4. Equivalent Circuit of IM

8/3/2019 4. Equivalent Circuit of IM

1/33

4. Equivalent circuit of IM.

Developing an equivalent circuit model is useful to study and predict the

performance characteristics of the induction machine with reasonable

accuracy.

In this section a steady-state per-phase equivalent circuit will be derived.

For convenience, consider a three-phase wound-rotor induction machine

as shown in fig 3.1.

Stator Rotor Air-gap

Fig. 4.1. 3 - phase induction machines equivalent circuit

model


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If currents flow in both stator and rotor windings,

resultant rotating magnetic fields will be produced in

the air gap.This resultant air gap field will induce voltages in both

stator windings (at supply frequency f1) and rotor

windings (at slip frequency f2).

It appears that the equivalent circuit may assume a formidentical to that of a transformer; therefore it can be

developed from the knowledge of a transformer

equivalent circuit and from knowledge of variation of

rotor frequency with speed in IM.

The only difference is on account of the fact that

the secondary winding (rotor winding) of an

induction motor rotates and mechanical power is

developed.


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The stator circuit model

As explained earlier, when a voltage V

1

is applied to

the stator terminals, the rotating flux generates

counter emf in the 3-phase of stator windings. The

applied voltage V1 has to overcome counter emf = -

E1, V1` and stator leakage impedance drop,I1(r1 + jx1).

Thus, 11111 jxrIVV

V1

E1

X1R

1

Rc

Xm

Fig. 4.2 - The stator circuit model.

V1 = per-phase terminal voltage

R1 = per-phase stator windingresistance

X1 = per-phase stator leakage

reactance

E1 = per-phase induced voltage inthe stator winding

Xm = per-phase stator magnetizingreactance

Rc = per-phase stator core loss resistance


4/33

The stator current consists of two components, theload component, I2` (-I2) and the exciting componentIe.

I2` counteracts the rotor mmf, I2N2. The exciting component Ie, whose function is to create

the resultant air gap flux and to provide the core loss,friction and windage loss and small stator and rotorI2R losses I0 is usually 30% - 50% of rated current of

IM. Thus, the exciting current can be resolved into twocomponents.

- magnetising component, Im lagging V1` by 900

and

- core loss component Ic which is in phase withV1`.

In the equivalent circuit, Ic and Im are accounted for bya shunt branch consisting of:-

- core loss resistance Rc = V1`/Ic

- magnetising reactance Xm = V1`/Im


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Since there is air-gap between stator and rotor of

induction motor, Xm (reactance on account of effect

of magnetizing current) will be much less than

transformer case. Thus, Im will be much greater than

in IM.

F,AT

,wb

Trans

forme

r

Induction motor

Note:- The graph indicates that, Reluctance of IM is much

greater than reluctance of transformers.

Xm = sLs =

2f1Ls

Ls =(Ns)2/RL

Fig. 4.3 Flux mmf relation in transformer and IM.


6/33

Rotor circuit model

Regarding the actual physical facts existing in the rotor circuit,

the per phase rotor current is:

Where, SE2 slip frequency emf

I2 slip frequency current

If the right hand side is divided by slip ,we get:- (for further analysis)

22

22

jSXr

SEI

2

2

2

2

jXS

rEI

R2

SX2

SE2

I2

S

R2

Fig. 4.4a Rotor equiv.

circuit at slip s.

Fig. 4.4b

I2

E2 per phase equi. Circuit at stator frequency.


7/33

To determine the complete equivalent circuit of the rotor, the line

frequency emf E2, the line frequency current I2 and the speed

dependant resistance r2/s must be referred to the stator primary

by the effective turn's ratio as in the transformer.

11

2

1

2. VEN

NE

2

2

2

2

1

2

2 jXS

r

N

NjX

S

r

2

1

22 IN

NI

Where, N1` = N1 Kw1

N2` = N2Kw2


8/33

V1

X1R1

Rc Xm

Pag

I1

Io

Ic

Im

a

II 2'

2

2

2'

2XaX

S

Ra

S

R2

2'

2 12

'

2EaEE E

1

1

2

1

2 EN

NE

2jX

Fig.4.5a Rotor equivalent circuit referred to the stator

S

r

2

fig.4.5b Exact equivalent circuit of IM

Where,

2

1

N

Na


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The equivalent circuit of IM is almost similar to the

transformer equivalent circuit. i.e.

- when S = 1, the exact equivalent circuit

becomes the equivalent circuit of a short circuited

transformer.

- In case S = 0, (at synchronous speed), the circuit

becomes the equivalent circuit of an open circuitedtransformer.


10/33

Analysis of the equivalent circuit of IM

The equivalent circuit shown in Fig.3.4b is not

convenient to use for the determination of the

performance of the induction machine .

As a result, several simplified versions have been

proposed.

Some of the commonly used versions of the equivalent

circuit are discussed here.


