Hey all,

The quadratic resistance in CFX-5.6 is stated as following a "generalized" Darcy's law, and uses the expression (for each component):

dP/dx = K*rho * |U| * u (ignoring the linear term)

Where |U| is the magnitude of the total velocity vector (speed), and K*rho is the R2 resistance constant. This would indicate, that the resistance in the x direction is a function of not just the u velocity, but the v, and w velocities as well (since |U|=sqrt(u^2+v^2+w^2)). This just doesn't feel right to me.

With unidirectional flow, I've always seen:

dP/dx = K*rho * |u| * u

using the absolute value of u, instead of the vector magnitude, |U|. This would give a smaller resistance for non-unidirectional flow than the CFX formulation.

Does someone know of a reference with a really good development of a generalized Darcy's law in 3D? One CFX paper references a "Convective Heat Transfer" by Bejan. I have this text, but the only reference I can find is a 1D formulation like the one I stated. Can anyone help?

Thanks, Jeff

Hi jeff, I think in CFX5, the Forchheimer-Brinkman-Darcy equations are in the vector form(3D) correct implemented.You can get The unidirectional flow formulation with simply setting v and w to 0. It always a Problem with thus empirical formulations and their Parameters!!In Bejan also are the equations in vector form (thik)! regards! hanni

hi Jeff & Hannibal,i m a new CFX 5.6 user so pls dont mind if my question seems trivial to u. i have problem with specifying BRINKMAN's equation in CFX 5.6 Brinkmans equation is-> -dp/dl = (myu)*v/k - (myu)*grad^2(v)

how to incorporate it into momentum source in the form S = Cr1*Ui - Cr2*|U|*Ui + Sspec

we will take Cr1 = (myu)/k but how to use the term [-(myu)*grad^2(v)] in the form [-Cr2*|U|*Ui + Sspec ] ?

Waiting for ur reply. Warm Regards, Paresh Jain

Paresh,

In order to simulate Brinkman's equation, you'll need to set momentum sources for each equation manually (i.e. your terms will have to be calculated in CEL and put into Sspec rather than Cr2).

You can get grad(u) from a CEL user FORTRAN function (see the CFX manual which shows how to do this). I suppose you can then put each of these into a user variable (one that's not solved) say ph1, phi2, phi3 and then call the grad user function again on on these new variables. Then combine everything into your Sspec term. I don't know if this will work, but it seems logical.

Jeff

Paresh,

Also, dont forget to compute the jacobian, dS/dU, and take the maximum value as the linearisation coefficient. This will give you much better convergence behaviour.

Neale

If each component of velocity independently generates resistance, doesn't that mean that the resistance is then a function of the coordinate system selected? If so, that doesn't seem to be the best choice of models for resistance to me.

EG assume rho and K are both = 1. If the flow is moving in a 2D flow field with velocity u,v = 0.5, 0.5 then the resistance in x and y directions would be proportional to (0.5*0.5,0.5*0.5) or (0.25,0.25) and the magnitude of resistance would be 0.353. If you then aligned the x' axis with the flow, velocity would be (0.707,0) and resistance would be (0.5,0) with a magnitude of 0.5. So depending on coordinate system, you get a different resistance to the flow. Let me know if you see something wrong with my math.

... to complete Martin's thought...

If you use multiply by the velocity magnitude, instead of individual components, the frame dependancy goes away.

The resistance in the x and y directions are then proportional to (.707*.5, .707*.5) and the magnitude is thus .5. Similary, if you align your x-axis with the flow you get (.707*.707, .707*0) and again a magnitude of .5.

Note that this also means that the pressure drop is in the right direction. If you reverse your original x-axis, you get (.707*(-.5), .707*.5) and the momentum source is now negative, keeping it in the right direction. By multiplying by velocity, a negative coefficient will always ensure that the source is resistive.

Regards, Robin

Robin/Martin,

Thanks for this. This has bugged me for a long time, but I see that the math does indeed work. I'll be able to sleep tonight.

Thanks, Jeff

please send informaion to me your hameed pakistan

to whom u want to join? for what ?

Paresh Jain