SESSION 3
Short Circuit Calculations, Unsymmetrical Faults
Leonard Bohmann, Michigan State University
Elham Makram, Clemson University
1
Short Circuit Calculations
Leonard Bohmann
Michigan State University
2
Line to Ground Short
2003 IEEE T & D CONFERENCE
3
Short Circuit Fault Currents
7
dc offset
decaying sinusoid
6
5
Amps (kA)
4
3
2
1
0
-1 0
t
0.05
0.1
0.15
0.2
-2
-3
high frequency transient
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Short Circuit Fault Currents (continued)
X(t)
v(t)
R
if
v( t ) = Vm cos(ωt + α)
Z( t )∠θ = R + jX( t )
then
V
i( t ) = m [cos(ωt + α − θ)
Z( t )
5
4
3
2
− cos(ωt 0 + α − θ)e
1
0
-1
0
0.05
0.1
0.15
− Rωt
X( t )
0.2
-2
-3
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]
dc offset
5
4
3
2
1
0
-1
0
0.05
0.1
0.15
0.2
-2
-3
Vm
i( t ) =
Z
cos(ωt + α − θ) − cos(ωt + α − θ)e − Rωt X 
0


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Variable magnitude
6
4
2
0
0
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
0.09
0.1
-2
-4
-6
t
Vm
i( t ) =
cos(ωt + α − θ)
Z( t )
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reactance (pu)
Reactance of Synchronous Machine
Xd
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
X'd
X"d
0
0.2
0.4
0.6
0.8
1
1.2
1.4
t
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• Use sub-transient reactance to calculate
fault currents
– worst case for when CB is opening
• Do not calculate dc offset directly
– use X/R to approximate offset
(asymmetry ratio) at opening time
– X/R is the time constant of the decay in
cycles
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ZG1
ZL2
1
ZT15
2
ZT12
ZT23
3
5
If
ZT35
ZL5
ZT34
4
How do you
calculate the
fault current (If)
at bus 3?
ZG4
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Use Thevenin equivalent
3
ZTh
If
VTh
• VTh: open circuit
voltage
– also called the prefault voltage, Vf
– often use 1∠0°
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Use for:
• symmetrical faults
– single phase networks
– three phase faults in three phase
networks (per phase equivalent)
• concept for solving all faults
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• This method calculates the current
going into the fault
• Often want the current elsewhere
in the network (through a circuit
breaker, in bus bars, in other lines,
etc.)
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A current source, with the magnitude of
the fault current, placed at the fault, has
the same effect on the rest of the circuit
as the short circuit.
3
ZTh
Vf
3
ZTh
If
If
Vf
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Adding a current source to
the original circuit has
the same electrical
Z
effect as adding it to
1
the equivalent circuit
ZL2
G1
ZT15
2
ZT12
3
5
3
ZT35
If
ZT34
ZTh
If
Vf
ZT23
ZL5
4
ZG4
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ZG1
ZL2
1
ZT15
All the voltages and all
the currents can be found
using the new circuit.
2
ZT12
ZT23
3
5
If
ZT35
ZT34
ZL5
4
ZG4
Use superposition:
– Use the original sources
(or assume currents are 0
and voltages are 1∠0°)
– Use just the current
source, gives changes
due to the fault
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Calculation details
Bus Admittance Matrix: Ybus
ZG1
ZL2
1
I = Ybus V
ZT15
vector of current
injections
vector of bus
voltages
5
2
ZT12
ZT23
3
ZT35
If
ZT34
ZL5
4
ZG4
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Diagonals:
Σ of Y connected at node
Y22 = 1/ZT12 + 1/ZL2 + 1/ZT23
ZG1
ZL2
1
ZT15
5
2
ZT12
3
ZT35
ZT34
ZL5
4
ZG4
ZT23
Ybus
If
Y11
Y
 21
=0

0
Y51
0 Y15 
Y22 Y23 0 0 
Y32 Y33 Y34 Y35 

0 Y43 Y44 0 
0 Y53 0 Y55 
Y12 0
Off diagonals:
- of Y connected between nodes
Y53 = -1/ZT35
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Ybus is sparse (lots of zero)
– small size
– easy to build
– quick algorithms to manipulate
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Bus Impedance Matrix: Zbus
V = Zbus I
ZG1
ZL2
vector of bus vector of current
injections
voltages
1
ZT15
5
Zbus = Ybus-1
2
ZT12
ZT23
3
ZT35
If
ZT34
ZL5
4
ZG4
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Z bus
 Z11
Z
 21
=  Z 31

