Lecture Objectives:

Explicit vs. Implicit

Residual, Stability, Relaxation

Simple algorithm

General Transport Equation unsteady-state 1-D

Fully explicit method:

Implicit method:

ΦΔττ

WΔττ

EΔττ

Peff Φ,Δττ

P(or W)Δττ

P)or E(x

τP

ΔττP SΦΦ2Φ

ΔxΔx

Γ ΦΦ

Δx

ρV

Δτ

ΦΦρ

ΦτW

τE

τP

eff Φ,τP(or W)

τP)or E(

xτP

ΔττP SΦΦ2Φ

ΔxΔx

Γ ΦΦ

Δx

ρV

Δτ

ΦΦρ

Value form previous time step (known value)

Make the difference between - Calculation for different time step- Calculation in iteration step

source1N1NNp

1-NNx

NN qTTT2xx

k/c TT

x

ρVTT

ρ

Explicit:

Unsteady state Advection diffusion equation, 1-D

Rarely used due to the problem with stability of calculation

To achieve stable calculation should be very small

NNN T/aa-f/aT/aaT/aaT NNoN1N1N1-NN1-N

NNN Ta-fTaTaTa No11N1-N1-NN

Unsteady state Advection diffusion equation, 1-D

source1N1NN1-NNx

NN qTTT2xx

k/cp TT

x

ρVTT

ρ

Implicit:

Iterative method:

NNN Ta-fTaTaTa No11N1-N1-NN

NNN T/aa-f/aT/aaT/aaT NNoN1N1N1-NN1-N

Explicit format (to solve system of equations)

2) Guess initial values: ..T ..,T ..,T ...,T 04

03

02

01

3) Substitute and calculate:

4….) Iterate for considered time stepIn iteration substitute only these values

Make the difference between iteration and calculation for next time step

ResidualExample:

x-exp(1/x)-2=0

Find x using iteration

Explicit form 1:

x=exp(1/x)+2

Explicit form 2:

x=1/(ln(x)-ln(2))

Solution process:

Guess x0

Iteration :

x1=exp(1/x0)+2 , R1=x1-x0

X2=exp(1/x1)+2 , R2=x2-x1

……..…….

Not all iteration process converge!

See the example for the same equation

Convergence example

Explicit form 2:

x=1/(ln(x)-ln(2)

Residual calculation for CFD

• Residual for the cell

Rijk=kijk-k-1

ijk

• Total residual for the simulation domain

Rtotal=Rijk|

• Scaled (normalized) residual R=Rijk|/F

iteration

cell positionVariable: p,V,T,…

For all cells

Flux of variable used for normalizationVary for different CFD software

RelaxationRelaxation with iterative solvers:

When the equations are nonlinearit can happen that you get divergency in iterative procedure for solving consideredtime step

Under-Relaxation is often required when you have nonlinear equations!

iteration

convergence

variabledivergence

solution

Solution is Under-Relaxation:

Y*=f·Y(n)+(1-f)·Y(n-1) Y – considered parameter , n –iteration , f – relaxation factor

For our example Y*in iteration 101=f·Y(100)+(1-f) ·Y(99)

f = [0-1] – under-relaxation -stabilize the iterationf = [1-2] – over-relaxation - speed-up the convergence

Value which is should be used for the next iteration

Example of relaxation(example from homework 3 assignment)

N1NNNN1-NN fTcTbTa

Example: Advection diffusion equation, 1-D, steady-state, 4 nodes

1NNN1-NNNNNN T/bcT/bafb/1T 1 2 3 4

1) Explicit format:

2) Guess initial values:

..T ..,T ..,T ...,T 04

03

02

01

3) Substitute and calculate:

20

111111 T/bcfb/1T

30

2211

222221 T/bcT/bafb/1T

40

3321

333331 T/bcT/bafb/1T

31

444441 T/bafb/1T

..T ..,T ..,T ...,T 14

13

12

11

Substitute and calculate:4) ..T ..,T ..,T ...,T 24

23

22

21

………………………….

