Mass Balance #3
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Hi Patil, Let's say that you start with 5 CD and 10 MEA (initial concentrations). Your mass conservation has to keep this into account. Let's try with the following: The issue here is that the total number of OZD and WATER (OZD + WATER) molecules produced in your reaction 3 is going to be double of whatever amount of CD (or MEA) was consumed to form them. Including the other reactions (1 and 2), I would then set the following mass conservations: We can now rewrite this problem in a form suitable for equipy: species_names = {'CD':0, 'MEA':1, 'CBNAC':2, 'CBMAC':3, 'OZD':4, 'WATER':5} Theoretically, since OZD and WATER are not taking part in any other reaction and you know that they are the same at equilibrium, you could make this problem more digestible, either using a placeholder instead of (OZD + WATER) or (they are formally the same) ignoring one of the two. This would lead to your second set of mass conservation equations. Please let me know if this makes sense to you, I have quickly gone through it so I may have misunderstood or missed something. Best, |
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Hi, there!
Thanks for such a nice implementation.
I take this opportunity to know more about mass Balance equations to be fed to the program. E.g.,
Consider the following system
CD + MEA = CBNAC
CD + MEA = CBMAC
CD + MEA = OZD + Water
What will be the mass Balance equations to be used here?
Using,
CD + CBNAC+CBMAC
MEA+CBNAC+CBMAC
Seems to give better in comparison when using
CD + CBNAC+CBMAC+OZD
MEA+CBNAC+CBMAC+OZD
Or
CD + CBNAC+CBMAC
MEA+CBNAC+CBMAC
OZD-WATER
It seems that I am failing to understand the logic
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