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
Ns = 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