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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
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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
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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..
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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
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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.
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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
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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
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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)
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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
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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
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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.
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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
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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
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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
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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
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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
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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
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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
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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
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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.
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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
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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