read, Purpose of Centrifugal Switch in Induction Motor As can be seen from the phasor diagram of Capacitor Split Phase Motor, there is an angle of β between the auxiliary winding current Ia and main winding current Im.
Mind that the angle between the auxiliary winding current Ia and main winding current Im is 180° if there is main winding alone and because of this Single Phase Induction Motor cannot start by itself. Also, the torque produced in any machine is directly proportional to IaImSinβ, therefore in this method of starting, there will be a net starting torque and the motor will start. The value of Capacitor used shall be selected based on the load starting torque requirement. If the starting torque requirement of load is more, then angle β shall be made more by selecting higher value of Capacitor. A maximum starting torque can be obtained by this method of starting if angle β = 90°. But to have β = 90°, the size and cost of Capacitor will increase. Therefore a compromise is made in between the load starting torque requirement and size & cost of Capacitor. It shall be noted that, auxiliary winding and Capacitor are in circuit for a short time only, and therefore these can be designed for a minimum cost. The torque speed curve for Capacitor Split Phase Motor is depicted in figure below.
It is clear from the figure that starting torque in this method is high. Capacitor Split Phase Motors have a typical power rating of 100 to 800 Watts. The value of starting Capacitor varies from 20 to 30 microF for 100 Watt Motors and 60 to 100 microF for 750 watt Motors. AC electrolytic capacitors are mostly used in this method of starting but Motors shall not be frequently started else electrolytic capacitors may get overheated and damage.
Why 3-Phase Induction Motors are Self-Starting but 3-Phase Synchronous Motors Not?. Let us consider a 3-phase induction motor first and see how it rotates. A 3- phase supply given to the armature of Induction Motor produces a rotating magnetic field. This rotating magnetic field rotates at synchronous speed Ns = (120xf)/P This rotating magnetic field links to the rotor coils and induces voltage which in turn produces current in the rotor. The current carrying rotor being placed in a magnetic field experiences a torque and hence begins to rotate in the direction of rotating magnetic field. Thus we see that Induction Motor is self-starting. It does not require nay external mean to rotate. Now we consider a 3-phase synchronous motor. A 3-phase supply is given to the armature of Synchronous Motor, produces a rotating magnetic field. However, in this case, the rotor has its own field produced by a DC current flowing through the rotor winding. This rotor field tends to align itself along with the rotating magnetic field produced by the stator i.e. armature winding. North pole of rotor tries to lock with the South pole of stator and South pole of Rotor tries to align along the North of Stator.
So what happens exactly? The North Pole of rotor tries to chase the South Pole of stator. But the stator magnetic field is rapidly rotating at synchronous speed, and before the North Pole of rotor could lock with South Pole of stator, the stator field has shifted position so that its North (stator) comes in the vicinity of North Pole of rotor and they repel as shown in figure below.
Because the rotor has certain inertia and the speed of the rotating magnetic field is too fast for it to catch up, it ends up vibrating. So Synchronous Motor fails to start. So what do you do for starting a Synchronous Motor?
We can give a reduced frequency supply to the stator, this will reduce the speed of rotation of the stator magnetic field and the rotor will easily catch up, once the rotor catches up we may increase the frequency. We can manually rotate the rotor till it catches speed near to synchronous speed and eventually locks the rotor field with the stator field. We can use Amortsieur Windings. The concept is to start the motor as an induction motor. Initially no DC field excitation is given and the motor operates as an induction motor. Once it attains some speed near to synchronous speed, DC excitation is given and the rotor field aligns itself with the stator field, and rotor attains synchronous speed.
Methods of Starting Single Phase Induction Motors. As discussed in earlier post Revolving Field Theory of Single Phase Induction Motors, a single phase induction motor with main stator winding alone has no inherent starting torque as the main stator winding and only produces stationary pulsating air gap flux wave. For the development of starting torque, rotating field at the starting must be produced. There are various methods of starting single phase induction motor, can be classified as below,
1) Split phase starting 2) Shaded pole starting 3) Repulsion motor starting 4) Reluctance starting Generally a single phase induction motor is known by the method employed for its starting. Basically the selection of a particular type and choice of starting single phase induction motor method depends upon the following factors: 1) Torque speed characteristics of load from starting to normal operating speed 2) The duty cycle 3) The starting and running line current limitation as imposed by the supply authorities In this post we will only discuss the Split Phase Starting method. Single Phase Induction Motor employing split phase starting method is known as Split Phase Motor. All the split phase motors have two winding, main winding and auxiliary winding. Both theses windings are connected in parallel but their magnetic axis are displaced by an angle 90°. Split phase starting method is further categorized into following: 1) Resistor split phase motors 2) Capacitor split phase motors 3) Capacitor start and run motors
Resistor split phase motors. A schematic diagram of the two stator winding in quadrature is shown if figure below. Subscript m and a stands for main and auxiliary winding of stator. CS is Centrifugal Switch. You may like to read, Purpose of Centrifugal Switch in Induction Motor.
As we know that, if the two winding currents are shifted in time phase, a rotating filed is created which is necessary for the production of starting torque. In order to achieve this main winding M is designed to have low resistance but higher reactance whereas the auxiliary winding is designed to have higher resistance (thin wire) but lower reactance. The use of thin wire for auxiliary winding is acceptable as auxiliary winding only remain in circuit during starting but the use of thick wire for main winding is necessary as main winding has to remain in circuit permanently. As the reactance is directly proportional to the square of number of turns, auxiliary winding has less number of turns as compared with main winding. In addition to the above mentioned points, leakage reactance of main winding is increases by placing it at the bottom of slot whereas auxiliary winding is placed at the top of slot to have low leakage reactance. As seen in above discussion, main winding has more reactive impedance as compared to the auxiliary winding, therefore main winding current Im lags behind the auxiliary winding current Ia as shown in the phasor diagram below.
Thus from the phasor we see that the angle between the two field produced by main and auxiliary winding is β. As we know that torque produced is directly proportional to torque angle which is β here, therefore a n net starting torque will be developed. The auxiliary winding is disconnected automatically by means of Centrifugal Switch CS at about 70-80% of synchronous speed. If the Centrifugal Switch fails to operate, auxiliary winding will remain in the circuit and noisy performance of single phase induction motor will result. Since auxiliary winding is short time rated, it must get overheated and consequently burn out.
Typical application of Resistor split phase induction motor is for fans, blowers, centrifugal pumps and refrigerator. Concept of Series and Shunt Faults. Electrical Faults can be classified into two categories:.
Shunt Faults and 2) Series Faults Shunt faults include power conductor or conductors to ground or short circuit between the conductors. Series type of fault is basically unbalance in system. Suppose we have used Fuse / Breaker to protect the circuit. If one or two phases open while the third phase remain in circuit, such kind of fault is called Series Fault. Notice that Series Fault may also occur in case of one or two Broken Conductor. Here broken conductor is like breaking of jumper on the tower of transmission line which is not touching the grounded tower body.
Shunt faults are characterized by increase in current and decrease in voltage and frequency whereas Series faults are characterized by increase in voltage and frequency and decrease in current in the faulted phase. Shunt faults are classified as: 1) Line-to-Ground Fault 2) Line-to-Line Fault 3) Double Line-to-Ground Fault 4) Three phase fault Of the above faults, first three faults are unsymmetrical fault as the symmetry is disturbed in one / two of the phases. The method of Symmetrical Components shall be applied for the analysis of such unbalance and fault.
Three phase fault is balanced fault which can also be analyzed using concept of symmetrical components. Series faults are classified as: 1) One Open Conductor
2) Two Open Conductors These faults also disturb the symmetry and therefore these faults are unbalanced faults and hence shall be analyzed using concept of symmetrical components. Neutral Voltage during Fault: The potential of neutral when it is grounded through some impedance or is isolated will not be at ground potential under unbalance condition as in unsymmetrical fault rather it will have some finite value with respect to ground. The potential of neutral is given as Vn = -InZn where Zn is neutral grounding impedance and In is neutral current. Notice the negative sign before the expression of neutral voltage Vn, it indicates the flow of current from ground to the neutral point and therefore the potential of neutral point will be less than the ground potential. For a three phase system we know that, Ia + Ib+ Ic = 3Ia0
You may also like to read Calculation of Symmetrical Components Therefore, Vn = -3Ia0Zn
Notice that only zero sequence current flows through the neutral and therefore voltage drop across neutral will be only due to zero sequence currents.
HRC Fuse – Construction, Working and Characteristics. HRC stands for High Rupturing Capacity. HRC Fuse has high rupturing capacity. Because of its high current rupturing capacity, a special method for extinguishing arc is required in the design of HRC Fuse.
Construction of HRC Fuse: HRC Fuse consists of heat resisting ceramic body having metal end caps on which silver current carrying element is welded in a special manner as shown in figure below.
As clear from the figure above, the fuse element have a portion of Tin Alloy, known as a Eutectic Material. This alloy is used to give the fuse specific operating characteristics. Also, constrictions in the fuse element are provided which play a very vital role in the operation of Fuse. The space between the body surrounding the Fuse element is filled filling powder such as with Silica Send, Chalk, plaster of peris etc.
Working Principle of HRC Fuse: Under normal operating conditions the current flowing through the Fuse element does not provide enough energy to melt the element. The heat produced is absorbed by the surrounding filling powder. If a large current flows the energy produced melts and vaporizes the fuse element before the fault current reaches the peak. The chemical reaction between the fuse element vapour and filling powder results into high resistance material which helps in extinguishing the arc. Now, will HRC Fuse blow off in case of overload condition?
Under overload condition the fuse element will not blow off but if the condition exists for prolonged period, the Eutectic Material will melt and break the fuse element. This is the purpose of providing Eutectic Material in the HRC Fuse.
Will Eutectic Material blow off during short circuit condition? Under high current short circuit conditions the smaller area constricted parts of the fuse element will melt rapidly and vaporize and will break before the Eutectic Material. That is why constrictions are provided in the HRC Fuse element.
Characteristics of HRC Fuse: A Fuse operates when its element melts due to heat produced by I2RF, where RFis Fuse resistance. This heat produced increases if the current flowing through the Fuse element increases. Therefore, we can conclude that a Fuse element will melt faster for large fault current while it will take some time for lower value of fault current. This timecurrent relationship of Fuse is known as Characteristics of Fuse and is very useful for proper selection of Fuse for a particular circuit and for coordination purpose. A typical Fuse characteristic is shown in figure below.
How to Interpret the Fuse Characteristics? In the above figure, curve for three Fuses of rating 60 A, 100 A and 200 A are give. We select Fuse of rating 60 A for the sake of understanding. See, if the current flowing through the Fuse element is around 350 A then the Fuse element will melt in .02 sec i.e. 20 ms while if the current is around 225 A then it will take 50 ms to melt. Thus we see that the Fuse characteristic is Inverse Time. Higher the current, lower will be the time to melt. Electrical Fuse – Types and Related Terminologies.
Fuse: A fuse is the small piece of wire connected between the two terminals of insulated mounted base. Fuse is always connected in the series of the circuit of low voltage equipment. It is the simplest and cheapest form of protection from overload and short circuit. The Fuse is expected to carry the normal current without heating and during overload / short circuit; Fuse gets overheated up to its melting point rapidly and thus breaking the circuit. The materials used for the Fuse are Tin, Lead, Silver, Zinc, Copper etc. For small value of current an alloy of Lead and Tin, in the ration of 37 & 63% are used. But for current more than 15A, this alloy is not used as the diameter of the wire will be large and after fusing the metal release will be excessive.
Important Terminologies Related to Fuse:
Minimum Fusing Current: It is the value of current flowing through the Fuse wire at which the Fuse wire will melt. Fuse Rating: Fuse rating is given in Ampere. It is basically that value of current at which the Fuse is expected to operate safely without melting. This value of current will definitely will be less than the Minimum Fusing Current. Fusing Factor: Fusing Factor is defined as the ratio of minimum fusing current to the fuse rating. Fusing Factor = Minimum Fusing Current / Fuse Rating The value of Fusing Factor is always greater than 1. Prospective Current: Prospective Current of Fuse is the value of current which will flow through it just before the melting of the fuse wire under Short Circuit condition. Melting Time / Pre-arcing Time: This is the time taken by a fuse wire to be broken by melting. It is counted from the instant; the over current starts to flow through fuse, to the instant when fuse wire is just broken by melting. Arcing Time:
After breaking of fuse wire there will be an arcing between both melted tips of the wire which will be extinguished at the current zero. The time from the instant of arc initiated to the instant of arc being extinguished is known as Arcing Time of fuse. Total Operating Time: Total Operating Time of Fuse is the sum of Pre-arcing and arcing time. Types of Fuses: There are basically two types of Fuses: AC Fuse DC Fuse This classification of Fuse arises because of arcing. In DC it is quite time taking to extinguish the arc. Therefore DC Fuses are made with longer wire so as to avoid arc. Hence DC Fuses are bigger in size. But in AC fuse as the current reduces to zero in every haft cycle (10 ms assuming 50 Hz frequency), arc is extinguished.
Other Types of Fuses: Cartridge Type Fuses: Cartridge fuses are used to protect electrical appliances such as motors air-conditions, refrigerator, pumps etc, where high voltage rating and currents required. They are available up to 600 A and 600 V AC and widely used in industries, commercial as well as home distribution panels.
Blade Type Fuses: This type of fuses, also called as spade or plug-in fuses comes in plastic body and two metal caps to fit in the socket. They are used in automobiles for wiring and short circuit protection.
Other Types of Fuses are SMD Fuses, Axial Fuses, Thermal Fuses, HRC (High Rupturing Capacity) fuse and High Voltage fuses. What is Endurance Test of Circuit Breaker?. Endurance Test of Circuit Breaker is conducted to check the healthiness of its mechanical parts i.e. operating mechanism. In this test, Circuit Breaker is operated several times and checked for any damage of its mechanical parts / contacts. The breaker should be in a position to open and close satisfactorily. This test is also called Mechanical Test. In mechanical tests, the circuit breaker is opened and closed several times (1000). Some operations (about 50) are conducted by energizing the relays, remaining are by closing the trip circuit by other means. Mechanical tests on high voltage AC circuit breakers are conducted without current and voltage in the main circuit. Out of the 1000 operations, about 100 operations are made by connecting the main circuit (contacts) in series with trip circuit. No adjustment or replacement of parts is permitted during the mechanical tests. However, lubrication is permitted as per manufacturer’s instructions.
After the Endurance Test, the contacts, linkages and all the other parts should be in good condition and should not show any permanent deformation or distortion. The dimensions should be within original limits. During repeated operations of the circuitbreaker, the weaker parts in the assembly may fail. The circuit-breaker is then considered to have failed in the mechanical test. The tests are then to be repeated after improvement in the design and manufacture. Successful performance in Mechanical Endurance Test proves the adequacy of design and also good quality of materials and manufacture. Though 1000 close-open cycles are specified in the standards, the manufacturer may conduct 10,000 or more operations to ascertain the reliability and for getting design data. What is the Need of Synchronizing two Different Power Sources?. For understanding the need of Synchronization of two Power Sources, first we shall consider the meaning of Synchronization. Suppose we have a trolley that can only drawn by either pushing or pulling it ,two workers are there to drive it if one of them is pushing in one direction but the other one is in another direction. What will happen? They can’t move it with different speed or different direction. Similar is going to happen with power source. Synchronization of two Power Sources means both the sources have the same
Phase Sequence Voltage Magnitude Frequency Phase Angle
There is a setting provided in the Synchronization Check Relay (25 SYN). There are two terms which are frequently used in Synchronization, Running Line and Incoming Line. The bus which is already charged and to which we are connecting a source is called Running Line as shown in figure below.
In above figure if we close the CB-1 then we are synchronizing S1 to the Bus, therefore S1 is Incoming Line and Bus is Running Line. Now, what will happen if we connect two sources in out of Synch.? When power sources are not synchronized, there are instances where there is a voltage difference at the same very node where the three sources are connected. Suppose Source S1 = 415V, 50 Hz, phase angle = 0° Source S2 = 415 V, 50 Hz, phase angle = -120° Thus, we are going to connect the R phase of Source S1 to Y phase of Source S2. Therefore from phase angle 0 to 90 degree, source S2 is stronger in magnitude and hence current will flow from source S2 to S12 but after 90 to 135 degree, source S1 is stronger in magnitude of voltage and hence current will flow from S1 to S2. In this manner a continuous circulation current will flow from one source to another whose magnitude depends on the system impedance.
This circulating current if high enough will burn the equipment connected. Therefore it is must that we should Synchronize the two sources. Working Principle of Petersen Coil. To better understand the working principle and need of Petersen Coil, let us have a look at the arcing ground phenomenon. We know that arcing ground phenomenon is observed in ungrounded 3 phase system. During arcing ground the voltage of healthy phase rises from phase voltage to line voltage i.e. it becomes √3Vph. Also, arcing in such phenomena is due to heavy capacitive charging current which is 3IC where IC = Vph / XC. Thus if there were any way to reduce this charging current then arcing ground phenomena could have been eliminated. Isn’t it?.
Well, you will say that we can connect a resistor in ground of system to minimize the capacitive charging current. Then why do we connect inductor in ground? Why Inductor is used to eliminate Arcing Ground?
To answer this question, let us consider a single line to ground fault and its phasor diagram for an ungrounded system as shown below.
From the phasor diagram it is can be easily observed that, the voltage of neutral point shifts from ground potential to phase voltage Vph but in opposite direction. This is the reason the direction of VC is reverse in figure above and shown by V’C. The fault current IC (IC = IA+IB) is perpendicular to the V’C. Thus if we want to eliminate the fault current then
we must connect an element which will take current in a direction opposite to I C. Carefully observe that IC is leading V’C by 90°.
Now as we are connecting an element in between the neutral point N and ground, therefore the voltage drop across that element will be V’C. Thus that element must take current equal to IC and shall lag from V’C by 90°. As inductor takes lagging current, therefore an inductor is connected in between neutral and ground to eliminate arcing ground. What is Petersen Coil?
Petersen Coil is nothing but an inductor used to connect ground of three phase system to the earth. In other words, the neutral of three phase system is grounded through Peterson Coil. Basically, such grounding is adopted to minimize the capacitive charging current during fault in the lines. This also eliminates the arcing ground. The inductor connected in figure above is Petersen Coil. This type of grounding is also known as Resonant Grounding. How does Petersen Coil Work?
As discussed earlier in this post, Petersen Coil must take current equal to the fault current ICso that it neutralizes the fault current. This is the reason, it is also known as fault neutralizer. Let us consider the figure shown above. The current through the Petersen Coil IL = Vph / ωL But the fault current IC = 3Vph / XC (how? Please read Arcing Ground) Therefore to neutralize the fault current, IL = IC
Hence, Vph / ωL = 3Vph / XC ⇒1/ωL = 3ωC ⇒L = 1/3ω2C Thus to neutralize the capacitive charging current, the value of inductance of Petersen coil shall be 1/3ω2C. Advantages of Resonant Grounding The use of Petersen coil reduces the line interruption due to transient line to ground fault. This is otherwise not possible with other kind of grounding. The tendency of developing three phase fault from single phase fault is reduces with the use of resonant grounding.
Working Principle of Earth Leakage Circuit Breaker ELCB and Residual Current Device RCD. An Earth Leakage Circuit Breaker (ELCB) is a safety device used to directly detect the leakage current to the Earth from an installation and cut the power supply. Basically ELCB is used where the earth impedance is high. Because of high earth impedance, the voltage difference between the Metallic part of the Installation and Earth will be quite high and dangerous from human safety point of view. It may strike in your smart mind that “What is the difference between Earth Fault current and Earth Leakage Current?” This is very important to know as we are going to discuss about safety device used to sense earth leakage current. Well, according to IEC 60947-2, Earth fault current is the current flowing to earth due to insulation fault and Earth leakage current is the current flowing from the live parts of the installation to earth in the absence of an insulation fault. In case of degradation of electrical insulation, the live conductor may get in touch with the metallic part of the equipment and because of high earth impedance; the potential difference between the body of equipment to the Earth will be high enough to result in shock to the working personnel. Earth Leakage Circuit Breaker (ELCB) detects the leakage current to the earth and trips the associated breaker to isolate the supply. There are two types of Earth Leakage Circuit Breaker (ELCB). One is Voltage Earth Leakage Circuit Breaker, also called Voltage ELCB and another is Current Earth Leakage Circuit Breaker, also known as Current ELCB. Working Principle of Voltage ELCB: Voltage ELCB is a voltage operated device. It has a coil and if the voltage across the coil exceeds a predetermined value such as 50 V, the current through the coil will be sufficient enough to trip the circuit.
Voltage ELCB is connected in between the metallic part of equipment and the Earth. If we take an example of insulation failure, then the voltage across the coil of Voltage ELCB will drive enough current to cut the power supply till the manually reset.
Working Principle of Current ELCB: The working of Current ELCB is quite interesting but easy. Current operated ELCB is also known as Residual Current Device, RCD. A Residual Current Device (RCD) has a toroidal iron core over which phase and neutral windings are wound. A search coil is also wound on the same iron core which in turn is connected to the trip coil. Figure below shows the constructional detail of RCD or Current ELCB.
Under normal operating condition, the current through the phase winding and neutral winding are same but both the windings are wound in such a manner to oppose the mmfs of each other, therefore net mmf in the toroidal iron core will be zero. Let us consider a condition where earth leakage current exists in the load side. In this case the current through the phase and neutral will no longer be equal rather phase current will be more than the neutral current. Thus mmf
produced by phase winding will be more than the mmf produced by neutral winding because of which a net mmf will exist in the toroidal iron core. Net mmf in Core = mmf by phase winding – mmf by neutral winding This net mmf in the core will link with the Search Coil and as the mmf is changing in nature (current is AC), an emf will be induced across the terminals of the Search Coil. This emf will in turn drive a current through the Trip Coil which will pull (because of current flow through the Trip Coil, it will behave as an electromagnet and hence will pull the lever to open contact) the supply contacts to isolate the power supply. Notice that Current ELCB works on Residual Current that is the reason it is also called Residual Current Device. A RCD / Current ELCB is also provided with test button to check the healthiness of the safety device. If you carefully observe the figure, you will notice that, when we press the Test Button, Load and phase winding are bypassed due to which only mmf because of neutral winding will exist in the core (as there is no opposing mmf as was the case with both the windings in service) which will cause RCD to trip to isolate the supply.
Working Principle of Residual Voltage Transformer. A Residual Voltage Transformer is used to measure the residual voltage of three phase system during single phase fault. During normal operating condition, the summation of three phase voltage is zero but in case of single phase fault, the scenario changes and there exists a residual voltage.
Let us first discuss residual voltage in case of single line to ground fault. Let us consider a solidly grounded system as shown in figure below.
Let us assume that a ground fault takes place in A phase (In many industries and numerical relays, normally the phases are said as A, B and C instead of R, Y and B, though they represent the same thing i.e. A phase means R phase, B means Y phase and C means B phase). Ea, Eb and Ec are the Generator terminal voltage per phase. Bold letters here represent vector form.
Because of ground fault in A phase, the voltage at the point of fault will become zero but the voltage of other two healthy phases will remain normal as the neutral is solidly grounded therefore the neutral potential will be maintained to earth potential. Va = 0 Vb = V ∠-120° Vc = V ∠120° Here V is per phase voltage under normal condition. Thus the residual voltage of system = Va+Vb+Vc = 0 + V ∠-120° + V ∠120° = V ∠-60° Thus we observe that, there exists a residual voltage in case of single line to ground fault. This residual voltage is measured by Residual Voltage Transformer. The primary of Residual Voltage Transformer is connected to three phase system and its secondary is connected in Broken Delta as shown in figure below.
The output of the secondary windings connected in broken delta is zero when balanced sinusoidal voltages are applied (as Va+Vb +Vc = 0), but under conditions of unbalance a residual voltage equal to three times the zero sequence voltage (V0) of the system will be developed. To measure this component i.e. 3V0, it is necessary for a zero sequence flux to be set up in the Residual Voltage Transformer (RVT), and for this to be possible there must be a return path for the resultant summated flux. Therefore, RVT core must have one or more unwound limbs linking the yokes in addition to the limbs carrying phase windings. Usually the core is made symmetrically, with five limbs, the two outermost ones being unwound. This two outermost unwound limbs provide return path for zero sequence flux. In case where three single phase transformer units are used to measure residual voltage, no extra limbs are requires as each single phase transformer has a core with closed magnetic path. It is very important to earth the primary winding neutral of Residual Voltage Transformer to provide return path for zero sequence current else zero sequence current cannot flow and hence the flux will contain 3rd harmonic component that is reflected in primary and secondary voltages of Residual Voltage Transformer. This voltage appearing at the secondary terminals of RVT is not the residual voltage of the system in any way. Core Balance Current Transformer. Core Balance Current Transformer or CBCT is a ring type current transformer through center of which a three core cable or three single core cables of three phase system passes. This type of
current transformer is normally used for earth fault protection for low and medium voltage system. A typical Core balance Current Transformer is shown in figure below.
Secondary of CBCT is connected to Earth Fault Relay. During normal operating condition as the vector sum of three phase current i.e. (Īa + Īb + Īc =0) is zero therefore no residual current in the primary will be present. Here residual current means zero sequence current. Therefore there will not be any flux developed in the CBCT core and hence no current in the secondary circuit of CBCT. Working Principle of CBCT: Let Īa, Īb and Īc be the three line currents and Φa, Φb and Φc be corresponding components of magnetic flux in the core. Assuming that the CT is operating in the linear region (Read B-H Curve to get idea of linearity), magnetic flux because of individual phase current will be directly proportional to the phase current and hence we can write as below, Φa = kIa Φb = kIb Φc = kIc where k is constant of proportionality. Mind here that same constant of proportionality is used as all the three phase current are producing magnetic flux in the same core i.e. magnetic material. Thus the resultant magnetic flux in the CBCT core, Φr = k(Īa + Īb + Īc) …………………..(1) But we know from theory of symmetrical components, Īa + Īb + Īc = 3Ī0 = Īn Where, Io is zero sequence current and In is neutral current. Hence we can write as Φr = kĪn …………………………(2) Now let us consider two cases: Case1: During normal condition Īa + Īb + Īc = 0 Hence from equation (1), Net resultant flux in the CBCT Core, Φr = 0 which means no secondary current and therefore the Earth Fault Relay won’t operate. Case2: During earth fault, three phase current passing through the center of Core Balance Current Transformer will not be balanced rather a zero sequence current will flow. For example for single line ground fault, If = 3Ia0 = In Thus from equation (2), Net magnetic flux in the CBCT core, Φr will have some finite value which in turn will induce current in the secondary circuit due to which earth fault relay will operate.
Because of this reason, a Core Balance Current Transformer or CBCT is also called Zero Sequence Current Transformer. Advantage of Core Balance Current Transformer:
The advantage of using CBCT for earth fault protectionis that only one CT core is used instead of three core as in conventional system where the secondary winding of three cores are connected residually. Thus the magnetizing current required for the production of a particular secondary current is reduced by one third which is a great advantage as the sensitivity of protection is increased. Also, the number of secondary turn does not need to be related to the cable rated current because no secondary current flows under normal operating condition as the currents are balanced. This allows the number of secondary turns to be chosen to optimize the effective primary pick-up current. Core Balance Current Transformer is normally mounted over a cable at a point close to the cable gland of the Switchgear. In case cables are already laid in a Switchgear, physically split core, which is also known as Slip-over type CT, are used. Knee Point Voltage of Current Transformer. According to IEC, Knee Point Voltage of a Current Transformer is defined as the voltage at which 10 % increase in voltage of CT secondary results in 50 % increase in secondary current. If you carefully read the definition, you will notice that it is something related to saturation of Current Transformer. Never mind, I will explain what exactly Knee Point Voltage mean and what is its significance. We know that CT core is made of CRGO (Cold Rolled Grain Oriented Silicon Steel). When the primary of CT is energized a working mmf is produced in the core. To produce a working mmf, excitation current Ie is taken. This mmf produces a flux inside the core of CT which links with the secondary winding and as per the Faraday’s Law of Electromagnetic Induction, an emf is generated across the terminals of CT secondary by Transformer action. The emf induced in the CT secondary terminal is given as E = 4.44fNØ Where f is frequency of supply, N is number of secondary turns and Ø is flux in the core of CT. But Ø is directly proportional to mmf and mmf in turn is directly proportional to current. Thus if we increase the current, flux Ø generated in the core will increase till the core saturates. Thus there must be a point where from the flux do not increase in the same proportion as the increase in current. This point is called Knee Point. After discussing this much, we can at least say that Knee Point is something related with Saturation of CT core. If there is something called saturation, then we must draw saturation curve of the CT core to have more insight.
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For drawing CT saturation curve, we apply voltage in the CT secondary (keeping primary open) in step of 10% of rated voltage till 120% and read the secondary current using Clamp Meter. All the readings are noted in table and a curve is drawn between applied voltage V and excitation current Ie. The curve drawn will look like as shown in figure below.
Carefully observe the saturation curve shown above. It is quite clear that beyond point K, we need to increase current to a larger extent to have some increase in voltage. This because the curve beyond point K becomes non-linear. The voltage at point K i.e. Vk is called Knee Point Voltage. This is the reason, in definition it is said that Knee Point Voltage of a Current Transformer is defined as the voltage at which 10 % increase in voltage of CT secondary results in 50 % increase in secondary current. This means that an increase in 50% current will lead to just an increase in 10% voltage. Therefore slope at Knee Point Voltage will be, Slope = Increase in Voltage / Increase in current = 0.1/0.5 = 0.2 Knee Point Voltage of Current Transformer is of importance in Protection Class CT i.e. where CT is used for protection purpose. Protection Class CT is normally specified as PS (Protection Special). PS is defined by knee point voltage of current transformer Vk and excitation current Ie
at Vk/2. The Burden of CT when used for protection purpose is quite high when compared with Metering Class CT, which means that voltage drop across theburden will be high. But voltage drop across the burden is equal to the voltage across the CT secondary and if the voltage across the CT secondary is high then it may drive the CT to saturate in normal condition. Therefore Knee Point voltage of Protection Class CT must be more than the voltage drop across the burden to maintain CT core in its linear zone.
Why Harmonic Current in Transformer Excitation Current?. Harmonics in the excitation current of Transformer is due to Hysteresis. As we know the relationship between Magnetic Flux Density, B and Magnetic Field Intensity, H is not linear as shown in figure below.
Also, B = Flux (Ø) / Area (A), and H = NI where N = Number of turns and I = Magnetizing Current Therefore, there exist hysteresis relationship between Flux and Magnetizing Current. The Transformer is preferably operated in saturation region which in turn means that for considerable increment in current causes a slight increase in flux, which gives flux wave sinusoidal shape, while magnetizing current is peaky. This explains why it is rich in 3rd harmonic component even though the supply is sinusoidal. If we observe the waveform of excitation current of Transformer, we see that the wave form of current is symmetrical which means absence of even harmonics. Remember that waveform will not be symmetrical if there is any even harmonic component.
The peaks of the magnetizing current and flux will occur simultaneously, while their zeros will not, due to hysteresis. Thus I hope it is clear that how harmonic component comes in the excitation current of Transformer. Sympathetic Inrush Current in Transformer. I would suggest to read Transformer Inrush Current before reading this article for better understanding of Sympathetic Inrush Current. Any event on the power system that causes a significant increase in the magnetizing voltage of the transformer core results in magnetizing inrush current flowing into the transformer. The three most common events are as follows: Energization of the Transformer. This is the typical event where magnetizing inrush currents are a concern. The excitation voltage on one winding is increased from 0 to full voltage. The transformer core typically saturates, with the amount of saturation determined by transformer design, system impedance, the remnant flux in the core, and the point on the voltage wave when the transformer is energized. The current needed to supply this flux may be as much as 40 times the full load rating of the transformer, with typical value for power transformers for 2 to 6 times the full load rating. Figure below shows the waveform during energization of a transformer.
Magnetizing Inrush Current during Fault Clearing. An external fault may significantly reduce the system voltage, and therefore reduce the excitation voltage of the transformer. When this fault is cleared, the excitation voltage returns to the normal system voltage level. The return of voltage may force a dc offset on the flux linkages, resulting in magnetizing inrush current. This magnetizing inrush current will be less than that of energization, as there is no remnant flux in the core. The current measured by the differential relay will be fairly linear due to the presence of load current, and may result in low levels of second harmonic current. Sympathetic inrush current. Energizing a transformer on the power system can cause sympathetic inrush currents to flow in an already energized parallel transformer. Energizing the second transformer causes a voltage drop across the resistance of the source line feeding the transformers. This voltage drop may cause a saturation of the already energized transformer in the negative direction. This saturation causes magnetizing inrush current to supply the flux. The magnitude of the magnetizing inrush current is generally not as severe as the other cases.
While charging a Transformer in Power System, harmonic restraining of other connected Transformers must be taken care as it may otherwise lead to the tripping of other connected Transformers.
Transformer Inrush Current. In this post we will discuss the Magnetizing Inrush Current in a Power Transformer. Magnetizing Inrush urrent in Transformers results from any abrupt changes of the magnetizing voltage. This current in transformer may be caused by energizing an unloaded transformer, occurrence of an external fault, voltage recovery after clearing an external fault and out-of phase synchronizing of connected Generator.Because the amplitude of inrush current can be as high as a short circuit current, a detailed analysis of the magnetizing inrush current under various conditions is necessary for the making required setting of protective system for the Transformers. First question which will come up in your smart mind, why a Power Transformer takes Inrush Current when energized in unloaded condition? When a power transformer is energized while keeping its secondary circuit open, it acts as an inductance. In normal condition of a Power Transformer, the flux produced in the core is in
quadrature with applied voltage i.e. Flux lags behind the applied voltage by 90° as shown in the figure below.
This means, Flux wave will reach its maximum value after 1/4 cycle or π/2 angle reaching maximum value of voltage wave. Hence as per the waves shown in the figure, at the instant when, the voltage is zero; the corresponding steady state value of flux should be negative maximum. But practically it is not possible to have flux at the instant of switching on the supply of Transformer. This is because, there will be no flux linked to the core prior to switching on the supply. The steady state value of flux will only reach after some finite time which in turn depends upon how fast the circuit can take energy. So the flux in the core also will start from its zero value at the time of switching on the transformer. As we know that, e = dφ/dt where φ is the Flux in the core Therefore assuming e = ESinwt,
Now suppose, Transformer is switched on when Voltage is zero. Therefore Flux will also start from zero. Therefore, total Flux at the end of first half cycle of voltage wave will be,
Where Øm = Maximum flux in the core in steady state or normal operating condition.
Therefore, the flux in the core of Transformer will be double the maximum value of flux in steady state condition. This phenomenon is also shown in figure below.
It is clear from the above graph that maximum flux in the core of Transformer will be 2Ø mwhen the applied voltage is at its zero. Now what will happen because of this higher value of flux in the core of Transformer? Transformer core is saturated just above the maximum steady state value of flux Øm. But when we switch on power supply to the Transformer’s primary, the maximum value of flux will jump to double of its steady state maximum value Øm. As, after steady state maximum value of flux Øm, the Transformer core becomes saturated, the current required to produced rest (2Øm-Øm = Øm) of flux will be very high. So transformer primary will draw a very high current from the source which is called Magnetizing Inrush Current in Transformer or Inrush Current in Transformer. The nature of Transformer Inrush Current is shown in figure below.
It should be noted that waveform of Transformer Inrush Current is asymmetric which means in Transformer Inrush current mainly 2 nd harmonic component will be present. It shall also be noted from the waveform that as time passes the Magnetizing Inrush Current of Transformer decays and becomes zero. Normally it takes few millisecond for Magnetizing Inrush Current to decay to zero.
The Magnetizing Inrush Current of Transformer may be up to 10 times higher than normal rated current of Transformer. Even though the magnitude of Magnetizing Inrush Currentis so high but it generally does not create any permanent fault in Transformer as it exists for few miliseconds. But still Magnetizing Inrush Current in Power Transformer is a problem, because during the time of Magnetizing Inrush Current the protection scheme of Transformer may operate and hence may trip the Primary side Circuit Breaker of Transformer which is not expected.
How do we prevent Tripping of Transformer due to Magnetizing Inrush Current? In Transformer Differential Protection an intestinal time delay of 20 milisecond is provided to prevent tripping of Transformer due to high Magnetizing Inrush Current. In modern Numerical Relay, 2nd Harmonics blocking feature is provided which blocks the 2ndHarmonics when it is more than set value, thereby don’t issue trip command to Transformer Primary side Breaker due to Magnetizing Inrush Current. Normally the setting of 2nd Harmonics blocking is set to 20% which means if 2ndHarmonic component is more than 20% of fundamental frequency value of current then Relay will understand that it is because of Magnetizing Inrush Current and hence won’t issue trip command but if it is less than 20% fundamental frequency value of current then Relay will treat it due to fault and will issue trip command to the primary side Circuit Breaker.
Is there no way to minimize Magnetizing Inrush Current of Transformer? Point on Wave Switch is used to minimize the Magnetizing Inrush Current of Transformer. If you want to know more about Point on Wave Switch, please write in comment box. I will post on Point on Wave Switch.
BDV Test in Transformer. BDV test means Breakdown Voltage Test. It is done for checking the dielectric strength of the oil of the Transformer. Dielectric strength means the maximum capacity to withstand voltage of insulating oil. This test shows the dielectric strength of Transformer oil.
In transformer oil has mainly two purposes, first for insulation, second as cooling of Transformer core and other winding. So while designing Transformer oil use in transformer depends on voltage rating. So testing of oil is done according to voltage rating. For the purpose of BDV test, oil sample from Transformer is taken in a Sample Bottle. While taking sample of oil from transformer, Sample bottle should be flushed well by Transformer oil and oil in Sample bottle should be vented properly so that atmospheric moisture could not ingress in the sample oil. A typical way of taking oil sample in Sample Bottle is shown in figure below.
Breakdown Voltage is measured by observing at what voltage, sparking straits between two electrodes emerged in the oil, separated by specific gap. Low value of BDV indicates presence of moisture content and conducting substances in the oil. For measuring BDV of transformer oil, portable BDV measuring kit is generally available at site. In this kit, oil is kept in a pot in which one pair of electrodes are fixed with a gap of 2.5 mm (in some kit it 4 mm) between them.
Now slowly rising voltage is applied between the electrodes. Rate of rise of voltage is generally controlled at 2 KV/s and observe the voltage at which sparking starts between the electrodes. That means at which voltage dielectric strength of transformer oil between the electrodes has been broken down. A typical value of BDV Test result for 220 / 6.6 kV Transformer oil is 65 kV and moisture content should be less than 10 ppm. SFRA Test of Transformer. SFRA stands for Sweep Frequency Response Analysis. This test is very reliable for condition monitoring of physical condition of transformer winding. Why to conduct SFRA test? The winding of Transformer may be subjected to mechanical stresses during transportation, heavy short circuit faults, transient switching impulses & lightening impulses etc. These mechanical stresses may cause displacement of transformer winding from their position and may also cause deformation of these winding. SFRA Test can efficiently detect: Displacement of Transformer core Deformation & displacement of winding Faulty core grounds Collapse of partial winding Broken or loosen clamp connections Short circuited turns Open winding conditions.
Principle of SFRA Test: As each of the electrical equipment is combination of R, L & C. In Transformer each winding turn is separated from other by paper insulation which acts as dielectric and windings themselves have inductance and resistance, a transformer can be considered as a complicated distributed network of resistance, inductance, and capacitance or in other words a Transformer is a complicated RLC circuit as shown in figure below. Because of this each winding of a transformer exhibits a particular frequency response.
In Sweep Frequency Response Analysis a sinusoidal voltage Vi is applied to one end of a winding and output voltage Vo is measured at the other end of the winding while keeping the other windings open. As the winding is itself a distributed RLC circuit it will behave like RLC filter and gives different output voltages at different frequencies. That means if we go on increasing the frequency of the input signal without changing its voltage level we will get different output voltages at different frequencies depending upon the RLC nature of the winding. If we plot these output voltages against the corresponding frequencies we will get a particular pattern for a particular winding as shown in figure below.
But after transportation, heavy short circuit faults, transient switching impulses and lightening impulses etc, if we do same Sweep Frequency Response Analysis test and superimpose the present signature with the earlier pattern and observe some deviation between the two graphs / signature. Thus we can conclude that there is mechanical displacement / deformation in the Winding / Core. Thus using SFRA test, we can say whether Transformer windings / core is OK or not. This method is simple yet reliable. Buchholz Relay- Transformer Protection. Buchholz relay is a gas actuated protection relay which is generally used in large oil immersed transformers of rating more than 500 kVA. It is used for the protection of a Transformer from the faults occurring inside the transformer. A typical Buchhloz Relay is shown below.
The gas actuated protective relay is designed to detect faults as well as to minimize the propagation of any damage, which might occur within oil-filled Transformers. The Buchholz relay is therefore particularly effective in case of:
Short-circuited core laminations Broken-down of core bolt insulation Overheating of some part of the windings Bad contacts Short circuits between phases, turns Earth faults-puncture of bushing insulators inside tank
Buchholz relay `can prevent the development of conditions leading to a fault in the transformer, such as the falling of the oil level due to leaks, or the penetration of air as a result of defects in the oil circulating system. The adoption of other forms of protection does not therefore exclude the use of the gasactuated Buchholz relay, as this device is the only means of detecting incipient faults, which if unnoticed, can cause heavy failures. Buchholz Relay is installed in between the Main Tank of Transformer and the Conservator as shown if figure below.
Construction of Buchholz Relay: Buchholz relay consists of an oil filled chamber. There are two hinged floats, one at the top and other at the bottom in the chamber. Each float is connected by a mercury switch. The mercury switch on the upper float is connected to an alarm circuit and that on the lower float is connected to an external circuit to cause breaker trip. A simplified construction diagram of Buchholz relay is shown in figure below.
Working Principle of Buchholz Relay: The operation of the Buchholz relay is based upon the fact that every kind of fault in an oil-filled transformer causes a decomposition of the insulating oil due to overheating in the fault zone or to the action of an intense electric field, and a generation of bubble of gas. These reach the relay which is normally filled with oil, through the pipe connecting the transformer to the conservator where the Buchholz relay is mounted. Whenever a minor fault occurs inside the transformer, heat is produced by the fault currents. The produced heat causes decomposition of transformer oil and gas bubbles are produced. These gas bubbles flow in upward direction and get collected in the Buchholz relay. The collected gas displaces the oil in Buchholz relay and the displacement is equivalent to the volume of gas collected. The displacement of oil causes the upper float to close the upper mercury switch which is connected to an alarm circuit. Hence, when minor fault occurs, the connected alarm gets activated. The collected amount of gas indicates the severity of the fault occurred. During minor faults the production of gas is not enough to move the lower float. Hence, during minor faults, the lower float is unaffected. During major faults, like phase to earth short circuit, the heat generated is high and a large amount of gas is produced. This large amount of gas will similarly flow upwards, but its motion is high enough to tilt the lower float in the Buccholz relay. In this case, the
lower float will cause the lower mercury switch which will trip the transformer from the supply i.e. transformer is isolated from the supply. Class-A, Class-B and Class-C Tripping Classification of Generator. Generator, Generator Transformer and Unit Transformer protections have been classified into Class-A, Class-B and Class-C. Class-A tripping is further classified into Class-A1 and ClassA2. Class-A, Class-B and Class-C Tripping Classification of Generator is based on the need of isolation of Generator on the basis of type of fault. In this post we will discuss each type of tripping classes and their significance.
Basis of Tripping Classification: The tipping classification of Generator is based on the need of isolation of Generator on the basis of type of fault. For example, there are some faults like Generator Differential Protection which calls for immediate tripping of Generator Breaker without delay whereas there are some fault like Loss of Excitation, Rotor Earth Fault etc. which do not call for immediate tripping of Generator. Class-A1 Trip: The protections for the faults in the Generator which need immediate isolation are grouped under this Class-A1. There are a list of faults which are kept under this class. They are as follows: a) Generator Differential Protection b) 100% Stator Earth Fault Protection c) Generator Over Voltage Protection d) Dead Machine Protection e) 95% Stator Earth Fault Protection f) Starting Over Current Protection In case of actuation of Class-A1 protection, Generator Circuit Breaker and Filed Circuit Breaker are opened along with turbine tripping. Class-A2 Trip: The protections for the faults in Generator Transformer (GT), Isolated Phase Bus Duct (IPBD), and Unit Transformer (UT) which need immediate isolation are grouped under this Class-A2. Normally following protections are kept under Class-A2: a) Over fluxing Protection of Generator b) Back up Impedance Protection of Generator
c) Differential Protection of GT d) Buchcholz Relay of GT e) PRD of GT f) Trip from OTI & WTI of GT g) Fire protection of GT h) Differential Protection of UT i) Buchcholz Relay & PRD of Main Tank of UT j) Trip from OTI & WTI of UT k) Fire protection of UT These protection when operated initiate tripping of Generator Circuit Breaker, Field Circuit Breaker, Generator Transformer Circuit Breakers & Unit Transformer LV Circuit Breakers and turbine. Class-B Trip: The protections for the faults in the Generator which do not need immediate isolation are grouped under this Class-B. The turbine is tripped first and Generator is allowed to run utilizing trapped steam in turbine. Let us suppose that there is some fault in the process side i.e. in steam cycle, under that condition also turbine will be tripped first while Generator will continue to run utilizing trapped steam till reverse power relay operates. Generator Circuit Breaker is tripped on initiation of reverse power. Normally, Loss of Excitation and Rotor Earth Fault of Generator are kept under this class. These protection when operated initiate tripping of Generator Circuit Breaker, Field Circuit Breaker and turbine. Class-C Trip: The protections for the faults / abnormal condition in the Grid which call for disconnection of the Generator from the Grid are grouped under this Class-C. In this case, Generator is isolated from the Grid by opening the suitable breaker i.e. Generator Transformer HV side Breaker. Mind that in this case only Generator is isolated from the Grid. Thus Generator continues to feed Station loads (also known as house load). Such scheme where generator is operated on house load at reduced power is known as Generator Islanding. Normally following protections of Generator are kept under this class: a) Unbalance or Negative Sequence Protection b) Back up Impedance Protection c) Under Frequency d) Over Frequency e) Pole Slipping Protection. Difference between Power Transformer and Distribution Transformer. The main and basic difference between a Power transformer and a Distribution Transformer is that Power transformers are used in transmission network of higher
voltages for step-up and step down application like 400 kV, 200 kV, 110 kV, 66 kV, 33kV and are generally rated above 200 MVA.
Distribution transformers are used for lower voltage distribution networks as a mean for end user connectivity. 11 kV, 6.6 kV, 3.3 kV, 440 V, 230 V and are generally rated less than 200 MVA.
On the basis of Transformer Size / Insulation Level, Power transformer is used for the transmission purpose at heavy load, high voltage greater than 33 KV and 100% efficiency. It also having a big in size as compare to distribution transformer, it used in generating station and Transmission substation at high insulation level. The distribution transformer is used for the distribution of electrical energy at low voltage as less than 33 kV in industrial purpose and 440 – 220 V in domestic purpose. It work at
low efficiency at 50-70%, small size, easy in installation, having low magnetic losses & it is not always fully loaded. On the basis of Iron Losses and Copper Losses, Power Transformers are used in Transmission network so they do not directly connect to the consumers, so load fluctuations are very less. These are loaded fully during 24 hours a day, so copper losses and Iron losses takes place throughout day. The average loads are nearer to full loaded or full load and these are designed in such a way that maximum efficiency occur at full load condition. Distribution Transformers are used in Distribution Network so directly connected to the consumer so load fluctuations are very high. Distribution Transformers are not loaded fully at all time so iron losses takes place 24 hr a day and copper losses takes place based on load cycle. The specific weight i.e. (iron weight)/(cu weight) is more. Average loads are about only 75% of full load and these are designed in such a way that maximum efficiency occurs at 75% of full load. As these are time dependent therefore All Day Efficiency is defined in order to calculate the efficiency. Power Transformers are used for transmission as a step up devices so that the I2rloss can be minimized for a given power flow. These transformers are designed to utilize the core to maximum and will operate very much near to the knee point of B-H curve. This brings down the mass of the core enormously. Naturally these transformers have the matched iron losses and copper losses at peak load i.e. the maximum efficiency point where both the losses match. Distribution transformers obviously cannot be designed like this. Hence the All-DayEfficiency comes into picture while designing it. It depends on the typical load cycle for which it has to supply. Definitely Core design will be done to take care of peak load and as well as all-day-efficiency. It is a bargain between these two points. Power transformer generally operated at full load. Hence, it is designed such that copper losses are minimal. However, a distribution transformer is always online and operated at loads less than full load for most of time. Hence, it is designed such that core losses are minimal.
In Power Transformer the flux density is higher than the distribution transformer. On the basis of Maximum Efficiency, The main difference between Power and Distribution Transformer is that Distribution Transformer is designed for maximum efficiency at 60% to 70% load as normally doesn’t operate at full load all the time. Its load depends on distribution demand. Whereas power transformer is designed for maximum efficiency at 100% load as it always runs at 100% load being near to generating station. Distribution Transformer is used at the distribution level where voltages tend to be lower .The secondary voltage is almost always the voltage delivered to the end consumer. Because of voltage drop limitations, it is usually not possible to deliver that secondary voltage over great distances. As a result, most distribution systems tend to involve many clusters of loads fed from distribution transformers, and this in turn means that the thermal rating of distribution transformers doesn’t have to be very high to support the loads that they have to serve. All Day Efficiency = (Output in KWhr) / (Input in KWhr) in 24 hrs which is always less than power efficiency. Dry Type Transformers versus Oil Filled Transformers. Before going to the mark difference between Dry Type Transformer and Oil Filled Transformer, it is worth to have some discussion on Dry Type Transformer.
Dry Type Transformer: Dry Type Transformers find use in locations where the use oil Filled Transformers increases the fire hazard such as shopping malls, Hospitals, residential complexes etc. In dry type transformers air is used as the cooling medium instead of oil.
The insulation used in dry type transformers are designed to withstand higher temperatures. Dry type transformers are more expensive than conventional transformers. Vacuum Pressure Impregnation, Epoxy Resin cast are some of the methods of Insulation adopted in Dry Type Transformer construction.
Comparison between Dry Type and Oil Filled Transformer:
Dry Type Transformers use air as the cooling medium. Oil Type Transformers are considered a potential fire and safety hazard for indoor application.
Dry Type Transformers can be located closer to the load unlike Oil Transformers which require special location and civil construction for safety reasons. Locating the Transformers near the loads may lead to savings in cable costs and reduced electrical losses.
Oil Type transformers may require periodic sampling of the oil and more exhaustive maintenance procedures.
Though Dry Type Transformers are advantageous, they are limited by size and voltage rating. Higher MVA ratings and voltage ratings may require the use of oil Transformers.
For outdoor applications, Oil Filled Transformers are cheaper than dry types.
Losses in Electrical Machine – Core Loss and Eddy Current Loss. There are two types of Losses in an Electrical Machine. They are
Core Loss or Iron Loss Ohmic Loss or Copper Loss
In this post we will discuss about Core Loss. Core Loss is again classified into two types:
Hysteresis Loss Eddy Current Loss
First we will have a look at how the core of a Transformer looks like. But the Core Loss take place in any electrical machine which face changing magnetic flux.
Hysteresis Loss: This loss is due to magnetic properties of iron part or core. When the magnetic field strength or the current is increased the flux increase, after a point when we further increase current the flux gets saturated. When we reduce the current from saturation to zero side the flux density starts to decrease. But when the current value reaches zero the flux density should also be zero but it is not zero. For zero current there is still some flux present in the material, this is known as Residual
Magnetic Flux or Remnant Magnetic Flux. Hence the amount of power is never recovered back. The power which gets trapped in the core of the material is lost in the form of heat.
Now we will consider the mathematical part of Hysteresis Loss. The Hysteris Loss in core is given as
Ph= KhfBmx Where Kh = Constant which depends on the volume and quality of core material. Bm = Maximum flux density in the core f = Frequency of Supply x = Steinmetz’s constant whose value varies from 1.5 to 2.5. Thus we see that Core Loss depend on Voltage as well as Frequency of Supply.
Eddy Current Loss: Eddy Current Loss takes place when a coil is wrapped around a core and alternating ac supply is applied to it. As the supply to the coil is alternating, the flux produced in the coil is also alternating.
By faradays law of electromagnetic induction, the change in flux through the core causes emf induction inside the core. Due to induction of emf eddy current starts to flow in the core. Due to this eddy current there will be an associated Ohmic loss which is called Eddy Current Loss.
Eddy current losses can be reduced by lamination in the core. Thin sheet steels must be used which are insulated from each other. Due to insulated sheets the amount of current which flows get reduced and hence the eddy current losses also reduces. Now we will take a look at the mathematical part of Eddy Current Loss. Eddy Current Loss is given as
Pe= Kef2Bm2 Where Ke = constant whose value depends on the volume and resistivity of the core material. Bm = Maximum flux density in the core f = Frequency of Supply It shall be noted that, from the equation of Eddy Current Loss it seems that Eddy Current Loss depends on the frequency of supply but it is not so rather it only depends on the Supply Voltage. How? As Pe = Kef2Bm2
But we know that
So,
Bm2f2= KE2 where K is constant Thus Eddy Current Loss Pe = KV2 Therefore, Eddy Current Loss only depends on the applied Voltage. Over Current Relay and Its Characteristics. Protection schemes can be divided into two major groupings: a) Unit schemes b) Non-unit schemes a) Unit Scheme: Unit type schemes protect a specific area of the system i.e. a transformer, transmission line, generator or bus bar. The unit protection schemes are based on Kirchhoff’s Current Law – the sum of the currents entering an area of the system must be zero. Any deviation from this must indicate an abnormal current path. In these schemes, the effects of any disturbance or operating condition outside the area of interest are totally ignored and the protection must be designed to be stable above the maximum possible fault current that could flow through the protected area. b) Non-unit scheme: The non-unit schemes, while also intended to protect specific areas, have no fixed boundaries. As well as protecting their own designated areas, the protective zones can overlap into other areas. While this can be very beneficial for backup purposes, there can be a tendency for too great an area to be isolated if a fault is detected by different non unit schemes.
The most simple of these schemes measures current and incorporates an inverse time characteristic into the protection operation to allow protection nearer to the fault to operate first. The non unit type protection system includes following schemes: Time graded over-current protection Current graded over-current protection Distance or Impedance Protection Over Current Protection: It finds its application from the fact that in the event of fault the current will increase to a value several times greater than maximum load current. A relay that operates or picks up when its current exceeds a predetermined value (setting value) is called Over-current Relay. Over-current protection protects electrical power systems against excessive currents which are caused by short circuits, ground faults, etc. Over-current relays can be used to protect practically any power system elements, i.e. transmission lines, transformers, generators, or motors. For feeder protection, there would be more than one over-current relay to protect different sections of the feeder. These over-current relays need to coordinate with each other such that the relay nearest fault operates first. Use time, current and a combination of both time and current are three ways to discriminate adjacent over-current relays. Over-current Relay gives protection against: Phase faults Earth faults Winding faults Short-circuit currents are generally several times (5 to 20) full load current. Hence fast fault clearance is always desirable on short circuits. Primary requirement of Over-current protection is that the protection should not operate for starting currents, permissible over-current, and current surges. To achieve this, the time delay is provided. Over-current Relay Ratings: In order for an over-current protective device to operate properly, over-current protective device ratings must be properly selected. These ratings include voltage, ampere and interrupting rating. Current limiting can be considered as another over-current protective device rating, although not all over-current protective devices are required to have this characteristic Voltage Rating: The voltage rating of the over-current protective device must be at least equal to or greater than the circuit voltage. The over-current protective device rating can be higher than the system voltage but never lower. Ampere Rating: The ampere rating of a over-current protecting device normally should not exceed the current carrying capacity of the conductors As a general rule, the
ampere rating of a over-current protecting device is selected at 125% of the continuous load current. Depending on the time of operation of relays, they are categorized as follows: a) Instantaneous Over-current Relay b) Inverse time over current Relay c) Inverse definite minimum time (IDMT) over-current Relay d) Very Inverse Relay e) Extremely Inverse Relay a) Instantaneous Over-current Relay: Instantaneous Over-current Relay is one in which no intentional time delay is provided for the operation. The time of operation of such Relay is approximately 100 ms. Instantaneous Over-current relay is employed where the impedance between the source and the Relay is small as compared with the impedance of the section to be provided.
Following are the important features of an Instantaneous Over-current Relay: 1) Operates in a definite time when current exceeds its Pick-up value. 2) Its operation criterion is only current magnitude. 3) Operating time is constant. 4) There is no intentional time delay. 5) Coordination of definite-current relays is based on the fact that the fault current varies with the position of the fault because of the difference in the impedance between the fault and the source 6) The relay located furthest from the source operate for a low current value 7) The operating currents are progressively increased for the other relays when moving towards the source. b) Inverse time over current Relay: Inverse time over-current Relay is one in which the time of actuation of Relay decreases as the fault current increases. The more the fault current the lesser will be the time of operation of the Relay. Normally it has more inverse characteristic near the pick-up value which in turn means that if fault current is equal to pick-up value then the relay will take infinite time to operate.
c) Inverse definite minimum time (IDMT) over-current Relay: Inverse definite minimum time (IDMT) over-current Relay is one in which the operating time is approximately inversely proportional to the fault current near pick-up value and then becomes constant above the pick-up value of the relay.
From the picture, it is clear that there is some definite time after which the Relay will operate. It is also clear that the time of operation at Pick-up value is nearly very high and as the fault current increases the time of operation decreases maintaining some definite time. d) Very Inverse Relay: Very Inverse Relay is one in which the range of operation is inverse with respect to fault current over a wide range. This happens so as the CT saturation occurs at a later stage but as soon as CT saturation occur there will not be any flux change and hence the current output of CT will become zero and hence the time of operation will nearly become constant. e) Extremely Inverse Relay: Extremely Inverse Relay is one in which CT saturation occur still at a later stage as compared with Very Inverse Relay and hence the characteristic remain inverse up to a larger range of fault current. The equation describing the Extremely Inverse Relay is I2t = K where I is operating current and t is time of operation of the Relay.
IEC (International Electrotechnical Commission) Standard Curve for Inverse Relays:
As per IEC, the time of operation of any Inverse relay can be calculated from the formula given below. Here,
K = Time of actuation α, β = Constant which depends on the type of Relay I = Fault Current I0 = Pick-up current Value of α and β for different types of Relay: Sr. No. Type of Relay α
β
1)
Inverse time over current Relay 0.02 / IDMT
0.14
2)
Very Inverse Relay
1.00
13.5
3)
Extremely Inverse Relay
2.00
80.00
Example: Suppose the pick-up current for an IDMT relay is set at 0.8 A and the fault current is 80 A then the time of actuation can be calculated as K = 0.14/[ (80/0.8)0.02– 1]. Unit and Non-Unit Protection Scheme. Protection schemes can be divided into two major groupings:
a) Unit schemes
b) Non-unit schemes a) Unit Protection Scheme: Unit type schemes protect a specific area of the system i.e. a transformer, transmission line, generator or bus bar. The unit protection schemes are based on Kirchhoff’s Current Law – the sum of the currents entering an area of the system must be zero. Any deviation from this must indicate an abnormal current path. In these schemes, the effects of any disturbance or operating condition outside the area of interest are totally ignored and the protection must be designed to be stable above the maximum possible fault current that could flow through the protected area.
In other words, it is possible to design protection systems that respond only to fault conditions occurring within a clearly defined zone. This type of protection system is known as unit protection. Certain types of unit protection are known by specific names, e.g.Restricted Earth Fault and Differential Protection. Unit protection can be applied throughout a power system and, since it does not involve time grading, is relatively fast in operation. The speed of response is substantially independent of fault severity.
Unit protection usually involves comparison of quantities at the boundaries of the protected zone as defined by the locations of the current transformers. This comparison may be achieved by direct hard-wired connections or may be achieved via a communications link. However certain protection systems derive their restricted property from the configuration of the power system and may be classed as unit protection, e.g. Earth Fault Protection applied to the high voltage delta winding of a power transformer. Whichever method is used, it must be kept in mind that selectivity is not merely a matter of relay design. It also depends on the correct coordination of current transformers and relays with a suitable choice of relaysettings, taking into account the possible range of such variables as fault currents, maximum load current, system impedances and other related factors, where appropriate.
b) Non-unit Protection Scheme: The non-unit schemes, while also intended to protect specific areas, have no fixed boundaries. As well as protecting their own designated areas, the protective zones can overlap into other areas. While this can be very beneficial for backup purposes, there can be a tendency for too great an area to be isolated if a fault is detected by different non unit schemes.
The most simple of these schemes measures current and incorporates an inverse time characteristic into the protection operation to allow protection nearer to the fault to operate first.
The non unit type protection system includes following schemes:
a) Time graded over-current protection
b) Current graded over-current protection
c) Distance or Impedance Protection. Programmable Scheme Logic (PSL) in Numerical Relays. Programmable Scheme Logic or PSL is a kind of feature provided in Numerical Relays to implement the protection scheme of a particular type. This feature of Numerical Relays makes it easier to implement many protection schemes in a single Numerical Relay for example, in Distance Relay we can configure Distance protection, over voltage protection, Over Current Protection, Earth Fault Protection etc. Now we will study about Programmable Scheme Logic, PSL. PSL is a logical block which is made from different but suitable DDB. Here DDB stand for Digital Data Bus. There are many DDBs offered in a Numerical Relay. Each DDB perform a unique function. Thus it is very important to have the knowledge of function of DDBs to implement a particular logic. Hope you got some idea of DDB but don’t worry I will go in detail with example to make it crystal clear. Lets us begin with an example. Let us assume that we have an Alstom Relay P442 and we want to implement a protection feature called Local Breaker Back-up (LBB) Protection in the Relay. So we need to finalize our logic for the operation of LBB. The generalized logic for LBB protection is Lock-out Relay Operated AND Current still existing Under the above logic the LBB Relay shall initiate its timer and shall give the tripping command to isolate the fault after a fixed time delay say 200 ms. Assuming the above logic for LBB, we will design a logic using Programmable Scheme Logic, PSL in the Numerical Relay. But before designing the PSL, we need to give the input to the Relay, in our case there are two inputs, one will be the contact of Lock-out Relay and another Current Transformer (CT) input. So our first step will be to assign the inputs to the Relay and label a name to each input. In the second step, we need to configure the output of Relay and label a name to the Relay Output contact. Let, Lock-out Relay contact Input is labeled as INPUT1. Mind that only digital inputs can be labeled. So we can not assign a name to CT input. Likewise let the Relay output contact to trip Breakers to isolate fault be RL1. Thus as per our logic when INPUT1 is high and over current still exists then RL1 shall get high after 200 ms to isolate the fault.
Carefully observe the figure above. We have used digital inputs and Over-current DDB I>1 (1stStage Over-Current) in an AND gate to start the timer and if the input status do not change for 200 ms then RL1 will change its status from low to high but in the time window of 200 ms, if the input status changes then the timer will reset and RL1 will remain low. Therefore, making a Programmable Scheme Logic, PSL in a Numerical Relay is just a logic building while having knowledge of function of each DDB to be used. This is just a brief of PSL to have some idea of PSL used in Numerical Relay. Hope you enjoyed this post.
Pick-up Current, Plug Setting Multiplier (PSM) and Time Setting Multiplier (TSM). Plug Setting Multiplier (PSM) and Time Setting Multiplier (TSM) are used only for Electromechanical Relays. These terms or parameters are not so used in Numerical Relays but they are conceptually used and incorporated in Numerical Relays too but the way of their implementation is quite different than that of Electromechanical Relays. In this post we will focus on the concept and implementation of Plug Setting Multiplier and Time Setting Multiplier for Electromechanical Relays. As we know that an Electromechanical Relay has a coil which when energized, operates the Relay to have contact changeover. But there shall be some minimum current which when flows through the Relay coil, produces enough magnetic force to pull the lever to make contact change over. Isn’t it? Yes, if you ever get a chance to see electromechanical relay, you will observe that there is a flapper kind of thing which is attached with the lever. The lever in turn is attached with contacts. Thus when a specified current flows through the relay coil, then only it will produce enough magnetic pull to attract the flapper and lever to operate the Relay. A simple picture of relay demonstrating its construction and operation is shown in figure below.
This minimum current in the Relay coil at which Relay starts to operate is called Pick-up Current. If the current through the Relay coil is less than the pick-up value then Relay won’t operate. On contrary, if the current through the Relay coil is more than the Pick-up current, Relay will operate. In industries, we normally perform Relay Pick-up and Drop-off Test to check the healthiness of relays. Hope your concept of Pick-up current of Relay is clear now. Now we will move on to Current Setting of electromechanical relays. Current Setting of Electromechanical Relays: Current Setting of relay is nothing but adjusting its pick-up value. Suppose we are using a CT of ratio 1000/1 A and the pick-up current needs to be set at 1.2 A. Then we will simply put the plug provided on relay coil to 120% or 1.2. Thus we can say that Pick-up current = Plug Position x Rated CT Secondary Current. The plug or tapping is provided on the Relay Coil so that changing the position of Plug changes the number of turns of the relay coil as shown in figure below.
As shown in figure above, the plug is kept at 5. This means that pick-up current of relay will be 5 times of rated CT Secondary current. Likewise, if we put the plug at 8.75 then pick-up current of relay will be 8.75 times of the rated CT Secondary current. Plug Setting Multiplier (PSM): Plug Setting Multiplier (PSM) is defined as the ratio of fault current to the pick-up current of the relay. Thus, Plug Setting Multiplier (PSM) = Fault Current / Pick-up Current = Fault Current / (Plug Position x Rated CT Secondary Current)
Suppose we are using CT of 100/5 A, a fault current of, say 250 A is flowing through the network protected by the relay. Assume that Current Setting or the position of plug is at 5 then Plug Setting Multiplier (PSM) = 250 / (5×5) =10 It shall be noted here that we shall not bother about PSM for instantaneous relay rather we shall consider PSM for relays having characteristics of Inverse Time, Very Inverse Time etc. For Detail on Relay Characteristics read Over Current Relay and Its Characteristics Time Setting Multiplier (TSM): Again it is worthwhile to mention that we shall not bother about TSM for instantaneous relay rather we shall consider TSM for relays having characteristics of Inverse Time, Very InverseTime etc. A Relay is generally provided with control to adjust the time of operation of the Relay. This adjustment is known as Time Setting Multiplier or TSM. Normally a Time Setting Dial is provided which is calibrated from 0 to 1 s in step of 0.05 s. For practical exposure, let us consider a relay as shown in figure below. Please Zoom the image to clearly view every part of the Relay for better understanding.
As can be seen from the figure, there is a Time Setting Dial which is rotated to set the time of operation of the relay. Lets say we want to set the time on Time Setting Dial to 0.5 s, then we need to rotate the dial till 0.5 s on the dial matches with the fixed mark provided. So our TSM is 0.5 here in the case. How to find the time of operation of Relay?
Well, assume that plug is set at 5 and TSM at 0.5 s. For finding the actual time of operation of relay we need to refer the Graph between the Operating Time and PSM which is generally provided on the Relay cover itself but in our figure it is not given. So we consider a graph between Operating Time and PSM as shown below.
For our case, PSM = 10 (Please see the calculation and case considered above in our discussion of PSM) and TSM = 0.5 s. From the Graph, the time of operation of Relay for PSM = 10 is 3 s. Therefore, Actual Time of Operation of Relay = 3s x TSM = 3 x0.5 s =1.5 s Thus we can say that actual time of operation of Relay is equal to the time obtained from the PSM & Operating Time Graph multiplied by TSM.
Why Zone-1 Setting 80% and Zone-2 150% in Distance Protection?. As discussed in my earlier post on Distance Protection Philosophy, Zone-1 setting is kept at 80% of the total impedance of the Protected Line. So question will come in your smart brain that why 80% even though Zone-1 is meant for primary protection?
Zone-1 is meant for protection of the primary line. Typically, it is set to cover 80% of the line length. Zone-1 provides fastest protection because there is no intentional time delay associated with it. Operating time of Zone-1 can be of the order of 1 cycle.
Zone 1 does not cover the entire length of the primary line because it is difficult to distinguish between faults which are close to bus B like fault at F1, F2, F3 and F4. In other words, if a fault is close to bus, one cannot ascertain if it is on the primary line, bus or on back up line. This is because of the following reasons:
CTs and PTs have limited accuracy. During fault, a CT may partially or complete saturate. The resulting errors in measurement of impedance seen by relay, makes it difficult to determine fault location at the boundary of lines very accurately. There are infeed and outfeed effects associated with working of distance relays. A distance relay scheme uses only local voltage and current measurements for a bus and transmission line. Hence, it cannot model infeed or outfeed properly. Because of infeed and outfeed effect the, the Relay may sense fault in 100% length of line even though the location of fault is actually not in 100% of line. Therefore a margin of 20% is given for the accuracy of measurement and infeed / outfeed effect. Next, Question which will definitely strike you that why do we keep the setting of Zone-2 150%? Why not more than 150%?
Zone-2 setting in Distance Relay is kept at 150 % to avoid Overlap Problem. See the picture below.
As clear from the picture above, if the reach of Zone-2 of a relay R1 is extended too much, then it can overlap with the Zone-2 of the relay R3. Under such a situation, there exists following conflict. If the fault is on line BC (and in Z2 of R3), relay R3 should get the first opportunity to clear the fault. Unfortunately, now both R1 and R3 compete to clear the fault. This means that Z2 of the relay R1 has to be further slowed down. As Zone-2 protection already have a time delay, due to overlapping we need to further introduce some time delay which will degrade the performance of Relay for Zone-2. Hence, a conscious effort is made to avoid overlaps of Z2 of relay R1 and R3. Setting Zone-2 of R1 to maximum of 150% of primary line impedance or primary line impedance plus 50% of smallest line impedance usually works out good compromise without getting into Z2 overlap problem.
Zones of Protection and Dead or Blind Zone in Power System. Zoning in Power system Protection is an important philosophy and must be done carefully so that no part of the system remains unprotected in any condition. To limit the extent of the power system that is disconnected when a fault occurs, protection is arranged in zones. The principle is shown in figure below.
Ideally, the zones of protection should overlap, so that no part of the power system is left unprotected. This is shown in figure below. As can be seen from the figure below, each Breaker is included in two different zones of protection to increase the reliability of protection scheme.
For practical physical and economic reasons, this ideal zoning in protection is not always possible to achieve because accommodation for current transformers being in some cases available only on one side of the circuit breakers, as shown in figure below.
This leaves a section between the Current Transformers and Circuit Breaker CB-A that is not completely protected against faults. As shown in figure above, a fault at F would cause the busbar protection to operate and open the circuit breaker but the fault may continue to be fed through the feeder. The feeder protection, if of the unit type, would not operate, since the fault is outside its zone. This problem is dealt with by intertripping or some form of zone extension, to ensure that the remote end of the feeder is tripped also.
The section of Power System which is not covered under any zone of protection is called Dead Zone or Blind Zone and special kind of protection shall be provided to take care of fault in Dead Zone. Normally overcurrent element is used for the protection of Dead Zone with some suitable logic interlock. The logic interlock depends on the configuration of power system and the condition in which Dead Zone is created. Let us take an example to have more insight. Carefully observe the figure below.
Transformer is fed by the Bus when the Breaker CB-A is close. Now suppose we want to take Transformer under maintenance, so for isolating the Transformer we will open CB-A and DS. After opening DS, it may be required to close the CB-A to feed some other connected feeder. As the Breaker is closed, a portion up to DS is charged. Now suppose a fault take place in between DS and CT-3. Assuming that CT-1, CT-2 and CT-3 are meant for protection of Zone in between the CTs using Differential protection, so a fault outside this zone will not be protected and hence zone in between CT-3 and DS is unprotected and called Dead Zone.
Read: Difference between Isolator and Breaker. Over Current Relay and Its Characteristics What will be the logic for implementing protection of this Dead Zone?
One may say, if the DS is open and CT-3 senses an overcurrent then Relay shall issue a tripping command to CB-A. That is all, Dead Zone is no more Dead rather it is protected. Notice that in this case Dead Zone or Blind Zone is created in a particular condition where the DS is open and CB-A is close.
Open Circuit and Short Circuit Characteristics of Synchronous Machine. Open Circuit Test and Short Circuit Test are performed on a Synchronous Machine to find out the parameters of Synchronous Machine and hence to have an idea of their performance. Open Circuit Test of Synchronous Machine is also called No Load, Saturation or Magnetizing Characteristics for the reason which will be clear after going through the post. For getting the Open Circuit Characteristics of Synchronous Machine, the alternator is first driven at its rated speed and the open terminal voltage i.e. voltage across the armature terminal is noted by varying the field current. Thus Open Circuit Characteristic or OCC is basically the plot between the armature terminal voltage Ef versus field current If while keeping the speed of rotor at rated value. It shall be noted that for OCC, the final value of Efshall be 125% of the rated voltage. Figure below shows the connection diagram for performing the Open Circuit Test of Alternator.
As clear from the figure above, an Ammeter is connected in series with the field circuit to measure the field current and a Voltmeter is connected across the armature terminals to note down the voltage generated. Figure (b) shows the plot between If and Ef. It can be seen from the graph that the relationship between the field current I fand no load generated voltage Ef is linear up to certain value of field current but as the the field current increases the relationship no
longer remains linear. The linear part of the relationship is because, at small value of filed current the whole mmf is required by the air gap to create magnetic flux but as the value of mmf exceeds some certain value, the iron parts get saturated and hence the relationship between the flux (No load generated emf is proportional to flux) and field current no longer remain linear. Next assume that if there were no saturation (assuming no iron part is present rather only air gap is present), the relationship between the field current and no load voltage would have been a straight line and that is why the straight line ob in the figure is called Air Gap Line. Thus we observe that because of saturation in iron parts of machine, the no load generated voltage Efdoes not increase in the same proportion as the increase in field current. Short Circuit Test of Synchronous Machine: For performing Short Circuit Test on an Alternator, the machine is driven at rated synchronous speed and the armature terminals are short circuited through an Ammeter as shown in figure below.
Now the field current If is gradually increased from zero until the armature short circuit current reaches its maximum safe value i.e. 125 to 150% of its rated current value. Readings of field current If and short circuit current are noted and plotted.
If you see the above plot of Short Circuit Test, you notice that the short circuit characteristics of a synchronous machine is a straight line. Why Short Circuit Characteristics of Synchronous Machine is Straight Line? For short circuit test, as the armature terminals are shorted, therefore terminal voltage Vt = 0. Therefore the air gap emf Er shall only be enough to provide the leakage impedance drop in the armature i.e. Er = Ia(Ra + jXal) where Xal = Armature Leakage Reactance As we know that, for a Synchronous machine the value of Xal is of the order of 0.1 to 0.2 per unit and Ra (Armature Resistance) is negligible thus we can write as Xal = 0.15 (Taking average value of 0.1 and 0.2) Ra = 0 then Er = Ia (Ra +jXal) = 0.15Ia Taking rated current of armature, Ia = 1 pu
Therefore, Er = 0.15 pu You must read Per Unit System in Electrical Engineering
Thus we observe that during short circuit test, the air gap generated emf Er is only 0.15 pu which mean that air gap flux must also be 0.15 pu. As the resultant air gap flux is only 0.15 of its rated value under normal voltage condition, such a low value of air gap flux does not saturate the iron parts of synchronous machine and hence the short circuit characteristics is a straight line. It shall also be noted here that, in case of short circuit test the armature mmf is almost entirely demagnetizing in nature which results in very low value of air gap flux. Transformer Testings. For the purpose of quality assurance and ensuring that the finished transformer conforms to customer requirements and is ready for service a barrage of tests are performed on the transformer. These can be broadly classified into the following 3 types:
1) Routine Tests: These are standard tests performed necessarily on all transformers. They are further divided into two subcategories: Dielectric Tests (for insulation testing) a) Separate Source Power Frequency Test on HV and LV b) Induced over voltage test c) Insulation Resistance of HV and LV winding d) Dielectric value of oil
Parametric Tests (for winding, loss and efficiency testing) a) Winding Resistance b) No load loss and No load Current c) Load Loss and Impedance
d) Turns ratio on all taps and all phases
2) Type Test: This test is done on one unit of a particular design. It verifies the response of the design to the expected boundary conditions of the design. Hence it is not necessary to perform it on all units manufactured. Generally the testing and certification for the validation of a design to the conformity of these tests is considered for a period of five years. There are two tests that come under this paradigm: a) Temperature Rise Test b) Impulse Test
3) Special Tests: These are performed at the request of the customer by third party testing organizations. These give an idea about the design integrity, manufacturing quality, resistance to fault currents, operation quality etc. These are the following: a) Short-Circuit Test b) Unbalanced Current Test c) Magnetic Balance Test d) Measurement of Zero Sequence Impedance e) Measurement of noise level As seen from the above list there are a total of 15 tests to be performed on a transformer.These tests are to be performed in the same sequence as it has been written above.