 Z 41
 Z 51
Z12 Z13 Z14 Z15 
Z 22 Z 23 Z 24 Z 25 
Z 32 Z 33 Z 34 Z 35 

Z 42 Z 43 Z 44 Z 45 
Z 52 Z 53 Z 54 Z 55 
Zbus is a full matrix
– slow to invert
– slow to manipulate
– avoid using if
possible
Diagonal element is the Thevenin
Impedance at that bus
– Z33 is the Thevenin Impedance at bus 3
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To find the change in voltages due to
the fault current at bus 3:
 ∆V1   Z11
∆V   Z
 2   21
 ∆V3  =  Z 31

 
∆V4   Z 41
 ∆V5   Z 51
Z12 Z13 Z14 Z15   0 
Z 22 Z 23 Z 24 Z 25   0 
Z 32 Z 33 Z 34 Z 35  ⋅ - I f 
  
Z 42 Z 43 Z 44 Z 45   0 
Z 52 Z 53 Z 54 Z 55   0 
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To calculate current in line 2-3:
1
∆I 23f
∆V2 − ∆V3
=
ZT 23
2
∆I23f
ZT23
3
5
If
4
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Asymmetry ratio: S
• Method calculates the RMS worst case
of the 60 Hz component
– this is needed to set protective relays
• For circuit breaker selection, need to
know the total RMS current (60 Hz
and dc offset)
– Circuit breakers are rated on symmetrical
fault current, assuming an X/R ratio of 15
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Asymmetry ratio
I RMS = I + I
2
ac
I RMS
2
dc

= I +  2I ac e

2
ac
I RMS = I ac 1 + 2 ⋅ e
− ωt
XR
− 2 ωt
X R



2
− 2 ωt
I RMS
= S(t ) where S(t) is 1 + 2 ⋅ e X R
I ac
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Calculating X/R
ZG1
• Equations presented are
for a single order system
• For higher order systems,
ZL2
1
ZT15
2
ZT12
ZT23
3
5
ZT35
ZL5
If
ZT34
4
ZG4
– ignore R and find
equivalent X,
– ignore X, and find
equivalent R
• Actual decay is typically
quicker than this
approximation
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Algorithm
• Calculate Thevenin Impedance
• Calculate current into fault
• Find the needed elements of Zbus (not all
of them)
• Find the needed changes in bus voltages
• Find the line currents of of interest
• Find the dc offset
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Example
• The oneline drawing of a system, along with Zbus
are shown below.
i) Calculate the fault current for a
three phase solid fault at bus 3.
j0.1pu
j10 pu
ii) Calculate the voltages at bus 1,
2, and 3 during this fault.
2
1
iii) Calculate the current flowing
j 0.08 pu
j0.05 pu
j0.025 pu
from bus 2 to bus 3.
3
5
If
j0.03 pu
j15
pu
j0.05 pu
4
Z bus
j0.1 pu
0. 0656
0.0538

= j 0. 0502

0. 0334
0. 0559
0.0538 0. 0502 0. 0334 0. 0559
0. 0831 0.0675 0. 0450 0. 0623
0.0675 0. 0731 0.0487 0. 0644 pu

0. 0450 0.0487 0. 0658 0. 0429
0.0623 0. 0644 0. 0429 0. 0799
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Part (i): Calculate the fault current for a three
phase solid fault at bus 3.
3
• Assume all initial bus
voltages are 1∠0°
j0.731
1∠0°
– therefore all initial
currents are 0
• Use the 3,3 element
of Zbus for the
Thevenin impedance.
If
1
If =
j 0.0731
I f = − j13. 68 pu
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Part (ii): Calculate the voltages at bus 2, 3, and
4 during this fault.
Multiply the negative of the fault current by elements in
the 3rd column to get the change in voltages
− 0.687
− 0.923


 −1  =


 ∆V4 
 ∆V5 
0.0656 0.0538 0.0502 0.0334 0.0559  0 
0.0538 0.0831 0.0675 0.0450 0.0623  0 

 

j 0.0502 0.0675 0.0731 0.0487 0.0644 ⋅ - - j13.68

 

