fixedGradient, zeroGradient?
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? 
fixedGradient bassically means that a specific value shall be assigned, for instance, noslip 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 
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. 
Quote:

Quote:

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 
I have provided a fairly selfexplanatory 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] http://www.cfdonline.com/Forums/ope...ssurebc.html 
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 steadystate solution (you can find this form in every fluid mechanic textbook): u(y)=1/(2*mu) dp/dx (hy)y which has roots at 0 and h. The derivative goes as follows u'(y)=1/(2*mu) dp/dx (h2*y) which is for u'(0)=1/(2*mu) dp/dx h != 0. All quatntites are nonzero, 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. 
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 Xdirection. Also, in the derivation, I have NOT assumed du/dy=0, which means that the wall shear stress is free to exist.

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 coordinate system as X and Y.

hey guys,
i think i have to revive this thread, because i dont quite understand the following statements: Firstly, Quote:
Quote:
Please correct me if I'm wrong, but zeroGradient for p at inlet doesn't seem to make sense to me. 
All times are GMT 4. The time now is 21:23. 