Need best way to measure Eff. for brushless DC motor

• Jun 21, 2001

• #2

Rich2001

Mechanical

Mar 23, 2001

896

US

Contact a local university or college that has a EE program, They generally has dynamometers that are used in the AC/DC machinery lab. Most shcools will support local industry and not charge a fee.

If, that is not an option, connect the motor to a centif. pump, measure the motor's input parameters and temperature and measure the flowrate and pressure developed by the pump. The pump manufacturer can give you the pump curve , to get the pump's efficiency from, know all these you can calculate the motor's/controller's efficiency.

 

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• Jun 22, 2001

• #3

don01

Electrical

Dec 29, 2000

216

AU

hi rich
You say you are looking at a 12 v 60 amp motor. so say around 720 watts output (yeh minus losses etc) go to your nearest technical library or physics dept of an eduction centre. If you grab a basic book on thermodynamics and look up horsepower or that man watts you will see some sketches of different and simple first principle devices to measure the output of yur motor.
2 that spring to mind is a calorimeter (water bath)type device with agitator and temp measurment or a lifting device
(you know that 1hp picture).
Another test we used at uni was to couple it up to a generating device and measure its output etc etc
mind you the dyno would make it real easy

Hope that helps a little
regards
Don

 

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• Jun 22, 2001

• #4

jbartos

Electrical

Jan 15, 2000

6,440

Suggestion: It appears that the second reply posting may result in more accurate efficiency value since the first reply posting is dependent on accurate measuring of the motor input power which may get difficult to measure accurately on the output of electronic controller circuit (considering the voltage and current waveforms).

 

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• Jul 3, 2001

• #5

electricpete

Electrical

May 4, 2001

16,774

US

To compute efficiency you need two out of three quantities: motor input power, motor output power, or motor losses.

efficieny= output/input = (input-losses)/input = output(output+losses).

jbartos seems to be suggesting that input power is the hardest to measure accurately. That part seems like cake to me. Capture voltage and current waveforms digitally, integrate the product numerically, divide by time interval. Simple and clean. In contrast, I suspect any attempt to measure motor losses will be complicated and ugly. DO you suggest we put the motor in a calorimeter? How do we manage heat loss conducted along the power leads? How do we manage friction associated with the shaft seal which must penetrate the calorimeter boundary to deliver the output power outide of the calorimeter. Or are you going to try to calculate it? I'm interested to know more.

 

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• Jul 4, 2001

• #6

jbartos

Electrical

Jan 15, 2000

6,440

Suggestions/questions to the previous posting:
1. How much is the test hardware for "Capture voltage and current waveforms digitally, integrate the product numerically, divide by time interval. Simple and clean."?
2. On which level is tought the integration?

 

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• Jul 4, 2001

• #7

electricpete

Electrical

May 4, 2001

16,774

US

Suggestions/questions to the previous posting:
1 - You can get a two-channel digital O-scope circuit card for several hundred bucks (requires you already have a laptop or PC). Digital capture storage capabilities is required for a large variety of experiements... so chances are if you buy the equipment you'd be able to use it again... and you shouldn't have a hard time borrowing the equipment.
2 - Allow me to restate the computation: Compute instantaneous power= voltage* current at each instant of time that samples are taken. Compute average power as the average of instantaneous power at all the sampled times.

But you are right, jbartos, that there are many ways to skin the cat. And it certainly depends on what you can get your hands on and what you are comfortable using.

 

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