I am trying to calculate the energy needed to bring water:MEA mixture from room temperature to 90 degree Celcius and then the evaporation of the mixtures at 90 degree Celcius
In principle you could obtain the latent heat of vaporization of the mixture by Differential Scanning Calorimetry (DSC). But it is needed a specific experimental skill to obtain good results.
Hope to have helped you
Regards
Federico
It depends how precise and accurate a result you need. First of all realise that what will vaporize will not be the same as the liquid. At 90 deg C what will come off will be about 3% MEA, so what will be vaporizing is water. Thus to a first approximation, the enthalpy of vaporization is that of water.
This is a well-known system, so you can easily find papers on its vapor-liquid equilibrium.
You can find the result in terms of the Clausisus-Clapeyron (CC) equation in the excel file based on the data of "J. Chem. Eng. Data 2009, 54, 2312–2316" and on the assumprtons that :
a) 7:3 means 0.3 MEA molar fraction;
b) 0.274 is close enough to 0.3, the desired mole fraction
c) DeltavapH is constant in the temp interval 283-363 K
The CC eq is ln pv= a0 -DvapH/R * 1/T therefore it is easy to get the value you want from the slope of this equation. Since the regression parameters are quite good I don't think you need to consider the variation of the vap. enthalpy with T.
If you mean 7:3 in terms of wheight then you need to get more data to transform the data I used into w/w fractions. You can find that in "Journal of Chemical and Engineering Data, Vol. 48, No. 3, 2003".
If you need to calculation of DeltavapH exactly fot a 0.3 mixture than you nee to first interpolate for 0.3 from the data in the spreadsheet.
What do you mean by "and then the evaporation..."?
I Hope this is what you need regarding the first part of your question.
Thank you for your question...
Next time don't expect to get my answer!
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