# DC Motors - Calculating Efficiency

Discussion in 'Homework Help' started by divinesage, Nov 5, 2010.

1. ### divinesage Thread Starter New Member

Sep 13, 2010
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A 10 hp, 200V DC shunt motor has an armature resistance of 1 ohm and shunt field resistance of 100 ohms. At no-load, the motor draws 3A from the 200V supply and and runs at 1500 r/min. Calculate the rotational losses under this operating condition. Ans: 199W

I've calculated power loss due to the shunt field and armature field, this works out to 0.09W lost in the shunt field and 9W lost in the armature field.

Power loss due to rotation = Power input - Power output - Power loss due to shunt and armature

I can find power input easily. But I have no idea on how to find the power output because torque is not given. Or am I applying the wrong concept here?

Feb 5, 2010
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3. ### divinesage Thread Starter New Member

Sep 13, 2010
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I do know how to convert the units to torque, that's fine. But there's absolutely no way I am able to do it for this question (as far as I know). So I'm guessing there must be some other way to obtain the info.

4. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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The Shunt field resistance is 100Ω.

Ishunt=200/100=2A

So 1A (3A-2A) flows in the armature. I_armature=1A.

The no-load rotational losses = Total armature power - Armature resistance power loss.

The Total armature power = V_DC_source*I_armature=200W.

Can you do the rest?

5. ### divinesage Thread Starter New Member

Sep 13, 2010
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0
Thank you very much for your help! I've gotten it now. I didn't realise you had to treat rotational inertia (moment of inertia) as a resistive force, I was getting a very small value for I_armature because I assumed the motor would have zero resistance.

6. ### t_n_k AAC Fanatic!

Mar 6, 2009
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I don't think it's true that rotational inertia is a source of losses (is that what you mean by "resistive force"?) in a machine.

The no-load losses are either electrical (+ magnetic) in nature or relate to mechanical losses such as windage and other frictional (e.g. bearing) losses. I would suggest inertia is used in the context of energy storage at constant speed or stored energy change during deceleration or acceleration. A motor running at constant speed, no-load condition will have stored energy which is maintained at a constant value by the electrical supply delivering enough energy to just overcome the aforementioned electrical and mechanical losses.

Certainly energy is required to overcome rotational inertia to take the machine from one stored energy state to another (e.g from a lower to a higher speed) but if the new state is constant, then the energy input required for the change is transient in nature. In contrast, losses persist even at steady state conditions.