In another thread, a poster asked about the how temperature affects tire pressure and was initially advised that there was not a temperature pressure type thing. In fact, Boyle’s Law does quantify the relationship between the temperature of a gas and the pressure of that gas when the volume remains constant (inside a tire).
Are any of our engineer/physicist type posters familiar enough with this law of physics to give us a reasonably accurate approximation for the change in tire pressure in psi in a road tire for every 10 degrees Fahrenheit of temperature change?
You are right and I stand corrected. Thanks. Just goes to show you what a broad base of knowledge is present on this forum. This is a compliment to all of you.
I am not smart enough to speak fully on this subject, but it has been played with a bit. Before a TT a few years back, Jan Ullrich had tyre warmers on his bike tyres before a TdF time trial. I don’t know how well it helped.
Some have it correct above but volume doesn’t matter if we assume that there is no change in volume during heating.
PV=nRT (as others have said). Or for two temperatures
P1V1=nRT1 and P2V2=nRT2
Assume constant volume so V1=V2 giving (nRT1)/P1 = (nRT2)/P2
cancel to get T1/P1 = T2/P2 solve for P2
P2 = (T2/T1)*P1
Lets say it is 60F and you pump your tires to 120psi in the morning and the temp goes up to 80F.
Note : Convert the temperature to Kelin
P2 = (300/289)*120 = 124psi
4 psi - not enough to blow off your tire
If the temp goes up to 100F we get
P2 = (321/289)*120 = 133psi
13 psi, could be a problem, not sure.
I could have screwed up somewear which is a possibility b/c my 6mo old was up at 4:15a and I took thermo 8yrs ago and I need to get to bed.
Andrew
So you all,what does that all mean for us idiots! Always told that the having 120psi and the heat caused your tires to blow was bull. (even without the laws given I thought the tolerance would not allow that). So to put it simply…Does a hot day make a difference?
Yeah, it is P1/T1 = P2/T2 and the temps are supposed to be in degrees Kelvin. In 1993, I was racing at World’s Toughest Tri in Lake Tahoe (2mile/100 mile/20 mile). Morning temps were 39 F or around 277K or so. Mid day, it was supposed to go up to ~80F, so around 297K. Based on this tire pressure would go up by around 8%. The RD told us that in years past, guys pumped up to over 160 psi at 7 am but their tired blew later in the day when temps went into the 80’s and tire pressure went up to the 175+ psi range, which is above the safe limit for many conti tubulars. You would not want to blow a tubular descending Monitor pass at 50+ mph. Add the additional heat from braking and you could see bad Beloki like outcomes. Fortunately, most competitors went with the much lower pressure given bumpy pavement along the course and no one had to blown tubulars due to increased pressure from heat during the day.
There were many other blowups on the course, but they were due to lack of respect/nutrition and the 8K ft of climbing on the bike or the 3K ft climbing on the run.
Thanks for the equation and the solution. I went looking for my Handbook of Chemistry and Physics but couldn’t find it. Turns out my son was using it to elevate his PC monitor. Pretty expensive phone book though.
I expected the relationship to be linear though. 60 degree to 80 degree change (20 degrees) equals 4 psi increase in pressure. 60 degree to 100 degree change (40 degrees) would equal 2 X 4 psi (20 degree change) would equal 8 psi rather than 13 psi.
Is this an incorrect expectation?
Young_Ironman, for the purpose of the solution of this problem, air will behave enough like an ideal gas to be considered the same.
The volume of the container (your tire) DOES increase, and the pressure is quite high and would then interact amonst itself. Ideal Gas Law does not then reliably predict results. I’ve measured a temperature difference of well over 100 degrees F with a pressure increase of less than 15 psi. Albeit not scientifically, this illustrates the temperature can be neglected
I don’t think temperature plays a roll in significant pressure differences at all. Unless you air up in sub zero temps and ride through an active lava field.
I expected the relationship to be linear though. 60 degree to 80 degree change (20 degrees) equals 4 psi increase in pressure. 60 degree to 100 degree change (40 degrees) would equal 2 X 4 psi (20 degree change) would equal 8 psi rather than 13 psi.
Is this an incorrect expectation?
You are correct, I had an error in my temp conversion
100F =~ 311K (not 321 that I had above)
This gives about P2=129 or 9psi (2x4.5).
Once your tire is inflated to 75-80 percent of working pressure, the volume changes as a result of increasing pressure are so small as to be considered insignificant in this situation. Equally so, at the modest pressures of a bicycle tire, air will behave enough like an ideal gas to be considered the same in this situation.
In fact, Boyle’s Law does quantify the relationship between the temperature of a gas and the pressure of that gas when the volume remains constant (inside a tire).
Actually, Boyle’s law states that the pressure and volume are inversly proportional at a constant temperature.