fluid mechanics – applications and numerical methodskrzyte/students/numerical_methods.pdf · k....

K. Tesch; Fluid Mechanics – Applications and Numerical Methods 1

Fluid Mechanics – Applications and

Numerical Methods

Krzysztof Tesch

Fluid Mechanics DepartmentGdansk University of Technology

Contents

Contents

Description of fluidat different scales

Turbulencemodelling

Finite differencemethod

Finite elementmethod

Finite volumemethod

Monte Carlo method

Lattice Boltzmannmethod

Other methods

References


Description of fluid at different scales

Turbulence modelling

Finite difference method

Finite element method

Finite volume method

Monte Carlo method

Lattice Boltzmann method

Other methods

References

Description of fluid at different

scales

Contents


Turbulencemodelling



Finite volumemethod

Monte Carlo method


Other methods

References


Descriptions

Contents


Turbulencemodelling



Finite volumemethod

Monte Carlo method


Other methods

References


Fluid motion may be described by three types ofmathematical models according to the observed scales:

Microscopic description (molecular mechanics anddynamics)

Mesoscopic description (kinetic theory, DPD) Macroscopic description (classical fluid mechanics)

Microscopic description

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Turbulencemodelling



Finite volumemethod

Monte Carlo method


Other methods

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Molecular mechanics takes advantage of classicalmechanics equations to model molecular systemswhereas molecular dynamics simulates movements ofatoms in the context of N-body simulation. Themotion of molecules is determined by solving theNewtons’s equation of motion

md2ri

dt2= Gi +

N∑

j=1 6=i

fij. (1)

The force exerted on a molecule consists of theexternal force such as gravity Gi and theintermolecular force fij = −∇V usually described bymans of the Lennard-Jones potential

Microscopic description

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Turbulencemodelling



Finite volumemethod

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V = 4ǫ

[

(

σ

‖r‖

)12

−(

σ

‖r‖

)6]

. (2)

In the above equations ‖r‖ is the distance betweenparticles, ǫ – the depth of the potential well thatcharacterises the interaction strength and σ – thefinite distance describing the interaction range.Further, the ensemble average makes it possible toobtain a macroscopic quantity from the correspondingmicroscopic variable. The disadvantage of moleculardynamics method is that the total number ofmolecules even in small volume is too large –proportional to 1023.

Molecular dynamics pseudocode

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Turbulencemodelling



Finite volumemethod

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t← 0;Calculate initial molecule position r;while not the end of calculations do

fij ← −∇V ;a← m−1fij;r← r+ v∆t+ 1

2a∆t2;

t← t+∆t;

Molecular dynamics - example

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Turbulencemodelling



Finite volumemethod

Monte Carlo method


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References


Mesoscopic description

Contents


Turbulencemodelling



Finite volumemethod

Monte Carlo method


Other methods

References


The key concept is the probability distributionfunction f (N) in the phase space. The phase space isconstituted of 3N spatial coordinates q1, . . . ,qN and3N momenta p1, . . . ,pN . The probability distributionfunction f (N) allows to express the probability to finda particle within the infinitesimal phase space

(q1,q1 + dq)× . . .× (qN ,qN + dq)×(p1,p1 + dp)× . . .× (pN ,pN + dp). (3)

The total number of molecules within the infinitesimalphase space is then

f (N) (q1, . . . ,qN ,p1, . . . ,pN ) dqN dpN . (4)


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The time evolution of the probability distributionfunction f (N) follows the Liouville equation

df (N)

dt=∂f (N)

∂t+

N∑

i=1

(

∂f (N)

∂pi· dpidt

+∂f (N)

∂qi· dqidt

)

= 0.

This means that the distribution function is constantalong any trajectory in phase space.The reduced probability distribution function isdefined as

Fs (q1, . . . ,qs,p1, . . . ,ps) =∫

R3(N−s)

∫

R3(N−s)

f (N) (q1, . . . ,qN ,p1, . . . ,pN ) dqN−s dpN−s.


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The above function is called the s-particle probabilitydistribution function. A chain of evolution equationsfor Fs for 1 ≤ s ≤ N is derived and called BBGKYhierarchy. This means that the sth equation for thes-particle distributions contains s+ 1 distribution.That hierarchy may be truncated. Truncating it at thefirst order results in Boltzmann equation

∂f

∂t+ v · ∇f = Ω(f) (5)

for the probability distribution function

f(r,v, t) = mNF1(q1,p1, t) (6)

for binary collisions with uncorrelated velocities beforethat collision.

Mesoscopic description – DPD

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The DPD (Dissipative Particle Dynamics) methodsimulate only a reduced number of degrees of freedom(coarse-grained models). The motion of particles isdetermined by solving the Newtons’s equation ofmotion

md2ri

dt2= Gi +

N∑

j=1 6=i

(

fCij + fDij + fRij)

where the interaction forces are the sum of

fCij conservative or repulsion forces fDij dissipative forces fRij random force

Macroscopic description – conservation

equations

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Conservation of mass

dρ

dt+ ρ∇ ·U = 0 (7)

Conservation of linear momentum

ρdU

dt= ρf +∇ · σ (8)

Decomposition of stress tensor σ = −ptδ+ τ.Another form of conservation of linear momentum

ρdU

dt= ρf −∇pt +∇ · τ (9)

Macroscopic description – energy equations

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Energy Equation

Kinetic ρdekdt

= ρf ·U+∇ · (σ ·U)−∇ · qTotal ρdec

dt= ∂pt

∂t+∇ · (τ ·U)−∇ · q

Mechanical ρdemdt

= ∇ · (σ ·U)− σ : DInternal ρde

dt= σ : D−∇ · q

Enthalpy ρdhdt

= τ : D−∇ · q+ dptdt

Macroscopic description – general transport

equations

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∂(ρf)

∂t+∇ · (ρUf) = Sf −∇ · k (10)

Left hand side represents transient and convectioneffects. It expresses rate of changeρdf

dt≡ ∂(ρf)

∂t+∇ · (ρUf). Right hand side represents

sources (positive and negative) and fluxes (transportdue to other mechanism than convection).

mass conservation equationf := 1, Sf := 0, k := 0

linear momentum conservation equationf ← U, Sf ← ρf , k← −σenergy conservation equationf := ek, Sf := ρf ·U, k := q− σ ·U

Macroscopic description – laws of

thermodynamics

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Second law of thermodynamics

ρds

dt≥ −∇ · q

T. (11)

Entropy balance

ρds

dt=φ

T−∇ · q

T. (12)

First law of thermodynamics

ρde

dt= τ : D− pt∇ ·U−∇ · q. (13)

Macroscopic description – constitutive

equations

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Finite volumemethod

Monte Carlo method


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References


Mechanical (rheological) constitutive equations Equations of state Fluxes

Macroscopic description – mechanical

constitutive equations

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Newtonian fluids τ = 2µD Non-Newtonian fluids

Generalised Newtonian fluids τ = 2µ(γ)D Differential type fluid τ = f(A1,A2, . . .)

Ai+1 =dAi

dt+Ai·

∂U

∂r+∇U·Ai, i = 1, 2, . . .

Integral type fluids

τ =t∫

−∞

f(t− τ) (δ−Ct(τ)) dτ

Rate type fluids τ = f(

τ,D, D)

τ+ λ1τ = 2µ(

D+ λ2D)

Generalised Newtonian fluids

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τ1n = τ

1n

0+ (kγ)

1m

µ1n =(

τ0|γ|

)1n

+ k1m |γ| 1m− 1n

Szulman

τ = τ0 + kγn + µ∞γ

µ = τ0|γ| + k|γ|n−1 + µ∞

Generalised Herschel

τ1n = τ

1n

0+ (kγ)

1n

µ1n =(

τ0|γ|

)1n

+ k1n

Generalised Casson

m := n

τ = τ0 + kγn

µ = τ0|γ| + k|γ|n−1

Herschel-Bulkley

n := 1,

m := 1n

µ∞ := 0

τ = τ0 + k√γ + µ∞γ

µ = τ0|γ| +k√|γ|+ µ∞

Luo-Kuang

n := 12

√τ =√τ0 +√kγ

√µ =√

τ0|γ| +

√k

Casson

n := 2

τ = τ0 + kγµ = τ0|γ| + k

Bingham

n := 1

τ = kγn

µ = k|γ|n−1

Ostwald-de Waele

n := 1 τ0 := 0

τ = kγµ = k

Newton

τ0 := 0 n := 1

τ0 := 0

k := 0,

µ∞ := k,

τ0 := 0

Macroscopic description – equations of state

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Monte Carlo method


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Fundamental equation of state e = f(s, ρ−1)

de

dt= T

ds

dt+ptρ2

dρ

dt(14)

Thermal equation of state pt = f(T, ρ−1)

pt = ρRT (15)

Caloric equation of state e = f(T, ρ−1)

de = cv dT +

(

T∂pt∂T− pt

)

dρ−1 (16)

Macroscopic description – fluxes

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General form w = −T · ∇ϕ due to assumptionw = f(∇ϕ). More precisely w depends only on ϕ and∇ϕ. Fourier’s law

q = −λ · ∇T (17) Fick’s law

ji = −ρDij · ∇gi (18) Darcy’s law

U = −µ−1K · ∇p (19)

In the case of isotropy T = α δ andw = −α δ · ∇ϕ = −α∇ϕ.

