In a 3-phase, 6-pole induction motor, The power crossing the air gap of a 60 Hz is 42.6 kW, and the power converted from electrical to mechanical form in the motor is 39.6 kW. If the total mechanical losses are 1000 W, find the load torque of this motor? Select one: O a. Tload = 330.4 N.m. O b. Tload = 318.6 N.m. O c. Tioad = 305.4 N.m. O d. None

In a 3-phase, 6-pole induction motor, The power crossing the air gap of a 60 Hz is 42.6 kW, and the power converted from electrical to mechanical form in the motor is 39.6 kW. If the total mechanical losses are 1000 W, find the load torque of this motor? Select one: O a. Tload = 330.4 N.m. O b. Tload = 318.6 N.m. O c. Tioad = 305.4 N.m. O d. None

Delmar's Standard Textbook Of Electricity

7th Edition

ISBN:9781337900348

Author:Stephen L. Herman

Publisher:Stephen L. Herman

Chapter35: Harmonics 916

Section: Chapter Questions

Problem 3RQ: Would a positive-rotating harmonic or a negative-rotating harmonic be more harmful to an induction...

See similar textbooks

Similar questions

Name three factors that determine the torque produced by an induction motor.
A 440-V-rms (line-to-line voltage) three-phase induction motor runs at 1150 rpm driving a load requiring 15 Nm of torque. The line current is 3.4 A rms at a power factor of 80 percent lagging. Find the output power, the power loss, and the efficiency.
A 220-V-rms (line-to-line), 60-Hz, three-phase induction motor operates at 3500 rpm while delivering its rated output power of 3 hp. The line current is 8 A rms and the losses are 300 W. Find the input power, power factor, and efficiency.
A 25-hp three-phase induction motor operates for a 440-Vrms (line-to-line) three phase source and draws a line current of 35 A rms at a power factor of 83 percent lagging under rated full load conditions. The full load speed is 1750 rpm. Under no load conditions the speed is 1797 rpm and the line current is 6.5 A rms at a power factor of 30 percent lagging. Find the power loss and efficiency with full load, the input power with no load and the speed regulation.
Draw the approximate equivalent (without Xm) circuit of an induction motor and derive the equation of the maximum torque..
IV. What are the various losses in an induction motor? On what actors do they depend?
electrical machines A 50-kW, 460-V, 50 Hz, two-pole induction motor has a slip of 5% when operating a full-loadconditions. At full-load conditions, the friction and windage losses are 700W, and the core losses are600 W. Find the following values for full-load conditions:a. The shaft speedb. The output power in wattsc. The load torque in N-m.d. The induced torque in N-me. The rotor frequency in Hz.
When drawing the steady state electrical model (for an AC induction motor), is the stator armaturre inductance considered? Why/Why not?
A three-phase induction motor is rated at 5 hp, 1760 rpm, with a line-to-line voltage of 220 V rms. The motor has a power factor of 80 percent lagging and an efficiency of 75 percent under full-load conditions. Find the electrical input power absorbed by the motor under full-load conditions. Also, find the rms line current.
The rotor of a 6-pole, 50-Hz induction motor is rotated by some means at 1000 r.p.m. Compute (i) rotor voltage (ii) rotor frequency (iii) rotor slip and (iv) torque developed. Can the rotor rotate at this speed by itself?
Consider the two-pole two-phase induction motor having two windings displaced 90° in space shown in Figure P16.8. The fields produced by the windings are given by Ba=Kia(t) cos(θ) and Bb=Kib(t) cos(θ−90°). The two-phase source produces currents given by ia(t)=Im cos(ωt) and ib(t)=Im cos(ωt−90°). Show that the total field rotates. Determine the speed and direction of rotation. Also find the maximum flux density of the rotating field in terms of K and Im.
Find the line current, shaft speed, load torque, induced torque, and rotor frequency of an induction motor given the following information: a) A total of six poles b) 67 hp c) 440V d) S of 6% e) Friction and gale losses of 300 W f) 600 W core losses g) Efficiency of 80% h) Power factor of 0.8