Question

4. A 3-phase, 340 V, Induction motor is supplied with 60 A at.85 pf lagging. Its stator and rotor copper losses are 4000 W and 800 W respectively, while its core losses are 2000 W, and its Friction, Windage and Stray losses are a combined 750 W. Determine: a. Draw a diagram depicting the energy flow in the motor (Similar to the one we had drawn in class) b The air gap power c. The mechanical power developed d. The shaft output power in Watts and HP e. Efficiency of the motor

Answer 1

Rating of induction motor 3 phi, v_L = 340 v(L - L), I_2 = 60 A @ 0.85 pf lagging stator copper loss = 4000 w Rotor copper loss = 800 w Total core loss = stator core loss + Rotor core loss Since the frequency in the rotor is slip times the supply frequency, hence rotor core loss are less & generally neglected. Therefore Total core loss = Stator core loss = 2000 w (given) Mechanical losses = Friction + windage + stray = 750 w (given) \r\n Energy Flow Diagram useful power p_out (HP Rating) or shaft power P_in = squareroot 3 v_L I_L cos phi = squareroot 3 times 340 times 60 times 0.85 = 30033.761 w P_in = 30.033 kw \r\n Air gap power P_2 = P_in - stator loss (copper + core) = 30.033 kw - (2 + 4) kw P_2 = 24.033 kw Gross Mechanical power P_in = P_2 - P_c where P_c = Rotor copper loss Therefore P_m = 24.033 kw - 0.8kw P_m = 23.233 kw Shaft o/p power P_out = P_m - Mechanical Losses = (23.233 - 0.75) kw