EPS2 Short Sem Report

7/26/2019 [PDF] EPS2 Short Sem Report

1/22

EEPB383 ELECTRICAL POWER SYSTEM II

SEMESTER 3 2015/2016

GROUP PROJECT

EVALUATION OF

UNBALANCED FAULT IN 11KV NETWORK

LECTURERS NAME:ASSOC. PROF. DR. JAGADEESH PASUPULETISECTION:01

GROUP MEMBERS:

1

KHAIRUL ANWAR SYAHMI BIN CHE ISMAIL EE090586

2 MUHAMMAD NAQIUDDIN BIN MOHD NORDIN EP092689

3 HANIF BIN ABDUL AZIZ EP093691

DATE OF SUBMISSION:3rd MAY 2016


2/22

2

TABLE OF CONTENTS

1. PART A .................................................................................................................................3

1.1 Positive Sequence ............................................................................................................. 3

1.2 Negative Sequence ............................................................................................................ 6

1.3 Zero Sequence .................................................................................................................. 6

1.4 Symmetrical Components of Bus Voltages During Fault ................................................ 9

1.5 Bus Voltages during Fault ................................................................................................ 9

1.6 Symmetrical Components of Fault Currents In Lines for Phase A ................................ 10

2. PART B .............................................................................................................................. 11

2.1 Positive Sequence ........................................................................................................... 11

2.2 Negative Sequence .......................................................................................................... 15

2.3 Zero Sequence ................................................................................................................ 15

2.4 Symmetrical Components of Bus Voltages During Fault .............................................. 18

2.5 Bus Voltages during Fault .............................................................................................. 18

2.6 Symmetrical Components of Fault Currents In Lines for Phase A ................................. 19

3. PART C .............................................................................................................................. 21


3/22

3

X1

XA1

= XA2

= j1.3716

X1

XA2= 9km x 0.1524 /km

=j1.3716

X YB1

= 3km x 0.0949/km

= j0.2847

X1

YB1

= X1YB1

= j0.2847

Fault Level (MVA) = Fault Level (p.u.) * Sbase

2500 = F.L.G1 x 100FLG1= 25 p.u.

XG1 = 1 / 25 = j0.04 p.u.

2000 = F.L.G2 x 100

FLG2= 20 p.u.

XG2 = 1 / 20 = j0.05 p.u.

BOLTED

FAULT1 2

5

43

6

1. PART A

1.1 Positive Sequence Network

X1

T1= j0.08

X1

T2=j0.08

X T3

= j0.10

X1T4

= j0.10

X G1

= j0.04

X G2

= j0.05

X1AB4

= 6km x 0.0949/km

= j0.5694

Though the operating conditions at the time

of fault are important, the loadscan usually

be neglected during short circuits, asvoltages dip very low so that currents drawn

by loads can be neglected in comparison

with short circuit currents. Thus, the 4MVA

and 7MVAloads are ignored in determining

the equivalent Thevenin network impedance

as seen from the faulted bus/point..


4/22

4

X

1

XA12= (j1.3716) (j1.3716)(j1.3716 +j1.3716)

= j0.0686

X1T34

= (j0.1) (j0.1)

(j0.1 +j0.1)

= j0.05

XYB

12

= (j0.2847) (j0.2847)(j0.2847 +j0.2847)

= j0.14235

X1T12

=(j0.08) (j0.08)

(j0.08 +j0.08)

= j0.04

X1G1

= j0.04

X G2

= j0.05

X AB4

= 6km x 0.0949/km

= j0.5694

1

3

5


5/22

5

1 2

1

j0.81175

j0.1486

X G1 + X T12 +XXA12

= j0.04+ j0.04+ j0.0686

= 0.1486

X1G2 + X

1T34 + X

1YB12

= j0.05+ j0.05 +j0.14235

= 0.24235

X1AB4

= j0.5694

X1(G2 + T34 + YB12+ AB4) =j(0.24235 + 0.5694)

=j0.81175

Paralleling the two equivalent impedances at the Faulted Bus No. 5:-

Z1 =(j0.1486) (j0.81175)

(j0.1486+j0.81175)

=j0.1256

In the next part, the impedances to the point of fault for both the

negative-sequence and zero-sequence (Z2and Z

0) are to be determined

before fault current in this unbalanced network can be found.