Macroscopic description – applications 1

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General form of the Navier-Stokes equation forNewtonian fluids

ρdU

dt= ρf −∇p+∇ ·

(

2µDD)

(20)

incompressible flow creeping flow inviscid flow Boussinesq approximation Oseen approximation filtration one-dimensional flows heat transfer


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– Incompressible flow (ρ = const)

ρdU

dt= ρf −∇p+∇ · (2µD) (21)

µ = const

ρdU

dt= ρf −∇p+ µ∇2U (22)

– creeping flow

ρ∂U

∂t= ρf −∇p+∇ · (2µD) (23)

2D creeping flow

∇4ψ = 0 (24)


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– inviscid flow (µ = 0)

ρdU

dt= ρf −∇p (25)

potential flows (∇×U = 0)

∇2ϕ = ∇2ψ = 0 (26)

– Boussinesq approximation– Oseen approximation U · ∇U ≈ U∞ · ∇U

ρ∂U

∂t+ ρU∞ · ∇U = ρf −∇p+∇ ·

(

2µDD)

(27)


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– filtration

ρdU

dt= ρf −∇p+ µ∇2U− R1U (28)

– one-dimensional flows

∇2U = a (29)


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– heat transfer The ‘fluid’ Fourier equation describesthe temperature field in the fluid.

cv

(

∂(ρT )

∂t+∇ · (ρTU)

)

= φµ +∇ · (λ∇T ) . (30)

For solids where U = 0 the above equation simplifiesto the ‘solid’ Fourier-Kirchhoff equation

c∂(ρT )

∂t= ∇ · (λ∇T ) + SE (31)

where internal energy sources are given by SE .

Comments

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Generally, the ‘fluid’ equation should be solvedtogether with ‘solid’ equation. This is called conjugateheat transfer. Not having to know the heat transfercoefficient is an advantage of this approach. Thedisadvantage is the necessity of increasing the totalnumber of mesh elements due to the additional solidvolume.It is not always possible because of storagelimitations. Then either the temperature or heat fluxmust be specified at the wall. Alternatives, throughboundary conditions, are discussed further such asspecified temperature, specified heat flux, specifiedtemperature and heat flux, adiabatic or specified heattransfer coefficient.

Macroscopic description – dimensionless form

of equations

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Mass conservation equation

∇+ ·U+ = 0 (32)Linear momentum conservation equation

ρ+(

Sh∂U+

∂t++U+ · ∇+U+

)

=

=ρ+f+

Fr− Eu∇+p+ +

µ+

Re∇2+U+ (33)

Fourier-Kirchhoff (internal energy)

ρ+c+v

(

Sh∂T+

∂t++U+ · ∇+T+

)

=

=Ec

Reφ+µ +

λ+

PrRe∇2+T+ (34)

Macroscopic description – dimensionless

numbers

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Sh := LUt0

= tcht0

= LfU

=Ut0U2

L

Fr := U2

f0L=

ρ0U2

L

ρ0f0

Re := LUρ0µ0

= LUν0

=ρ0U

2

Lµ0U

L2

Eu := p0ρ0U2 =

p0L

ρ0U2

L

Ec := U2

cv0T0Pr := cv0µ0

λ0= ν0

λ0cv0ρ0

Sc := µ0ρ0D0

= ν0D0

Da := K0

L2

De := λ0t0

Wi := λ0γ0

Macroscopic description – compatibility

conditions

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General transport equation

∂(ρf)

∂t+∇ · (ρUf) = Sf −∇ · k. (35)

From Reynolds’ transport theorem arises generalcompatibility condition n · [ρUf + k] = 0.

– Mass conservation: f := 1, Sf := 0, k := 0. C.C.takes form n · [ρU] = 0 or n · [U] ≡ [Un] = 0.– Linear momentum: f ← U, Sf ← ρf , k← −σ andC.C. n · [ρUU− σ] = 0 or n · [σ] = [σn] = 0.– Energy conservation: f := ek, Sf := ρf ·U,k := q− σ ·U and C.C. n · [ρUek − σ ·U+ q] = 0or n · [q] or [λ∂T

∂n] = 0.

Macroscopic description – boundary

conditions 1

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Compatibility conditions are insufficient! Furtherconditions are needed:

adhesion l ·U = Ul = 0 thermal equilibrium on surfaces [T ] = 0

Boundary condition related to heat transfer (arisefrom C.C.)

Dirichlet: T = f1(x, y, z, t) Neumann: qn := n · q or qn = f2(x, y, z, t) mixed: αT − λ∂T

∂n= f3(P, t)

Macroscopic description – boundary

conditions 2

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Other surfaces than walls

inlet: n− 1 conditions where n stands for thenumber of equations

outlet: Generally, σn = n · σ plus T distribution.Usually p distribution due to σn ≈ −pn plus∂T∂n

= 0

symmetry: ∂ϕ∂n

= 0 for all scalar variables ϕ periodicity (translation and rotation):

ϕ(P1) = ϕ(P2) where P1 and P2 arecorresponding points on periodic surfaces

Mathematical classification

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The Navier-Stokes equations are second ordernonlinear partial differential equations. In general,they cannot be classified. However they possesproperties of quasi-linear, semi-linear and linear secondorder partial differential equations. Sometimes theycan be simplified to those and can be divided into:

hyperbolic, parabolic, elliptic.

Semi-linear second order PDEs

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For two independent variables x, y:

A(x, y)∂2U

∂x2+B(x, y)

∂2U

∂x∂y+ C(x, y)

∂2U

∂y2+

F

(

x, y, U,∂U

∂x,∂U

∂y

)

= 0 (36)

for all (x, y) over a domain Ω the above equation is:

hyperbolic if B2 − 4AC > 0, parabolic if B2 − 4AC = 0, elliptic if B2 − 4AC < 0.

Important elliptic second order PDEs

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Laplace equation

∂2U

∂x2+∂2U

∂y2= 0. (37)

Poisson equation

∂2U

∂x2+∂2U

∂y2= f(x, y). (38)

Helmholtz equation

∂2U

∂x2+∂2U

∂y2+ k2U = 0. (39)

Important parabolic second order PDEs

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Heat equation

∂U

∂t− α∂

2U

∂x2= 0, (40)

∂U

∂t− α∂

2U

∂x2= f(x, t). (41)

Important hyperbolic second order PDEs

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Wave equation

∂2U

∂t2− a2∂

2U

∂x2= 0. (42)

Telegraph equations

∂2U

∂x2− a∂

2U

∂t2− b∂U

∂t− c U = 0. (43)

Important mixed type second order PDEs

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Euler-Tricomi equation

∂2U

∂x2− x∂

2U

∂y2= 0. (44)

It is of hyperbolic type for x > 0, parabolic atx = 0 and elliptic for x < 0.

Generalised Euler-Tricomi equation

∂2U

∂x2− f(x)∂

2U

∂y2= 0. (45)

Important mixed type second order PDEs

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Potential gas flow

(1−Ma2∞)∂2ϕ′

∂x2+∂2ϕ′

∂y2= 0. (46)

It is of hyperbolic type for Ma2∞ > 1, parabolic atMa2∞ = 1 and elliptic for Ma2∞ < 1.The velocity potential for the x axis dominatedflow is

ϕ(x, y) = U∞x+ ϕ′(x, y). (47)

Velocity components are then given as

Ux(x, y) =∂ϕ

∂x= U∞ + U ′x(x, y), (48a)

Uy(x, y) =∂ϕ

∂y= U ′y(x, y). (48b)


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Turbulence features

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Irregularity Unsteadiness 3-D in terms of space and vortex structures Diffusivity Dissipation Energy cascade Need for constant energy supply

Kolmogorov scales 1

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Kolmogorov scales are the smallest vorticity scaleswhere nearly the whole dissipation takes place. Thereare three scales, for velocity UK = (νε)1/4, length

LK = (ν3ε−1)1/4

and time tK = (νε−1)1/2

.The Reynolds number for these scales

ReK =UKLKν

=(νε)1/4 (ν3ε−1)

1/4

ν= 1. (49)

It means that at this level the inertial forces are of thesame order as the viscous forces.The dissipation intensity of the kinetic energy offluctuation can also be estimated in terms of a lengthscale for large scale motion (vorticity) as

Kolmogorov scales 2

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ε ∼ U2

t=U2

L/U =U3

L . (50)

It means that the energy U2 of the large scales isdissipated proportionally to time L/U . Substitutingthe dissipation in equation for Lk with that for ε wehave

LK =

(

ν3LU3

)1/4

. (51)

Introducing a Reynolds number for large scalesReL = UL

νit is possible to find a relation for the ratio

of length scales by means of this Reynolds number inthe form of

LLK∼( U3

ν3L

)1/4

L =

(ULν

)3/4

= Re3/4L . (52)

Turbulence glossary

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DNS – Direct Numerical Simulation LES – Large Eddy Simulation RANS – Raynolds Averaged Navier-Stokes URANS – Unsteady Raynolds Averaged

Navier-Stokes DES – Detached Eddy Simulation SST – Shear Stress Transport RNG – ReNormalisation Group EARSM – Explicit Algebraic Reynolds Stress

Models RST – Reynolds Stress Transport


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DNS LES RANS

Models based on the Boussinesq hypothesis(0-eq, 1-eq, 2-eq models)

Models which do not take advantage of theBoussinesq hypothesis

RST models EARSM

Decompositions and averages

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Decomposition f(r, t) = f(r) + f ′(r, t)

Realisations f(r) := limN→∞

1N

N∑

i=1

fi(r, t)

Time fτ (r) := limτ→∞

1τ

τ∫

0

f(r, t) dt

Time ft(r, t) :=1∆t

t+∆t∫

t

f(r, t) dt

Spatial fV (r, t) :=1|V |

∫∫∫

Vf(r, t) dV

Comments

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In practice it is usually enough to know what theaverage velocity is (not the fluctuation). The velocityvector field is be decomposed into average andfluctuation components U = U+U′. The time ofaveraging ∆t should be chosen to be greater than thefluctuation range and smaller than the function that isgoing to be averaged. The averaging process of theNavier-Stokes equation introduces a number of newunknown functions.