.... ,f)T-(1fTT ,f)T-(1fTT 02

12

1r2

01

11

1r1

.... ,f)T-(1fTT ,f)T-(1fTT 12

22

2r2

11

21

2r1 Substitute and calculate:

Navier Stokes Equations

0z

v

y

v

x

v zyx

)(Sz

vμ

y

vμ

x

vμ

y

p)

z

vv

y

vv

x

vv

τ

vρ( yM2

y2

2

y2

2

y2

yz

yy

yx

y

TTgρ

xM2x

2

2x

2

2x

2x

zx

yx

xx S

z

vμ

y

vμ

x

vμ

x

p)

z

vv

y

vv

x

vv

τ

vρ(

zM2z

2

2z

2

2z

2z

zz

yz

xz S

z

vμ

y

vμ

x

vμ

z

p)

z

vv

y

vv

x

vv

τ

vρ(

In order to use linear equation solver we need to solve two problems:

1) find velocities that constitute in advection coefficients2) link pressure field with continuity equation

This velocities that constitute advection coefficients: F=V

Pressure is in momentum equations which already has one unknown

Continuity equation

Momentum x

Momentum y

Momentum z

Pressure and velocities in NS equations

How to find velocities that constitute advection coefficients?

xM2x

2

2x

2

2x

2x

zx

yx

xx S

z

vμ

y

vμ

x

vμ

x

p)

z

vv

y

vv

x

vv

τ

vρ(

fVaVaVaVaVaVaVaLPx,LHx,HNx,NSx,SWx,WEx,EPx,P

................................

a ,V

a

VVV6a

2Wx

2E

zyx

2P

xxx

xx

For the first step use Initial guessAnd for next iterative steps usethe values from previous iteration

Pressure and velocities in NS equations

How to link pressure field with continuity equation?

SIMPLE (Semi-Implicit Method for Pressure-Linked Equations ) algorithm

The momentum equations can be solved only when the pressure field is given or is somehow estimated. Use * for estimated pressure and the corresponding velocities

P EW

x

x x

xxx

)/2P– (P

)/2P (P– )/2P (P

P– P

x

p EWEPPWew

xM2x

2

2x

2

2x

2x

zx

yx

xx S

z

vμ

y

vμ

x

vμ

x

p)

z

vv

y

vv

x

vv

τ

vρ(

sideEW

LxLHxHNxNSxSWxWExEPxP

)/2P– (P fVaVaVaVaVaVaVa A

x

AeAw

Aw=Ae=Aside

We have two additional equations for y and x directions

SIMPLE algorithmGuess pressure field: P*W, P*P, P*E, P*N , P*S, P*H, P*L

1) For this pressure field solve system of equations:

Solution is:

sideEW

LxLHxHNxNSxSWxWExEPxP

)/2P– (P fVaVaVaVaVaVaVa A

xx:

y:

z:

………………..………………..

LxHxNxSxWxExPx *V,*V,*V,*V,*V,*V,*V

P = P* + P’

2) The pressure and velocity correction

P’ – pressure correction

V = V* + f(P’)

For all nodes E,W,N,S,…

V’ – velocity correction

Substitute P=P* + P’ into momentum equations (simplify equation) and obtain

3) Substitute V = V* + f(P’) into continuity equation solve P’ and then V

V’=f(P’)

V = V* + V’

4) Solve T , k , equations

SIMPLE algorithm

Step1: solve V* from momentum equations

Step2: introduce correction P’ and express V = V* + f(P’)

Step3: substitute V into continuity equation solve P’ and then V

Step4: Solve T , k , e equations

Guess p*

start

end

Converged (residual check)

yes

no

p=p*

Other methods

SIMPLERSIMPLEC variation of SIMPLEPISO

COUPLED - use Jacobeans of nonlinear velocity functions to form linear matrix ( and avoid iteration )