Stevenson J. Power system analysis
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6 CHAPTER 12 UNSYMMETRICAL FAULTS
+
V{ = 1.0�
FIGU 12.10
jO.1696
I
(l)
f
A
I
®
It +
jO. 1696
jO.1999
[(0)
fA
V
(
2
)
3,1
®
,
Series connection of the Thevenin equivalents of the sequence networks for the single line-to-ground
fault of Example 12.2.
transformers are solidly grounded on both sides, the zero-sequence network is fully
connected, as shown in Fig. 12.9(b), and has the bus impedance matrix
@
i
O.lSS3
iO.1407
jO.0493
jO. 0347
Z(O) -
jO.1407
jO. 1999 jO.0701
jO.0493
hll
-
jO.0493
jO.070 I
jO. 1099
jO.1407
iO.O]47
jO .049]
jO.1407
iO.1SS]
Since the line-to-ground fault is at bus , we must connect the Thevenin
equivalent circuits of the sequence networks in series, as shown in Fig. 12.10. From
this gure we can calculate the symmetrical components of the current If
A
out of
the system and into the fault,
](0)
-
1(1)
-
1(2)
-
fA
-
fA
-
fA
-
Z
(I)
+
Zm +
Z
«l)
33 3
3
33
1.0
j(0.1696 + 0.1696 + 0.1999)
-
-
j1.8549 per unit
12.2 SINGLE LlNE-TO-GROUND FAULTS 487
The total current in the fault is
I
jA
=
3
I} = -j5.5643
per unit
and since the base current in the high-voltage transmission line is 100,000/ X
345 = 167.35 A, we have
IrA
=
-j5.5643 X 167. 35 =
931/2700
A
The phasc-a sequence voltages at bus
@
, the terminals of machine 2, are
calcu lated from Eqs. (12.6) with k
=
3 and j = 4,
-Z((J)
/
(IJ) =
-(
JO.140
7
)(-
J·
I.8549) = -0.261 0per unit
�'I� J /1
V
(I) V
-
Z(1)/(
I
)
=
1
-
(J·
0.
12
11)( -
J·1
.8549)
=
0.7754 per unit
-I II J � J /1
-z
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