kcme2005
Mechanical
- Sep 8, 2005
- 16
We are trying to calculate the vapor pressure of an Ethylene Gylcol/Water mixture.
We are using pure ethylene glycol (no additives), and our mixture is 60% EG and 40% H2O.
The Critical Pressure and Temperature of EG is:
Pressure - 1116.8 psia
Temperature - 733.3°F
The Critical Pressure and Temeprature of H2O is:
Pressure - 3199.2 psia
Temperature - 705.2°F
We want to know what the Critical Pressure and Temperature of the mixture will be. Does anyone know how to go about doing this?
We used Raoults Law assuming that the mixture is an ideal mixture, but we did not get the results we were expecting.
See our calculations below:
EG
Molecular Weight - 62.07 g/mol
Density - 1.11 g/cc
Boiling Point - 387.7°F
H2O
Molecular Weight - 18.02 g/mol
Density - 1.00 g/cc
At a volume of 10cc
Weight EG = 6cc*1.11 g/cc = 6.66g
Weight H2O = 4cc*1.00 g/cc = 4.0g
Number of Moles
EG = 6.66g/62.07g/mol = 0.1073 moles
H2O = 4.00g/18.02g/mol = 0.2220 moles
Mole Fractions
EG = 0.1073mol/0.3293mol = .326
H2O = 0.2220mol/0.3293mol = .674
Vapor Pressure of EG at Critical Temp of H2O
Assuming vapor pressure of EG is linear.
EG = (705.2°F-387.7°F)/(733.3°F-387.7°F)*1116.8psia = 1026psia
Combined Vapor Pressure
EG = 0.326*1026psia = 334psia
H2O = 0.674*3199.2psia = 2156psia
EG + H2O = 334 + 2156 = 2490 psia @ 705.2°F
Any help would be greatly appreciated.