Question

A synchronous machine operating on bus bar with $V=1.0p.u,X_s=1.0p.u,R_a=0,P=0.6p.u$ at unity power factor. If the mechanical power input is increased such that the active power delivered is increased to 0.8 p.u., keeping field excitation same. Then the reactive power is

• A. 0.151 p.u.
• B. -0.151 p.u.
• C. -0.302p.u.
• D. 0.302 p.u.

Answer 1

The correct option is B -0.151 p.u.

$P=V{I}_{a}cos\varphi$

$0.6=1\times I_a\times 1$

$I_a=0.6p.u.$

$E=\sqrt{(Vcos\varphi +I_a{R}_a{)}^2+(Vsin\varphi +I_a{X}_s{)}^2}$

$=\sqrt{(1\times 1+0{)}^2+(0+0.6\times 1{)}^2}$

$=1.166p.u$

$P=\frac{EV}{X_s}\sin \delta$

$0.6=\frac{1.166\times 1}{1}\sin \delta$

$\delta ={\sin }^{-1}\left(\frac{0.6}{1.166}\right)={30.9}^{\circ }$

$\frac{P_2}{P_1}=\frac{\sin {\delta }_2}{\sin {\delta }_1}$

$\frac{0.8}{0.6}=\frac{\sin {\delta }_2}{\sin 30.96}$

$\delta ={43.3}^{\circ }$

$Q=\frac{V}{X_s}[E\cos \delta -V]$

$=\frac{1}{1}\left[1.166\cos \right({43.3}^{\circ })-1]$

$=-0.151p.u$