Question

The armature winding of a 200V, 4 pole, and series motor is lap-connected. There are 280 slots and each slot has 4 conductors. The current is 45A and flux per pole 48 mWb. The field and armature resistance is 0.32 and .052. The iron and friction loss is 800W. calculate

I. Back EMF

II. Total output of motor​

Answer 1

Answer:Given:

V = 200V, P = 4, A = P, 280 slots 4 conductors / slot, series motor, I = 45 A , Ф = 18mWb, Rs = 0.3 Ω, Ra = 0.5 Ω.

To find:

The pull in newton at the rim of the pulley.

Solution:

We know that, Eb = V - I (Ra + Rs)

                                    = 200-45(0.3 + 0.5)

                                    = 164V

Also, Eb = ФPNZ / 60A                        (Z = 280 x 4)

Therefore, 164 = (18 x 10⁻³ x 4 x N x 1120) / (60 x 4)

                            N = 488.0952 rpm

Now, Ta = (Eb x Ia) / (2πn / 60)

Substituting the values, we get, Ta = 144.3853 Nm     (Torque developed)

Also, T(lost) = 800 / (2πn / 60) = 15.6515 Nm

The net torque exerted, Tsh = 144.3853 Nm - 15.6515 Nm  =  128.7337

Now, Tsh = Net pull x Radius

      128.7337  = net pull x 0.41/2

        Net pull = 627.969N

Thus, the net pull in newton at the rim of the pulley is 627.969N.

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