6 replies to this topic

[trainspotter124]

I've come across a problem in a research paper where they've calculated the equilibrium state for the adiabatic mixing of 108kg of air at 0degC & humidity of 100%, and 115kg of methanol at 14degC. The answer they've come up with (without showing the methodology) is 3.5%w/w in the gas phase - I'm trying to figure out how they've come to this answer.

 

Trying to get approximately the same answer, my working as follows:

 

Vapour pressure of methanol at 14degC = 0.0931 bar

 

Molar concentration = 0.0931 bar/1.01325 bar = 9%mol

 

Molar mass of air = 29 g/mol

 

Molar mass of methanol = 32 g/mol

 

9%mol * 32/29 = 10%w/w

 

This is nowhere near the 3.5%w/w they have calculated?

 

Also tried at 0degC for methanol (assuming the air cools the methanol to 0degC) and using the same method above, I get slightly closer to the answer of 4.1%w/w - any ideas?

 

Thanks

[latexman]

First, do the adiabatic mixing calculation to determine the total mixture temperature.  It's not 14^oC; it's between 0^oC and 14^oC.  My guess is, it's about 9^oC.  The water/MeOH non-idealities may affect the results.  Try using activity coefficients.

[Pilesar]

Humidity of 100% means water is in your system. Did you account for that? The resulting system may be non-ideal in the liquid phase.

[MrShorty]

I'm not sure there's enough water in the system to substantially impact the partial pressure/activity coefficient of methanol. Certainly, if you want the calculations to be rigorous, you would need to consider the liquid phase activity coefficients of methanol and water. For quick guestimates, it seems to me that you can ignore non-idealities between methanol and water in the liquid.

 

What I notice is that your "problem statement" doesn't mention what pressure they used for the calculations, but your calculations all seem to assume a pressure of 1 atm. If the system pressure is higher (say about 2 atm), that would put you in the ballpark where P(MeOH)/P(system) is about 3.5%

 

Of course, if they did specify 1 atm pressure and you just didn't carry that over into the problem statement, then this is not the answer. I would double check and make sure that 1 atm pressure is the actual pressure being used for the calculations.

[PaoloPemi]

even with the regular / ideal model (as for your calc's) you can observe a relatively large amount of CH4O which evaporates and the final temperature decreases as suggested by others...

for example mixing a Stream 40 flow 108 Kg/s  and molar fractions H2O 0.001, N2  0.7992 N2 0.1998 Kg/s) @ 0 C, 1 Atm  and stream 41 flow 115 Kg/s (CH4O) @ 14 C, 1 Atm the predicted temperature is about -0.5 C and the CH4O fraction (in vapor phase) is not far from suggested value (see attached image)

Attached Files

•  mixer-air-methanol.jpg   99.94KB   0 downloads

Edited by PaoloPemi, 05 October 2023 - 02:28 AM.

[latexman]

My guess was off, so I cranked up Aspen+.  I selected NRTL with binary interactions (it's so easy, so why not).  1 atm assumption.  Results obtained are adiabatic mixed temperature = -2.2^oC, and MeOH vapor concentration = 3.7% w/w.

 

Very similar to PaoloPemi.

 

 Capture.JPG   56.29KB   1 downloads

 Capture 2.JPG   48.23KB   0 downloads

[trainspotter124]

Ah great, thanks both