simulación numérica de flujos reactivos

An introduction to the numerical simulation of reacting flows

Pascal Bruel Laboratoire de Mathmatiques et de leurs Applications CNRS - Pau University France (Cordoba 08/2007)1


Pascal BRUEL

LABORATOIRE DE MATHMATIQUES APPLIQUES UMR 5142 CNRS-UPPA Pau France.

[email protected]



Main objectives

Understanding where do the zero Mach number Navier-Stokes (NS) equations come from.

Understanding the basic structure of an isobaric planarpremixed laminar flame.

Being able to construct and understand a diagram ofturbulent premixed combustion.

Understanding a simple model of turbulent premixedcombustion.



Main objectives (continued)

Understanding a numerical method specificallydeveloped to deal with zero Mach number reacting flows.

Miscellaneous: 1. Other jet engine related situations for which the zero

Mach NS equations cannot be used: the accidental boringof a combustion chamber.

2. On the need of experimental data to compare with: presentation of an experiment dedicated to the test ofnumerical simulations.



INTRODUCTION

Why it is important to improve the predictivecapabilities of numerical simulations ofreacting flows: an example for jet propulsion.



PROPULSION DEVICES: challenges

FOR MORE ENVIRONMENTALLY FRIENDLY AIRCRAFT'S

CLEANER ENGINES ARE NEEDED

AIRBUS A380

CFM56-5B



PROPULSION DEVICES: challenges

CLEANER ENGINES: how ?

Fuel replacement : may be in the long term !

With fossil fuels: Better efficiency and new combustor design.

EXAMPLES OF NEW TECHNOLOGIES ALREADY IN OPERATION

DOUBLE ANNULAR COMBUSTOR (DAC) PROPOSED BY CFM INTERNATIONAL SINCE 1995 (CFM56-5B for A320 and CFM56-7B for

B737): reduction of 40 % of Nox (Responsible in particular of theproduction of ozone through a photolytic reaction ).



Examples of ICAO engine exhaust emission for the clean enginesof today (source: www.qinetiq.com/aviation_emission_databank/index.asp)



CLEANER ENGINES: what is under development ?

Improvement of the injection system (Twin-annular Pre-Swirl by CFMI) to optimize the premixing air-fuel in any situation in order to control the combustion regime. Basically, one tries to burn in a lean premixed prevaporised regime.

LPP concepts



CLEANER ENGINES: what is under development ?

LPP concepts: problems

Mixing hot air + fuel can not be perfect because of risk of auto-ignition: there still exists inhomogeneities of equivalenceratio.

Combustion instabilities are more likely to occur.

Where do they come from ?



A source of coherent motion: the thermo-acoustic instability( Reference: G. Searby, Ecole de combustion 2002, La Londe les Maures, France)

Consider a combustion chamber of volume V. By combining the various governing equations (continuity, momentum, energy and equation of state) and by using a linear development around a mean state (index 0), it is then possible to obtain the following equation:

tn

np

tqpc

tp

+

= && 022

2

2 ')1(''

===

=

==

i0

02

v

p

i

0

0

; c ; CC

production mole n

number mole nnfluctuatio releaseheat '

nfluctuatio pressure '

ii

i

i

hqp

W

qqqppp

&&

&&

&&&




If we can neglect the second term in the right hand side (diluted reactants for instance) then the equation reads as:

tqpc

tp

=

')1('' 2222 &

Using the linearised momentum equation, the above equation can be rewritten as:

tq

tuc

tp

=

')1('.' 202

2 &




Integrating once, multiplying by

and adding the linearised momentum equation multiplied by uyields:

'')1()''.()'21'

21( 2

0

202

0

2qp

cupu

cp

t&

=++

The first term represents the time derivative of the acoustic energy E, the second is its flux divergence and the third one is a source term.




Integrating over the chamber volume V delimited by the surface Aand over one acoustic period T:

'')1(. 20

qpc

nFEdtd

TVAV

& =+ If the flux through A is zero, one recovers the well-known Rayleigh criterion: depending on the sign of the integral of the RHS, the energy can be amplified. If there exists a relation between the pressure fluctuation and the heat release fluctuations one may have an unstable unstable systemsystem.



Understanding some aspects of the asymptoticbehavior of the Navier-Stokes (NS) equations: wheredo the zero Mach number NS equations come from ?



Zero Mach number N-S equations

In many systems of practical interest, the pressure is said to bethermodynamically constant e.g. the density variations are thenunivoquely linked to the temperature variations, namely:

Example: Imagine a system in which a plane reacting wavepropagates at speed Vf with respect to a reactants medium andwhich converts them into products

cstT =

VfProducts

Temperature Tp

Reactants R

Temperature Tr




r f rr rp p

2an d is th e so u n d sp eed in th e reactan ts : r rRa T

War =

fr

r

VM

a=

The pressure change through the wave can be evaluated from the momentum equation (neglecting the diffusive terms) by:

22 2and so r f

Vpp V M =

where Mr is the Mach number defined by:

As a consequence, the equation of state leads to:

In the case Vf




There exists a more formal way to deal with this kind of flow, based on the regular pertubation of the Navier-Stokes equations. We shall address that in the way proposed by Mller (Mller, B., Low Mach number asymptotics of the Navier-Stokes equations and numerical implications, von Karman Institute for Fluid Dynamics, Lecture Series 1999-03, March 1999).

Beforehands, it is necessary to introduce the notion of the asymptotic development of a function (taken from Joulin, G., Mthodes asymptotiques, Ecole de Combustion, Collonges, France, May 1994).




0lim

C onsider the real function ( ; ) ( ) o f real variab le x param eterized by .

G iven a series o f N +1 functions ( ) such that:

(0 ) (1) (N )( ) ( ) ...... ( ) w hen 0

(i+1or equ ivalen tly:

f x f x

=

0 (or ) (i+1) (i)( ), Landau no tation)(i)

(0) ( )If there ex ists a series of functions ( ), ......... ( ) such that:N (i) ( ) (N )( ; ) ( ) ( ) ( ( )), w hen 0

0

o

Nf x f x

if x f x oi

= =

= =

(n )




Note that

) /

)

N (i) ( )Then, ( ) ( ) is said to be an asymptotic development of ( ; )0

(N)at order ( ) and for 0.

