A conservative FE-discretisation of the Navier-Stokes equation

transcript

JASS 2005, St. PetersburgThomas Satzger

Overview

• Navier-Stokes-Equation– Interpretation– Laws of conservation

• Basic Ideas of FD, FE, FV

• Conservative FE-discretisation of Navier-Stokes-Equation

Navier-Stokes-Equation

• The Navier-Stokes-Equation is mostly used for the numerical simulation of fluids.

• Some examples are- Flow in pipes- Flow in rivers- Aerodynamics- Hydrodynamics

The Navier-Stokes-Equation writes:

gupuuut

Equation of momentum

0 u Continuity equation

with u Velocity field

Pressure fieldp

Density

Dynamic viscosity

The interpretation of these terms are:

gupuuut

Derivative of velocity field

Convection

Pressure gradientDiffusion

Outer Forces

The corresponding for the components is:

for the momentum equation, and

for the continuity equation.

With the Einstein summation jiji

jjiji xayxay

and the abbreviation we get:tt

iijjiijjii gupuuu

)2,1( i

for the momentum equation, and

for the continuity equation.

Now take a short look to the dimensions:

iijjiijjii gupuuu

Navier-Stokes-Equation - Interpretation

We see that the momentum equations handles with accelerations. If we rewrite the equation, we get:

iijjiijjii gupuuu

This means:

Total acceleration is the sum of the partial accelerations.

Interpretation of the Convection

fluid particle

iiu pi

1ijj u

igijj uu

Transport of kinetic energy by moving the fluid particle

Interpretation of the pressure Gradient

fluid particle

iiu ijj uu ijj u

Acceleration of the fluid particle by pressure forces

Interpretation of the Diffusion

iiu ijj uu pi

1igijj u

fluid particle

Distributing of kinetic Energy by friction

Interpretation of the continuity equation 0 jju

• Conservation of mass in arbitrary domain

11u 12u

this means: influx = out flux

huhu 11 21 huhu

01212 2211

for 0h we get

02211 uu

Navier-Stokes-Equation - Laws of conservation

Conservation of kinetic energy:

We must know that the kinetic energy doesn't increase, this means:

Proof:

duuduE iikin 212

duuuuduuduudtdE

iititiiitiikin 21

0kinEdtd

With the momentum equation it holds

Using the relations (proof with the continuity equation)

puuuu iijjijjii

puuuuuu

puuuuduuEdtd

iiijjiijji

iijjijjiitikin

ijjiji uuuu

ijjiijijijji

ijijijii

product

ruleijii

uuuuuuuuu

uuuuuuuuuuuu

Additionally it holds

Therefore we get

Due to Greens identity we have

puuuuuuE

iiijjiijjikin

continuity

equationjjjj

product

rulejj

upupuppuup0

0 ijij

ijji uuuu

This means in total

We have also seen that the continuity equation is very important for energy conservation.

0kinEdtd

Basic Ideas of FD, FE, FV

We can solve the Navier-Stokes-Equations only numerically.Therefore we must discretise our domain. This means, we regard our Problem only at finite many points.There are several methods to do it:

•Finite Difference (FD)One replace the differential operator with the difference operator, this mean you approximate by

or an similar expression.

xfhxfxf

• Finite Volume (FV)- You divide the domain in disjoint subdomains- Rewrite the PDE by Gauß theorem- Couple the subdomains by the flux over the

boundary• Finite Elements (FE)

- You divide the domain in disjoint subdomains- Rewrite the PDE in an equivalent variational

problem- The solution of the PDE is the solution of the

variational problem

Comparison of FD, FE and FV

Finite Difference

Finite Element

Finite Volume

Advantages and DisadvantagesFinite Difference:

+ easy to programme- no local mesh refinement- only for simple geometries

Finite Volume:+ local mesh refinement+ also suitable for difficult geometries

Finite Element:+ local mesh refinement+ good for all geometries

BUT:Conservation laws aren't always complied by the discretisation. This can lead to problems in stability of the solution.

Conservative FE-Elements

We use a partially staggered grid for our discretisation.

