Calculating Efficiency and Speed of a DC Motor at Half-Load Torque

[Numbskull]

Homework Statement

A 20kW DC shunt motor supplied with 500V, draws a full-load current of 45A and rotates at 600rpm. The no-load current is 5A. The field resistance is 220 Ohms and the armature resistance is 0.3 Ohms. Calculate the armature current, speed and efficiency of the motor when the torque falls to half of its full-load value, assuming that flux per-pole does not change. These are ALL of the given values.

Relevant Equations

V = E + IaRa, P = I^2R

First some basic figures which are very rounded as I'm interested in the approach to the problem rather than accuracy of the answer at this stage:

The field current will be 500/220 = 2.2727 Amps, and so power is a fixed loss at 1136.36 Watts
The armature current is 45 - 2.2727 = 42.72 Amps. At full load, the power loss in the armature is 547 Watts. At no-load it's negligible at a couple of watts.

Calculate the armature current, speed and efficiency of the motor when the torque falls to half of its full-load value, assuming that flux per-pole does not change.

The question is exactly as it is posed, so I'm not sure if this suggests that half-load is half of the armature power or current, or half of the total power consumed by the machine under full load (which is 22.5kW and thus 11.25kW). I've made further calculations but is there something I've overlooked?

 

[anorlunda]

half-load is half of the armature power or current, or half of the total power consumed by the machine under full load

I think you're a bit hasty in saying that. Would you predict zero power at no load?

You can learn more about DC shunt motor efficiency here:http://www.engineeringenotes.com/el...tics-performance-electrical-engineering/36558

 

[Numbskull]

Would you predict zero power at no load?

Was just checking that I'm on the right lines. So (approximate figures):

Field winding loss is fixed at 1136 Watts, where the current is 2.27 Amps.

Motor current at no-load is (5 - 2.27) = 2.73 Amps. Therefore, excluding the field winding current, we have 42.72 - 2.72 = 40 Amps in the armature, then divided by two (mid point / half-load) is 20 Amps, it being a linear relationship. With an armature current of 20 Amps, and a resistance of 0.3Ω, we have an energy loss of 120 Watts.

Now I'm a bit stuck because I'm not sure how to calculate the efficiency. I know it's the output power minus the losses divided by the input power, but if so, what is the input power at half? Is it 10kW or is it (500*45)/2 which is 11.25kW?

Thanks

 

[anorlunda]

What did that linked paper say about efficiency?

More important, if this is your homework, you must have covered the subject in the textbook or lectures. What did they say?

 

[Numbskull]

Unfortunately the notes only provide a single example which applies to a generator. No mention of half-load or motors.

 

[Numbskull]

I surprisingly got a rapid email reply from my tutor who confirmed that in this case, it was the armature current half way between no-load and full load.

 

[anorlunda]

I surprisingly got a rapid email reply from my tutor who confirmed that in this case, it was the armature current half way between no-load and full load.

OK, then you have two kinds of losses, Field losses which do not vary with load and armature losses which do vary with armature current.

 

[Numbskull]

Yes, I've got the answer with little effort to be honest. As far as I can see, the need to mention 20kW in the question appears immaterial, which is what confused me.

 

[Babadag]

However, you have and another formulas such these:

Emf=ke*Φ*rpm/60

T=kT*Φ*Ia

where Ia=armature current and T=torque

 

[Babadag]

And, of course, Pm=km.T.rpm/60 as output [on motor shaft] power.

 

[Numbskull]

And, of course, Pm=km.T.rpm/60 as output [on motor shaft] power.

Unfortunately I am not provided with a number of motor parameters so am unable to use those formulas.

 

[Babadag]

You don't have to know any parameter.Take this proportional only.

Told/Tnew=Iaold/Ianew

Eold=V-Ra*Iaold

Enew==V-Ra*Ianew

Eold/Enew=rpmold/rpmnew

Pmold/Pmnew=Told/Tnew*rpmold/rpmnew

Pinput=V*(Ia+Iex); eff=Pm/Pinput

 

[Numbskull]

Thank you Babadag. I'm on to the next part of the question now which asks what value resistor would need to be placed in the field circuit (in addition to the 220Ω) to increase the motor speed to 1000rpm with the same load torque.

I know that reducing the current in the field winding, and thus the flux, will increase the speed of the motor, I just have to figure out which equations to use to relate it all

 

[Babadag]

If the load will be the same then E remains the same

First find ke*Φnew=E*60/RPMnew

Then ke*Φnew/ke*Φold=Iexnew/Iexold [field current]