RANS 1

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Averaged mass conservation equation

∇ · U = 0 (53)

Averaged Navier-Stokes equation

ρ∂U

∂t+ ρ∇·

(

UU)

= ρf −∇p+µ∇2U−ρ∇·U′U′ (54)

Reynolds stress tensor R := −ρU′U′ and thetotal stress tensor σ := −pδ+ 2µD+R makes itpossible to obtain the averaged momentumequation

ρdU

dt= ρf +∇ · σ (55)

RANS 2

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Averaged Fourier-Kirchhoff equation

cv

(

∂(ρT )

∂t+∇ ·

(

ρT U)

)

=

= 2µD2 +∇ · (λ∇T )− cv∇ ·(

ρT ′U′)

+ ρε.(56)

The averaging process of the Navier-Stokes equationintroduces six unknown (because of the symmetry)components of the Reynolds stress tensor. Theaveraged Fourier-Kirchhoff equations gives a furtherthree of the vector T ′U′.

Comments

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It is important to realise that the closure of system ofthe mass conservation and Navier-Stokes equationshas been lost. Further modelling is required.Formulating additional relationships for unknownfunctions to achieve closure of equations is calledturbulence modelling. Any additional closure equationmust fulfil a few basic criteria such as coordinateinvariance. This is fulfilled by proper tensorformulation of the exact and modelled equations.Another criterion is called realisability meaning that asolution must be physical.Practically, however, it is difficult to achieve all theserequirements. This is because some parts of the exacttransport equations are modelled or even dropped.

Closure

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There are two main approaches to achieve closure.The models may be divided between those whichassume the eddy viscosity hypothesis and those whichdo not

Models not assuming the eddy viscosity hypothesis

Reynolds stress transport equation Algebraic stress tensor models

Boussinesq hypothesis assumed

Reynolds stress transport equation

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∂R

∂t+∇ ·

(

UR)

= −∇U · (RT +R)+

+∇ ·((

Ck2ε−1 + ν)

∇R)

−Π+2

3ρεδ (57)

The left hand side represents unsteadiness andconvection. On the right hand side the two first termsrepresent production. The two terms under divergenceare responsible for diffusion.The right hand side fourth term is the secondunknown tensor Π need to be modelled. The lastright hand side term 2

3ρεδ is the so called dissipation

tensor for isotropic turbulence.

Algebraic stress tensor models

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Without using the eddy viscosity hypothesis twotransport equations for k and a second variable areformulated. Instead of the linear Boussinesqhypothesis – algebraic, non-linear relationships areformulated between the stress anisotropy tensor a andthe average flow properties. The tensor a is related tothe Reynolds stress tensor by:

a :=R

k− 2

3δ. (58)

Typically, relationships depend on average strain rateand spin tensors

a = f(D, Ω). (59)

Boussinesq hypothesis

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The turbulence stresses is related to the mean flowRxy = µt

∂Ux

∂y. This linear relationship is

R = a0δ+ 2µtD. (60)

The trace of this relations allows to find a constant−2ρk = 3a0 which gives

R = −2

3ρkδ+ 2µtD. (61)

The Reynolds equation becomes

∂(ρU)

∂t+∇·

(

ρUU)

= ρf −∇pe+∇·(

2µeD)

(62)

where µe := µt + µ, pe := p+ 23ρk.

Eddy viscosity and diffusivity hypothesis

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The eddy diffusivity hypothesis is introduced by directanalogy with the eddy viscosity hypothesis. It reducesthe number of unknown functions in theFourier-Kirchhoff equation −cvρT ′U′ = λt∇T .The Fourier-Kirchhoff equation then becomes

cv

(

∂(ρT )

∂t+∇ ·

(

ρT U)

)

= 2µD2+∇·(

λe∇T)

+ρε,

(63)where λt can be estimated by means of the turbulentPrandtl number λt =

µtcvPrt

. Effective conductivity isintroduced by means of the definitionλe := λt + λ = µtcv

Prt+ λ.

Trace of RST equation

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Calculating the trace of the Reynolds stress transportequation

∂R

∂t+∇ ·

(

UR)

= −∇U · (RT +R)+

+∇ ·((

Ck2ε−1 + ν)

∇R)

−Π+2

3ρεδ (64)

results in kinetic energy k transport equation which isused in the preceding one- and two-equationturbulence models trR = −2ρk for µt = Cµρk

2ε−1.The traces of Π by definition trΠ = 0 and thetransport equation for k takes the following form

∂(ρk)

∂t+∇·(ρkU) = ∇U : R+∇·

((

µtσ−1k + µ

)

∇k)

−ρε(65)

Zero-equation model

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Zero k and ε are assumed. It allows the Boussinesqequation to be reduced to R = 2µtD.Eddy viscosity µt is modelled by means of thePrandtl-Kolmogorov hypothesis. This hypothesiscomes directly from dimensionless analysis µt = cρUL.The velocity scale U is often approximated by meansof the maximal velocity |U|max and length scale L bythe volume of the flow domain |V | by U ∼ |U|max,L ∼ 3

√

|V |.The Boussinesq hypothesis takes the following form

R = Cρ 3√

|V ||U|maxD (66)

No new unknown functions! However, zero-equationmodels are not as accurate but they are robust (firstapproximation for more complex models).

One-equation model

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One transport equation for k is introduced

∂(ρk)

∂t+∇·(ρkU) = ∇U : R+∇·

((

µtσ−1k + µ

)

∇k)

−ρε

where ‘production’ ∇U : R = 2µtD2. According to

Prandl-Kolmogorov hypothesis U =√k and ε ∼ U3

Lso

ε = k3/2L−1. Finally, the k transport equation arrives

ρdk

dt= 2µtD

2 +∇ ·((

µtσ−1k + µ

)

∇k)

− ρk3/2L−1

(67)where eddy viscosity is estimated as µt = ρ

√kL by

means of another Prandtl hypothesis.

Two-equation k-ε model

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Two additional equations have to be formulated. Thefirst, for the kinetic energy k, comes from theReynolds stress transport equation

ρdk

dt= 2µtD

2 +∇ ·((

µtσk

+ µ

)

∇k)

− ρε

and that for the dissipation ε is analogous to it

ρdε

dt= Cε1

ε

k2µtD

2 +∇ ·((

µtσε

+ µ

)

∇ε)

−Cε2ρε2

k.

(68)Both of them are transport equations for a scalarfunction. The eddy viscosity depends on both k and εand is postulated, as previously, to have the formµt = Cµρ

k2

ε.

Comments

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The five constants in equations are empirical, that is,should be deduced from experiment for a specificgeometry. This ‘standard’ set is given by

σk = 1, σε = 1.3,

Cµ = 0.09, Cε1 = 1.44, Cε2 = 1.92. (69)

Two-equation k-ω model

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The turbulent frequency ω is proportional to the ratioof dissipation and kinetic energy ω ∼ ε

kand using the

constant Cµ, they are then related by ε = Cµkω. Theeddy viscosity takes the form µt = ρ k

ω. The two

transport equations take the following form

ρdk

dt= 2µtD

2+∇·((

µtσk1

+ µ

)

∇k)

−Cµρkω (70)

ρdω

dt= α1

ω

k2µtD

2 +∇ ·((

µtσω1

+ µ

)

∇ω)

− β1ρω2

(71)This ‘standard’ set is constant is σk1 = 2, σω1 = 2,Cµ = 0.09, α1 =

59, β1 =

340.

Two equations SST model

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The shear stress model combines the k−ω model nearthe wall with the k − ε far from it. Firstly, the k − εmodel has to be transformed to the k − ω formulationby means of relation ε = Cµkω. This results in

∂(ρk)

∂t+∇·

(

ρkU)

= 2µtD2+∇·

((

µtσk2

+ µ

)

∇k)

−Cµρkω,

∂(ρω)

∂t+∇ ·

(

ρωU)

= α2ω

k2µtD

2 +∇ ·((

µtσω2

+ µ

)

∇ω)

+

−β2ρω2 + 2ρ

ωσω2∇k · ∇ω.

Additional cross-diffusion terms now appear. The‘standard’ set of constants is different from that forthe original k − ε σk2 = 1, σω2 = 0.856, Cµ = 0.09,α2 = 0.44, β2 = 0.0828.

Two equations SST model

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Secondly, the equations for the k − ω model aremultiplied by a blending function F1 and thetransformed k − ε equations by (1− F1). Theequations then are added. This results in

∂(ρk)

∂t+∇·

(

ρkU)

= 2µtD2+∇·

((

µtσk3

+ µ

)

∇k)

−Cµρkω,

∂(ρω)

∂t+∇ ·

(

ρωU)

= α3ω

k2µtD

2 +∇ ·((

µtσω3

+ µ

)

∇ω)

+

−β3ρω2 + (1− F1)2

ωρσω3∇k · ∇ω.

Constants marked with the subscript ‘3’, namely σk3,σω3, α3, β3 are linear combinations of constants fromthe component models C3 = F1C1 + (1− F1)C2.

Additional passive transport equation

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The additional variable transport equation may beadded to the closed system after averaging, as

∂(ρf)

∂t+∇·

(

ρfU)

= −∇·(

ρf ′U′)

−∇·k+Sf . (72)The first term of the right hand side can be modelledby means of the eddy diffusivity hypothesis and theturbulent diffusivity coefficient Γ , −ρf ′U′ = Γ∇fand the additional transport equation takes the form

∂(ρf)

∂t+∇ ·

(

ρfU)

= ∇ ·((

µtSct

+ ρD

)

∇f)

+ Sf

(73)where the turbulent diffusivity coefficient Γ may berepresented as a function of the eddy viscosity and theturbulent Schmidt number Γ := µt

Sctwhere Sc := µ

ρD.

