University of Jordan Faculty of Engineering & Technology Electric Power Engineering Department EE471: Electrical Machines-II Tutorial # 2: 3-ph Induction Motor/Generator Question #1 A 100 hp, 60-Hz, three-phase induction motor has a slip of 0.03 at full load. Compute the efficiency of the motor at full load when the friction and windage losses are 900W, the stator core loss is 4200W, and the stator copper loss is 2700W. Question #2 A three-phase, 480-volt, 60-Hz, 12-pole, induction motor has the following parameters: R 1 = 1.0 Ω; R`2= 0.5 Ω; X eq = 10. Ω; X c = 100 Ω; R c = 800 Ω Calculate the following: a. The slip at maximum torque. b. The current at maximum torque c. The speed at maximum torque d. The maximum torque.
Question #3 A three-phase, 30-hp, 480-V, 4-pole, 60-Hz induction motor has the following equivalent circuit parameters in per phase referred to stator: R 1 = 0.25 R 2 = 0.20 X 1 = 1.30 X 2 = 1.20 X m = 35 The total core, friction, and windage losses may be assumed constant at 1250 W. The motor is connected directly to a 440-V source, and it runs at a slip of 3.5%. Compute the following: a) Motor speed, input current, and power factor and b) Power input, shaft output torque, and efficiency of the motor. Question #4 A three-phase, 6-pole, 60-Hz induction motor is operating at a speed of 1152 rpm. The power input to the motor is 44 kw, the rotational losses are 500 W, and the stator copper loss is 1600 W. Find the following: a) Slip, airgap power, and rotor copper loss b) Developed torque, developed horsepower, output torque, and output horsepower Question #5 A 3-phase, 10-hp, 230-V, 4-pole, 60-Hz, Y-connected induction motor develops its full-load torque at a slip of 4.5% when operating at 230 V and 60-Hz. The per-phase equivalent circuit impedances of the motor are R 1 = 0.35 X 1 = 0.5 X 2 = 0.5 X m = 15.0 The mechanical and core losses are assumed to be negligible. Determine the following: a) Rotor resistance R 2, and slip S, b) Starting torque, maximum torque, and rotor speed at maximum torque. Question #6 A 3-phase, 20-hp, 440-V, 6-pole, 60-Hz, Y-connected induction motor has a starting torque of 98 N-m and a full-load torque of 72 N-m. Calculate: a) The starting torque when the applied voltage is reduced to 300 V. b) The applied voltage in order to develop a starting torque equal to the full-load torque. Question #7 A 3-phase, 440-V, 25-hp, 60-Hz, 1750-rpm, Y-connected, wound-rotor induction motor has the following equivalent circuit parameters. R 1 = 0.20 R 2 = 0.15 X 1 = 1.0 X 2 = 0.80 X m = 30 Determine the resistance of the rheostat in the rotor circuit such that the maximum torque occurs at starting. Calculate the starting torque before and after adding the resistance. 2
Question #8 A 480V, 60 Hz, 6-pole, three-phase, delta-connected induction motor with the following parameters: R 1 =0.461 Ω, R 2 =0.258 Ω, X 1 =0.507 Ω, X 2 =0.309 Ω, X m =30.74 Ω Rotational losses are 2450W. Now, suppose that the machine is being driven by a mechanical system such that it is rotating at 1224 rpm. Calculate the following information: a. slip b. Line Current c. Power and Reactive Power at the terminals d. Airgap Power e. Torque Developed f. Mechanical Power g. Efficiency h. Much of the theory for this problem is identical to a motor problem: Solution: Slip is given by Using the rpm equation, s = (1200-1224)/1200 = -0.02 Now, phase current is given by where phase impedance is given by Using the above equation, Zin = -10.3 + j5.29 Ω And noting that the machine is delta connected, V 1 = V LL = 480V, I 1 = -36.7 - j18.8 A. I 1 =41.4 A. Therefore, I L = 3 41.4 = 71.7 A Note that the real part of phase current is negative, indicating that real power flows out of the terminals. Also, the imaginary part of the phase current, which indicates that reactive power must flow into the terminals. An induction generator cannot operate without a reactive power supply, either from a power grid or capacitor bank Using complex notation, S=3VI * or P elec =3 V Re{I} = -53.0kW, Q elec =- V Im{I} = +27.2kVAR Airgap Power is given by This approach requires rotor current to be found. With no core loss resistance: 3
Giving I 2 = 37.8 A. Substituting into the power equation P gap = -55.4kW Torque developed can be found from where synchronous speed in radians per second is given by giving τ = -441 Nm Mechanical power can be found using and Therefore mechanical power in kw is: P out = -55.4 (1.0+0.02) - 2.45, P mech = -58.9 kw Efficiency is given by Therefore, η = 53.0/58.9 = 89.9 % Summary It can be seen from the above analysis that the equations for an induction motor can all be applied to an induction generator. (As long as output and input power are correctly described as either electrical or mechanical).. Induction generation used to be relatively rare. However, it is becoming increasingly common as induction generators are the generator of choice for large wind turbines. The requirement of an induction generator to have an independent reactive power supply has caused significant research into the impact of large wind farms of power system stability. 4
Question #9 As an example, consider the use of a 10 hp, 1760 r/min, 440 V, three-phase induction motor as an asynchronous generator. The full-load current of the motor is 10 A and the full-load power factor is 0.8. Required capacitance per phase if capacitors are connected in delta: Apparent power S = 3 E I = 1.73 440 10 = 7612 VA Active power P = S cos θ = 7612 0.8 = 6090 W Reactive power Q = = 4567 VAR For a machine to run as an asynchronous generator, capacitor bank must supply minimum 4567 / 3 phases = 1523 VAR per phase. Voltage per capacitor is 440 V because capacitors are connected in delta. Capacitive current I c = Q/E = 1523/440 = 3.46 A Capacitive reactance per phase X c = E/I c = 127 Ω Minimum capacitance per phase: C = 1 / (2 π f X c ) = 1 / (2 3.141 60 127) = 21 microfarads. If the load also absorbs reactive power, capacitor bank must be increased in size to compensate. Prime mover speed should be used to generate frequency of 60 Hz: Typically, slip should be similar to full-load value when machine is running as motor, but negative (generator operation): if N s = 1800, one can choose N=N s +40 rpm Required prime mover speed N = 1800 + 40 = 1840 rpm. 5