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

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Discussion Overview

The discussion revolves around calculating the efficiency and speed of a DC motor when operating at half-load torque. Participants explore various aspects of motor performance, including armature current, power losses, and the impact of field current on motor speed. The conversation includes theoretical considerations and practical calculations related to motor efficiency and operation under different load conditions.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Homework-related
  • Debate/contested

Main Points Raised

  • One participant presents initial calculations for field current and armature current, questioning the interpretation of "half-load" in terms of power or current.
  • Another participant challenges the assumption that power would be zero at no load, prompting further clarification on the relationship between load and power consumption.
  • A participant calculates the armature current at no-load and proposes a method to find the efficiency, expressing uncertainty about the input power at half-load.
  • One participant mentions receiving clarification from a tutor that "half-load" refers to the armature current halfway between no-load and full load.
  • Discussion includes various formulas related to motor performance, including those for electromotive force (Emf) and output power.
  • Another participant notes the lack of specific motor parameters needed to apply certain formulas effectively.
  • One participant expresses a need to determine the resistor value required to increase motor speed while maintaining the same load torque, indicating a focus on practical application.

Areas of Agreement / Disagreement

Participants generally agree on the definition of half-load as the armature current halfway between no-load and full load. However, there is ongoing uncertainty regarding the calculations of efficiency and input power, with multiple interpretations presented without consensus.

Contextual Notes

Participants express limitations in their understanding based on the provided materials, noting that the example in their notes applies to generators rather than motors. There is also a lack of specific motor parameters necessary for applying certain formulas, which may affect the calculations discussed.

Who May Find This Useful

This discussion may be useful for students and practitioners interested in the performance characteristics of DC motors, particularly in understanding efficiency calculations and the impact of load conditions on motor operation.

Numbskull
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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?
 
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anorlunda said:
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
 
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?
 
Unfortunately the notes only provide a single example which applies to a generator. No mention of half-load or motors.
 
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.
 
Numbskull said:
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.
 
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.
 
However, you have and another formulas such these:

Emf=ke*Φ*rpm/60

T=kT*Φ*Ia

where Ia=armature current and T=torque
 
  • #10
And, of course, Pm=km.T.rpm/60 as output [on motor shaft] power.
 
  • #11
Babadag said:
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.
 
  • #12
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
 
  • #13
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 :confused:
 
  • #14
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]
 

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