6/22

6

X0

XA1

= X0XA2

= j0.2808

XXA2

= 9km x 0.0312 /km

=j0.2808

X0YB1

= 3km x 0.0303/km

= j0.0909

XYB1

= X0YB1

= j0.0909

BOLTED

FAULT1 2

6

21

5

1.2Negative Sequence Network for Part A

Since negative-sequence reactances are the same as positive-sequence reactances,

Z2= Z

1 =j0.1256

1.3 Zero Sequence Network for Part A

X T1

= j0.08

X T2

=j0.08

X T3

= j0.10

X0T4

= j0.10

G1 G2

X AB4

= 6km x 0.0303/km

= j1.818


7/22

7

X

0

XA12= (j0.2808) (j0.2808)(j0.2808+ j0.2808)

= j0.01404

X0T34

= (j0.1) (j0.1)

(j0.1 +j0.1)

= j0.0500

XYB

12

= (j0.0909) (j0.0909)(j0.0909 +j0.0909)

= j0.04545

3

1

4

6

2

5

1

X0T12

= (j0.08) (j0.08)

(j0.08 +j0.08)

= j0.04000

G1 G2

X AB4

= j1.818


8/22

8

X T34 + X

YB12= j0.05 +j0.04545

= j0.09545

X AB4

= j1.818

1 2

X0AB4

= j0.0907

COPYRIGHT -JOHN-HANIF-SYAHMI-NAQIUBOGARD

Z1bus=

0474.00021.00374.00077.0

0021.00383.00121.00338.0

0374.0010.01812.00375.0

0077.00338.00375.01256.0

jjj

jjjj

jjjj

jjjj

Z0bus=

25.0000

020.000

000907.00026.0

000026.00525.0

j

j

jj

jj

X T12 +X XA12

= j0.04+ j0.01404

= j0.05404

X AB4 = (j0.09545 *j1.818) =j0.0907

(j0.09545 +j1.818)

X

0

= Z

0

= (j0.05404) (j1.91345)(j0.05404+j1.91345)

= j0.0525

I10(F) = I1

1(F) = I1

2(F )= 1.0

(X0+ X1+ X2)

= (1.0 p.u.) = 3.29-90 puj(0.1256 + 0.1256 + 0.0525)

= -j3.29 pu

I3abc(F )= 3I1

0= 3(V5(F))

(X0+ X1+ X2)

= 3(1.0 p.u.) = 9.87-90 puj(0.1256 + 0.13256 + 0.0525)

= -j9.87pu

j0.05404

The bus impedance matrices

(Zero Sequence reactance for generator assumed to be half of the

Positive Sequences reactance)


9/22

9

The symmetrical components of bus voltages during fault:

=

3.29)(-1256.00

3.29)j0.1256(-1

3.29)(-0526.00

jj

j

jj

=

4132.0

587.0

173.0

=

3.29)j0.0375(--0

3.29)(-0375.01

3.29)j0.0026(-0

j

jj

j

=

1234.0

88.0

1055.8 3

x

=

3.29)j0.0338(--0

3.29)(-0338.01

3.29)j0(--0

j

jj

j

=

111.0

89.0

0

=

3.29)j0.0077(--0

3.29)j0.0077(--1

3.29)j0(--0

j

j

j

=

0253.0

974.0

0

Bus voltages during fault:

435.0

564.0

173.0

=

10694.0

10694.0

0

1210.0

88.0

1055.8 3x

=

11495.0

11495.0

748.0

111.0

89.0

0

=

11394.0

11394.0

779.0


10/22

10

0253.0

974.0

0

=

11999.0

11999.0

9497.0

Symmetrical components of fault currents in lines for phase a:

=

5694.0

)41727.0(1234.0

5694.0

587.088.0

818.1

)16578.0(55.8

j

j

j

m

=

9051.0

9051.0

900867.0

=

1086.0

)4132.0(1314.01086.0

58273.089.0

)16578.0(0

j

j=

9078.2

9078.2

0

=

19325.0

)1234.0(0253.0

19235.0

879.0974.0

)55.8(0

j

j

m

=

905.0

905.0

0


11/22

11

9051.0

9051.0

900867.0

=

90433.0

90433.0

901.1

9078.2

9078.2

0

=

9078.2

9078.2

56.5

2

2

42

1

1

111

aa

aa(F)Iabc

905.0

905.0

0

=

9042.0

9042.0

9001.1

Finally; questions asked for fault currents flowing in the 11 kV underground cables.

