Ac Motor and power

Question:A 10-hp six-pole 60 Hz three-phase induction motor runs at 1150 rpm under full-load conditions. Determine the slip and the fequency of the rotor currents at full load. Also estimate the speed if the load torque drops in half.

Answer:slip = 4.167%, f_slip = 2.5 Hz at full load;slip = 2.083%, speed = 1175 rpm at half power

Question:Explain why induction motor develops zero torque at synchronous speed.

Answer:The voltage induced in the rotor conductors is:v_c = B l u.When the rotor turns at synchronous speed, the relative speed between the field and conductors is u = 0. Hence, v_c = 0.Since the induced voltage is zero, the rotor currents are zero. The force acting on the rotor conductors: f = i l x B, is zero. There is zero torque at synchronous speed.

Question:A 60-Hz induction motor is needed to drive a load at approximately 850 rpm. How many poles should the motor have? What is the slip of this motor for a speed of 850 rpm?Answer: P = 8; s = 5.55%

Example: DC to AC power conversion for 3-phase induction motorA 4-pole induction motor drives a load at 2500 rpm. This is accomplished byusing an electronic converter to convert a 400-V DC source into a set of 3-phaseAC voltages. Given that the frequency of the AC voltages is 86.8 Hz assuming that the slip is4%. The load is 2 hp. If the DC-to-AC converter has a power efficiency of 88%and the motor has a power efficiency of 80%, estimate the current taken fromthe DC source.

Solution:

Pout,motor = 2 hp x 746 = 1492 WThe input power to the motor is:Pin,motor = Pout,motor/ηmotor = 1492/0.80 = 1865 WPout,converter = Pin,motorThe input power to the converter is:Pin,converter = Pout,converter/ηconvertor = 1865/0.88 = 2119.32 WFinally, the current taken from the 400-V source is:I = Pin,converter/400 = 2119.32/400 = 5.298 A