Boundary condition in viscous fluid
 

Hi all,

I read on a website that the boundary condition between a flow and a wall is both no slip condition (fluid and wall at the same velocity) and null pressure gradient.

I dont really understand the reason of the null pressure gradient. As the velocity gradient can be high in a boundary layer, I would believe that there could be a pressure gradient too.

Could someone tell me about the physical reason ?

By the way, here is the page which says that the pressure gradient is null (see "Viscous wall") http://www.grc.nasa.gov/WWW/winddocs...c.html#bc:wall

Re: Boundary condition in viscous fluid
 

There is no pressure boundary condition for the Navier-Stokes equations on the wall - the no-slip condition is enough.

In numerical methods, if projection methods are used, then the derived pressure equation requires "extra" boundary conditions which are chosen for compatability; e.g. take the momentum equations in vector form and take the dot product with the unit normal vector to the wall. This gives an equation for the normal derivative of the pressure on the wall.

For example on a flat surface (y=0), ignoring bouyancy forces etc, the boundary condition becomes,

dP/dy = nu.d^2u/dy^2.

Note this is not generally zero.

Re: Boundary condition in viscous fluid
 

Hi, and thank you for answering...

"Note this is not generally zero."

I understand that the theorical result is non zero.

But some graphics in books shows that the velocity varies almost linearly near to the surface.

So with the formula

dP/dy = nu.d^2u/dy^2

It could explain that in most cases, the pressure gradient is very close to zero at the surface.

Is it right ?

Re: Boundary condition in viscous fluid
 

Another way to view this is under the boundary layer approximation the pressure does not vary normal to the wall. Basically for a boundary layer of thickness R^{-1/2} (R is the Reynolds number) then the velocity normal to the wall, for an attched boundary layer, will be O(R^{-1/2}) which implies that P is indendent of the normal direction to O(1/R). This argument of course requires R->infinity.

This argument works for curved walls, provided the curvature is not too great, but it is not always true! What happens if the boundary-layer separates?

Re: Boundary condition in viscous fluid
 

hi,

just would like to check, it is stated tt

dP/dy = nu.d^2u/dy^2.

what is "nu"? is tt the dynamic viscosity?

hence, in most projection mtds such as fractional step, the BC for pressure is usually dp/dn=0. so is tt an approx?

i'm wondering if this approx is still valid for a moving boundary?

tks

Re: Boundary condition in viscous fluid
 

projecting the momentum equation in direction normal to the surface, as pointed out, by Tom, he have:

dP/dn = nu d^2u/dy^2 (eq. 1) This condition in non dimensional form became:

dP/dn = (1/Re)d^2u/dy^2 (eq. 2)

So if Re--> infinity it become

dP/dn=0 (eq.3)

In most practical situations, such as aeronautical flows, the approximation given by equation 3 is reasonable.

If the wall is moving, you have to project the corresponding momentum equation on the normal direction and obtain the corresponding boundary condition. This condition depends by the equation you are using. For example if you are using the N-S equation in the relative frame of reference the RHS of eq.3 will contain the acceleration of the wall.

Re: Boundary condition in viscous fluid
 

And if we have, i.e. LES model? Then it would be: 1. in case of a subgrid model: dP/dn=(1/Re+SGS)*d^2u/dn^2, or, hence, SGS=f(t), dP/dn=f(t) if Re->infinity dP/dn->f(t) 2. in case of a tensor model: dP/dn=(1/Re)*d^2u/dn^2+dq/dn, where q - stress tensor. In this case dP/dn->dq/dn.

In both cases we have, that dP/dn=f(t), and depends on turbulence.

Re: Boundary condition in viscous fluid
 

In that case, does it mean that if we are solving the NS eqns in ALE (Arbitrary Lagrangian Eulerian) form in an inertial reference frame, it 'll become dP/dn= a, where a is the known acceleration of wall?

same can also be applied for immersed boundary/cartesian cut cell method?

thanks

Re: Boundary condition in viscous fluid
 

Actually it would be the normal component of the acceleration.

Re: Boundary condition in viscous fluid
 

Tks ag!

Btw, can someone recommend me some references which explains how to project the momentum eqn in the normal direction. I would like to learn to proof it for myself.

Tks in advance.

Re: Boundary condition in viscous fluid
 

to project the equation you have only to make the dot (scalar) product of your momentum equation (which is a vectorial one) and the direction normal to the wall taking into account the boundary condition (which are Vn=0 for Euler system of equation and V=0 for N-S). By the way if you look at the following paper:

Investigation of the Flow Structure Around a Rapidly Pitching Airfoil Miguel R. Visbal* and J. S. Shang

AIAA JOURNAL VOL. 27, NO. 8

you will see explicitely which are the boundary condition implemented there in.

You can also search on the WEB a PhD thesys which probably will contain more details....

Re: Boundary condition in viscous fluid
 

hi mar, i just went over to the library to look at the reference you gave.

however, i found that it doesn't use the fractional step scheme and solve the poisson eqn. It was a compressible code. So is that valid?

Also, i've looked thru quite a few papers on ALE/moving grid but none seems to explicitly state that dp/dn=a(normal) where a= acceleration of wall.

Why is this so? Or is it too simple that everyone already know it...

