Current-transformer.ppt

GRID Technical Institute

This document is the exclusive property of Alstom Grid and shall not be transmitted by any means, copied, reproduced or modified without the prior written consent of Alstom Grid Technical Institute. All rights reserved.

Current Transformers

GRID Technical Institute

This document is the exclusive property of Alstom Grid and shall not be transmitted by any means, copied, reproduced or modified without the prior written consent of Alstom Grid Technical Institute. All rights reserved.

Current Transformer Function

 Reduce power system current to lower value for measurement.  Insulate secondary circuits from the primary.  Permit the use of standard current ratings for secondary equipment.

REMEMBER : The relay performance DEPENDS on the C.T which drives it !

3

s

Instrument Transformer Standards IEC

IEC 185:1987

CTs

IEC 44-6:1992

CTs

IEC 186:1987

VTs

BS 7625

VTs

BS 7626

CTs

BS 7628

CT+VT

BS 3938:1973

CTs

BS 3941:1975

VTs

AMERICAN

ANSI C51.13.1978

CTs and VTs

CANADIAN

CSA CAN3-C13-M83

CTs and VTs

AUSTRALIAN

AS 1675-1986

CTs

EUROPEAN

BRITISH

4

s

Current Transformer Construction

BAR PRIMARY

WOUND PRIMARY

Primary

Secondary

5

s

Bar Primary Type Current Transformer(s)

Primary Conductor Primary Insulation

Core Secondary Winding

6

s

Ring Type Current Transformer

Typical Protection Bar-Primary Current Transformer Laminated „strip‟ wound steel toroidal core Insulation to stop flash-over from HV primary to core & secondary circuit

1000 turns sec. ? „Feeder‟ or „Bus-bar‟ forming 1 turn of primary circuit

RELAY

1000A ? Generator, or system voltage source

7

s

1A ? Insulation covered wire, giving inter-turn insulation & secondary to core insulation

Polarity

Is

P2

P1 Ip

S1

S2

Inst. Current directions :P1  P2 S1  S2 Externally 8

s

Differential Protection (1)

Operation for Internal Faults P1

P2

IFP1

IFP2

P2

P1

S1

S2

IFS1

IFS2

S2

S1

IFS

R

9

s

Differential Protection (2)

Stability for External Faults P1

P2

S1

S2

IFP

IFS

IFS

R

10

s

P2

P1

S2

S1

Differential Protection (3)

Maloperation due to Incorrect Connections IFP

S1

S2

IFS

S2

S1

IFP IFS

R 2IFS

11

s

Basic Theory

12

s

Basic Theory (1) IS IP

R

1 Primary Turn N Secondary Turns

For an ideal transformer :PRIMARY AMPERE TURNS = SECONDARY AMPERE TURNS  IP = N x IS

13

s

Basic Theory (2) IS IP

ES

R

For IS to flow through R there must be some potential ES = the E.M.F. ES = IS x R ES is produced by an alternating flux in the core. ES  dØ dt 14

s

Basic Theory (3) NP IP NS IS

EK ZCT

ZB

VO/P 15

s

=

ISZB = EK - ISZCT

Basic Formulae

Circuit Voltage Required : ES = IS (ZB + ZCT + ZL) Volts where :-

IS

=

Secondary Current of C.T. (Amperes)

ZB

=

Connected External Burden (Ohms)

ZCT

=

C.T Winding Impedance (Ohms)

ZL

=

Lead Loop Resistance (Ohms)

Require EK > ES

16

s

Basic Formulae (1) Maximum Secondary Winding Voltage : EK = 4.44 x B A f N Volts …. 1 where :EK

=

Secondary Induced Volts

(knee-point voltage) B

=

Flux Density (Tesla)

A

=

Core Cross-sectional Area

(square metres)

17

s

f

=

System Frequency (Hertz)

N

=

Number of Turns

Simple Selection Example Example Calculation : C.T. Ratio = 2000 / 5A T

Max Flux Density = Core C.S.A = 20 cm2

RS = 0.31 Ohms IMAX Primary = 40 kA

Find maximum secondary burden permissible if no saturation is to occur. Solution :

