Question

The speed of a separately excited dc motor is controlled by two single phase full converters, one in armature circuit and other in the field circuit. Both converters are fed from same $1-\varphi ,230V,50Hz$ source. Armature resistance is $0.5\Omega$ and field circuit resistance is $200\Omega$ . Firing angle for converter connected to field is zero and motor constant is 0.8 V-s/rad. Armature and field current are continuous and ripple free. If the armature current is 30 A and motor speed is 25.25 rps, then firing angle of armature converter will be______ degree.

Answer 1

For field controlled converter,

$\alpha =0^{\circ }$

$\therefore \text{Field voltage},V_f=\frac{2V_m}{\pi }=\frac{2\sqrt{2}\times 230}{\pi }=207.072V$

$\text{Field current},I_f=\frac{V_f}{R_f}=\frac{207.072}{200}=1.035A$

For armature of motor we can write KVL,

$V_a=V_t=E_a+I_a{R}_a$

$\frac{2\sqrt{2}\times 230}{\pi }\cos \alpha =(K.I_f)\frac{2\pi N}{60}+30\times 0.5$

Where, Motor speed , n = 25.5 rps

$\text{Motor speed in rpm,}N=25.25\times 60=1515rpm$

$\frac{2\sqrt{2}\times 230}{\pi }\cos \alpha =(0.8\times 1.035)\times \frac{2\pi \times 1515}{60}+15$

$207.072\cos \alpha =131.362+15$

$207.072\cos \alpha =146.362$

$\cos \alpha =\frac{146.362}{207.072}=0.7068$

$\alpha ={45.02}^{\circ }$