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 Andrew Hayes December 19, 2005 11:25

stupid question

If I am doing steady-state 2-d flow between plates and look at the solution matrix for the final velocity each column should be the final steady-state velocity - be the same thing, or should it show the development of the flow field? I have a symmetry line down the middle in the x-direction, and don't know if I should set the boundary condition for this as 1.5*Vin or use the stream function and solve for it. I am using the stream function-velocity equations.

thanks

 Juergen Kertz December 19, 2005 12:19

Re: stupid question

Hello Andrew,

I am a very Beginner, so my answer might be very wrong.

As far as I understand your flow setup, the result depends on how you create the flow. I did some similar calc with a penalty approach, and these were my observations, which match the analytic results, and "reality":

1.) If flow is caused by gravity, you should get a uniform (in flow direction) and parabolic velocity distribution.

2.) If flow is caused by a velocity boundary condition, (maybe you have set velocities to a value at inflow) then you should notice a development of the distribution in flow direction, and after some lenght there should be a parabolic distribution.

I hope this answer is correct so far.

I can't help you about the symmetry as I never implemented it, but usually a velocity symmetry is setting the velocity derivate to zero. So if flow is u in x-direction, symmetry is : dv/dy=0

 Andrew Hayes December 19, 2005 12:25

Re: stupid question

my flow is created with an intial uniform velocity in the x-dir. The Re=100. What I meant by 'the same' was a half-parabolic curve - since I have a symmetry line (with Umax in the x-dir located at the symmetry line) at each of the columns in the resulting solution matrix being the same parabola or should I see a development of the parabola. Your second answer is what I was looking for. I am doing steady-state, so I wasn't sure if I should see a development of the parabola or a final parabola and no development. thanks

 diaw December 19, 2005 21:41

Re: stupid question

The shape of the velocity profile will change from the inlet 'shape' to the flow-governed final shape eg. parabolic (if laminar flow). The shape development will depend on the shape of the inlet velocity profile.

eg. if a linear inlet profile, then this will gradually change to the parabolic-type shape. If inlet profile is already parabolic, then the shape changes will be slightly different - but should still end up parabolic. The final profile will be governed by the plate b/c's, fluid properties & entry mass-flow.

Don't be to quick to impose symmetry on the solution - model the whole flow domain & let the physics work itself out.

I hope that this helps.

diaw...

 Andrew Mettler Hayes December 19, 2005 22:40

Re: stupid question

I am having trouble getting my inlet velocity to translate through the flow field. I am using the streamfunction-velocity equations. My BC's are at the y=h; Ux=Uy=0, and Psi=0. At the exit Uy=Uy(evaluated at the node before it), as is Ux and Psi, since I am assuming fully developed flow at the exit. At the entrance Uy=0, Ux=Uin(uniform flowfield), the streamfunction I wasn't too sure about, but I integrated d(Psi)= Uxdy to get a value of the streamfunction, but I had to disregard the constant of integration. (Psi = Ux(deltaY)), and along the symmetry line, Psi = 0, Ux= 2*Psi (evaluated at one grid point above the symmetry line)/2*dy, Uy=0. I am using an 8 point finite-element method, but I immediately lose my inlet velocity at the first node, and the centerline velocity never develops in a positive manner.

Any suggestions would be great.

thanks

 diaw December 19, 2005 23:08

Re: stupid question

Andrew, I'm looking through one of my text books at a similar problem to yours for Streamfunction (Ideal flow):

B/C's: @ y=0 => Psi=0 @ y=h => Psi = U*h !!! @ x=0 => Psi = U*y @ x=L => nothing specified

The y=0 boundary is a symmetry boundary.

I hope that helps you on your way.

diaw...

 Juergen Kertz December 20, 2005 03:36

Re: stupid question

This looks like the trouble I had with my first try (few days ago), though I used a different approach (viscous flow without inertia)

I got rid of it by reducing to "very simple", so I could check each step by hand.

First of all I would recommend to throw out the symmetry, and bring it back when the rest of the calc preforms well. I used linear (3-node) triangles on a rectangular 10x10 node grid, b.c. about the same as yours. You should be able to solve this problem for low Reynolds by hand (laminar flow) with usual analytic formulas (just simplify and integrate the x-momentum-conservation, integration constans is where your b.c. come in). Try setting gravity instead of velocity, then you should get a constant velocity distribution along lenght. If not, you know that there is an error in the code.

I experienced a huge loss of velocity and continuity (as I understand your problem) if the inflow velocity near the wall-nodes were too high.

 Andrew Hayes December 20, 2005 09:26

Re: stupid question

Diaw,

Is that 'y' @ x=0 the distance from symmetry line for each node?

thanks

 Andrew Hayes December 20, 2005 12:16

Re: stupid question

that BC at x=0 is sort of odd. Especially given that the BC at y=h for Psi is Psi=U*h. It seems that @x=o it would be Psi=U*(distance from h)-especially with the no slip BC at the wall. Does that make sense?

 diaw December 20, 2005 12:25

Re: stupid question

B/C's: @ y=0 => Psi=0 @ y=h => Psi = U*h !!! @ x=0 => Psi = U*y @ x=L => nothing specified

The y=0 boundary is a symmetry boundary.

----------

Sorry, the text became strung together once the message was posted.

Symmetry boundary (pipe centrline): @ y=0 => Psi=0

Top boundary: @ y=h => Psi = U*h !!!

Inlet boundary (left) : @ x=0 => Psi = U*y (linear)

Outlet boundary : @ x=L => nothing specified

Hopefully that prints out better.

diaw...

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