(0) (0)( ) lim ( ; ( )0

(1) (0) (0) (1)( ) lim ( ( ; ( ) ( ))/ ( ),.........0

if x f xi

f x f x

f x f x f x

=

= =




(i)

(i)

For a given series , the AD of ( ; ) does not necessarily exists, so all the difficulty is to guess at

the form of the !

f x




2

0

(0)

(1)

(2)

(0) (1) (2)

Exercise 1: Consider the following series( ) log( )( ) 1( )

we have log( ) 1 when 0

determine ( ), ( ) and ( ) such that(i) ( )( ) ( ) is an asymptotic expansion o

i

i

f x f x f xif x

==

= ==

>> >>

xf log(1+ )




Regular and singular perturbationsConsider two "problems" P(y,x, ) and P(y,x,0) and their related solutions y(x, ) and y(x,0). If there exists an asymptotic expansion of y(x, )of the form y(x,0) (1) when 0, valid for all x, then y(o

+ x, ) y(x,0)

is said to be a regular perturbation of y(x,0) when 0.Sometimes it is said that P(y,x, ) is a regular perturbation of P(y,x,0).A perturbation which is not regular is said to be singular,

for instance, when there is no single asymptotics expansions valid for the entire domain of x.




Example of regular perturbation: the low Mach numberasymptotic analysis of the Navier-Sokes equations. They read(perfect gas, no buoyancy forces, no chemical reactions):

2 2

.( ) 0

.( ) .

.( ) .( .( )

1 1 ;2 2v

t

ptE H + T qt

pE e c T H = E

p RT

.

+ = + = + + = +

= + = + +=

u

u u u

u u)

u u

( ) ( ) ( )2with (newtonian fluid) : .3

T = + u u u I




In dimensionless form (Mller, 1999):*

* **

* ** * * * *

* 2

* * 2* * * * * * * *

* * * * *

*

.( ) 0

1 1.( ) . Re

.( ) .( .( )Re 1 Pr Re

:

/ ; / ; / ; / ; / ;

/( / );ref ref ref ref ref ref

ref ref

t

pt ME MH + T qt

with

p p p u L

t t L u E

.

+ = + = + 1+ = +

= = = = = ==

u

u u u

u u )

u u

* * *

*2* * 2 * * * * * * * * **

/( / ); /( / ); /( / )

1 ; ; ; ( 1) ; /( )2

Re ;Pr ;/

ref ref ref ref ref ref

ref ref

ref ref

ref ref ref ref p refref

ref ref ref ref

E p e e p H H p

u ppE e M H = E p T T e Q QL

u L c uM M

p

= = =

= + + = = =

= = = =

u




From now on, we shall drop the superscript * for denoting the dimensionless variables and it is assumed that the low Mach number asymptotic analysis can be considered as a regular perturbation.

(i) So each independent variables is expanded in terms of a series ( )where is the small parameter, for instance:

MM

(0) (0) (1) (1) (2) (2)

(0) (0) (1) (1) (2) (2)

(0) (0) (1) (1) (2) (2)

( , ; ) ( ) ( , ) ( ) ( , ) ( ) ( , ) ...... ( , ; ) ( ) ( , ) ( ) ( , ) ( ) ( , ) ..... ( , ; ) ( ) ( , ) ( ) ( , ) ( ) ( ,

p t M M p x t M p x t M p x t

u t M M u x t M u x t M u x t

t M M x t M x t M x

= + + += + + += + +

x

x

x

( )

) ......the scaling functions are chosen such that:

( )i i

t

M M

+

=




0 1 2

(0) (0) (0)

(1) (0) (1) (1) (0)

(2) (0) (2) (1) (1) (2) (0)

(0)(0)

(0)

Exercise 2: if one retains an AD with three terms in M namely, M 1, M and M , show that:

(

(

(

)

)

)

pT

=

== += + +

=

u u

u u u

u u u u




(0)(0) (0)

(1)(1) (1)

(2)(2) (2)

If one proceeds similarly for the governing equations.The continuity equation yields:

.( ) 0

.( ) 0

.( ) 0

t

t

t

+ = + = + =

u

u

u




(0) (0) (0)

(1) (1) (1)

(0) (0)(0) (0) (0) (2)

Exercise 3: Show that the expansion of the momentum equation yields:

( , ) 0 e.g. ( , ) ( )( , ) 0 e.g. ( , ) ( )

1.( ) .Re

with:

p x t p x t p tp x t p x t p t

pt

(0)

(0) (0)

= == =

+ = +=

u u u

( ) ( ) ( )2 .3T(0) (0) (0) + u u u I




(0)(0) (0) (0) (0)

(1)(1) (1) (1) (1)

(2)(2) (0) (0) (2) (2) (2)

For the energy equation, we have:( ) .( ) .( )

1 Pr Re( ) .( ) .( )

1 Pr Re( ) 1.( ) .( .( )

Re 1 Pr Re

E H T qtE H T qtE H T qt

.

1+ = + 1+ = + 1+ = +

u

u

u u )+

(0) (0) (0) (0) (0)

(1) (0) (1) (1) (0) (1) (1)

the following relations also hold:

( ) ( ) with ( )1

( ) ( ) ( ) with ( )1

H H H p

H H H H p

= = = + =

u u

u u u




(0)(0) (0) (0) (0) (0)

(1)(0) (1) (1) (0) (1) (1) (1)

Thus, at orders 0 and 1, the energy equations read:

. .( ) ( 1)Pr Re

. . .( ) ( 1)Pr Re

by employing the continuity equation

dp p T qdt

dp p p T qdt

1+ = + 1+ + = +

u

u u

(0) (0)(0) (0) (0) (0) (0) (0)

and the state equation at order 0, the energy equation at order 0 can be expressed as:

. . .( )1 1 Pr Re

T dpT T qt dt

1+ = +

u




(0)(0) (0)

(0) (0)(0) (0) (0) (2)

(0) (0)(0) (0) (0) (0)

Thus, the system of zeroth-order Navier-Stokes equations reads as follows:

.( ) 0

1.( ) .Re

. . .( )1 Pr Re

t

pt

T dpT Tt dt

(0)

+ = + = +

1+ =

u

u u u

u

(0) (0)

(0) (0) (0)

(2)

( , ) ( , ) ( )Since the second order pressure is decoupled from the density and temperature fluctuations, acoustic waves are absent from the flow described by such a system.

q

x t T x t p tp

+=

It is such a system that we shall consider in the following unless stated explicitly otherwise.



Asymptotics of the N-S equations

But if one is interested in the acoustics for slow flows two-timescale development. One time scale (fast) is related to acoustics while the second (slow) is, as before, related to the reference conv

* *

*

ective time scale, ie:Reference flow time scale: ( / )

Reference acoustic time scale: /( )

The dimensionless acoustic time scale is defined by / /

where is the d

ref ref ref

refref ref

ref

ref

t L u

pL

t t M

t

=

=

= =imensionless flow time scale used previously.




