I am sure you are going to get many answers to your question, but it might be helpful for the subsequent discussion to note the following: In fluid mechanics the traction vector t ( also called the stress vector, the force per unit area) at any point in space is defined as the scalar product of t and the unit normal n of a surface that passes through the point in question. That surface is said to be traction free if t=0. The traction vector for a Newtonian fluid is defined as: t=T.n, where T is the total stress ( a tensor) : T=-pI +2mu D, where D is the rate of deformation tensor , and I is the unit tensor. Now consider fully developed laminar flow is a channel of height 2b. The velocity field for this 2-D flow is given by u_z=-dp/dx(2b y-y^2)/2mu, u_y=0. Now we can use the flow field to compute the traction at the inlet and exit planes of the flow. At the inlet t=T.n has two components:
t_z=(T.e_z).e_z=p_inlet, t_y=(T.e_z).e_y=dp/dx(b-y)
At the exit the values are
t_z=(T.e_z).e_z=p_exit, t_y=(T.e_z).e_y=dp/dx(b-y)
If we define the pressure level at the exit to be p_exit=0, then it is clear that the exit condition for a fully developed flow is NOT traction free unless dp/dx=0, which is a good approximation for a long channel. One can show that for large Reynolds numbers mu
I am sure you are going to get many answers to your question, but it might be helpful for the subsequent discussion to note the following: In fluid mechanics the traction vector t ( also called the stress vector, the force per unit area) at any point in space is defined as the scalar product of t and the unit normal n of a surface that passes through the point in question. That surface is said to be traction free if t=0. The traction vector for a Newtonian fluid is defined as: t=T.n, where T is the total stress ( a tensor) : T=-pI +2mu D, where D is the rate of deformation tensor , and I is the unit tensor. Now consider fully developed laminar flow is a channel of height 2b. The velocity field for this 2-D flow is given by u_z=-dp/dx(2b y-y^2)/2mu, u_y=0. Now we can use the flow field to compute the traction at the inlet and exit planes of the flow. At the inlet t=T.n has two components:
t_z=(T.e_z).e_z=p_inlet, t_y=(T.e_z).e_y=dp/dx(b-y)
At the exit the values are
t_z=(T.e_z).e_z=p_exit, t_y=(T.e_z).e_y=dp/dx(b-y)
If we define the pressure level at the exit to be p_exit=0, then it is clear that the exit condition for a fully developed flow is NOT traction free unless dp/dx=0, which is a good approximation for a long channel. One can show that for large Reynolds numbers mu
Hi Brian
Its an old thread, but thank you for the explanation. It cleared the doubts. If an own code is written, I think it can be implemented.
But still looking for how to implement traction free condition in a commercial code like Fluent.
@Ranjith G - how you implemented it in Fluent. ?
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