Question 7.6: For the system of Example 7.5 find the voltages and currents...
For the system of Example 7.5 find the voltages and currents at the fault point for a double line-to-ground fault. Assume
Z_{f}=j0.05 p.u.
Z_{g}=j0.033 p.u.
The sequence network connection is as shown in Figure 7.27. Steps of the network reduction are also shown. From the figure, sequence currents are as follows:
I_{+} =\frac{1\angle 0}{0.45\angle 90° }= 2.24\angle – 90°
I_{-}=-I_{+} \left\lgroup\frac{0.29}{0.29+ 0.2585}\right\rgroup
= -1.18\angle -90°
I_{0} =-1.06\angle -90°
The sequence voltages are calculated as follows.
V_{+}=E_{+}-I_{+}Z_{+}
=1\angle 0-\left(2.24\angle -90°\right)\left(0.26\angle -90° \right)
=0.42
V_{-}=-I_{-}Z_{-}
=+\left(1.18\right) \left(0.2085\right) =0.25
V_{0}=-I_{0}Z_{0}
=\left(1.06\right) \left(0.14\right) =0.15
The phase currents are obtained as
I_{A}=0
I_{B} =\alpha ^{2} I_{+} +\alpha I_{-} +I_{0}
=\left(1\angle 240\right) \left(2.24\angle -90 °\right) +\left(1\angle 120\right) \left(-1.18\angle -90°\right) +\left(-1.06\angle 90°\right)
=3.36\angle 151.77°
I_{C} =\alpha I_{+} +\alpha ^{2} I_{-} +I_{0}
=\left(1\angle 120\right) \left(2.24\angle -90 °\right) +\left(1\angle 240\right) \left(-1.18\angle -90°\right) +\left(-1.06\angle 90°\right)
=3.36\angle 28.23°
The phase voltages are found as
V_{A}=V_{+}+V_{-}+V_{0}
=0.42+0.25+0.15
=0.82
V_{B} =\alpha ^{2} V_{+} +\alpha V_{-} +V_{0}
=\left(1\angle 240°\right) \left(0.42\right) +\left(1\angle 120°\right) \left(0.25\right) +\left(0.15\right)
=0.24\angle -141.49°
V_{C} =\alpha V_{+} +\alpha ^{2} V_{-} +V_{0}
=\left(1\angle 120°\right) \left(0.42\right) +\left(1\angle 240°\right) \left(0.25\right) +\left(0.15\right)
=0.24\angle 141.49°
