Answer to Question #203134 in Electrical Engineering for Anusha

Question

A 3 phase, 400 V, 50 Hz, 4 pole, star connected SCIM has the following per phase constants

in ohms referred to stator:

r1= 0.15, x1=0.45, x2=0.45, Xm= 28.5

Rotational losses= 400W.

Compute stator current, rotor speed, output torque and efficiency when the motor is operated at

rated voltage and frequency at 4% slip.

Answer 1

$I'_2=\frac{v_{pn}}{(R_s+\frac{R_r}{S})+ j(X_{is}+X_{ir})}=\frac{230.94 \angle0}{(0.15+\frac{0.12}{0.04})+ j(0.45+0.45)} =70.494 \angle -15.95^0$

$I_m=\frac{V_{pm}}{jX_m}=\frac{230.94 \angle0}{j28.5} A = 8.103 \angle -90^0$

a. State current, $I_s=I_m+I'_s=73.137 \angle -22.06^0$

b. Synchronuos speed $N_s=\frac{120f}{P}=\frac{120*50}{4}=1500 rpm$

$N_r=N_s(1-s)=1500(1-0.04)=1440 rpm$

c. Air gap power, $=3 I_2'^2* \frac{R_r}{s} \implies P_g=3(70.414)^2*(\frac{0.12}{0.04})=44.7246 kW$

Rotor power developed =(1-s)P_g =(1-0.04)(44.7246)kW=42.936 kW

Rotational loss =400 W = 0.4 kW

Shaft power = 42.936 -0.4 =42.536 kW

Output torque, $T_{sh}=\frac{60}{2 \pi N_s}* Shaft power output=\frac{60}{2 \pi *1440}*42.536*1000=282.07 Nm$

d. Total power output = 42.536 kW

Total power input = $3v_{pm}I_s cos\phi_s=3(230.94)(73.137) cos 22.06 W= 46.961 kW$

$Efficiency = \frac{42.536}{46.961}*100=90.577 \%$