3 Unsymmetrical Faults Eet308 Updated Sv

7/30/2019 3 Unsymmetrical Faults Eet308 Updated Sv

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CHAPTER 5

UNSYMMETRICALFAULT

REVISED BY : PN ZETTY NURAZLINDA ZAKARIAPREPARED BY: EN NASRUL HELMEI BIN HALIM

SCHOOL OF ELECTRICAL SYSTEM ENGINEERINGUniMAP


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IntroductionMost of the faults occurs on power systems areunsymmetrical faults, which may consist of unsymmetrical short circuits, unsymmetrical faultsthrough impedance, or open conductors.

Unsymmetrical faults occur as:Single line-to-ground faultsLine-to-line-faultsDouble line-to-ground faults

The path of the fault current from line to line or line toground may or may not contain impedance.


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Introduction (cont..)

One or two open conductors result inunsymmetrical fault, through either the breakingof one or two conductors e.g. the breaking of fuse

Since unsymmetrical fault causes unbalancedcurrents to flow in the system, the method of

symmetrical components is very useful in ananalysis to determine the currents and voltagesin the system after the occurrence of the fault.


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Unsymmetrical Faults in Power

SystemsConsider a general power network in Fig. 1.

It is assumed that a faultoccurs at point F in thesystem, as a result of which currents I a , I b , I c

flow out of the system.V a , V b , V c are voltages of line a, b, c with respect toground. Fig. 1


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Systems (cont..)

Thevenin equivalents circuit : (a) zero, (b) positive and(c) negative sequence.

+

-

E a V a1

Z 1 I a1 +

-

V a2

Z 2 I a2 +

-

V a0

Z 0 I a0

Fig. 2

(a) (b) (c)


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Systems (cont..)The positive sequence voltages of

synchronous machines will be identicaland will equal the pre-fault voltage at faultpoint, F .

This voltage labeled as E a (refer Fig 2(b)).


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Systems (cont..)The sequence/symmetrical components of voltages at faultpoint can be expressed in terms of sequence currents andThevenin sequence impedance as (refer Fig 2):

2

1

0

2

1

0

2

1

0

0000

00

0

0

a

a

a

a

a

a

a

I I

I

Z Z

Z

E V V

V


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And the phase voltages during fault;

2

1

0

2

2

1

1

111

a

a

a

c

b

a

V

V

V

aa

aa

V

V

V


Systems (cont..)

1201

1201

;where

2a

a


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Systems (cont..)The sequence/symmetrical components of currents can beexpressed in terms of phase currents:-

c

b

a

a

a

a

I

I

I

aa

aa

I

I

I

2

2

2

1

0

1

1

111

31

1201

1201

;where

2a

a


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Systems (cont..)The phase currents are;

2

1

0

2

2

1

1111

a

a

a

c

b

a

I

I I

aa

aa

I

I I

1201

1201

;where

2

a

a


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How to do fault analysis???Single-line-Ground Fault

Line-Line FaultDouble-Line-Ground Fault

What should we analyse???Component currents & voltagesPhase currents & voltages during fault


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Single Line-to-Ground Faults

Caused by lightning or by conductorsmaking contact with grounded structures.

Fig. 3 shows a line-to-ground fault at F ina power system through a fault impedanceZ F .

The phases are so labeled that the faultoccurs on phase a .

REMEMBER : ZF Zn


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At the fault point F , the currents out of the power systemand the line to ground voltages are constrained as follows:

Fig. 3

I b = 0 I c = 0 V a = I a Z f

i f Z f = 0; V a = 0


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The symmetrical components of the fault currents are:

00

11

111

31

2

2

2

1

0 a

a

a

a I

aaaa

I I

I

From which it is easy to see that

aaaa I I I I 31

021


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Fig 4: Connection of sequence network for a single line-to-ground fault

V a = 3I a1 Z F = I a Z F

3 Z F

aaaa I I I I 31

021


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Phase a voltage in terms of symmetrical components is

1210

210

)( aa

aaaa

I Z Z Z E

V V V V

From the Thevenin equivalent sequence network;

f

aa Z Z Z Z

E I

30211 or

0021

1

f

aa

Z if

Z Z Z E

I


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2

1

0

2

2

1

1

111

a

a

a

c

b

a

I

I

I

aa

aa

I

I

I

The fault currents is

1

021

3 a

aaa

a F

I

I I I

I I


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case: syn. generator

Fig. 6


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Exercise 1 A 100 MVA, 13. 8 kV, wye connected synchronousgenerator has a subtransient reactance X of 0.2 pu, anegative reactance X 2 of 0.25 pu, a zero sequencereactance of 0.1 pu and negligible resistance as in Fig. 6.

