Question

A 100 kW, 500 V, 2000 rpm separately excited dc motor is energized from 400 V, 50 Hz, $3-\varphi$ source through a $3-\varphi$ full converter. The voltage drop in conducting thyristor is 1 V. The dc motor has following parameters:
$r_a=0.2\Omega K_m=1.6\text{V-s/rad}{L}_a=8mH$
Rated armature current = 210 A.
The value of firing angle of converter for speed of 2000 rpm so that motor draws rated armature current and its respective supply side power factor will be

• A. ${45.57}^{\circ },0.85$
• B. ${68.43}^{\circ },0.67$
• C. ${45.57}^{\circ },0.67$
• D. ${68.43}^{\circ },0.85$

Answer 1

The correct option is C ${45.57}^{\circ },0.67$

At rated armature current and speed of 2000 rpm
$V_0=V_t=K_m{\omega }_m+I_a{r}_a+$ drop across thyristor
$\frac{3\sqrt{2}\times 400}{\pi }.\cos \alpha =\frac{1.6\times 2\pi \times 2000}{60}+210\times 0.2+1V$
$=335.10+42+1=378.10$
$\cos \alpha =\frac{\left(378.10\right)\times \pi }{3\sqrt{2}\times 400}=0.699$
$\alpha ={45.57}^{\circ }$
as frring angle $\alpha <{60}^{\circ }\left({45.57}^{\circ }\right)$
$\therefore$ rms value of source current
$I_{sr}=I_a\sqrt{\frac{2}{3}}=210\sqrt{\frac{2}{3}}=171.464A$
supply power factor $=\frac{V_i}{\sqrt{3V_s}.I_{sr}}=\frac{378.10\times 210}{\sqrt{3}\times 400\times 171.464}=0.669\text{lag}$