Question

A 220 V DC shunt motor has an armature resistance, Re-0.1 S2, a field resistance, Ry=100 12, and negligible rotational losses. Under certain load conditions, the shunt motor runs at 1600 rpm while drawing a total current, I = 100 A. Determine: a. the output power of the motor; b. the output torque in N-m; and, c. the efficiency Without making any changes to the applied voltage or the field, the motor operates with a new M mechanical load that requires a mechanical developed torque of 100 N-m. Determine: d. the total current that the motor consumes; e. the operating speed; and, f. the efficiency.

Answer 1

Given ve 220 V , Ra= 0.18. Rf = 100-2. Shunt motor speed N = 1600 rpm load Wrent I = 100 A Armature current In Load current - field current a I ap = It - If 144 1 - 4 = 100- 220 Tag = 97.8 A output power Pout = Exia - (V-Ia Ra) Ia = (220-97.880.1) x97.8 = 20560 W Pourt = 20.56 kW output torque To= Pout - b. Pout 2TN 60 20560 80 x 1600 60 Torque Ti = 122.7 N-m.

Poutput c. Efficiency n - K100 Pinput C. Eficiency n - = Postput 100 Ponte x 100 Pout X 100 - Vxτι 20560 .x100 220 X 100 n = 93.5% New developed torque aq=100 Nmi We know that for shunt motor a a la Iaz - Inix The - 99.8 x 100m 122.1 Armature current Iar = 79.7 A 79. d. Total current Ite= Iaz + I field - Iaz + ut a = 74.7 + 3200 i n It2 = 811A

e. for constant field of shunt motor Speed Na E - Na (V+ - Inha) N₂ = V - I agra 1. Vt - Idi Ra N₂ = N, x Vt - Iaz Ra Vį - Idi Ra 220- 79.7x0.1 220-97.8 X0.1 New speed Nz a 1613.1 rpm f. Efficiency n Efficiency (V+ - Iaz Ra)x Iaz - x100 VtX Itz - (220- 79.7X0.1)*79.7 X 100 220 X 81.1 ng - 94.7%