Question

A 3-phase alternator has synchronous reactance, $X_d=1.7241p.u.$ and is connected to a very large system. The terminal voltage is $1.0\angle 0^{\circ }p.u.$ and the generator is supplying to the system a current of 0.8 p.u. at 0.9 power factor lagging. If the real power output of the generator remains constant but the excitation of the generator is increased by 20%, then the Q delivered to the bus by the generator is ____p.u. (Answer upto three decimals).

• A. 0.632

Answer 1

The correct option is A 0.632

By applying KVL in the loop,
$-E\angle \delta +j{I}_0{X}_d+V_R=0$

$E\angle \delta =1\angle 0^{\circ }+\left(j1.7241\right)(0.8\angle -{25.84}^{\circ }$

$E\angle \delta =2.026\angle {37.78}^{\circ }p.u.$

$P=\frac{|E||V|}{|X|}\sin \delta =\frac{2.026\times 1}{1.7241}\sin \left({37.78}^{\circ }\right)$

$P=0.73p.u.$

Power angle ${\delta }^{\prime },$ after increasing the excitation to 20% is

$0.72=\frac{1.20\times 2.026\times 1}{1.7241}\sin {\delta }^{\prime }$

${\delta }^{\prime }={\sin }^{-1}\left[\frac{0.72\times 1.7241}{1.20\times 2.026}\right]={30.70}^{\circ }$

If excitation is increased by 20% then,

$Q_R=\frac{|E||V|}{|X|}\cos {\delta }^{\prime }-\frac{|V{|}^2}{X}$

Where ${\delta }^{\prime }$ is the new power angle

Reactive power delivered to bus,

$Q_R=\frac{1.2\times 2.026\times 1}{1.7241}\cos ({30.70}^{\circ }-\frac{1^2}{1.7241})$

$=0.632p.u$