11/33

Approximate Equivalent Circuit

If the voltage drop across R1 and X1 is small and the

terminal voltage V1 does not appreciably differ from the

induced voltage E1, the magnetizing branch (i.e. Rc andXm), can be moved to the machine terminals as shown

in the fig. 3.6 below. (this is done in transformer equivalent circuit

parameters determination)

V1

X1R1 X2

S

R'

2Rc

Xm

I2

Pag

I1

Io

fig. 4.6 approx. equiv. circuit


12/33

( f )2


13/33

Therefore, if a machine operates from a constant

voltage and constant-frequency source, the sum of

core losses and friction and windage losses remains

essentially constant at all operating speeds. These losses can thus be lumped together and termed

the constant rotational losses of the inductionmachine.

If the core loss is lumped with the windage andfrication loss, Rc can be removed from the equivalent

circuit, as shown in the fig. below.

V1

X1R1 X2

S

R'

2Xm

I2

Pag

I1

Io

Fig. 4.7


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IEEE RECOMMENDED EQUIVALENT CIRCUITIEEE RECOMMENDED EQUIVALENT CIRCUIT

In the induction machine , because of its air gap, the exciting current Ie

is high of the order of 30 to 50 percent of the full-load current.

The leakage Reactance X1 is also high .The IEEE recommends that, in such situation, the magnetizing reactance Xm

can not be moved to the machine terminals (as is done in transformers.), but

be retained at its appropriate place , as shown in the fig.3.8 below.

The resistance RC is however, omitted, and the core loss is lumped

with the windage and friction losses.

This equivalent circuit is to be preferred for situation in whichthe induced voltage E1 differs appreciably from the terminal voltage V1.

Fig. 4.8 a. Equivalent circuit of IM without core loss.

V1

X1

R1

Xm

Pag

I1

Io

'

2I

'

2X

S

R '2Zf

a

b


15/33

Steady state performance parameters of IM, such as current,

speed, torque, losses, etc. can be computed from the circuit of

fig. 3.8.

In this fig., Zfis the per phase impedance offered by the rotatingair-gap field.

Note that Zfincludes mXjXs

r,2

2

m

m

fff

XXjs

r

jXjXsr

jXRZ

22

22

V1

X1

R1

Xm

Pag

I1

Io

'

2I

'

2X

S

R'

2Zf

a

Fig 4.8b


16/33

Impedance Z1 as seen by stator voltage V1 is:-

And

The total power lost in

The power lost in

therefore,

fZjXrZ 111

1

1

1

Z

VI

:,&2

2

isjXjXs

r

m

s

rIRIZ ff

22

2

2

1

fg RIs

rIP2

1

22

2


17/33

Determining the Equivalent circuit model parameters of

IM. The response of an IM to changes in load can be

determined from the equivalent circuit of the motor. The parameters R1,X1, R2, X2, Xm can be determined

by performing a series of tests on the IM. i.e.

- No-load test

- DC test for stator resistance- Blocked rotor test.

These tests need precision because the resistances

vary with temperature and the rotor resistance also

varies with rotor frequency.


18/33

The No-load test

The induction motor is made

to run at no load at rated

voltage and frequency.

Per phase values of applied

stator voltage Vnl

, input

current Inl and input power

Pnl are recorded ( by two

wattmeter method).

W1A1

A2

A3W2

3-Phase

IM

A

C

B

V

Fig. 4.9 Circuit diagram for

no-load and blocked-rotor

Test

V1

X1

R1

Xm

Pag

I1

Io

'

2I

'

2X

S

R'

2Fig. 4.10 IEEE RECOMMENDEDFig. 4.10 IEEE RECOMMENDED

EQUIVALENT CIRCUITEQUIVALENT CIRCUIT


19/33

Since slip Snl at no-load is very small, r2/s at no-load

is very large as compared to Xm. Thus, the total impedance of parallel branches

consisting of jXm and (r2/s + jX2) is almost equal tojXm.

Then, the no-load reactance Xnl seen from the stator

terminals is:-

jXnl = jX1 + jXm = X1 stator selfreactance.

VNL

X1

R1

Xm

INL

Fig. 4.11 Equivalent circuit under no-load test.


20/33

From no-load test readings we can get;

nl

nl

nl I

V

Z 2

nl

nl

nl

I

PR

22

nlnlnl

RZX , ,

During no-load condition, as stated earlier, the rotational

losses ( friction, windage and core losses are usually

assumed constant and can be obtained from the relation,

12rIPmP nlnlrot

Rotational losses or fixed losses are:- Stator and

rotorCore losses and friction and windage losses.

mnlXXX

1

If X1 is known, Xm can be

determined.


21/33

Thus,

The no-load test measures the rotational losses and

gives information about its magnetisation current.

i.e., no-load test gives Xnl and Prot Note that, the only load during no-load test is friction

and windage loss.

The mechanical power(Pm = Pg - Prcul) isconsumed by mechanical losses.


22/33

P

M

eout TP

Fig. 4.12 Power flow diagram of IM.

StatorI2

Rlo

ss

Stato

rcorelo

ss

Roto

rI2

Rloss

Rotorcore

loss

Frictionandwindageloss

Pg

Rotor input

powerShaft power

Mech. powerdeveloped

Input power

(3IVcos)


23/33

DC test for stator resistanceThe rotor resistance r2 plays an extremely critical role

in the operation of an IM. i.e.- It determines the shape of the speed torque curve.

