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January 24, 2004, 09:51 
Output Boundary in Simple Method

#1 
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In the Versteeg Book titled Computatiomal Fluid Dynamics,page 197,and page 198,
There is the explanation on the special treatment against the Output boundary condition in the Simple Method. If there is boundary at the (NI,J) position, To make all variable(except pressure) zero, We make UniCj=Uni1,j But,during the iteration cycles of the SIMPLE algorithm there is no gurantee these velocity will conserve mass over thecpmputational domain as awhole. So,we must make Uni,j= Uni1,jxMin/Mout .........  I can not understand the reason why there is no gurantee these velocity will conserve mass over thecpmputational domain as awhole only making UniCj=Uni1,j. Onother question is, If we make Uni,j= (Uni1,j)X Min/Mout, (Min is inflow mass and Mout is outflow mass) The wake status after the obstale is drasticaly changed compared with real physical model even if the obstacle is not located at near the Output boundary. 

January 24, 2004, 12:02 
Re: Output Boundary in Simple Method

#2 
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OK,
Imagine a domain with inflow and outflow and a pressure equation: div(1/a_P grad(p)) = div(flux). The pressure equation has beed derived from the div(flux) = 0 condition for every cell. In order for that to be possible, div(flux) over the whole domain should also be zero (what goes in, comes out). Now, let's have a look at the boundary conditions: if you specify the velocity (flux) on a boundary, you do not want the continuity to mess with it. I can safely assume that you have fixed the flux at the inlet, but at the outlet you've got the choice (I am talking incompressible flows!). 1) If you fix the outlet pressure, you will have a pressure gradient at the outlet and this will be adjusted by the solution to guarantee global continuity (div(flux) = 0 over the whole domain). All is well, no adjustments are required. 2) If you specify zero gradient both on the velocity and the pressure at the outlet, the the pequation will have zero gradient boundary conditions all the way around (inlet, outlet and walls!). Also, you have just said there will be no pressure gradient at the outlet boundary. Thus, you've got two results:  the pressure level in the solution is undefined (it does not matter, boundary conditions only deal with the gradient of p).  there is no mechanism for the pressure equation to change the flow condition at any boundary (zero gradient!).  if global continuity is not satisfied, the pressure equation does NOT have a solution: it is impossible to satisfy div(flux) = 0 for every cell if this is not the case for the domain as a whole. What you need to do is easy: 1) sum up all the fluxes that are going in 2) extrapolate the velocity at the outlet boundary from the cells next to the boundary. 3) adjust the boundary flux such that it matches the incoming flux. They then remain fixed as far as the pressure equation is concerned. ... and all is well! (you may need to fix the pressure in one point of the domain to stop your solver going pressure bananas, but that's another story. Hope this was useful, Hrv 

January 27, 2004, 08:58 
Re: Output Boundary in Simple Method

#3 
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Dear Hrvoje Jasak ,
Thank you for your clear explanation on my quetion. I have still 2 questions on this issue. No1. If we have 1.0 meter sized cube domain. and have same size of rectangle shaped inlet and outlet at East and West with the same size and same location at each side. If we asign 1.0 meter/sec air speed at West inlet. According to the continuity equetion rule, we get 1.0 meter/sec at the East outlet. Is this a realistic flow as the CFD result ? No 2. If we have obstacle such as 0.3 meter sized cube at the center of domain, we should have a wake with turbulence at the back side of the cube. But ,we will get no wake at the backside of cube after the extapolation on the outlet flux. Regards, Yoshi 

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