Turbulent boundary layer

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1 2 5 10 20 50 100 2000

5

10

15

20

25

y+

U+

1 2 5 10 20 50 100 200

0

5

10

15

20

25

Friction velocity Uτ :=√

τ0ρ

Characteristic length l := νUτ

dimensionless distance y+ := yl

dimensionless velocity U+ := Ux

Uτ

y+ < 11 laminar sub-layer

5 < y+ < 30 bufferregion

11 < y+ < 250 turbu-lent sublayer (log-lawlayer)

y+ < 250 inner turbu-lent boundary layer

y+ > 250 outer turbu-lent boundary layer

Filtration of N-S equations 1

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Filtration of the N-S equations is associated with LESmethod. Small scales are removed by means offiltering

f(r, t) =

∫∫∫

R3

+∞∫

−∞

f(r ′′, t′′)G(r− r ′′, t− t′′) dt′′ dV ′′

where G is a filter. Typically it is a product

G(r− r ′′, t− t′′) := Gt(t− t′′)3∏

i=1

Gvi(xi−x′′i ). (74)

For Gt(t− t′′) := τ−1H(t′′) andGvi(xi − x′′i ) := δ(xi − x′′i ) we have time average

fτ (r) := limτ→∞

1τ

τ∫

0

f(r, t) dt.

Filtration of N-S equations 2

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Filtration of the N-S equations results in

ρ∂U

∂t+ρ∇·

(

UU)

= ρf−∇p+∇·(

2µD− ρ (L+C+R))

(75)where Leonard’s decomposition

L := UU− UU, C := UU′ +U′U, R := U′U′

(76)represents the cross stress tensor C (interactionsbetween large and small scales), Reynolds subgridtensor R (interactions among subgrid scales) andLeonard tensor L (interactions among the largescales).For L = 0 and C = 0 we have Reynolds equations.

Finite difference method

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The method

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The finite difference method (introduced by Euler inXVIII century) replaces the region by a finite mesh ofpoints at which the dependent variable isapproximated.

All partial derivativesat each mesh pointare approximated fromneighbouring valuesby means of Taylor’stheorem. Thismeans that derivativesat each pointare approximated bydifference quotients.

Taylor’s theorem

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Assuming that f has continuous derivatives overcertain interval the Taylor expansion is used

f(x0 +∆x) =m−1∑

n=0

dnf(x0)

n!+

dmf(c)

m!. (77)

where x := x0 +∆x, c = x0 + θ∆x and θ ∈]0; 1[.The above equation may also be written as

f(x0 +∆x) = f(x0) + f ′(x0)∆x+1

2f ′′(x0)∆x

2

+1

6f ′′′(x0)∆x

3 + . . . +1

m!f (m)(c)∆xm. (78)

Taylor’s theorem

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Instead of f (m) at unknown point c it is rewritten interms of another unknown quantity of order ∆xm

f(x0 +∆x) = f(x0) + f ′(x0)∆x+ f ′′(x0)∆x2

2

+ . . . + f (m−1)(x0)∆xm−1

(m− 1)!+O(∆xm) (79)

Discarding (truncating) O(∆xm) one gets anapproximation to f . The error in this approximation isO(∆xm). Roughly speaking it says that knowing thevalue of f and the values of its derivatives at x0 it ispossible to write down the equation for its value atthe point x0 +∆x.

First order finite difference

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Taking under consideration the Taylor expansion up tothe first derivative

f(x0 +∆x) = f(x0) + f ′(x0)∆x+O(∆x2) (80)

then neglecting O(∆x) and rearranging gives the firstorder finite difference approximation to f ′(x0)

f ′(x0) ≈f(x0 +∆x)− f(x0)

∆x. (81)

This approximation is called a forward approximation.Replacing ∆x by −∆x in Taylor expansion one getsbackward approximation

f ′(x0) ≈f(x0)− f(x0 −∆x)

∆x. (82)

Second order finite difference

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Taking under consideration the Taylor expansion up tothe second derivative

f(x0+∆x) = f(x0)+f′(x0)∆x+f

′′(x0)∆x2

2+O(∆x3)

(83)then neglecting O(∆x2). Doing the same for −∆xand combining the two above we have

f ′(x0) ≈f(x0 +∆x)− f(x0 −∆x)

2∆x(84)

after neglecting O(∆x2). This is so called the secondorder central difference approximation to f ′(x0).

Second order finite difference

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Higher order approximation to derivatives is alsopossible. This can be done by taking more terms inthe Taylor expansion. Doing so up to the third we get

f(x0 +∆x) = f(x0) + f ′(x0)∆x+1

2f ′′(x0)∆x

2

+1

6f ′′′(x0)∆x

3 +O(∆x4). (85)

Replacing ∆x for −∆x and combing the results thendropping O(∆x4) gives the second order symmetricdifference approximation to f ′′

f ′′(x0) ≈f(x0 +∆x)− 2f(x0) + f(x0 −∆x)

∆x2.

(86)

Differences

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Selected finite differences approximation to first andsecond derivatives are given in the following table.These can be used to solve ordinary differentialequations by replacing derivatives by theirapproximations.

Approximation Type Order

f ′(x0)f(x0+∆x)−f(x0)

∆xforward 1st

f ′(x0)f(x0)−f(x0−∆x)

∆xbackward 1st

f ′(x0)f(x0+∆x)−f(x0−∆x)

2∆xcentral 2nd

f ′′(x0)f(x0+∆x)−2f(x0)+f(x0−∆x)

∆x2symmetric 2nd

Differences

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Equations for f ′ approximate the slope of the tangentin x0 by means of chords (backward, forward andcentral finite difference).

Differences

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The typical subscript notation is

f(x0 +mh, y0 + nh) ≡ fi+mj+n. (87)

Now it is possible to express selected finite differencesapproximations to derivatives in somewhat simplermanner

Approximation Type Order

f ′ifi+1−fi

hforward 1st

f ′ifi−fi−1

hbackward 1st

f ′ifi+1−fi−1

2hcentral 2nd

f ′′ifi+1−2fi+fi−1

h2symmetric 2nd

Poisson equation

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Poisson equation is ∇2Uz = a. Two dimensionalversions of this equation is written as

∂2Uz∂x2

+∂2Uz∂y2

= a. (88)

The next step would be to replace second orderderivatives by symmetric finite differenceapproximation

Ui+1j − 2Uij + Ui−1jh2

+Uij+1 − 2Uij + Uij−1

h2= a.

(89)It can be rewritten to give Uij as a function ofsurrounding variables

Uij =Ui+1j + Ui−1j + Uij+1 + Uij−1 − ah2

4. (90)

Poisson equation - mesh and boundary

conditions

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The domain is discretised in the x and y directions bymeans of constant mesh size h (figure on the left). Uzis unknown at black mesh points and known at whitepoints from the boundary condition.

For instance the Dirichletboundary conditionspecifies the values of Uzdirectly. In this case Uz = 0meaning no slip wall. If theboundary values are knownthen discrete Poissonequation gives a systemof linear equations for Uij.

Poisson equation - solution methods

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The accuracy of results depends on the size of themesh represented here by h. Mesh size should bedecreased until there is no significant influence onnumerical results.The set of linear equations can be solved eitherdirectly by means of an appropriate method (Gausselimination for instance) or indirectly by means ofiterative solution methods or the relaxation method(point-Jacobi iteration)

Un+1ij =

Uni+1j + Un

i−1j + Unij+1 + Un

ij−1 − ah24

(91)

or point-Gauss-Seidel (faster than point-Jacobi)

Un+1ij =

Uni+1j + Un+1

i−1j + Unij+1 + Un+1

ij−1 − ah24

. (92)

Poisson equation - solution methods

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Another indirect method is so called SuccessiveOver-Relaxation method

Un+1ij = (1− w)Un

ij+

wUni+1j + Un+1

i−1j + Unij+1 + Un+1

ij−1 − ah24

(93)

where w is a relaxation parameter. For w ∈]1, 2[ wehave over-relaxation and for w := 1 this methodcorresponds to the point-Gauss-Seidel method. Onecan also consider under-relaxation method forw ∈]0, 1[.The best choice of w value needs numericalexperiments. It also depends on specific problems.

Poisson FDM pseudocode

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Data: Read input variables and BCsw ← 1; n← 1;repeat

R← 0;for i← 1 to imax do

for j ← 1 to jmax do

if not boundary(Unij) then

Un+1ij ← Un

i+1j+Un+1i−1j+U

nij+1+U

n+1ij−1−ah

2

4;

R← max(

|Un+1ij − Un

ij|, R)

;

Un+1ij ← (1− w)Un

ij + wUn+1ij ;

n← n+ 1;

until n ≤ nmax and R > Rmin;

Results - Poisson equation

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24

68

102

4

6

810

0

3

6

2 4 6 8 10

2

4

6

8

10

x

y

Laplace equation

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Laplace equation is ∇2ϕ = 0. Two dimensionalversions of this equation is written as

∂2ϕ

∂x2+∂2ϕ

∂y2= 0. (94)

Replacing second order derivatives by symmetric finitedifference approximation

ϕi+1j − 2ϕij + ϕi−1jh2

+ϕij+1 − 2ϕij + ϕij−1

h2= 0.

