Fault In Power Systems

EEP 4113 Fault in Electric Power System

Dr. Mohd Taufiq Ishak

Module Outline

1. Introduction 2. Symmetrical/ Balanced Faults 3. Unsymmetrical/ Unbalanced Faults

EEP 4113 – Advanced Power Systems - Fault in Power System

2

Introduction • Analysis types: power flow - evaluate normal operating conditions fault analysis - evaluate abnormal operating conditions

• Fault analysis is also known as short circuit study. • In normal condition, a power system is operating at balanced 3-phase AC system. • Whenever a fault occurred, the bus voltages and flow of current in the network elements get affected. • Faults can cause over current at certain point of power system. EEP 4113 – Advanced Power Systems - Fault in Power System

3

Introduction • Faults occur in power system due to: insulation failure in the equipments flashover of lines initiated by lightning mechanical damage to conductors and towers accidental faulty operation

• Faults can be classified as Shunt faults and Series faults. Series faults are not so severe as compared to shunt faults.

EEP 4113 – Advanced Power Systems - Fault in Power System

4

Introduction • Fault types: Symmetrical/ balanced faults (3-phase) Unsymmetrical/ unbalanced faults single-line to ground and double-line to ground line-to-line faults

• The relative frequency of occurrence of various faults in the order of severity are as follows: balanced 3-phase fault double line to ground fault line to line fault single line to ground fault

5% 10% 15% 70%

EEP 4113 – Advanced Power Systems - Fault in Power System

5

Introduction • When a fault occurs in a power system, bus voltages reduces and large current flows in the lines. • This may cause damage to the equipments. • The magnitude of the fault currents depends on: the impedance of the network the internal impedances of the generators the resistance of the fault (arc resistance)

EEP 4113 – Advanced Power Systems - Fault in Power System

6

Introduction • Faulty section should be isolated from the rest of the network immediately. • This can be achieved by providing relays and circuit breakers. • The protective relays sense the occurrence of the fault and send signals to circuit breakers to open the circuit under faulty condition. • Proper relay setting and relay coordination are required for effective protection.

EEP 4113 – Advanced Power Systems - Fault in Power System

7

Introduction • The main purposes of fault analysis: specifying ratings for circuit breakers and fuses protective relay settings specifying the impedance of transformers and generators

• Network impedances are governed by generator impedances transformer connections and impedances transmission line impedances Load impedances grounding connections and resistances EEP 4113 – Advanced Power Systems - Fault in Power System

8

Subtransient and transient • Generator behavior is divided into three periods sub-transient period, lasting for the first few cycles transient period, covering a relatively longer time steady state period

• Sub-transient reactances, XG = Xd” determine the interrupting capacity of HV circuit breakers determine the operation timing of the protective relay system for high-voltage networks

• Transient reactances, XG = Xd’ determine the interrupting capacity of MV circuit breakers determine the operation timing of the protective relay system for medium-voltage networks EEP 4113 – Advanced Power Systems - Fault in Power System

9

Symmetrical Three-phase Fault • The balanced fault is a phenomenon where the three phases are short circuited simultaneously. • Since the network is balanced, it is solve on per phase basis. • A fault represents a structural network change equivalent to the addition of an impedance at the place of the fault if the fault impedance is zero, the fault is referred to as a bolted fault or solid fault

• For small networks, it can be solved by the Thévenin’s method, and for large networks, it is solved by the Bus Impedance Matrix method. EEP 4113 – Advanced Power Systems - Fault in Power System

10

Balanced Fault – Thévenin’s Method • Assumptions: generators are modeled as an emf behind the subtransient or transient reactance network resistances are neglected shunt capacitances are neglected system is considered as having no-load

• The fault is simulated by switching a fault impedance at the faulted bus. • The change in the network voltages is equivalent to adding the pre-fault bus voltage with all other sources short circuited. EEP 4113 – Advanced Power Systems - Fault in Power System

11

Balanced Fault – Thévenin’s Method • Example 1:

G1

G2

EEP 4113 – Advanced Power Systems - Fault in Power System

12

Balanced Fault – Thévenin’s Method • 3-phase fault with fault impedance, Zf = j0.16 on bus 3.

EEP 4113 – Advanced Power Systems - Fault in Power System

13

Balanced Fault – Thévenin’s Method • Convert the impedances of Z12, Z23, and Z31 from  to Y.

