Transcript SIMPLE Semi-Implicit Method for Pressure

Lecture Objectives
• Review
• Define Residual and Relaxation
• SIMPLE CFD Algorithm
SIMPLE Semi-Implicit Method for Pressure-Linked Equations
Review
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Conservation equations
Turbulent flow and turbulence modeling
RANS Equation
Discretization
System of equation and solution methods
– Accuracy
– Numerical stability of solution procedure
(new)
• Solution algorithm (new today)
Residual
Example:
x-exp(1/x)-2=0
Find x using iteration
Explicit form 1:
Explicit form 2:
x=exp(1/x)+2
x=1/(ln(x)-ln(2))
Solution process:
Guess x0
Not all iteration
process converge!
Iteration :
x1=exp(1/x0)+2 ,
X2=exp(1/x1)+2 ,
……..
…….
R1=x1-x0
R2=x2-x1
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
RFijk=Fkijk-Fk-1ijk
Variable: p,V,T,…
iteration
cell position
• Total residual for the simulation domain
RFtotal=S|RFijk|
For all cells
• Scaled (normalized) residual
RF=S|RFijk|/FF
Flux of variable F used for normalization
Vary for different CFD software
Relaxation
Relaxation with iterative solvers:
divergence
variable
When the equations are nonlinear
it can happen that you get divergency
in iterative procedure for solving considered
time step
solution
convergence
Solution is Under-Relaxation:
iteration
Y*=f·Y(n)+(1-f)·Y(n-1)
Y – considered parameter , n –iteration , f – relaxation factor
Value which is should be used for the next iteration
For our example Y*in iteration 101=f·Y(100)+(1-f) ·Y(99)
f = [0-1] – under-relaxation -stabilize the iteration
f = [1-2] – over-relaxation - speed-up the convergence
Under-Relaxation is often required when you have nonlinear equations!
Example of relaxation
(example from homework 3 assignment)
Example: Advection diffusion equation, 1-D, steady-state, 4 nodes
a NTN-1  b NTN  c NTN1  f N
1) Explicit format:
TN  1/ b Nf N  a N /b NTN-1  c N /b NTN1
1
2
3
4
2) Guess initial values:
T10  ..., T20  .., T30  .., T40  ..
3) Substitute and calculate:
T11  1/ b1f1  c1/b1T0 2
T12  1/ b 2 f 2  a 2 /b 2 T11  c 2 /b 2 T 0 3
T13  1/ b3f 3  a 3 /b3T12  c3 /b3T 0 4
T14  1/ b 4 f 4  a 4 /b 4 T13
4) Substitute and calculate:
Substitute and calculate:
………………………….
T11  ..., T21  .., T31  .., T41  ..
T11r  fT11  (1- f)T10 , T21r  fT21  (1- f)T20 , ....
T12  ..., T22  .., T32  .., T42  ..
T12r  fT12  (1- f)T11 , T22r  fT22  (1- f)T21 , ....
Navier Stokes Equations
Continuity equation
v x v y v z


0
x
y z
This velocities that constitute advection coefficients: F=rV
Momentum x
v x
v x
v x
v x
p
2vx
 2vx
2vx
ρ(
 vx
 vy
 vz
)    μ 2  μ 2  μ 2  SM x
τ
x
y
z
x
x
y
z
Momentum y
v y
v y
v y
v y
2vy
2vy
2vy
p
ρ(
 vx
 vy
 vz
)    μ 2  μ 2  μ 2  SM y  ρ g (T  T )
τ
x
y
z
y
x
y
z
Momentum z
v z
v z
v z
v z
p
 2 vz
 2 vz
 2 vz
ρ(
 vx
 vy
 vz
)    μ 2  μ 2  μ 2  SM z
τ
x
y
z
z
x
y
z
Pressure is in momentum equations
which already has one unknown
In order to use linear equation solver we need to solve two problems:
1) find velocities that constitute in advection coefficients
2) link pressure field with continuity equation
Pressure and velocities in NS
equations
How to find velocities that constitute in advection coefficients?
v x
v x
v x
v x
p
2vx
 2vx
2vx
ρ(
 vx
 vy
 vz
)    μ 2  μ 2  μ 2  SM x
τ
x
y
z
x
x
y
z
a P Vx,P  a E Vx, E  a WVx, W  aSVx,S  a NVx, N  a HVx, H  a LVx,P L  f
aP  6

x
2
x
Vx

,
a


W
x 2
x
x 2
................................
aE  

r
Vx  Vy  Vz
r
For the first step use Initial guess
And for next iterative steps use
the 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
v x
v x
v x
v x
p
2vx
 2vx
2vx
ρ(
 vx
 vy
 vz
)   μ
 μ 2  μ 2  SM x
τ
x
y
z
x
x 2
y
z
x
W

p Pw – Pe (PW  PP )/2 – (PP  PE )/2 (PW – PE )/2



x
x
x
x
a P Vx P  a E Vx E  a W Vx W  a S Vx S  a N Vx N  a H Vx H  a L Vx L  f 
Aw
x
P
x
Ae
Aw=Ae=Aside
(PW – PE )/2
Aside
x
We have two additional equations for y and x directions
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
E
SIMPLE algorithm
Guess 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:
x:
a P Vx P  a E Vx E  a W Vx W  a S Vx S  a N Vx N  a H Vx H  a L Vx L  f 
y:
………………..
………………..
z:
Solution is:
(PW – PE )/2
Aside
x
V *x P , V *x E , V *x W , V *x S , V *x N , V *x H , V *x L
2) The pressure and velocity correction
P = P* + P’
V = V* + V’
P’ – pressure correction
V’ – velocity correction
For all nodes E,W,N,S,…
Substitute P=P* + P’ into momentum equations (simplify equation) and obtain
V’=f(P’)
V = V* + f(P’)
3) Substitute V = V* + f(P’) into continuity equation solve P’ and then V
4) Solve T , k , e equations
SIMPLE algorithm
start
p=p*
Guess p*
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
no
Converged
(residual check)
yes
end
Other methods
SIMPLER
SIMPLEC
PISO
variation of SIMPLE
COUPLED - use Jacobeans of nonlinear velocity functions to form
linear matrix ( and avoid iteration )