How hot does the air in a PCP get? Probably Hotter than you think.

[lloyd-ss]

Please bear with me as I ramble on about filling PCPs.

You know how when fill a PCP from a tank that flows really fast, that the valve on the tank gets cold and the reservoir on the gun gets warm? Did you ever give it much more than just a passing thought? Until a few months ago, I didn’t either.
To make a long story short, the air filling that reservoir  gets REALLY HOT, and probably hotter than you might think.

I had purchased one of those YongHeng compressors and although I was happy with how fast it filled, wasn’t too thrilled with the cleanliness of the air. So even  though you can buy a decent purpose-built moisture and oil filter for the HPA, I was going to make something out of what I had on hand, because I can't stop myself.    I had a 12” piece of 7/8” O.D. DOM steel tube, internal threaded  at both ends, some fill adapters, and some felt carpet pad about ¼” thick. I rolled up a 2” x 3” piece of the pad like a cinnamon roll and stuffed it into the tube, and then threaded a pair of fill adapters, without the check valves,  into both ends of the tube. I hooked a fill line from a CF tank that had about 2,000 psi in it,  to the steel tube and gave the valve a sharp twist to blast a little air thru the tube just to see how the air would flow thru the felt.

WELL, plenty of sound of the air rushing into the tube, but nothing rushing out. And the tube instantly got so hot that I had to put it down and turn the valve off as fast as I could.  (Yes, I know better, please don’t remind me.)

So what happened? Well, the exit fitting actually DID have a check valve in it, so that tube was filled almost instantaneously to 2000 psi. Can anybody say adiabatic compression?
After I checked my pants and took a little walk outside and regained my composure, I came back in to figure out just what had happened.  The tube was back at room temp. I slowly released the remaining  pressure and threaded out the end plugs.The picture at the end pretty much shows it. Apparently, the air rushing into the 7/8” dia x 12”  long tube was re-compressed to 2000 psi so quickly that the compression was almost fully adiabatic (all the heat stays in the compressed air and is not transferred to the vessel) and that air was hot enough to melt, char, and burn that synthetic felt carpet pad. That black chunk of stuff in the picture is about half of what was left of the carpet pad, but it was all charred, no un-burned pad was left at all. Even though the compressed air is only 20% oxygen, that was still plenty to support the combustion process.
I have since done some analysis and tried measuring some temps with thermocouples, with limited success. If you do the temperature calcs for the compression, you might say, aww, it can’t get that hot. Hmmm, maybe it does. More to follow.
Lloyd

[Insanity]

Interesting reminds me of pyrokinesis.

[lloyd-ss]

Interesting reminds me of pyrokinesis.

Ha ha, I had to look that up. I don't know too many people I'd trust with that one.

[Insanity]

Not quite someone creating fire lol.

[mobilehomer]

Not nearly as hot as the HPA tubes on the Yong Heng!

[rsterne]

OUCH!!!.... Lloyd told me about this quite a while ago.... but I failed to comprehend how serious it was.... 

Bob

[MJP]

This makes you wonder how hot does the air momentarily get behind the pellet before it starts to move and the air starts to cool from pressure drop?
You are raising the pressure very rapidly behind the pellet when the valve opens.

Marko

[subscriber]

If you compare the volume (stored energy of HPA) of your steel pipe, Lloyd, to a springer; where 50 CC of ambient air is compressed to only 0.5 CC of hot HPA; then the magnitude of the potential problem comes into focus.

The potential for an exothermic bang was significant, simply because the energy in that volume of HPA had heat of combustion superimposed on top.  If the combustion reaction had been more brisk, the resultant pressure rise might not have been able to vent back to your air source fast enough...

The fact that the pipe was so full of felt probably reduced the temperature rise and rate to the point where it was not enough to run away.  If the amount of felt had been reduced, there would have been more air and more oxygen, to speed along the reaction and increase the temperature more dramatically.

Food for thought, indeed.

[lloyd-ss]

Thanks for the comments everyone, the incident was eye opening for me and warranted some research. But what I found regarding compression of air was quite confusing, often conflicting, and by its very nature, difficult to pin down to exact values. Too many variables in the air compression process, along with non-linear changes.
But I really am quite interested in knowing what kind of heat we are dealing with. About all I can say is that I came up with some good generalizations, but no concrete values. Here is why (in my opinion), but some of you probably know more about this subject or have different opinions. The subject is too “mushy”, not like measuring a block of steel where you get one right answer, LOL.
Here is what I think I learned. Compression of air, a “real” gas, can range from adiabatic (constant entropy) where all heat of compression stays in the air, to isothermal (constant temperature) where the gas is compressed very slowly with no temperature increase. Or the process can be somewhere in between, called polytropic. And the process varies along the length of the airtube.
Two equations can define the pressure and temperature changes when the air is compressed.
T2=T1*(V1/V2)^(k-1)   and  P2=P1*(V1/V2)^k     note:compression ratio can be instead of V1/V2
K, or gamma, is the ratio of specific heats of air, constant pressure Cp=1.00 and constant volume Cv=.718. Therefore, k=1.4, but only for pure adiabatic processes, and can go as low as k=1 for pure isothermal processes, but k is usually somewhere in between.
So, by varying the value of k from 1 to 1.4 in the 2 equations above, you can get wildly varying results for temp and pressure. (remember to use the same value for k for each pair of calculations.
Looking at subscribers comment likening my ¾” x 10” air tube to an extra long springer cylinder, lets try an adiabatic calc for a ¾” x 3” springer, compressing down to .1” (30 to 1).
T2=297*30^(1.4-1) = 1158K or 1624F
P2=14.7*30^1.4 = 1719psi
The temp in particular looks too high, but I don’t know? Let’s re-do the calcs using a value of 1.0 for k, pure isothermal.
T2=297*30^(1.0-1) = 297K or 75F
P2=14.7*30^1.0 = 441psi
Well, that answer doesn’t look correct either. So it must be somewhere between adiabatic and isothermal. I am only half joking when I say this, but these formulas were written so that if you don’t like the answer, you can just say, “No, the answer isn’t wrong, you just chose the wrong process.”
For the 10” long tube that I burned the felt in, the process is even more complicated. When that tube is filled through a fill hose, the air expanding into the 10” tube gets chilled, but then it packs up at the far end and is recompressed and reheated. So, what is the compression ratio in the 10” tube? Is the heat lost during the expansion equal to the heat energy that is converted during the re-compression?
I was able to measure a temperature spike over 500F in that 10” tube with a couple of thermocouples that Shorty sent me. More to come about that. And if the air gets that hot, where does all the heat go? More to come.

