166.MET February 2022 Q.6(b)

 An 18.65 KW,4-pole,50HZ, 3 phase induction motor has friction and windage losses of 2.5 percent of the output. The full load slip is 4% compute for full load (a) the rotor Cu loss (b) the rotor input (c) the shaft torque (d) the gross electromagnetic torque.

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Given

P = 18650 W

F = 50 Hz

Poles P = 4

Friction and windage loss = 2.5 % = 18650 X 0.025 = 466 KW

Slip s = 4 % = 0.04

To find

     a)      Rotor copper loss

     b)      Rotor input

     c)       Shaft torque

     d)      Gross electromagnetic torque

Solution

Losses = 466 KW

So total output PT = 18650 + 466 = 19116 KW

      a)      Rotor copper loss / Rotor total output = s / (1-s)

      Rotor copper loss / 19116 = 0.04 / (1 – 0.04)

                   Rotor copper loss = 796.6 W

      b)      Rotor input = Rotor total output + Rotor copper loss

                          = 19116 + 796.6

         Rotor input = 19912.6 W

      c)        Ns = 120 f / P

             = (120 x 50) / 4

             = 1500 rpm

Speed N = ( 1 – slip ) x Ns

               = (1 – 0.04) X 1500

               = 1440 rpm

Shaft torque TSH = (9.55 x P) / N

                            = (9.55 x 18650) / 1440

                            = 123.7 Nm

       d)      Gross torque TTOTAL  = (9.55 x PT) / N

                                            = (9.55 x 19116) / 1440

                                            = 126.8 Nm