EEP 4113 – Advanced Power Systems - Fault in Power System

14

Balanced Fault – Thévenin’s Method • Combine the parallel branches to get the Thévenin’s impedance at faulted bus, i.e. Z33.

EEP 4113 – Advanced Power Systems - Fault in Power System

15

Balanced Fault – Thévenin’s Method • Combine the parallel branches to get the Thévenin’s impedance at faulted bus, i.e. Z33. V3 (0) I3 (F )  Z 33  Z f V1 (0)  V2 (0)  V3 (0)  1.0 pu

V3 (0) 1.0 I3 (F )     j 2.0 Z 33  Z f j 0.34  j 0.16

EEP 4113 – Advanced Power Systems - Fault in Power System

16

Balanced Fault – Thévenin’s Method • The current flow for the two generators: j 0.6 I G1  I 3 ( F )   j1.2 pu j 0.4  j 0.6 IG2

j 0.4  I 3 ( F )   j 0.8 pu j 0.4  j 0.6

• Bus voltage changes, V1  0  ( j 0.2)(  j1.2)  0.24 pu V2  0  ( j 0.4)(  j 0.8)  0.32 pu V3  ( j 0.16)(  j 2)  1.0  0.68 pu EEP 4113 – Advanced Power Systems - Fault in Power System

17

Balanced Fault – Thévenin’s Method • The fault voltages: V1 ( F )  V1 (0)  V1  1.0  0.24  0.76 pu V2 ( F )  V2 (0)  V2  1.0  0.32  0.68 pu V3 ( F )  V3 (0)  V3  1.0  0.68  0.32 pu

• The short circuit currents in the lines: V1 ( F )  V2 ( F ) 0.76  0.68 I12 ( F )     j 0.1 pu z12 j 0.8 V1 ( F )  V3 ( F ) 0.76  0.32 I13 ( F )     j1.1 pu z13 j 0.4 V2 ( F )  V3 ( F ) 0.68  0.32 I 23 ( F )     j 0.9 pu z23 j 0.4 EEP 4113 – Advanced Power Systems - Fault in Power System

18

Balanced Fault – Thévenin’s Method • For more accurate solutions use the pre-fault bus voltages which can be obtained from the results of a power flow solution include loads - to preserve linearity, convert loads to constant impedance model Thévenin’s theorem allows the changes in the bus voltages to be obtained bus voltages are obtained by superposition of the prefault voltages and the changes in the bus voltages current in each branch can be solved

EEP 4113 – Advanced Power Systems - Fault in Power System

19

Balanced Fault – Short Circuit Capacity • Measures the electrical strength of the bus • Stated in MVA. • Determines the dimension of busbars and the interrupting capacity of circuit breakers • Definition: SCC  3VLL,k (0) I k ( F ) x 103

where VLL ,k (0) is the line-to-line voltage at bus k in kV I k (F ) is the faulted current in A EEP 4113 – Advanced Power Systems - Fault in Power System

20

Balanced Fault – Short Circuit Capacity • In per unit (pu), I k ( F ) pu

Vk (0)  X kk

• The approximate formula for SCC or short-circuit MVA: SCC 

SB MVA X kk

where SB is the base MVA and Xkk is the p.u. reactance to the point of fault. EEP 4113 – Advanced Power Systems - Fault in Power System

21

Balanced Fault – Short Circuit Capacity • Example 2 of SCC calculation: Z 33  j 0.34 S B  100MVA S B 100MVA SCC3    294MVA Z 33 0.34

EEP 4113 – Advanced Power Systems - Fault in Power System

22

Balanced Fault – Bus Impedance Matrix • Network reduction by Thévenin’s method is not efficient and not applicable to large networks. • Consider the following system: operating under balanced conditions each generator represented by a constant emf behind a proper reactance (Xd, Xd’, or Xd”) lines represented by their equivalent  model EEP 4113 – Advanced Power Systems - Fault in Power System

23

Balanced Fault – Bus Impedance Matrix • Place the pre-fault voltages into a vector: V1 (0)       Vbus (0)  Vk (0)      Vn (0)

• Replace the loads by a constant impedance model using the pre-fault bus voltages: Z iL 

Vi (0)

2

S L*

EEP 4113 – Advanced Power Systems - Fault in Power System

24

Balanced Fault – Bus Impedance Matrix • The change in the network voltage caused by the fault is equivalent to placing a fault voltage at the faulted bus with all the other sources short-circuited:

EEP 4113 – Advanced Power Systems - Fault in Power System

25

Balanced Fault – Bus Impedance Matrix • Using superposition, the fault voltages are calculated from the pre-fault voltages by adding the change in bus voltages due to the fault: Vbus ( F )  Vbus (0)  Vbus

where

 V1       Vbus  Vk       Vn  EEP 4113 – Advanced Power Systems - Fault in Power System

26

Balanced Fault – Bus Impedance Matrix • The injected bus currents: I bus  YbusVbus

• Current entering every bus is zero except at the faulted bus. Since the current is leaving the bus k, it is taken as negative current.  0  Y11          I k ( F )  Yk1          0  Yn1

 Y1k 



 Ykk 



 Ynk

 Y1n   V1         Ykn  Vk          Ynn  Vn 

=

EEP 4113 – Advanced Power Systems - Fault in Power System

I bus ( F )  YbusVbus

27

Balanced Fault – Bus Impedance Matrix • Solving for Vbus, Vbus  ZbusI bus (F )

where Zbus = Y-1bus is the bus impedance matrix. Vbus ( F )  Vbus (0)  ZbusI bus ( F )

V1 ( F )  V1 (0)   Z11              Vk ( F )  Vk (0)   Z k1              Vn ( F ) Vn (0)  Z n1

 Z1k 



 Z kk 



 Z nk

 Z1n   0           Z kn   I k ( F )         Z nn   0 

EEP 4113 – Advanced Power Systems - Fault in Power System

28

Balanced Fault – Bus Impedance Matrix • Since there is only one nonzero element in the current vector, Vk ( F )  Vk (0)  Z kk I k ( F )

• From the Thévenin’s circuit, Vk ( F )  Z f I k ( F )

• Combining the both equations, the fault current becomes, Vk (0) Ik (F )  Z kk  Z f

EEP 4113 – Advanced Power Systems - Fault in Power System

29

Balanced Fault – Bus Impedance Matrix • For ith equation, Vi ( F )  Vi (0)  Zik I k ( F )

• Substituting Ik(F), Z ik Vi ( F )  Vi (0)  Vk (0) Z kk  Z f

• For fault current in the line, I ij ( F ) 

Vi ( F )  V j ( F ) zij

Small letter of z, it is taken from the original network and not from the Zbus!

EEP 4113 – Advanced Power Systems - Fault in Power System

30

Balanced Fault – Bus Impedance Matrix • Example 3: Using the previous example, Zf = j0.16 pu. The bus admittance matrix, j 2.5   j8.75 j1.25 Ybus   j1.25  j 6.25 j 2.5   j 2.5 j 2.5  j5.0

The impedance matrix,  j 0.16 1 Z bus  Ybus   j 0.08  j 0.12

j 0.08 j 0.24 j 0.16

j 0.12 j 0.16 j 0.34

EEP 4113 – Advanced Power Systems - Fault in Power System

31

Balanced Fault – Bus Impedance Matrix • The fault current, V3 (0) 1.0 I3 (F )     j 2.0 pu Z 33  Z f j 0.34  j 0.16

• The bus voltages during fault, V1 ( F )  V1 (0)  Z13I 3 ( F )  1.0  ( j 0.12)(  j 2.0)  0.76 pu V2 ( F )  V2 (0)  Z 23I 3 ( F )  1.0  ( j 0.16)(  j 2.0)  0.68 pu V3 ( F )  V3 (0)  Z 33I 3 ( F )  1.0  ( j 0.34)(  j 2.0)  0.32 pu

EEP 4113 – Advanced Power Systems - Fault in Power System

32

Balanced Fault – Bus Impedance Matrix • The fault currents in the lines, V1 ( F )  V2 ( F ) 0.76  0.68 I12 ( F )     j 0.1 pu z12 j 0.8 V1 ( F )  V3 ( F ) 0.76  0.32 I13 ( F )     j1.1 pu z13 j 0.4 I 23 ( F ) 

V2 ( F )  V3 ( F ) 0.68  0.32    j 0.9 pu z23 j 0.4

EEP 4113 – Advanced Power Systems - Fault in Power System

33

Unsymmetrical Fault • Unsymmetrical faults are analyzed using symmetrical components. • The symmetrical components method allows unbalanced three-phase phasor quantities to be replaced by the sum of three separate but balanced symmetrical components. • The balanced 3-phase phasors consists of: positive sequence negative sequence zero sequence EEP 4113 – Advanced Power Systems - Fault in Power System

