The following are the results obtained from the no-load, blocked rotor and DC resistancetests on a 30 40 hp, 400 V, 1470 rpm, 4-pole, 50 Hz, Class C wye-connected squirrel-cageinduction motor with 2% slip.No load TestVNL= 500 VINL = 40 APNL = 800 WBlocked Rotor TestVBR= 50 VIBR = 25 APBR = 100 WDC Resistance TestRpc = 0.02 OhmDetermine:i.The parameters of approximate equivalent circuitii.Draw its approximate equivalent circuit with proper labeliii.Determine the windage and friction loss of the machinePnl = V3VNL!nL Cos 0nlPBr = V3VBR Br Cos OBrNema class CX, =0.3X RX = 0.7X,PeuNL = 31 N*R,P, = PNL. – P.eRBR

Given V=400V phase voltage =4003=231 V

The following are the results obtained from the no-load, blocked rotor and DC resistance tests on a 30 40 hp, 400 V, 1470 rpm, 4-pole, 50 Hz, Class C wye-connected squirrel-cage induction motor with 2% slip. No load Test VNL= 500 V INL = 40 A PNL = 800 W Blocked Rotor Test VBr= 50 V IBR = 25 A PBR = 100 W DC Resistance Test RDc = 0.02 Ohm Determine: i. The parameters of approximate equivalent circuit ii. Draw its approximate equivalent circuit with proper label iii. Determine the windage and friction loss of the machine

The following are the results obtained from the no-load, blocked rotor and DC resistance tests on a 30 40 hp, 400 V, 1470 rpm, 4-pole, 50 Hz, Class C wye-connected squirrel-cage induction motor with 2% slip. No load Test VNL= 500 V INL = 40 A PNL = 800 W Blocked Rotor Test VBr= 50 V IBR = 25 A PBR = 100 W DC Resistance Test RDc = 0.02 Ohm Determine: i. The parameters of approximate equivalent circuit ii. Draw its approximate equivalent circuit with proper label iii. Determine the windage and friction loss of the machine

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 armature resistance of a shunt motor fed from a 220 Volt DC source is 5 ohms and the resistance of the excitation winding is R ohm and it rotates stably. Then, a resistance of R ohm is added to the excitation circuit (a resistor with the same value as the winding resistance) to control the speed. How many RPM is the motor in the last state in steady state.Mechanical losses will be neglected, motor magnetically linear, no saturation, temperature change will be neglected.Armature current = Ia = 3 ALoad torque=Tload= 8 Nma) 143 rpmb) 734 rpmc) 1361 rpm d) 77 rpm e) 1333 rpm
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d) A 2-pole induction motor is fed with a 230 V, 50 Hz supply and is running at no load. i. Calculate its angular speed in rpm. ii. Explain how can we slow down the speed of the motor above, at no load? iii. Explain what will happen to the speed of the motor if we slowly add amechanical load to the motor.
1. A 50-hp, 250-V 1200 r/min dc shunt motor with compensating windings has an armature resistance (including the brushes, compensating windings, and interpoles) of 0.06 2. Its field circuit has a total resistance Radj + Rr of 50 22, which produces a no-load speed of 1200 r/min. There are 1200 turns per pole on the shunt field winding (see figure). Find the speed of this motor when its input current is A. 100 A. B. 200 A. C. 300A. I EA RA www 0.06 Ω 50 Ω Radj 12 RF LF IL |¹F NF = 1200 turns VT = 250 V
Q2. A 230 V single-phase induction motor, 50 Hz, has the following winding impedances (at rest): Zm=28 <70° Q. Za=42 <45°Q A) Calculate the input current at start-up. B) Obtain the amount of capacitor in series with the auxiliary winding for which the torque Maximum startup. Then compare the input current, auxiliary winding current and torque all at start-up in two modes with and without starter capacitor.
Developed torque??
what are the disadvantages of slip ring induction motor?
Answer it right..
1- To control the speed of an induction motor, the supply frequency is reduced by 10%. For the same magnetizing current to remain constant, the supply voltage must be....... a) Reduced by 10%, b) Reduced by 20%, c) Increased by 10%, d) Increased by 20% 2. When maximum starting torqua ic ohtoinod
Reducing the supply voltage so that Vnew = Vold / V3 reduces the maximum torque of the motor so that Tmaxnew= ...Tmaxold . * .... ...