N = 2000 / 5 = 400 Turns IS MAX = 40,000 / 400 = 100 Amps From equation 1 the knee point voltage is :VK = (4.44 x 1.6 x 20 x 50 x 400) = 284 Volts 104 Therefore Maximum Burden = 284 / 100 = 2.84 Ohms Hence Maximum CONNECTED burden : 2.84 - 0.31 = 2.53 Ohms

18

s

Mag Curves Flux Density (B) Tesla (Wb/m2)

Material : Cross

1.8 1.6

Knee Point

1.4 1.2 1.0 0.8

0.6 0.4 0.2 0

0

0.02 0.04 0.06 0.08 0.1 0.12

Magnetizing Force (H) Ampere-Turns / mm (Magnetic Field Strength : H, A/m) 19

s

Mag Curves Cont...

ES ES = 4.44 N f A B = Kv B

(Secondary E.M.F.)

where :- Kv = 4.44 N f A

(B in Wb/m2 ; A in m2) Ie = H . L = K i . H N where :Ki = L/N L = mean magnetic path in metres H = amp. Turns / metre Ie = Amps

(Magnetising Current) Ie 20

s

Mag Curves Cont... B

Multiply By KV to obtain Volts

Units :

KV = E = 4.44 N f A (m2 x turns) B Ki = Ie = L (m / turns) H

Note :

N

L = Mean Magnetic Path

ro

Mean Magnetic Path = 2r r

ri

r = r o - ri 2

H Multiply By Ki to obtain Amps

KV x B = Volts (E) Ki x H = Amps 21

s

+ ri

Magnetisation Characteristic (1)

C.T Ratio 100/5 A connected to CDG11. Relay Burden = 2 VA (at 10% rated I) Problem :-

Calculate the necessary values of Kv and Ki to provide the necessary output at ten times setting (Assume maximum flux density before saturation = 1.6 Tesla)

22

s

Magnetisation Characteristic (2) (i)

Bar Type Primary CDG current setting = 10% = 0.5 A Volts required to operate relay = 2 / 0.5 = 4 V Therefore voltage required at 10 times setting = 10 x 4 = 40 V Hence for a flux density of 1.6T the C.T. should have an output of at

least 40 V. With bar primary, number of turns = 20. Ek = 4.44 B A f N = 4.44 x 1.6 x A x 10-4 x 20 x 50

A = 40 / 0.71 = 56.3 cm2 Assume a stacking factor of 0.92 => Total C.S.A. = 61.2 cm2 Assume I.D. = 18cm, O.D. = 30cm & Depth = 10.2cm

KV = AN / 45 = (56.3 x 20) / 45 = 25 Ki = L / N = 24TT / 20 = 3.77 cm / turn 23

s

Magnetisation Characteristic (3)

(ii)

Wound Type Primary Assume primary has 5 turns - Therefore secondary has 100 turns Ek = 4.44 B A f N = 4.44 x 1.6 x A x 10-4 x 100 x 50 A = 40 / 3.55 = 11.26 cm2 Assume a stacking factor of 0.92 => Total C.S.A. = 12.24 cm2 Assume I.D. = 18cm, O.D. = 30cm & Depth = 2.04cm

KV = AN / 45 = (11.26 x 100) / 45 = 25 Ki = L / N = 24TT / 100 = 0.754 cm/turn

24

s

Low Reactance Design

With evenly distributed winding the leakage reactance is very low and usually ignored. Thus ZCT ~ RCT

25

s

Exciting Voltage (VS)

Knee-Point Voltage Definition

+10% Vk Vk %50+ Iek

Iek Exciting Current (Ie) 26

s

C.T. Equivalent Circuit Ip

ZCT Is

P1

Ip/N S1 N

27

Ie

Ze

Vt

Es

Ip = Primary rating of C.T.

