EEL303: Power Engineering I - Tutorial 9
1. For the given power system shown in Figure 1, draw the positive, negative and zero
sequence networks.
Figure 1: Given Power System
Figure 2: Positive Sequence Network
Figure 3: Negative Sequence Network
Figure 4: Zero Sequence Network
2. Two synchronous machines are connected through three-phase transformers to the transmission line shown in Figure. 5(a). The ratings and reactances of the machines and
transformers are
′′
Machines 1 and 2: 100MVA, 20kV; Xd = X1 = X2 = 20%, X0 =4%, Xn = 5%
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EEL303: Power Engineering I - Tutorial 9
Transformers T1 and T2 : 100MVA, 20Star/345Star kV; X = 8%
Both transformers are solidly grounded on two sides. On a chosen base of 100MVA,
345kV in the transmission line circuit the line reactances are X1 = X2 = 15% and
X0 = 50%. The system is operating at nominal voltage without prefault currents when
a bolted (Zf = 0) single line to ground fault occurs on phase A at bus 3. Using the bus
impedance matrix for each of the three sequence networks, determine the subtransient
current to ground at the fault, the line-to-ground voltages at the faulted bus and at the
terminals of machine 2.
[Ans: I(f A) = 9316 270 A;V3a = 0kV ; V3b = 11.8856 − 122.71kV ; V3c = 11.8856 122.71kV ;
V4a = 3.3466 0kV ; V4b = 11.7636 − 121.8kV ; V4c = 11.7636 121.8kV ]
Figure 5:
Figure 6:
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EEL303: Power Engineering I - Tutorial 9
Solution:
The transformers are solidly grounded on both sides, the zero-sequence network is
fully connected, as shown in Figure. 5(d). Since fault takes place on bus 3, third
column of ZBus of every sequence network is required. These can be obtained from
the respective YBus . Third column of each of the sequence network is given as follows:

0
ZBus−3



=








j0.0493
j0.0789
j0.0789





 j0.1104
 j0.1104
j0.0701

 1

 2

 ; ZBus−3 = 
 ; ZBus−3 = 

 j0.1696
 j0.1696
j0.1999










j0.1407
j0.1211
j0.1211
Since, the LG fault is at bus 3, we must connect the Thevenin equivalent circuits
of the sequence networks in series, as shown in Figure. 6. From the figure we can
calculate the symmetrical components of the current If A out of the system and into
the fault using prefault voltage Vf as
I(f A)(0) = I(f A)(1) = I(f A)(2) =
Vf
(1)
Z33
+
(2)
Z33
+
(0)
Z33
=
16 0
= −j1.8549p.u
j(0.1696 + 0.1696 + 0.1999)
The total current in the fault is
(0)
I(f A) = 3I(f A) = −j5.5648p.u
√
and since the base current in the HV transmission line is 100000/ 3×345 = 167.35A,
we have The total current in the fault is
I(f A) = −j5.5648 × 167.35 = 9316 270A
The phase-a sequence voltages at bus 3 are
(0)
(0) (0)
V3a = 0 − Z33 If A = −(j0.1999)(−j1.8549) = −0.3708p.u
(1)
(1) (1)
V3a = Vf − Z33 If A = 1 − (j0.1696)(−j1.8549) = 0.6854p.u
(2)
(2) (2)
V3a = 0 − Z33 If A = −(j0.1696)(−j1.8549) = −0.3146p.u
From the above symmetrical components we can calculate a − b − c LG voltages at
bus 3 as follows:
 
 

  

0
0
−0.3708
1 1 1
V3a
 
 

  

 V3b  1 a2 a  0.6854  −0.5562 − j0.8660 1.02936 − 122.71
=
=

 =







  

−0.5562 + j0.8660
1 a a2 −0.3146
V3c
1.02936 122.71
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EEL303: Power Engineering I - Tutorial 9
The phase-a sequence voltages at bus 4, the terminals of machine 2, are
(0)
(0) (0)
V4a = −Z43 If A = −(j0.1407)(−j1.8549) = −0.2610p.u
(1)
(1) (1)
V4a = Vf − Z43 If A = 1 − (j0.1211)(−j1.8549) = 0.7754p.u
(2)
(2) (2)
V4a = −Z43 If A = −(j0.1211)(−j1.8549) = −0.2246p.u
Note that subscripts A and a denote voltages and currents in the high voltage and
low voltage circuits respectively, of the Star-Star connected transformer. No phase
shift is involved. From the above symmetrical components we can calculate a − b − c
LG voltages at bus 4 as follows:

 
 


0.28986 0
1 1 1
0.2898 + j0.0
−0.2610

 
 
  

