90.MET February 2020 Q.10

The total input power to a 500 V, 50 Hz, 6 pole, 3 phase induction motor running at 975 rpm is 40 Kw. The stator losses are 1 Kw. Friction and windage losses are 2 Kw. Calculate

a)Slip

b)Rotor Cu loss

c)Efficiency

                  If you are noticing some error in problems kindly comment below.Thanks

GIVEN

V = 500

f = 50 Hz

p = 6

N = 975

P1 = 40 Kw

TO FIND

a)Slip , s
b)Rotor Cu loss
c)Efficiency

SOLUTION

We know that Ns = 120 f / p
                             = (120 x 50) / 6
                             = 1000 rpm

a) Slip, s = (Ns - N) / Ns
               = (1000 - 975) /  1000

               = 0.025
               = 2.5 %

b) Motor input P1 = 40 Kw

    Stator loss = 1 Kw
 
    Rotor input P2 = Motor input - stator loss
                             = 40 - 1
                             = 39 Kw

    Rotor Cu loss = slip x Rotor input
                           = s x P2
                           = 0.025 x 39
                           = 0.975 Kw

c) Power output, Pout = Rotor input - Rotor Cu loss - Friction & windage losses
                                    = 39 - 0.975 - 2
                                    = 36.025 Kw
 
    Mechanical efficiency = Pout / Pin
                                         = 36.025 / 40
                                         = 0.9
                                         = 90 %