Protection Application Handbook (англ. яз.)
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UNBALANCED FAULT CALCULATION
Fault Calculation
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Equations (13) to (17) gives for a two phase fault the following
block diagram:
Insert 13, 17 into 7, 8 and 9 and you will get:
and:
The unknown phase currents and voltages can be calculated by
inserting 11,13 and 14 into 17 into the basic equations:
Before the fault occurred there were only positive sequence volt-
age.
E
a
= E
1
E
b
= a
2
E
1
E
c
= aE
1
I
1
I
2
Z
2
I
1
= -I
2
I
0
V
1
E
1
1
Z
1
Z
1
Z
2
Z
bc
++
-----------------------------------
–
(18)=
V
2
E
1
Z
2
Z
1
Z
2
Z
bc
++
-----------------------------------
(19)=
V
0
0=
(20)
I
b
I
c
– a
2
a–
E
1
Z
1
Z
2
Z
bc
++
-----------------------------------==
(21)
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“E
a
“, “E
b
” and “E
c
” are the voltages, when the system is in a
balanced condition.
This gives:
and:
and:
In reality, we have “Z
1
= Z
2
= Z”. The following network is
achieved:
The following equations are achieved:
V
a
= E
a
(27)
I
b
I
c
–
E
b
E
c
–
Z
1
Z
2
Z
bc
++
-----------------------------------
==
(22)
V
a
E
a
1
Z
1
Z
2
–
Z
1
Z
2
Z
bc
++
-----------------------------------
–
=
(23)
V
b
E
b
E
b
Z
1
E
c
Z
2
–
Z
1
Z
2
Z
bc
++
-----------------------------------
–=
(24)
V
c
E
c
E
c
Z
1
E
b
Z
2
–
Z
1
Z
2
Z
bc
++
-----------------------------------
–=
(25)
I
b
I–
c
E
b
E
c
–
2Z Z
bc
+
-----------------------
==
(26)
UNBALANCED FAULT CALCULATION
Fault Calculation
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BA THS / BU Transmission Systems and Substations
LEC Support Programme
V
b
= E
b
- ZI
b
(28)
V
c
= E
c
- ZI
c
(29)
For a two phase fault without fault resistance “Z
bc
” is set to 0.
Impedance between phase and earth
The general and the basic equations will still be the same as in
“Impedance between two phases” but the following special equa-
tions are achieved:
The special equations:
I
b
= 0 (30)
I
c
= 0 (31)
V
a
= I
a
Z
a
(32)
E
1
E
2
Z
1
Z
2
Z
0
Z
a
V
C
E
3
V
B
V
A
I
a
= 0
= 0
0
UNBALANCED FAULT CALCULATION
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Transformation of the equations 30, 31, 5, 6, as in ”Impedance
between two phases” gives the following result:
This gives the following figure:
I
0
I
1
I
2
==
(33)
I
1
I
2
I
0
E
1
3Z
a
Z
0
Z
1
Z
2
+++
-------------------------------------------------
===
(34)
V
1
E
1
E
1
Z
1
3Z
a
Z
0
Z
1
Z
2
+++
-------------------------------------------------
–=
(35)
V
2
E
1
Z
2
3Z
a
Z
0
Z
1
Z
2
+++
-------------------------------------------------
=
(36)
V
0
E
1
Z
0
3Z
a
Z
0
Z
1
Z
2
+++
-------------------------------------------------
=
(37)
I
a
3E
a
2Z Z
0
3Z
a
++
-------------------------------------
=
(38)
I
b
I
c
0==
(39)
UNBALANCED FAULT CALCULATION
Fault Calculation
for worldwide cooperation
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BA THS / BU Transmission Systems and Substations
LEC Support Programme
This simple network is achieved:
Example 1
Generator: X
g
= 24.2 Ω/ph, (X
1
= X
2
)
Transformers: X
k1
= 12.1 Ω/ph, X
k2
= 10 Ω/ph
(Zero and positive sequence impedances equal).
Network: X
n
= 8.3 Ω/ph
Line: X
1
= 40 Ω/ph, X
0
= 120 Ω/ph
V
a
E
a
1
2Z Z
0
+
2Z Z
0
3Z
a
++
-------------------------------------
–
=
(40)
V
b
E
b
E
a
Z
0
Z–
2Z Z
0
3Z
a
++
-------------------------------------
–=
(41)
V
c
E
c
E
a
Z
0
Z–
2Z Z
0
3Z
a
++
-------------------------------------
–=
(42)
E
C
Z
V
C
0
Z
a
1/3(Z
0
-Z)
V
B
V
A
E
B
E
A
Z
Z
UNBALANCED FAULT CALCULATION
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Calculate the fault current for:
a) Two phase fault with zero fault resistance.
b) Single phase solid earth fault.
a) Two phase short circuit
This scheme is valid when “Z
1
=Z
2
=Z”.
