An 8-pole, 50 Hz, three-phase induction motor is running at 705 rpm and has a rotor copper loss of 5 kW. Its rotor input is

Option 4 : 83.33 kW

CT 1: Network Theory 1

11532

10 Questions
10 Marks
10 Mins

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:

Given that, number of poles (P) = 8

Frequency (f) = 50 Hz

Rotor speed (Nr) = 705 rpm

Synchronous speed \( = \frac{{120 \times 50}}{8} = 750\;rpm\)

Slip, \(s = \frac{{750 - 705}}{{750}} = 0.06\)

Copper losses = 5 kW

**Rotor input = 5 / 0.06 = 83.33 kW**