A 3-phase, 240-V, 50-Hz, 6-pole, Y-connected induction motor is operating with a per-unit slip of 0.08, 105 A line current, PF=0.9 lagging,and 2400 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. PRCL = 2.09 KW and Pout = 26.88 KWO b. PRCL = 4.5 KW and Pout = 28.14 KWO c. PRCL = 2.95 KW and Pout = 30.43 KWO d. NoneClear my choice

Using the power flow diagram of the induction motor, The input power of the induction motor is,…

Answered: A 3-phase, 240-V, 50-Hz, 6-pole,…

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...

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The full load slip of a 4-pole, 60Hz, 440V (line-to-line) three phase induction motor is 0.05. (i). What is the speed of the rotating stating? What is the frequency (in Hz) of the rotor current? (ii). When the output power of the motor is 90 hp at a speed of 1732 rmp, rotational losses (which include frictional and core losses) are 10,100W, while copper losses for both the rotor and stator total 3700W. Determine the motor efficiency and line current, assuming the motor has a 0.8pf lagging under the conditions.
a. A three-phase, 220-V, 60-Hz induction motor runsat 1,140 r/min. Determine the number of poles (forminimum slip), the slip, and the frequency of therotor currents.b. To reduce the starting current, a three-phasesquirrel cage induction motor is started byreducing the line voltage to Vs/2. By what factorare the starting torque and the starting currentreduced?
A three-phase induction motor, operating at rated voltage and frequency, has a startingtorque of 1.35 times and a maximum torque of 2.2 times, both with respect to its rated-loadtorque. Ignoring the effects of the stator resistance and rotational losses and assumingconstant rotor resistance, determine the slip rated load, and the rotor current at starting (as a percentage of the rotor current at rated load). The slip at maximum torque is 0.343 if that helps.
A 6-pole, 60-Hz, Y-connected, three-phase induction motor develops a torque (Td) of 250 N.m at a speed of 720 rpm. The rotor resistance is 0.4 Ohm/phase, and the stator winding impedance is negligible. A) What is the air-gap power? B) What is the rotor current refered to the stator? C) If the rotational loss is 1kW, what is the motor output power in hp?
Three phase 460 hp,415v,4pole,50hz induction motor is operating at Over voltage of 480v and reduced load of 80% of the rated value and draws a line current of 624.06A.The losses at these operating conditions are as follows .The stator conductor loss is 12.6 kw sketch the power flow diagram ,enter the known values and determine (a) rotor speed,(b) shaft output torque ,(c)developed mechanical torque in Nm (d)input power to the stator (f)The input power factor
A 80 kW, 480-V, 60-Hz, six pole induction motor has a slip of 5 percent when operating at fullload conditions. At full-load conditions, the friction and windage losses are 300 W, and the core losses are 400 W. For full-load conditions, compute the induced torque,T-ind, in N-m.
A 3300-V, 50-Hz, 3-phase, 4-pole, star-connected induction motor has identical primary and referred secondary impedances of value 3 + j9 ohm per phase. The turns-ratio per phase is 3/I (stator/rotor), and the rotor winding is connected in delta and brought out to slip rings. Calculate: the full-load torque at rated slip of 5%; the maximum torque at normal voltage and frequency; the supply voltage reduction which can be withstood without the motor stalling; the maximum torque if the supply voltage and frequency both fall to half normal value; the increase in rotor-circuit resistance which, at normal voltage and frequency, will permit maximum torque to be developed at starting. Express this: as a fraction of normal R2and as (3) ohmic values to be placed in series with each of the slip-ring terminals and star connected.
A 230-V, 60-Hz, 6-pole, Y-connected, three-phase induction motorhas the following parameters in ohms/phase as referred to thestator: R1 = 12.5, X1 = 21.3, R2 = 28.6, X2 = 13.6, Rc = 4200, andXm = 1800. The friction and windage loss is 3% of the powerdeveloped. If the rotor speed is 1125 rpm, and using the exactequivalent circuit diagram determine the following:(i) the input power,(ii) the stator copper loss,(iii) the rotor copper loss,(iv) the air-gap power,(v) the power developed,(vi) the torque developed,(vii) the efficiency, and(viii) draw the power flow diagram
Three-phase 460-hp, 415 V, 4-pole, 50 Hz induction motor is operating at overvoltage of 480 V and reduced load of 80% of the rated value and draws a line current of 624.06A. The losses at these operating conditions are as follows: The stator conductor loss is 12.6kW, the rotor conductor loss is 10.4kW, the core loss is 11.00kW; and rotational losses is 4.129 kW. Sketch the power flow diagram, enter the known values, and determine (a) rotor speed (b) shaft output torque (c) developed mechanical torque in Nm. (d) input power to the stator; (f) the input power factor. Do only part (a), (b) & (c)
A three-phase induction motor, at rated voltageand frequency, has a starting torque of 140 percentand a maximum torque of 210 percent of full-loadtorque. Neglect stator resistance and rotational lossesand assume constant rotor resistance. Determinea. The slip at full load.b. The slip at maximum torque.c. The rotor current at starting as a percentage offull-load rotor current.
A 440 V , 25 hp , 4 - pole , Y - connected three - phase wound - rotor induction motor has the following circuit parameters : R1 = 0.460 , R2 = 0.33 0 , x1 = 1.1 0 , x2 = 0.46 02 , and Xm = 25 22. ( stator referred ) Rotational losses ( including core loss ) = 500 W ( constant ) . If the terminal voltage and supply frequency are simultaneously reduced to 50 % , calculate the stator current , the developed air - gap torque , and the maximum torque at a slip of 0.04 .
A series dc motor 20-hp, 500-V, is having armature, series field and shunt field winding resistances of 0.08 Ω, 0.3 Ω and 125 Ω, respectively. If the back – emf is 478 V, determine the motor current when the connection is a) long – shunt. b) short – shunt c) series d) shunt

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