Cfm's & Psi- Egg on My Face

[Stan Smith]

Yep, got a little egg on my face. I was trying to figure out what the cfm's would be with a given psi to see if I could use my 1.5hp emglo for an hvlp conversion gun that I'm getting from Jeff Jewitt. At first I thought simultaneous equations would give the answer. Wrong! Sometimes it's a bear getting old. Daughter popped in to visit. She's a 4th grade teacher. I asked her to check what I did and she immediately said I needed to use Boyle's Law. Blew me away that she could remember that from college over 10 years ago right off the top of her head. So here it is FWIW:

With the temp constant, the volume of a given mass of gas is inversely proportional to the pressure exerted on it.

Initial Pressure Final Volume
Final Pressure = Initial Volume

Applying that to my little 1.5hp emglo compressor:
124psi/29psi = X/3.6cfm
= 15.5cfm

Cfm's required for my new hvlp conversion gun are 6.6cfm per specs..

[Mark Rios]

That's okay. You know what they say, "When you get egg on your face... Make an omelet". Or something like that.

[JayStPeter]

I don't think that math works at all. If you go purely by specs, there's a sears 60 gal compressor that claims 10.2scfm at 90psi and 12.4 at 40.
Not even close to 90/40=12.4/10.2. Using your math it would output almost 23 scfm at 40. I don't think you want to really get into the equations to figure it out.
I'm guessing that if you don't just hold the trigger down it'll probably work. You can hear when your compressor kicks in and stop for a little bit.

Jay

[Dean Baumgartner]

I don't think that math works at all. If you go purely by specs, there's a sears 60 gal compressor that claims 10.2scfm at 90psi and 12.4 at 40.
Not even close to 90/40=12.4/10.2. Using your math it would output almost 23 scfm at 40. I don't think you want to really get into the equations to figure it out.
I'm guessing that if you don't just hold the trigger down it'll probably work. You can hear when your compressor kicks in and stop for a little bit.

Jay

The actual math for converting volume at one pressure to volume at another pressure is P1a (absolute)*V1=P2a*V2 unfortunatly that's not how most compressors work, there is slippage which is air lost past the rings. There are also thermal effects on capacity ie. the higher the pressure and the higher the speed the lower the efficiency of the system. These losses in a single stage compressor can be as high as 40%. In other words the actual air delivered could only be 60% of the actual displacement. Probably the best thing to do is use the manufacturer's numbers and hope they don't lie much.

[Steve Clardy]

Stan. When I spray on a jobsite doing a staircase, I use a Porter Cable PSH1 with a Porter Cable 6 gallon panckae compressor. I have never ran out of air. I run 15#s at the gun, and yes the compressor runs almost constant. As long as it will keep the 15#'s, thats all thats necessary. As far as the math on this, I didn't make it that far in college.

[Dale Thompson]

Stan,
Your daughter has Boyle's Law down pat. Just a thought though. Boyle's Law assumes NO temperature change. This only happens when the volume is reduced VERY slowly or never. In a REAL adiabatic process, such as with a compressor, there IS a temperature change. Fortunately, the temperatures that are used in the IDEAL GAS LAW are ABSOLUTE. In other words, absolute temperature is equal to temperature in degrees Fahrenheit plus 460 degrees. Therefore the change in pressure is probably minimal because the differential divisor is quite small. Just a thought. I kind of like physics - even at my advanced age. Bottom line?? You're fine!!

Dale T.

[Stan Smith]

Well, thank you Dale. Physics are fascinating, but not my forte' by a long shot. I actually looked up Boyle's law to make sure my daughter was correct. I realize that there are a lot of variables including temp and another one of which would include an in-line air filter, which I haven't even gotten yet. I guess my main concern is that I didn't want the compressor to be constantly on. For the amount of spraying, that I'm planning, I don't think that it will be. Jeff Jewitt said it would be fine but he didn't have the stats for my compressor. The manual only shows the 1 stat and that's it. From what I'm reading you can't be sure about what stats the manufacturers provide either. I guess that's just a quirk of the times we live in (i.e. they could at least provide ranges).

Stan

[Rob Russell]

I don't think that math works at all. If you go purely by specs, there's a sears 60 gal compressor that claims 10.2scfm at 90psi and 12.4 at 40.
Not even close to 90/40=12.4/10.2. Using your math it would output almost 23 scfm at 40. I don't think you want to really get into the equations to figure it out.

Jay

Jay,

The reason it works out that way doesn't have that much to do with Boyle's law. It's a function of the air displaced per cycle of the compressor pump and the extra HP required to achieve higher pressure.

Sort of like how many places rate their dust suckers.

If you had the compressor running, not into a tank and against pressure, but open into normal atmospheric pressure, it'll achieve a certain CFM. Call that "free air" output (like a DC running with no piping on it). Now, run the output into a tank at 40 PSI, and it'll do 12.4 CFM. That "free air" CFM figure isn't going to be that much higher than 12.4. Now, increase the pressure to 90 PSI and your output drops to 10.2 cfm. That's because the motor is working harder to overcome the increased load of hte higher pressure and can't spin the pump as fast, so your output drops.

Assume, for discussion sake, that 12.4 CFM is the most the pump head can put out at its rated RPM (i.e., that the "free air" and 40 CFM output is the same). If you put a larger motor on the compressor, you'd be able to get 90 psi output closer to 12.4 cfm. That's because the additional HP will allow the pump to spin closer to rated RPM as the additional HP overcomes the resistance of the higher output pressure.

Rob