196.MET September 2022 Q.7(b)

  A 230 V dc shunt motor runs at 1000 rpm and takes 5A. The armature resistance of the motor is 0.025 Ω and shunt field resistance is 230 Ω. Calculate the drop in speed when the motor is loaded and takes the line current of 41 A. Neglect armature reaction.

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Given

V= 230 

N1 = 1000 

Im = 5A

Ra = 0.025

Rf = 230

Il = 41 A

 

To find 

Drop in speed

Solution

Field current, If = V / Rf 

                          = 230 / 230

                          = 1 A

Armature current, Ia1 =  Im - If

                                   = 5 - 1

                                   = 4 A

 

When motor is loaded Armature current ,  Ia2 = Il - If

                                                                          = 41 - 1

                                                                          = 40 A

We know that Back emf Eb = V - Ia x Ra

So Eb1 = 230 - (4 x 0.025)

             = 229.9 V

Eb2 = 230 - (40 x 0.025) 

        = 229 V

We know that (Eb2 / Eb1) = (N2 / N1) x (φ2 / φ1)

Assuming that φ1 = φ2

 So (Eb2 / Eb1) = (N2 / N1)

N2 = (Eb2 x N1) / Eb1

      = (229 x 1000) / 229.9

      = 996.08 rpm

Drop in speed = N1 - N2

                        = 1000 - 996.08

                        = 3.92 rpm