Re: Boundary condition in viscous fluid
 

I've also found a reference which discuss about the boundary condition for pressure. it says that there are 2 possibilities - 1. just multiply by the unit normal vector at the boundary (as stated in earlier post) or 2. since u(n+1) - u(*)= -del(t)(dp(n+1)/dn), at boundary u(n+1)=u(*) so dp/dn=0. it is valid at regions where velocity is prescribed.

hence, as stated in the reference, these 2 conditions 'll give 2 different BC.

in other words, even for moving body, dp/dn=a(n) or 0.

Can someone confirm this?

Re: Boundary condition in viscous fluid
 

You've forgotten the nonlinear inertial terms - there is an acceleration due to the motion of the wall => dP/dn is nonzero; e.g. if n=i, then u.u_x is nonzero.

Re: Boundary condition in viscous fluid
 

>hi mar, i just went over to the library to look at the >reference you gave.

>however, i found that it doesn't use the fractional step >scheme and solve the poisson eqn. It was a compressible >code.

YES IT'S A COMPRESSIBLE CODE

>So is that valid?

YES IT IS VALID

>Also, i've looked thru quite a few papers on ALE/moving >grid but none seems to explicitly state that dp/dn=
:(normal) where a= acceleration of wall.
:Why is this so? Or is it too simple that everyone already >know it...

IT DEPENDS BY THE SYSTEM OF REFERENCE AND THE FLOW VELOCITIES YOU ARE USING; DEPENDING ON THIS TWO CHOICES YOU WILL HAVE DIFFERENT SYSTEMS OF EQUATIONS.

PROJECTING THESE DIFFERENT EQUATIONS ON THE NORMAL TO THE WALL YOU WILL OBTAIN DIFFERENCT WALL-BC.

for example: if you use relative velocities in the relative frame of reference ---> dp/dn= inertial terms.

if you use absolute velocities in the relative frame of reference--> dp/dn = a_n + inertial terms

Hi

Re: Boundary condition in viscous fluid
 

let me try to prove this:

for NS eqn in absolute frame ie standard form,

projecting the inviscid momentum equation in a direction normal (outward) to the boundary (assume wall is horizontal ie n=-j),

dp/dn=-d(u.n)/dt-(u.(del u)).n where u=(u,v), n=(0,1)

d(u.n)/dt=a(y) acceleration of wall in y dir

(u.(del u))=u(du/dx)+v(du/dy) (x comp) u(dv/dx)+v(dv/dy) (y comp)

(u.(del u)).n=-(u(dv/dx)+v(dv/dy))

since due to viscosity dv/dy=0, dv/dx=0 therefore dp/dy=a(y). the result is also the same for problems with inclined wall, just that dp/dn=a(n).

this is applicable to formulation such as ALE form and also methods such as immersed boundart/cartesian grid for moving problems.

So, is this correct? I'm not sure about the assumption that dv/dx=0. thanks in advance.

Re: Boundary condition in viscous fluid
 

As I told you before you have to specify 1) the frame of reference 2) the "nature" of the velocities

You said that the reference system is the absolute but you didn't specified if the velocities are relative or absolute ( a popular choice for example is absolute velocities in the relative frame)

Supposing that the velocities are absolute the convective fluxes will contain the grid velocities (vg)

rho(v-vg) rho v(v-vg) rho E (v-vg)

This is clearly understood if you consider that your equations represent a conservation in a volume.

Suppose for the sake of simplicity that this volume is moving with velocity vg in the x direction.

_______

| | x

1| -> |2 ----------->

|_______|

If the fluid entering your volume has an absolute velocities equal to vg What is the net flux entering your volume??

IT IS ZERO because the fluid and the control-volume are moving at the same velocity.

Once it'is clear that the use of absolute velocities implies the appearance of grid velocities in the convective fluxes; it's evident that when you project this momentum equation on the normal direction and evaluate this projection at the wall the convective term disappear because the boundary condition at the wall is v=vg.

I hope that now things are clearer...

If you are a student I hope I will have a mention on your thesis even if you don't know who I am...

Obviously I'm joking.

Hi

Re: Boundary condition in viscous fluid
 

hi,

so u are referring to the ALE formulation, right? but what about for immersed boundary/cartesian grid methods? In those schemes, the grid is stationary. so does it mean the convective term will give rho u(b)*U(b) where u(b)= interpolated x or y velocity, U(b) = face center normal velocity, b=boundary(known) ?

moreover, the poisson eqn of the fractional step is given as summation(dp/dn)*area=summation(U(*)*area/(delta t)), where U(*) is the face normal interpolated velocity. if dp/dn(b)=-rho*a(n)(b) (as given in the AIAA paper), then the eqn will become

(summation exclude boundary)(dp/dn)*area=summation(U(*)*area/(delta t))+rho*a(n)(b)*area

thank you!

Hope to clear this up once and for all.

Re: Boundary condition in viscous fluid
 

The way Nekton (Fluent Technologies?) does it is NOT to have any boundary condition on pressure!

To 'solve' pressure, we

(a) GUESS it (yes, anything you come-up with will be a guess)

(b) correct it, but this will cause velocity to be non-solenoidal

(c) correct velocity to make it divergence-free.

Pressure is really a different animal. Velocity is always continuous (most numerical methods force C0 continuity) but there is no assumption of continuity for pressure!