dumb question on compressor displacement cfm

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.

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

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).

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.

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.

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

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

Real compressors operate somewhere between isentropic and adiabatic. Smaller single-stage units like this are closer to adiabatic than isentropic.

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 someth> However, is your rating in CFM at some pressure?

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.

Don't worry about it. I've seen ratings on machines at the stores that have been done by high dollar engineers, and they're farther off than yours.

Steve

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

astonishing

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.

Swept volume (not compressed volume) would be 1/2 half of Grant's calculation. Assuming a single acting piston there would be only 1/2 active stroke per cyclinder per revolution or 900 total strokes per minute.

Don't worry about it. I've seen ratings on machines at the stores that have been done by high dollar engineers, and they're farther off than yours.

Steve

Assuming a single

revolution or 900 total

How do you figure that? I think you are confusing a compressor with a 4 cycle engine.

How do you make a crankshaft that only moves the piston up every other rev?

Or is it that they both go up each rev but the valves only close on one or the other?

Actually, a factor of 2 is probably about close. Back in the mid-90s I was told that this compressor is capable of about 18 CFM IIRC. And half of 45 would be

Grant

Robert Sw> Swept volume (not compressed volume) would be 1/2 half of Grant's calculation.

Assuming a single

revolution or 900 total

I believe he is confusing a compressor with a 4 cycle engine, where the four cycle engine requires two crankshaft revolutions to complete the intake-compress-power-exhaust cycle, unlike a compressor which requires only one revolution for the intake-output cycle.

astonishing

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.

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.