A three phase four pole induction motor is operating on an input frequency of 75 Hz and slip of 4%.

If the rotor resistance of the machine is 1 Ω. The torque developed by the motor is given by: (Assume the operating voltage is 415 V. assuming stator voltage drop = 0)

This question was previously asked in

UGC NET Paper 2: Electronic Science Nov 2020 Official Paper

Option 1 : 6889 syn.watts

Official Paper 1: Held on 24 Sep 2020 Shift 1

12478

50 Questions
100 Marks
60 Mins

__Concept__:-

Torque under running condition of induction motor is given by

\(\frac{{{T_r}}}{{phase}} = \frac{{60}}{{2\pi {N_S}}} \cdot \frac{{sE_2^2{R_2}}}{{R_2^2 + {{\left( {S{X_2}} \right)}^2}}}\)

Where R_{2} = Rotor Resistance

X_{2} = Rotor reactance

E_{2} = Rotor induced emf

S = slip

N_{s }= synchronous speed (In rpm)

\({N_s} = \frac{{120f}}{p}\)

Where f = frequency

P = no. of poles

Solution:- given E_{2} = 415 V, S = 4%=0.04, R_{2} = 1 Ω, X_{2} = 0, f = 75 hz, p = 4

\({N_s} = \frac{{120f}}{p} = \frac{{120\; \times\; 75}}{4} = 2250\;rpm\)

\(\frac{{{T_r}}}{{phase}} = \frac{{60}}{{2\pi {N_S}}} \cdot \frac{{sE_2^2{R_2}}}{{R_2^2 + {{\left( {S{X_2}} \right)}^2}}}\)

\(\frac{{{T_r}}}{{phase}} = \frac{{60}}{{2\pi \; \times \;2250}}\left\{ {\frac{{.04\; \times\; {{\left( {415} \right)}^2}\; \times\; 1}}{{{1^2}}}} \right\}\)

\(\frac{{{T_r}}}{{phase}} = 29.238\) N-m

To convert N-m into watt we use following formula

Power in watt = 0.105 * T _{N-m} * N (rpm)

Power in watt = 0.05 × 29.238 × 2250

≈ 6900 W