The asymptotic expansion of the NS equations in terms of the Mach number is carried out by considering a single space scale and the two time scales defined above, for instance for the pressure (dropp

(0) (1) 2 (2)

,

ing the * as before):

( , ; ) ( , , ) ( , , ) ( , , ) .....

The time derivative at constant and yields:1

M

p t M p x t M p x t M p x t

x M

t t M

= + + +

+

x

x




(0)(0) (0)

(1) (0)(0)

(2) (1)(1)

(0) (0) (

If one proceeds similarly as before, the zeroth, first and second order equations yields:Continuity

0 ( , )

.( ) 0

.( ) 0

Momentum0

t

t

t

p p p

= = + = + =

= =

+

+

x

u

u

0)

(0) (0)(1) (1)

(0)

(1) (0)(0) (2)

( , )( 1

( ( 1.( ) .Re

t) p p

) ) pt

(0)

= = + + = +

xu u

u u u u




(0)

(1) (0)(0) (0) (0)

(2) (1)(1) (1) (1)

(0) (0) (0)

(0)

Energy equation( ) 0

( ) ( ) .( ) .( ) ( )1 Pr Re

( ) ( ) .( ) .( ) ( )1 Pr Re

State equations( , ) ( , ) ( )( ) ( 1

E

E E H T qt

E E H T qt

t x T t x p tp t

= 1+ = + 1+ = +

==

+

+

u

u

(0))( ) ( )E t




(1) (0)(0) (0) (0) (0)

(0)

The first order energy equation can be also expressed as:

.( ) .( ) ( 1)( )Pr Re

Deriving the above equation with respect to and substracting timesthe diverg

p dpp T qdt

p

1+ = + u

2 22 (1) (0) (0)

(0) (1) (0) (0)2 (0)

ence of the first order momentum equation yields:( ) ( ).( ) ( 1) with

( , )This is a wave equation and its source is to the change over acoustic time ofth

p q p tc p ct

= = x

(0)e leading order heat release rate. If one approximates by the ambient speed of sound taken as reference one recoves the equation presented in the introduction !

c



Understanding the basic structure of an isobaric planarpremixed laminar flame.



Premixed laminar flame: chemical kinetics

Some (rapid) reminders To have chemical reactions, molecules have to collide.

But even if they do collide, there is not necessarilychemical reactions.

So this the probability of collision times the probabibilityof success of the collision which controls the rate ofproduction of species (Svante Arrhenius, Nobel Prize, 1903).




2

1 , molecule mean free path ie mean distance covered by a molecule 2

between two successive collisions.with : effective collision cross section (d is the "diameter" of a molecule) and the

cl n

d n

=

=7

00

-23 -1 0

averagenumber of molecules per unit volume (typically 10 )

8 , average thermal velocity, with mass of the individual molecule and

Boltzmann constant ( 1,3806 10 J.K ). But

c

T

l m

k Tv m km

k R

=

=N23

where is the Avogadro number ( 6,022 10 ) and R perfect gas constant ( 8,314472 J/ mole/K), so

8=

Note: the ratio between the thermal velocity and the sound speed is of order one:

c8

T

T

RTvM

RTM

v RT

=

N

8

M

=




-9

It is then possible to define an average collision time ie the average time between two successive collisions by:

(typically 10 )

Notes: These average quant

cc

T

lt sv

=

ities results from the statistical analysis of the behavior of a large number of molecules (Maxwell-Boltzmann statistics).

When establishing the (macroscopic) governing equations describing a fluid evolution (Navier-Sokes equations), the fluid is supposed to be continuous and the scales and of space and time variations that these equations can "capture" is such that:

L t

L and t .c cl t




' ", ,

1 1

Considering now a chemical system involving species reacting according to elementary reactions:

, 1, (1)j N j N

j jj r j r

j j

NM

v A v A r M= =

= = =

,1

3

,

For each species , the following relation holds:

o

: molecular mass of species [kg/mole]

: molar concentration of species [mole/m ]

: mass production of spec

j

j j j

j

j

j

j

r MA

A A A rr

jA

jA

A r

AdC

M W Wdt

M A

C A

W

=

== =

3ies due to reaction [kg/m /s]jA r




For each elementary reaction, the changes of concentration of the involvedspecies are related one to each other. Indeed, consider two species and ,their changes of concentration are such that:

k lA A

" ', ," ' " ' " ', , , , , ,

" ', , ,

and in mass : ( )

k k l

l

j j

k r k rA A Ar

l r l r k r k r l r l rA

A r A j r j r r

C C CC

W M

= = = =

' ", ,

1 1

is given by:

where quantify the "efficiency" of the collisions and thus are function of the temperature.

direct inversej r j r

j j

direct inverse

rj N j N

v vr r A r A

j j

r r

K C K C

K et K

= =

= ==




" ', ,

,

1

For instance: ( ) exp

If equilibrium is reached for reaction : 0

Then: / ( )

is th

direct

j

direct inversej r j r

j

directdirect

r

A r r

j Nv v equil

r r A rj

equilr

EK B TRTr

W

K K C K T

K

=

=

=

= =

= =e equilibrium constant of the elementary reaction .r




is called the activation energy of the reaction, not to be confused with the energy released bythe reaction. How can we evaluate the latter ?Consider an isolated system containing initially re

directE

actants at temperature T whichundergo complete chemical reactions and yields products which are brought back to the initial temperature either by cooling or heating the system. The amount of energy Q requiredby this process (heating or cooling) is called the heat of reaction.The first law of thermodynamic states that:

where is the total internal energy, the pressure, the sys

dQ dU PdV dH VdP UP V

= + = tem volume and its enthalpy.

If combustion is isobaric, then

If combustion develops at constant volume, then

finalfinal

initialinitial

finalfinal

initialinitial

H U PV

Q dH H

Q dU U

= +

= =

= =




Exercise 4: Consider a one-step irreversible exothermic reaction irrversible R P at constant pressure, with no heat and mass exchange with the surrounding environment, suppose that the heat capaciti

es are similar

and do not depend on temperature ie ( ) ( ) . Calculate the

final temperature of the products as a function of the heat of reaction .

R Pp p PC T C T C

Q= =



The importance of mixing

To have combustion, one has to bring together reactants and products, how ?: I put the species ( , ) in a box and I mix:

1- perfectly

2- partially

3- not at all!

Premixed flame

Diffusion flame

Partially premixedflame



Premixed laminar flame: the basic planar structure

0lS

flPerfectlypremixed mixture for exampleair+propane

products

x

0Rl

P

S

p

Hypotheses: planar wave, stationary in the laboratory coordinate systemirreversible reaction R P, large activation energy / 1C , and constant.

