Question 7.7: For the system of Example 7.5, find the voltages and current...
For the system of Example 7.5, find the voltages and currents at the fault point for a line-to-line fault through an impedance Z_{f} = j0.05 p.u.
The sequence network connection is as shown in Figure 7.30. From the diagram,
I_{+}=- I_{-}=\frac{1\angle 0}{0.5185\angle 90°}
=1.93\angle – 90° p.u.
I_{0}=0
The phase currents are thus
I_{A}=0
I_{B}=- I_{C}
=\left(\alpha ^{2} – \alpha \right)I_{+}
=\left(1\angle 240°- 1\angle 120°\right) \left(1.93\angle – 90°\right)
=3.34\angle – 180° p.u.
The sequence voltages are
V_{+}=E_{+}-I_{+}Z_{+}
=1\angle 0-\left(1.93\angle – 90°\right)\left(0.26\angle 90° \right)
=0.5 p.u.
V_{-}=-I_{-}Z_{-}
=-\left(1.93\angle – 90°\right) \left(0.2085\angle 90°\right)
=0.4 p.u.
V_{0}=-I_{0}Z_{0}
=0
The phase voltages are obtained as shown below:
V_{A}=V_{+}+V_{-}+V_{0}
=0.9 p.u.
V_{B} =\alpha ^{2} V_{+} +\alpha V_{-} +V_{0}
=\left(1\angle 240°\right) \left(0.5\right) +\left(1\angle 120°\right) \left(0.4\right)
=0.46\angle- 169.11°
V_{C} =\alpha V_{+} +\alpha ^{2} V_{-} +V_{0}
=\left(1\angle 120°\right) \left(0.5\right) +\left(1\angle 240°\right) \left(0.4\right)
=0.46\angle 169.11°
As a check, we calculate
V_{B}- V_{C}=0.17\angle – 90°
I_{B}Z_{f}=\left(3.34\angle – 180°\right) \left(0.05\angle 90°\right)
=0.17\angle – 90°
Hence,
V_{B}- V_{C}=I_{B}Z_{f}
