Volumetric Ratio Calculation

[MMcCawley]

Hi,

I'm trying to determine the proper equation required to calculate volumetric ratios between compressed, liquified gases and the same gas at atmospheric pressure. For example, let's assume that I have 10cc of liquified CO2 in a pressure vessel at room temperature ( Vapor pressure = 849 psig @ R.T.) and I release it to the atmosphere (also at room temperature). What volume of gas would I have at atmospheric pressure? Any assistance with the proper equation would be greatly appreciated.

Thanks

 

[BigInch]

Convert your volume of CO2 to mass of liquid CO2, find its density when at vessel conditions from NIST table for CO2 liquid range. Convert that same mass of liquid to its gas volume at whatever temperature and pressure condition you want PV = ZRT, or again see the same tables in the gas range.

"People will work for you with blood and sweat and tears if they work for what they believe in......" - Simon Sinek

 

[zdas04]

I would approach it a tiny bit differently. I'd get the mass. Then calculate the theoretical density at room temperature and atmospheric pressure. Divide the mass by the density at room temperature and local atmospheric pressure.

David Simpson, PE
MuleShoe Engineering

"Belief" is the acceptance of an hypotheses in the absence of data.
"Prejudice" is having an opinion not supported by the preponderance of the data.
"Knowledge" is only found through the accumulation and analysis of data.

 

[MMcCawley]

BigInch-
CO2 Liquid Density @ 20degC: .77g/cc
CO2 Molecular Weight: 44.01g/mol
Assume 1cc compressed liquid volume to keep things simple

Mass=DV=(.77g/cc)(1cc)=.77g
n=Mass/Molecular Weight=(.77g/44.01g/mol)=.0175mol
V=nRT/P=(.0175mol)(82.057cm^3atm/Kmol)(293.15K)/1atm=420.9cm^3

Does this look correct? The reason I ask is that its approximately half of what is listed on AirLiquide's website HERE so I'm trying to figure out why there's a discrepancy.

 

[MMcCawley]

zdas04-
CO2 Liquid Density @ 20degC: .77g/cc
Assume 1cc compressed liquid volume to keep things simple

Mass=DV=(.77g/cc)(1cc)=.77g

Gas Density @ 1atm, 20degC: .0018394g/cc

V=M/D=.77g/.0018394g/cc=418.6cc (basically the same as above).

So I'm assuming the data on Air Liquide's website is incorrect or could I be failing to account for something?

 

[zdas04]

In RefProp if I put in 20C and 59 bara I get that it is a subcooled liquid with a density of 779.39 kg/m^3. So with 1 cc you have 0.838 gm.

RefProp says that at 20C and 1 bara the fluid is superheated and the density is 1.8152 kg/m^3. Dividing mass by density gives me 429 cm^3. I think you gave Air Liquide an incorrect parameter.

David Simpson, PE
MuleShoe Engineering

"Belief" is the acceptance of an hypotheses in the absence of data.
"Prejudice" is having an opinion not supported by the preponderance of the data.
"Knowledge" is only found through the accumulation and analysis of data.

 

[Montemayor]

I have done this type of calculation many times in the past and, like David, always resorted to simple, direct available thermo data. Please refer to the attached Excel Workbook for the explanation and calculation of the coresponding gas volume to 10 cm[sup]3[/sup] of liquid CO[sub]2[/sub].

I also have found various mistakes in the AirLiquide compressed gases website some years back, so I wouldn't be surprised if their algorithm for detrmining the gas/liquid ratio was inaccurate.

The main point I would draw attention to in this thread is that the liquid CO[sub]2[/sub] - as defined by McCawley (the OP)- is definitely in the saturated state. This is the normal way you would find liquid CO[sub]2[/sub] if it is pure. And the OP has stated that it is CO[sub]2[/sub], so we must assume that it has pure properties. The OP has failed to state this fact, as well as identifying his atmospheric pressure and so-called "room" temperature. But note that since the liquid CO[sub]2[/sub] has to be saturated at the stated vapor pressure, this then identifies what the OP's "room" temperature is: 21.6 [sup]o[/sup]C.

The calculation is simple; what is harder is understanding the phase equilibria taking place when one has a pure, liquefied gas. There is no supercooling.