0195136047.pdf

572 ROTATING MACHINES

EXAMPLE 13.2.2

For the motor specified in Example 13.2.1, compute the following:

(a) The load componentI 2 ′of the stator current, the internal torqueT, and the internal power

Pmfor a slip of 0.02.

(b) The maximum internal torque, and the corresponding slip and speed.

(c) The internal starting torque and the corresponding stator load currentI 2 ′.

Solution

Let us first reduce the equivalent circuit of Figure 13.2.6 to its Thévenin-equivalent form shown

in Figure 13.2.8. With the aid of Equations (13.2.9) and (13.2.10), we obtain

V ̄ 1 a=V ̄ 1 jXm

R 1 +j(Xl 1 +Xm)

∼=V ̄ 1 Xm

Xl 1 +Xm

=

220

√

3

15

0. 5 + 15

0°= 122. 9  0° V

R 1 ′′+jX′′ 1 =

( 0. 3 +j 0. 5 )j 15

0. 3 +j( 0. 5 + 15 )

= 0. 281 +j 0. 489

(a) Corresponding to a slip of 0.02,R 2 ′/S= 0. 15 / 0. 02 = 7 .5. From Equation (13.2.11), we

get

I 2 ′=

122. 9

√

7. 82 + 0. 6892

=

122. 9

7. 83

= 15 .7A

The internal torqueTcan be calculated from either Equation (13.2.5) or (13.2.12),

T=

1

125. 7

3 × 15. 72 × 7. 5 =

5546

125. 7

= 44 .12 N·m

From Equation (13.2.6), the internal mechanical power is calculated as

Pm= 3 × 15. 72 × 7. 5 × 0. 98 =5435 W

which is also the same asTωm=Tωs( 1 −S)= 44. 12 × 125. 7 × 0 .98. In Example 13.2.1,

this value is calculated as 5460 W; the small discrepancy is due to the approximations.

(b) From Equation (13.2.14) it follows that

SmaxT=

0. 15

√

0. 2812 + 0. 6892

=

0. 15

0. 744

= 0. 202

or the corresponding speed atTmaxis( 1 − 0. 202 ) 1200 =958 r/min. From Equation

(13.2.15), the maximum torque can be calculated as

Tmax=

1

125. 7

0. 5 × 3 × 122. 92

0. 281 +

√

0. 2812 + 0. 6892

=

1

125. 7

0. 5 × 3 × 122. 92

0. 281 + 0. 744

= 175 .8N·m

(c) Assuming the rotor-circuit resistance to be constant, withS=1 at starting, from Equations

(13.2.11) and (13.2.12) we get