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HOW TO SOLVE THE NAVIER-STOKES EQUATION

Benk Janos Department of Informatics, TU München

JASS 2007, course 2: Numerical Simulation: From Models to Software

Based on:

ON PRESSURE BOUNDARY CONDITIONS FOR THE

INCOMPRESSIBLE NAVIER-STOKES EQUATION

Phlilp M. Gresho and Robert L. Sani

International Journal For Numerical Methods In Fluids, Vol 7, 1111-1145(1987)

Content

Short introduction

Analysis of the continuum equation

Pressure Poisson equation

Boundary conditions

Discrete approximation to the continuum equation

Short introduction

- Important field of application of the numerical simulation

- The flow is a result of different physical processes - Numerical flow simulation has a various fields of

application, real scenario simulations.

http://www.cfd-online.com/Links/misc.html#picts


( ) gugradPugradut

u+∆=+••+

∂∂

Re

1

0=udiv

The second equation is the continuity equation

The momentum equation for incompressible fluids

If it would be compressible fluid

tudiv

∂∂

=ρ

(1)

(2)

(3)


t

u

∂∂

Each part from the (1) equation has a contribution to the momentum

The velocity change describing the acceleration of a infinite mass point. It must be in balance with …

( ) ugradu •• the convective term describing the frictionless acceleration induced by the velocity filed.

gradP the gradient of the pressure. (by definition is an acceleration)


u∆Re

1 This component reflects the interior drag of the fluid.

Re is the Reynolds number.

The drag appears between two layers of fluid with different

velocity.

- The friction force is acting against the velocity gradient

- This laminar flow, which opposes turbulent flow


- At gas flow this term can be neglected, but by fluids not (e.g. honey)

-The internal friction is also called viscosity, which characterize each fluid.

vs - mean fluid velocity, L - characteristic length,

µ - (absolute) dynamic fluid

viscosity, ν - cinematic fluid viscosity:

ν = µ / ρ, ρ - fluid density.

source:http://en.wikipedia.org/wiki/Reynolds_number


Ω∂≡Γ

We need boundary conditions so that the problem will be solvable

- Ω is the fluid domain , and is bounded by Γ

gExternal accelerations. ex: gravity

This component contains other forces which

are not represented by other terms


- This leads to the BC for the continuity equation (mass conservation) n is the normal vector on the boundary

∫Γ

=• 0wn (4)

w is the velocity on the boundary (Dirichlet BC) u = w(x,t) on Γ

( ) nxwnu ⋅=⋅ 0,0

We consider the initial situation t=0


In 2D we have the following vectors on the boundary

n -> normal vector

τ -> tangential component

00 =⋅∇ uAccordingly to equation (2), this holds also for the initial

velocity field in Ω


Calculate the pressure field from a velocity field.

Want to have a relation between pressure and velocity, in discrete and in continuum time.

Show one way to get the Poisson equation


( ) ugradPugradut

u∆=+••+

∂∂

Re

1

( )( ) ( )uPuutu22

Re

1/ ∇⋅∇=∇+∇⋅+∂∂⋅∇

We start with equation (1), and we neglect external influences

First we apply the divergence operator, so we get:

We use the following expression to process further the equation:

( ) uuu ×∇×∇−⋅∇∇=∇ 2


( ) 0=rotAdiv

( ) )/)()(Re

1( 22 tuuPuu ∂⋅∇∂−⋅∇∇=∇+∇⋅⋅∇

And because:

for any (differentiable) vector field.

We obtain:

We obtain the pressure Poisson equation:

∇⋅−∇⋅∇=∇ uuuP22

Re

1


Puuut

uu∇−∇•−∇=

∆−+ 20

Re

1

A slightly modified system has to be solved for the discrete time dependent solution.

Discretizing the first(1) equation:

Expressing u+ and insert the solution in the second(2) equation, we can calculate the pressure

0=⋅∇ +u

Boundary condition

To complete the specification of the problem for the pressure, we must set BC’s (boundary conditions) on Γ

It is important how we include the boundary conditions in our system.

For the boundary condition we need a scalar value, so we have 2 choice: either the normal or the tangential projection

Boundary condition

( )nnn uutuunPPn ∇⋅+∂∂−∇=∂∂≡∇⋅ )/(Re

1/ 2

on Γ

The BC with the tangential component (Dirichlet):

First we choose the normal component: (Neumann)

( )τττττ uutuuPP ∇⋅+∂∂−∇=∂∂≡∇⋅ )/(Re

1/ 2

It has been proven that these 2 conditions are equivalent.

We can calculate the pressure up to an arbitrary

additive constant!!!

on Γ

Boundary condition

( ) fuuuPtu ≡∇⋅−∇=∇+∂∂ 2

Re

1/

( ) 0/ =∂∂⋅∇ tu

The question is: When the original NS problem is well-posed, so is the associated Poisson/Neumann problem?

Here is the proof: We start with the first equation

and in Ω

also ( ))/(/ tufnnP ∂∂−⋅=∂∂ on Γ

Boundary condition

∫ ∫Ω Ω

⋅∇=∇ fP2

∫ ∫Γ Γ

⋅=∂∂ fnnP /

( ) ∫∫ΓΓ

⋅=∂∂−⋅ fntufn )/(

Apply Green formula, and we obtain an integration on the boundary

Substituting the left side, we obtain the following equation

Boundary condition

∫∫ ∫Γ

•

Γ Γ

⋅

=⋅=∂∂⋅= ),(),()/(0 txwndt

dtxwntun

Finally we obtain the following equation

w is representing the velocity on the boundary.

We obtained the mass conservation law (4), which is

satisfied.

This means that our transformed problem is also well posed.


Build the system of equations

We want to calculate the real pressure, on the boundary

Check whether the system has a correct boundary result


We can write the momentum and the continuity equation in the following discretized general way:

( ) ( ))(tgDu

tfKuGPuuAuM

=

+=++•

M – mass matrix, (for equidistant discretization is the unit matrix)

A – advection matrix

G – gradient matrix

K – diffusion matrix

D – divergence matrix


TGD

uuAtfKub

=

−+≡ )()(

f(t) , g(t) - represents the effects of the boundary Dirichlet conditions on velocity.

We use the following notations and equalities:

Denoting G with C we get:

=

••

g

b

u

P

C

CMT 0


( ) 0=tgP TH

We consider the hydrostatic pressure P=(1->1),

the system has a solution if g(t) satisfies the following

condition.

From the above define system we can derive the discrete pressure Poisson equation:

( ) ( )•

−− −=⋅⋅⋅ gtubMCPCMCTT

,11


We use the following staggered mesh:

We apply the continuity equation for cell with

P0(on the boundary)

0=−

+−

••••

h

vv

l

uu snwe


( )nnnuutuunPPn ∇⋅+∂∂−∇=∂∂≡∇⋅ )/(

Re

1/ 2

Expressing all 3 terms (ue is known on the boundary) from the momentum equation.

We use Taylor expansion to express the velocity at the “fictive” nodes, and we use the boundary velocities

With l, h ->0 (on the limit) we get the real boundary condition.

)()/(4

3/)32( 222 lOxvlvvv

nmn+∂∂=+−Γ

Thank you for your attention!

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