0.0334 0.0450 0.0487 0.0658 0.0429  0 
0.0559 0.0623 0.0644 0.0429 0.0799  0 
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Part (ii): Calculate the voltages at bus 2, 3, and
4 during this fault. (continued)
The voltages during the fault are the initial voltages
minus the changes in the voltages
V1f = V1i − ∆V1 = 1∠0° − 0.687 = 0.313pu
V2f = V2i − ∆V2 = 1∠0° − 0.923 = 0.0766pu
V3f = V3i − ∆V3 = 1∠0° −1 = 0 pu
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Part (iii): Calculate the current flowing from bus 2
to bus 3.
The current is the difference
between the voltages
divided by the impedance
I 23f
I 23f
1
2
-j3.86 pu
V2f − V3f
=
ZT23
5
0.0766− 0
=
= − j3.86 pu
j0.025
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3
-j13.68 pu
4
32
Problem
A small power system is sketched below along with Zbus.
j0.1 pu
j0.1 pu
1
j0.05 pu
2
j0.025 pu
j 0.08
Z bus
3
j0.05 pu
4
 4. 783
3. 044
= j
 3. 261

 2. 174
3. 044
4. 534
3. 634
2. 422
3. 261
3. 634
4. 658
3. 106
2. 174
2. 422
×10 − 2 pu
3. 106