E RTD




2

2

2

p 2

Governing equationsu = cst =

dY du ( , ) (Y is the mass fraction of propane)dx dx

dT d TuC ( , )dx dx

= cstwith ( , ) ( ) exp( / ), / 1

( ) ; ( ) ; ( ) 0, ( )

R R P P

R R P

u uYD w T Y

Qw T Y

Tw T Y B T Y E RT E RT

T T Y Y Y T T

==

= +

= = = + = + =

Estimation of the flame thickness and flame speed?




Corresponding profiles

T Yp

0 0

Yp

T

TpTp

TRX (-) X (+)

0l




2p R

p R2

Eliminating between the energy equation and the species equation yields:

1d C (T-T ) ( )d C (T-T ) ( )u

dx dx

/( )where the Lewis is defined by , ratio of the thermal

RR

p

p

w

Q Y YQ Y Y LeC

CLe Le

D

+ + =

=

p R p

diffusivity

anf the species diffusion coefficient.If 1,C (T-T ) ( ) can be cast under the form a + bexp( uC / ) but

since T et Y are bounded and because of the boundary conditions in the reactanRLe Q Y Y x = +

ts,-a and b are equal to zero. So, for 1, one obtains 1-

with .

R

P R P

P R Rp

T T YLeT T Y

QT T YC

= =

= +



Premixed laminar flame: the basic planar structure2

p 2

p 0 0 2

dT d TuC ( ) exp( / )dx dx

In terms of order of magnitude, one can write that:

uC ( ) exp( / )( )

P R P RP R P

l l

QB T Y E RT

T T T T QB T Y E RT

= +

00l

R l pS C

0

0R

0

0 6 2.11

exp( / )

The rate of conversion of reactants into products is therefore

given by .

Methane+air: For an initial temperature between 150 and 600 K: 0.08 1.610

l PR p

l

l

l R

S B E RTC

S

S T

+



Premixed planar laminar flame: detailed structure

SL0

Tp

TR

Reaction rate

TemperatureFlamethickness

Reactants Products

Preheating zone Reactionzone

For the detail of the asymptotic analysis seeClavin (1985)



Flame stability



Premixed laminar flame:hydrodynamic instability(Darrieus-Landaus analysis)

Wave length of front perturbation

b

'

'

'

( , , ) ( , , )

( , , ) ( , , )

( , , ) ( , , )

unpP p

unpP p

unpP p

P

u x z t u u x z t

v x z t v v x z t

p x z t p p x z t

= += += +

0

unp Rp l

Punp

p

unpP

unpP

u u S

v v

p p

= == ===

'

'

'

( , , ) ( , , )

( , , ) ( , , )

( , , ) ( , , )

unpR R

unpR R

unpR r

R

u x z t u u x z tv x z t v v x z tp x z t p p x z t

= += += +

0lS

0lS

x

z

x

Reactantsside

Productsside

Productsside

Reactantsside z

Pertubated flamefront xf=F(z,t)

0lS

0

0

unpR l

unpp

unpR

unpR

u u Sv v

p p

= == ===

0lS

x=0 x=0

Perturbed state flame front witha small periodic corrugation >> flame thickness

Unpertubed state (unp): Planar flame at xf=0




x=0

x

( )Front location : , . 0 corresponds to the location of the unperturbed front.x F t z F= =

2

,1 Unit tangent vector to F curve

1

TFz

Fz

= +

t

2

1, Unit normal vector to F curve

1

TFz

Fz

= +

n

tn

z-




( , ) Flow velocity vectoru v=U

( ,0) Absolute front velocity vector ( projected on the x direction):Ft

= D

2. Flow velocity component normal to the front

1

Fu vzFz

= +

U n

2. Absolute front velocity component normal to the front

1

FtD

Fz

= =

+

D n

2. Relative flow-front velocity component normal to the front

1

F Fu vt z

Fz

= +

U n D.n




Hypotheses:1 The front is considered as a discontinuity2 The perturbations are of small amplitude and of large wave length (with respect to the flame thickness) and F(z,t) exp( t i

= + ) with 1, initial amplitude of the perturbation3 . for any front shape F4 remains piecewise constant ( or )5 Euler equations on both sides of the front ie:

.( ) 0 ;

l

u b

kz

S

=

=

U n D.n

uu 1.( ) ;p T cstt

+ = = u u




such that:( , , ) '( , , )( , , ) '( , , )( , , ) '( , , )

unp

unp

unp

(u,v)u x z t u u x z tv x z t v v x z tp x z t p p x z t

= += += +

T

We shall seek solutions of the form :u =




By injecting the sought form of solution in the Euler equations and dropping the quadratic terms, one gets:

' 1 '

' 1 '

' ' 0

valid on both sides, wit

unpunp

unpunp

u' u put x xv' v put x z

u vx z

+ = + =

+ = h matching of the solution at the pertubed flame front.




2 2

2 2

0 0

' 1 '' ' ' ' 1 ' '

' 1 '

Thus, the field of pressure per

unpunp

unpunp

unpunp

u' u pux t x x u v u v p psum u

t x z x x z x zv' v puy t x z

+ = + + + = + + = 14243 14243

2 2

2 2

turbation satisfies a Laplace equation:' ' 0

The solution is sought under the form: ' exp( )exp: growth rate of the perturbation in tim

p px z

p A t ikz nx

+ = = +

2 2

e (unstable if Re( ) 0)( 0) : wave number of the perturbation along the z axis: growth rate of the perturbation in space.

Injecting this in the Laplace equation yields The damping of the pertu

kn

n k n k

>>

= = bation at infinity in space implies that:

reactants side: x0n k n k = + =




' exp( )exp( )

' exp( )exp( )

Convention: the upper (resp. lower) sign is for the reactants (resp. products) side Solution to this is obtained

unpunp

unpunp

u' u ku A t ikz kxt xv' v iku A t ikz kxt x

+ = + + = +

m

by i) solving the homogeneous system and ii) adding a particular solution

exp( ) exp exp( )( )

exp( ) exp exp( )( )

unp unp unp

unp unp unp

Aku'(x,z,t)= B x kx t ikzu u k

iAkv'(x,z,t)= C x kx t ikzu u k

+ +

m




An unstable solution corresponds to Re( )>0. In such a case, the solution has still to satisfy the condition that there is no perturbation at infinity so the integration constant and must be set tB C

o zero in the reactants

side ie for x




Consequently, there are four unknowns which are , , and . To determine them, one has to use the matching at the front between the gasdynamicspertubations in the reactants and in the products. T

R P Pa a b

-

2

he pertubed flame front is supposed to propagate normally to itself at the same speedas the plane flame At the reactants side of the flame front ie for x=0

.

