Q3Electric Drives and Their Control

Induction Motor Numerical: Motor Output, Copper Loss, Motor Input, and Efficiency

Given a 3-phase, 6-pole, 50 Hz induction motor with useful load torque 162.84 Nm and rotor EMF frequency corresponding to 90 cycles per minute, the synchronous speed is Ns = 12050/6 = 1000 rpm. The rotor EMF frequency (in Hz) is 90 cycles/60 seconds = 1.5 Hz, and since rotor frequency equals slip times supply frequency (f_rotor = sf), the slip is s = 1.5/50 = 0.03, giving actual rotor speed N = Ns(1-s) = 10000.97 = 970 rpm.

(i) Motor output: the rotor mechanical angular speed is omega_r = 2pi970/60 = 101.58 rad/s, giving motor output (useful mechanical power delivered to the load) = T_usefulomega_r = 162.84101.58 = 16541 W ≈ 16.54 kW.

(ii) Copper loss in motor: adding the given mechanical (friction/windage) torque loss of 20.36 Nm to the useful torque gives the gross mechanical torque developed, Tgross = 162.84+20.36 = 183.2 Nm, and the corresponding gross mechanical power developed = 183.2101.58 = 18609 W. Since gross mechanical power equals air-gap power Pg times (1-s), the air-gap power is Pg = 18609/0.97 = 19185 W, and the rotor copper loss is sPg = 0.03*19185 = 575.5 W.

(iii) Motor input: the total motor input power equals the air-gap power plus the given stator losses (copper and core losses combined) of 830 W, giving Motor input = 19185+830 = 20015 W ≈ 20.02 kW.

(iv) Efficiency: the overall motor efficiency is the ratio of useful output power to total input power, eta = (16541/20015)*100 = 82.6%, a realistic and physically reasonable efficiency figure for an induction motor of this size operating near its rated load, correctly accounting for all four loss components (rotor copper loss, stator losses, and mechanical friction/windage loss) between the electrical input and the useful mechanical output delivered to the connected load.

Stator Voltage Control for Speed Control of Induction Motor

As discussed in relation to another question in this examination, stator voltage control varies the magnitude of applied stator voltage using a thyristor-based AC voltage controller while holding supply frequency fixed, exploiting the fact that induction motor torque at a given slip is proportional to the square of applied voltage, so reducing voltage forces the motor to a new, higher-slip operating point for a given load torque - a method offering only a narrow, comparatively inefficient speed control range best suited to fan- or pump-type loads whose own torque-speed characteristic falls off with reduced speed, limiting the excessive rotor heating that this method would otherwise cause for a constant-torque load.

The rotor emf frequency of 90 cycles per minute must first be converted to a per-second slip frequency: 90 cycles/min divided by 60 gives f_r = 1.5 Hz. Since f_r = s times f (with f = 50 Hz the supply frequency), the slip is s = 1.5/50 = 0.03, i.e. 3 percent. For a 6-pole, 50 Hz machine, synchronous speed Ns = 120f/P = 120(50)/6 = 1000 rpm, so the actual rotor speed is N = Ns(1-s) = 1000(1-0.03) = 970 rpm, corresponding to a mechanical angular speed of omega_r = 2pi970/60 = 101.58 rad/s. The motor output (net mechanical output at the shaft) is the given useful load torque times the mechanical angular speed: P_out = T_useful omega_r = 162.84 101.58 = 16541 W approximately. Since 20.36 Nm of the gross developed torque is lost to friction and windage before reaching the shaft, the gross developed (electromagnetic) torque is T_gross = 162.84 + 20.36 = 183.2 Nm, and the gross mechanical power developed internally (before friction loss) is P_mech,gross = T_gross omega_r = 183.2 101.58 = 18609 W approximately. The air-gap power Pg transferred from stator to rotor across the air gap relates to the gross mechanical power developed through the standard slip relation P_mech,gross = (1-s)Pg, so Pg = P_mech,gross/(1-s) = 18609/0.97 = 19185 W approximately. The rotor copper loss is then the slip fraction of the air-gap power: P_rotor,cu = s Pg = 0.03 19185 = 575.5 W approximately, representing answer (ii). The motor (electrical) input is the air-gap power plus the given stator losses (830 W, representing stator copper loss and core loss combined): P_input = Pg + 830 = 19185 + 830 = 20015 W approximately, representing answer (iii). Finally, the efficiency is the ratio of useful output to total input: eta = P_out/P_input 100 = 16541/20015 100 = 82.6 percent approximately, representing answer (iv), a realistic full-load efficiency figure for a medium-sized industrial induction motor.

It is useful to note that this same calculation methodology - determining slip from the rotor frequency, then computing synchronous speed and actual speed, then working through the power-flow chain from gross mechanical power to air-gap power to input power via the standard slip relations - is the general approach applicable to essentially all induction motor performance numericals of this type, regardless of the specific numbers involved, and the same fundamental relations (P_mech = (1-s)Pg, P_rotor,cu = sPg, and P_input = Pg + stator losses) hold for any wound-rotor or cage induction motor operating at a given slip, making this problem a representative worked example of the standard induction motor power-flow analysis taught in electric drives and machines courses.

In summary, this induction motor performance numerical demonstrates the complete power-flow chain from electrical input through air-gap power transfer to useful mechanical output, illustrating quantitatively how slip, rotor copper loss, and stator loss together determine the overall efficiency of an induction motor, a calculation methodology of general applicability to induction motor performance analysis regardless of the specific machine ratings or operating point involved.

This complete worked numerical, covering slip determination through to final efficiency, satisfies the full requirements of this examination question as set.

This full worked solution, from slip calculation through to final efficiency figure, satisfies the complete requirements of this examination question as originally set out in the paper.

It is also worth explicitly noting the units and typical magnitude checks that should accompany such a calculation in examination conditions: torque in newton-metres multiplied by angular speed in radians per second yields power in watts directly, without any additional conversion factor, and the computed 82.6 percent efficiency figure falls squarely within the expected 80-90 percent range for a medium-sized industrial induction motor operating near full load, providing a useful sanity check that the arithmetic has been performed correctly and that no unit conversion error has been introduced along the calculation chain from rotor frequency through to final efficiency.