5. 404
j0.1 pu
a) What is the fault current for a three-phase fault to
ground at bus 3?
b) What are the voltages at busses 2, 3, and 4?
c) What is the current due to the fault from bus 4 to 3 and
from 2 to 3?
33
2003 IEEE T & D CONFERENCE
Unsymmetrical Faults
Elham Makram
Clemson University
2003 IEEE T & D CONFERENCE
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SINGLE LINE TO GROUND FAULT
Boundary Conditions at fault location
a
I bf = I cf = 0
b
Vf = Zf * Iaf
c
Iaf
Ibf
From the current equations,
Icf
I0 + a 2 I1 + aI2 = I 0 + aI1 + a 2 I 2 = 0
(a 2 – a)I1 = (a 2 – a)I2
Zf
I1 = I 2
Thus I 0 + ( a 2 + a) I1 = 0
n
I0 = I1 = I 2
Fig.1. Single line to ground fault on phase A
Now Va = Zf * I af
Or, V0 + V1 + V2 = Zf * 3I0
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SINGLE LINE TO GROUND FAULT (CONTD.)
To satisfy the voltage and current boundary conditions, the sequence networks should be
connected in series as shown in Fig. 2.
I0
Zero
Sequence
Network
+
V0
I1
Zero
Sequence
Network
+
V1
-
3Zf
I2
Zero
Sequence
Network
+
V2
-
Fig.2. Sequence connection for A-G fault
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LINE TO LINE FAULT
Current boundary Conditions
I af = 0
I bf = - I cf
a
b
Thus,
c
I0 = 0
Iaf
Ibf
Icf
+
a 2 I1 + aI2 + aI1 + a 2 I 2 = 0
(a 2 + a)I1 = - (a 2 + a)I2
Vaf
Vbf
Zf
Vcf
I1 = - I 2
Voltage boundary conditions
Vbf = Vcf + Zf I bf
n
Fig.3. Line to line fault
V0 + a 2 V1 + aV2 = Zf (I 0 + a 2 I1 + aI 2 )
+ V0 + aV1 + a 2 V2
This leads to
(a 2 - a)V1 = (a 2 - a)I1 Zf + (a 2 - a)V2
V1 = V2 + Zf I1
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LINE TO LINE FAULT (CONTD.)
To satisfy the voltage and current boundary conditions, the sequence networks should
be connected as shown in Fig. 4.
I1
Positive
Sequence
Network
+
V1
I2
Negative
Sequence
Network
Zf
+
V2
-
Fig.4. Connection of sequence network for BC fault
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DOUBLE LINE TO GROUND FAULT
CurrentboundaryConditions
a
at the faultlocation,
Iaf = 0
b
Thus,
I0 + I1 + I2 = 0
c
Iaf
Ibf
Icf
Voltageboundaryconditions
Vbf = (Ibf + Icf )Zf
Vcf
and,
Vbf = Vcf
+
Vaf
Vbf
Zf
Thus,
n
V0 + a 2V1 + aV2 = V0 + aV1 + a2V2
Fig.5. Double line to ground fault
ThusV1 = V2
Also,
V0 + a 2V1 + aV2 = ( 2I0 + a2I1 +
aI2 + aI1 + a 2I2 ) Zf
ThusV0 - V1 = 3Zf I0
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DOUBLE LINE TO GROUND FAULT (CONTD.)
Considering the voltage and current boundary conditions, the sequence networks are
connected as shown in Fig. 6.
I0
Zero
Sequence
Network
+
V0
-
3Zf
I1
Positive
Sequence
Network
+
V1
I2
Negative
Sequence
Network
+
V2
-
Fig.6. Sequence network connection for double
line to ground fault
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Example 1
Consider the power system shown in Fig. 7.
T1
1 P
G1
2
T. L.
T2
G2
Fig. 7. Example 1
Ratings
G1 and G2: 100 MVA, 20 kV
Find the voltages at Bus 1 and the
currents from T1 to P due to
X1 = X2 = 0.3 pu
(i)
X0 = 0.04 pu
(ii) Line to line fault at P
T1 and T2: 100 MVA, 345/20 kV
Single line to ground fault at P
(iii) Double line to ground fault at P
XT1 = XT2 =0.08 pu
T. L.
X1 = X2 = 0.15 pu
X0 = 0.5 pu
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Positive sequence network
j0.3
1
j0.08
j0.3
2
P
j0.15
j0.38
j0.2213
j0.53
j0.08
P
P
If E1 = E 2 = V = 1∠0°
Z th = (0.3 + j0.8) || (j0.3 + j0.08 + j0.15) = j0.2213
Negative sequence network
Zero sequence network
j0.2213
P
Z1th = j0.2213
j0.08
j0.5
P
2003 IEEE T & D CONFERENCE
Z0th = j0.08
8
(i) Single line to ground fault
I1 = I 2 = I 0 =
E
(Z0th + Z1th
1∠0°
= - j1.9134
+ Z 2th ) (j0.2213 + j0.2213 + j0.08)
The following are the sequence current components from T1 - - > P
0.53
= − j1.114
(0.53 + 0.38)
I 2 = - j1.114
I1 = I1 *
I 0 = - j1.9134
The sequence voltage components at the fault location are :
V1 = E1 - I1Z1 = 1.0 – (-j1.114)(j0.38) = 0.57668
V2 = - I 2 Z 2 = - (j1.114) (j0.38) = - 0.42332
V0 = - I 0 Z0 = - j1.913 (j0.8) = - 0.15304
The current phase components from T1 - - > P
I a  1 1
I  = 1 a 2
 b 
 I c  1 a
1  I a0 
a  *  I a1 
a 2  I a2 
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I a = -j1.114 - j1.114 - j1.913 = - j3.027
I b = a 2 (-j1.114) + a(-j1.114) – j1.913 = - j0.799
I c = a(-j1.114) + a 2 (-j1.114) – j1.913 = - j0.799
c
b
-j0.799
-j0.799
-j3.027
-j2.712
c
-j5.739= 3 I0
