Download - 279 39 Solutions Instructor Manual Chapter 10 Symmetrical Components Unsymmetrical Fault Analyses
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Chapter 10
10.1 Input:
» a=exp(i*2*pi/3);
» A=[1 1 1;1 a^2 a;1 a a^2];
» I=[15*exp(0);20*exp(i*120*pi/180);30*exp(i*(-120)*pi/180)];
The zero, positive and negative sequence currents are given by
» IS=inv(A)*I
I ao = -3.3333 - 2.8868i = 4.4096 ∠-139.11 0 A
I a1 = -3.3333 + 2.8868i = 4.4096∠
139.110
AI a2 = 21.6667 + 0.0000i = 21.6667 ∠00 A
10.2 Input:
» a=exp(i*2*pi/3);
» A=[1 1 1;1 a^2 a;1 a a^2];
» VS=[50*exp(i*180*pi/180);100*exp(i*90*pi/180);50*exp(i*0)];
Computation of phase to neutral voltages
» VP=A*VS
Van = 0.0000 + 100.00i = 100 ∠900 V
V bn = 11.6000 – 6.7000i = 13.3975 ∠ –30 0 V
Vcn = -161.600 – 93.300i = 186.6025 ∠ –150 0 V
10.3 Assume that the neutral is un-earthed. Hence, there is no path for the zero sequence
currents to flow. Therefore, there is no zero sequence component of voltages in the line
voltages.
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If V c a is assumed to have an angle of 180 0, the phase angles of the other line voltages are
calculated using the cosine law which yields the following
V a b = 200 ∠70.53 0 V, V b c = 220 ∠-58.99 0 V, and V c a = 180 ∠180 0 V
Input data:
» a=exp(i*2*pi/3);
» Vab=200*exp(i*70.53*pi/180);Vbc=220*exp(i*(-58.99)*pi/180);
» Vca=180*exp(i*180*pi/180);
Computation of symmetrical components of line voltages
» Vab1=(Vab+a*Vbc+a^2*Vca)/3 = (8.7762e+001 +1.7896e+002i) V= 199.3220 ∠63.8767 0 V
» Vab2=(Vab+a^2*Vbc+a*Vca)/3 = (-21.1011 + 9.6002i) V
= 23.1823 ∠155.5361 0 V
Computation of sequence components of voltages to neutral
» Van1=i*Vab1 = (-1.7896e+002 +8.7762e+001i) V
» Van2=-i*Vab2 = (9.6002 +21.1011i) V
» Van0=0;
Computation phase ‘a’ to neutral voltage
» Van=Van1+Van2+Van0 = (-1.6936e+002 +1.0886e+002i)
= 201.3313 ∠147.2675 0 V
Computation of phase current in line ‘a’
» Ian=Van/10 = -16.9361 +10.8863i = 20.1331 ∠147.2675 0 A
Computation phase ‘b’ to neutral voltage
» Vbn=a^2*Van1+a*Van2+Van0 = (1.4241e+002 +1.0887e+002i)
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= 179.2564 ∠37.3964 0 V
Computation of phase current in line ‘b’
» Ibn=Vbn/10 = 14.2411 +10.8867i = 17.9256 ∠37.3964 0 A
Computation phase ‘c’ to neutral voltage
» Vcn=a*Van1+a*2*Van2+Van0 = (-3.2672e+001 -2.0334e+002i)
= 205.9470 ∠-99.1282 0 V
Computation of phase current in line ‘c’
» Icn=Vcn/10 = -3.2672 -20.3339i = 20.5947 ∠-99.1282 0 A
10.4 Computation of phase sequence currents» Ian1=Van1/10 = (-17.8961 + 8.7762i) A
» Ian2=Van2/10 = (0.9600 + 2.1101i) A
» Ian0=Van0/10 = 0 A
Computation of power from sequence voltages and currents
» S=3*(Van0*conj(Ian0)+Van1*conj(Ian1)+Van2*conj(Ian2))
= 1.2080e+004 W
Verification by computing power fro line currents and resistance
» (abs(Ian)^2+abs(Ibn)^2+abs(Icn)^2)*10 = 1.1508e+004
The slight difference is due to round off error in computations
10.5 When the neutral is grounded, it provides a return path for the line currents to flow.
Thus if I n is the neutral and I a, I b, and I c are the line currents,
I n = I a + I b + I c
Also, I a0 =31
(I a + I b + I c).