The routine tests need to be performed on all transformers that are manufactured, whereas type and special tests have to be performed under the conditions that have been mentioned. It is necessary to perform the above tests in the same order because; each test can cause slight changes in the mechanical and electrical characteristics of the transformer. For example it can cause some of the insulated material to come under excessive levels of dielectric stress which may not be a regular occurrence under normal working conditions. They can cause modifications to the geometry of the core coil assembly thereby affecting the building factor. Also, if the tests are performed in a haphazard fashion without taking into account the parametric variations induced due to the previous tests, then probably the test results are not indicating the correct or operational values. It is recommended to perform tests on Transformer as per the relevant BIS and IEEE standards. Transformer Winding Temperature Indicator. The winding is the component with the highest temperature within the transformer and it is the component which is subjected to the fastest temperature increases as the load increases. Therefore for the control of the temperature parameter within the transformer, the temperature of the winding, as well as top oil, is measured. The temperature of winding of Transformer is measured using Winding Temperature Indicator (WTI) and the temperature of Transformer Oil is measured using Oil Temperature Indicator.
The purpose of WTI is to indicate the winding temperature of HV and LV winding of the Transformer and operates the alarm, trip and cooler control contacts (For detail on Cooler Unit visit Transformer Cooling Classes).
As can be seen from the above figure, Black needle shows the current winding temperature while the Red needle shows the highest winding temperature reached for a particular day. This Red needle needs to be reset manually.
Also, for setting winding temperature for Alarm and Trip, two knobs are provided. The Green knob shows the setting of winding temperature for Alarm and Red knob shows the Trip temperature setting as shown in figure.
Construction Detail of Winding Temperature Indicator, WTI:
Winding temperature indicator consists of a sensor bulb placed in an oil filled pocket in the transformer tank cover. The bulb is connected to the Instrument housing by means of two flexible capillary tubes. One capillary is connected to the measuring bellow of the instrument and the other to a compensating bellow.
Working Principle Winding Temperature Indicator:
An indirect system is used to measure winding temperature, since it is dangerous to place a sensor close to the winding due to the high voltage. The indirect measurement is done by means of a Thermal Image.
The measuring system is filled with a liquid which changes its volume with rising of temperature. Inside the instrument is fitted a heating resistance which is fed by a current proportionate to the current flowing through the transformer winding. To do this we connect the terminal of the heating resistance with the Bushing Current Transformer so that reflection of change in load is reflected in the Winding Temperature Indicator, WTI.
The Winding Temperature Indicator, WTI is provided with a maximum temperature indicator. The heating resistance is fed by a current transformer associated to the loaded winding of the transformer. The increase in the temperature of the resistance is proportionate to that of the winding. The sensor bulb of the instrument is located in the hottest oil of the transformer, therefore, the winding temperature indicates a temperature of hottest oil plus the winding temperature rise above hot oil Level the hotspot temperature.
Winding Temperature of Transformer may rise due to increased loading of Transformer or due to some internal fault. Normally the Winding Temperature Indicator, WTI gives alarm at 85 °C and Trip signal at 95 °C in India.
What is SOTF Protection?. SOTF Protection stands for Switch On to Fault. This protection is provided for high speed clearance of detected fault immediately after Manual Closure of Circuit Breaker. Let us suppose that a fault is existing and we gave a closing command to Breaker, then in that case SOTF protection will immediately trip the Breaker. SOTF Protection is helpful in the sense that while taking maintenance on a Line, it may happen so that Earth Switch of a particular Circuit Breaker is close. In that case if we give closing command to the Circuit Breaker, it will immediately trip on SOTF Protection. But it is normal to strike in your smart brain that how can we close a Breaker when a fault is already existing? Thing is that you can close the Breaker by bypassing all the required logics, so in that case SOTF protection will come into picture. Suppose while maintenance Earthing Rod was used for discharging a particular section but after competition of maintenance the Earth Rod is still connected to that section. In such case there is nothing or no logic to prevent closing of Breaker and as soon as one closes the Circuit Breaker, it will trip on SOTF protection. SOTF Protection gets enabled whenever all the three poles of a Circuit Breaker is open for more than a settable time say 110s (This time can be changed and configured as per the design.) SOTF protection is enabled in two cases. They are mentioned below.
Case1: If no Closing command to the Breaker is Present. When all the three poles of Circuit Breaker is open for more than a particular time say 110 s, as soon as we give a closing command to the Circuit Breaker, the SOTF protection is enabled for 500 ms and then gets reset. Case2: When an external closing command is present. Here external closing means closing the Circuit Breaker either manually or remotely. The SOTF protection logic is activated immediately. As soon as all the poles are closed SOTF protection is enable for 500 ms and then is reset. During this SOTF time frame of 500 ms, individual distance protection can be enabled or disabled. When a particular distance zone say Zone-2 is enabled, Zone-2 will actuate immediately without waiting for Zone-2 time delay if there exists any fault in this zone. How SOTF Relay Senses a Fault? Basically there are two elements in relay providing SOTF feature. They are Voltage Level Detector and Current Level Detector. The purpose of Voltage Level Detector is to ensure Dead Pole situation and Current Level Detector ensures that a fault occurred i.e. Current Level Detector is used for detection of fault. Therefore SOTF Relay element must have two settings, one for Voltage Level Detector and another for Current Level Detector. Typical setting for both the elements are as follows: Vph < 75% VN
Iph > 5% IN
Here VNand IN are Nominal Voltage and Nominal Current respectively. Thus if the phase voltages are normal and if phase current increases from the setting then SOFT protection will sense it a fault and will issue three phase trip command provided all other conditions as mentioned in case1 and case2 for SOTF are met.
Distance Protection Philosophy. Distance protection is a non-unit system of protection offering considerable economic and technical advantages. Unlike phase and neutral over current protection, the key advantage of distance protection is that its fault coverage of the protected circuit is independent of source
impedance variations. Let us take an example of this to understand how distance protection is independent of source impedance. Consider the figure below.
In the figure above, R1 is an over current relay which is used for the protection of Transmission Line. If there is a fault at F1,
Equivalent source impedance Zs = 10×10/20 = 5 Ω
Impedance up to the point of Fault = 5+4 = 9 Ω
Fault current IF1= 220×103/1.732*9 = 220×103/15.588 = 14113.5 A
Therefore the setting of over current Relay should be more than 14113.5 A.
Now consider the case,
Here fault is not on the Transmission Line but it is assumed to be inside Switchyard and only one source is feeding the power to the network. Proceeding in the similar manner, Fault Current IF2= 220×103/1.732*10 = 12702A
Therefore for the protection of Transmission Line, the setting of Relay shall be kept less than 12702 A. But for earlier case we saw that setting of Relay R1 shall be more than 14113.5 A thus overall the setting shall be > 14113.5 but <12702 A which is impractical. Therefore over current Relay is not suitable here and it depends on the source impedance.
Distance protection is therefore used for the protection of Transmission Line. It is simple to apply and fast in isolating the faulty section from the healthy network. Distance Protection provides primary as well as back-up protection to the protected line. I will show this back-up protection function latter in this post.
PRINCIPLES OF DISTANCE RELAYS:
Since the impedance of a transmission line is proportional to its length, for distance measurement it is justified to use a relay capable of measuring the impedance of a line up to a predetermined point. This predetermined point is called Reach of the Relay.
Such a relay is described as a distance relay and is designed to operate only for faults occurring between the relay location and the selected reach point, thus giving discrimination for faults that may occur in different line sections. The basic principle of distance protection involves the division of the voltage at the relaying point by the measured current. The apparent impedance so calculated is compared with the reach point impedance which is settable in the Relay. If the
measured impedance is less than the reach point impedance, it is assumed that a fault exists on the line between the relay and the reach point and issues trip command to the concerned Breaker Trip Coil either through Master Trip Relay or directly (in case of single pole tripping of breaker, assuming single pole Auto Reclosure is allowed).
If measured value of impedance V/I is less than setting z then Relay assumes a fault as clear from the above diagram.
ZONE CONCEPT IN DISTANCE PROTECTION:
Consider the figure below and carefully observe.
Here there are three sub-stations namely A, B and C. For sub-station A, the distance protection is divided into three zones Z1a, Z2a and Z3a which are called Zone-1, Zone-2 and Zone-3 protection. Similarly for sub-station D the three zones will be Z1d, Z2d and Z3d. Zone-1 is normally set to 80% of total length of Line (here line length is AB between two consecutive substation). Zone-2 is set to 150 % of total line length and Zone-3 set at 120% of (100% line length + 100% of Longest Line from Remote substation i.e. B). It should be noted that all Zones are setting is done in terms of impedance.
Assume the distance between A and B = 200 KM
Total Impedance of Line AB = 61 Ohm
Current Transformer ratio = 1000/1A
Potential Transformer ratio = 400 kV / 110 V
So for Zone-1 Impedance setting = 80 % of Total Line Impedance = 80% of 61
= 0.8×61 = 48.8 Ohm ????? (Will it be????)
It won’t be…..because you need to consider CT & PT ratio for calculating the impedance as the Relay is sensing current and voltage through CT and PT only.
PT/CT Ratio = 1000/(400×103/110) = 1000×110/400,000 = 0.28
So the required setting for Zone-1 = 48.8×0.28 = 13.66 ohm. Which means if the distance Relay senses Impedance less than 13.66 Ohm then it will pick-up for Zone-1.
In the same manner, Setting for Zone-2 = (150% of 61) × CT/PT ratio
= 1.5×61×0.28 = 25.62 ohm
Which means if the distance Relay senses Impedance less than 25.62 Ohm then it will pick-up for Zone-2.
Setting for Zone-3 = 120% of (Impedance of Line AB+ Impedance of Longest Line from substation B)
Assume the Longest Line from substation B is having an impedance of 61 Ohm.
Therefore Setting for Zone-3 = (120% of (61+61)) × CT/PT ratio
= 1.2×122×0.28 = 41 Ohm
Which means if the distance Relay senses Impedance less than 41 Ohm then it will pickup for Zone-3.
So we now know how to calculate the setting for different Zones of Distance Protection. Now suppose our substation is A and we are providing distance protection so Relay is located at A. For fault in Zone-1, obviously we need to isolate the fault without any time delay. Now say our breaker at A opened but as we are connected to the substation B so their breaker at B shall also trip so as to isolate the fault completely otherwise fault will be feed from substation B side even though our breaker at A opened. Thus if fault in Zone-1 occurs then Distance Relay shall trip Breaker at A and send a signal to Remote Substation B by receiving which Remote substation B shall trip their breaker at B. This signal is called Carrier Signal which is sent through Power Line Carrier Communication (PLCC) Line. This is the purpose of PLCC. I will post on PLCC latter so be there.
Thus for Zone-1, time delay = 0. Got it? (If no then write in comment box I will be happy to clear your doubt)
Next, suppose there is a fault in Zone-2 then our breaker at A shall not trip rather Remote Substation breaker at C shall trip (If fault is in section CD in figure above) as it will be in their Zone-1. So we need to introduce some time delay in our Distance Relay to operate for Zone-2 fault. This time delay is usually kept around 350 ms. If within 350 ms Remote substation breaker at B trips then our Breaker at A won’t trip but if suppose because of any Reason Remote Substation breaker at C fail to trip then our breaker at A will definitely trip.
See how Zone-2 is working as Back-up protection for line CD. Got it friend?
Now if there is a fault in the remaining 20% of line which is protected by Zone-1 at our substation A then it will be sensed by our Relay at A in Zone-2 but for Remote substation B it will be Zone-1 so their breaker at B will instantaneously trip but our breaker at A also need to trip otherwise our substation will continue to feed the fault by receiving carrier signal.
Now coming to Zone-3, if there is a fault in Zone-3 then our breaker at A is not supposed to trip rather Remote substation breaker at C &D is supposed to trip. Therefore we introduce some time delay for the operation of Zone-3 which is typically of the order of 1s. If because of any reason breaker at C & D fail to trip within 1s then our Distance relay will operate to open our Breaker at A.
There is one more Zone in modern Distance Relay which is called Reverse Zone or Zone-4. As the name Reverse Zone implies it is back-up protection of the Substation where Distance Relay is installed, in our case to the substation A. The setting for zone is normally 10% of the impedance of protected line.
Distance Relay Zone Characteristics on R-X Plane:
The reach point of a relay is the point along the line impedance locus that is intersected by the boundary characteristic of the relay.
Philosophy of Primary and Back-up Protection. The protection provided by the protective relaying equipment can be categorized into two types as:
a) Primary protection
b) Back-up protection
The primary protection is the first line of defense and is responsible to protect all the power system elements from all the types of faults. The backup protection comes into play only when the primary protection fails.
In the event of failure or non-availability of the Primary Protection some other means of ensuring that the fault is isolated must be provided. These secondary systems are referred to as Back-up Protection. Back-up protection may be considered as either being local or remote. Local backup protection is achieved by protection which detects an un-cleared primary system fault at its own location and which then trips its own circuit breakers, e.g. Time Graded Overcurrent Relays.
What is Time Grading? Protection systems in successive zones are arranged to operate in times that are graded through the sequence of equipment so that upon the occurrence of a fault, although a number of protection equipment respond, only those relevant to the faulty zone complete the tripping function. The others make incomplete operations and then reset. The speed of response will often depend on the severity of the fault, and will generally be slower than for a unit system.
As shown in figure above, if a fault occurs then Relay C is supposed to trip instantaneously, but in case of failure of operation of Relay C to isolate the fault, Relay B shall issue trip command after a time delay of 350 ms. In the worst case when both theRelays B & C fails, Relay A shall operate after 1 sec. This is Time Grading where a time is provided for main Relay to operate. Remote back-up protection is provided by protection that detects an un-cleared primary system fault at a remote location and then issues a local trip command, e.g. the second or third zones of a distance relay. In both cases the main and back-up protection systems detect a fault simultaneously, operation of the back-up protection being delayed to ensure that the primary protection clears the fault if possible. Normally being unit protection, operation of the primary protection will be fast and will result in the minimum amount of the power system being disconnected. Operation of the back-up protection will be, of necessity, slower and will result in a greater proportion of the primary system being lost.
The extent and type of back-up protection applied will be related to the failure risks and relative economic importance of the system. For distribution systems where fault clearance times are not critical, time delayed remote back-up protection may be adequate. For EHV systems, where system stability is at risk unless a fault is cleared quickly, multiple primary protection systems, operating in parallel and possibly of different types (e.g. distance andunit protection), will be used to ensure fast and reliable tripping. Back-up overcurrent protection may then optionally be applied to ensure that two separate protection systems are available during maintenance of one of the primary protection systems. Back-up protection systems should, ideally, be completely separate from the primary systems. For example a circuit protected by a current differential relay may also have time graded overcurrent and earth fault relays added to provide circuit breaker tripping in the event of failure of the main primary unit protection. To maintain complete separation and thus integrity, current transformers, voltage transformers, relays, circuit breaker
trip coils and d.c. supplies would be duplicated. This ideal is rarely attained in practice. The following compromises are typical:
a) Separate current transformers(cores and secondary windings only) are provided. This involves little extra cost or accommodation compared with the use of common current transformers that would have to be larger because of the combined burden. This practice is becoming less common when digital or numerical relays are used, because of the extremely low input burden of these relay types.
b) Voltage transformers are not duplicated because of cost and space considerations. Each protection relay supply is separately protected (fuse or MCB) and continuously supervised to ensure security of the VT output. An alarm is given on failure of the supply and, where appropriate, prevents an unwanted operation of the protection.
c) Trip supplies to the two protections should be separately protected (fuse or MCB). Duplication of tripping batteries and of circuit breaker tripping coils may be provided. Trip circuits should be continuously supervised.
d) It is desirable that the main and back-up protections (or duplicate main protections) should operate on different principles, so that unusual events that may cause failure of the one will be less likely to affect the other.
Numerical relaysmay incorporate suitable back-up protection functions (e.g. a distance relay may also incorporate time-delayed overcurrent protection elements as well). A reduction in the hardware required to provide back-up protection is obtained, but at the risk that a common relay element failure (e.g. the power supply) will result in simultaneous loss of both main and back-up protection.
Autoreclosing Philosophy in Distance Protection. Autoreclosing is a feature which is provided in the Line Circuit Breaker so that Single Pole of a Breaker may trip and close when command is given to do so. Based on the time duration of fault existing in the Power System, faults can be classified into three categories as
Transient Fault Semi-Transient Fault Permanent Fault.
Transient fault exists only for very short duration and these can the removed faster if the line is disconnected from the system momentarily so that arc extinguishes. After the arc is deionized, line can be reclosed to resume the service. Thus, in this way the interruption in the Power Supply is reduced and loss of revenue is also saved. It has been found that 80% of the fault in Power System are Transient in nature, 12% are Semitransient and remaining 8% are only permanenet fault. Semi-transient fault are those fault which take some finite time to clear by itself. For example, suppose a bord spanning the two lines sit then it will cause a fault which will clear by itself after the burning of cause of fault,after some time say 1sec. Thus we will expect, Autorecloser to take place for 1 sec i.e. Breaker shall close after a time delay of 1 sec.Here note that the time after which fault clears by itself is called DEAD TIME. Therefore, in our example DEAD TIME = 1 sec. But for Permanent Fault, Autorecloser will not help as the cause of fault continuously exists so if we incorporate the Autorecloser the Breaker will again trip after the Autorecloser. So how many attempts will the Realy take to Autoreclose and after how much time it will take another consecutive attempt to Autoreclose? “Here we come to another concept, called RECLAIM TIME. RECLAIM TIME is the time after which Relay will take another consecutive attempt to Autoreclose. This RECLAIM TIME is typically set at 25 sec. The number of attempt for Autorecloser is set in the Relay which is 4 for MiCOM P444 Distance Protection Relay. This means that Relay will take four Autorecloser shots and at the end of fourth shot, if still fault is existing, the Line will be taken out.” In figure below, a Numerical Relay is shown.
Thus we see that for permanent fault Autorecloser won’t help as we need to attend the fault and rectify it. Autorecloser can be Single pole or Three pole. Here Pole means Breaker of any of the three phase i.e. either R, Y or B phase. Single Pole Autorecloser take place during Line to ground fault. It shall be noted that Autorecloser facility is provided only in Line Breaker and that to by Distance Protection Relay. Single Autorecloser take place in the following conditions: Zone-1 protection operated AND Carrier Channels are healthy AND Three Pole Tripping has not taken place. OR Zone-2 protection operated with Carrier Signal Received. This seems surprising that only a single pole of Breaker trips during Zone-1 fault. But it’s true. The phase, say B phase, in which Line to ground fault has taken place will only trip and reclose after the DEAD TIME. If within the RECLAIM TIME, another fault take place then Three Phase trip will occur. During the DEAD TIME period, power is fed to the system via the two healthy phases. In case of Three Pole Autorecloser, all the three phases are opened independently irrespective of type of fault be it Single L-G or L-L or L-L-L fault and reclosed after the DEAD TIME. During the DEAD TIME period, no power can be transmitted and therefore system is liable to operate unstably.
Fig. G65: Example of single-line diagram
Calculation using software Ecodial General network characteristics
Number of poles and protected poles
4P4d
Earthing system
TN-S
Tripping unit
Micrologic 2.3
Neutral distributed
No
Overload trip Ir (A)
510
Voltage (V)
400
Short-delay trip Im / Isd (A)
5100
Frequency (Hz)
50
Cable C3
Upstream fault level (MVA)
500
Length
20
Resistance of MV network (mΩ)
0.0351
Maximum load current (A)
509
Reactance of MV network (mΩ)
0.351
Type of insulation
PVC
Ambient temperature (°C)
30
Transformer T1
Rating (kVA)
630
Conductor material
Copper
Short-circuit impedance voltage (%)
4
Single-core or multi-core cable
Single
Transformer resistance RT (mΩ)
3.472
Installation method
F
Transformer reactance XT (mΩ)
10.64
Phase conductor selected csa (mm2)
2 x 95
3-phase short-circuit current Ik3 (kA)
21.54
Neutral conductor selected
2 x 95
csa (mm2)
Cable C1
PE conductor selected csa (mm2)
1 x 95
Length (m)
5
Cable voltage drop ΔU (%)
0.53
Maximum load current (A)
860
Total voltage drop ΔU (%)
0.65
Type of insulation
PVC
3-phase short-circuit current Ik3 (kA)
19.1
Ambient temperature (°C)
30
1-phase-to-earth fault current Id (kA)
11.5
Conductor material
Copper
Switchboard B6
Single-core or multi-core cable
Single
Reference
Linergy 800
Installation method
F
Rated current (A)
750
Number of layers
1
Circuit-breaker Q7
Phase conductor selected csa (mm2)
2 x 240
Load current (A)
255
Neutral conductor selected csa (mm2)
2 x 240
Type
Compact
PE conductor selected csa (mm2)
1 x 120
Reference
NSX400F
Voltage drop ΔU (%)
0.122
Rated current (A)
400
3-phase short-circuit current
21.5
Number of poles and
3P3d
Ik3 (kA)
Courant de défaut phase-terre Id (kA)
protected poles
15.9
Circuit-breaker Q1
Tripping unit
Micrologic 2.3
Overload trip Ir (A)
258
2576
Load current (A)
860
Short-delay trip Im / Isd (A)
Type
Compact
Cable C7
Reference
NS1000N
Length
5
Rated current (A)
1000
Maximum load current (A)
255
Number of poles and protected poles
4P4d
Type of insulation
PVC
Tripping unit
Micrologic 5.0
Ambient temperature (°C)
30
Overload trip Ir (A)
900
Conductor material
Copper
Short-delay trip Im / Isd (A)
9000
Single-core or multi-core cable
Single
Tripping time tm (ms)
50
Installation method
F
Phase conductor selected csa (mm2)
1 x 95
Switchboard B2
Reference
Linergy 1250
Neutral conductor selected csa (mm2)
-
Rated current (A)
1050
PE conductor selected csa
1 x 50
(mm2)
Circuit breaker Q3
Cable voltage drop ΔU (%)
0.14
Load current (A)
509
Total voltage drop ΔU (%)
0.79
Type
Compact
3-phase short-circuit current Ik3 (kA)
18.0
Reference
NSX630F
1-phase-to-earth fault current Id (kA)
10.0
Rated current (A)
630
Fig. G66: Partial results of calculation carried out with Ecodial software (Schneider Electric). The calculation is performed according to Cenelec TR50480
The same calculation using the simplified method recommended in this guide Dimensioning circuit C1 The MV/LV 630 kVA transformer has a rated no-load voltage of 420 V. Circuit C1 must be suitable for a current of:
per phase Two single-core PVC-insulated copper cables in parallel will be used for each phase.These cables will be laid on cable trays according to method F. Each conductor will therefore carry 433A. Figure G21a indicates that for 3 loaded conductors with PVC isolation, the required c.s.a. is 240mm2. The resistance and the inductive reactance, for the two conductors in parallel, and for a length of 5 metres, are:
(cable resistance: 23.7 mΩ.mm2/m) (cable reactance: 0.08 mΩ/m)
Dimensioning circuit C3 Circuit C3 supplies two 150kW loads with cos φ = 0.85, so the total load current is:
Two single-core PVC-insulated copper cables in parallel will be used for each phase. These cables will be laid on cable trays according to method F. Each conductor will therefore carry 255A. Figure G21a indicates that for 3 loaded conductors with PVC isolation, the required c.s.a. is 95mm2. The resistance and the inductive reactance, for the two conductors in parallel, and for a length of 20 metres, are:
(cable resistance: 23.7 mΩ.mm2/m) (cable reactance: 0.08 mΩ/m) Dimensioning circuit C7 Circuit C7 supplies one 150kW load with cos φ = 0.85, so the total load current is:
One single-core PVC-insulated copper cable will be used for each phase. The cables will be laid on cable trays according to method F. Each conductor will therefore carry 255A. Figure G21a indicates that for 3 loaded conductors with PVC isolation, the required c.s.a. is 95mm2. The resistance and the inductive reactance for a length of 20 metres is:
(cable resistance: 23.7 mΩ.mm2/m) (cable reactance: 0.08 mΩ/m) Calculation of short-circuit currents for the selection of circuit-breakers Q1, Q3, Q7 (see Fig. G67) Circuit components
R (mΩ)
X (mΩ)
Z (mΩ)
Ikmax (kA)
Upstream MV network, 500MVA fault level (see Fig. G34)
0,035
0,351
Transformer 630kVA, 4% (see Fig. G35)
2.9
10.8
Cable C1
0.23
0.4
Sub-total
3.16
11.55
Cable C3
2.37
1.6
Sub-total
5.53
13.15
Cable C7
1.18
0.4
Sub-total
6.71
13.55
11.97
20.2
14.26
17
15.12
16
Fig. G67: Example of short-circuit current evaluation
The protective conductor When using the adiabatic method, the minimum c.s.a. for the protective earth conductor (PE) can be calculated by the formula given in Figure G58:
For circuit C1, I = 20.2kA and k = 143. t is the maximum operating time of the MV protection, e.g. 0.5s This gives:
A single 120 mm2 conductor is therefore largely sufficient, provided that it also satisfies the requirements for indirect contact protection (i.e. that its impedance is sufficiently low).
Generally, for circuits with phase conductor c.s.a. Sph ≥ 50 mm2, the PE conductor minimum c.s.a. will be Sph / 2. Then, for circuit C3, the PE conductor will be 95mm2, and for circuit C7, the PE conductor will be 50mm2. Protection against indirect-contact hazards For circuit C3 of Figure G65, Figures F41 and F40, or the formula given TN system - Protection against indirect contact may be used for a 3-phase 4-wire circuit. The maximum permitted length of the circuit is given by:
(The value in the denominator 630 x 11 is the maximum current level at which the instantaneous short-circuit magnetic trip of the 630 A circuit-breaker operates). The length of 20 metres is therefore fully protected by “instantaneous” over-current devices. Voltage drop The voltage drop is calculated using the data given in Figure G28, for balanced three-phase circuits, motor power normal service (cos φ = 0.8). The results are summarized on Figure G68 The total voltage drop at the end of cable C7 is then: 0.77%. C1
C3
C7
2 x 240mm2
2 x 95mm2
1 x 95mm2
0.22
0.43
0.43
Load current (A)
866
509
255
Length (m)
5
20
5
Voltage drop (V)
0.48
2.19
0.55
Voltage drop (%)
0.12
0.55
0.14
c.s.a.
∆U per conductor(V/A/km) see Fig. G28
Fig. G68: Voltage drop introduced by the different cables.
3-phase short-circuit current (Isc) at any point within a LV installation. In a 3-phase installation Isc at any point is given by:
where U20 = phase-to-phase voltage of the open circuited secondary windings of the power supply transformer(s). ZT = total impedance per phase of the installation upstream of the fault location (in Ω)
Method of calculating ZT Each component of an installation (MV network, transformer, cable, busbar, and so on...) is characterized by its impedance Z, comprising an element of resistance (R) and an inductive reactance (X). It may be noted that capacitive reactances are not important in short-circuit current calculations. The parameters R, X and Z are expressed in ohms, and are related by the sides of a right angled triangle, as shown in the impedance diagram of Figure G33.
Fig. G33: Impedance diagram The method consists in dividing the network into convenient sections, and to calculate the R and X values for each. Where sections are connected in series in the network, all the resistive elements in the section are added arithmetically; likewise for the reactances, to give RT and XT.
The impedance (ZT) for the combined sections concerned is then calculated from Any two sections of the network which are connected in parallel, can, if predominantly both resistive (or both inductive) be combined to give a single equivalent resistance (or reactance) as follows: Let R1 and R2 be the two resistances connected in parallel, then the equivalent resistance R3 will be given by:
or for reactances It should be noted that the calculation of X3 concerns only separated circuit without mutual inductance. If the circuits in parallel are close togother the value of X3 will be notably higher.
Determination of the impedance of each component Network upstream of the MV/LV transformer (see Fig. G34) The 3-phase short-circuit fault level PSC, in kA or in MVA[1] is given by the power supply authority concerned, from which an equivalent impedance can be deduced. Psc
Uo (V)
Ra (mΩ)
Xa (mΩ)
250 MVA
420
0.07
0.7
500 MVA
420
0.035
0.351
Fig. G34: The impedance of the MV network referred to the LV side of the MV/LV transformer
A formula which makes this deduction and at the same time converts the impedance to an equivalent value at LV is given, as follows:
where Zs = impedance of the MV voltage network, expressed in milli-ohms Uo = phase-to-phase no-load LV voltage, expressed in volts Psc = MV 3-phase short-circuit fault level, expressed in kVA
The upstream (MV) resistance Ra is generally found to be negligible compared with the corresponding Xa, the latter then being taken as the ohmic value for Za. If more accurate calculations are necessary, Xa may be taken to be equal to 0.995 Za and Ra equal to 0.1 Xa. Figure G36 gives values for Ra and Xa corresponding to the most common MV[2] short-circuit levels in utility power-supply networks, namely, 250 MVA and 500 MVA. Transformers (see Fig. G35) The impedance Ztr of a transformer, viewed from the LV terminals, is given by the formula:
where: U20 = open-circuit secondary phase-to-phase voltage expressed in volts Sn = rating of the transformer (in VA) Usc = the short-circuit impedance voltage of the transformer expressed in % The transformer windings resistance Rtr can be derived from the total load-losses as follows:
so that
in milli-ohms
where Pcu = total load-losses in watts In = nominal full-load current in amps Rtr = resistance of one phase of the transformer in milli-ohms (the LV and corresponding MV winding for one LV phase are included in this resistance value).
Note: for an approximate calculation, in the absence of more precise information on transformer characteristics, Cenelec 50480 suggests to use the following guidelines:
if U20 is not known, it may be assumed to be 1.05 Un
in the absence of more precise information, the following values may be used: Rtr = 0.31 Ztr and Xtr = 0.95 Ztr Example: for a transformer of 630kVA with Usc=4% / Un = 400V, approximate calculation gives:
U20 = 400 x 1.05 = 420V
Ztr = 4202 / 630000 x 4% = 11 mΩ
Rtr = 0.31 x Ztr = 3.5 mΩ and Xtr = 0.95 x Ztr = 10.6 mΩ Rated Power kVA)
Oil-immersed
Cast-resin
Usc (%)
Rtr (mΩ)
Xtr (mΩ)
Ztr (mΩ)
Usc (%)
Rtr (mΩ)
Xtr (mΩ)
Ztr (mΩ)
100
4
37.9
59.5
70.6
6
37.0
99.1
105.8
160
4
16.2
41.0
44.1
6
18.6
63.5
66.2
200
4
11.9
33.2
35.3
6
14.1
51.0
52.9
250
4
9.2
26.7
28.2
6
10.7
41.0
42.3
315
4
6.2
21.5
22.4
6
8.0
32.6
33.6
400
4
5.1
16.9
17.6
6
6.1
25.8
26.5
500
4
3.8
13.6
14.1
6
4.6
20.7
21.2
630
4
2.9
10.8
11.2
6
3.5
16.4
16.8
800
6
2.9
12.9
13.2
6
2.6
13.0
13.2
1,000
6
2.3
10.3
10.6
6
1.9
10.4
10.6
1,250
6
1.8
8.3
8.5
6
1.5
8.3
8.5
1,600
6
1.4
6.5
6.6
6
1.1
6.5
6.6
2,000
6
1.1
5.2
5.3
6
0.9
5.2
5.3
Fig. G35: Resistance, reactance and impedance values for typical distribution 400 V transformers (no-load voltage = 420 V) with MV windings ≤ 20 kV
Busbars The resistance of busbars is generally negligible, so that the impedance is practically all reactive, and amounts to approximately 0.15 mΩ/metre[3] length for LV busbars (doubling the spacing between the bars increases the reactance by about 10% only). In practice, it's almost never possible to estimate the busbar length concerned by a short-circuit downstream a switchboard. Circuit conductors The resistance of a conductor is given by the formula:
where ρ = the resistivity of the conductor material at the normal operating temperature ρ has to be considered:
at cold state (20°C) to determine maximum short-circuit current,
at steady state (normal operating temperature) to determine minimum short-circuit current. L = length of the conductor in m S = c.s.a. of conductor in mm2 20 °C
PR/XLPE 90 °C
PVC 70 °C
Copper
18.51
23.69
22.21
Alu
29.41
37.65
35.29
Fig. G35b: Values of ρ as a function of the temperature, cable insulation and cable core material, according to IEC60909-0 and Cenelec TR 50480 (in mΩ mm2/m).
Cable reactance values can be obtained from the manufacturers. For c.s.a. of less than 50 mm2 reactance may be ignored. In the absence of other information, a value of 0.08 mΩ/metre may be used (for 50 Hz systems) or 0.096 mΩ/metre (for 60 Hz systems). For busways (busbar trunking systems) and similar pre-wired ducting systems, the manufacturer should be consulted.
Motors At the instant of short-circuit, a running motor will act (for a brief period) as a generator, and feed current into the fault. In general, this fault-current contribution may be ignored. However, if the total power of motors running simultaneously is higher than 25% of the total power of transformers, the influence of motors must be taken into account. Their total contribution can be estimated from the formula: Iscm = 3.5 In from each motor i.e. 3.5m In for m similar motors operating concurrently. The motors concerned will be the 3-phase motors only; single-phase-motor contribution being insignificant. Fault-arc resistance Short-circuit faults generally form an arc which has the properties of a resistance. The resistance is not stable and its average value is low, but at low voltage this resistance is sufficient to reduce the fault-current to some extent. Experience has shown that a reduction of the order of 20% may be expected. This phenomenon will effectively ease the current-breaking duty of a CB, but affords no relief for its fault-current making duty. Recapitulation table (see Fig. G36) Parts of power-supply system
R (mΩ)
X (mΩ) Xa = 0.995 Za
Supply network Figure G34 Transformer where
Figure G35
Rtr is often negligible compared to Xtr for transformers > 100 kVA Circuit-breaker
Not considered in practice
Busbars
Negligible for S > 200 mm2 in the
XB = 0.15 mΩ/m
[a]
formula: Circuit conductors[b]
Cables: Xc = 0.08 mΩ/m [a]
Motors
See 3-phase short-circuit current (Isc) at any point within a LV installation Motors (often negligible at LV)
Three-phase maximum circuit current in kA [a] ρ = resistivity at 20°C [b] If there are several conductors in parallel per phase, then divide the resistance of one conductor by the number of conductors. The reactance remains practically unchanged. U20: Phase-to-phase no-load secondary voltage of MV/LV transformer (in volts). Psc: 3-phase short-circuit power at MV terminals of the MV/LV transformers (in kVA). Pcu: 3-phase total losses of the MV/LV transformer (in watts). Sn: Rating of the MV/LV transformer (in kVA). Usc: Short-circuit impedance voltage of the MV/LV transfomer (in %). RT : Total resistance. XT: Total reactanc Fig. G36: Recapitulation table of impedances for different parts of a power-supply system
Example of short-circuit calculations (see Fig. G37) LV installation
R (mΩ)
X (mΩ)
0.035
0.351
2.35
8.5
MV network Psc = 500 MVA Transformer 20 kV / 420 V Pn = 1000 kVA Usc = 5% Pcu = 13.3 x
RT (mΩ)
XT (mΩ)
103 watts Single-core cables 5 m copper 4 x 240 mm2/phase
Xc = 0.08 x 5 = 0.40
2.48
9.25
Isc1 = 25 kA
Three-core cable 100 m 95 mm2copper
Xc = 100 x 0.08 = 8
22
17.3
Isc3 = 8.7 kA
Three-core cable 20 m 10 mm2copper final circuits
Xc = 20 x 0.08 = 1.6
59
18.9
Isc4 = 3.9 kA
Main circuitbreaker
Not considered in practice
Busbars 10 m
Not considered in practice
RT : Total resistance. XT: Total reactance. Isc : 3-phase maximum short-circuit current Calculations made a G36 Fig. G37: Example of maximum short-circuit current calculations for a LV installation supplied at 400 V (nominal) from a 1000 kVA MV/LV transformer
Notes 1. ^ Short-circuit MVA:
EL Isc where:
EL = phase-to-phase nominal system voltage expressed in kV (r.m.s.)
Isc = 3-phase short-circuit current expressed in kA (r.m.s.)
2. ^ up to 36 kV 3. ^ For 50 Hz systems, but 0.18 mΩ/m length at 60 Hz.
Location of protective devices. General rule (see Fig. G7a) A protective device is necessary at the origin of each circuit where a reduction of permissible maximum current level occurs.
Possible alternative locations in certain circumstances (see Fig. G7b) The protective device may be placed part way along the circuit:
If AB is not in proximity to combustible material, and
If no socket-outlets or branch connections are taken from AB Three cases may be useful in practice: Consider case (1) in the diagram
AB ≤ 3 metres, and
AB has been installed to reduce to a practical minimum the risk of a short-circuit (wires in heavy steel conduit for example) Consider case (2)
The upstream device P1 protects the length AB against short-circuits in accordance
Consider case (3)
The overload device (S) is located adjacent to the load. This arrangement is
convenient for motor circuits. The device (S) constitutes the control (start/stop) and overload protection of the motor while (SC) is: either a circuit-breaker (designed for motor protection) or fuses type aM The short-circuit protection (SC) located at the origin of the circuit conforms with the
principles of Calculation of minimum levels of short-circuit current .
Circuits with no protection (see Fig. G7c) Either The protective device P1 is calibrated to protect the cable S2 against overloads and short-
circuits Or Where the breaking of a circuit constitutes a risk, e.g.
Excitation circuits of rotating machines
circuits of large lifting electromagnets
the secondary circuits of current transformers
No circuit interruption can be tolerated, and the protection of the cabling is of secondary importance.
a.
b.
c.
Short-circuit current at the secondary terminals of a MV/LV distribution transformer. The case of one transformer In a simplified approach, the impedance of the MV system is assumed to be negligibly small,
so that:
where
and:
S = kVA rating of the transformer U20 = phase-to-phase secondary volts on open circuit In = nominal current in amps Isc = short-circuit fault current in amps Usc = short-circuit impedance voltage of the transformer in %. Typical values of Usc for distribution transformers are given in Figure G31 Transformer rating (kVA)
Usc in %
Oil-immersed
Cast-resin dry type
50 to 750
4
6
800 to 3,200
6
6
Fig. G31: Typical values of Usc for different kVA ratings of transformers with MV windings ≤ 20 kV
Example 400 kVA transformer, 420 V at no load Usc = 4%
The case of several transformers in parallel feeding a busbar The value of fault current on an outgoing circuit immediately downstream of the busbars (see Fig. G32) can be estimated as the sum of the Isc from each transformer calculated separately. It is assumed that all transformers are supplied from the same MV network, in which case the values obtained from Figure G31when added together will give a slightly higher fault-level value than would actually occur. Other factors which have not been taken into account are the impedance of the busbars and of the cable between transformers and circuit breakers.
The conservative fault-current value obtained however, is sufficiently accurate for basic installation design purposes. The choice of circuit breakers and incorporated protective devices against shortcircuit and fault currents is described in Selection of a circuit-breaker .
Fig. G32: Case of several transformers in parallel.
Conductor sizing: methodology and definition. Methodology (see Figure G1) Following a preliminary analysis of the power requirements of the installation, as described in The consumer substation with LV metering, a study of cabling[1] and its electrical protection is undertaken, starting at the origin of the installation, through the intermediate stages to the final circuits. The cabling and its protection at each level must satisfy several conditions at the same time, in order to ensure a safe and reliable installation, e.g. it must:
Carry the permanent full load current, and normal short-time overcurrents
Not cause voltage drops likely to result in an inferior performance of certain loads, for example: an excessively long acceleration period when starting a motor, etc. Moreover, the protective devices (circuit-breakers or fuses) must:
Protect the cabling and busbars for all levels of overcurrent, up to and including short-circuit currents
Ensure protection of persons against indirect contact hazards, particularly in TN- and ITearthed systems, where the length of circuits may limit the magnitude of short-circuit currents,
thereby delaying automatic disconnection (it may be remembered that TT- earthed installations are necessarily protected at the origin by a RCD, generally rated at 300 mA). The cross-sectional areas of conductors are determined by the general method described in Practical method for determining the smallest allowable cross-sectional area of circuit conductors of this Chapter. Apart from this method some national standards may prescribe a minimum cross-sectional area to be observed for reasons of mechanical endurance. Particular loads (as noted in Chapter Characteristics of particular sources and loads) require that the cable supplying them be oversized, and that the protection of the circuit be likewise modified.
Fig. G1: Flow-chart for the selection of cable size and protective device rating for a given circuit
Definitions Maximum load current: IB
At the final circuits level, this design current (according to IEV "International Electrotechnical Vocabulary" ref 826-11-10) corresponds to the rated kVA of the load. In the case of motor-starting, or other loads which take a high in-rush current, particularly where frequent starting is concerned (e.g. lift motors, resistance-type spot welding, and so on) the cumulative thermal effects of the overcurrents must be taken into account. Both cables and thermal type relays are affected.
At all upstream circuit levels this current corresponds to the kVA to be supplied, which takes account of the diversity and utilization factors, ks and ku respectively, as shown in Figure G2.
Fig. G2: Calculation of maximum load current IB Maximum permissible current: Iz Current carrying capacity Iz is the maximum permissible that the cabling for the circuit can carry indefinitely, without reducing its normal life expectancy. The current depends, for a given cross sectional area of conductors, on several parameters:
Constitution of the cable and cable-way (Cu or Alu conductors; PVC or EPR etc. insulation; number of active conductors)
Ambient temperature
Method of installation
Influence of neighbouring circuits Overcurrents An overcurrent occurs each time the value of current exceeds the maximum load current IB for the load concerned. This current must be cut off with a rapidity that depends upon its magnitude, if permanent damage to the cabling (and appliance if the overcurrent is due to a defective load component) is to be avoided. Overcurrents of relatively short duration can however, occur in normal operation; two types of overcurrent are distinguished:
Overloads These overcurrents can occur in healthy electric circuits, for example, due to a number of small short-duration loads which occasionally occur co-incidentally: motor starting loads, and so on. If either of these conditions persists however beyond a given period (depending on protective-relay settings or fuse ratings) the circuit will be automatically cut off.
Short-circuit currents These currents result from the failure of insulation between live conductors or/and between live conductors and earth (on systems having low-impedance-earthed neutrals) in any combination, viz: 3 phases short-circuited (and to neutral and/or earth, or not) 2 phases short-circuited (and to neutral and/or earth, or not) 1 phase short-circuited to neutral (and/or to earth).
Overcurrent protection principles. A protective device is provided at the origin of the circuit concerned (see Fig. G3 and Fig. G4 ).
Acting to cut-off the current in a time shorter than that given by the I2t characteristic of the circuit cabling
But allowing the maximum load current IB to flow indefinitely The characteristics of insulated conductors when carrying short-circuit currents can, for periods up to 5 seconds following short-circuit initiation, be determined approximately by the formula: I2t = k2 S2
which shows that the allowable heat generated is proportional to the squared cross-sectional-area of the condutor. where t = Duration of short-circuit current (seconds) S = Cross sectional area of insulated conductor (mm2) I = Short-circuit current (A r.m.s.) k = Insulated conductor constant (values of k2 are given in Figure G52) For a given insulated conductor, the maximum permissible current varies according to the environment. For instance, for a high ambient temperature (θa1 > θa2), Iz1 is less than Iz2 (see Fig. G5). θ means “temperature”. Note: ISC: 3-phase short-circuit current ISCB: rated 3-ph. short-circuit breaking current of the circuit-breaker Ir (or Irth)[1]: regulated “nominal” current level; e.g. a 50 A nominal circuit-breaker can be regulated to have a protective range, i.e. a conventional overcurrent tripping level (see Fig. G6) similar to that of a 30 A circuit-breaker.
Fig. G3: Circuit protection by circuit breaker
Fig. G4: Circuit protection by fuses
Fig. G5: I2t characteristic of an insulated conductor at two different ambient temperatures.
Practical values for a protective scheme. General rules A protective device (circuit-breaker or fuse) functions correctly if:
Its nominal current or its setting current In is greater than the maximum load current IB but less than the maximum permissible current Iz for the circuit, i.e. IB ≤ In ≤ Iz corresponding to zone “a” in Figure G6
Its tripping current I2 “conventional” setting is less than 1.45 Iz which corresponds to zone “b” in Figure G6 The “conventional” setting tripping time may be 1 hour or 2 hours according to local standards and the actual value selected for I2. For fuses, I2 is the current (denoted If) which will operate the fuse in the conventional time.
Its 3-phase short-circuit fault-current breaking rating is greater than the 3-phase short-circuit current existing at its point of installation. This corresponds to zone “c” in Figure G6.
Fig. G6: Current levels for determining circuir breaker or fuse characteristics IB ≤ In ≤ Iz zone a I2 ≤ 1.45 Iz zone b ISCB ≥ ISC zone c
Applications
Protection by circuit-breaker Criteria for circuit-breakers: IB ≤ In ≤ Iz and ISCB ≥ ISC. By virtue of its high level of precision the current I2 is always less than 1.45 In (or 1.45 Ir) so that the condition I2 ≤ 1.45 Iz (as noted in the “general rules” above) will always be respected.
Particular case If the circuit-breaker itself does not protect against overloads, it is necessary to ensure that, at a time of lowest value of short-circuit current, the overcurrent device protecting the circuit will operate correctly. This particular case is examined in Calculation of minimum levels of short-circuit current. Protection by fuses Criteria for fuses: IB ≤ In ≤ Iz/k3 and ISCF ≥ ISC. The condition I2 ≤ 1.45 Iz must be taken into account, where I2 is the fusing (melting level) current, equal to k2 x In (k2 ranges from 1.6 to 1.9) depending on the particular fuse concerned.
A further factor k3 has been introduced
such that I2 ≤ 1.45 Iz
will be valid if In ≤ Iz/k3. For fuses type gG: In < 16 A → k3 = 1.31 In ≥ 16 A → k3 = 1.10 Moreover, the short-circuit current breaking capacity of the fuse ISCF must exceed the level of 3phase short-circuit current at the point of installation of the fuse(s). Association of different protective devices The use of protective devices which have fault-current ratings lower than the fault level existing at their point of installation are permitted by IEC and many national standards in the following conditions:
There exists upstream, another protective device which has the necessary short-circuit rating, and
The amount of energy allowed to pass through the upstream device is less than that which
can be withstood without damage by the downstream device and all associated cabling and appliances. In pratice this arrangement is generally exploited in:
The association of circuit-breakers/fuses
The technique known as “cascading” or “series rating” in which the strong current-limiting performance of certain circuit-breakers effectively reduces the severity of downstream shortcircuits Possible combinations which have been tested in laboratories are indicated in certain manufacturers catalogues.
General method for cable sizing. Possible methods of installation for different types of conductors or cables The different admissible methods of installation are listed in Figure G8, in conjonction with the different types of conductors and cables.
Conductors cables
Method of installation
Without fixings
Bare conductors
[b]
Clipped direct
-
Condu it syste ms
-
Cable trunking systems (including skirting trunking,
Cable ducting systems
Cable ladder,
tray, Cable
trunking)
rackets
[a]
Sup port wire
Cable
flush floor
-
On insulat ors
-
-
+
-
+
-
+
-
Insulated conductors
-
-
+
+
Sheathed cables
Multicore
+
+
+
+
+
+
0
+
(including
Single -core
0
+
+
+
+
+
0
+
armoured and mineral insulated)
+ : Permitted. – : Not Permitted. 0 : Not applicable, or not normally used in practice. [a] Insulated conductors are admitted if the cable trunking systems provide at least he degree of protection IP4X or IPXXD and if the cover can only be removed by means of a tool or a deliberate action. [b] Insulated conductors which are used as protective conductors or protective bonding conductors may use any appropriate method of installation and need not be laid in conduits, trunking or ducting systems. Fig. G8: Selection of wiring systems (table A.52.1 of IEC 60364-5-52)
Possible methods of installation for different situations: Different methods of installation can be implemented in different situations. The possible combinations are presented in Figure G9. The number given in this table refer to the different wiring systems considered. Situations
Method of installation
Without fixings
Clipped direct
Conduit System s
Cable trunking (including skirting trunking, flush floor trunking)
Cable ducting systems
Cable ladder, cable tray, cable brackets
On insulators
Support wire
Accessible
40
33
41, 42
6, 7, 8, 9,12
43, 44
30, 31, 32, 33, 34
-
0
Not accessible
40
0
41, 42
0
43
0
0
0
Cable channel
56
56
54, 55
0
30, 31, 32, 34
-
-
Buried in ground
72, 73
0
70, 71
-
70, 71
0
-
-
Embedded in structure
57, 58
3
1, 2, 59, 60
50, 51, 52, 53
46, 45
0
-
-
Building voids
Surface mounted
-
20, 21, 22, 23, 33
4, 5
6, 7, 8, 9, 12
6, 7, 8, 9
30, 31, 32, 34
36
-
Overhead/free in air
-
33
0
10, 11
10, 11
30, 31, 32, 34
36
35
Window frames
16
0
16
0
0
0
-
-
Architrave
15
0
15
0
0
0
-
-
Immersed 1
+
+
+
-
+
0
-
-
– : Not permitted. 0 : Not applicable or not normally used in practice. + : Follow manufacturer’s instructions. Note: The number in each box, e.g. 40, 46, refers to the number of the method of installation in Table A.52.3. Fig. G9: Erection of wiring systems (table A.52.2 of IEC 60364-5-52)
Examples of wiring systems and reference methods of installations An illustration of some of the many different wiring systems and methods of installation is provided in Figure G10. Several reference methods are defined (with code letters A to G), grouping installation methods having the same characteristics relative to the current-carrying capacities of the wiring systems. Item No.
Methods of installation
1
Room
Description
Reference method of installation to be used to obtain current-carrying capacity
Insulated conductors or single-core cables in conduit in a thermally insulated wall
A1
2
Multi-core cables in conduit in a thermally insulated wall
A2
4
Insulated conductors or single-core cables in conduit on a wooden, or masonry wall or spaced less than 0,3 x conduit diameter from it
B1
5
Multi-core cable in conduit on a wooden, or mansonry wall or spaced less than 0,3 x conduit diameter from it
B2
20
Single-core or multi-core cables: fixed on, or sapced less than 0.3 x cable diameter from a wooden wall
C
30
Single-core or multi-core cables:
Room
C
On unperforated tray run horizontally or vertically
31
Single-core or multi-core cables:
E or F
On perforated tray run horizontally or vertically
36
Bare or insulated conductors on insulators
G
70
Multi-core cables in conduit or in cable ducting in the ground
D1
71
Single-core cable in conduit or in cable ducting in the ground
D1
Fig. G10: Examples of methods of installation (part of table A.52.3 of IEC 60364-5-52)
Maximum operating temperature: The current-carrying capacities given in the subsequent tables have been determined so that the maximum insulation temperature is not exceeded for sustained periods of time. For different type of insulation material, the maximum admissible temperature is given in Figure G11. Type of insulation
Temperature limit °C
Polyvinyl-chloride (PVC)
70 at the conductor
Cross-linked polyethylene (XLPE) and ethylene propylene rubber
90 at the conductor
(EPR)
Mineral (PVC covered or bare exposed to touch)
70 at the sheath
Mineral (bare not exposed to touch and not in contact with combustible material)
105 at the seath
Fig. G11: Maximum operating temperatures for types of insulation (table 52.1 of IEC 60364-5-52)
Correction factors In order to take environment or special conditions of installation into account, correction factors have been introduced. The cross sectional area of cables is determined using the rated load current IB divided by different correction factors, k1, k2, ...:
I’B is the corrected load current, to be compared to the current-carrying capacity of the considered cable. Ambient temperature The current-carrying capacities of cables in the air are based on an average air temperature equal to 30 °C. For other temperatures, the correction factor is given in FigureG12 for PVC, EPR and XLPE insulation material. The related correction factor is here noted k1. Ambient temperature °C
Insulation
PVC
XLPE and EPR
10
1.22
1.15
15
1.17
1.12
20
1.12
1.08
25
1.06
1.04
30
1
1
35
0.94
0.96
40
0.87
0.91
45
0.79
0.87
50
0.71
0.82
55
0.61
0.76
60
0.50
0.71
65
-
0.65
70
-
0.58
75
-
0.50
80
-
0.41
Fig. G12: Correction factors for ambient air temperatures other than 30 °C to be applied to the current-carrying capacities for cables in the air (from table B.52.14 of IEC 60364-5-52)
The current-carrying capacities of cables in the ground are based on an average ground temperature equal to 20 °C. For other temperatures, the correction factor is given in Figure G13 for PVC, EPR and XLPE insulation material. The related correction factor is here noted k2.
Ground temperature °C
Insulation
PVC
XLPE and EPR
10
1.10
1.07
15
1.05
1.04
20
1
1
25
0.95
0.96
30
0.89
0.93
35
0.84
0.89
40
0.77
0.85
45
0.71
0.80
50
0.63
0.76
55
0.55
0.71
60
0.45
0.65
65
-
0.60
70
-
0.53
75
-
0.46
80
-
0.38
Fig. G13: Correction factors for ambient ground temperatures other than 20 °C to be applied to the current-carrying capacities for cables in ducts in the ground (from table B.52.15 of IEC 60364-5-52)
Soil thermal resistivity The current-carrying capacities of cables in the ground are based on a ground resistivity equal to 2.5 K•m/W. For other values, the correction factor is given in Figure G14. The related correction factor is here noted k3.
Thermal resistivity, K•m/W
0.5
0.7
1
1.5
2
2.5
3
Correction factor for cables in buried ducts
1.28
1.20
1.18
1.1
1.05
1
0.96
Correction factor for direct buried cables
1.88
1.62
1.5
1.28
1.12
1
0.90
Note 1: The correction factors given have been averaged over the range of conductor sizes and types of installation included in Tables B.52.2 to B.52.5. The overall accuracy of correction factors is within ±5 %. Note 2: The correction factors are applicable to cables drawn into buried ducts; for cables laid direct in the ground the correction factors for thermal resistivities less than 2.5 K•m/W will be higher. Where more precise values are required they may be calculated by methods given in the IEC 60287 series. Note 3: The correction factors are applicable to ducts buried at depths of up to 0.8 m. Note 4: It is assumed that the soil properties are uniform. No allowance had been made for the possibility of moisture migration which can lead to a region of high thermal resistivity around the cable. If partial drying out of the soil is foreseen, the permissible current rating should be derived by the methods specified in the IEC 60287 series. Fig. G14: Correction factors for cables in buried ducts for soil thermal resistivities other than 2.5 K.m/W to be applied to the current-carrying capacities for reference method D (table B.52.16 of IEC 60364-5-52)
Based on experience, a relationship exist between the soil nature and resistivity. Then, empiric values of correction factors k3 are proposed in Figure G15, depending on the nature of soil. Nature of soil
k3
Very wet soil (saturated)
1.21
Wet soil
1.13
Damp soil
1.05
Dry soil
1.00
Very dry soil (sunbaked)
0.86
Fig. G15: Correction factor k3 depending on the nature of soil
Grouping of conductors or cables The current-carrying capacities given in the subsequent tables relate to single circuits consisting of the following numbers of loaded conductors: Two insulated conductors or two single-core cables, or one twin-core cable (applicable to single-phase circuits); Three insulated conductors or three single-core cables, or one three-core cable (applicable to three-phase circuits). Where more insulated conductors or cables are installed in the same group, a group reduction factor (here noted k4) shall be applied. Examples are given in Figures G16 to G18 for different configurations (installation methods, in free air or in the ground). Figure G16 gives the values of correction factor k4 for different configurations of unburied cables or conductors, grouping of more than one circuit or multi-core cables. Arrangement (cables touching)
Number of circuits or multi-core cables
1
2
3
4
5
6
Reference methods
7
8
9
12
16
20
Bunched in air, on a surface, embedded orenclosed
1.00
0.8 0
0.7 0
0.6 5
0.6 0
0.5 7
0.54
0.5 2
0.5 0
0.4 5
0.41
0.38
Single layer on wall, floor or unperforated tray
1.00
0.8 5
0.7 9
0.7 5
0.7 3
0.7 2
0.72
0.7 1
0.7 0
No further reduction factor for more than nine circuits or multi-core cables
Single layer fixed directly under a wooden ceiling
0.95
0.8 1
0.7 2
0.6 8
0.6 6
0.6 4
0.63
0.6 2
0.6 1
Single layer on a perforated horizontal or vertical tray
1.00
0.8 8
0.8 2
0.7 7
0.7 5
0.7 3
0.73
0.7 2
0.7 2
Single layer on ladder support or cleats etc.
1.00
0.8 7
0.8 2
0.8 0
0.8 0
0.7 9
0.79
0.7 8
0.7 8
Methods A to F
Method C
Methods E and F
Fig. G16: Reduction factors for groups of more than one circuit or of more than one multi-core cable (table B.52.17 of IEC 60364-5-52)
Figure G17 gives the values of correction factor k4 for different configurations of unburied cables or conductors, for groups of more than one circuit of single-core cables in free air. Method of installation
Perforated trays
Number of tray
31
Number of three-phase circuits
1
2
3
1
0.98
0.91
0.87
2
0.96
0.87
0.81
3
0.95
0.85
0.78
Use as a multiplier to rating for
Three cables in horizontal formation
Vertical perforated trays
Ladder supports, cleats, etc... formation
Perforated trays
Vertical perforated trays
Ladder supports, cleats, etc...
31
1
0.96
0.86
2
0.95
0.84
32
1
1.00
0.97
0.96
33
2
0.98
0.93
0.89
34
3
0.97
0.90
0.86
31
1
1.00
0.98
0.96
2
0.97
0.93
0.89
3
0.96
0.92
0.86
1
1.00
0.91
0.89
2
1.00
0.90
0.86
32
1
1.00
1.00
1.00
33
2
0.97
0.95
0.93
34
3
0.96
0.94
0.90
31
Three cables in vertical formation
Three cables in horizontal formation
Three cables in trefoil formation
Fig. G17: Reduction factors for groups of more than one circuit of single-core cables to be applied to reference rating for one circuit of single-core cables in free air - Method of installation F. (table B.52.21 of IEC 60364-5-52)
Figure G18 gives the values of correction factor k4 for different configurations of cables or conductors laid directly in the ground. Number of circuits
Cable to cable clearance(a)
Nil (cables touching)
One cable diameter
0.125 m
0.25 m
0.5 m
2
0.75
0.80
0.85
0.90
0.90
3
0.65
0.70
0.75
0.80
0.85
4
0.60
0.60
0.70
0.75
0.80
5
0.55
0.55
0.65
0.70
0.80
6
0.50
0.55
0.60
0.70
0.80
7
0.45
0.51
0.59
0.67
0.76
8
0.43
0.48
0.57
0.65
0.75
9
0.41
0.46
0.55
0.63
0.74
12
0.36
0.42
0.51
0.59
0.71
16
0.32
0.38
0.47
0.56
0.38
20
0.29
0.35
0.44
0.53
0.66
(a) for Multicore cables
(a) for Singlecore cables
Fig. G18: Reduction factors for more than one circuit, single-core or multi-core cables laid directly in the ground. Installation method D. (table B.52.18 of IEC 60364-5-52)
Harmonic current The current-carrying capacity of three-phase, 4-core or 5-core cables is based on the assumption that only 3 conductors are fully loaded. However, when harmonic currents are circulating, the neutral current can be significant, and even higher than the phase currents. This is due to the fact that the 3rd harmonic currents of the three phases do not cancel each other, and sum up in the neutral conductor. This of course affects the current-carrying capacity of the cable, and a correction factor noted here k5 shall be applied. In addition, if the 3rd harmonic percentage h3 is greater than 33%, the neutral current is greater than the phase current and the cable size selection is based on the neutral current. The heating effect of harmonic currents in the phase conductors has also to be taken into account. The values of k5 depending on the 3rd harmonic content are given in Figure G19. Third harmonic content of phase current %
Correction factor
Size selection is based on phase current
0 - 15
1.0
15 - 33
0.86
Size selection is based on neutral current
33 - 45
0.86
> 45
1.0[a]
[a] If the neutral current is more than 135 % of the phase current and the cable size is selected on the basis of the neutral current then the three phase conductors will not be fully loaded. The reduction in heat generated by the phase conductors offsets the heat generated by the neutral conductor to the extent that it is not necessary to apply any reduction factor to the current carrying capacity for three loaded conductors. Fig. G19: Correction factors for harmonic currents in four-core and five-core cables (table E.52.1 of IEC 60364-5-52)
Admissible current as a function of nominal cross-sectional area of conductors IEC standard 60364-5-52 proposes extensive information in the form of tables giving the admissible currents as a function of cross-sectional area of cables. Many parameters are taken into account, such as the method of installation, type of insulation material, type of conductor material, number of loaded conductors. As an example, Figure G20 gives the current-carrying capacities for different methods of installation of PVC insulation, three loaded copper or almunium conductors, free air or in ground. Nominal cross-sectional area of conductors(mm2)
1
Installation methods of Table B.52.1
A1
A2
B1
B2
C
D1
D2
2
3
4
5
6
7
8
13.5
13
15.5
15
17.5
18
19
Copper
1.5
2.5
18
17.5
21
20
24
24
24
4
24
23
28
27
32
30
33
6
31
29
36
34
41
38
41
10
42
39
50
46
57
50
54
16
56
52
68
62
76
64
70
25
73
68
89
80
96
82
92
35
89
83
110
99
119
98
110
50
108
99
134
118
144
116
130
70
136
125
171
149
184
143
162
95
164
150
207
179
223
169
193
120
188
172
239
206
259
192
220
150
216
196
262
225
299
217
246
185
245
223
296
255
341
243
278
240
286
261
346
297
403
280
320
300
328
298
394
339
464
316
359
Aluminium
2.5
14
13.5
16.5
15.5
18.5
18.5
4
18.5
17.5
22
21
25
24
6
24
23
28
27
32
30
10
32
31
39
36
44
39
16
43
41
53
48
59
50
53
25
57
53
70
62
73
64
69
35
70
65
86
77
90
77
83
50
84
78
104
92
110
91
99
70
107
98
133
116
140
112
122
95
129
118
161
139
170
132
148
120
149
135
186
160
197
150
169
150
170
155
204
176
227
169
189
185
194
176
230
199
259
190
214
240
227
207
269
232
305
218
250
300
261
237
306
265
351
247
Note: In columns 3, 5, 6, 7 and 8, circular conductors are assumed for sizes up to and including 16 mm2. Values for larger sizes relate to shaped conductors and may safely be applied to circular conductors. Fig. G20: Current-carrying capacities in amperes for different methods of installation, PVC insulation, three loaded conductors, copper or aluminium, conductor temperature: 70 °C, ambient temperature: 30 °C in air, 20 °C in ground (table B.52.4 of IEC 60364-5-52).
3-phase short-circuit current (Isc) at any point within a LV installation. In a 3-phase installation Isc at any point is given by:
where U20 = phase-to-phase voltage of the open circuited secondary windings of the power supply transformer(s). ZT = total impedance per phase of the installation upstream of the fault location (in Ω)
Method of calculating ZT Each component of an installation (MV network, transformer, cable, busbar, and so on...) is characterized by its impedance Z, comprising an element of resistance (R) and an inductive reactance (X). It may be noted that capacitive reactances are not important in short-circuit current calculations. The parameters R, X and Z are expressed in ohms, and are related by the sides of a right angled triangle, as shown in the impedance diagram of Figure G33.
282
Fig. G33: Impedance diagram The method consists in dividing the network into convenient sections, and to calculate the R and X values for each. Where sections are connected in series in the network, all the resistive elements in the section are added arithmetically; likewise for the reactances, to give RT and XT. The impedance (ZT) for the combined sections concerned is then calculated from Any two sections of the network which are connected in parallel, can, if predominantly both resistive (or both inductive) be combined to give a single equivalent resistance (or reactance) as follows: Let R1 and R2 be the two resistances connected in parallel, then the equivalent resistance R3 will be given by:
or for reactances It should be noted that the calculation of X3 concerns only separated circuit without mutual inductance. If the circuits in parallel are close togother the value of X3 will be notably higher.
Determination of the impedance of each component Network upstream of the MV/LV transformer (see Fig. G34) The 3-phase short-circuit fault level PSC, in kA or in MVA[1] is given by the power supply authority concerned, from which an equivalent impedance can be deduced. Psc
Uo (V)
Ra (mΩ)
Xa (mΩ)
250 MVA
420
0.07
0.7
500 MVA
420
0.035
0.351
Fig. G34: The impedance of the MV network referred to the LV side of the MV/LV transformer
A formula which makes this deduction and at the same time converts the impedance to an equivalent value at LV is given, as follows:
where Zs = impedance of the MV voltage network, expressed in milli-ohms Uo = phase-to-phase no-load LV voltage, expressed in volts Psc = MV 3-phase short-circuit fault level, expressed in kVA The upstream (MV) resistance Ra is generally found to be negligible compared with the corresponding Xa, the latter then being taken as the ohmic value for Za. If more accurate calculations are necessary, Xa may be taken to be equal to 0.995 Za and Ra equal to 0.1 Xa. Figure G36 gives values for Ra and Xa corresponding to the most common MV[2] short-circuit levels in utility power-supply networks, namely, 250 MVA and 500 MVA. Transformers (see Fig. G35) The impedance Ztr of a transformer, viewed from the LV terminals, is given by the formula:
where: U20 = open-circuit secondary phase-to-phase voltage expressed in volts Sn = rating of the transformer (in VA) Usc = the short-circuit impedance voltage of the transformer expressed in % The transformer windings resistance Rtr can be derived from the total load-losses as follows:
so that
in milli-ohms
where Pcu = total load-losses in watts In = nominal full-load current in amps Rtr = resistance of one phase of the transformer in milli-ohms (the LV and corresponding MV winding for one LV phase are included in this resistance value).
Note: for an approximate calculation, in the absence of more precise information on transformer characteristics, Cenelec 50480 suggests to use the following guidelines:
if U20 is not known, it may be assumed to be 1.05 Un
in the absence of more precise information, the following values may be used: Rtr = 0.31 Ztr and Xtr = 0.95 Ztr Example: for a transformer of 630kVA with Usc=4% / Un = 400V, approximate calculation gives:
U20 = 400 x 1.05 = 420V
Ztr = 4202 / 630000 x 4% = 11 mΩ
Rtr = 0.31 x Ztr = 3.5 mΩ and Xtr = 0.95 x Ztr = 10.6 mΩ Rated Power kVA)
Oil-immersed
Cast-resin
Usc (%)
Rtr (mΩ)
Xtr (mΩ)
Ztr (mΩ)
Usc (%)
Rtr (mΩ)
Xtr (mΩ)
Ztr (mΩ)
100
4
37.9
59.5
70.6
6
37.0
99.1
105.8
160
4
16.2
41.0
44.1
6
18.6
63.5
66.2
200
4
11.9
33.2
35.3
6
14.1
51.0
52.9
250
4
9.2
26.7
28.2
6
10.7
41.0
42.3
315
4
6.2
21.5
22.4
6
8.0
32.6
33.6
400
4
5.1
16.9
17.6
6
6.1
25.8
26.5
500
4
3.8
13.6
14.1
6
4.6
20.7
21.2
630
4
2.9
10.8
11.2
6
3.5
16.4
16.8
800
6
2.9
12.9
13.2
6
2.6
13.0
13.2
1,000
6
2.3
10.3
10.6
6
1.9
10.4
10.6
1,250
6
1.8
8.3
8.5
6
1.5
8.3
8.5
1,600
6
1.4
6.5
6.6
6
1.1
6.5
6.6
2,000
6
1.1
5.2
5.3
6
0.9
5.2
5.3
Fig. G35: Resistance, reactance and impedance values for typical distribution 400 V transformers (no-load voltage = 420 V) with MV windings ≤ 20 kV
Busbars The resistance of busbars is generally negligible, so that the impedance is practically all reactive, and amounts to approximately 0.15 mΩ/metre[3] length for LV busbars (doubling the spacing between the bars increases the reactance by about 10% only). In practice, it's almost never possible to estimate the busbar length concerned by a short-circuit downstream a switchboard. Circuit conductors The resistance of a conductor is given by the formula:
where ρ = the resistivity of the conductor material at the normal operating temperature ρ has to be considered:
at cold state (20°C) to determine maximum short-circuit current,
at steady state (normal operating temperature) to determine minimum short-circuit current. L = length of the conductor in m S = c.s.a. of conductor in mm2 20 °C
PR/XLPE 90 °C
PVC 70 °C
Copper
18.51
23.69
22.21
Alu
29.41
37.65
35.29
Fig. G35b: Values of ρ as a function of the temperature, cable insulation and cable core material, according to IEC60909-0 and Cenelec TR 50480 (in mΩ mm2/m).
Cable reactance values can be obtained from the manufacturers. For c.s.a. of less than 50 mm2 reactance may be ignored. In the absence of other information, a value of 0.08 mΩ/metre may be used (for 50 Hz systems) or 0.096 mΩ/metre (for 60 Hz systems). For busways (busbar trunking systems) and similar pre-wired ducting systems, the manufacturer should be consulted. Motors At the instant of short-circuit, a running motor will act (for a brief period) as a generator, and feed current into the fault. In general, this fault-current contribution may be ignored. However, if the total power of motors running simultaneously is higher than 25% of the total power of transformers, the influence of motors must be taken into account. Their total contribution can be estimated from the formula: Iscm = 3.5 In from each motor i.e. 3.5m In for m similar motors operating concurrently. The motors concerned will be the 3-phase motors only; single-phase-motor contribution being insignificant. Fault-arc resistance Short-circuit faults generally form an arc which has the properties of a resistance. The resistance is not stable and its average value is low, but at low voltage this resistance is sufficient to reduce the fault-current to some extent. Experience has shown that a reduction of the order of 20% may be expected. This phenomenon will effectively ease the current-breaking duty of a CB, but affords no relief for its fault-current making duty. Recapitulation table (see Fig. G36) Parts of power-supply system
R (mΩ)
X (mΩ) Xa = 0.995 Za
Supply network Figure G34 Transformer with Figure G35
where
Rtr is often negligible compared to Xtr for transformers > 100 kVA Circuit-breaker
Not considered in practice
Busbars
Negligible for S > 200 mm2 in the
XB = 0.15 mΩ/m
[a]
formula: Circuit conductors[b]
Cables: Xc = 0.08 mΩ/m [a]
Motors
See 3-phase short-circuit current (Isc) at any point within a LV installation Motors (often negligible at LV)
Three-phase maximum circuit current in kA [a] ρ = resistivity at 20°C [b] If there are several conductors in parallel per phase, then divide the resistance of one conductor by the number of conductors. The reactance remains practically unchanged. U20: Phase-to-phase no-load secondary voltage of MV/LV transformer (in volts). Psc: 3-phase short-circuit power at MV terminals of the MV/LV transformers (in kVA). Pcu: 3-phase total losses of the MV/LV transformer (in watts). Sn: Rating of the MV/LV transformer (in kVA). Usc: Short-circuit impedance voltage of the MV/LV transfomer (in %). RT : Total resistance. XT: Total reactanc Fig. G36: Recapitulation table of impedances for different parts of a power-supply system
Example of short-circuit calculations (see Fig. G37) LV installation
R (mΩ)
X (mΩ)
RT (mΩ)
XT (mΩ)
0.035
0.351
2.35
8.5
MV network Psc = 500 MVA Transformer 20 kV / 420 V Pn = 1000 kVA Usc = 5% Pcu = 13.3 x 103 watts Single-core cables 5 m copper 4 x 240 mm2/phase
Xc = 0.08 x 5 = 0.40
Main circuitbreaker
Not considered in practice
Busbars 10 m
Not considered in practice
2.48
9.25
Isc1 = 25 kA
Three-core cable 100 m 95 mm2copper
Xc = 100 x 0.08 = 8
22
17.3
Isc3 = 8.7 kA
Three-core cable 20 m 10 mm2copper final circuits
Xc = 20 x 0.08 = 1.6
59
18.9
Isc4 = 3.9 kA
RT : Total resistance. XT: Total reactance. Isc : 3-phase maximum short-circuit current Calculations made as described in figure G36 Fig. G37: Example of maximum short-circuit current calculations for a LV installation supplied at 400 V (nominal) from a 1000 kVA MV/LV transformer
Notes 1. ^ Short-circuit MVA:
EL Isc where:
EL = phase-to-phase nominal system voltage expressed in kV (r.m.s.)
Isc = 3-phase short-circuit current expressed in kA (r.m.s.)
2. ^ up to 36 kV 3. ^ For 50 Hz systems, but 0.18 mΩ/m length at 60 Hz.
Isc at the receiving end of a feeder as a function of the Isc at its sending end. The network shown in Figure G38 typifies a case for the application of Figure G39 , derived by the «method of composition» (mentioned in chapter Protection against electric shocks and electric fires ). These tables give a rapid and sufficiently accurate value of short-circuit current at a point in a network, knowing:
The value of short-circuit current upstream of the point considered
The length and composition of the circuit between the point at which the short-circuit current level is known, and the point at which the level is to be determined It is then sufficient to select a circuit-breaker with an appropriate short-circuit fault rating immediately above that indicated in the tables. If more precise values are required, it is possible to make a detailed calculation or to use a software package, such as Ecodial. In such a case, moreover, the possibility of using the cascading technique should be considered, in which the use of a current limiting circuit-breaker at the upstream position would allow all circuit-breakers downstream of the limiter to have a short-circuit current rating much lower than would otherwise be necessary (See chapter LV switchgear: functions and selection ).
Method Select the c.s.a. of the conductor in the column for copper conductors (in this example the c.s.a. is 47.5 mm2). Search along the row corresponding to 47.5 mm2 for the length of conductor equal to that of the circuit concerned (or the nearest possible on the low side). Descend vertically the column in which the length is located, and stop at a row in the middle section (of the 3 sections of the Figure) corresponding to the known fault-current level (or the nearest to it on the high side). In this case 30 kA is the nearest to 28 kA on the high side. The value of short-circuit current at the downstream end of the 20 metre circuit is given at the intersection of the vertical column in which the length is located, and the horizontal row corresponding to the upstream Isc (or nearest to it on the high side). This value in the example is seen to be 14.7 kA. The procedure for aluminium conductors is similar, but the vertical column must be ascended into the middle section of the table.
In consequence, a DIN-rail-mounted circuit-breaker rated at 63 A and Isc of 25 kA (such as a NG 125N unit) can be used for the 55 A circuit in Figure G38. A Compact rated at 160 A with an Isc capacity of 25 kA (such as a NS160 unit) can be used to protect the 160 A circuit.
