I can't seem to get my head around the relative humidity of air inside a pressurized tire.
What I'd like to know is if water condenses out of the air inside a tire when it gets cold. I'm trying to figure out the implications of using nitrogen instead of air as an inflation medium.
For example, if water doesn't condense out of the pressurized air inside a tire when it gets cold then it doesn't matter if one uses air or nitrogen, the gas medium reacts the same.
Welcome!
The water vapor contained in trapped atmospheric air changes volume much more dramatically than air with changes in tire temperature.
For each combination of internal pressure and temperature, humid air absorbs and condensates different amounts of water.
Water may affect metal parts of the rim.
Pure nitrogen does not have those problems.
What I'd like to know is if water condenses out of the air inside a tire when it gets cold. I'm trying to figure out the implications of using nitrogen instead of air as an inflation medium.
I don't know much about the subject, but it seems like dehumidifying the air before compressing it would be simpler than having to stock a bunch of ##N_2## cylinders. No?
I'm trying to figure out the implications of using nitrogen instead of air as an inflation medium.
The primary reason why nitrogen is used, is to prevent fire supported by compressed air, in the wheel well of an aircraft. Nitrogen also prevents aircraft wheel imbalance, due to internal ice formation at high altitude, low temperatures.
The problem with humid air in a tire is that, if you inflate the tire to the correct pressure on a hot and humid day, when the tire cools, the tire pressure will fall as the water condenses to a liquid in the tire.
If you pour an excess of liquid water, into the wheel before inflation, then with changing temperature, the tire pressure will vary as a function of the changing vapour pressure of water, also, the rim and valve will probably corrode.
How bad could it get? There is a code table below showing how gauge pressure changes in a tire with temperature, for dry air or nitrogen, and for saturated wet air. Those are my computations from a year ago when I was considering truck tires. The effect will be proportionally greater at lower pressures, subtract a constant. They have not been checked by others, so take them with a grain of salt, (which may accelerate corrosion).
As tire pressure varies with temperature due to moisture, the contact patch and the tread wear across the tire will change. At lower pressures, the contact patch will increase in area, so the sidewall will also deflect more, reducing tire life.
The tension and the speed of sound in the belt and tread will be greater at greater pressures. Low pressure at high speeds can destroy a tire when road speed approaches the sound velocity in the belt, as a noise front builds up at the first point of road contact, a sound barrier.
Of note: A Californian study on permeable pavements found that asphalt pavements, with a typical albedo of 0.1, can produce surface temperatures of 70-80°C on a hot and sunny summer day in cities such as Phoenix, Arizona.
(Li et al. 2013).
We can significantly exceed that in Australia.
During a stock car race, we need to maintain a proper tire diameter once the tire is brought up to proper operating temperatures. usually 180 to 210 degrees F. depending upon the weight of vehicle and particular tire compound. in order to maintain the proper diameter, we use nitrogen instead of air because this gas does not have moisture in it, has larger molecules and will not bleed through the tire wall. The lack of moisture means the tire will expand at a more consistent rate vs. moisture rich air which could steam up on you. Yes the moisture ( water) turns to steam and you can grow tire diameter up to an inch or more. Two things happen when diameter grows. The tire contact patch is reduced on all four tires and stagger grows again on all four tires. Stagger is used to assist the rearend turning when a locked differential is used. Both of these impact handling dramatically.
Just about all air compressors will use moisture latent air and unless you use air filtered through an expensive air dryer, $1100, you will have moisture in the air. I sell expensive machines requiring instrument quality compressed air. A bottle of nitrogen from local welding supply shop is cheap and lasts the entire season.
Go to a junk yard and buy an old refrigerator and take out the motorized pump. Suck the air out of the mounted and balanced tire and replace it with nitrogen. Will decrease your lap time!
If you can draw air into your compressor from a refrigerated cold store, there will be significantly less water dissolved in the air.
The partial pressure of water in saturated air, halves approximately, for each 10°C drop in temperature.
At +20°C, ppH2O = 2337 Pa.
At +10°C, ppH2O =1227 Pa.
At +0°C, ppH2O = 611 Pa.
At -5°C, ppH2O = 421 Pa.
At -10°C, ppH2O = 286 Pa.
At -20°C, ppH2O = 125 Pa.
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#9
CapriRacer
6
0
Thanks for all the replies. it's going to take me a while to sort through it all.
