Answer to Question #108685 in Mechanical Engineering for beiken

Question #108685
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.
1
Expert's answer
2020-04-10T13:39:45-0400

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

we know that relation for Torque

T=k (V- Eb)/Ra

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




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