User Name Remember Me Password
 Register Blogs Members List Search Today's Posts Mark Forums Read

 August 2, 2010, 20:45 fixedGradient, zeroGradient? #1 Member   Join Date: Dec 2009 Posts: 33 Rep Power: 9 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: 11 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 miladrakhsha, soheil_r7, wenxu and 2 others like this.

 August 3, 2010, 04:31 #3 Member   Join Date: Dec 2009 Posts: 33 Rep Power: 9 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: Germany
Posts: 1,267
Rep Power: 23
Quote:
 Originally Posted by idrama 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).

August 3, 2010, 06:51
#5
Senior Member

Anton Kidess
Join Date: May 2009
Location: Germany
Posts: 1,267
Rep Power: 23
Quote:
 Originally Posted by smillion 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.

 August 3, 2010, 09:50 #6 Senior Member   Claus Meister Join Date: Aug 2009 Location: Wiesbaden, Germany Posts: 241 Rep Power: 11 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: 14 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] http://www.cfd-online.com/Forums/ope...ssure-b-c.html solefire, saba_saeb, Pirlu and 3 others like this.

 August 7, 2010, 04:35 #8 Senior Member   Claus Meister Join Date: Aug 2009 Location: Wiesbaden, Germany Posts: 241 Rep Power: 11 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.

 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: 14 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.

 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: 14 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.

August 30, 2012, 10:03
#11
Member

Join Date: Jul 2012
Posts: 31
Rep Power: 7
hey guys,

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

Firstly,

Quote:
 Originally Posted by idrama 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 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.

 Thread Tools Display Modes Linear Mode

 Posting Rules You may not post new threads You may not post replies You may not post attachments You may not edit your posts BB code is On Smilies are On [IMG] code is On HTML code is OffTrackbacks are On Pingbacks are On Refbacks are On Forum Rules

 Similar Threads Thread Thread Starter Forum Replies Last Post Bedotto OpenFOAM Running, Solving & CFD 4 April 22, 2013 06:02 daniel_mills OpenFOAM Running, Solving & CFD 44 February 17, 2011 18:08 andrea.pasquali OpenFOAM 7 May 30, 2010 12:45 Stylianos OpenFOAM 3 March 23, 2010 13:28 kassiotis OpenFOAM Running, Solving & CFD 0 July 22, 2009 09:28

All times are GMT -4. The time now is 14:25.

 Contact Us - CFD Online - Privacy Statement - Top