Question

The terminal voltage of the generator is held at 1.20 p.u. and the voltage of the infinite bus is 1.0 p.u. The value of steady state power limit of a system consisting of a generator equivalent reactance 0.5 p.u. connected to an infinite bus through a series reactance of 1.0 p.u. is

• A. 2.512 p.u.
• B. 0.591 p.u
• C. 1.151 p.u.
• D. 1.615 p.u.

Answer 1

The correct option is C 1.151 p.u.

$I=\left(\frac{V_t-V}{j1}\right)=\left(\frac{1.2\angle \theta -1\angle 0^{\circ }}{1\angle 90{}^{\circ }}\right)=\left(1.2\angle \right(\theta -90{}^{\circ })-1\angle -{90}^{\circ })$

$E\angle \delta =V_t+j0.5I$

$E\angle \delta =1.2\angle \theta +\left(1.2\angle \right(-90+\theta )+j)0.5j$

$=1.2\angle \theta +1.2\times 0.5\angle \theta -0.5$

$=1.2\angle \theta +0.6\angle \theta -0.5$

At steady state power, $\delta =90{}^{\circ }$

$1.2\cos \theta +0.6\cos \theta -0.5=0$

$1.8\cos \theta =0.5$

$\theta =73.87{}^{\circ }$

$I=\frac{1.2\angle 73.87{}^{\circ }-1}{j1}=1.3316\angle 30.1{}^{\circ }A$

$E=V+j1.5I$

$=1+j1.5\times 1.3316\angle 30.1{}^{\circ }$

$=1.727\angle 90{}^{\circ }p.u.$

$P_{max}=\frac{1.727\times 1}{1.5}=1.151p.u.$