SKEE2413 on Induction Motor [Compatibility Mode]

Chapter 4.

Three-phase Induction Machines

Introduction The induction machine is the most rugged and

the most widely used machine in industry. Both stator and rotor winding carry alternating

currents. The alternating current (ac) is supplied to the

stator winding directly and to the rotor winding by induction – hence the name induction machine.

Application (1-phase): washing machines, ceiling fans, refrigerators, blenders, juice mixers, stereo turntables, etc.

2-phase induction motors are used primarily as servomotors in a control system.

Application 3-phase: pumps, fans, compressors, paper mills, textile mills, etc.

Induction Machine

Construction

Construction

Unlike dc machines, induction machines have a uniform air gap. The stator is composed of laminations of high-grade sheet steel.

A three-phase winding is put in slots cut on the inner surface of the stator frame.

The rotor also consists of laminated ferromagnetic material, with slots cut on the outer surface.

Squirrel-cage Rotor

Wound Rotor

Wound rotor

Slip ring

Construction

The three-phase winding are displaced from each other by 120 electrical degrees in space

Current flows in a phase coil produce a sinusoidally distributed mmf wave centered on the axis of the coil.

Alternating current in each coil produces a pulsating mmf wave. Mmf waves are displaced by 120 degrees in space from each other. Resultant mmf wave is rotating along the air gap with constant peak.

Cross-sectional view Y-connected stator ∆-connected stator

Induction Motor Operation

RMF – rotating magnetic field

Rotating Magnetic Field – consider 2-pole machine

a) Three phase stator winding, aa’, bb’ and cc’ displaced by 120o.

b) Mmf (pulsating) in space at various instants due to a.c current in coil aa’

c) Instantaneous 3 phase current

a. Graphical Method – Resultant mmf (magnitude and direction)

Resultant mmf

Mmf phase a

at t = t0= t4

Graphical Method

Constant amplitude, move around the air gap

n = synchronous speed

f = f1= supply freq.,

p = no of polesrpm

pairs pole p ; radsf2

rad/sin speed sSynchronou

1-11 ===

ppe πωω

b. Analytical Method

Motion of the resultant mmf

N = effective number of turns

ia= current in phase ‘a’

Analytical Method

Induced Voltages

where r = radius of the stator; ι = axial length of stator

A

Induced Voltages

V per phase

At Standstill operation

E1 = 4.44f1N1ΦpKw1

E2 = 4.44f2N2ΦpKw2 ; f1 = f2

E2 = 4.44f1N2ΦpKw2

Running Operation

120

pnf =

at slip s

* E2 – induced rotor voltage at standstill

frequency slip :

pair pole :

frequencyrotor :

frequencysupply :

frequency ssynchronou :

:where

)1(

; s ;

radsin edreprensent beCan

2

2

-1

ω

ωωω

ωω

ωωωω

ωωω

p

s

ps

m

e

s

sm

es

ss

ms

−=

==−=

Example 1

A three-phase, 100 hp, 460 V, four-pole, 60 Hz induction machine delivers rated output power at a slip of 0.05. Determine the

(a) Synchronous speed and motor speed.(b) Speed of the rotating air gap field.(c) Frequency of the rotor circuit.(d) Slip frequency (in rpm).(e) Speed of the rotor field relative to the

(i) rotor structure(ii) stator structure(iii) stator rotating field

(f) Rotor induced voltage at the operating speed, if the stator-to-rotor turns ratio is 1:0.5

Pg 219: 1800 &1710rpm, 1800rpm, 3 Hz, 90 rpm, (90rpm, 1800rpm, 0rpm), 6.64 V/ph)

Sol_pg21

Equivalent Circuit Model

• To study and predict the performance of the induction machine

Induction Motor Drives SEE4433 Dr Zainal / Dr Awang

25

Stator voltage equation:

V1 = R1 I1 + j(2πf)LlI1 + Eag;

Eag – air gap voltage or back e.m.f

Eag = E1 = k f1 φag

Rotor voltage equation:

E2 = R2 I2 + js(2πf)Ll2

E2 = k f2 φag = k sf1 φag = sE1

E2 – induced emf in rotor circuit ; E1=R2/sI2+j2πfLI2



sE2 – rotor voltage at standstill


This model is not convenient to use to predict

circuit performance

Example 2

A three-phase, 15 hp, 460 V, four-pole, 60 Hz, 1728 rpm induction motor delivers full output power to a load connected to its shaft. The windage and friction loss of the motor is 750 W. Determine the

(a) mechanical power(b) air gap power(c) rotor copper loss.

Pg 226: 11940 W, 12437.5 W, 497.5 W

Sol_pg29


Assume small volt

drop across R1and X1 – ease computation of IΦ and I2’,

V1 = E1

Due to machine air gap, IΦ is high- 30-50% of full –load

current, X1 is high, core loss (Rc) is lumped into the

mechanical losses


R12<<(X1+Xm)2

X1 << Xm

For simplification, replace V1, R1,X1, Xm with Vth, Rth, Xth ( at terminal Pag)

Equivalent Circuit ParametersRc, Xm, R1, X1, X2, R2

No-Load Test The parameters of the equivalent circuit, Rc, Xm, R1, X1,

X2, and R2 can be determined from the results of a no-load test, a blocked-rotor test and from measurement of the dc resistance of the stator winding.

The no-load test, like the open circuit test on a transformer, gives information about exciting current and rotational losses.

This test is performed by applying balanced polyphase voltages (415V) to the stator windings at the rated frequency(50Hz).

The rotor is kept uncoupled from any mechanical load.