Fig. G38: Determination of downstream short-circuit current level Isc using Figure G39 Copper 230 V / 400 V
c.s.a.of phase conductor s (mm2)
Length of circuit (in metres)
1.5
2.5
4
1.2
1.3
1.8
2.6
3. 6
5.2
7.3
1 0 3
1.1
1.5
2.1
3.0
4.3
6. 1
8.6
12. 1
1 7 2
1.7
2.4
3.4
4.9
6.9
9. 7
13. 7
19. 4
2 7
6
10
16
1.8
2.6
3.6
5.2
7.3
10. 3
1 4. 6
21
29
4 1
2. 2
3.0
4.3
6.1
8.6
12.2
17. 2
2 4
34
49
6 9
1. 7
2 . 4
3. 4
4.9
6.9
9.7
13. 8
19.4
27
3 9
55
78
1 1 0
25
1 . 3
1. 9
2. 7
3 . 8
5. 4
7.6
10. 8
15. 2
21
30
43
6 1
86
121
1 7 2
35
1 . 9
2. 7
3. 8
5 . 3
7. 5
10. 6
15. 1
21
30
43
60
8 5
120
170
2 4 0
3 2 6
47.5
1 . 8
2 . 6
3. 6
5. 1
7 . 2
10 .2
14. 4
20
29
41
58
82
1 1 5
163
231
70
2 . 7
3 . 8
5. 3
7. 5
1 0 . 7
15 .1
21
30
43
60
85
120
1 7 0
240
340
2. 6
3 . 6
5 . 1
7. 2
10 .2
1 4 . 5
20
29
41
58
82
115
163
2 3 1
326
461
95
120
150
1.2
1. 6
2. 3
3. 2
4 . 6
6 . 5
9. 1
12 .9
1 8 . 3
26
37
52
73
103
146
206
2 9 1
412
1. 8
2. 5
3. 5
5 . 0
7 . 0
9. 9
14 .0
1 9 . 8
28
40
56
79
112
159
224
3 1 7
448
185
1.5
2. 1
2. 9
4. 2
5 . 9
8 . 3
11 .7
16 .6
2 3
33
47
66
94
133
187
265
3 7 4
529
240
1.8
2. 6
3. 7
5. 2
7 . 3
1 0 . 3
4. 6
21
2 9
41
58
83
117
165
233
330
4 6 6
659
300
2.2
3. 1
4. 4
6. 2
8 . 8
1 2 . 4
17 .6
25
3 5
50
70
99
140
198
280
396
5 6 1
2x120
2.3
3. 2
4. 6
6. 5
9 . 1
1 2 . 9
18 .3
26
3 7
52
73
103
146
206
292
412
5 8 3
2x150
2.5
3. 5
5. 0
7. 0
9 . 9
1 4 . 0
20
28
4 0
56
79
112
159
224
317
448
6 3 4
2x185
2.9
4. 2
5. 9
8. 3
1 1 . 7
1 6 . 6
23
33
4 7
66
94
133
187
265
375
530
7 4 9
3x120
3.4
4. 9
6. 9
9. 7
1 3 . 7
1 9 . 4
27
39
5 5
77
110
155
219
309
438
619
3x150
3.7
5. 3
7. 5
10 .5
1 4 . 9
2 1
30
42
6 0
84
119
168
238
336
476
672
3x185
4.4
6. 2
8. 8
12 .5
1 7 . 6
2 5
35
50
7 0
10 0
141
199
281
398
562
Isc upstream (in kA)
Isc downstream (in kA)
100
93
9 0
87
82
7 7
7 0
62
54
4 5
37
29
22
17. 0
12. 6
9.3
6.7
4. 9
3.5
2.5
1 8
90
84
8 2
79
75
7 1
6 5
58
51
4 3
35
28
22
16. 7
12. 5
9.2
6.7
4. 8
3.5.
2.5
1 8
80
75
7 4
71
68
6 4
5 9
54
47
4 0
34
27
21
16. 3
12. 2
9.1
6.6
4. 8
3.5
2.5
1 8
70
66
6 5
63
61
5 8
5 4
49
44
3 8
32
26
20
15. 8
12. 0
8.9
6.6
4. 8
3.4
2.5
1 8
60
57
5 6
55
53
5 1
4 8
44
39
3 5
29
24
20
15. 2
11. 6
8.7
6.5
4. 7
3.4
2.5
1 8
50
48
4 7
46
45
4 3
4 1
38
35
3 1
27
22
18. 3
14. 5
11. 2
8.5
6.3
4. 6
3.4
2.4
1 7
40
39
3 8
38
37
3 6
3 4
32
30
2 7
24
20
16. 8
13. 5
10. 6
8.1
6.1
4. 5
3.3
2.4
1 7
35
34
3 4
33
33
3 2
3 0
29
27
2 4
22
18. 8
15. 8
12. 9
10. 2
7.9
6.0
4. 5
3.3
2.4
1 7
30
29
2 9
29
28
2 7
2 7
25
24
2 2
20
17. 3
14. 7
12. 2
9.8
7.6
5.8
4. 4
3.2
2.4
1 7
25
25
2 4
24
24
2 3
2 3
22
21
1 9 . 1
17 .4
15. 5
13. 4
11. 2
9.2
7.3
5.6
4. 2
3.2
2.3
1 7
20
20
2 0
19 .4
19 .2
1 8 . 8
1 8 . 4
17 .8
17 .0
1 6 . 1
14 .9
13. 4
11. 8
10. 1
8.4
6.8
5.3
4. 1
3.1
2.3
1 7
15
14. 8
4 1. 8
14 .7
14 .5
1 4 . 3
1 4 . 1
13 .7
3. 3
1 2 . 7
11 .9
11. 0
9.9
8.7
7.4
6.1
4.9
3. 8
2.9
2.2
1 6
10
9.9
9. 9
9. 8
9. 8
9 . 7
9 . 6
9. 4
9. 2
8 . 9
8. 5
8.0
7.4
6.7
5.9
5.1
4.2
3. 4
2.7
2.0
1 5
7
7.0
6. 9
6. 9
6. 9
6 . 9
6 . 8
6. 7
6. 6
6 . 4
6. 2
6.0
5.6
5.2
4.7
4.2
3.6
3. 0
2.4
1.9
1 4
5
5.0
5. 0
5. 0
4. 9
4 . 9
4 . 9
4. 9
4. 8
4 . 7
4. 6
4.5
4.3
4.0
3.7
3.4
3.0
2. 5
2.1
1.7
1 3
4
4.0
4. 0
4. 0
4. 0
4 . 0
3 . 9
3. 9
3. 9
3 . 8
3. 7
3.6
3.5
3.3
3.1
2.9
2.6
2. 2
1.9
1.6
1 2
3
3.0
3. 0
3. 0
3. 0
3 . 0
3 . 0
2. 9
2. 9
2 . 9
2. 9
2.8
2.7
2.6
2.5
2.3
2.1
1. 9
1.6
1.4
1 1
2
2.0
2. 0
2. 0
2. 0
2 . 0
2 . 0
2. 0
2. 0
1 2 . 0
1. 9
1.9
1.9
1.8
1.8
1.7
1.6
1. 4
1.3
1.1
1 0
1
1.0
1. 0
1. 0
1. 0
1 . 0
1 . 0
1. 0
1. 0
1 . 0
1. 0
1.0
1.0
1.0
0.9
0.9
0.9
0. 8
0.8
0.7
0 6
Note: for a 3-phase system having 230 V between phases, divide the above lengths by Fig. G39: Isc at a point downstream, as a function of a known upstream fault-current value and the length and c.s.a. of the intervening conductors, in a 230/400 V 3-phase system.
Calculation of minimum levels of short-circuit current. If a protective device in a circuit is intended only to protect against short-circuit faults, it is essential that it will operate with certainty at the lowest possible level of short-circuit current that can occur on the circuit In general, on LV circuits, a single protective device protects against all levels of current, from the overload threshold through the maximum rated short-circuit current breaking capability of the device. The protection device should be able to operate in a maximum time to ensure people and circuit safety, for all short-circuit current or fault current that may occur. To check that behavior, calculation of minimal short-circuit current or fault current is mandatory. In addition, in certain cases overload protective devices and separate short-circuit protective devices are used.
Examples of such arrangements Figures G40 to G42 show some common arrangements where overload and short-circuit protections are achieved by separate devices.
Fig. G40: Circuit protected by aM fuses
Fig. G41: Circuit protected by circuit breaker without thermal overload relay
Fig. G42a: Circuit breaker D provides protection against short-circuit faults as far as and including the load As shown in Figures G40 and G41, the most common circuits using separate devices control and protect motors. Figure G42a constitutes a derogation in the basic protection rules, and is generally used on circuits of prefabricated bustrunking, lighting rails, etc. Variable speed drive Figure G42b shows the functions provided by the variable speed drive, and if necessary some additional functions provided by devices such as circuit-breaker, thermal relay, RCD.
Protection to be provided
Protection generally provided by the variable speed drive
Additional protection if not provided by the variable speed drive
Cable overload
Yes
CB / Thermal relay
Motor overload
Yes
CB / Thermal relay
Downstream shortcircuit
Yes
Variable speed drive overload
Yes
Overvoltage
Yes
Undervoltage
Yes
Loss of phase
Yes
Upstream short-circuit
Circuit-breaker (short-circuit tripping)
Internal fault
Circuit-breaker (short-circuit and overload tripping)
Downstream earth fault (indirect contact)
(self protection)
Direct contact fault
RCD ≥ 300 mA or CB in TN earthing system RCD ≤ 30 mA
Fig. G42b: Protection to be provided for variable speeed drive applications
Conditions to be fulfilled The protective device must fulfill:
instantaneous trip setting Im < Iscmin for a circuit-breaker
fusion current Ia < Iscmin for a fuse
The protective device must therefore satisfy the two following conditions: Its breaking capacity must be greater than Isc, the 3-phase short-circuit current at its point of
installation Elimination of the minimum short-circuit current possible in the circuit, in a time tc compatible
with the thermal constraints of the circuit conductors, where:
(valid for tc < 5 seconds) where S is the cross section area of the cable, k is a factor depending of the cable conductor material, the insulation material and initial temperature. Example: for copper XLPE, initial temperature 90 °C, k = 143 (see IEC60364-4-43 §434.3.2 table 43A). Comparison of the tripping or fusing performance curve of protective devices, with the limit curves of thermal constraint for a conductor shows that this condition is satisfied if:
Isc (min) > Im (instantaneous or short timedelay circuit-breaker trip setting current level), (see Fig. G43 )
Isc (min) > Ia for protection by fuses. The value of the current Ia corresponds to the crossing point of the fuse curve and the cable thermal withstand curve (see Fig. G44 and Fig. G45)
Fig. G43: Protection by circuit breaker
Fig. G44: Protection by aM-type fuses
Fig. G45: Protection by gG-type fuses
Practical method of calculating Lmax In practice this means that the length of circuit downstream of the protective device must not exceed a calculated maximum length: The limiting effect of the impedance of long circuit conductors on the value of short-circuit currents must be checked and the length of a circuit must be restricted accordingly. The method of calculating the maximum permitted length has already been demonstrated in TN- and IT- earthed schemes for single and double earth faults, respectively. Two cases are considered below: 1 - Calculation of Lmax for a 3-phase 3-wire circuit
The minimum short-circuit current will occur when two phase wires are short-circuited at the remote end of the circuit (see Fig. G46).
Fig. G46: Definition of L for a 3-phase 3-wire circuit Using the “conventional method”, the voltage at the point of protection P is assumed to be 80% of the nominal voltage during a short-circuit fault, so that 0.8 U = Isc Zd, where: Zd = impedance of the fault loop Isc = short-circuit current (ph/ph) U = phase-to-phase nominal voltage
For cables ≤ 120 mm2, reactance may be neglected, so that
[1]
where: ρ = resistivity of conductor material at the average temperature during a short-circuit, Sph = c.s.a. of a phase conductor in mm2 L = length in metres The condition for the cable protection is Im ≤ Isc with Im = magnetic trip current setting of the CB.
This leads to
which gives
with U = 400 V ρ = 0.023 Ω.mm2/m[2] (Cu) therefore
with Lmax = maximum circuit length in metres In general, the value of Im is given with +/- 20% tolerance, so Lmax should be calculated for Im+20% (worst case).
k factor values are provided in the following table, taking into account these 20%, and as a function of cross-section for Sph > 120 mm2[1] Cross-section (mm2)
≤ 120
150
185
240
300
k (for 400 V)
5800
5040
4830
4640
4460
2 - Calculation of Lmax for a 3-phase 4-wire 230/400 V circuit The minimum Isc will occur when the short-circuit is between a phase conductor and the neutral at the end of the circuit. A calculation similar to that of example 1 above is required, but for a single-phase fault (230V).
If Sn (neutral cross-section) = Sph Lmax = k Sph / Im with k calculated for 230V, as shown in the table below
Cross-section (mm2)
≤ 120
150
185
240
300
k (for 400 V)
3333
2898
2777
2668
2565
If Sn (neutral cross-section) < Sph, then (for cable cross-section ≤ 120mm2)
Tabulated values for Lmax Figure G47 below gives maximum circuit lengths (Lmax) in metres, for:
3-phase 4-wire 400 V circuits (i.e. with neutral) and
1-phase 2-wire 230 V circuits protected by general-purpose circuit-breakers. In other cases, apply correction factors (given in Figure G51) to the lengths obtained. In general, the value of Im is given with +/- 20% tolerance.
Lmax values below are therefore calculated for Im+20% (worst case). For the 50 mm2 c.s.a., calculation are based on a 47.5 mm2 real c.s.a. Operating current level Im of the instantaneous magnetic tripping element (in A)
c.s.a. (nominal cross-sectional-area) of conductors (in mm2)
1.5
2.5
4
6
10
16
25
35
50
100
167
26 7
400
63
79
133
21 2
317
80
63
104
16 7
250
41 7
100
50
83
13 3
200
33 3
125
40
67
10 7
160
26 7
427
160
31
52
83
125
20 8
333
200
25
42
67
100
16 7
267
417
250
20
33
53
80
13 3
213
333
46 7
320
16
26
42
63
10 4
167
260
36 5
50
495
70
95
1
400
13
21
33
50
83
133
208
29 2
396
500
10
17
27
40
67
107
167
23 3
317
560
9
15
24
36
60
95
149
20 8
283
41 7
630
8
13
21
32
53
85
132
18 5
251
37 0
700
7
12
19
29
48
76
119
16 7
226
33 3
452
800
6
10
17
25
42
67
104
14 6
198
29 2
396
875
6
10
15
23
38
61
95
13 3
181
26 7
362
4
1000
5
8
13
20
33
53
83
11 7
158
23 3
317
4
1120
4
7
12
18
30
48
74
10 4
141
20 8
283
3
1250
4
7
11
16
27
43
67
93
127
18 7
253
3
1600
5
8
13
21
33
52
73
99
14 6
198
2
2000
4
7
10
17
27
42
58
79
11 7
158
2
5
8
13
21
33
47
63
93
127
1
2500
3200
4
6
10
17
26
36
49
73
99
1
4000
5
8
13
21
29
40
58
79
1
5000
4
7
11
17
23
32
47
63
8
6300
5
8
13
19
25
37
50
6
8000
4
7
10
15
20
29
40
5
10000
5
8
12
16
23
32
4
12500
4
7
9
13
19
25
3
Fig. G47: Maximum circuit lengths in metres for copper conductors (for aluminium, the lengths must be multiplied by 0.62)
Figures G48 to G50 give maximum circuit length (Lmax) in metres for:
3-phase 4-wire 400 V circuits (i.e. with neutral) and
1-phase 2-wire 230 V circuits protected in both cases by domestic-type circuit-breakers or with circuit-breakers having similar tripping/current characteristics. In other cases, apply correction factors to the lengths indicated. These factors are given in Figure G51. Circuit-breaker rating (A)
c.s.a. (nominal cross-sectional-area) of conductors (in mm2)
1.5
2.5
4
6
10
16
25
6
200
333
533
800
10
120
200
320
480
800
16
75
125
200
300
500
800
20
60
100
160
240
400
640
25
48
80
128
192
320
512
800
32
37
62
100
150
250
400
625
40
30
50
80
120
200
320
500
50
24
40
64
96
160
256
400
63
19
32
51
76
127
203
317
80
15
25
40
60
100
160
250
100
12
20
32
48
80
128
200
125
10
16
26
38
64
102
160
Fig. G48: Maximum length of copper-conductor circuits in metres protected by B-type circuit-breakers
Circuit-breaker rating (A)
6
c.s.a. (nominal cross-sectional-area) of conductors (in mm2)
1.5
2.5
4
6
10
100
167
267
400
667
16
25
10
60
100
160
240
400
640
16
37
62
100
150
250
400
625
20
30
50
80
120
200
320
500
25
24
40
64
96
160
256
400
32
18.0
31
50
75
125
200
313
40
15.0
25
40
60
100
160
250
50
12.0
20
32
48
80
128
200
63
9.5
16.0
26
38
64
102
159
80
7.5
12.5
20
30
50
80
125
100
6.0
10.0
16.0
24
40
64
100
125
5.0
8.0
13.0
19.0
32
51
80
Fig. G49: Maximum length of copper-conductor circuits in metres protected by C-type circuit-breakers
Circuit-breaker rating (A)
c.s.a. (nominal cross-sectional-area) of conductors (in mm2)
1.5
2.5
1
429
714
2
214
357
4
6
571
857
10
16
25
3
143
238
381
571
952
4
107
179
286
429
714
6
71
119
190
286
476
762
10
43
71
114
171
286
457
714
16
27
45
71
107
179
286
446
20
21
36
57
86
143
229
357
25
17.0
29
46
69
114
183
286
32
13.0
22
36
54
89
143
223
40
11.0
18.0
29
43
71
114
179
50
9.0
14.0
23
34
57
91
143
63
7.0
11.0
18.0
27
45
73
113
80
5.0
9.0
14.0
21
36
57
89
100
4.0
7.0
11.0
17.0
29
46
71
125
3.0
6.0
9.0
14.0
23
37
57
Fig. G50: Maximum length of copper-conductor circuits in metres protected by D-type circuit-breakers
Circuit detail
3-phase 3-wire 400 V circuit or 1-phase 2-wire 400 V circuit (no neutral)
1-phase 2-wire (phase and neutral) 230 V circuit
3-phase 4-wire 230/400 V circuit or 2-phase 3-wire 230/400 V circuit (i.e with neutral)
Sph / S neutra
Sph / S neutra
Fig. G51: Correction factor to apply to lengths obtained from Figures G47 to G50
Note: IEC 60898 accepts an upper short-circuit-current tripping range of 10-50 In for type D circuitbreakers. European standards, and Figure G50 however, are based on a range of 10-20 In, a range which covers the vast majority of domestic and similar installations.
Examples Example 1 In a 3-phase 3-wire 400 V installation the protection is provided by a 50 A circuit-breaker type NS80HMA, the instantaneous short-circuit current trip, is set at 500 A (accuracy of ± 20%), i.e. in the worst case would require 500 x 1,2 = 600 A to trip. The cable c.s.a. = 10 mm2 and the conductor material is copper. In Figure G47, the row Im = 500 A crosses the column c.s.a. = 10 mm2 at the value for Lmax of 67 m. The circuit-breaker protects the cable against short-circuit faults, therefore, provided that its length does not exceed 67 metres. Example 2 In a 3-phase 3-wire 400 V circuit (without neutral), the protection is provided by a 220 A circuitbreaker type NSX250N with an instantaneous short-circuit current trip unit type MA set at 2,000 A (± 20%), i.e. a worst case of 2,400 A to be certain of tripping. The cable c.s.a. = 120 mm2 and the conductor material is copper. In Figure G47 the row Im = 2,000 A crosses the column c.s.a. = 120 mm2 at the value for Lmax of 200 m. Being a 3-phase 3-wire 400 V circuit (without neutral), a correction factor from Figure G51 must be applied. This factor is seen to be 1.73. The circuit-breaker will therefore protect the cable against short-circuit current, provided that its length does not exceed 200 x 1.73 = 346 metres.
Notes 1. ^ a b For larger c.s.a.’s, the resistance calculated for the conductors must be increased to account for the non-uniform current density in the conductor (due to “skin” and “proximity” effects Suitable values are as follows:
150 mm2: R + 15 %
185 mm2: R + 20 %
240 mm2: R + 25 %
300 mm2: R + 30 %
2. ^ Resistivity for copper EPR/XLPE cables when passing short-circuit current, eg for the max temperature they can withstand = 90°C (cf Figure G35b).
Verification of the withstand capabilities of cables under short-circuit conditions. Thermal constraints When the duration of short-circuit current is brief (several tenths of a second up to five seconds maximum) all of the heat produced is assumed to remain in the conductor, causing its temperature to rise. The heating process is said to be adiabatic, an assumption that simplifies the calculation and gives a pessimistic result, i.e. a higher conductor temperature than that which would actually occur, since in practice, some heat would leave the conductor and pass into the insulation. For a period of 5 seconds or less, the relationship I2t = k2S2 characterizes the time in seconds during which a conductor of c.s.a. S (in mm 2) can be allowed to carry a current I, before its temperature reaches a level which would damage the surrounding insulation. The factor k is given in Figure G52 below. Conductor insulation
PVC ≤ 300 mm2
PVC > 300 mm2
EPR XLPE
Rubber 60 °C
Initial temperature °C
70
70
90
60
Final temperature °C
160
140
250
200
Material of conductor
Copper
115
103
143
141
Aluminium
76
68
94
93
Fig. G52: Value of the constant k according to table 43A of IEC 60364-4-43
The method of verification consists in checking that the thermal energy I2t per ohm of conductor material, allowed to pass by the protecting circuit-breaker (from manufacturers catalogues) is less than that permitted for the particular conductor (as given in Figure G53 below). S (mm2)
PVC
XLPE
Copper
Aluminium
Copper
Aluminium
1.5
0.0297
0.0130
0.0460
0.0199
2.5
0.0826
0.0361
0.1278
0.0552
4
0.2116
0.0924
0.3272
0.1414
6
0.4761
0.2079
0.7362
0.3181
10
1.3225
0.5776
2.0450
0.8836
16
3.3856
1.4786
5.2350
2.2620
25
8.2656
3.6100
12.7806
5.5225
35
16.2006
7.0756
25.0500
10.8241
50[a]
29.839
13.032
46.133
19.936
[a] For 50mm2 cable, the values are calculated for the actual cross-section of 47.5mm2 Fig. G53: Maximum allowable thermal stress for cables I2t (expressed in ampere2 x second x 106)
Example Is a copper-cored XLPE cable of 4 mm2 c.s.a. adequately protected by a iC60N circuit-breaker? (see Fig. G53b) Figure G53 shows that the I2t value for the cable is 0.3272 x 106, while the maximum “let-through” value by the circuit-breaker, as given in the manufacturer’s catalogue, is considerably less ( < 0.1.106 A2s). The cable is therefore adequately protected by the circuit-breaker up to its full rated breaking capability.
Electrodynamic constraints For all type of circuit (conductors or bus-trunking), it is necessary to take electrodynamic effects into account. To withstand the electrodynamic constraints, the conductors must be solidly fixed and the connection must be strongly tightened. For bus-trunking, rails, etc. it is also necessary to verify that the electrodynamic withstand performance is satisfactory when carrying short-circuit currents. The peak value of current, limited by the circuit-breaker or fuse, must be less than the busbar system rating. Tables of coordination ensuring adequate protection of their products are generally published by the manufacturers and provide a major advantage of such systems.
Fig. G53b: Example of energy limitation of a MCB for different ratings.
Connection and choice for protective earthing conductor. Protective (PE) conductors provide the bonding connection between all exposed and extraneous conductive parts of an installation, to create the main equipotential bonding system. These conductors conduct fault current due to insulation failure (between a phase conductor and an exposed conductive part) to the earthed neutral of the source. PE conductors are connected to the main earthing terminal of the installation. The main earthing terminal is connected to the earthing electrode (see Chapter E) by the earthing conductor (grounding electrode conductor in the USA). PE conductors must be:
Insulated and coloured yellow and green (stripes)
Protected against mechanical and chemical damage In IT and TN-earthed schemes it is strongly recommended that PE conductors should be installed in close proximity (i.e. in the same conduits, on the same cable tray, etc.) as the live cables of the
related circuit. This arrangement ensures the minimum possible inductive reactance in the earth-fault current carrying circuits. It should be noted that this arrangement is originally provided by bus-trunking.
Connection PE conductors must:
Not include any means of breaking the continuity of the circuit (such as a switch, removable links, etc.)
Connect exposed conductive parts individually to the main PE conductor, i.e. in parallel, not in series, as shown in Figure G54
Have an individual terminal on common earthing bars in distribution boards.
Fig. G54: A poor connection in a series arrangement will leave all downstream appliances unprotected TT scheme The PE conductor need not necessarily be installed in close proximity to the live conductors of the corresponding circuit, since high values of earth-fault current are not needed to operate the RCDtype of protection used in TT installations. IT and TN schemes
The PE or PEN conductor, as previously noted, must be installed as close as possible to the corresponding live conductors of the circuit and no ferro-magnetic material must be interposed between them. A PEN conductor must always be connected directly to the earth terminal of an appliance, with a looped connection from the earth terminal to the neutral terminal of the appliance (see Fig. G55).
TN-C scheme (the neutral and PE conductor are one and the same, referred to as a PEN conductor) The protective function of a PEN conductor has priority, so that all rules governing PE conductors apply strictly to PEN conductors
TN-C to TN-S transition The PE conductor for the installation is connected to the PEN terminal or bar (see Figure G56) generally at the origin of the installation. Downstream of the point of separation, no PE conductor can be connected to the neutral conductor.
Fig. G55: Direct connection of the PEN conductor to the earth terminal of an appliance
Fig. G56: The TN-C-S scheme
Types of materials Materials of the kinds mentioned below in Figure G57 can be used for PE conductors, provided that the conditions mentioned in the last column are satisfied. Type of protective earthing conductor(PE)
IT scheme
TN scheme
TT scheme
Conditions to be respected
Supplementary conductor
In the same cable as the phases, or in the same cable run
Strongly ecommended
Strongly recommended
Correct
Independent of the phase conductors
Possible[a]
Possible[a][b]
Correct
The PE conductor must be insulated to the same level as the phases
The PE conductor may
be bare or insulated[b] The electrical
[c]
Metallic housing of bus-trunking or of other prefabricated prewired ducting[e]
Possible
External sheath of extruded, mineralinsulated conductors (e.g. «pyrotenax» type systems)
Possible[c]
Certain extraneous conductive elements[f] such as:
Possible[d]
[c]
PE possible PEN possible[h]
Correct
PE possible[c] PEN not recommended[b][c]
Possible
PE possible[d]
Possible
Steel building structure
Machine frames
Water pipes[g] Metallic cable ways, such as, conduits[i]ducts, trunking, trays, ladders, and so on…
by protection against deterioration by mechanical, chemical and electrochemical hazards
PEN forbidden
continuity must be assured
Their conductance
must be adequate
Possible[d]
PE possible[d] PEN not recommended[b][d]
Possible
Forbidden for use as PE conductors, are: metal conduits[i], gas pipes, hot-water pipes, cable-armouring tapes[i]or wires[i] [a] In TN and IT schemes, fault clearance is generally achieved by overcurrent devices (fuses or circuit-breakers) so that the impedance of the fault-current loop must be sufficiently low to assure positive protective device operation. The surest means of achieving a low loop impedance is to use a supplementary core in the same cable as the circuit conductors (or taking the same route as the circuit conductors). This solution minimizes the inductive reactance and therefore the impedance of the loop. [b] The PEN conductor is a neutral conductor that is also used as a protective earth conductor. This means that a current may be flowing through it at any time (in the absence of an earth fault). For this reason an insulated conductor is recommended for PEN operation. [c] The manufacturer provides the necessary values of R and X components of the impedances (phase/PE, phase/PEN) to include in the calculation of the earth-fault loop impedance. [d] Possible, but not recomended, since the impedance of the earth-fault loop cannot be known at the design stage. Measurements on the completed installation are the only practical means of assuring adequate protection for persons. [e] It must allow the connection of other PE conductors. Note: these elements must carry an indivual green/yellow striped visual indication, 15 to 100 mm long (or the letters PE at less than 15 cm from each extremity). [f] These elements must be demountable only if other means have been provided to ensure uninterrupted continuity
of protection. [g] With the agreement of the appropriate water authorities. [h] In the prefabricated pre-wired trunking and similar elements, the metallic housing may be used as a PEN conductor, in parallel with the corresponding bar, or other PE conductor in the housing. [i] Forbidden in some countries only. Universally allowed to be used for supplementary equipotential conductors. Fig. G57: Choice of protective conductors (PE).
Choice of earthing method – implementation. After consulting applicable regulations, Figure E16 and Figure E17 can be used as an aid in deciding on divisions and possible galvanic isolation of appropriate sections of a proposed installation.
Division of source This technique concerns the use of several transformers instead of employing one high-rated unit. In this way, a load that is a source of network disturbances (large motors, furnaces, etc.) can be supplied by its own transformer. The quality and continuity of supply to the whole installation are thereby improved. The cost of switchgear is reduced (short-circuit current level is lower). The cost-effectiveness of separate transformers must be determined on a case by case basis.
Network islands The creation of galvanically-separated “islands” by means of LV/LV transformers makes it possible to optimise the choice of earthing methods to meet specific requirements (see Fig. E18 and Fig. E19).
Fig. E18: TN-S island within an IT system
Fig. E19: IT islands within a TN-S system
Conclusion The optimisation of the performance of the whole installation governs the choice of earthing system. Including:
Initial investments, and
Future operational expenditures, hard to assess, that can arise from insufficient reliability, quality of equipment, safety, continuity of service, etc. An ideal structure would comprise normal power supply sources, local reserve power supply sources (see Selection criteria for the TT, TN and IT systems ) and the appropriate earthing arrangements.
Installation and measurements of earth electrodes. A very effective method of obtaining a low-resistance earth connection is to bury a conductor in the form of a closed loop in the soil at the bottom of the excavation for building foundations.
The resistance R of such an electrode (in homogeneous soil) is given (approximately) in ohms by: where L = length of the buried conductor in metres ρ = soil resistivity in ohm-metres The quality of an earth electrode (resistance as low as possible) depends essentially on two factors:
Installation method
Type of soil
Installation methods Three common types of installation will be discussed: Buried ring (see Fig. E20) This solution is strongly recommended, particularly in the case of a new building. The electrode should be buried around the perimeter of the excavation made for the foundations. It is important that the bare conductor be in intimate contact with the soil (and not placed in the gravel or aggregate hard-core, often forming a base for concrete). At least four (widely-spaced) vertically arranged conductors from the electrode should be provided for the installation connections and, where possible, any reinforcing rods in concrete work should be connected to the electrode. The conductor forming the earth electrode, particularly when it is laid in an excavation for foundations, must be in the earth, at least 50 cm below the hard-core or aggregate base for the concrete foundation. Neither the electrode nor the vertical rising conductors to the ground floor, should ever be in contact with the foundation concrete. For existing buildings, the electrode conductor should be buried around the outside wall of the premises to a depth of at least 1 metre. As a general rule, all vertical connections from an electrode to above-ground level should be insulated for the nominal LV voltage (600-1,000 V). The conductors may be:
Copper: Bare cable ( ≥ 25 mm2) or multiple-strip (≥ 25 mm2) and (≥ 2 mm thick)
Aluminium with lead jacket: Cable ( ≥ 35 mm2)
Galvanised-steel cable: Bare cable ( ≥ 95 mm2) or multiple-strip ( ≥ 100 mm2 and ≥ 3 mm thick) The approximate resistance R of the electrode in ohms: where L = length of conductor in metres ρ = resistivity of the soil in ohm-metres (see Influence of the type of soil )
Fig. E20: Conductor buried below the level of the foundations, i.e. not in the concrete Earthing rods (see Fig. E21)
For n rods: Vertically driven earthing rods are often used for existing buildings, and for improving (i.e. reducing the resistance of) existing earth electrodes. The rods may be: Copper or (more commonly) copper-clad steel. The latter are generally 1 or 2 metres long
and provided with screwed ends and sockets in order to reach considerable depths, if necessary (for instance, the water-table level in areas of high soil resistivity) Galvanised[1] steel pipe ≥ 25 mm diameter or rod ≥ 15 mm diameter, ≥ 2 metres long in each
case.
Fig. E21: Earthing rods connected in parallel
It is often necessary to use more than one rod, in which case the spacing between them should exceed the depth to which they are driven, by a factor of 2 to 3. The total resistance (in homogeneous soil) is then equal to the resistance of one rod, divided by the number of rods in question.
The approximate resistance R obtained is:
if the distance separating the rods > 4L
where L = the length of the rod in metres ρ = resistivity of the soil in ohm-metres (see Influence of the type of soil) n = the number of rods Vertical plates (see Fig. E22)
For a vertical plate electrode: Rectangular plates, each side of which must be ≥ 0.5 metres, are commonly used as earth electrodes, being buried in a vertical plane such that the centre of the plate is at least 1 metre below the surface of the soil. The plates may be:
Copper of 2 mm thickness
Galvanised[1] steel of 3 mm thickness The resistance R in ohms is given (approximately), by: where L = the perimeter of the plate in metres ρ = resistivity of the soil in ohm-metres (see Influence of the type of soil)
Fig. E22: Vertical plate - 2 mm thickness (Cu)
Influence of the type of soil Measurements on earth electrodes in similar soils are useful to determine the resistivity value to be applied for the design of an earth-electrode system Type of soil
Mean value of resistivity in Ωm
Swampy soil, bogs
1 - 30
Silt alluvium
20 - 100
Humus, leaf mould
10 - 150
Peat, turf
5 - 100
Soft clay
50
Marl and compacted clay
100 - 200
Jurassic marl
30 - 40
Clayey sand
50 - 500
Siliceous sand
200 - 300
Stoney ground
1,500 - 3,000
Grass-covered-stoney sub-soil
300 - 500
Chalky soil
100 - 300
Limestone
1,000 - 5,000
Fissured limestone
500 - 1,000
Schist, shale
50 - 300
Mica schist
800
Granite and sandstone
1,500 - 10,000
Modified granite and sandstone
100 - 600
Fig. E23: Resistivity (Ωm) for different types of soil
Type of soil
Average value of resistivity in Ωm
Fertile soil, compacted damp fill
50
Arid soil, gravel, uncompacted non-uniform fill
500
Stoney soil, bare, dry sand, fissured rocks
3000
Fig. E24: Average resistivity (Ωm) values for approximate earth-elect
Measurement and constancy of the resistance between an earth electrode and the earth The resistance of the electrode/earth interface rarely remains constant Among the principal factors affecting this resistance are the following: Humidity of the soil
The seasonal changes in the moisture content of the soil can be significant at depths of up to 2 meters. At a depth of 1 metre the resistivity and therefore the resistance can vary by a ratio of 1 to 3 between a wet winter and a dry summer in temperate regions Frost
Frozen earth can increase the resistivity of the soil by several orders of magnitude. This is one reason for recommending the installation of deep electrodes, in particular in cold climates Ageing
The materials used for electrodes will generally deteriorate to some extent for various reasons, for example: Chemical reactions (in acidic or alkaline soils) Galvanic: due to stray DC currents in the earth, for example from electric railways, etc. or due to dissimilar metals forming primary cells. Different soils acting on sections of the same conductor can also form cathodic and anodic areas with consequent loss of surface metal from the latter areas. Unfortunately, the most favourable conditions for low earthelectrode resistance (i.e. low soil resistivity) are also those in which galvanic currents can most easily flow. Oxidation
Brazed and welded joints and connections are the points most sensitive to oxidation. Thorough cleaning of a newly made joint or connection and wrapping with a suitable greased-tape binding is a commonly used preventive measure.
Measurement of the earth-electrode resistance There must always be one or more removable links to isolate an earth electrode so that it can be tested.
There must always be removable links which allow the earth electrode to be isolated from the installation, so that periodic tests of the earthing resistance can be carried out. To make such tests, two auxiliary electrodes are required, each consisting of a vertically driven rod. Ammeter method (see Fig. E25)
Fig. E25: Measurement of the resistance to earth of the earth electrode of an installation by means of an ammeter
When the source voltage U is constant (adjusted to be the same value for each test) then:
In order to avoid errors due to stray earth currents (galvanic -DC- or leakage currents from power and communication networks and so on) the test current should be AC, but at a different frequency to that of the power system or any of its harmonics. Instruments using hand-driven generators to make these measurements usually produce an AC voltage at a frequency of between 85 Hz and 135 Hz. The distances between the electrodes are not critical and may be in different directions from the electrode being tested, according to site conditions. A number of tests at different spacings and directions are generally made to cross-check the test results.
Use of a direct-reading earthing-resistance ohmmeter These instruments use a hand-driven or electronic-type AC generator, together with two auxiliary electrodes, the spacing of which must be such that the zone of influence of the electrode being tested should not overlap that of the test electrode (C). The test electrode (C) furthest from the electrode (X) under test, passes a current through the earth and the electrode under test, while the second test electrode (P) picks up a voltage. This voltage, measured between (X) and (P), is due to the test current and is a measure of the contact resistance (of the electrode under test) with earth. It is clear that the distance (X) to (P) must be carefully chosen to give accurate results. If the distance (X) to (C) is increased, however, the zones of resistance of electrodes (X) and (C) become more remote, one from the other, and the curve of potential (voltage) becomes more nearly horizontal about the point (O). In practical tests, therefore, the distance (X) to (C) is increased until readings taken with electrode (P) at three different points, i.e. at (P) and at approximately 5 metres on either side of (P), give similar values. The distance (X) to (P) is generally about 0.68 of the distance (X) to (C).
[a] the principle of measurement is based on assumed homogeneous soil conditions. Where the zones of influence of electrodes C and X overlap, the location of test electrode P is difficult to determine for satisfactory results.
[b] showing the effect on the potential gradient when (X) and (C) are widely spaced. The location of test electrode P is not critical and can be easily determined.
Fig. E26: Measurement of the resistance to the mass of earth of electrode (X) using an earth-electrode-testing ohmmeter
Notes 1. ^ a b Where galvanised conducting materials are used for earth electrodes, sacrificial cathodic protection anodes may be necessary to avoid rapid corrosion of the electrodes where the soil is aggressive. Specially prepared magnesium anodes (in a porous sack filled with a suitable “soil”) are available for direct connection to the electrodes. In such circumstances, a specialist should be consulted.
Selection criteria for the TT, TN and IT systems. Selection does not depend on safety criteria. The three systems are equivalent in terms of protection of persons if all installation and operating rules are correctly followed. The selection criteria for the best system(s) depend on the regulatory requirements, the required continuity of service, operating conditions and the types of network and loads In terms of the protection of persons, the three system earthing arrangements (SEA) are equivalent if all installation and operating rules are correctly followed. Consequently, selection does not depend on safety criteria. It is by combining all requirements in terms of regulations, continuity of service, operating conditions and the types of network and loads that it is possible to determine the best system(s) (see Fig. E16). Selection is determined by the following factors:
Above all, the applicable regulations which in some cases impose certain types of SEA
Secondly, the decision of the owner if supply is via a private MV/LV transformer (MV subscription) or the owner has a private energy source (or a separate-winding transformer) If the owner effectively has a choice, the decision on the SEA is taken following discussions with the network designer (design office, contractor). The discussions must cover:
First of all, the operating requirements (the required level of continuity of service) and the operating conditions (maintenance ensured by electrical personnel or not, in-house personnel or outsourced, etc.)
Secondly, the particular characteristics of the network and the loads (see Fig. E17 ).
TNS
TNC
IT1[a]
IT2[b]
Comments
Fault current
-
--
--
+
--
Fault voltage
-
-
-
+
-
Touch voltage
+/-
-
-
+
-
Protection of persons against indirect contact
+
+
+
+
+
Protection of persons with emergency generating sets
+
-
-
+
-
TT
Electrical characteristics
Protection
Protection against fire (with an RCD)
+
+
Not allowed
+
+
Continuous overvoltage
+
+
+
-
+
Transient overvoltage
+
-
-
+
-
Overvoltage if transformer breakdown(primary/secondary)
-
+
+
+
+
-
+
+
+
+
Overvoltages
Electromagnetic compatibility Immunity to nearby lightning strikes
Immunity to lightning strikes on MV lines
-
-
-
-
-
Continuous emission of an electromagnetic field
+
+
-
+
+
Transient non-equipotentiality of the PE
+
-
-
+
-
Interruption for first fault
-
-
-
+
+
Voltage dip during insulation fault
+
-
-
+
-
Continuity of service
Installation
Special devices
-
+
+
-
-
Number of earth electrodes
-
+
+
-/+
-/+
Number of cables
-
-
+
-
-
Cost of repairs
-
--
--
-
--
Installation damage
+
-
-
++
-
Maintenance
[a] IT-net when a first fault occurs. [b] IT-net when a second fault occurs. Fig. E16: Comparison of system earthing arrangements
Type of network
Advised
Very large network with high-quality earth electrodes for exposed conductive parts (10 Ω max.)
Very large network with low-quality earth electrodes for exposed conductive parts (> 30 Ω)
TN
Disturbed area (storms) (e.g. television or radio transmitter)
TN
Network with high leakage currents (> 500 mA)
TN[d]
Network with outdoor overhead lines
TT[e]
Emergency standby generator set
IT
Type of loads
Loads sensitive to high fault currents (motors, etc.)
IT
Loads with a low insulation level (electric furnaces,welding machines, heating elements, immersion heaters, equipment in large kitchens)
TN[i]
Numerous phase-neutral single-phase loads (mobile, semi-fixed, portable)
TT[j] TN-S
Loads with sizeable risks (hoists, conveyers, etc.)
TN[k]
Numerous auxiliaries (machine tools)
TN-S
Miscellaneous
Supply via star-star connected power transformer[n]
TT
Premises with risk of fire
IT[o]
Increase in power level of LV utility subscription, requiring a private substation
TT[p]
Installation with frequent modifications
TT[q]
Installation where the continuity of earth circuits is uncertain (work sites, old installations)
TT[s]
Electronic equipment (computers, PLCs)
TN-S
Machine control-monitoring network, PLC sensors and actuators
IT[t]
[a] When the SEA is not imposed by regulations, it is selected according to the level of operating characteristics (continuity of service that is mandatory for safety reasons or desired to enhance productivity, etc.). Whatever the SEA, the probability of an insulation failure increases with the length of the network. It may be a good idea to break up the network, which facilitates fault location and makes it possible to implement the system advised above for each type of application. [b] The risk of flashover on the surge limiter turns the isolated neutral into an earthed neutral. These risks are high for regions with frequent thunder storms or installations supplied by overhead lines. If the IT system is selected to ensure a higher level of continuity of service, the system designer must precisely calculate the tripping conditions for a second fault. [c] Risk of RCD nuisance tripping.
[d] Whatever the SEA, the ideal solution is to isolate the disturbing section if it can be easily identified. [e] Risks of phase-to-earth faults affecting equipotentiality. [f] Insulation is uncertain due to humidity and conducting dust. [g] The TN system is not advised due to the risk of damage to the generator in the case of an internal fault. What is more, when generator sets supply safety equipment, the system must not trip for the first fault. [h] The phase-to-earth current may be several times higher than In, with the risk of damaging or accelerating the ageing of motor windings, or of destroying magnetic circuits. [i] To combine continuity of service and safety, it is necessary and highly advised, whatever the SEA, to separate these loads from the rest of the installation (transformers with local neutral connection). [j] When load equipment quality is not a design priority, there is a risk that the insulation resistance will fall rapidly. The TT system with RCDs is the best means to avoid problems. [k] The mobility of this type of load causes frequent faults (sliding contact for bonding of exposed conductive parts) that must be countered. Whatever the SEA, it is advised to supply these circuits using transformers with a local neutral connection. [l] Requires the use of transformers with a local TN system to avoid operating risks and nuisance tripping at the first fault (TT) or a double fault (IT). [l2] With a double break in the control circuit. [m] Excessive limitation of the phase-to-neutral current due to the high value of the zero-phase impedance (at least 4 to 5 times the direct impedance). This system must be replaced by a star-delta arrangement. [n] The high fault currents make the TN system dangerous. The TN-C system is forbidden. [o] Whatever the system, the RCD must be set to Δn ≤ 500 mA. [p] An installation supplied with LV energy must use the TT system. Maintaining this SEA means the least amount of modifications on the existing network (no cables to be run, no protection devices to be modified). [q] Possible without highly competent maintenance personnel. [r] This type of installation requires particular attention in maintaining safety. The absence of preventive measures in the TN system means highly qualified personnel are required to ensure safety over time. [s] The risks of breaks in conductors (supply, protection) may cause the loss of equipotentiality for exposed conductive parts. A TT system or a TN-S system with 30 mA RCDs is advised and is often mandatory. The IT system may be used in very specific cases. [t] This solution avoids nuisance tripping for unexpected earth leakage. Fig. E17: Influence of networks and loads on the selection of system earthing arrangements.
Characteristics of TT, TN and IT systems. TT system
(see Fig. E12) The TT system:
Technique for the protection of persons: the exposed conductive parts are earthed and residual current devices (RCDs) are used
Operating technique: interruption for the first insulation fault
Fig. E12: TT system Note: If the exposed conductive parts are earthed at a number of points, an RCD must be installed for each set of circuits connected to a given earth electrode. Main characteristics
Simplest solution to design and install. Used in installations supplied directly by the public LV distribution network.
Does not require continuous monitoring during operation (a periodic check on the RCDs may be necessary).
Protection is ensured by special devices, the residual current devices (RCD), which also prevent the risk of fire when they are set to ≤ 500 mA.
Each insulation fault results in an interruption in the supply of power, however the outage is limited to the faulty circuit by installing the RCDs in series (selective RCDs) or in parallel (circuit selection).
Loads or parts of the installation which, during normal operation, cause high leakage currents, require special measures to avoid nuisance tripping, i.e. supply the loads with a separation transformer or use specific RCDs (see TT system - Protective measures).
TN system
see (Fig. E13 and Fig. E14) The TN system: Technique for the protection of persons:
Interconnection and earthing of exposed conductive parts and the neutral are mandatory
Interruption for the first fault using overcurrent protection (circuit-breakers or fuses) Operating technique: interruption for the first insulation fault
Fig. E13: TN-C system
Fig. E14: TN-S system Main characteristics Generally speaking, the TN system:
Requires the installation of earth electrodes at regular intervals throughout the
installation
Requires that the initial check on effective tripping for the first insulation fault be carried out by calculations during the design stage, followed by mandatory measurements to confirm tripping during commissioning
Requires that any modification or extension be designed and carried out by a qualified electrician
May result, in the case of insulation faults, in greater damage to the windings of
rotating machines May, on premises with a risk of fire, represent a greater danger due to the higher
fault currents In addition, the TN-C system:
At first glance, would appear to be less expensive (elimination of a device pole and of
a conductor)
Requires the use of fixed and rigid conductors
Is forbidden in certain cases: Premises with a risk of fire For computer equipment (presence of harmonic currents in the neutral) In addition, the TN-S system:
May be used even with flexible conductors and small conduits
Due to the separation of the neutral and the protection conductor, provides a clean PE (computer systems and premises with special risks)
IT system (see Fig. E15) IT system: Protection technique:
Interconnection and earthing of exposed conductive parts
Indication of the first fault by an insulation monitoring device (IMD)
Interruption for the second fault using overcurrent protection (circuit-breakers or fuses) Operating technique:
Monitoring of the first insulation fault
Mandatory location and clearing of the fault
Interruption for two simultaneous insulation faults
Fig. E15: IT system Main characteristics
Solution offering the best continuity of service during operation
Indication of the first insulation fault, followed by mandatory location and clearing, ensures systematic prevention of supply outages
Generally used in installations supplied by a private MV/LV or LV/LV transformer
Requires maintenance personnel for monitoring and operation
Requires a high level of insulation in the network (implies breaking up the network if it is very large and the use of circuit-separation transformers to supply loads with high leakage currents)
The check on effective tripping for two simultaneous faults must be carried out by calculations during the design stage, followed by mandatory measurements during commissioning on each group of interconnected exposed conductive parts
Protection of the neutral conductor must be ensured as indicated in Protection of the neutral conductor inside chapter Sizing and protection of conductors.
Definition of standardised earthing schemes. he different earthing schemes (often referred to as the type of power system or system earthing arrangements) described characterise the method of earthing the installation downstream of the secondary winding of a MV/LV transformer and the means used for earthing the exposed conductive-parts of the LV installation supplied from it The choice of these methods governs the measures necessary for protection against indirect-contact hazards. The earthing system qualifies three originally independent choices made by the designer of an electrical distribution system or installation:
The type of connection of the electrical system (that is generally of the neutral conductor) and of the exposed parts to earth electrod (s)
A separate protective conductor or protective conductor and neutral conductor being a single conductor
The use of earth fault protection of overcurrent protective switchgear which clear only relatively high fault currents or the use of additional relays able to detect and clear small insulation fault currents to earth In practice, these choices have been grouped and standardised as explained below. Each of these choices provides standardised earthing systems with three advantages and drawbacks:
Connection of the exposed conductive parts of the equipment and of the neutral conductor to the PE conductor results in equipotentiality and lower overvoltages but increases earth fault currents
A separate protective conductor is costly even if it has a small cross-sectional area but it is much more unlikely to be polluted by voltage drops and harmonics, etc. than a neutral conductor is. Leakage currents are also avoided in extraneous conductive parts
Installation of residual current protective relays or insulation monitoring devices are much more sensitive and permits in many circumstances to clear faults before heavy damage occurs (motors, fires, electrocution). The protection offered is in addition independent with respect to changes in an existing installation
TT system (earthed neutral) (see Fig. E3) One point at the supply source is connected directly to earth. All exposed- and extraneousconductive-parts are connected to a separate earth electrode at the installation. This electrode may or may not be electrically independent of the source electrode. The two zones of influence may overlap without affecting the operation of protective devices.
Fig. E3: TT System
TN systems (exposed conductive parts connected to the neutral) The source is earthed as for the TT system (above). In the installation, all exposed- and extraneousconductive-parts are connected to the neutral conductor. The several versions of TN systems are shown below. TN-C system (see Fig. E4) The neutral conductor is also used as a protective conductor and is referred to as a PEN (Protective Earth and Neutral) conductor. This system is not permitted for conductors of less than 10 mm2 or for portable equipment. The TN-C system requires an effective equipotential environment within the installation with dispersed earth electrodes spaced as regularly as possible since the PEN conductor is both the neutral conductor and at the same time carries phase unbalance currents as well as 3rd order harmonic currents (and their multiples). The PEN conductor must therefore be connected to a number of earth electrodes in the installation. Caution: In the TN-C system, the “protective conductor” function has priority over the “neutral function”. In particular, a PEN conductor must always be connected to the earthing terminal of a load and a jumper is used to connect this terminal to the neutral terminal.
Fig. E4: TN-C system TN-S system (see Fig. E5) The TN-S system (5 wires) is obligatory for circuits with cross-sectional areas less than 10 mm2 for portable equipment. The protective conductor and the neutral conductor are separate. On underground cable systems where lead-sheathed cables exist, the protective conductor is generally the lead sheath. The use of separate PE and N conductors (5 wires) is obligatory for circuits with cross-sectional areas less than 10 mm2 for portable equipment.
Fig. E5: TN-S system TN-C-S system (see Fig. E6 and Fig. E7) The TN-C and TN-S systems can be used in the same installation. In the TN-C-S system, the TN-C (4 wires) system must never be used downstream of the TN-S (5 wires) system, since any accidental interruption in the neutral on the upstream part would lead to an interruption in the protective conductor in the downstream part and therefore a danger.
Fig. E6: TN-C-S system
Fig. E7: Connection of the PEN conductor in the TN-C system
IT system (isolated or impedance-earthed neutral) IT system (isolated neutral) No intentional connection is made between the neutral point of the supply source and earth (see Fig. E8).
Fig. E8: IT system (isolated neutral) Exposed- and extraneous-conductive-parts of the installation are connected to an earth electrode. In practice all circuits have a leakage impedance to earth, since no insulation is perfect. In parallel with this (distributed) resistive leakage path, there is the distributed capacitive current path, the two paths together constituting the normal leakage impedance to earth (see Fig. E9).
Fig. E9: IT system (isolated neutral) Example (see Fig. E10) In a LV 3-phase 3-wire system, 1 km of cable will have a leakage impedance due to C1, C2, C3 and R1, R2 and R3 equivalent to a neutral earth impedance Zct of 3,000 to 4,000 Ω, without counting the filtering capacitances of electronic devices.
Fig. E10: Impedance equivalent to leakage impedances in an IT system IT system (impedance-earthed neutral) An impedance Zs (in the order of 1,000 to 2,000 Ω) is connected permanently between the neutral point of the transformer LV winding and earth (see Fig. E11). All exposed- and extraneousconductive-parts are connected to an earth electrode. The reasons for this form of power-source earthing are to fix the potential of a small network with respect to earth (Zs is small compared to the leakage impedance) and to reduce the level of overvoltages, such as transmitted surges from the MV windings, static charges, etc. with respect to earth. It has, however, the effect of slightly increasing the first-fault current level.
Fig. E11: IT system (impedance-earthed neutral).
Earthing connections. In a building, the connection of all metal parts of the building and all exposed conductive parts of electrical equipment to an earth electrode prevents the appearance of dangerously high voltages between any two simultaneously accessible metal parts
Definitions National and international standards (IEC 60364) clearly define the various elements of earthing connections. The following terms are commonly used in industry and in the literature. Bracketed numbers refer to Figure E1
Fig. E1: An example of a block of flats in which the main earthing terminal (6) provides the main equipotential connection; the removable link (7) allows an earth-electrode-resistance check
Earth electrode (1): A conductor or group of conductors in intimate contact with, and providing an electrical connection with Earth (cf details in section 1.6 of Chapter E.)
Earth: The conductive mass of the Earth, whose electric potential at any point is conventionally taken as zero
Electrically independent earth electrodes: Earth electrodes located at such a distance from
one another that the maximum current likely to flow through one of them does not significantly affect the potential of the other(s)
Earth electrode resistance: The contact resistance of an earth electrode with the Earth
Earthing conductor (2): A protective conductor connecting the main earthing terminal (6) of an installation to an earth electrode (1) or to other means of earthing (e.g. TN systems); Exposed-conductive-part: A conductive part of equipment which can be touched and which
is not a live part, but which may become live under fault conditions Protective conductor (3): A conductor used for some measures of protection against electric
shock and intended for connecting together any of the following parts:
Exposed-conductive-parts
Extraneous-conductive-parts
The main earthing terminal
Earth electrode(s)
The earthed point of the source or an artificial neutral Extraneous-conductive-part: A conductive part liable to introduce a potential, generally earth potential, and not forming part of the electrical installation (4). For example: Non-insulated floors or walls, metal framework of buildings Metal conduits and pipework (not part of the electrical installation) for water, gas, heating, compressed-air, etc. and metal materials associated with them
Bonding conductor (5): A protective conductor providing equipotential bonding
Main earthing terminal (6): The terminal or bar provided for the connection of protective conductors, including equipotential bonding conductors, and conductors for functional earthing, if any, to the means of earthing.
Connections The main equipotential bonding system The bonding is carried out by protective conductors and the aim is to ensure that, in the event of an incoming extraneous conductor (such as a gas pipe, etc.) being raised to some potential due to a fault external to the building, no difference of potential can occur between extraneous-conductiveparts within the installation.
The bonding must be effected as close as possible to the point(s) of entry into the building, and be connected to the main earthing terminal (6). However, connections to earth of metallic sheaths of communications cables require the authorisation of the owners of the cables. Supplementary equipotential connections These connections are intended to connect all exposed-conductive-parts and all extraneousconductive-parts simultaneously accessible, when correct conditions for protection have not been met, i.e. the original bonding conductors present an unacceptably high resistance. Connection of exposed-conductive-parts to the earth electrode(s) The connection is made by protective conductors with the object of providing a low-resistance path for fault currents flowing to earth.
Components (see Fig. E2) Effective connection of all accessible metal fixtures and all exposed-conductive-parts of electrical appliances and equipment, is essential for effective protection against electric shocks. Component parts to consider:
as exposed-conductive-parts
as extraneous-conductive-parts Elements used in building construction
Cableways Metal or reinforced concrete (RC):
Conduits
Impregnated-paper-insulated lead-covered cable, armoured or unarmoured Mineral insulated metal-sheathed cable (pyrotenax,
Reinforcement rods
Prefabricated RC panels Surface finishes: Floors and walls in reinforced concrete without further
cradle of withdrawable switchgear Appliances Exposed metal parts of class 1 insulated appliances
Switchgear
Steel-framed structure
etc.)
surface treatment
Tiled surface Metallic covering:
Metallic wall covering Building services elements other than electrical
Non-electrical elements metallic fittings associated with cableways (cable
trays, cable ladders, etc.) Metal objects:
Close to aerial conductors or to busbars
In contact with electrical equipment.
Metal pipes, conduits, trunking, etc. for gas,water
and heating systems, etc. Related metal components (furnaces,
tanks,reservoirs, radiators) Metallic fittings in wash rooms, bathrooms, toilets,
etc.
Metallised papers
Component parts not to be considered: as exposed-conductive-parts Diverse service channels, ducts, etc.
Conduits made of insulating material
Mouldings in wood or other insulating material
Conductors and cables without metallic sheaths Switchgear
Wooden-block floors
Rubber-covered or linoleum-covered floors
Dry plaster-block partition
Brick walls
Carpets and wall-to-wall carpeting
Enclosures made of insulating material Appliances
as extraneous-conductive-parts
All appliances having class II insulation regardless of the type of exterior envelope
Fig. E2: List of exposed-conductive-parts and extraneous-conductive-parts.
Cables and busways. Distribution and installation methods (see Fig. E36) Distribution takes place via cableways that carry single insulated conductors or cables and include a fixing system and mechanical protection.
Fig. E36: Radial distribution using cables in a hotel
Busbar trunking (busways) Busways, also referred to as busbar trunking systems, stand out for their ease of installation, flexibility and number of possible connection points Busbar trunking is intended to distribute power (from 20 A to 5000 A) and lighting (in this application, the busbar trunking may play a dual role of supplying electrical power and physically holding the lights).
Busbar trunking system components A busbar trunking system comprises a set of conductors protected by an enclosure (see Fig. E37). Used for the transmission and distribution of electrical power, busbar trunking systems have all the necessary features for fitting: connectors, straights, angles, fixings, etc. The tap-off points placed at regular intervals make power available at every point in the installation.
Fig. E37: Busbar trunking system design for distribution of currents from 25 to 4000 A
The various types of busbar trunking: Busbar trunking systems are present at every level in electrical distribution: from the link between the transformer and the low voltage switch switchboard (MLVS) to the distribution of power sockets and lighting to offices, or power distribution to workshops.
Fig. E38: Radial distribution using busways We talk about a distributed network architecture. There are essentially three categories of busways.
Transformer to MLVS busbar trunking Installation of the busway may be considered as permanent and will most likely never be modified. There are no tap-off points. Frequently used for short runs, it is almost always used for ratings above 1,600 /2,000 A, i.e. when the use of parallel cables makes installation impossible. Busways are also used between the MLVS and downstream distribution switchboards. The characteristics of main-distribution busways authorize operational currents from1,000 to 5,000 A and short-circuit withstands up to 150 kA.
Sub-distribution busbar trunking with low or high tap-off densities Downstream of main-distribution busbar trunking , two types of applications must be supplied: Mid-sized premises (industrial workshops with injection presses and metalwork machines or large supermarkets with heavy loads). The short-circuit and current levels can be fairly high (respectively 20 to 70 kA and 100 to 1,000 A) Small sites (workshops with machine-tools, textile factories with small machines,supermarkets with small loads). The short-circuit and current levels are lower (respectively 10 to 40 kA and 40 to 400 A) Sub-distribution using busbar trunking meets user needs in terms of: Modifications and upgrades given the high number of tap-off points Dependability and continuity of service because tap-off units can be connected under energized conditions in complete safety The sub-distribution concept is also valid for vertical distribution in the form of 100 to 5,000 A risers in tall buildings.
Lighting distribution busbar trunking Lighting circuits can be distributed using two types of busbar trunking according to whether the lighting fixtures are suspended from the busbar trunking or not.
busbar trunking designed for the suspension of lighting fixtures These busways supply and support light fixtures (industrial reflectors, discharge lamps, etc.). They are used in industrial buildings, supermarkets, department stores and warehouses. The busbar trunkings are very rigid and are designed for one or two 25 A or 40 A circuits. They have tap-off outlets every 0.5 to 1 m. busbar trunking not designed for the suspension of lighting fixtures Similar to prefabricated cable systems, these busways are used to supply all types of lighting fixtures secured to the building structure. They are used in commercial buildings (offices, shops, restaurants, hotels, etc.), especially in false ceilings. The busbar trunking is flexible and designed for one 20 A circuit. It has tap-off outlets every 1.2 m to 3 m. Busbar trunking systems are suited to the requirements of a large number of buildings.
Industrial buildings: garages, workshops, farm buildings, logistic centers, etc.
Commercial areas: stores, shopping malls, supermarkets, hotels, etc.
Tertiary buildings: offices, schools, hospitals, sports rooms, cruise liners, etc.
Standards Busbar trunking systems must meet all rules stated in IEC 61439-6. This defines the manufacturing arrangements to be complied with in the design of busbar trunking systems (e.g.: temperature rise characteristics, short-circuit withstand, mechanical strength, etc.) as well as test methods to check them. The new standard IEC61439-6 describes in particular the design verifications and routine verifications required to ensure compliance. By assembling the system components on the site according to the assembly instructions, the contractor benefits from conformity with the standard.
The advantages of busbar trunking systems Flexibility
Easy to change configuration (on-site modification to change production line configuration or extend production areas).
Reusing components (components are kept intact): when an installation is subject to major modifications, the busbar trunking is easy to dismantle and reuse.
Power availability throughout the installation (possibility of having a tap-off point every meter).
Wide choice of tap-off units. Simplicity
Design can be carried out independently from the distribution and layout of current consumers.
Performances are independent of implementation: the use of cables requires a lot of derating coefficients.
Clear distribution layout
Reduction of fitting time: the trunking system allows fitting times to be reduced by up to 50% compared with a traditional cable installation.
Manufacturer’s guarantee.
Controlled execution times: the trunking system concept guarantees that there are no unexpected surprises when fitting. The fitting time is clearly known in advance and a quick solution can be provided to any problems on site with this adaptable and scalable equipment.
Easy to implement: modular components that are easy to handle, simple and quick to connect. Dependability
Reliability guaranteed by being factory-built
Fool-proof units
Sequential assembly of straight components and tap-off units making it impossible to make any mistakes Continuity of service
The large number of tap-off points makes it easy to supply power to any new current consumer. Connecting and disconnecting is quick and can be carried out in complete safety even when energized. These two operations (adding or modifying) take place without having to stop operations.
Quick and easy fault location since current consumers are near to the line
Maintenance is non existent or greatly reduced Major contribution to sustainable development
Busbar trunking systems allow circuits to be combined. Compared with a traditional cable
distribution system, consumption of raw materials for insulators is divided by 4 due to the busbar trunking distributed network concept (see Fig. E39).
Reusable device and all of its components are fully recyclable.
Does not contain PVC and does not generate toxic gases or waste.
Reduction of risks due to exposure to electromagnetic fields.
Fig. E39: Example of a set of 14 x 25A loads distributed along 34 meters (for busway, Canalis KS 250A)
New functional features for Canalis Busbar trunking systems are getting even better. Among the new features we can mention:
Increased performance with a IP55 protection index and new ratings of 160 A through to 1000 A (Ks).
New lighting offers with pre-cabled lights and new light ducts.
New fixing accessories. Quick fixing system, cable ducts, shared support with “VDI” (voice, data, images) circuits. Busbar trunking systems are perfectly integrated with the environment:
white color to enhance the working environment, naturally integrated in a range of electrical distribution products.
conformity with European regulations on reducing hazardous materials (RoHS).
Examples of Canalis busbar trunking systems
Fig. E41: Rigid busbar trunking able to support light fittings: Canalis KBA or KBB (25 and 40 A)
Fig. E43: A busway for medium power distribution: Canalis KN (40 up to 160 A)
Fig. E44: A busway for medium power distribution: Canalis KS (100 up to 1000 A)
Fig. E45: A busway for high power distribution: Canalis KT (800 up to 5000 A).
The architecture design. The architecture design considered in this document starts at the preliminary design stage (see Fig. D3 step1). It generally covers the levels of MV/LV main distribution, LV power distribution, and exceptionally the terminal distribution level. (see Fig. D2). In buildings all consumers are connected in low voltage. It means that MV distribution consists in:
connection to utility,
distribution to MV/LV substation(s),
MV/LV substation(s) itself.
Fig. D2: Example of single-line diagram The design of an electrical distribution architecture can be described by a 3-stage process, with iterative possibilities. This process is based on taking account of the installation characteristics and criteria to be satisfied.
Internal MV circuits. Internal MV circuits are dedicated to the supply of the secondary MV/LV substations dispersed in the installation. They are three typical principles commonly used for this purpose Fig. D11:
Single feeder
Dual feeder
Open ring.
Comparison of these three typical principles of internal distribution is given Fig. D12. MV circuit configuration
Characteristic to consider
Single feeder
Open ring
Dual feeder
Site topology
Any
Single or several buildings
Single or several buildings
Power demand
Any
> 1250kVA
> 2500kVA
Disturbance sensitivity
Long interruption acceptable
Short interruption acceptable
Short interruption not acceptable
Fig. D12: Comparison of the typical internal circuits.
Number of MV/LV transformers. For every MV/LV substation, the definition of the number of MV/LV transformers takes into account the following criteria:
Total power supplied by the substation
Standardization of the rated power to reduce the number of spare transformers
Limit of the rated power. It is recommended to set this limit at 1250 kVA in order to facilitate the handling and the replacement of the transformers
Scalability of the installation
Need to separate the loads having a high level of sensitivity to the electrical perturbations
Need to dedicate a transformer to the load generating a high level of perturbation such as voltage dips, harmonics, flicker
Need for partial or total redundancy. When required, two transformers each sized for the full load and equipped with an automatic change-over are installed
Loads requiring a dedicated neutral system. IT for example to ensure the continuity of operation in case of phase to earth fault.
LV distribution - centralized or distributed layout. Layout Position of the main MV and LV equipment on the site or in the building. This layout choice is applied to the results of stage 1. Selection guide As recommended in IEC60364-8-1 §6.3, MV/LV substation location can be determined by using the barycenter method:
taking into account service conditions: in dedicated premises if the layout in the workshop is too restrictive (temperature, vibrations, dust, etc.)
Placing heavy equipment (transformers, generators, etc.) close to walls or to main exits for ease of maintenance. A layout example is given in the following diagram (Fig. D13):
Fig. D13: The position of the global load barycentre guides the positioning of power sources
Centralized or distributed layout of LV distribution In centralized layout, each load is connected directly to the power source. (Fig. D14):
Fig. D14: Example of centralized layout with point to point links In distributed layout, loads are connected to sources via a busway. This type of distribution is well adapted to supply many loads that are spread out, where easy change is requested or future new connection (need of flexibility) (Fig. D15):
Fig. D15: Example of distributed layout, with busway Factors in favour of centralized layout (see summary table in Fig. D16):
Installation flexibility: no,
Load distribution: localized loads (high unit power loads). Factors in favor of distributed layout:
Installation flexibility: "Implementation" flexibility (moving of workstations, etc…),
Load distribution: uniform distribution of low or medium unit power loads
Load distribution
Flexibility (see Installation flexibilityfor definition of the flexibility levels)
Localized loads
No flexibility
Centralized
Intermediate distribution loads
Uniformly distributed loads
Decentralized
Flexibility of design
Implementation flexibility
Centralized
Decentralized
Operation flexibility
Fig. D16: Recommendations for centralized or distributed layout
Centralized distribution gives greater independence of circuits, reducing the consequences of a failure from power availability point of view. The use of decentralized distribution with busway is a way to merge all the circuits in one: it makes it possible to take into account the diversity factor (ks), which means cost savings on conductor sizing (See Fig. D17). The choice between centralized and decentralized solutions, according to the diversity factor, allows to find an economic optimum between investment costs, installation costs and operating costs. These two distribution modes are often combined.
Fig. D17: Example of a set of 14 x 25A loads distributed along 34 meters (for busway, Canalis KS 250A).
Presence of LV back-up generators. LV backup-up generator is the association of an alternator mechanically powered by a thermal engine. No electrical power can be delivered until the generator has reached its rated speed. This type of device is therefore not suitable for an uninterrupted power supply. Depending, if the generator is sized to supply power to all or only part of the installation, there is either total or partial redundancy. A back-up generator runs generally disconnected from the network. A source changeover and an interlocking system is therefore required (see Fig. D18). The generator back-up time depends on the quantity of available fuel.
Fig. D18: Connection of a back-up generator The main characteristics to consider for implementing LV back-up generator:
Sensitivity of loads to power interruption (see Voltage Interruption Sensitivity for definition),
Availability of the public distribution network (see Service reliability for the definition),
Other constraints (e.g.: generators compulsory in hospitals or high buildings) In addition the presence of generators can be decided to reduce the energy bill or due to the opportunity for co-generation. These two aspects are not taken into account in this guide. The presence of a back-up generator is essential if the loads cannot be shed (only short interruption acceptable) or if the utility network availability is low. Determining the number of back-up generator units is in line with the same criteria as determining the number of transformers, as well as taking account of economic and availability considerations (redundancy, start-up reliability, maintenance facility). Determining the generator apparent power, depends on:
installation power demand of loads to be supplied,
transient constraints that can occur by motors inrush current for example.
Configuration of LV circuits. Single feeder configuration
Fig. D20 This is the reference configuration and the most simple. A load is connected to one single source. This configuration provides a minimum level of availability, since there is no redundancy in case of power source failure.
Fig. D20: Single feeder configuration
Parallel transformers configuration Fig. D21 The power supply is provided by more than 1 transformer generally connected in parallel to the same main LV switchboard.
Fig. D21: Parallel transformers configuration
Variant: Normally open coupled transformers Fig. D22 In order to increase the availability it is possible to split the main LV switchboard into 2 parts, with a normally open bus-coupler (NO). This configuration may require an Automatic Transfer Switch between the coupler and transformer incomers. These 2 configurations are more often used when power demand is greater than 1 MVA.
Fig. D22: Normally open coupled transformers
Main LV switchboard interconnected by a busway Fig. D23 Transformers are physically distant, and operated in parallel. They are connected by a busway, the load can always be supplied in the case of failure of one of the sources. The redundancy can be:
Total: each transformer being able to supply all of the installation,
Partial: each transformer only being able to supply part of the installation. In this case, part of the loads must be disconnected (load-shedding) in the case of one of transformer failure.
Fig. D23: Main LV switchboard interconnected by a busway
LV ring configuration Fig. D24 This configuration can be considered as an extension of the previous configuration with interconnection between switchboards. Typically, 4 transformers connected in parallel to the same MV line, supply a ring using busway. A given load is then supplied by several transformers. This configuration is well suited to large sites, with high load density (in kVA/m2). If all of the loads can be
supplied by 3 transformers, there is total redundancy in the case of failure of one of the transformers. In fact, each busbar can be fed by one or other of its ends. Otherwise, downgraded operation must be considered (with partial load shedding). This configuration requires special design of the protection plan in order to ensure discrimination in all of the fault circumstances. As the previous configuration this type of installation is commonly used in automotive industry or large site manufacturing industry.
Fig. D24: Ring configuration
Double-ended power supply Fig. D25 This configuration is implemented in cases where maximum availability is required. The principle involves having 2 independent power sources, e.g.:
2 transformers supplied by different MV lines,
1 transformer and 1 generator,
1 transformer and 1 UPS.
An automatic transfer switch (ATS) is used to avoid the sources being parallel connected. This configuration allows preventive and curative maintenance to be carried out on all of the electrical distribution system upstream without interrupting the power supply.
Fig. D25: Double-ended configuration with automatic transfer switch
Configuration combinations Fig. D26 An installation can be made up of several sub-asssemblies with different configurations, according to requirements for the availability of the different types of load. E.g.: generator unit and UPS, choice by sectors (some sectors supplied by cables and others by busways).
Fig. D26: Example of a configuration combination 1: Single feeder, 2: Main LV switchboard interconnected by a busway, 3: Double-ended For the different possible configurations, the most probable and usual set of characteristics is given in the following table:
Characteristic to be considered
Configuration
Single feeder (fig. D20)
Parallel transformer or transformers connected via a coupler (fig. D21-D22)
Main LV switchboard interconnected by a busway (fig D24)
LV ring
Site topology
Any
Any
1 level 5000 to 25000 m2
1 level 5000 to m2
Power demand
< 2500kVA
Any
≥ 2500kVA
> 2500kVA
Location latitude
Any
Any
Medium or high
Medium or high
Load distribution
Localized loads
Localized loads
Intermediate or uniform load distribution
Intermediate or uniform load distribution
Maintainability
Minimal
Standard
Standard
Standard
Disturbances sensitivity
Low sensitivity
High sensitivity
High sensitivity
High sensitivity
Fig. D27: Recommendations for the configuration of LV circuits.
Example: electrical installation in a printworks. Brief description Printing of personalized mailshots intended for mail order sales.
Installation characteristics Characteristic
Category
Activity
Mechanical
Site topology
single storey building 10000m2 (8000m2 dedicated to the process,
2000m2 for ancillary areas)
Layout latitude
High
Service reliability
Standard
Maintainability
Standard
Installation flexibility
No flexibility planned:
HVAC
Process utilities
Office power supply Possible flexibility:
finishing, putting in envelopes
special machines, installed at a later date
rotary machines (uncertainty at the draft design stage)
Power demand
3500kVA
Load distribution
Intermediate distribution
Power interruptions sensitivity
Sheddable circuits:
offices (apart from PC power sockets)
air conditioning, office heating
social premises
maintenance premises long interruptions acceptable:
printing machines
workshop HVAC (hygrometric control
Finishing, envelope filling
Process utilities (compressor, recycling of cooled water) No interruptions acceptable:
servers, office PCs
Disturbance sensitivity
Average sensitivity:
motors, lighting
High sensitivity:
IT
No special precaution to be taken due to the connection to the EdF network (low level of disturbance) Disturbance capability Other constraints
Non disturbing Building with lightning classification: lightning surge
arresters installed Power supply by overhead single feeder line
Technological characteristics Criteria
Service conditions
Category
IP: standard (no dust, no water protection)
IK: standard (use of technical pits, dedicated premises) °C: standard (temperature regulation)
Required service index
211
Offer availability by country
No problem (project carried out in Europe)
Other criteria
Not applicable
Architecture assessment criteria Criteria
Category
On-site work time
Standard
Environmental impact
Minimal: compliance with European standard regulations
Preventive maintenance costs
Standard
Power supply availability
Tier 1
Step 1: Architecture fundamentals Choice
Main criteria
Solution
Connection to upstream network
Isolated site, 3500 kVA
MV single-line service
MV Circuits
Layout + criticality
single feeder
Number of transformers
Power > 2500kVA
2 x 2000kVA
Number and distribution of substations
Surface area and power distribution
2 possible solutions: 1 substation or 2 substations
if 1 substations: Normaly open bus-coupler
between MLVS if 2 substations: Main LV switchboard
interconnected by a busway MV Generator
Site activity
No
Fig. D31: Two possible single-line diagrams Step 2: Architecture details Choice
Main criteria
Solution
Layout
Service conditions
Dedicated premises
LV circuit configuration
2 transformers, requested by the power demand
Solution from fig.D22 or D23 are possible
Centralized or distributed layout
Uniform loads, distributed power, scalability possibilities Non-uniform loads, direct link from MLVS
Decentralized with busbar trunking: finishing sector, envelope filling Centralized with cables:
special machines, rotary machines, HVAC, process utilities, offices (2 switchboards), office air conditioning, social premises, maintenance
Presence of backup generator
Criticality ≤ low
No back-up generator
Network availability: standard Presence of UPS
Criticality
UPS unit for IT devices and office workstations
Fig. D32: Detailed single-line diagram (1 substation based on fig.D22)
Fig. D33: Detailed single-line diagram (2 substation based on fig.D24)
Choice of technological solutions Choice
Main criteria
Solution
MV/LV substation
Service conditions
indoor (dedicated premises)
MV switchboard
Offer availability by country
SM6 (installation in Europe)
Transformers
Service conditions
cast resin transfo (avoids constraints related to oil)
LV switchboard
Service conditions, service index for LV switchboards
MLVS: Prisma P Sub-distribution: Prisma
Busway
Load distribution
Canalis KS (fig.D32 or D33) Canalis KT for main distribution (fig D33)
UPS units
Installed power to be supplied, back-up
Galaxy PW
time Power factor correction
Reactive power to provide for the minimum up to the full load without harmonic (see chapter Power Factor Correction for more information), presence of harmonics.
