How to calculate large DC motor efficiency?

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Discussion Overview

The discussion centers on methods for calculating the efficiency of large DC motors, exploring both experimental and analytical approaches. Participants delve into specific techniques and tests applicable to this context.

Discussion Character

  • Technical explanation
  • Experimental/applied
  • Debate/contested

Main Points Raised

  • Some participants suggest that efficiency can be calculated both experimentally and analytically.
  • One participant outlines a method for estimating efficiency at full power rating, which involves recording motor nameplate values, calculating electrical power for the shunt field and armature, and dividing mechanical output power by total electrical power.
  • Another participant reiterates the same estimation method, providing a detailed example with specific values for a 400 HP motor.
  • Questions arise regarding the applicability of Swinburne’s Test and Hopkinson’s Test for calculating large machine efficiency, with one participant affirming that both tests can be used, noting that Hopkinson’s Test is more accurate than Swinburne’s Test.

Areas of Agreement / Disagreement

Participants generally agree on the two primary methods for calculating efficiency but express differing views on the accuracy and applicability of specific tests like Swinburne’s and Hopkinson’s Tests.

Contextual Notes

Participants mention the need for specific equipment and conditions for accurate measurements, such as the requirement for identical DC machines in Hopkinson’s Test and the limitations of no-load conditions in Swinburne’s Test.

Ana Mido
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What are the two methods for calculating large DC motor efficiency ?
 
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Experimentally and analytically.
 
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billy_joule said:
Experimentally and analytically.
In detail, please.
 
For a reasonable estimate at full power rating:
  • Record motor nameplate values.
  • Calculate electrical power required by the shunt field, and armature power. Add them together.
  • Divide mechanical output (shaft) power into the electrical power required to generate it.
Example: 400 HP. 1750 RPM. 500V/633A armature. 300V/4.84A shunt field.
316.5 kW + 1452 watts = 317.952 kW
400 HP = 298.28 kW
298.28/317.95 = 93.8%

Experimentally, design and build a suitably sized dynamometer (or haul the motor to a rewind shop large enough to have the requisite equipment), and couple it to the motor. Measure output shaft power while simultaneously measuring and recording armature and field voltage and current. Do the math.
 
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Asymptotic said:
For a reasonable estimate at full power rating:
  • Record motor nameplate values.
  • Calculate electrical power required by the shunt field, and armature power. Add them together.
  • Divide mechanical output (shaft) power into the electrical power required to generate it.
Example: 400 HP. 1750 RPM. 500V/633A armature. 300V/4.84A shunt field.
316.5 kW + 1452 watts = 317.952 kW
400 HP = 298.28 kW
298.28/317.95 = 93.8%

Experimentally, design and build a suitably sized dynamometer (or haul the motor to a rewind shop large enough to have the requisite equipment), and couple it to the motor. Measure output shaft power while simultaneously measuring and recording armature and field voltage and current. Do the math.
What about Swinburne’s Test and Hopkinson’s Test ?. Can they be used for calculating large Machine efficiency?
 
Ana Mido said:
What about Swinburne’s Test and Hopkinson’s Test ?. Can they be used for calculating large Machine efficiency?

Sure.
To do a Hopkinson you need two identical DC machines, but it's more accurate than a Swinburne (which is run no-load, and doesn't account for several significant effects).
 
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