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### Transcript of Chapter 15: Induction Motor Equivalent Circuit frick/EE4220-EM_Dynamics/lecture1¢ ...

• 10/29/2003 Electromechanical Dynamics 1

Chapter 15: Induction Motor Equivalent Circuit

• 10/29/2003 Electromechanical Dynamics 2

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

• 10/29/2003 Electromechanical Dynamics 3

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

• 10/29/2003 Electromechanical Dynamics 4

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

• 10/29/2003 Electromechanical Dynamics 5

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

1 2 jsxR

Es I

+ =

• 10/29/2003 Electromechanical Dynamics 6

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

1 21 jsxR

Es II

+ ==

2

1

22

1 1 Z

E

jxsR

E I =

+ =

• 10/29/2003 Electromechanical Dynamics 7

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 2 2 21

2 1

2 ++=

2 2 21

2 1

2 xIxIXVQ mT ++=

22 QPS +=

SP=θcos

sRIPr 2 2 2=

rjr PsRIP == 2 2 1

( )sPPPP rjrrm −=−= 1

Tp VSI =

PPm=η s

rm

n

P

n

P T

55.955.9 ==

• 10/29/2003 Electromechanical Dynamics 8

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 = 22 cos

2 cos 11

2 1

TVZI s

R I == αα

• 10/29/2003 Electromechanical Dynamics 9

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

55.9 αZn

V T

s

T b = ( )1arctan rx=α

• 10/29/2003 Electromechanical Dynamics 10

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

• 10/29/2003 Electromechanical Dynamics 11

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

• 10/29/2003 Electromechanical Dynamics 12

Torque-speed Curve

• 10/29/2003 Electromechanical Dynamics 13

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

( ) NLNLmvfNLm NLNLvf

NLNLNLNLNLNL

QEXPPER

rIPPP

PSQIVS

22

1 2

22

3

3

=+=

−=+

−==

21 LLRr =

• 10/29/2003 Electromechanical Dynamics 14

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

−=

=

−==

• 10/29/2003 Electromechanical Dynamics 15

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

• 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

• 10/29/2003 Electromechanical Dynamics 16

Induction Motor Equivalent Circuit

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