MCB’s. Instantaneous Trip Setting. 6, 8, 10, 13, 16, 20, 25, 32, 40, 50, 63, 80, 100 and 125 A. Instantaneous tripping – less than 0.1 sec. for tripping. B.
3 – 5 of the rated continuous currents.
C.
5 – 10 of the rated continuous currents.
D.
10 – 20 of the rated continuous currents.
Suitable for resistive loads, heaters, lighting etc. Suitable for the reactive loads – motors, general lighting etc. Suitable for high reactive loads – transformers, motors etc.
Permissible I2t (let-through energy) in A2s for circuit-breakers type B with rated current up to and including 63A.
Permissible I2t (let-through energy) in A2s for circuit-breakers type C with rated current up to and including 63A.
Cables withstanding capacities.
The application of fault current protective devices to cable protection is detailed in BS 7671 and is given by:
I2t k2S2 Where:
I2t is the energy let-through value of the protective device k2S2 is the energy withstand of the cable.
The code allows the breaking capacity of a circuit-breaker to be less than its associated prospective fault current when back-up protection is employed in cascading through a suitable upstream protective devices. Back-up protection consists of an upstream short-circuit protective device (SCPD) that helps a downstream circuit-breaker to break fault currents greater than its maximum breaking current. However, where an MCCB, MCB or fuse is the upstream SCPD, and the downstream SCPD is an MCB,
coordination tests can be used to validate that the I2t of the specific combination will not exceed the I2t value of the downstream MCB at its maximum breaking capacity. The I 2t of the upstream SCPD “A” and downstream MCB “B” operating together at 20 kA, will be equal to or less than the I 2t of MCB “B” at 10 kA. The I 2t to be used in the conductor fault current assessment would be that of MCB “B” at 10 kA.
MCCB’s. (MCCBs) may have fixed or adjustable protection settings, normally a three position toggle operating handle giving on-off-tripped indication plus reset function, and a performance level relative to the incoming supply such that they can be installed at a point close to the supply transformer. Ratings: 16A to 1600A (may be upto 3200A), with the short circuit withstanding capacities upto 100kA in selections.
Rated Short-Time Withstand Current. Circuit-breakers of Selectivity Category B have a short-time delay (STD) allowing timegraded selectivity between circuit-breakers in series.
Icw is the fault current the circuit-breaker will withstand for the maximum short-time delay time. Preferred times are: 0.05, 0.1, 0.25, 0.5 and 1.0 second.
In order to provide protection against electric shock in accordance with the Wiring, it is necessary to determine the maximum value of Zs that will give the required disconnection time. This will give the max. length of the circuits downstream of the device for achieving the protections at around 80% of the phase – to – N voltages. No tolerance may be required for the IT – trippings, but a 20% may be considered in the case of instantaneous types.
Calculation Methodology
This calculation is based on IEEE Std 80 (2000), "Guide for safety in AC substation grounding". There are two main parts to this calculation:
Earthing grid conductor sizing
Touch and step potential calculations
IEEE Std 80 is quite descriptive, detailed and easy to follow, so only an overview will be presented here and IEEE Std 80 should be consulted for further details (although references will be given herein).
Prerequisites The following information is required / desirable before starting the calculation:
A layout of the site
Maximum earth fault current into the earthing grid
Maximum fault clearing time
Ambient (or soil) temperature at the site
Soil resistivity measurements at the site (for touch and step only)
Resistivity of any surface layers intended to be laid (for touch and step only)
Earthing Grid Conductor Sizing Determining the minimum size of the earthing grid conductors is necessary to ensure that the earthing grid will be able to withstand the maximum earth fault current. Like a normal power cable under fault, the earthing grid conductors experience an adiabatic short circuit temperature rise. However unlike a fault on a normal cable, where the limiting temperature is that which would cause permanent damage to the cable's insulation, the temperature limit for earthing grid conductors is the melting point of the conductor. In other words, during the worst case earth fault, we don't want the earthing grid conductors to start melting! The minimum conductor size capable of withstanding the adiabatic temperature rise associated with an earth fault is given by re-arranging IEEE Std 80 Equation 37:
Where (mm2)
is the minimum cross-sectional area of the earthing grid conductor
is the energy of the maximum earth fault (A2s) is the maximum allowable (fusing) temperature (ºC) is the ambient temperature (ºC) is the thermal coefficient of resistivity (ºC - 1) is the resistivity of the earthing conductor (μΩ.cm)
is is the thermal capacity of the conductor per unit volume(Jcm - 3ºC - 1) The material constants Tm, αr, ρr and TCAP for common conductor materials can be found in IEEE Std 80 Table 1. For example. commercial hard-drawn copper has material constants:
Tm = 1084 ºC
αr = 0.00381 ºC - 1
ρr = 1.78 μΩ.cm
TCAP = 3.42 Jcm - 3ºC - 1.
As described in IEEE Std 80 Section 11.3.1.1, there are alternative methods to formulate this equation, all of which can also be derived from first principles). There are also additional factors that should be considered (e.g. taking into account future growth in fault levels), as discussed in IEEE Std 80 Section 11.3.3.
Touch and Step Potential Calculations When electricity is generated remotely and there are no return paths for earth faults other than the earth itself, then there is a risk that earth faults can cause dangerous voltage gradients in the earth around the site of the fault (called ground potential rises). This means that someone standing near the fault can receive a dangerous electrical shock due to:
Touch voltages - there is a dangerous potential difference between the earth and a metallic object that a person is touching
Step voltages - there is a dangerous voltage gradient between the feet of a person standing on earth
The earthing grid can be used to dissipate fault currents to remote earth and reduce the voltage gradients in the earth. The touch and step potential calculations are performed in
order to assess whether the earthing grid can dissipate the fault currents so that dangerous touch and step voltages cannot exist.
Step 1: Soil Resistivity The resistivity properties of the soil where the earthing grid will be laid is an important factor in determining the earthing grid's resistance with respect to remote earth. Soils with lower resistivity lead to lower overall grid resistances and potentially smaller earthing grid configurations can be designed (i.e. that comply with safe step and touch potentials). It is good practice to perform soil resistivity tests on the site. There are a few standard methods for measuring soil resistivity (e.g. Wenner four-pin method). A good discussion on the interpretation of soil resistivity test measurements is found in IEEE Std 80 Section 13.4. Sometimes it isn't possible to conduct soil resistivity tests and an estimate must suffice. When estimating soil resistivity, it goes without saying that one should err on the side of caution and select a higher resistivity. IEEE Std 80 Table 8 gives some guidance on range of soil resistivities based on the general characteristics of the soil (i.e. wet organic soil = 10 Ω.m, moist soil = 100 Ω.m, dry soil = 1,000 Ω.m and bedrock = 10,000 Ω.m).
Step 2: Surface Layer Materials Applying a thin layer (0.08m - 0.15m) of high resistivity material (such as gravel, blue metal, crushed rock, etc) over the surface of the ground is commonly used to help protect against dangerous touch and step voltages. This is because the surface layer material increases the contact resistance between the soil (i.e. earth) and the feet of a person standing on it, thereby lowering the current flowing through the person in the event of a fault. IEEE Std 80 Table 7 gives typical values for surface layer material resistivity in dry and wet conditions (e.g. 40mm crushed granite = 4,000 Ω.m (dry) and 1,200 Ω.m (wet)). The effective resistance of a person's feet (with respect to earth) when standing on a surface layer is not the same as the surface layer resistance because the layer is not thick enough to have uniform resistivity in all directions. A surface layer derating factor needs to be applied in order to compute the effective foot resistance (with respect to earth) in the presence of a finite thickness of surface layer material. This derating factor can be approximated by an empirical formula as per IEEE Std 80 Equation 27:
Where
is the surface layer
derating factor is the soil resistivity (Ω.m) is the resistivity of the surface layer material (Ω.m) is the thickness of the surface layer (m) This derating factor will be used later in Step 5 when calculating the maximum allowable touch and step voltages.
Step 3: Earthing Grid Resistance A good earthing grid has low resistance (with respect to remote earth) to minimise ground potential rise (GPR) and consequently avoid dangerous touch and step voltages. Calculating the earthing grid resistance usually goes hand in hand with earthing grid design - that is, you design the earthing grid to minimise grid resistance. The earthing grid resistance mainly depends on the area taken up by the earthing grid, the total length of buried earthing conductors and the number of earthing rods / electrodes. IEEE Std 80 offers two alternative options for calculating the earthing grid resistance (with respect to remote earth) - 1) the simplified method (Section 14.2) and 2) the Schwarz equations (Section 14.3), both of which are outlined briefly below. IEEE Std 80 also includes methods for reducing soil resistivity (in Section 14.5) and a treatment for concrete-encased earthing electrodes (in Section 14.6). Simplified Method IEEE Std 80 Equation 52 gives the simplified method as modified by Sverak to include the effect of earthing grid depth:
Where
is the earthing grid resistance with respect to remote earth (Ω)
is the soil resistivitiy (Ω.m) is the total length of buried conductors (m) is the total area occupied by the earthing grid (m2) is the depth of the earthing grid (m)
Schwarz
Equations
The Schwarz equations are a series of equations that are more accurate in modelling the effect of earthing rods / electrodes. The equations are found in IEEE Std 80 Equations 53, 54, 55(footnote) and 56, as follows:
Where
is the earthing grid resistance with respect to remote earth (Ω)
is the earth resistance of the grid conductors (Ω) is the earth resistance of the earthing electrodes (Ω) is the mutual earth resistance between the grid conductors and earthing electrodes (Ω) And the grid, earthing electrode and mutual earth resistances are:
Where
is the soil resistivity (Ω.m)
is the total length of buried grid conductors (m) is radius
for conductors buried at depth metres, or simply
metres and with cross-sectional
for grid conductors on the surface
is the total area covered by the grid conductors (m2) is the length of each earthing electrode (m) is number of earthing electrodes in area is the cross-sectional radius of an earthing electrode (m) and
are constant coefficients depending on the geometry of the grid
The coefficient
(1) For depth
(2) For depth
can be approximated by the following: :
:
(3) For depth
The coefficient
(1) For depth
(2) For depth
(3) For depth
: can be approximated by the following: :
:
: Where in both cases,
is
the length-to-
width ratio of th earthing grid.
Step 4: Maximum Gri Current The maximum
grid current is t
worst case eart
fault current tha would flow via
the earthing gri back to remote earth. To calculate the maximum grid current, you firstly need to calculate the worst case symmetrical earth fault current at the
facility that wou
have a return path through
remote earth (c this
). This
can be found
from the power
systems studies
or from manual calculation. Generally speaking, the
highest relevan
earth fault level will be on the primary side of the largest distribution
transformer (i.e either the
terminals or the
delta windings).
Current Division Factor Not all of the earth fault
current will flow back through
remote earth. A portion of the earth fault
current may ha
local return pat (e.g. local generation) or there could be
alternative retu
paths other tha remote earth (e.g. overhead earth return cables, buried
pipes and cable etc). Therefore
current division factor
must
be applied to account for the
proportion of th fault current flowing back
through remote earth. Computing the
current division factor is a task
that is specific t
each project an
the fault locatio and it may
incorporate som
subjectivity (i.e "engineeing
judgement"). In any case, IEEE Std 80 Section
15.9 has a good discussion on calculating the
current division factor. In the
most conservative case, a current division factor of
can
be applied, meaning that 100% of earth fault current flows back
through remote earth.
The symmetrica grid current calculated by:
Decrement Fact
The symmetrica
current is not th
maximum grid c
because of asym
in short circuits
namely a dc cur
offset. This is ca
by the decreme
factor, which ca
calculated from
Std 80 Equation
Where
is t
decrement facto is the duration of the fault (s)
is the dc time offset constant (see below)
The dc time offs
derived from IE Equation 74:
Where
is th
fault location is the system frequency (Hz)
The maximum g calculated by:
Step 5: Touch
One of the goal
protect people a the event of an
ac electric curre
human body ca
range of 60 to 1
fibrillation and h
duration of an e
the risk of mort
faults are cleare
need to prescrib touch and step lethal shocks.
The maximum t
touch scenarios
from IEEE Std S
50kg and 70kg:
Touch voltage li
difference betw
the potential of
during a fault (d
50kg person:
70kg person:
Step voltage lim
surface potentia distance of 1m
earthed object:
50kg person:
70kg person: Where is the step voltage limit (V) is the surface layer derating factor (as calculated in Step 2) is the soil resistivity (Ω.m) is the maximum fault clearing time (s)
The choice of bo
expected weigh
women are exp
to choose 50kg.
Step 6: Groun
Normally, the p around the site they are at the
(where the faul flow of current
gradients in and
difference betw
ground potentia
a maximum po
potentials aroun fault.
The maximum G
Where is the maximum grid current found earlier in Step 4 (A) is the earthing grid resistance found earlier in Step 3 (Ω)
Step 7: Earth
Now we just ne
and step potent
exceed either o the grid design
However if it do
further analysis
of the maximum 16.5.
Mesh Voltage C
The mesh volta
earthing grid an
Where ::
is
is the maximum grid current found earlier in Step 4 (A) is the geometric spacing factor (see below) is the irregularity factor (see below) is the effective buried length of the grid (see below)
Geometric Sp
The geometric s
Where
is th
is the depth of buried grid conductors (m) is the cross-sectional diameter of a grid conductor (m) is a weighting factor for depth of burial = is a weighting factor for earth electrodes /rods on the corner mesh for grids with earth electrodes along the grid perimeter or corners
for grids with no earth electrodes on the corners or on the perimeter
is a geometric factor (see below)
Geometric Fa
The geometric f
With
for square grids, or otherwise
for square and rectangular grids, or otherwise
for square, rectangular and L-shaped grids, or otherwise Where
is th
is the length of grid conductors on the perimeter (m) is the total area of the grid (m2) and
are the maximum length of the grids in the x and y directions (m)
is the maximum distance between any two points on the grid (m)
Irregularity Fa
The irregularity
Where
is the
Effective Buri
The effective bu
For grids wi
Where
is th
is the total length of earthing electrodes / rods (m)
For grids wi
Where
is th
is the total length of earthing electrodes / rods (m) is the length of each earthing electrode / rod (m) and
are the maximum length of the grids in the x and y directions (m)
Step Voltage Ca
The maximum a
Where ::
is
is the maximum grid current found earlier in Step 4 (A) is the geometric spacing factor (see below) is the irregularity factor (as derived above in the mesh voltage calculation) is the effective buried length of the grid (see below)
Geometric Sp
The geometric s
Where is the depth of buried grid conductors (m) is a geometric factor (as derived above in the mesh voltage calculation)
is th
Effective Buri
The effective bu
Where
is th
is the total length of earthing electrodes / rods (m) What Now?
Now that the m
, and
then the earthin
If not, however
Redesign the earthing grid to lower the grid resistance (e.g. more grid conductors, more earthing electrodes, increasing cross-sectional area of conductors, etc). Once this is done, re-compute the earthing grid resistance (see Step 3) and re-do the touch and step potential calculations.
Limit the total earth fault current or create alternative earth fault return paths
Consider soil treatments to lower the resistivity of the soil
Greater use of high resistivity surface layer materials
Worked Ex
In this example
line and a delta
Step 1: Soi
The soil resistiv
Step 2: Sur
A thin 100mm l
Step 3: Ear
Figure 1. Propose
A rectangular e
Length of 90m and a width of 50m
6 parallel rows and 7 parallel columns
Grid conductors will be 120 mm2 and buried at a depth of 600mm
22 earthing rods will be installed on the corners and perimeter of the grid
Each earthing rod will be 3m long
Using the simpl
Step 4: Ma
Suppose that th
The X/R ratio at
The decrement
Fianlly, the max
kA
Step 5: Tou
Based on the av
The maximum a
V
The maximum a
V
Step 6: Gro
The maximum g
V
The GPR far exc
Step 7: Ear
Mesh Voltage
The component
and
is the number of parallel rows and columns respectively (e.g. 6 and 7)
m
V
m
V.
ntroduction Number of Earthing Electrode and Earthing Resistance depends on the resistivity of soil and time for fault current to pass through (1 sec or 3 sec). If we divide the area for earthing required by the area of one earth plate gives the number of earth pits required. There is no general rule to calculate the exact number of earth pits and size of earthing strip, but discharging of leakage current is certainly dependent on the cross section area of the material so for any equipment the earth strip size is calculated on the current to be carried by that strip. First the leakage current to be carried is calculated and then size of the strip is determined. For most of the electrical equipment like transformer, diesel generator set etc., the general concept is to have 4 number of earth pits. 2 no’s for body earthing with 2 separate strips with the pits shorted and 2 nos for Neutral with 2 separate strips with the pits shorted. The Size of Neutral Earthing Strip should be capable to carry neutral current of that equipment. The Size of Body Earthing should be capable to carry half of neutral Current.
For example for 100kVA transformer, the full load current is around 140A. The strip connected should be capable to carry at least 70A (neutral current) which means a strip of GI 25x3mm should be enough to carry the current and for body a strip of 25×3 will do the needful. Normally we consider the strip size that is generally used as standards. However a strip with lesser size which can carry a current of 35A can be used for body earthing. The reason for using 2 earth pits for each body and neutral and then shorting them is to serve as back up. If one strip gets corroded and cuts the continuity is broken and the other leakage current flows through the other run thery by completing the circuit.
Similarly for panels the no of pits should be 2 nos. The size can be decided on the main incomer circuit breaker. For example if main incomer to breaker is 400A, then body earthing for panel can have a strip size of 25×6 mm which can easily carry 100A.
Number of earth pits is decided by considering the total fault current to be dissipated to the ground in case of fault and the current that can be dissipated by each earth pit. Normally the density of current for GI strip can be roughly 200 amps per square cam. Based on the length and dia of the pipe used the number of earthing pits can be finalized.
1. Calculate numbers of pipe earthing A. Earthing resistance and number of rods for isolated earth pit (without buried earthing strip) The earth resistance of single rod or pipe electrode is calculated as per BS 7430:
R=ρ/2×3.14xL (loge (8xL/d)-1) Where: ρ = Resistivity of soil (Ω meter), L = Length of electrode (meter), D = Diameter of electrode (meter)
Example: Calculate isolated earthing rod resistance. The earthing rod is 4 meter long and having 12.2mm diameter, soil resistivity 500 Ω meter. R=500/ (2×3.14×4) x (Loge (8×4/0.0125)-1) =156.19 Ω. The earth resistance of single rod or pipe electrode is calculated as per IS 3040:
R=100xρ/2×3.14xL (loge(4xL/d)) Where: ρ = Resistivity of soil (Ω meter), L = Length of electrode (cm), D = Diameter of electrode (cm)
Example: Calculate number of CI earthing pipe of 100mm diameter, 3 meter length. System has fault current 50KA for 1 sec and soil resistivity is 72.44 Ω-Meters. Current Density At The Surface of Earth Electrode (As per IS 3043):
Max. allowable current density I = 7.57×1000/(√ρxt) A/m2 Max. allowable current density = 7.57×1000/(√72.44X1) = 889.419 A/m2 Surface area of one 100mm dia. 3 meter Pipe = 2 x 3.14 x r x L = 2 x 3.14 x 0.05 x3 = 0.942 m2 Max. current dissipated by one Earthing Pipe = Current Density x Surface area of electrode Max. current dissipated by one earthing pipe = 889.419x 0.942 = 837.83 A say 838 Amps Number of earthing pipe required = Fault Current / Max.current dissipated by one earthing pipe. Number of earthing pipe required = 50000/838 = 59.66 Say 60 No’s. Total number of earthing pipe required = 60 No’s. Resistance of earthing pipe (isolated) R = 100xρ/2×3.14xLx(loge (4XL/d)) Resistance of earthing pipe (isolated) R = 100×72.44 /2×3.14x300x(loge (4X300/10)) = 7.99 Ω/Pipe Overall resistance of 60 no of earthing pipe = 7.99/60 = 0.133 Ω. Top
B. Earthing resistance and number of rods for isolated earth pit (with buried earthing strip) Resistance of earth strip (R) As per IS 3043:
R=ρ/2×3.14xLx (loge (2xLxL/wt)) Example: Calculate GI strip having width of 12mm , length of 2200 meter buried in ground at depth of 200mm, soil resistivity is 72.44 Ω-meter.
Resistance of earth strip(Re) = 72.44/2×3.14x2200x(loge (2x2200x2200/.2x.012)) = 0.050 Ω From above calculation overall resistance of 60 no of earthing pipes (Rp) = 0.133 Ω. And it connected to bury earthing strip. Here net earthing resistance = (RpxRe)/(Rp+Re) Net eatrthing resistance = (0.133×0.05)/(0.133+0.05) = 0.036 Ω.
C. Total earthing resistance and number of electrode for group (parallel) In cases where a single electrode is not sufficient to provide the desired earth resistance, more than one electrode shall be used. The separation of the electrodes shall be about 4 m. The combined resistance of parallel electrodes is a complex function of several factors, such as the number and configuration of electrode the array. The total resistance of group of electrodes in different configurations as per BS 7430:
Ra=R (1+λa/n) where a=ρ/2X3.14xRxS Where: S = Distance between adjustment rod (meter), λ = Factor given in table below, n = Number of electrodes, ρ = Resistivity of soil (Ω meter), R = Resistance of single rod in isolation (Ω) Factors for parallel electrodes in line (BS 7430) Number of electrodes (n)
Factor (λ)
2
1.0
3
1.66
4
2.15
5
2.54
6
2.87
7
3.15
8
3.39
9
3.61
10
3.8
For electrodes equally spaced around a hollow square, e.g. around the perimeter of a building, the equations given above are used with a value of λ taken from following table.
For three rods placed in an equilateral triangle, or in an L formation, a value of λ = 1.66 may be assumed. Factors for electrodes in a hollow square (BS 7430) Number of electrodes (n)
Factor (λ)
2
2.71
3
4.51
4
5.48
5
6.13
6
6.63
7
7.03
8
7.36
9
7.65
10
7.9
12
8.3
14
8.6
16
8.9
18
9.2
20
9.4
For Hollow square total number of electrodes (N) = (4n-1). The rule of thumb is that rods in parallel should be spaced at least twice their length to utilize the full benefit of the additional rods. If the separation of the electrodes is much larger than their lengths and only a few electrodes are in parallel, then the resultant earth resistance can be calculated using the ordinary equation for resistances in parallel. In practice, the effective earth resistance will usually be higher than calculation.
Typically, a 4 spike array may provide an improvement 2.5 to 3 times. An 8 spike array will typically give an improvement of maybe 5 to 6 times. The Resistance of Original Earthing Rod will be lowered by Total of 40% for Second Rod, 60% for third Rod,66% for forth rod.
Example: Calculate Total Earthing Rod Resistance of 200 Number arranges in Parallel having 4 Meter Space of each and if it connects in Hollow Square arrangement. The Earthing Rod is 4 Meter Long and having 12.2mm Diameter, Soil Resistivity 500 Ω. First Calculate Single Earthing Rod Resistance:
R = 500/ (2×3.14×4) x (Loge (8×4/0.0125)-1) =136.23 Ω.
Now calculate total resistance of earthing rod of 200 number in parallel condition:
a = 500/(2×3.14x136x4) =0.146 Ra (Parallel in Line) =136.23x (1+10×0.146/200) = 1.67 Ω.
If earthing rod is connected in Hollow square than rod in each side of square is 200 = (4n-1) so n = 49 No. Ra (in hollow square) =136.23x (1+9.4×0.146/200) = 1.61 Ω.
(1) Demand factor (in IEC, Max.Utilization factor (Ku)):
The word “demand” itself says the meaning of Demand Factor. The ratio of the maximum coincident demand of a system, or part of a system, to the total connected load of the system.
Demand Factor = Maximum demand / Total connected load
For example, an over sized motor 20 Kw drives a constant 15 Kw load whenever it is ON. The motor demand factor is then 15/20 =0.75= 75 %. Demand Factor is express as a percentage (%) or in a ratio (less than 1).
Demand factor is always < =1. Demand Factor is always change with the time to time or hours to hours of use and it will not constant. The connected load is always known so it will be easy to calculate the maximum demand if the demand factor for a certain supply is known at different time intervals and seasons. The lower the demand factor, the less system capacity required to serve the connected load.
Calculation: (1) A Residence Consumer has 10 No’s Lamp of 400 W but at the same time It is possible that only 9 No’s of Bulbs are used at the same time. Here Total Connected load is 10×40=400 W. Consumer maximum demand is 9×40=360 W. Demand Facto of this Load = 360/400 =0.9 or 90%. (2) One Consumer have 10 lights at 60 Kw each in Kitchen, the load is 60 Kw x 10 = 600 KW. This will be true only if All lights are Turns ON the same time (Demand factor=100% or 1) For this Consumer it is observed that only half of the lights being turned ON at a time so we can say that the demand factor is 0.5 (50%). The estimated load = 600 Kw X 0.5 = 300 Kw.
Use of demand factors: Feeder conductors should have sufficient Ampere Capacity to carry the load. The Ampere Capacity does not always be equal to the total of all loads on connected branch-circuits. This factor must be applied to each individual load, with particular attention to electric motors, which are very rarely operated at full load. As per National Electrical Code (NEC) demand factor may be applied to the total load. The demand factor permits a feeder ampearcity to be less than 100 percent of all the branch-circuit loads connected to it. Demand factor can be applied to calculate the size of the sub-main which is feeding a Sub panel or a fixed load like a motor etc. If the panel have total load of 250 kVA , considering a Demand factor of 0.8, we can size the feeder cable for 250 x 0.8= 200 kVA. Demand factors for buildings typically range between 50 and 80 % of the connected load. In an industrial installation this factor may be estimated on an average at 0.75 for motors.
For incandescent-lighting loads, the factor always equals 1.
Demand Factor For Industrial Load Text Book of Design of Elect. Installation- Jain Electrical Load
Demand Factor
1 No of Motor
1
Up to 10 No’s of Motor
0.75
Up to 20 No’s of Motor
0.65
Up to 30 No’s of Motor
0.6
Up to 40 No’s of Motor
0.5
Up to 50 No’s of Motor
0.4
Demand Factor Text Book of Design of Elect. Installation- Jain Utility
Demand Factor
Office ,School
0.4
Hospital
0.5
Air Port, Bank, Shops,
0.6
Restaurant, Factory,
0.7
Work Shop, Factory (24Hr Shift)
0.8
Arc Furnace
0.9
Compressor
0.5
Hand tools
0.4
Inductance Furnace
0.8
Demand Factor Saudi Electricity Company Distribution Standard Utility
Demand Factor
Residential
0.6
Commercial
0.7
Flats
0.7
Hotel
0.75
Mall
0.7
Restaurant
0.7
Office
0.7
School
0.8
Common Area in building
0.8
Public Facility
0.75
Street Light
0.9
Indoor Parking
0.8
Outdoor Parking
0.9
Park / Garden
0.8
Hospital
0.8
Workshops
0.6
Ware House
0.7
Farms
0.9
Fuel Station
0.7
Factories
0.9
Demand Factor Text Book of Principal of Power System-V.K.Mehta Utility
Demand Factor
Residence Load (<0.25 KW)
1
Residence Load (<0.5 KW)
0.6
Residence Load (>0.1 KW)
0.5
Restaurant
0.7
Theatre
0.6
Hotel
0.5
School
0.55
Small Industry
0.6
Store
0.7
Motor Load (up to 10HP)
0.75
Motor Load (10HP to 20HP)
0.65
Motor Load (20HP to 100HP)
0.55
Motor Load (Above 100HP)
0.50
(2) Diversity factor:
Diversity Factor is ratio of the sum of the individual maximum demands of the various sub circuit of a system to the maximum demand of the whole system.
Diversity Factor = Sum of Individual Maximum Demands / Maximum Demand of the System.
Diversity Factor = Installed load / Running load.
The diversity factor is always >= 1. Diversity Factor is always >1 because sum of individual max. Demands >Max. Demand. In other terms, Diversity Factor (0 to 100%) is a fraction of Total Load that is particular item contributed to peak demand. 70% diversity means that the device operates at its nominal or maximum load level 70% of the time that it is connected and turned ON.
It is expressed as a percentage (%) or a ratio more than 1. If we use diversity value in % than it should be multiply with Load and if we use in numerical value (>1) than it should be divided with Load. Diversity occurs in an operating system because all loads connected to the System are not operating simultaneously or are not simultaneously operating at their maximum rating. The diversity factor shows that the whole electrical load does not equal the sum of its parts due to this time Interdependence (i.e. diverseness). In general terms we can say that diversity factor refers to the percent of time available that a machine. 70% diversity means that the device operates at its nominal or maximum load level 70% of the time that it is connected and turned ON. Consider two Feeders with the same maximum demand but that occur at different intervals of time. When supplied by the same feeder, the demand on such is less the sum of the two demands. In electrical design, this condition is known as diversity. Diversity factor is an extended version of demand factor. It deals with maximum demand of different units at a time/Maximum demand of the entire system. Greater the diversity factor, lesser is the cost of generation of power. Many designers prefer to use unity as the diversity factor in calculations for planning conservatism because of plant load growth uncertainties. Local experience can justify using a diversity factor larger than unity, and smaller service entrance conductors and transformer requirements chosen accordingly. The diversity factor for all other installations will be different, and would be based upon a local evaluation of the loads to be applied at different moments in time. Assuming it to be 1.0 may, on some occasions, result in a supply feeder and equipment rating that is rather larger than the local installation warrants, and an overinvestment in cable and equipment to handle the rated load current. It is better to evaluate the pattern of usage of the loads and calculate an acceptable diversity factor for each particular case.
Calculation: One Main Feeder have two Sub feeder (Sub Feeder A and Sub Feeder B), Sub Feeder-A have demand at a time is 35 KW and Sub Feeder-B have demands at a time is 42 KW, but the maximum demand of Main Feeder is 70 KW. Total individual Maximum Demand =35+42=77 KW. Maximum Demand of whole System=70 KW So Diversity factor of The System= 77/70 =1.1 Diversity factor can shoot up above 1.
Use of diversity factor: The Diversity Factor is applied to each group of loads (e.g. being supplied from a distribution or subdistribution board). Diversity factor is commonly used for a complete a coordination study for a system. This diversity factor is used to estimate the load of a particular node in the system. Diversity factor can be used to estimate the total load required for a facility or to size the Transformer Diversity factors have been developed for main feeders supplying a number of feeders, and typically 1.2 to 1.3 for Residence Consumer and 1.1 to 1.2 for Commercial Load. 1.50 to 2.00 for power and lighting loads. Note: Reciprocal of the above ratio (will be more than 1) also is used in some other countries. Diversity factor is mostly used for distribution feeder size and transformer as well as to determine the maximum peak load and diversity factor is always based on knowing the process. You have to understand what will be on or off at a given time for different buildings and this will size the feeder. Note for typical buildings diversity factor is always one. You have to estimate or have a data records to create 24 hours load graph and you can determine the maximum demand load for node then you can easily determine the feeder and transformer size. The diversity factor of a feeder would be the sum of the maximum demands of the individual consumers divided by the maximum demand of the feeder. In the same manner, it is possible to compute the diversity factor on a substation, a transmission line or a whole utility system. The residential load has the highest diversity factor. Industrial loads have low diversity factors usually of 1.4, street light practically unity and other loads vary between these limits.
Diversity Factor in distribution Network (Standard Handbook for Electrical Engineers” by Fink and Beaty) Elements of System
Residential
Commercial
General Power
Large Industrial
Between individual users
2.00
1.46
1.45
Between transformers
1.30
1.30
1.35
1.05
Between feeders
1.15
1.15
1.15
1.05
Between substations
1.10
1.10
1.10
1.10
From users to transformers
2.00
1.46
1.44
From users to feeder
2.60
1.90
1.95
1.15
From users to substation
3.00
2.18
2.24
1.32
From users to generating station
3.29
2.40
2.46
1.45
Diversity Factor for Distribution Switchboards Number of circuits
Diversity Factor in % (ks)
Assemblies entirely tested 2 and 3
90%
4 and 5
80%
6 to 9
70%
10 and more
60%
Assemblies partially tested in every case choose
100%
Diversity Factor as per IEC 60439 Circuits Function
Diversity Factor in % (ks)
Lighting
90%
Heating and air conditioning
80%
Socket-outlets
70%
Lifts and catering hoist
For the most powerful motor
100%
For the second most powerful motor
75%
For all motors
80%
Diversity Factor for Apartment block Apartment
Diversity Factor in % (ks)
2 To 4
1
5To 19
0.78
10To 14
0.63
15To 19
0.53
20To 24
0.49
25To 29
0.46
30 To 34
0.44
35 To 39
0.42
40To 40
0.41
50 To Above
0.40
Diversity Factor Text Book of Principal of Power System-V.K.Mehta Area
Residence Ltg
Commercial Ltg
Ind. Ltg
Between Consumer
3
1.5
1.5
Between Transformer
1.3
1.3
1.3
Between Feeder
1.2
1.2
1.2
Between S.S
1.1
1.1
1.1
(3) Load factor:
The ratio of the Actual Load of equipment to Full load of equipment.
Load Factor=Actual Load / Full Load
It is the ratio of actual kilowatt-Hours used in a given period, divided by the total possible kilowatt -hours that could have been used in the same period at the peak KW level. Load Factor = ( energy (kWh per month) ) / ( peak demand (kW) x hours/month ) In other terms Load factor is defined as the ratio of Average load to maximum demand during a given period. Load Factor= Average Load / Maximum Demand during given Time Period
The Load factor is always <=1. Load Factor is always less than 1 because maximum demand is always more than average demand. Load Factor can be calculated for a single day, for a month or for a year. Load factor in other terms of efficiency. It is used for determining the overall cost per unit generated. Higher the load factor is GOOD and it will more Output of Plan, lesser the cost per unit which means an electricity generator can sell more electricity at a higher spark spread, Fixed costs are spread over more kWh of output. A power plant may be highly efficient at High load factors. Low load factor is a BED. A low load factor will use electricity inefficiently relative to what we could be if we were controlling our peak demand. A power plant may be less efficient at low load factors. For almost constant loads, the load factor is close to unity. For Varying Load Factor is closed Zero. Load Factor is a measure of the effective utilization of the load and distribution equipment, i.e. higher load factor means better utilization of the transformer, line or cable. A high load factor means power usage is relatively constant. Low load factor shows that occasionally a high demand is set. To service that peak, capacity is sitting idle for long periods, thereby imposing higher costs on the system. Electrical rates are designed so that customers with high load factor are charged less overall per kWh. Sometimes utility companies will encourage industrial customers to improve their load factors. Load factor is term that does not appear on your utility bill, but does affect electricity costs. Load factor indicates how efficiently the customer is using peak demand. Calculation: Motor of 20 hp drives a constant 15 hp load whenever it is on. The motor load factor is then 15/20 = 75%.
Demand Factor & Load Factor
Introduction to Power Requirement for Building – J. Paul Guyer, Utility
Demand Factor (%)
Load Factor (%)
Communications – buildings
60-65
70-75
Telephone exchange building
55-70
20-25
Air passenger terminal building
65-80
28-32
Aircraft fire and rescue station
25-35
13-17
Aircraft line operations building
65-80
24-28
Academic instruction building
40-60
22-26
Applied instruction building
35-65
24-28
Chemistry and Toxicology Laboratory
70-80
22-28
Materials Laboratory
30-35
27-32
Physics Laboratory
70-80
22-28
Electrical and electronics laboratory
20-30
3-7
Cold storage warehouse
70-75
20-25
General warehouse
75-80
23-28
Controlled humidity warehouse
60-65
33-38
Hazardous/flammable storehouse
75-80
20-25
Disposal, salvage, scrap building
35-40
25-20
Hospital
38-42
45-50
Laboratory
32-37
20-25
K-6 schools
75-80
10-15
7-12 schools
65-70
12-17
Churches
65-70
5-25
Post Office
75-80
20-25
Retail store
65-70
25-32
Bank
75-80
20-25
Supermarket
55-60
25-30
Restaurant
45-75
15-25
Auto repair shop
40-60
15-20
Hobby shop, art/crafts
30-40
25-30
Bowling alley
70-75
10-15
Gymnasium
70-75
20-45
Skating rink
70-75
10-15
Indoor swimming pool
55-60
25-50
Theatres
45-55
8-13
Library
75-80
30-35
Golf clubhouse
75-80
15-20
Museum
75-80
30-35
(4) Coincidence factor (in IEC, Factor of simultaneity (ks)):
The reciprocal of diversity factor is coincidence factor The coincidence factor is the ratio of the maximum demand of a system, or part under consideration, to the sum of the individual maximum demands of the subdivisions
Coincidence factor = Maximum demand / Sum of individual maximum demands
Expressed as a percentage (%) or a ratio less than 1.
The Confidence Factor is always <=1. Usually Confidence Factor will decrease as the number of connected customer’s increases. The factor ks is applied to each group of loads (e.g. distribution or sub-distribution board). The determination of these factors is the responsibility of the designer, since it requires a detailed knowledge of the installation and the conditions in which the individual circuits are to be exploited. For this reason, it is not possible to give precise values for general application.
(5) Maximum demand:
The maximum demand of an installation is the maximum rate of consumption expressed in amperes, kW or kVA. It is generally taken as the average rate of consumption over a period of time. Example the 15-minute maximum kW demand for the week was 150 kW. Maximum demand does not include motor starting currents or other transient effects. Fault currents and overload currents are also excluded. Maximum demand in KW is relevant only for metering/tariff purposes. Maximum demand (often referred to as MD) is the largest current normally carried by circuits, switches and protective devices. It does not include the levels of current flowing under overload or short circuit conditions. Maximum Demand is a greatest of all demands that occur during a specific time The major disadvantage of allocating load using the diversity factors is that most utilities will not have a table of diversity factors and sometime it is not viable to determine accurate Diversity Factor. In this situation Maximum Demand is very helpful to calculate size of Feeder or TC. The kVA rating of all distribution transformers is always known for a feeder. The metered readings can be taken to each transformer based upon the transformer rating. An “allocation factor” (AF) can be calculate. Allocation Factor= Metered Demand (KVA) / Total KVA. Equipment Demand= AF x Total KVA of Equipments Calculation: Actual Loading or Size of TC-1 and TC-2.
Total Load on TC-1 =10+11+12+08= 41 KW. Maximum Diversity Demand of TC-1= 41 / 1.1 = 37.3 KW. Total Load on TC-2 =4+3+12+02= 21 KW. Maximum Diversity Demand of TC-2= 21 / 1.2 = 17.5 KW. Total Load= 37.3 + 17.5 =54.8 KW. Allocating Factor (AF)= M.D / Total Load Allocating Factor (AF)= 0.27.
Actual Load on TC-1=0.27×37.3 = 1.20 KW. Actual Load on TC-2=0.27×17.5 = 4.8 KW. Assessment of maximum demand is very easy for Resistive Load , For example, the maximum demand of a 240 V single-phase 8 kW shower heater can be calculated by dividing the power (8 kW) by the voltage (240 V) to give a current of 33.3 A. This calculation assumes a power factor of unity, which is a reasonable assumption for such a purely resistive load. Lighting circuits pose a special problem when determining MD. Discharge lamps are particularly difficult to assess, and current cannot be calculated simply by dividing lamp power by supply voltage. The reasons for this are Control gear losses result in additional current, the power factor is usually less than unity so current is greater, and Chokes and other control gear usually distort the waveform of the current so that it contains harmonics which are additional to the fundamental supply current. So long as the power factor of a discharge lighting circuit is not less than 0.85, the current demand for the circuit can be calculated from: current (A) = (lamp power x 1.8) / supply voltage (V) For example, the steady state current demand of a 240 V circuit supplying ten 65 W fluorescent lamps would be: I = 10X65X1.8A / 240 = 4.88A Switches for circuits feeding discharge lamps must be rated at twice the current they are required to carry, unless they have been specially constructed to withstand the severe arcing resulting from the switching of such inductive and capacitive loads.
Where to use Demand and Diversity factor:
There is generally confusion between Demand factor and Diversity factor. Demand factors should be ideally applied to individual loads and diversity factor to a group of loads. When you talk about ‘diversity’, there are naturally more than one or many loads involved. Demand factor can be applied to calculate the size of the sub-main, which is feeding a Sub panel or a fixed load like a motor etc, individual Load. Demand factors are more conservative and are used by NEC for service and feeder sizing. If the Sub panel have total load is 250 kVA , considering a Demand factor of 0.8, we can size the feeder cable for 250 x 0.8= 200 kVA. The Diversity Factor is applied to each group of loads (e.g. being supplied from a distribution or subdistribution board), size the Transformer. Demand factors and diversity factors are used in design. For example, the sum of the connected loads supplied by a feeder is multiplied by the demand factor to determine the load for which the feeder must be sized. This load is termed the maximum demand of the feeder. The sum of the maximum demand loads for a number of sub feeders divided by the diversity factor for the sub feeders will give the maximum demand load to be supplied by the feeder from which the sub feeders are derived.
Calculate Size of Electrical Switchgear by Demand & Diversity Factor:
The estimated electrical demand for all feeders served directly from the service entrance is calculated by multiplying the total connected loads by their demand factors and then adding all of these together. This sum is divided by the diversity factor (frequently assumed to be unity) to calculate the service entrance demand which is used to determine ampacity requirements for the service entrance conductors. When used Diversity and Demand Factor in an electrical design it should be applied as follows, the sum of the connected loads supplied by a feeder-circuit can be multiplied by the demand factor to determine the load used to size the components of the system. The sum of the maximum demand loads for two or more feeders is divided by the diversity factor for the feeders to derive the maximum demand load. Example-1: Calculate Size of Transformer having following details:
Feeder Breaker-1 Demand Load= Feeder Breaker-1xDemand Factor. Feeder Breaker-1 Demand Load=2000×0.7=1400 KVA Feeder Breaker-2 Demand Load= Feeder Breaker-2xDemand Factor. Feeder Breaker-2 Demand Load=1500×0.6=900 KVA Feeder Breaker-3 Demand Load= Feeder Breaker-3xDemand Factor. Feeder Breaker-2 Demand Load=1000×0.5=500 KVA Total Feeder Breaker Demand=1400+900+500=2800KVA Transformer Demand Load= Total Feeder Breaker Demand / Diversity Factor. Transformer Demand Load=2800/1.1 =2545 KVA If we Calculated Total Load on Transformer without any Demand & Diversity=2000+1500+1000=4500KVA. But after Calculating Demand & Diversity Factor Total Load on Transformer =2545 KVA Example-2: Calculate Size of Main Feeder of Main Transformer having following Details:
Sum of Maximum Demand of Customer on TC-1 =10 KWx0.65 =6.5 KW Sum of Maximum Demand of Customer on TC-2 =20 KWx0.75 =15 KW Sum of Maximum Demand of Customer on TC-3 =30 KWx0.65 =19.5 KW As Diversity of Consumer Connected on TC-1 is 1.5 so, Maximum Demand on TC-1 =6.5 KW/1.5 = 4 KW. As Diversity of Consumer Connected on TC-2 is 1.1 so, Maximum Demand on TC-2 =15 KW/1.1 = 14 KW As Diversity of Consumer Connected on TC-3 is 1.5 so, Maximum Demand on TC-3 =19.5 KW/1.5 = 13 KW.
Individual Maximum Demand on Main Transformer =04+14+13= 31 KW. Maximum Demand on Main Feeder =04+14+13 / 1.3 =24 KW
Significance of Load Factor and Diversity Factor
Load factor and diversity factor play an important part in the cost of the supply of electrical energy. Higher the values of load factor and diversity factors, lower will be the overall cost per unit generated. The capital cost of the power station depends upon the capacity of the power station. Lower the maximum demand of the power station, the lower is the capacity required and therefore lower is the capital cost of the plant. With a given number of consumers the higher the diversity factor of their loads, the smaller will be the capacity of the plant required and consequently the fixed charges due to capital investment will be much reduced. Similarly higher load factor means more average load or more number of units generated for a given maximum demand and therefore overall cost per unit of electrical energy generated is reduced due to distribution of standing charges which are proportional to maximum demand and independent of number of units generated. Thus the suppliers should always try to improve the load factor as well as diversity factor by inducing the consumers to use the electrical energy during off peak hours and they may be charged at lower rates for such schemes.
Main-Tie-Main.
Load Configuration. Both Bus#1 & Bus#2 are supplying normal loads that mean interruption for PT1 or PT2 is accepted for fault located between M2-PT2 and main supply. No critical load (instantaneous interruption is not allowed) connected on Bus #1 and Bus #2. It shall be supplied from UPS. Load on Bus #1 has a standby load on Bus#2 or vice versa, so if the bus #2 fail, load on bus #1 is operated. Basic Operation. This diagram may be useful for our discussion. The basic M1-T-M2 configuration is shown. During normal operation M1 & M2 breakers are closed and tie breaker T is opened. Supply coming from PT-1 and PT-2. This drawing indicate when M2 open, T and M1 CBs are closed (abnormal condition). This condition is done for maintenance purpose for equipment located between M2 to upstream (main source). Load transfer from bus #2 to bus #1 can be carried out without interruption done by ATS scheme. If fault located on bus #2 to tie breaker T or bus #1 to tie breaker T, load transfer is prohibited by ATS scheme. But for fault located from M2 to upstream load transfer is allowed with deenergizing bus #2 loads first, then tie-breaker T closed by ATS scheme. Loads may be in service after this tranfer, if the loads is set in auto position. We cannot maintain the supply on fault bus (e.g. bus #2 or bus #1) before correction is made. A redundant bus tie or switch isolator acting as maintenance bypass to ATS operation. Based on discussion above, I do not know, where we have to install those equipment to maintain supply for fault on respectively bus. Normally Closed Tie Breaker Operation. It is possible to operate tie breaker in closed position, but we have to consider a short-circuit level on that bus. By calculation (Short-circuit study), a fault on bus #1 or bus #2 the magnitude become double. So, we
have to ensure the equipment s.c. rating (buse, breakers, feeder loads, feeder breakers, and etc) meet the requirements for tie-in in closed position. Note: Temporary closing three breakers for maintenance purpose is allowed within 3 cycles to 1 (one) second is accepted. Relay application. 1). Bus differential for bus#1 and bus #2 may be applied (we apply on 4.16 kV systems). 2). Directional relay may be applied on incoming breaker M1 & M2 if the NC for tie breaker T is applied. 3). Restrictive earth fault is applied for transformer with low resistance grounding. 4). Please consider to provide better coordination for instantaneous relay between incoming breaker and load breakers as well as ground fault protection. 5). Syncheck relay is required for synchronising bus # & bus #2 before closing tie breaker T. We provide permissive closed for ATS schecme. ATS can only be operated if the upstream system is in synchronising condition (Generating buses are in remote but located closed to each other). Conclusions. 1). We cannot maintain load on bus faulted before repairing is made. 2). I do not know the location for instaling redudant bus tie breaker or isolator to prevent faulted bus total failure. 3). Comprehensive study shall be caried out to operate tie breaker in NC. Especially in selecting electrical equipment and relay coordination. 4). Pay more attention on safety aspect in establishing the ATS scheme. http://www.sayedsaad.com/substation/index_SF6circuitbreaker.html.
Basic electrical design of a PLC panel (Wiring diagrams).
Building the PLC panel It is uncommon for engineers to build their own PLC panel designs (but not impossible of course). For example, once the electrical designs are complete, they must be built by an electrician. Therefore, it is your responsibility to effectively communicate your design intentions to the electricians through drawings. In some factories, the electricians also enter the ladder logic and do debugging. This article discusses the design issues in implementation that must be considered by the designer.
Electrical wiring diagrams of a PLC panel In an industrial setting a PLC is not simply “plugged into a wall socket”. The electrical design for each machine must include at least the following components. 1. 2.
Transformers – to step down AC supply voltages to lower levels Power contacts – to manually enable/disable power to the machine with e-stop buttons
3. Terminals – to connect devices 4. Fuses or circuit breakers – will cause power to fail if too much current is drawn 5. Grounding – to provide a path for current to flow when there is an electrical fault 6. Enclosure – to protect the equipment, and users from accidental contact A control system of a PLC panel will normally use AC and DC power at different voltage levels. Control cabinets are often supplied with single phase AC at 220/440/550V, or two phase AC at 220/440V AC, or three phase AC at 330/550V.
This power must be dropped down to a lower voltage level for the controls and DC power supplies. 110Vac is common in North America, and 220 V AC Is common in Europe and the Commonwealth countries. It is also common for a control cabinet to supply a higher voltage to other equipment, such as motors.
Motor controller example An example of a wiring diagram for a motor controller is shown in Figure 1. Note that symbols are discussed in detail later). Dashed lines indicate a single purchased component. This system uses 3 phase AC power (L1, L2 and L3) connected to the terminals. The three phases are then connected to a power interrupter. Next, all three phases are supplied to a motor starter that contains three contacts, M, and three thermal overload relays (breakers).
Figure 1 – A Motor Controller Schematic
The contacts, M, will be controlled by the coil, M. The output of the motor starter goes to a three phase AC motor. Power is supplied by connecting a step down transformer to the control electronics by connecting to phases L2 and L3. The lower voltage is then used to supply power to the left and right rails of the ladder below. The neutral rail is also grounded. The logic consists of two push buttons:
Start push button is normally open, so that if something fails the motor cannot be started. Stop push button is normally closed, so that if a wire or connection fails the system halts safely.
The system controls the motor starter coil M, and uses a spare contact on the starter, M, to seal in the motor starter. The diagram also shows numbering for the wires in the device. This is essential for industrial control systems that may contain hundreds or thousands of wires. These numbering schemes are often particular to each facility, but there are tools to help make wire labels that will appear in the final controls cabinet.
Figure 2 – A Physical Layout for the Control Cabinet Once the electrical design is complete, a layout for the controls cabinet is developed, as shown in Figure 2. The physical dimensions of the devices must be considered, and adequate space is needed to run wires between components. In the cabinet the AC power would enter at the terminal block, and be connected to the main breaker.
It would then be connected to the contactors and overload relays that constitute the motor starter. Two of the phases are also connected to the transformer to power the logic. The start and stop buttons are at the left of the box (note: normally these are mounted elsewhere, and a separate layout drawing would be needed). The final layout in the cabinet might look like the one shown in Figure 1.
Figure 3 – Final PLC Panel Wiring
When being built the system will follow certain standards that may be company policy, or legal requirements. This often includes items such as;
Hold downs – the will secure the wire so they don’t move Labels – wire labels help troubleshooting Strain reliefs – these will hold the wire so that it will not be pulled out of screw terminals Grounding – grounding wires may be needed on each metal piece for safety
A photograph of an industrial controls cabinet is shown in Figure 4:
Figure 4 – An industrial control cabinet with wire runs, terminal strip, buttons on PLC panel front, etc.
When including a PLC in the ladder diagram still remains. But, it does tend to become more complex. Figure 5 below shows a schematic diagram for a PLC based motor control system, similar to the previous motor control example. This figure shows the E-stop wired to cutoff power to all of the devices in the circuit, including the PLC. All critical safety functions should be hardwired this way.
Figure 5 – An Electrical Schematic with a PLC.
Measurement of N–PE loop resistance in TN, TT and IT systems.
N–PE loop resistance Up-to-date test instruments, with built in modern electronics, can measure loop resistance even between the neutral N and protection PE conductors in spite of
possible high currents in the neutral conductor. The current, which is driven by phase voltages through different linear and non-linear loads, causes voltage drops of extremely irregular (non sine wave) shape. The voltage drops interfere with the test voltage and thereby disturb the measurement. Internal test voltage (approx. 40V, AC, <15 mA) is used, as there is no mains voltage between neutral and protection conductors. Important advantage of this measurement against Fault Loop test (L–PE) is, that the RCD does definitly not trip during the measurement, this is due to the low test current.
Used test instrument Eurotest 61557 uses special (patented) measurement principle to filter the test signal and therefore assures correct measurement results.
What can we deduce from the measurement? The following conclusions can be reached on the basis of the measurement result: 1. 2. 3.
Type of used protection conductor connection (TN, TT or IT-system) Earth Resistance value for TT-system In case of TT or TN-system, the result is quite similar to the Fault Loop
Resistance value, this is why the test instrument can also calculate the Fault Loop Prospective Short circuit current.
Generally about the measurement principle As there is no mains voltage between the N and PE terminals which could be used as a test voltage the instrument must generate an internal one. This voltage may be either DC or AC. Used instrument uses AC test voltage, measurement is done following the U-I method according to the figure below.
Fi gure 1 – N–PE loop resistance measurement principle
Result = Ut / It = RN-PE Where:
Ut – Test voltage measured by the V-meter. It – Test current measured by the A-meter. RN-PE – Resistance of N-PE loop.
1. Measurement of N–PE loop resistance in TNsystem The test instrument measures the resistance of the neutral and the protection conductors from the power transformer to the measurement site (the loop is marked with a bold line on upper figure). The test result in this case is quite low (maximum a couple of ohms), showing that a TN-system is involved.
Figure 2 – Resistance measurement between neutral and protection conductor in TN-system
Result 1 = RN + RPE Result 2 = Ipsc = 230V × 1,06 / (RN + RPE) Where:
RN – Resistance of neutral conductor (marked with bold line) RPE – Resistance of protection conductor (marked with bold dotted line) Ipsc – Prospective short circuit current of fault loop
2. Measurement of N–PE loop resistance in TTsystem The test instrument measures the resistance in the following loop – Neutral conductor from power transformer to measurement site (mains outlet), protection conductor from the mains outlet to earth electrode and then back to the power transformer via soil and the transformer’s earthing system (the loop is marked with a bold line on the figure 3 below). The test result in this case is quite high (in excess of ten ohms), showing that a TTsystem is involved.
Figure 3 – Resistance measurement between the neutral and the protection conductor in a TT-system
Result 1 = RN + RPE + RE + RO Result 2 = Ipsc = 230V × 1,06 / (RN + RPE + RE + RO) As it could be presumed, that resistance R E is much higher than the sum of all other resistances, the following can be noted:
Result 1 ≈ RE Result 2 = Ipsc = 230V × 1,06 / RE Where:
RN – Resistance of neutral conductor from power transformer to measurement site (mains outlet) RPE – Resistance of protection conductor from the mains outlet to earthing electrode RE – Earth resistance of protection earth electrode RO – Earth resistance of transformer’s earthing system Ipsc – Prospective short circuit fault loop current
3. Measurement of N–PE loop resistance in ITsystem As can be seen from the Figure 4, there is no hard wired connection between the neutral and protection conductor in an IT-system. The test result is therefore very high (it can even be out of display range), showing that an IT-system is involved.
Figure 4 – Resistance measurement between neutral and protection conductor in IT-system Attention! – A high test result in itself is not sufficient evidence that an IT-system is involved (it could be just an interrupted protection conductor in a TN or TT-system).
Detailed calculation of currents and power according to the type of load.
Calculation of currents & power Currents and power analysis are key factors in any design or redesign of an installation they will enable the source(s) to be sized according to the purpose of the installation, the intended use of the circuits and the receivers to be supplied. The current consumed Ia corresponds to the nominal current consumed by a receiver independently of the utilisation factor and the coincidence factor, but taking into account the aspects of efficiency (η factor), displacement factor or phase shift (cos φ) for motors or other inductive or capacitive loads. For non-linear (or distorting) loads, the quadratic sum of the fundamental current and the harmonic currents must be calculated in order to obtain the actual rms current.
Let’s break the calculation of the power into few parts, so we can easily follow:
1. Purely resistive load The current consumed Ia of a purely resistive load is calculated by simply applying the formulas. For single phase:
and for the three-phase:
But beware, very few loads are totally resistive. Incandescent lighting is losing ground to solutions that offer higher performance levels, but which are on the other hand less “pure” from an electrical viewpoint.
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2. Non-distorting load that is not purely resistive The nominal power (Pn) of a motor corresponds to the mechanical power available on its shaft. The actual power consumed (Pa) corresponds to the active power carried by the line. This is dependent on the efficiency of the motor:
The current consumed (Ia) is given by the following formulae. For single-phase:
and for the three-phase:
Where:
Ia – rms current consumed (in A) Pn – nominal power (in W; this is the useful power)
U – voltage between phases in three-phase, and between phase and neutral in singlephase (in V) η – efficiency cosφ – displacement factor.
3. Calculation of the current consumed by several receivers The example described below shows that the current and power calculations must be carried out in accordance with precise mathematical rules in order to clearly distinguish the different components.
Example of asynchronous motors A group of circuits consists of two three-phase asynchronous motors M1 and M2 connected to the same panel (mains supply: 400V AC – 50 hz). The nominal power of the motors are respectively: Pn1 = 22 kW and Pn2 = 37 kW. The displacement factors are cosφ1 = 0.92 for M1 and cosφ2 = 0.72 for M2 the efficiencies are η1 = 0.91 and η2 = 0.93 respectively.
Calculation of the power consumed:
The reactive power can in this case be calculated by determining the value of tanφ from cosφ. the relationship with the tangent is given by the formula:
Calculation of the reactive power:
Calculation of the apparent power:
Calculation of the total current consumption for M1, M2, M1 + M2 and the corresponding power factor:
The active power (in W) and the reactive power (in VAr) can be added together algebraically, while the apparent power and currents can only be added together geometrically.
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Presentation of the results All power analyses must show, as in the table below, at least for each group of:
Active power circuits which corresponds (to the nearest efficiency) to the energy supplied, Reactive power so that the compensation devices (capacitors) can be sized, Apparent power so that the power of the source can be determined and Current consumed so that the trunking and protection devices can be calculated. M1 M2 M1 + M2 (Total t)
Active power: P [kW]
Pa1 = 24.18
Pa2 = 39.78
Pt = 63.96
Reactive power: Q [kVAR]
Q1 = 10.30
Q2 = 38.35
Qt = 48.65
Apparent power: S [kVA]
S1 = 26.28
S2 = 55.26
St = 80.36
Current consumed: Ia [A]
Ia1 = 38
Ia2 = 80
Iat = 116
0.92
0.72
0.80
cosφ
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4. Overloads on conductors according to the total harmonic distortion The current circulating in each phase is equal to the quadratic sum of the fundamental current (referred to as 1st harmonic order) and all the harmonic currents (of the following orders):
The THDi (Total Harmonic Distortion) expresses the ratio between the share of all the harmonic currents and the total current as a percentage.
I1 being the rms value of the fundamental and in In the rms value of the nth order harmonic. The principle is to apply a current reduction factor that can be calculated based on the THDi. For a permissible THDi value of 33%, the current must thus in theory be reduced in each phase by a factor K:
If the factor is not applied, the current will then be increased by:
This remains acceptable and explains why the standard does not recommend any derating or oversizing of cross-sections up to 33% THDi. Above 33%. the standard recommends an increase in the current IB which results in necessary oversizing of the neutral conductor.
Reduction of the current or oversizing of multi-core cables may also be necessary for the phase conductors. It should be noted that the standard recommends a reduction factor of 0.84. which in fact corresponds to a pessimistic THDi of 65%.
Related to the neutral conductor, it is considered that if all the harmonics are 3rd order and its multiples, they will be added together and the current due to the harmonics in the neutral will then be IN = 3 × Iph, which can be expressed using an equivalent notation, THDn = 3 THDi. Devices whose load is said to be non-linear do not consume a current that is a reflection of the voltage applied. This leads to unnecessary power consumption: the distorting power that generates an additional current, the consequences of which must not be overlooked. But this current is never expressed directly because it involves a fairly complex mathematical calculation, the fourier transform, to ascertain its relative overall part (THDi: total harmonic distortion) or the value order by order: ih2, ih3, ih4, ih5,..ihn.
With no precise measurements, it is difficult to know exactly the current level that corresponds to each harmonic order. It is therefore preferable to simply increase the cross-section of the neutral conductor as a precaution, since it is known that the main 3rd order harmonics and their multiples are added together in the neutral. and to adapt the protection of this conductor. Standard IEC 60364 indicates the increasing factors to be applied to the cross-section of the neutral conductor according the percentage of 3rd order harmonics.
In principle, the neutral must be the same cross-section as the phase conductor in all single-phase circuits. In three-phase circuits with a cross-section greater than 16 mm2 [25 mm2 aluminium]. The cross-section of the neutral can be reduced to crosssection/2. However this reduction is not permitted if:
The loads are not virtually balanced The total 3rd order harmonic currents are greater than 15%
If this total is greater than 33%, the cross-section of the active conductors of multi-core cables is chosen by increasing the current In by a fixed multiplication factor of 1.65. For single-core cables, only the cross-section of the neutral is increased. In practice, the increase of the current Ia in the neutral is compensated by an increase of its cross-section. When the neutral is loaded, a reduction factor of 0.86 is applied to the permissible current of cables with 3 or 1 conductors.
The current reduction factor KN or rather its inverse which will be used to oversize the neutral conductor will then be:
With a total 3rd order harmonic distortion of 65%, the current of the phase conductors must be increased by 119% and that in the neutral conductor by 163%. If the THDi were to reach 100%, 1/KN would theoretically reach 2.12. This value would be impossible to reach as it would mean that the harmonic had totally replaced the fundamental. The theoretical overcurrent limit for the neutral in relation to the phases is:
These calculations demonstrate that the harmonic currents above all must not be ignored both in terms of “hidden” power consumption and in terms of sizing the conductors which may be overloaded. The relative complexity of the calculations leads to the use of generic derating values which normally cover most cases, just as software is used elsewhere.
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Example of following the standards for defining a protection device with neutral overloaded by harmonics For a 3P+N circuit, intended for 170 A, with TNS system, with total 3rd order harmonic distortion of more than 33%. When sizing the phase cables, the reduction factor of 0.84 (loaded neutral, see above) must be included. This requires a minimum cross-section of 70 mm2 per phase. The neutral conductor must be sized to withstand a current of 1.45 × 170 A = 247 A, i.e. a cross-section of 95 mm2. A circuit breaker must therefore be chosen that is capable of withstanding the current that may cross the neutral:
In device ≥ IB neutral ⇒ In = 250 A But the device must be set according to the current that may flow in the phases:
Ir ≥ IB phases ⇒ Ir ≥ 170 A (and < 206 A, limit of the cable) A 250 A unprotected interrupted neutral circuit breaker, set to 0.7 is therefore suitable for this application.
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5. Distorting load that is not purely resistive The current consumed (Ia) is given by the following formulae:
where:
Ia – rms current consumed (in A) Pn – nominal power (in W; this is the useful power) U – voltage between phases in three-phase, and between phase and neutral in singlephase (in V) η – efficiency PF – power factor
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Example of a fluorescent luminaire and electronic ballast The nominal active power consumed by the luminaire is 9 W, and the measured apparent power is 16 VA. The measured displacement factor is cosφ = 0.845 and the power factor PF = 0.56. The measured current consumed Ia is 0.07 A. As cosφ and the power factor are different, it is not possible to calculate the value of the tanφ or that of the reactive power Q (VAR) for the receiver in question. The measured cosφ and power Q which would be calculated can only be calculated for the reactive power part connected with the sinusoidal component of the signal, in fact the current of the fundamental at 50Hz: 0.045 A measured in this case.
The powers relative to this linear and sinusoidal part of the load can be calculated as follows
S = 230 × 0.045 = 10.3 VA P = S × cosφ = 10.3 × 0.85 = 8.7 W Q = 5.5 VAR which is confirmed by the calculation of the power triangles Q2 = P2 – S2 or by the tanφ: Q = P × tanφ = 8.7 × 0.63 = 5.5 VAR Therefore not all the apparent power consumed is linear as there is a significant difference between the measured total apparent power S (16 VA) and the calculated theoretical sinusoidal power (10.3 VA).
It can also be seen that the sinusoidal active power of the device 8.7 W is very similar to the measured total active power 9 W. It can therefore be deduced that a large part of power S (16 – 10.3 = 5.7 VA) is consumed without producing any active power. The fluorescent luminaire and electronic ballast in the example consumes unproductive power in the form of harmonic currents. The total harmonic distortion is easy to calculate and represents expressed as a rate.
The spectral decomposition of the signal carried out on this luminaire shows that the main harmonic is 3rd order (34 mA) but that all the following odd-order harmonics are present and decaying. The main purpose of the above example is to demonstrate that active power information (in W) only for a non-linear receiver is very inadequate. The cosφ has no real relevance or meaning as it is only applicable to the fundamental signal. Only the apparent power and power factor (PF or ?\.,) information can really quantify and qualify the power that must be supplied by the source.
In the example given, it can be seen that an active power of approximately 9 W corresponds to a consumed power of 16 VA. Many modern devices (light bulbs, computer equipment, domestic appliances and electronic equipment) have this particular feature of consuming non-linear currents. For domestic use, where only the power in W is billed (sic), the power savings shown for these products is attractive. In practice, the currents consumed are higher than it seems and the energy distributor is supplying wasted energy. In large commercial or industrial installations the situation is different. A poor power factor results in consumption of reactive power that is billed. Compensation of non-linear loads thus becomes meaningful and useful here, but also at the design stage when it prevents oversizing of the energy sources, which it must be remembered supply VA (volt-amperes) and not W (watts). Important: Unlike linear loads (page 29), for non-linear loads the active powers (in W) can be added together algebraically, the apparent powers must only be added together geometrically, and likewise the currents which must be the same order. The reactive powers Q must not be added together except to as certain the relative part of the power associated with the sinusoidal fundamental signal and the part connected with the harmonic signals. Go back to currents and power calculations ↑
Reference // Power balance and choice of power supply solutions by Legrand. Pick Up Current | Current Setting | Plug Setting Multiplier and Time Setting Multiplier of Relay. Pick Up Current of Relay In all electrical relays, the moving contacts are not free to move. All the contacts remain in their respective normal position by some force applied on them continuously. This force is called controlling force of the relay. This controlling force may be gravitational force, may be spring force, may be magnetic force. The force applied on the relay’s moving parts for changing the normal position of the contacts, is called deflecting force. This deflecting force is always in opposition of controlling force and presents always in the relay. Although the deflecting force always presents in the relay directly connected to live line, but as the magnitude of this force is less than controlling force in normal condition, the relay does not operate. If the actuating current in the relay coil increases gradually, the deflecting force in electro mechanical relay, is also increased. Once, the deflecting force crosses the controlling force, the moving parts of the relay initiate to move to change the position of the contacts in the relay. The current for which the relay initiates it operation is called pick up current of relay. Current Setting of Relay The minimum pick up value of the deflecting force of an electrical relay is constant. Again the deflecting force of the coil is proportional to its number of turns and current flowing through the coil. Now, if we can change the number of active turns of any coil, the required current to reach at minimum pick value of the deflecting force, in the coil also changes. That means if active turns of the relay coil is reduced, then proportionately more current is required to produce desired relay actuating force. Similarly if active turns of the relay coil is increased, then proportionately reduced current is required to produce same desired deflecting force. Practically same model relays may be used in different systems. As per these systems requirement the pick up current of relay is adjusted. This is known as current setting of relay. This is achieved by providing required number of tapping in the coil. These taps are brought out to a plug bridge. The number of active turns in the coil can be changed by inserting plug in different points in the bridge. The current setting of relay is expressed in percentage ratio of relay pick up current to rated secondary current of CT.
That means, For example, suppose, you want that, an over current relay should operate when the system current just crosses 125% of rated current. If the relay is rated with 1 A, the normal pick up current of the relay is 1 A and it should be equal to secondary rated current of current transformer connected to the relay. Then, the relay will be operated when the current of CT secondary becomes more than or equal 1.25 A. As per definition, The current setting is sometimes referred as current plug setting. The current setting of over current relay is generally ranged from 50 % to 200 %, in steps of 25 %. For earth fault relay it is from 10% to 70% in steps of 10%. Plug Setting Multiplier of Relay Plug setting multiplier of relay is referred as ratio of fault current in the relay to its pick up current.
Suppose we have connected on protection CT of ratio 200/1 A and current setting is 150%. Hence, pick up current of the relay is, 1 × 150 % = 1.5 A Now, suppose fault current in the CT primary is 1000 A. Hence, fault current in the CT secondary i.e. in the relay coil is, 1000 × 1/200 = 5A Therefore PSM of the relay is, 5 / 1.5 =3.33 Time Setting Multiplier of Relay The operating time of an electrical relay mainly depends upon two factors : 1.
How long distance to be traveled by the moving parts of the relay for closing relay contacts and 2. How fast the moving parts of the relay cover this distance. So far adjusting relay operating time, both of the factors to be adjusted. The adjustment of travelling distance of an electromechanical relay is commonly known as time setting. This adjustment is commonly known as time setting multiplier of relay. The time setting dial is calibrated from 0 to 1 in steps 0.05 sec. But by adjusting only time setting multiplier, we can not set the actual time of operation of an electrical relay. As we already said, the time of operation also depends upon the speed of operation. The speed of moving
parts of relay depends upon the force due to current in the relay coil. Hence, it is clear that, speed of operation of an electrical relay depends upon the level of fault current. In other words, time of operation of relay depends upon plug setting multiplier. The relation between time of operation and plug setting multiplier is plotted on a graph paper and this is known as time / PSM graph. From this graph one can determine, the total time taken by the moving parts of an electromechanical relay, to complete its total travelling distance for different PSM. In time setting multiplier, this total travelling distance is divided and calibrated from 0 to 1 in steps of 0.05. So when time setting is 0.1, the moving parts of the relay has to travel only 0.1 times of the total travelling distance, to close the contact of the relay. So, if we get total operating time of the relay for a particular PSM from time / PSM graph and if we multiply that time with the time setting multiplier, we will get, actual time of operation of relay for said PSM and TSM. For getting clear idea, let us have a practical example. Say a relay has time setting 0.1 and you have to calculate actual time of operation for PSM 10. From time / PSM graph of the relay as shown below, we can see the total operating time of the relay is 3 seconds. That means, the moving parts of the relay take total 3 seconds to travel 100 % travelling distance. As the time setting multiplier is 0.1 here, actually the moving parts of the relay have to travel only 0.1 × 100% or 10% of the total travel distance, to close the relay contacts. Hence, actual operating time of the relay is 3 × 0.1 = 0.3 sec. i.e. 10% of 3 sec. Time vs PSM Curve of Relay This is relation curve between operating time and plug setting multiplier of an electrical relay. The x-axis or horizontal axis of the Time / PSM graph represents, PSM and Y-axis or vertical axis represents time of operation of the relay. The time of operation represents in this graph is that, which required to operate the relay when time setting multiplier set at 1. From the Time / PSM curve of a typical relay shown below, it is seen that, if PSM is 10, the time of operation of relay is 3 sec. That means, the relay will take 3 seconds to complete its operation, with time setting 1. It is also seen from the curve that, for lower value of plug setting multiplier, i.e. for lower value of fault current, the time of operation of the relay is inversely proportional to the fault current. But when PSM becomes more than 20, the operating time of relay becomes almost constant. This feature is necessary in order to ensure discrimination on very heavy fault current flowing through sound feeders. Calculation of Relay Operation Time For calculating actual relay operating time, we need to know these following operation.
1. Current setting. 2. Fault current level. 3. Ratio of current transformer. 4. Time / PSM curve. 5. Time setting. Step-1 From CT ratio, we first see the rated secondary current of CT. Say the CT ratio is 100 / 1 A, i.e. secondary current of CT is 1 A. Step-2 From current setting we calculate the trick current of the relay. Say current setting of the relay is 150 % therefore pick up current of the relay is 1 × 150% = 1.5 A. Step-3 Now we have to calculate PSM for the specified faulty current level. For that, we have to first divide primary faulty current by CT ratio to get relay faulty current. Say the faulty current level is 1500 A, in the CT primary, hence secondary equivalent of faulty current is 1500/(100/1) = 15
A Step-4 Now, after calculating PSM, we have to find out the total time of operation of the relay from Time / PSM curve. From the curve, say we found the time of operation of relay is 3 second for PSM = 10. Step-5 Finally that operating time of relay would be multiplied with time setting multiplier, in order to get actual time of operation of relay. Hence say time setting of the relay is 0.1. Therefore actual time of operation of the relay for PSM 10, is 3 × 0.1 = 0.3 sec or 300 ms. Working Principle of Over Current Relay In an over current relay, there would be essentially a current coil. When normal current flows through this coil, the magnetic effect generated by the coil is not sufficient to move the moving element of the relay, as in this condition the restraining force is greater than deflecting force. But when the current through the coil increased, the magnetic effect increases, and after certain level of current, the deflecting force generated by the magnetic effect of the coil, crosses the restraining force, as a result, the moving element starts moving to change the contact position in the relay. Types of Over Current Relay Depending upon time of operation, there are various types of Over Current relays, such as, 1. Instantaneous over current relay. 2. Definite time over current relay. 3. Inverse time over current relay.
Inverse time over current relay or simply inverse OC relay is again subdivided as inverse definite minimum time (IDMT), very inverse time, extremely inverse time over current relay or OC relay. Instantaneous Over Current Relay Construction and working principle of instantaneous over current relay quite simple. Here generally a magnetic core is wound by current coil. A piece of iron is so fitted by hinge support and restraining spring in the relay, that when there is not sufficient current in the coil, the NO contacts remain open. When current in the coil crosses a present value, the attractive force becomes sufficient to pull the iron piece towards the magnetic core and consequently the no contacts are closed. The preset value of current in the relay coil is referred as pick up setting current. This relay is referred as instantaneous over current relay, as ideally, the relay operates as soon as the current in the coil gets higher than pick up setting current. There is no intentional time delay applied. But there is always an inherent time delay which can not be avoided practically. In practice the operating time of an instantaneous relay is of the order of a few milliseconds. Fig. Definite Time Over Current Relay This relay is created by applying intentional time delay after crossing pick up value of the current. A definite time over current relay can be adjusted to issue a trip output at definite amount of time after it picks up. Thus, it has a time setting adjustment and pick up adjustment. Inverse Time Over Current Relay Inverse time is a natural character of any induction type rotating device. This means the speed of rotation of rotating art of the device is faster if input current is increased. In other words, time of operation inversely varies with input current. This natural characteristic of electromechanical induction disc relay in very suitable for over current protection. This is because, in this relay, if fault is more severe, it would be cleared more faster. Although time inverse characteristic is inherent to electromechanical induction disc relay, but the same characteristic can be achieved in microprocessor based relay also by proper programming. Inverse Definite Minimum Time Over Current Relay or IDMT O/C Relay Ideal inverse time characteristics can not be achieved, in an over current relay. As the current in the system increases, the secondary current of the current transformer is increased proportionally. The secondary current is
fed to the relay current coil. But when the CT becomes saturated, there would not be further proportional increase of CT secondary current with increased system current. From this phenomenon it is clear that from trick value to certain range of faulty level, an inverse time relay shows exact inverse characteristic. But after this level of fault, the CT becomes saturated and relay current does not increase further with increasing faulty level of the system. As the relay current is not increased further, there would not be any further reduction in time of operation in the relay. This time is referred as minimum time of operation. Hence, the characteristic is inverse in the initial part, which tends to a definite minimum operating time as the current becomes very high. That is why the relay is referred as inverse definite minimum time over current relay or simply IDMT relay. Motor Protection Relay for High Voltage Induction Motor. Above 90% of motors used in an industry are induction motors, because they are cheap, robust & easy to maintain. For higher HP (>250HP) motors we prefer high voltage, because it will reduce operating current and the size of the motor. Why we require protection of motors? To understand this we need to know the cost associated with the failure of motor, i.e. Loss of production (Cost of production) Replacement of motor (Replacement cost) Cost of repair Cost of man hours due to this emergency The basic function of a protective relay is to identify the fault and isolate it from the healthy part of the system. This will improve the reliability of power system. For protection of motor, we have to identify the various causes of failure and to address the same. The various causes of failure are as below.
