# species conservation problem

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 October 10, 2006, 17:47 species conservation problem #1 H Patel Guest   Posts: n/a Hi I am simulating a mixing tank with impeller in stady state.My flow domain invoves two species toluene and styrene.I dont have any inlet or outlet boundary condition. i have only wall boundary conditions.To specify initial condition( two seperate layer) i used initialise and patch panel to specify two separate layers for species.for example i created a cell register by selacting cells in top 15% volume. and patched value of 1 for tol in that register.i swithched off species equation and i solved flow equation for 1st order and then swithched on species equation and both species and flow equations are solved. My quation is when i used fluent's Report menu to calculate vol weighted avg. for mass fraction of toluene,It was showing 0.15% before solving but after solving it was showing 0.08%. Why individual species are not conserved? Any help will be appreciated. thanks

 October 12, 2006, 09:25 Re: species conservation problem #2 RoM Guest   Posts: n/a Define a custom function as "specie_mass_fraction*density" and calculate the volume intergral of this function. This is equal to the total mass of the specie in the domain and should be constant. RoM

 October 13, 2006, 10:37 Re: species conservation problem #3 H patel Guest   Posts: n/a thank you very much for your reply.My question is why vol weighted avg. mass fraction is not conserved.i will appreciate any help H Patel

 October 17, 2006, 04:37 Re: species conservation problem #4 RoM Guest   Posts: n/a Mass fraction is not conservative because its related to density and density will change druing mixing. Just take the following example: We have two volumes: Volume1 is 1m3 and filled with specie1 which has a density of 1kg/m3. Volume2 is 2m3 and filed with specie2 which has a density of 2kg/m3. The volume averaged mass fraction of specie 1 bevor mixing is 1/3. After mixing we have the same density in both volumes (5/3 kg/m3). So the new mass fraction of specie 1 is 1/5 for each volume and ofcourse volume averaged too. bevor mixing |-------------------------------- | V1=1m3, roh_1=1 kg/m3, x_1=1 | | x_2=0 | |-------------------------------- | | | V2=2m3, roh_2=2 kg/m3, x_1=0 | | x_2=1 | |-------------------------------| after mixing |-------------------------------- | roh=5/3 kg/m3 , x_1=1/5 | | x_2=4/5 | |-------------------------------- | | | roh=5/3 kg/m3 , x_1=1/5 | | x_2=4/5 | |-------------------------------| RoM

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