4-20-A three-phase,60-Hz, six pole wound-rotor induction motor draws 10KW when driving its normal load.it draw 700W ,when the load is disconnected The rotor and stator copper loses under normal load are 295 W and 310 W respectively. Calculate the efficiency, the specd, and the shaft torque of this motor under normal load condition, Assume equal rotational and core losses and negligible no-load copper Isses.

4-20-A three-phase,60-Hz, six pole wound-rotor induction motor draws 10KW when driving its normal load.it draw 700W ,when the load is disconnected The rotor and stator copper loses under normal load are 295 W and 310 W respectively. Calculate the efficiency, the specd, and the shaft torque of this motor under normal load condition, Assume equal rotational and core losses and negligible no-load copper Isses.

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

Section: Chapter Questions

Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...

See similar textbooks

Similar questions

Developed torque??
Q2: A single-phase induction motor with a 4-pole, 50-Hz, 130-V, is rotating in the clockwise direction at a speed of 1450 rpm. Determine its per-unit slip (a) in the direction of rotation and (b) in the opposite direction. At standstill the rotor resistance is 15 Q, determine the effective rotor resistance in each branch.
what are the disadvantages of slip ring induction motor?
22. A) Why is the power factor of a lightly loaded 3-phase squirrel-cage induction motor very low? a) most of the power supplied to the motor at light loads is magnetization or inductive power b) most of the power supplied to the motor at light loads is resistive power c) most of the power supplied to the motor at light loads is capacitive power d) none of the above B) What happens to the power factor as load is increased on the motor? a) the power factor increases b) the power factor deceases c) remains the same
Q4) What is the effect in series motor if you assume V = Eb , with low resistance fielding winding ? a) b) The given information about series motor on full – load (with that the flux is proportional to the field current) is a ) takes ( 78A) and runs at (850 r.p.m) , the series winding resistance is (0.04 2) , with diverter in parallel with the field winding is a (0.18 2) and the armature resistance is (0.09 S2). ( 220 V Neglected the iron, friction and windage losses. Determine the efficiency , speed when developing full load torque ,the maximum efficiency .
Q. Given the vector field Ē = xâ, +x²yâ, + xy°â, A (0,0,0)| В (0,1,0) y a) Find the line integral E-dl along the path ABCDA. D (1,0,0) C (1,1,0) b) Evaluate [(V×E)·â¸ds over the surface S.
On full-load a 300 V series motor takes 90A and runs at 15 rev/s. The armature resistance is 0.1 ohm and the series winding resistance is 50 m.Determine the speed when developing full load torque but with a 0.2 ohm diverter in parallel with the field winding. (Assume that the flux is proportional to the field current.
Reducing the supply voltage so that Vnew = Vold / V3 reduces the maximum torque of the motor so that Tmaxnew= ...Tmaxold . * .... ...
On full load, a 300 V series motor takes 90A and runs at 15 rev/s. The armature resistance is 0.1 ohm, and the series winding resistance is 50 mohm. Determine the speed when developing full load torque but with a 0.2-0hm diverter parallel to the field winding. (Assume that the flux is proportional to the field current.
A 460 V, 60 Hz, Star connected, 870 r.p.m., Eight poles, Three-Phase Induction Motor has the following per phase parameters: R1 = 0.07, R2 = 0.05, X1 = X2 = 0.2, Xm = 6.5, Draw the approximate equivalent circuit and determine the following:I. Starting TorqueII. Maximum TorqueIII. Rated TorqueIV. Stator CurrentV. Power FactorVI. Input PowerVII. Magnetizing Current
A 3-phase, 220-V, 50-Hz, 6-pole, Y-connected induction motor is operating with a per-unit slip of 0.07, 95 A line current, PF=0.8 lagging, and 2200 W stator copper losses. The total friction, windage, and core losses are lumped and assumed to be constant at 3.5 KW. Find the rotor copper losses and the output power of this motor? Select one: O a. None O b. PRCI = 1.08 KW and Pout = 18.8 KW O c. PRCL = 1.87 KW and Pout = 21.4 KW O d. PRCL = 10.5 KW and Pout = 29.14 KW