Thermal stress on winding Single phasing Earth fault Short circuit Locked rotor Number of hot starts Bearing failure Thermal Stress on Winding – If a motor runs continuously more than its rated capacity then this will over heat the winding and insulation. Subsequently deteriorate the winding insulation resulting in failure of
motor. If the voltage is less than the designed value then also it will over heat the winding at rated load and fail the motor. Single Phasing – Loss of one phase supplied to the motor (in case of 3-phase motor) leads to single phasing. If we start the motor on load, then the motor will fail due to unbalance. Earth Fault – If any part of winding comes in contact with the ground then we can say the motor is earthed. If we start the motor then it will lead to failure of motor. Short Circuit – If there is a contact between two phases of a three phase winding or between the turns of a phase, then this will termed as short circuit. Locked Rotor – If the driven equipment is in jammed condition or the motor shaft is jam, then this is known as locked rotor. If we start the motor then it will fail. Number of Hot Start – Each motor is designed to withstand a certain number of hot starts. Consider a motor is in running condition, if we stop the motor & immediately start the same, then this is called as hot start. Depending upon the thermal curve of a motor we have to give certain time to bring down the temperature of winding. Bearing Failure – If bearing fails then rubbing of rotor on stator will occur, resulting physical damage of insulation and winding. The bearing failure can be avoided by monitoring the bearing temperature. Bearing temperature detector (BTD) is used for monitoring and tripping of motor in case of abnormality. All motor protection relays operate on the basis of current taken by the motor. Motor protection relay is used for high voltage area having the following features Thermal overload protection Short circuit protection Single phasing protection Earth fault protection Locked rotor protection Number of start protection For setting of the relay we require the CT ratio & full load current of the motor. The setting of different element is listed below
Thermal over Load Element - To set this element we have to identify the % of Full load current on which the motor is running continuously. Thermal setting = (Full load current x % of Full load) / CT ratio. Short circuit Element - The range available for this element is 1 to 5 times of starting current. Time delay is also available. We normally set at 2 times of I starting with a time delay of 0.1 second.
Single Phasing Element - This element will operate, if there is an unbalance in current of three phases. It is also called as unbalance protection. The element is set for 1/3rd of starting current. If it tripped during starting, then the parameter will changed to 1/2 of starting current. Earth Fault Protection - This element measures the neutral current of star connected CT secondary. The range available for this element is 0.02 to 2 times of CT primary current. Time delay is also available. We normally set at 0.1 times of CT primary current with a time delay of 0.2 seconds. If tripped during starting of motor, then the time setting can be raised to 0.5 sec. Locked rotor protection - The range available for this element is 1 to 5 times of full load current. Time delay is also available. We normally set at 2 times of FLC. The time delay will be more than the starting time of the motor. "Starting time means the time require by the motor to reach its full speed." Number of hot start protection - Here we will provide the number of start allowed in specified time duration. By this we will limit the number of hot start given to the motor. The schematic diagram to connect a motor protection relay is as below
Modern digital motor protection relays are having some extra features, i.e. protection against no load running of a motor and thermal protection. In case of no load running, the relay senses the motor current. If it is less than the specified value then it will trip the motor. We can also connect the temperature probe to the relay, which will monitor the bearing and winding temperature and trip the motor if it exceeds the specified value. Feeder Protection Relays. Distance Protection Relay One of the important protections in Power System Protection is Feeder Protection. Different types of relays were used for feeder protection like electromagnetic type relays, static type relays etc. But now a day we are using Numerical relays for all protections. The benefits of Numerical relays are, 1. Accurate tripping, 2. Less tolerance, 3. Fault events and counter storage 4. Display of fault parameters on screen (Fault parameters means current, voltage, resistance and reactance values during fault and fault distance, Numerical relays can store thousands of tripping events). The main inputs needed for distance protection is Voltage and Current from corresponding feeder PT and CT. According to site condition we will set certain impedance values in relay settings (i.e., R and X values) for fault detection. Relay will monitor Current & Voltage in the feeder line (PT and CT secondary), and from these values, the relay will calculate Impedance value Z. i.e., Z=V/I. In normal load condition the impedance values on line will be high. But when fault comes on the feeder line, the impedance will decrease and becomes less than the impedance setting in the relay, then the distance relay will trip with in 40 ms in zone 1 (different zones are there and that will be explained later) and isolate power equipments from fault. I.e., during fault, relay will trip and shows the fault parameters like fault current, voltage, reactance, resistance and fault distance on relay screen. Suppose if the fault is on 25km, then relay will show Fault Distance (FD) = 25km, and thus it becomes easy to identify the location where there is fault. For Distance protection now a day Quadrilateral characteristics is used. We already discussed that for fault identification we have to set different parameters in relay. i.e., 1. Forward and Backward Resistance (RF, RB), 2. Forward and backward Reactance (XF, XB), 3. RCA (Relay Characteristics Angle) and 4. Line Impedance per km.
These parameters are used for making Quadrilateral characteristics. Suppose if RCA=70° and we are using parallelogram characteristics (Quadrilateral), we can plot the graph by setting Forward Resistance (RF) value in positive X axis, Backward Resistance (RB) in negative X axis, Forward Reactance (XF) value in positive Y axis, and Backward Reactance (XB) in negative Y axis. And plot parallelogram with a slope of RCA angle. Thus we will get a parallelogram graph and the protection zone is inside the parallelogram. Means during fault the impedance will reach inside the parallelogram then the relay will trip. In graph there are 4 quadrants of operation 1. First quadrant (R and X values + ve) If the load is inductive and the fault is in forward direction from Relay, then the relay will trip in this quadrant values. 2. Second quadrant (R – ve and X + ve) If the load is capacitive and the fault is in reverse direction from Relay, then the relay will trip in this quadrant values. 3. Third quadrant (R – ve and X - ve) If the load is inductive and the fault is in reverse direction from Relay, then the relay will trip in this quadrant values. 4. Fourth quadrant (R + ve and X - ve) If the load is capacitive and the fault is in forward direction from Relay, then the relay will trip in this quadrant values. Different Zones of operation, fault distance calculation and other feeder protection relays etc. will be explained in next article... Remark: A model of
quadrilateral characteristics is shown below….
Types of Electrical Protection Relays or Protective Relays. Definition of Protective Relay A relay is automatic device which senses an abnormal condition of electrical circuit and closes its contacts. These contacts in turns close and complete the circuit breaker trip coil circuit hence make the circuit breaker tripped for disconnecting the faulty portion of the electrical circuit from rest of the healthy circuit. Now let’s have a discussion on some terms related to protective relay. Pickup Level of Actuating Signal: The value of actuating quantity (voltage or current) which is on threshold above which the relay initiates to be operated. If the value of actuating quantity is increased, the electromagnetic effect of the relay coil is increased and above a certain level of actuating quantity the moving mechanism of the relay just starts to move. Reset Level: The value of current or voltage below which a relay opens its contacts and comes in original position. Operating Time of Relay: Just after exceeding pickup level of actuating quantity the moving mechanism (for example rotating disc) of relay starts moving and it ultimately close the relay contacts at the end of its journey. The time which elapses between the instant when actuating quantity exceeds the pickup value to the instant when the relay contacts close. Reset Time of Relay: The time which elapses between the instant when the actuating quantity becomes less than
the reset value to the instant when the relay contacts returns to its normal position. Reach of Relay: A distance relay operates whenever the distance seen by the relay is less than the pre-specified impedance. The actuating impedance in the relay is the function of distance in a distance protection relay. This impedance or corresponding distance is called reach of the relay. Power system protection relays can be categorized into different types of relays. Types of Relays Types of protection relays are mainly based on their characteristic, logic, on actuating parameter and operation mechanism. Based on operation mechanism protection relay can be categorized as electromagnetic relay, static relay and mechanical relay. Actually relay is nothing but a combination of one or more open or closed contacts. These all or some specific contacts the relay change their state when actuating parameters are applied to the relay. That means open contacts become closed and closed contacts become open. In electromagnetic relay these closing and opening of relay contacts are done by electromagnetic action of a solenoid. In mechanical relay these closing and opening of relay contacts are done by mechanical displacement of different gear level system. In static relay it is mainly done by semiconductor switches like thyristor. In digital relay on and off state can be referred as 1 and 0 state. Based on Characteristic the protection relay can be categorized as1. Definite time relays 2. Inverse time relays with definite minimum time(IDMT) 3. Instantaneous relays. 4. IDMT with inst. 5. Stepped characteristic. 6. Programmed switches. 7. Voltage restraint over current relay. Based on of logic the protection relay can be categorized as1. Differential. 2. Unbalance. 3. Neutral displacement. 4. Directional. 5. Restricted earth fault. 6. Over fluxing. 7. Distance schemes. 8. Bus bar protection. 9. Reverse power relays. 10. Loss of excitation. 11. Negative phase sequence relays etc. Based on actuating parameter the protection relay can be categorized as-
1. Current relays. 2. Voltage relays. 3. Frequency relays. 4. Power relays etc. Based on application the protection relay can be categorized as1. Primary relay. 2. Backup relay. Primary relay or primary protection relay is the first line of power system protection whereas backup relay is operated only when primary relay fails to be operated during fault. Hence backup relay is slower in action than primary relay. Any relay may fail to be operated due to any of the following reasons, 1. 2. 3. 4.
The protective relay itself is defective. DC Trip voltage supply to the relay is unavailable. Trip lead from relay panel to circuit breaker is disconnected. Trip coil in the circuit breaker is disconnected or defective. 5. Current or voltage signals from CT or PT respectively is unavailable. As because backup relay operates only when primary relay fails, backup protection relay should not have anything common with primary protection relay. Some examples of Mechanical Relay are1. o o o 2. o o o o 3. 4. 5.
Thermal OT trip (Oil Temperature Trip) WT trip (Winding Temperature Trip) Bearing temp trip etc. Float type Buchholz OSR PRV Water level Controls etc. Pressure switches. Mechanical interlocks. Pole discrepancy relay.
List Different Protective Relays are used for Different Power System Equipment Protection Now let’s have a look on which different protective relays are used in different power system equipment protection schemes. Relays for Transmission & Distribution Lines Protection SL
Lines to be
Relays to be used
protected 400 KV Transmission Line
Main-I: Non switched or Numerical Distance Scheme Main-II: Non switched or Numerical Distance Scheme
220 KV Transmission Line
Main-I : Non switched distance scheme (Fed from Bus PTs) Main-II: Switched distance scheme (Fed from line CVTs) With a changeover facility from bus PT to line CVT and vice-versa.
3
132 KV Transmission Line
Main Protection : Switched distance scheme (fed from bus PT). Backup Protection: 3 Nos. directional IDMT O/L Relays and 1 No. Directional IDMT E/L relay.
4
33 KV lines
Non-directional IDMT 3 O/L and 1 E/L relays.
5
11 KV lines
Non-directional IDMT 2 O/L and 1 E/L relays.
1
2
Relays for Transformer Protection Voltage Ratio and SL Capacity of Transformer 1
11/132 KV Generator Transformer
Relays on HV Side
Relays on LV Side
3 nos Non-Directional O/L Relay 1 no Non-Directional E/L Relay and/or standby E/F + REF Relay
--
Common Relays
Differential Relay or Overall differential Relay Overflux Relay Buchholz Relay OLTC Buchholz Relay PRV Relay
OT Trip Relay WT Trip Relay
2
3
4
13.8/220 KV 15.75/220 KV 18/400 KV 21/400 KV Generator Transformer
3 nos Non-Directional O/L Relay 1 no Non-Directional E/L Relay and/or standby E/F + REF Relay
220 /6.6KV Station Transformer
3 nos Non-Directional O/L Relay 1 no Non-Directional E/L Relay and/or standby E/F + REF Relay
Gen-volt/6.6KV UAT
3 nos Non-Directional O/L Relay
--
Differential Relay or Overall differential Relay Overflux Relay Buchholz Relay OLTC Buchholz Relay PRV Relay OT Trip Relay WT Trip Relay
3 nos NonDirectional O/L Relay
Differential Relay Overflux Relay Buchholz Relay OLTC Buchholz Relay PRV Relay OT Trip Relay WT Trip Relay
3 nos NonDirectional O/L Relay
Differential Relay Overflux Relay Buchholz Relay OLTC Buchholz
Relay PRV Relay OT Trip Relay WT Trip Relay
5
6
7
8
132/33/11KV upto 8 MVA
132/33/11KV above 8 MVA & below 31.5 MVA
3 nos O/L Relay 1 no E/L Relay
3 nos O/L Relay 1 no Directional E/L Relay
2 nos O/L Relays 1 no E/L Relay
Buchholz Relay OLTC Buchholz Relay PRV Relay OT Trip Relay WT Trip Relay
3 nos O/L Relay 1 no E/L Relay
Differential Relay Buchholz Relay OLTC Buchholz Relay PRV Relay OT Trip Relay WT Trip Relay Differential Relay Overflux Relay Buchholz Relay OLTC Buchholz Relay PRV Relay OT Trip Relay WT Trip Relay
132/33KV, 31.5 MVA & above
3 nos O/L Relay 1 no Directional E/L Relay
3 nos O/L Relay 1 no E/L Relay
220/33 KV, 31.5MVA &
3 nos O/L Relay 1 no Directional E/L
3 nos O/L Relay Differential 1 no Directional E/L Relay
50MVA 220/132KV, 100 Relay MVA
9
400/220KV 315MVA
3 nos Directional O/L Relay (with dir.highset) 1 no Directional E/L relay. Restricted E/F relay 3 nos Directional O/L Relay for action
Relay
Overflux Relay Buchholz Relay OLTC Buchholz Relay PRV Relay OT Trip Relay WT Trip Relay
Differential Relay Overflux Relay Buchholz 3 nos Directional Relay O/L Relay OLTC (with dir.highset) Buchholz 1 no Directional E/L Relay relay. PRV Relay Restricted E/F relay OT Trip Relay WT Trip Relay Over Load (Alarm) Relay
Points to be remembered in respect of protection of transformers 1.
2. 3.
4.
5. 6. 7. 8.
No Buchholz relay for transformers below 500 KVA capacity. Transformers up to 1500 KVA shall have only Horn gap protection. Transformers above 1500 KVA and upto 8000 KVA of 33/11KV ratio shall have one group control breaker on HV side and individual LV breakers if there is more than one transformer. Transformers above 8000 KVA shall have individual HV and LV circuit breakers. The relays indicate above shall be provided on HV and LV. LAs to be provided on HV and LV for transformers of all capacities and voltage class. OLTC out of step protection is to be provided where master follower scheme is in operation. Fans failure and pumps failure alarms to be connected.
9.
Alarms for O.T., W.T., Buchholz (Main tank AND OLTC) should be connected.
Equivalent Circuit of Transformer referred to Primary and Secondary. Equivalent Circuit of Transformer Equivalent impedance of transformer is essential to be calculated because the electrical power transformer is an electrical power system equipment for estimating different parameters of electrical power system which may be required to calculate total internal impedance of an electrical power transformer, viewing from primary side or secondary side as per requirement. This calculation requires equivalent circuit of transformer referred to primary or equivalent circuit of transformer referred to secondary sides respectively. Percentage impedance is also very essential parameter of transformer. Special attention is to be given to this parameter during installing a transformer in an existing electrical power system. Percentage impedance of different power transformers should be properly matched during parallel operation of power transformers. The percentage impedance can be derived from equivalent impedance of transformer so, it can be said that equivalent circuit of transformer is also required during calculation of % impedance. Equivalent Circuit of Transformer Referred to Primary For drawing equivalent circuit of transformer referred to primary, first we have to establish general equivalent circuit of transformer then, we will modify it for referring from primary side. For doing this, first we need to recall the complete vector diagram of a transformer which is shown in the
figure below. consider the transformation ratio be,
Let us
In the figure above, the applied voltage to the primary is V1 and voltage across the primary winding is E1. Total current supplied to primary is I1. So the voltage V1 applied to the primary is partly dropped by I1Z1 or I1R1 + j.I1X1 before it appears across primary winding. The voltage appeared across winding is countered by primary induced emf E1. So voltage equation of this portion of the transformer can be written as,
The equivalent circuit for that equation can be drawn as below,
From the vector diagram above, it is found that the total primary current I1 has two components, one is no load component Io and the other is load component I2′. As this primary current has two components or branches, so there must be a parallel path with primary winding of transformer. This parallel path of current is known as excitation branch of equivalent circuit of transformer. The resistive and reactive branches of the excitation circuit can be represented as
The load component I2′ flows through the primary winding of transformer and induced voltage across the winding is E1 as shown in the figure right. This induced voltage E1transforms to secondary and it is E2 and load component of primary current I2′ is transformed to secondary as secondary current I2. Current of secondary is I2. So the voltage E2 across secondary winding is partly dropped by I2Z2 or I2R2 + j.I2X2 before it appears across load. The load voltage is V2.
The
complete
equivalent
circuit
of
transformer
is
shown
below.
Now if we see the voltage drop in secondary from primary side, then it would be ′K′ times greater and would be written as K.Z2.I2. Again I2′.N1 = I2.N2
Therefore,
From above equation, secondary impedance of transformer referred to primary is,
So, the complete equivalent circuit of transformer referred to primary is shown in the figure below,
Approximate Equivalent Circuit of Transformer Since Io is very small compared to I1, it is less than 5% of full load primary current, Iochanges the voltage drop insignificantly. Hence, it is good
approximation to ignore the excitation circuit in approximate equivalent circuit of transformer. The winding resistance and reactance being in series can now be combined into equivalent resistance and reactance of transformer, referred to any particular side. In this case it is side 1 or primary side.
Equivalent Circuit of Transformer Referred to Secondary In similar way, approximate equivalent circuit of transformer referred to secondary can be drawn. Where equivalent impedance of transformer referred to secondary, can be derived as.
Knee Point Voltage of Current Transformer PS Class. Current Transformer PS Class Before understanding Knee Point Voltage of Current Transformer and current transformer PS class we should recall the terms instrument security factor of CT and accuracy limit factor. Instrument Security Factor or ISF of Current Transformer Instrument security factor is the ratio of instrument limit primary current to the rated primary current. Instrument limit current of a metering current transformer is the maximum value of primary current beyond which current transformer core becomes saturated. Instrument security factor of CT is the significant factor for choosing the metering instruments which to be connected to the secondary of the CT. Security or Safety of the measuring unit is better, if ISF is low. If we go through the example below it would be clear to us. Suppose one current transformer has rating 100/1 A and ISF is 1.5 and another current transformer has same rating with ISF 2. That means, in first CT, the metering core would be saturated at 1.5 × 100 or 150 A, whereas is second CT, core will be saturated at 2 × 100 or 200 A. That means whatever may be the primary current of both CTs, secondary current will not increase further after 150 and 200 A of primary current of the CTs respectively. Hence maximum secondary current of the CTs would be 1.5 and 2.0 A. As the maximum current can flow through the instrument connected to the first CT is 1.5 A which is less than the maximum value of current can flow through the instrument connected to the second CT i.e. 2 A. Hence security or safety of the instruments of first CT is better than later. Another significance of ISF is during huge electrical fault, the short circuit current, flows through primary of the CT does not affect destructively, the measuring instrument attached to it as because, the secondary current of the CT will not rise above the value of rated secondary current multiplied by ISF. Accuracy Limit Factor or ALF of Current Transformer
For protection current transformer, the ratio of accuracy limit primary current to the rated primary current. First we will explain, what is rated accuracy limit primary current?. Broadly, this is the maximum value of primary current, beyond which core of the protection CT or simply protection core of of a CT starts saturated. The value of rated accuracy limit primary current is always many times more than the value of instrument limit primary current. Actually CT transforms the fault current of the electrical power system for operation of the protection relays connected to the secondary of that CT. If the core of the CT becomes saturated at lower value of primary current, as in the case of metering CT, the system fault will not reflect properly to the secondary, which may cause, the relays remain inoperative even the fault level of the system is large enough. That is why the core of the protection CT is made such a way that saturation level of that core must be high enough. But still there is a limit as because, it is impossible to make one magnetic core with infinitely high saturation level and secondly most important reason is that although the protection care should have high saturation level but that must be limited up to certain level otherwise total transformation of primary current during huge fault may badly damage the protection relays. So it is clear from above explanation, rated accuracy limit primary current, should not be so less, that it will not at all help the relays to be operated on the other hand this value must not be so high that it can damage the relays. So, accuracy limit factor or ALF should not have the value nearer to unit and at the same time it should not be as high as 100. The standard values of ALF as per IS-2705 are 5, 10, 15, 20 and 30. Knee Point Voltage of Current Transformer This is the significance of saturation level of a CT core mainly used for protection purposes. The sinusoidal voltage of rated frequency applied to the secondary terminals of current transformer, with other winding being open circuited, which when increased by 10% cause the exiting current to increase 50%. The CT core is made of CRGO steel. It has its won saturation level. The EMF induced in the CT secondary windings is E2 = 4.44φfT2 Where, f is the system frequency, φ is the maximum magnetic flux in Wb. T2 is the number of turns of the secondary winding. The flux in the core, is produced by excitation current Ie. We have a non-liner relationship between excitation current and magnetizing flux. After certain value of excitation current, flux will not further increase so rapidly with increase in excitation current. This non-liner relation curve is also called B - H curve. Again from the equation above, it is found that, secondary voltage of acurrent
transformer is directly proportional to flux φ. Hence one typical curve can be drawn from this relation between secondary voltage and excitation current as shown below. It is clear from the curve that, linear relation between V and Ie is maintained from point A and K. The point ′A′ is known as ′ankle point′ and point ′K′ is known as ′Knee Point′.
In differential and restricted earth fault (REF) protection scheme, accuracy class and ALF of the CT may not ensure the reliability of the operation. It is desired that, differential and REF relays should not be operated when fault occurs outside the protected transformer. When any fault occurs outside the differential protection zone, the faulty current flows through the CTs of both sides of electrical power transformer. The both LV and HV CTs have magnetizing characteristics. Beyond the knee point, for slight increase in secondary emf a large increasing in excitation current is required. So after this knee point excitation current of both current transformers will be extremely high, which may cause mismatch between secondary current of LV & HV current transformers. This phenomena may cause unexpected tripping of power transformer. So the magnetizing characteristics of both LV & HV sides CTs, should be same that means they have same knee point voltage Vk as well as same excitation current Ieat Vk/2. It can be again said that, if both knee point voltage of current transformer and magnetizing characteristic of CTs of both sides of power transformer differ, there must be a mismatch in high excitation currents of the CTs during fault which ultimately causes the unbalancing between secondary current of both groups of CTs and transformer trips. So for choosing CT for differential protection of transformer, one should consider current transformer PS class rather its convectional protection
class. PS stands for protection special which is defined by knee point voltage of current transformer Vk and excitation current Ie at Vk/2. Why CT Secondary Should Not Be Kept Open? The electrical power system load current always flows through current transformer primary; irrespective of whether the current transformer is open circuited or connected to burden at its secondary.
If CT secondary is open circuited, all the primary current will behave as excitation current, which ultimately produce huge voltage. Every current transformer has its won non-linear magnetizing curve, because of which secondary open circuit voltage should be limited by saturation of the core. If one can measure the rms voltage across the secondary terminals, he or she will get the value which may not appear to be dangerous. As the CT primary current is sinusoidal in nature, it zero 100 times per second.(As frequency of the current is 50 Hz). The rate of change of flux at every current zero is not limited by saturation and is high indeed. This develops extremely high peaks or pulses of voltage. This high peaks of voltage may not be measured by conventional voltmeter. But these high peaks of induced voltage may breakdown the CT insulation, and may case accident to personnel. The actual open-circuit voltage peak is difficult to measure accurately because of its very short peaks. That is why CT secondary should not be kept open. Theory of Transformer We have discussed about the theory of ideal transformer for better understanding of actual elementary theory of transformer. Now we will go through the practical aspects one by one of an electrical power transformer and try to draw vector diagram of transformer in every step.
As we said that, in an ideal transformer; there are no core losses in transformer i.e. loss free core of transformer. But in practical transformer, there are hysteresis and eddy current losses in transformer core. Theory of Transformer on No-Load Theory of Transformer On No-load, and Having No Winding Resistance and No Leakage Reactance of Transformer Let us consider one electrical transformer with only core losses, which means, it has only core losses but no copper loss and no leakage reactance of transformer. When an alternating source is applied in the primary, the source will supply the current for magnetizing the core of transformer. But this current is not the actual magnetizing current, it is little bit greater than actual magnetizing current. Actually, total current supplied from the source has two components, one is magnetizing current which is merely utilized for magnetizing the core and other component of the source current is consumed for compensating the core losses in transformer. Because of this core loss component, the source current in transformer on noload condition supplied from the source as source current is not exactly at 90° lags of supply voltage, but it lags behind an angle θ is less than 90°. If total current supplied from source is I o, it will have one component in phase with supply voltage V1 and this component of the current I w is core loss component. This component is taken in phase with source voltage, because it is associated with active or working losses in transformer. Other component of the source current is denoted as I μ. This component produces the alternating magnetic flux in the core, so it is watt-less; means it is reactive part of the transformer source current. Hence I μ will be in quadrature with V1 and in phase with alternating flux Φ. Hence, total primary current in transformer on no-load condition can be represented as
Now you have seen how simple is to explain the theory of transformer in no-load.
Theory of Transformer on Load Theory of Transformer On Load But Having No Winding Resistance and Leakage Reactance Now we will examine the behavior of above said transformer on load, that means load is connected to the secondary terminals. Consider, transformer having core loss but no copper loss and leakage reactance. Whenever load is connected to the secondary winding, load current will start to flow through the load as well as secondary winding. This load current solely depends upon the characteristics of the load and also upon secondary voltage of the transformer. This current is called secondary current or load current, here it is denoted as I2. As I2 is flowing through the secondary, a self mmf in secondary winding will be produced. Here it is N2I2, where, N2 is the number of turns of the secondary winding of transformer.
This mmf or magneto motive force in the secondary winding produces flux φ2. This φ2 will oppose the main magnetizing flux and momentarily weakens the main flux and tries to reduce primary self induced emf E1. If E1 falls down below the primary source voltage V1, there will be an extra current flowing from source to primary winding. This extra primary current I2′ produces extra flux φ′ in the core which will neutralize the secondary counter flux φ2. Hence the main magnetizing flux of core, Φ remains unchanged irrespective of load. So total current, this transformer draws from source can be divided into two components, first one is utilized for magnetizing the core and compensating the core loss i.e. Io. It is no-load component of the primary current. Second one is utilized for compensating the counter flux of the secondary winding. It is known as load component of the primary current. Hence total no load primary current I1 of a electrical power transformer having no winding resistance and leakage reactance can be represented as follows Where, θ2 is the angle between Secondary Voltage and Secondary Current of transformer. Now we will proceed one further step toward more practical aspect of a transformer.
Theory of Transformer On Load, With Resistive Winding, But No Leakage Reactance Now, consider the winding resistance of transformer but no leakage reactance. So far we have discussed about the transformer which has ideal windings, means winding with no resistance and leakage reactance, but now we will consider one transformer which has internal resistance in the winding but no leakage reactance. As the windings are resistive, there would be a voltage drop in the windings.
We have proved earlier that, total primary current from the source on load is I1. The voltage drop in the primary winding with resistance, R1 is R1I1. Obviously, induced emf across primary winding E1, is not exactly equal to source voltage V1. E1 is less than V1 by voltage drop I1R1.
Again in the case of secondary, the voltage induced across the secondary winding, E2 does not totally appear across the load since it also drops by an amount I2R2, where R2 is the secondary winding resistance and I2 is secondary current or load current. Similarly, voltage equation of the secondary side of the transformer will be
Theory of Transformer On Load, With Resistance As Well As Leakage Reactance in Transformer Windings Now we will consider the condition, when there is leakage reactance of transformer as well as winding resistance of transformer.
Let leakage reactances of primary and secondary windings of the transformer are X1 and X2respectively. Hence total impedance of primary and secondary winding of transformer with resistance R1 and R2 respectively, can be represented as,
We have already established the voltage equation of a transformer on load, with only resistances in the windings, where voltage drops in the windings occur only due to resistive voltage drop. But when we consider leakage reactances of transformer windings, voltage drop occurs in the winding not only because of resistance, it is because of impedance of transformer windings. Hence, actual voltage equation of a transformer can
easily be determined by just replacing resistances R1 & R2 in the previously established voltage equations by Z1 and Z2. Therefore, the voltage equations are,
Resistance drops are in the direction of current vector but, reactive drop will be perpendicular to the current vector as shown in the above vector diagram of transformer. Small and Large Motor Protection Scheme. The abnormalities in motor or motor faults may appear due to mainly two reasons 1. Conditions imposed by the external power supply network, 2. Internal faults, either in the motor or in the driven plant. Unbalanced supply voltages, under-voltage, reversed phase sequence and loss of synchronism (in the case of synchronous motor) come under former category. The later category includes bearing failures, stator winding faults, motor earth faults and overload etc. The degree of motor protection system depends on the costs and applications of the electrical motor. Small Motor Protection Scheme Generally motors up to 30 hp are considered in small category. The small motor protection in this case is arranged by HRC fuse, bimetallic relay and under voltage relay - all assembled into the motor contractor - starter itself. Most common cause of motor burn outs on LV fuse protected system is due to single phasing. This single phasing may remain undetected even if the motors are protected by conventional bimetallic relay. It can not be detected by a set of voltage relays connected across the lines. Since, even when one phase is dead, the motor maintains substantial back emf on its faulty phase terminal and hence voltage across the voltage relay is prevented from dropping - off. The difficulties of detecting single phasing can be overcome by employing a set of three current operated relays as shown in the small
motor protection circuit given below. The current operated relays are very simple instantaneous relays. There are mainly two parts in this relay one is a current coil and other is one or more normally open contacts (NO Contacts). The NO contacts are operated by the mmf of the current coil. This relay is connected in series with each phase of the supply and backup by HRC fuse. When the electrical motor starts and runs the supply current passes through the current coil of the protective relay. The mmf of the current coil makes the NO contacts closed. If suddenly a single phasing occurs the corresponding current through the current coil will falls and the contacts of the corresponding relay will become to its normal open position. The NO contacts of the all three relays are connected in series to hold - in the motor contractor. So if any one relay contact opens, results to release of motor contractor and hence motor will stop running.
Large Motor Protection Large motor specially induction motors require protection against1. 2. 3. 4.
Motor bearing failure, Motor over heating, Motor winding failure, Reverse motor rotation.
Motor Bearing Failure
Ball and roller bearings are used for the motor up to 500 hp and beyond this size sleeve bearings are used. failure of ball or roller bearing usually causes the motor to a standstill very quickly. Due to sudden mechanical jamming in motor bearing, the input current of the motor becomes very high. Current operated protection, attached to the input of the motor can not serve satisfactorily. Since this motor protection system has to be set to override the high motor starting current. The difficulty can be over come by providing thermal over load relay. As the starting current of the motor is high but exists only during starting so for that current the there will be no over heating effect. But over current due mechanical jamming exists for longer time hence there will be a over heating effect. So stalling motor protection can be offered by the thermal overload relay. Stalling protection can also be provided by separate definite time over current relay which is operated only after a certain predefined time if over current persists beyond that period. In the case of sleeve bearing, a temperature sensing device embedded in the bearing itself. This scheme of motor protection is more reliable and sensitive to motor bearing failure since the thermal withstand limit of the motor is quite higher than that of bearing. If we allow the bearing over heating and wait for motor thermal relay to trip, the bearing may be permanently damaged. The temperature sensing device embedded in the bearing stops the motor if the bearing temperature rises beyond its predefined limit. Motor Over Heating The main reason of motor over heating that means over heating of motor winding is due to either of mechanical over loading, reduced supply voltage, unbalanced supply voltage and single phasing. The over heating may cause deterioration of insulation life of motor hence it must be avoided by providing proper motor protection scheme. To avoid over heating, the motor should be isolated in 40 to 50 minutes even in the event of small overloads of the order of 10 %. The protective relay should take into account the detrimental heating effects on the motor rotor due to negative sequence currents in the stator arising out of unbalance in supply voltage. The motor should also be protected by instantaneous motor protection relay against single phasing such as a stall on loss of one phase when running at full load or attempting to start with only two of three phases alive. Motor Winding Failure The motor protection relay should should have instantaneous trip elements to detect motor winding failure such as phase to phase and phase to earth faults. Preferably phase to phase fault unit should be energized from positive phase sequence component of the motor current and
another instantaneous unit connected in the residual circuit of the current transformers be used for earth faults protection. Reverse Motor Rotation Specially in the case of conveyor belt, the reverse motor rotation must be avoided. The reverse rotation during starting can be caused due to inadvertent reversing of supply phases. A comprehensive motor protection relay with an instantaneous negative sequence unit will satisfy this requirement. If such relay has not been provided, a watt-meter type relay can be employed. NB: However, we have to provide some additional motor protection system for synchronous motor which is discussed in details in synchronous motor protection topic. Motor Thermal Overload Protection. For understanding motor thermal overload protection in induction motor we can discuss the operating principle of three phase induction motor. There is one cylindrical stator and a three phase winding is symmetrically distributed in the inner periphery of the stator. Due to such symmetrical distribution, when three phase power supply is applied to the stator winding, a rotating magnetic field is produced. This field rotates at synchronous speed. The rotor is created in induction motor mainly by numbers solid copper bars which are shorted at both ends in such a manner that they form a cylinder cage like structure. This is why this motor is also referred as squirrel cage induction motor. Anyway let's come to the basic point of three phase induction motor - which will help us to understand clearly about motor thermal overload protection.As the rotating magnetic flux cuts each of the bar conductor of rotor, there will be an induced circulating current flowing through the bar conductors. At starting the rotor is stand still and stator field is rotating at synchronous speed, the relative motion between rotating field and rotor is maximum. Hence, the rate of cuts of flux with rotor bars is maximum, the induced current is maximum at this condition. But as the cause of induced current is, this relative speed, the rotor will try to reduce this relative speed and hence it will start rotating in the direction of rotating magnetic field to catch the synchronous speed. As soon as the rotor will come to the synchronous speed this relative speed between rotor and rotating magnetic field becomes zero, hence there will not be any further flux cutting and consequently there will not be any induced current in the rotor bars. As the induced current becomes zero, there will not be any further need of maintaining zero relative speed between rotor and rotating magnetic field hence rotor speed falls. As soon as the rotor speed falls the relative speed between rotor and rotating magnetic field again acquires a non zero value which again causes induced
current in the rotor bars then rotor will again try to achieve the synchronous speed and this will continue till the motor is switch on. Due to this phenomenon the rotor will never achieve the synchronous speed as well as it will never stop running during normal operation. The difference between the synchronous speed with rotor speed in respect of synchronous speed, is termed as slip of induction motor. The slip in a normally running induction motor typically varies from 1 % to 3 % depending upon the loading condition of the motor. Now we will try to draw speed current characteristics of induction motor – let’s have an
example of large boiler fan. In the characteristic Y axis is taken as time in second, X axis is taken as % of stator current. When rotor is stand still that is at starting condition, the slip is maximum hence the induced current in the rotor is maximum and due to transformation action, stator will also draw a heavy current from the supply and it would be around 600 % of the rated full load stator current. As the rotor is being accelerated the slip is reduced, consequently the rotor current hence stator current falls to around 500 % of the full load rated current within 12 seconds when the rotor speed attains 80 % of synchronous speed. After that the stator current falls rapidly to the rated value as the rotor reaches its normal speed. Now we will discuss about thermal over loading of electrical motor or over heating problem of electric motor and the necessity of motor thermal overload protection. Whenever we think about the overheating of a motor, the first thing strikes in our mind is over loading. Due to mechanical over loading of the motor draws higher current from the supply which leads to excessive over heating of the motor. The motor can also be excessively over heated if the rotor is mechanically locked i.e. becomes stationary by any external mechanical force. In this situation the motor will draw excessively high current from the supply which also leads to thermal over loading of
electrical motor or excessive over heating problem. Another cause of overheating is low supply voltage. As the power id drawn by the motor from the supply depends upon the loading condition of the motor, for lower supply voltage, motor will draw higher current from mains to maintain required torque. Single phasing also causes thermal over loading of motor. When one phase of the supply is out of service, the remaining two phases draw higher current to maintain required load torque and this leads to overheating of the motor. Unbalance condition between three phases of supply also causes over heating of the motor winding, as because unbalance system results to negative sequence current in the stator winding. Again, due to sudden loss and reestablish of supply voltage may cause excessive heating of the motor. Since due to sudden loss of supply voltage, the motor is deaccelerated and due to sudden reestablishment of voltage the motor is accelerated to achieve its rated speed and hence for that motor draws higher current form the supply. As the thermal over loading or over heating of the motor may lead to insulation failure and damage of winding, hence for proper motor thermal overload protection, the motor should be protected against the following conditions 1. Mechanical over loading, 2. Stalling of motor shaft, 3. Low supply voltage, 4. Single phasing of supply mains, 5. Unbalancing of supply mains, 6. Sudden Loss and rebuilding of supply voltage. The most basic protection scheme of the motor is thermal over load protection which primarily covers the protection of all the above mentioned condition. To understand the basic principle of thermal over load protection let’s examine the schematic diagram of basic motor control scheme.
In the figure above, when START push is closed, the starter coil is energized through the transformer. As the starter coil is energized, normally open (NO) contacts 5 are closed hence motor gets supply voltage at its terminal and it starts rotating. This start coil also closes contact 4 which makes the starter coil energized even the START push button contact is released from its close position. To stop the motor there are several normally closed (NC) contacts in series with the starter coil as shown in the figure. One of them is STOP push button contact. If the STOP push button is pressed, this button contact opens and breaks the continuity of the starter coil circuit consequently makes the starter coil de-energized. Hence the contact 5 and 4 come back to their normally open position. Then, in absence of voltage at motor terminals it will ultimately stop running. Similarly any of the other NC contacts (1, 2 & 3) connected in series with starter coil if open; it will also stop the motor. These NC contacts are electrically coupled with various protection relays to stop operation of the motor in different abnormal conditions. Let’s look at the thermal over load relay and its function in motor thermal overload protection. The secondary of the CTs in series with motor supply circuit, are connected with a bimetallic strip of the thermal over load relay (49). As shown in the figure below, when current through the secondary of any of the CTs, crosses it’s predetermined values for a predetermined time, the bimetallic strip is over heated and it deforms which ultimately causes to operate the relay 49. As soon as the relay 49 is operated, the NC contacts 1 and 2 are opened which de-energizes the starter coil and hence stop the
motor. An other thing we have to remember during providing motor thermal overload protection. Actually every motor does have some predetermined overload tolerance value. That means every motor may run beyond its rated load for a specific allowable period depending on its loading condition. How long a motor can run safely for a particular load is specified by the manufacturer. The relation between different loads on motor and corresponding allowable periods for running the same in safe condition is referred as thermal limit curve of the motor. Let’s look at the curve of a particular motor, given
below. Here Y axis or vertical axis represents the allowable time in seconds and X axis or horizontal axis represents percentage of overload. Here it is clear from the curve that, motor can run safely without any damage due to overheating for prolonged period at 100 % of the rated load. It can run safely 1000 seconds at 200 % of normal rated load. It can run safely 100 seconds at 300 % of normal rated load. It can run safely 15 seconds at 600 % of normal rated load. The upper portion of the curve represents the normal running condition of the rotor and the lower most portion represents the mechanical locked condition of the rotor. Now the operating time Vs actuating current curve of the chosen thermal over load relay should be situated below the thermal limit curve of the motor for satisfactory and safe operation. Let’s have a discussion on more details-
Remember the characteristics of starting current of the motor – During start up of the induction motor, the stator current goes beyond 600 % of normal rated current but it stays up to 10 to 12 seconds after that stator current suddenly falls to normal rated value. So if the thermal overload relay is operated before that 10 to 12 seconds for the current 600 % of normal rated then the motor cannot be started. Hence, it can be concluded that the operating time Vs actuating current curve of the chosen thermal over load relay should be situated below the thermal limit curve of the motor but above the starting current characteristics curve of the motor. Probable position of the thermal current relay characteristics is bounded by these two said curves as shown in the graph by highlighted area. Another thing has to be remembered during choosing of thermal overload relay. This relay is not an instantaneous relay. It has a minimum delay in operation as the bimetallic strip required a minimum time to be heated up and deformed for maximum value of operating current. From the graph it is found that the thermal relay will be operated after 25 to 30 seconds if either the rotor is suddenly mechanically blocked or motor is fail to start. At this situation the motor will draw a huge current from the supply. If the motor is not isolated sooner, severer damage may occur.
This problem is overcome by providing time over current relay with high pickup. The time current characteristics of these over current relays are so chosen that for lower value of over load, the relay will not operate since thermal overload relay will be actuated before it. But for higher value of overload and for blocked rotor condition time over load relay will be operated instead of thermal relay because former will actuate much before the latter. Hence both the bimetallic over load relay and time over current relay are provided for complete motor thermal overload protection. There is one main disadvantage of bimetallic thermal over load relay, as the rate of heating and cooling of bi-metal is affected by ambient temperature, the performance of the relay may differ for different ambient temperatures. This problem can be overcome by using RTD or resistance temperature detector. The bigger and more sophisticated motors are protected against thermal over load more accurately by using RTD. In stator slots, RTDs are placed along with stator winding. Resistance of the RTD changes with changing temperature and this changed resistive value is sensed by a Wheatstone bridge circuit. This motor thermal overload protection scheme is very simple. RTD of stator is used as one arm of balanced Wheatstone bridge. The amount of current through the relay 49 depends upon the degree of unbalancing of the bridge. As the temperature of the stator winding is increased, the electrical resistance of the detector increases which disturbs the balanced condition of the bridge. As a result current start flowing through the relay 49 and the relay will be
actuated after a predetermined value of this unbalanced current and ultimately starter contact will open to stop the supply to the motor.
Electrical Fault Calculation | Positive Negative Zero Sequence Impedance. Before applying proper electrical protection system, it is necessary to have through knowledge of the conditions of electrical power system during faults. The knowledge of electrical fault condition is required to deploy proper different protective relays in different locations of electrical power system. Information regarding values of maximum and minimum fault currents, voltages under those faults in magnitude and phase relation with respect to the currents at different parts of power system, to be gathered for proper application of protection relay system in those different parts of the electrical power system. Collecting the information from different parameters of the system is generally known as electrical fault calculation. Fault calculation broadly means calculation of fault current in any electrical power system. There are mainly three steps for calculating faults in a system. 1. Choice of impedance rotations. 2. Reduction of complicated electrical power system network to single equivalent impedance. 3. Electrical fault currents and voltages calculation by using symmetrical component theory. Impedance Notation of Electrical Power System If we look at any electrical power system, we will find, these are several voltage levels. For example, suppose a typical power system where electrical power is generated at 6.6 kV then that 132 kV power is transmitted to terminal substation where it is stepped down to 33 kV and 11 kV levels and this 11 kV level may further step down to 0.4kv. Hence from this example it is clear that a same power system network may have different voltage levels. So calculation of fault at any location of the said system becomes much difficult and complicated it try to calculate impedance of different parts
of the system according to their voltage level. This difficulty can be avoided if we calculate impedance of different part of the system in reference to a single base value. This technique is called impedance notation of power system. In other wards, before electrical fault calculation, the system parameters, must be referred to base quantities and represented as uniform system of impedance in either ohmic, percentage, or per unit values. Electrical power and voltage are generally taken as base quantities. In three phase system, three phase power in MVA or KVA is taken as base power and line to line voltage in KV is taken as base voltage. The base impedance of the system can be calculated from these base power and base voltage, as
follows, Per unit is an impedance value of any system is nothing but the radio of actual impedance of the system to the base
impedance value.
Percentage impedance value can be
calculated by multiplying 100 with per unit value. Again it is sometimes required to convert per unit values referred to new base values for simplifying different electrical fault calculations. In that case,
The choice of impedance notation depends upon the complicity of the system. Generally base voltage of a system is so chosen that it requires minimum number of transfers. Suppose, one system as a large number of 132 KV over head lines, few numbers of 33 KV lines and very few number of 11 KV lines. The base voltage of the system can be chosen either as 132 KV or 33 KV or 11 KV, but here the best base voltages 132 KV, because it requires minimum number of transfer during fault calculation. Network Reduction After choosing the correct impedance notation, the next step is to reduce network to a single impedance. For this first we have to convert the impedance of all generators, lines, cables, transformer to a common base value. Then we prepare a schematic diagram of electrical power system showing the impedance referred to same base value of all those generators, lines, cables and transformers. The network then reduced to a common
equivalent single impedance by using star/delta transformations. Separate impedance diagrams should be prepared for positive, negative and zero sequence networks. There phase faults are unique since they are balanced i.e. symmetrical in three phase, and can be calculated from the single phase positive sequence impedance diagram. Therefore three phase fault current
is obtained by, Where, I f is the total three phase fault current, v is the phase to neutral voltage z 1 is the total positive sequence impedance of the system; assuming that in the calculation, impedance are represented in ohms on a voltage base. Symmetrical Component Analysis The above fault calculation is made on assumption of three phase balanced system. The calculation is made for one phase only as the current and voltage conditions are same in all three phases. When actual faults occur in electrical power system, such as phase to earth fault, phase to phase fault and double phase to earth fault, the system becomes unbalanced means, the conditions of voltages and currents in all phases are no longer symmetrical. Such faults are solved by symmetrical component analysis. Generally three phase vector diagram may be replaced by three sets of balanced vectors. One has opposite or negative phase rotation, second has positive phase rotation and last one is co-phasal. That means these vectors sets are described as negative, positive and zero sequence, respectively.
The equation between phase and sequence quantities are,
Therefore,
Where all quantities are referred to the reference phase r. Similarly a set of equations can be written for sequence currents also. From, voltage and current equations, one can easily determine the sequence impedance of the system. The development of symmetrical component analysis depends upon the fact that in balanced system of impedance, sequence currents can give rise only to voltage drops of the same sequence. Once the sequence networks are available, these can be converted to single equivalent impedance. Let us consider Z 1, Z2 and Z0 are the impedance of the system to the flow of positive, negative and zero
sequence current respectively. For earth fault
Phase to phase faults
Double phase to earth
faults Three phase faults If fault current in any particular branch of the network is required, the same can be calculated after combining the sequence components flowing in that branch. This involves the distribution of sequence components currents as determined by solving the above equations, in their respective network according to their relative impedance. Voltages it any point of the network can also be determine once the sequence component currents and sequence impedance of each branch are known. Sequence Impedance Positive Sequence Impedance The impedance offered by the system to the flow of positive sequence current is called positive sequence impedance. Negative Sequence Impedance The impedance offered by the system to the flow of negative sequence current is called negative sequence impedance. Zero Sequence Impedance The impedance offered by the system to the flow of zero sequence current is known as zero sequence impedance. In previous fault calculation, Z1, Z2 and Z0 are positive, negative and zero sequence impedance respectively. The sequence impedance varies with the type of power system components under consideration:1.
In static and balanced power system components like transformer and lines, the sequence impedance offered by the system are the same for positive and negative sequence currents. In other words, the positive sequence impedance and negative sequence impedance are same for transformers and power lines.
2.
But in case of rotating machines the positive and negative sequence impedance are different. 3. The assignment of zero sequence impedance values is a more complex one. This is because the three zero sequence current at any point in a electrical power system, being in phase, do not sum to zero but must return through the neutral and /or earth. In three phase transformer and machine fluxes due to zero sequence components do not sum to zero in the yoke or field system. The impedance very widely depending upon the physical arrangement of the magnetic circuits and winding. 1. The reactance of transmission lines of zero sequence currents can be about 3 to 5 times the positive sequence current, the lighter value being for lines without earth wires. This is because the spacing between the go and return(i.e. neutral and/or earth) is so much greater than for positive and negative sequence currents which return (balance) within the three phase conductor groups. 2. The zero sequence reactance of a machine is compounded of leakage and winding reactance, and a small component due to winding balance (depends on winding tritch). 3. The zero sequence reactance of transformers depends both on winding connections and upon construction of core. External and Internal Faults in Transformer. External Faults in Power Transformer External Short - Circuit of Power Transformer The short - circuit may occurs in two or three phases of electrical power system. The level of fault current is always high enough. It depends upon the voltage which has been short - circuited and upon the impedance of the circuit up to the fault point. The copper loss of the fault feeding transformer is abruptly increased. This increasing copper loss causes internal heating in the transformer. Large fault current also produces severe mechanical stresses in the transformer. The maximum mechanical stresses occurs during first cycle of symmetrical fault current. High Voltage Disturbance in Power Transformer High voltage disturbance in power transformer are of two kinds, 1. 2.
Transient Surge Voltage Power Frequency Over Voltage
Transient Surge Voltage High voltage and high frequency surge may arise in the power system due to any of the following causes,
Arcing ground if neutral point is isolated. Switching operation of different electrical equipment. Atmospheric Lightening Impulse. Whatever may be the causes of surge voltage, it is after all a traveling wave having high and steep wave form and also having high frequency. This wave travels in the electrical power system network, upon reaching in the power transformer, it causes breakdown the insulation between turns adjacent to line terminal, which may create short circuit between turns. Power Frequency Over Voltage There may be always a chance of system over voltage due to sudden disconnection of large load. Although the amplitude of this voltage is higher than its normal level but frequency is same as it was in normal condition. Over voltage in the system causes an increase in stress on the insulation of transformer. As we know that, voltage V = 4.44Φ.f.T ⇒ V ∝ Φ, increased voltage causes proportionate increase in the working flux. This therefore causes, increased in iron loss and dis - proportionately large increase in magnetizing current. The increase flux is diverted from the transformer core to other steel structural parts of the transformer. Core bolts which normally carry little flux, may be subjected to a large component of flux diverted from saturated region of the core alongside. Under such condition, the bolt may be rapidly heated up and destroys their own insulation as well as winding insulation. Under Frequency Effect in Power Transformer As, voltage V = 4.44Φ.f.T ⇒ V ∝ Φ.f as the number of turns in the winding is fixed. Therefore, Φ ∝ V/f From, this equation it is clear that if frequency reduces in a system, the flux in the core increases, the effect are more or less similar to that of the over voltage. Internal Faults in Power Transformer The principle faults which occurs inside a power transformer are categorized as, 1. 2. 3. 4.
Insulation breakdown between winding and earth Insulation breakdown in between different phases Insulation breakdown in between adjacent turns i.e. inter - turn fault Transformer core fault
Internal Earth Faults in Power Transformer
Internal Earth Faults in a Star Connected Winding with Neutral Point Earthed through an Impedance In this case the fault current is dependent on the value of earthing impedance and is also proportional to the distance of the fault point from neutral point as the voltage at the point depends upon, the number of winding turns come under across neutral and fault point. If the distance between fault point and neutral point is more, the number of turns come under this distance is also more, hence voltage across the neutral point and fault point is high which causes higher fault current. So, in few words it can be said that, the value of fault current depends on the value of earthing impedance as well as the distance between the faulty point and neutral point. The fault current also depends up on leakage reactance of the portion of the winding across the fault point and neutral. But compared to the earthing impedance,it is very low and it is obviously ignored as it comes in series with comparatively much higher earthing impedance. Internal Earth Faults in a Star Connected Winding with Neutral Point Solidly Earthed In this case, earthing impedance is ideally zero. The fault current is dependent up on leakage reactance of the portion of winding comes across faulty point and neutral point of transformer. The fault current is also dependent on the distance between neutral point and fault point in the transformer. As said in previous case the voltage across these two points depends upon the number of winding turn comes across faulty point and neutral point. So in star connected winding with neutral point solidly earthed, the fault current depends upon two main factors, first the leakage reactance of the winding comes across faulty point and neutral point and secondly the distance between faulty point and neutral point. But the leakage reactance of the winding varies in complex manner with position of the fault in the winding. It is seen that the reactance decreases very rapidly for fault point approaching the neutral and hence the fault current is highest for the fault near the neutral end. So at this point, the voltage available for fault current is low and at the same time the reactance opposes the fault current is also low, hence the value of fault current is high enough. Again at fault point away from the neutral point, the voltage available for fault current is high but at the same time reactance offered by the winding portion between fault point and neutral point is high. It can be noticed that the fault current stays a very high level throughout the winding. In other word, the fault current maintain a very high magnitude irrelevant to the position of the fault on winding. Internal Phase to Phase Faults in Power Transformer
Phase to phase fault in the transformer are rare. If such a fault does occur, it will give rise to substantial current to operate instantaneous over current relay on the primary side as well as the differential relay. Inter Turns Fault in Power Transformer Power Transformer connected with electrical extra high voltage transmission system, is very likely to be subjected to high magnitude, steep fronted and high frequency impulse voltage due to lightening surge on the transmission line. The voltage stresses between winding turns become so large, it can not sustain the stress and causing insulation failure between inter - turns in some points. Also LV winding is stressed because of the transferred surge voltage. Very large number of Power Transformer failure arise from fault between turns. Inter turn fault may also be occurred due to mechanical forces between turns originated by external short circuit. Core Fault in Power Transformer In any portion of the core lamination is damaged, or lamination of the core is bridged by any conducting material causes sufficient eddy current to flow, hence, this part of the core becomes over heated. Some times, insulation of bolts (Used for tightening the core lamination together) fails which also permits sufficient eddy current to flow through the bolt and causing over heating. These insulation failure in lamination and core bolts causes severe local heating. Although these local heating, causes additional core loss but can not create any noticeable change in input and output current in the transformer, hence these faults can not be detected by normal electrical protection scheme. This is desirable to detect the local over heating condition of the transformer core before any major fault occurs. Excessive over heating leads to breakdown of transformer insulating oil with evolution of gases. These gases are accumulated in Buchholz relay and actuating Buchholz Alarm. Backup Protection of Transformer | Over Current and Earth Fault. Over Current and Earth Fault Protection of Transformer Backup protection of electrical transformer is simple Over Current and Earth Fault protection applied against external short circuit and excessive over loads. These over current and earth Fault relays may be of Inverse Definite Minimum Time (IDMT) or Definite Time type relays. Generally IDMT relays are connected to the in-feed side of the transformer. The over current relays can not distinguish between external short circuit, over load and internal faults of the transformer. For any of the above fault, backup protection i.e. over current and earth fault protection connected to in-feed side of the transformer will operate.
Backup protection is although generally installed at in feed side of the transformer, but it should trip both the primary and secondary circuit
breakers of the transformer. Over Current and Earth Fault protection relays may be also provided in load side of the transformer too, but it should not inter trip the primary side circuit breaker like the case of backup protection at in-feed side. The operation is governed primarily by current and time settings and the characteristic curve of the relay. To permit use of over load capacity of the transformer and coordination with other similar relays at about 125 to 150 % of full load current of the transformer but below the minimum short circuit current. Backup protection of transformer has four elements, three over current relays connected each in each phase and one earth fault relay connected to the common point of three over current relays as shown in the figure. The normal range of current settings available on IDMT over current relays is 50 % to 200 % and on earth fault relay 20 to 80 %.
Another range of setting on earth fault relay is also available and may be selected where the earth fault current is restricted due to insertion of impedance in the neutral grounding. In the case of transformer winding with neutral earthed, unrestricted earth fault protection is obtained by connecting an ordinary earth fault relay across a neutral current transformer. The unrestricted over current and earth fault relays should have proper time lag to co-ordinate with the protective relays of other circuit to avoid indiscriminate tripping.
Transformer Protection and Transformer Fault. There are different kinds of transformers such as two winding or three winding electrical power transformers, auto transformer, regulating transformers, earthing transformers, rectifier transformers etc. Different transformers demand different schemes of transformer protection depending upon their importance, winding connections, earthing methods and mode of operation etc.It is common practice to provide Buchholz relay protection to all 0.5 MVA and above transformers. While for all small size distribution transformers, only high voltage fuses are used as main protective device. For all larger rated and important distribution transformers, over current protection along with restricted earth fault protectionis applied. Differential protection should be provided in the transformers rated above 5 MVA. Depending upon the normal service condition, nature of transformer faults, degree of sustained over load, scheme of tap changing, and many other factors, the suitable transformer protection schemes are chosen. Nature of Transformer Faults Although an electrical power transformer is a static device, but internal stresses arising from abnormal system conditions, must be taken into
consideration. A transformer generally suffers from following types of transformer fault1. Over current due to overloads and external short circuits, 2. Terminal faults, 3. Winding faults, 4. Incipient faults. All the above mentioned transformer faults cause mechanical and thermal stresses inside the transformer winding and its connecting terminals. Thermal stresses lead to overheating which ultimately affect the insulation system of transformer. Deterioration of insulation leads to winding faults. Some time failure of transformer cooling system, leads to overheating of transformer. So the transformer protection schemes are very much required. The short circuit current of an electrical transformer is normally limited by its reactance and for low reactance, the value of short circuit current may be excessively high. The duration of external short circuits which a transformer can sustain without damage as given in BSS 171:1936. Transformer % reactance
Permitted fault duration in seconds
4%
2
5%
3
6%
4
7 % and over
5
The general winding faults in transformer are either earth faults or interturns faults. Phase to phase winding faults in a transformer is rare. The phase faults in an electrical transformer may be occurred due to bushing flash over and faults in tap changer equipment. Whatever may be the faults, the transformer must be isolated instantly during fault otherwise major breakdown may occur in the electrical power system. Incipient faults are internal faults which constitute no immediate hazard. But it these faults are over looked and not taken care of, these may lead to major faults. The faults in this group are mainly inter-lamination short circuit due to insulation failure between core lamination, lowering the oil level due to oil leakage, blockage of oil flow paths. All these faults lead to overheating. So transformer protection scheme is required for incipient transformer faults also. The earth fault, very nearer to neutral point of transformer star winding may also be considered as an incipient fault. Influence of winding connections and
earthing on earth fault current magnitude. There are mainly two conditions for earth fault current to flow during winding to earth faults, 1. A current exists for the current to flow into and out of the winding. 2. Ampere-turns balance is maintained between the windings. The value of winding earth fault current depends upon position of the fault on the winding, method of winding connection and method of earthing. The star point of the windings may be earthed either solidly or via a resistor. On delta side of the transformer the system is earthed through an earthing transformer. Grounding or earthing transformer provides low impedance path to the zero sequence current and high impedance to the positive and negative sequence currents. Star Winding with Neutral Resistance Earthed In this case the neutral point of the transformer is earthed via a resistor and the value of impedance of it, is much higher than that of winding impedance of the transformer. That means the value of transformer winding impedance is negligible compared to impedance of earthing resistor. The value of earth current is, therefore, proportional to the position of the fault in the winding. As the fault current in the primary winding of the transformer is proportional to the ratio of the short circuited secondary turns to the total turns on the primary winding, the primary fault current will be proportional to the square of the percentage of winding short circuited. The variation of fault current both in the primary and secondary winding is shown below. Star Winding with Neutral Solidly Earthed In this case the earth fault current magnitude is limited solely by the winding impedance and the fault is no longer proportional to the position of the fault. The reason for this non linearity is unbalanced flux linkage. Protection of Lines or Feeder. As the length of electrical power transmission line is generally long enough and it runs through open atmosphere, the probability of occurring fault in electrical power transmission line is much higher than that of electrical power transformers and alternators. That is why a transmission line requires much more protective schemes than a transformer and an alternator.Protection of line should have some special features, such asDuring fault, the only circuit breaker closest to the fault point should be tripped. 2. If the circuit breaker closest the faulty point, fails to trip the circuit breaker just next to this breaker will trip as back up. 1.
3.
The operating time of relay associated with protection of line should be as minimum as possible in order to prevent unnecessary tripping of circuit breakers associated with other healthy parts of power system.
These above mentioned requirements cause protection of transmission line much different from protection of transformer and other equipment of power systems. The main three methods of transmission line protection are 1. Time graded over current protection. 2. Differential protection. 3. Distance protection. Time Graded Over Current Protection This may also be referred simply as over-current protection of electrical power transmission line. Let' discuss different schemes of time graded over current protection. Protection of Radial Feeder In radial feeder, the power flows in one direction only, that is from source to load. This type of feeders can easily protected by using either definite time relays or inverse time relays. Line Protection by Definite Time Relay This protection scheme is very simple. Here total line is divided into different sections and each section is provided with definite time relay. The relay nearest to the end of the line has minimum time setting while time setting of other relays successively increased, towards the source. For example, suppose there is a source at point A, in the figure below
At point D the circuit breaker CB-3 is installed with definite time of relay operation 0.5 sec. Successively, at point C an other circuit breaker CB-2 is installed with definite time of relay operation 1 sec. The next circuit breaker CB-1 is installed at point B which is nearest of the point A. At point B, the relay is set at time of operation 1.5 sec. Now, assume a fault occurs at point F. Due to this fault, the faulty current flow through all the current transformers or CTs connected in the line. But as the time of operation of relay at point D is minimum the CB-3, associated with this relay will trip first to isolate the faulty zone from rest part of the line. In case due to any reason, CB-3 fails to trip, then next higher timed relay will operate the associated CB to trip. In this case, CB-2 will trip. If CB2 also fails to trip, then next circuit breaker i.e. CB-1 will trip to isolate major portion of the line. Advantages of Definite Time Line Protection The main advantage of this scheme is simplicity. The second major advantage is, during fault, only nearest CB towards the source from fault point will operate to isolate the specific position of the line. Disadvantage of Definite Time Line Protection If the number of sections in the line is quite large, the time setting of relay nearest to the source, would be very long. So during any fault nearer to the source will take much time to be isolated. This may cause severe destructive effect on the system.
Over Current Line Protection by Inverse Relay The drawback as we discussed just in definite time over current protection of transmission line, can easily be overcome by using inverse time relays. In inverse relay the time of operation is inversely proportional to fault current.
In the above figure, overall time setting of relay at point D is minimum and successively this time setting is increased for the relays associated with the points towards the point A. In case of any fault at point F will obviously trip CB-3 at point D. In failure of opening CB-3, CB-2 will be operated as overall time setting is higher in relay at point C. Although, the time setting of relay nearest to the source is maximum but still it will trip in shorter period, if major fault occurs near the source, as the time of operation of relay is inversely proportional to faulty current. Over Current Protection of Parallel Feeders For maintaining stability of the system it is required to feed a load from source by two or more than two feeders in parallel. If fault occurs in any of the feeders, only that faulty feeder should be isolated from the system in order to maintain continuity of supply from source to load. This requirement makes the protection of parallel feeders little bit more complex than simple non direction over current protection of line as in the case of radial feeders. The protection of parallel feeder requires to use directional relays and to grade the time setting of relay for selective tripping.
There are two feeders connected in parallel from source to load. Both of the feeders have non-directional over current relay at source end. These relays should be inverse time relay. Also both of the feeders have directional relay or reverse power relay at their load end. The reverse power relays used here should be instantaneous type. That means these relays should be operated as soon as flow of power in the feeder is reversed. The normal direction of power from source to load. Now, suppose a fault occurs at point F, say the fault current is If. This fault will get two parallel paths from source, one through circuit breaker A only and other via CB-B, feeder-2, CB-Q, load bus and CB-P. This is clearly shown in figure below, where IA and IB are current of fault shared by feeder-1 and feeder-2 respectively.
As per Kirchoff's current law, IA + IB = If. Now, IA is flowing through CB-A, IB is flowing through CB-P. As the direction of flow of CB-P is reversed it will trip instantly. But CB-Q will not trip as flow of current (power) in this circuit breaker is not reversed. As soon as CB-P is tripped, the fault current IB stops flowing through feeder and hence there is no question of further operating of inverse time over current relay. I A still continues to flow even CB-P is tripped. Then because of over current I A, CBA will trip. In this way the faulty feeder is isolated from system. Differential Pilot Wire Protection This is simply a differential protection scheme applied to feeders. Several differential schemes are applied for protection of line but Mess Price Voltage balance system and Translay Scheme are most popularly used. Merz Price Balance System The working principle of Merz Price Balance system is quite simple. In this scheme of line protection, identical CT is connected to each of the both ends of the line. The polarity of the CTs are same. The secondary of these current transformer and operating coil of two instantaneous relays are formed a closed loop as shown in the figure below. In the loop pilot wire is used to connect both CT secondary and both relay coil as shown.
Now, from the figure it is quite clear that when the system is under normal condition, there would not be any current flowing through the loop. As the secondary current of one CT will cancel out secondary current of other CT. Now, if any fault occurs in the portion of the line between these two CTs, the secondary current of one CT will no longer equal and opposite of secondary current of other CT. Hence there would be a resultant circulating current in the loop. Due this circulating current, the coil of both relays will close the trip circuit of associate circuit breaker. Hence, the faulty line will be isolated from both ends. Busbar Protection | Busbar Differential Protection Scheme. In early days only conventional over current relays were used for busbar protection. But it is desired that fault in any feeder or transformer connected to the busbar should not disturb busbar system. In viewing of this time setting of busbar protection relays are made lengthy. So when faults occurs on busbar itself, it takes much time to isolate the bus from source which may came much damage in the bus system.In recent days, the second zone distance protection relays on incoming feeder, with operating time of 0.3 to 0.5 seconds have been applied for busbar protection. But this scheme has also a main disadvantage. This scheme of protection can not discriminate the faulty section of the busbar. Now days, electrical power system deals with huge amount of power. Hence any interruption in total bus system causes big loss to the company. So it becomes essential to isolate only faulty section of busbar during bus fault.
Another drawback of second zone distance protection scheme is that, sometime the clearing time is not short enough to ensure the system stability. To overcome the above mentioned difficulties, differential busbar protection scheme with an operating time less than 0.1 sec., is commonly applied to many SHT bus systems. Differential Busbar Protection Current Differential Protection The scheme of busbar protection, involves, Kirchoff’s current law, which states that, total current entering an electrical node is exactly equal to total current leaving the node. Hence, total current entering into a bus section is equal to total current leaving the bus section. The principle of differential busbar protection is very simple. Here, secondaries of CTs are connected parallel. That means, S1 terminals of all CTs connected together and forms a bus wire. Similarly S 2 terminals of all CTs connected together to form another bus wire. A tripping relay is connected across these two bus wires.
Here, in the figure above we assume that at normal condition feed, A, B, C, D, E and F carries current IA, IB, IC, ID, IE and IF. Now, according to Kirchoff’s current law, Essentially all the CTs used for differential busbar protection are of same current ratio. Hence, the summation of all secondary currents must also be equal to zero. Now, say current through the relay connected in parallel with
all CT secondaries, is iR, and iA, iB, iC, iD, iE and iF are secondary currents. Now, let us apply KCL at node X. As per KCL at node X,
So, it is clear that under normal condition there is no current flows through the busbar protection tripping relay. This relay is generally referred as Relay 87. Now, say fault is occurred at any of the feeders, outside the protected zone. In that case, the faulty current will pass through primary of the CT of that feeder. This fault current is contributed by all other feeders connected to the bus. So, contributed part of fault current flows through the corresponding CT of respective feeder. Hence at that faulty condition, if we apply KCL at node K, we will still get, iR = 0.
That means, at external faulty condition, there is no current flows through relay 87. Now consider a situation when fault is occurred on the bus itself.
At this condition, also the faulty current is contributed by all feeders connected to the bus. Hence, at this condition, sum of all contributed fault current is equal to total faulty current. Now, at faulty path there is no CT. (in external fault, both fault current and contributed current to the fault by different feeder get CT in their path of flowing).
The sum of all secondary currents is no longer zero. It is equal to secondary equivalent of faulty current. Now, if we apply KCL at the nodes, we will get a non zero value of iR. So at this condition current starts flowing through 87 relay and it makes trip the circuit breakercorresponding to all the feeders connected to this section of the busbar. As all the incoming and outgoing feeders, connected to this section of bus are tripped, the bus becomes dead. This differential busbar protection scheme is also referred as current differential protection of busbar. Differential Protection of Sectionalized Bus During explaining working principle of current differential protection of busbar, we have shown a simple non sectionalized busbar. But in moderate high voltage system electrical bus sectionalized in than one sections to increase stability of the system. It is done because, fault in one section of bus should not disturb other section of the system. Hence during bus fault, total bus would be interrupted.
Let us draw and discuss about protection of busbar with two sections.
Here, bus section A or zone A is bounded by CT 1, CT2 and CT3 where CT1 and CT2 are feeder CTs and CT3 is bus CT. Similarly bus section B or zone B is bounded by CT4, CT5 and CT6where CT4 is bus CT, CT5 and CT6 are feeder CT. Therefore, zone A and B are overlapped to ensure that, there is no zone left behind this busbar protection scheme. ASI terminals of CT1, 2 and 3 are connected together to form secondary bus ASI BSI terminals of CT 4, 5 and 6 are connected together to form secondary bus BSI. S 2 terminals of all CTs are connected together to form a common bus S 2. Now, busbar protection relay 87A for zone A is connected across bus ASI and S 2. Relay 87B for zone B is connected across bus BSI and S 2. This section busbar differential protection scheme operates in some manner simple current differential protection of busbar. That is, any fault in zone A, with trip only CB 1, CB2and bus CB. Any fault in zone B, will trip only CB 5, CB6 and bus CB. Hence, fault in any section of bus will isolate only that portion from live system. In current differential protection of busbar, if CT secondary circuits, or bus wires is open the relay may be operated to isolate the bus from live system. But this is not desirable. DC Circuit of Differential Busbar Protection
A typical DC circuit for busbar differential protection scheme is given
below. Here, CSSA and CSSB are two selector switch which are used to put into service, the busbar protection system for zone A and zone B respectively. If CSSA is in “IN” position, protection scheme for zone A is in service. If CSSB is in “IN” position, protection for zone B is in service. Generally both of the switches are in “IN’ position in normal operating condition. Here, relay coil of 96A and 96B are in series with differential busbar protection relay contact 87A-1 and 87B-1 respectively. 96A relay is multi contacts relay. Each circuit breaker in
zone A is connected with individual contact of 96A. Similarly, 96B is multi contacts relay and each circuit breaker in zone-B is connected with individual contacts of 96B. Although here we use only one tripping relay per protected zone, but this is better to use one individual tripping relay per feeder. In this scheme one protective relay is provided per feeder circuit breaker, whereas two tripping relays one for zone A and other for zone B are provided to bus section or bus coupler circuit breaker. On an interval fault in zone A or bus section A, the respective bus protection relay 87A, be energized whereas during internal fault in zone B, the respective relay 87B will be energized. As soon as relay coil of 87A or 87B is energized respective no. contact 87A-1 or 87B-1 is closed.Hence, the tripping relay 96 will trip the breakers connected to the faulty zone. To indicate whether zone A or B busbar protection operated, relay 30 is used. For example, if relay 87A is operated, corresponding “No” contact 87A-2 is closed which energized relay 30A. Then the No contact 30A-1 of relay 30A is closed to energized alarm relay 74. Supervision relay 95 of respective zone is also energized during internal fault, but it has a time delay of 3 second. So, it reset as soon as the fault is cleared and therefore does not pick up zone bus wire shorting relay 95x which in turn shorts out the bus wires. An alarm contact is also given to this auxiliary 95x relay to indicate which CT is open circuited. No volt relay 80 is provided in both trip and non-trip section of the DC circuit of differential busbar protection system to indicate any discontinuity of D. C. supply. Voltage Differential Protection of Busbar The current differential scheme is sensitive only when the CTs do not get saturated and maintain same current ratio, phase angle error under maximum faulty condition. This is usually not 80, particularly, in the case of an external fault on one of the feeders. The CT on the faulty feeder may be saturated by total current and consequently it will have very large errors. Due to this large error, the summation of secondary current of all CTs in a particular zone may not be zero. So there may be a high chance of tripping of all circuit breakers associated with this protection zone even in the case of an external large fault. To prevent this maloperation of current differential busbar protection, the 87 relays are provided with high pick up current and enough time delay. The greatest troublesome cause of current transformer saturation is the transient dc component of the short circuit current. This difficulties can be overcome by using air core CTs. This current transformer is also called linear coupler. As the core of the CT does not use iron the secondary characteristic of these CTs, is straight line. In voltage differential busbar protection the CTs of all incoming and outgoing feeders are connected in series instead of connecting them in parallel.
The secondaries of all CTs and differential relay form a closed loop. If polarity of all CTs are properly matched, the sum of voltage across all CT secondaries is zero. Hence there would be no resultant voltage appears across the differential relay. When a buss fault occurs, sum of the all CT secondary voltage is no longer zero. Hence, there would be current circulate in the loop due to the resultant voltage. As this loop current also flows through the differential relay, the relay is operated to trip all the circuit beaker associated with protected bus zone. Except when ground fault current is severally limited by neutral impedance there is usually no selectivity problem when such a problem exists, it is solved by use of an additional more sensitive relaying equipment including a supervising protective relay. Protection of Capacitor Bank. Like other electrical equipments, shunt capacitor may also be subjected to internal and external electrical faults. Hence this equipment is also to be protected from internal and external faults. There are numbers of schemes available for protection of capacitor bank, but during applying any of the schemes, we should remember the initial investment on that capacitor for economical point of view. We should compare the initial investment in the capacitor and the cost of the protection applying on it. There are mainly 3 types of protection arrangement are applied to a capacitor bank. 1. 2.
3.
Element Fuse. Unit. Fuse. Bank Protection.
Element Fuses Manufacturers of capacitor unit commonly provide inbuilt fuse in each element of the unit. In this case, if any fault occurs in any element itself, it is automatically disconnected from rest of the unit. In this case, the unit still serves its purpose, but with smaller output. In smaller rated capacitor bank only these inbuilt protection scheme is applied to avoid the expenditure of other special protective equipments. Unit Fuse The unit fuse protection is generally provided to limit the duration of arc inside a faulty capacitor unit. As the arc duration is limited, there is less chance of major mechanical deformation and huge production of gas in the faulty unit, and hence the neighborhood units of the bank are saved. If each unit of a capacitor bank is individually protected against fuse, then in case of failure of one unit, the capacitor bank can still be running without interruption before removing and replacing the faulty unit. Another major advantage of providing fuse protection to each unit of the bank is that, it indicates the exact location of the faulty unit. But during choosing the size of the fuse for this purpose, it should be taken into consideration that the fuse element must withstand the excessive loading due to harmonics in the system. In the view of that the current rating of the fuse element for this purpose is taken as 65 % above the full load current. Whenever the individual unit of capacitor bank is protected by fuse, it is necessary to provide discharge resistance in each of the units. Bank Protection Although in general fuse protection is provided with each of the capacitor units, but when a capacitor unit is under fault and the associated fuse element is blown out, the voltage stress increases to the other capacitor units connected in series in same row. Generally, each capacitor unit is designed for withstanding 110% of its normal rated voltage. If any other capacitor unit further becomes out of service, in the same row where previously one unit is damaged, the voltage stress upon other healthy units of that row will increase further and easily crosses the limit of maximum allowable, voltage of these units. Hence it is always desirable to replace damaged capacitor unit from the bank as soon as possible to avoid excess voltage stress on the other healthy units. Hence, there must be some indicating arrangement to identify the exact faulty unit. As soon as the faulty unit is identified in a bank, the bank should be removed from the service for replacing the faulty unit. There are several methods of sensing unbalance voltage caused by failure of capacitor unit. The figure below is showing the
most common arrangement of capacitor bank protection. Here, the capacitor bank is connected in star formation. Primary of a potential transformer is connected across each phase. The secondary of all three potential transformers are connected in series to form an open delta and a voltage sensitive relay is connected across this open delta. In exact balanced condition there must not be any voltage appears across the voltage sensitive relay because summation of balanced 3 phase voltages is zero. But when there would be any voltage unbalancing due to failure of capacitor unit, the resultant voltage will appear across the relay and the relay will be actuated for providing an alarm and trip signals. The voltage sensitive relay can be so adjusted that up to a certain voltage unbalancing only alarm contacts would be closed and for certain higher voltage level the trip contacts along with alarm contacts would be closed. The potential transformer connected across the capacitors of each phase also serves for discharging of the bank after being switched off.
