Tutorial 3 - July 2010

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1. Three 15 MVA, 30 kV synchronous generators A, B, and C are connected via threereactors to a common bus bar, as shown in Figure 1. The neutrals of generators Aand B are solidly grounded, and the neutral of generator C is grounded through a

reactor of 2.0 . The generator data and the reactance of the reactors are tabulatedbelow. A line-to-ground fault occurs on phase a of the common bus bar. Neglect

prefault currents and assume generators are operating at their rated voltage.Determine the fault current in phasea.

Figure 1

The generator base impedance is

== 6015

(30)Z

2

B

The reactor per-unit reactance, and the per-unit generator C neutral reactor are

pu1.0==60

6XR

pu3333.0==602Xn

The positive-sequence impedance network is shown in Figure (a), and the zerosequence impedance network is shown in Figure (b).

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The positive-sequence impedance is

3.0

1

3.0

1

35.0

1++=

1X

1 pu105.0X1 =

The negative-sequence impedance network is the same as the positive-sequenceimpedance network, except for the value of the generator negative-sequencereactance. Therefore, the negative-sequence impedance is

255.0

1

255.0

1

255.0

1++=

2X

1 pu085.0X 2 =

The zero-sequence impedance is

26.0

1

156.0

1

156.0

1++=

0X

1 pu06.0X 0 =

The line-to-ground fault current in phasea is

pu12j)06.0085.0105.0(j

3I3I )0(aa =

++==

2. The reactance data for the power system shown in Figure 2 in per unit on a commonbase is as follows:

X1 = 10 %

X0 = 5 %

X1 = 25 %

X0 = 25 %

X1 = 25 %

X0 = 25 %

XG1 = 10 %

XG0 = 5 %

X1 = 30 %

X0 = 50 %

Figure 2

Obtain the Thevenin sequence impedances for the fault at bus 1 and compute thefault current in per unit for the following faults:(a) A bolted three-phase fault at bus 1.(b) A bolted single line-to-ground fault at bus 1.(c) A bolted line-to-line fault at bus 1.(d) A bolted double line-to-ground fault at bus 1.

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The positive-sequence impedance network is shown below, and impedanceto the point of fault is

pu2275.0j65.035.0

)65.0)(35.0(jZ1 =

+=

Since negative-sequence reactances are the same as positive-sequence reactances,X2=X1= 0.2275.

The zero-sequence impedance network is shown below, and impedance to the pointof fault is

pu1875.0j75.025.0

)75.0)(25.0(jZ0 =

+=

(a) For a bolted three-phase fault at bus 1, the fault current is

pu3956.4j2275.0

1jI f ==

(b) For a bolted single-line to ground fault at bus 1, the fault current is

pu669.4j)1875.02275.02275.0(j

3I3I )0(af =

++==

(c) For a bolted line-to-line fault at bus 1, the fault current in phase b is

pu1978.2j)2275.02275.0(j

1I )1(a =

+=

pu3jI f 3.8067-I(1)

a ==

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(d) For a bolted double line-to-line fault at bus 1, we have

pu02767.3j

)1875.02275.0(

)1875.0)(2275.0(j2275.0j

1I )1(a =

++

=

pu6597.1j02767.3jI )0(a =+

=0.1875)(0.2275

0.2275

pu9792.4jI3I )0(af ==

3. The zero-, positive-, and negative-sequence bus impedance matrices for athree-bus power system are

jZ0 = pu

0.300.080.12

0.080.100.050.120.050.20

1Z = jZ2 = pu

0.250.120.15

0.120.200.100.150.100.16

Determine the per unit fault current and the bus voltages during fault for(a) A bolted three-phase fault at bus 2.(b) A bolted single line-to-ground fault at bus 2.(c) A bolted line-to-line fault at bus 2.(d) A bolted double line-to-ground fault at bus 2.

(a) The symmetrical components of fault current for a bolted balanced three-phasefault at bus 2 is given by

(1)

afI = (1)22

f

Z

V=

j0.20

1= -j5 ; (2)afI = 0 ;

(0)

afI = 0

The fault current

fc

fb

fa

I

I

I

=

2

2

aa1

aa1

111

0

j5-

0

=

0

0

0

305

1505

90-5

For balanced fault we only have the positive-sequence component of voltage. Thus,bus voltages during fault for phasea are

0.50.2

0.11.0

Z

Z1.0V

1

22

1

12)F(1 === p.u.

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00.2

0

Z

ZV

22

F)F(2 === p.u.

0.4

0.2

0.121.0

Z

Z1.0V

1

22

1

32)F(3 === p.u.

(b) The symmetrical components of fault current for a single line-to-groundfault at bus 2 is given by

(1)

afI =(2)

afI =(0)

afI = (0)22

(2)

22

(1)

22

f

ZZZ

V

++

Let0

f 01.0V =

Then (1)afI =(2)

afI =(0)

afI =0.10)0.20(0.20j

01.0 0

++

= 2j

The fault current (1)af(2)

af(1)

af(0)

afaf I3IIII =++= = - j 6; bfI = cfI = 0

The sequence components of voltage at bus 2 are calculated as

0.20)j()0.10j(IZV(0)

af

(0)

22

(0)

a2 === 2

0.60)j2()0.20j(1.0IZVV (1)af(1)

22f

(1)

a2 ===

0.40)j2()j0.20(IZV (2)af(2)

