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June 11, 2020, 06:47 
Partial pressure of a species

#1 
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AnneSophie
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Hi all,
I wonder how I can get the partial pressure of a species to use it in a UDF? Is there a macro available to do this? Thanks for the advice! Kind regards, AnneSophue 

June 11, 2020, 09:20 

#2 
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Khan
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for ideal gas,
density = (PM/RT). M is molecular weight. P = (density*RT)/M I guess the right hand side in above equation when multiplied by mass fraction of a species would give partial pressure of that species. species mass fraction is accessible in Fluent. Check Fluent manual on the particular macro available for the purpose. 

June 15, 2020, 05:01 
Partial Pressure

#3 
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Partial pressure of a species is determined using pressure in the domain and mole fractions. Do note that partial pressure is affected by moles and not by mass. So, you have to convert mass fractions to mole fractions.
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June 15, 2020, 09:37 

#4 
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Khan
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mass divided by molecular weight gives molar quantities.


June 15, 2020, 09:53 
Solution?

#5 
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AnneSophie
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Dear Khan and Vinerm,
Thank you so much for your help! This is what I conclude: molar mass of the mixture based on ideal gas law (with R=universal gass constant): real mm_mix=(C_R(c,t)*R*C_T(c,t))/C_P(c,t); mole fraction of water (with mm_h2o= molar mass of h2o): real molfr_h2o=C_YI(c,t,0)*(mm_mix/mm_h2o); partial pressure of h2o: real pw=molfr_h2o*C_P(c,t); Does that seem ok? 

June 15, 2020, 10:30 

#6 
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Khan
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I request you to verify this before using it, through simple example hand calculations:
Partial pressure of species 'i' in an ideal gas mixture = (density_mixture*R*Temp_mixture*mass fraction_species_i)/(Molecular weight of species_i) 

June 15, 2020, 11:01 
Partial pressure

#7 
Senior Member

If you have a binary mixture, then you can calculate it as
where is mass fraction of species named 2 and P is absolute pressure. This will return partial pressure for the species 2. M denotes molecular weight.
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June 15, 2020, 11:26 

#8  
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AnneSophie
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Quote:
Hi Khan, Thanks, but I don't really understand where this equation comes from. Could you explain this? And also, do you mean my conclusion was incorrect? Thanks for your time! 

June 15, 2020, 11:29 

#9  
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AnneSophie
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Quote:
Hi Vinerm, Thanks for this usefull equation. Unfortunately, my mixture contains CO2, h2o and air, so I have more than 2 components. In which way could I adapt the equation? And also, do you think my conclusion from the previous reply was incorrect? Thanks for your time! 

June 15, 2020, 11:49 
Partial Pressure

#10 
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Then it won't work.
Yes, your calculation is correct. It would work. You can use your calculation for binary and you will find that it fits the equation I mentioned.
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June 15, 2020, 11:50 

#11 
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AnneSophie
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Ok, thank you very much for all the help, Vinerm!


June 15, 2020, 21:17 

#12  
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Khan
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Quote:
If you plug in values of first and second equations of yours into third one, what you get is my equation. So your conclusion is correct...! (caution: For the sake of confidence in your simulation, I urge you to verify this equation by simple examples) (mixture density * mass fraction of i) = mass of i / mixture volume; divide right hand side by molecular mass of i and you get (moles of i / mixture volume). multiply by RT (R is J/mol.K of mixture = Pa*m3/mol.K of mixture) and you get (moles of i / moles of mixture)* mixture pressure which is nothing but partial pressure of i. Regards. 

June 16, 2020, 03:42 

#13  
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AnneSophie
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Quote:
Yes, now I can see it too. Thank you so much for the help, Khan! 

Tags 
macro, partial pressure, porous media, transient 2d 
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