low pressure in tires with cold weather

[Curt Harms]

I've never understood this suggestion (in theory or realistic terms). Air is already 78% nitrogen. The remaining 22% has negligible effect at the pressure we run our daily drivers.

Reminds of my wife's trip to the local tire shop to top up a tire originally installed by Costco (they use Nitrogen). Shop initially refused to top up with shop air (they didn't have nitrogen) because it would "contaminate" the nitrogen in the tires. After informing them of the above percentage, I was still mystified by their reluctance to add 3-4psi of shop air, adding a whopping 2-3% of "contaminants".

I'm not sure how relevant this is to cars but most aircraft manufacturers recommend nitrogen in tires. Not because nitrogen is good for the tires but because compressed air can contain moisture, nitrogen from a bottle does not. Get enough moisture in a tire then leave it outside in freezing weather (or cold soaked for a few hours at -60^oF at altitude) and you have a frozen puddle, resulting in an unbalanced tire. Also sorta related, there is sometimes a table to be found for high pressure accumulators. I noticed the same phenomon, pressure drops when it gets cold and I'd get the nitrogen added. Later I discovered a table that gave differing "full" values depending on temperature.

[Pat Barry]

This whole discussion prompted me to dust off my memory related to the old PV=nRT formula. I did the calculations and here is a resulting table showing the pressure versus temperature. Note: It assumes a constant gas (air/nitrogen) volume (probably not true for a tire or a football that can inflate), also assumes that the gas is always a gas and not a condensing material. None the less the table probably is a pretty good first order estimate:


TempVsPressure.jpg
Also, for you NFL fans, I also did the calculations for footballs

tempVsPressure2.jpg
.

[Jason Roehl]

One of the reasons for nitrogen in airplane tires is that if one bursts during braking, you have a non-flammable gas flooding the area as bits of hot rubber come into contact with even hotter brake parts. Pointless in car tires.

[Jason Roehl]

This whole discussion prompted me to dust off my memory related to the old PV=nRT formula. I did the calculations and here is a resulting table showing the pressure versus temperature. Note: It assumes a constant gas (air/nitrogen) volume (probably not true for a tire or a football that can inflate), also assumes that the gas is always a gas and not a condensing material. None the less the table probably is a pretty good first order estimate:


TempVsPressure.jpg
Also, for you NFL fans, I also did the calculations for footballs

tempVsPressure2.jpg
.

I had to double check with a friend who is a nuclear chemist to make sure I was using Charles' Law correctly, but I found that if a football (assuming no change in volume) is inflated to 12.5 PSI at 70ºF, and the temp drops to 50ºF, the pressure will drop about 1.1 PSI--more than double the drops your chart shows for 80ºF to 60ºF to 40ºF, interestingly enough (but your chart is right). Supposedly the Patriots' balls were 2 PSI low...

[Art Mann]

A friend of mine who is an electrical engineer at Boeing came over last night. I asked him why airliner tires are inflated with nitrogen. He didn't know for sure but he knows some people who would know the correct answer and he said he would ask. If I hear back from him, I will post what he said.

[Pat Barry]

I had to double check with a friend who is a nuclear chemist to make sure I was using Charles' Law correctly, but I found that if a football (assuming no change in volume) is inflated to 12.5 PSI at 70ºF, and the temp drops to 50ºF, the pressure will drop about 1.1 PSI--more than double the drops your chart shows for 80ºF to 60ºF to 40ºF, interestingly enough (but your chart is right). Supposedly the Patriots' balls were 2 PSI low...

Jason, thanks for the sanity check. I now see that I used PSIG for the calculations, not absolute pressure. Using absolute pressure (correct application) the pressure drops will be more than what I calculated in the tables above. I did a quick check and got 1 PSI for the football example at the temperatures you identified.

[Jason Roehl]

Jason, thanks for the sanity check. I now see that I used PSIG for the calculations, not absolute pressure. Using absolute pressure (correct application) the pressure drops will be more than what I calculated in the tables above. I did a quick check and got 1 PSI for the football example at the temperatures you identified.

LOL! I heard back from my nuclear chemist friend, and he went back on what he originally said--that we SHOULD use PSIG, not absolute pressure. The atmospheric pressure wouldn't change. However it just occurred to me that what's tripping everything up is that the volume of the football is not constant. Footballs are neither perfectly rigid where the volume never changes, nor are they "perfect balloons" that can collapse completely. With an increase in pressure, their volume probably follows some sort of inverse exponential curve that approaches some theoretical limit until they burst.

I may need to do an experiment...if I can find a needle...

[Anthony Whitesell]

Both are incorrect. The ideal gas law and Boyle's law both use temperature in Kelvin and pressure in Pascals. When using 70 deg F (294K) and 40 deg F (277K) and a starting pressure of 13 psig (89631 pascal) all converted to Kelvin and pascals, then perform the Boyle's law calculation (T2/T1*P1) and convert the resulting pressure back to psig, the footballs are at 12.2 psig (84448 pascal). As soon as you calculate the ratio of temperatures in Kelvin, you already there is going a 6% drop in pressure. Also remember that this is for an IDEAL gas, which there aren't any and void of water vapor. One aspect I have not heard mention of is the source to fill the balls. A bicycle pump, air compressor, or concentrated gas cylinder?

