A 3-phase induction motor draws 50 kW from a 220 V, 50 Hz mains. The rotor emf makes 100 oscillations/minute. If the stator losses are 2 kW the rotor copper loss would be

Concept:

Power flow in the Induction motor is as shown below.

Rotor input or air gap power \({{P}_{in}}=\frac{3I_{2}^{2}{{R}_{2}}}{s}\)

Rotor copper losses \({{P}_{cu}}=s\times {{P}_{in}}=3I_{2}^{2}{{R}_{2}}\)

Gross mechanical power output \({{P}_{g}}={{P}_{in}}-{{P}_{cu}}=\frac{3I_{2}^{2}{{R}_{2}}}{s}-3I_{2}^{2}{{R}_{2}}=3I_{2}^{2}{{R}_{2}}\left( \frac{1-s}{s} \right)\)

The relation between the rotor air gap power, rotor copper losses and gross mechanical power output is,

\({{P}_{in}}:{{P}_{cu}}:{{P}_{g}}=\frac{3I_{2}^{2}{{R}_{2}}}{s}:3I_{2}^{2}{{R}_{2}}:3I_{2}^{2}{{R}_{2}}\left( \frac{1-s}{s} \right)\)

\(\Rightarrow {{P}_{in}}:{{P}_{cu}}:{{P}_{g}}=\frac{1}{s}:1:\left( \frac{1-s}{s} \right)=1:s:\left( 1-s \right)\)

Calculation:

Frequency of rotor emf (f_r) = 100 oscillations/minute

= 100/60 oscillations/second

= 1.66 Hz

Supply frequency (f) = 50 Hz

Slip (s) = f_r/f = 1.66/50 = 0.033

Motor input = 50 kW

Stator losses = 2 kW

Rotor input = stator output = 50 – 2 = 48 kW

Rotor copper losses = 0.033 × 48 = 1.6 kW