dumb question on compressor displacement cfm

One of my compressors has bore 4" stroke 3-1/2" two cylinders (single stage) and can run at 900 rpm max.
I am trying to figure the displacement CFM and I'm coming up with an astonishing
and quite unrealistic 45.8 CFM.
I think the actual number is more like 14 to 18. But I can't see what I'm doing wrong.
Help!
3.14 * 2^2 * 3.5 * 2 / 12^3 * 900 = 45.82-ish
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Grant Erwin wrote:

45.82 CFM at 0 PSIG? Like a muffin fan?
I think you need to adjust from raw displacement at 0 PSIG to the portion of the displacement that remains after the air charge has been compressed to the output pressure. After all, until the pressure in the cylinder reaches the pressure in the tank, no air will flow through the check valves into the tank.
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"Pete C." wrote:

If I have this correct, assuming 90 PSIG output pressure you need 7:1 compression to get to output pressure where you actually feed into the tank. 3.5" stroke / 7 = .5" output stroke * 2 cylinders = 1" effective output stroke * 12.56 square inches cylinder bore = 12.56 CFM @ 90 PSIG (with some rounding error and assuming minimal dead air space between the cylinder and the valves).
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"Pete C." wrote:

Bah! Forgot the conversion from CIM to CFM and RPM in the equation, but you get the idea. 6.56 CFM @ 90 PSIG I think is the result.
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"Pete C." wrote:

3.14 * 2^2 = 12.56 square inch cylinder bore
15 PSIA to 105 PSIA = 7:1
3.5" stroke / 7 = 0.5" output stroke
0.5" output stroke * 2 cyl = 1" output stroke
1" output stroke * 12.56 square inch bore = 12.56 cubic inches output per revolution
12.56 cubic inches output per revolution * 900 RPM = 11,304 cubic inches output per minute
1 cubic foot = 1728 cubic inches
11,304 cubic inches output per minute / 1728 = 6.54 cubic feet output per minute
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"Pete C." wrote:

3.14 * 2^2 = 12.56 square inch cylinder bore
15 PSIA to 105 PSIA = 7:1
3.5" stroke / 7 = 0.5" output stroke
0.5" output stroke * 2 cyl = 1" output stroke
1" output stroke * 12.56 square inch bore = 12.56 cubic inches output per revolution
12.56 cubic inches output per revolution * 900 RPM = 11,304 cubic inches output per minute
1 cubic foot = 1728 cubic inches
11,304 cubic inches output per minute / 1728 = 6.54 cubic feet output per minute
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"Pete C." wrote:

3.14 * 2^2 = 12.56 square inch cylinder bore
15 PSIA to 105 PSIA = 7:1
3.5" stroke / 7 = 0.5" output stroke
0.5" output stroke * 2 cyl = 1" output stroke
1" output stroke * 12.56 square inch bore = 12.56 cubic inches output per revolution
12.56 cubic inches output per revolution * 900 RPM = 11,304 cubic inches output per minute
1 cubic foot = 1728 cubic inches
11,304 cubic inches output per minute / 1728 = 6.54 cubic feet output per minute
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wrote:

Pete, your calcs neglect thermodynamic effects so they're only valid for isentropic comression -- no temperature change during compression. That is only approached with very slow compression, certainly not at 900 RPM.
Real compressors operate somewhere between isentropic and adiabatic. Smaller single-stage units like this are closer to adiabatic than isentropic.
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Don Foreman wrote:

It neglects a lot like dead space between the swept volume of the cylinder of the valves, temperature, etc. and has rounding error too, but it's a lot closer to the correct number than the original 45.8 CFM.
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wrote:

The only difference between your calculation and Grant's is that you figured the flow at 105 PSIA, i.e., a 7:1 compression ratio.
45.8/6.54 = 7.00
Compressors are rated in SCFM -- at a standardized pressure, temperature and humidity close to room temp and atmospheric pressure -- so in that sense, Grant's numbers are more appropriate.
What you both neglected, as Don pointed out, is the thermodynamic losses.
--
Ned Simmons

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Ned Simmons wrote:

SCFM would be one rating on a compressor, but it is not the only one. The volume delivered at a useable pressure is also normally listed.
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wrote:

Not normally. Even though a compressor may be rated 20 CFM @ 90 PSI, those cubic feet are measured at atmospheric pressure.
--
Ned Simmons