For XA1 and XA2, they are between the designated Bus 1 and Bus 3. So we divide

the fault current by half:

2

121 (F)I

abc

XAXA

9078.2

9078.2

56.5

=

9039.1

9039.1

78.2

pu

For YB1 and YB2, they are located between bus 2 and bus 4. Again, divided by half:

5.021 (F)Iabc

XBXB

9051.0

9051.0

9001.1

=

90255.0

90255.0

9051.0

pu

As for XAB4, its the current which flows between bus 1 and 2, which is:

=

9042.0

9042.0

9011.1

pu.


12/22

12

X1

XA1

= X1 XA2

= j1.3716

X1

XA2= 9km x 0.1524 /km

=j1.3716

X1YB1

= 3km x 0.0949/km

= j0.2847

X1

YB2

= X1YB1

= j0.2847

Fault Level (MVA) = Fault Level (p.u.) * Sbase

2500 = F.L.G1 x 100

FLG1= 25 p.u.

XG1 = 1 / 20 = j0.04 p.u.

2000 = F.L.G2 x 100

FLG2= 20 p.u.

XG2 = 1 / 20 = 0.05 .u.

BOLTED

FAULT

1

2. PART B

2.1 Positive Sequence Network

X1

T1= j0.08

X1

T2=j0.08

X1T3= j0.10

X1T4

= j0.10

X G1

= j0.04

X G2

= j0.05

X1AB4

= 6km x 0.0949/km

= j0.5694

Though the operating conditions at the time

of fault are important, the loadscan usually

be neglected during short circuits, as

voltages dip very low so that currents drawn

by loads can be neglected in comparison

with short circuit currents. Thus, the 4MVA

and 7MVAloads are ignored in determining

the equivalent Thevenin network impedance

as seen from the faulted bus/ oint..

3

56


13/22

13

X XA1= X

1XA2

= j1.3716

X1T34

= (j0.1) (j0.1)

(j0.1 +j0.1)

= j0.05

1

X1T12

=(j0.08) (j0.08)

(j0.08 +j0.08)

= j0.04

X1G1

= j0.04

X G2

= j0.05

X AB4

= 6km x 0.0949/km

= j0.5694

X1

YB2= X

1YB1

= j0.2847

2

34

6


14/22

14

1 2

1

j0.9541

j1.3616

X G1 + X T12 +XXA2

= j0.04+ j0.04+ j1.3716

= 1.3616

X1G2 + X

1T34 + X

1YB1

= j0.05+ j0.05 +j0.2847

= 0.3847

X1AB4

= j0.5694

X1(G2 + T34 + YB1+ AB4) =j(0.3847+ 0.5694)

=j0.9541

Paralleling the two equivalent impedances at the Faulted Bus No. 5:-

Z1 =(j1.3616) (j0.9541)

(j1.3616+j0.9541)

=j0.56


15/22

15

X0

XA1

= X0XA2

= j0.2808

XXA2

= 9km x 0.0312 /km

=j0.2808

X0YB1

= 3km x 0.0303/km

= j0.0909

XYB1

= X0YB1

= j0.0909

BOLTED

FAULT1 2

6

43

5

2.2Negative Sequence Network for Part B

Since negative-sequence reactances are the same as positive-sequence reactances,

Z2= Z

1 =j0.56

2.3 Zero Sequence Network for Part B

X T1

= j0.08

X T2

=j0.08

X T3

= j0.10

X0T4

= j0.10

G1 G2

X AB4

= 6km x 0.0303/km

= j1.818


16/22

16

X

0

XA2= j0.2808

X0T34

= (j0.1) (j0.1)

(j0.1 +j0.1)

= j0.05

XYB

1

= j0.0909

3

1

4

6

2

5

1

X0T12

= (j0.08) (j0.08)

(j0.08 +j0.08)

= j0.04

G1 G2

X AB4

= j1.818


17/22

17

X0T34 + X

0YB1

= j0.05 +j0.0909

= j0.1409

X AB4

= j1.818

1

2

X AB4

= j1.9589

1

Z1bus=

049.00008.00420.00302.0

0008.00393.00064.00159.0

042.00064.03232.02321.0

0302.00159.02321.05757.0

jjjj

jjjj

jjjj

jjjj

Z0bus=

25.0000

020.000

001322.00198.0

000198.002757.0

j

j

jj

jj

X T12 +XXA2

= j0.04+ j0.2808

= j0.3208

X0AB4= (j0.1409 +j1.818) =j1.9589

X0= Z55

0= (j0.3208) (j1.9589)

