Braking a Separately Excited DC Motor: Calculation and Analysis

[khedira]

Homework Statement

A separately excited DC motor with 220V, 970rpm, 100A and Ra=0.05ohms. It is braked by plugging from an initial speed of 1000rpm.

I need to calculate i)the resistance to be placed in the armature circuit to limit braking current to twice the full load value, ii)the braking torque and iii)torque when speed fallen to zero.

for i),
braking current limit = 200A
Ea(rated) = Va(rated) - [Ia(rated)*Ra]
= 220 - (100*0.05)
= 215V

Ea = Ke*wm(rated)
215 = Ke*(970*2*pi/60)
Ke = 2.1166 V/(rad/s)

For braking by plugging, Va = -220V
Va = Ea + (Ia*Ra)
-220 = (2.1166*1000rpm*2*pi/60) + [-200*(0.05+Rext)]
Rext = 2.158ohms

ii),
braking torque = Ke*Ia
= 2.1166*-200
= -423.23Nm

Are my answers correct for part i and part ii?

For part iii, i have no idea how i can approach to solve the problem. Since torque = Ke*Ia, but when speed fallen to zero, Ia = 0? If Ia = 0, then torque will be 0 too??

Please advise. Any help is very much appreciated. Thank you.

 

[LightHero]

Homework Equations Ea = Ke*wm(rated) Va = Ea + (Ia*Ra) braking torque = Ke*IaThe Attempt at a Solution For part i, braking current limit = 200AEa(rated) = Va(rated) - [Ia(rated)*Ra] = 220 - (100*0.05) = 215VEa = Ke*wm(rated)215 = Ke*(970*2*pi/60)Ke = 2.1166 V/(rad/s)For braking by plugging, Va = -220VVa = Ea + (Ia*Ra)-220 = (2.1166*1000rpm*2*pi/60) + [-200*(0.05+Rext)]Rext = 2.158ohmsii),braking torque = Ke*Ia = 2.1166*-200 = -423.23Nm