0195136047.pdf

568 ROTATING MACHINES

Pm=Pg( 1 −S)=Tωm=m 1 (I 2 ′)^2

R′ 2 ( 1 −S)

S

(13.2.6)

This much power is absorbed by a resistance ofR 2 ′( 1 −S)/S, which corresponds to the load.

For this reason, the resistance termR 2 ′/Shas been split into two terms, as in Equation (13.2.4)

and shown in the equivalent circuit of Figure 13.2.6. From Equation (13.2.6) we can see that of

the total power delivered to the rotor, the fraction 1−Sis converted to mechanical power and

the fractionSis dissipated as rotor copper loss. We can then conclude that an induction motor

operating at high slip values will be inefficient.

The total rotational losses, including the core losses, can be subtracted fromPmto obtain the

mechanical power outputPothat is available in mechanical form at the shaft for useful work,

Po=Pm−Prot=Toωm (13.2.7)

The per-unit efficiency of the induction motor is then given by

η=Po/Pi (13.2.8)

Let us now illustrate this procedure and the analysis of the equivalent circuit in the following

example.

Electric

power input Pi

Mechanical power

output Po

= m 1 V 1 I 1 cos φ 1 = m 1 E 1 I' 2 cos φ 2

= Tωs

Power transferred

across air gap Pg

= rotor input power

= m 1 (I' 2 )^2 R' 2 /S

Mechanical power

developed Pm

= Pg(1 − S)

= Tωm

Stator copper loss

= m 1 I^21 R 1

S

1 − S

= m 1 (I' 2 )^2 R' 2

Rotor copper loss

= m 1 (I' 2 )^2 R' 2

= SPg

Rotational losses

(core losses will be

included here) Prot

= Pm − Prot = Toωm

Figure 13.2.7Power flow in an induction motor.

EXAMPLE 13.2.1

The parameters of the equivalent circuit shown in Figure 13.2.6 for a three-phase, wye-connected,

220-V, 10-hp, 60-Hz, six-pole induction motor are given in ohms per phase referred to the stator:

R 1 =0.3,R 2 ′= 0 .15,Xl 1 = 0. 5 ,X′ 12 = 0 .2, andXm=15. The total friction, windage, and core

losses can be assumed to be constant at 400 W, independent of load. For a per-unit slip of 0.02,

when the motor is operated at rated voltage and frequency, calculate the stator input current, the

power factor at the stator terminals, the rotor speed, output power, output torque, and efficiency.

Solution

From the equivalent circuit of Figure 13.2.6, the total impedance per phase, as viewed from the

stator input terminals, is given by