Question

The generator having a synchronous reactance of $X_d=1.7241$ p.u. is connected to a very large system. The terminal voltage is $1.0\angle 0°$ p.u. and generator is supplying a current of 0.8 p.u. at 0.9 p.f. lagging to the system. If the real power output and terminal voltage of the generator remains constant but the excitation of the generator is increased by 20% then the reactive power delivered to the bus is_______p.u.

• A. 0.6326

Answer 1

The correct option is A 0.6326
Synchronous internal voltage = $|E|\angle {\delta }^o=\overrightarrow{V_t}+j\overrightarrow{X_d}\overrightarrow{I}$

$|E|\angle \delta =1.0\angle 0°+\left(1.7241\right)(0.8\angle -25.84°)$
$=2.026\angle 37.78°$ p.u.

Active power from the generator,
$P_e=\frac{|E||V_t|}{|X|}sin\delta =\frac{2.026\times 1}{1.7241}sin\left({37.78}^o\right)=0.72p.u.$

Increasing excitation by 20% with P and V constant gives,

$P_c=0.72=\frac{1.0\times 1.2\times 2.026}{1.7241}sin\delta$

$\delta ={\sin }^{-1}\left(0.51\right)={30.70}^o$

$Q_{new}=\frac{|V_t|}{X_d}(|E_i|\cos \delta -|V_t|)$

$Q_{new}=\frac{1.0}{1.7241}[1.20\times 2.026\cos (30.70)-1.0]$

$Q_{new}=0.632p.u.$