Engineering Calculating Power Loss and Input for a Permanent Magnet DC Motor

[Name15]

Homework Statement

A permanent magnet DC motor has an armature resistance of 0.4 Ohms. When a voltage of 110 V is applied to the motor it reaches a steady-state speed of rotation of 126 rad/s and draws 39 A.
(a) Find the power loss in the armature
(b) the power input to the motor

I really don't know what I'm doing. I just used the equations I already know. What is the significance of steady state speed being reached? What do i do with the 126 rad/s?

Homework Equations

P=IV (power = current x voltage)
P=(I^2)(R) (power = current squared x resistance)

The Attempt at a Solution

(a) P = (39^2) x (0.4) = 608.4 W
(b) P = 39 x 110 = 4290 W

 

[Hesch]

What is the significance of steady state speed being reached?

The armature current will be steady, thus the self induction of the armature can be ignored.

What do i do with the 126 rad/s?

The motor constant, K_m, can be calculated.
Back EMF = 110V - 39A * 0.4Ω = 94.4V →
K_m = 94.4V / 126 rad/s = 0.749 Vs.
You are just not asked about that, but maybe the the missing questions could be:

c) Find the steady state torque at 100 rad/s
d) Find the steady state current at 100 rad/s

 

[Name15]

Hi, thanks for your help!
But why would I find Torque and current at 100 rad/s and not 126 rad/s?
Continuing from your calculations: I find Torque = (0.749 Vs)x(39A) = 29.2 Nm. So using P=Tw; Power = 29.2 x 126 rad/s = 3681 W.
Power input to motor = (39 x 110) = 4290 W. So power loss is (4290 - 3681) = 609 W

Is this correct?

 

[Hesch]

But why would I find Torque and current at 100 rad/s and not 126 rad/s?

That's just an example, how the exercise could be continued.

I find Torque = (0.749 Vs)x(39A) = 29.2 Nm. So using P=Tw; Power = 29.2 x 126 rad/s = 3681 W.
Power input to motor = (39 x 110) = 4290 W. So power loss is (4290 - 3681) = 609 W

Yes, that's correct.