Answer to Question #108685 in Mechanical Engineering for beiken

Question

A dc shunt motor is driving a constant-torque load at the rated speed and rated terminal voltage. The motor has the following rated data:

Terminal voltage = 115 V

Speed = 312 rpm

Field constant K(phi) = 3 V sec

If the terminal voltage of the motor is reduced by 10%, what is the motor speed? Assume that the field voltage is also reduced by the same ratio.

Answer 1

Here, Terminal voltage = 115 V and speed = 312 rpm

we know that relation for Torque

T=k (V- E_b)/R_a

here torque is constant for both condition so we can take

$k_1 \frac{( V_1- E_b)}{R_a}= k_2 \frac{( V_2- E_b)}{R_a}$

$3 \frac{( V- E)}{R_a}= k_2 \frac{( 0.9 V- 0.9E)}{R_a}$

$k_2= \frac {3}{0.9}=3.33$

So, here speed becomes

$N_2= 313 \times \frac{3}{3.33}=282 rpm$

So the motor speed be 282 rpm