A 220V, 1000rpm, 60A separately excited DC motor has an armature resistance of 0.1Ω . It is fed from a single phase full converter with an AC source voltage of 230V, 50Hz. Assuming continuous conduction,

compute:

a) Firing angle for rated motor torque at 600 rpm

b) Firing angle for rated motor torque at (-500) rpm

c) Motor speed for α = 150° and half rated torque

Mumbai University > Electronics Engineering > Sem7 > Power Electronics 2

Marks: 10M

Year: May 2013

$200=K_m=\frac{2\pi \times 1000}{60}+60 \times 0.1$ OR $K_m=\big[\frac{220-6}{2\pi \times 1000}\big] \times 60=2.044 V-s/rad$ OR 2.044Nm/A (a) for rated motor, armature current =60A $V_0=V_tK_m \omega _m+I_ar_a \ \frac{2 \sqrt 2 \times 230}{\pi}cos\alpha=2.044 \frac{2\pi \times 600}{60}+60 \times 0.1=134.43V \ \alpha = cos^{-1}\big[\frac{134.43 \times \pi}{2\sqrt2 \times 230}\big] $ (b) at (-500) rpm, $$\frac{2\sqrt2 \times 230}{\pi}cos\alpha=2.044\frac{2\pi \times (-500)}{60}+60 \times 0.1 \ \frac{2\sqrt2 \times 230}{\pi}cos\alpha=-107.024+6=-101.024V \ $$