(Sorry this is so dry. I keep telling myself it is always good to learn something new. I say the same thing to my grandkids, LOL.)

[lloyd-ss]

Measuring the temperature rise inside the 10" long tube will take a lot more money and brains than I have. But I did manage to get a rough measurement of greater than 500F for the temperature flash as the tube is filled.
Shorty had sent me some extremely fine thermocouples over a year ago. Thinner than 40 gauge wire and about 36" long. My wife was in the shop a while back and wanted to know what those long hairs were that I had taped to the end of a cabinet. Awkward and creepy.
The picture shows one of the thermocouple "hairs" epoxied into an 8-32 brass screw. A regular thermocouple is also show for comparison. The regular one has to much mass for a quick response. It was difficult trying to calibrate the T-couple. Between an ice cube and a heat gun and a piece of copper and some other thermometers and thermocouples, I was at least able to find out where a good "hot" reading was. The T-couple had to be hooked to a compensator and then thru a little instrument amp IC to get the output voltage high enough to track, over 30mV.
There was a thermocouple at each end of the tube. Catching the temperature spike during filling was difficult, but it was very interesting to see how the temp varied inside the tube as the tube was bled of all air pressure. The air stream got very cold. Max temp was definitely greater than 500F and occurred at the closed end of the tube.

[Chickenthief]

Reminds me of a fire piston!
Combustibles, heat and lots of oxygen = Poof

[lloyd-ss]

So Where Does all of the Compression Heat Go?
---------------
EDIT - 10-17-19 @3:23
I just realized that I slipped a couple of decimal points in a calculation and have to re-visit this.
Sorry. Lloyd
--------------------------
 EDIT -10-17-19 @ 4:38
I corrected the slipped decimal places, but I am not 100% sure about this calculation.
Lloyd
----------------------------------------------
So Where Does all of the Compression Heat Go?

Using the 3/4 dia x 10" long tube again. If the temp in that tube spikes up by 500 degrees F, what happens?

Heat (joules) = mass * delta T (C) * specific heat (Joules/Kg-K) (.83 for air)
Heat = .013 kg (3/4 I.D. x 10" long tube of 2000 psi air)(weight of air only) * delta T of 315 deg C * 830 Joules
Heat = 3399 Joules of heat = .013 * 315 * 830
3399 Joules of new heat energy in the air.

So, the air has 3399 joules of heat and is 315 deg C above its original room temp. Now, if all that 3399 Joules of heat is transferred to the steel air tube and end caps that weigh 10.7 ounces, how hot does the tube become?

3399 Joules of heat = .3Kg of steel * delta T temp change in degree C * 502 (specific heat of steel)
3399 J/(.3 kg * 502 J/kg-K) = 22 deg C

So what that means is that after that hot air inside the tube transfers all of its heat energy into the steel tube, the temperature of the tube rises 22 degrees C.  Enough to be obvious. But the tube is also loosing heat to the surrounding atmosphere at the same time.

But back to the Yong Heng pump, if you do that compression 3600 times per minute, the heat builds up REALLY FAST.

[subscriber]

Reminds me of a fire piston!
Combustibles, heat and lots of oxygen = Poof

I wonder what pressure and temperature fire pistons reach.  Obviously enough to start combustion; so some might say at least 451 F:  https://en.wikipedia.org/wiki/Fahrenheit_451. 

I think it is well over 300 degrees C.

[lloyd-ss]

I had not seen those videos of the fire piston. It sure seems to follow along with the adiabatic calculations.
I was curious about how much combustion might be supported in one of those tubes.  Burning wood takes, BY WEIGHT, about 11 pounds of air per pound of wood. Wood might be 45 lbs per cubic foot and air .0807 lbs per cubic foot.
One pound of wood = .0222 cuft
One pound of air = 12.4cuft
It takes 11 pounds of air (136.4 cuft) to burn one pound of wood (.0222 cuft), so the ratio of air to wood BY VOLUME is 6,144 air to one wood. So luckily, there doesn't appear to be enough air in the little glass tubes to support much combustion.

That 3/4" x 10" air tube at 2,000 psi in the previous example, however, has .024 pounds of air, and that would burn .0022 pounds of wood at the 11 to 1 by weight ratio. That is almost exactly one gram of wood with complete combustion. I just weighed a piece of the felt the same size as what burned in the tube, and it weighed about 2 grams. Playing with fire! Ignorance is not bliss, LOL.