In another scheme, the
capacitors in each phase are divided into two equal parts connected in series. Discharge coil is connected across each of the parts as shown in the figure. In between the secondary of discharge coil and the sensitive voltage unbalance the relay an auxiliary transformer is connected which serves to regulate the voltage difference between secondary voltages of discharge coil under normal conditions.
Here the capacitor bank is connected in star and the neutral point is connected to the ground through a potential transformer. A voltage sensitive relay is connected across the secondary of the potential transformer. As soon as there is any unbalance between the phases, the resultant voltage will appear across the potential transformer and hence the voltage sensitive relay will be actuated
beyond a preset value. Here, the capacitor bank of each phase is divided into two equal parts connected in parallel and the star points of both parts are interconnected through a current transformer. The secondary of the current transformer are connected across a current sensitive relay. In case any misbalancing occurs between the two parts of the bank, there would be a unbalance current flowing through the current transformer and hence the current sensitive relay will actuate. In this scheme for discharging the bank after switching off, discharge coil may be connected across the capacitors in each phase.
In another scheme of protection of capacitor bank, the star point of a three phase capacitor bank is connected to the ground through a current transformer and a current sensitive relay is connected across the secondary of the current transformer. As soon as there is any unbalancing between the phases of capacitor bank, there must be a current flowing to the ground through the current transformer and hence the current sensitive relay will be actuated to trip the circuit breaker associated with the capacitor bank.
Basic electrical design of a PLC panel (Wiring diagrams).
Building the PLC panel It is uncommon for engineers to build their own PLC panel designs (but not impossible of course). For example, once the electrical designs are complete, they must be built by an electrician. Therefore, it is your responsibility to effectively communicate your design intentions to the electricians through drawings. In some factories, the electricians also enter the ladder logic and do debugging. This article discusses the design issues in implementation that must be considered by the designer.
Electrical wiring diagrams of a PLC panel In an industrial setting a PLC is not simply “plugged into a wall socket”. The electrical design for each machine must include at least the following components. 1. 2.
Transformers – to step down AC supply voltages to lower levels Power contacts – to manually enable/disable power to the machine with e-stop buttons 3. Terminals – to connect devices
4. Fuses or circuit breakers – will cause power to fail if too much current is drawn 5. Grounding – to provide a path for current to flow when there is an electrical fault 6. Enclosure – to protect the equipment, and users from accidental contact A control system of a PLC panel will normally use AC and DC power at different voltage levels. Control cabinets are often supplied with single phase AC at 220/440/550V, or two phase AC at 220/440V AC, or three phase AC at 330/550V.
This power must be dropped down to a lower voltage level for the controls and DC power supplies. 110Vac is common in North America, and 220 V AC Is common in Europe and the Commonwealth countries. It is also common for a control cabinet to supply a higher voltage to other equipment, such as motors.
Motor controller example An example of a wiring diagram for a motor controller is shown in Figure 1. Note that symbols are discussed in detail later). Dashed lines indicate a single purchased component. This system uses 3 phase AC power (L1, L2 and L3) connected to the terminals. The three phases are then connected to a power interrupter. Next, all three phases are supplied to a motor starter that contains three contacts, M, and three thermal overload relays (breakers).
Figure 1 – A Motor Controller Schematic
The contacts, M, will be controlled by the coil, M. The output of the motor starter goes to a three phase AC motor. Power is supplied by connecting a step down transformer to the control electronics by connecting to phases L2 and L3. The lower voltage is then used to supply power to the left and right rails of the ladder below. The neutral rail is also grounded. The logic consists of two push buttons:
Start push button is normally open, so that if something fails the motor cannot be started. Stop push button is normally closed, so that if a wire or connection fails the system halts safely.
The system controls the motor starter coil M, and uses a spare contact on the starter, M, to seal in the motor starter. The diagram also shows numbering for the wires in the device. This is essential for industrial control systems that may contain hundreds or thousands of wires. These numbering schemes are often particular to each facility, but there are tools to help make wire labels that will appear in the final controls cabinet.
Figure 2 – A Physical Layout for the Control Cabinet Once the electrical design is complete, a layout for the controls cabinet is developed, as shown in Figure 2. The physical dimensions of the devices must be considered, and adequate space is needed to run wires between components. In the cabinet the AC power would enter at the terminal block, and be connected to the main breaker.
It would then be connected to the contactors and overload relays that constitute the motor starter. Two of the phases are also connected to the transformer to power the logic. The start and stop buttons are at the left of the box (note: normally these are mounted elsewhere, and a separate layout drawing would be needed). The final layout in the cabinet might look like the one shown in Figure 1.
Figure 3 – Final PLC Panel Wiring
When being built the system will follow certain standards that may be company policy, or legal requirements. This often includes items such as;
Hold downs – the will secure the wire so they don’t move Labels – wire labels help troubleshooting Strain reliefs – these will hold the wire so that it will not be pulled out of screw terminals Grounding – grounding wires may be needed on each metal piece for safety
A photograph of an industrial controls cabinet is shown in Figure 4:
Figure 4 – An industrial control cabinet with wire runs, terminal strip, buttons on PLC panel front, etc.
When including a PLC in the ladder diagram still remains. But, it does tend to become more complex. Figure 5 below shows a schematic diagram for a PLC based motor control system, similar to the previous motor control example. This figure shows the E-stop wired to cutoff power to all of the devices in the circuit, including the PLC. All critical safety functions should be hardwired this way.
Figure 5 – An Electrical Schematic with a PLC.
Electrical Distribution Architecture In Water Treatment Plants.
Water treatment plants For both drinking water and wastewater treatment, 4 different sizes of plants have been distinguished. The size of plants can be expressed in quantity of treated water per day, or in corresponding number of inhabitants. Four different types of (waste) water treatment plants have been distinguished, depending on destination and size:
T1 – Autonomous water treatment plant // See single-line diagram T2 – Small water or wastewater treatment plant // See single line diagram T3 – Medium sized water or wastewater treatment plant See single line diagrams // – Double-radial architecture and – Open medium voltage loop T4 – Large water or wastewater treatment plant See single line diagrams // – Radial-double feed – Open medium voltage loop Characteristics T1 T2 T3
T4
m3/day (drinking water or waste water)
1K-5K
5K-50K
50K-200K
200K-1000K
Inhabitants
1K-10K
10K-100K
100K-500K
500K-1000K
25-125 kVA
125-1250 kVA
1.25-5 MVA
5-25 MVA
Power demand
Electrical Distribution Guidance is given for the selection of Electrical Distribution architecture in water treatment plants. This includes the selection between different possible configurations of MV and LV circuits and the implementation of back-up power sources. The most relevant characteristics of the electrical installation are taken into account, such as typology, power demand, sensitivity to power interruptions, …
Connection to the utility network MV circuit configuration Configurations of LV circuits Backup generators Presence of uninterruptible power supply (UPS).
Commissioning tests of protection relays at site (before set to work).
Installation of protection relays Installation of protection relays at site creates a number of possibilities for errors in the implementation of the scheme to occur. Even if the scheme has been thoroughly tested in the factory, wiring to the CTs and VTs on site may be incorrectly carried out, or the CTs/VTs may have been incorrectly installed. The impact of such errors may range from simply being a nuisance (tripping occurs repeatedly on energisation, requiring investigation to locate and correct the errors) through to failure to trip under fault conditions, leading to major equipment damage, disruption to supplies and potential hazards to personnel.
The strategies available to remove these risks are many, but all involve some kind of testing at site. Commissioning tests at site are therefore invariably performed before protection equipment is set to work. The aims of commissioning tests are: 1. 2. 3.
To ensure that the equipment has not been damaged during transit or installation To ensure that the installation work has been carried out correctly To prove the correct functioning of the protection scheme as a whole
The tests carried out will normally vary according to the protection scheme involved, the relay technology used, and the policy of the client. In many cases, the tests actually conducted are determined at the time of commissioning by mutual agreement between the client’s representative and the commissioning team. The following tests are invariably carried out, since the protection scheme will not function correctly if faults exist.
Wiring diagram check, using circuit diagrams showing all the reference numbers of the interconnecting wiring General inspection of the equipment, checking all connections, wires on relays terminals, labels on terminal boards, etc. Insulation resistance measurement of all circuits [details] Perform relay self-test procedure and external communications checks on digital/numerical relays [details] Test main current transformers Polarity check Magnetisation Curve Test main voltage transformers Polarity check Ratio check
Phasing check Check that protection relay alarm/trip settings have been entered correctly [details] Tripping and alarm circuit checks to prove correct functioning In addition, the following checks may be carried out, depending on the factors noted above (not covered in this technical article):
Secondary injection test on each relay to prove operation at one or more setting values Primary injection tests on each relay to prove stability for external faults and to determine the effective current setting for internal faults (essential for some types of electromechanical relays) Testing of protection scheme logic
Insulation resistance tests All the deliberate earth connections on the wiring to be tested should first be removed, for example earthing links on current transformers, voltage transformers and DC supplies. Some insulation testers generate impulses with peak voltages exceeding 5kV. In these instances any electronic equipment should be disconnected while the external wiring insulation is checked. The insulation resistance should be measured to earth and between electrically separate circuits. The readings are recorded and compared with subsequent routine tests to check for any deterioration of the insulation. The insulation resistance measured depends on the amount of wiring involved, its grade, and the site humidity. Generally, if the test is restricted to one cubicle, a reading of several hundred megohms should be obtained. If long lengths of site wiring are involved, the reading could be only a few megohms. Protection relay self-test procedure Digital and numerical relays will have a self-test procedure that is detailed in the appropriate relay manual. These tests should be followed to determine if the relay is operating correctly.
This will normally involve checking of the relay watchdog circuit, exercising all digital inputs and outputs and checking that the relay analogue inputs are within calibration by applying a test current or voltage. For these tests, the relay outputs are normally disconnected from the remainder of the protection scheme, as it is a test carried out to prove correct relay, rather than scheme, operation.
To shorten testing and commissioning times of SIPROTEC relays, extensive test and diagnostic functions are available to the user in DIGSI 5
Unit protection schemes involve relays that need to communicate with each other. This leads to additional testing requirements. The communications path between the relays is tested using suitable equipment to ensure that the path is complete and that the received signal strength is within specification. Numerical relays may be fitted with loopback test facilities that enable either part of or the entire communications link to be tested from one end.
After completion of these tests, it is usual to enter the relay settings required. This can be done manually via the relay front panel controls, or using a portable PC and suitable software. Whichever, method is used, a check by a second person that the correct settings have been used is desirable, and the settings recorded. Programmable scheme logic that is required is also entered at this stage.
SIPROTE C relay wiring test editor for monitoring and testing of binary inputs, binary outputs and LED (click to expand)
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Current transformer tests
The following tests are normally carried out prior to energisation of the main circuits: checking of polarity and current transformer magnetisation curve.
Polarity checks. Each current transformer should be individually tested to verify that the primary and secondary polarity markings are correct (see Figure 1). The ammeter connected to the secondary of the current transformer should be a robust moving coil, permanent magnet, centre-zero type. A low voltage battery is used, via a single-pole push-button switch, to energise the primary winding. On closing the push-button, the DC ammeter, A, should give a positive flick and on opening, a negative flick.
Figure 1 – Current transformer polarity check
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Checking of magnetisation curve Several points should be checked on each current transformer magnetisation curve. This can be done by energising the secondary winding from the local mains supply through a variable auto-transformer while the primary circuit remains open. See Figure 2.
The characteristic is measured at suitable intervals of applied voltage, until the magnetising current is seen to rise very rapidly for a small increase in voltage. This indicates the approximate knee-point or saturation flux level of the current transformer.
The magnetising current should then be recorded at similar voltage intervals as it is reduced to zero.
Figure 2 – Testing current transformer magnetising curve
Care must be taken that the test equipment is suitably rated. The short-time current rating must be in excess of the CT secondary current rating, to allow for measurement of the saturation current. This will be in excess of the CT secondary current rating. As the magnetising current will not be sinusoidal, a moving iron or dynamometer type ammeter should be used. It is often found that current transformers with secondary ratings of 1A or less have a kneepoint voltage higher than the local mains supply. In these cases, a step-up interposing transformer must be used to obtain the necessary voltage to check the magnetisation curve.
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Voltage transformer tests Voltage transformers require testing for polarity, ratio and phasing.
Polarity check of voltage transformer The voltage transformer polarity can be checked using the method for CT polarity tests. Care must be taken to connect the battery supply to the primary winding, with the polarity ammeter connected to the secondary winding. If the voltage transformer is of the capacitor type, then the polarity of the transformer at the bottom of the capacitor stack should be checked. Go back to Commissioning tests ↑
Ratio check of VT This check can be carried out when the main circuit is first made live. The voltage transformer secondary voltage is compared with the secondary voltage shown on the nameplate.
Nam plate of a single phase voltage transformer (photo credit: emadrlc.blogspot.com)
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Phasing check of VT The secondary connections for a three-phase voltage transformer or a bank of three single-phase voltage transformers must be carefully checked for phasing. With the main circuit alive, the phase rotation is checked using a phase rotation meter connected across the three phases, as shown in Figure 3 below. Provided an existing proven VT is available on the same primary system, and that secondary earthing is employed, all that is now necessary to prove correct phasing is a voltage check between, say, both ‘A’ phase secondary outputs. There should be nominally little or no voltage if the phasing is correct. However, this test does not detect if the phase sequence is correct, but the phases are displaced by 120o from their correct position, i.e. phase A occupies the position of phase C or phase B in Figure 3.
This can be checked by removing the fuses from phases B and C (say) and measuring the phase-earth voltages on the secondary of the VT. If the phasing is correct, only phase A should be healthy, phases B and C should have only a small residual voltage.
Figure 3 – Voltage transformer phasing check
Correct phasing should be further substantiated when carrying out ‘on load’ tests on any phase-angle sensitive relays, at the relay terminals. Load current in a known phase CT secondary should be compared with the associated phase to neutral VT secondary voltage.
The phase angle between them should be measured, and should relate to the power factor of the system load. If the three-phase voltage transformer has a broken-delta tertiary winding, then a check should be made of the voltage across the two connections from the broken delta VN and VL, as shown in Figure 3 above. With the rated balanced three- phase supply voltage applied to the voltage transformer primary windings, the broken-delta voltage should be below 5V with the rated burden connected. Go back to Commissioning tests ↑
Protection relay setting checks (alarm and trip settings) At some point during commissioning, the alarm and trip settings of the relay elements involved will require to be entered and/or checked. Where the complete scheme is engineered and supplied by a single contractor, the settings may already have been entered prior to despatch from the factory, and hence this need not be repeated. The method of entering settings varies according to the relay technology used. For electromechanical and static relays, manual entry of the settings for each relay element is required. This method can also be used for digital/numerical relays. However, the amount of data to be entered is much greater, and therefore it is usual to use appropriate software, normally supplied by the manufacturer, for this purpose. The software also makes the essential task of making a record of the data entered much easier.
The maximum allowable step potential is 5,664V, which exceeds the step voltage calculated above and the earthing system passes the step potential criteria. Having passed both touch and step potential criteria, we can conclude that the earthing system is safe. Once the data has been entered, it should be checked for compliance with the recommended settings as calculated from the protection setting study. Where appropriate software is used for data entry, the checks can be considered complete if the data is checked prior to download of the settings to the relay. Otherwise, a check may required subsequent to data entry by inspection and recording of the relay settings, or it may be considered adequate to do this at the time of data entry. The recorded settings form an essential part of the commissioning documentation provided to the client.
FAULT CURRENT COMPUTATION USING SEQUENCE NETWORKS In this section we shall demonstrate the use of sequence networks in the calculation of fault currents using sequence network through some examples.
Example 8.4 Consider the network shown in Fig. 8.10. The system parameters are given below Generator G : 50 MVA, 20 kV, X" = X1 = X2 = 20%, X0 = 7.5%
Motor M : 40 MVA, 20 kV, X" = X1 = X2 = 20%, X0 = 10%, Xn = 5%
Transformer T1 : 50 MVA, 20 kV Δ /110 kVY, X = 10%
Transformer T2 : 50 MVA, 20 kV Δ /110 kVY, X = 10%
Transmission line: X1 = X2 = 24.2 Ω , X0 = 60.5 Ω
We shall find the fault current for when a (a) 1LG, (b) LL and (c) 2LG fault occurs at bus-2.
Fig. 8.10 Radial power system of Example 8.4. Let us choose a base in the circuit of the generator. Then the per unit impedances of the generator are:
The per unit impedances of the two transformers are
The MVA base of the motor is 40, while the base MVA of the total circuit is 50. Therefore the per unit impedances of the motor are
For the transmission line
Therefore
Let us neglect the phase shift associated with the Y/ Δ transformers. Then the positive, negative and zero sequence networks are as shown in Figs. 8.11-8.13.
Fig. 8.11 Positive sequence network of the power system of Fig. 8.10.
Fig. 8.12 Negative sequence network of the power system of Fig. 8.10.
Fig. 8.13 Zero sequence network of the power system of Fig. 8.10. From Figs. 8.11 and 8.12 we get the following Ybus matrix for both positive and negative sequences
Inverting the above matrix we get the following Zbus matrix
Again from Fig. 8.13 we get the following Ybus matrix for the zero sequence
Inverting the above matrix we get
Hence for a fault in bus-2, we have the following Thevenin impedances
Alternatively we find from Figs. 8.11 and 8.12 that
(a) Single-Line-to-Ground Fault : Let a bolted 1LG fault occurs at bus-2 when the system is unloaded with bus voltages being 1.0 per unit. Then from (8.7) we get
per unit
Also from (8.4) we get
per unit Also I fb = I fc = 0. From (8.5) we get the sequence components of the voltages as
Therefore the voltages at the faulted bus are
(b) Line-to-Line Fault : For a bolted LL fault, we can write from (8.16)
per unit
Then the fault currents are
Finally the sequence components of bus-2 voltages are
Hence faulted bus voltages are
(c) Double-Line-to-Ground Fault : Let us assumes that a bolted 2LG fault occurs at bus-2. Then
Hence from (8.24) we get the positive sequence current as
per unit The zero and negative sequence currents are then computed from (8.25) and (8.26) as
per unit
per unit Therefore the fault currents flowing in the line are
Furthermore the sequence components of bus-2 voltages are
Therefore voltages at the faulted bus are
Example 8.5 Let us now assume that a 2LG fault has occurred in bus-4 instead of the one in bus-2. Therefore
Also we have
Hence
per unit Also
per unit
per unit Therefore the fault currents flowing in the line are
We shall now compute the currents contributed by the generator and the motor to the fault. Let us denote the current flowing to the fault from the generator side by Ig , while that flowing from the motor by Im . Then from Fig. 8.11 using the current divider principle, the positive sequence currents contributed by the two buses are
per unit
per unit
Similarly from Fig. 8.12, the negative sequence currents are given as
per unit
per unit Finally notice from Fig. 8.13 that the zero sequence current flowing from the generator to the fault is 0. Then we have
per unit Therefore the fault currents flowing from the generator side are
and those flowing from the motor are
It can be easily verified that adding Ig and Im we get If given above. In the above two examples we have neglected the phase shifts of the Y/ Δ transformers. However according to the American standard, the positive sequence components of the high tension side lead
those of the low tension side by 30° , while the negative sequence behavior is reverse of the positive sequence behavior. Usually the high tension side of a Y/ Δ transformer is Y-connected. Therefore as we have seen in Fig. 7.16, the positive sequence component of Y side leads the positive sequence component of the Δ side by 30° while the negative sequence component of Y side lags that of the Δ side by 30° . We shall now use this principle to compute the fault current for an unsymmetrical fault. Let us do some more examples.
Example 8.6 Let us consider the same system as given in Example 8.5. Since the phase shift does not alter the zero sequence, the circuit of Fig. 8.13 remains unchanged. The positive and the negative sequence circuits must however include the respective phase shifts. These circuits are redrawn as shown in Figs. 8.14 and 8.15. Note from Figs. 8.14 and 8.15 that we have dropped the √3 α vis-à-vis that of Fig. 7.16. This is because the per unit impedances remain unchanged when referred to the either high tension or low tension side of an ideal transformer. Therefore the per unit impedances will also not be altered.
Fig. 8.14 Positive sequence network of the power system of Fig. 8.10 including transformer phase shift.
Fig. 8.15 Negative sequence network of the power system of Fig. 8.10 including transformer phase shift. Since the zero sequence remains unaltered, these currents will not change from those computed in Example 8.6. Thus
and
per unit
Now the positive sequence fault current from the generator Iga1 , being on the Y-side of the Y/ Δ transformer will lead I ma1 by 30° . Therefore per unit
per unit Finally the negative sequence current I ga2 will lag I ma2 by 30° . Hence we have per unit per unit
Therefore
Also the fault currents flowing from the motor remain unaltered. Also note that the currents flowing into the fault remain unchanged. This implies that the phase shift of the Y/ Δ transformers does not affect the fault currents.
Example 8.7 Let us consider the same power system as given in Example 1.2, the sequence diagrams of which are given in Figs. 7.18 to 7.20. With respect to Fig. 7.17, let us define the system parameters as: Generator G1 : 200 MVA, 20 kV, X" = 20%, X0 = 10% Generator G2 : 300 MVA, 18 kV, X" = 20%, X0 = 10% Generator G3 : 300 MVA, 20 kV, X " = 25%, X0 = 15% Transformer T1 : 300 MVA, 220Y/22 kV, X = 10% Transformer T2 : Three single-phase units each rated 100 MVA, 130Y/25 kV, X = 10% Transformer T3 : 300 MVA, 220/22 kV, X = 10% Line B-C : X1 = X2 = 75 Ω , X0 = 100 Ω
Line C-D : X1 = X2 = 75 Ω , X0 = 100 Ω Line C-F : X1 = X2 = 50 Ω , X0 = 75 Ω Let us choose the circuit of Generator 3 as the base, the base MVA for the circuit is 300. The base voltages are then same as those shown in Fig. 1.23. Per unit reactances are then computed as shown below.
Generator G1 :
, X0 = 0.15
Generator G2 : , X0 = 0.0656 Generator G3 :
, X0 = 0.15
Transformer T1 :
Transformer T2 :
Transformer T3 :
Line B-C :
,
Line C-D :
,
Line C-F :
,
Neglecting the phase shifts of Y/ Δ connected transformers and assuming that the system is unloaded, we shall find the fault current for a 1LG fault at bus-1 (point C of Fig. 7.17). From Figs. 7.18 and 7.19, we can obtain the positive and negative sequence Thevenin impedance at point C as (verify)
X1 = X2 = j 0.2723 per unit Similarly from Fig. 7.20, the Thevenin equivalent of the zero sequence impedance is
X0 = j 0.4369 per unit Therefore from (8.7) we get
per unit
Then the fault current is Ifa = 3 Ifa0 = 3.0565 per unit.
Double- Line -to Ground Fault The faulted segment for a 2LG fault is shown in Fig. 8.7 where it is assumed that the fault has occurred at node k of the network. In this the phases b and c got shorted through the impedance Zf to the ground. Since the system is unloaded before the occurrence of the fault we have the same condition as (8.8) for
(8.17)
the phase-a current. Therefore
Fig. 8.7 Representation of 2LG fault. Also voltages of phases b and c are given by
(8.18)
(8.19)
Therefore We thus get the following two equations from (8.19)
(8.20)
(8.21)
Substituting (8.18) and (8.20) in (8.21) and rearranging we get Also since I fa = 0 we have
(8.22)
(8.23)
(8.24)
The Thevenin equivalent circuit for 2LG fault is shown in Fig. 8.8. From this figure we get
(8.25)
The zero and negative sequence currents can be obtained using the current divider principle as
(8.26)
Fig. 8.8 Thevenin equivalent of a 2LG fault.
Example 8.3
Let us consider the same generator as given in Examples 8.1 and 8.2. Let us assume that the generator is operating without any load when a bolted 2LG fault occurs in phases b and c. The equivalent circuit for this fault is shown in Fig. 8.9. From this figure we can write
Fig. 8.9 Equivalent circuit of the generator in Fig. 8.4 for a 2LG fault in phases b and c. Combining the above three equations we can write the following vector-matrix form
Solving the above equation we get
Hence
We can also obtain the above values using (8.24)-(8.26). Note from Example 8.1 that
Then
Now the sequence components of the voltages are
Also note from Fig. 8.9 that
and Vb = Vc = 0. Therefore
which are the same as obtained before.
Line-to-Line Fault The faulted segment for an L-L fault is shown in Fig. 8.5 where it is assumed that the fault has occurred at node k of the network. In this the phases b and c got shorted through the impedance Zf . Since the system is unloaded before the occurrence of the fault we have
(8.8)
Fig. 8.5 Representation of L-L fault. Also since phases b and c are shorted we have
(8.9)
(8.10)
Therefore from (8.8) and (8.9) we have We can then summarize from (8.10)
(8.11)
Therefore no zero sequence current is injected into the network at bus k and hence the zero sequence remains a dead network for an L-L fault. The positive and negative sequence currents are negative of each other. Now from Fig. 8.5 we get the following expression for the voltage at the faulted point Again
(8.12)
(8.13)
Moreover since I fa0 = I fb0 = 0 and I fa1 = - I fb2 , we can write
(8.14)
Therefore combining (8.12) - (8.14) we get
(8.15)
Equations (8.12) and (8.15) indicate that the positive and negative sequence networks are in parallel. The sequence network is then as shown in Fig. 8.6. From this network we get
(8.16)
Fig. 8.6 Thevenin equivalent of an LL fault.
Example 8.2 Let us consider the same generator as given in Example 8.1. Assume that the generator is unloaded when a bolted ( Zf = 0) short circuit occurs between phases b and c. Then we get from (8.9) I fb = - I fc . Also since the generator is unloaded, we have I fa = 0. Therefore from (7.34) we get
Also since V bn = V cn , we can combine the above two equations to get
Then
We can also obtain the above equation from (8.16) as
Also since the neutral current I n is zero, we can write V a = 1.0 and
Hence the sequence components of the line voltages are
Also note that
which are the same as obtained before.
Single-Line-to-Ground Fault Let a 1LG fault has occurred at node k of a network. The faulted segment is then as shown in Fig. 8.2 where it is assumed that phase-a has touched the ground through an impedance Zf . Since the system is unloaded before the occurrence of the fault we have
(8.1)
Fig. 8.2 Representation of 1LG fault. Also the phase-a voltage at the fault point is given by
(8.2)
(8.3)
From (8.1) we can write Solving (8.3) we get
(8.4)
This implies that the three sequence currents are in series for the 1LG fault. Let us denote the zero, positive and negative sequence Thevenin impedance at the faulted point as Z kk0 , Z kk1 and Z kk2respectively. Also since the Thevenin voltage at the faulted phase is Vf we get
(8.5)
three sequence circuits that are similar to the ones shown in Fig. 7.7. We can then write
Then from (8.4) and (8.5) we can write
(8.6)
Again since
We get from (8.6)
(8.7)
The Thevenin equivalent of the sequence network is shown in Fig. 8.3.
Fig. 8.3 Thevenin equivalent of a 1LG fault.
Example 8.1 A three-phase Y-connected synchronous generator is running unloaded with rated voltage when a 1LG fault occurs at its terminals. The generator is rated 20 kV, 220 MVA, with subsynchronous reactance of 0.2 per unit. Assume that the subtransient mutual reactance between the windings is 0.025 per unit. The neutral of the generator is grounded through a 0.05 per unit reactance. The equivalent circuit of the generator is shown in Fig. 8.4. We have to find out the negative and zero sequence reactances.
Fig. 8.4 Unloaded generator of Example 8.1. Since the generator is unloaded the internal emfs are
Since no current flows in phases b and c, once the fault occurs, we have from Fig. 8.4
Then we also have
From Fig. 8.4 and (7.34) we get
Therefore
From (7.38) we can write Z1 = j ω ( Ls + Ms ) = j 0.225. Then from Fig. 7.7 we have
Also note from (8.4) that
Therefore from Fig. 7.7 we get
Comparing the above two values with (7.37) and (7.39) we find that Z 0 indeed is equal to j ω ( Ls - 2 Ms ) and Z2 is equal to j ω ( Ls + Ms ). Note that we can also calculate the fault current from (8.7) as
Introduction The sequence circuits and the sequence networks developed in the previous chapter will now be used for finding out fault current during unsymmetrical faults.
Three Types of Faults
Calculation of fault currents
Let us make the following assumptions:
The power system is balanced before the fault occurs such that of the three sequence networks only the positive sequence network is active. Also as the fault occurs, the sequence networks are connected only through the fault location. The fault current is negligible such that the pre-fault positive sequence voltages are same at all nodes and at the fault location. All the network resistances and line charging capacitances are negligible. All loads are passive except the rotating loads which are represented by synchronous machines.
Based on the assumptions stated above, the faulted network will be as shown in Fig. 8.1 where the voltage at the faulted point will be denoted by Vf and current in the three faulted phases are Ifa , I fb and I fc . We shall now discuss how the three sequence networks are connected when the three types of faults discussed above occur.
Fig. 8.1 Representation of a faulted segment. Calculation of the cables capacitances & inductance values.
Capacitance of Cables, Charging Current, and Charging Reactive Power.
.
. The reactances of the insulated cables.
.
.
.
.
Three branches in an electrical network can be connected in numbers of forms but most common among them is either star or delta form. In delta connection, three branches are so connected, that they form a closed loop. As these three branches are connected nose to tail, they form a triangular closed loop, this configuration is referred as delta connection. On the other hand, when either terminal of three branches is connected to a common point to form a Y like pattern is known as star connection. But these star and delta connections can be transformed from one form to another. For simplifying complex network, delta to star or star to delta transformation is often required.
Delta - Star Transformation
The replacement of delta or mesh by equivalent star connection is known as delta - star transformation. The two connections are equivalent or identical to each other if the impedance is measured between any pair of lines. That means, the value of impedance will be the same if it is measured between any pair of lines irrespective of whether the delta is connected between the lines or its equivalent star is connected between that lines.
Consi der a delta system that's three corner points are A, B and C as shown in the figure. Electrical resistance of the branch between points A and B, B and C and C and A are R1, R2and R3 respectively. The resistance between
the
points
A
and
B
will
be,
Now, one star system is connected to these points A, B, and C as shown in the figure. Three arms R A, RB and RC of the star system are connected with A, B and C respectively. Now if we measure the resistance value between points A and B, we will get, Since the two systems are identical, resistance measured between terminals A and B in both systems must be equal.
Similarly, resistance between points B and C being equal in the two systems, points C
and
A
being
equal
And resistance between in the two systems,
Adding equations (I), (II) and
(III) we get, Subtracting equations (I), (II) and (III) from equation (IV) we get,
The relation of delta - star transformation can be expressed as follows. The equivalent star resistance connected to a given terminal, is equal to the product of the two delta resistances connected to the same terminal divided by the sum of the delta connected resistances. If the delta connected system has same resistance R at its three sides then equivalent star resistance r will be,
Star - Delta Transformation For star - delta transformation we just multiply equations (v), (VI) and (VI), (VII) and (VII), (V) that is by doing (v) × (VI) + (VI) × (VII) + (VII) × (V) we get,
Now dividing equation (VIII) by equations (V), (VI) and equations (VII)
separately we get,
.
Concept of Subtransient, Transient & Steady State. The concept of Subtransient, Transient and Steady State arises in case of fault in an Alternator. Let us assume a sudden short circuit in three phase of alternator. The fault current will flow in all the three phases of alternator and its waveform will be as shown in figure below.
.
When the alternator is short-circuited, the currents in all the three-phases rise rapidly to a high value of about 10 to 18 times of full load current, during the first quarter cycle. The flux crossing the air gap is large during a first couple of cycles. The reactance during these first two or three cycle is least and the short circuit current is high. This reactance is called subtransient reactance and is denoted by X”. The first few cycles come under subtransient state. After a first few cycles, the decrement in the r.m.s. value of short circuit current is less rapid than the decrements during the first few cycles. This state is called the Transient State and the reactance in this state is called transient reactance X’. The circuit breaker contacts separate in the transient state. Finally the transient dies out and the current reaches a steady sinusoidal state called the Steady State. The reactance in this state is called steady state reactance Xd. Since the short circuit current of the alternator lags behind the voltage by 90 degree, the reactance involved are direct axis reactance. As clear from the figure above, the d.c. components in the three phases are different; hence the waveforms of the three phases are not identical. If voltage of phase, say, Y, is maximum at the instant of short circuit, the DC component of short circuit current is zero. Hence the waveform is symmetrical as shown in figure below.
.
The currents and reactance are given by the following expressions,.
.
Where I = Steady state current, r.m.s. value I’ = Transient current, r.m.s. value I” = Sub-transient current, r.m.s. value Ea = Induced e.m.f. per phase Xd = Direct axis synchronous reactance Xd’ = Direct axis transient reactance Xd” = Direct axis sub-transient reactance As the short circuit occurs, the short-circuit current attains high value. The circuit breakercontact starts separating after the operation of the protective relay. The contacts of thecircuit breaker separate during ‘transient state.’ The r.m.s. value of the current at the instant of the contact separation is called the breaking current of the circuit-breaker and is expressed in kA. If a circuit-breaker closes on existing fault, the current would increase to a high value during the first, half cycle. The highest peak value of the current is reached during the peak of the first current loop. This peak value is called making current of the circuit breaker and is expressed in kA. This is the reason making current of Circuit Breaker is higher than the Breaking Current.
Now we will go into the discussion back while taking an example of Electrical circuit. Consider the circuit given below.
If the battery is ON at t=0 and kept ON. Inductor has no energy stored initially. So it has no voltage across it.
Inductor voltage = 0 Rate of change of current = 0 Initially current in inductor = 0 Energy stored in inductor = 0 As we know that Inductor does not allow sudden change in currents. So just after the moment when switch is ON, the current in inductor is zero. That means at t=0+ inductor acts as open circuit. But as the time passes inductor allows current. This happens until coil voltage drop is equal to applied voltage.
But after some time, i.e. when coil voltage drop is equal to applied voltage, then current flowing through the circuit is constant as inductor voltage is zero which means rate of change of current is zero. This state is called Steady State.
So if we solve the differential equation of the circuit, we’re actually finding the response as a function of time, which includes both transient and steady state response. Basic Principle of Relay Operation. Relay is a switch which senses fault in a system and once fault is sensed by the Relay, it issues trip command to the Circuit Breaker, CB to isolate the faulty section of the network from the healthy section.
The Relay detects the abnormal condition by continuously monitoring electrical quantities which are different for healthy and faulty condition. The electrical quantities which may change during fault condition are voltage, current, frequency and phase angle. If one or more of the above electrical quantities change, that signals the presence, type and location of the fault to the Relay. After detecting the fault condition, Relay pick-up, its contact will change from NO to NC or vice versa. So we can wire up a particular kind of Relay contact to Breaker tripping circuit. So whenever, the Relay picks up, the tripping of Breaker will take place. You
may
like
to
Read, Why
CT
Secondary
Shall
Never
be
Kept
Open?
A simplified Relay circuit is shown in figure below. Figure below shows one of the three phase system for simplicity.
As shown in the figure above, Current Transformer CT secondary winding is directly connected to the Relay coil. Under normal condition, the current through the Relay coil is not sufficient enough to pull the plunger and close the circuit of Breaker Tripping Coil. Notice here that Breaker Tripping coil is solely responsible for the tripping of Circuit Breaker. If trip coil of breaker fails, then tripping of Breaker will not take place. This is the reason, two trip coils are normally provided in Circuit Breaker to get reliable operation of Breaker. Not only two Trip Coils are provided in CB rather a Trip Coil monitoring Relay is also used. If case of fault i.e. if it happens to be any open circuit in Trip Coil, then the Trip Coil Supervision Relay will be flagged to attract the attention of the operator.
In case of fault, the current through the CT secondary will go up which will cause increased current through the Relay coil. If it happens that the current through the Relay coil exceeds the setting value or pick-up value then the coil will get produce sufficient magnetic pull to the plunger and thus plunger will complete the CB trip circuit. As soon as the CB trip circuit is complete, current will start flowing in the Trip Coil which in turn will pull a lever to trip the Circuit Breaker CB. In the above figure, it is shown that Relay coil is directly pulling the plunger to complete the Breaker Trip Coil circuit but in actual practice, Relay coil when picked up will change its contact status. Let us say Relay Normally Open (NO) contact is wired to the Breaker Trip Coil Circuit. Therefore when the Relay coil is in de-energized state, the circuit of Trip Coil of CB is not complete and hence no tripping of the CB. During fault condition as the current through the Relay coil exceeds the pick-up value, the Relay coil will get actuated which in turn will force its
contact to change over i.e. NO contact will change to Normally Close (NC) thereby closing the Trip Coil circuit of the Breaker.
Since Trip Coil circuit of Breaker is complete, current will flow through the Trip Coil causing CB to trip.
Introduction and Architecture of Numerical Relay. Most of us are aware of Electromagnetic Relays and Static Relays but most us may not be well acquainted with Numerical Relay. If I define a Numerical Relay, honestly speaking it will seem to be quite tough but in reality they are very user friendly and easy to implement different types of protection scheme. However I am going to define Numerical Relays.
Numerical Relay is a device in which measured electrical quantities are sequentially sampled and then converted into numerical data which are mathematically or logically processed to take decision for issuing trip command.
Numerical Relay is basically Digital Relay for which manufacturers have developed specified hardware which can be used in conjunction with suitable Software o meet different protection needs. A Digital Relay comprises both Hardware and Software. The Hardware part is briefly described below. CPU: CPU stand for Central Processing Unit which is responsible for the processing of protection algorithms and digital filtering. Memory:Memory is of two types. One is RAM (Random Access Memory) and ROM (Read Only Memory). RAM serves for the purpose of retaining the input data to the Relay and processing the data during the compilation of algorithm. ROM is used to store Software needed for the working of Relay. ROM is also needed for storing Event and Disturbance data. Event and Disturbance Recording is a must feature for a digital relay because these data are used for troubleshooting any event. A typical Numerical Relay can store as much as 520 Events and 50 Disturbances. The most attractive feature of such relay is that it works on FIFO (First In, First Out). Suppose if it happens to be the number of disturbances exceeds 50 then the Relay will delete the last Disturbance and will register new disturbance. Input Module: The analog single from the Power System is stepped down using Current Transformer and Potential Transformer and then fed to the Numerical Relay using low pass filter. Low pass filter is incorporated in the input module to eliminate any noise single induced in the line due to corona or induction effect of nearby high voltage line. The output from the Filter is then fed to Sample and Hold (S/H)circuit.
A Sample and Hold (S/H) circuit is used to keep the rapidly changing instantaneous value constant during the period of conversion for processing. In addition to the analog input, Numerical Relay is designed to accept digital input too. Separate terminals are provided for the analog and digital inputs. Multiplexer and Analog to Digital Converter: The CPU accepts the input in digital form but the input from Current Transformer CT and Potential Transformer PT are analog in nature. Therefore and A/D converter is used to convert the analog signal to digital signal. In case more than one analog quantity is to be converted into digital form, Multiplexer is used for selecting any analog input at a time to convert into digital form.
Output Module: Output module provided in Numerical Relay is digital contacts which are actuated when a trip decision is taken by the CPU. These output digital contacts are a pulse which is generated as a response signal. The timing of pulse can be changed by the user. Digital Input / Communication Module: Numerical Relay is provided with serial and parallel ports for the interconnection with control and communication system of the substation. Digital output contacts of Numerical Relay is used for wiring with the Auxiliary Relays to extend tripping command to the Circuit Breaker.
Software: Numerical Relay is equipped with software to communicate with external device to program to Relay or one can program by navigating through the Relay Menu. Hardware for Metering: In principle, the hardware setup discussed above can be used for both measurement and protection function. However, considering the order of difference between current magnitudes in case of fault and load, there can be loss of accuracy during metering applications. Consider a hypothetical case where in maximum load current is 100 A and maximum fault current is 20 times this load current i.e. 2000 A. Let a 12 bit unipolar A/D converter be used for sampling current signal. This implies that resolution of A/D converter is 2000/(212-1)=0.488 A. This resolution may be inadequate for metering purposes. One solution is to increase resolution i.e. the number of bits in A/D converter. For example, one may use 16 bit A/D converter in place of 12 bit A/D converter. However, increasing the number of bits of A/D converter also affects the selection of processor. A good design guideline is to choose a processor with double the number of bits of A/D converter. This ensures that truncation and numerical precision problems associated with finite precision arithmetic do not cause significant loss of accuracy. For example, with 16 bit A/D converter, 32 bit processor is the natural choice. Alternatively, a variable gain amplifier can be used along with the A/D converter. At low currents, high gain setting is used and at high currents low gain setting is preferred. However, during the change from one setting to another, loss of information can take place. Therefore, a simple solution would be to keep metering and protection functionality separate. In the next post we will be discussing about some interesting features of Numerical Relay. So be there and follow ELECTRICAL CONCEPTS.
PLC Step By Step: Your Complete Guide. This course will help know TIMERS, Counters, Relays, Coils and Ladder Logic and build you FIRST PLC Program From Scratch. Click below to register. Why CT Secondary Shall Never be Kept Open?. The most important precaution which shall be taken care while working with a CT is that its secondary should never be kept open. In this post we will discuss this aspect of CT. As we know that Current Transformers (CTs) are always used with secondary winding connected with Ammeter, Relays or Wattmeter Coils. A precaution which shall always be taken is that Never open the secondary winding circuit of a Current Transformer while its Primary Winding is energized. If the secondary winding circuit of a CT is kept open then it will lead to severe consequence to the personnel opening the CT secondary and to the CT itself. The question arises why?
To understand this, first we should know the basic difference between a Power Transformer and a CT. The basic difference between a Power Transformer and a CT is that, in Power Transformer the primary current is the reflection of the secondary current by N1I1 = N2I2while in CT the primary current is dependent on the load current or line current as Ct is connected in series with the line. So primary current of CT (assuming constant line current) is constant irrespective of whether the secondary of CT is connected with burden or not. During normal operation of CT, the primary and secondary winding produces mmf which by lenze’s law opposes each other. As the secondary mmf is slightly less than the primary mmf, the net mmf is small. This net mmf is the working / magnetizing mmf of the core of CT. Now, in case secondary winding is kept open then secondary current will be zero while the primary current of CT will remain same. Therefore the opposing mmf of secondary will no longer exist. Hence the net mmf is due to primary current only i.e. N1I1 which is very large. This large mmf will produce large flux in the core and will saturate the core. Again, due to large flux in the core the flux linkage of secondary winding will be large which in turn will produce a large voltage across the secondary terminals of the CT. This large voltage across the secondary terminals will be very dangerous and will lead to the insulation failure and there is a good chance that the person who is opening the CT secondary while primary is energized will die due to shock. Also, because of excessive core flux, the hysteresis and eddy current loss will be very high and the CT will get overheated. As CT is oil filled, because of overheating, the oil of CT will get boil and start to vaporize. Because of vaporization of CT oil, the CT housing will get pressurized and blast. This blasting will lead to fire and smoke. Again, it is not the end here but due to smoke, the nearby lines will trip due to earth fault which may trip the Power Generating Station.
Difference between Isolators and Sectionalizers in Power System. An isolator is a switch used in power lines, to completely open (physically) a specific section of line, when maintenance or modification work is to be done. The isolator can be operated manually, or in some cases, opened by energizing a motor. With a threephase system, the isolator opens all three lines. A picture of Isolator is shown below to visualize.
A sectionalizer is an automatic switch, installed downstream of a recloser. The recloser is a feature provided in a circuit breaker, which opens when a short circuit occurs on the line such as a falling tree branch. Ordinarily, such short circuits clear themselves, for example by the branch falling to the ground. Thus the recloser opens the circuit when the fault occurs, and after a short interval, re-closes itself and restores power to the system. A picture of sectionalizer is shown below to visualize.
However, not all short-circuit faults clear themselves. For example, a fallen branch may lie across the lines and not fall to the ground. A sectionalizer counts the successive openings and closings of the recloser, and after a pre-set number, it opens. Thus it isolates a particular section of line where a continuing fault exists, leaving most of the larger area protected by the recloser still with power.
So both of these are devices that isolate a portion of a power distribution system. But the first is a safety device, to protect people working on the lines. The second is an automatic device, to cut off power to a section where a continuing fault has occurred. Circuit Breaker and Arc Phenomenon. What is Circuit Breaker?
Circuit Breaker is switch capable of making or breaking the circuit under no-load as well as onload condition. It can make or break circuit either manually or by remote control. A Circuit Breaker in conjunction with Relay can break the circuit under fault condition.
You may also like to read, Basic Principle of Relay Operation
Operating Principle of Circuit Breaker:
A Circuit Breaker CB consists of two contacts which are called electrodes, one of which remain fixed, called fixed contact and another moving contact. Under normal operating condition, this contact will remain closed to supply power but as soon as fault is sensed by the Relay, trip coil of Circuit Breaker energizes and the moving contact of CB is pulled apart by some mechanism to open the CB.
When contacts of CB are separated under fault condition, an arc is stuck between the fixed and moving contacts. The current is thus able to continue till the arc persists. The production of arc not only delays the current interruption but it also produces huge amount of heat which if exceeds a limit may damage the system or CB itself. Therefore, the design of CB is done in such a way to minimize the arcing period so that
1) Heat produced during arcing may not exceeds the dangerous value.
2) To have fast fault clearing.
It is worth here to mention that a typical Breaker opening and closing time remain around 30-35 ms and 60-70 ms respectively. Notice that CB opening time is less than the closing time to ensure fast fault clearing.
Arcing Phenomenon in Circuit Breaker:
When a short circuit occurs, heavy current flows through the contacts of circuit breaker before they are opened by the protective system. At the instant when the contacts begin to open after getting trip command from the Relay, the contact area decreases rapidly and large fault current causes increased current density and hence rise in temperature. The heat produced in the medium in between the contacts is sufficient enough to ionize the medium. This ionized medium acts as a conductor and arc is stuck in between the contacts of the circuit breaker. It shall be noted here that the potential difference between the fixed and moving contacts is quite small and just enough to maintain the arc. This arc provides a low resistance path to the current and thus due to arcing the current in the circuit remain uninterrupted as long as arcing persists.
During the arcing period the current flowing through the contacts of circuit breaker deepens upon the arc resistance. The greater the arc resistance the smaller will be the current flowing through the contacts of CB. The arcing resistance depends upon the following factors:
Degree of Ionization:
The more the ionization of medium between the contacts, the less will be the arcing resistance.
Length of Arc:
The arc resistance increases as the length of arc increases i.e. as the separation between the contacts of Breaker increases the arcing resistance also increases.
Cross Section of the Arc:
The arcing resistance increases with decrease in the cross sectional area of the arc.
Principle of Arc Extinction:
As we discussed earlier in this post, ionization of medium in between the contacts and potential difference across the contacts are responsible for the production and maintenance of arc. Thus for arc extinction, we can increase the separation between the contacts to such an extent that potential difference across the contacts is not sufficient enough to maintain the arc. But this philosophy is impractical as in EHV (Like 220 kV, 400 kV, 765 kV etc.) system; the separation between the contacts to extinguish the arc will be many meters which is not practically achievable.
Another way for extinction of arc is to demonize the medium in between the contacts. If the arc path is demonized the arc extinction will definitely be facilitated. This may be achieved by cooling the arc or by quickly removing the ionized particles from the space in between the contacts. This principle of arc extinction is used in all modern Circuit Breakers.
SF6 Circuit Breaker – Construction, Working Principle and Types. In Sulpher Hexafluoride or SF6 Circuit Breaker, sulpher hexafluoride (SF6) gas is used as an insulating and arc quenching medium. SF6 gas has many superior properties which makes it perfect for arc quenching.Sulpher Hexafluoride or SF6 Circuit Breaker is most popular and widely used breaker. This type of breaker is mostly used for EHV systems like 220 kV, 400 kV and 765 kV. It is suggested to read “Why SF6 gas used in HV/EHV Circuit Breakers“.
Construction of SF6 Circuit Breaker Like other circuit breaker viz. Vacuum Circuit Breaker, Air Blast Circuit Breaker etc., SF6Circuit Breaker has fixed contact as well as moving contact. Theses fixed and moving contacts are known as MAIN CONTACT. There exists one another contact which is known as ARCING CONTACT. Arcing Contact is part of fixed contact. Basically, Arcing contacts are only designed to withstand arcing. It is not designed for carrying load current. In spite, main contacts are designed to carry load current and not the arcing. Therefore, it can be said that, while closing of SF6 circuit breaker, first Arcing Contact will close. Thereafter arcing contact will close. Similarly while opening, first main contact will open and then arcing contact will open. Notice here that, during opening operation of SF6Circuit Breaker, the order of opening main and arcing contact is revered as that in closing operation. This is because, while opening if the main contact opens first, there will not be any arcing as the current is getting path through the arcing contact. But if the arcing contact open first then during opening of main contact there will be arcing and as discussed main contacts are not meant to withstand arcing. Apart from Fixed contact and moving contacts, SF6 Circuit Breaker has following main components:
Interrupter Insulating Nozzle SF6 Gas Chamber
An interrupt is a chamber which encloses the breaker contacts, insulating nozzle, SF6 gas chamber. Interrupter is made of porcelain. Figure below shows the basic parts of SF6 circuit breaker.
Figure-1 Carefully observe the figure and notice the different parts, though some parts like SF6 gas chamber, nozzle, valve etc are not shown in the above figure but they will be shown while discussing the working principle. So, please be patient till then and read further. Working principle of SF6 Circuit Breaker The contacts of SF6 Circuit Breaker are surrounded in an environment of SF6 gas at some pressure. Actually, the dielectric strength of SF6 gas is directly proportional to its
pressure. In 220 kV, 400 kV and 765 kV applications, the gas pressure is maintained at 6.5 bar. Let’s consider breaker opening operation for better understanding of operating mechanism. First have a look at the contacts when the breaker is in fully close position as shown in Figure-1. Now we will open the breaker and will observe its mechanism step by step. Step-1: Main Contact Open
As discussed earlier in the post, main contact will open first. This is shown in the figure above. Observe in figure that, though main contacts are open, arcing contacts are still close. As main contacts open, the piston in the cylinder moves causes the SF6 gas to compress due to reduction of volume Vp. Step-2: Arcing Contacts Open
As soon as arcing contacts separates from contact 1, an arc is strikes. Due to this arcing, heat is produced. This heat of arc further increases the pressure of SF 6 in the chamber Vt. Mind that, the pressure of arc extinguisher i.e. SF 6 is increased by the heat of arc. This is the reason; such breaker is called self compensating type. Here self compensating means that, the capacity of breaker to interrupt the fault is proportional to fault current. Step-3: Arcing Contact separates from Nozzle
When arcing contact separates from the insulating Nozzle, the pressurized SF6 gas in volume Vt is released in the arc. This causes the arc to extinguish at the moment the current passes though the natural zero. Thus, the pressurized SF6 gas extinguishes the arc and hence circuit is interrupted. In case of small current like in unloaded transformer or reactor, the thermal energy of arc is not enough to pressurize the SF6 gas. In such case the pressure developed in the SF6 gas chamber Vp in Step-1 is extinguishes the arc. Types of SF6 Circuit Breaker As discussed, in 220 kV, 400 kV and 765 kV applications, the SF6 gas pressure is maintained at 6.5 bar. You will be amazed that, even though voltage level is increasing, same pressure of SF6 i.e. 6.5 bar is used for 220, 400 and 765 kV applications. Actually as we go up at higher voltage level, the number of contacts increases in SF6 circuit Breaker. Based on this philosophy, SF6 circuit breaker can be classified into following types: Single
Single Breaker Circuit Breaker Double Break Circuit Breaker Multi Break Circuit Breaker Break SF6 Circuit Breaker
In Single Break Circuit Breaker, only one moving and fixed contacts are present. This means that, there will only be one interrupter unit in such breaker. Single break SF6 circuit breaker is used for 220 kV applications. Double Break SF6 Circuit Breaker
In such type of breaker, there are two set of moving and fixed contacts connected in series. Therefore, to enclose two set of contacts, there must be two interrupt unit in series. This type of breaker is used in 400 kV applications. In double break circuit breaker, grading capacitorsare used to equalize the voltage distribution across each contact. Thus for 400 kV application, the voltage across each contact will be 200 kV. Therefore it is logical to use SF6 gas at a pressure same as used in 200 kV application. Ha ha..got it? Multi Break SF6 Circuit Breaker
In multi break circuit break, more than two set of fixed and moving contacts are used. Such type of breaker is used in 765 kV applications.
Vacuum Circuit Breaker- Construction and Working. Vacuum Circuit Breaker Vacuum Circuit Breaker (VCB) is one where vacuum of the order of 10 -6 to 10-10 torr is used as an arc quenching medium. 1 torr is equivalent to a pressure represented by a barometric head of 1 mm mercury. Vacuum Circuit Breaker is used for low and medium voltage applications. Construction of Vacuum Circuit Breaker
Vacuum Circuit Breaker consists of Enclosure, Contacts, Vapor Condensing Shield, Metallic Bellows and Seal. Enclosure. The enclosure is made of impermeable insulating material like glass. The enclosure must not be porous and should retain high vacuum of the order of 10-7 torr. Contacts. There are two types of contacts, moving and fixed. The moving contact is connected with large stem connected to operating mechanism of breaker. Contacts of Vacuum Circuit Breaker have generally disc shaped faces. The disc is provided with symmetrical grooves in such a way that the segments of the two contacts are not in the same line. The magnetic field set-up by the components of currents with such geometry causes the plasma of the arc to move rapidly over the contacts instead of remaining stable at one point. The concentration of the arc is thus prevented and the arc remains in diffused state. The sintered material used for contact tip are generally copperchromium or copper bismuth alloy. Vapor Condensing Shield. These metallic shields are supported on insulating housing such that they cover the contact region. The metal vapor released from the contact surface during arcing is condensed on these shields and is prevented from condensing on the insulting enclosure. Metallic Bellows. One end of the bellows is welded to the enclosure. The other end is welded to the moving contact. The bellows permit the sealed construction of the interrupter and yet permit movement of the contact. Stainless steel bellows are generally used in vacuum interrupters. Carefully observe every component of Vacuum Circuit Breaker as shown in figure below.
Arc Extinction in Vacuum Circuit Breaker (VCB)
The arc interruption process in Vacuum Circuit Breaker interrupter is quite different from that in other types of circuit breakers. The vacuum as such is a dielectric medium and arc cannot persist in ideal vacuum. However, the separation of current carrying contacts causes the vapor to be released from the contacts. Thus, as the contacts separate, the contact space is filled with vapor of positive ions liberated from the contact material. The vapor density depends on the current in the arc. During the decreasing mode of the current wave the rate of release of the vapor reduces and after the current zero, the medium regains the dielectric strength provided vapor density around contacts has substantially reduced. While interrupting a current of the order of a few hundred amperes by separating flat contacts under high vacuum, the arc generally has several parallel paths. Thus the total current is divided in several parallel arcs. The parallel arcs repel each other so that the arc tends to spread over the contact surface. Such an arc is called diffused arc. The diffused arc can get interrupted easily. At higher values of currents of the order of a few thousand amperes, the arc gets concentrated on a small region and becomes self-sustained arc. The concentrated arc around a small area causes rapid vaporization of the contact surface. The transition from diffused arc to the concentrated arc depends upon the material and shape of contact, the magnitude of current and the condition of electrodes. The interruption of arc is possible when the vapor density varies in phase with the current and the arc remains in the diffused state. The arc does not strike again if the metal vapor is quickly removed from the contact zone. Thus the arc extinction process in vacuum circuit breaker is related to a great extent to the material and shape of the contacts and the technique adopted in condensing the metal vapor. The contact geometry is so designed that the root of the arc keeps on moving so that the temperature at one point on the contact does not reach a very high value. The rapid building up of dielectric strength after final arc extinction is a unique advantage of vacuum circuit breaker. They are ideally suitable for capacitor switching as they can give restrike free performance. Degree of Vacuum in VCB Interrupters
The breakdown voltage of contact gap varies with the absolute pressure in the vacuum circuit breaker interrupter unit. As the absolute pressure is reduced from 10-1 Torr to 103 Torr, the dielectric strength (kV/mm) increases but above 10-4 Torr, the breakdown strength and pressure characteristic becomes almost flat as shown in figure below.
The dielectric strength in this region is above 12 kV /mm. In vacuum interrupters vacuum level of the order of 10-6 to 10-10 Torr is used. This is called high vacuum range. During the passage of time and after arc interruptions, the vacuum level goes on reducing. However it remains in the range of 10– 5 Torr and 10-8 Torr. Vacuum in the range of 10-3 is sufficient for interruption. Gas Insulated Switchgear or GIS. Gas Insulated Switchgear (GIS) is a compact metal encapsulated switchgear consisting of high voltage components such as circuit breakers, disconnectors, Current Transformer, Earth Switch, Bus bar etc. which can be safely operated in confined spaces. GIS is used where space is limited, for example, extensions, in city buildings, on roofs, on offshore platforms etc.
In Gas Insulated Switchgear SF6 gas is used as an insulating medium and as an arc quenching medium in Circuit Breaker. But there are some GIS where Clean Air is used an insulating medium and Vacuum as an interrupting medium. For example, in Siemens 8VN1GIS (up to 145 kV), vacuum is used as an interrupting medium while clean air is used as insulating medium. Due to advancement of technology, today Gas Insulated Switchgear or GIS are available for voltage ranging from 12 kV to 1200 kV. GIS is basically modular switchgear as shown in figure below.
Siemens press picture
The various modules are factory assembled and are filled with SF 6 gas. Thereafter, they are taken to site for final assembly. Such sub-stations are compact and can be installed conveniently on any floor of a multi-storied building or in an underground sub-station. As the units are factory assembled, the installation time is substantially reduced. Such installations are preferred in cosmopolitan cities, industrial townships, hydro-stations where land is very costly and higher cost of SF6 insulated switchgear is justified by saving due to reduction in floor-area requirement. SF6 insulated switchgear or GIS is also preferred in heavily polluted areas where dust chemical fumes and salt layers can cause frequent flashovers in conventional outdoor substations. Types of Gas Insulated Switchgear
Based on the constructional feature, GIS can be classified into following types: Isolated Phase GIS Basically in such kind of GIS, the modules are assembled phase wise i.e. Breaker, CT, Isolators of individual phases are assembled separately. Modules for each phase of a bay are separate. That is why such GIS are called Isolated Phase GIS. It is quite obvious that the space requirement for Isolated Phase GIS is more. Integrated Three Phase GIS As the name suggests, all three phases’ equipment like Breaker, CT, Isolators etc. are kept in a single module filled with gas. The space requirement for such GIS is 1/3rd of space requirement in Isolated Phase GIS. Hybrid GIS Hybrid GIS is an optimal combination of Isolated Phase and Integrated Three Phase GIS. Highly Integrated Switchgear (HIS) In HIS type of Gas Insulated Switchgear, total substation equipments are encapsulated together in single enclosure. With the HIS, the circuit breakers, disconnectors, earthing switches and instrument transformers are accommodated in compressed gas tight enclosures, which make the switchgear extremely compact. HIS, the gas insulated switchgear can be used for indoor and outdoor purpose as HIS requires less than half the space of a comparable air insulated switchgear. Figure below shows the HIS.
Siemens press picture
Notice that, the GIS in above figure is used for outdoor purpose but the same could be used for indoor purpose. I used this picture to just show you that GIS does not mean
that every equipment is kept in a room filled with SF6 gas rather GIS mean equipment kept in a module filled with SF6 gas. Advantages of SF6 Gas Insulated Switchgear The space occupied by Gas Insulated Switchgear installation is only about 10% of that of conventional outdoor sub-station. Thus the high cost of GIS is partly compensated by saving in cost of space. Protection from pollution. The moisture, pollution, dust etc., have little influence on Gas Insulated Switchgear. However, to facilitate installation and maintenance, such substations are generally housed inside a small building. The construction of the building need not be very strong like convention power houses. Reduced Switching overvoltages. The overvoltages while closing and opening line, cables, motors, capacitors etc. are low. Reduced Installation Time. The principle of modular construction reduces the installation time to a few weeks. Conventional sub-stations require a few months for installation. The gas pressure (4 kg/cm2) is relatively low and does not pose serious leakage problems. Increased Safety. As the enclosures are at earth potential, there is no possibility of accidental contact by service personnel to live parts. Disadvantage of SF6 Gas Insulated Switchgear High cost compared to conventional outdoor sub-station. Excessive damage in case of internal fault. Long outage periods as repair of damaged part at site may be difficult. Requirement of cleanliness are very stringent. Dust or moisture can cause internal flashovers. Procurement of gas and supply of gas to site is problematic. Adequate stock of gas must be maintained.
Directional Earth Fault Protection. Generally earth faults are Single Line to Ground (SLG) and Line-Line to Ground (LLG) faults. Earth faults are characterized by the presence of Zero Sequence Curren I 0. Since, except for unbalance, normal system operation is not having Zero Sequence Current I 0, much more sensitive pick-up is possible for earth fault by using zero sequence current component I 0 = (Ia + Ib + Ic) / 3 and declaring a fault if I0 exceeds a threshold.
We know that I0 = (Ia + Ib + Ic) / 3
However, in a system with multiple sources or parallel paths, we require earth fault relays to be directional as discussed in earlier post How to Incorporate Directional Featurein a Relay.
As we discussed in earlier post How to Incorporate Directional Feature in a Relay, that for making a Relay directional we need Reference Phasor. The reference phasor is called asPolarizing Quantity. For ground fault relaying both Voltage and Current Polarization can be used.
We will consider each Voltage and Current Polarization separately for Earth Fault Protection.
Voltage Polarization:
Let the system be initially unloaded and a ground fault occur on phase A. Therefore Ib = Ic = 0 and Ia = 3I0. For Single Line to Ground fault there is a drop in voltage of phase A while phase B and C voltages remain unchanged. Phasor diagram for Voltage and current for SLG fault can be drawn as below.
Voltage and Current Phasor under Single Line to Ground Fault:
In the phasor diagram only 3I0 is shown as Ib = Ic =0 and Ia = 3I0 for Single Line to Ground fault.
Now we will find the Zero Sequence Voltage under the fault.
As V0 = (Vag+Vbg+Vcg)/3, phasor sum of Va,Vb and Vc is to be taken. 3V0 = Vag+Vbg+Vcg, phasor sum
From phasor diagram it is clear that Zero Sequence Voltage 3V 0 is in phase opposition with Vag (Phase Voltage of A). Therefore it is appropriate to take -3V0 as a reference phasor. In normal power system V0is not present but available only during the fault. Let the maximum torque be drawn at 60 degrees lag with respect to -3V0 phasor as shown in figure below.
As we know that Zero Torque Line is perpendicular to the Maximum Torque Line, therefore we draw Zero Torque Line as shown in figure above.
It is then clear that zero Torque Line which separates the plane into Operate and Do Not Operate zone leads -3V0by 30 degrees. Thus, for fault in the correct region, 3I0 lags -3V0 hence falls in operate region. If fault is behind the relay, 3I 0 will lead -3V0 by about 45 to 60 degrees and hence will lie in do not operate region. Hence, earth fault directional unit will not pick-up.
Current Polarization:
For providing direction feature in earth fault relay we can also use current as refrenec phasor which is called current polarization. It is an alternative for voltage polarization. It does not require an additional Potential Transformer (PT).
For balanced system, Ia+Ib+Ic = 0, phasor sum is taken here.
Therefore, I0 = (Ia+Ib+Ic)/3 = 0 which means absence of Zero Sequence Current in balanced system.
During ground fault say at phase A, 3I0 flows from ground to neutral of a Wye connection of Transformer. If we assume for simplicity that Ib = Ic = 0, then 3I0 and Ia are in phase. This indicates that directional unit for ground relay should pick-up as Ia is in phase with 3I 0. Thus we place maximum torque line at zero degrees with respect to I0 phasor. The correspondingOperate and Do Not Operate zones are marked in figure below.
If fault is behind the relay, then the Ia will fall in Do Not Operate region and hence relay will not pickup as Zero sequence Current through the neutral of Wye connection and Relay will be in phase opposition.
Sensitive Earth Fault Protection. Sensitive Earth fault Protection scheme is used for the detection of earth fault. Protection against phase to ground faults can be a difficult problem since ground fault currents vary within a large range, becoming almost negligible in some situations. The ground fault current magnitude depends on the power system grounding which can vary from solidly grounded to ungrounded neutral. The sensitive earth fault protection is usually used in alternators and
transformers with high resistance grounding. High resistance grounding restricts the earth fault current to a very less value and permits the operation of equipment. Also Read, Concept of Neutral Grounding As we know that, the earth fault current magnitude is given by, IF = 3VLN/ (Z0 + Z1 + Z2 + 3ZE) Where Z0 = Zero sequence impedance Z1 = Positive sequence impedance Z2 = Negative sequence impedance ZE = Impedance of earth fault ZE represents the impedance of the ground return circuit including the fault arc, the grounding circuit, and the intentional neutral impedance, if present. If we consider a solidly grounded system, then ZE = 0 and if Z0 = Z1= Z2 then earth fault current will be given as shown below. IF = VLN/ Z1
The Sensitive Earth Fault protection scheme works by measuring the residual current across the three phases in a system. Measurement of three phase residual current is done either by using Core Balance Current Transformer (CBCT) or three CTs connected in parallel. In the ideal condition, the residual current will be zero as all the currents flow through the three phases. Here Residual current means current flowing through neutral or zero sequence current.
As we know that IN = 3I0= IR + IY +IB
But during normal operation, IR + IY+IB = 0 therefore, no residual current will flow. The most important part in SEF or Sensitive Earth Fault Protection is to make proper setting of the Relay. The protection setting should take into consideration that the three CTs do not have identical characteristics and will perform differently for heavy phase-to-phase faults or for initial asymmetrical motor starting currents. This can produce false residual currents. The setting should also be above the line maximum unbalance current. The above conditions must be satisfied to avoid nuisance tripping. In addition, the ground fault protection must be sensitive to minimum ground fault current at the end of the line. Sensitive earth Fault protection scheme is very sensitive to detection of earth fault in the sense that its setting can be as low as 0.2%.
Concept of Neutral Grounding. The concept of system grounding is extremely important, as it affects the susceptibility of the system to voltage transients, determines the types of loads the system can accommodate, and helps to determine the system protection requirements. The system grounding arrangement is determined by the grounding of the power source. For commercial and industrial systems, the types of power sources generally fall into four broad categories:
Utility Service – The system grounding is usually determined by the secondary winding configuration of the upstream utility substation transformer. Generator – The system grounding is determined by the stator winding configuration. Transformer– The system grounding on the system fed by the transformer is determined by the transformer secondary winding configuration. Neutral grounding is generally of three types:
Solid Grounding Resistance Grounding Reactance Grounding
Each of the grounding method serves a specific purpose and based on the suitability of our need, we use any one of the grounding method.
Solidly Grounded Systems: The solidly grounded system is the most common system arrangement, and one of the most used. The most commonly used configuration is the solidly grounded star, because it support single-phase phase to neutral loads. In this type of grounding method, the star point is directly connected to the ground. The figure below, shows the relationship between the phase and line voltage for Solidly Grounded System.
It can be seen from the above figure that the system voltage with respect to ground is fixed by the phase-to-neutral winding voltage. It means that the line-to-ground insulation level of equipment need only be as large as the phase-to-neutral voltage, which is 57.7% (100/1.732 = 57.7 %) of the phase-to-phase voltage. It also means that the system is less susceptible to phase-to-ground voltage transients. This is very important benefit of Solidly Ground System. A common characteristic of solidly-grounded system is that a short circuit to ground will cause a large amount of short circuit current to flow. This condition is known as a ground fault. As can be seen from figure below the voltage on the faulted phase is depressed, and large current flows in the faulted phase since the phase and fault impedance are small.
The voltage and current on the other two phases are not affected. Thus a solidly grounded system supports a large ground fault current. Statistically, 90-95% of all system short-circuits are ground faults. The occurrence of a ground fault on a solidly grounded system necessitates the removal of the fault as quickly as possible. This is the major disadvantage of the solidly-grounded system as compared to other types of system grounding.
A solidly-grounded system is very effective at reducing the possibility of line-to-ground voltage transients. However, to do this the system must be effectively grounded. One measure of the effectiveness of the system grounding is the ratio of the available ground-fault current to the available three-phase fault current. For effectively grounded systems this ratio is usually at least 60%. To summarize, The solidly grounded system is the most popular, is required where single-phase phaseto-neutral loads must be supplied, and has the most stable phase-to-ground voltage characteristics. However, the large ground fault current is a disadvantage and can be hindrance to system reliability.
Resistance Grounded Systems: In Resistance Grounding method, the neutral point is connected to the ground by using a Resistor. The resistor is sized to allow 1-10 A to flow continuously if a ground fault occurs.
The resistor is sized to be less than or equal to the magnitude of the system charging capacitance to ground. If the resistor is thus sized, the high-resistance grounded system is usually not susceptible to the large transient overvoltages that an ungrounded system can experience. If no ground fault current is present, the phasor diagram for the system is the same as for a solidly grounded system. However, if a ground fault occurs on one phase the system response is as shown in figure below. As can be seen from figure below, the ground fault current is limited by the grounding resistor.
If the approximation is made that ZA (impedance of winding) and ZF (Fault impedance) are very small compared to the ground resistor resistance value R, then the ground fault current is approximately equal to the phase-to-neutral voltage of the faulted phase divided by R. The faulted phase voltage to ground in that case would be zero and the unfaulted phase voltages to ground would be 173% of their values without a ground fault present. The ground fault current is not large enough to force its removal by taking the system off-line. Therefore, the high resistance grounded system has the same operational advantage in this respect as the ungrounded system.
Reactance Grounding: A Reactance Grounded system is one in which the neutral point is grounded through an impedance which is highly inductive. Reactance Grounding lies between the effective grounding and Resonant Grounding (will be discussed in next post). Reactance is provided to keep the fault current within safe limit. This method of grounding is used where the charging current is high like in capacitor bank, line reactors used for voltage control of transmission line etc. Restricted Earth Fault Protection of Transformer. Restricted Earth Fault (REF) protection is basically a Differential Protection. The only difference in between the Differential Protection and REF Protection is that, latter protection is more sensitive as compared to the former protection scheme. In earlier posts we have already discussed Differential Protection of Transformer and various characteristics of Differential Protection. In this post we will focus on Restricted earth Fault protection. Also Read,
Transformer Differential Protection Percentage Differential Protection – Slope in Differential Protection Harmonic Restraining in Differential Protection For the sake of understanding REF Protection, we take a Transformer of configuration DYn i.e. HV side of Transformer is Delta connected while the LV side is Start connected and neutral is grounded solidly.
As shown in figure above, there are a total of four Current Transformers (CTs), three CTs connected in each phase i.e. R, Y and B and one CT connected in neutral. The secondary of these four CTs are connected in parallel. The parallel connected CT secondary are then connected to REF Relay Coil. Basically REF protection Relay element is an over current element. Under balanced condition i.e. under normal operation the sum of currents through the secondary of CTs will be zero and current in neutral CT will also be zero. But as soon as a fault takes place in the secondary winding of Transformer, the current in R, Y and B phase will no longer be balanced. Also under earth fault a current will flow through the neutral CT. Because of this unbalance, the summation of current will not be zero but it will have some finite value and hence the relay will pick up. It shall be noted that for a fault outside the Transformer i.e. for through fault Restricted Earth Fault Protection will not operate as in this case of through fault, the vector sum of currents in CT secondary will be zero. This is the reason; such kind of protection scheme is for restricted zone and hence called Restricted Earth Fault Protection.
Now, it is normal to ask that Differential Protection is also a zone protection and it shall operate for any internal fault in Transformer, then why do we need extra Restricted Earth Fault Protection? This is really a very smart question. See, what happens is, the setting of differential protection is normally kept at 20%. So, differential relay shall pick if the differential current exceeds 0.2 A. Now let us consider a case where earth fault occurs just near the neutral point as shown in figure below.
Since the location of fault is very near to the neutral point, the voltage driving the fault current will be very less and hence the reflection of such a low current in primary side of transformer will also will be low. Thus in such case, Transformer differential protection may not operate as its setting is quite high at 20%. Therefore for protection of Transformer from such a fault we need more sensitive protection scheme which is implemented by using Restricted Earth Fault Protection. The sensitivity of REF protection is superior as compared to Differential Protection. Normally the setting of REF protection is kept as low as 5%. Basically the sensitivity of REF protection increases as we are using CT in neutral of transformer and whenever an earth fault take place it is damn sure that current will complete its path through the neutral and hence increasing the sensitivity of REF protection.
Difference between Earthing, Grounding and Neutral. Earthing means connecting the dead part, it means the part which does not carries current under normal condition to the earth. For example electrical equipment’s frames, enclosures, supports etc.
While Grounding means connecting the live part (it means the part which carries current under normal condition) to the earth. For example neutral of power transformer. The purpose of Earthing is to minimize risk of receiving an electric shock if touching metal parts when a fault is present. While the purpose of Grounding is the protections of power system equipment and to provide an effective return path from the machine to the power source. For example grounding of neutral point of a star connected transformer. Ground is a source for unwanted currents and also as a return path for main current some times. While Earthing is done not for return path but only for protection of delicate equipment. Harmonic Restraining in Differential Protection. In this post we will discuss about the feature of Harmonic Restraining feature of Differential Protection. First we should know why Harmonic Restraining is needed?
Suppose we are going to energize the Transformer, obviously the Transformer will have an Inrush current which is around 6 times of full load current. Refer “Transformer Inrush Current” for detail on Transformer Inrush Current. Figure given below shows waveform of Inrush Current of Transformer.
Therefore, the Differential protection will operate. Thus we won’t ever be able to energize the Transformer or we need to bypass the Transformer Differential Protection when we are going to energize. Is it a good practice to Bypass Transformer Protection? You will say NO.So what we need to do for preventing the Differential Protection operation due to Inrush Current?
This requirement calls for Harmonic Restraining in Differential Protection of Transformer. The inrush current of a Transformer, if analyzed, is rich in 2 nd harmonic component. So we can use this fact to prevent the operation of Differential Protection. Therefore a 2nd harmonic Restraining is provided in Transformer Differential Protection. Normally the setting of 2ndHarmonic Restraining is kept at 20% which means that if the 2nd harmonic component in the differential current of Relay is more than 20% of differential current Idthen Differential Protection Relay will not operate as it will think that it is because of Transformer Inrush Current but if the 2nd harmonic component in the differential current of Relay is less than 20% of differential current Id then Differential Protection Relay will operate.
Similarly, during overfluxing of a Transformer the Transformer current is rich in 5th harmonic current. But it may happen so that Transformer is overfluxed for some short time period say around 2 sec. So for this short period of overfluxing should not cause the operation of Differential Protection. Note that during overluxing, the differential current Id will increase which will operate the Differential protection Relay.
Therefore to prevent the operation of Differential protection Relay due to overfluxing, 5thharmonic Restraining is provided in such a manner that if the 5th harmonic component in the differential current of Relay is more than 25% (say) of differential current Id then Differential
Protection Relay will not operate but if 5th harmonic component in the differential current of Relay is less than 25% of differential current Id then Differential Protection Relay will operate.
