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Temperature affect on propane meter

Post questions and info about combustion (flammable vapor) powered cannons here. This includes discussion about fuels, ratios, ignition systems, safety, and anything else relevant.
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Temperature affect on propane meter

Unread postAuthor: homedepotpro » Wed Nov 21, 2007 11:58 pm

As many of you know the weather is changing and the air is getting colder. I would still like to shoot my advance comb., but im not sure if/how to account the the temperature's affect on the propane. My logic says lower temp, lower vapor pressure, that means less psi it the meter to get the same amount of gas, but i don't really know. I was wondering if i could get some help, im looking for a table, some equations or maybe just some educated estimates.

thanks
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Unread postAuthor: hi » Thu Nov 22, 2007 12:05 am

thats true, but you are forgetting something, the air it self...

the air will condense as well as the propane, so you wont have to change anything. thats my theory...
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Unread postAuthor: SpudBlaster15 » Thu Nov 22, 2007 12:43 am

As hi said, the density of both atmospheres (Chamber and meter) will rise equally with a fall in temperature. Your air/fuel proportions will not alter, assuming you use the same meter pressure as you did prior to winter.
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Unread postAuthor: homedepotpro » Thu Nov 22, 2007 12:44 am

yeah hi, but im pretty sure they don't condense at the same rate. at 0*F propane's vapor pressure is 11psi and oxygen is, idk, way the hell up there
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Unread postAuthor: SpudBlaster15 » Thu Nov 22, 2007 2:14 am

Sigh... I could show you this in a much simpler sequence of calculations (canceling out variables that divide to become the number 1), but with comments like "My logic says lower temp, lower vapor pressure, that means less psi it the meter to get the same amount of gas" I doubt you would understand unless I put it into real world perspective.

The ideal gas law: PV = nRT. P = pressure, V = volume, n = number of moles of the gas, R = the gas constant (8.314 472), T = temperature.

For ease of calculation, we will assume that the meter volume is 0.0001m<sup>3</sup>, and the chamber volume is 0.01m<sup>3</sup>. The chamber pressure will be assumed to be constant at absolute atmospheric level, about 101,352.932 pascals, and the meter pressure will be assumed to be 300000.00000 pascals, about 43.5psia. Temperature will be the only variable.

First, rearrange to n = PV/RT

We will assume a temperature of 300 kelvin, probably common summer shooting conditions. (It doesn't get that warm around here too often, but oh well)

n<sub>chamber</sub> = ((101352.932)(0.010000))/((8.314472)(300.00000))

n<sub>meter</sub> = ((300000.00000)(0.000100))/((8.314472)(300.00000))

This is a ratio of (((101352.932)(0.010000))/((8.314472)(300.00000)))/(((300000.00000)(0.000100))/((8.314472)(300.00000))) = 33.7843:1

Now we will decrease the temperature to 272 kelvin, or 0*C.

n<sub>chamber</sub> = ((101352.932)(0.010000))/((8.314472)(272.00000))

n<sub>meter</sub> = ((300000.00000)(0.000100))/((8.314472)(272.00000))

This is a ratio of (((101352.932)(0.010000))/((8.314472)(272.00000)))/(((300000.00000)(0.000100))/((8.314472)(272.00000))) = 33.7843:1

The ratios are identical, meaning that temperature will effect all gases in an identical manner.

And this is exactly why we don't overcomplicate meter pressure calculations by using PV = nRT.
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Unread postAuthor: psycix » Thu Nov 22, 2007 9:10 am

Its still the same 4.2%, but yes, there is more fuel in your meter, aswell, more oxygen in your chamber.
Thats why combustion cannons are said to have more power in cold weather (but your pvc gets brittle!)
When temperature get VERY cold, you might have your propane turning liquid when exceeding the vapor pressure.
Then you should have OR a bigger meter, or measure two shots with only half amount of the fuel (and thus pressure).
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Unread postAuthor: homedepotpro » Thu Nov 22, 2007 10:41 am

thank you spudblaster, you are a very knowledgeable spudgunner.
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