Ie

= Secondary excitation current

N

Is

= Secondary current

= C.T. ratio

Zb = Burden of relays in ohms

Es = Secondary excitation voltage

(r+jx) ZCT = C.T. secondary winding impedance in ohms (r+jx) Ze = Secondary excitation impedance in ohms (r+jx)

Vt = Secondary terminal voltage across the C.T. terminals

s

Zb

Phasor Diagram 

Ip/N

Ie Ie

Is Es

28

Ep

Im Ic

Ep =

Primary voltage

Im =

Magnetising current

Es =  = Ic =

Secondary voltage Flux Iron losses (hysteresis & eddy currents)

Ie = Ip = Is =

Excitation current Primary current Secondary current

s

Saturation

29

s

Steady State Saturation (1)

E= 100V

100A

100A

1A

1A 100/1

E

100/1

1 ohm

E

100 ohm

E= 1V

100A 1A 100/1

30

E=?

100A E

s

10 ohm

1A E= 10V

100/1

E

1000 ohm

Steady State Saturation (2) 

+V

0V

A

-V

Time

Assume :- Zero residual flux Switch on at point A 31

s

Steady State Saturation (3) +V

0V

C

 -V

Time

Assume :- Zero residual flux Switch on at point C 32

s

Steady State Saturation (4) +V 

0V

B

-V

Time

Assume :- Zero residual flux Switch on at point B 33

s

Steady State Saturation (5)

+V Mag Core Saturation 0V

Mag Core Saturation -V

Time 34

s

Steady State Saturation (6) +V

Prospective Output Voltage 

Mag Core Saturation

0V Mag Core Saturation

-V

+V Output lost due to steady state saturation

0V Actual Output Voltage

-V Time 35

s

Transient Saturation v = VM sin (wt + ) L1

R1 Z1

i1

v = VM sin (wt + ) i1  

VM V sin (wt   -  )  M sin ( -  ) . e Z1 Z1

  ˆ1 sin (wt   -  ) - ˆ1 sin ( -  ) . e 

36

STEADY STATE

s



-R1t / L1

-R1t / L1

TRANSIENT

where : -

Z1 

R12  w 2L12

  tan-1 V ˆ1  M Z1

wL1 R1

Transient Saturation : Resistive Burden

Required Flux ØSAT

FLUX Actual Flux Mag Current

0

Primary Current Secondary Current CURRENT

0

37

10

s

20 30

40 50

60

70

80

M

CT Types

38

s

Current Transformer Function

Two basic groups of C.T. 

Measurement C.T.s

 Limits well defined 

Protection C.T.s

 Operation over wide range of currents Note : They have DIFFERENT characteristics

39

s

Measuring C.T.s Measuring C.T.s  Require good accuracy up to approx 120% rated current.  Require low saturation level to protect instruments, thus use nickel iron alloy core with low exciting current and knee point at low flux density.

B Protection C.T.

Protection C.T.s  Accuracy not as important as above.  Require accuracy up to many times rated current, thus use grain orientated silicon steel with high saturation flux density. 40

s

Measuring C.T.

H

CT Errors

41

s

Current Transformer Errors ZS

Is

Ip Ie Ze

Zb

Es

 Ip/N Ie

I Iq p/N Ie Ic’ I s Es 42

Ie Is Ep

s

Current Transformer Error Transformer Error vs Primary Current

Error

Current Error Phase Error Primary Current

Rated Current

Errors can be reduced by :1. 2. 3. 43

s

Using better quality magnetic material Shortening the mean magnetic path Reducing the flux density in the core

Current Transformer Errors Current Error Definition. Error in magnitude of the secondary current, expressed as a percentage, given by :Current error % = 100 (kn IS - Ip) Ip kn Ip

= =

Rated Transmission Ratio Actual Primary Current

Is

=

Actual Secondary Current

Current Error is :-

+ve : When secondary current is HIGHER than the rated nominal value. -ve : When secondary current is LOWER than the rated nominal value. 44

s

Current Transformer Errors

Phase Error Definition. The displacement in phase between the primary and secondary current vectors, the direction of the vectors being chosen so the angle is zero for a perfect transformer. Phase Error is :-

+ve : When secondary current vector LEADS the primary current vector. -ve : When secondary current vector LAGS the primary current vector.