That is, I n = 3 I a0
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10.6 Computation of generator 1 reactance
» Xg1=0.12*(10.6/11)^2*(100/50) = 0.2229 pu
Computation of generator 2 reactance
» Xg2=0.10*(6.6/6.6)^2*(100/30) = 0.3333 pu
Computation of transformer 1 reactance
» Xt1=0.15*(220/220)^2*(100/75) = 0.2000 pu
Computation of transformer 2 reactance
» Xt2=0.15*(220/220)^2*(100/50) = 0.3000 pu
Computation of transformer 3 reactance» Xt3=0.15*(220/220)^2*(100/30) = 0.5000 pu
Computation of motor reactance
» Xm=0.08*(10/11)^2*(100/25) = 0.2645 pu
» Zbase=(220^2)/100 = 484 Ω
Computation of line 1 impedance
» Zl1=(50+i*100)/Zbase = (0.1033 + 0.2066i) pu
Computation of line 2 impedance
» Zl2=(25+i*50)/Zbase = (0.0517 + 0.1033i) pu
Computation of line 3 impedance
» Zl3=(45+i*60)/Zbase = (0.0930 + 0.1240i) pu
Representation of the load as a parallel R- X combination
» S=30*(0.8+i*0.6) = (24.0000 +18.0000i) MVA
» Rload=(11/11)^2*100/real(S) = 4.1667 pu
» Xload=(11/11)^2*100/imag(S) = 5.5556 pu
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The per unit single line diagram of the network is shown below:
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10.7 Here 0.Also.and,, b ====== cabcabcbbaa ZZZZZZZZZ Hence, the
sequence impedance is given by
[Z 012 ] =⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
012
201
120
sss
sss
sss
ZZZ
ZZZ
ZZZ
(I)
where
( ) ( )
( ) [ ]( ) ( )
( ) [ ]( ) ( )babacbas
bababbas
babbas
ZZZaaZaZZaZZ
ZZZaaZZaaZZZ
ZZZZZZ
−=++=++=
−=++=++=
+=++=
31
31
31
31
31
31
231
31
222
221
0
Substitution for the sequence impedances in (I) leads to
[Z 012 ] =
( ) ( ) ( )( ) ( ) ( )( ) ( ) ( )⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
+−−−+−−−+
=⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
bababa
bababa
bababa
ZZZZZZ
ZZZZZZ
ZZZZZZ
Z
Z
Z
2
2
2
31
2
1
0
10.8 Connection of the neutral point to ground provides a path for the zero sequence
currents to flow. Staring from first principle, the sequence impedance matrix is written as
[ ] [ ]( )
( )( )
[ ]AZZ
ZZ
ZZ
AZ⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
+
+
+
= −
gb
gb
ga
00
00
001
012
On performing the indicated matrix multiplications leads to the final result
[Z 012 ] =( ) ( ) ( )( ) ( ) ( )( ) ( ) ( )⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
++−−
−++−−−++
=⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
gbababa
bagbaba
babagba
ZZZZZZZ
ZZZZZZZZZZZZZZ
Z
ZZ
32
3232
31
2
1
0
The expression for the neutral voltage is given by
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gan ZIV 30 ×=
10.9 Assume a system base of 100 MVA. The voltage base for the various parts of the
network is as per the transformation ratios of the transformers.
Generator 1 per unit reactance: 5 .0;25.0 021 === X X X
Generator 2 per unit reactance: 3333.0;25.0 021 === X X X
Transformer 1-2 per unit reactance: 2 .0021 === X X X
Transformer 1-5 per unit reactance: 1143.0021 === X X X
Transformer 3-4 per unit reactance: 2 .0021 === X X X
Transformer 6-4 per unit reactance: 3 .0021 === X X X
Transmission line 2-3 per unit impedance:
( )( )
5165.01722.0132
10090302
j j
Z +=×+=
Transmission line 5-6 per unit impedance:
( )( )
1653.01033.0220
10080502
j j
Z +=×+=
Series representation of load
( ) ( )( )( ) ( )( ) ( )( )( ) ( ) pu2.110301040
1030101000.1
pu6.110301040
1040101000.1
0.300.40)6.08.0(50
2626
662
2626
662
=×+×
××=
=×+×
××=
+=+×=
X
R
j jS
pu05.0= f Z
Assembly of positive and negative sequence networks:
The positive sequence network is shown in the figure below.
