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Old   August 2, 2010, 20:45
Default fixedGradient, zeroGradient?
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Hey guys,

I've been reading through the Versteeg CFD book and I'm having some trouble understanding the boundary conditions.
My question is, if you don't know the value of the pressure or velocity at the boundary then do you simply put zeroGradient or is it more complex than that?
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Old   August 3, 2010, 00:53
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fixedGradient bassically means that a specific value shall be assigned, for instance, no-slip condition which means U=0 at wall (Def. of fluid). Moreover, at the inlet usually a velcotiy is prescirbed.

zeroGradient means that the gradient of respective quantity is zero, meaning that the actual value is constant. Ususally, pressure is taken to be zeroGradient at a inlet, because the pressure is assumed to be constant there.

Hopfully that will help
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Old   August 3, 2010, 04:31
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idrama thanks. But I have another question.
What boundary conditions would you put for the outlet? Because we wouldn't have any idea of the velocity and pressure values.
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Old   August 3, 2010, 06:50
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Quote:
Originally Posted by idrama View Post
fixedGradient bassically means that a specific value shall be assigned, for instance, no-slip condition which means U=0 at wall (Def. of fluid). Moreover, at the inlet usually a velcotiy is prescirbed.
U=0 would be fixedValue, not fixedGradient! (Dirichlet vs Neumann).
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Old   August 3, 2010, 06:51
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Quote:
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idrama thanks. But I have another question.
What boundary conditions would you put for the outlet? Because we wouldn't have any idea of the velocity and pressure values.
Since you are not interested in the exact pressure value for an incompressible flow in a channel, but only pressure differences, you'd set the pressure to be 0 at the outlet and zeroGradient on the inlet.
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Old   August 3, 2010, 09:50
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Firstly, you are right; i wrote under pressure (ain't got no time).

Actually, you can set every constant value you want at the outlet. Fact is, that in the NS equations only the alteration of the pressure is presented, e.g. dp/dx, which means that only for pressure differences are being solved. In postprocessing is it just easier to handle the pressure distribution about zero. Check it out: Make a easy pipe case with 100000 Pa and 0 Pa at the outlet. Note, my statements are only for incompressible flows. I've never dealt compressible or something else.

Good luck
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Old   August 6, 2010, 19:21
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I have provided a fairly self-explanatory derivation[1] for why the zero gradient B/C is applied for pressure at walls. The same logic can be extended to inlets as well.

References:
[1] zero gradient pressure B/C
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Old   August 7, 2010, 04:35
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Hello,

I think there is something wrong! You wrote:

u_A = 0 => du_A/dn = 0 [1]

Should I have understood everything correctly, then is [1] wrong. Just consider the steady-state solution (you can find this form in every fluid mechanic textbook):

u(y)=-1/(2*mu) dp/dx (h-y)y

which has roots at 0 and h. The derivative goes as follows

u'(y)=-1/(2*mu) dp/dx (h-2*y)

which is for u'(0)=-1/(2*mu) dp/dx h != 0. All quatntites are non-zero, even dp/dx due to the pressure drop.
Furthremore, du/dn = 0 implies you would have no wall shear stresses.
Note, the presents of root of a function f at a postion t, i.e. f(t)=0, does not implie that the slope is zero.
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Old   August 7, 2010, 12:09
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I think du/dx = 0 and dv/dx =0 are vaild as both u and v at the bottom wall do not change in the X-direction. Also, in the derivation, I have NOT assumed du/dy=0, which means that the wall shear stress is free to exist.
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Old   August 7, 2010, 12:15
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Ah, I think I know what happened. You have confused my "x" for "n". Just for clarification, nowhere in the derivation have I ever assumed/stated that du/dn = 0 at the wall. There are no derivatives with respect to 'n' in my derivation. That's why I clearly marked out the co-ordinate system as X and Y.

Last edited by msrinath80; August 7, 2010 at 14:10.
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Old   August 30, 2012, 10:03
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hey guys,

i think i have to revive this thread, because i dont quite understand the following statements:

Firstly,

Quote:
Originally Posted by idrama View Post
Ususally, pressure is taken to be zeroGradient at a inlet, because the pressure is assumed to be constant there.
How can the pressure be assumed to be constant at the inlet? In the simple case of stationary flow in a cylindrical tube, e.g., the pressure gradient is non-zero for every point in the tube (dp/dz = -delta P / L).


Quote:
Originally Posted by msrinath80 View Post
I have provided a fairly self-explanatory derivation[1] for why the zero gradient B/C is applied for pressure at walls. The same logic can be extended to inlets as well.l
You assumed zero velocity at the wall (no-slip). clearly, this cannot be assumed for the inlet, right?


Please correct me if I'm wrong, but zeroGradient for p at inlet doesn't seem to make sense to me.
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