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Chapter 6

Induction Motors

Part 2

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Induced Torque in an Induction Motor

• The induced torque in an induction motor is

• To find rotor current I2, the stator circuit is replaced with

its Thevenin equivalent circuit.

Per-phase equivalent circuit of an induction motor.

2

conv AG

ind

m sync sync

P P RI

sτ

ω ω ω

= = =

2 2

2

3

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(a) The Thevenin equivalent voltage of the stator circuit. (b) The Thevenin impedance.(c) The resulting simplified equivalent circuit of an induction motor

3 4

( ) ( )

( ) ( )

M

TH

1 1 M

M 1 1

TH TH TH

1 1 M

TH

2

TH 2 TH 2

TH

2 2 2

TH 2 TH 2

2

AG TH 2

ind 2 2

sync sync TH 2 TH 2

jX=

R + j(X + X )

jX (R + jX )Z = R + jX =

R + j(X + X )

=R + R /s+ j(X + X )

VI =

R + R /s + X + X

P 3V R /s= =

R + R /s + X + X

φ

τω ω

V V

VI

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2

A typical induction motor torque-speed characteristic curve

5Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

No-Load Torque

Linear region

Maximum (Pullout) Torque in an Induction Motor

6

( ) ( )

( )

( )

2

AG TH 2

ind 2 2

sync sync TH 2 TH 2

ind

2

max 22

TH TH 2

2

TH

max22

sync TH TH TH 2

P 3V R /s= =

R + R /s + X + X

d= 0

ds

Rs =

R + X + X

3V=

2w R + R + X + X

τω ω

τ

τ

• Slip at maximum torque can be varied by changing rotor resistance while the corresponding maximum torque is independent of R2

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EQ. 1

EQ. 2

EQ. 3

The effect of varying rotor resistance on the torque-speed characteristic of a wound-rotor induction motor.

7Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

Per EQ.2Per EQ.2Per EQ.2Per EQ.2 �� ������, →��� ������, →�� = 1 − �������

8

3

9 10

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11

��� =1

1� !

+1

�# + � #

=1

1�26.3 +

10.641 + �1.106

= 0.590 + j1.10 = ��� + � ��

Note:

12Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

=3 (255.2)�

2 1800 ∙2/60 [0.59 + 0.59� + (1.10 + 0.464)�

= 229 12

2

4

13Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

b) The starting torque of this motor is found by setting s =1 in

EQ. 1

345�65 =3 7��

���

849:;[ ��� + ��� + �� + �

�]

345�65 =3 (255.2�) 0.332

188.5[ 0.59 + 0.332 � + 1.106 + 0.464 �]

345�65 = 104 12

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Induction Motor Testing

• The DC Test

• The No-Load Test

• The Locked (blocked) Rotor Test

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• The DC Test: to obtain stator resistance, R1. An adjusted dc voltage is applied between two terminals of the stator

circuit such that rated armature current flows.

DC

1

DC

VR =

2I

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Induction Motor Testing

5

Induction Motor Testing

• The No-Load Test:

By measuring the no load voltage 7:=, the no load current

>:=, and the no load power ?:= we can compute.

�@A =7@A

3>@A & �@A =

?@A

3 (>@A)� & @A = (�@A)� − (�@A)�

Because the slip s is small at no load , CD4⁄ is very large, so

the rotor circuit reduces to:

� !||(�� G + � 4) ≈ � !

So finally:

:= ≅ # + !

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• The Locked-Rotor (or Blocked-Rotor) Test:

For R2, X1+X2, and XM (using the no-load test results).

Again we measure 7J= , >J= , ��� ?J= , and compute:

�JA =7JA

3>JA & �JA =

?JA

3 (>JA)� & JA = (�JA)� − (�JA)�

Since s =1, >� ≫ >! and thus J= = # + � and �J= = �# + ��.

Now using �#from the dc test � �� = �J= − �#

Next we assume # = � = LMN�G

And finally: ! = := − #

Induction Motor Testing

19

• The Locked-Rotor (or Blocked-Rotor) Test:

To find �; note: ?:= = ?4;= + ?6;= + ?O&P + ?;Q6R

=0 (s =0 at no load)

Solving for ?;Q6R = ?:= − ?O&P − ?4;= − ?6;=

= ?:= − ?OP − 3(>#)��#

Then:

�S =T(UVN)D

WXYZ[

Induction Motor Testing

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Induction Motor Testing

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Induction Motor Testing

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Induction Motor Testing

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Induction Motor Testing