03/ 37 Kw, 300V, 50-Hz, 6-pole, 3-4 , an induction motor has an efficiency of 0.9, slip 3%, speed is 970 r.p.m., windage and fricition loss is 600w, stator losses is 1200w, power factor 0.85. Calculate (1) Mechanical power (2) Rotor Cu loss Pcu (3) In put power. (4) The shaft torque (5) The phase current

03/ 37 Kw, 300V, 50-Hz, 6-pole, 3-4 , an induction motor has an efficiency of 0.9, slip 3%, speed is 970 r.p.m., windage and fricition loss is 600w, stator losses is 1200w, power factor 0.85. Calculate (1) Mechanical power (2) Rotor Cu loss Pcu (3) In put power. (4) The shaft torque (5) The phase current

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

Would a positive-rotating harmonic or a negative-rotating harmonic be more harmful to an induction motor? Explain your answer.
i need the answer quickly
I need the answer as soon as possible
6- Which of the following supply is given to the rotor winding of 3-Phase Induction Motor: a) DC Supply b) AC Supply c) No Supply d) Pulsating Supply 7- 3-Phase Induction Motor will have Starting Torque and Starting Current a) High, High b) Low, Low. c) High, Low d) Low, High 8- Which of the following method is used for controlling the speed of an induction motor: a) Changing number of stator poles b) Supply voltage control c) Frequency control d) All the options are correct 9- Which of the following method cannot be used for starting of 3-Phase Squirrel Induction Motor: a) Star-Delta Starting b) Line resistor startling c) transformer Starter d) Rotor Resistance Starting 10- The rotor of a stepper motor has no: a) windings b) commutator c) brushes d) all of the above
An 8 pole, 50 Hz slip ring induction motor runs at a slip of 2.67% on full load. The rotor resistance per phase of the winding is 0.2 ohm. What external resistance per phase must be added in the rotor circuit so that the motor now runs ate a slip of 17.33% such that the torque remains same in both the cases. (Ans: 1.1 ohm)
A 400 V, 50 Hz 30 hp, three-phase induction motor is drawing 50 A current at 0.8 power factor lagging. The stator and rotor copper losses are 1.5 kW and 900 W respectively. The friction and windage losses are 1050 W and the core losses are 1200 W. The air-gap power of the motor will be
1.why the slip of induction motor is never zero, 2.what happens if the slip of induction motor is one, 3.what is the value of the slip for an induction motor at on load and full load
0-4:-A three phase induction motor has 4-pole runs at 1440 rpm on 500 V, s0 Hz. The mechanical power developed by the rotor is 20.3 hp. The mechanical losses are 2.23 hp. Determine (a) the slip (b) the rotor copper losses, and (c) the efficiency.
B/ Fill in the blanks: TV ezur 1- In a three-phase induction motor iron losses in stator will be..... that in rotor. 2- In three-phase induction motors sometimes copper bars are placed deep in the rotor to........... 3- Blocked rotor test on induction motors is used to find out. R..., ke, X. 4- The maximum torque in an induction motor depends on .............and .....
A three-phase, 60-Hz induction motor runs at 890 r/min at no load and at 840 r/min at full load. (a) How many poles does this motor have? (b) What is the slip at rated load? (c) What is the speed at one-quarter of the rated load? (d) What is the rotor’s electrical frequency at one-quarter of the rated load?
WITH DIAGRAM FOR UPVOTE COMPLETE WORK
tørque in new Draw the power flow diagram of induction motor and explain the different power losses, the air gap power and the converted pòwer. Explain how by solving the equivalent circuit of three phase induction motor the input current, input power factor, air gap power, rotor copper loss, converted power, output power and induced torque are calculated. (b)A 440 V, 50 Hz, 2-pole, Y-connected induction motor is rated 75 KW. The equivalent circuit parameters are: R, = 0.0752 R, = 0.0652 X = 1.22 Pare = 1.1kW X, = 0.172 X, = 0.17Q Pe =150W PEAW = 1.0kW For a slip of 0.04 find the line-eurrent, the stator"power factor, the stator copper_loss, the air gap power, the converted power the induced torque and the Toad torque. MISC