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Thus, )3937.01133.0(21 j Z Z +==
Computation of zero sequence network:
Based on the types of transformer connections, the zero sequence network is
shown below.
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From the diagram, it is seen that )4506.01595.0(0 j Z +=
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The network below depicts the inter-connection of the sequence networks for a
SLG fault.
Take the node 4 voltage (11 kV) as the fault voltage pu 0.1= f V (reference
phasor) for fault calculations.
( ) pu j j j j Z Z Z Z
V I
f
f a
°−∠=−=
++++++=
+++=
5856.667412.06802.02946.0
15.0)4506.01595.0()3937.01133.0()3937.01133.0(0.1
30211
Base current kA I Base 2486.510113
101003
6=
×××=
021 aaa III ==
Fault current pu pu I I aa °−∠=−∠×== 5856.662236.25856.667412.0330
1
kA°−∠=−∠×= 5856.666708.115856.662236.22486.5 0
Computation of sequence voltages:
» Va0=-Ia0*Z0 = (–0.3535 – 0.0242i) pu = 0.3542 ∠ –176.0837 ° pu
» Va1=Vf-Ia1*Z1 = (0.6988 – 0.0389i) pu = 0.6999 ∠ –3.1862 ° pu
» Va2=-Ia2*Z2 = (–0.3012 – 0.0389i) pu = 0.3037 ∠ –17.6410 ° pu
Computation of line voltages:
» A=[1 1 1;1 a^2 a;1 a a^2];
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» VS=[Va0;Va1;Va2];
» Vabc=A*VS
Vabc = =⎥
⎥⎥
⎦
⎤
⎢
⎢⎢
⎣
⎡
c
b
a
VV
V
⎥
⎥⎥
⎦
⎤
⎢
⎢⎢
⎣
⎡
°∠°−∠
°∠
=⎥
⎥⎥
⎦
⎤
⎢
⎢⎢
⎣
⎡
+ 122.09321.0395973.1221.0148
66.5603-0.1112
0.8807i0.5523-0.8514i-0.5523-
0.1020i-0.0442
Line voltage aV is also given by
°−∠×=×= 5856.667412.015.01a f a I Z V = 0.1112 ∠ –66.5856 °
The above verifies the computations.
10.10 (i) 2LL fault:
The interconnection of the sequence networks is drawn below:
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» Zf=0.05;
Computation of sequence currents:
» Ia1=Vf/(Z1+Z2+Zf) = (0.3971 - 1.1305i) = 1.1982 ∠ –70.6445 ° pu
» Ia2=-Ia1 = (–0.3971 + 1.1305i) = 1.1982 ∠109.3555 ° pu
Computation of sequence voltages
» Va0=0;
» Va1=Vf–Ia1*Z1 = (0.5099 – 0.0283i) pu
» Va2=-Ia2*Z2 = (0.4901 + 0.0283i) pu
Computation of phase voltages:
» VS=[Va0;Va1;Va2];
» Vabc=A*VS
Vabc = =⎥
⎥⎥
⎦
⎤
⎢
⎢⎢
⎣
⎡
c
b
a
VV
V
⎥
⎥⎥
⎦
⎤
⎢
⎢⎢
⎣
⎡
°∠°−∠
°∠
=⎥
⎥⎥
⎦
⎤
⎢
⎢⎢
⎣
⎡
+−−
177.81670.4514178.20580.5492
01.0000
0.0172i0.45100.0172i-0.5490
1.0000
Computation of line voltages
» Vab=Vabc(1) –Vabc(2) = (1.5490 + 0.0172i) = 1.5490 ∠0.6361 ° pu
» Vbc=Vabc(2) –Vabc(3) = (–0.0979 – 0.0344i) = 0.1038 ∠ –160.6445 ° pu
» Vca=Vabc(3) –Vabc(1) = (–1.4510 + 0.0172i) = 1.4511 ∠179.3210 ° pu
Computation of line currents
» Ib=–i*sqrt(3)*Ia1 = (–1.9581 – 0.6878i) = 2.0754 ∠ –160.6445 ° pu
» Ic=-Ib = (1.9581 + 0.6878i) = 2.0754 ∠19.3555 ° pu
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Verification of computations
» Vbc=Zf*Ib = (–0.0979 – 0.0344i) = 0.1038 ∠ –160.6445 ° pu
Computation of actual line currents
» IBase*abs(Ib) = 10.8929 kA
(ii) 2LG fault