1l

F Fu vt z cst S

Fz

= = = +

U n D.n

+

linearization ' (1)

Doing the same at the products side of the flame front ie for x=0 yields ' = (2)

l R

P

F Fu S u Ft t

Fu Ft

=




2

The velocity component tangential to the front must besimilar on both sides of the flame front

1

' ' (3)Finally, since the flame propagation thro

l R l P

Fu v Fz u vzF

zS ikF v ES ikF v

+ = + +

+ = +

u.t

ugh the reactants isconstant, so is the pressure drop through the front

' ' (4)

R Pp p

=




Injecting the solutions into Eqs. (1)-(4) yields an homogeneous systemin , , and . To get a non trivial solution, the determinant of that systemhas to be equal to zero. After (long !) calculati

R P Pa a b

2 2

- +

ons yields the following dispersion relation:

1 1 2 (1 )( ) 0

there exists two real roots, one negative and one positive the front is (always

l lS k E S kE

+ + + =

1/ 22+

) unstable !

1( ) with ( ) 11l

E E EkS E EE E

+ = = +




z

x=0

Products

p-

p-

p+

p+

p-

p-

p+

p+

lu S>

lu S

lu S >

Reactants

x




But we know by experience that there exists stable premixed flames ! there exists stabilizing phenomena

Drawback of the Darrieus-Landau model: it is not valid for the shor

0l

t wavelength perturbations.For these short wavelengths ( ), the diffusional-thermal structure of the flame is affectedand so is the flame propagation velocity in the reactants.

Reactants Products

0l

concentration

temperatureProduction term

Reaction zone

Heat flux

Reactants flux

0l

0l



Premixed laminar flame:hydrodynamic instability(Darrieus-Landaus analysis+Marksteins correction)

The front structure is affected by the result of a competition between the transverse fluxes of heat and species (Lewis number).Markstein proposed to account for this through a dependency of thepropag

020 0 0

2

ation velocity of the curved front to its curvature radius :

(1 ) (1 ) (1 )

is called the Markstein number. By carrying out the same kind of analysis as DL, one obtains t

ll l l l

RFS S S S Ma

z R RMa

= = + = +LL

20 0 0

he following expression for :

1( ) ( 2 ) 11

1Using asymptotic analysis in the joint limit (1- ) (1) and , the expressionLe

of the Markstein number is give

l l lE E EE Ma k Ma k E Ma k

E E

O

+ = + +

n by:

2 1 2 2 1(1- )Le 1 21 1

E EMa LogEE E

+= + + +



Premixed laminar flame:hydrodynamic instability(Darrieus-Landaus analysis+Marksteins correction)

Curves corresponding to the dispersion relation for the growth rate of the perturbation

Ma = 0Ma < 0

Ma > 0

kn0 'kmax

'0

Wave number

17



Premixed laminar flame:hydrodynamic instability(Darrieus-Landaus analysis+Marksteins correction+gravity effects)

Rayleigh-Taylor instability comes into play!

Reactants

Upwards propagation: destabilizing effect

SL0SL0g

Products




Reactants

g SL0SL0

ProductsDownwards propagation: stabilizing effect




20 0

By carrying out the same kind of analysis as DL+Markstein, and taking into account the boyancy effect at the font, one obtains the following expression for :

1( ) ( 2 )1 l l

E E EE Ma k Ma k EE E

+ = + +

2

20

0

0

0

1 1 1

where the Froude number is defined by:

g is taken positive (resp. negative) for a downwards (resp. upwards) flame propagation.

lr l

lr

l

E Ma kE F k

SFg

=




Ma > 0

0

Fr < 0| Fr |

Markstein(| Fr | )

kn0

Darrieus-Landau

increases

0

kn1'

Ma > 0

0

Fr > 0Markstein(Fr )

kn0'

Darrieus-Landau

kn2'

kn1'

Zoom

increases

Upwards propagation Downwards propagation



Diagram of turbulent premixed combustion



Diagram of turbulent premixed combustion: turbulence scales

uA(t) uB(t)| |A B

d

'

( ) '( ), ( ) '( ), with '( ) '( ) 0

denotes the time (Reynolds) average of . For flow with variable density

one introduces the density weighted (Favre) average .

we call ,

A A A B B B A B

d

U t U u t U t U u t u t u t

u

= + = + = =

=

'

the average of the velocity difference at scale

ie ( ) ( )d A B

d

u U t U t




( ) ( )2 2A B

A one-time two-point correlation coefficient can also be defined as:

' 'r( )u' u'

if the two signals are perfectly correlated r( ) 1 and perfectly uncorrelated r( ) 0.

The integral length scale

A Bu u=

= =

d

d d

0'

can therefore be defined as: r( )

Thus, will designate the average velocity fluctuations at the integral length scale

l dl

u

+

=




There exists in a turbulent flow a spectrum of scales of fluctuation from the integral length scale to the Kolmogorov length scale at whichfluctuation cannot "survive" because of the viscous di

d

ssipation.

)1(Re'

Ou =

Scale at which the kineticenergy of the velocityfluctuations is transformedinto heat chaleur

4/3Re

=Ideal energy cascade: transfer without losses)//()//( '2''2' uuuu



Diagram of turbulent premixed combustion: premixed flame scales0 0

00

Velocity scale: time scale , transit time through

Space scale: the flame front ie. the time required by the flame front to propagate over a distance equal to its thickness.

The dia

l lf

ll

SS

=

gram is established by comparing the time scales of turbulence and the transit time through the premixed flame.

premixed flame one time scaletwo dimensionless numbers, th

Turbulence two time scales

e Damkhler

number and the Karlovitz number ff

Da Ka

= =



Diagram of turbulent premixed combustion: premixed flame scales

These two numbers are supplemented by the turbulent Reynolds number'Re . If Re (1), we are back to a laminar flow !

Exercise 5: show that the following expression holds:

u O

=

1

0 0

3/ 2 1/ 2

0 0

1

0 0

'

'

' Re

and plot the

l l

l l

l l

uDaS

uKaS

uS

= = =

0 0

'iso-curves of , and Re in the plane , and in log-log coordinatesl l

uDa KaS



Thickened wrinkled flamesPreheating zone

SL0

Reaction zone

Preheating zone

SL0 SL0

reactants products

SL0

Wrinkled flame, small vortex

reactants products

Preheating zone

Reaction zone

reactants

Wrinkled flame, larger vortex

products

Reaction zone

reactants products

Reaction zonePreheatingzone

Diagram of turbulent premixed combustion: schematic of flame-vortex interaction

Thickened flames



Diagram of turbulent premixed combustion

0

Exercise 6: show that the line 1 corresponds to the situationwhere .l

Ka

==

Now we shall consider an example of estimation of the regimesof combustion potentially present in a nuclear reactor, in case of

hydrogen release



Diagram of turbulent premixed combustion: practical application

Hydrogen risk in a PWR in case of water leakage in the primary water circuit of the reactor




If H2 gets into the confinement.