34

Unsymmetrical Fault – Sym. Components • Representation of symmetrical components: I c1

I b2 120°

120° 120°

I a0 0 I bI 0

120°

I a1

I b1

120° 120°

I a2

c

I c2

+ve Sequence A balanced 3-phase system in the normal a-b-c sequence

-ve Sequence A balanced 3-phase system of reversed sequence a-c-b sequence

Zero Sequence Three phasors equal in magnitude and phase revolving in the +ve phase rotation

EEP 4113 – Advanced Power Systems - Fault in Power System

35

Unsymmetrical Fault – Sym. Components • Positive sequence phasors: I a1  I a1 0  I a1 I b1  I a1 240  a 2 I a1 I c1  I a1 120  aI a1

• Operator a identities a  1120  0.5  j 0.866 a 2  1240  0.5  j 0.866 a 3  1360  1  j 0 1 a  a2  0 EEP 4113 – Advanced Power Systems - Fault in Power System

36

Unsymmetrical Fault – Sym. Components • Negative sequence phasors: I a2  I a2 0  I a2 I b2  I a2 120  aI a2 I c2  I a2 240  a 2 I a2

• Zero sequence phasors I a0  I a0 0  I a0 I b0  I a0 0  I a0 I c0  I a0 0  I a0

EEP 4113 – Advanced Power Systems - Fault in Power System

37

Unsymmetrical Fault – Sym. Components • Relating unbalanced phasors to symmetrical components: I a  I a0  I a1  I a2  I a0  I a1  I a2 I b  I b0  I b1  I b2  I a0  a 2 I a1  aI a2 I c  I c0  I c1  I c2  I a0  aI a1  a 2 I a2

• In matrix notation  I a  1 1  I   1 a 2  b   I c  1 a

1   I a0   1  a  Ia  a 2   I a2 

EEP 4113 – Advanced Power Systems - Fault in Power System

38

Unsymmetrical Fault – Sym. Components • [A] is known as the symmetrical components transformation matrix: 1 1 A  1 a 2 1 a

I abc  AI a012

1 a  a 2 

• Solving for the symmetrical components leads to: I a012  A1I abc

1 1 1 1 A  1 a 3 1 a 2

1 1 * 2 a  A 3 a 

EEP 4113 – Advanced Power Systems - Fault in Power System

39

Unsymmetrical Fault – Sym. Components • In component form, the calculation for symmetrical components are: 1  I a  I b  I c  3 1 I a1  I a  aI b  a 2 I c 3 1 2 I a  I a  a 2 I b  aI c 3

I a0









• Similar expressions exist for voltages: V abc  AVa012 Va012  A1V abc EEP 4113 – Advanced Power Systems - Fault in Power System

40

Unsymmetrical Fault – Sym. Components • The apparent power (S) may also be expressed in terms of symmetrical components: S (3 )  V S (3 ) 

abcT

I



abc*



012 T AVa



012 * AI a

012T  Va

T

A

* 012* A Ia

Since AT  A, then AT A*  3



S (3 )  3 V

012T

I

012*



0 0*  3Va I a

1 1*  3Va I a

2 2*  3Va I a

EEP 4113 – Advanced Power Systems - Fault in Power System

41

Unsymmetrical Fault – Sym. Components • Example 4: Obtain the symmetrical components of a set of unbalanced currents: I a  1.625, I b  1.0180, I c  0.9132

Solution: 1.625  1.0180  0.9132   0.4596.5 3 2 1 . 6  25   a ( 1 . 0  180  )  a (0.9132) 1 Ia   0.94  0.1 3 2 1 . 6  25   a (1.0180)  a(0.9132) 2 Ia   0.6022.3 3

I a0

EEP 4113 – Advanced Power Systems - Fault in Power System

42

Unsymmetrical Fault – Sym. Components • Solution: Ic

I a0 I b0 I c0

Ia

1320

96.50

250

Ib

a-b-c set

Zero-sequence set

I c1

I b2

-0.1

22.30

I a1 0

I b1 Positive-sequence set

I a2

I c2 Negative-sequence set

EEP 4113 – Advanced Power Systems - Fault in Power System

43

Unsymmetrical Fault – Sym. Components • Example 5: The symmetrical components of a set of unbalanced voltages are Va0  0.690, Va1  1.030, Va2  0.8  30

