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Technische Universität München

Janos Benk: Computational fluid dynamics (CFD) www5.in.tum.de/wiki/index.php/Lab_Course_Computational_Fluid_Dynamics_-_Summer_10

9th SIMLAB Course, Belgrade, October 7, 2010

Computational fluid dynamics (CFD)9th SIMLAB Course

Janos Benk

October 3-9, 2010




Overview

• Introduction

• Potential flow

• Stokes equation and discretization

• Boundary Conditions

• Navier-Stokes equation and its dicretization

• (Parallelization)




Introduction

• What defines a flow?

• What are the quantities in such a incompressible flow field?

Velocity vector

pressure scalar




Introduction

• We are looking for a relation between the velocity vector and the pressure

• We note vel = (u,v) the velocity vector in 2D

• The pressure is noted by p

• For the case of simplicity we consider only stationary scenarios though the whole

tutorial

• We use a regular structured grid (as the simplest grid)

• The cells form a mesh

x

y One cell




Introduction

• Grid based method

• Finite difference (reuse some knowledge from the previous lecture)

• Other discretization techniques are more favorable in practice, due to the limitations

of the finite difference method.

• “Finite volume”

• “Finite element”




Potential flow

• The first try is the “Potential flow”, the simples flow equation

• The velocity is directly derived from the pressure (potential) flow

• First we specify that per cell no matter can be gained or disappear.

( vel = (u,v) )

0

0

=∂∂+

∂∂

=⋅∇

y

v

x

u

vel

v1

v2

u1 u2

p

x

y




Potential flow

• Replace the “?” with values

3

?

-2

p

x

y

0

0

=∂∂+

∂∂

=⋅∇

y

v

x

u

vel

?

X

-X+1




Potential flow

• Replace the “?” with values

-2

4

3

p

x

y

0

0

=∂∂+

∂∂

=⋅∇

y

v

x

u

vel

?

-3




Potential flow

• The potential is a Poisson equation, with zero right hand side.

• The velocity is directly derived from the pressure (potential) flow

• The different boundary conditions can be implemented through the potential

• To be the solution uniquely determined, we use a Dirichlet boundary condition for

0

0

02

==∇⋅=∇⋅=∇−

p

ppn

pn

p

out

Inlet

Outlet

Walls




Potential flow

• Is the velocity still divergence free?

• Is the following equation still satisfied? vel=(u,v)

• With the following equations:

pvel ∇=

0=⋅∇ vel

02 =∇− p

02 =∇=⋅∇ pvel




Potential flow

• How does the potential field looks for a channel flow?

• And the velocity field (u,v)

p=0

Inlet

p=-1

Outlet

Walls

pvu ∇=),(

02 =∇− p

x

y

p

u




Stokes equation

• There is a “complete” new second equation:

0=⋅∇ vel

extfvelpt

vel +∇=∇+∂

∂ 2

Re

1

0=∂

∂t

vel

• Since we consider only stationary problems:

extfvelp +∇=∇ 2

Re

1




Stokes equation

extfvelp +∇=∇ 2

Re

1

• What does the second equation mean ?

• This is the so called “impulse equation”, at each point the sum of the acting forces

must equal zero

Grad(p)

vel




Stokes equation

• Reformulate the equation in terms of u,vand p instead of vel,p

• Vel=(u,v)0=⋅∇ vel

extfvelp +∇=∇ 2

Re

1

• The equations are

0=∂∂+

∂∂

y

v

x

u

xextfy

u

x

u

x

p,2

2

2

2

Re

1 +

∂∂+

∂∂=

∂∂

yextfy

v

x

v

y

p,2

2

2

2

Re

1 +

∂∂+

∂∂=

∂∂




Stokes equation

• Calculate the parabolic profile of a channel flow:

0)( =Hu

02

2

Re

1c

y

u

x

p =

∂∂=

∂∂

32

20

2Re)( cycy

cyu ++=

xextfy

u

x

u

x

p,2

2

2

2

Re

1 +

∂∂+

∂∂=

∂∂

yextfy

v

x

v

y

p,2

2

2

2

Re

1 +

∂∂+

∂∂=

∂∂

x

y

10)( cxcxp +=

0)0( =u

x

pc

∆∆=0




Stokes equation

• Finite difference discretization:

• “Cell wise” view of the cont. eq.

xextfy

u

x

u

x

p,2

2

2

2

Re

1 +

∂∂+

∂∂=

∂∂

yextfy

v

x

v

y

p,2

2

2

2

Re

1 +

∂∂+

∂∂=

∂∂

0=∂∂+

∂∂

y

v

x

u

• The operators in the velocity points, since the impulse is point wise satisfied (x and y).




Stokes equation xextfy

u

x

u

x

p,2

2

2

2

Re

1 +

∂∂+

∂∂=

∂∂

yextfy

v

x

v

y

p,2

2

2

2

Re

1 +

∂∂+

∂∂=

∂∂

0=∂∂+

∂∂

y

v

x

u




Stokes equation

Boundary conditions:

• No-Slip:

)0,0(),( =vu

• Free-Slip:

• Inflow:

• Outflow: 02

=+= iar

vvv

)0,(),( uvu = ),0(),( vvu =

),(),( 00 vuvu =

0/),( =∂∂ nvu




Stokes equation

• Driven cavity




Stokes equation

• Which are the unknowns?

