Gas Ratios
Gas Ratios
(OP)
Please forgive my ignorance on this subject but it is not my field. I have a question concerning Gas mixtures. I need a mixture of 2-1/2% Methane with 97-1/2% Nitrogen. Is this a simple ratio or do densities and such come into play? For example if I have 10 psi of Methane does this mean that I need 390 psi Nitrogen to equal 2-1/2% of 400 psi?
Thanks in advance
Ray Havermahl
Independent Engineering Labs
Thanks in advance
Ray Havermahl
Independent Engineering Labs





RE: Gas Ratios
It sounds like you are thinking along the lines of partial pressure, in which case you need the temperature of the gases too.
RE: Gas Ratios
Secondly, there are compressibility issues with gases at higher pressures where they no longer act purely as ideal gases. This can be avoided as a factor if, as I think you imply, you intend to make this mixture by filling the cylinder first with 10psi of methane (methane acts as an ideal gas at this pressure) and then pressuring it up to 400 psi with nitrogen (nitrogen still essentially acts as an ideal gas at this pressure). And to top the whole thing off, there may be some temperature effects to think about depending how accurate of a mixture you need.
[Interesting question for everyone. What drives the compressiblity factor for the methane in the mixture at 400 psi? The system pressure or the partial pressure of methane?]
Finally, how accurate of a blend do you need and how do you plan to make this blend (eg, flow control both streams into the cylinder, pressure the tank first with one component and then top up the tank with the other to the final 400 psi, etc?).
RE: Gas Ratios
As you've touched on.
A quick way to get mixtures(bottle not continuous flow) is to evacuate a cylinder or start with a cylinder that has been precharged, add N2 to a pressure of 97.5% of the final absolutute pressure, then add CH4 to increase the pressure to 100% again on an absolute pressure basis, letting the cylinder temperature equilibriate at each stage. You'll still need an assay to confirm and verify the final result, but it is pretty good.
To obtain weight basis, then you use a good scale. Weight and volume ratios not identical, but are convertable from one to the other.
At low pressure, the volume ratios are taken to be equal to the partial pressure ratios. It is pretty accurate for temperatures well above the critical temperature of the individual components,and one of the components is only present in small amounts (as in your case).
When you mix gases at high pressures, you have to develop the pseudo-critical properties of the mixture. There are several procedures developed for this. Some work better for certain gases (and mixtures) than others. Again, you will not see much departure from the ideal gas case (partial pressure ratios equal to the partial volume ratios) for samll mixing ratios.
RE: Gas Ratios
Again thanks.
Ray
RE: Gas Ratios
In response to your question;
The compressibility factor is a pure component or mixture property and not a partial property such as the vapor fugacity of methane in a mixture.
Greg
RE: Gas Ratios
I would like to expand on my previous remark after reviewing my thermo book.
The partial compressibility factor is the partial derivative of Z wrt n1 keeping T & n2 constant.
If we use the simplest form of the virial equation to calculate Z (i.e. Z=1+BP/RT), the partial compressibility factor is then equal to;
1+P/RT*(B11+y2^2*D12)
where
D12 = 2B12-B11-B22
B11,B12&B22 are virial coefficients
B=y1^2*B11+2y1*y2*B12+y2^2*B22
yi=mole fraction of component i
The Virial coefficients are functions of T only and therefore at constant T the above equation is equal to:
1 + constant1*P+ constant2*P*(y2)^2
Therefore as y2 approaches 0 (i.e. almost pure methane), the partial compressibility will depend on the system pressure.
Greg