Obtain the original unbalanced voltages. Solution:

Va  0.690  1.030  0.8  30  1.708824.2 Vb  0.690  a 2 1.030  a0.8  30  0.490

Vc  0.690  a1.030  a 2 0.8  30  1.7088155.8

EEP 4113 – Advanced Power Systems - Fault in Power System

44

Unsymmetrical Fault – Sym. Components • Solution: Vc

Vb

Va0 Vb0 Vc0

Va 900 24.20

a-b-c set

Vc1

Zero-sequence set

Vb2

Va1

Vb1

Positive-sequence set

Vc2

Va2 Negative-sequence set

EEP 4113 – Advanced Power Systems - Fault in Power System

45

Unsymm. Fault – Sequence Impedances • The impedance offered to the flow of a sequence current creating sequence voltages: positive sequence impedance, Z1 negative sequence impedance, Z2 zero sequence impedance, Z0

• Augmented network models: wye-connected balanced loads transmission line 3-phase transformers generators

EEP 4113 – Advanced Power Systems - Fault in Power System

46

Unsymm. Fault – Sequence Impedances • Balanced Y-connected load

Line-to-ground voltages Va  Z s I a  Z m I b  Z m I c  Z n I n Vb  Z m I a  Z s I b  Z m I c  Z n I n Vc  Z m I a  Z m I b  Z s I c  Z n I n

Neutral current I n  I a  Ib  Ic

EEP 4113 – Advanced Power Systems - Fault in Power System

47

Unsymm. Fault – Sequence Impedances • Balanced Y-connected load Rewriting in matrix form, Va   Z s  Z n V   Z  Z n  b  m Vc  Z m  Z n

Zm  Zn Zs  Zn Zm  Zn

Zm  Zn  I a  Z m  Z n   I b  Z s  Z n   I c 

V abc  Z abcI abc

Writing Vabc and Iabc in symmetrical components, AVa012  Z abc AI a012

Multiplying by A-1, Va012  A1Z abc AI a012  Z 012I a012 where Z 012  A1Z abc A EEP 4113 – Advanced Power Systems - Fault in Power System

48

Unsymm. Fault – Sequence Impedances • Balanced Y-connected load 1 1 1 012 Z  1 a 3 1 a 2

1  Zs  Zn a 2   Z m  Z n a   Z m  Z n

Zm  Zn Zs  Zn Zm  Zn

Z m  Z n  1 1 Z m  Z n  1 a 2 Z s  Z n  1 a

1 a  a 2 

0 0   Z s  3Z n  2Z m   0 Zs  Zm 0   0 0 Z s  Z m   Z s  3Z n 0 0    0 Z s 0  when Z m  0  0 0 Z s  EEP 4113 – Advanced Power Systems - Fault in Power System

49

Unsymm. Fault – Sequence Impedances • Transmission Lines

A

B

Positive B Sequence

C

C

A

EEP 4113 – Advanced Power Systems - Fault in Power System

Negative Sequence

50

Unsymm. Fault – Sequence Impedances • Transmission Lines Va1

Va1  Z s I a  Z n I n  Va2

Va2

Vb1

Vb1  Z s I b  Z n I n  Vb2

Vb2

Vc1

Vc2

Vn  0  Z n I n I n  I a  Ib  Ic  0

Vn Va1   Z s  Z n  1  Vb    Z n Vc1   Z n   

Vc1  Z s I c  Z n I n  Vc2

Zn Zs  Zn Zn

2   I a  Va   2    Z n   I b   Vb  Z s  Z n   I c  Vc2 

Zn

EEP 4113 – Advanced Power Systems - Fault in Power System

51

Unsymm. Fault – Sequence Impedances • Transmission Lines 1 2 Vabc  Z abcI abc  Z abc 1 2 AV012  Z abc AI 012  AV012

Z 012  A1Z abc A 1 1 1  1 a 3 1 a 2  Z s  3Z n   0  0

1  Z s  Z n a 2   Z n a   Z n 0 Zs 0

Zn Zs  Zn Zn

Z n  1 1 Z n  1 a 2 Z s  Z n  1 a

1 a  a 2 

0 0  Z s 

EEP 4113 – Advanced Power Systems - Fault in Power System

52

Unsymm. Fault – Sequence Impedances • Synchronous Generators Similar modeling of impedances to sequence impedances Typical values for common generators remember that the transient fault impedance is a function of time positive sequence values are the same as Xd, Xd’, and Xd” negative sequence values are affected by the rotation of the rotor X2  Xd” zero sequence values are isolated from the airgap of the machine. the zero sequence reactance is approximated to the leakage reactance X0 XL

