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GATE EE 2021 Official Paper

CT 1: Ratio and Proportion

2672

10 Questions
16 Marks
30 Mins

__Concept:__

**Power Stages in an Induction Motor:**

Stator iron loss (eddy current loss and hysteresis losses) considered as constant loss and it depends on the supply frequency and magnetic flux density in the iron core.

The iron loss of the rotor is negligible because the frequency of rotor currents under normal running conditions is always small.

\( {{P}_{2}}:{{P}_{cu}}:{{P}_{m}}=1:s:\left( 1-s \right)\)

Where,

s is the slip of motor and P_{cu} is rotor copper loss.

**Calculation:**

Given,

Stator input (P_{1}) = 40 kW

Stator loss = 1 kW

Rotor input (P_{2}) = 40 kW - 1 kW = 39 kW

f = 50 Hz

P = 6

\({N_s} = \frac{{120f}}{P} = \frac{{120 × 50}}{6} = 1000\;RPM\)

N_{r} = 975 RPM

\(s = \frac{{{N_s} - {N_r}}}{{{N_s}}} = \frac{{1000 - 975}}{{1000}} = 0.025\)

∴ P_{m} = (1 - s) × P_{2} = (1 - 0.025) × 39 = 38.025 kW

Total friction and windage losses = 2.025 kW

From power flow diagram,

Rotor output (P_{o}) = (P_{m}) - (windage friction loss)

P_{o} = 38.025 kW - 2.025 kW = 36 kW

\(\eta = \frac{{{P_o}}}{{{P_1}}} = \frac{{36}}{{40}} = 0.9\)

**Hence, the efficiency of the motor is 90%.**