
[Sponsors] 
August 2, 2010, 20:45 
fixedGradient, zeroGradient?

#1 
Member
Join Date: Dec 2009
Posts: 33
Rep Power: 8 
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? 

August 3, 2010, 00:53 

#2 
Senior Member
Claus Meister
Join Date: Aug 2009
Location: Wiesbaden, Germany
Posts: 241
Rep Power: 10 
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 

August 3, 2010, 04:31 

#3 
Member
Join Date: Dec 2009
Posts: 33
Rep Power: 8 
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. 

August 3, 2010, 06:50 

#4 
Senior Member
Anton Kidess
Join Date: May 2009
Location: Delft, Netherlands
Posts: 1,139
Rep Power: 20 
U=0 would be fixedValue, not fixedGradient! (Dirichlet vs Neumann).


August 3, 2010, 06:51 

#5 
Senior Member
Anton Kidess
Join Date: May 2009
Location: Delft, Netherlands
Posts: 1,139
Rep Power: 20 
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.


August 3, 2010, 09:50 

#6 
Senior Member
Claus Meister
Join Date: Aug 2009
Location: Wiesbaden, Germany
Posts: 241
Rep Power: 10 
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 

August 6, 2010, 19:21 

#7 
Senior Member
Srinath Madhavan (a.k.a pUl)
Join Date: Mar 2009
Location: Edmonton, AB, Canada
Posts: 703
Rep Power: 13 
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] zero gradient pressure B/C 

August 7, 2010, 04:35 

#8 
Senior Member
Claus Meister
Join Date: Aug 2009
Location: Wiesbaden, Germany
Posts: 241
Rep Power: 10 
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. 

August 7, 2010, 12:09 

#9 
Senior Member
Srinath Madhavan (a.k.a pUl)
Join Date: Mar 2009
Location: Edmonton, AB, Canada
Posts: 703
Rep Power: 13 
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.


August 7, 2010, 12:15 

#10 
Senior Member
Srinath Madhavan (a.k.a pUl)
Join Date: Mar 2009
Location: Edmonton, AB, Canada
Posts: 703
Rep Power: 13 
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.
Last edited by msrinath80; August 7, 2010 at 14:10. 

August 30, 2012, 10:03 

#11  
Member
Join Date: Jul 2012
Posts: 31
Rep Power: 6 
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. 

Thread Tools  
Display Modes  


Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
fixedGradient BC type with the channelOodles solver  Bedotto  OpenFOAM Running, Solving & CFD  4  April 22, 2013 06:02 
Pressure instability with rhoSimpleFoam  daniel_mills  OpenFOAM Running, Solving & CFD  44  February 17, 2011 18:08 
interFoam  andrea.pasquali  OpenFOAM  7  May 30, 2010 12:45 
ZeroGradient pressure Outlet BC prob (adjustPhi.C)  Stylianos  OpenFOAM  3  March 23, 2010 13:28 
zeroGradient pressure and moving mesh  kassiotis  OpenFOAM Running, Solving & CFD  0  July 22, 2009 09:28 