Answer to Question #203136 in Electrical Engineering for Renoir

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.


1
Expert's answer
2021-06-07T05:46:27-0400

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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