# DC Shunt Motor

Discussion in 'Homework Help' started by Jess_88, Oct 19, 2011.

1. ### Jess_88 Thread Starter Member

Apr 29, 2011
174
1
Hey guys

I keep getting a different answer to a question. Can anyone explain where I'm going wrong?

Question:
220V DC shunt motor has an armature resistance of 0.2Ω and field resistance of 110Ω. At no-load, the motor runs at 1000rmp, and it draws a line current of 7A. At full-load, the input to the motor is 11kW. Consider the air gap flux remains fixed at its value at no load; that is armature reaction

a) fined Ea, speed and speed regulation at full load

b) Developed torque at full load

c) Starting torque if the starting armature current is limited to 150% of full load current.

Ea = 210.4V, N2 = 960.7, %SR = 41%
Pd = 100.4N.m, Tst = 150.6N.m

I can't get the first part.... Can anyone help me get started?
My attempt
a)
Ra = 0.2, Rf = 110
V = 220

If = V/Rf = 220/110 = 2A
Ia = IL -If = 7A - 2A = 5A
Ea = V - Ia*Ra = 220 - 5*0.2 = 219
...wrong

thanks guys

2. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
782

With 7A draw the armature current is 7A minus the field current [2A]

Ia = 7 - 220/110=7-2=5A

So the Ra voltage drop is Ra*Ia=0.2*5=1V

The motor internal emf at 1000rpm is 220-1=219V

The emf per rpm=219/1000=0.219 V/rpm