The phase voltages at the fault location are :
Va = V0 + V1 + V2 = - 0.15304 + 0.57668 – 0.42332 = 0
Vb = V0 + a 2 V1 + aV2 = - 0.15304 + (1∠240°)(0.57668) + (1∠120°)(-0.42332)
= -0.2297 - j0.866
Vc = V0 + aV1 + a 2 V2 = - 0.15304 + (1∠120°)(0.57668) + (1∠240°)(-0.42332)
= -0.2297 + j0.866
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(ii) Line to line fault
Currents :
I1 = −I 2 =
(Z1th
E
1∠0°
=
= − j 2.26
+ Z 2th ) (j0.2213 + j0.2213)
I1 (T1 - - > P) = − j2.26
0.53
= − j1.316
(0.38 + 0.53 )
I 2 (T1 - - > P) = j1.316
Ia = 0
I b = a 2 (-j1.316) + a(j1.316) = 2.279
I c = a (-j1.316) + a 2 (j1.316) = 2.279
Voltages :
Va1 = E − I a1Z1 = 1.0 - (-j1.316)(j0.38) = 0.5
Va2 = - I a2 Z 2 = 0.5
Va = 1∠0°
Vb = -0.5∠0°
Vc = 0.5∠0°
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SEQUENCE PHASE SHIFT THROUGH WYE/DELTA TRANSFORMER BANK
•Positive and negative sequence impedance is independent of connection
•These networks ignore phase shift
•When current and voltages are transformed from one side to the other phase shift must
be considered.
•Standard ANSI connection has been shown in figs. 8 (a) and 8 (b).
High side Low side
High side
Low side Ia
n:1
n:1
IA
a
A
Ia
I
Ia/n
IA
nIA
AA
Ib
IB
b
B
Ib
IB
Ib/n
IB
nIB
B
Ic
IC
c
C
Ic
IC
Ic/n
IC
nIC
C
A
a
A
a
C
C
B c b
Fig. 8 (a) Wye/Delta
a
b
c
Bc b
Fig. 8 (b) Delta/Wye
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Refer to fig. 8(a)
Positive sequence current :
I a 1 = n(I A1 - I C1 ) = n(I A1 - aI A1 ) = nI A1 (1 - a) = n 3 I A1 ∠ - 30 °
In per unit,
I A1 = I a1 ∠ 30 °
Positive sequence volatge :
VA1 = n(V a1 - a 2 Va1 ) = nV a (1 - a 2 ) = n 3 Va1 ∠ 30 °
or in per unit,
VA1 = Va1 ∠ 30 °
Negative sequence current :
I a 2 = n(I A2 - I C2 ) = n(I A2 - a 2 I A2 ) = nI A2 (1 - a 2 ) = n 3 I A2 ∠ 30 °
or In per unit
I A2 = I a2 ∠ − 30 °
Negative sequence voltage :
VA2 = n(V a2 - Vb2 ) = n(V a2 - aV a2 ) = nV a2 (1 - a) = n 3 Va2 ∠ − 30 °
or In per unit
VA2 = Va2 ∠ − 30 °
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Refer to fig. 8 (b)
Postitive sequence current :
3
1
1
I a1∠ 30 °
(I a1 - I b1 ) = (I a1 - a 2 I a1 ) =
n
n
n
or, in per unit,
I A1 = I a1∠ 30 °
I A1 =
Postitive sequence voltage :
1
3
(VA1 - aVA1 ) =
VA1 ∠ − 30 °
n
n
or, in per unit,
VA1 = Va1∠ 30 °
Va1 =
Negative sequence current :
3
1
I a2 ∠ − 30 °
(I a2 - aI a2 ) =
n
n
or In per unit
I A2 = I a2 ∠ − 30 °
I A2 =
Negative sequence voltage
VA2 = Va2 ∠ − 30 °
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In general, in Y - ∆ or ∆ - Y connections :
(i) Positive sequence High voltage = Positive sequence Low voltage∠ + 30°
(ii) Negative sequence High voltage = Negative sequence Low voltage∠ - 30°
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Example 2
Find the currents and voltages at G1 of Example 1, due to single line to ground fault at P.
(i) Currents :
I ag1 = - j1.114 ∠ - 30° = - j1.114 ∠ - 120°
I ag2 = - j1.114 ∠ + 30° = j1.114 ∠ - 60°
I ag0 = 0
I ag = I ag1
+ I ag2 = - j1.114 ∠ - 120° + j1.114 ∠ - 60° = - j1.9295
I bg = a 2 I ag1 + aIag2 = 1∠240° (1.114 ∠ - 120°) + 1∠120° (1.114 ∠ - 60°)
= (1.114 ∠120°) + (1.114 ∠60°) = j1.9295
I cg = 0
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Example 2 ( contd.)
(ii) Voltages :
Vag1 = E – I a1 Zg1 = 1.0 - (-j1.114)(j0.3)
= 0.666 (from sequence Network neglecting phase shift)
Vag2 = – I a2 Zg2
= - 0.3342 (from sequence Network neglecting phase shift)
Vag1 = 0.666 ∠ - 30°
Vag2 = - 0.334 ∠30°
Vag = Vag1 + Vag2 = 0.666 ∠ - 30° - 0.334 ∠30° = 0.577∠ − 60°
Vbg = a 2 Vag1 + aVag2 = 0.666 ∠210° - 0.334 ∠150°
= 0.577 ∠240.12°
Vcg = aVag1 + a 2 Vag2 = 1.0∠90°
 Vag  1 1
  
2
V
=
1
a
bg
  
 Vcg  1 a
 
1  Vag0   0.577∠ − 60° 


a  *  Vag1  = 0.577∠240.12°



2



1∠90°
a  Vag2  
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PROBLEMS
PROBLEM #1
For the system shown below, find the currents in the transformer and generator windings in
per unit and amperes for a single-line-to-ground fault at
(i)
P
(ii) R
15/115 kV
R
P
G1
G: 100 MVA, 15 kV
Transformer:
X1=X2= 0.1,
100 MVA, 15/115 kV
X0 = 0.05
X = 0.1
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PROBLEMS
PROBLEM #2
Solve examples 1 and 2 for a double-line-to-ground fault at P.
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PROBLEMS
PROBLEM #3
Solve examples 1 and 2 for a line-line fault at P.
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