# outflow boundary in a steady state 1D convection diffusion eq.

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 September 30, 2012, 03:52 outflow boundary in a steady state 1D convection diffusion eq. #1 New Member   hans Join Date: Sep 2012 Posts: 7 Rep Power: 4 Hi all, I've just plunged myself into cfd, so quite new to the game. I've started with writing, what i assumed to be a simple, code for an advection diffusion equation. Im not sure what to do with an outflow boundary, or at least how to understand the results. Im looking at the steady state solution of a 1D problem. I want to have an outflow where a concentration can pass though, like say the open end of a pipe. For this i have used an end boundary of dC/dx=0 (correct if this is wrong). At the start of the pipe (x=0) i have a set value for the boundary, say C=0.1. If i have a positive convective term (or 0), from start towards the end of the pipe, the steady state solution gives a concentration equal to the set cocentration, 0.1 (regardless of the value of the diffusive term). If i use a negative convetive term the outlet will beome an inlet and i assume will have a set value of 0 (i'm not so sure about this one) So the outlet boundary can only take on 2 values (in the steady state solutiuon) 0.1 or 0, regardsless of the diffusion term. I'm not sure how to proceed, is the solution correct and are these the only 2 states the outflow boundary can be in? seems logical that the diffusive term should have some role to play and that the outlow boundary can have a range of concentrations, or not? All help, or comments are welcome.

 October 9, 2012, 06:57 #2 Senior Member   Rami Ben-Zvi Join Date: Mar 2009 Posts: 148 Rep Power: 8 If I correctly understand your problem (steady, 1D, and I also assumed constant coefficients and no source, with Dirichlet BC at x=0 and homogeneous Neumann BC at x=L), the equation has the trivial solution C(x) = C(x=0). Am I missing something?

 October 14, 2012, 17:39 #3 New Member   hans Join Date: Sep 2012 Posts: 7 Rep Power: 4 So it seems. After thinking on it for a while, seems logical. I was expecting another solution to be possible, so it took me a while to fully understand. Taanks for your reply

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