EEP 4113 – Advanced Power Systems - Fault in Power System

53

Unsymm. Fault – Sequence Impedances • Synchronous Generators Zero Sequence

Positive Sequence

Negative Sequence

X0

X1

X2

VT0

VT1

VT2

EEP 4113 – Advanced Power Systems - Fault in Power System

54

Unsymm. Fault – Sequence Impedances • Loaded Synchronous Generators

EEP 4113 – Advanced Power Systems - Fault in Power System

55

Unsymm. Fault – Sequence Impedances • Loaded Synchronous Generators  Z s  3Z n Z 012   0  0 Va0   1 Va  Va2   

0  Z 0   0 0 Z s   0

0 Zs 0

 0  Z      Ea    0  0   0 

0

0 Z1 0

0 Z1 0

0  0 Z 2 

0   I a0   1  0  Ia  Z 2   I a2 

EEP 4113 – Advanced Power Systems - Fault in Power System

Z0

Z1

Z2

Va0

Va1

Va2 56

Unsymm. Fault – Sequence Impedances • Transformers • Series Leakage Impedance the magnetization current and core losses represented by the shunt branch are neglected (they represent only 1% of the total load current) the transformer is modeled with the equivalent series leakage impedance

• Three single-phase units & five-legged core three-phase units the series leakage impedance is the same for all the sequences

Z 0  Z1  Z 2  ZL • Three-legged core three-phase units the series leakage impedance is the same for the positive and negative sequence only

Z1  Z 2  ZL EEP 4113 – Advanced Power Systems - Fault in Power System

57

Unsymm. Fault – Sequence Impedances • Transformers • the positive sequence line voltage on the HV side leads the corresponding line voltage on the LV side by 30° • consequently, for the negative sequence voltages the corresponding phase shift is -30° • Zero-sequence network connections of the transformer depends on the winding connection

EEP 4113 – Advanced Power Systems - Fault in Power System

58

Unsymm. Fault – Sequence Impedances • Transformers • Representation circuits for Zero-sequence:

EEP 4113 – Advanced Power Systems - Fault in Power System

59

Unsymm. Fault – Sequence Impedances • Transformers • Representation circuits for Zero-sequence:

EEP 4113 – Advanced Power Systems - Fault in Power System

60

Unsymm. Fault – Single Line-to-ground Fault

Va  Z f I a Ib  Ic  0  I a0  1 1  1 1  I a   3 1 a  I a2  1 a 2   I a0



I a1

EEP 4113 – Advanced Power Systems - Fault in Power System



I a2

1  I a  a 2   0  a   0 

Ia  3

61

Unsymm. Fault – Single Line-to-ground Fault

Since,

Va0   0   Z 0  1    Va    Ea    0 Va2   0   0     

Va0  0  Z 0 I a0 Va1  Ea  Z 1I a1 Va2  0  Z 2 I a2

0 Z1 0

0   I a0   1  0  Ia  Z 2   I a2 

Va  Va  V  Va  Ea  (Z 1  Z 2  Z 0 ) I a0 0

1 a

2

The fault current, 3Z f I a0  Ea  ( Z 1  Z 2  Z 0 ) I a0 I f  Ia 

3I a0

3 Ea  1 Z  Z 2  Z 0  3Z f

EEP 4113 – Advanced Power Systems - Fault in Power System

62

Unsymm. Fault – Line-to-line Fault

Vb  Vc  Z f I b Ib  Ic  0 Ia  0  I a0  1 1  1 1  I a   3 1 a  I a2  1 a 2  

EEP 4113 – Advanced Power Systems - Fault in Power System

1  0  a 2   I b  a   I b 

63

Unsymm. Fault – Line-to-line Fault I a1   I a2 Vb  Va0  a 2Va1  aVa2

 I a  1 1  I   1 a 2  b   I c  1 a

Vc  Va0  aVa1  a 2Va2

The fault current,

Vb  Vc  (a

Ib  I c  (a 2  a) I a1   j 3I a1

Va  Va0  Va1  Va2

2

 a)(Va1

 Va2 )

