Question 4.12: No-load test and blocked-rotor test were performed on a 10 h...
No-load test and blocked-rotor test were performed on a 10 hp, four-pole, 400 V, 50 Hz, three-phase induction motor to determine its efficiency. The test data are given as follows:
no-load test: V = 400 V, I = 6 A, P = 300 W
blocked-rotor test: V = 40 V, I = 24 A, P = 700 W
Calculate the efficiency of the motor on full load.
Losses under blocked-rotor test is considered equal to I^{2} R losses in the two windings. If R_{e} is the equivalent resistance of the two windings referred to stator side, then total copper loss
3 I^{2} R^{′} _{e} = 700 W
R^{′} _{e}=\frac{700}{3 ×24 ×24}=0.4 Ω
At no load, I^{2} R loss in the windings
= 3 I_{0} ^{2} R^{′} _{e}= 3 × 6^{2}× 0.4= 43.2 W
No-load power input = 300 W
Core loss + Frictional and Windage loss = 300 – 43.2
= 256.8 W
Output = 10 hp = 10 × 735.5 = 7355 W
Efficiency, η=\frac{Output}{Output + losses}=\frac{10 × 735.5 ×100 }{10 × 735.5 + 700 + 256.8}
= 88.5 per cent