(95)It can be rewritten to give ϕij as a function ofsurrounding variables

ϕij =ϕi+1j + ϕi−1j + ϕij+1 + ϕij−1

4. (96)

Laplace eq. – Neumann boundary condition

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Neumann boundary condition specifies values of thederivative ∂

∂nof a solution ϕ on boundary ∂Ω to fulfil

∂ϕ

∂n= N(x, y) (97)

where the normal derivatives is defined as

∂ϕ

∂n= n · ∇ϕ = nx

∂ϕ

∂x+ ny

∂ϕ

∂y(98)

and (x, y) ∈ ∂Ω. If U = ∇ϕ we get

∂ϕ

∂n= n ·U = nxUx + nyUy. (99)

Laplace eq. – Neumann boundary condition

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We have two equation for apoint located on boundary ‘2’

∂2f

∂x2=fi+1j − 2fij + fi−1j

h2,

∂f

∂x=fi+1j − fi−1j

2h= Nij .

Point fi−1j is located outsidethe Ω area. Eliminating it weget

∂2f

∂x2=

2fi+1j − 2Nijh− 2fijh2

and

fij =2fi+1j + fi+1j + fij−1 − 2Nijh

4.

Laplace FDM pseudocode

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if boundary(ϕnij) 6= 0 then

switch ϕ(Unij) do

case 1: ϕn+1ij ←

ϕni+1j+ϕ

n+1

i−1j+ϕn

ij+1+ϕn+1

ij−1

4;

;

case 2: ϕn+1ij ←

2ϕni+1j+ϕn

ij+1+ϕn+1

ij−1−2hNij

4;

;...

case 6: ϕn+1ij ←

2ϕni+1j+2ϕn

ij+1−2hNij

4;

;...

R← max(

|ϕn+1ij − ϕn

ij |, R)

;

ϕn+1ij ← (1− w)ϕn

ij + wϕn+1ij ;

n← n+ 1;


Results - Laplace equation

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2 4 6 810 12 14 16 2

46810

−10−50

Biharmonic equation

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The biharmonic equation is ∇4ψ ≡ ∇2 · ∇2ψ = 0.Two dimensional versions of this equation is written as

∂4ψ

∂x4+ 2

∂4ψ

∂x2∂y2+∂4ψ

∂y4= 0. (101)

It is a fourth-order elliptic partial differential equationthat describes creeping flows in terms of a streamfunction ψ where the velocity components areUx =

∂ψ∂y

and Uy = −∂ψ∂x.

The Dirichlet boundary condition specifies both: astream function ψ and its normal derivative ∂ψ

∂n. Two

conditions are needed due to the fourth order of thebiharmonic equation.

Biharmonic equation - approximation to

derivatives

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The finite difference approximations to ∂4ψ∂x4

, ∂4ψ∂y4

are

∂4ψ

∂x4=ψi+2j + ψi−2j − 4ψi+1j − 4ψi−1j + 6ψij

h4,

(102a)

∂4ψ

∂y4=ψij+2 + ψij−2 − 4ψij+1 − 4ψij−1 + 6ψij

h4.

(102b)

The fourth order mixed derivative is approximated as

∂4ψ

∂x2∂y2=ψi+1j+1 + ψi−1j−1 + ψi−1j+1 + ψi+1j−1

h4

+4ψij − 2ψi+1j − 2ψi−1j − 2ψij+1 − 2ψij−1

h4. (103)

Biharmonic equation - discrete equation

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From the discrete biharmonic equations ψij can beexpressed as a function of surrounding variables

ψij =−ψi+2j − ψi−2j − ψij+2 − ψij−2 + 4ψij

20

+ 8ψi−1j + ψij+1 + ψij−1 + ψi+1j

20

− 2ψi+1j+1 + ψi−1j−1 + ψi−1j+1 + ψi+1j−1

20. (104)

Biharmonic equation - boundary conditions

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From the below figure two purely geometricrelationships arise ∂ψ

∂n= n · ∇ψ = −Ul,

∂ψ∂l

= l · ∇ψ = Un. For an impermeable boundary one

gets Un = 0⇒ ∂ψ∂l

= 0. The general relationshipbetween volumetric flow rate and the stream functionsis

V =

∫

L

U · n dL =

∫

L

∂ψ

∂ldL =

∫

L

dψ = ψA − ψB.

(105)

Biharmonic FDM pseudocode

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if not boundary(Unij) then

Un+1ij ← −Un

i+2j−Uni−2j−U

nij+2−U

nij−2+4Un

ij

20+

8Uni−1j+U

nij+1+U

nij−1+U

ni+1j

20−

2Uni+1j+1+U

ni−1j−1+U

ni−1j+1+U

ni+1j−1

20;

R← max(

|Un+1ij − Un

ij|, R)

;

Un+1ij ← (1− w)Un

ij + wUn+1ij ;

n← n+ 1;


Results - biharmonic equation

Contents


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Finite volumemethod

Monte Carlo method


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References


15

1015

20 14

812

0

0.5

1

Complex geometry

Contents


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References


Point 1 is located insidethe Ω area

f1 =h f0 + d f2h+ d

(106)

Point 1 is located outsidethe Ω area

f1 =h f0 − d f2h− d (107)

Navier-Stokes equations

Contents


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Typical numerical approaches for the incompressibleNavier-Stokes equations:

Ωz-ψ (vorticity-stream function) formulationmethod

Artificial compressibility method Pressure/velocity correction (operator splitting

methods)

Projection methods MAC (Marker-and-Cell) Fractional step method SIMPLE (Semi-Implicit Method for Pressure

Linked Equations), SIMPLER (SIMPLERevisited), SIMPLEC

Bad idea

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The incompressible Navier-Stokes equations

∂U

∂t+U · ∇U = −1

ρ∇p+ ν∇2U. (108)

Explicit forward difference in time

Un+1 −Un

∆t+Un ·∇Un = −1

ρ∇pn+ν∇2Un. (109)

Problems:

∇ ·Un+1 6= 0,pn+1 =?

Better idea

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∂U

∂t+U · ∇U = −1

ρ∇p+ ν∇2U, (110a)

∇ ·U = 0. (110b)

Un+1 −Un

∆t+Un · ∇Un = −1

ρ∇pn+1 + ν∇2Un, (111a)

∇ ·Un+1 = 0. (111b)

Problems:

∇2pn+1 = ρ∆t∇ · (Un −∆tUn · ∇Un +∆t ν∇2Un)

BCs?

Semi implicit

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∂U

∂t+U · ∇U = −1

ρ∇p+ ν∇2U, (112a)

∇ ·U = 0. (112b)

Semi implicit approach

Un+1 −Un

∆t+Un · ∇Un+1 = −1

ρ∇pn+1 + ν∇2Un+1, (113a)

∇ ·Un+1 = 0. (113b)

Fully implicit

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∂U

∂t+U · ∇U = −1

ρ∇p+ ν∇2U, (114a)

∇ ·U = 0. (114b)

Fully implicit approach

Un+1 −Un

∆t+Un+1 · ∇Un+1 = −1

ρ∇pn+1 + ν∇2Un+1,

(115a)

∇ ·Un+1 = 0. (115b)

Artificial compressibility method

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∂U

∂t+U · ∇U = −1

ρ∇p+ ν∇2U, (116a)

∇ ·U = 0. (116b)

Explicit forward difference in time

Un+1 −Un

∆t+Un · ∇Un = − 1

ρ0∇pn + ν∇2Un, (117a)

βpn+1 − pn

∆t+∇ ·Un = 0. (117b)

Projection method

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Ut −Un

∆t= −Un · ∇Un + ν∇2Un, (118)

∇ ·Ut 6= 0, BC

Un+1 −Ut

∆t= −1

ρ∇pn+1, (119)

∇ ·Un+1 = 0, ¬BC

∇2pn+1 =ρ

∆t∇ ·Ut (120)

Square cylinder

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Vorticity-stream function formulation

Contents


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References


The incompressible 2D Navier-Stokes equations∂U

∂t+U · ∇U = −1

ρ∇p+ ν∇2U, (121a)

Ωz =∂Uy∂x− ∂Ux

∂y. (121b)

2D Helmholtz equation (Ux =∂ψ∂y, Uy = −∂ψ

∂x)

∂Ωz

∂t+U · ∇Ωz = ν∇2Ωz, (122a)

∇2ψ = −Ωz. (122b)

Ωn+1z − Ωn

z

∆t+Un · ∇Ωn

z = ν∇2Ωnz , (123a)

∇2ψn+1 = −Ωn+1z . (123b)

Finite element method

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Method of weighted residuals

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The mathematical foundation of the finite elementmethod is in the method of weighted residuals.Imagine ordinary differential equation

y′′(x) + 20x = 0 (124)

subjected to boundary conditions y(0) = y(1) = 0.The exact general solution of this equation is

y(x) := −10

3x3 + C1x+ C2 (125)

and a specific solution subjected to boundaryconditions

y(x) := −10

3(x3 − x). (126)


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The method seeks an approximate solution y in thegeneral form

y(x) :=N∑

i=1

CiNi(x) (127)

where Ni are known trial functions which should becontinuous and fulfilled boundary conditions. Theconstants Ci are unknown and they will bedetermined. A residual R appears when substitutingapproximate solution y into the differential equations

R(x) := y′′(x) + 20x 6= 0. (128)

The unknown Ci constant are determined fori = 1, . . . N from

∫ 1

0

Wi(x)R(x) dx = 0. (129)


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∫ 1

0

Wi(x)R(x) dx = 0 i = 1, . . . , N (130)

Choices for the weighting functions Wi

Collocation method Wi(x) := δ(x− xi) Subdomain method

Wi(x) := H(x− xi−1)−H(x− xi) Galerkin’s method Wi(x) := Ni(x)

∫ 1

0

Ni(x)R(x) dx = 0 i = 1, . . . , N (131)

Least Squares Method Wi(x) :=∂R∂Ci

∫ 1

0

∂R

∂CiR(x) dx = 0 i = 1, . . . , N (132)

Method of weighted residuals - example

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A polynomial trial functions can be assumed

N(x) := xr(x− 1)s. (133)

It is continuous and fulfils boundary conditions. Justone trial function for r = s = 1 is the simplest case

N1(x) := x(x− 1). (134)

The approximate solution y(x) :=∑N

i=1CiNi(x)where N := 1 takes the following form

y(x) = C1N1(x) = C1(x2 − x). (135)

Residual may now be expressed as

R(x) = 2C1 + 20x 6= 0. (136)


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The unknown constant C1 may be determined uponintegrating (Galerkin’s method of weighted residuals)

∫ 1

0

x(x− 1)(2C1 + 20x) dx = 0. (137)

This gives −13(5 + C1) = 0 and allows to determine

C1 = −5. The approximate solution is now

y(x) = −5x(x− 1) (138)

and can be compared with the exact solution

y(x) := −10

3(x3 − x). (139)


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The simplest case with just one trial functionapproximates the exact solution more or lessacceptably. Better agreement is possible with morethan one trial functions.