As for the car tires, the same 6% drop applies so a 30 psig tire at 70 deg F drops to 28.2 psig at 40 deg F. The TPMS in my car goes off just under 28 psig.

[Jason Roehl]

Both are incorrect. The ideal gas law and Boyle's law both use temperature in Kelvin and pressure in Pascals. When using 70 deg F (294K) and 40 deg F (277K) and a starting pressure of 13 psig (89631 pascal) all converted to Kelvin and pascals, then perform the Boyle's law calculation (T2/T1*P1) and convert the resulting pressure back to psig, the footballs are at 12.2 psig (84448 pascal). As soon as you calculate the ratio of temperatures in Kelvin, you already there is going a 6% drop in pressure. Also remember that this is for an IDEAL gas, which there aren't any and void of water vapor. One aspect I have not heard mention of is the source to fill the balls. A bicycle pump, air compressor, or concentrated gas cylinder?

As for the car tires, the same 6% drop applies so a 30 psig tire at 70 deg F drops to 28.2 psig at 40 deg F. The TPMS in my car goes off just under 28 psig.

We both did that. As long as the reference point is correct (absolute zero), units don't matter since the equation is T2/T1 = P2/P1. Try it. You can do it in Fahrenheit gradations above absolute zero and the difference will still be the same percentage. But neither of us did 70ºF to 40ºF. I calculated 70ºF and 12.5 PSIG (minimum pressure for the NFL) starting conditions dropped to 50ºF, which yields a pressure of about 11.4 PSIG.

[Anthony Whitesell]

I'm sorry Jason, I don't follow. I made the mistake of using Deg F first time I thought of using the gas law. Then I made the mistake of using Deg C. Neither compute the same as Kelvin. T2/T1 = 40/70 deg F = 0.57, 4.4/21.1 deg C = 0.21; and in Kelvin 277/294 =0.94. To reproduce the calculation using 70 and 50 deg F and 12.5 psig starting pressure. 12.5 psig=86184 pascal; 283K (50degF)/294K(70degF) = 0.96 (4% loss) * 86184 pascal (12.5 psig) = 82959 pascal or 12.0psig.

[Jason Roehl]

Anthony, you're right that you can't use 70ºF/50ºF for the equation. However, if you add 460ºF to each of those, it will work, since absolute zero is -459.67ºF. That sets absolute zero as the reference point. The temperature units cancel. The same would be true for Celsius if you add 273.15 to the appropriate Celsius temperatures (which is how you get the Kelvin temperatures).

The reason I had come up with a different number is that I used absolute pressure (PSIG + PSIA, which is the gauge reading plus atmospheric pressure). However, since footballs aren't rigid, atmospheric pressure has an effect that changes the volume of the football. Probably not a large effect, but it does complicate the calculation, and we don't (can't--it's not ever measured by the NFL) know what that volume change was on the field. Anyways, I was multiplying that 96% by 27.2 PSI (absolute, or 12.5 PSIG + 14.7 PSIatmos).

[Pat Barry]

I used deg K for the calaculations I did both times (PSIG and abs pressure). The pressure unit of measure doesn't matter since gong pascal to PSI is a simple proportion. I suppose PSIG is actually OK to use since a deflated football would read 0 psig and go up from there to whatever. In absolute you just need to add 14.7 psi. Anyway, the pressure goes down with temperature as we all know. I'm looking forward to actual test data if anyone can produce that.

[Pat Barry]

I found someone doing the air pressure vs temperature football test
13.3 psig to 11.9 psig (76 deg vs 52 deg)

[Frank Trinkle]

I have nitrogen in all four of my new large Michelins on my Armada SUV. I drive in Hot areas, Cold areas, and Freezing areas. What I have found with the nitrogen fill is that my tire pressure never strays more than 1-2 psi no matter what conditions I'm driving in. This is a far cry from my prior air filled tires that would give me the tire alerts regularly when the temperature outside dropped.
+1 all the way for Nitrogen. Oh, I am also a pilot, and all of our tires on our Helicopters are nitrogen filled, much for the same reason as my SUV.

Cheers,
Frank

[Jason Roehl]

Okay, so here's the thing with nitrogen. Air is 78% nitrogen, almost 21% oxygen and about 1% all kinds of other stuff. Just for round numbers, let's say it's 80/20 nitrogen/oxygen. You fill your tires with air to 35 PSI. By the law of partial pressures (Dalton's Law, if you need to look it up), that means that 1/5 of that is oxygen (7 PSI). Assuming the pro-nitrogen-fill guys are correct, then the oxygen will leak out, but the nitrogen won't.

So, in a little while, the tires will be at 28 PSI, all nitrogen. Then then get topped off with more air (80/20 nitrogen/oxygen). Now the tires are at 35 PSI again, but 80% of that 7 PSI addition was nitrogen, or about 5.6 PSI. This means the tires are now 33.6 PSI nitrogen and 1.4 PSI oxygen.

The tires leak down again (only the oxygen, to 33.6 PSI), and get topped off again (to 35 PSI). The nitrogen is then 34.8 PSI...