# Basic boundary condition question

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 September 17, 2008, 15:04 Hi all, Given that the outw #1 Senior Member   Srinath Madhavan (a.k.a pUl|) Join Date: Mar 2009 Location: Edmonton, AB, Canada Posts: 698 Rep Power: 12 Hi all, Given that the outward drawn unit normal to inlet and outlet boundaries point parallel to the x-direction, are the following the only conditions that need to be satisfied at these boundaries? For a fixed velocity inlet, u = fixed value v = fixed value (dp/dx) = 0.0 For a fixed pressure outlet, p = fixed value du/dx = 0.0 dv/dx = 0.0 Where: In 2D: U = [u,v] (u and v are the two components of velocity vector U) and assume that 'd' represents the partial derivative symbol. I would appreciate if someone can point out if I am incorrect or missing anything here. Thanks for your help.

 September 18, 2008, 13:02 Anyone? Even a simple Yes or N #2 Senior Member   Srinath Madhavan (a.k.a pUl|) Join Date: Mar 2009 Location: Edmonton, AB, Canada Posts: 698 Rep Power: 12 Anyone? Even a simple Yes or No will suffice.

 September 18, 2008, 13:15 For incompressible laminar flo #3 Senior Member   Xiaofeng Liu Join Date: Mar 2009 Location: State College, PA, USA Posts: 118 Rep Power: 8 For incompressible laminar flow, Yes __________________ Xiaofeng Liu, Ph.D., P.E., Assistant Professor Department of Civil and Environmental Engineering Penn State University 223B Sackett Building University Park, PA 16802 Web: http://water.engr.psu.edu/liu/

 September 19, 2008, 10:59 Thanks Xiaofeng. #4 Senior Member   Srinath Madhavan (a.k.a pUl|) Join Date: Mar 2009 Location: Edmonton, AB, Canada Posts: 698 Rep Power: 12 Thanks Xiaofeng.

 October 1, 2008, 17:05 For a plain channel or pipe fl #5 Senior Member   Srinath Madhavan (a.k.a pUl|) Join Date: Mar 2009 Location: Edmonton, AB, Canada Posts: 698 Rep Power: 12 For a plain channel or pipe flow, dp/dx = 0 at the inlet boundary does not seem to make sense. The pressure should be allowed to change in the X-direction. Can anyone comment on that?

 October 1, 2008, 17:55 Hi Srinath Yes, I would lik #6 Senior Member   Niels Gjoel Jacobsen Join Date: Mar 2009 Location: Deltares, Delft, The Netherlands Posts: 1,607 Rep Power: 25 Hi Srinath Yes, I would like to give my two cents: Ideally you would say, that the only thing you know for sure is the pressure at the inlet and at the outlet, i.e. the energy gradient. Thus the intuitive way to set the boundary conditions would be to apply the pressure at both ends, which is equivalent of applying a body force. This solution converges asymptotically, thus far to slow to be interesting if you start with initial field 0 (unfortunately I cannot recall the reference on the asymptotic part). This would give a correct pressure gradient, i.e. constant, along your uniform channel. Another method is to apply the velocity at the inlet and a non-zero pressure gradient, but that implies that you _know_ the solution to the pressure field in advance, thus people have settled with the zero pressure gradient. I have read another article (again, cannot recall the exact reference), which did compare the results between a zeroGradient and the correct gradient solution, and the differences was small, at least from an engineer-point-of-view. Best regards, Niels __________________ Please note that I do not use the Friend-feature, so do not be offended, if I do not accept a request.

 October 1, 2008, 19:56 Thanks very much for that resp #7 Senior Member   Srinath Madhavan (a.k.a pUl|) Join Date: Mar 2009 Location: Edmonton, AB, Canada Posts: 698 Rep Power: 12 Thanks very much for that response Niels. You've brought some interesting facts to my attention. I'll try and dig up more information on this.

 October 14, 2008, 00:25 Hello People, I am have a s #8 Member   Mahendra Join Date: Mar 2009 Location: Pune, Maharashtra, India Posts: 65 Rep Power: 8 Hello People, I am have a small doubt regarding pressure boundary condition. In Fluent i have specified -1500 Pa gauge (negative) boundary condition at outlet. In OpenFOAM what boundary condition should i give. like this or something different? outlet { type fixedValue; value -1225; // (-1500/1.225) pressure/density since units of pressure are [0 2 -2 0 0 0 0]. } --

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