Radioactive products release in the

atmosphere

Energy deposit, spark,

Mixture H2/Air/H2O in the confinement

Damage to the airtightness of the confinement

Pressure

Combustion




28 March 1979 : Three Mile Island accident

Between 100 et 174 : the reactor core wasuncovered

Between 174 et 200 : emergency cooling systemsstopped the uncovering process

Between 200 et 930 : the water level was back to normal

320 kg H2 (8 % in volume) releasedPressure rise of 2 bars due to a deflagration

2 2 22 2586,6 /

Zr H O ZrO H QQ kJ mole

+ + +=




Two preliminary steps are necessary:

The determination of the basic properties of lean H2-Air mixtures (laminar flame speed).

the determination of the turbulence length scales in the reactor.




Laminar flame speed ? Collecting data in the litterature.

Calculations with Chemkin II (1986) / Premix (1985) withdetailed chemistry or global scheme.




Reactants

Flame front

electrodes

Flamefront

reactants

Spheric bomb(Dowdy et al, Lamoureux et al)

Double kernels(Andrews et Bradley, Koroll et al)

Bunsen burner(Liu et MacFarlane, Lewis et von Elbe)

Opposed jets(Egolfopoulos et Law)

reactants

reactants

Flame front

electrodes

Flame front

reactants

Experimental set-up s to determine the laminar flame velocity




*

*

*

Concentration en hydrogne [%]

V

i

t

e

s

s

e

d

e

f

l

a

m

m

e

l

a

m

i

n

a

i

r

e

[

c

m

/

s

]

8 10 12 14 16 18 20 22 24 26 28 300

25

50

75

100

125

150

175

200

225

250 Andrews et Bradley (1973)Liu et MacFarlane (1983)Lewis et von Elbe (1987)Egolfopoulos et Law (1990)Dowdy et al (1990)Law (1992)Koroll et al (1993)Lamoureux et al (2000)

*




+

+

+

+

++

x xxxxx xx

x

x

x

x

C o n cen tra tio n en h y d ro g n e [% ]

V

i

t

e

s

s

e

d

e

f

l

a

m

m

e

l

a

m

i

n

a

i

r

e

[

c

m

/

s

]

8 1 0 1 2 1 4 1 6 1 8 2 0 2 2 2 4 2 6 2 8 3 00

2 5

5 0

7 5

1 0 0

1 2 5

1 5 0

1 7 5

2 0 0

2 2 5

2 5 0 K ee e t a l (1 9 8 5 )W es tb ro o k e t D ry e r (1 9 8 4 )M aas e t W arn a tz (1 9 8 8 )B a lak ris h n an e t W illiam s (1 9 9 4 )G as R es ea rch In s titu te (1 9 9 9 )R ac tio n g lo b a le (1 q u a tio n )G ttg en s , M au ss e t P e te rs (1 9 9 2 )L am o u reu x e t a l (2 0 0 0 )

+x

calculations

In thelitterature




Comparison between experiments and numerical results

*

*

*

+

+

+

+

++

x xxxxx xx

x

x

x

x

C o n ce n tra tio n en h y d ro g n e [% ]

V

i

t

e

s

s

e

d

e

f

l

a

m

m

e

l

a

m

i

n

a

i

r

e

[

c

m

/

s

]

8 1 0 1 2 1 4 1 6 1 8 2 0 2 2 2 4 2 6 2 8 3 00

2 5

5 0

7 5

1 0 0

1 2 5

1 5 0

1 7 5

2 0 0

2 2 5

2 5 0




Plausible domain

Concentration en hydrogne [% ]

V

i

t

e

s

s

e

d

e

f

l

a

m

m

e

l

a

m

i

n

a

i

r

e

[

c

m

/

s

]

8 10 12 14 16 18 20 22 24 26 28 300

20406080

100120140160180200220240260 V itesse moyenne numrique

V itesse moyenne exprimentale

Concentration en hydrogne [%]

V

i

t

e

s

s

e

d

e

f

l

a

m

m

e

l

a

m

i

n

a

i

r

e

[

c

m

/

s

]

8 9 10 11 12 13 14 15 160

10

20

30

40

50

60

70

80

90 Average numerical valuesAverage experimental values




Overall view of the reactor

0 m

6



Diagram of turbulent premixed combustion: practical application Generic flow configuration retained

Jet Grid turbulenceWake

0.03 - 10.02 - 0.05Grid0.22 - 5.60.05 - 2Wake

1 - 150.05 - 0.8Jet

[m/s][m]Configuration 'u

0.02 2 m0.03 15 m /s

'u



Resulting estimated locations of the related regimes ofturbulent premixed combustion

10-2 10-1 100 101 102 103 104 105 106 10710-2

10-1

100

101

102

103

104

105u' /SL

0

lt / L0

Ret = 1

Ka = 1

Ka = 100

Da = 1 Da = 100

Ret = 0.01

Ret = 100

Da < 1Ka > 100

Ka < 1Da >> 1

Da >> 1Ka > 1

Flamelet regime



A minimal model of turbulent combustion in the

Flamelet regime



Premixed flamelet model

Pioneered by Bray-Moss-Libby in the 1980 sDa>>1, Ret>>1, Ka



Premixed flamelet model

R

P R

T TcT T

= Progress variable

0 1

P(c;x,t)

(x,t) (x,t)

cP(c;x,t) = (x,t)(c)+(x,t)(1-c) + f(c) [H(c)-H(1-c)]



Premixed flamelet model: equations for a planar 1D turbulent flame

( )1 1uE c = + %

productsreactants

0

4 '' '' 03

'' ''

ut x

u u u p u u ut x x x

c u c cD u ct x x

+ = + + = + =

%

% % %

% % % &

+

Closure

required



Premixed flamelet model: ?Under the model hypotheses, the signal c(t) at a given

point has the following shape (telegraphic signal)

c

Flamelet crossings

1 1 with ,

i i

i N i NP

P p R Ri iR P

tc t t t tt t

= =

= =< >= = =+

fw iR

tiP

t1iP

t + 2iPt +

Average reaction rate per flamelet crossing

1iRt + Mean crossing frequency

2Exercise 7: show that R Pt t

= +



Premixed flamelet model: ?c

c

Pt

RtReconstructed regular instantaneous signalof same duration, same number of crossingsobtained by duplicating the same pattern

Instantaneous signal

Two crossings for a 2duration P R

P R

t tt t

+ = +

Pt Pt Pt

Rt Rt Rt

Solution Ex. 7:



Premixed flamelet model: ?Let us call p0(x,t) the probability that

c(x,t)=0,

and derive a differential equation for it.

