Question

An alternator with a synchronous impedance of (0 + j1.25) p.u. delivers rated current to infinite bus bar at pf 0.8 lagging. For the same excitation power factor just before falling out of step would be

• A. 0.785 lagging
• B. 0.895 leading
• C. 0.785 leading
• D. 0.895 lagging

Answer 1

The correct option is B 0.895 leading

Take, $V_t=1\angle 0^o$ p.u.

So, $I_a=1\angle -co{s}^{-1}\left(0.8\right)$p.u.

Alternator excitation emf, ${\overrightarrow{E}}_f={\overrightarrow{V}}_t+{\overrightarrow{I}}_a{\overrightarrow{Z}}_s$

${\overrightarrow{E}}_f=1\angle 0^o+[1\angle -co{s}^{-1}(0.8\left)\right]\times 1.25\angle {90}^o$

${\overrightarrow{E}}_f=1+1.25\angle {53.13}^o$

$|{\overrightarrow{E}}_f|=\sqrt{(1+1.25cos{53.13}^{\circ }{)}^2+(1.25sin{53.13}^{\circ }{)}^2}$

$=2.01p.u.$

When motor just fall out of step,
$\delta ={90}^{\circ }$

Now for same excitation,
$2.01\angle {90}^{\circ }=1\angle 0^{\circ }+I_a\times 1.25\angle {90}^{\circ }$

$\overrightarrow{I_a}=\frac{j2.01-1}{j1.25}=1.608+j0.8$

$\overrightarrow{I_a}=1.8\angle {26.45}^o$ p.u.

Power factor, $cos\left({26.45}^o\right)=0.895$ leading