Question

The speed of a 50 kW, 500 V, 120 A, 1500 rpm separately excited dc motor is controlled by a 3-$\varphi$ full converter fed from 400 V, 50 Hz supply. Motor armature resistance is 0.1 $\Omega$. If the firing angle (degree) is varied from ${\alpha }_1$ to ${\alpha }_2$ to obtain speed between 1000 rpm and - 1000 rpm and rated torque, then the value of ${\alpha }_1+$${\alpha }_2$ in degrees is

• A. 186.84${}^o$
• B. 180${}^o$
• C. 176.84${}^o$
• D. 156.84${}^o$

Answer 1

The correct option is C 176.84${}^o$

Back emf, $E=K_m\omega$

$K_m=\frac{500-120\times 0.1}{2\pi \times \left(\frac{1500}{60}\right)}=3.106$

For speed of 1000 rpm at rated load torque,

$V_t=K_m{\omega }_s+I_a{T}_a$

$\frac{3V_{ml}}{\pi }cos\alpha =3.106\times 2\pi \times \frac{1000}{60}+120\times 0.1$

$cos{\alpha }_1=\frac{337.26\times \pi }{3\times 400\sqrt{2}}=0.624$
${\alpha }_1={51.39}^o$
For speed of - 1000 rpm at rated torque

$\frac{3\sqrt{2}\times 400}{\pi }cos{\alpha }_2=3.106\times 2\pi \times \frac{(-1000)}{60}+120\times 0.1$
$cos{\alpha }_2=\frac{-313.26\times \pi }{1200\sqrt{2}}=-0.58$

${\alpha }_2={125.44}^o$
${\alpha }_1+{\alpha }_2={51.39}^o+{125.44}^o$
$={176.83}^o$