Ee251 Note 1 - Introduction To Fault Studies

Introduction to Fault Studies Prof. Artemio P. Magabo Prof. Rowaldo R. del Mundo

Department of Electrical and Electronics Engineering University of the Philippines - Diliman

Per-Unit Quantities Definition:

Per

Unit Value

Actual Value Base Value

Note: 1. The per-unit value is a dimensionless quantity. 2. The actual value may be complex but the base value is always real. 3. Percent Value = Per-Unit Value x 100 Department of Electrical and Electronics Engineering

Advantages of Per-Unit Calculations 1. Manufacturers specify the impedances of their equipment in percent (or per-unit) using the nameplate rating of the equipment.

2. The per-unit impedances of electrical equipment of the same type but different ratings usually lie within a narrow range. This makes the detection of an erroneous impedance data easy. Also, if the impedance of a particular equipment is not known, it is acceptable for most studies to select from a range of tabulated typical values. Department of Electrical and Electronics Engineering

Advantages of Per-Unit Calculations 3. In many instances, the base voltages can be selected so that the per-unit turns ratio of any transformer in the system is 1:1. This makes the primary and secondary currents (and voltages) of any transformer identical. 4. When per-unit values are used, the connection of a three-phase transformer does not affect its per-unit equivalent impedance. The transformer connection, however, determines the ratio of the base voltages on the two sides of the transformer. Department of Electrical and Electronics Engineering

Advantages of Per-Unit Calculations 5. Per-unit representation yields more relevant information and easily correlated data. 6. Network calculations are the same for singlephase and three-phase systems. There is less chance of mix-up between phase and line voltages, single-phase and three-phase powers, and primary and secondary voltages.

7. Per-unit calculation is more convenient to use when the solution requires a digital computer. Department of Electrical and Electronics Engineering

Per-Unit Values Need to define base values for: 1. Voltage, V 1. Current, I 3. Impedance, Z 4. Volt-Ampere, VA Note: 1. V, I, Z and VA are not independent of each other.The base values must be chosen so that basic electrical laws are satisfied .

2. The rating of electrical equipment specify KVA (or MVA) and voltage (volts or KV). Department of Electrical and Electronics Engineering

Choice of Per-Unit Values 1. Choose any two of the electrical parameters. In general, the base Volt-Ampere and base Voltage are chosen. Note: For actual power systems, volt and voltampere are small units. The bases are expressed in kV and MVA. 2. Calculate the Base Impedance and Base Current. Note: The base MVA will also serve as base for power and reactive power. The base Z will also be used as base for resistance and reactance. Department of Electrical and Electronics Engineering

Single-Phase System Given a base voltage Vp and a base volt-ampere VAp, find the base current I and base impedance Z. Since VAp

VpI , then

Base I Also, since Vp

Base Z

Base VAp Base Vp

in Amps

IZ, then Base Vp

(Base Vp )2

Base I

Base VAp

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in

Example: Given base Vp = 20,000 volts and base VAp = 2,000,000 volt-amperes

Base I

Base Z

Base VAp Base Vp

100 A

(Base Vp )2 Base VAp

200

Note: 1. If base Vp=20,000 volts, then base kV=20 kV. 2. If base VAp=2,000,000 , then base kVAp=2,000 and base MVAp=2. Department of Electrical and Electronics Engineering

Single-Phase System Other Formulas:

Base I

Base VAp Base Vp Base kVAp Base kVp

in Amps

in Amps

Base MVAp x 1000 Base kVp

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in Amps

Single-Phase System Other Formulas:

Base Z

(Base Vp )2 Base VAp

in

(Base kVp )2 x 1000 Base kVAp (Base kVp )2 Base MVAp

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in

in

Example: For the single-phase system shown, find the voltage, power and reactive power at the source. 100+j300

 Vs

 I

Load

 VL

100 KVA @ 0.9 PF lag VL=20kV

Solution using actual values:

 Let VL

20 0o kV , the reference phasor.