 V4b  1 a2 a  0.7754  −0.5364 − j0.8660 1.01876 − 121.8

=
=
 =


 
 
  

2
1.01876 121.8
−0.5364 + j0.8660
1 a a
−0.2246
V4c

V4a

To express the line-to-ground voltages at faulted bus
√ and at terminals of machine 2
in kV, we multiply the respective p.u values by 20/ 3, which gives
V3a = 0; V3b = 11.8856 − 122.71kV ; V3c = 11.8856 122.71kV ;
V4a = 3.3466 0kV ; V4b = 11.7636 − 121.8kV ; V4c = 11.7636 121.8kV ;
3. The system shown in Figure 7. is operating at nominal system voltage without prefault currents when a bolted LL fault occurs at bus 3 on lines B and C. Using the
bus impedance matrices of the sequence networks for subtransient conditions, determine
the currents in the fault, the line-to-line voltages at the fault bus, and the line-to-line
voltages at the terminals of machine 2.
Figure 7:
Solution: From Figure 8 the sequence currents are calculated as follows:
Vf
16 90
(1)
(2)
I(f A) = −I(f A) = (1)
= −j2.9481p.u
=
(2)
j(0.1696 + 0.1696)
Z33 + Z33
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EEL303: Power Engineering I - Tutorial 9
Figure 8:
Uppercase A is used here because the fault is in the high voltage transmission line
circuit. Since, I(f A)(0) = 0, the components of currents in the fault are calculated
from
(1)
(2)
I(f A) = I(f A) + I(f A) = −j2.9481 + j2.9481 = 0
(1)
(2)
I(f B) = a2 I(f A) + aI(f A) = −j2.9481(−0.5 − j0.866) + j2.9481(−0.5 + j0.866)
= −5.1061 + j0p.u
I(f C) = −I(f B) = 5.1061 + j0p.u
The base current in the transmission line is 167.35A, and so
I(f A) = 0
I(f B) = −5.1061 × 167.35 = 8556 180A
I(f C) = −5.1061 × 167.35 = 8556 0A
Symmetrical components of phase-A voltage to ground at bus 3 are
0
V3A
=0
1
2
1 1
V3A
= V3A
= 1 − Zkk
If A = 1 − (j0.1696)(−j2.9481) = 0.5 + j0p.u
Line-to-ground voltages at fault bus 3 are
(0)
(1)
(2)
V(3A) = V(3A) + V(3A) + V(3A) = 0 + 0.5 + 0.5 = 16 0p.u
(0)
(1)
(2)
V(3B) = V(3A) + a2 V(3A) + aV(3A) = 0 + a2 0.5 + a0.5 = 0.56 180p.u
V(3C) = V(3B) = 0.56 180p.u
Line-to-line voltages at fault bus 3 are
V(3,AB) = V(3A) − V(3B) = (1 + j0) − (−0.5 + j0) = 1.56 0p.u
V(3,BC) = V(3B) − V(3C) = (−0.5 + j0) − (−0.5 + j0) = 0
V(3,CA) = V(3C) − V(3A) = (−0.5 + j0) − (−1.0 + j0) = 1.56 180p.u
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In volts, the line-to-line voltages are
345
V(3,AB) = 1.56 0 × √ = 2996 0kV
3
EEL303: Power Engineering I - Tutorial 9
V(3,BC) = 0
345
V(3,CA) = 1.56 0 × √ = 2996 180kV
3
The sequence voltages of phase A at bus 4 are
(0) (0)
0
V4A
= −Z43 If A = 0
(1) (1)
1
V4A
= Vf − Z43 If A = 1 − (j0.1211)(−j2.9481) = 0.643p.u
(2) (2)
2
V4A
= −Z43 If A = −(j0.1211)(−j2.9481) = 0.357p.u
At machine 2 terminals, indicated by lower case a, the voltages are
0
V4a
=0
1
1 6
V4a
= V4A
− 30 = 0.6436 − 30 = 0.5569 − j0.3215p.u
2
2 6
30 = 0.3576 30 = 0.3092 + j0.1785p.u
V4a
= V4A
This is according to ASA where HT side positive sequence voltage leads corresponding LT side voltage by 30o . The opposite holds good for negative sequence voltage.
0
1
2
V4a = V4a
+ V4a
+ V4a
= 0.8661 − j0.1430 = 0.87786 − 9.4p.u
For phase-b of machine 2,
0
V4b0 = V4a
=0
1
V4b1 = a2 V4a
= −0.5569 − j0.3215p.u
2
2
V4a
= aV4a
= −0.3092 + j0.1785p.u
V4b = V4b0 + V4b1 + V4b2 = −0.8661 − j0.143 = 0.87786 − 170.6p.u