The inter-connected sequence networks are shown below:
Computation of sequence currents
» Ia1=Vf/((Z1+Zf)+(Z2+Zf)*(Z0+Zf)/(Z2+Z0+2*Zf)) = (0.5939 –1.4048i) pu
» Ia2=–((Z0+Zf)/(Z2+Z0+2*Zf))*Ia1 = (–0.3050 + 0.7624i) pu
» Ia0=–((Z2+Zf)/(Z2+Z0+2*Zf))*Ia1 = (–0.2889 + 0.6424i) pu
» Ia0+Ia1+Ia2 = 0; which verifies the computations
» IS=[Ia0;Ia1;Ia2];
» Iabc=A*IS
Iabc = =⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
c
b
a
I
I
I
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
++−
−
1.7420i1.4435
0.1851i2.3101
0.0000i0.0000
» 3*Ia0 = (–0.8666 + 1.9271i) pu
» Iabc(2)+Iabc(3) = (–0.8666 + 1.9271i)
Since 03 acb III =+ , the computations are again verified.
Computation of sequence voltages
» Va0=–Ia0*Z0 = (0.3355 + 0.0277i) pu
» Va1=1.0–Ia1*Z1 = (0.3797 – 0.0747i) pu
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» Va2=-Ia2*Z2 = (0.3347 + 0.0337i) pu
Computation of phase voltages
» VS=[Va0;Va1;Va2];
» Vabc=A*VS
Vabc = =⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
c
b
a
V
V
V
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
°∠°∠°−∠
=⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
++−
−
3532.500.1131
4195.1750.1159
7226.01.0500
0.0871i0.0722
0.0093i0.1155
0.0132i1.0499
pu
Computation of line voltages
» Vab=Vabc(1) –Vabc(2) = (1.1654 – 0.0225i) = 1.1656 ∠ –1.1058 ° pu
» Vbc=Vabc(2) –Vabc(3) = (–0.1877 – 0.0778i) = 0.2032 ∠ –157.4722 ° pu
» Vca=Vabc(3) –Vabc(1) = (–0.9777 + 0.1003i) = 0.9829 ∠174.1404 ° pu
Computation of actual line currents
» IBase*abs(Iabc)
kA
I
I
I
c
b
a
⎥⎥
⎥
⎦
⎤
⎢⎢
⎢
⎣
⎡
°∠°∠°−∠
=⎥⎥
⎥
⎦
⎤
⎢⎢
⎢
⎣
⎡
3532.5011.8743
4195.17512.1637
9638.750.0000
(iii) Three phase fault
The sequence networks for a three-phase fault are drawn below.
Computation of sequence currents
» Ia1=Vf/(Z1+Zf) = (0.8989 – 2.1672i) = 2.3462 ∠ –67.4722 ° pu
» Ia=Ia1;
Computation of actual current
» IBase*abs(Ia) = 12.3142 ∠ –67.4722 ° kA
» angleIb=angleIa+(angle(a^2)*180/pi) = ∠ –187.4722 °
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» angleIc=angleIa+(angle(a)*180/pi) = ∠52.5278 °
Computation of voltages
» Va=Zf*Ia = (0.0449 – 0.1084i) = 0.1173 ∠ –67.4722 ° pu
» angleVb=angleVa+angle(a^2)*180/pi = ∠ –187.4722 °
» angleVc=angleVa+angle(a)*180/pi = ∠52.5278 °
10.11 Input data from problem 10.9
» Ia1=0.2946–i*0.6802;Ia2=Ia1;Ia0=Ia1;
» Va0=-0.3535–i*0.0242;
» Va1=0.6988–i*0.0389;
» Va2=-0.3012–i*0.0389;
» a=exp(i*2*pi/3);
» A=[1 1 1;1 a^2 a;1 a a^2];
Computation of positive and negative sequence currents flowing in the circuit on the
generator side
» Ia1g=((1.6+i*1.2)/(i*0.125+(0.0646+i*0.3551)+1.6+i*1.2))*Ia1
= (0.1625 – 0.6054i) pu
» Ia2g=Ia1g;
Computation of zero sequence current flowing from node 5
» Ia05=((1.6+i*1.2)/(i*0.1143+(0.1033+i*0.1653)+i*0.3+1.6+i*1.2))*Ia0
= (0.1459 – 0.5839i) pu
Computation of positive and negative sequence currents flowing from node 5.