0 0.2 0.4 0.6 0.8 10

0.5

1

1.5

x

y

y

y


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The two polynomial trial functions can be assumed

N1(x) := x(x− 1), N2(x) := x2(x− 1). (140)

Both are continuous and fulfil boundary conditions.The approximate solution y(x) :=

∑Ni=1CiNi(x)

where N := 2 takes the following form

y(x) = C1N1(x)+C2N2(x) = C1(x2−x)+C2(x

3−x2).(141)

Residual may now be expressed as

R(x) = 2C1 + 2C2(3x− 1) + 20x 6= 0. (142)


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The unknown constants C1, C2 may be determinedupon integrating

∫ 1

0

x(x− 1)(2C1 + 2C2(3x− 1) + 20x) dx = 0,

∫ 1

0

x2(x− 1)(2C1 + 2C2(3x− 1) + 20x) dx = 0.

This gives 10 + 2C1 + C2 = 0 and1 + 1

6C1 +

215C2 = 0 and allows to determine

C1 = C2 = −103. The approximate solution is now

y(x) = −10

3x(x− 1)(x+ 1). (143)


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The case with two trial function approximates theexact solution very well. It is far better that theprevious case. There is no visible difference. In fact, itis even the exact solution

−10

3x(x− 1)(x+ 1) = −10

3(x3 − x). (144)

Method of weighted residuals - comments

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The method of weighted residuals constitutesfoundation of the final element method

The method exploits an integral formulation tominimise residual errors

Trail functions of this method are global. It isusually difficult task to find a proper one thatsatisfies boundary conditions. The moredimensions the worse.

Finite Element method

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The approximate solution is defined as

y(x) =N+1∑

i=1

yini(x) (145)

over the interval [x1;xN+1] which is divided into Nsubintervals (elements). The unknown values of yi atpoint xi correspond to constants Ci of the weightedresiduals method. However, the trial function is localand defined as

ni(x) =

x−xi−1

xi−xi−1xi−1 ≤ x ≤ xi,

xi+1−xxi+1−xi

xi ≤ x ≤ xi+1,

0 otherwise

(146)

Trial functions

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Trial function

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For instance, if x ∈ [x3;x4] we have a linearinterpolation

y(x) = y3n3(x) + y4n4(x) = y3x4 − xx4 − x3

+ y4x− x3x4 − x3

.

(147)The above definition arises from the equation of theline passing through two different points (x3, y3) and(x4, y4)

(y − y3)(x4 − x3) = (y4 − y3)(x− x3) (148)

or knowing that y = a+ bx we solve for a and b from

y3 = a+ bx3, (149a)

y4 = a+ bx4. (149b)

Method

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Let us consider the same differential equationy′′ + 20x = 0 as previously. The approximate solutionis in the form y(x) =

∑N+1i=1 yini(x). The idea of

minimising the residual R(x) = y′′ + 20x 6= 0 isadapted from the weighted residuals to the finiteelement method∫ xN+1

x1

nj(x)R(x) dx = 0 j = 1, . . . , N + 1.

The unknown values yj may be determined uponintegrating (Galerkin’s finite element method)

∫ xN+1

x1

nj(x)

(

N+1∑

i=1

(yini(x))′′ + 20x

)

dx = 0

j = 1, . . . , N + 1. (150)

Method

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For the three trial functions nj we have threeequations. However, in any sub interval only two trialfunctions are nonzero. If so∫ x2

x1

n1(x) ((y1n1(x) + y2n2(x))′′ + 20x) dx = 0

∫ x2

x1

n2(x) ((y1n1(x) + y2n2(x))′′ + 20x) dx+

∫ x3

x2

n2(x) ((y2n2(x) + y3n3(x))′′ + 20x) dx = 0

∫ x3

x2

n3(x) ((y2n2(x) + y3n3(x))′′ + 20x) dx = 0

Method

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We can decouple the second equation for the timebeing and couple it latter at the global assemblyprocess. We have then

∫ xj+1

xj

nk(x) ((yjnj(x) + yj+1nj+1(x))′′ + 20x) dx = 0

j = 1, . . . , N ; k = j, j + 1. (151)

The above equation is valid for each subinterval(element) and suggest the so called ‘element’formulation. It is also well known that

∫ xN+1

x1

y(x) dx =N∑

j=1

∫ xj+1

xj

y(x) dx. (152)

‘Element’ formulation

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The approximate solution is expressed as

ye = yjN1(x) + yj+1N2(x) = N · ye (153)

where the known local trial functions N and theunknown nodal values ye are collected as vectors

N := (N1, N2), (154a)

ye := (yj, yj+1). (154b)

The local trial functions are simply a linearinterpolation

N1 :=xj+1 − xxj+1 − xj

xj ≤ x ≤ xj+1, (155a)

N2 :=x− xj

xj+1 − xjxj ≤ x ≤ xj+1. (155b)


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For each element we have the Galerkin residualcondition

∫ xj+1

xj

NR dx = 0 j = 1, . . . , N. (156)

Taking under consideration our differential equationy′′ + 20x = 0 and the approximate solution ye it isnow possible to express the residual as

∫ xj+1

xj

N

(

d2yedx2

+ 20x

)

dx = 0. (157)

The second derivative has to be replaced. This is dueto linear nature of the trial functions.


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Integration by parts makes it possible to replace thesecond derivative

Ndyedx

∣

∣

∣

∣

xj+1

xj

−∫ xj+1

xj

dN

dx

dyedx

dx+

∫ xj+1

xj

N20x dx = 0.

Finally, matrix form of the Galerkin residual conditionfor each element is now

∫ xj+1

xj

dN

dx

dN

dxdx·ye =

∫ xj+1

xj

N20x dx+Ndyedx

∣

∣

∣

∣

xj+1

xj

j = 1, . . . , N.

Element equations

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The so called ‘stiffness’ matrix for each element e maybe introduced

Ke :=

∫ xj+1

xj

dN

dx

dN

dxdx. (158)

The above matrix is symmetric. The so called‘displacement’ vector is also introduced

Fe :=

∫ xj+1

xj

N20x dx+ Ndyedx

∣

∣

∣

∣

xj+1

xj

. (159)

The Galerkin residual condition for each element maynow be written as

Ke · ye = Fe. (160)

Element equations

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It is possible to simplify the matrices even further. Forthe linear trial functions one gets the ‘stiffness’ matrix

Ke :=

∫ xj+1

xj

(

dN1

dxdN1

dxdN1

dxdN2

dxdN1

dxdN2

dxdN2

dxdN2

dx

)

dx =1

xj+1 − xj

(

1 −1−1 1

)

and the ‘displacement vector’

Fe :=

∫ xj+1

xj

(

N120xN220x

)

dx+

(

N1dyedx

∣

∣

xj+1

xj

N2dyedx

∣

∣

xj+1

xj

)

.

If the gradients are dropped, as discussed further, wehave

Fe = −10

3(xj − xj+1)

(

2xj + xj+1

xj + 2xj+1

)

.

Element equations - example

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For the interval [0; 1] divided equally into 3 elementswe have the element matrices

K1 = K2 = K3 =

(

3 −3−3 3

)

.

The global assembly process (coupling):

K =

K111 K12

1 0 0K12

1 K221 +K11

2 K122 0

0 K122 K22

2 +K113 K12

3

0 0 K123 K22

3

results in

K =

3 −3 0 0−3 6 −3 00 −3 6 −30 0 −3 3

.

Element equations - example

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The ‘displacement’ vector for equally divided interval[0; 1] take form

F1 =

(

1027− dy(0)

dx2027

+dy( 1

3)

dx

)

,F2 =

(

4027− dy( 1

3)

dx5027

+dy( 2

3)

dx

)

,F3 =

(

7027− dy( 2

3)

dx8027

+ dy(1)dx

)

.

After the global assembly process one finally gets

F =

F 11

F 21 + F 1

2

F 22 + F 1

3

F 23

=

1027− dy(0)

dx602712027

8027

+ dy(1)dx

.

System of equations - example

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The global (assembled) system of linear equations is

3 −3 0 0−3 6 −3 00 −3 6 −30 0 −3 3

·

y1y2y3y4

=

1027− dy(0)

dx602712027

8027

+ dy(1)dx

.

It cannot, however, be solved yet. This is due tonecessity of applying the global boundary conditions.These are y1 = y4 = 0. Two typical methods ofapplying them are discussed further.

