Question 7.5: Consider a system with sequence impedances given by Z+ = j0....
Consider a system with sequence impedances given by Z_{+} = j0.2577, Z_{-} = j0.2085, and Z_{0} = j0.14; find the voltages and currents at the fault point for a single line-to-ground fault.
The sequence networks are connected in series for a single line-to-ground fault.
The sequence currents are given by
I_{+}=I_{-}=I_{0}=\frac{1}{j\left(0.2577 +0.2085 +0.14\right) }
=1.65\angle -90° p.u.
Therefore,
I_{A} =3I_{+}=4.95\angle -90° p.u.
I_{B} =I_{C}=0
The sequence voltages are as follows:
V_{+}=E_{+}-I_{+}Z_{+}
=1\angle 0-\left(1.65\angle -90°\right)\left(0.2577\angle 90°\right)
=0.57 p.u.
V_{-}=-I_{-}Z_{-}
=-\left(1.65\angle -90°\right)\left(0.2085\angle 90°\right)
=-0.34 p.u.
V_{0}=-I_{0}Z_{0}
=-\left(1.65\angle -90°\right)\left(0.14\angle 90°\right)
=-0.23p.u.
The phase voltages are thus
V_{A}=V_{+}+V_{-}+V_{0}=0V_{B}=\alpha ^{2}V_{+}+\alpha V_{-}+ V_{0}
=\left(1\angle 240°\right)\left(0.57\right)+ \left(1\angle 120°\right)\left(-0.34\right) + \left(-0.23\right)
=0.86\angle -113.64° p.u.
V_{C}=\alpha V_{+}+\alpha ^{2}V_{-}+V_{0}
=\left(1\angle 120°\right)\left(0.57\right)+ \left(1\angle 240°\right)\left(-0.34\right) +\left(-0.23\right)
=0.86\angle 113.64° p.u.