Question

A 14,92kw , 400 v , 400 r.p.m . Dc shunt motor draws a current of 40A when running at full-load. The moment of inertia of the roating system is 7,5kg-m^2. If the starting current is 1.2 times full load current. CALCULATE
a) full-load torque.
b) the time required for the motor to attain the reted speed against full-load.

Answer 1

First we determine the motor's efficiency - a ratio of the mechanical output power and the electrical input power:

$E = \frac{P_out}{P_in}=\frac{14920W}{U\cdot I}=\frac{14920W}{400V\cdot 40A}=0.93$

The output mechanical power is directly proportional to torque and angular speed. From this relationship we can isolate the torque:

$P_{out}=\tau \cdot\omega=E\cdot P_{in}=E\cdot I \cdot U\\\implies\tau = \frac{I \cdot U \cdot E}{\omega}=\frac{I\cdot U\cdot E\cdot 60}{2\pi\cdot \omega_{rpm}}$

whereby the last expression uses the angular speed in rpm units.

(a) we calculate the torque, given that the motor draws 1.2 times the rated current, i.e., 48 Amps:

$\tau = \frac{I\cdot U\cdot E\cdot 60}{2\pi\cdot \omega_{rpm}}=\frac{48A\cdot 400V\cdot0.93\cdot 60}{2\pi \cdot 400 rpm} = 426.3 Nm$

(b) use formula for the acceleration time:

$t_{start} = \frac{j\cdot \omega_{rpm}}{\frac{60}{2\pi}\cdot \tau}=\frac{7.5\frac{kg}{m^2}\cdot 400rpm}{9.55\cdot 426.3kg\frac{m^2}{s^2}}\approx 0.74s$

The startup time will be 0.74s