Boundary conditions

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Extracting only these equations that are related tounknown functions y2 and y3 for y1 = y4 = 0 results in

· · · ·· 6 −3 ·· −3 6 ·· · · ·

·

·y2y3·

=

·602712027

·

or simpler in

(

6 −3−3 6

)

·(

y2y3

)

=

(

602712027

)

The above system may now be solved to obtain theunknown values y2, y3.

Boundary conditions

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The second method does not change the layout of thematrices. However, it involves modification of specificelements by multiplying them by a ‘large’ number.These elements are located on the diagonal of the‘stiffness’ matrix and corresponding positions of the‘displacement’ vector (if non-zero)

3 · 107 −3 0 0−3 6 −3 00 −3 6 −30 0 −3 3 · 107

·

y1y2y3y3

=

0602712027

0

.

ODE FEM pseudocode

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Data: Read N elements, nodes and BCsCreate global matrix K and vectors F, y;for e← 1 to N do

Ke ←∫

edNdx

dNdx

dx;Fe ←

∫

eN20x dx;

Add Ke to K;Add Fe to F;

Apply BCs;Solve linear system K · y = F;

Results - ODE

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0 0.2 0.4 0.6 0.8 10

0.5

1

1.5

x

y

3 elements, 4 nodes

Results - ODE

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0 0.2 0.4 0.6 0.8 10

0.5

1

1.5

x

y

6 elements, 7 nodes

Results - ODE

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0 0.2 0.4 0.6 0.8 10

0.5

1

1.5

x

y

9 elements, 10 nodes

Results - ODE

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0 0.2 0.4 0.6 0.8 10

0.5

1

1.5

x

y


Linear and quadratic interpolation

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Considering line equation ye = a+ bx and utilising itfor two different points (xj, yj) and (xj+1, yj+1) wecan get the following system of equations

(

yjyj+1

)

=

(

1 xj1 xj+1

)

·(

ab

)

. (161)

It can be easily solved for a and b(

ab

)

=

(

1 xj1 xj+1

)−1

·(

yjyj+1

)

. (162)

Keeping in mind that ye = N · ye whereN = (N1, N2) and ye = (yj, yj+1) we can utilise thesolution for a and b to get

ye = a+bx =xj+1 − xxj+1 − xj

yj+x− xj

xj+1 − xjyj+1 = N1yj+N2yj+1.


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Introducing L1 and L2 for a one-dimensional element

L1 := N1 =xj+1 − xxj+1 − xj

, (163a)

L2 := N2 =x− xj

xj+1 − xj(163b)

one can formulate similar system of equation forquadratic interpolation ye = a+ bx+ cx2 through thepoints (xj, yj), (xj+ 1

2, yj+ 1

2) and (xj+1, yj+1)

yjyj+ 1

2

yj+1

=

1 xj x2j1 xj+ 1

2x2j+ 1

2

1 xj+1 x2j+1

·

abc

. (164)


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0 0.2 0.4 0.6 0.8 1

0

0.5

1

L1 L2

0 0.2 0.4 0.6 0.8 1

0

0.5

1

N1 N2 N3

The quadratictrial functionscan be alsoexpressed interms of lineartrial functions

N1 := L1 (2L1 − 1) ,

N2 := 4L1L2,

N3 := L2 (2L2 − 1)

Quadratic interpolation

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The ‘stiffness’ matrix for each element e may now becalculated as

Ke :=

∫ xj+1

xj

dN

dx

dN

dxdx =

1

3(xj+1 − xj)

7 −8 1−8 16 −81 −8 7

.

The ‘displacement’ vector is now

Fe :=

∫ xj+1

xj

N20x dx =10

3(xj+1−xj)

xj2(xj + xj+1)

xj+1

.

The Galerkin residual condition for each element isthe same as previously

Ke · ye = Fe. (166)

ODE quadratic FEM pseudocode

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Data: Read N linear elements, n nodes and BCsInsert midpoints xj+ 1

2← xj+xj+1

2;

n← 2n− 1 ;Create global matrix K and vectors F, y;for e← 1 to N do

Ke ←∫

edNdx

dNdx

dx;Fe ←

∫

eN20x dx;



ODE - linear vs quadratic interpolation

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0 0.2 0.4 0.6 0.8 10

0.5

1

1.5

x

y

0 0.2 0.4 0.6 0.8 10

0.5

1

1.5

x

y

6 elements, 7 nodes. 3 elements, 7 nodes.

Mesh refinement

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The generalised p-norm is given by

‖f‖p :=(∫

L

|f(x)|p dx) 1

p

(167)

where for p := 2 we have a special case

‖f‖2 :=√

∫

L

f2(x) dx. (168)

The error E := y − y of a finite element solution ymay now be defined by means of 2-norm. It may takethe following form

‖E ′‖22 ≤ CN∑

e=1

r2e . (169)

Mesh refinement

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The element residue re is defined as

re := |Le| ‖f + y′′e‖2 (170)

but due to the linear form of trial functions N ′′i = 0 itis true that y′′e = 0. This means that the elementresidue is re := |Le| ‖f‖2 and solution error

‖E ′‖22 ≤ CN∑

e=1

|Le|2‖f‖22. (171)

Element’s length is |Le| = xj+1 − xj and utilising thetrapezoidal rule we can express the element residue as

re := |Le|√

∫

Le

f(x)2 dx ≈ (xj+1 − xj)32

√

f2j + f2

j+1

2.

The above approximation is used for mesh refinement.

Mesh refinement - results

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Monte Carlo method


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References


0.00

0.50

1.00

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Residue


Contents


Turbulencemodelling



Finite volumemethod

Monte Carlo method


Other methods

References


0.00

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0.75

1.00

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Turbulencemodelling



Finite volumemethod

Monte Carlo method


Other methods

References


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Turbulencemodelling



Finite volumemethod

Monte Carlo method


Other methods

References


0.00

0.25

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0.87

1.00

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Residue


Contents


Turbulencemodelling



Finite volumemethod

Monte Carlo method


Other methods

References


0.00

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0.63

0.75

0.87

1.00

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y

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2

3

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Residue


Contents


Turbulencemodelling



Finite volumemethod

Monte Carlo method


Other methods

References


0.00

0.25

0.38

0.50

0.63

0.75

0.87

1.00

0

0.5

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y

0 0.2 0.4 0.6 0.8 10

1

2

3

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Residue

FEM for Poisson equation

Contents


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References


Letthe two dimensional formof Poisson equation on Ω

∂2U

∂x2+∂2U

∂y2= −a (172)

be subjected to the Dirichlet boundary conditionU(x, y) = 0 for every (x, y) ∈ ∂Ω. It is true that

∫∫

Ω

f(x, y) dx dy =∑

e

∫∫

Ωe

f(x, y) dx dy. (173)


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Ue = N ·Ue (174)

where the known local trial functions N and theunknown nodal values Ue are

N := (N1, N2, N3), (175a)

Ue := (U1, U2, U3). (175b)


∫∫

Ωe

NR dx = 0. (176)


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Taking under consideration Poisson equation and theapproximate solution Ue it is now possible to expressthe residual as

∫∫

Ωe

N

(

∂2Ue∂x2

+∂2Ue∂y2

+ a

)

dx dy = 0. (177)

The second derivative has to be replaced (due tolinear nature of the trial functions). This can be doneby means of Green’s first identity∫∫

S

(

ψ∂2ϕ

∂x2+ ψ

∂2ϕ

∂y2

)

dx dy =

∫

∂S

ψ∂ϕ

∂ndL

−∫∫

S

(

∂ψ

∂x

∂ϕ

∂x+∂ψ

∂y

∂ϕ

∂y

)

dx dy. (178)


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Integration by means of Green’s identity makes itpossible to replace the second derivative

∫∫

Ωe

(

∂N

∂x

∂Ue∂x

+∂N

∂y

∂Ue∂y

)

dx dy

−∫

∂Ωe

N∂Ue∂n

dL−∫∫

Ωe

Na dx dy = 0. (179)

The matrix form of the Galerkin residual condition foreach element can now be expressed

∫∫

Ωe

(

∂N

∂x

∂N

∂x+∂N

∂y

∂N

∂y

)

dx dy ·Ue =

∫∫

Ωe

Na dx dy.

Element equation

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Introducing the ‘stiffness’ matrix for each element e

Ke :=

∫∫

Ωe

(

∂N

∂x

∂N

∂x+∂N

∂y

∂N

∂y

)

dx dy (180)

and the ‘displacement’ vector

Fe :=

∫∫

Ωe

Na dx dy (181)

one may obtain the Galerkin residual condition foreach element in the form

Ke ·Ue = Fe. (182)

Element equation

Contents


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The expanded version of the ‘stiffness’ matrix is

Ke :=

∫∫

Ωe

(

∂N1

∂x∂N1

∂x+∂N1

∂y∂N1

∂y∂N1

∂x∂N2

∂x+∂N1

∂y∂N2

∂y∂N1

∂x∂N3

∂x+∂N1

∂y∂N3

∂y∂N1

∂x∂N2

∂x+∂N1

∂y∂N2

∂y∂N2

∂x∂N2

∂x+∂N2

∂y∂N2

∂y∂N2

∂x∂N3

∂x+∂N2

∂y∂N3

∂y∂N1

∂x∂N3

∂x+∂N1

∂y∂N3

∂y∂N2

∂x∂N3

∂x+∂N2

∂y∂N3

∂y∂N3

∂x∂N3

∂x+∂N3

∂y∂N3

∂y

)

dx dy.

