Question

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)

1. 6889 syn.watts
2. 7898 syn.watts
3. 8998 syn.watts
4. 9998 syn.watts

Answer 1

Correct Answer - Option 1 : 6889 syn.watts

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 R₂ = Rotor Resistance

X₂ = Rotor reactance

E₂ = 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 E₂ = 415 V, S = 4%=0.04, R₂ = 1 Ω, X₂ = 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