KevinC SJ,CA

Soapy water would be bad. But "moist air" is just air mixed with water vapor - a gas.

Mark Dalen

If you have saturated air at 250 def F and cool it, you will have a puddle of water. The N in the equation has changed.

If you start with 70 degree saturated air, then you are correct.

s/c 6

Thanks for all your comments.... How about with Khumo victoracer 700 on a B5 platform?

s4wood

It was at a Tracquest event a few years ago. I probably started with 40 cold....the thing was I didn't check pressures before I went out on my first session. When I released the pressure..the hot air was burning my hand...it was VERY hot.

Mark Dalen

I have seen this too, so here is a long explaination. I have tried to make this as non-technical as possible so everyone can follow along. Kevin clearly understands the physics - I think he simply overlooked one aspect of what can happen with water present.

Here is what is going on - maybe this will also dispel the myth that Nitrogen is used by race teams because the pressure hardly changes with temp. It can also give everyone a rule of thumb about what will happen to temps vs. pressures.

The ideal gas law is PV=NRT
P =Pressure
V =Volume
N = No of Moles of Gas (It is a way of talking about the amount of gas present.)
R = Gas Constant
T = Temperature
For the environment of the tire, N does not change as the tire heats, nor does R, the gas constant. V can be taken as unchanged (the actual change in volume is very, very small).
When you distill all this down you see that P2/P1=T2/T1 (The two conditions we are interested in.)
NOTE - There is no factor for different gas in this, nor should there be. Using air, Nitrogen, Helium, Argon or whatever else DOES NOT MATTER. You can even use Hydrogen in your tires and it will follow the same rules (of course I will bring the marshmellows to the track if you want to do that). Even a moist gas behaves the same.

So if you go out at 30 psi and 70 deg F and come back in at 40 psi, we can calculate the temperature (absolute, not F). Add ~460 to deg F for Deg R (absolute) and we can see that P2/P1 is 40/30=1.333. 70+460=530. 530*1.33=705 then convert back to Deg F (subtract 460) and you can see that the gas within the tire had to be 245 deg F. A very typical and believable value.
I use the example of starting around 30 and getting to 60psi. Can't happen within practical limits using this equation. That would imply that the temp doubled for 530 deg R ambient to 1060, which converts to 600 deg F for the internal tire temp.

Then how is it possible that this kind of pressure increase actually happens. Realize that a gas (air in this case) can hold more water when hot than at lower temps. This is what causes dew as overnight temps drop. If there is liquid water in the tire at 70 deg, but it gets vaporized as the tire heats to 250 deg +, then out equation can not be simplified the way I did at the start. N has changed between the two states, as has R. The pressures go nuts. So Kevin was absolutely correct in what he said ASSUMING there is not enough water to drop out of vapor at the lower temperature.
Note to Kevin - All my old textbooks with the constants, etc are in storage, so I can not do the math to estimate this. Clearly you understand the math and physics. It would be interesting (at least to me) to calculate what the pressure would be by going from 70 to ~250 deg with a water content that gives saturation at the higher temp. It would end up being a complex equation, as the amount of water would increase as the pressure grows along with temp, but an estimate would be pretty close.

Hence my statement about that being impossible (with the assumption of ~30 starting pressure) was just playing with words. It is very possible but it indicates that there is a lot of water in the tire that needs to be purged.

Hope this helps clarify what is going on, and what I meant with my other posts.

s4wood

I like them a lot. The old ones IMHO were better than the new version though.

KevinC SJ,CA

Mark, I don't think I'm smart enough to figure that one out. ;o) But the link below might give you the ammunition needed.

However, it shouldn't be too hard to estimate how much water would need to be converted to water vapor in order to double the expected pressure at a given temperature. If the pressure doubles, it would simply mean that the number of moles of water vapor added to the gas mixture was equal to the number of moles of air (or to simplify, we could just assume that the tire only contained N2 and H2O.) From the ideal gas law, N = PV/RT. If I was a little less lazy, I could figure out how many moles of N2 was in that tire, and therefore how many moles of H20 were added to the gas mixture, and therefore how many cc's of liquid water had to be in the tire. How did our prof's put it? Excercise left to the student. ;o)<ul><li><a href="http://www.cgd.ucar.edu/cms/wcollins/ATOC_4710_5710/Lectures/Lecture_6.htm">Gas Laws Notes</a></li></ul>

Mark Dalen

The problem being left to the student is that it has been way too many years since I could find my way around all those charts.

I think my ONLY chance of actually solving this would be with an experiment. Hmmm -maybe my wife's pressure cooker has a pressure gauge - I promise I won't blow anything up this time.