PL

KVA(PF)

90 kW

QL

PL tan(cos 1 0.9)

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43.59 kVars

Solve for the current

 I

PL

jQL

 VL

From KVL,

 Vs

(100

90

j43.59 20 4.5 j2.18 5  j300)I

 VL

(0.32 71.56o )(5 21.1 We get

Ps

jQs

j1.13

25.84 A

25.84)

20.0

o

21.13 3.07 kV

  o o Vs I (21.13)(5) (3.07 25.84 ) 105.67 28.91o 92.5 j51.09

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Solution using per-unit values: Assume

Base KVAp=100

We get

Base I

Base Z

 Let VL

Base KVAp Base kVp

Base kVp=20

100 20

(Base KVp )2 (1000) Base kVAp

5A

4k

1.0 0o p.u, the reference phasor.

PL

KVA(PF)

90 kW

QL

PL tan(cos 1 0.9)

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0.9 p.u. 0.44 p.u.

Solve for the line impedance in per-unit

Z

100 j300 4000

0.025

j0.075 p.u.

0.025+j0.075

 I

 Vs

 I

PL

jQL

 VL

Load

 VL

PL=0.9 QL=0.44 VL=1.0

0.9

j0.44 1.0 0.9 j0.44 1.0

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25.84 p.u.

 Vs

Ps

  (0.025 j0.075)I VL (0.08 71.56o )(1.0 25.84) 1.0 1.06 j0.05 1.06 3.07o p.u.   jQs Vs I (1.06)(1.0) (3.07o 25.84o ) 1.06 28.91o

0.925

j0.5109 p.u.

In actual values,

 Vs

1.06(20)

Ps

0.925(100)

Qs

0.5109(100)

21.13 kV 92.5 kW 51.09 kVar

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Three-Phase System For a three-phase system,

VA3

3VAp

VL

3 Vp

Define:

Base VA3 Base VL

3 x Base VAp 3 x Base Vp

Note: The base current and base impedance will still be I and Z, respectively. Department of Electrical and Electronics Engineering

Three-Phase System Base I

Base kVAp Base kVp

in Amps

Base kVA3

in Amps

3 Base kVL Base MVA3 x 1000 in Amps 3 Base kVL

Base Z

(Base kVp )2 Base MVAp

(Base kVL )2 in Base MVA3

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Example: Given Base MVA3 =30 and Base kVL=120, find the base current and base impedance.

Base I

Base Z

Base MVA 3 x 1000 3 Base kVL 30 x 1000 144.34 A 3 (120) (Base kVL )2 Base MVA3

(120)2 30

480

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If we use single-phase analysis, we get Base MVAp=10 and Base kVp=69.28

Base I

Base Z

Base MVAp x 1000 Base kVp 10 x 1000 144.34 A (69.28)

(Base kVp )2 Base MVAp

(69.28)2 10

480

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Example: Determine the per-unit equivalent circuit for the three-phase network shown. G1

0.381+j0.508

12 MW 0.9 PF lag 13.8 kV

0.571+j0.762

G2

18 MW 0.95 PF lag

Single-phase equivalent circuit 0.381+j0.508

P1=4 MW PF=0.9 lag

V1=7.97 kV

0.571+j0.762

 V1

 V2 PL=6 MW PF=0.95 lag

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Define single-phase bases. Let Base MVAp=5 and Base kVp=7.97.

Base I

Base Z

5 x 1000 627.6 A 7.97 (7.97)2 12.7 5

Per-unit single-phase equivalent circuit 0.03+j0.04 P1=0.8 p.u. PF=0.9 lag

V1=1.0 p.u.

0.045+j0.06

 V1

 V2 PL=1.2 p.u. PF=0.95 lag

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Define three-phase bases. Let Base MVA3 =15 and Base kVL=13.8.

Base I

Base Z

15 x 1000

627.6 A

3(13.8) (13.8)2 12.7 15

Per-unit single-phase equivalent circuit 0.03+j0.04 P1=0.8 p.u. PF=0.9 lag

V1=1.0 p.u.