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Ia15=((0.1722+i*0.5165+i*0.4)/(i*0.1143+0.1033+i*0.1653+i*0.3+0.1722+i*0.5165
+i*0.4))*Ia1g = (0.0983 – 0.3715i) pu
» Ia25=Ia15;
Computation of phase currents flowing from node 5
» IS5=[Ia05;Ia15;Ia25];
» Iabc5=A*IS5
Iabc5 =
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
°−∠
°−∠°−∠
=
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
−
−−
=
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
3481.772177.0
3481.772177.0
5286.753703.1
0.2124i0.0477
0.2124i0.0477
1.3268i0.3424
c
b
a
I
I
I
pu
Computation of sequence voltages at node 5
» Va15=1.0+Ia15*(0.1033+i*0.1653+i*0.3) = (1.1830 + 0.0073i) pu
» Va25=Ia25*(0.1033+i*0.1653+i*0.3) = (0.1830 + 0.0073i) pu
» Va05=(0.1033+i*0.1653+i*0.3)*Ia05 = (0.2867 + 0.0076i)
Computation of phase voltages at node 5» VS5=[Va05;Va15;Va25];
» Vabc5=A*VS5
Vabc5 =⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
°∠°−∠
°∠=⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
+−−−
+=⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
5805.1149526.0
5927.1149522.0
7721.06529.1
0.8663i0.3963
0.8658i0.3963
0.0223i1.6527
c
b
a
V
V
V
pu
Computation of line voltages at node 5
» Vab=Vabc5(1) –Vabc5(2) = (2.0490 + 0.8881i) = 2.2332 ∠23.4324 ° pu
» Vbc=Vabc5(2) –Vabc5(3) = (–0.0000 – 1.7321i) = 1.7321 ∠ –90.0000 ° pu
» Vca=Vabc5(3) –Vabc5(1) = (–2.0490 + 0.8440i) = 2.2160 ∠157.6129 ° pu
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10.12 Computation of positive sequence impedance
» Zp=(0.1722+i*0.9165)*(0.1033+i*0.5796)/(0.1722+i*0.9165+0.1033+i*0.5796)
= (0.0646 + 0.3551i) pu
» Z1=i*0.125*(Zp+1.6+i*1.2)/(i*0.125+Zp+1.6+i*1.2)
= (0.0046 + 0.1203i) pu
The positive sequence network for a 3-phase fault at node-1 is shown in the
figure below.
