0195136047.pdf

13.2 INDUCTION MACHINES 569

Zt=R 1 +jXl 1 +

jXm

(

R′ 2

S

+jX′l 2

)

R 2 ′

S

+j(XM+X′l 2 )

= 0. 3 +j 0. 5 +

j 15 ( 7. 5 +j 0. 2 )

7. 5 +j( 15 + 0. 2 )

=( 0. 3 +j 0. 5 )+( 5. 87 +j 3. 10 )= 6. 17 +j 3. 60

= 7. 14  30 .26°

Phase voltage= 220 /

√

3 =127 V

Stator input current= 127 / 7. 14 = 17 .79 A

Power factor=cos 30.26°= 0. 864

Synchronous speed= 120 × 60 / 6 =1200 r/min

Rotor speed=( 1 − 0. 02 ) 1200 =1176 r/min

Total input power=

√

3 × 220 × 17. 79 × 0. 864 = 5856 .8W

Stator copper loss= 3 × 17. 792 × 0. 3 = 284 .8W

Power transferred across air gapPg= 5856. 8 − 284. 8 =5572 W

Pgcan also be obtained as follows:

Pg=m 1 (I 2 ′)^2 R′ 2 /S=m 1 I 12 Rf

whereRfis the real part of the parallel combination ofjXmandR 2 ′/S+jX′l 2. Thus,

Pg= 3 × 17. 792 × 5. 87 =5573 W

Internal mechanical power developed=Pg( 1 −S)= 0. 98 × 5572 =5460 W

Total mechanical power output= 5460 − 400 =5060 W, or 5060/ 745. 7 = 6 .8hp

Total output torque=output powerωm =output power( 1 −S)ωs

Since,

ωs=

4 πf

poles

=

4 π× 60

6

= 40 π= 125 .7 mechanical rad/s

it follows that

Total output torque=

5060

0. 98 × 125. 7

= 41 .08 N·m

Efficiency=

5060

5856. 8

= 0. 864 ,or 86.4%

The efficiency may alternatively be calculated from the losses:

Total stator copper loss=284.8 W

Rotor copper loss=m 1 (I 2 ′)^2 R′ 2 =SPg= 0. 02 × 5572 = 111 .4W

Friction, windage, and core losses=400 W

Total losses= 284. 8 + 111. 4 + 400 = 796 .2W

Output=5060 W

Input= 5060 + 796. 2 = 5856 .2W

Efficiency= 1 −

losses

input

= 1 −

796. 2

5856. 2

= 1 − 0. 136 = 0 .864, or 86.4%

Torque-Speed Characteristics of 3-Phase Induction Motors

Because the torque–slip characteristic is one of the most important aspects of the induction motor,
we will now develop an expression for torque as a function of slip and other equivalent circuit
parameters. Recalling Equation (13.2.8) and the equivalent circuit of Figure 13.2.6, we will obtain