The neutral of generator is solidly grounded. Assume thatthe generator is initially unloaded and operating at ratedvoltage, and that a single line-to-ground fault occurs onphase a at the terminals of generator.

Evaluate the voltages and currents in each phase duringthe subtransient period immediately after the fault occurs.

Also, determine the line-to-line voltages at that time.Base : 100 MVA, 13.8 kV


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Exercise 1: ANSWERS

Line currentsIa = 5.45590 o = 22.82 kA

Ib = 0Ic = 0

Line VoltagesVa = 0Vb = 0.983 -106 o

Vc = 0.983+106 o

Line-to-line voltagesVab = Va Vb = 0.98674 o

Vbc = Vb Vc = 1.890 -90o

Vca = Vc Va = 0.986 106 o


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Example 5.1The one-line diagram of a simple power system is shown in Fig. 5.1. Theneutral of each generator is grounded through a current-limiting reactor of 0.25/3 per-unit on a 100 MVA base. The system data expressed in per-unit ona common 100 MVA base is tabulated below. The generators are running onno-load at their rated voltage and rated frequency with their rated e.m.f. inphase.

Fig. 5.1


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Example 5.1 (cont..)

Item BaseMVA

VoltageRating

X 1 X 2 X 0

G 1 100 20kV 0.15 0.15 0.05G 2 100 20kV 0.15 0.15 0.05T 1 100 20/220kV 0.10 0.10 0.10

T 2 100 20/220kV 0.10 0.10 0.10L12 100 220kV 0.125 0.125 0.30L13 100 220kV 0.15 0.15 0.35

L23 100 220kV 0.25 0.25 0.7125


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Example 5.1 (cont..)

Determine:a) The fault current when a single line-to-

ground fault occurs at bus 3 through afault impedance Z F = j0.1p.u.b) Phase Voltages at all buses during fault

at bus 3


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Solution:

Step 1:Draw the positive, negative and zero sequence

networks of the system.

Step 2:

Reduce the networks to their Thevenin equivalentcircuit viewed from the faulted bus (bus 3)


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Solution (cont..):

Positive sequence network


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Solution (cont..):

Using to Y conversion


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Solution (cont..):

0357.0

25.015.0125.0)15.0)(125.0(

1

j

j j j j j

Z s

0595.0

25.015.0125.0)25.0)(125.0(

2

j

j j j j j

Z s

0714.0

25.015.0125.0)25.0)(15.0(

3

j

j j j j j

Z s


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Solution (cont..):

Negative sequence network


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Solution (cont..):

Negative sequence network

22.0

0714.03095.02857.0

)3095.0)(2857.0(2

j

j j j j j

Z


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Solution (cont..):

Zero sequence network


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Solution (cont..):

0771.0

7125.035.03.0)35.0)(3.0(

1

j

j j j j j

Z s

183.0

25.015.0125.0)7125.0)(35.0(

3

j

j j j j j

Z s

1569.0

7125.035.03.0)7125.0)(3.0(

2

j

j j j j j

Z s


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Solution (cont..):

Zero sequence network

35.0

183.02569.04771.0

)2569.0)(4771.0(0

j

j j j j j

Z


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Solution (cont..):

The symmetrical components of the faultcurrents are:

p.u.9174.0)1.0(335.022.022.0

0.1

3021021

j j j j j

Z Z Z Z E

I I I f

aaaa


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Solution (cont..):The fault current,

0

0

7523.2

9174.0

9174.0

9174.0

1

1

1111

1

111

2

2

2

1

0

2

2

j

j

j

j

aa

aa

I

I

I

aa

aa

I

I

I

a

a

a

c

b

a

..7523.2 u p j I I a F

A

kV

MVA I F

29.722

7523.243.262

7523.22203

100


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Solution (cont..):

The symmetrical components of bus voltageduring fault are

2018.0

7982.0

3211.0

)9174.0(22.000

0)9174.0(22.00

00)9174.0(35.0

0

0.1

0

00

0000

0

0

2

1

0

2

1

0

2

1

0

j j

j j

j j

I

I

I

Z

Z Z

E

V

V

V

a

a

a

a

a

a

a


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Solution (cont..):Voltage at bus 3 during fault

56.1250647.1

56.1250647.1

02752.02018.0

7982.0

3211.0

1

1

111

1

1

111

2

2

2

1

0

2

2

aa

aa

V

V

V

aa

aa

V

V

V

a

a

a

c

b

a

kV234.234

kV2200647.1

p.u.56.1250647.1

kV234.234

kV2200647.1

p.u.56.1250647.1

60.544kVkV2202752.0

p.u.02752.0

c

c

b

b

a

a

V

V

V

V

V

V


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Line-to-Line FaultsFig. 7 shows a line-to-linefault at F in a power systemthrough a fault impedanceZ F . The phases can alwaysbe relabeled, such that thefault is on phases b and c.

