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Unformatted text preview: ECE 341 Homework #5 Solution Due: Mar. 2, Feb. ‘12 Problem 1 : An eight-pole, 60 Hz, 3-phase, wye-connected induction motor has mechanical losses of 620 W at s=0.04. At the same time, the motor delivers 30 HP at its shaft. Calculate (a) the input power to the rotor (P AG ) (2 points) (b) the load torque and total torque (2 points) Solution: ns=900 rpm, Pout=30 HP8760=22.38 kW a. Pconv=Pout+Pmech_loss=22.38+0.62=23 kW P AG = Pconv/(1-s)=23/0.96=23.96 kW b. 35 . 247 900 96 . 2 22380 m out load P Nm 21 . 254 900 96 . 2 23000 m conv total P Nm Problem 2: A six-pole, 480 V, 3-phase, 60 Hz induction motor takes 200 A of current in each phase at starting, and 30 A/Phase when it runs with full load. The starting torque is 1.8 times the torque at the full load at 480 V. If it is desired that the starting torque be the same as the full load torque, determine (a) The applied line-to-line voltage (2 points) (b) The corresponding line current (phase) of the motor...
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- Winter '08
- Electric motor, iL, induction motor, load torque