Question

P4-17) A 240 V separately exited dc motor has an armature resistance of Ra = 006 Ω, when it is connected to a 240 V supply, it draws 90 A from the supply and rotates at 1200 rpm. Find the torque developed by the motor at this operating condition. If the torque developed is 280 Nm for the same excitation and same supply voltage, then find the speed and armature current. a) b)

Answer 1

(a) v = 240v, I = 90A, N_1 = 1200 rpm E_1 = v - IR = 240 - (90 times 0.06) E_1 = 234.6v o/p power = E_1 times I = 21,114W Torque times speed = o/p power Torque = o/p power/angular speed = 21114/2 pi N/60 = 21114/2 pi times 1200/60 = 168.02 N - M (b) NEW TORQUE = 280 N - M T proportionalto phi I_a implies T proportionalto I_a T_1/T_2 = I_a_1/I_a_2 implies 168.02/280 = 90/I_a_2 I_a_2 = 150A Therefore E_2 = 240 - I_a_2 times 0.06 E_2 = 231V \r\n Therefore o/p power = E_2 times I_a_2 = 34650W Therefore T = P/2 pi N/60 implies T = 60p/2 pi N N = 60 p/2 pi times T = 60 times 34650/2 pi times 280 N = 1181.72 rpm