Traction boundary conditions

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 January 11, 2004, 11:04 Traction boundary conditions #1 m malik Guest   Posts: n/a I have the following question on 3-dimensional FEM solution of momentum-continuity equations. Consider a simple capillary flow problem with only the pressure being specified at the two ends; velocities not specified. How does one apply the traction conditions: ,i,e, tx = (mu)(du/dz+dw/dx), ty = (mu)(dv/dz+dw/dy), and tz = -p+2(mu)(dw/dz) when velocity field is not know a-priori? (z = coordinate along the capillary axis). Thanks.

 January 12, 2004, 19:40 Re: Traction boundary conditions #2 xueying Guest   Posts: n/a The traction equals t = -p n + mu n.(du/dx + transpose(du/dx)) where t, n, u, x are vectors In my code, I specify the traction as follows, If the flow is fully developed, then n.(du/dx)=0, so t = -p n + mu n.(transpose(du/dx)) The traction is encoded thru the weak form integral, du/dx is computed from the finite element basis functions and coefficients. You can also refer to papers related with open flow boundaries.

 January 13, 2004, 09:37 Re: Traction boundary conditions #3 m malik Guest   Posts: n/a I am dealing with the polymer flows, and in my code I apply the traction boundary condition in the following manner. The nonlinear finite element equations are solved using the Picard iteration scheme. In each iteration, the traction vector is calculated using the external pressure and the mu.(dui/dxj + duj/dxi) terms from the velocity field of the previous iteration. No assumption of fully developed flow is imposed. Is this the right approach?

 January 13, 2004, 12:56 Re: Traction boundary conditions #4 xueying Guest   Posts: n/a If it's not fully developed flow and it's a viscous flow, your imposing method should be correct, at least we impose exactly the same traction. Since your flow is polymer flow, do you have elastic stress contribution from polymer? In our code, if it's polymer flow, then it's viscoelastic flow, the traction has contributions from pressure, viscous stress, and elastic stress.

 January 13, 2004, 13:49 Re: Traction boundary conditions #5 m malik Guest   Posts: n/a Thanks Xueying for your response. No, I do not have elastic stress contribution. You are probably talking about the elastic dumbbell models. I am just considering steady state viscoplatic flows. However, irrespective of the model, why should one impose fully developed flow condition? For example in tube flow, the solution must approach to fully developed with increasing aspect (length-to-diameter) ratio. (This is one of the things that is bothering me.)

 January 14, 2004, 05:15 Re: Traction boundary conditions #6 Rami Guest   Posts: n/a Re: Traction boundary conditions

I think a formal and general answer would be the simplest way to answer your question. Tensor notation is used below, in which a repeated index implies summation. In addition, "," stands for covariant derivative (which, for Cartesian coordinates, degenerates to partial derivative).

The traction vector, tk, in a given direction ni is defined as

tk = sik ni

where sik is the stress tensor.

The constitutive relation for a fluid with no elasticity is

sik = -p dik + mv dik emm + 2m (eik – 1/3 dik emm)

where p is the pressure, m and mv are the viscosity and the bulk viscosity, respectively, dik is Kronecker's delta and the strain-rate tensor, eik , is related to the velocity vector, uk, by

eik = 1/2 (ui,k + uk,i)

In the finite element method, the shape-functions relate the nodal values to the element distributions. Therefore, having the nodal values (at any iteration level), the traction may be calculated by the above equations.

A developed flow assumption is not required in a pipe flow. You may definitely solve for a developing flow. For a long enough domain you will notice the flow approaches the fully developed profile asymptotically.

I hope this helps,

Rami

 March 29, 2012, 12:46 #7 Member   Hesam Moghaddam Join Date: Mar 2012 Posts: 49 Rep Power: 6 Hi, How can I apply free traction BC in CFX? my BC is: du/dn=0 thank you so much Hesam

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