The currents and the voltages at the fault can be expressed

as

I a = 0 I b = - I c V b - V c = I b Z F

i f Z F = 0; V b = V c

Fig. 7Z F


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Line-to-Line Faults

The symmetrical components of the fault currents are:

b

b

a

a

a

I I

aaaa

I I

I 0

11

111

31 22

2

1

0

From which it is easy to see that

120 ;0 aaa I I I


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Line-to-Line Faults

Fig. 8: Connection of sequence networks for a line-to-line faults

Z F

12 aa I I V a = I a1 Z F = -I a2 Z F


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Line-to-Line Faults

Phase a voltage in terms of symmetrical components is

2111

21

Z I Z I E

V V V

aaa

aaa

From the Thevenin equivalent sequence network;

f

aa Z Z Z

E I

211 or

021

1

f

aa

Z if

Z Z E

I


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Line-to-Line Faults

2

1

2

2

0

1

1

111

a

a

c

b

a

I

I

aa

aa

I

I

I

Hence, the fault currents is;

Since the phase currents are as follows;

since I a0 = 0

cb F I I I


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Line-to-Line FaultsTherefore, phase voltages during fault are;

2

12

2

0

11

111

a

a

c

b

a

V V

aaaa

V V

V

since V a0 = 0

2

2

1

21

2

21

aac

aab

aaa

V aaV V

aV V aV

V V V Hence;


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Line-Line Faults

case: syn. generator

Fig 9


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The neutral of generator is solidly grounded. Assume thatthe generator is initially unloaded and operating at ratedvoltage, and that a line-to-line fault occurs between phaseb an d c at the terminals of generator.

Evaluate the voltages and currents in each phase duringthe subtransient period immediately after the fault occurs. Also, determine the line-to-line voltages at that time.Base : 100 MVA, 13.8 kV


48/74

Exercise 2: ANSWERS

Line currentsIa = 0

Ib= 3.849180 o

Ic = 3.8490 = 16.1 kA

Line VoltagesVa = 1.1120 o

Vb = 0.556180 o

Vc = 0.556180 o

Line-to-line voltagesVab = Va Vb = 1.6680 o

Vbc = Vb

Vc= 00 o

Vca = Vc Va = 1.668180 o


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Example 5.2

Repeat Example 5.1 when a line-to-line fault occur at bus 3 through a fault impedance Z F = j0.1 p.u.


50/74

Solution:zero sequence impedance, Z 0 = j0.35 p.u.positive sequence impedance, Z 1 = j0.22 p.u.negative sequence impedance, Z 2 = j0.22 p.u.

The symmetrical components of fault currents

00a I

p.u.8519.1

1.022.022.01

21

21

j

j j j

Z Z Z

E I I

F

aaa


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Solution (cont..)

The phase currents

2075.3

2075.3

0

8519.1

8519.1

0

1

1

111

11

111

2

2

2

1

0

2

2

j

j

aa

aa

I I

I

aaaa

I I

I

a

a

a

c

b

a

The fault current,

..2075.3 u p j I I I cb F

A

kV

MVA I F

74.8412075.343.262

2075.32203

100


52/74

Solution (cont..):

The symmetrical components of bus voltageduring fault are

4074.0

5926.0

0

)8519.1(22.000

0)8519.1(22.00

00)0(35.0

0

0.1

0

00

00

00

0

0

2

1

0

2

1

0

2

1

0

j j

j j

j

I

I

I

Z

Z

Z

E

V

V

V

a

a

a

a

a

a

a


53/74


54/74

Double Line-to-Ground FaultsFig. 10 shows a doubleline-to-ground fault at F in a power system. The

fault may in generalhave an impedance Z F as shown.