0.045+j0.06

 V1

 V2 PL=1.2 p.u. PF=0.95 lag

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Generator Impedances Manufacturers provide the following impedances in percent or per-unit: 1. Armature Resistance: Ra

2. Direct-Axis Reactances: Xd”, Xd’ and Xd 3. Quadrature-Axis Reactances: Xq”, Xq’ and Xq 4. Negative-Sequence Reactance: X2 5. Zero-Sequence Reactance: X0

Note: The impedances are based on the machine’s nameplate rating: MVA (or kVA) and kV. Department of Electrical and Electronics Engineering

Per Unit Values for Transformers The ohmic values of the resistance and leakage reactance of a transformer depend on whether the impedances are measured on the high- or low-voltage side of the transformer. The impedance of a transformer is expressed in percent (or per-unit) using its Rated MVA and Rated kVs as bases, respectively. The per-unit impedance is the same whether it is referred to the high-voltage or low-voltage side. The per-unit impedance of a three-phase transformer bank, from identical single-phase units, is the same regardless of the connection. Department of Electrical and Electronics Engineering

Example: Consider a single-phase transformer with the following nameplate rating: 10 MVA

127-69 kV

From

Base Z we get

Base Z

Z=8%

(Base kVp )2 Base MVAp

1272 10 692 10

1,612.9 476.1

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at the HV side

at the LV side

We get the actual impedance of the transformer

Z

0.08(1612.9) 129 0.08(476.1) 38.1

at the HV side at the LV side

129

Equivalent Circuit at the HV side

 IH

38.1

 IH

 IX

Equivalent Circuit at the LV side

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 IX

Assuming the current is rated

Rated MVA x 1000 Rated I Rated kV  10000 IH 78.74 A at the HV side 127  10000 IX 144.93 A at the LV side 69 Per-unit equivalent circuit: Let Base MVAp=10 Base kVp=127 at the HV side Base kVp=69 at the LV side Department of Electrical and Electronics Engineering

We get

Base Z=1612.9 =476.1

We also get

at the HV side at the LV side

Base I=78.74 A at the HV side =144.93 A at the LV side

The transformer impedance is 0.8 p.u. regardless of which side is chosen. The current is 1.0 p.u. regardless of side. 0.08 Per-Unit Equivalent Circuit

1.0

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Example: Given 3 single-phase transformers each rated 10 MVA, 127-69 kV, Z=8%. The transformers are connected to form a three-phase bank, wyegrounded at the HV side and delta-connected at the LV side. What is the rating of the bank? Volt-Ampere Rating=3x10=30 MVA Voltage Rating

Base Z

127 3

220 kV at the HV side 69 kV at the LV side

2202 at the HV side 1,612.9 30 692 at the LV side 158.7 30

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From the previous example, we get the actual impedance of the transformer. At the HV side,

Z At the LV side,

129 , connected in wye 129 0.08 p.u. 1612.9

Z

38.1

Z

12.7 , connected in wye 12.7 0.08 p.u. 158.7

, connected in delta

The bank rating: 30 MVA, 220Y-69 kV, Z=8% Department of Electrical and Electronics Engineering

Change in Base Values Given a per-unit impedance Zpu1 which is based on kV1 and MVA1. Suppose the bases are changed to kV2 and MVA2. Find Zpu2, the new per-unit value. The actual value can be expressed as

Z in

(kV1 )2 Zpu1 MVA1

(kV2 )2 Zpu2 MVA2

Solving for Zpu2, we get

Zpu2

Zpu1

kV1 kV2

2

MVA 2 x MVA1

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Example: Consider a three-phase transformer unit with the following nameplate rating: 50 MVA, 220Y-69 Z=10% Find the percent Z when the bases are 100 MVA, 230 kV at the HV side. From

Zpu2 we get

Zpu1

kV1 kV2

2

MVA 2 x MVA1

2

Zpu2

220 100 10 x 230 50

18.3%

Note: At the LV side, Base kV=72.1, Zpu2=18.3%. Department of Electrical and Electronics Engineering

