Solution Electrical Numerical 5

Numerical 5: The total input power to a 500 V, 50Hz, 6-Pole phase induction motor running at 975 rpm is 40kW. The stator losses are 1 kW. Friction and windage losses are 2 kW total. Calculate (i) Slip, (ii) Rotor Cu Loss, (iii) Efficiency.

Solution:(i) Synchronous Speed: $\displaystyle \mathrm{N_s = \frac{120f}{P} =\ \frac{120\times 50}{6}=\ 1000\ RPM}$ 

Slip; $\displaystyle \mathrm{s = \frac{N_s -N}{N_s} =\ \frac{1000-975}{1000}=\ 0.025 }$

(ii) Input power = 40 kW.

Stator loss  = 1 kW.

Rotor input = 40 - 1 = 39 kW.

Rotor cu-loss = slip x rotor input = 0.025 x 39 = 0.975 kW.

(iii)  Generated power in rotor = Rotor input - Rotor cu-loss =39 - 0.975 = 38.025 kW.

Output power from rotor = Generated power in rotor - Friction and windage losses = 38.025 - 2 = 36.025 kW

Efficiency:  

$\displaystyle \mathrm{\eta = \frac{Output\ power\ from\ rotor}{total\ input\ power }\times 100}$ 

$\displaystyle \mathrm{\eta =\ \frac{36.025}{40}\times100}$ = 90%