Question

Two similar star connected alternator having $X_1=X_2=3X_0=j0.6p.u.$ are connected in parallel. One generator is solidly grounded while other has isolated ground. Prefault voltage is $1\angle 0p.u.$ and a single line to ground fault occurs on generator terminals at phase R when phase sequence is RBY. Assume generator is unloaded, then select the correct statement.

• A. $I_f=3.75\angle {90}^{o}p.u.$
• B. $V_{BN}=0.9437\angle {113.41}^{o}p.u.$
• C. $V_{YN}=0.9437\angle {113.41}^{o}p.u.$
• D. $V_{RY}=0.9437\angle -{66.58}^{o}p.u.$

Answer 1

The correct option is D $V_{RY}=0.9437\angle -{66.58}^{o}p.u.$

$X_1=X_2=3X_0=j0.6$
$X_1=X_2=j0.6p.u$
$X_0=j0.2p.u$
As two generators are connected in parallel
$X_{1eq}=\frac{j0.6}{2}=j0.3p.u$
$X_{2eq}=\frac{j0.6}{2}=j0.3p.u$
As one generator is solidly grounded and other is isolated
$X_{0eq}=j0.2p.u$
For single line to ground fault
$I_{RO}=I_{R1}=I_{R2}=\frac{E_f}{X_1+X_2+X_0}=\frac{1\angle 0}{j0.3+j0.3+j0.2}=-j1.25$
$I_F=3I_{R1}=-j3.75p.u$
$V_{R1}=E_f-I_{a1}{X}_1=1\angle 0-(-j1.25)\times j0.3=0.625p.u$
$V_{R2}=-I_{a2}{X}_2=-0.375p.u$
$V_{R0}=-I_{a0}{X}_0=-0.25p.u$
As phase RBY
$\left[\begin{array}{c}{V}_{RN}\\ V_{BN}\\ V_{YN}\end{array}\right]=\left[\begin{array}{ccc}1& 1& 1\\ 1& {\alpha }^2& \alpha \\ 1& \alpha & {\alpha }^2\end{array}\right]\left[\begin{array}{c}{V}_{R0}\\ V_{R1}\\ V_{R2}\end{array}\right]$
$V_{RN}=(0.625-0.325-0.25)=0V$
$V_{BN}=V_{R0}+{\alpha }^{2}V{R}_1+\alpha V_{R2}$
$=-0.25+1\angle 240\times 0.625+1\angle 120(-0.375)$
$=-0.375-j0.08660p.u$
$=0.9437\angle -{113.41}^o$
$V_{YN}=V_{R0}+{\alpha }^{2}V{R}_1+\alpha V_{R2}$
$=-0.25+-.625\times 1\angle 120+1\angle 240\times -0.375$
$=0.9437\angle {113.41}^o$
$V_{RY}=V_{RN}-V_{YN}=0.9437\angle -66.58$