Question

The armature winding of a 200 V, 4-pole series motor is lap-connected. There are 280 slots and each slot and each slot has 4 conductors. The current is 45 A and the flux per pole is 18 mWb. The field resistance is 0.3 ohm ; the armature resistance 0.5 ohm and the iron and friction losses total 800 W. The pulley diameter is 0.41 m. Find the pull in newton at the rim of the pulley.​

Answer 1

samj me nahi aaya please hindi me lilho fir pucho ok

Answer 2

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.