Percentage Differential Protection – Slope in Differential Protection. As discussed in the post “Transformer Differential Protection”, differential protection is supposed to operate for the internal faults or for the zone of protection it is intended for. Differential protection is not supposed to operate for a through fault. Through fault means a fault outside the zone of protection. Thus as discussed in earlier post, for a through fault the differential current through the overcurrent element of the Differential Protection Relay is zero while there is some definite value of Differential Current for internal faults. But actually there are many limitations due to which a differential current flows the Differential Relay in normal operation also.
A practical transformers and CTs pose some challenge to Differential Protection. They are as follows:
The primary of transformer will carry no load current even when the secondary is open circuited. This will lead to differential current on which the protection scheme should not operate. It is not possible to exactly match the CT ratio as per equation. This would also lead to differential currents under healthy conditions. If the transformer is used with an off nominal tap, then differential currents will arise as the CT ratio calculated for a particular Tap (Nominal Tap) will be different for different Tap, even under healthy conditions.
Thus we see that because of the above reasons a differential current will flow through the Differential Protection Relay. So Differential Protection will operate which is not expected to operate for the above said reasons. So
what to do to prevent tripping because of the differential current caused by the above mentioned reasons?
To prevent the Differential Protection scheme from picking up under such conditions, aPercentage Differential Protectionscheme is used. It improves security at the cost of sensitivity.
In Percentage Differential Protection, we provide a slope feature to the Differential Protection Relay. In modern Numerical Differential protection Relay two slopes are provided. A typical Slope characteristic is shown in figure below.
Notice an offset of to account for the no load current of Transformer. If we don’t provide this offset then the Differential protection will operate during no load of Transformer and will trip the Transformer Primary side Breaker which is not desired.
The current on the X-axis is the average current of primary and secondary winding referred to primary. It indicates the restraining current called the Biasing Current, I bwhile the corresponding difference on Y-axis represents the differential current. The Differential Protection Relay will pick up if magnitude of differential current is more than a fixed percentage of the restraining current.
Let for differential Protection to operate, Id should be greater than the x% of Ib. Therefore,
Id/Ib> 0.0x
But Id/Ib = Slope of the curve
Thus Differential Protection will operate if the Slope is greater than some fixed value which is set in the Differential Protection Relay. Carefully observe the operating zone in the Slope characteristic of Differential Protection Relay. Consider the figure below.
Suppose, the current in the secondary of CT is 1A at normal operating condition. Therefore the Biasing Current Ib = (1+1)/2 = 1A
While the Differential Current Id = (1-1) = 0A
Thus as discussed above, the restraining current is more than the differential current, Differential Protection Relay will not operate.
Now assume a through fault, so the primary side CT current will be 2 A (say) and secondary side CT gets saturated so current in secondary side CT = 0 A.
Thus, Differential current Id = (2-0) = 2A
and Biasing Current Ib = 2/2 = 1A
Thus we see that Differential protection will operate but it is not expected to operate as the fault is through fault. Thus to prevent tripping on through fault we provide a slope so that Differential current increases as the Biasing current increases and Differential protection will operate if the slope exceeds a particular value (which can be set in the Relay).
Calculation of Stabilizing Resistor in High Impedance Differential Protection. Before going into the calculation part of Stabilizing Resistor, I will first explain the purpose of Stabilizing Resistor in High Impedance Differential Protection.
Stabilizing Resistor in High Impedance Differential Protection is used to prevent the operation of Relay in case of through fault. Through fault is a fault outside the zone of protection. Lets us assume that High Impedance Differential protection is used to protect a Bus bar as shown in figure.
It shall be noted here that, in High Impedance Differential Protection, all the CTs are connected in parallel and then the four wires i.e. R, Y, B and N are connected with the Relay as shown in figure above. If there is any fault in the bus, the according to Kirchhoff’s current law, the summation of current will not be zero and a net current will flow through the Relay coil to operate it. In normal condition, the summation of current will be zero and hence no current will flow through the Relay coil and hence the Relay will be stable. Mathematically under normal condition, I1+ I2 + I3 = 0 As Relay sees only summation of current hence we normally employ an overcurrent element in High Impedance Differential Protection. This is the main difference between a high impedance and low impedance differential protection.
Let us consider a through fault i.e. fault outside the zone of protection. To be more specific, let a fault take place after the CT of any feeder. If all the CT’s maintain the same nominal ratio for all external faults the assumed scheme is perfectly valid since no current can flow in the relay coil. However, when the instantaneous overcurrent relay is set low enough to give useful sensitivity to internal faults the Relay may in practice operate falsely on external faults due to a reduction of the nominal ratio of the fault CT resulting from fault CT core saturation. This reduction of the fault CT nominal ratio results in a “false” differential relay current that may operate the instantaneous overcurrent relay. The wort condition will be when a CT gets completely saturated. Thus we need to make Relay insensitive for through fault. To do this we use Stabilizing Resistor. How Stabilizing Resistor makes Relay Insensitive to through Fault? Well, the main cause for the flow of current through the Relay coil in High Impedance Differential Protection is the Voltage across the terminals of CT. We consider the worst case here when a CT gets completely saturated for through fault. When a CT gets completely saturated, it will no longer will be a source of current rather it will behave purely as a Resistor having a value equal to the CT secondary winding. Thus the fault current will not go toward the relay rather it will go circulate through the saturated CT secondary only as current always chooses a path having least resistance. Let the fault current be IF and the resistance of CT secondary be RCT. Therefore the voltage developed across the saturated CT will be, Vs = IFRCTwhen looping of CT secondaries are done at CT Junction Box only. Or, Vs = IF(RCT+ 2RL) when looping of CT is done at Panel or near Relay end. Here looping of CT secondaries means parallel connection of CT secondaries. It may happen so that we are doing the paralleling at the CT Junction Box (JB) or at Panel (Relay end). If paralleling is done at Relay end then lead resistance of saturated CT up to panel shall be considered for the calculation of driving voltage across the common point of CTs but if paralleling is done at CT JB only then lead resistance of saturated CT from CT core to CT JB shall only be considered which is very less and can be ignored. Now, let us assume that the setting of High Impedance Differential Relay for internal fault be Is. So to make Relay insensitive for through fault, the voltage developed shall not drive a current Is through the Relay, hence we put a Stabilizing Resistor R stbin series with the Relay Coil and the value of Stabilizing Resistor Rstbis given as Rstb= Vs / Is
So, Rstb = IFRCT / Is when paralleling is done at CT JB
Or Rstb= Vs = IF(RCT + 2RL) / Iswhen looping of CT is done at Panel or near Relay end. Thus during through fault, for the worst condition of CT saturation, the current through the Relay coil will not be enough to cross setting value of Is and thus will not operate. Rated Normal, Short Circuit Making & Breaking, Short Time Current Rating & Operating Duty Cycle of Circuit Breaker. Rated Normal Current:
The rated normal current of a circuit-breaker is the r.m.s. value of the current which the circuit breaker can carry continuously and with temperature rise of the various parts within specified limits.
The design of contacts and other current carrying parts in the interrupter of the circuit breaker are generally based on the limits of temperature rise. For a given cross section of the conductor and a certain value of current, the temperature rise depends upon the conductivity of the material. Hence, high conductivity material is preferred for current carrying parts. The crosssection of the conductors should be increased for materials with lower conductivity.
The use of magnetic materials in close circuits should be avoided to prevent heating due to hysteresis loss and eddy currents. The rated current of a circuit-breaker is verified by conducting temperature-rise tests.
Rated Short Circuit Breaking Current:
The rated short-circuit-breaking-current of a circuit-breaker is the highest value of short circuit current which a circuit-breaker is capable of breaking under specified conditions of transient recovery voltage and power frequency voltage. It is expressed in kA r.m.s. at contact separation.
Rated Short Circuit Making Current:
It may so happen that circuit-breaker may close on an existing fault. In such cases the current increase to the maximum value at the peak of first current loop. The circuit breaker should be able to close without hesitation as contacts touch. The circuit breaker should be able to withstand the high mechanical forces during such a closure. These capabilities are proved by carrying out making current test. The rated short-circuit making current of a circuit-breaker is the peak value of first current loop of short-circuit current (I pk) which the circuit-breaker is capable of making at its rated voltage.
Rated making current = 1.8 x √2 x Rated short circuit breaking
= 2.5 x Rated short circuit breaking current.
In the above equation, the factor √2 converts the r.m.s. value to peak value. Factor 1.8 takes into account the doubling effect of short-circuit-current with considerations to slight drop in current during the first quarter cycle.
You may like to read, Why Making Current of Circuit Breaker is more than Breaking Current?
Rated Short Time Current:
The short time current of a circuit-breaker is the r.m.s. value of current that the circuit breaker can carry in a fully closed position during a specified time under prescribed conditions of use and behavior. It is normally expressed in terms of kA for a period of one second. Adjacent poles experience mechanical force during the test.
The rated duration of short circuit is generally 1 second and the circuit breaker should be able to carry short-circuit current equal to its rated breaking-current for one second. During the shorttime current test, the contacts should not get damaged or welded.
The current carrying parts and insulation should not get deteriorated. Generally, the crosssection of conductors based on normal current rating requirements is quite adequate for carrying the rated short-circuit current for the duration of 1 second.
Operating Duty Cycle:
The operating sequence denotes the sequence of opening and closing operations which the circuit-breaker can perform under specified conditions. The operating mechanism experiences severe mechanical stresses during the auto reclosure duty. As per IEC, the circuit-breaker should be able to perform the operating sequence as per one of the following two alternatives:
(i) O-t-CO-T-CO
where, O=opening operation
C=closing opeartion
CO=closing followed by opening
t=3 mintes for circuit-breaker not to be used for rapid auto reclosure
t=0.3 second for circuti-breaker to be used for rapid auto reclosure.
T=3 minutes.
(ii) CO-t’-CO
where t’-15 second for circuit-breaker not to be used for rapid auto reclosure.
You may like to read, AutoreclosingPhilosophy in Distance Protection.
Why Transformer Open Circuit Test Conducted on LV side and Short Circuit on HV Side?. As we know that Transformer Open Circuit Test is conducted on LV side and that Short Circuit Test on HV side. To understand the reason behind this, we will consider a Single Phase Transformer of rating 3300 / 220 Volt, 33 kVA. Thus the voltage at the LV side of this Transformer is 220 Volt. Therefore, for Open Circuit Test on LV side the range of Voltmeter will be 220 V.
Now, Full Load Current = (33×1000) / 220 = 150 A
So, excitation current = 4% of Full Load Current
= 0.06×150 = 6 A
Here note that Excitation Current is taken as 4% of the Full Load Current as the range of excitation current is 2-6%.
Also, the range of Wattmeter will be 220 V and 6A.
We see that the rating of instruments required for the testing are standard and easily available. Furthermore, using standard instruments, more accurate results can be obtained. If the Open Circuit Test is conducted on HV side then a source of 3300 Volt may not be readily available. At the same time the instrument ranges required will be 3300 V, 0.4 A [4% of(33×1000) / 3300 = 4% of 10 = 0.4 A] and 3300 V & 0.4A which are not within the range of ordinary instruments and hence result obtained may not be accurate. Also, it is not safe to work on HV side from safety point of view.
Now, coming to Short CircuitTest. For a Short Circuit Test conducted on HV side the range of voltmeter required will be,
Range of Voltmeter = 5% of Rated Voltage
= 0.05×3300
= 165 V
Note that voltage required to circulate rated current is around 2-12% of the rated voltage, that is why we have considered 5% for selection of range of voltmeter.
Range of Ammeter = Rated Current
= 33×1000 / 3300
= 10 A
Range of Wattmeter = 165 V , 10A
Thus we observe that the range of instruments required to perform the test fall within the range of standard instruments which are easily available and accurate.
At the same time, if we conduct the Short Circuit Test on LV side the
Range of Voltmeter = 5% of Rated Voltage
= 0.05×220
= 11 V
Range of Ammeter = Rated Current
= 33×1000 / 220
= 150 A
Range of Wattmeter = 11 V , 150A
Instruments of such range and Auto Transformer capable of handling 150 A may not be readily available and at the same time result may not be much accurate.
It is for these reasons, Open Circuit Test is conducted on LV side and Short Circuit Test on HV side.
Short Circuit Test of Transformer. Open Circuit Test and Short Circuit Test are two important tests which are carried out on a Transformer to determine its equivalent circuit parameters, Voltage Regulation and Efficiency.
Short Circuit Test Short Circuit Test of Transformer is performed on HV side and the supply voltage is so adjusted that rated current flows through the shorted secondary. As rated current is flowing through the shorted secondary that means that rated current will also flow in the primary because of Transformer action.
The supply voltage required for the flow of rated current in the shorted secondary is around 212% of rated voltage. Thus the supply voltage i.e. primary voltage is very less which in turn means that core losses during short circuit test will be negligible (as core loss is directly proportional to the square of primary voltage.). Thus short circuit test gives us information of Ohmic loss of Transformer and the power measured by the Wattmeter is Ohmic loss. An equivalent circuit of the Short Circuit Test referred to secondary side is shown in figure below. Mind that shunt branch in the equivalent circuit is not shown as the core loss taking place during Short Circuit Test is negligible.
Let us assume that, Vsc, Iscand Psc be the Voltage, Current and Power measured by the Voltmeter, Ammeter and Wattmeter respectively. Therefore from the equivalent circuit diagram, Equivalent leakage impedance referred to HV side, ZeH = (R1+R2) + j(X1+X2) = Vsc / Isc
But Equivalent Resistance referred to HV side, ReH = R1+R2
= Psc / Isc2
Hence Equivalent Leakage Reactance referred to HV side, XeH = (X1+X2)
But in case leakage impedance parameters for both primary and secondary side is required then usually, R1 = R2 = ReH / 2 and X1 = X2 = XeH / 2 referred to same side is taken.
Related Posts: Why Transformer Open Circuit Test Conducted on LV side and Short Circuit on HV Side? 2. Open Circuit Test of Transformer 3. Transformer Load Test or Back to Back Test or Sumpner’s Test 4. Rated Normal, Short Circuit Making & Breaking, Short Time Current Rating & Operating Duty Cycle of Circuit Breaker.
1.
Open Circuit Test of Transformer. Open Circuit Test and Short Circuit Tests are two important tests which are carried out on a Transformer to determine its equivalent circuit parameters, Voltage Regulation and Efficiency. Though theses parameters can also be found using the physical dimension of Core and Winding detail but using Open Circuit and Short Circuit Tests are quite easy and simple.
The circuit diagram for Open Circuit Test is shown below. As clear from the figure below, Voltmeter, Ammeter and Wattmeter are connected in LV side of the Transformer and HV side is left open. Rated LV voltage is applied to the LV side of the Transformer and the reading of Voltmeter, Ammeter & Wattmeter is noted for further analysis.
As the Transformer Secondary i.e. HV side is kept open therefore Transformer will only take excitation current to set up magnetic flux in the core. Therefore Ammeter A will
read Excitation Current Ie which is around 2-6% of the full load current. As Ie is very less therefore, Primary leakage impedance drop is negligible and we can say that applied voltage V1 to LV side is equal to the voltage induced in the Primary winding i.e. E1. Therefore, the equivalent circuit when referred to Primary side reduces to as shown in figure below.
In input power as metered by Wattmeter consists of two components, one is Core Loss and another Ohmic Loss. The exciting current being 2 to 6% of the full load current, the ohmic loss in the primary (Ie2r1) varies from 0.04 to 0.36% of full load Primary ohmic loss. In view of this, the ohmic loss in Primary is negligible when compared to the normal core loss which is being directly proportional to square of applied Voltage. Therefore wattmeter reading can directly be taken as the core loss in the Transformer. Thus we see that Open Circuit Test gives us Core Loss of Transformer. The phasor diagram for Open Circuit Test of Transformer is shown below.
Let, V1 = Applied Rated Voltage on Primary Ie = Excitation Current Pc = Core Loss The Core Loss, Pc= V1IeCosƟ So, no load power factor CosƟ = Pc / (V1Ie) Also from phasor diagram, Ic = IeCosƟ Im =IeSinƟ But Pc = V1Ic
So, Ic = Pc/ V1
Therefore, Core Loss Resistance Rc= V1 / Ic
= V1/ IeCosƟ = V12/ V1IeCosƟ
= V12/ Pc Now, Magnetizing Reactance Xm = V1 / Im
= V1 / IeSinƟ Here Rc and Xmare values referred to LV side. Sometimes a voltmeter is placed in the open HV side of the Transformer to measure the Secondary voltage to get turn ratio. Thus, Open Circuit Test gives us the following information: Transformer Core Loss at rated voltage and frequency. Transformer Shunt Branch Parameters like Rc and Xm. Turn Ratio of the Transformer. Transformer Load Test or Back to Back Test or Sumpner’s Test. Transformer Load Test is basically carried out for determining the maximum temperature rise of the Transformer. However is not viable to conduct this test by connecting the secondary of the Transformer to rated load. In case of small Transformer, rated load can be connected to the secondary of Transformer but for large Transformer, rated load capable of consuming rated power is not easily available and also this will lead to wastage of energy. Therefore the best and smart way to load a Transformer is to conduct Back to Back Test or Sumpner’s Test. It shall be noted here that, Back to Back Test or Sumpner’s test can also be conducted for calculating the efficiency of a Transformer but it is better to calculate the efficiency of Transformer using Open Circuit Test and Short Circuit Test of Transformer as this will give more accurate result. In this post we will focus on Back to Back Test or Sumpner’s Test of single phase Transformer. The Back to back Test on single phase transformer requires two identical Transformers. In this Test, the primaries of two Transformers are connected in parallel and are energized at rated voltage and frequency as shown in figure below.
With Secondary open, the wattmeter W1 records the total core loss occurring in both the Transformers. Now, secondary of both the Transformers are connected in series with their polarity in phase opposition so that the voltage across the terminals ab is zero as indicated by the voltmeter V2. Here it shall be noted that, the basis for selection of range of voltmeter V2 shall be the double of rated secondary voltage of the Transformers. In case, the reading of voltmeter V2 is not zero, this means that the series connection of secondary of Transformer are not in phase opposition and therefore we need to change the connection by just connecting the terminals ad together and check that voltage across bc is zero as indicated by voltmeter V2. Now, if we short the secondary of Transformers, no current will flow as Vab = 0 and therefore the reading of wattmeter W1 will not change. Now we will inject voltage in the secondary circuit by means of voltage regulator, fed from the source connected to the primary or from a separate source. The injected voltage is regulated till rated current flow in the secondary circuit. By Transformer Action, rated current now will start flowing in primary circuit too. Note here that, the full load or rated current in primary completes its path as shown by the dotted line in figure above. As this full load current in the primary is not flowing through the current coil of the wattmeter, therefore the reading of wattmeter W1will remain unaltered. The reading of voltmeter V2will now show the voltage drop due to leakage impedance of both the Transformers. As full load current is flowing in the secondary circuit, the wattmeter W2 will read the total full load ohmic loss of Transformers. Let us suppose that, Pc = Core Loss in individual Transformer
Poh = Ohmic Loss in individual Transformer then, Reading of wattmeter W1=2Pc Reading of wattmeter W2= 2Poh
Therefore, from the above two reading efficiency of each Transformer can be determined. From the above discussion, it is quite clear that, even though Transformer is not connected to any load in Sumpner’s Test, rated current is flowing in the primary as well as in the secondary of Transformers and hence full load ohmic and core loss is taking place. If temperature rise of the two Transformers is to be measured, then two Transformers are kept under rated loss condition for several hours till maximum stable temperature is reached.
Types of Relay Output Contact. We know that Relay is a switch which senses fault in a system and once fault is sensed by the Relay, it issues trip command to the Circuit Breaker to isolate the faulty section of the network from the healthy section. The Relay detects the abnormal condition by continuously monitoring electrical quantities which are different for healthy and faulty condition. The electrical quantities which may change during fault condition are voltage, current, frequency and phase angle. If one or more of the above electrical quantities change, that signals the presence, type and location of the fault to the Relay. After detecting the fault condition, Relay pick-up, its contact will change from NO to NC or vice versa. So we can wire up a particular kind of Relay contact to Breaker tripping circuit. So whenever, the Relay picks up, the tripping of Breaker will take place. Relay coil when picked up will change its contact status. Let us say Relay Normally Open (NO) contact is wired to the Breaker Trip Coil Circuit. Therefore when the Relay coil is in de-energized state, the circuit of Trip Coil of CB is not complete and hence no tripping of the CB. During fault condition as the current through the Relay coil exceeds the pick-up value, the Relay coil will get actuated which in turn will force its contact to change over i.e. NO contact will change to Normally Close (NC) thereby closing the Trip Coil circuit of the Breaker. Relays may be fitted with a variety of contact systems for providing electrical outputs for tripping and remote indication purposes. The most common types of relay Output contacts are as follows: a) Self-reset Type. The contacts remain in the operated condition only while the controlling quantity is applied, returning to their original condition when it is removed
b) Hand or Electrical Reset. These contacts remain in the operated condition after the controlling quantity is removed. They can be reset either by hand or by an auxiliary electromagnetic element. The majority of protection relay elements have self-reset contact systems, which, if so desired, can be modified to provide hand reset output contacts by the use of auxiliary elements. Hand or electrically reset relays are used when it is necessary to maintain a signal or lockout condition. Contacts are generally shown on engineering diagrams in the position corresponding to the un-operated or de-energized condition, regardless of the continuous service condition of the equipment. For example, an under-voltage relay, which is continually energized in normal circumstances, would still be shown in the deenergized condition. A ‘make’ contact is one that closes when the relay picks up, whereas a ‘break’ contact is one that is closed when the relay is de-energized and opens when the relay picks up. Examples of these conventions and variations are shown in figure below.
.
How does Trip Circuit Supervision Work?. Trip Circuit Supervision Circuit senses any fault either in the trip coil of breaker or trip circuit. On sensing a fault, Trip Circuit Supervision Relay changes its contact status to window annunciation on the panel. Here fault in the circuit means any break or open circuit. In this post we will be discussing the working principle of trip circuit supervision.
Basically, a breaker is equipped with two trip coils. Both the trip coil energizes if relay issues a tripping command. On energization of trip coil, breaker mechanism opens the circuit breaker. Therefore, it is very important to monitor the trip coil healthiness otherwise during the requirement the breaker may not open to clear the fault. Figure below, shows the simplified diagram of trip circuit supervision.
Trip Circuit Supervision Relay is provided to monitor the healthiness of trip coil. As there are two trip coils therefore, there will be two Trip Circuit Supervision Relays and hence two different circuits. This Relay has three coils namely RLA, RLB and main coil (TCS) of the Relay as shown in the figure above. Now, breaker may either be open or close, therefore we need to monitor the healthiness in both the state of the breaker. Thus, trip circuit supervision is divided as pre-close and post close supervision. As shown in the figure, contacts A and B are breaker auxiliary contacts. Mind that theses contacts are connected to the breaker mechanism and therefore their status depends on the breaker position. If the breaker is open, contact A will be OPEN and contact B will be CLOSE. Similarly if breaker is close, contact B will be OPEN and contact A will be CLOSE. Now we will consider pre-close and post close supervision separately. Pre-close Trip Circuit Supervision:
Pre-close means that breaker is open. Thus the status of contact A will be OPEN and that of B will be CLOSE. If you observe the circuit carefully, you will notice that, in this case current will be flowing through both the coils RLA and RLB. As coils RLA and RLB are energized, their contact will be close.
Now carefully observe the circuit of TCS Relay. Since RLA and RLB are close, DC supply will be extended and hence TCS relay will be energized. Therefore, its output contact 1-2 will be open and hence no window will appear. This means that trip circuit is healthy. Assume there is any open circuit, in such case as no current could flow through the coils RLA and RLB, hence relay TCS will not be energized. Because of this its output contact 1-2 will be close for window annunciation . This means to the operator that either DC supply has failed or there is some problem in the trip circuit. Post Close Trip Circuit Supervision:
Post close means that breaker is close. Thus the status of contact A will be CLOSE and that of B will be OPEN. If you observe the circuit carefully, you will notice that, in this case current will be flowing through both the coils RLA.
Now carefully observe the circuit of TCS Relay. Since output contact of RLA is close, DC supply will be extended to TCS relay and hence TCS relay will be energized. Therefore, itsoutput contact 1-2 will be open and hence no window will appear. This
means that trip circuit is healthy. Assume there is any open circuit, in such case as no current could flow through the coils RLA, hence relay TCS will not be energized. Because of this its output contact 1-2 will be close to annunciate window. This means to the operator that either DC supply has failed or there is some problem in the trip circuit. It shall be noticed that, as the resistance of relay coil is very less therefore a high resistanceshall be connected in series so that less current flows through the circuit to operate the trip coil of the breaker. It shall also be noticed that, in case of protection trip, separately positive DC voltage is extended to the trip coil of circuit breaker so that full current flows through the trip coil to operate the breaker as clear from the figure. Basic Criteria for Selection of 33 kV/ 11 kV Or 66kV/11kV Substation. Basically 33 kV / 11 kV is used for distribution of power at the substation and 66 kV / 11 kV is also used for the distribution. But whether to use 66/11 kV or 33/11 kV depends on many factors. Following are some of the factors which are taken care for the selection of voltage of substation.
a) The substation rating is defined as per the power handling capacity, location and purpose of substation. b) Thumb rule for the economical voltage rating has been categorized for different power range to be received as below:
Load up to 150 MVA – voltage rating of 132 kV. Load up to 80 MVA – voltage rating of 66 kV. Load up to 5 MVA – Voltage rating of 33 kV.
Thus, when a substation is of rating 66/11 kV, means substation has been designed to receive 80 MVA on 66 kV and it will distribute the received power on 11 kV. Again, substation of rating 33/11kV means, the substation has been designed to receive 5 MVA of power at 33 kV and it will distribute the same on 11 kV.
Advantages & disadvantages of Harmonics in Power System. Harmonic voltages and currents in a Power System are a result of non-linear electric loads. In a normal AC Power System, the current varies sinusoidally at a specific frequency, usually 50 or 60 Hertz. When a linear electrical load is connected to the system, it draws a sinusoidal current at the same frequency as the voltage, though usually not in phase with the voltage.
Current harmonics are caused by non-linear loads. When a non-linear load, such as a rectifier / inverter, is connected to the system, it draws a current that is not necessarily sinusoidal. The current waveform can become quite complex, depending on the type of load and its interaction with other components of the system. Regardless of how complex the current waveform becomes, as described through Fourier Series analysis, it is possible to decompose it into a series of simple sinusoids, which start at the power system fundamental frequency and occur at integer multiples of the fundamental frequency. Further examples of non-linear loads include common office equipment such as computers and printers, Fluorescent lighting, battery chargers, electronic ballasts, variable frequency drives, and switching mode power supplies. Total Harmonic Distortion or THD is a common measurement of the level of harmonic distortion present in power systems. THD is defined as the ratio of total harmonics to the value at fundamental frequency.
where Vn is the RMS voltage of nth harmonic and n = 1 is the fundamental frequency.
Effects of Harmonics: One of the major effects of power system harmonics is to increase the current in the system. This is particularly the case for the third harmonic, which causes a sharp increase in the zero sequence current, and therefore increases the current in the neutral conductor. Electric motors experience losses due to hysteresis and losses due to eddy currents set up in the iron core of the motor. These are proportional to the frequency of the current. Since the harmonics are at higher frequencies, they produce higher core losses in a motor than the power frequency would. This results in increased heating of the motor core, which (if excessive) can shorten the life of the motor. The 5th harmonic causes a Counter Electromotive Force in large motors which acts in the opposite direction of rotation. The Counter Electromotive Force is not large enough to counteract the rotation; however it does play a small role in the resulting rotating speed of the motor. How Harmonics Eliminated from Alternator Generated Voltage?. In this post we will discuss about how to eliminate or suppress the harmonics from the emf waveform of an Alternator. In an Alternator, the primary source of harmonics in the emf waveform is due to non sinusoidal field waveform. If the field waveform would have been sinusoidal then there have been no harmonics in generated emf of an Alternator. Therefore, first attempt is made to make the field waveform sinusoidal as far as possible and them means of reducing or suppressing the harmonics is adopted. Field waveform of an Alternator can be made sinusoidal by the following methods:
Small air gap at the pole centre and large air gap at the pole end in an Salient Pole Synchronous machine tens to make the field flux sinusoidal. Skew pole faces if possible.
In Turbo-Alternator or cylindrical pole synchronous machine, the air gap is uniform and hence field winding is distributed in slots in such a manner to make the field waveform sinusoidal. Figure below shows the Rotor of an Alternator.
Having adopted all the above mentioned methods, the filed waveform along the air gap periphery is still not purely sinusoidal but it is flat topped. As a result, harmonic emf is always generated in the Alternator. These harmonics can however be eliminated / suppressed by the following methods:
The distribution of armature winding along the air gap periphery tends to make the generated emf waveform sinusoidal. With chorded coil, harmonics can be eliminated. If the ξ be the chording angle for fundamental flux wave, then for nth space harmonics the chording angle becomes nξ electrical. Therefore, pitch factor for nth harmonics,
Kp= Cos(nξ/2) If we want to eliminate 5th harmonics then, ξ = 36° as Kp = Cos(180/2) = 0 Therefore, generated emf E = Where Kp = Pitch factor Kd = Winding Distribution Factor f =Frequency Nph = Number of turns per phase Ø =Flux
KpKdπfNphØ
A chording angle of 30° is most useful in an Alternator as it gives the following pitch factors, 0.966 for fundamental, 0.707 for 3rd harmonic, 0.259 for 5th and 7th harmonics and 0.7.7 for 9th harmonics. Mind that triplen harmonics is eliminated from the generated line emf by Star connection, though 5thand 7th harmonics of reduced magnitude are present in the line emf. By Skewing the Armature Slot tooth / slot harmonics are eliminated. Thus, using the above mentioned ways, harmonics in the generated emf of a synchronous generator is reduced.
Why Output Voltage of Alternator can’t have Even Harmonics?. Output voltage of a Synchronous Generator cannot contain even harmonics. For understanding the reason, carefully observe the figure below.
As shown in the figure, one full pitched coil a and a’ are placed. The fundamental component of field flux wave induce maximum emf in coil sides a and a’ as these are cutting the maximum filed flux Øm1. If the rms value of emf in each coil be E1 then the resultant emf of fundamental frequency across the coil ends A and B = 2E1. Now, the second harmonics component of field flux wave also induces emf in coil sides a and a’. The induced emf because of second harmonics in coil sides a and a’ are maximum as because these are cutting the maximum field flux Øm2. But here it shall be noted that in both the coil sides the emf induced will oppose each other as both coil sides are cutting maximum positive filed flux wave of second harmonic component. Therefore the net / resultant emf because of second harmonic component across the coil sides A and B = 0 This shows that second / even though even / second harmonic is present in the field flux wave, second harmonic cannot be there in the output voltage of an Alternator. It shall however be noted that field flux waveform of an Alternator is symmetrical and hence it do not have any even harmonics and hence there will not be any even harmonics in the generated voltage of an Alternator. Suppression of Harmonics in Transformer. In this post we will focus on various methods adopted to suppress harmonics in a Transformer. Harmonics in a Transformer can be suppressed by flowing methods: Lower Flux Density: Harmonics in Transformer can be suppressed by having lower flux density in core but it has one disadvantage that, this method of reducing flux density in core will lead to higher cross sectional area of the core and more number of turn for maintain voltage ratio. Because V = 1.414πfNØ So if B is reduced then for having a constant value of V, we need to increase cross sectional area A to maintain flux Ø (as Ø = BA) and number of turns N. Thus this method of reducing harmonics is uneconomical and is not used.
Type of Connection: Triplen harmonics are those which are having frequency multiple of 3. Triplen harmonic current and voltage in the line are suppressed by using Star or Delta connection of the winding in the Transformer. Primary or Secondary Winding in Delta: As we know that harmonic voltages have more nuisance effect than harmonic current. Thus every effort is made to suppress the third harmonic voltages in the Transformer. When any of the Primary or Secondary Winding is connected in Delta, triplen harmonic voltages are suppressed considerably. In view of this, one of the Transformers winding either Primary or Secondary must be connected in Delta to suppress triplen harmonics. Use of Tertiary Winding: As seen in previous paragraph, we need to connect either Primary or Secondary must be connected in Delta to suppress triplen harmonics but in case it is not possible to connect either of the Primary or Secondary in Delta then three phase Transformer must be designed to have an extra i.e. tertiary winding connected in Delta. This closed Delta formed by the tertiary winding provides a path for the third harmonic currents and therefore the EMF and flux becomes approximately sinusoidal in nature. It shall be noted that tertiary winding has no any effect on the fundamental frequency component voltages because the phasor sum of three EMFs mutually displaced by an angle of 120° is zero in the closed Delta connected tertiary winding. Star Delta Earthing Transformer: The third harmonic voltages in the two winding Star-Star Transformer can also be suppressed by using Star-Delta Earthing Transformer as shown in figure below.
The closed Delta provides a path for the third harmonic currents and hence the EMF and flux remains sinusoidal. For Star-Star Transformer, the third harmonic voltages can also be suppressed by using 4 wire supply or earthing Transformer Neutral point if Alternator / Generator neutral is grounded. The neutral wire in this case provide a path for the flow of third harmonic current and hence the flux & EMF remain sinusoidal. Transformer Overflux Protection. Transformer Overflux Protection is provided to protect the Transformer core from overfluxing. A Transformer is designed to operate at a particular flux level. In case the flux in the core of Transformer exceeds a certain level, the core loss increases which may lead to overheating of components which in turn may result into internal fault. Therefore, overflux protection is provided.
A transformer is designed to operate at or below a maximum magnetic flux density in the transformer core. Above this design limit the eddy currents in the core and nearby conductive components cause overheating which within a very short time may cause severe damage. The magnetic flux in the core is proportional to the voltage applied to the winding divided by the impedance of the winding. The flux in the core increases with either increasing voltage or decreasing frequency. During start-up or shutdown of generator-connected transformers, or following a load rejection, the transformer may experience an excessive ratio of Volts to Hertz (V/f), that is, become overexcited. When a transformer core is overexcited, the core is operating in a non-linear magnetic region, and creates
harmonic components in the exciting current. A significant amount of current at the 5th harmonic is characteristic of overexcitation.
Assuming Number of turns constant, Flux is directly proportional to V/f. Here V is supply voltage and f is frequency of supply. In case of any Transformer, signal for supply voltage V is taken from PT. Let us assume that Transformer Primary is connected with 220 kV. Thus normal voltage of primary of Transformer will be 220 kV at a frequency of 50 Hz. Also assume that the PT ratio is 220 kV/110 V. Therefore, V/f ratio = 110/50 = 2.2 Thus at a V/f ratio of 2.2 the Transformer will operate satisfactorily. So the question arises which V/f ratio may cause the overfluxing. For answering this we need to have a look at the Hysteresis curve of the core material and from the curve we can judge at which flux level Transformer can be subject for a particular time safely. Normally the setting of overfluxing is kept 110% of nominal value or 1.1 pu. This means at a flux level of 1.1×2.2 = 2.42 the Transformer will operate safely but above 2.42 the Transformer core will be subjected to overflux. Does this mean that at a V/f ratio of 2.5 Transformer shall be tripped instantaneously? No it doesn’t mean so. Because Transformer core may tolerate such an overflux for some short time duration and hence instantaneous tripping is not required. Therefore, wise decision is to give anINVERSEcharacteristics to the tripping which mean more the ratio of V/f less will be time of tripping. Now we consider two cases: Case1: Transformer Primary voltage rises to 247 kV while frequency is 50.1 Hz As primary of Transformer rises to 247 kV at a frequency f = 50.1 Hz The PT secondary Voltage = 247×110/220 = 123.5 V Hence, V / f = 123.5/50.1 = 2.465 Thus the Relay will pick-up and as the characteristics is inverse, the relay will trip after some time because we have kept the setting 2.42. If the Primary Voltage is maintaining at 247 kV , then we can do nothing and the Relay will definitely trip. Case2: Transformer is provide with Tap Changer Suppose the Transformer is provided with Tap Changer. As the Transformer is provided with Tap Changer in the primary side, we can increase the Tap position from the nominal value which will result in increase in the value of N1 (Primary number of turns) and hence,
But this is not going to help us as we have taken the voltage signal from the PT which is connected to the Primary side i.e. and primary side voltage is maintained at 247 kV, hence V/f will be same. Thus we observe that, even though we have Tap Changer, in the present scenario we can do nothing to prevent tripping of Transformer on overfluxing though the Transformer is not actually in overflux condition (as we have increased the number of turns in the primary side.) Therefore, to take advantage of Tap Changer, we can make a provision of taking voltage signal from the secondary side PT of Transformer Relay. In such case, if the primary turn of Transformer is
increased then its reflection on secondary side PT will be observed proportionally and tripping on Overflux protection can be prevented. In case of no load operation of Transformer, we can give voltage signal to the Relay from the Primary side PT. In this way the purpose will be served without compromising the overflux protection. Thus we see, how important is tap changer in preventing tripping of Transformer from Overfluxing.
Concept of Tap-Changers in Transformer. The modern loads are designed to operate at satisfactorily at one voltage level. It therefore of great importance to keep the consumers terminal voltage within prescribed limits. The Transformers output voltage and hence the terminal voltage the consumer can be controlled by using tap either on primary or secondary side Transformer.
is a of of
As we know that V1 / V2= N1 / N2 Where V1 = Primary Voltage V2 = Secondary Voltage N1 = Primary Number of Turns N2 = Secondary Number of Turns Thus V2 = V1(N2 / N1) Thus,
If we increase the Primary Number of Turns N , output voltage of Transformer V decreases. If we decrease the Primary Number of Turns N , output voltage of Transformer V increases. If we increase the Secondary Number of Turns N , output voltage of Transformer V increases. If we decrease the Secondary Number of Turns N , output voltage of Transformer V decreases. 1
2
1
2
2
2
2
2
Now, whether will we change Primary Number of Turns N1 or Secondary Number of Turns N2depends whether we have provided Tap on Primary side or Secondary side. The choice of providing tap on Primary side or Secondary side is based on maintaining voltage per turn constant as far as possible. The flux in the core of Transformer depends on voltage applied in the primary, V1 = 1.414πfN1Ø So, Ø = V1 / 1.414 πfN1 Thus the flux in the core shall be maintained constant. If the Primary voltage per turn i.e. Flux decrease, which means poor utilization of core while in case the Primary voltage per turn increases that means overflux which may cause heating and saturation of the core. Let us take an example of Generating Transformer. As the primary of the Generating Transformer is connected with the Generator output terminals therefore variation in the Primary voltage will be very less. Therefore, the flux in the core of Transformer will be constant and hence the wise decision will be to put the Taps on the secondary side. Other factors which shall also be taken care while deciding upon the side Taps should be provided are:
Transformer Taps are provided on HV side as in this case Tag changing Gera will handle low current and chance of sparking will be less. If we see the construction of Transformer, we will observe that LV winding are placed just after the core to limit the insulation requirement to be provided and HV winding are placed on the LV winding. Thus it is quite difficult to provide the Taps on the LV winding of the Transformer.
Now after deciding the side where Tap is to be provided in Transformer, next question is that whether Tap shall be provided in the center of the winding or at the end of the winding? A general sense says that Tap shall be provided in the middle of the winding because in this case the forces on the winding will be less. Since the current flowing in the Primary and Secondary coils are in opposite direction, these currents interact with the leakage flux in between the two windings and produce a radial force repelling each other as shown in the figure below.
Now, suppose the winding is tapped at one end. When some of the winding is cut out by tap changer, axial force in addition to radial force is also developed as shown in figure below.
Under short circuit condition, the axial force tending to compress the winding against the core is very large which may damage the winding insulation. In order to eliminate this, physical position of the Tapped winding should be in the middle of the Transformer winding so that no axial force arises after some of the turns are cut out.
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Advantage & Disadvantage of Auto Transformer over Two Winding Transformer. The weight of conductor for any winding depends upon two parameters:
Current carried by the winding Number of turns required in the winding.
Thus we can say that weight of conductor in a winding is directly proportional to its Ampere Turns i.e. MMF. Now, we will focus our attention on the weight of conductor required for Auto Transformer and a Two Winding Transformer.
As discussed in earlier post Basic principle of Auto Transformer, the current carried by winding AC i.e. IAC= I1 and that of winding BC i.e. IBC = (I2-I1). Also the number of turns for winding AC is (N1-N2) and that for winding BC is N2. Therefore, The weight of conductor required for Auto Transformer, WCAT ∝Ampere turns of winding AC + Ampere turns of winding BC ∝ (N1-N2)xI1+ N2x(I2-I1) ∝ N1I1+N2I2-2N2I1 But N2 / N1= I1 / I2 So, N2I2= N1I1 Therefore, The weight of conductor required for Auto Transformer, WCAT
∝ N1I1+N2I2-2N2I1 ∝ 2N1I1 – 2N2I1 ∝ 2(N1 – N2)xI1 …………………….(1) If we want to compare the weight of conductor required for Auto Transformer and Two Winding Transformer, then both the Transformers must have same voltage ratio (V2 / V1), current ratio (I2 / I1), input VA i.e. V1I1and output VA i.e. V2I2. Assuming all the above parameters to be same for Two Winding and Auto Transformer, Weight of conductor in Two Winding Transformer WCTWT, ∝Ampere Turns of Primary + Ampere Turns of Secondary ∝ N1I1+ N2I2 But N1I1= N2I2 So, Weight of conductor in Two Winding Transformer WCTWT, ∝ 2N1I1 ……………………………………….(2) Therefore from equation (1) and (2), WCAT / WCTWT= [2(N2 – N1)xI1] / 2N1I1 = 1 – N2/N1 = (1 – k)
As for step down Auto Transformer k<1, hence the weight of conductor required for Auto Transformer is less than that required for Two Winding Transformer. Saving in Conductor = 1 – WCAT/ WCTWT = 1 – (1 – k) = k Hence there is net saving of conductor in case of Auto Transformer. Let us assume that k =0.1, thus the saving in conductor for Auto Transformer will be only 10% but if k = 0.9 then saving in conductor will be 90% which is quite lucrative. Thus we can conclude that Auto Transformer is more economical when the voltage ratio k is more near to unity. Another important aspect is core. As the conductor required for Auto Transformer is less than that required for Two Winding Transformer that means for Auto Transformer lower window dimension will be required. Thus by using Auto Transformer, there is net saving in Core material as well as conductor material, the saving will be more and more as we tend toward voltage ratio unity. Owing to reduction in conductor and core material, the Ohmic loss in conductor and core loss is reduced considerably. Therefore an Auto Transformer has higher efficiency than Two Winding Transformer of same output. Reduction in conductor material means lower value of winding resistance. Since a part of winding is common to both the Primary and Secondary circuit in Auto Transformer, leakage reactance will be less. Because of lower value of leakage reactance, a superior voltage regulation is achieved with Auto Transformer. Disadvantage of Auto Transformer: As seen earlier in this post that saving of conductor in Auto Transformer is k so saving in conductor decreases as the value of voltage ratio k decreases. Another biggest disadvantage of Auto Transformer is direct electrical connection between the Primary and Secondary circuit. If Primary is supplied with high voltage then any open circuit in common winding will lead to dangerously high voltage on LV side which may damage load as well as this dangerously high voltage will be very
harmful of working personnel. Thus special protection must be provided to prevent such an occurrence. Auto-Transformer Starting of an Induction Motor. It can easily be seen from the slip torque characteristics of an induction motor that, there is some finite torque when the slip s=1 i.e. speed is zero. This simply means that Induction Motor is a self-starting motor and begins to rotate on its own when connected to a 3 phase supply. At the instant of starting, a three phase Induction Motor behaves like a Transformer with its secondary winding shorted. Therefore, Induction Motor during starting takes a high current from the supply mains. To limit this high starting current of Induction Motor, different starting methods are used. In this post we will have a look at the AutoTransformer Starting Method of Induction Motor. The main philosophy of starting any Induction Motor is to start it at a reduced voltage and as soon as the motor reaches its rated speed, full supply voltage is applied to the terminals of Induction Motor. A schematic diagram for AutoTransformer Starting of an Induction Motor is given below.
It shall be observed that, using Auto-Transformer we are only applying a reduced voltage xV1 to the Stator terminal of Induction Motor. Here x is less than 1. As soon as the Induction Motor reaches its rated speed, full supply voltage is applied to the terminals of stator. Therefore, per phase starting current of Motor = xV1/Zsc = xIsc Here Isc is the current through the stator during direct switching of motor.
Thus we observe that starting current of Motor has reduced and is x times that of current during DOL (Direct Online) starting. Again, Input VA of Auto-Transformer = Output VA of the Auto-Transformer Ist.V1= xV1(xIsc) Therefore, Per phase starting current from the Supply Mains Ist = x2Isc Thus per phase starting current from Supply Mains has now became x2 times that of DOL current. Mind that it has reduced as x is less than 1 so x2 will be much less than 1. Thus the main advantage of using Auto-Transformer is that it reduces the starting current from the Supply Mains by x2 times. Note that starting current is the motor winding is x times while the starting current from the supply mains has became x2 time of DOL starting current. Now we will have a look at Torque of Induction Motor. As the torque of an Induction Motor is directly proportional of square of applied voltage at the stator terminals, therefore Starting Torque with Auto-Transformer Test = K(xV1)2 where K is constant of proportionality. Starting Torque with DOL Test = KV12 where K is constant of proportionality. Therefore, Test with Auto-Transformer/ Test with DOL = x2 Thus starting torque with Auto Transformer is less than the starting torque with DOL starting by a factor of x2.
Why Star Delta Starter Preferred in Induction Motor?. The main purpose of any starter is to reduce the requirement of high starting current. Normally the starting current of an induction motor is 6 to 7 times of the full load current. If one has an induction motor with a DOL starter, drawing a high current from the line, which is higher than the current for which this line is designed. This will cause a drop in the line voltage, all along the line, both for the consumers between the substation and this consumer, and those, who are in the line after this consumer. This is the reason for which a starter is to be used. In a squirrel cage induction motor, the starter is used only to decrease the input voltage to the motor so as to decrease the starting current.
It is T.P.D.T switch used to first start the motor with the winding connected in star and then switch for delta connection in running position. TPDT stands for Triple Pole, Double Throw.
Why Star Delta Starter Preferred in Induction Motor? If the winding of Induction Motor is connected in star, the voltage per phase supplied to each winding is reduced by 0.577.In general the voltage per phase in delta connection is Vs, the phase current in each stator winding is (Vs/Zs), where Zs represents the impedance per phase of the motor at standstill or start.
The line current or the input current to the motor is Ist (starting current) = (1.732*Vs)/Zs which is the current when it has to be started by DOL starter. Now, if the stator winding is connected as star, the phase or line current drawn from supply at start (standstill) = (Vs/Zs)/(1.732) which is 0.333= (0.577*0.577) of the starting current, if DOL starter is used. The voltage per phase in each stator winding is now Vs/1.732. So the starting current is reduced by 33.3%. Because of the reduction in starting current, starting torque reduces. Therefore we can conclude that by using Star Delta starter, the starting current is reduced to approximately two-thirds. Since starting current is reduced, the voltage drops during the starting of motor in systems are reduced. Why Induction Motor Star Point not Grounded in Industries?. In any electrical system, we do the neutral grounding at the power source e.g. the starpoints of generators or transformers. By keeping the grounded neutrals at the power source, earth fault current will have a return path from the point of short-circuit at downstream to the source. In this way the direction of earth fault current flow can be easily identified and the earth fault protection relays in the distribution system can easily be coordinated.
As clear from the figure, there will not be any earth fault current in the case where there is no neutral grounding of the source because of absence of return path. But as we can
see, if the source is having neutral grounding then earth fault current will have a return path and earth fault current will flow from the point of fault to the source. After reading the above paragraph, I am sure that a question will strike your smart brain, why do we do neutral grounding as there won’t be any earth fault current in absence of neutral grounding? Yes, it is correct that there will not be any earth fault current in absence of neutral grounding but neutral grounding has many advantages, they are as follows:
Voltage of the phases is limited to phase to Ground Voltage The high voltage due to arching ground or transient line to ground fault are eliminated. Sensitive protective Relays for earth fault protection can be used. Over voltage due to lightening are discharged to ground otherwise there would have been a positive reflection at the isolated neutral of the system.
Hope your doubt is clear now. This is the reason Neutral grounding is done in a system. Now coming to grounding of Star Point of Induction Motor, Grounding a motor star point will create an earth path for earth fault current to flow through that motor’s star point. If there are 10 motors in a process plant and their star points are all grounded then obviously there are 10 additional paths for earth fault currents to flow through. If all the motors’ star points are grounded in this way the earth fault current detections by the protection relays will be complicated and it is most likely that Relay will trip at the incorrect locations because earth fault currents are flowing in many directions toward multiple grounded neutral points. Therefore the electrical consumers i.e. the load, including the capacitor banks, even if they are star connected are not to be grounded. Motor is a balanced 3-phase load. However when the system supply voltage is unbalanced caused by unbalanced loads somewhere else or due to network conductors problem, the motor operating under unbalance voltage will result in unbalance current in the 3 windings. The same is true for the generator windings under that condition. The
design engineer may then decide that individual machines should be fixed with negative phase sequence current protection. Even if there is a neutral voltage shift in the induction motor, we should not ground the motor’s neutral point. If we ground the induction Motor, it may create nuisance trip on earth fault protection relays. Why LT Motors are Delta connected but HT Motors are Star?. In industries Low Tension i.e. low voltage Motors like Motors supplied by three phase 415 V are normally have Stator connected in Delta while High Tension i.e. Motors supplied by high voltage like 6.6 kV have Stator connected in Star configuration. The reason behind this is technical while making the Motor economical. Following are the main reasons due to which high voltage Motors stator are connected in Star:
As the Stator winding of Motor is to be connected with high voltage, it is better to configure the Stator in STAR as in this configuration, the phase current remains the same as the line current but the phase voltage reduces by to Vph = Vline/1.732 which means that insulation requirement from phase winding will be less. The second most important reason is that, starting current for Motors is 6 to 7 times of full load current. So start-up power will be large if HT motors are delta connected. It may cause instability i.e. voltage dip in case of small Power System. In STAR connected HT Motors starting current will be less compared to delta connected motor as voltage is V and current is line current. So starting power and starting torque are reduced. As current is less in STAR configuration, copper (Cu) required for winding will be less. ph
Following are the main reasons due to which low voltage Motors stator are connected in Delta:
In Delta connection, the insulation requirement will not be problem as voltage level is less in LT Motors. Starting current will not be problem as starting power in all will be less. So no problem of voltage dips. Starting torque should be large, as motors are of small capacity and hence Stator should be connected in Delta to have more current and hence more starting torque.
Difference between an Induction Motor and a Synchronous Motor. The basic difference is that an induction motor is an asynchronous machine whereas the other one, as the name suggests is a synchronous machine. Following are some important differences between a Induction Motor and a Synchronous Motor:
Synchronous motors operate at synchronous speed (RPM=120f/p) while induction motors operate at less than synchronous speed (RPM=120f/p – slip). Slip is nearly zero at zero load torque and increases as load torque increases. Synchronous motors require a DC power source for the rotor excitation. Synchronous motors require slip rings and brushes to supply rotor excitation. Induction motors don’t require slip rings, but some induction motors have them for soft starting or speed control. Synchronous motors require rotor windings while induction motors are most often constructed with conduction bars in the rotor that are shorted together at the ends to form a “squirrel cage.”
Synchronous motors require DC excitation to be supplied to the rotor windings; induction motors don’t.
Synchronous motors require a starting mechanism in addition to the mode of operation that is in effect once they reach synchronous speed. Three phase induction motors can start by simply applying power, but single phase motors require an additional starting circuit. The power factor of a synchronous motor can be adjusted to be lagging, unity or leading while induction motors must always operate with a lagging power factor. Synchronous motors are generally more efficient that induction motors. Synchronous motors can be constructed with permanent magnets in the rotor eliminating the slip rings, rotor windings, DC excitation system and power factor adjustability. Synchronous motors are usually built only is sizes larger than about 1000 Hp (750 kW) because of their cost and complexity. However, permanent magnet synchronous motors and electronically controlled permanent synchronous motors called brushless DC motors are available in smaller sizes.
Difference between Squirrel Cage and Slip Ring Induction Motor. Please have a look at the below figures of Squirrel Cage Induction and Wound Rotor / Slip Ring Induction Motor. Squirrel Cage Induction Motor:
Slip Ring / Wound Rotor Induction Motor:
Sr. No Squirrel Cage Induction Motor . 1) 1. In Squirrel cage induction motors the rotor is simplest and most rugged in construction.
Slip Ring / Wound Rotor Induction Motor In slip ring induction motors the rotor is wound type. In the motor the slip rings, brushes are provided. Compared to squirrel cage rotor the rotor construction is not simple.
2)
Cylindrical laminated core rotor with Cylindrical laminated core rotor is heavy bars or copper or Aluminum wound like winding on the stator. or alloys are used for conductors.
3)
Rotor conductors or rotor bars are At starting the 3 phase windings are short circuited with end rings. connected to a star connected rheostat and during running condition, the windings are short circuited at the slip rings.
4)
Rotor bars are permanently short It is possible to insert additional circuited and hence it is not possible resistance in the rotor circuit. to connect external resistance in the Therefore it is possible to increase
circuit in series with the rotor the torque; the additional series conductors. resistance is used for starting purposes. 5)
Cheaper cost.
Cost is slightly higher.
6)
No moving contacts in the rotor.
Carbon brushes, slip rings etc are provided in the rotor circuit.
7)
Higher efficiency.
Comparatively less efficiency.
8)
Low starting torque. It has 1.5 time High starting torque. It can be full load torque. obtained by adding external resistance in the rotor circuit.
9)
Speed control by rotor resistance is Speed control by rotor resistance is not possible. possible.
10)
Starting current is 5 to 7 times the Less starting current compared to full load current. squirrel cage Induction Motor.
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Difference between Synchronous Motors with Damper Winding and Induction Motor. A Squirrel Cage Induction Motor (SCIM) only has a cage rotor winding, whereas a synchronous machine has both a cage (damper) and a wound DC field winding as shown in figure below. Synchronous Motor with Damper Winding:
Squirrel Cage Induction Motor:
In both cases, and when using only the cage winding, both motors will run close to synchronous speed just like a normal SCIM. However, when the synchronous motor field winding is supplied with DC, it will lock into synchronism and its cage or damper winding will carry zero current as the relative speed between the damper winding and rotating flux is zero. Under load disturbances the damper winding will have current induced in it, develop additional torque, and thus help to damp out transient disturbances. In smaller synchronous motors the damper winding is used to start the motor just like a normal SCIM and when close to synchronous speed the DC field is applied and the rotor locks into and runs at synchronous speed. Capacitor Split Phase Motors or Starting of Single Phase Induction Motor by Capacitor. We have already discussed about staring methods of Single Phase Induction Motor. If you miss that please read here, 1) Revolving Field Theory of Single Phase Induction Motors 2) Starting Methods of Single Phase Induction Motors The schematic diagram of Capacitor Split Phase Motor is shown in figure below.
Like in resistor split phase motor, there are two windings, Main and Auxiliary winding but the basic difference between the two method is that in Capacitor Split Phase Motors a capacitor of suitable value is connected in series with the auxiliary winding. Capacitor is connected in series with the auxiliary winding to obtain the desired time phase displacement between the auxiliary winding current Ia and main winding current Im. A centrifugal Switch is also provided the cut out auxiliary winding when the speed of Single Phase Induction Motor reaches 70 to 80% of synchronous speed. You may like to read, Purpose of Centrifugal Switch in Induction Motor As can be seen from the phasor diagram of Capacitor Split Phase Motor, there is an angle of β between the auxiliary winding current Ia and main winding current Im.
Mind that the angle between the auxiliary winding current Ia and main winding current Im is 180° if there is main winding alone and because of this Single Phase Induction Motor cannot start by itself. Also, the torque produced in any machine is directly proportional to IaImSinβ, therefore in this method of starting, there will be a net starting torque and the motor will start. The value of Capacitor used shall be selected based on the load starting torque requirement. If the starting torque requirement of load is more, then angle β shall be made more by selecting higher value of Capacitor. A maximum starting torque can be obtained by this method of starting if angle β = 90°. But to have β = 90°, the size and cost of Capacitor will increase. Therefore a compromise is made in between the load starting torque requirement and size & cost of Capacitor. It shall be noted that, auxiliary winding and Capacitor are in circuit for a short time only, and therefore these can be designed for a minimum cost. The torque speed curve for Capacitor Split Phase Motor is depicted in figure below.
It is clear from the figure that starting torque in this method is high. Capacitor Split Phase Motors have a typical power rating of 100 to 800 Watts. The value of starting Capacitor varies from 20 to 30 microF for 100 Watt Motors and 60 to 100 microF for 750 watt Motors. AC electrolytic capacitors are mostly used in this method of starting but Motors shall not be frequently started else electrolytic capacitors may get overheated and damage.
Why 3-Phase Induction Motors are Self-Starting but 3-Phase Synchronous Motors Not?. Let us consider a 3-phase induction motor first and see how it rotates. A 3- phase supply given to the armature of Induction Motor produces a rotating magnetic field. This rotating magnetic field rotates at synchronous speed Ns = (120xf)/P This rotating magnetic field links to the rotor coils and induces voltage which in turn produces current in the rotor. The current carrying rotor being placed in a magnetic field experiences a torque and hence begins to rotate in the direction of rotating magnetic field. Thus we see that Induction Motor is self-starting. It does not require nay external mean to rotate. Now we consider a 3-phase synchronous motor. A 3-phase supply is given to the armature of Synchronous Motor, produces a rotating magnetic field. However, in this case, the rotor has its own field produced by a DC current flowing through the rotor winding. This rotor field tends to align itself along with the rotating magnetic field produced by the stator i.e. armature winding. North pole of rotor tries to lock with the South pole of stator and South pole of Rotor tries to align along the North of Stator.
So what happens exactly? The North Pole of rotor tries to chase the South Pole of stator. But the stator magnetic field is rapidly rotating at synchronous speed, and before the North Pole of rotor could lock with South Pole of stator, the stator field has shifted position so that its North (stator) comes in the vicinity of North Pole of rotor and they repel as shown in figure below.
Because the rotor has certain inertia and the speed of the rotating magnetic field is too fast for it to catch up, it ends up vibrating. So Synchronous Motor fails to start. So what do you do for starting a Synchronous Motor?
We can give a reduced frequency supply to the stator, this will reduce the speed of rotation of the stator magnetic field and the rotor will easily catch up, once the rotor catches up we may increase the frequency. We can manually rotate the rotor till it catches speed near to synchronous speed and eventually locks the rotor field with the stator field. We can use Amortsieur Windings. The concept is to start the motor as an induction motor. Initially no DC field excitation is given and the motor operates as an induction motor. Once it attains some speed near to synchronous speed, DC excitation is given and the rotor field aligns itself with the stator field, and rotor attains synchronous speed.
Methods of Starting Single Phase Induction Motors. As discussed in earlier post Revolving Field Theory of Single Phase Induction Motors, a single phase induction motor with main stator winding alone has no inherent starting torque as the main stator winding and only produces stationary pulsating air gap flux wave. For the development of starting torque, rotating field at the starting must be produced. There are various methods of starting single phase induction motor, can be classified as below,
1) Split phase starting 2) Shaded pole starting 3) Repulsion motor starting 4) Reluctance starting Generally a single phase induction motor is known by the method employed for its starting. Basically the selection of a particular type and choice of starting single phase induction motor method depends upon the following factors: 1) Torque speed characteristics of load from starting to normal operating speed 2) The duty cycle 3) The starting and running line current limitation as imposed by the supply authorities In this post we will only discuss the Split Phase Starting method. Single Phase Induction Motor employing split phase starting method is known as Split Phase Motor. All the split phase motors have two winding, main winding and auxiliary winding. Both theses windings are connected in parallel but their magnetic axis are displaced by an angle 90°. Split phase starting method is further categorized into following: 1) Resistor split phase motors 2) Capacitor split phase motors 3) Capacitor start and run motors
Resistor split phase motors. A schematic diagram of the two stator winding in quadrature is shown if figure below. Subscript m and a stands for main and auxiliary winding of stator. CS is Centrifugal Switch. You may like to read, Purpose of Centrifugal Switch in Induction Motor.
As we know that, if the two winding currents are shifted in time phase, a rotating filed is created which is necessary for the production of starting torque. In order to achieve this main winding M is designed to have low resistance but higher reactance whereas the auxiliary winding is designed to have higher resistance (thin wire) but lower reactance. The use of thin wire for auxiliary winding is acceptable as auxiliary winding only remain in circuit during starting but the use of thick wire for main winding is necessary as main winding has to remain in circuit permanently. As the reactance is directly proportional to the square of number of turns, auxiliary winding has less number of turns as compared with main winding. In addition to the above mentioned points, leakage reactance of main winding is increases by placing it at the bottom of slot whereas auxiliary winding is placed at the top of slot to have low leakage reactance. As seen in above discussion, main winding has more reactive impedance as compared to the auxiliary winding, therefore main winding current Im lags behind the auxiliary winding current Ia as shown in the phasor diagram below.
Thus from the phasor we see that the angle between the two field produced by main and auxiliary winding is β. As we know that torque produced is directly proportional to torque angle which is β here, therefore a n net starting torque will be developed. The auxiliary winding is disconnected automatically by means of Centrifugal Switch CS at about 70-80% of synchronous speed. If the Centrifugal Switch fails to operate, auxiliary winding will remain in the circuit and noisy performance of single phase induction motor will result. Since auxiliary winding is short time rated, it must get overheated and consequently burn out.
Typical application of Resistor split phase induction motor is for fans, blowers, centrifugal pumps and refrigerator. Concept of Series and Shunt Faults. Electrical Faults can be classified into two categories:.
Shunt Faults and 2) Series Faults Shunt faults include power conductor or conductors to ground or short circuit between the conductors. Series type of fault is basically unbalance in system. Suppose we have used Fuse / Breaker to protect the circuit. If one or two phases open while the third phase remain in circuit, such kind of fault is called Series Fault. Notice that Series Fault may also occur in case of one or two Broken Conductor. Here broken conductor is like breaking of jumper on the tower of transmission line which is not touching the grounded tower body.
Shunt faults are characterized by increase in current and decrease in voltage and frequency whereas Series faults are characterized by increase in voltage and frequency and decrease in current in the faulted phase. Shunt faults are classified as: 1) Line-to-Ground Fault 2) Line-to-Line Fault 3) Double Line-to-Ground Fault 4) Three phase fault Of the above faults, first three faults are unsymmetrical fault as the symmetry is disturbed in one / two of the phases. The method of Symmetrical Components shall be applied for the analysis of such unbalance and fault.
Three phase fault is balanced fault which can also be analyzed using concept of symmetrical components. Series faults are classified as: 1) One Open Conductor
2) Two Open Conductors These faults also disturb the symmetry and therefore these faults are unbalanced faults and hence shall be analyzed using concept of symmetrical components. Neutral Voltage during Fault: The potential of neutral when it is grounded through some impedance or is isolated will not be at ground potential under unbalance condition as in unsymmetrical fault rather it will have some finite value with respect to ground. The potential of neutral is given as Vn = -InZn where Zn is neutral grounding impedance and In is neutral current. Notice the negative sign before the expression of neutral voltage Vn, it indicates the flow of current from ground to the neutral point and therefore the potential of neutral point will be less than the ground potential. For a three phase system we know that, Ia + Ib+ Ic = 3Ia0
You may also like to read Calculation of Symmetrical Components Therefore, Vn = -3Ia0Zn
Notice that only zero sequence current flows through the neutral and therefore voltage drop across neutral will be only due to zero sequence currents.
HRC Fuse – Construction, Working and Characteristics. HRC stands for High Rupturing Capacity. HRC Fuse has high rupturing capacity. Because of its high current rupturing capacity, a special method for extinguishing arc is required in the design of HRC Fuse.
Construction of HRC Fuse: HRC Fuse consists of heat resisting ceramic body having metal end caps on which silver current carrying element is welded in a special manner as shown in figure below.
As clear from the figure above, the fuse element have a portion of Tin Alloy, known as a Eutectic Material. This alloy is used to give the fuse specific operating characteristics. Also, constrictions in the fuse element are provided which play a very vital role in the operation of Fuse. The space between the body surrounding the Fuse element is filled filling powder such as with Silica Send, Chalk, plaster of peris etc.
Working Principle of HRC Fuse: Under normal operating conditions the current flowing through the Fuse element does not provide enough energy to melt the element. The heat produced is absorbed by the surrounding filling powder. If a large current flows the energy produced melts and vaporizes the fuse element before the fault current reaches the peak. The chemical reaction between the fuse element vapour and filling powder results into high resistance material which helps in extinguishing the arc. Now, will HRC Fuse blow off in case of overload condition?
Under overload condition the fuse element will not blow off but if the condition exists for prolonged period, the Eutectic Material will melt and break the fuse element. This is the purpose of providing Eutectic Material in the HRC Fuse.
Will Eutectic Material blow off during short circuit condition? Under high current short circuit conditions the smaller area constricted parts of the fuse element will melt rapidly and vaporize and will break before the Eutectic Material. That is why constrictions are provided in the HRC Fuse element.
Characteristics of HRC Fuse: A Fuse operates when its element melts due to heat produced by I2RF, where RFis Fuse resistance. This heat produced increases if the current flowing through the Fuse element increases. Therefore, we can conclude that a Fuse element will melt faster for large fault current while it will take some time for lower value of fault current. This timecurrent relationship of Fuse is known as Characteristics of Fuse and is very useful for proper selection of Fuse for a particular circuit and for coordination purpose. A typical Fuse characteristic is shown in figure below.
How to Interpret the Fuse Characteristics? In the above figure, curve for three Fuses of rating 60 A, 100 A and 200 A are give. We select Fuse of rating 60 A for the sake of understanding. See, if the current flowing through the Fuse element is around 350 A then the Fuse element will melt in .02 sec i.e. 20 ms while if the current is around 225 A then it will take 50 ms to melt. Thus we see that the Fuse characteristic is Inverse Time. Higher the current, lower will be the time to melt. Electrical Fuse – Types and Related Terminologies.
Fuse: A fuse is the small piece of wire connected between the two terminals of insulated mounted base. Fuse is always connected in the series of the circuit of low voltage equipment. It is the simplest and cheapest form of protection from overload and short circuit. The Fuse is expected to carry the normal current without heating and during overload / short circuit; Fuse gets overheated up to its melting point rapidly and thus breaking the circuit. The materials used for the Fuse are Tin, Lead, Silver, Zinc, Copper etc. For small value of current an alloy of Lead and Tin, in the ration of 37 & 63% are used. But for current more than 15A, this alloy is not used as the diameter of the wire will be large and after fusing the metal release will be excessive.
Important Terminologies Related to Fuse:
Minimum Fusing Current: It is the value of current flowing through the Fuse wire at which the Fuse wire will melt. Fuse Rating: Fuse rating is given in Ampere. It is basically that value of current at which the Fuse is expected to operate safely without melting. This value of current will definitely will be less than the Minimum Fusing Current. Fusing Factor: Fusing Factor is defined as the ratio of minimum fusing current to the fuse rating. Fusing Factor = Minimum Fusing Current / Fuse Rating The value of Fusing Factor is always greater than 1. Prospective Current: Prospective Current of Fuse is the value of current which will flow through it just before the melting of the fuse wire under Short Circuit condition. Melting Time / Pre-arcing Time: This is the time taken by a fuse wire to be broken by melting. It is counted from the instant; the over current starts to flow through fuse, to the instant when fuse wire is just broken by melting. Arcing Time:
After breaking of fuse wire there will be an arcing between both melted tips of the wire which will be extinguished at the current zero. The time from the instant of arc initiated to the instant of arc being extinguished is known as Arcing Time of fuse. Total Operating Time: Total Operating Time of Fuse is the sum of Pre-arcing and arcing time. Types of Fuses: There are basically two types of Fuses: AC Fuse DC Fuse This classification of Fuse arises because of arcing. In DC it is quite time taking to extinguish the arc. Therefore DC Fuses are made with longer wire so as to avoid arc. Hence DC Fuses are bigger in size. But in AC fuse as the current reduces to zero in every haft cycle (10 ms assuming 50 Hz frequency), arc is extinguished.
Other Types of Fuses: Cartridge Type Fuses: Cartridge fuses are used to protect electrical appliances such as motors air-conditions, refrigerator, pumps etc, where high voltage rating and currents required. They are available up to 600 A and 600 V AC and widely used in industries, commercial as well as home distribution panels.
Blade Type Fuses: This type of fuses, also called as spade or plug-in fuses comes in plastic body and two metal caps to fit in the socket. They are used in automobiles for wiring and short circuit protection.
Other Types of Fuses are SMD Fuses, Axial Fuses, Thermal Fuses, HRC (High Rupturing Capacity) fuse and High Voltage fuses. What is Endurance Test of Circuit Breaker?. Endurance Test of Circuit Breaker is conducted to check the healthiness of its mechanical parts i.e. operating mechanism. In this test, Circuit Breaker is operated several times and checked for any damage of its mechanical parts / contacts. The breaker should be in a position to open and close satisfactorily. This test is also called Mechanical Test. In mechanical tests, the circuit breaker is opened and closed several times (1000). Some operations (about 50) are conducted by energizing the relays, remaining are by closing the trip circuit by other means. Mechanical tests on high voltage AC circuit breakers are conducted without current and voltage in the main circuit. Out of the 1000 operations, about 100 operations are made by connecting the main circuit (contacts) in series with trip circuit. No adjustment or replacement of parts is permitted during the mechanical tests. However, lubrication is permitted as per manufacturer’s instructions.
After the Endurance Test, the contacts, linkages and all the other parts should be in good condition and should not show any permanent deformation or distortion. The dimensions should be within original limits. During repeated operations of the circuitbreaker, the weaker parts in the assembly may fail. The circuit-breaker is then considered to have failed in the mechanical test. The tests are then to be repeated after improvement in the design and manufacture. Successful performance in Mechanical Endurance Test proves the adequacy of design and also good quality of materials and manufacture. Though 1000 close-open cycles are specified in the standards, the manufacturer may conduct 10,000 or more operations to ascertain the reliability and for getting design data. What is the Need of Synchronizing two Different Power Sources?. For understanding the need of Synchronization of two Power Sources, first we shall consider the meaning of Synchronization. Suppose we have a trolley that can only drawn by either pushing or pulling it ,two workers are there to drive it if one of them is pushing in one direction but the other one is in another direction. What will happen? They can’t move it with different speed or different direction. Similar is going to happen with power source. Synchronization of two Power Sources means both the sources have the same
Phase Sequence Voltage Magnitude Frequency Phase Angle
There is a setting provided in the Synchronization Check Relay (25 SYN). There are two terms which are frequently used in Synchronization, Running Line and Incoming Line. The bus which is already charged and to which we are connecting a source is called Running Line as shown in figure below.
In above figure if we close the CB-1 then we are synchronizing S1 to the Bus, therefore S1 is Incoming Line and Bus is Running Line. Now, what will happen if we connect two sources in out of Synch.? When power sources are not synchronized, there are instances where there is a voltage difference at the same very node where the three sources are connected. Suppose Source S1 = 415V, 50 Hz, phase angle = 0° Source S2 = 415 V, 50 Hz, phase angle = -120° Thus, we are going to connect the R phase of Source S1 to Y phase of Source S2. Therefore from phase angle 0 to 90 degree, source S2 is stronger in magnitude and hence current will flow from source S2 to S12 but after 90 to 135 degree, source S1 is stronger in magnitude of voltage and hence current will flow from S1 to S2. In this manner a continuous circulation current will flow from one source to another whose magnitude depends on the system impedance.
This circulating current if high enough will burn the equipment connected. Therefore it is must that we should Synchronize the two sources. Working Principle of Petersen Coil. To better understand the working principle and need of Petersen Coil, let us have a look at the arcing ground phenomenon. We know that arcing ground phenomenon is observed in ungrounded 3 phase system. During arcing ground the voltage of healthy phase rises from phase voltage to line voltage i.e. it becomes √3Vph. Also, arcing in such phenomena is due to heavy capacitive charging current which is 3IC where IC = Vph / XC. Thus if there were any way to reduce this charging current then arcing ground phenomena could have been eliminated. Isn’t it?.
Well, you will say that we can connect a resistor in ground of system to minimize the capacitive charging current. Then why do we connect inductor in ground? Why Inductor is used to eliminate Arcing Ground?
To answer this question, let us consider a single line to ground fault and its phasor diagram for an ungrounded system as shown below.
From the phasor diagram it is can be easily observed that, the voltage of neutral point shifts from ground potential to phase voltage Vph but in opposite direction. This is the reason the direction of VC is reverse in figure above and shown by V’C. The fault current IC (IC = IA+IB) is perpendicular to the V’C. Thus if we want to eliminate the fault current then
we must connect an element which will take current in a direction opposite to I C. Carefully observe that IC is leading V’C by 90°.
Now as we are connecting an element in between the neutral point N and ground, therefore the voltage drop across that element will be V’C. Thus that element must take current equal to IC and shall lag from V’C by 90°. As inductor takes lagging current, therefore an inductor is connected in between neutral and ground to eliminate arcing ground. What is Petersen Coil?
Petersen Coil is nothing but an inductor used to connect ground of three phase system to the earth. In other words, the neutral of three phase system is grounded through Peterson Coil. Basically, such grounding is adopted to minimize the capacitive charging current during fault in the lines. This also eliminates the arcing ground. The inductor connected in figure above is Petersen Coil. This type of grounding is also known as Resonant Grounding. How does Petersen Coil Work?
As discussed earlier in this post, Petersen Coil must take current equal to the fault current ICso that it neutralizes the fault current. This is the reason, it is also known as fault neutralizer. Let us consider the figure shown above. The current through the Petersen Coil IL = Vph / ωL But the fault current IC = 3Vph / XC (how? Please read Arcing Ground) Therefore to neutralize the fault current, IL = IC
Hence, Vph / ωL = 3Vph / XC ⇒1/ωL = 3ωC ⇒L = 1/3ω2C Thus to neutralize the capacitive charging current, the value of inductance of Petersen coil shall be 1/3ω2C. Advantages of Resonant Grounding The use of Petersen coil reduces the line interruption due to transient line to ground fault. This is otherwise not possible with other kind of grounding. The tendency of developing three phase fault from single phase fault is reduces with the use of resonant grounding.
Working Principle of Earth Leakage Circuit Breaker ELCB and Residual Current Device RCD. An Earth Leakage Circuit Breaker (ELCB) is a safety device used to directly detect the leakage current to the Earth from an installation and cut the power supply. Basically ELCB is used where the earth impedance is high. Because of high earth impedance, the voltage difference between the Metallic part of the Installation and Earth will be quite high and dangerous from human safety point of view. It may strike in your smart mind that “What is the difference between Earth Fault current and Earth Leakage Current?” This is very important to know as we are going to discuss about safety device used to sense earth leakage current. Well, according to IEC 60947-2, Earth fault current is the current flowing to earth due to insulation fault and Earth leakage current is the current flowing from the live parts of the installation to earth in the absence of an insulation fault. In case of degradation of electrical insulation, the live conductor may get in touch with the metallic part of the equipment and because of high earth impedance; the potential difference between the body of equipment to the Earth will be high enough to result in shock to the working personnel. Earth Leakage Circuit Breaker (ELCB) detects the leakage current to the earth and trips the associated breaker to isolate the supply. There are two types of Earth Leakage Circuit Breaker (ELCB). One is Voltage Earth Leakage Circuit Breaker, also called Voltage ELCB and another is Current Earth Leakage Circuit Breaker, also known as Current ELCB. Working Principle of Voltage ELCB: Voltage ELCB is a voltage operated device. It has a coil and if the voltage across the coil exceeds a predetermined value such as 50 V, the current through the coil will be sufficient enough to trip the circuit.