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On Tue, 23 Jun 2009 16:46:46 -0700, Grant Erwin

Your math for swept volume looks o.k. However, is your rating in CFM at some pressure? Also, you're not accounting for compression ratio. The less compression ratio, the less efficiency. The volume you swept loaded at ~14.7 psia at ambient temp. It exhausts at the tank back pressure and a higher temperature, and that volume between the piston and the reed valve at tdc re-expands, limiting the amount of air sucked in. (Based on that, I guess good design would minimize that volume by placing the valves very close to the pistons at tdc.)
No, I didn't calculate it. I'd have to dig out thermo texts and I think I chunked them. Just thinking about it makes my head hurt.
Pete Keillor
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When the crankshaft speed and piston displacement volume are calculated, that result is generally referred to as: free air displacement, meaning pressure doesn't enter into the calculation, (essentially the same as the cubic inch or liter, or CC displacement in engines).
If the pump was used to inflate a very thin, empty/collapsed plastic bag without reaching the point of stretching the bag, the results would be similar.
--
WB
.........
metalworking projects
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On Tue, 23 Jun 2009 16:46:46 -0700, Grant Erwin

Displacement is displacement, delivery at 0 psig pressure if the valves can work with near 0 psig pressure differential. Your numbers look correct.
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Grant Erwin writes:

Your displacement CFM calculation looks correct.
Does it really run 900 rpm? That would take like a 10 or 15 HP motor to deliver 90 psi. Maybe you have something like a 1725 rpm 2 HP motor and a 10:1 pulley, so you're only running at 172 rpm?
Pete Keillor writes:

No, no, no. Please don't start up this old canard. Compressor CFM is measured in FREE AIR, not compressed. When a compressor pumps one "CFM" (cubic foot per minute), that means the intake port inhales one cubic foot of "free air" (air at atmospheric pressure, which is 0 psig) every minute.
CFM is a unit of mass flow per time, not volume.
http://www.truetex.com/aircompressors.htm
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Richard J Kinch wrote:

It's run by an 18hp gas engine.
The driving pulley is 5-1/2" and the driven pulley is 19" and the engine runs between 2200 and 3000 rpm depending on how the variable speed control is set. So yes, I can drive the pump at 900 rpm, and yes, 900 rpm is the max speed that this air pump (a Quincy model 244) can run.
Grant
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Grant Erwin writes:

Then 45 cfm at 90 psi is entirely within reason; 1 hp being roughly equivalent to 3 to 4 cfm at 90 psi.
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Richard J Kinch wrote:

Ah, but there is something even you may not know yet, Mr. Kinch. One horsepower from a gas engine doesn't equal one horsepower from an electric motor (assuming this is not something sold at Home Depot, in which case the horsepower rating is largely fictional). The good people at Quincy a long time ago told me that the electric equivalent of an 18 hp gas engine is a 7.5hp electric motor.
My compressor's specs are now known:
770 rpm -> 21.4 CFM @ 100psi 900 rpm -> 25.4 CFM @ 100psi
So to get 25.4 CFM you would indeed need most of a 7.5hp electric motor.
Why is a gas engine horsepower different from an electric motor horsepower?
Oh, boy. <ducking>
Grant
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Lots of ways to describe the difference but one is that the gasoline engine generally has a throttle and is therefore economically not designed to deliver its maximum horsepower constantly. They generally have a fairly constant torque over a wide RPM range so the horsepower is pretty much proportional to the speed. Onan and probably others have built nearly identical engines rated at 1800 and at 3600 RPM, with the higher speed giving about twice the horsepower. The electric motor can deliver its rated horsepower (or slightly more) constantly but can not deliver a lot more even for a short time. The gasoline motor is commonly rated for the maximum power it can deliver, but it can not deliver that amount constantly and have a satisfactory service life. There are industrial internal combustion engines which are rated for continuous horsepower.
A useful rating system would give both peak and continous horsepower and motors are sometimes so rated (and over-rated). Actually, the horsepower to drive a compressor at a certain speed and pressure is exactly the same regardless of the type of motor. But a 5HP electric motor can deliver that 5HP constantly with good life but a 5HP gasoline motor would be running "wide-open" and would not last long.
Lots of other ways to look at it; electric motors can withstand short term torque overload without stalling better than a gasoline engine, for example. Just quite different speed, torque, and other characteristics for the two types of motors.
Don Young
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