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Computation of base current
» IBase=100*10^6/(sqrt(3)*6.6*10^3) = 8.7477 kA
» Zf=0.05;
Computation of voltage (pre-fault) at node-1
» Vnode=1.0+(1.0/(1.6+i*1.2))*Zp = (1.1324 + 0.1226i) = 1.1390 ∠6.1815 °
Set the fault voltage Vf equal to the magnitude of voltage at node-1, that is,
Vf = 1.1390 ∠0°
» Vf=abs(Vnode);
Computation of positive sequence current
» Ia1=Vf/(Z1+Zf) = (3.5650 – 7.8479i) = 8.6197 ∠ –65.5699 ° pu
» Ia=Ia1;
Computation of actual current
» IBase*abs(Ia) = 75.403 kA
» angleIb=angleIa+(angle(a^2)*180/pi) = ∠ –185.5699 °
» angleIc=angleIa+(angle(a)*180/pi) = ∠ 54.4301 °
Computation of voltages
» Va=Zf*Ia = (0.1782 – 0.3924i) = 0.4310 ∠ –65.5699 ° pu
» angleVb=angleVa+angle(a^2)*180/pi = ∠ –185.5699 °
» angleVc=angleVa+angle(a)*180/pi = ∠54.4301 °
10.13 Input data:
» Zl2=i*0.2+0.1722+i*0.5165+i*0.2;
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» Zl1=i*0.1143+0.1033+i*0.1653+i*0.3;
» Zlo=1.6+i*1.2;Zg=i*0.125;
» IBase=8.7477;Zf=0.05; Va=0.1782–i*0.3924;
Ia = (3.5650 – 7.8480i);
Parallel combination of lines 1 and 2
» Zp=Zl1*Zl2/(Zl1+Zl2) = (0.0646 + 0.3551i);
Component of current in the load side of the network
» Ialo=(Zg/(Zg+Zp+Zlo))*Ia = (0.4258 – 0.1620i) pu
= 3.9853 ∠ –20.8347 ° KAComponent of current in the generator side
» Iag=((Zp+Zlo)/(Zg+Zp+Zlo))*Ia = (3.1392 – 7.6860i) pu
= 72.6266 ∠ –67.7833 0 KA
Voltage at node-4
» Van4=Va+Ialo*Zp = (0.2632 – 0.2517i) = 0.3642 ∠ –43.7142 ° pu
Computation of current in line-1
» Ial1=(Zl2/(Zl1+Zl2))*Ialo = (0.2606 – 0.1003i) pu
= 2.4430 ∠ –21.0420 ° KA
Computation of current in line-2
» Ial2=(Zl1/(Zl1+Zl2))*Ialo = (0.1651 – 0.0618i) pu
= 1.5423 ∠ –20.5063 KA
Computation of voltage at node-2
» Van2=Va+Ial2*(i*0.2) = (0.1906 – 0.3594i) = 0.4068 ∠ –62.0658 ° pu
Computation of voltage at node-3
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» Van3=Va+Ial2*(i*0.2+0.1722+i*0.5165) = (0.2509 – 0.2847i) pu
= 0.3795 ∠ –48.6134 ° pu
The computations are verified by computing the voltage at node-4 via the line-2 circuit as
under
» Van4=Va+Ial2*Zl2 = (0.2632 – 0.2517i) = 0.3642 ∠ –43.7142 ° pu
Computation of voltage at node-5
» Van5=Va+Ial1*(i*0.1143) =(0.1897 – 0.3626i) = 0.4092 ∠ –62.3883 ° pu
Computation of voltage at node-6
» Van6=Va+Ial1*(i*0.1143+0.1033+i*0.1653) = (0.2332 – 0.3299i)
= 0.4040 ∠-54.7471 ° pu
10.14 Computation of positive sequence network (reference figs.(a) and (b))
» Z14=0.1722+i*0.9165;
» Z4F=0.0517+i*0.3826;
» ZF1=0.0517+i*0.1970;
» ZFs=Z4F*ZF1/(Z14+Z4F+ZF1) = (0.0107 + 0.0506i) pu
» Z1s=ZF1*Z14/(Z14+Z4F+ZF1) = (0.0321 + 0.1206i) pu
» Z4s=Z14*Z4F/(Z14+Z4F+ZF1) = (0.0325 + 0.2344i) pu
» Z1=ZFs+(Z1s+i*0.125)*(Z4s+1.6+i*1.2)/(Z1s+Z4s+i*0.125+1.6+i*1.2)
= (0.0557 + 0.2737i)
» Z2=Z1;
Computation of zero sequence network
» Z0=(Z4F+1.6+i*1.2)*(ZF1)/(Z4F+ZF1+1.6+i*1.2) = (0.0559 + 0.1807i)
Computation of fault and ground impedances
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» Zf=5*100/(220^2) = 0.0103 pu
» Zg=10*100/(220^2) = 0.0207 pu
The interconnection of the sequence networks is shown below
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Computation of sequence fault currents
» Vf=1.0;
» Ia1=Vf/((Z1+Zf)+(Z2+Zf)*(Z0+Zf+3*Zg)/(Z2+Z0+2*Zf+3*Zg))
