Question

A \(311 \mathrm{~V}, 60 \mathrm{~Hz}\), six pole, Y-connected, \(25 \mathrm{hp}\), design class B induction motor was tested in the laboratory and the following results were obtained.

- Idle; \(311 \mathrm{~V}, 22 \mathrm{~A}, 1200 \mathrm{~W}, 60 \mathrm{~Hz}\)

- Locked rotor; \(28.6 \mathrm{~V}, 64.5 \mathrm{~A}, 2200 \mathrm{~W}, 60 \mathrm{~Hz}\)

- DA experiment; \(14.5 \mathrm{~V}, 64.5 \mathrm{~A}\) Draw the equivalent circuit of this motor and calculate the rotor resistance and stator reactance values.

Answer 1

Step 1 Given, = Line voltage VL = 311 V Frequency (f) = 60 Hz 311 = 179.55 V Vph V3 From DC analysis, V = 14.5 V I = 64.5 A 14.5 2R1 64.5 R1 = 0.1122

Step 2 From Idle or No load condition: VL Vph 311 VL = u = Vph 311 Vph 179.55 V V3 = 179.55 V No load current = 22 A 179.55 X1 + Xm- 22 A X1 + Xm = 8.161 2

Step 3 From the locked rotor test, Vph = = VL = 28. 6 V 28.6 28:19 16.51 V Iph = 64. 5 A Blocked rotor impedance |Zlr' = |Rlr + jXL. IZLR'I = = 0.256 22 16.51 64.5 Olr = cos PLR -1 V3xVurxIur 2200 Olr = cos -1 V3x28.6X64.5 Olr = cos -' [0.688] Olr = 46.484 °

= : RLR ZLR X cos OLR RiR = 0.256 x cos 46.484° RIR 0.256 X 0.688 RiR = 0.176 22 Now, RR = R1 + R2 :. R2 = 0.176 – 0.112 :. R2 0.064 22 Step 4 XLR = ZLR X sin Ole XLR 0.256 x sin 46.484° XLR 0.1856 22 = =

XLR X1 + X2 = 2 0.1856 X = X2 = 2 X = X2 = 0.092 X1 + Xm 8. 161 12 Xm (8. 161 – 0.092) 12 8. 069 12 = Xm Step 5 The equivalent circuit of the induction motor is:

R,+ jX, jX2 Xom R Ο. R] = 0. 112 Ω R2 = 0. 064 Ω X = Χ2 = 0. 092 Ω Χn = 8. 069 Ω