Question

The power input to a 415v, 50 hz, 6 pole, 3-phase induction motor running at 975 rpm is 40 kw. The stator losses are 1kw and friction and windage losses total 2 kw. The efficiency of the motor is

Answer 1

Answer:

Explanation:power i/p = 40 kw

Slip = 0.025

Rotor i/p = 40 – 1 = 39 kw

GMPO = (1 - S) 39 = 38.025 kw

Shaft o/p = 38.025 – 2

= 36.025 kw

∴ η = 36.025/40 = 90%

Answer 2

Answer:  η= 90%

Explanation:

• Given, power input($P_{1}$) = 40 kw, f=50 hz, P=6, $N_{r}=975 \ rpm$

       Rotor input ($P_{2}$)= (40 – 1) = 39 kw

• $N_{s}=\frac{120f}{P}$

             =$\frac{120*50}{6}$

            =1000 rpm

         S=$\frac{N_{s}-N_{r} }{N_{s} }$

      ⇒S=$\frac{1000-975}{1000}$

      ∴ S= 0.025

• $P_{m}$ = (1 - S) $P_{2}$

             = (1-0.025)*39

              = 38.025 kw

• Total friction and windage losses=2.025 kW

• Shaft output($P_{o}$) = 38.025 – 2.025

                             = 36 kW

• Efficiency ( η)= $\frac{P_{o} }{P_{1} }$

                           ⇒η=$\frac{36}{40}$

                           ⇒ η=0.9

                           ∴ η= 90%

• Hence, the required efficiency is 90%.

  #SPJ2