Question

(d) A 100 V, 4.9 A, 50000 rpm motor used for an application that requires a continuous torque of 0.75 Nm at a speed of 2750 rpm. The peak torque required for acceleration is 6 Nm. The motor has speed constant, Kg = 0.183 V/rad/s, armature resistance, Ra= 1.42, and rated torque, T, = 0.86 Nm. The torque-speed characteristic is also shown as in Figure Q1. i. Determine minimum armature voltage required for this application.

Answer 1

I la ha 3 ^ Rsn Vs Eylm - 5000 rpm Vs = loov Ra = 1.90 Ir = 4.9 V Nmua Tp = 6Nm Ke = 0.183 / rad's T = 0.75Nm Nc=2750RPM 그 Vg - In Ra - Eg Vs - Igra = Eg For Eg to be minimum, I should be maximum as Vs and Ra should be constant. So; KE is also equivalent for Torque - Amb relationship KE = 0.183 Nm/A V So, Ia is maximum at peak torque. Tp =ke la 6 = Tp = 0.18 x la I a = 6 - 32.788 A I I 0.183 Ia ~32.77A Eg = Vs - In Ra 32.78*1.4 =100 - 45.9 Eg = 54.10 - Loo

Eg 354.1V → in all cases. :) So, minimum armature voltage is 54.10 Also, when Eg < Vs then it works motor Eg > Vs - then it works generator

60 N = 2750 RPM We = 2 and = 2X3.19 X 2750 = 287.833 rad's 60 E = KEX WC - 0173x287.833 Eg = 52.67 U