In this case “Z
bc
” is set to 0.
The value of “Z” is calculated:
Then “Z” = 22.4 Ω/ph and “I
b
” = 1.23 kA/ph
b) Single phase solid earth fault
This scheme is also valid when “Z
1
=Z
2
=Z” and “Z” = 22.4 Ω/ph
Z
X
g
X
k1
+()X
L
X
k2
X
n
++()
X
g
X
k1
X
L
X
k2
X
n
++++
--------------------------------------------------------------------
=
UNBALANCED FAULT CALCULATION
Fault Calculation
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This gives “X
0tot
” = 11.1 Ω/ph and
It should be noted that the current at single phase fault is higher
than the fault current at three phase fault.
Example 2:
Transformer:
20MVA, 16/77 kV, x
k
=8%, Yd11
Generator: 20MVA, 16 kV, x(transient) =25%, X
2
= X
1
Line: X
1
= 84 Ω/ph, X
0
= 300 Ω/ph
The line is considered unloaded before the fault and all resis-
tance and capacitance is neglected. The voltage at the fault po-
sition is 75 kV before the fault.
Calculate the fault current through the earth connection of the
transformer, the phase currents on both sides of the transformer
and the voltages (to earth) in the HV terminals of the transformer.
I
a
3E
a
2Z Z
0
3Z
a
++
-------------------------------------
=
X
0tot
X
k1
X
0L
X
k2
+()
X
k1
X
0L
X
k2
++
-----------------------------------------
=
I
a
55 3×
3 11.1+2 22.4×()
-------------------------------------------------
1.71 kA
=
=
UNBALANCED FAULT CALCULATION
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LEC Support Programme
Calculation of the 77 kV impedances.
Generator:
Transformer:
Line:
X
1
= X
2
= 84 Ω/ph, X
0
= 300 Ω/ph
The following block diagram is achieved:
X
1
= X
2
= 181.7 Ω/ph
X
0
= 323.7 Ω/ph
The current through the earth connection of the transformer.
I
g
= 3xI
0
= 0.189 kA
The component voltages at the HV side:
U
1
= E
1
+dU
1
= 43.3-(j97.7)(-j0.063) = 37.15 kV
U
2
= dU
2
= -j97.7(-j0.063) = -6.15 kV
U
0
= dU
0
= -j23.7(-j0.063) = -1.49 kV
Phase voltages:
X
1
X
2
0.25
16
2
20
---------
×
77
16
------
2
74Ω/ph=×==
X
1
X
2
0.08
77
20
------
2
23.7Ω/ph=×==
I
1
I
2
I
0
=
75
3 2x181 7, 323 7,+()
----------------------------------------------------------- 0 .063 kA/ph===
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U
a
= U
0
+U
1
+U
2
= 29.5/0° kV
U
b
= U
0
+a
2
U
1
+aU
2
= 41.3/-114.4° kV
U
c
= U
0
+aU
1
+a
2
U
2
= 41.3/114.4° kV
Phase currents:
I
a
= I
0
+I
1
+I
2
= 3x0.063 = 0.19 kA
I
b
= I
0
+a
2
I
1
+aI
2
= 0.063(1+a
2
+a) = 0 kA
I
c
= I
0
+aI
1
+a
2
I
2
= 0.063(1+a
2
+a) = 0 kA
Phase currents on the LV side:
Connection:
Figure 2. The positive sequence current is turned +30°, while the negative
sequence current is turned -30°.
At the 16 kV side there is no zero sequence current as the trans-
former is Yd connected
“I
0
”=0.
The phase currents at the 16 kV side:
I
a
77
16
------
0 0.063e
j30°
0.063e
j– 30°
++()==
77
16
------
= 3 0,063× 0.52 kA=
I
b
77
16
------
0 0.063e
j30°
a
2
0.063e
j30°–
a++()==
77
16
------
0.063e
j30°
e
j120°–
0.063e
j30°–
e
j120°
+()= 0kA=
UNBALANCED FAULT CALCULATION
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I
c
77
16
------
0 0.063e
j30°
a 0.063e
j30°–
a
2
++()==
77
16
------
0.063e
j30°
e
j120°
0.063e
j30°–
e
j120°–
+()=
77
16
------
3 0,063× 0.52 kA==