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BC for the stream function in the computing domain

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Old   July 26, 2000, 13:21
Default BC for the stream function in the computing domain
Pierre Forges
Posts: n/a
July 26, 2000

Dear colleagues,


(x,y) Cartesian coordinates.

(r,s) Curvilinear coordinates.

P Stream Function

P is defined as follow:

u = dP/dy = dP/dr dr/dy + dP/ds ds/dy [1]

v =-dP/dx =-dP/dr dr/dx - dP/ds ds/dx [2]


For x=0 and y in [0,ymax] u=1 v=0

Now if I go to the computing domain, I have:

1=dP/dr dr/dy + dP/ds ds/dy

0=dP/dr dr/dx + dP/ds ds/dx

Two equations with two unknowns: dP/dr and dP/ds

Solving for dP/ds and integrating from s=0 to s>0 yields:

P(r=0,s)= P(0,0) + Int_{0}^{y(s)}

[dr/dx] /[dr/dx ds/dy - dr/dy ds/dx] dy

I have used my 2 BC, namely u=1 and v=0, to get a

Dirichlet BC for the stream function P. I think

it's the way to do it. Let's see now another

boundary condition.


For x=xmax and y in [0,ymax]



In the computing domain I have:

du/dr dr/dx + du/ds ds/dx=0

dv/dr dr/dx + dv/ds ds/dx=0

Using eqs [1] and [2] yields:

[d^2P/dr^2 dr/dy + d^2P/drds ds/dy] dr/dx +

[d^2P/drds dr/dy + d^2P/ds^2 ds/dy] ds/dx=0 [3]

[d^2P/dr^2 dr/dx + d^2P/drds ds/dx] dr/dx +

[d^2P/drds dr/dx + d^2P/ds^2 ds/dx] ds/dx=0 [4]

From these two equations, how can I get my

boundary condition for the stream function P?

I cannot isolate dP/ds from eqs [3] and [4]! If it was possible I would have been able to integrate from s=0 to s>0 as I did before.

Same remark applies when I have the

following BC: du/dy=0 and v=0. It's a

BC for a symmetry line.

Thank you so much in advance for

replying to my question.

My very best regards,

Dr. Pierre Forges

UAE University Mech. Eng. Dept.
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Old   July 26, 2000, 14:47
Default Re: BC for the stream function in the computing do
John C. Chien
Posts: n/a
(1). The only reason of using the coordinate transformation is that after the transformation, the boundary condition specification will be easier. (2). For example, initially the wall is a curve, that is y=y(x). After the coordinate transformation, this becomes Eta=Eta1=a constant, using the body aligned coordinates. (3). In other words, the original wall boundary condition is Psi=a constant along y=y(x), now it becomes Psi=a constant along Eta=Eta1=a constant. (3). The same is true for the inlet condition. If u=1 along yinlet=yinlet(x), then after transformation, it becomes u=1 along Xi=0. And if you can set v_eta=0, then the flow will be uniform at Xi=0 in Eta direction. (4). So, the principle of using the coordinate transformation is to make the specification of the boundary conditions easier in the transformed coordinates. (5). In terms of the simple English, the inlet boundary condition is: at the inlet plane, velocity is uniform and is along the streamwise direction. (velocity has only Xi component) This inlet plane could be positioned originally at an angle to the x-y coordinate system and was difficult to specify the inlet boundary condition. After the transformation, it becomes easier. (6). In your case, you can keep the transformation the same as the original coordinates in the inlet and the exit region!!!! And use general transformation elsewhere. In other words, there is no transformation in the inlet and the exit regions. That is in those regions, Xi=x, Eta=y. Then those coordinate transformation factors will have value of either 0 or 1. In this way, u=1 at x=0, is the same as u=1 at Xi=0 or =Xi_zero (can be non-zero value). (7). the idea is: don't make the simple problem (u=1 at x=0) complicated through the coordinate transformation. as what you have just illustrated in your message. You are making the problem more difficult, not easier to solve.
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