Per-Unit Representation Procedure: 1. Establish the Base MVA and the Base kVs a. Assume a Base MVA for the entire system

b. Assume a Base Voltage for any power system component (Generator, Transformer or transmission line) c. Compute the Base Voltages for the remaining power system components using the voltage ratios of the transformers

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Per-Unit Representation 2. Using the base MVA and the base voltages, compute the base current and base impedance of each subsystem. 3. Determine the per-unit impedance of every system component. a. If the impedance is expressed in ohms, divide the actual value by the base value.

b. If the impedance is expressed in percent, use the formula for changing base values. Department of Electrical and Electronics Engineering

Example: Draw the per-unit reactance diagram using 50 MVA and 13.8 kV as bases in the circuit of G1. T1 T2 G1

L2

L1

G2

T3

G3

G1: 20 MVA, 13.8 kV X=20% G2: 30 MVA, 18 kV X=20% G3: 30 MVA, 20 kV X=20% T1: 25 MVA, 220Y-13.8 kV X=10% T2: Single-phase each 10 MVA, 127-18 kV X=10% T3: 50 MVA, 220Y-22Y kV X=10% L1: X=80 L2: X=100 Department of Electrical and Electronics Engineering

Three-phase rating of transformer T2: 30 MVA, 220Y-18 kV X=10% Base MVA = 50 Base kV = 13.8 in generator G1 = 220 in lines L1 and L2 = 18 in generator G2 = 22 in generator G3 Convert all reactances to the new bases: G1: X=0.2 x 50/20=0.5 p.u. G2: X=0.2 x 50/30=0.33+ p.u. G3: T1:

X=0.2 x (20/22)2 x 50/30=0.275 p.u. X=0.1 x 50/25=0.2 p.u.

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T2: T3:

X=0.1 x 50/30=0.166+ p.u. X=0.1 p.u.

In lines L1 and L2,

Base Z

2202 50

968

L1:

X=80/968=0.083 p.u.

L1:

X=100/968=0.103 p.u. 0.2

 + EG1

0.5

0.083

 + EG3

0.103

0.375

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0.167

0.333 + 

EG2

Symmetrical Components In a three-phase system, a given set of unbalanced phasors may be replaced by three sets of balanced phasors which are referred to as the symmetrical components of the original unbalanced phasors. The positive-sequence phasors are three-phase balanced and have the same phase sequence as the original set of unbalanced phasors. The negative-sequence phasors are three-phase balanced but having a phase sequence opposite to that of the original set of unbalanced phasors. The zero-sequence phasors are single-phase, equal in magnitude and in the same direction. Department of Electrical and Electronics Engineering

Unbalanced Conditions Causes of unbalanced conditions:

1. Non-transposition of transmission lines 2. Single-phase transformers used in threephase banks are not identical 3. Single-phase loads are not distributed equally among the phases

4. Faults Note: Single-phase analysis cannot be used when the system is unbalanced. Department of Electrical and Electronics Engineering

Assume a sequence abc for the original set of  unbalanced phasors.  Vb2 Vc1  Va1

 Vb1

PositiveSequence (abc)

 Vc2

ZeroSequence

 Va0

 Vb0

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 Va2 NegativeSequence (acb)

 Vc0

The vector sum of the sequence components is shown below to get the phase voltages.    Vc2 Vc1 Va2     Va1 Va Vb Vc   Vb2 Vb1 We get

 Va  Vb  Vc

 Va0  Vb0  Vc0

 Va1  Vb1  Vc1

 Va2  Vb2  Vc2

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Express all sequence components using phase a.

 Va0

 Vc1

 Vb0

 Vc0

 Va1

 Vb1

Note: a

 Vb1  Vc1

1 120o

 a Va1  aVa1 2

 Vb2  Vc2

 Vb2

 aVa2  2 a Va2

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 Vc2

 Va2

Substitution  gives 

Va  Vb  Vc

Va0  Va0  Va0

  Va1 Va2   2 a Va1 aVa2   aVa1 a2 Va2

Using matrix notation, we get

 Va 1 1  2 = V 1 a b Vc 1 a or

 Vabc

1 a a2

 Va0  V  a1 Va2

 A V012

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We can show by matrix inversion that A-1 is

1 1 A = 1 a2 1

1 a

A-1 =

a2

a

1 3

The inverse equations are

 Va0  1 = V 3  a1 Va2 or

 V012

A

1 1

1

1 a

1 a2

1 a2

a

 Vabc

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 Va  V b Vc

1 1

1 a

1 a2

1 a2

a

The transformation also applies to unbalanced currents.

