a question for you mathmaticians...

[billybob_81067]

How many ounces of liquid CO2 will it take to fill a standard 15 inch tire? I don't know the volume of a tire, or the expansion rate of CO2, so if you know that or where to find it that'd be cool

[Target Practice]

I would use the Shell Method for finding Volumes of Revolution. This would give you the volume of the tire at what ever your inside radius would be...it's flexible. It is labor intensive, and requires more than the average bear's level of calculus. But hell, it sure can impress people! Any body higher in math want to do this? I got a Taylor/Maclaurin Series test coming up I have to study for...

Newton, L'Hopital, Leibniz...Oh, My!

[johnny_chase]

well, if i remember my calculus of last sememster...
post the inside and outside radius of the tire, i'll work out an approximate number.

btw, taylor polynomials are the most fun thing in the world![sarcasm...]

have you done taylor's inequality yet?

[vf-xx]

Depends on how accurate you need to be. If you're just looking for a rough estimate just find the volume of a cylinder equal to the outer radius, then subract a cylindrical section for the hub. After taking that volume, use material propery sheets to figure out the volume of one ounce of Co2. Not too hard really.


Example: Assume the following:
Mag rims with touring tires (ie the closest you can get to true cylindrical tires)

the outer radius of the tire (r0) is 15.in and has a wall thickness of .5.in.

the radius of the rim (r2) is 11.in

The width (h) of the tire is 8.in with a sidewall thickness of .25.in

Soln:

r0-.5 = r1 = 14.5.in
pi*(r1)^2 = a1 = 660.52 in^2
pi*(r2)^2 = a2 = 380.13 in^2
a1 - a2 = a3 = 280.39 in^2

a3*(h-2*.25) = V = 2102.9 in^3

Then just figure out how much volume one ounce of liquid co2 fills and divide that into the total volume to get ounces

V/(voume of one oz of co2)

[billybob_81067]

Approximation

Well, I don't need to get really THAT accurate. I just want to know approximately how many tire fills I could get out of my 12 oz co2 tank, AKA my "portable air compressor."

I used my pickup tire for some measurements just a bit ago and it's approximately 31" tall and 9.5" wide, so minus about a half inch tread thickness and quarter inch sidewall thickness would make the internal dimensions 30" and 9" on a 16" rim.

So therefore it would be [pi*15^2 minus pi*8^2]*9. This works out to [706.5 - 200.96]*9 or 4549.86 cubic inches.

Now that seems like a really large amount, but I dunno! Am I overlooking something here?

Also if anyone knows where to get the expansion info for the CO2 that'd be awesome. Oh and lets say the tire needs to be filled up to 45 psi.

[vf-xx]

hrm.... Well how about this.

The density of Co2 is about 1.977 (taken from google search)

Estimate that your tank is about 850 psi

You're shooting for 45 psi final (which seems a bit high IMO for a car, but whatever....).

Assume Ideal gas conditions (big assumption but good for estimates).

P*V = n*R*T

P = pressure
V = volume
R = Gas Constant
T = Tempurature
n = moles of Gas

Give yourself a standard temperature (average outside temperature).

You already have a tire volume and you can calculate the Co2 tank volume.

Then you just calculate the required moles to fill the tire to the desired pressure and compair that to the moles in the co2 tank.


Edit: That tire volume sounds reasonable. It takes a LOT of air to fill one up. I wouldn't expect too much. A 12 gram co2 cylinder can fill maybe two street bike tires to around 100psi. Those tires have MUCH less volume than a car tire.

[billybob_81067]

Ok, so using your P*V=n*R*T formula what goes in for the R on both the co2 tank formula and tire formula? Is that where the co2 density number goes?

Tire:

45psi*4550cu. in. = n*R*80
=
204,750 = 80nR
=
nR = 2560


Tank:
850psi*22cu. in. = n*R*80
=
18,700 = 80nR
=
nR = 234


I think and hope that I did this right. Now I just need to know what to put in for the R! Oh and 45psi is my way of kinda averaging between passenger car tires and like my pickup tires. I usually run 30-35psi in car tires and I run 60psi in my pickup fronts and 80psi in the rears.