45

s

Current Transformer Ratings

46

s

Current Transformer Ratings (1) Rated Burden  Value of burden upon which accuracy claims are based  Usually expressed in VA  Preferred values :2.5, 5, 7.5, 10, 15, 30 VA

Continuous Rated Current  Usually rated primary current

Short Time Rated Current  Usually specified for 0.5, 1, 2 or 3 secs  No harmful effects  Usually specified with the secondary shorted

Rated Secondary Current  Commonly 1, 2 or 5 Amps 47

s

Current Transformer Ratings (2) Rated Dynamic Current

Ratio of :IPEAK : IRATED (IPEAK = Maximum current C.T. can withstand without suffering any damage). Accuracy Limit Factor - A.L.F. (or Saturation Factor) Ratio of :IPRIMARY : IRATED up to which the C.T. rated accuracy is maintained. e.g. 200 / 1A C.T. with an A.L.F. = 5 will maintain its accuracy for IPRIMARY < 5 x 200 = 1000 Amps 48

s

Choice of Ratio Clearly, the primary rating IP  normal current in the circuit if thermal (continuous) rating is not to be exceeded. Secondary rating is usually 1 or 5 Amps (0.5 and 2 Amp are also used).

If secondary wiring route length is greater than 30 metres - 1 Amp secondaries are preferable. A practical maximum ratio is 3000 / 1. If larger primary ratings are required (e.g. for large generators), can use 20 Amp secondary together with interposing C.T. e.g. 5000 / 20 - 20 / 1 49

s

Current Transformer Designation

Class “P” Specified in terms of :-

i) Rated burden ii) Class (5P or 10P) iii) Accuracy limit factor (A.L.F.) Example :15 VA 10P 20 To convert VA and A.L.F. into useful volts Vuseful  VA x ALF IN

50

s

BS 3938 Classes :-

5P, 10P. „X‟

Designation (Classes 5P, 10P) (Rated VA) (ALF)

(Class)

Multiple of rated current (IN) up to which declared accuracy will be maintained with rated burden connected. 5P or 10P. Value of burden in VA on which accuracy claims are based. (Preferred values :- 2.5, 5, 7.5, 10, 15, 30 VA) ZB = rated burden in ohms 51

s

= Rated VA IN2

Protection Current Transformers

Table 3 - Limits of Error for Accuracy Class 5P and 10P

Accuracy Current Error at Phase Displacement at Composite Error Class rated primary rated primary current (%) at rated current (%) Minutes Centiradians accuracy limit primary current 5P 5 1 60 1.8 10P

52

s

3

10

Interposing CT

53

s

Interposing CT

LINE CT

NP

NS

ZCT

Burden presented to line CT = ZCT + ZB x NP2 NS2 54

s

ZB

NEG.

5A

0.5

1A

R 500/5

0.1

0.4

1VA @ 1A  1.0

„Seen‟ by main ct :- 0.1 + (1)2 (2 x 0.5 + 0.4 + 1) = 0.196 (5) Burden on main ct :- I2R = 25 x 0.196 = 4.9VA Burden on a main ct of required ratio :0.5

R 500/1

1.0

Total connected burden = 2 x 0.5 + 1 = 2 Connected VA = I2R = 2  The I/P ct consumption was about 3VA. 55

s

Current Transformer Designation

56

s

Current Transformer Designation Class “X” Specified in terms of :-

57

s

i)

Rated Primary Current

ii)

Turns Ratio (max. error = 0.25%)

iii)

Knee Point Voltage

iv)

Mag Current (at specified voltage)

v)

Secondary Resistance (at 75°C)

Choice of Current Transformer  Instantaneous Overcurrent Relays

 Class P Specification  A.L.F. = 5 usually sufficient  For high settings (5 - 15 times C.T rating) A.L.F. = relay setting

 IDMT Overcurrent Relays

 Generally Class 10P  Class 5P where grading is critical Note : A.L.F. X V.A < 150  Differential Protection

 Class X Specification  Protection relies on balanced C.T output 58

s

Selection Example

59

s

Burden on Current Transformers

1. Overcurrent : RCT + RL + Rr

2. Earth : RCT + 2RL + 2Rr

RCT

RCT

RCT RCT

Rr

60

Rr

s

RCT

IF RL

RL

RCT

IF

RL

Rr

RL

Rr

RL

Rr

IF

IF

RL

Rr

RL

Rr

RL

Rr

Overcurrent Relay VK Check Assume values :

