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Consider a 3-phase, 4 Pole, AC induction motor being driven at 75 Hz. Initially, the motor is connected to a mechanical load that requires 1,816 W of power. While connected to this load, the motor spins at 1,996 rpm. Now a new load is connected that requires 2,334 W of power. What is the new slip on the motor for the new load assuming that the motor is operating in its normal operating range?

Respuesta :

Answer:

New slip = 0.145

Explanation:

Revolutions per minute rpm = 1996

Synchronous speed  Ns= 120 * f / p

= 120 * 75 / 4 = 2250

Hence initial slip S1 = 2250 - 1996 / 2250 = 0.113

From output power = [tex] 3*S*E^2*R2 [/tex]

Hence at constant E and R, output power depends on slip

So for new power and new slip to new initial power and initial slip,

Hence 2334/1816 = S2/ S1

New slip S2 = 2334 * 0.113 / 1816 = 0.145

Hence, new speed = Ns ( 1 - S2 )

= 2250*(1 - 0.145)

= 1924 rpm