Tire Pressure Loss After Five Months

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I've read (tire rack, and other auto sources) it's normal for tires to lose 1 psi per month or so.....but also the weather/temperature changes affects the pressure too.

This winter, my TPMS light went on when we got our first "cold" days...I was like oh great, I got a nail.....turned out, nope, just needed air pressure. Filled up to 38 psi and good to go.
 
Originally Posted By: Spector
remember for every 10 degree change in outdoor temperature the tire pressure changes 1PSI. so if the temp goes from 50degrees to 40 the tire pressure will drop 1 PSI. In addition, as noted, tires lose pressure at varying rates over the course of a month.

Nitrogen filled tires supposedly lose less air over time
Nitrogen is dry.
 
Originally Posted By: LS2JSTS
Do changes in temp affect Nitrogen filled tires at the same rate as "regular" air filled tires??


Yes, there is only a 3% difference in the expansion rates between air and nitrogen. After all air is 78% nitrogen, so you'd expect them to behave very similarly.
 
Originally Posted By: Carbon
The water vapor level is important in how much the pressure in the tire drops with temperature drop.


I've done a little research to see if the presence of water in a closed chamber has a major effect on pressure. What I found was that the boiling point of water at 30 psi is 273°F - way, way higher than anything that could be experienced in normal operation. The only time you would experience that would be if there was something terribly wrong - like a fire or a long sustained brake application - and likely the brakes would fade before the tire could build up that amount of temperature.

Recognizing that normal pressure build in a passenger car tire is 3 psi and 5 psi is the absolute extreme, then that only represents a temperature build up of about 60°F - well below the boiling point.

Other than the fact that a supply of water inside a tire would keep the air fully saturated, I don't think this represents much of a pressure differential.

Other than that, the water vapor behaves very much like any other gas.
 
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Interesting topic. I don't know the answers, but I suspect it is not all that simple.

Think dew. Water can covert from vapor to liquid at temperatures below the boiling point.
Graph of water vapor pressure versus temperature. Note that at the normal boiling point of 100°C, the vapor pressure equals the standard atmospheric pressure of 760 Torr or 101.325 kPa.
220px-Water_vapor_pressure_graph.jpg


http://en.wikipedia.org/wiki/Vapor_pressure

I understand that the graph does not go to high enough pressures to correspond to tires, it seems clear that this is not going to act as an ideal gas. So at some amount of water in the tire, not all of the water is maintained as vapor. As vapor converts to liquid as the temperature drops, I expect a drop in pressure faster than the drop in the kelvin temperature.
 
Those chartts are basically for 100% water vapour, in a closed vessel.

When muckeing around with gas mixes, you have to work with the mole fraction of the gasses to find out how much that gas contributes to the whole.

Take the saturation curves below
1101.jpg


At 25C, each cubic metre of air (1.17Kg) can hold around 20g of water, or 1.7% by mass. Mole fraction is around 1:35, so water has around 1/35th of the total effect.

Assuming that the compressor managed to take all of the air/water, and pump it into your tyre with no water removal...

if the tyre were to go from 25C to 100C, the air would go from 30psi to 37.55 psi (assuming a rigid, fixed pressure tyre), and the water from about nothing to 14.7psi.

The tyre pressure at 100C would be 30 +(35x7.55+1x14.7)/36 = 30+7.83, or 37.83, a difference of around 3/4%
 
I am trying to follow the calculation.

At 25C =298K lets assume 30PSIG of ideal gas. That's 44.7 PSI absolute.

At 100C = 373K, I would expect 30*(373/298) = 55.95 PSI absolute.

That is 41.25 PSIG. That's less than your calculation that adds water to the mix.
 
Sorry,
was distracted and using guage for the tyre, and absolute for the water.

Good pick-up.

Makes the water effect even less.

30+(35x11.25 + 1x14.7)/36 = 30+11.34

41.34 wet, versus 41.25 dry.
 
I think Carbon has a point about the calculation.

But if I use a starting point of 30 psig water saturated air and compare that to 30 psig dry air:

1) I think that graph says the water vapor at 25ºC is 24 g making the percentage of water 2%
2) Normal pressure build up in a tire is 5 psi max.
3) That means the air chamber temperature would climb from 77ºF (25ºC) to 137ºF (58ºC) for dry air.
4) The 30 psig saturated air is 98% air = 29.1 psig and for the same temperature buildup would climb to 34.0 psig
5) And the additional water that the air could take on would be … oh… let’s call it 115 g, so now the saturated air is 1170g+115g = 1285g or 10% or 38.9 psig or 4.9 psi more.

Unfortunately, this graph is for atmospheric air (14.7 psia) and we are dealing with pressures 3 times that. Those pressures do not hold water vapor as well and not only do we start off with more air and less water, the affect of water vapor at the higher temperature will be less as well.

Does anyone have any graphs for saturated air at elevated pressures?
 
I feel that when the temperature drops, and the water vapor approaches the vapor point, there may be a larger pressure drop as some water converts to liquid or solid.

Effects do seem like they are going to be less than I had suspected.
 
Free water in a tyre would be way worse than water vapour in the air, I reckon, as there's a source to keep "feeding" the air as it warms up.
 
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