lecture5 cfd

8/10/2019 Lecture5 CFD

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From conservation laws to Navier-Stokes equations

Finite Volume formulation of Navier-Stokes equations

Discretization of RANSE

Outline of lecture 5

From conservation laws to Navier-Stokes equations,

Finite-Volume formulation of Navier-Stokes equations,

Mass and Momentum equations,

Volume and surface integral evaluation

ECN-CNRS A generalized unstructured finite volume discretisation - Part 1
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F i l N i S k i
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Derivation according to an arbitrary velocity field


Let

Ud be an arbitrary velocity field and g(x, t) a scalar function.

g

t=g

t+g

Ud

For a domain Ddwith a boundaryDdmoving with the velocity

Ud,

we have :

t

DdgdV=

Dd

g

t dV+

Ddg

Ud

n dS


F ti l t N i St k ti
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This relation can be interpreted as follows :

Variation of the inte-

gral ofgin the moving

domain Dd

=

Integral of the tem-

poral variation of g

on Dd

+

Convective flux

convectif of g

across the bound-

aryD

d


From conservation laws to Navier Stokes equations
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Let us recall the theorem establishing the expression of the material

derivative of a volume integral.Let Dmbe a material domain. With some regularity hypothesis on the

fields, we have :

d

dt

Dm

gdV= Dm

g

t

dV+Dm

gU

n dS (1)

For the material domain Dmcoinciding with Ddat timet, we get the

following relation :

d

dt

Dm

g dV=

t

Dd

g dV+

Dd

g

U

Udn dS (2)


From conservation laws to Navier-Stokes equations RANSE formulation
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RANSE formulation

Reynolds Averaged Navier-Stokes Equations

Mass conservation

For isothermal viscous flows, conservation laws are reduced to

momentum and mass conservation. Let us consider a domain Dd ofboundaryDdmoving with the velocity

Ud and Dm, a material domain

coinciding with the material domain at time t.

Mass conservation :d

dt

DmdV= 0

Momentum conservation :

d

dt

D

m

U dV=

D

m

fv dV+

D

m

T dS

fv : volumic force (for us, gravity)T : Surface constraint

T =

n ,

: constraint tensor


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RANSE formulation


Mass conservation

For a Newtonian fluid:

= (p+2

3div(

U))I+ 2D

wherepis the pressure, the dynamic viscosity, I the identity tensor

and D the rate-of-strain tensor.

Using this relation leads to:

d

dt

Dm

U dV =

Dm

fv dV

Dm

p+

2

3div(

U)

n dS+

Dm

2Dn dS

(3)




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RANSE formulation


Mass conservation

Using relation (2) and the Gauss theorem, we get the new expression

of the mass and momentum conservation equations:

t

Dd

dV+

Dd

(U

Ud) n

dS= 0 (4)

t

Dd

U dV+

Dd

U

(U

Ud) n

dS=

Dd

fv dV

Dd

p+

2

3div(

U)

dV +

Dd

div(2D) dV

(5)

wherefv includes the gravity

g and additional source terms to be

specified later.


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q




Mass conservation

Notations

Let us consider an incompressible and isothermal flow of a fluid

defined by its density and dynamic viscosity . LetUibe the

cartesian components in directionsiof velocity,fithe components ofthe volume forces andpthe pressure.

Momentum and mass conservation laws may be written under an

integral formulation for a volume Vbounded by surfaces S, moving at

velocity

Ud.n is an outbound normalized normal-to-the-face vector.


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Mass conservation

From equations (4)and (5), one gets :

t

Dd

dV+

Dd

(U

Ud) n

dS= 0 (6)

t

Dd

UidV +

Dd

Ui

(U

Ud) n

dS=

Dd

fvidV

D

d

xi

p+

2

3div(

U)

dV +

D

d

2

xj

DijdV

(7)




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Mass conservation

RANSE formulation




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Mass conservation

Mean and fluctuating parts decomposition 1/3

The instantaneous velocity components and pressure are

decomposed into mean and fluctuating parts:

Ui = Ui + u

i

p = p+ p (8)

which leads to the modified mass and momentum conservation

equations (also called Reynolds Averaged Navier-Stokes Equations(RANSE)).



Fi i V l f l i f N i S k i

RANSE formulation

R ld A d N i S k E i
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Mass conservation


One gets the mean mass and momentum conservation equations:

t

Dd

dV+

Dd

(UjUd

j )njdS= 0 (9)

t

Dd

UidV+

Dd

Ui

(UjUd

j )nj

dS+

Dd

uiu

jnjdS=

Dd

fvi dV

Dd

xip

+

2

3div

(

U)dV

+

Dd

xj

2Dij

dV

(10)



Fi it V l f l ti f N i St k ti

RANSE formulation

R ld A d N i St k E ti
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Mass conservation


The Reynolds stress tensor uiu

jis defined by :

uiu

j= 2

3kij + 2tDij + f

EASM (11)

whereij is the Kronecker symbol.

Terms t,kandfEASM are the turbulent viscosity coefficient, turbulent

kinetic energy and source terms associated with the non-linear EASM

turbulence closure, respectively. For isotropic linear turbulenceclosures,fEASM = 0.



