Answer to Question #203136 in Electrical Engineering for Renoir

Question

A 208 V, 60 Hz, 4 poles, star connected SCIM gave the following test results

N.L. test: 450W, 208V, 1.562A, B.R. Test : 59.4W, 27V, 2.77A

The stator winding resistance between any two terminals=2.4Ω.

a. Compute the equivalent circuit parameters of the motor.

b. Determine the shaft torque and the efficiency of the motor if it is running at its rated speed of

1710rpm.

Answer 1

No-load test

$cos \phi _o= \frac{1200}{\sqrt{3}*208*22}=0.151 \implies \phi _o =cos^{-1}0.151 =81.29^0$

$I_w=I_{oc}cos \phi_o=22*0.151=3.33A$

$I_{\mu}=I_{oc}sin \phi_o=22*sin 81.29=21.746 A$

$R_c=\frac{V_{oc}(ph)}{I_w}=\frac{208}{\sqrt{3}*3.33}=36.053 \Omega$

$X_m=\frac{V_{oc}ph}{I_{\mu}}=\frac{208}{\sqrt{3}*21.746}=5.22 \Omega$

Locked rotor test

$Z_{eq}=\frac{V_{sc}ph}{I_{sc}}=\frac{24.6}{\sqrt{3}*64.5}=0.22 \Omega$

$R_{eq}=\frac{2200}{3*64.5^2}=0.176 \Omega$

$X_{eq}=\sqrt{Z_{eq}^2-R_{eq}^2}=\sqrt{0.22^2-0.176^2}=0.131 \Omega$

$X_1=X_2'= \frac{0.131}{2}=0.0655 \Omega$

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

d. Total power output = 42.536 kW

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

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