(we drop out the x for convenience)We have:

p0(t+dt)= P(c(t)=0).P(c remains at 0 during dt) + P(c(t)=1).P(c switches from 1 to 0 during dt)

So:

p0(t+dt)= p0(t) . P(c remains at 0 during dt) + (1- p0(t)).(1- P(c remains at 1 during dt))



Suppose that the statistics of tr and tbfollow a Poisson law namely:

P(c remains at 0 during dt)= e-adt=

P(c remains at 1 during dt)= e-dt=

rtdte /btdte /

Premixed flamelet model: ?

p0(t+dt)= p0(t) . e-adt +(1- p0(t)).(1- e- dt)

in the limit dt tends to 0: p0(t)=-(a+ ) p0(t)+ and the solution is given by:

tt etpetp )(0)(

0 )0(]1[)(

++ =++=



Premixed flamelet model: ?

and

+==+= )(1)()( 010 tptptpFor +t

It is now possible to calculate the auto-correlation signal of c, namely R()= to extract its

integral time scale

))1(/1)(()1)(()()()1)()((1)()(

==+=>=+=+=+=+>==+



Premixed flamelet model: ?>>=+



Premixed flamelet model: ? is the statistical average carried out over a large

number of flamelet crossings. Each flamelet is characterized

by its conversion rate given by and by its crossing time

at the point of observ

f

lc

l

w

S t

flamelets

00 0

flamelets

ation, so its related production rate is

If are supposed to be statistically independent, then:

(with 1)l

R lc

l

R lf c

l

l

R l c cf c R l R

l l l

S t

Sw t

S

S t tw t S S I I

=

=



Premixed flamelet model: ?0

01For low turbulence intensity can be considered as a

constant (Bray et al., 1988). Non constant expressions have been also proposedby Bray (1990). Thus, the mean reaction rate can be writ

l

cR

l

tS It

00

00 2

ten as:

(1- )2

(1 )since (two-delta pdf) it can be reexpressed in Favre average:1

(1 )2 (1 ) (1 )

with

l

l

cR

l

cR

l

P R

R

t c cS It

ccc

t c cS Ic

T TT

< > < >+< >= +

+ +=

%%

% %%



Premixed flamelet model: ?This demonstrates the proportionality of the

mean reaction rate to (1-) that reads in the end:

)~1(~),,,,,( ccITSf ooLf = L

This typically the source term considered by Kolmogorov, Petrovskii and Piskounov (KPP,1937) but for

a constant density - constant diffusion case.



Premixed flamelet model: " " and " " ?u u u c The simplest approach gradient approximations

2" "3

" " (no possibility of so-called counter-gradient diffusion)

where the turbulent

ji li j t t ij

j i l

ti

t j

uu uu u kx x x

cu cSc x

= + + + =

%% %

%

dynamic viscosity has to be calculated by useof a turbulence model ( - model for instance) and the turbulentSchmidt number has to be set. More elaborated approaches rely on the resolution of dedicat

k

ed

equations for either " " or " " or both. (See Pope, 2000)i j iu u u c



Premixed flamelet model: propagation properties ?We use the KPP technique to analyse the propagation properties of an 1-D

turbulent planar flame modelled as (frozen turbulence):

0

2 4( ) 03 3

( )

with a "generic" source term given by:(1 ) where is the heat release parameter

(1 )

and

w

rt

t

w D

b r

r

ut x

ku uu p ut x t x x xc uc uD

t x x x

c cAc

T TT

+ = + + + =

+ =

= +=

%

% % % %

% % % %

% %%

is the turbulence kinetic energy

prevailing in the reactants.

rk



Premixed flamelet model: propagation properties ?

We consider two different situations:

Variable density Constant Dt Regular mean reaction rate

Situation 1

0 10

=c(1-c)

c

Situation 2

10

(c)

cc*0

Or Variable density Constant Dt Quenched mean reaction rate




In both situations one can use the KPP

technique to study the steady regime of propagation

= 21

~ PcddPP

0

1

tf DAm =

Dimensionless velocity:

with cstSmdxcd

mcDcP trt ===

~)~()~(

:as written be can equation- steady The c~

)0~(*~0

*~)~1(

)~1(~)~()~(

0

01

==

>+

= +

cDDcc

ccc

ccD

cDc

tt

Dt

tw

No quenching: 0* =c




Without quenching c*=0: KPP scenarioCharacteristic equation near the singular points:

Reactants side:

Products side:

(node) )0('2 if

roots real ,0)0('122

KPP

ss

=>=+

point) (saddle roots real

,0)1('122

=++ ss

There exists a continuous spectrum of

propagation velocity limited from below by

KPPt

r

twt S

DAS =

)0('2 0

0 0.2 0.4 0.6 0.8 10

0.01

0.02

0.03

0.04

0.05

< KPPKPP = 2 > KPP

P(c)

c

noeud

pointdeselle




),

~)~(~

],0[

**

*

cc

ccPPcd

dPP

c

(at branchlinear above the intersectsthat one

the is trajectory selected the ,1],[c interval the On

is solution the and

:to reduced is equation-P the interval the On

*

==

There is only one possible trajectory and consequently only one propagation velocity which

is smaller than !KPP

tS

Integration from the vicinity of the saddle point in the direction of the point (c*, c*) yields the trajectory in the phase space usable to get the flame structure in the physical space (Sabelnikov et al., 1998). Then, the resulting profiles can be compared with those produced by simulations in the physical space.

With quenchingc*0

Tangent to PKPP




Asymptotic behavior of the P-equation when * 0c

Development at first order

*

~

cc=c~Outer solution : Inner solution :

)0('2lim )0('2lim

matching

)0('2lim =

So when c* 0 (c*) min



A numerical method for Mach zero reacting flows: the artificialcompressibility method

(in cooperation with Catherine Corvellec, Wladimyr Dourado Vladimir Sabelnikov)



Artificial compressibility approach

Developing a numerical method to deal with unsteadyreacting flows at "zero" Mach number

This method should:

be simple, versatile and easy to implement

lead to a program structure familiar to people usuallycalculating inert compressible flows (unlike specificschemes as SIMPLE(R) or PISO approaches )




Basic idea: introducing a finite-sound speed in the systemContinuity equation equation for the pressure

Proposed by Chorin (1967) for inert flows Unsteady inert flows (Soh et al. 1988, McHugh et al. 1995 Extended to steady zero Mach number reacting flows by Bruel et al. (1996). Extended to unsteady zero Mach number reacting flows by Corvellec et al. (1999).