Voltage ELCB is connected in between the metallic part of equipment and the Earth. If we take an example of insulation failure, then the voltage across the coil of Voltage ELCB will drive enough current to cut the power supply till the manually reset.
Working Principle of Current ELCB: The working of Current ELCB is quite interesting but easy. Current operated ELCB is also known as Residual Current Device, RCD. A Residual Current Device (RCD) has a toroidal iron core over which phase and neutral windings are wound. A search coil is also wound on the same iron core which in turn is connected to the trip coil. Figure below shows the constructional detail of RCD or Current ELCB.
Under normal operating condition, the current through the phase winding and neutral winding are same but both the windings are wound in such a manner to oppose the mmfs of each other, therefore net mmf in the toroidal iron core will be zero. Let us consider a condition where earth leakage current exists in the load side. In this case the current through the phase and neutral will no longer be equal rather phase current will be more than the neutral current. Thus mmf
produced by phase winding will be more than the mmf produced by neutral winding because of which a net mmf will exist in the toroidal iron core. Net mmf in Core = mmf by phase winding – mmf by neutral winding This net mmf in the core will link with the Search Coil and as the mmf is changing in nature (current is AC), an emf will be induced across the terminals of the Search Coil. This emf will in turn drive a current through the Trip Coil which will pull (because of current flow through the Trip Coil, it will behave as an electromagnet and hence will pull the lever to open contact) the supply contacts to isolate the power supply. Notice that Current ELCB works on Residual Current that is the reason it is also called Residual Current Device. A RCD / Current ELCB is also provided with test button to check the healthiness of the safety device. If you carefully observe the figure, you will notice that, when we press the Test Button, Load and phase winding are bypassed due to which only mmf because of neutral winding will exist in the core (as there is no opposing mmf as was the case with both the windings in service) which will cause RCD to trip to isolate the supply.
Working Principle of Residual Voltage Transformer. A Residual Voltage Transformer is used to measure the residual voltage of three phase system during single phase fault. During normal operating condition, the summation of three phase voltage is zero but in case of single phase fault, the scenario changes and there exists a residual voltage.
Let us first discuss residual voltage in case of single line to ground fault. Let us consider a solidly grounded system as shown in figure below.
Let us assume that a ground fault takes place in A phase (In many industries and numerical relays, normally the phases are said as A, B and C instead of R, Y and B, though they represent the same thing i.e. A phase means R phase, B means Y phase and C means B phase). Ea, Eb and Ec are the Generator terminal voltage per phase. Bold letters here represent vector form.
Because of ground fault in A phase, the voltage at the point of fault will become zero but the voltage of other two healthy phases will remain normal as the neutral is solidly grounded therefore the neutral potential will be maintained to earth potential. Va = 0 Vb = V ∠-120° Vc = V ∠120° Here V is per phase voltage under normal condition. Thus the residual voltage of system = Va+Vb+Vc = 0 + V ∠-120° + V ∠120° = V ∠-60° Thus we observe that, there exists a residual voltage in case of single line to ground fault. This residual voltage is measured by Residual Voltage Transformer. The primary of Residual Voltage Transformer is connected to three phase system and its secondary is connected in Broken Delta as shown in figure below.
The output of the secondary windings connected in broken delta is zero when balanced sinusoidal voltages are applied (as Va+Vb +Vc = 0), but under conditions of unbalance a residual voltage equal to three times the zero sequence voltage (V0) of the system will be developed. To measure this component i.e. 3V0, it is necessary for a zero sequence flux to be set up in the Residual Voltage Transformer (RVT), and for this to be possible there must be a return path for the resultant summated flux. Therefore, RVT core must have one or more unwound limbs linking the yokes in addition to the limbs carrying phase windings. Usually the core is made symmetrically, with five limbs, the two outermost ones being unwound. This two outermost unwound limbs provide return path for zero sequence flux. In case where three single phase transformer units are used to measure residual voltage, no extra limbs are requires as each single phase transformer has a core with closed magnetic path. It is very important to earth the primary winding neutral of Residual Voltage Transformer to provide return path for zero sequence current else zero sequence current cannot flow and hence the flux will contain 3rd harmonic component that is reflected in primary and secondary voltages of Residual Voltage Transformer. This voltage appearing at the secondary terminals of RVT is not the residual voltage of the system in any way. Core Balance Current Transformer. Core Balance Current Transformer or CBCT is a ring type current transformer through center of which a three core cable or three single core cables of three phase system passes. This type of
current transformer is normally used for earth fault protection for low and medium voltage system. A typical Core balance Current Transformer is shown in figure below.
Secondary of CBCT is connected to Earth Fault Relay. During normal operating condition as the vector sum of three phase current i.e. (Īa + Īb + Īc =0) is zero therefore no residual current in the primary will be present. Here residual current means zero sequence current. Therefore there will not be any flux developed in the CBCT core and hence no current in the secondary circuit of CBCT. Working Principle of CBCT: Let Īa, Īb and Īc be the three line currents and Φa, Φb and Φc be corresponding components of magnetic flux in the core. Assuming that the CT is operating in the linear region (Read B-H Curve to get idea of linearity), magnetic flux because of individual phase current will be directly proportional to the phase current and hence we can write as below, Φa = kIa Φb = kIb Φc = kIc where k is constant of proportionality. Mind here that same constant of proportionality is used as all the three phase current are producing magnetic flux in the same core i.e. magnetic material. Thus the resultant magnetic flux in the CBCT core, Φr = k(Īa + Īb + Īc) …………………..(1) But we know from theory of symmetrical components, Īa + Īb + Īc = 3Ī0 = Īn Where, Io is zero sequence current and In is neutral current. Hence we can write as Φr = kĪn …………………………(2) Now let us consider two cases: Case1: During normal condition Īa + Īb + Īc = 0 Hence from equation (1), Net resultant flux in the CBCT Core, Φr = 0 which means no secondary current and therefore the Earth Fault Relay won’t operate. Case2: During earth fault, three phase current passing through the center of Core Balance Current Transformer will not be balanced rather a zero sequence current will flow. For example for single line ground fault, If = 3Ia0 = In Thus from equation (2), Net magnetic flux in the CBCT core, Φr will have some finite value which in turn will induce current in the secondary circuit due to which earth fault relay will operate.
Because of this reason, a Core Balance Current Transformer or CBCT is also called Zero Sequence Current Transformer. Advantage of Core Balance Current Transformer:
The advantage of using CBCT for earth fault protectionis that only one CT core is used instead of three core as in conventional system where the secondary winding of three cores are connected residually. Thus the magnetizing current required for the production of a particular secondary current is reduced by one third which is a great advantage as the sensitivity of protection is increased. Also, the number of secondary turn does not need to be related to the cable rated current because no secondary current flows under normal operating condition as the currents are balanced. This allows the number of secondary turns to be chosen to optimize the effective primary pick-up current. Core Balance Current Transformer is normally mounted over a cable at a point close to the cable gland of the Switchgear. In case cables are already laid in a Switchgear, physically split core, which is also known as Slip-over type CT, are used. Knee Point Voltage of Current Transformer. According to IEC, Knee Point Voltage of a Current Transformer is defined as the voltage at which 10 % increase in voltage of CT secondary results in 50 % increase in secondary current. If you carefully read the definition, you will notice that it is something related to saturation of Current Transformer. Never mind, I will explain what exactly Knee Point Voltage mean and what is its significance. We know that CT core is made of CRGO (Cold Rolled Grain Oriented Silicon Steel). When the primary of CT is energized a working mmf is produced in the core. To produce a working mmf, excitation current Ie is taken. This mmf produces a flux inside the core of CT which links with the secondary winding and as per the Faraday’s Law of Electromagnetic Induction, an emf is generated across the terminals of CT secondary by Transformer action. The emf induced in the CT secondary terminal is given as E = 4.44fNØ Where f is frequency of supply, N is number of secondary turns and Ø is flux in the core of CT. But Ø is directly proportional to mmf and mmf in turn is directly proportional to current. Thus if we increase the current, flux Ø generated in the core will increase till the core saturates. Thus there must be a point where from the flux do not increase in the same proportion as the increase in current. This point is called Knee Point. After discussing this much, we can at least say that Knee Point is something related with Saturation of CT core. If there is something called saturation, then we must draw saturation curve of the CT core to have more insight.
You may like to read
Why CT Secondary Shall Never be Kept Open?
Difference between Current Transformer & Potential Transformer
For drawing CT saturation curve, we apply voltage in the CT secondary (keeping primary open) in step of 10% of rated voltage till 120% and read the secondary current using Clamp Meter. All the readings are noted in table and a curve is drawn between applied voltage V and excitation current Ie. The curve drawn will look like as shown in figure below.
Carefully observe the saturation curve shown above. It is quite clear that beyond point K, we need to increase current to a larger extent to have some increase in voltage. This because the curve beyond point K becomes non-linear. The voltage at point K i.e. Vk is called Knee Point Voltage. This is the reason, in definition it is said that Knee Point Voltage of a Current Transformer is defined as the voltage at which 10 % increase in voltage of CT secondary results in 50 % increase in secondary current. This means that an increase in 50% current will lead to just an increase in 10% voltage. Therefore slope at Knee Point Voltage will be, Slope = Increase in Voltage / Increase in current = 0.1/0.5 = 0.2 Knee Point Voltage of Current Transformer is of importance in Protection Class CT i.e. where CT is used for protection purpose. Protection Class CT is normally specified as PS (Protection Special). PS is defined by knee point voltage of current transformer Vk and excitation current Ie
at Vk/2. The Burden of CT when used for protection purpose is quite high when compared with Metering Class CT, which means that voltage drop across theburden will be high. But voltage drop across the burden is equal to the voltage across the CT secondary and if the voltage across the CT secondary is high then it may drive the CT to saturate in normal condition. Therefore Knee Point voltage of Protection Class CT must be more than the voltage drop across the burden to maintain CT core in its linear zone.
Why Harmonic Current in Transformer Excitation Current?. Harmonics in the excitation current of Transformer is due to Hysteresis. As we know the relationship between Magnetic Flux Density, B and Magnetic Field Intensity, H is not linear as shown in figure below.
Also, B = Flux (Ø) / Area (A), and H = NI where N = Number of turns and I = Magnetizing Current Therefore, there exist hysteresis relationship between Flux and Magnetizing Current. The Transformer is preferably operated in saturation region which in turn means that for considerable increment in current causes a slight increase in flux, which gives flux wave sinusoidal shape, while magnetizing current is peaky. This explains why it is rich in 3rd harmonic component even though the supply is sinusoidal. If we observe the waveform of excitation current of Transformer, we see that the wave form of current is symmetrical which means absence of even harmonics. Remember that waveform will not be symmetrical if there is any even harmonic component.
The peaks of the magnetizing current and flux will occur simultaneously, while their zeros will not, due to hysteresis. Thus I hope it is clear that how harmonic component comes in the excitation current of Transformer. Sympathetic Inrush Current in Transformer. I would suggest to read Transformer Inrush Current before reading this article for better understanding of Sympathetic Inrush Current. Any event on the power system that causes a significant increase in the magnetizing voltage of the transformer core results in magnetizing inrush current flowing into the transformer. The three most common events are as follows: Energization of the Transformer. This is the typical event where magnetizing inrush currents are a concern. The excitation voltage on one winding is increased from 0 to full voltage. The transformer core typically saturates, with the amount of saturation determined by transformer design, system impedance, the remnant flux in the core, and the point on the voltage wave when the transformer is energized. The current needed to supply this flux may be as much as 40 times the full load rating of the transformer, with typical value for power transformers for 2 to 6 times the full load rating. Figure below shows the waveform during energization of a transformer.
Magnetizing Inrush Current during Fault Clearing. An external fault may significantly reduce the system voltage, and therefore reduce the excitation voltage of the transformer. When this fault is cleared, the excitation voltage returns to the normal system voltage level. The return of voltage may force a dc offset on the flux linkages, resulting in magnetizing inrush current. This magnetizing inrush current will be less than that of energization, as there is no remnant flux in the core. The current measured by the differential relay will be fairly linear due to the presence of load current, and may result in low levels of second harmonic current. Sympathetic inrush current. Energizing a transformer on the power system can cause sympathetic inrush currents to flow in an already energized parallel transformer. Energizing the second transformer causes a voltage drop across the resistance of the source line feeding the transformers. This voltage drop may cause a saturation of the already energized transformer in the negative direction. This saturation causes magnetizing inrush current to supply the flux. The magnitude of the magnetizing inrush current is generally not as severe as the other cases.
While charging a Transformer in Power System, harmonic restraining of other connected Transformers must be taken care as it may otherwise lead to the tripping of other connected Transformers.
Transformer Inrush Current. In this post we will discuss the Magnetizing Inrush Current in a Power Transformer. Magnetizing Inrush urrent in Transformers results from any abrupt changes of the magnetizing voltage. This current in transformer may be caused by energizing an unloaded transformer, occurrence of an external fault, voltage recovery after clearing an external fault and out-of phase synchronizing of connected Generator.Because the amplitude of inrush current can be as high as a short circuit current, a detailed analysis of the magnetizing inrush current under various conditions is necessary for the making required setting of protective system for the Transformers. First question which will come up in your smart mind, why a Power Transformer takes Inrush Current when energized in unloaded condition? When a power transformer is energized while keeping its secondary circuit open, it acts as an inductance. In normal condition of a Power Transformer, the flux produced in the core is in
quadrature with applied voltage i.e. Flux lags behind the applied voltage by 90° as shown in the figure below.
This means, Flux wave will reach its maximum value after 1/4 cycle or π/2 angle reaching maximum value of voltage wave. Hence as per the waves shown in the figure, at the instant when, the voltage is zero; the corresponding steady state value of flux should be negative maximum. But practically it is not possible to have flux at the instant of switching on the supply of Transformer. This is because, there will be no flux linked to the core prior to switching on the supply. The steady state value of flux will only reach after some finite time which in turn depends upon how fast the circuit can take energy. So the flux in the core also will start from its zero value at the time of switching on the transformer. As we know that, e = dφ/dt where φ is the Flux in the core Therefore assuming e = ESinwt,
Now suppose, Transformer is switched on when Voltage is zero. Therefore Flux will also start from zero. Therefore, total Flux at the end of first half cycle of voltage wave will be,
Where Øm = Maximum flux in the core in steady state or normal operating condition.
Therefore, the flux in the core of Transformer will be double the maximum value of flux in steady state condition. This phenomenon is also shown in figure below.
It is clear from the above graph that maximum flux in the core of Transformer will be 2Ø mwhen the applied voltage is at its zero. Now what will happen because of this higher value of flux in the core of Transformer? Transformer core is saturated just above the maximum steady state value of flux Øm. But when we switch on power supply to the Transformer’s primary, the maximum value of flux will jump to double of its steady state maximum value Øm. As, after steady state maximum value of flux Øm, the Transformer core becomes saturated, the current required to produced rest (2Øm-Øm = Øm) of flux will be very high. So transformer primary will draw a very high current from the source which is called Magnetizing Inrush Current in Transformer or Inrush Current in Transformer. The nature of Transformer Inrush Current is shown in figure below.
It should be noted that waveform of Transformer Inrush Current is asymmetric which means in Transformer Inrush current mainly 2 nd harmonic component will be present. It shall also be noted from the waveform that as time passes the Magnetizing Inrush Current of Transformer decays and becomes zero. Normally it takes few millisecond for Magnetizing Inrush Current to decay to zero.
The Magnetizing Inrush Current of Transformer may be up to 10 times higher than normal rated current of Transformer. Even though the magnitude of Magnetizing Inrush Currentis so high but it generally does not create any permanent fault in Transformer as it exists for few miliseconds. But still Magnetizing Inrush Current in Power Transformer is a problem, because during the time of Magnetizing Inrush Current the protection scheme of Transformer may operate and hence may trip the Primary side Circuit Breaker of Transformer which is not expected.
How do we prevent Tripping of Transformer due to Magnetizing Inrush Current? In Transformer Differential Protection an intestinal time delay of 20 milisecond is provided to prevent tripping of Transformer due to high Magnetizing Inrush Current. In modern Numerical Relay, 2nd Harmonics blocking feature is provided which blocks the 2ndHarmonics when it is more than set value, thereby don’t issue trip command to Transformer Primary side Breaker due to Magnetizing Inrush Current. Normally the setting of 2nd Harmonics blocking is set to 20% which means if 2ndHarmonic component is more than 20% of fundamental frequency value of current then Relay will understand that it is because of Magnetizing Inrush Current and hence won’t issue trip command but if it is less than 20% fundamental frequency value of current then Relay will treat it due to fault and will issue trip command to the primary side Circuit Breaker.
Is there no way to minimize Magnetizing Inrush Current of Transformer? Point on Wave Switch is used to minimize the Magnetizing Inrush Current of Transformer. If you want to know more about Point on Wave Switch, please write in comment box. I will post on Point on Wave Switch.
BDV Test in Transformer. BDV test means Breakdown Voltage Test. It is done for checking the dielectric strength of the oil of the Transformer. Dielectric strength means the maximum capacity to withstand voltage of insulating oil. This test shows the dielectric strength of Transformer oil.
In transformer oil has mainly two purposes, first for insulation, second as cooling of Transformer core and other winding. So while designing Transformer oil use in transformer depends on voltage rating. So testing of oil is done according to voltage rating. For the purpose of BDV test, oil sample from Transformer is taken in a Sample Bottle. While taking sample of oil from transformer, Sample bottle should be flushed well by Transformer oil and oil in Sample bottle should be vented properly so that atmospheric moisture could not ingress in the sample oil. A typical way of taking oil sample in Sample Bottle is shown in figure below.
Breakdown Voltage is measured by observing at what voltage, sparking straits between two electrodes emerged in the oil, separated by specific gap. Low value of BDV indicates presence of moisture content and conducting substances in the oil. For measuring BDV of transformer oil, portable BDV measuring kit is generally available at site. In this kit, oil is kept in a pot in which one pair of electrodes are fixed with a gap of 2.5 mm (in some kit it 4 mm) between them.
Now slowly rising voltage is applied between the electrodes. Rate of rise of voltage is generally controlled at 2 KV/s and observe the voltage at which sparking starts between the electrodes. That means at which voltage dielectric strength of transformer oil between the electrodes has been broken down. A typical value of BDV Test result for 220 / 6.6 kV Transformer oil is 65 kV and moisture content should be less than 10 ppm. SFRA Test of Transformer. SFRA stands for Sweep Frequency Response Analysis. This test is very reliable for condition monitoring of physical condition of transformer winding. Why to conduct SFRA test? The winding of Transformer may be subjected to mechanical stresses during transportation, heavy short circuit faults, transient switching impulses & lightening impulses etc. These mechanical stresses may cause displacement of transformer winding from their position and may also cause deformation of these winding. SFRA Test can efficiently detect: Displacement of Transformer core Deformation & displacement of winding Faulty core grounds Collapse of partial winding Broken or loosen clamp connections Short circuited turns Open winding conditions.
Principle of SFRA Test: As each of the electrical equipment is combination of R, L & C. In Transformer each winding turn is separated from other by paper insulation which acts as dielectric and windings themselves have inductance and resistance, a transformer can be considered as a complicated distributed network of resistance, inductance, and capacitance or in other words a Transformer is a complicated RLC circuit as shown in figure below. Because of this each winding of a transformer exhibits a particular frequency response.
In Sweep Frequency Response Analysis a sinusoidal voltage Vi is applied to one end of a winding and output voltage Vo is measured at the other end of the winding while keeping the other windings open. As the winding is itself a distributed RLC circuit it will behave like RLC filter and gives different output voltages at different frequencies. That means if we go on increasing the frequency of the input signal without changing its voltage level we will get different output voltages at different frequencies depending upon the RLC nature of the winding. If we plot these output voltages against the corresponding frequencies we will get a particular pattern for a particular winding as shown in figure below.
But after transportation, heavy short circuit faults, transient switching impulses and lightening impulses etc, if we do same Sweep Frequency Response Analysis test and superimpose the present signature with the earlier pattern and observe some deviation between the two graphs / signature. Thus we can conclude that there is mechanical displacement / deformation in the Winding / Core. Thus using SFRA test, we can say whether Transformer windings / core is OK or not. This method is simple yet reliable. Buchholz Relay- Transformer Protection. Buchholz relay is a gas actuated protection relay which is generally used in large oil immersed transformers of rating more than 500 kVA. It is used for the protection of a Transformer from the faults occurring inside the transformer. A typical Buchhloz Relay is shown below.
The gas actuated protective relay is designed to detect faults as well as to minimize the propagation of any damage, which might occur within oil-filled Transformers. The Buchholz relay is therefore particularly effective in case of:
Short-circuited core laminations Broken-down of core bolt insulation Overheating of some part of the windings Bad contacts Short circuits between phases, turns Earth faults-puncture of bushing insulators inside tank
Buchholz relay `can prevent the development of conditions leading to a fault in the transformer, such as the falling of the oil level due to leaks, or the penetration of air as a result of defects in the oil circulating system. The adoption of other forms of protection does not therefore exclude the use of the gasactuated Buchholz relay, as this device is the only means of detecting incipient faults, which if unnoticed, can cause heavy failures. Buchholz Relay is installed in between the Main Tank of Transformer and the Conservator as shown if figure below.
Construction of Buchholz Relay: Buchholz relay consists of an oil filled chamber. There are two hinged floats, one at the top and other at the bottom in the chamber. Each float is connected by a mercury switch. The mercury switch on the upper float is connected to an alarm circuit and that on the lower float is connected to an external circuit to cause breaker trip. A simplified construction diagram of Buchholz relay is shown in figure below.
Working Principle of Buchholz Relay: The operation of the Buchholz relay is based upon the fact that every kind of fault in an oil-filled transformer causes a decomposition of the insulating oil due to overheating in the fault zone or to the action of an intense electric field, and a generation of bubble of gas. These reach the relay which is normally filled with oil, through the pipe connecting the transformer to the conservator where the Buchholz relay is mounted. Whenever a minor fault occurs inside the transformer, heat is produced by the fault currents. The produced heat causes decomposition of transformer oil and gas bubbles are produced. These gas bubbles flow in upward direction and get collected in the Buchholz relay. The collected gas displaces the oil in Buchholz relay and the displacement is equivalent to the volume of gas collected. The displacement of oil causes the upper float to close the upper mercury switch which is connected to an alarm circuit. Hence, when minor fault occurs, the connected alarm gets activated. The collected amount of gas indicates the severity of the fault occurred. During minor faults the production of gas is not enough to move the lower float. Hence, during minor faults, the lower float is unaffected. During major faults, like phase to earth short circuit, the heat generated is high and a large amount of gas is produced. This large amount of gas will similarly flow upwards, but its motion is high enough to tilt the lower float in the Buccholz relay. In this case, the
lower float will cause the lower mercury switch which will trip the transformer from the supply i.e. transformer is isolated from the supply. Class-A, Class-B and Class-C Tripping Classification of Generator. Generator, Generator Transformer and Unit Transformer protections have been classified into Class-A, Class-B and Class-C. Class-A tripping is further classified into Class-A1 and ClassA2. Class-A, Class-B and Class-C Tripping Classification of Generator is based on the need of isolation of Generator on the basis of type of fault. In this post we will discuss each type of tripping classes and their significance.
Basis of Tripping Classification: The tipping classification of Generator is based on the need of isolation of Generator on the basis of type of fault. For example, there are some faults like Generator Differential Protection which calls for immediate tripping of Generator Breaker without delay whereas there are some fault like Loss of Excitation, Rotor Earth Fault etc. which do not call for immediate tripping of Generator. Class-A1 Trip: The protections for the faults in the Generator which need immediate isolation are grouped under this Class-A1. There are a list of faults which are kept under this class. They are as follows: a) Generator Differential Protection b) 100% Stator Earth Fault Protection c) Generator Over Voltage Protection d) Dead Machine Protection e) 95% Stator Earth Fault Protection f) Starting Over Current Protection In case of actuation of Class-A1 protection, Generator Circuit Breaker and Filed Circuit Breaker are opened along with turbine tripping. Class-A2 Trip: The protections for the faults in Generator Transformer (GT), Isolated Phase Bus Duct (IPBD), and Unit Transformer (UT) which need immediate isolation are grouped under this Class-A2. Normally following protections are kept under Class-A2: a) Over fluxing Protection of Generator b) Back up Impedance Protection of Generator
c) Differential Protection of GT d) Buchcholz Relay of GT e) PRD of GT f) Trip from OTI & WTI of GT g) Fire protection of GT h) Differential Protection of UT i) Buchcholz Relay & PRD of Main Tank of UT j) Trip from OTI & WTI of UT k) Fire protection of UT These protection when operated initiate tripping of Generator Circuit Breaker, Field Circuit Breaker, Generator Transformer Circuit Breakers & Unit Transformer LV Circuit Breakers and turbine. Class-B Trip: The protections for the faults in the Generator which do not need immediate isolation are grouped under this Class-B. The turbine is tripped first and Generator is allowed to run utilizing trapped steam in turbine. Let us suppose that there is some fault in the process side i.e. in steam cycle, under that condition also turbine will be tripped first while Generator will continue to run utilizing trapped steam till reverse power relay operates. Generator Circuit Breaker is tripped on initiation of reverse power. Normally, Loss of Excitation and Rotor Earth Fault of Generator are kept under this class. These protection when operated initiate tripping of Generator Circuit Breaker, Field Circuit Breaker and turbine. Class-C Trip: The protections for the faults / abnormal condition in the Grid which call for disconnection of the Generator from the Grid are grouped under this Class-C. In this case, Generator is isolated from the Grid by opening the suitable breaker i.e. Generator Transformer HV side Breaker. Mind that in this case only Generator is isolated from the Grid. Thus Generator continues to feed Station loads (also known as house load). Such scheme where generator is operated on house load at reduced power is known as Generator Islanding. Normally following protections of Generator are kept under this class: a) Unbalance or Negative Sequence Protection b) Back up Impedance Protection c) Under Frequency d) Over Frequency e) Pole Slipping Protection. Difference between Power Transformer and Distribution Transformer. The main and basic difference between a Power transformer and a Distribution Transformer is that Power transformers are used in transmission network of higher
voltages for step-up and step down application like 400 kV, 200 kV, 110 kV, 66 kV, 33kV and are generally rated above 200 MVA.
Distribution transformers are used for lower voltage distribution networks as a mean for end user connectivity. 11 kV, 6.6 kV, 3.3 kV, 440 V, 230 V and are generally rated less than 200 MVA.
On the basis of Transformer Size / Insulation Level, Power transformer is used for the transmission purpose at heavy load, high voltage greater than 33 KV and 100% efficiency. It also having a big in size as compare to distribution transformer, it used in generating station and Transmission substation at high insulation level. The distribution transformer is used for the distribution of electrical energy at low voltage as less than 33 kV in industrial purpose and 440 – 220 V in domestic purpose. It work at
low efficiency at 50-70%, small size, easy in installation, having low magnetic losses & it is not always fully loaded. On the basis of Iron Losses and Copper Losses, Power Transformers are used in Transmission network so they do not directly connect to the consumers, so load fluctuations are very less. These are loaded fully during 24 hours a day, so copper losses and Iron losses takes place throughout day. The average loads are nearer to full loaded or full load and these are designed in such a way that maximum efficiency occur at full load condition. Distribution Transformers are used in Distribution Network so directly connected to the consumer so load fluctuations are very high. Distribution Transformers are not loaded fully at all time so iron losses takes place 24 hr a day and copper losses takes place based on load cycle. The specific weight i.e. (iron weight)/(cu weight) is more. Average loads are about only 75% of full load and these are designed in such a way that maximum efficiency occurs at 75% of full load. As these are time dependent therefore All Day Efficiency is defined in order to calculate the efficiency. Power Transformers are used for transmission as a step up devices so that the I2rloss can be minimized for a given power flow. These transformers are designed to utilize the core to maximum and will operate very much near to the knee point of B-H curve. This brings down the mass of the core enormously. Naturally these transformers have the matched iron losses and copper losses at peak load i.e. the maximum efficiency point where both the losses match. Distribution transformers obviously cannot be designed like this. Hence the All-DayEfficiency comes into picture while designing it. It depends on the typical load cycle for which it has to supply. Definitely Core design will be done to take care of peak load and as well as all-day-efficiency. It is a bargain between these two points. Power transformer generally operated at full load. Hence, it is designed such that copper losses are minimal. However, a distribution transformer is always online and operated at loads less than full load for most of time. Hence, it is designed such that core losses are minimal.
In Power Transformer the flux density is higher than the distribution transformer. On the basis of Maximum Efficiency, The main difference between Power and Distribution Transformer is that Distribution Transformer is designed for maximum efficiency at 60% to 70% load as normally doesn’t operate at full load all the time. Its load depends on distribution demand. Whereas power transformer is designed for maximum efficiency at 100% load as it always runs at 100% load being near to generating station. Distribution Transformer is used at the distribution level where voltages tend to be lower .The secondary voltage is almost always the voltage delivered to the end consumer. Because of voltage drop limitations, it is usually not possible to deliver that secondary voltage over great distances. As a result, most distribution systems tend to involve many clusters of loads fed from distribution transformers, and this in turn means that the thermal rating of distribution transformers doesn’t have to be very high to support the loads that they have to serve. All Day Efficiency = (Output in KWhr) / (Input in KWhr) in 24 hrs which is always less than power efficiency. Dry Type Transformers versus Oil Filled Transformers. Before going to the mark difference between Dry Type Transformer and Oil Filled Transformer, it is worth to have some discussion on Dry Type Transformer.
Dry Type Transformer: Dry Type Transformers find use in locations where the use oil Filled Transformers increases the fire hazard such as shopping malls, Hospitals, residential complexes etc. In dry type transformers air is used as the cooling medium instead of oil.
The insulation used in dry type transformers are designed to withstand higher temperatures. Dry type transformers are more expensive than conventional transformers. Vacuum Pressure Impregnation, Epoxy Resin cast are some of the methods of Insulation adopted in Dry Type Transformer construction.
Comparison between Dry Type and Oil Filled Transformer:
Dry Type Transformers use air as the cooling medium. Oil Type Transformers are considered a potential fire and safety hazard for indoor application.
Dry Type Transformers can be located closer to the load unlike Oil Transformers which require special location and civil construction for safety reasons. Locating the Transformers near the loads may lead to savings in cable costs and reduced electrical losses.
Oil Type transformers may require periodic sampling of the oil and more exhaustive maintenance procedures.
Though Dry Type Transformers are advantageous, they are limited by size and voltage rating. Higher MVA ratings and voltage ratings may require the use of oil Transformers.
For outdoor applications, Oil Filled Transformers are cheaper than dry types.
Losses in Electrical Machine – Core Loss and Eddy Current Loss. There are two types of Losses in an Electrical Machine. They are
Core Loss or Iron Loss Ohmic Loss or Copper Loss
In this post we will discuss about Core Loss. Core Loss is again classified into two types:
Hysteresis Loss Eddy Current Loss
First we will have a look at how the core of a Transformer looks like. But the Core Loss take place in any electrical machine which face changing magnetic flux.
Hysteresis Loss: This loss is due to magnetic properties of iron part or core. When the magnetic field strength or the current is increased the flux increase, after a point when we further increase current the flux gets saturated. When we reduce the current from saturation to zero side the flux density starts to decrease. But when the current value reaches zero the flux density should also be zero but it is not zero. For zero current there is still some flux present in the material, this is known as Residual
Magnetic Flux or Remnant Magnetic Flux. Hence the amount of power is never recovered back. The power which gets trapped in the core of the material is lost in the form of heat.
Now we will consider the mathematical part of Hysteresis Loss. The Hysteris Loss in core is given as
Ph= KhfBmx Where Kh = Constant which depends on the volume and quality of core material. Bm = Maximum flux density in the core f = Frequency of Supply x = Steinmetz’s constant whose value varies from 1.5 to 2.5. Thus we see that Core Loss depend on Voltage as well as Frequency of Supply.
Eddy Current Loss: Eddy Current Loss takes place when a coil is wrapped around a core and alternating ac supply is applied to it. As the supply to the coil is alternating, the flux produced in the coil is also alternating.
By faradays law of electromagnetic induction, the change in flux through the core causes emf induction inside the core. Due to induction of emf eddy current starts to flow in the core. Due to this eddy current there will be an associated Ohmic loss which is called Eddy Current Loss.
Eddy current losses can be reduced by lamination in the core. Thin sheet steels must be used which are insulated from each other. Due to insulated sheets the amount of current which flows get reduced and hence the eddy current losses also reduces. Now we will take a look at the mathematical part of Eddy Current Loss. Eddy Current Loss is given as
Pe= Kef2Bm2 Where Ke = constant whose value depends on the volume and resistivity of the core material. Bm = Maximum flux density in the core f = Frequency of Supply It shall be noted that, from the equation of Eddy Current Loss it seems that Eddy Current Loss depends on the frequency of supply but it is not so rather it only depends on the Supply Voltage. How? As Pe = Kef2Bm2
But we know that
So,
Bm2f2= KE2 where K is constant Thus Eddy Current Loss Pe = KV2 Therefore, Eddy Current Loss only depends on the applied Voltage. Over Current Relay and Its Characteristics. Protection schemes can be divided into two major groupings: a) Unit schemes b) Non-unit schemes a) Unit Scheme: Unit type schemes protect a specific area of the system i.e. a transformer, transmission line, generator or bus bar. The unit protection schemes are based on Kirchhoff’s Current Law – the sum of the currents entering an area of the system must be zero. Any deviation from this must indicate an abnormal current path. In these schemes, the effects of any disturbance or operating condition outside the area of interest are totally ignored and the protection must be designed to be stable above the maximum possible fault current that could flow through the protected area. b) Non-unit scheme: The non-unit schemes, while also intended to protect specific areas, have no fixed boundaries. As well as protecting their own designated areas, the protective zones can overlap into other areas. While this can be very beneficial for backup purposes, there can be a tendency for too great an area to be isolated if a fault is detected by different non unit schemes.
The most simple of these schemes measures current and incorporates an inverse time characteristic into the protection operation to allow protection nearer to the fault to operate first. The non unit type protection system includes following schemes: Time graded over-current protection Current graded over-current protection Distance or Impedance Protection Over Current Protection: It finds its application from the fact that in the event of fault the current will increase to a value several times greater than maximum load current. A relay that operates or picks up when its current exceeds a predetermined value (setting value) is called Over-current Relay. Over-current protection protects electrical power systems against excessive currents which are caused by short circuits, ground faults, etc. Over-current relays can be used to protect practically any power system elements, i.e. transmission lines, transformers, generators, or motors. For feeder protection, there would be more than one over-current relay to protect different sections of the feeder. These over-current relays need to coordinate with each other such that the relay nearest fault operates first. Use time, current and a combination of both time and current are three ways to discriminate adjacent over-current relays. Over-current Relay gives protection against: Phase faults Earth faults Winding faults Short-circuit currents are generally several times (5 to 20) full load current. Hence fast fault clearance is always desirable on short circuits. Primary requirement of Over-current protection is that the protection should not operate for starting currents, permissible over-current, and current surges. To achieve this, the time delay is provided. Over-current Relay Ratings: In order for an over-current protective device to operate properly, over-current protective device ratings must be properly selected. These ratings include voltage, ampere and interrupting rating. Current limiting can be considered as another over-current protective device rating, although not all over-current protective devices are required to have this characteristic Voltage Rating: The voltage rating of the over-current protective device must be at least equal to or greater than the circuit voltage. The over-current protective device rating can be higher than the system voltage but never lower. Ampere Rating: The ampere rating of a over-current protecting device normally should not exceed the current carrying capacity of the conductors As a general rule, the
ampere rating of a over-current protecting device is selected at 125% of the continuous load current. Depending on the time of operation of relays, they are categorized as follows: a) Instantaneous Over-current Relay b) Inverse time over current Relay c) Inverse definite minimum time (IDMT) over-current Relay d) Very Inverse Relay e) Extremely Inverse Relay a) Instantaneous Over-current Relay: Instantaneous Over-current Relay is one in which no intentional time delay is provided for the operation. The time of operation of such Relay is approximately 100 ms. Instantaneous Over-current relay is employed where the impedance between the source and the Relay is small as compared with the impedance of the section to be provided.
Following are the important features of an Instantaneous Over-current Relay: 1) Operates in a definite time when current exceeds its Pick-up value. 2) Its operation criterion is only current magnitude. 3) Operating time is constant. 4) There is no intentional time delay. 5) Coordination of definite-current relays is based on the fact that the fault current varies with the position of the fault because of the difference in the impedance between the fault and the source 6) The relay located furthest from the source operate for a low current value 7) The operating currents are progressively increased for the other relays when moving towards the source. b) Inverse time over current Relay: Inverse time over-current Relay is one in which the time of actuation of Relay decreases as the fault current increases. The more the fault current the lesser will be the time of operation of the Relay. Normally it has more inverse characteristic near the pick-up value which in turn means that if fault current is equal to pick-up value then the relay will take infinite time to operate.
c) Inverse definite minimum time (IDMT) over-current Relay: Inverse definite minimum time (IDMT) over-current Relay is one in which the operating time is approximately inversely proportional to the fault current near pick-up value and then becomes constant above the pick-up value of the relay.
From the picture, it is clear that there is some definite time after which the Relay will operate. It is also clear that the time of operation at Pick-up value is nearly very high and as the fault current increases the time of operation decreases maintaining some definite time. d) Very Inverse Relay: Very Inverse Relay is one in which the range of operation is inverse with respect to fault current over a wide range. This happens so as the CT saturation occurs at a later stage but as soon as CT saturation occur there will not be any flux change and hence the current output of CT will become zero and hence the time of operation will nearly become constant. e) Extremely Inverse Relay: Extremely Inverse Relay is one in which CT saturation occur still at a later stage as compared with Very Inverse Relay and hence the characteristic remain inverse up to a larger range of fault current. The equation describing the Extremely Inverse Relay is I2t = K where I is operating current and t is time of operation of the Relay.
IEC (International Electrotechnical Commission) Standard Curve for Inverse Relays:
As per IEC, the time of operation of any Inverse relay can be calculated from the formula given below. Here,
K = Time of actuation α, β = Constant which depends on the type of Relay I = Fault Current I0 = Pick-up current Value of α and β for different types of Relay: Sr. No. Type of Relay α
β
1)
Inverse time over current Relay 0.02 / IDMT
0.14
2)
Very Inverse Relay
1.00
13.5
3)
Extremely Inverse Relay
2.00
80.00
Example: Suppose the pick-up current for an IDMT relay is set at 0.8 A and the fault current is 80 A then the time of actuation can be calculated as K = 0.14/[ (80/0.8)0.02– 1]. Unit and Non-Unit Protection Scheme. Protection schemes can be divided into two major groupings:
a) Unit schemes
b) Non-unit schemes a) Unit Protection Scheme: Unit type schemes protect a specific area of the system i.e. a transformer, transmission line, generator or bus bar. The unit protection schemes are based on Kirchhoff’s Current Law – the sum of the currents entering an area of the system must be zero. Any deviation from this must indicate an abnormal current path. In these schemes, the effects of any disturbance or operating condition outside the area of interest are totally ignored and the protection must be designed to be stable above the maximum possible fault current that could flow through the protected area.
In other words, it is possible to design protection systems that respond only to fault conditions occurring within a clearly defined zone. This type of protection system is known as unit protection. Certain types of unit protection are known by specific names, e.g.Restricted Earth Fault and Differential Protection. Unit protection can be applied throughout a power system and, since it does not involve time grading, is relatively fast in operation. The speed of response is substantially independent of fault severity.
Unit protection usually involves comparison of quantities at the boundaries of the protected zone as defined by the locations of the current transformers. This comparison may be achieved by direct hard-wired connections or may be achieved via a communications link. However certain protection systems derive their restricted property from the configuration of the power system and may be classed as unit protection, e.g. Earth Fault Protection applied to the high voltage delta winding of a power transformer. Whichever method is used, it must be kept in mind that selectivity is not merely a matter of relay design. It also depends on the correct coordination of current transformers and relays with a suitable choice of relaysettings, taking into account the possible range of such variables as fault currents, maximum load current, system impedances and other related factors, where appropriate.
b) Non-unit Protection Scheme: The non-unit schemes, while also intended to protect specific areas, have no fixed boundaries. As well as protecting their own designated areas, the protective zones can overlap into other areas. While this can be very beneficial for backup purposes, there can be a tendency for too great an area to be isolated if a fault is detected by different non unit schemes.
The most simple of these schemes measures current and incorporates an inverse time characteristic into the protection operation to allow protection nearer to the fault to operate first.
The non unit type protection system includes following schemes:
a) Time graded over-current protection
b) Current graded over-current protection
c) Distance or Impedance Protection. Programmable Scheme Logic (PSL) in Numerical Relays. Programmable Scheme Logic or PSL is a kind of feature provided in Numerical Relays to implement the protection scheme of a particular type. This feature of Numerical Relays makes it easier to implement many protection schemes in a single Numerical Relay for example, in Distance Relay we can configure Distance protection, over voltage protection, Over Current Protection, Earth Fault Protection etc. Now we will study about Programmable Scheme Logic, PSL. PSL is a logical block which is made from different but suitable DDB. Here DDB stand for Digital Data Bus. There are many DDBs offered in a Numerical Relay. Each DDB perform a unique function. Thus it is very important to have the knowledge of function of DDBs to implement a particular logic. Hope you got some idea of DDB but don’t worry I will go in detail with example to make it crystal clear. Lets us begin with an example. Let us assume that we have an Alstom Relay P442 and we want to implement a protection feature called Local Breaker Back-up (LBB) Protection in the Relay. So we need to finalize our logic for the operation of LBB. The generalized logic for LBB protection is Lock-out Relay Operated AND Current still existing Under the above logic the LBB Relay shall initiate its timer and shall give the tripping command to isolate the fault after a fixed time delay say 200 ms. Assuming the above logic for LBB, we will design a logic using Programmable Scheme Logic, PSL in the Numerical Relay. But before designing the PSL, we need to give the input to the Relay, in our case there are two inputs, one will be the contact of Lock-out Relay and another Current Transformer (CT) input. So our first step will be to assign the inputs to the Relay and label a name to each input. In the second step, we need to configure the output of Relay and label a name to the Relay Output contact. Let, Lock-out Relay contact Input is labeled as INPUT1. Mind that only digital inputs can be labeled. So we can not assign a name to CT input. Likewise let the Relay output contact to trip Breakers to isolate fault be RL1. Thus as per our logic when INPUT1 is high and over current still exists then RL1 shall get high after 200 ms to isolate the fault.
Carefully observe the figure above. We have used digital inputs and Over-current DDB I>1 (1stStage Over-Current) in an AND gate to start the timer and if the input status do not change for 200 ms then RL1 will change its status from low to high but in the time window of 200 ms, if the input status changes then the timer will reset and RL1 will remain low. Therefore, making a Programmable Scheme Logic, PSL in a Numerical Relay is just a logic building while having knowledge of function of each DDB to be used. This is just a brief of PSL to have some idea of PSL used in Numerical Relay. Hope you enjoyed this post.
Pick-up Current, Plug Setting Multiplier (PSM) and Time Setting Multiplier (TSM). Plug Setting Multiplier (PSM) and Time Setting Multiplier (TSM) are used only for Electromechanical Relays. These terms or parameters are not so used in Numerical Relays but they are conceptually used and incorporated in Numerical Relays too but the way of their implementation is quite different than that of Electromechanical Relays. In this post we will focus on the concept and implementation of Plug Setting Multiplier and Time Setting Multiplier for Electromechanical Relays. As we know that an Electromechanical Relay has a coil which when energized, operates the Relay to have contact changeover. But there shall be some minimum current which when flows through the Relay coil, produces enough magnetic force to pull the lever to make contact change over. Isn’t it? Yes, if you ever get a chance to see electromechanical relay, you will observe that there is a flapper kind of thing which is attached with the lever. The lever in turn is attached with contacts. Thus when a specified current flows through the relay coil, then only it will produce enough magnetic pull to attract the flapper and lever to operate the Relay. A simple picture of relay demonstrating its construction and operation is shown in figure below.
This minimum current in the Relay coil at which Relay starts to operate is called Pick-up Current. If the current through the Relay coil is less than the pick-up value then Relay won’t operate. On contrary, if the current through the Relay coil is more than the Pick-up current, Relay will operate. In industries, we normally perform Relay Pick-up and Drop-off Test to check the healthiness of relays. Hope your concept of Pick-up current of Relay is clear now. Now we will move on to Current Setting of electromechanical relays. Current Setting of Electromechanical Relays: Current Setting of relay is nothing but adjusting its pick-up value. Suppose we are using a CT of ratio 1000/1 A and the pick-up current needs to be set at 1.2 A. Then we will simply put the plug provided on relay coil to 120% or 1.2. Thus we can say that Pick-up current = Plug Position x Rated CT Secondary Current. The plug or tapping is provided on the Relay Coil so that changing the position of Plug changes the number of turns of the relay coil as shown in figure below.
As shown in figure above, the plug is kept at 5. This means that pick-up current of relay will be 5 times of rated CT Secondary current. Likewise, if we put the plug at 8.75 then pick-up current of relay will be 8.75 times of the rated CT Secondary current. Plug Setting Multiplier (PSM): Plug Setting Multiplier (PSM) is defined as the ratio of fault current to the pick-up current of the relay. Thus, Plug Setting Multiplier (PSM) = Fault Current / Pick-up Current = Fault Current / (Plug Position x Rated CT Secondary Current)
Suppose we are using CT of 100/5 A, a fault current of, say 250 A is flowing through the network protected by the relay. Assume that Current Setting or the position of plug is at 5 then Plug Setting Multiplier (PSM) = 250 / (5×5) =10 It shall be noted here that we shall not bother about PSM for instantaneous relay rather we shall consider PSM for relays having characteristics of Inverse Time, Very Inverse Time etc. For Detail on Relay Characteristics read Over Current Relay and Its Characteristics Time Setting Multiplier (TSM): Again it is worthwhile to mention that we shall not bother about TSM for instantaneous relay rather we shall consider TSM for relays having characteristics of Inverse Time, Very InverseTime etc. A Relay is generally provided with control to adjust the time of operation of the Relay. This adjustment is known as Time Setting Multiplier or TSM. Normally a Time Setting Dial is provided which is calibrated from 0 to 1 s in step of 0.05 s. For practical exposure, let us consider a relay as shown in figure below. Please Zoom the image to clearly view every part of the Relay for better understanding.
As can be seen from the figure, there is a Time Setting Dial which is rotated to set the time of operation of the relay. Lets say we want to set the time on Time Setting Dial to 0.5 s, then we need to rotate the dial till 0.5 s on the dial matches with the fixed mark provided. So our TSM is 0.5 here in the case. How to find the time of operation of Relay?
Well, assume that plug is set at 5 and TSM at 0.5 s. For finding the actual time of operation of relay we need to refer the Graph between the Operating Time and PSM which is generally provided on the Relay cover itself but in our figure it is not given. So we consider a graph between Operating Time and PSM as shown below.
For our case, PSM = 10 (Please see the calculation and case considered above in our discussion of PSM) and TSM = 0.5 s. From the Graph, the time of operation of Relay for PSM = 10 is 3 s. Therefore, Actual Time of Operation of Relay = 3s x TSM = 3 x0.5 s =1.5 s Thus we can say that actual time of operation of Relay is equal to the time obtained from the PSM & Operating Time Graph multiplied by TSM.
Why Zone-1 Setting 80% and Zone-2 150% in Distance Protection?. As discussed in my earlier post on Distance Protection Philosophy, Zone-1 setting is kept at 80% of the total impedance of the Protected Line. So question will come in your smart brain that why 80% even though Zone-1 is meant for primary protection?
Zone-1 is meant for protection of the primary line. Typically, it is set to cover 80% of the line length. Zone-1 provides fastest protection because there is no intentional time delay associated with it. Operating time of Zone-1 can be of the order of 1 cycle.
Zone 1 does not cover the entire length of the primary line because it is difficult to distinguish between faults which are close to bus B like fault at F1, F2, F3 and F4. In other words, if a fault is close to bus, one cannot ascertain if it is on the primary line, bus or on back up line. This is because of the following reasons:
CTs and PTs have limited accuracy. During fault, a CT may partially or complete saturate. The resulting errors in measurement of impedance seen by relay, makes it difficult to determine fault location at the boundary of lines very accurately. There are infeed and outfeed effects associated with working of distance relays. A distance relay scheme uses only local voltage and current measurements for a bus and transmission line. Hence, it cannot model infeed or outfeed properly. Because of infeed and outfeed effect the, the Relay may sense fault in 100% length of line even though the location of fault is actually not in 100% of line. Therefore a margin of 20% is given for the accuracy of measurement and infeed / outfeed effect. Next, Question which will definitely strike you that why do we keep the setting of Zone-2 150%? Why not more than 150%?
Zone-2 setting in Distance Relay is kept at 150 % to avoid Overlap Problem. See the picture below.
As clear from the picture above, if the reach of Zone-2 of a relay R1 is extended too much, then it can overlap with the Zone-2 of the relay R3. Under such a situation, there exists following conflict. If the fault is on line BC (and in Z2 of R3), relay R3 should get the first opportunity to clear the fault. Unfortunately, now both R1 and R3 compete to clear the fault. This means that Z2 of the relay R1 has to be further slowed down. As Zone-2 protection already have a time delay, due to overlapping we need to further introduce some time delay which will degrade the performance of Relay for Zone-2. Hence, a conscious effort is made to avoid overlaps of Z2 of relay R1 and R3. Setting Zone-2 of R1 to maximum of 150% of primary line impedance or primary line impedance plus 50% of smallest line impedance usually works out good compromise without getting into Z2 overlap problem.
Zones of Protection and Dead or Blind Zone in Power System. Zoning in Power system Protection is an important philosophy and must be done carefully so that no part of the system remains unprotected in any condition. To limit the extent of the power system that is disconnected when a fault occurs, protection is arranged in zones. The principle is shown in figure below.
Ideally, the zones of protection should overlap, so that no part of the power system is left unprotected. This is shown in figure below. As can be seen from the figure below, each Breaker is included in two different zones of protection to increase the reliability of protection scheme.
For practical physical and economic reasons, this ideal zoning in protection is not always possible to achieve because accommodation for current transformers being in some cases available only on one side of the circuit breakers, as shown in figure below.
This leaves a section between the Current Transformers and Circuit Breaker CB-A that is not completely protected against faults. As shown in figure above, a fault at F would cause the busbar protection to operate and open the circuit breaker but the fault may continue to be fed through the feeder. The feeder protection, if of the unit type, would not operate, since the fault is outside its zone. This problem is dealt with by intertripping or some form of zone extension, to ensure that the remote end of the feeder is tripped also.
The section of Power System which is not covered under any zone of protection is called Dead Zone or Blind Zone and special kind of protection shall be provided to take care of fault in Dead Zone. Normally overcurrent element is used for the protection of Dead Zone with some suitable logic interlock. The logic interlock depends on the configuration of power system and the condition in which Dead Zone is created. Let us take an example to have more insight. Carefully observe the figure below.
Transformer is fed by the Bus when the Breaker CB-A is close. Now suppose we want to take Transformer under maintenance, so for isolating the Transformer we will open CB-A and DS. After opening DS, it may be required to close the CB-A to feed some other connected feeder. As the Breaker is closed, a portion up to DS is charged. Now suppose a fault take place in between DS and CT-3. Assuming that CT-1, CT-2 and CT-3 are meant for protection of Zone in between the CTs using Differential protection, so a fault outside this zone will not be protected and hence zone in between CT-3 and DS is unprotected and called Dead Zone.
Read: Difference between Isolator and Breaker. Over Current Relay and Its Characteristics What will be the logic for implementing protection of this Dead Zone?
One may say, if the DS is open and CT-3 senses an overcurrent then Relay shall issue a tripping command to CB-A. That is all, Dead Zone is no more Dead rather it is protected. Notice that in this case Dead Zone or Blind Zone is created in a particular condition where the DS is open and CB-A is close.
Open Circuit and Short Circuit Characteristics of Synchronous Machine. Open Circuit Test and Short Circuit Test are performed on a Synchronous Machine to find out the parameters of Synchronous Machine and hence to have an idea of their performance. Open Circuit Test of Synchronous Machine is also called No Load, Saturation or Magnetizing Characteristics for the reason which will be clear after going through the post. For getting the Open Circuit Characteristics of Synchronous Machine, the alternator is first driven at its rated speed and the open terminal voltage i.e. voltage across the armature terminal is noted by varying the field current. Thus Open Circuit Characteristic or OCC is basically the plot between the armature terminal voltage Ef versus field current If while keeping the speed of rotor at rated value. It shall be noted that for OCC, the final value of Efshall be 125% of the rated voltage. Figure below shows the connection diagram for performing the Open Circuit Test of Alternator.
As clear from the figure above, an Ammeter is connected in series with the field circuit to measure the field current and a Voltmeter is connected across the armature terminals to note down the voltage generated. Figure (b) shows the plot between If and Ef. It can be seen from the graph that the relationship between the field current I fand no load generated voltage Ef is linear up to certain value of field current but as the the field current increases the relationship no
longer remains linear. The linear part of the relationship is because, at small value of filed current the whole mmf is required by the air gap to create magnetic flux but as the value of mmf exceeds some certain value, the iron parts get saturated and hence the relationship between the flux (No load generated emf is proportional to flux) and field current no longer remain linear. Next assume that if there were no saturation (assuming no iron part is present rather only air gap is present), the relationship between the field current and no load voltage would have been a straight line and that is why the straight line ob in the figure is called Air Gap Line. Thus we observe that because of saturation in iron parts of machine, the no load generated voltage Efdoes not increase in the same proportion as the increase in field current. Short Circuit Test of Synchronous Machine: For performing Short Circuit Test on an Alternator, the machine is driven at rated synchronous speed and the armature terminals are short circuited through an Ammeter as shown in figure below.
Now the field current If is gradually increased from zero until the armature short circuit current reaches its maximum safe value i.e. 125 to 150% of its rated current value. Readings of field current If and short circuit current are noted and plotted.
If you see the above plot of Short Circuit Test, you notice that the short circuit characteristics of a synchronous machine is a straight line. Why Short Circuit Characteristics of Synchronous Machine is Straight Line? For short circuit test, as the armature terminals are shorted, therefore terminal voltage Vt = 0. Therefore the air gap emf Er shall only be enough to provide the leakage impedance drop in the armature i.e. Er = Ia(Ra + jXal) where Xal = Armature Leakage Reactance As we know that, for a Synchronous machine the value of Xal is of the order of 0.1 to 0.2 per unit and Ra (Armature Resistance) is negligible thus we can write as Xal = 0.15 (Taking average value of 0.1 and 0.2) Ra = 0 then Er = Ia (Ra +jXal) = 0.15Ia Taking rated current of armature, Ia = 1 pu
Therefore, Er = 0.15 pu You must read Per Unit System in Electrical Engineering
Thus we observe that during short circuit test, the air gap generated emf Er is only 0.15 pu which mean that air gap flux must also be 0.15 pu. As the resultant air gap flux is only 0.15 of its rated value under normal voltage condition, such a low value of air gap flux does not saturate the iron parts of synchronous machine and hence the short circuit characteristics is a straight line. It shall also be noted here that, in case of short circuit test the armature mmf is almost entirely demagnetizing in nature which results in very low value of air gap flux. Transformer Testings. For the purpose of quality assurance and ensuring that the finished transformer conforms to customer requirements and is ready for service a barrage of tests are performed on the transformer. These can be broadly classified into the following 3 types:
1) Routine Tests: These are standard tests performed necessarily on all transformers. They are further divided into two subcategories: Dielectric Tests (for insulation testing) a) Separate Source Power Frequency Test on HV and LV b) Induced over voltage test c) Insulation Resistance of HV and LV winding d) Dielectric value of oil
Parametric Tests (for winding, loss and efficiency testing) a) Winding Resistance b) No load loss and No load Current c) Load Loss and Impedance
d) Turns ratio on all taps and all phases
2) Type Test: This test is done on one unit of a particular design. It verifies the response of the design to the expected boundary conditions of the design. Hence it is not necessary to perform it on all units manufactured. Generally the testing and certification for the validation of a design to the conformity of these tests is considered for a period of five years. There are two tests that come under this paradigm: a) Temperature Rise Test b) Impulse Test
3) Special Tests: These are performed at the request of the customer by third party testing organizations. These give an idea about the design integrity, manufacturing quality, resistance to fault currents, operation quality etc. These are the following: a) Short-Circuit Test b) Unbalanced Current Test c) Magnetic Balance Test d) Measurement of Zero Sequence Impedance e) Measurement of noise level As seen from the above list there are a total of 15 tests to be performed on a transformer.These tests are to be performed in the same sequence as it has been written above.
The routine tests need to be performed on all transformers that are manufactured, whereas type and special tests have to be performed under the conditions that have been mentioned. It is necessary to perform the above tests in the same order because; each test can cause slight changes in the mechanical and electrical characteristics of the transformer. For example it can cause some of the insulated material to come under excessive levels of dielectric stress which may not be a regular occurrence under normal working conditions. They can cause modifications to the geometry of the core coil assembly thereby affecting the building factor. Also, if the tests are performed in a haphazard fashion without taking into account the parametric variations induced due to the previous tests, then probably the test results are not indicating the correct or operational values. It is recommended to perform tests on Transformer as per the relevant BIS and IEEE standards. Transformer Winding Temperature Indicator. The winding is the component with the highest temperature within the transformer and it is the component which is subjected to the fastest temperature increases as the load increases. Therefore for the control of the temperature parameter within the transformer, the temperature of the winding, as well as top oil, is measured. The temperature of winding of Transformer is measured using Winding Temperature Indicator (WTI) and the temperature of Transformer Oil is measured using Oil Temperature Indicator.
The purpose of WTI is to indicate the winding temperature of HV and LV winding of the Transformer and operates the alarm, trip and cooler control contacts (For detail on Cooler Unit visit Transformer Cooling Classes).
As can be seen from the above figure, Black needle shows the current winding temperature while the Red needle shows the highest winding temperature reached for a particular day. This Red needle needs to be reset manually.
Also, for setting winding temperature for Alarm and Trip, two knobs are provided. The Green knob shows the setting of winding temperature for Alarm and Red knob shows the Trip temperature setting as shown in figure.
Construction Detail of Winding Temperature Indicator, WTI:
Winding temperature indicator consists of a sensor bulb placed in an oil filled pocket in the transformer tank cover. The bulb is connected to the Instrument housing by means of two flexible capillary tubes. One capillary is connected to the measuring bellow of the instrument and the other to a compensating bellow.
Working Principle Winding Temperature Indicator:
An indirect system is used to measure winding temperature, since it is dangerous to place a sensor close to the winding due to the high voltage. The indirect measurement is done by means of a Thermal Image.
The measuring system is filled with a liquid which changes its volume with rising of temperature. Inside the instrument is fitted a heating resistance which is fed by a current proportionate to the current flowing through the transformer winding. To do this we connect the terminal of the heating resistance with the Bushing Current Transformer so that reflection of change in load is reflected in the Winding Temperature Indicator, WTI.
The Winding Temperature Indicator, WTI is provided with a maximum temperature indicator. The heating resistance is fed by a current transformer associated to the loaded winding of the transformer. The increase in the temperature of the resistance is proportionate to that of the winding. The sensor bulb of the instrument is located in the hottest oil of the transformer, therefore, the winding temperature indicates a temperature of hottest oil plus the winding temperature rise above hot oil Level the hotspot temperature.
Winding Temperature of Transformer may rise due to increased loading of Transformer or due to some internal fault. Normally the Winding Temperature Indicator, WTI gives alarm at 85 °C and Trip signal at 95 °C in India.
What is SOTF Protection?. SOTF Protection stands for Switch On to Fault. This protection is provided for high speed clearance of detected fault immediately after Manual Closure of Circuit Breaker. Let us suppose that a fault is existing and we gave a closing command to Breaker, then in that case SOTF protection will immediately trip the Breaker. SOTF Protection is helpful in the sense that while taking maintenance on a Line, it may happen so that Earth Switch of a particular Circuit Breaker is close. In that case if we give closing command to the Circuit Breaker, it will immediately trip on SOTF Protection. But it is normal to strike in your smart brain that how can we close a Breaker when a fault is already existing? Thing is that you can close the Breaker by bypassing all the required logics, so in that case SOTF protection will come into picture. Suppose while maintenance Earthing Rod was used for discharging a particular section but after competition of maintenance the Earth Rod is still connected to that section. In such case there is nothing or no logic to prevent closing of Breaker and as soon as one closes the Circuit Breaker, it will trip on SOTF protection. SOTF Protection gets enabled whenever all the three poles of a Circuit Breaker is open for more than a settable time say 110s (This time can be changed and configured as per the design.) SOTF protection is enabled in two cases. They are mentioned below.
Case1: If no Closing command to the Breaker is Present. When all the three poles of Circuit Breaker is open for more than a particular time say 110 s, as soon as we give a closing command to the Circuit Breaker, the SOTF protection is enabled for 500 ms and then gets reset. Case2: When an external closing command is present. Here external closing means closing the Circuit Breaker either manually or remotely. The SOTF protection logic is activated immediately. As soon as all the poles are closed SOTF protection is enable for 500 ms and then is reset. During this SOTF time frame of 500 ms, individual distance protection can be enabled or disabled. When a particular distance zone say Zone-2 is enabled, Zone-2 will actuate immediately without waiting for Zone-2 time delay if there exists any fault in this zone. How SOTF Relay Senses a Fault? Basically there are two elements in relay providing SOTF feature. They are Voltage Level Detector and Current Level Detector. The purpose of Voltage Level Detector is to ensure Dead Pole situation and Current Level Detector ensures that a fault occurred i.e. Current Level Detector is used for detection of fault. Therefore SOTF Relay element must have two settings, one for Voltage Level Detector and another for Current Level Detector. Typical setting for both the elements are as follows: Vph < 75% VN
Iph > 5% IN
Here VNand IN are Nominal Voltage and Nominal Current respectively. Thus if the phase voltages are normal and if phase current increases from the setting then SOFT protection will sense it a fault and will issue three phase trip command provided all other conditions as mentioned in case1 and case2 for SOTF are met.
Distance Protection Philosophy. Distance protection is a non-unit system of protection offering considerable economic and technical advantages. Unlike phase and neutral over current protection, the key advantage of distance protection is that its fault coverage of the protected circuit is independent of source
impedance variations. Let us take an example of this to understand how distance protection is independent of source impedance. Consider the figure below.
In the figure above, R1 is an over current relay which is used for the protection of Transmission Line. If there is a fault at F1,
Equivalent source impedance Zs = 10×10/20 = 5 Ω
Impedance up to the point of Fault = 5+4 = 9 Ω
Fault current IF1= 220×103/1.732*9 = 220×103/15.588 = 14113.5 A
Therefore the setting of over current Relay should be more than 14113.5 A.
Now consider the case,
Here fault is not on the Transmission Line but it is assumed to be inside Switchyard and only one source is feeding the power to the network. Proceeding in the similar manner, Fault Current IF2= 220×103/1.732*10 = 12702A
Therefore for the protection of Transmission Line, the setting of Relay shall be kept less than 12702 A. But for earlier case we saw that setting of Relay R1 shall be more than 14113.5 A thus overall the setting shall be > 14113.5 but <12702 A which is impractical. Therefore over current Relay is not suitable here and it depends on the source impedance.
Distance protection is therefore used for the protection of Transmission Line. It is simple to apply and fast in isolating the faulty section from the healthy network. Distance Protection provides primary as well as back-up protection to the protected line. I will show this back-up protection function latter in this post.
PRINCIPLES OF DISTANCE RELAYS:
Since the impedance of a transmission line is proportional to its length, for distance measurement it is justified to use a relay capable of measuring the impedance of a line up to a predetermined point. This predetermined point is called Reach of the Relay.
Such a relay is described as a distance relay and is designed to operate only for faults occurring between the relay location and the selected reach point, thus giving discrimination for faults that may occur in different line sections. The basic principle of distance protection involves the division of the voltage at the relaying point by the measured current. The apparent impedance so calculated is compared with the reach point impedance which is settable in the Relay. If the
measured impedance is less than the reach point impedance, it is assumed that a fault exists on the line between the relay and the reach point and issues trip command to the concerned Breaker Trip Coil either through Master Trip Relay or directly (in case of single pole tripping of breaker, assuming single pole Auto Reclosure is allowed).
If measured value of impedance V/I is less than setting z then Relay assumes a fault as clear from the above diagram.
ZONE CONCEPT IN DISTANCE PROTECTION:
Consider the figure below and carefully observe.
Here there are three sub-stations namely A, B and C. For sub-station A, the distance protection is divided into three zones Z1a, Z2a and Z3a which are called Zone-1, Zone-2 and Zone-3 protection. Similarly for sub-station D the three zones will be Z1d, Z2d and Z3d. Zone-1 is normally set to 80% of total length of Line (here line length is AB between two consecutive substation). Zone-2 is set to 150 % of total line length and Zone-3 set at 120% of (100% line length + 100% of Longest Line from Remote substation i.e. B). It should be noted that all Zones are setting is done in terms of impedance.
Assume the distance between A and B = 200 KM
Total Impedance of Line AB = 61 Ohm
Current Transformer ratio = 1000/1A
Potential Transformer ratio = 400 kV / 110 V
So for Zone-1 Impedance setting = 80 % of Total Line Impedance = 80% of 61
= 0.8×61 = 48.8 Ohm ????? (Will it be????)
It won’t be…..because you need to consider CT & PT ratio for calculating the impedance as the Relay is sensing current and voltage through CT and PT only.
PT/CT Ratio = 1000/(400×103/110) = 1000×110/400,000 = 0.28
So the required setting for Zone-1 = 48.8×0.28 = 13.66 ohm. Which means if the distance Relay senses Impedance less than 13.66 Ohm then it will pick-up for Zone-1.
In the same manner, Setting for Zone-2 = (150% of 61) × CT/PT ratio
= 1.5×61×0.28 = 25.62 ohm
Which means if the distance Relay senses Impedance less than 25.62 Ohm then it will pick-up for Zone-2.
Setting for Zone-3 = 120% of (Impedance of Line AB+ Impedance of Longest Line from substation B)
Assume the Longest Line from substation B is having an impedance of 61 Ohm.
Therefore Setting for Zone-3 = (120% of (61+61)) × CT/PT ratio
= 1.2×122×0.28 = 41 Ohm
Which means if the distance Relay senses Impedance less than 41 Ohm then it will pickup for Zone-3.
So we now know how to calculate the setting for different Zones of Distance Protection. Now suppose our substation is A and we are providing distance protection so Relay is located at A. For fault in Zone-1, obviously we need to isolate the fault without any time delay. Now say our breaker at A opened but as we are connected to the substation B so their breaker at B shall also trip so as to isolate the fault completely otherwise fault will be feed from substation B side even though our breaker at A opened. Thus if fault in Zone-1 occurs then Distance Relay shall trip Breaker at A and send a signal to Remote Substation B by receiving which Remote substation B shall trip their breaker at B. This signal is called Carrier Signal which is sent through Power Line Carrier Communication (PLCC) Line. This is the purpose of PLCC. I will post on PLCC latter so be there.
Thus for Zone-1, time delay = 0. Got it? (If no then write in comment box I will be happy to clear your doubt)
Next, suppose there is a fault in Zone-2 then our breaker at A shall not trip rather Remote Substation breaker at C shall trip (If fault is in section CD in figure above) as it will be in their Zone-1. So we need to introduce some time delay in our Distance Relay to operate for Zone-2 fault. This time delay is usually kept around 350 ms. If within 350 ms Remote substation breaker at B trips then our Breaker at A won’t trip but if suppose because of any Reason Remote Substation breaker at C fail to trip then our breaker at A will definitely trip.
See how Zone-2 is working as Back-up protection for line CD. Got it friend?
Now if there is a fault in the remaining 20% of line which is protected by Zone-1 at our substation A then it will be sensed by our Relay at A in Zone-2 but for Remote substation B it will be Zone-1 so their breaker at B will instantaneously trip but our breaker at A also need to trip otherwise our substation will continue to feed the fault by receiving carrier signal.
Now coming to Zone-3, if there is a fault in Zone-3 then our breaker at A is not supposed to trip rather Remote substation breaker at C &D is supposed to trip. Therefore we introduce some time delay for the operation of Zone-3 which is typically of the order of 1s. If because of any reason breaker at C & D fail to trip within 1s then our Distance relay will operate to open our Breaker at A.
There is one more Zone in modern Distance Relay which is called Reverse Zone or Zone-4. As the name Reverse Zone implies it is back-up protection of the Substation where Distance Relay is installed, in our case to the substation A. The setting for zone is normally 10% of the impedance of protected line.
Distance Relay Zone Characteristics on R-X Plane:
The reach point of a relay is the point along the line impedance locus that is intersected by the boundary characteristic of the relay.
Philosophy of Primary and Back-up Protection. The protection provided by the protective relaying equipment can be categorized into two types as:
a) Primary protection
b) Back-up protection
The primary protection is the first line of defense and is responsible to protect all the power system elements from all the types of faults. The backup protection comes into play only when the primary protection fails.
In the event of failure or non-availability of the Primary Protection some other means of ensuring that the fault is isolated must be provided. These secondary systems are referred to as Back-up Protection. Back-up protection may be considered as either being local or remote. Local backup protection is achieved by protection which detects an un-cleared primary system fault at its own location and which then trips its own circuit breakers, e.g. Time Graded Overcurrent Relays.
What is Time Grading? Protection systems in successive zones are arranged to operate in times that are graded through the sequence of equipment so that upon the occurrence of a fault, although a number of protection equipment respond, only those relevant to the faulty zone complete the tripping function. The others make incomplete operations and then reset. The speed of response will often depend on the severity of the fault, and will generally be slower than for a unit system.
As shown in figure above, if a fault occurs then Relay C is supposed to trip instantaneously, but in case of failure of operation of Relay C to isolate the fault, Relay B shall issue trip command after a time delay of 350 ms. In the worst case when both theRelays B & C fails, Relay A shall operate after 1 sec. This is Time Grading where a time is provided for main Relay to operate. Remote back-up protection is provided by protection that detects an un-cleared primary system fault at a remote location and then issues a local trip command, e.g. the second or third zones of a distance relay. In both cases the main and back-up protection systems detect a fault simultaneously, operation of the back-up protection being delayed to ensure that the primary protection clears the fault if possible. Normally being unit protection, operation of the primary protection will be fast and will result in the minimum amount of the power system being disconnected. Operation of the back-up protection will be, of necessity, slower and will result in a greater proportion of the primary system being lost.
The extent and type of back-up protection applied will be related to the failure risks and relative economic importance of the system. For distribution systems where fault clearance times are not critical, time delayed remote back-up protection may be adequate. For EHV systems, where system stability is at risk unless a fault is cleared quickly, multiple primary protection systems, operating in parallel and possibly of different types (e.g. distance andunit protection), will be used to ensure fast and reliable tripping. Back-up overcurrent protection may then optionally be applied to ensure that two separate protection systems are available during maintenance of one of the primary protection systems. Back-up protection systems should, ideally, be completely separate from the primary systems. For example a circuit protected by a current differential relay may also have time graded overcurrent and earth fault relays added to provide circuit breaker tripping in the event of failure of the main primary unit protection. To maintain complete separation and thus integrity, current transformers, voltage transformers, relays, circuit breaker
trip coils and d.c. supplies would be duplicated. This ideal is rarely attained in practice. The following compromises are typical:
a) Separate current transformers(cores and secondary windings only) are provided. This involves little extra cost or accommodation compared with the use of common current transformers that would have to be larger because of the combined burden. This practice is becoming less common when digital or numerical relays are used, because of the extremely low input burden of these relay types.
b) Voltage transformers are not duplicated because of cost and space considerations. Each protection relay supply is separately protected (fuse or MCB) and continuously supervised to ensure security of the VT output. An alarm is given on failure of the supply and, where appropriate, prevents an unwanted operation of the protection.
c) Trip supplies to the two protections should be separately protected (fuse or MCB). Duplication of tripping batteries and of circuit breaker tripping coils may be provided. Trip circuits should be continuously supervised.
d) It is desirable that the main and back-up protections (or duplicate main protections) should operate on different principles, so that unusual events that may cause failure of the one will be less likely to affect the other.
Numerical relaysmay incorporate suitable back-up protection functions (e.g. a distance relay may also incorporate time-delayed overcurrent protection elements as well). A reduction in the hardware required to provide back-up protection is obtained, but at the risk that a common relay element failure (e.g. the power supply) will result in simultaneous loss of both main and back-up protection.
Autoreclosing Philosophy in Distance Protection. Autoreclosing is a feature which is provided in the Line Circuit Breaker so that Single Pole of a Breaker may trip and close when command is given to do so. Based on the time duration of fault existing in the Power System, faults can be classified into three categories as
Transient Fault Semi-Transient Fault Permanent Fault.
Transient fault exists only for very short duration and these can the removed faster if the line is disconnected from the system momentarily so that arc extinguishes. After the arc is deionized, line can be reclosed to resume the service. Thus, in this way the interruption in the Power Supply is reduced and loss of revenue is also saved. It has been found that 80% of the fault in Power System are Transient in nature, 12% are Semitransient and remaining 8% are only permanenet fault. Semi-transient fault are those fault which take some finite time to clear by itself. For example, suppose a bord spanning the two lines sit then it will cause a fault which will clear by itself after the burning of cause of fault,after some time say 1sec. Thus we will expect, Autorecloser to take place for 1 sec i.e. Breaker shall close after a time delay of 1 sec.Here note that the time after which fault clears by itself is called DEAD TIME. Therefore, in our example DEAD TIME = 1 sec. But for Permanent Fault, Autorecloser will not help as the cause of fault continuously exists so if we incorporate the Autorecloser the Breaker will again trip after the Autorecloser. So how many attempts will the Realy take to Autoreclose and after how much time it will take another consecutive attempt to Autoreclose? “Here we come to another concept, called RECLAIM TIME. RECLAIM TIME is the time after which Relay will take another consecutive attempt to Autoreclose. This RECLAIM TIME is typically set at 25 sec. The number of attempt for Autorecloser is set in the Relay which is 4 for MiCOM P444 Distance Protection Relay. This means that Relay will take four Autorecloser shots and at the end of fourth shot, if still fault is existing, the Line will be taken out.” In figure below, a Numerical Relay is shown.
Thus we see that for permanent fault Autorecloser won’t help as we need to attend the fault and rectify it. Autorecloser can be Single pole or Three pole. Here Pole means Breaker of any of the three phase i.e. either R, Y or B phase. Single Pole Autorecloser take place during Line to ground fault. It shall be noted that Autorecloser facility is provided only in Line Breaker and that to by Distance Protection Relay. Single Autorecloser take place in the following conditions: Zone-1 protection operated AND Carrier Channels are healthy AND Three Pole Tripping has not taken place. OR Zone-2 protection operated with Carrier Signal Received. This seems surprising that only a single pole of Breaker trips during Zone-1 fault. But it’s true. The phase, say B phase, in which Line to ground fault has taken place will only trip and reclose after the DEAD TIME. If within the RECLAIM TIME, another fault take place then Three Phase trip will occur. During the DEAD TIME period, power is fed to the system via the two healthy phases. In case of Three Pole Autorecloser, all the three phases are opened independently irrespective of type of fault be it Single L-G or L-L or L-L-L fault and reclosed after the DEAD TIME. During the DEAD TIME period, no power can be transmitted and therefore system is liable to operate unstably.
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