How to calculate large DC motor efficiency?

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SUMMARY

The discussion outlines two primary methods for calculating the efficiency of large DC motors: experimental and analytical approaches. The analytical method involves recording motor nameplate values, calculating the electrical power required by the shunt field and armature, and dividing the mechanical output power by the total electrical power. An example calculation is provided, demonstrating a 93.8% efficiency for a 400 HP motor. Additionally, the discussion mentions Swinburne’s Test and Hopkinson’s Test, noting that while both can be used for efficiency calculations, the Hopkinson Test is more accurate due to its methodology involving two identical DC machines.

PREREQUISITES
  • Understanding of DC motor specifications and nameplate values
  • Knowledge of electrical power calculations for shunt fields and armatures
  • Familiarity with experimental setups, including dynamometers
  • Awareness of Swinburne’s Test and Hopkinson’s Test methodologies
NEXT STEPS
  • Research the design and construction of dynamometers for motor testing
  • Learn about Swinburne’s Test and its limitations in efficiency measurement
  • Study Hopkinson’s Test and its application in comparing DC machine efficiencies
  • Explore advanced electrical power measurement techniques for DC motors
USEFUL FOR

Electrical engineers, motor technicians, and anyone involved in the design, testing, or optimization of large DC motors will benefit from this discussion.

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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