Fig. 10

The currents and the voltages conditions at the fault areexpressed as

I a = 0or

I a0 +I a1 +I a2 = 0

V b = V c = (I b + I c )Z F

i f Z F = 0; V b = V c = 0

Z F


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Double Line-to-Ground FaultsThe symmetrical components of voltages at F are:

b

b

a

a

a

a

V

V

V

aa

aa

V

V

V

2

2

2

1

0

1

1

111

3

1

))((31 2

21 baaa V aaV V V

Where V c = V b

Therefore;


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Double Line-to-Ground Faults

f a0

f a2a1a0

f a2a12

a0a22

a1a0

f c b b

Z3I

]ZII[2I)]ZaIIa(I)IaaI[(I

)ZI(IV

a2a1

a1a12

a0 b

VVwhere

aVVaVV


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;ISolve

VVZ3I

a)V(aVZ3I

a0

a1a0f a0

a12

a0f a0

F

aaa Z Z Z I E

I 32110


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;ISolve

ZIZIE

VV

a2

2a21a1a

a2a1

2

112 Z

Z I E I

aaa


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;ISolve

)II(I

0III

a1

a0a2a1

a0a2a1

)3()3(

))3//((

02

021

021

1

F

F

a

F

aa

Z Z Z Z Z Z

Z

E Z Z Z Z

E I


60/74


Fig. 11: Connection of sequence network for a double line-to-groundfault

3 Z F

3 Z F


61/74


The symmetrical components of fault currents isgiven by;

)3()3(

))3//((

02

021

0211

F

F

a

F

aa

Z Z Z Z Z Z Z

E

Z Z Z Z E

I 2

112 Z

Z I E I aaa

F

aaa Z Z

Z I E I 32

110


62/74


The fault currents is

0

210

212

022

10

3

2

)()(

a

aaa

aaaaaa

cb F

I

I I I

aI I a I I aaI I I I I


63/74

Double Line-to-Ground FaultsPhase currents during fault ;

Phase voltages during fault ;

2

2

10

21

2

0

210

aaac

aaab

aaaa

V aaV V V

aV V aV V

V V V V

22

10

21

2

0

210

aaac

aaab

aaaa

I aaI I I aI I a I I

I I I I


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case:syn. generator

Fig 12


65/74


The neutral of generator is solidly grounded. Assume thatthe generator is initially unloaded and operating at ratedvoltage, and that a double line to ground fault occursbetween phase b an d c to g ro u n d at the terminals of generator.Evaluate the voltages and currents in each phase duringthe subtransient period immediately after the fault occurs.

Also, determine the line-to-line voltages at that time.Base : 100 MVA, 13.8 kV


66/74

Exercise 3: ANSWERS

Line currentsIa = ?Ib = ?Ic = ?

Line VoltagesVa = ?Vb = ?Vc = ?

Line-to-line voltagesVab = Va Vb = ?Vbc = Vb Vc = ?Vca = Vc Va = ?


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Example 5.3

Repeat Example 5.1 when a double line-to-ground faultoccur at bus 3 through a fault impedance Z F = j0.1 p.u.


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Solution:

zero sequence impedance, Z 0 = j0.35 p.u.positive sequence impedance, Z 1 = j0.22 p.u.negative sequence impedance, Z 2 = j0.22 p.u.

The symmetrical components of fault currents

..6017.2

))1.0(335.0(22.0))1.0(335.0(22.0

22.0

1

))3//(( 0211

u p j

j j j j j

j

Z Z Z Z E

I F

aa

..9438.1

22.0)22.0)(6017.2(1

2

112

u p j

j j j

Z Z I E

I aaa

6579.0

3.035.0)22.0)(6017.2(1

3211

0

j

j j j j

Z Z Z I E

I F

aaa


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70/74


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Solution (cont..):Voltage at bus 3 during fault

1801974.0

1801974.0

00855.14276.0

4276.0

2303.0

1

1

111

1

1

111

2

2

2

1

0

2

2

aa

aa

V

V

V

aa

aa

V

V

V

a

a

a

c

b

a

kV43.43

kV2201974.0

p.u.1801974.0

kV43.43

kV2201974.0

p.u.1801974.0

8.81kV32kV2200855.1

p.u.00855.1

c

c

b

b

a

a

V

V

V

V

V

V


72/74


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Exercise 4

Item MVA Voltage X X1 X2 X0Generator (Y-grounded)

25 11 kV 15% 15% 10%

Transformer T1 &T2( -Ygrounded)

25 11/66 kV 10% 10% 5%

Motor (Y grounded)

25 11 kV 15% 15% 10%

Line 25 66 kV 10% 10% 5%

Figure 13


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3 Unsymmetrical Faults Eet308 Updated Sv