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INF5620 Lecture: Numerical solution of the Navier-Stokes equations

INF5620 Lecture: Numerical solution of the Navier-Stokes equations

Dec 6, 2012

Table of contents

The physical and mathematical problem
      The Navier-Stokes equations
      Boundary conditions
The classical splitting method
      A simple, naive approach
      A working scheme
            Summary
      Boundary conditions
      Spatial discretization by the finite element method
      Increasing the implicitness
Methods based on slight compressibility
Applications

The physical and mathematical problem

Applications involving fluid flow:

The Navier-Stokes equations

Assumptions:

Primary unknowns:


Figure 1: Flow around a cylinder.

Momentum balance (Newton's 2nd law):

uut+(uu)uu=1ϱp+ν2uu+ff,

Mass balance (eq. of continuity):

uu=0.

Boundary conditions

The classical splitting method

Idea: split the N-S equations into simpler problems (operator splitting).

A simple, naive approach

The equation for uu looks like a diffusion equation...why not a Forward Euler scheme?

uut+(uu)uu=1ϱp+ν2uu+ff

uun+1uunΔt+(uun)uun=1ϱpn+ν2uun+ffn,

uun+1=uunΔt(uun)uunΔtϱpn+Δtν2uun+Δtffn.

Two fundamental problems:

A working scheme

Idea: Forward Euler in time, but evaluate p at tn+1 and enforce uun+1=0.

uun+1=uunΔt(uun)uunΔtϱpn+1+Δtν2uun+Δtffn,uun+1=0

Intermediate velocity (Forward Euler):

uu=uunΔt(uun)uunβΔtϱpn+Δtν2uun+Δtffn

Seek correction δuu such that

uun+1=uu+δuu, fulfills

uun+1=0.

Subtract uu equation from original uun+1 equation to find δuu:

δuu=uun+1uu=ΔtϱΦ, where

Φ=pn+1βpn.

The oldest methods had β=0, but β0 gives in general better speed and accuracy.

uun+1=0 implies

δuu=uu, which gives

2Φ=ϱΔtuu.

When Φ is computed,

uun+1=uuΔtϱΦ, and

pn+1=Φ+βpn.

Summary

  1. Compute the intermediate velocity uu from (???)
  2. Solve the Poisson equation (???) for Φ
  3. Update the velocity: uun+1=uuΔtϱΦ
  4. Update the pressure: pn+1=Φ+βpn
Basically, we have u=f approximation problems (1, 3, 4) and a Poisson equation to solve.

Boundary conditions

Problem: p condition at one point only in the original N-S equations. Now we need boundary conditions for Φ along the whole boundary (Poisson equation).

Spatial discretization by the finite element method

Ω(uuvv(u)+Δt((uu1)uu1)vv(u)Δtϱpvv(u)+Δtνuu1vv(u)Δtf1)dx+ΩN,u(νuunpnn)vv(u)ds, vv(u)V(u).

Natural boundary condition:

νuunpnn(=0) Usually uu/n=0 and p=0 at outlets.

Pressure Poisson equation:

ΩΦv(Φ)dx=ϱΔtΩuuv(Φ)dx+ΩN,pΦnv(Φ)ds,v(Φ)V(Φ).

Velocity update:

Ωuuvv(u)dx=Ω(uuΔtϱΦ)vv(u)dx,vv(u)V(u).

Pressure update:

Ωpv(Φ)dx=Ω(Φ+βp1)v(Φ)dx,v(Φ)V(Φ).

Increasing the implicitness

Stability (due to Forward Euler-style scheme):

Δth22ν+Uh. h: minimum element size, U: typical velocity.

Better stability by a Backward Euler scheme:

uun+1=uunΔt(uun+1)uun+1Δtϱpn+1+Δtν2uun+1+Δtffn+1,uun+1=0.

Intermediate velocity (pn+1βpn):

uu=uunΔt(uu)uuβΔtϱpn+1+Δtν2uu+Δtffn+1

Deal with nonlinearity in a simple way (1 Pickard it.):

(uu)uu(uun)uu.

Then we have a linear problem for uu:

uu=uunΔt(uun)uuβΔtϱpn+Δtν2uu+Δtffn+1

Correction (assume uun+1uu small):

δuu=Δt((uun+1)uun+1(uun)uu)ΔtϱΦ+Δtν(2(uun+1uu)ΔtϱΦ.

So, as before,

2Φ=ϱΔtuu

Methods based on slight compressibility

uu=0 is problematic. Allow slight compressibility in the fluid:

pt+c2uu=0. c: speed of sound.

Now we have evolution equations for uu and p:

uut=(uu)uu1ϱp+ν2uu+ff,pt=c2uu.

Forward Euler:

uun+1=uunΔt(uun)uunΔtϱpn+Δtν2uun+Δtffn,pn+1=pnΔtc2uun.

Applications


Figure 2: Flow in a channel.


Figure 3: Flow in a half a channel with a symmetry line.


Figure 4: Flow over a backward facing step.


Figure 5: Flow around a cylinder.