If max C.T

= =

7226 A 1000 / 5 A 7.5 VA 10P 20

RCT = Rr = RL =

0.26  0.02  0.15 

Check to see if VK is large enough : Required voltage = VS = IF (RCT + Rr + RL) = 7226 x 5 (0.26 + 0.02 + 0.15) = 36.13 x 0.43 = 15.54 Volts 1000 Current transformer VK approximates to :VK ~ VA x ALF + RCT x IN x ALF In = 7.5 x 20 + 0.26 x 5 x 20 = 56 Volts 5 VK > VS therefore C.T VK is adequate 61

s

Earth Fault Relay VK Check Assume values : As per overcurrent. Note

For earth fault applications require to be able to pass 10 x relay setting.

Check to see if VK is large enough : Total load connected = 2RL + RCT + 2Rr = 2 x 0.15 + 0.26 + 2 x 0.02 

Maximum secondary current = 56 = 93.33A 0.6

Typical earth fault setting

= =

30% IN 1.5A

Therefore C.T can provide > 60 x setting C.T VK is adequate 62

s

VK = 56 Volts

Voltage Transformers

63

s

Voltage Transformers

64

s



Provides isolation from high voltages



Must operate in the linear region to prevent accuracy problems - Do not over specify VT



Must be capable of driving the burden, specified by relay manufacturer



Protection class VT will suffice

Typical Working Points on a B-H Curve Flux Density ‘B’

Saturation Cross C.T.’s & Power Transformers

1.6

Tesla 1.0

0.5

V.T.’s

Protection C.T. (at full load) 1000

‘H’ 2000

3000 Magnetising Force AT/m

65

s

Types of Voltage Transformers

Two main basic types are available:  Electromechanical VT`s

 Similar to a power transformer  May not be economical above 132kV  Capacitor VT`s (CVT)

 Used at high voltages  Main difference is that CVT has a capacitor divider on the front end.

66

s

Electromagnetic Voltage Transformer

NP / NS = Kn

LP

RP IP

EP = ES

67

s

IS

Ie LM

VP

LS

RS

IM

Re

IC

VS

ZB

(burden)

Basic Circuit of a Capacitor V.T.

C1

L T

VP C2

68

s

ZB VC2

Vi

VS

Ferro-resonance

 The exciting impedance of auxiliary transformer T and the capacitance of the potential divider form a resonant circuit.  May oscillate at a sub normal frequency  Resonant frequency close to one-third value of system frequency  Manifests itself as a rise in output voltage, r.m.s. value being 25 to 50 per cent above normal value  Use resistive burden to dampen the effect

69

s

VT Earthing

 Primary Earthing

 Earth at neutral point  Required for phase-ground measurement at relay  Secondary Earthing  Required for safety  Earth at neutral point  When no neutral available - earth yellow phase (VERY COMMON)  No relevance for protection operation

70

s

VT Construction



5 Limb

 Used when zero sequence measurement is required (primary must also be earthed)



Three Single Phase

 Used when zero sequence measurement is required (primary must also be earthed)



3 Limb

 Used where no zero sequence measurement is required



V Connected (Open Delta)

   

71

s

No yellow phase Cost effective Two phase-phase voltages No ground fault measurement

VT Connections

Broken Delta A

B

da

a

72

s

C

N

V Connected a

b

c

dn

b

c n

a

b

c

VT Construction - Residual

 Used to detect earthfault  Useful where current operated protection cannot be used  Connect all secondary windings in series  Sometimes referred to as `Broken Delta`  Residual Voltage is 3 times zero sequence voltage  VT must be 5 Limb or 3 single phase units

 Primary winding must be earthed

73

s

Voltage Factors Vf

 Vf is the upper limit of operating voltage.

74



Important for correct relay operation.



Earthfaults cause displacement of system neutral, particularly in the case of unearthed or impedance earthed systems.

s

Protection of VT’s

75



H.R.C. Fuses on primary side



Fuses may not have sufficient interrupting capability



Use MCB

s

GRID Technical Institute

This document is the exclusive property of Alstom Grid and shall not be transmitted by any means, copied, reproduced or modified without the prior written consent of Alstom Grid Technical Institute. All rights reserved.