(j0.3208+j1.9589)

= j0.2757

If = 3Ia0= 3(V5(F))

(X0+ X1+ X2)

= 3(1.0 p.u.)

j(0.5736 + 0.5736 + 0.2757)

= -j2.1084 pu

I11(F) = -j2.1084/3

= -j0.703 pu

Bus impedance matrix for case B

j0.3208


18/22

18

Symmetrical components of bus voltages during fault:

=

)703.0(-.5757.00

)703.0j0.5757(-1

)703.0(-2757.00

jj

j

jj

=

405.0

595.0

194.0

=

)703.0j0.2321(-0

)703.0(2321.01

)703.0j0.0198(0

j

jj

j

=

.16.0

837.0

0139.0

=

)703.0j0.0159(-0

)703.0(0159.01

)703.0j0(-0

j

jj

j

=

011.0

988.0

0

=

)703.0j0.0302(-0

)703.0j0.0302(-1

)703.0j0(-0

j

j

j

=

0212.0

98.0

0

Bus voltages during fault:

405.0

595.0

194.0

=

10694.0

10694.0

0

.16.0

837.0

0139.0

=

11494.0

11494.0

663.0

011.0

988.0

0

=

11399.0

11399.0

997.0


19/22

19

0212.0

98.0

0

=

11999.0

11999.0

96.0

Symmetrical components of fault currents in lines for phase a:

=

5694.0

)405.0(16.05694.0

595.0837.0818.1

)194.0(0139.0

j

j

j

=

90425.0

90425.0

90099.0

=

4116.1

)405.0(011.04116.1

595.0988.0

)194.0(0

j

j=

90278.0

90278.0

0

=

3347.0

)16.0(0212.0

3347.0

837.098.0

)0139.0(0

j

j=

90427.0

90427.0

0


20/22

20

90425.0

90425.0

90099.0

=

90326.0

90326.0

9095.0

90278.0

90278.0

0

=

90278.0

90278.0

556.0

2

2

42

1

1

111

aa

aa(F)Iabc

90427.0

90427.0

0

=

9042.0

9042.0

90854.0

For XA1, it is between the designated Bus 1 and Bus 3. The current does not flow

through XA2 because its disconnected by the circuit breaker A0. So the fault current

flowing through XA1 is equal to (F)IabcXA1

90326.0

90326.0

9095.0

pu.

For YB1, no current flow through the cable due to closed circuit breaker B0,

effectively disconnecting it from the bus network. Whereas the cable YB2 carries

current equal to the current from cable AB4 which is:

(F)I(F)I(F)I abcABabc

YB

abc

4242 =

9042.0

9042.0

90854.0

pu.


21/22

21

XXA1

= XA2

= j1.3716

XXA2

= 9km x 0.1524 /km

=j1.3716

X1YB1

= 3km x 0.0949/km

= j0.2847

X1

YB1

= X1YB1

= j0.2847

1 2 (11kV)

6

43

5

7 (0.4kV)

3. PART C

X T1

= j0.08

X T2

=j0.08X

1T3

= j0.10

X T4

= j0.10

X1G1

= j0.04

X1G2

= j0.05

X1AB4

= 6km x 0.0949/km

= j0.5694


22/22

22

1 2

7 (0.4kV)

2

7 (0.4kV)

X G1 + X T12 +XXA12

= j0.04+ j0.04+ j0.0686

= 0.1486

If= 1.0 / Z

= 1.0 / j0.33= 3.03 -90 pu

IB= SB/ VB

= 100MVA / (3*0.4kV)= 144.34kA

Iactual= 144.34K * 3.06 = 437.35kA (very high and dangerous)

Scc= (3)(400)(437.35)/1M= 303MVA

X1AB4

= j0.5694

X1G2 + X

1T34 + X

1YB12

= j0.05+ j0.05 +j0.14235

= 0.24235

X T5

=j0.15

(X1G1 + X

1T12 +X

1XA12) + X

1AB4

=j0.1486+j0.5694

=j0.718

Zs= (j0.728) (j0.23235) / (j0.728+j0.23235)

=j0.18

Z= j0.18+j0.15

=j0.33

j0.18

j0.15

EPS2 Short Sem Report

Documents

Transcript of EPS2 Short Sem Report