But I still can't get my head around this whole thing. I think part of this confusion is the business of the Ideal Gas Law. If water vapor doesn't follow it, then there should be some test data that shows that the pressure buildup during operation is different between nitrogen and air, but I haven't seen any. You would think a manufacturer of nitrogen generating equipment would have widely circulated such a test as proof of nitrogen's effectiveness. I can only conclude that it doesn't show that difference.
If anyone has seen such a test, please post a link. It would be very much appreciated.
If water vapor doesn't follow it, then there should be some test data that shows that the pressure buildup during operation is different between nitrogen and air, but I haven't seen any.
Nitrogen behaves the same as dry air. In both those cases, the number, n, of molecules of gas in the tire remains constant.
Given an ideal gas; P · V = n · R · T . Assuming a fixed volume, V, tire, pressure, P, will vary in linear proportion only to absolute temperature, T. P = T · ( n · R / V ).
When there is excess liquid water present in the tire, some water will evaporate, until the contained air becomes saturated with water. The number, n, of molecules of gas in the tire has therefore changed, due to the water vapour. That will then change the tire pressure, P, in a way determined by the solubility of water in air, the partial pressure, ppH2O, which is exponentially proportional to absolute temperature. P = T · n · ( R / V ).
If the mass of water is limited, then at some specific temperature, all the liquid water will have evaporated, so there will be no increasing, n, at higher temperatures.
#11
Dullard
565
469
Possibly noteworthy: The water content of compressed air is typically 'saturated.' This can be misleading. The vapor pressure of water is a function of temperature (only). A tank of 120 PSIG compressed air has as much water (per unit volume) as a tank of 60 PSIG compressed air (equilibrium, same temp). Use of the 60 PSIG source would result in a tire (30 PSIG) with twice as much water as the 120 PSIG source (not 'saturated' in either case). The compressed air source matters.
Possibly noteworthy: The water content of compressed air is typically 'saturated.' This can be misleading. The vapor pressure of water is a function of temperature (only). A tank of 120 PSIG compressed air has as much water (per unit volume) as a tank of 60 PSIG compressed air (equilibrium, same temp). Use of the 60 PSIG source would result in a tire (30 PSIG) with twice as much water as the 120 PSIG source (not 'saturated' in either case). The compressed air source matters.
If I assume the air in a compressed air tank is saturated (a pretty good assumption), then filling a tire at a lower pressure should result in the tire NOT having saturated air - and using that chart, I could calculate the temperature where condensation inside the tire occurs.
#13
Dullard
565
469
I'll stand by my post; The 'amount of water' and 'vapor pressure' are not the same thing. Your last paragraph is correct (and is exactly what I was trying to convey), so I think we agree.
#14
CapriRacer
6
0
Dullard said:
I'll stand by my post; The 'amount of water' and 'vapor pressure' are not the same thing. Your last paragraph is correct (and is exactly what I was trying to convey), so I think we agree.
I see your point. I have to be very careful about which words I choose to describe this. I should know better as I frequently have to deal with such things.
What I was trying to point out was that the air within the tire was not at 100% humidity, and that's not always how folks think this works. I'm still working on the issue of whether or not water vapor behaves like an Ideal gas in the absence of condensation and boiling. I'm working on a webpage to help me sort this all out.
#15
Dullard
565
469
As I understand it: For the purposes of your question, water vapor (no condensation, no evaporation) is sufficiently close to ideal.
In true Dunning-Kruger style, I found that I did not know what I did not know, until I tried to compute it. It took a while to find a robust equation for the partial pressure of water in saturated air.
This is the one I use now.
You will note that there is a temperature term, but no pressure term.
Code:
' Compute the partial pressure in Pa, of H2O in saturated air
' given temperature in Tc = °C
Function ppH2o( Byval Tc As Double ) As Double
Static As Double c0 = +0.99999683, c1 = -0.90826951e-2,_
c2 = +0.78736169e-4, c3 = -0.61117958e-6, c4 = +0.43884187e-8,_
c5 = -0.29883885e-10, c6 = +0.21874425e-12, c7 = -0.17892321e-14,_
c8 = +0.11112018e-16, c9 = -0.30994571e-19
Dim As Double pT = c0 + Tc*(c1 + Tc*(c2 + Tc*(c3 + _
Tc*(c4 + Tc*(c5 + Tc*(c6 + Tc*(c7 + Tc*(c8 + Tc * C9 ))))))))
pT *= pT ' ^2
pT *= pT ' ^4
pT *= pT ' ^8
Return 610.78 / pT ' pp of H2O in pascals
End Function
The table appended to post 17 is probably correct, but it may leave a reader with the wrong impression. Helium is difficult to contain. Helium is a small molecule - That's more about the mass than the diameter. At a given temperature, a smaller molecule moves at a higher velocity (MV^2). It soccer terms (football for you Europeans): It's the number of shots on goal, not the size of the ball.
