Question

A 3-$\varphi$, 11 kV generator feeds power to a constant power unity power factor load of 100 MW through a 3-$\varphi$ transmission line. The line-to-line voltage at the terminals of machine is maintained at 11 kV. The per unit positive sequence impedance of line based on 100 MVA and 11 kV is j0.25. The line-to-line voltages at the load terminal is measured to be less than 11 kV. The total lagging var to be injected at the terminals of load to increase the line-to-line voltage at the load terminal to 11 kV is

• A. -100 MVAR
• B. -12.68 MVAR
• C. 12.68 MVAR
• D. 100 MVAR

Answer 1

The correct option is C 12.68 MVAR

$S_{base}=100MVA$

$V_{base}=11kV$


At receiving end bus,

$Q_R+Q_{injected}=Q_L$

$P_R=\frac{V_S.V_R}{X}.sin\delta$

$\Rightarrow$ ${\delta }_0=si{n}^{-1}\left(0.25\right)={14.47}^o$

$\Rightarrow$ $Q_R=\frac{V_S.V_R}{X}.cos\delta -\frac{V_S.V_R}{X}$

$Q_R=\frac{V_S.V_R}{X}\left[cos\right({14.47}^{\circ })-1]=\frac{1\times 1}{0.25}\left[cos\right({14.47}^{\circ })-1]$

$Q_R=-0.1268p.u.$

From equation (i),

$Q_R+Q_{injected}=Q_L$ $[\because Q_L=0,loadisupf]$

$Q_{injected}=-Q_R=-(-0.1268)p.u.$

$Q_{injected}=0.1268p.u.$

$Q_{injected}\left(actual\right)=Q_{C(p.u.)}.S_{Base}$

$Q_{injected}=0.1268\times 100$

$=12.68MVAR$