5.14.           The per-phase parameters of the equivalent circuit, Fig. 5-8(a), for a 400-V, 60-Hz, 3-pbase, wye­ connected. 4-pole induction motor are:

/

R1  =  2R;  = 0.2 Q         X 1    = 0.5 Q       x; = 0.2 Q        X:n = 20 Q

If the total mechanical and iron losses at 1755 rpm are 800 W, compute (a) input current, (b) input power, (c) output power, (d) output torque, and (e) efficiency (all at 1755  rpm).

 

n   =   120(60)   =  1800 rpm

 

1800 - 1755         1

 

s                     4             s  =---,1-"8"0'""0",---                40

From the given circuit, the equivalent impedance per phase is

z  = (0.2  +  '0.5)          (j20)(4   + j0.2)

e                              J           4  + j(20  + 0.2)

 

= (0.2  + j0.5)  + (3.77  + j0.944)  = 4.223L20° Q

..

and the phase voltage is 400/.J!= 231 V.

 

 

(a)                                               input current = ..E.!.... = 54.65 A 4.223

 

 

(b)                      total  input power  =  [3 (400)(54.65)(cos 20°)  = 35.58 kW

 

 

8 Â

(c)            The total power crossing the airgap, P , is the power    in the three 3.77 0 resistances (see the

expression for z.above).   Thus,

P   =  3(54.65)2(3 77)  =  33.789 kW

.K

[Or, by subtraction of the stator losses, P, = 35 580 - 3(54.65)2(0.2) = 33.788 kW.] The total developed power is then                                                                                     ·

 

Pd  = (1  - s)P

8

 

=  (0.975)(33.79)   =  32.94   kW

 

and the total output power is

 

 

= Â

P   = Pd  - (800 W)       32.14  kW

0

 

 

 

(d)                             output  torque  =

 

P0       =  ........ 32140.,..,,...,...  =  174.9 N  ·m

 

.,................ Â

. com        21t(l755)/60

 

 

(e)                                                   efficiecy  =   3z.14   = 90.3% 35.58