It is well suited to transform a compressible solver based code intoa code able to deal with zero Mach number reacting flows




artificial compressibility factor: controls the magnitude of the artificial sound speed that distributes the pressure throughout the computational domain

pseudo-time term: is brought to zerobetween two physical time steps

1 .( ) 0pt

+ + = U original equation

real unsteady term: treated as a source term during convergence loop in pseudo-time.




2c

c 2

The artificial sound speed is given by a

and the pseudo-Mach number is M 1

but its value can be controlled through the setting of .

u

uu

= +

=



Artificial compressibility approach for 1D turbulent flame: a concrete example

2

( )

regroups the convective terms and the other fluxes.is the source vector, namely:

02 4;3 3

i v

i v

i vR t

t

t x

uuu u p kx

c uc cDx

+ =

= + + =

%%% %

% % % %

q F F S

F FS

q = ; F F ;

(1 )with a generic flamet model source term ie: (1 ) ww D

00

c cAc

= +% %

%

S =




2

( )

/ 0 0 0 0with: ; and with: 0 1 0 ; 0 1 0

0 0 1 0 0 1

giving:

ac v

ac i

ac

t xp

upu u u pc c

+ + = = = = =

= + +

%

% % %M% %

ac

ac2 1 2 1 2

ac

q q F F S

q = I q q I q F I F I I

q = q = F 23

Since we deal with both the physical time t and the pseudo-time ,there are two nested loops, that we shall index by n (physical time) and (pseudo time,so from the solution at

Rk

uc

% %

1

1, 1 1, 1 1, 11, 1

physical time , the solution at time is obtained

when the term tends to zero. The corresponding system reads as:

( )

n n n

n n nac vn

t t t t

x t

+

+ + + + + ++ +

= +

+ =

ac

ac

q

q F F qS




1, 1 1, 1 1, 11, 1

1, 1 11, 12

1,

Discretisation of the physical time deivative and treatment of the source term:

( )

3 ( )

2 2

The term

n n nac vn

n n n nn

n

x t

O tt t t

+ + + + + +

+ +

+ + + +

+ +

+ =

= +

acq F F qS

q q q qq

q

1,

1,

1

1,1, 1 1,

1, 1 1,

is treated implicitly in pseudo time ie:

.

with: is the increment brought to zero during the

iteration cyle in . Thus:

n

nn

n n

n n

t

+

+++ + +

+ + +

= + =

acac

ac ac ac

qq q qq

q q q

q 1,1, 11,1, 1

2 3 . 3 ( )2 2 2

nn n n nnn

O tt t t

+ + ++ + = + + ac

ac

q q q qq qq




1, 1 1, 1 1, 1

The source term is decomposed into its positive and negative contribution. The former istreated explicitly in pseudo-time while the latter is treated implicitly, namely:

win n n + + + + + += +- +S S S

1,1, 1 1, 2

1, 1 1,

1, 1 1,1, 1

1,

th:

( )

( )

The unsteady terms in pseudo-time is expressed with an implicit Euler expression, ie:

(

nn n

n n

n nn

nO

O

O

++ + +

+ + +

+ + ++ +

+ = + + = +

= +

ac-- - ac

+ +

ac acac

SS S qq

S S

q qq

.

)




1, 1 1, 1,2

1,

An implicit treatment in pseudo-time is applied for the convective and diffusive terms

( ) ( ) ( ) ( )

So, the system of equation c

n n nac v ac v ac vac

ac

nO

x x x

+ + + ++ = + +

F F F F F Fqq

.

1,1,1,

1

1,

an be written as:

1 3 ( )+ - +2

3

nn ac vnac

n

n

x x t x

++

+

+

+ = +

F FI A P C G q S

q

. . . . .

, 1

1, 1, 1, 1,1, 1, 1, 1,

2 2

The different matrices are defined by:

n n n

n n n nac vn n n n

t t

+ + + ++ + + +

= = -

ac ac ac ac

q q q

SF F qA = P C = Gq q q q




1,

1, 1,

I is the 3x3 unit matrix and after some algebra (long but not difficult), one gets for theother matrices:

0 0 0 0 021 2 0 0 03

(2 (2 ) )0 (1 ) 0 0(1

n

n nR

w w

R

u u kA c D cc c u

+

+ +

+ + +

% %% %% % %

%

A = C =

1,

1

1,

1,

1, 1,

1,

)

0 0 00

4 1 4 0 0 1 0 3 3

0 0 110 0

In , the indicates that the derivative ha

w

n

D

n

n

n nt t

t

t

n

c

ux x

cSc x

+

+

+

+ +

+

= = +

%

%

P G

P

. .

.

. s to be made performing the matrix productafter .




Mesh stencil and spatial discretisation

1,

and n

ac

i

up

c

+ = %%q

x( 1)ux i ( )px i ( )ux i ( 1)px i +( 1)px i ( 2)px i +( 1)ux i +

Non uniform physicalmesh

( 1)i + ( 2)i + Uniform computational mesh( 3)i +( )i( 3)i ( 2)i ( 1)i

1 =




[ ][ ][ ]

1,

1,1,1/ 21/ 2

notation 1,1,

1, 1,

The numbering for the unknowns on a -node mesh is such that:

=

with

n

nacn1 ii

nnac ac2ii i

n naci 3 i

nx

p

u

c

+

++++

+ +

=

%%

M

q

q q

q

2, -1 , the boundary conditions are applied at nodes 1 and .The space derivatives will involve the jacobian of the mesh transformation ie:

For and derivatives at a pressure

i nx i i nx

x x

u c

= = =

% % 2node: ( )( ) ( 1)

2For derivatives at a velocity/scalar node: ( )( 1) ( )

4For and derivatives at a velocity/scalar node: ( )( 1) ( 1)

p

u

xu u

xp p

xu u

ix x i x i

p ix x i x i

u c ix x i x i

= = = = +

= = + % %




1,1, 1,1, 1,

12 12 12 1 11/ 2

21

With such a choice the space derivatives can be expressed as:Implicit convective terms

. ( )

.

p

nn nn nac ac ac

2 x 2 2i ii ii

ac1

A i A Ax

Ax

+ + ++ +

q q q

q1,

1, 1,1, 1,21 211/ 2 1/ 21/ 2 1/ 2

1,1, 1,1, 1,

1 11 11/ 2

( )

( ).2

for ,

u

nn nn nac ac

x 1 1i ii iin

n nn nac ac acxlm 1 lm lm lm 1mi ii i

simulación numérica de flujos reactivos

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