Why does a small air motor require more CFM than the compressor's rated watts?

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Mike_In_Plano
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I'm a sparky, so forgive my ignorance regarding many things mechanical...

I'm puzzled why a small air motor that's rated to put out about 3.5 shaft watts requires 4.3 CFM @ 60PSI while the compressor on the other end is rated to consume roughly 1200 watts to maintain the flow at the stated pressure.

Is it likely that I'm reading the specs incorrectly, are the CFM values quantified differently at the two ends, or is this process so horridly inefficient?

Best Regards,

- Mike
 
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Doesn't sound quite right. Why not show us what you're looking at so we have at least some clue how to help you?
 
When small compressor manufacturers use CFM, they're talking about the volumetric displacement of their machine, so if local temperature and pressure are equal to standard conditions, the CFM is equal to SCFM. Air motor manufacturers do basically the same thing, except here they're assuming that if the inlet temperature and discharge pressure are equal to standard conditions, the CFM is equal to SCFM. Note that what is meant by "standard conditions" can vary depending on what industry you are in. For the industrial gas industry, we generally use 14.7 psia and 70 F in the US.

The air motor you quote (60 psig inlet, 4.3 CFM, 3.5 W output) gives an efficiency of only 1.3%. I don't know about air motors this small, but I'd have expected a whole lot better. Larger air motors typically run around 50% from what I've seen, though the smaller ones (about 10 times larger than what you're referencing) may be as low as 10%. These are ballpark numbers, so you'll need to check individual units for efficiency.

The air compressor you quote (60 psig discharge, 4.3 CFM, 1200 W input) has an efficiency of about 39%, which isn't very good but for a machine this small, it's not that bad.
 
Thank you for the prompt reply.

The motor in question is a Micro Motors MMR-5000. I found the performance curves at:
http://www.airoil.com/micromotor.pdf
For the app, I was targeting about 25,000 RPM at 1.1oz-in of torque, which seems to be a sweet spot - halfway to unloaded speed.

I need about 3.5 shaft watts, but I think I messed up on the computation. I simply had:

P(watts) = torque * rev/sec

I think it should have been:

P = torque * rev/sec * 2pi

That gives me 20.3 shaft watts (more than needed / wanted), but it still doesn't look very efficient...

- Mike