1  0  a   I a1  a 2   I a1 

 Z f Ib

I a1

Ea  1 Z  Z2  Zf

EEP 4113 – Advanced Power Systems - Fault in Power System

64

Unsymm. Fault – Double Line-to-ground Fault Vb  Vc  Z f ( I b  I c )

I a  I a0  I a1  I a2  0 From previous section, Vb  Va0  a 2Va1  aVa2 Vc  Va0  aVa1  a 2Va2

Since Vb  Vc

Va1  Va2 Vb  Z f ( I a0  a 2 I a1  aI a2  I a0  aI a1  a 2 I a2 )  Z f (2 I a0  I a1  I a2 )  3Z f I a0 EEP 4113 – Advanced Power Systems - Fault in Power System

65

Unsymm. Fault – Double Line-to-ground Fault

I a0

Va0  Va1 Ea  Z 1I a1   0 3Z f Z  3Z f

Va0  0  Z 0 I a0

Since Va1  Ea  Z 1I a1 Va2  0  Z 2 I a2

1 1 E  Z Ia 2 a Ia   Z2

I a1

 Z  1

Ea Z 2 ( Z 0  3Z f ) Z 2  Z 0  3Z f

I f  I b  I c  3I a0

By using I a0  I a1  I a2  0

Fault current

EEP 4113 – Advanced Power Systems - Fault in Power System

66

Conclusion Sequence Impedance Circuit:

EEP 4113 – Advanced Power Systems - Fault in Power System

67

Example

• The neutral of each generator is grounded through a current limiting resistor of 25/3 % on a 100 MVA base • All network data is expressed on a 100 MVA base • Find the fault current for 3-phs, 1-phs, L-L, L-L-G bolted faults at bus 3 through a fault impedance Zf = j0.1 pu. EEP 4113 – Advanced Power Systems - Fault in Power System

68

Example

• Positive sequence impedance network:



Y

EEP 4113 – Advanced Power Systems - Fault in Power System

69

Example

• Negative sequence impedance network:



Y

EEP 4113 – Advanced Power Systems - Fault in Power System

70

Example

• Zero sequence impedance network:

Open circuited



Open circuited

Y

EEP 4113 – Advanced Power Systems - Fault in Power System

71

Example • 3-phase fault V3a( 0)

1.0 I (F )  1    j3.125 pu  902.11  90 Z 33  Z f j 0.22  j 0.1 a 3

• Single Line-to-ground fault a V 0 1 2 3 (0) I3  I3  I3  1   j 0.9174 pu 2 0 Z 33  Z 33  Z 33  3Z f

Fault current

 I 3a  1 1  b  2 I  1 a  3   I 3c  1 a   

1   I 30  3I 30   j 2.7523  0      pu a  I3    0    0   a 2   I 30   0   0

EEP 4113 – Advanced Power Systems - Fault in Power System

72

Example

• Line-to-line fault I 30  0 I 31   I 32 

V3a( 0) 1 Z 33



2 Z 33

Zf

  j1.8519 pu

Fault current  I 3a  1 1 1   0   0   b    j1.8519   3.2075 pu 2 I  1 a a  3       I 3c  1 a a 2   j1.8519   3.2075        

EEP 4113 – Advanced Power Systems - Fault in Power System

73

Example

• Double Line-to-ground fault Positive sequence component

V3a( 0)

I 31  1 Z 33 

2 0 Z 33 ( Z 33  3Z f ) 2 0 Z 33  Z 33  3Z f

  j 2.6017 pu

Negative sequence component

I 32



1 1 V3a(0)  Z 33 I3 2 Z 33

1  ( j 0.22)(  j 2.6017)   j1.9438 pu j 0.22

EEP 4113 – Advanced Power Systems - Fault in Power System

74

Example

• Double Line-to-ground fault Zero sequence component I 30  

1 1 V3a( 0)  Z 33 I3 0 Z 33

 3Z f

 j 0.6579 pu

Phase currents  I 3a  1 1 1   j 0.6579   0   b    j 2.6017  4.058165.93 pu 2 I  1 a a  3       I 3c  1 a a 2   j1.9438   4.05814.07         Fault current I3 ( F )  I3b  I3c  1.973290 EEP 4113 – Advanced Power Systems - Fault in Power System

75