- It determines the speed at which the pull-out torque occurs.

To determine r2, it is necessary to find r1 of the stator.- Thus, a DC voltage is applied to the stator windings of an IM.

- The current in the stator is adjusted to the rated value & the voltage is

measured.

If the supply dc voltage is connected between the two

phases, ( the 3rd phase being open) then, the dc

current flows through the two phase windings and therequired per phase stator resistance is:-

,2 1dc

dc

I

Vr

dc

dc

I

Vr

21


24/33

DC

w

A

V

+

_

R

S

T

Fig 4.13 Circuit for Dc test to determine stator

resistance


25/33

Blocked rotor test

During this test;

- An Ac voltage is applied to the stator; and the current

is adjusted to the rated value.- The rotor is blocked so that it can not rotate.

- The current, voltage and power are measured.

Since the rotor is kept stationary, the slip = 1,and the rotor

resistance r2/s is just equal to r2 which is quite a small value.

V1

X1

R1

Xm

Pag

I1

Io

'

2I

'

2X

S

R'

2

Fig. 4.14 IEEE RECOMMENDED EQUIVALENT CIRCUITFig. 4.14 IEEE RECOMMENDED EQUIVALENT CIRCUIT


26/33

Compared with Xm, r2 & X2 are small, so that the whole input

current flows through them. The equivalent circuit under this

condition looks a series combination of r1, X1, X2, r2.

cos3 LLin IVP

Ibr

Fig.4.15 equivalent circuit during blocked rptor test

LL

in

IV

P

3cos

arccos


27/33

Impedance of the IM during block rotor test is:-

sincos brbrbrbrbr ZZjXRZ

21 rrRbr

12 rRr br

21 XXXbr

;3 L

L

br I

VZ

Where, Xbris the reactance of stator & rotor

In practice, it is difficult to get separately stator and

rotor reactance, and are usually taken from

experimental data from tables.

22

2

brbrbr

br

br

br

RZX

I

PR

br

br

br I

V

Z


28/33

However, for wound rotor machines x1 is assumed to be equal to x2.i.e. x1 = x2= Xbr

For squirrel cage induction machines, total leakage reactance Xbr

(=x1+ x2) can be distributed between stator and rotor as per thefollowing table:

Empirical distribution of leakage reactance Xbr

0.5

0.6

0.70.5

0.5

0.4

0.30.5

1.Class A (normal Tst , normal Ist and low

slip)

2.Class B (normal Tst, low Ist and low slip)3.Class C (high Tst , low Ist and low slip)

4.Class D (high Tst , low Ist and high slip)

X2X1

Fraction of XbrClass of motor

For design classes, please refer Stephen J. Chapman, 2nd ed.


29/33

0.5

0.6

0.7

0.5

0.5

0.4

0.3

0.5

1.Class A (normal Tst , normal Istand low slip)

2.Class B (normal Tst, low Ist and

low slip)

3.Class C (high Tst , low Ist and lowslip)

4.Class D (high Tst , low Ist and high

slip)

X2X1

Fraction of XbrClass of motor


30/33

Thus, the block rotor testhelps to determineequivalent circuit parameters r2, Xbr(X1+X2).

If X1 & X2 are taken from tables, then Xm can bedetermined. i. e.; Xnl = X1 + Xm

In general, by performing these three tests and usingexperimental values of X1 and X2 from tables, we can

determine the equivalent circuit parameters of IM.

S i f f i i d i d l


31/33

Separation of friction and windage loss

from the no-load test result The power input to the induction motor at no-load has to

supply the stator copper loss, core loss and

friction and windage loss.

The dc resistance of the stator winding is measured and its perphase effective value r1 for AC is calculated from the relation:-

r1 = (1.1 - 1.3) (dc resistance per phase)

For computing the friction and windage loss, the appliedvoltage to the unloaded induction motor is varied from 20% toabout 1.25 of the rated voltage.

The input power, current and voltage are recorded so that agraph can be plotted.

The speed with reduced voltage, will fall only slightly so thatthe friction and windage loss remains substantially constant.


32/33

From the input-power readings, the corresponding stator ohmic

loss is subtracted to obtain the core loss and friction and

windage loss ( rotational loss); i.e.

Where, Pnl - is the per phase power input,

Inl - is the per phase stator current and

r1 - is the effective per phase stator resistance.

The plot of the rotational loss Prot with variable stator voltage is

shown in fig below.

PC

Pf&W

Prot

VVrated

)rIP(mP 12nlnlrot

Fig. 4.16

Core loss at

rated voltage

Fric. &

Windage loss

Prot VS V


33/33

The intercept of the extraplotted Prot curve with the ordinategives the friction and windage loss, because the core loss iszero for zero applied voltage.

In order to get a motor accurate value of mechanical loss

(friction and windage loss), rotational loss Prot should beplotted against (Voltage)2.

This plot of Prot with (voltage)2 is almost linear and, therefore,the extrapolation is easier.

P

rot

V2Vrat.

Friction &windage loss

(Pf&w)

Core loss at

rated voltage (Pc)

Prot Vs V2

Fig. 4.17

4. Equivalent Circuit of IM

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