22

(2)

a2 ===

Phase components of line to ground voltage of bus 2 are computed as

c2

b2

a2

V

V

V

=

2

2

aa1

aa1

111

0.40

0.60

0.20

=


0.10)j()0.05j(IZV (0)af(0)

12

(0)

a1 === 2

0.80)j()0.10j(1.0IZVV (1)af(1)

12f

(1)

a1 === 2

0.20)j2()j0.10(IZV (2)af(2)12(2)a1 ===


0

0.91650

109.11

0.9165 0109.11

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c1

b1

a1

V

V

V

=

2

2

aa1

aa1

111

0.20

0.80

0.10

=


0.16)j()0.08j(IZV (0)af(0)

32

(0)

a3 === 2

0.76)j()0.12j(1.0IZVV(1)

af

(1)

32f

(1)

a3 === 2

0.24)j2()j0.12(IZV (2)af(2)

32

(2)

a3 ===


c3

b3

a3

V

V

V

=

2

2

aa1

aa1

111

0.24

0.76

0.16

=

(c) The symmetrical components of fault current for line-to-line fault at bus 2 are

(0)

afI = 0

(1)

afI = -(2)

afI = (2)22

(1)

22

f

ZZ

V

+=

)0.20(0.20j

01.00

+

= -j2.5

The phase components of fault currents in the fault are

0IIII (2)af(1)

af

(0)

afaf =++=

==+= (1)af(2)

af

(1)

af

2

bf I3jIaIaI -4.33

== bfcf II 4.33


0)()0.10j(IZV (0)af(0)

22

(0)

a2 === 0

0.50)j2.5()0.20j(1.0IZVV (1)af(1)22f(1)a2 ===

0.50)j2.5()j0.20(IZV(2)

af

(2)

22

(2)

a2 ===


0.50 00

0.95390

114.79

0.9539 0114.79

0.36 00

0.96250

115.87

0.9625 0115.87

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c2

b2

a2

V

V

V

=

2

2

aa1

aa1

111

0.50

0.50

0

=


0)()0.05j(IZV (0)af(0)

12

(0)

a1 === 0

0.75)j()0.10j(1.0IZVV(1)

af

(1)

12f

(1)

a1 === 5.2

0.25)j2.5()j0.10(IZV (2)af(2)

12

(2)

a1 ===


c1

b1

a1

V

V

V

=

2

2

aa1

aa1

111

0.25

0.75

0

=


0)()0.08j(IZV (0)af(0)

32

(0)

a3 === 0

0.70)j()0.12j(1.0IZVV(1)

af

(1)

32f

(1)

a3 === 5.2

0.30)j2.5()j0.12(IZV (2)af(2)

32

(2)

a3 ===


c3

b3

a3

V

V

V

=

2

2

aa1

aa1

111

=

0.30

0.70

0

(d) The symmetrical components of fault current for a double line-to-ground fault atbus 2 is given by

.75j

0.10j0.20j

)0.10j()0.20j(0.20j

1.0

ZZ

ZZZ

VI

(0)

22

(2)

22

(0)

22

(2)

22(1)

22

f(1)

af 3=

++

=

++

=

1

0

0

0.6614 0139.11

0.6140130.11

10

0

0.5 0180

0.50

180

1 00

0.60830

145.285

0.6083 0145.285

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j1.25j0.10j0.20

j0.10)j3.75(

ZZ

ZII

(0)

22

(2)

22

(0)

22(1)

af

(2)

af =+

=+

=

j2.5

j0.10j0.20

j0.20)j3.75(

ZZ

ZII

(0)

22

(2)

22

(2)

22(1)

af

(0)

af =

+

=

+

=

Phase components of current at the fault bus are

=++= (2)af(1)

af

(0)

afaf IIII 0

00(2)

af

(1)

af

2(0)

afbf 21025150j2.5IaIaII ++=++= .175.3

= 5.7282 0139.11 0.175.3 302530j2.5IaIaII 0(2)af

2(1)

af

(0)

afcf ++=++=

= 5.7282 0.8940

Fault current =+= cfbff III 7.5090


0.25)j()0.10j(IZV (0)af(0)

22

(0)

a2 === 5.2

0.25)j3.75()0.20j(1.0IZVV (1)af(1)

22f

(1)

a2 ===

0.25)j1.25()j0.20(IZV (2)af(2)

22

(2)

a2 ===


c2

b2

a2

V

V

V

=

2

2

aa1

aa1

111

0.25

0.25

0.25

=


0.125)j()0.05j(IZV (0)af(0)

12

(0)

a1 === 5.2

0.625)j3()0.10j(1.0IZVV (1)af(1)

12f

(1)

a1 === 75.

0.125)j1.25()j0.10(IZV (2)af(2)12

(2)a1 ===


0.750

0 0

0

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c1

b1

a1

V

V

V

=

2

2

aa1

aa1

111

0.125

0.625

0.125

=


0.20)j()0.08j(IZV (0)af(0)

32

(0)

a3 === 5.2

0.55)j()0.12j(1.0IZVV (1)af(1)

32f

(1)

a3 === 75.3

0.15)j1.25()j0.12(IZV (2)af(2)

32

(2)

a3 ===


c3

b3

a3

V

V

V

=

2

2

aa1

aa1

111

0.15

0.55

0.20

=

0.87500

0.500

120

0.50 0120

0.90 00

0.37750

113.413

0.3775 0113.413

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Tutorial 3 - July 2010