and, for phase c of machine 2
0
V4c0 = V4a
=0
1
V4c1 = aV4a
= 0.6436 90p.u
2
V4c2 = a2 V4a
= 0.3576 − 90p.u
V4c = V4c0 + V4c1 + V4c2 = j0.286p.u
Line-to-line voltages at the terminals of machine 2 are
V4,ab = V4,a − V4,b = 1.7322 + j0p.u
V4,bc = V4,b − V4,c = −0.8661 − j0.429 = 0.96656 − 153.65p.u
V4,ca = V4,c − V4,a = −0.8661 + j0.429 = 0.96656 153.65p.u
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EEL303: Power Engineering I - Tutorial 9
Line-to-line voltages at machine 2 terminals are
20
V4,ab = 1.73226 0 × √ = 206 0kV
3
20
V4,bc = 0.96656 − 153.65 × √ = 11.26 − 153.65kV
3
20
V4,ca = 0.96656 153.65 × √ = 11.26 153.65kV
3
4. Find the sub-transient currents and the line-to-line voltages at the fault under subtransient conditions when a double line-to-ground fault, with phases B and C involved and
Zf = 0 occurs at the terminals of machine 2 in the system of Figure 9.Assume that the
(0)
system is unloaded and operating at rated voltage when the fault occurs. [Z44 = j0.19,
(1)
(2)
Z44 = Z44 = j0.1437.]
[Ans:If a = 0; If b = 19, 2626 154.6o A; If c = 19, 2626 25.4o A;If = 16, 5386 90o A;
V4,ab = 12.5686 0o kV ; V4,bc = 0; V4,ca = 12.5686 180okV ]
Figure 9:
Figure 10:
Solution: To simulate the double line-to-ground fault at bus 4, we connect the
Thevenin equivalents of all three sequence networks in parallel as shown in Figure
10, from which we obtain
Vf
1 + j0
(1)
If a =
= −j4.4342p.u
=
(2) (0)
(j0.1437)(j0.19)
Z44 Z44
(1)
j0.1437 +
Z44 + (2)
(0)
(j0.1437 + j0.19)
Z44 + Z44
Electrical Engineering Dept - IIT Delhi
EEL303: Power Engineering I - Tutorial 9
Therefore, the sequence voltages at the fault are
(1)
(2)
(0)
(1)
(1)
V4a = V4a = V4a = Vf − If a Z44 = 1 − (−j4.4342)(j0.1437) = 0.3628p.u
Current injections into the negative- and zero-sequence networks at the fault bus are
calculated by current division as follows:
(0)
(2)
If a
=
(1)
−If a
=
(1)
−If a
Z44
(2)
Z44
+
(0)
Z44
= j4.4342 ×
j0.19
= j2.5247p.u
j(0.1437 + 0.19)
= j4.4342 ×
j0.1437
= j1.9095p.u
j(0.1437 + 0.19)
(2)
(0)
If a
Z44
(2)
Z44
(0)
Z44
+
The currents out of the system at the fault point are
(0)
(1)
(2)
If a = If a + If a + If a = j1.9095 − j4.4342 + j2.5247 = 0
(0)
(1)
(2)
If b = If a +a2 If a +aIf a = 1.90956 90o +(16 240o)(4.43426 −90o )+(16 120o)(2.52476 90o )
= −6.0266 + j2.8642 = 6.67266 154.6op.u
(0)
(1)
(2)
If c = If a +aIf a +a2 If a = 1.90956 90o +(16 120o )(4.43426 −90o )+(16 240o )(2.52476 90o )
= 6.0266 + j2.8642 = 6.67266 25.4o p.u
and the current If into the ground is
(0)
If = If b + If c = 3If a = j5.7285p.u
Calculating a-b-c voltages at the fault bus, we find that
(0)
(1)
(2)
(1)
V4a = V4a + V4a + V4a = 3V4a = 3 × 0.3628 = 1.0884p.u
V4b = V4c = 0
V4,ab = V4a − V4b = 1.0884p.u
V4,bc = V4b − V4c = 0
V4,ca = V4c − V4a = −1.0884p.u
√
Base current equals 100 × 103 /( 3 × 20) = 2887 A in the circuit of machine 2, and
so we find that
If a = 0
If b = 2887 × 6.67266 154.6o = 19, 2626 154.6o A
If b = 2887 × 6.67266 25.4o = 19, 2626 25.4o A
If c = 2887 × 5.72856 90o = 16, 5386 90o A
√
The base line-to-neutral voltage in machine 2 is 20/ 3 kV, and so
√
V4,ab = 1.0884 × 20/ 3 = 12.5686 0o kV
V4,ca
V4,bc = 0
√
= −1.0884 × 20/ 3 = 12.5686 180o kV
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