A 220V, 1500rpm, 10A separately excited DC motor has an armature resistance of 1 Ω . It is fed from a single phase fully controlled bridge rectifier with an AC source voltage of 230V, 50Hz. Assuming continuous load current, compute

a) Motor speed at firing angle of 30° and torque of 5NM.

b) Developed torque at firing angle of 45° and speed of 1000rpm.

Mumbai University > Electronics Engineering > Sem7 > Power Electronics 2

Marks: 10M

Year: May 2015

Under rated operating conditions of the separately .excited dc motor.

$V_t=E_a+I_ar_a=K_m+I_ar_a \\ 220=K_m\frac{2\pi \times 1500}{60}+10 \times 1=50.\pi .K_m+10 $

Therefore, Motor constant

Or 1.337Nm/A

(a) for a torque of 5 Nm ,motor armature current,

$I_a=\frac{5}{1.337}=3.74A$

The equation giving the operation of converter – motor is

$V_0=V_t=E_a+I.r_a \\ \frac{2V_m}{\pi}cos\alpha=K_m.\omega_m+I_a.r_a \\ \frac{2\sqrt 2 \times 230}{\pi}cos30^0=1.337 \omega_m+3.74 \times 1 \\ \omega_m=\frac{179.3-3.74}{1.337}=131.31 \ \ \ rad/sec \\ \frac{2\pi.N}{60}=131.31 \ \ \ rad/sec $