10-07-2012, 06:04 PM
Join Date: Feb 2008
My Ride: M3 & ZHP
No one said that it needs to be 20°C outside for the pressure to be 1 atm. It's just a common temperature and easy to use for calculations. You made an assumption about the number of moles while WDE46 and I instead chose to make an assumption about temperature. Also, 1 mole is only 22.4L at STP. So your math essentially incorporated that assumption. That's why your numbers are similar to ours. They'd probably be even closer (or exact, I didn't check) if we used 25°C
Originally Posted by SeanC
Apparently my posts are not long enough for some people to understand. lol.
Let me put this into context for those who are still using 2 sets of P and T or having difficulty understanding what the law says. No need to start making assumptions once you have the temperature reading. For example, no need to assume 1 atm at 20 degrees, as the atmospheric pressure depends mostly on the altitude, not temperature. You may have a situation where you fill up the system at -40 degrees near the sea level, in which case, atmospheric pressure will still be 1 atm.
P is unknown.
V is arbitrary, assume a realistic starting volume, let's say 0.5 L. It will cancel out regardless.
n is the amount of substance in moles.
R is the gas constant = 8.314 J/(K*mol), meaning it takes 8.314 Joules of energy to raise 1 mole of gas's temperature by 1 degrees Kelvin.
T is the temperature reading from the coolant temperature sensor (thanks for the correction on the location). Assume 96 degrees Celcius = 369.15 Kelvin (the latter is the unit used in the equation).
We also know that 1 mole of air occupies 22.4 L of air. 0.5 L air will have 0.02223 moles of air in it (n=0.5/22.4).
P = (0.5/22.4) * 8.314 * 369.15 / 0.5 = 137.014 Joules/L
Using 1 L = 1 dm^3 = 0.001 m^3, we have:
P = 137.014 J / 0.001 m^3 = 137,014 Pa = 137.014 kPa = 1.37014 bar...
This is the pressure of the air in the expansion tank without taking into account the expansion of the coolant.
Assuming the coolant will expand so as to fill half of the empty space in the ET, it follows, from P1/V1 = P2/V2 (special form of the ideal gas law that is valid after the maximum operating temperature is reached, after this point on, I assume T will be constant at 96 degrees Celcius), that this pressure will double to 2.74 bars...
If one does not allow the necessary space for the coolant to expand, it again follows, from P1/V1 = P2/V2 that arbitrarily large pressure can be reached. Of course, the pressure will be relieved whenever it reaches 3 bars as long as there is air left in the system, but if there is not air left, the the cap or ET will give up at some point due to high pressures of the liquid pushing against the cap.
I hope the math, and my conclusions, are clear now.
Last edited by TerraPhantm; 10-07-2012 at 06:08 PM.