Question

A 230 V, 250 rpm, 100 A separately-excited dc motor has an armature resistance of 0.5 $\Omega$. The motor is connected to 230 V dc supply and rated dc voltage applied to the field winding. It is driving a load whose torque-speed characteristic is given by $T_L=500-10\omega$. The steady state speed at which the motor will drive the load is

• A. 300.75 rpm
• B. 428 rpm
• C. 157 rpm
• D. 600 rpm

Answer 1

The correct option is A 300.75 rpm

At rated load,
Motor counter emf, $E_a=V_t-I_a{r}_a$
or
$K_m.{\omega }_r=230-100\times 0.5=180V$

Where,
${\omega }_r$= rated motor speed in rad.sec

$\therefore$ Motor constant,
$K_m=\frac{180\times 60}{2\pi \times 20},V-s/rad$

Armature current at any speed $\omega$ is given by,
$I_a=\frac{V_t-E_a}{r_a}=\frac{230-K_m\omega }{0.5}$

$\therefore$ Motor torque,
$T_e=K_m{I}_a=\frac{K_m}{0.5}[230-K_m\omega ]$

Under steady state,
Motor torque, $T_s=$ load torque, $T_L$
(or)
$\frac{K_m}{0.5}[230-k_m\omega ]=500-10\omega$
(or)
$\frac{230}{0.5}\times \frac{180\times 60}{2\pi \times 250}-\frac{1}{0.5}{\left[\frac{180\times 60}{2\pi \times 250}\right]}^2.\omega =500-10\omega$
(or)
$3162.73-94.545\omega =500-10\omega$
(or)
$\omega =\frac{3162.73-500}{84.545}=31.495$ rad/sec

Motor speed
$N_m=\frac{31.495\times 60}{2\pi }=300.75$ rpm