I just completed getting my 1953 Tecumseh 2 cyl in line compressor running and in service.

It fill a 60 gallon tank.
I timed how long it took it to get to the following pressures with no air being used--just filling the tank.
20 PSI in 40 seconds
40 Psi in 1: 40 ( one minute, 40 seconds)
60 PSI in 2:40
80 PSI in 3:50
100 PSI in 5:10
110 PSI in 6:00

Can anyone show me how to calculate the free air delivery, or the delivery at 40 PSI and,say 90 PSI?

__________________
1982 300SD " Wotan" ..On the road as of Jan 8, 2007 with Historic Tags

Go to Harbor Freight and get a little black book called Pocket Reference by Thomas J. Glover. In there are air tables and formula from which you can calculate the cfm. I know these tables because they are from my great great grandfather's company the B. F. Sturtevant Fan Co.

There is another companion book also by Glover called The Handyman in Your Pocket which is worth the effort.

I can help get you started - 60 gallons = 8 cubic feet.

No personal expertise here, but if one searches a bit on-line, one can find some cookbook formulas purported to be useful for calculating free air delivery based on the sort of measurements the OP made.

I did not get into this problem as I thought some information might be lacking. As soon as I started to break the problem down that is.

The right way to calculate air delivery perhaps is by speed of the compressor including its cylinderical volume. If two stage will make a differance as well.

I was thinking about 6-8 cubic feet a minute as a general range though at say ninety pounds pressure. That may be too optimistic though. Some comparison with simular setups by compressor manufactureres might give a better guideline.

Losses by an older head unit will have to be estimated as well. There may be little or a lot depending on it's condition.

If I wanted a pretty accurate estimate. I would take an air tool with a known rate of consumption. Run it off the comoressor tank while watching the pressure gauge. If the unit could maintain pressure for say a minute or more then use the next tool that has a higher rate of consumption. Or lesser if the first one tried was too much for it.

Actual delivery would be found well enough that way at a given pressure. Most air tools have their consmption posted somewhere at a given pressure.

How about inflating a rubber bag of known volume?

Thanks for the replies. I did do some searches, and found no useful help. It seems simple issue. I know the volume of the tank. I can give times to any pressure,or the time between any two pressures. I can even put an orifice of a known size in the discharge line with the compressor running, and time what happens then.
I can even find the ideal gas formula. I just don't know how to use that to tell what I want to know.

__________________
1982 300SD " Wotan" ..On the road as of Jan 8, 2007 with Historic Tags

To sustain a given pressure may need more information than presented was my point of view. The time to fill that tank is not all that useful I believe.

To sustain a given pressure may need more information than presented was my point of view. The time to fill that tank to a certain pressure is not all that useful I believe. Or the math to enable a result very complex.

As noted earlier, no personal expertise here, but how about thinking about it this way?

Your compressor took your 60 gallon receiver from 80 PSI to 100 PSI in 80 seconds. As noted earlier in the thread, 60 gallons is 8 cubic feet.

The air that was added to the receiver during that time occupied 8 cubic feet at 20 PSI. How much volume did that air occupy when it was at atmospheric pressure (call that 14.7 PSI) when it was "free air" at the inlet?

8 cubic feet x 20 / 14.7 = 10.9 cubic feet at atmospheric pressure (Yes, I'm pretending that Boyle's law is good enough here. Yes, I'm ignoring any change in temperature. Yes, I'm ignoring condensation of water.)

How long did it take your compressor to do this? 80 seconds is 1.33 minutes.

10.9 cubic feet at atmospheric pressure / 1.33 minutes = 8.2 cubic feet of air at atmospheric pressure per minute. (That's what I understand "free air delivery" to mean; someone please step in here and correct me if that's wrong.)

A similar 20 PSI rise from 40 to 60 PSI required only one minute. Applying the same sort of thinking there would show 10.9 CFM when working in that pressure range.

A 10 PSI rise from 100 to 110 PSI required 50 seconds. Applying the same sort of thinking would show 6.5 CFM in that pressure range.

Eskimo--THANKS--That was the information I needed--and the results seem reasonable.

__________________
1982 300SD " Wotan" ..On the road as of Jan 8, 2007 with Historic Tags

Flawed a little but helpful. I would estimate the charge time calculation to apply to the lower pressure as the useful cfm output. It would be biased slightly in your favour. Sounds to me like about a two horsepower single stage setup in decent shape.

My large v twin compressor is tired now and I will have to rebuild the head to make it really useful anymore.

I have a spare modern vertical tank compressor as well that is junk compared to the past compressors. Head runs at too great a speed and the horspower rating is deceptive. They call it a five horse I think but at best it is a three horse. I would never work it hard or it would throw in the towel.

At least the head is cast iron. That may be the only good thing.The aluminium heads seem to crack etc.