1

1




Stokes equation• What to do at the boundary?




Stokes equation

p0 p1

p2 p3

v0

u0

v1

u1

• What to do at the boundary?




Stokes equation

p0 p1

p2 p3

v0

u0

v1

u1

-v0

2-u0

-v1

-u1




Stokes equation• Continuity equation:

0=∂∂+

∂∂

y

v

x

u

p0 p1

p2 p3

v0

u0

v1

u1

-v0

2-u0

-v1

-u1

v0 – u0 = 0

u0 + v1 = 0

-v0 - u1 = 0

-v1 + u1 = 0

1

1




Stokes equation

p0 p1

p2 p3

v0

u0

v1

u1

-v0

2-u0

-v1

-u1

(1/Re)(5v0– v1)+p0-p2 = 0

(1/Re)(5u0-u1)-p0+p1 =(1/Re) 2

(1/Re)(-v0 +5v1)+p1-p3 = 0

(1/Re)(-u0 + 5u1)-p2+p3 = 0

1

1

0Re

12

2

2

2

=

∂∂+

∂∂−

∂∂

y

u

x

u

x

p

0Re

12

2

2

2

=

∂∂+

∂∂−

∂∂

y

v

x

v

y

p




Stokes equation

v0 – u0 = 0

u0 + v1 = 0

-v0 - u1 = 0

-v1 + u1 = 0

(1/Re)(5v0– v1)+p0-p2 = 0

(1/Re)(5u0-u1)-p0+p1 =(1/Re) 2

(1/Re)(-v0 +5v1)+p1-p3 = 0

(1/Re)(-u0 + 5u1)-p2+p3 = 0

=

−−−−

−−−−

−−−

−

0

0

2

0

0

0

0

0

3

2

1

0

1

1

0

0

1151

1151

1115

1115

11

11

11

11

p

p

p

p

u

v

u

v

Write the system of equation with Re=1

With unknown vector [v0,u0,v1,u1,p0,p1,p2,p3] , first cont. eq. then momentum equation




Stokes equation

• Due to the singularity we set p0=0 and delete the fourth line from the system

=

−−−−

−−−

−

0

0

2

0

0

0

0

3

2

1

1

1

0

0

1151

1151

115

115

11

11

11

p

p

p

u

v

u

v

• We calculate the solution with Octave




Stokes equation

0 1.5

0.5

1.0

0.083

0.083 -0.083

-0.083

• The solution vector is:[0.083333 0.083333 - 0.083333 - 0.083333 1.5 0.5 1.0]

• “Visualize” these data on the grid




Stokes equation

• 3D Example:




Navier-Stokes equation

• Is the velocity still divergence free?

0=⋅∇ vel

( ) extfvelpvelvelt

vel +∇=∇+∇⋅+∂

∂ 2

Re

1

0=∂

∂t

vel

• Since we consider only stationary problems:

( ) extfvelpvelvel +∇=∇+∇⋅ 2

Re

1





• The new term in the equation is the so called “convective” or “transport” term

( )velvel ∇⋅

∂∂+

∂∂

∂∂+

∂∂

=

∂∂∂∂

⋅

y

vv

x

vu

y

uv

x

uu

v

u

y

x

v

u

• Which in more detailed form is (see the non-linearity)

• The velocity field transports the

“velocity”.

• The diffusion “spreads” the velocity in

each direction equally

• This transports in the flow direction

velocityGrad(p)

vel





• Which model to use when?

• Would you use Navier-Stokes for “slow” , viscous flow (diffusion term is dominant)?

• Would you use Navier-Stokes for non viscous flow?

• (Péclet number, is a good indicator for this)





• We reformulate the equation in terms

of u,v and p instead of vel,p

• Vel=(u,v)0=⋅∇ vel

• The equations are (the convection term is transformed slightly)

0=∂∂+

∂∂

y

v

x

u

( ) ( )xextf

y

uv

x

u

y

u

x

u

x

p,

2

2

2

2

2

Re

1 +∂

∂−∂

∂−

∂∂+

∂∂=

∂∂

( ) ( )yextf

y

v

x

uv

y

v

x

v

y

p,

2

2

2

2

2

Re

1 +∂

∂−∂

∂−

∂∂+

∂∂=

∂∂

( ) extfvelvelvelp +∇⋅−∇=∇ 2

Re

1





• Convection term, x component





• Convection term, y component





Outline of the solving method:

• Coupled approach:

Using non-linear solvers (fix point or Newton method)

• Partitioned approach only for time dependent problem

b

p

v

u

vuA =

),(

...2 =∇ p

...),(11 =++ tt

vu





• 2D Example:




Parallelization

• Shared memory systems

• The basics of parallelization on the matrix level on distributed memory system

• Distribute the unknown vector to processes

• Distribute the corresponding lines of the matrix and the right hand side

bAx =

=

p0

p1

p2

p0

p1

p2

p0

p1

p2




Parallelization

• Let’s think in terms of iterative processes

• How to divide among processors?




Parallelization

• We need additional cells in the same way as we need boundary cells




Parallelization

• (Ghost cells)




Parallelization

• Communication needed (?)




Thank you for your attention!

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