AC motor2

Lecture 24 Lecture 24 -- EE743EE743

Induction MachinesTorque Speed characteristics

2

Importance of T-speed characteristics

Torque speed characteristic of a motor is important from

the point of view of its applications to specific situations.

To calculate the torque produced by the machine, first, we

compute the motor power. The motor power, or the

mechanical power supplied to the load is

The electrical power inputted to the machine can be

calculated from the eq. circuit given below

lossesinem PPP =

3

Importance of T-speed characteristics

4

Machine Torque

Power dissipated is

Motor power Pem is

Motor torque can be calculated from

{ }

+==

s

RRIIVP

'2

1

2

1*

111 3~~

Re3

extrotor RRR

XaX

RaR

+=

=

=

2

22'

2

22'

2

( ) ( )'212

122

1

2

1 33 RRIRaRIP +=+=

( ) 21'2'21'2

1

2

1

133 IR

s

sRR

s

RRIPem

=

+

+=

memem TP =2

1'2

13IR

s

sT

m

em

=

5

Machine Torque

Since Tm can be written as

Substituting for I1 in equation (3), we will have

( )ssyncm = 1

2

1

'23 I

s

RT

sync

em

=

( )2'212'

21

2

12

1

XXs

RR

VI

++

+

=

( )2'212'

21

2

1'23

XXs

RR

V

s

RT

sync

em

++

+

=

6

Power calculations

Figure 2 shows the plot of equation (3) for the values of slip

from zero to unity. In Fig.1 . This

corresponds to the normal range of the speed of an induction

motor from starting (m=0, s=1) to the synchronous speed

(m=sync, s=0)

( )extRRaR += 22'

2

7

smax and Tmax calculation

To find smax first set derivative of (4) with respect to s

equal to zero.

Which will results in

substituting smax in (4) will result in Tmax

0=ds

dTm

( )2'2121

'2

max

XXR

Rs

++

=

( )

+++

=2'

21211

2

max2

3

XXRR

VT

sync

8

Torque-speed characteristic

Equation (5) shows that the slip at which the maximum torque

occurs is proportional to the rotor resistance. Equation (6) shows

that the maximum torque is independent of the rotor resistance

9


The machine is operating as a motor for the range of s for

which the torque-speed curves are shown in fig.1 and fig.2.

In this range, torque is positive and sync is greater than

rotor speed m. Note that the torque is zero at slip equal to

zero. As was state before, slip is given by

Let us consider three different cases:

sync

msyncs

=

10


Case 1: sync and m are rotating in the same direction and

sync is rotating faster than m.


sync is rotating slower than m.

Case 3: sync and m are rotating in different directions.

11

Torque-speed characteristic. Case 1.


sync is rotating faster than m.

This case is the normal operation of the induction machine. Machine

operates as a motor. Note also in equation (7) sync and m are in

the same direction and the slip is positive for motor operation

12



sync is rotating slower than m.

If the speed of the machine is increased beyond its synchronous

speed by an external prime motor, but still rotated in the same

direction as the stator field, the slip will be negative (s

13


Case 3: sync and m are rotating in different directions.

Suppose an induction motor is operating under normal conditions at

the same value of positive slip in stable region (0

14

Example 1

On no-load a 3-phase delta-connected induction motor

takes 6.8A and 390W at 220V line to line. R1=0.1/phase,

friction and windage losses are 120W. Determine Xm,and

Rm of the motor equivalent circuit.

AI

I linephase 926.33

8.6

3===

LLphase VVV == 220 WRI phasephase 54.12 =

m

phase

R

VWlossesCore

2

46.8854.13

120390==

=

== 14.54746.88

2202

mR ==

110828.1

1 3

m

mR

G

15

Example 1

===

11084.17

220

926.3 3

V

IY mm

===

110746.17

1 322mm

m

m GYX

B = 35.56mX

16

Example 2

The motor of the previous example take 30A and 480W at

36V line-to-line, when the rotor is blocked. Determine the

complete equivalent circuit of the motor. Assume that

X1=X2.

( ) eqeqeq XRZXX ==

==+ 0.2533.0

3/30

36 22

22'21

( )( )

eqRRR ===+ 533.03/30

3/4802

'21

== 433.01.0 '21 RsoR

== 12

21

eqXXX

17

Example 3

An induction motor has an output of 30kW at =0.86. For

this operating condition Pcoil,1=Pcoil,2=Pcore=Prot.Determine the slip

3,

0

inP

PEff =

WEff

PPin 884,34

86.0

000,3003, ===

WPloss 884,4000,30884,343, ==

M

18

Example 3

WPPPPP rotcoreRAG 663,33'20=+++=

3,

3,'2

AG

R

P

Ps = WP

R221,1

4

884,4'2

==

%6.3036.0663,33

221,1ors ==

19

Example 4

A wound rotor six-pole 60Hz ind. motor has R2=0.8 and

runs at 1152 rpm (s=0.04) at a given load. The load torque

remains constant at all speeds. How much resistance must

be inserted in the rotor circuit to change the speed to 960

rpm (s=0.2). Neglect the motor leakage reactance, X1 and

X2.

The air gap power function is:

( )2'212'

21

21

'2

'2

2'2 33

XXs

RR

V

s

R

s

RIPg

++

+

=

20

Example 4

If (see eq. circuit ), the voltage, current,

air-gap power, and torque conditions remain the same, i.e

tconss

Rtan

'2 =

..'2 ConstTthenConst

s

R==

2.004.0

''2

'2 insertRRR +=

=== 2.38.0)8.0(504.0

2.0 '2

'2

'RRRinsert

21

Example 5

A 400V, 3-phase WYE connected motor has

, and

using the approx. eq. circuit. Determine the max.

electromagnetic power, Pd

( ) phasejZ /2.16.01 += ( ) phasejZ /3.15.0'2 +=

( )2'212'

21

1

21

1'2

XXs

RR

V

ZZ

VI

++

+

=+

= VV 2313

4001 ==

( )22

'2

5.25.0

6.0

231

+

+

=

s

I

22

Example 5

( ) ( )sfunctionIRs

sPsRIPP ggd =

===

2'2

'2

'2

2'2

11

( )CBss

ssAPd

++

=

2

2

( ) ( )( )

( )BsCBss

ssA

CBss

sA

ds

dPd +++

++

== 2

210

22

2

2155.0

max=

dPats

( )( )

WCBss

ssAP

s

phased 962,6

155.0

22

2

max,, =++

=

=

WP phased 886,20962,633max,, ==

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