[vf-xx]

R is a gas constant. I believe it's universal, but I don't remember for sure. I"m about dead for the night, try stealing a look at a theromdynamics or a Chemistry book.

[bjjb99]

VF-XX is on the right track. However, the CO2 in the tank is not entirely gas and not entirely liquid at ambient temperatures. The two have completely different densities. You can't use the volume of the tank to determine the amount of CO2 present without including some means of figuring out the ratio of gas to liquid in the tank.

However, vf-xx touched on the solution. Regardless of how
much CO2 is liquid and how much is gas, there are the same number of molecules of CO2 present.

Those molecules have mass. How do we fill CO2 tanks? By weight (mass).

Fair warning, I'm going metric on you folks:

Weight of CO2 in tank: 12 ounces
Mass of CO2 in tank: 341 grams

You can use the molecular weight of CO2 to compute the number of moles of gas present.

Molecular weight of CO2: 44 grams/gram-mole
Amount of CO2 in tank: (341 grams) / (44 grams/gram-mole) - 7.75 gram-moles of CO2

Now we can start looking at PV=nRT

P = tire pressure = 45 psi = 411504 Pascals
n = amount of CO2 = 7.75 gram-moles
T = gas temperature = about 80F = about 300 kelvin
R = universal gas constant = 8.314 Joules/gram-mole/kelvin

V = n * R * T / P
V = 7.75 * 8.314 * 300 / 411504
V = 0.0470 cubic meters

So the volume that the CO2 will occupy at 45 psi and at about 80 degrees F is right around 0.047 cubic meters.

From billybob's post, the tire volume is about 4550 cubic inches. That's 0.0746 cubic meters, which is more volume than the CO2 will occupy at 45 psi.

You will get about 0.63 tire fills from a 12 ounce tank of CO2. You'd need a 20 ounce tank to fill the tire completely to 45 psi.

BJJB

[dudedude]

Wouldn't there be the small problem of the CO2-filled tire heating up when driving - followed by tire going "pop", the car going "screech", and you going "aaaaaaaaaaaaaaaaaaaahhh"?

[bjjb99]

Probably not, actually. The reason you have blowouts in CO2 tanks when they get warm is because the CO2 is stored in the form of a gas and in the form of a liquid, the proportions of each being determined by pressure and temperature. When the temperature goes up, the amount of gas increases and the amount of liquid decreases. Since CO2 gas wants to take up much more space than the same mass of CO2 liquid, the pressure goes up in the tank. Eventually there's no liquid in the tank, leaving just a gas under high pressure. If that pressure is too high, then the burst disk opens and vents the tank.

With the car tire, there is very, very little liquid CO2 present at all. A pressure of 45psi just isn't high enough to force much CO2 into liquid form at room-temperature. For the car tire, you can consider all the CO2 to be in the form of a gas, and it will behave pretty much like an ideal gas under everyday driving conditions. If the temperature goes up a couple of degrees then the gas pressure will go up by about 1 percent.

No "pop", "screech", or "aaaaaaaaaaaaaaaaaaaahhh".

BJJB

[billybob_81067]

Originally posted by bjjb99


You will get about 0.63 tire fills from a 12 ounce tank of CO2. You'd need a 20 ounce tank to fill the tire completely to 45 psi.

Sweet... and seeing as how this was calculated for a a pickup tire on a 16 inch rim, then there would be no problem at all with filling up a regular passenger car tire to 30-35 psi with 12 oz co2... I got myself a nice little air tank for those roadside emergencies!

[bjjb99]

Keep in mind that rapidly depressurizing the CO2 tank will cause it to chill down very quickly. If it gets too cold it might start sputtering liquid CO2 into your tire. Now, that liquid CO2 won't stay liquid for long once it hits the inner walls of the tire, but in the time it takes to boil off it might make the spot where it was in contact with the tire a bit brittle. Fill slowly, and use a long coil of remote hose to give the CO2 time to warm up nicely.

BJJB

[vf-xx]

Thanks bjjb99. I was getting there eventually, but I never posted while fully awake. Stupid night job.....

hehe.