Finite Volume formulation of Navier Stokes equations

RANSE formulation

Reynolds Averaged Navier Stokes Equations
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Mass conservation


Once the definition of the Reynolds stresses is included in theequations, one finally gets :

t

D

d

dV+

D

d

(UjUd

j )njdS= 0 (12)

t

Dd

UidV+

Dd

Ui

(UjUd

j )nj

dS=

Dd

(fvi + fEASMi )dV

Dd

pT

xi

dV+

Dd

xj2effDijdV

(13)

with:

pT = p+2

3div(

U) +2k

3eff = +t

(14)




RANSE formulation

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Mass conservation

Mass conservation for a fluid with uniform density


From conservation laws to Navier-Stokes equationsFinite Volume formulation of Navier-Stokes equations

RANSE formulationReynolds Averaged Navier-Stokes Equations
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Mass conservation

If the density is uniform and constant, then the mass conservation

equation becomes:

t

V

dV+

S(U

Ud) n dS= 0 (15)

The displacement of a surface limiting a volume control should verify

the following spatial conservation formula (geometrical identity) :

t

V

dV

S

Ud

n dS= 0 (16)

And, then, even for a moving domain, mass conservation equation

reads as follows :

S

U

n dS= 0 (17)

which gives, once discretized :

f

Uf

Sf= 0 (18)



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Mass conservation

Reynolds Averaged Navier-Stokes Equations 1/2

This leads to further simplifications of the mean momentum equations.

xj(2effDij) =

xj( +t)(Ui

xj+Uj

xi)

=

xj

( +t)(

Ui

xj)

+t

xj

Uj

xi

(19)

where the reduced incompressibility condition div(

U) = 0 has beenused to simplify this formulation.



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y g q

Mass conservation

Reynolds Averaged Navier-Stokes Equations 2/2

Using the incompressibility condition and Gauss theorem leads to the

final formulation of the momentum conservation :

t

Dd

UidV+Dd

Ui(UjUdj )njdS= Dd

(fvi + fEASMi )dV

Dd

pTnidS+

Dd

eff

Ui

xj

njdS+

Dd

t

xj

Uj

xidV

(20)

with:

pT = p+2k

3eff = +t

(21)



Finite volume discretisationEvaluation of volume integrals
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g

Evaluation of surface integrals

Principles of finite volume discretisation

The computational domain is discretized with an unstructured grid and

each individual cell volumeVis considered as a control volume where

the integral formulation of the Navier-Stokes equations has to be

satisfied.

All variables are located at the cell geometric centers of control volume

V andno hypothesis is made concerning the shape of this control

volumei.e. a control volume is made of an arbitrary number of

constitutive faces noted S.This peculiarity is fundamental if one wants to implement local mesh

adaption strategies later on.



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Discretization of RANSE Evaluation of surface integrals

Principles of finite volume discretisation

Typical unstructured control volume



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Volume integrals

If we postulate a spatial linear variation of Qleading to a second order

discretisation in space:

Q(x) = QP + (xxP) (Q)P (22)

withQP= Q(xP). The integral of a functionQover a domain V isapproximated by:

V

QdV =

V(QP + (xxP) (Q)P)dV

= QP

VdV+

V

(xxP) (Q)P)dV

= QPV

(23)



Di i i f RANSE


E l i f f i l
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Surface integrals

The surface integral splits into a sum of faces of the discrete volume ofintegrationV:

S

QndS = f

S

nQfdSf

= f

Sn(Qf + (xxP) (Q)f)dSf

= f

SfQf

(24)

where

Sf =SndSf is the surface-oriented vector andQfis the valueofQat the center of the face. All variablesQbeing located at the

center of cells, one has first to rebuild the value of function at the

center of the face (noted here Qf) from the cell-centered values of the

function (QL etQR) from each side of the face.



Di ti ti f RANSE


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Discretization of the momentum equations

Transport equation for a generic variable Qfor a cellV

of centerCand limited by an arbitrary number of faces fis given by :

(VQ)C +

t(VQ)C +

f

(FcfFdf) = (SVQ ) +

f

(SfQ) (25)

Fcf =

SQ

(UjU

dj )nj

dS

= .

mfQf

Fdf =

SeffQ

xjnjdS

.mf = (UjUd

j )njSf

(26)

Terms Fcf and Fdfare respectively convection and diffusion fluxes

across the facef,.

mfbeing the mass flux across this face. SVQ andS

fQ

are surface and volume source terms.







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Discretization of the momentum equations

For momentum equations,Q = Uiand the pressure term associatedto the i momentum equation will be included in the surface source term

as:

SfQ =

SpTnidS (27)

Temporal derivatives are evaluated by upwind second-order

discretisation :A

t ecAc + epAp+ eqAq (28)

Subscriptcstands for the current time step, and petqrefer toprevious time steps.

Remark These coefficients will be notedeq,ep,ec oreq,ep,ec inthese lectures.





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The first term of the left-hand side of equation (25) corresponds to a

pseudo-steady term needed to stabilize the solution procedure forsteady flows. The corresponding derivative is evaluated by :

A

= (AcAc0)/ (29)

Ac0 is the previous estimation ofAc within the non-linear loop. Finally,

a generic discrete transport equation reads :

(ec + 1/)(VQ)cC +f

(FcfFdf) = (SVQ ) +

f

(SfQ)

(eVQ)pC (eVQ)qC + (VQ)

c0C/

(30)

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lecture5 cfd

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