Question #203136

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.


Expert's answer

No-load test

cosϕo=1200320822=0.151    ϕo=cos10.151=81.290cos \phi _o= \frac{1200}{\sqrt{3}*208*22}=0.151 \implies \phi _o =cos^{-1}0.151 =81.29^0

Iw=Ioccosϕo=220.151=3.33AI_w=I_{oc}cos \phi_o=22*0.151=3.33A

Iμ=Iocsinϕo=22sin81.29=21.746AI_{\mu}=I_{oc}sin \phi_o=22*sin 81.29=21.746 A

Rc=Voc(ph)Iw=20833.33=36.053ΩR_c=\frac{V_{oc}(ph)}{I_w}=\frac{208}{\sqrt{3}*3.33}=36.053 \Omega

Xm=VocphIμ=208321.746=5.22ΩX_m=\frac{V_{oc}ph}{I_{\mu}}=\frac{208}{\sqrt{3}*21.746}=5.22 \Omega

Locked rotor test

Zeq=VscphIsc=24.6364.5=0.22ΩZ_{eq}=\frac{V_{sc}ph}{I_{sc}}=\frac{24.6}{\sqrt{3}*64.5}=0.22 \Omega

Req=2200364.52=0.176ΩR_{eq}=\frac{2200}{3*64.5^2}=0.176 \Omega

Xeq=Zeq2Req2=0.2220.1762=0.131ΩX_{eq}=\sqrt{Z_{eq}^2-R_{eq}^2}=\sqrt{0.22^2-0.176^2}=0.131 \Omega

X1=X2=0.1312=0.0655ΩX_1=X_2'= \frac{0.131}{2}=0.0655 \Omega

Tsh=602πNsShaftpoweroutput=602π144042.5361000=282.07NmT_{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 = 3vpmIscosϕs=3(230.94)(73.137)cos22.06W=46.961kW3v_{pm}I_s cos\phi_s=3(230.94)(73.137) cos 22.06 W= 46.961 kW

Efficiency=42.53646.961100=90.577%Efficiency = \frac{42.536}{46.961}*100=90.577 \%

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