We write: N for the number of grid pointsiu

ivfor the horizontal velocity in the i-th grid pointfor the vertical velocity in the i-th grid point

The FE-approximation is an element of an finite-dimensional function space with the basis

The approximation has the representation

Nff 21,..., 2: IRfi whereby

iNiiiih fvfuu

If we use a Nodal basis, this means

we can rewrite the approximation

point grid th-iif

0points gridother allNif

0point grid th-iNif

points gridother allif

Every approximation should have the following properties:continuousconservative

In the continuous case the continuity equation was very important for the conservation of mass and energy.

If the approximation complies the continuity pointwise in the whole area, e.g. , then the approximation preserves energy.

Now we search for a conservative interpolation for the velocities in a box.

We also assume that the velocities complies the discrete continuity equation.

1 1, u v 2 2, u v

3 3, u v 4 4, u v

1 1, u v 2 2, u v

3 3, u v 4 4, u v

hvvhuu

1 1, u v 2 2, u v

3 3, u v 4 4, u v

hvvhuu

4331 hvvhuu

We also assume that the velocities complies the discrete continuity equation:

1 1, u v 2 2, u v

3 3, u v 4 4, u v

hvvhuu

4331 hvvhuu

043314321 vvvvuuuu (1)

The bilinear interpolation isn't conservative1 1, u v 2 2, u v

3 3, u v 4 4, u v

1 2 3 4

, 1 1 1 1

x y x y x y x yu x y u u u u

h h h h h h h hx y x y x y x y

v x y v v v vh h h h h h h h

3 3, u v 4 4, u v

1 2 3 4

, 1 1 1 1

x y x y x y x yu x y u u u uh h h h h h h hx y x y x y x y

v x y v v v vh h h h h h h h

It is easy to show that

0general in

3 3, u v 4 4, u v

, , ,,1, 3, 4,2,

,2 4,,1 4,

, 1 1 1 1

, 1 1 1

f x y f x y f x yf x yu u uu

f x yvxv

x y x y x y x yu x y u u u uh h h h h h h hx y x y x yv x y v v vh h h h h h

, ,4, 4 4,

x y f x yv v

x yvh h

Basis on the box

These basis function for the bilinear interpolation are calledPagoden.

The picture shows the function on the whole support.

, i iu v

Now we are searching a interpolation of the velocities which complies the continuity equation on the box.

How can we construct such an interpolation?

1 1, u v 2 2, u v

3 3, u v 4 4, u v

1 1, u v 2 2, u v

3 3, u v 4 4, u v

Divide the box in four triangles.

51u52u53u

51v52v53v

Divide the box in four triangles.Make on every triangle an linear interpolation.

What's the right velocity in the middle?

51u52u53u

51v52v53v

51u52u53u

51v52v53v

We must have at every point in the box the following relations:

, , 0 , ,

u x y v x y u x y v x yx y x y

v x y u x yy x

, , 0 , ,

v x y u x yy x

51u52u53u

51v52v53v

, , 0 , ,

v x y u x yy x

51u52u53u

51v52v53v

, , 0 , ,

v x y u x yy x

51u52u53u

51v52v53v

, , 0 , ,

v x y u x yy x

51u52u53u

51v52v53v

, , 0 , ,

v x y u x yy x

51u52u53u

51v52v53v

, , 0 , ,

v x y u x yy x

51u52u53u

51v52v53v

51u52u53u

51v52v53v

5 33u u

51u52u53u

51v52v53v

1 35 33 2

u uhu uh

51u52u53u

51v52v53v

1 3 1 35 33 2 2

u u v vh hu uh h

51u52u53u

51v52v53v

1 3 1 35 3 3 1 3 1 3 1 3 1 33

1 12 2 2 2u u v vh hu u u u u v v u u v vh h

51u52u53u

51v52v53v

1 3 1 35 3 3 1 3 1 3 1 3 1 33

5 42u u

51u52u53u

51v52v53v

1 3 1 35 3 3 1 3 1 3 1 3 1 33

2 45 42 2

h u uu u

51u52u53u

51v52v53v

1 3 1 35 3 3 1 3 1 3 1 3 1 33

2 4 2 45 42 2 2

h u u h v vu u

51u52u53u

51v52v53v

1 3 1 35 3 3 1 3 1 3 1 3 1 33

2 4 2 45 4 4 2 4 2 4 2 4 2 42

1 12 2 2 2h u u h v v

u u u u u v v u u v vh h

51u52u53u

51v52v53v

1 3 1 35 3 3 1 3 1 3 1 3 1 33

2 4 2 45 4 4 2 4 2 4 2 4 2 42

1 12 2 2 2h u u h v v

u u u u u v v u u v vh h

free are and

41 55 uu

51u52u53u

51v52v53v

4 3 4 35 3 3 4 3 4 3 3 4 3 44

1 12 2 2 2v v u uh hv v v v v u u u u v vh h

51u52u53u

51v52v53v

4 3 4 35 3 3 4 3 4 3 3 4 3 44

1 12 2 2 2v v u uh hv v v v v u u u u v vh h

2 1 2 15 1 1 2 1 2 1 1 2 1 21

1 12 2 2 2h v v h u u

v v v v v u u u u v vh h

free are and

23 55 uv

Till now we have: 5 1 3 1 33

u u u v v 5 2 4 2 4212

u u u v v

5 3 4 3 4412

v u u v v 5 1 2 1 2112

v u u v v free are and 23 55 uv

free are and 41 55 uu

Till now we have:

With the discrete continuity equation

we get

5 1 3 1 3312

u u u v v 5 2 4 2 4212

u u u v v

5 3 4 3 4412

v u u v v 5 1 2 1 2112

1 2 3 4 1 2 3 4 0 (1)u u u u v v v v

5 53 2u u 5 54 1v v

Till now we have:

With the discrete continuity equation

we get

Therefore we choose

5 1 3 1 3312

u u u v v 5 2 4 2 4212

u u u v v

5 3 4 3 4412

v u u v v 5 1 2 1 2112

1 2 3 4 1 2 3 4 0 (1)u u u u v v v v

5 53 2u u 5 54 1v v

5 5 5 5 5 1 2 3 4 1 2 3 41 2 3 4

1: : : : :4

1: : : : :

u u u u u u u u u v v v v

v v v v v u u u u v v v v

5 1 2 3 4 1 2 3 4

1:41:4

u u u u u v v v v

v u u u u v v v v

Now we calculate the basis., ,

Ni u i N uh

i ii v i N vh i

f fuu v

1 1, u v 2 2, u v

3 3, u v 4 4, u v

5 5, u v

5 1 2 3 4 1 2 3 4

1:41:4

u u u u u v v v v

v u u u u v v v v

Now we calculate the basis.

1 1, u v 2 2, u v

3 3, u v 4 4, u v

5 5, u v

5 1 2 3 4 1 2 3 4

1:41:4

u u u u u v v v v

v u u u u v v v v

Ni u i N u

ii v i N vh i

f fuu v

,i uf ,i vf

Ni u i N uh

ii i N vih v

1 1, u v 2 2, u v

3 3, u v 4 4, u v

5 5, u v

5 1 2 3 4 1 2 3 4

1:41:4

u u u u u v v v v

v u u u u v v v v

,i uf ,i vf

Now we calculate the basis.