R1 X1

Xm

I1

cctopen 0

R

s

R

0;N

NNs

NN load, noAt

22

s

rs

sr

→∞==∴

=−=

≈

Blocked-Rotor Test The blocked-rotor test, like the short-circuit test on a

transformer, gives information about leakage impedances. In this test the rotor is blocked so that the motor cannot rotate,

and balanced polyphase voltages are applied to the stator terminal ( increases voltage until stator current reaches its rated value).

The blocked-rotor test should be performed under the same conditions of rotor current and frequency that will prevail in the normal operating conditions.

The IEEE recommends a frequency of 25% of the rated frequency for the blocked-rotor test. However, for normal motors of less than 20 hp rating, the effects of frequency are negligible and the blocked-rotor test can be performed directly at the rated frequency

cctopen , 1

R

s

R

1;N

NNs

;0N ),rotor test(block load fullAt

222

s

rs

r

2 mm XXRR ⟨⟨→==∴

=−=

=

R1 X1 X2

R2

Equivalent Circuit Parameters Measurement of average dc resistance per stator

phase : R1

No load test :VNL

INL

PNL

Blocked-rotor test:VBL

INL

PNL

Example 3 The following test results are obtained from a three-phase, 60 hp, 2200 V,

six-pole, 60 Hz squirrel-cage induction motor. No-load test:

supply frequency = 60 HZline voltage = 2200 Vline current = 4.5 Ainput power = 1600 W

Blocked-rotor test:frequency = 15 Hzline voltage = 270 Vline current = 25 Ainput power = 9000 W

Average DC resistance per stator phase:R1 = 2.8 ohm

(a) Determine the no-load rotational loss.(b) Determine the parameters of the IEEE-recommended equivalent circuit.(c) Determine the parameters (Vth, Rth, Xth) for the thevenin equivalent circuit.

Pg: 230: 1429.9 W Sol_pg38(IM)

Performance Characteristics

Induction Motor Drives SEE4433 Dr Zainal / Dr Awang

40

Performance calculation using SPEC

22

2

22

2

21

12

1

11

11

3 :losscopper Rotor

3 :gap-air theacrossPower

3 :loss Core

3 :losscopper Stator

current and voltagephase bemust and :Note

cos3 :PowerInput

RIPPPP

s

RIP

R

VP

RIP

IV

IVP

lr

lclsin

g

mlc

ls

in

=−−=

=

=

=

= φ

I 1 R1 L 1

V 1 Rm L m

L 2

R2

S

I 2

s

RsIPT

PPPP

sPs

sRII

s

RI

PPP

mm

oe

FWFWosh

g

lrgo

ωω2

22

22

22

22

22

2

)1(3

: torquegnetic)(electroma Developed

loss. windageandfriction : ;

:shaft at thePower

)1()1(3

R33

:poweroutput Gross

−==

−=

−=−=−=

−=

s

RpIT

p

s

RIT

ss

ee

ee

e

mm

22

2

1

2

1

22

11

1

3

Then,

frequency.supply theis ;But

3

,)1(

Since

ω

ωωω

ω

ωωω

ωω

=⇒

=

=∴

−=⇒−=

ExampleA single phase equivalent circuit of a 6-pole SCIM thatoperates from a 220 V line voltage at 60 Hz is given below.Calculate the stator current, input power factor, outputpower, torque and efficiency at a slip of 2.5%. The fixedwinding and friction losses is 350 W. Neglect the core loss.Also calculate the starting current.

Solution

I1 R1 X1

V1 Xm

X2

R2

I2

0.2Ω 0.5Ω

20Ω 0.1Ω

0.2Ω

1V 220V line-to line 3

220V 127V

3 2.5% 0.025

= ÷

= =

= =

mN

sp

PPT

WPPP

WsPP

W

PPP

WRIP

W

IVP

AZ

VI

jj

jjj

jXs

RjXjXRZ

XXX

e

o

m

oe

FWosh

go

lsing

ls

oin

oo

in

o

min

m

.4.78

)025.01)(3/60(2

9611

)1()(

Torque neticElectromag

%3.8910758

9611

powerInput

powerOutput Efficiency

611,9350961,9

shaft at thePower

961,9)025.01(216,10)1(

power Gross

216,10 540758,10

loss) core g(neglectinrotor tosferredPower tran

540)2.0)(30(33

758,10

)20)(cos30)(3220(3cos3

200.30202.4

3220

202.4202.0

025.01.0

2.0025.01.0

205.02.0

//)(

20,2.0,5.0

1

21

21

11

11

22

11

21

=−

=−

==

===

=−=−=

=−=−=

=−=

−=

===

===

−∠=∠

==

Ω∠=

++

+++=

+++=

Ω=Ω=Ω=

πωω

φInput power

factor

motors.induction starting taken whenbe should emotors.Carinduction for common is This

current. load full than times5about iscurrent starting that theNote

16776.0

322076.0

202.01.02.01.0

205.02.0

1 ,standstillAt :Solution

11 A

Z

VI

jj

jjjZ

s

in

in

===

Ω=

++

+++==

=

The following results were obtained on a 3 phase, starconnected stator, 75 kW, 3.3 kV, 6-pole, 50 Hz squirrel-cageinduction motor.

No-load (NL) test: Rated frequency, 50 HzVNL = 3.3 kV (line), INL = 5A, PNL = 2500 W

Blocked-rotor (BR) test: Frequency 50 HzVBR = 400 V (line), IBR = 27 A, PBR = 15000 W

DC test on stator resistance per phase = 3.75Ω.

i) Determine the parameters of the IEEE recommended equivalent circuit.

ii) Find the parameters of the Thevenin equivalent circuit as seen from the rotor circuit.

iii) For a slip of 4%, calculate the stator current, power factor and efficiency of the motor.

Example

Sol_pg46

SKEE2413 on Induction Motor [Compatibility Mode]

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Transcript of SKEE2413 on Induction Motor [Compatibility Mode]