Are you sure you measured this correctly? 19 CFM needs a mighty big hose and fittings. Doesn't a jackhammer only use something like 30-40 CFM? They use HUGE hoses and fittings for those. I have a much smaller compressor (2 Hp, about 6.3 CFM at 90 PSI) and some tools like air chisel, die grinder, paint sprayer, all use air at a rate that seems very much in line with their ratings.
I doubt you could get 20 CFM from the standard 3/8" shop air hoses at 90 PSI, without tremendous pressure drop at the tool.
I was messing with my compressor the other day to figure out just how much air it actually delivers, and also to see how much air my air tools actually use. The compressor (C-H 5 hp, 2 stage, 80 gal. tank) is rated at
13.7 cfm at 90 psi and 12.5 cfm at 175 psi.
It turns out that pumping from 100 to 140 psi takes 2 min. 40 secs., which gives an average rate of about 11 cfm. Resetting the pressure switch for a range of 145 to 175 psi makes it take 3 mins., for an average rate of a little under 10 cfm.
So, that wasn't all that disappointing, though I did spend a certain amount of time wondering if the previous owner has swapped in a too-small motor pulley, among other wishful thoughts. Unfortunately, the pump was running at the correct speed, as a quick calculation and an e-mail to Campbell-Hausfeld confirmed.
The real shocker came when I measured air consumption by my drill and air shears: 18.6 cfm and 19.8 cfm, respectively, when run continuously at 90 psi. These things are rated at 4 cfm each, which either implies a duty cycle of about 20% or it's simply a huge lie. Criminy.
Anyway, this probably isn't a big surprise to most of you, but I thought it was interesting to get some actual numbers.
I measured it the same basic way I measured the pumping rate, i.e. I measured the pressure drop and time, then calculated the volume using the ideal gas law assuming isothermal conditions. For the drill, for example, it took 1 min., 14 secs. to drop the pressure from 132 to
100 psi, so:
32 psi = 2.18 atm of air pressure lost
2.18 atm x 80 gals. = 174.15 gals. of air used
174.15 gals. / 7.48 = 23.28 cubic feet of air used
23.28 cu.ft / 74 secs. = 0.315 cu. ft./sec.
0.315 cu.ft./sec. * 60 = 18.9 cu.ft./min.
The numbers make sense compared to the pumping rate, anyway. It takes the compressor 160 secs. to add 40 psi, so that's 0.25 psi/sec. It takes the tool 74 secs. to use 32 psi, so that's 0.43 psi/sec. To me, that says the tool can use air almost twice as fast as the pump can deliver it, with both running continuously. Incidentally, I was using a 25' 1/4" hose to the tool, although the manufacturer recommends 3/8".
I didn't let the temperature equilibrate after releasing pressure, so the isothermal assumption isn't strictly true, but the temperature of the tank didn't change much, so I think the numbers are still in the ballpark.
But then, I'm so unversed on the subject that I'm not certain if the units of measure of air tool ratings are cubic feet per minute at standard temperature and pressure (14.7 psi/25C) or cubic feet at the rated running pressure. That's about a
1:6 difference for 90 psi, isn't it?
Never having done this kind of test myself, I'm curious about the technique you used to measure the tools' air consumption.
I'd probably have pumped the tank up as high as safely possible, closed it off from the compressor, set the pressure regulator to 90 psi, and run the tool for a measured time, which would reduce the tank pressure enough so that you could calculate the volume of air which must have exited it during that time.
Did you have to factor in the temperature changes of the compressed air which may have occurred when filling or emptying the storage tank?
Were the tools doing any work or just free running?
But it is food for thought...
Jeff
-- Jeff Wisnia (W1BSV + Brass Rat '57 EE)
"If you can keep smiling when things go wrong, you've thought of someone to place the blame on."
It's probably also worth mentioning that these tools are Enco cheapies, not top of the line by any means. I'd be interested to know if better quality tools use less air.
I'm afraid I don't know how the ratings are applied, but I'd be interested to know also. My calculations would be for room temp/pressure.
Yes, that's exactly what I did, except I didn't close the tank off from the compressor. I just stopped the tool before the pressure reached the cut-in pressure.
To be more accurate, I should have done that or let the temperature re-equilbrate with the room before recording the pressure. I didn't bother to do either, so the amount of air actually used is probably a bit less than my calculations suggest.
Not to be a doubting Thomas, but often, "One test is worth a thousand expert opinions."
Wot happens if you let the compressor run and keep the trigger on the air tool pulled? Does the air pressure at the compressor eventually drop below the regulated pressure, or can the compressor keep up with the tool? If it can, what's the compressor on-off duty cycle like.
I've got a feeling that maybe your math gave you SCFM (Standard CFM) but I think air tools are rated at CFM at the operating (compressed) pressure.
Now that I've gotten interested in this, I hope someone with knowledge of the right units of measure can educate me/us.
"Tom Young" wrote in news: snipped-for-privacy@attbi.com:
Those manufacturer ratings are for ideal conditions, with NEW tools and NEW compressor. Run em 20 minutes each...all bets are off.
Your calculations seem appropriate for the conditions you saw, but, I did not see a correction for air density (barometric pressure) in the compressor calculations. Suggestions to conserve air and get more work out of the air it uses:
1: De-gum the air tool motor blades and oil it. (Blaster air tool oil works wonders for removing varinish and gum, and oiling the tool)
Checked the compressor for blow-by/leak-down.
