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Chapter 15: Induction Motor Equivalent Circuit

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Introduction

• To gain better understanding of the properties and behaviors of the induction motor, it is necessary to examine the equivalent circuit– develop an equivalent circuit from the basic principles

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Wound-rotor Induction Motor• A 3-phase wound-rotor induction

motor is very similar in construction to a 3-phase transformer– the motor has 3 identical primary

windings, 3 identical secondary windings, and perfect symmetry

• single phase analysis is possible

– at standstill, the motor acts exactly like a conventional transformer

• model begins with the transformer’s equivalent circuit

– circuit parameters• Rm = iron, windage, and friction losses

• Xm = magnetizing reactance

• r1 = stator winding resistance

• x1 = stator leakage reactance

• r2 = rotor winding resistance

• x2 = rotor leakage reactance

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Wound-rotor Induction Motor

– the magnetizing branch can not be neglected

• I0 may be 40% of Ip due to the air gap

• for motors greater than 2 hp, the magnetization branch can be shifted to the input terminals

– for a motor running at a slip s, the rotor speed is ns(1-s)

• this will modify the values of E1, I1, E2 and I2

• the frequency in the secondary winding will become sf

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Wound-rotor Induction Motor– the induced rotor voltage

– the rotor leakage reactance

– rotor resistance is not frequency dependent

– the model of the rotor circuit is

• the effective value of I1 is equal to the effective value of I2 even though their frequencies are different

• the effective value of E1 is equal to the effective value of E2 divided by the slip

• the phase angle between E2 and I2 is the same as the angle between E1 and I1

12 EsE =

22 jxsxj =′

xRrR += 22

22

12 jsxR

EsI

+=

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Wound-rotor Induction Motor

– on the primary side

– therefore

• Without affecting the value of the stator current, I1, the impact of the slip is shifted from the rotor voltage and reactance to the rotor resistance– greatly simplifies the analysis

22

121 jsxR

EsII

+==

2

1

22

11 Z

E

jxsR

EI =

+=

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Power Relationships• The equivalent circuit enables the

deduction of basic electromechanical power relationships

• By inspection– active power absorbed

– reactive power absorbed

– apparent power absorbed

– power factor

– line current

– active power supplied to the rotor

– I2R losses dissipated in the rotor

– developed mechanical power

– developed torque

– efficiency

sRIrIRVP mT 2221

21

2 ++=

2221

21

2 xIxIXVQ mT ++=

22 QPS +=

SP=θcos

sRIPr 222=

rjr PsRIP == 221

( )sPPPP rjrrm −=−= 1

Tp VSI =

PPm=ηs

rm

n

P

n

PT

55.955.9 ==

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Breakdown Torque & Speed

• The developed torque is given by the power delivered to the rotor and the synchronous speed

– the power delivered is a function of the resistance R2/s

• Maximum torque occurs at maximum power transfer– maximum power transfer is

found when R2/s is equal to the absolute value of Z1

– the voltage drops are equal

• The phasor diagram of this special case:

from geometry:– AB and BC have the same length

– the angle ABC = (180°-α)– angle CAB = angle ACB = α/2– hence

12 ZsR =

sr nPT 55.9=

1121 ZIsRI = 22cos

2cos 11

21

TVZIs

RI == αα

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Breakdown Torque & Speed

– the slip at breakdown torque is

– the current at the breakdown torque is

– the breakdown torque is

• note that the magnitudes of both the breakdown torque and breakdown current are independent of the rotor current resistance R2

12 ZRsb =

( )211 cos2 αZVI Tb =

( )( )221

2

cos4

55.9αZn

VT

s

Tb = ( )1arctan rx=α

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Breakdown Torque Calculation

• Example– determine the maximum torque and corresponding speed of a

440 V, 1800 rpm, 60 Hz, 5-hp, squirrel cage induction motor• r1 = 1.5 ohms, r2 = 1.2 ohms,jx = 6 ohms,jXm = 110 ohms,

Rm = 900 ohms

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Torque-speed Curve

• The complete torque-speed curve of a motor can be determined by selecting various values of slip and solving the circuit equations

• Example– determine the torque and speed of a 440 V, 1800 rpm, 60 Hz,

5-hp, squirrel cage induction motor• r1 = 1.5 ohms, r2 = 1.2 ohms,jx = 6 ohms,jXm = 110 ohms,

Rm = 900 ohms

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Torque-speed Curve

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Test to Determine the Equivalent Circuit

• No-load and locked-rotor tests provide approximate values of r1, r2, Xm, Rm, and x

• No-load test– small slip leads to high value of R2/s with I1 being negligible compared to I0– no-load circuit essentially consist of Xm and Rm

• measure the stator resistance RLL, and assume a wye connection

• run the motor at no-load and at rated line voltage, measuring the no-load current and the total three-phase active power

( ) NLNLmvfNLmNLNLvf

NLNLNLNLNLNL

QEXPPER

rIPPP

PSQIVS

22

12

22

3

3

=+=

−=+

−==

21 LLRr =

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Test to Determine the Equivalent Circuit

• Locked-rotor test– under locked-rotor conditions

• stator current is almost 6 pu, slip is unity, and R2/s equals r2/1 = r2• magnetization branch can be neglected

– locked-rotor circuit consist of x, r1, and r2• apply a reduced 3-phase voltage

such that the stator current is about equal to its rated value

• measure the line voltage, line current, and 3-phase power

( )( ) 122

2

22

3

3

3

rIPr

IQX

PSQIVS

LRLR

LRLRm

LRLRLRLRLRLR

−=

=

−==

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Test to Determine the Equivalent Circuit

• Example– a no-load test conducted on a 30 hp, 835 rpm, 440 V, 3-phase, 60 Hz,

squirrel-cage induction motor yields the following results

• no-load line-to-line voltage: 440 V

• no-load current: 14 A

• no-load power: 1470 W

• resistance measured between two terminals: 0.5 Ω– the results of locked-rotor test, conducted at reduced voltage

• locked-rotor line-to-line voltage: 163 V

• locked-rotor power: 7200 W

• locked-rotor current: 60 A

– determine the equivalent circuit of the motor

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Induction Motor Equivalent Circuit

• Homework– Problems: 15-2, 15-4, 15-5, 15-7