= (0.7345 – 2.3521i) pu
» Ia2=-((Z0+Zf+3*Zg)/(Z2+Z0+2*Zf+3*Zg))*Ia1 = (–0.0989 + 1.1002i) pu
» Ia0=-((Z2+Zf)/(Z2+Z0+2*Zf+3*Zg))*Ia1 = (–0.6356 + 1.2519i)Verification of computations
» Ia0+Ia1+Ia2 = 0
Computation of line fault currents
» IS=[Ia0;Ia1;Ia2];
» Iabc=A*IS
Iabc =⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
°∠°∠
°∠=⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
++
+=⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
9268.513.3023
6605.1634.1093
00.0000
2.5996i2.0364
1.1560i3.9433-
0.0000i0.0000
c
b
a
I
I
I
Computation of fault sequence voltages
» Va0=-Z0*Ia0 = (0.2617 + 0.0449i) pu
» Va1=Ea-Z1*Ia1 = (0.3153 – 0.0699i) pu
» Va2=-Z2*Ia2 = (0.3067 – 0.0343i) pu
Computation of voltages at fault point
» VS=[Va0;Va1;Va2];
» Vabc=A*VS
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Vabc =⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
∠
∠
−∠=⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
++
−=⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
0
0
0
9696.990.1061
8273.1310.1202
8380.30.8856
1045001840
0895008010
0593088360
i..-
i..-
i..
V
V
V
c
b
a
Computation of line voltages
» Vab=Vabc(1) –Vabc(2) = (0.9638 – 0.1488i) = 0.9752 ∠ –8.7779 ° pu
» Vbc=Vabc(2) –Vabc(3) = (–0.0618 – 0.0149i) = 0.0635 ∠ –166.4276 ° pu
» Vca=Vabc(3) –Vabc(1) = (–0.9020 + 0.1637i) = 0.9167 ∠169.7115 ° pu
If Z g is set equal to ∞ , it is seen from the interconnected sequence network that the zero
sequence network is on open circuit. Therefore, there will be no zero sequence currents
and the boundary conditions are similar to a double line fault.
10.15 Computation of positive sequence network (reference figs.(a) and (b))
» Z1F=i*0.2+(0.1722+i*0.5165)/4 = (0.0430 + 0.3291i) pu
» Z4F=i*0.2+3*(0.1722+i*0.5165)/4 = (0.1291 + 0.5874i) pu
» Z14=i*0.1143+0.1033+i*0.1653+i*0.3 = (0.1033 + 0.5796i) pu
» Z1s=Z14*Z1F/(Z1F+Z4F+Z14) = (0.0159 + 0.1275i) pu
» ZFs=Z1F*Z4F/(Z1F+Z4F+Z14) = (0.0215 + 0.1295i) pu
» Z4s=Z4F*Z14/(Z1F+Z4F+Z14) = (0.0487 + 0.2276i) pu
» Z1=ZFs+(i*0.125+Z1s)*(1.6+i*1.2+Z4s)/(i*0.125+Z1s+1.6+i*1.2+Z4s)
= (0.0539 + 0.3605i) pu
» Z2=Z1;
Computation of fault impedance
» Zf=5*100/(132^2) = 0.0287 pu
The inter-connection of the sequence networks is as under
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» Vf=1.0;
Computation of sequence fault currents
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» Ic=–Ib = (2.3195 + 0.4390i) = 2.3606 ∠10.7185 ° pu
» Zf*Ib = (–0.0666 – 0.0126i) (II)
Comparing (I) and (II) shows that Vbc = Zf*Ib which verifies the computations.
In case Zf is set equal to zero, it is a dead short circuit between two lines. The fault
currents and voltages at the fault point can be determined by following the procedure
outlined above by setting Zf = 0.
A couple of problems in the book for this chapter had a few errors. Herewith, pleasefind the errata:
Problem As printed in book Correction______________
10.3 V ab = 200∠0° V ab = 200V bc = 220∠120 ° V bc= 220V ca = 180∠ –60 ° V bc= 180
Add the text at the end of the problem:[Note: Assume V ca = V ca ∠120 °, use thecosine law for the triangle to calculate the
phase angles of the other line voltages]
10.9 SLG occurs at node 3 SLG occurs at node 4___________________________________________________________________-