I think I've figured out what I needed to figure out. Thanks to all who contributed. Here it is:
If I have an air compressor with a full tank at 120 psi, the process of filling that tank resulted in water condensing out, and the air inside that tank is saturated = 100% relative humidity. If I use that air to fill a tire to 30 psi, then the air entering the tire is NOT saturated, and no liquid water condenses out. Please note I am assuming both the tank and the tire are at room temperature.
2 possibilities:
1) There was no liquid water in the uninflated tire's air chamber - in which case, the pressurized tire has a gaseous mixture that can be treated like an ideal gas. This would be no different if the gas were nitrogen or any other gas.
Please note that if the tire is allowed to cool below freezing, it is possible that water will condense out of the air, but it will be frost - snow-like. But once the tire heats up due to being operated, then the frost will evaporate and return to the condition before it cooled.
2) If there was liquid water in the air chamber, it is possible for the inflation medium to be completely saturated and there to be additional liquid water in the chamber. In this case, the liquid water will add additional pressure when the tire is operated due to the heat generated. At worst, this will add about 1/2 a psi to the pressure buildup. While this is nothing to worry about in a street tire, in a race tire, this is enough to affect the handling of the car - hence, why race tires benefit from using nitrogen as an inflation medium.
In this case, the liquid water will add additional pressure when the tire is operated due to the heat generated. At worst, this will add about 1/2 a psi to the pressure buildup.
At what temperature was the tire inflated, and at what temperature was it then operated?
Was the excess liquid water in the compressed air reservoir, or in the tire?
For saturated air, with excess liquid water in the tire;
Inflated at 20°C, the ppH2O = 2.337 kPa.
Operated at 70°C, the ppH2O = 31.168 kPa.
The increase then is ( 31.168 - 2.337 ) kPa = 28.831 kPa.
28.831 kPa = 4.182 psi.
When increasing the temperature from 20°C to 70°C, that is the increase in pressure due to ppH2O alone.
#21
CapriRacer
6
0
Baluncore,
Thanks for replying. I posted because I wanted to see if I missed something - and as you'll see, I did!
Baluncore said:
At what temperature was the tire inflated,
As stated, room temperature. Let's assume that is 20°C = 68°F
Baluncore said:
and at what temperature was it then operated?
And here's what I left out. A rule of thumb we tire engineers use is that a properly loaded and inflated tire will experience about a 10% pressure buildup. Anything more than that indicates an overloaded tire. Overloaded (underinflated) tires fail in the long term due to deterioration of the rubber (Arrhenius rule!)
Please note: This does not apply to racing tires, because racing tires have a short lifespan - although you will see tires fail when pushed too far. Both F1 and NASCAR have experienced such failures.
Having added that, a tire starting at 68°F (20°C), 30 psi (207 kPa) will experience an air chamber pressure rise to 33 psi (228 kPa) and a corresponding temperature rise from 68°F (20°C) to 106°F (41°C).
Baluncore said:
Was the excess liquid water in the compressed air reservoir, or in the tire?
In the tire. What was in the tank stayed in the tank,
Baluncore said:
For saturated air, with excess liquid water in the tire;
Inflated at 20°C, the ppH2O = 2.337 kPa.
= 0.34 psi
Baluncore said:
Operated at 70°C, the ppH2O = 31.168 kPa.
And this where that tidbit of information comes home to roost. A tire's air chamber would never see 70°C (158°F). So the rest of that calculation doesn't apply.
And this where that tidbit of information comes home to roost. A tire's air chamber would never see 70°C (158°F). So the rest of that calculation doesn't apply.
What worst case would it see in Arizona or the Australian outback?
Inflated at 20°C, the water vapour will provide 0.339 psi of the internal pressure. Then we often see 40°C outback air temperatures, with 70°C road surface temperatures. Black body radiation of heat by the tire, (heat from the road), will limit the rise to, say, about 20°C above ambient air temperature. That internal tire temperature is then 60°C, +40°C above the inflation temperature of 20°C. At 60°C the partial pressure of water will be 2.890 psi, an increase of 2.551 psi over the inflation temperature.
