Question

A 20 MVA, 3-phase star connected alternator with an impedance of 5 $\Omega$ and a resistance of 0.5 $\Omega$ is operating in parallel with constant voltage 11 kV bus-bars. If its field current is adjusted to give an excitation voltage of 12 kV. The value of armature current and power factor under maximum power conditions will be respectively

• A. 1784 A, 0.705 lagging
• B. 1879.57 A, 0.669 lagging
• C. 1879.57 A, 0.669 leading
• D. 1784 A, 0.705 leading

Answer 1

The correct option is D 1784 A, 0.705 leading

At maximum power conditions,
$\delta ={\theta }_s$

$\delta =co{s}^{-1}\left[\frac{0.5}{5}\right]={84.260}^{\circ }$

$E_f\angle \delta =V_t\angle 0+I_a{Z}_s$

$E_f\text{per phase =}\frac{12000}{\sqrt{3}}=6928.20V$

$V_t\text{per phase}=\frac{11000}{\sqrt{3}}=6350.852V$

$E_{f}cos\delta +j{E}_{f}sin\delta =V_t+I_a\angle -\varphi .Z_s\angle {\theta }_s$

$=V_t+I_a{Z}_s\angle {\theta }_s-\varphi$
$E_{f}cos\delta +j{E}_{f}sin\delta =V_t+I_a{Z}_{s}cos({\theta }_s-\varphi )+j{I}_a{Z}_{s}sin({\theta }_s-\varphi )$

$E_{f}cos\delta =V_t+I_a{Z}_{s}cos({\theta }_s-\varphi )$

$692.98=6350+5I_{a}cos({\theta }_s-\varphi )$

$5I_{a}cos({\theta }_s-\varphi )=-5657.015$

$I_{a}cos({\theta }_s-\varphi )=-1131.403$ ...(i)

Again, $5I_{a}sin({\theta }_s-\varphi )=6893.46193$ ...(ii)
From (i) and (ii) squaring and adding we get,
$I_a=1783.5A$

$cos({\theta }_s-\varphi )=-0.6343$

${\theta }_s-\varphi =129.373$

$\varphi =-45.11$

$cos\varphi =0.705leading$

$\text{Alternate Solution:}$

For $P_{max}$,
${\theta }_s=\delta ={84.26}^{\circ }$ and leading pf

$P_{max}=\frac{11}{5}[12-11\cos {84.26}^{\circ }]=\sqrt{3}\times 11\times I\cos \varphi$

$Icos\varphi =1.2586$ ....(i)

$Q=\frac{11}{5}(-11sin84.26)=\sqrt{3}\times 11\times Isin\varphi$

$Isin\varphi =-1.2638$ ....(iii)

From equation (i) and (iii),
$\varphi ={45.11}^{\circ }\left(leading\right)$

$pf=\cos {45.11}^{\circ }=0.705\left(leading\right)$

$I=\frac{1.2586}{0.705}=1.7836kA=1783.6A$