5 1 2 3 4 1 2 3 4

1:41:4

u u u u u v v v v

v u u u u v v v v

Ni u i N u

ii v i N vh i

f fuu v

1 1, u v 2 2, u v

3 3, u v 44, u v

5 5, u v

,i uf ,i vf

Ni u i N u

ii v i N vh i

f fuu v

1 1, u v 2 2, u v

3 3, u v 44, u v

5 5, u v

,i uf ,i vf

45 1 2 3 1 2 3 4

5 1 2 3 4 1 2 3 4

u u u u v v v v

Ni u i N u

ii v i N vh i

f fuu v

1 1, u v 2 2, u v

3 3, u v 44, u v

5 5, u v

,i uf ,i vf

45 1 2 3 1 2 3 4

5 1 2 3 4 1 2 3 4

u u u u v v v v

Ni u i N u

ii v i N vh i

f fuu v

5 1 2 3 4 1 2 3 4

1:41:4

u u u u u v v v v

v u u u u v v v v

1 1, u v 2 2, u v

33, u v 4 4, u v

5 5, u v

,i uf ,i vf

Ni u i N u

ii v i N vh i

f fuu v

1 1, u v 2 2, u v

33, u v 4 4, u v

5 5, u v

,i uf ,i vf

5 1 2 4 1 2 3 4

5 1 2 3 4 1 2

u u u u v v v v

v u u u u

v v v v

Ni u i N u

ii v i N vh i

f fuu v

1 1, u v 2 2, u v

33, u v 4 4, u v

5 5, u v

,i uf ,i vf

5 1 2 4 1 2 3 4

5 1 2 3 4 1 2

u u u u v v v v

v u u u u

v v v v

Ni u i N u

ii v i N vh i

f fuu v

5 1 3 4 1 2 3 4

5 1 2 3 4 1 4

u u u u v v v v

u u v v v v

1 1, u v 22, u v

3 3, u v 4 4, u v

5 5, u v0

,i uf ,i vf

Ni u i N u

ii v i N vh i

f fuu v

5 2 3 4 1 2 3 4

5 1 2 3 4 1 2 3 4

u u u u v v v v

v u u u v

u v v v

11, u v 2 2, u v

3 3, u v 4 4, u v

5 5, u v

,i uf ,i vf

Ni u i N uh

ii i N vih v

5 1 2 3 4 1 2 3 4

1:41:4

u u u u u v v v v

v u u u u v v v v

1 1, u v 2 2, u v

3 3, u v 4 4, u v

5 5, u v

,i uf ,i vf

Ni u i N uh

ii i N vih v

5 1 2 3 4 1 2 3 4

5 1 2 3 1 2 3 44

u u u u u v v v v

v u u u v v v vu

1 1, u v 2 2, u v

3 3, u v 44, u v

5 5, u v0

,i uf ,i vf

Ni u i N uh

ii i N vih v

5 1 2 3 4 1 2 3 4

5 1 2 4 13 2 3 4

u u u u u v v v v

v u u u v v v vu

1 1, u v 2 2, u v

33, u v 4 4, u v

5 5, u v

,i uf ,i vf

Ni u i N uh

ii i N vih v

1 1, u v 22, u v

3 3, u v 4 4, u v

5 5, u v

,i uf ,i vf

5 1 2 3 4 1 2 3 4

5 1 3 4 1 2 32 4

u u u u u v v v v

v u u u v v vu v

Ni u i N uh

ii i N vih v

5 1 2 3 4 1 2 3 4

5 2 3 4 1 31 2 4

u u u u u v v v v

v u u u v v vu v

11, u v 2 2, u v

3 3, u v 4 4, u v

5 5, u v

,i uf ,i vf

Ni u i N uh

ii v i N vh i

5 1 2 3 4 1 2 3 4

1:41:4

u u u u u v v v v

v u u u u v v v v

1 1, u v 2 2, u v

3 3, u v 4 4, u v

5 5, u v

,i N uf ,i N vf

Ni u i N uh

ii v i N vh i

5 1 2 3 4 1 2 3 4

1:41:4

u u u u u v v v v

v u u u u v v v v

1 1, u v 2 2, u v

3 3, u v 4 4, u v

5 5, u v

,i N uf ,i N vf

Ni u i N uh

ii v i N vh i

5 1 2 3 4 1 2 3 4

1:41:4

u u u u u v v v v

v u u u u v v v v

1 1, u v 2 2, u v

3 3, u v 4 4, u v

5 5, u v

, ,i N u i vf f , ,i N v i uf f

, ,i u i N vf f , ,i v i N uf f

Linear interpolation providesthe basis.

View on conservative elements in 3D

Partially staggered grid in 3D

, , u v w

We also search for a conservative interpolation of the velocities.

, , u v w

We also search for a conservative interpolation of the velocities.

, , u v w

Divide every box into 24 tetrahedrons, on which you make a linear interpolation

A conservative FE-discretisation of the Navier-Stokes equation

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