Q sol:eA3 phase, 4 pole, 50 HZ, 380v, IM, starConcted has z1=0.1-4j and z2= 0.2+1.2johm/phase and no-lode crent of 1-10j AThe fall Lord speed is 1410 rpmFin1-motor efficiency and power factor2 if theTorque decreased 10% findefficiency and power factorGiven data.A 30, 4P, 50t12, 380V, Star Connected IM2₁= 0·1+joy 222₂= 0·2+j1-2-2/phaa.No-load (ument Is = 1-j₁o Afull load speed Nv = 1410Rotor speed Ny = 14-10 vpm120f=statov speed Ns =PPower fa% ship =Motor efficiency=120x50 1500 vpm4NS-NyNgNvNsX100=X1001410L 500X1001500-14101500X100I know that the law 0.2efficiency √ (@0-1)² + (0-06×1.2) ²0.94€ lag== pout /pin x 100%way using the law?Ns/Nrx100%

Problem relating to the formula of motor efficiency

Answered: Motor efficiency Power fa 11 = = Ns Nv…

Motor efficiency Power fa 11 = = Ns Nv Ns x100 14-10 L 500 X100 1500 6% 84 I know that the law 02 efficiency√3)² + (0-06×12) ² 0.94€ lag = pout/pin x 100% way using the law? Ns/Nrx100%

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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Concept explainers

Winding and Constructions

Alternating Current (AC) winding is generally of the 3-phase type due to the inherent advantages of the 3-phase machine. Armature coils must be connected to the yield balance (equivalent in magnitude and 2/3 of the series phase difference of the rod) 3-phase electromotive forces (emf). The armature should be divided into step-bands to start with slots around the perimeter.In the construction of AC machines, the frame is used as an external protective shield to protect it from environmental conditions. The frame also serves as the outer perimeter, allowing the interior to be easily placed. The static state section of the devices is the stator winding attached stator.

Question

Q sol:
e
A3 phase, 4 pole, 50 HZ, 380v, IM, star
Concted has z1=0.1-4j and z2= 0.2+1.2j
ohm/phase and no-lode crent of 1-10j A
The fall Lord speed is 1410 rpm
Fin
1-motor efficiency and power factor
2 if theTorque decreased 10% find
efficiency and power factor
Given data.
A 30, 4P, 50t12, 380V, Star Connected IM
2₁= 0·1+joy 22
2₂= 0·2+j1-2-2/phaa.
No-load (ument Is = 1-j₁o A
full load speed Nv = 1410
Rotor speed Ny = 14-10 vpm
120f=
statov speed Ns =
P
Power fa
% ship =
Motor efficiency
=
120x50 1500 vpm
4
NS-Ny
Ng
Nv
Ns
X100=
X100
1410
L 500
X100
1500-1410
1500
X100
I know that the law 0.2
efficiency √ (@0-1)² + (0-06×1.2) ²
0.94€ lag
=
= pout /pin x 100%
way using the law?
Ns/Nrx100%

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