Variations in the output of a compressor can come from a wide range of causes. The rated CFM is at a specific barometric pressure and air temperature. (Big difference in pumping air in Miami vs. Denver.) Perpendicularity of the piston rings to the bore, variations in the ring gap, groove widths, etc can make a huge difference in how efficient the pump is on a standard air compressor.
Ahh, that makes a difference. Running unloaded, the air consumption is maximum, but the work output is zero. Put a load on the tool, the motor slows down, air consumption is reduced, and the efficiency goes from zero to some small number. (Try a die grinder, then imagine you have a 2 HP motor at the same speed attached to the grinding wheel. You couldn't hold the thing it would leap out of your hand. In other words, air tools are cheap to make, lightweight and smaller than electrically powered tools, but nowhere as efficient as direcly powering them by electricity.)
I have a cheap, hardware store die grinder that definitely keeps my 2 Hp compressor busy, but the compressor can keep about 75 PSI in the tank, when it is working on actual grinding. And, this is often very light grinding with small Cratex wheels, and spinning quite fast. So, that is in the range of 6 - 7 CFM. I have an air chisel that works out pretty similarly, maybe just a little more air consumption. If these tools gobbled
19 CFM, my compressor would eventually allow the tank to run down to about 10 PSI, where the tool would just stop working completely. So, I can tell you that my cheap tools don't use anywhere near the air you report.
Note also that the adiabatic effects can be considerable, and are working against you in your tests. Ie, adiabatic heating raises tank pressure faster when the pump is running, and lowers tank pressure faster when the tool is emptying it. So, it makes you think your compressor puts out maybe up to
20% more than your isothermal assumptions, and that the tools use maybe 20% more air than you computed. That still doesn't account for your huge numbers.
First of all, our air tools (like everyone else's) seem to run down the compressor faster than it can recover... but I think the math you use is wrong...
Boyle's Law says that at a constant temperature the volume of a given quantity of any gas varies inversely as the pressure to which the gas is subjected. It's the famous P1*V1=P2*V2 equation.
If we assume that your 80 gallon tank at 132 PSI is our P1V1 factor (and adding 15psi to the pressure since 0 gauge pressure is 15psi absolute)
80 gallons * 147 pounds gives us a value of 11760 thingamabobs
If the tank pressure drops to 100 psi gauge pressure, from Boyles Law that will occupy a larger volume, right? so we divide our 11760 by the new pressure 115 pounds ( gauge plus atmospheric)
11760 thingamabobs / 115 pounds = 102.3 gallons
so the drop of 32 psi changed the volume of the same amount of gas from 80 gallons to 102.3 gallons... Since the tank didn't magically get any bigger, that missing (102.3-80 ) 22.3 gallons must have been what went through the tool, right?
22.3 gallons / 7.48 gallons per cubic foot = about 3 cubic feet in 74 seconds or about 2.4 cfm...
Even after doing the math, the air tools in our shop still seem to outrun the compressor...
It's a mystery, yes indeed... Reminds me of the peak half cycle RMS power amplifier ratings ...
No. The 80 gallon tank is said to have lost 32 psi of pressure. Call it
2 atm. That's 160 gallons at 1 atm, or about 21 cu-ft. This was said to take 74 seconds, so 21 cu-ft / 74 seconds * 60 seconds/minute = 17 CFM, not your 2.4.
Your error is that for CFM (or SCFM), air volume must be understood as "free air", that is, volumes of air at atmospheric pressure, even though the delivered pressure is higher and the delivered volume lower. Think of it as measuring the final exhaust, not the input, of the tool. The "S" in "SCFM" just means a standardized ambient pressure, temperature, and humidity, which is a negligible consideration in most situations.
there s/b someone here that can give us a power input to air output formula. that would be useful. or maybe just hp to air output?
746w = 1hp, that's input or output, and the compressor loses about
40% of the input power.
my napkin scribble says 15.5amp = 1hp output (at 120vac). now just what is that in cu.ft/min (at say, 90psi)? any chemistry or physics majors out there (i know there are ;-). --Loren
I have a 1.5hp Sanborn unit, old. I don't know anything for sure, but it takes ALL of a 15A 115V circuit, and it isn't particularly efficent, and it delivers about 4 cfm (into 90 psi) as far as I can tell just from running a whole bunch of different things with it. So I would guesstimate about 4 cfm per 15A @ 115V.
Ah, that's a sensible explanation. Thanks for the clarity.
I tested my drill again, only this time I chucked a 1-3/8" Forstner bit and drilled through a 2x4. The drill was close to stalling most of the time, and it still used a hefty 15 cfm. Maybe that's what I should expect from a $30 drill :-)
Yes, I should have let the temperature equilibrate before reading the pressure (or measured the tank air temperature at each end and calculated the effect), but I really just wanted a general sense of things.
The tools are new, albeit cheap. I bought the compressor used, but the tank says it was manufactured in 1995, so I assume it's 8 years old or so.
Fair enough. I'm in Berkeley, CA, which is maybe a couple of hundred feet above sea-level, so pretty much 1 atm.
I will look into Blaster -- thanks. I'm currently using Mobil Almo 525. As for recommendation 2, there is a bit of air that puffs out of the oil filler tube; it feels about like it would if you pursed your lips and blew with moderate force. Probably (?) something like 1/4 cfm (I realize I'm getting into murky water here).
I tried this today, and, yes, the pressure drops below the cut-in pressure and keeps dropping even after the compressor kicks in. The regulator is set at 90 psi and I let it go to about 70 psi or so. I'm afraid I didn't measure the rate at which it continued to drop, but it was definite.
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