 Ia 1 1  Ib = 1 a2 Ic 1 a

 Ia0  1 = Ia1 3 Ia2 or

 Ia bc

1 1

 Ia0  Ia1 Ia2

1 a a2

1 a

1 a2

1 a2

a

 A I012 and

 Ia  Ib Ic

 I012

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 A Ia bc 1

Example: Determine the symmetrical components of the following unbalanced voltages:

 Va

 Vb

4 0o

We get 

Va0

 Va1

90o

3

 Vc

8 143.1o

   1 (Va Vb Vc ) 3 1 (4 j3 6.4 j4.8) 3 0.8 j0.6 1.0 143.05o    2 1 ( V a V a Vc ) a b 3 1 3

(4

o

3 30

o

8 23.1 )

o

4.9 18.38

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 Va2

 1 (Va 3

 a2 Vb

(4

3 150o

1 3

 aVc ) 8 263.1o )

2.15 86.08o  We also get  Vb0 Va0 1.0 143.05o   2 Vb1 a Va1 4.9 258.38o   Vb2 aVa2 2.15 33.92o   o Vc0 Va0 1.0 143.05   o Vc1 aVa1 4.9 138.38   2 o Vc2 a Va2 2.15 153.92 Department of Electrical and Electronics Engineering

The sequence components are shown below

 Vc1

 Vb2

 Vc2

 Va1 Positive Sequence

 Va2    Va0 Vb0 Vc0

 Vb1

Negative Sequence

Zero Sequence

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The sum of the sequence components is shown below. 

 Vc Vc2

 Va0

 Vb0

 Vc1

 Va1

 Vc0  Vb1

 Va2  Va

 Vb2

 Vb

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Zero-Sequence Current Zero-sequence current cannot flow in or out of a delta connection. In a wye-connection with neutral return, the neutral carries the in-phase zero-sequence currents. a  I0  3I0 n c b

 I0  I0

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Zero-Sequence Current Zero-sequence current can circulate in the delta– connected winding of a transformer. Balancing ampere turns must be satisfied.

 I0

a

 I0  I0

b c

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Example: Single Line-to-Ground Fault

 Ia  Ic

 Given: Ia  Ia0  Ia1  Ia2

 Ib

a

b

c   1500 0o A Ib Ic 0    o 1 ( I I I ) 500 0 A a b c 3    2 o 1 ( I a I a I ) 500 0 A a b c 3    2 o 1 ( I a I a I ) 500 0 A a b c 3

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Example: Line-to-Line Fault

 Ia  Ic

 Given: Ia  Ia0  Ia1  Ia2

0  1 (Ia 3  1 (Ia 3  1 (Ia 3

 Ib

a

b

c   Ib Ic 1500 0o A   Ib Ic ) 0   a Ib a2 Ic ) 866 90o A   a2 Ib a Ic ) 866 90o A

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Sequence Networks When the system is unbalanced, currents and voltages will contain positive-, negative- and zerosequence components. The ratio of a sequence voltage to its corresponding sequence current represents a sequence impedance. Thus, the unbalanced power system can be replaced by three sequence networks. 1. Positive-Sequence Network 2. Negative-Sequence Network 3. Zero-Sequence Network Note: A balanced three-phase system is modeled using the positive-sequence network. Department of Electrical and Electronics Engineering

The Fault Point The system is assumed to be balanced except at the fault point. a b

Line-to-    ground Va Vb Vc voltages

 Ia

 Ib

 Ic

c Fault Currents

Ground

Note: An equivalent network at the fault point can be determined using Thevenin’s theorem. Department of Electrical and Electronics Engineering