Question 9.12: 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 I²R 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}^{\prime}=700 WR_{e}^{\prime}=\frac{700}{3 \times 24 \times 24}=0.4 \Omega
At no load, I^{2} R loss in the windings
\begin{aligned}&=3 I_{0}^{2} R_{e}^{\prime}=3 \times 6^{2} \times 0.4 \\&=43.2 W\end{aligned}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, \eta =\frac{\text { Output }}{\text { Output }+\text { Losses }}=\frac{10 \times 735.5 \times 100}{10 \times 735.5+700+256.8}
= 88.5 per cent