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# What motor to use on a large compressor

## What motor to use on a large compressor

#1
05-16-10, 03:06 PM
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Join Date: May 2010
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What motor to use on a large compressor

I need help with the HP and RPM requirements for an electric motor.

I have been given a "The United States Air Compressor Co." Model GFA-1000-4-CS (the dashes are dots on the metal plate) air compressor.The original motor was removed by a previous owner.

The pulley on the compressor is about 18" in diameter, 3 groove, and indicates it spins CCW when the pulley is between you and the compressor head. The tank is a 120 gallon, ASME rated 200PSI tank, in very good condition, no significant internal corrosion. The pressure switch is on at 140PSI and off at 175PSI. The motor control box has been removed, so I can't use it to determine the maximum amperage.

The pump has four cylinders arranged as a pair of two stages.

Any help is greatly appreciated, thanks!

#2
05-16-10, 03:14 PM
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Join Date: Mar 2006
Location: Wet side of Washington state.
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You really need the diameter and stroke of the low pressure (larger) cylinder to calculate the displacement. From this you can calculate the delivered air and you need about one horsepower per four cubic feet delivered. With that large of a compressor pulley I suspect that you will turn the compressor a maximum of 850 to 900 rpm, maybe slower.

#3
05-16-10, 04:03 PM
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Join Date: May 2010
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First and foremost thank you for answering!

Without taking it apart getting an accurate answer to those questions is tough.

It certainly isn't as large as I thought it was. The outer casing of the cylinder (after discounting the fins) is about 4" in diameter. I'd guess it would be somewhere between 3 and 3.5" in diameter for the piston. The entire length of the cylinder (not counting the head) is 6".

Assuming I can do the math and guess work right, I would place the volume per cylinder (larger ones) to be between .008 cubic feet and .013 cubic feet. With a lot of assumptions that gives me somewhere between 14 and 23cfm. Using your numbers that would put it at roughly somewhere between a 4HP and 6HP motor.

The 3-groove pulley certainly seems overkill for that sized motor, or am I missing something?

Being curious, I decided to scrounge and see what I could do with what was on hand. I drained and replaced the oil (used 30W non-detergent). Turned the compressor by hand for a few minutes and from the sound it appeared to be working. I put a small amount of oil in each intake and turned it by hand a few more minutes. I rigged a 1HP, 3450RPM motor with a 4" pulley on it and fired it up. Until it starts building pressure it runs fine, around 40psi the motor stalls (even with the relief valve open it was still building pressure fairly quickly).

I measured the mounting bracket for the motor. It complies with a NEMA 224, which of course is discontinued, so I can't cross reference motors available with that base and their HP.

Does my math seem plausible? For the smaller end I used 3" cylinder, 3" stroke, 850RPM, for the larger I used 3.5" cylinder, 4" stroke, 900RPM.

Last edited by RichInTN; 05-16-10 at 04:03 PM. Reason: Got over excited and forgot to say thank you!
#4
05-16-10, 06:10 PM
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Join Date: Mar 2006
Location: Wet side of Washington state.
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When figuring the displacement of a multi-stage compressor you only work with the volume of the low-pressure cylinder because the subsequent stages work with the air compressed by the low pressure cylinders.

I'd be surprised if the stroke is four inches but I'll do the arithmetic using several different cylinder volumes. For a 3-inch low pressure cylinder and 3-inch stroke the volume is diameter squared times 1/4 Pi (0.7854) or 9 x 0.7854 =7.07 times the length of the stroke (3) = 21.21 cubic inches displacement per stroke. Since there are two low-pressure pistons you multiply the volume of one cylinder by two and get 42.41 cubic inches per revolution.

Multiply the cylinder displacement per revolution times the rotational speed to get displacement per minute, 42.41 times 800 = 33,929.28 cubic inches per minute displacement. Divide by 1728 (number of cubic inches in a cubic foot) and get 19.64 cubic feet displacement per minute.

Doing it for a 3-inch bore and 4-inch stroke would yield 26.18 cubic feet displacement per minute at 800 rpm.

I may have overstated the speed but I think 850 rpm would be the fastest you would want to turn that machine. It is possible that 450 to 600 rpm would be more in tune with the original speed. I do know that with that much displacement a one-horsepower motor would never be able to turn it at even 450 rpm. So going just by the guesstimate of a 3-inch bore and 3-inch stroke I would say you need at least a 5 horsepower motor and a 7-1/2 horsepower wouldn't be oversizing it by much.

Note that I have stated the output as displacement and not "free air" which is what pneumatic tools require. To get the "free air" output you multiply the displacement by a "fudge factor" which for two-stage compressors is 80% at all discharge pressures. If it were a single-stage machine the fudge factor is 70% at 100 psi and drops rapidly after that as the pressure rises.

#5
05-31-10, 07:06 AM
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Fort Allen Antique Farm Equipment Association, Inc.
check this 1
go to rotigels engine link theres a 1926 U S air comp.