Similarly, the same for the ‘displacement’ vector

Fe := a

∫∫

Ωe

N1

N2

N3

dx dy. (183)

The actual form of matrices depends on the trialfunctions. Linear form of these are discussed further.

Linear trial function

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Considering plane equation

Ue = a+ bx+ cy (184)

one can formulate the following system of equations

U1 = a+ bxi + cyi (185a)

U2 = a+ bxj + cyj (185b)

U3 = a+ bxk + cyk (185c)

for three different points (xi, yi), (xj, yj), (xk, yk).Solving these for a, b and c results in

Ue =U1

2Se(ai+bix+ciy)+

U2

2Se(aj+bjx+cjy)+

U3

2Se(ak+bkx+cky).

Linear trial function

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The linear trial functions then are

N1 =1

2Se(ai + bix+ ciy),

N2 =1

2Se(aj + bjx+ cjy),

N3 =1

2Se(ak + bkx+ cky)

where

ai = xjyk − xkyj ; aj = xkyi − xiyk; ak = xiyj − xjyi;bi = yj − yk; bj = yk − yi; bk = yi − yj;ci = xk − xj; cj = xi − xk; ck = xj − xi;

Se =1

2|ckbj − cjbk|;

Linear interpolation

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00.5

1 0

0.5

1

0

0.5

1

x

y

z

00.5

1 0

0.5

1

0

0.5

1

x

y

z

00.5

1 0

0.5

1

0

0.5

1

x

y

z

00.5

1 0

0.5

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0

0.5

1

x

y

z

Integrals and derivatives

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Now it is possible to calculate necessary derivativesappearing in the ‘stiffness’ matrix

∂N1

∂x=

bi2Se

,∂N2

∂x=

bj2Se

,∂N3

∂x=

bk2Se

,

∂N1

∂y=

ci2Se

,∂N2

∂y=

cj2Se

,∂N3

∂y=

ck2Se

.

The same concerns integrals appearing in the‘displacement’ vector

∫∫

Se

Nα1 N

β2N

γ3 dx dy = 2Se

α!β!γ!

(α+ β + γ + 2)!. (188a)

Element equation

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Now it is possible to simplify the matrices evenfurther. For the linear trial functions one gets the‘stiffness’ matrix

Ke =1

4Se

b2i + c2i bibj + cicj bibk + cickbibj + cicj b2j + c2j bjbk + cjckbibk + cick bjbk + cjck b2k + c2k

(189)

and the ‘displacement vector’

Fe =Se3

aiajak

. (190)

Four element example

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1(0, 0) 2(10, 0)

3(10, 10)4(0, 10)

5(5, 5)

1

2

3

4Ω

∂Ω

U = 0

U = 0

U=0

U=0

Global matrix

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The global assembly process (coupling) for theconsidered four element case:

K=

K111 +K11

4 K121 0 K12

4 K131 +K13

4K12

1 K221 +K11

2 K122 0 K23

1 +K132

0 K122 K22

2 +K113 K12

3 K232 +K13

3K12

4 0 K123 K22

3 +K224 K23

3 +K234

K131 +K13

4 K231 +K13

2 K232 +K13

3 K233 +K23

4 K331 +K33

2 +K333 +K33

4

.

The element matrices are identical

K1 = K2 = K3 = K4 =1

2

1 0 −10 1 −1−1 −1 1

.

Finally, the global ‘stiffness’ matrix is

K =

1 0 0 0 −10 1 0 0 −10 0 1 0 −10 0 0 1 −1−1 −1 −1 −1 4

.

Global vector

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The ‘displacement’ element vector for the consideredfour elements

F1 = F2 = F3 = F4 =25

3

111

.

After the global assembly process one finally gets

F =

F 11 + F 1

4

F 21 + F 1

2

F 22 + F 1

3

F 23 + F 2

4

F 31 + F 3

2 + F 33 + F 3

4

=50

3

11112

.

Poisson FEM pseudocode

Contents


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Ke ←∫∫

Ωe

(

∂N∂x

∂N∂x

+ ∂N∂y

∂N∂y

)

dx dy;

Fe ←∫∫

ΩeNa dx dy;



Results - PDE

Contents


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Finite volumemethod

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24

68

10

24

6810

0

3

6

4 elements, 5 nodes

Results - PDE

Contents


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Finite volumemethod

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Other methods

References


24

68

10

24

6810

0

3

6

8 elements, 9 nodes

Results - PDE

Contents


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Finite volumemethod

Monte Carlo method


Other methods

References


24

68

10

24

6810

0

3

6


FEM for Laplace equation

Contents


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Let the two dimensional form of Laplace equation

∂2ϕ

∂x2+∂2ϕ

∂y2= 0 (191)

or ∇2ϕ = 0 on Ω be subjected to both boundaryconditions on ∂Ω:

Neumann

∂ϕ

∂n= n·∇ϕ = n·(Ux, Uy) = nxUx+nyUy =: −fN ,

Dirichlet (as previously)

ϕ = const =: fD. (192)


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As previously, FEM formulation for Poisson equationsubjected to Dirichlet and Neumann BCs is∫∫

Ωe

(

∂N

∂x

∂Ue∂x

+∂N

∂y

∂Ue∂y

)

dx dy

−∫

∂Ωe

N∂Ue∂n

dL−∫∫

Ωe

Na dx dy = 0.

Poisson ∇2ϕ = −a with Dirichlet BC

∫∫

Ωe

(

∂N

∂x

∂N

∂x+∂N

∂y

∂N

∂y

)

dx dy·Ue =

∫∫

Ωe

Na dx dy,

Laplace ∇2ϕ = 0 with Dirichlet + Neumann BC∫∫

Ωe

(

∂N

∂x

∂N

∂x+∂N

∂y

∂N

∂y

)

dx dy·ϕe = −∫

∂Ωe

NfN dL.

Element equation

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Introducing the ‘stiffness’ matrix as previously foreach element e

Ke :=

∫∫

Ωe

(

∂N

∂x

∂N

∂x+∂N

∂y

∂N

∂y

)

dx dy (193)

and the ‘displacement’ vector

Fe := −∫

∂Ωe

NfN dL (194)

one may obtain the Galerkin residual condition foreach element in the form

Ke ·ϕe = Fe. (195)

Element equation

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The expanded version of the ‘stiffness’ matrix look thesame as previously but the ‘displacement’ vector isnow

Fe := fN

∫

∂Ωe

N dL = −1

2|L|fN1. (196)

The vector 1 may take of the three following forms

110

,

101

,

011

. (197)

|L| stands for element side length.

Laplace FEM pseudocode

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Ke ←∫∫

Ωe

(

∂N∂x

∂N∂x

+ ∂N∂y

∂N∂y

)

dx dy;

Fe ← −∫

∂ΩeNfN dL;



Geometry and mesh - Laplace equation

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∂ϕ∂x=Ux

∂ϕ∂y

= 0

∂ϕ∂y

= 0

∂ϕ∂n

= 0

ϕ=const

456 nodes and 818 elements


Contents


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Other methods

References


−10

12 0

1

−2

1

−10

12 0

1

1

2

Geometry and mesh - Laplace equation

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∂ϕ∂x=Ux

∂ϕ∂y

= 0

∂ϕ∂y

= 0

∂ϕ∂n

= 0∂ϕ∂y

= 0

ϕ=const

351 nodes and 607 elements


Contents


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Finite volumemethod

Monte Carlo method


Other methods

References


−10

12 0

1

−2

1

−10

12 0

1

0

1

2

Creeping flow – Stokes equations

Contents


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References


Re≪ 1

0 = ρg −∇p+ µ∇2U,

∇ ·U = 0.

0 = ρgx −∂p

∂x+ µ

(

∂2Ux∂x2

+∂2Ux∂y2

)

,

0 = ρgy −∂p

∂y+ µ

(

∂2Uy∂x2

+∂2Uy∂y2

)

,

∂Ux∂x

+∂Uy∂y

= 0.

FEM formulation

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Ue = N ·Ue

where the quadratic trial functions N and theunknown nodal values Ue are

N := (N1, N2, N3, N4, N5, N6),

Ue := (U1, U2, U3, U4, U5, U6).


∫∫

Ωe

NR dx = 0.

Quadratic interpolation

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The quadratictrial func-tions can beexpressed interms of lineartrial functions

N1 := L1 (2L1 − 1) ,

N2 := L2 (2L2 − 1) ,

N3 := L3 (2L3 − 1) ,

N4 := 4L1L2,

N5 := 4L2L3,

N6 := 4L1L3.

Momentum conservation equation

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The Galerkin residual condition

∫∫

Ωe

N

(

ρgx −∂pe∂x

+ µ

(

∂2Uxe∂x2

+∂2Uxe∂y2

))

dx dy = 0

by means of Green’s first identity

∫∫

S

(

ψ∂2ϕ

∂x2+ ψ

∂2ϕ

∂y2

)

dx dy =

∫

∂S

ψ∂ϕ

∂ndL

−∫∫

S

(

∂ψ

∂x

∂ϕ

∂x+∂ψ

∂y

∂ϕ

∂y

)

dx dy


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is

ρgx

∫∫

Ωe

N dx dy −∫∫

Ωe

N∂N

∂xdx dy · pe

− µ∫∫

Ωe

(

∂N

∂x

∂N

∂x+∂N

∂y

∂N

∂y

)

dx dy ·Uxe = 0

orKpxe · pe +Kxye ·Uxe = gxFe.

Similarly

Kpye · pe +Kxye ·Uye = gyFe.


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Kpxe :=

∫∫

Ωe

N∂N

∂xdx dy

Kpye :=

∫∫

Ωe

N∂N

∂ydx dy

Kxye := µ

∫∫

Ωe

(

∂N

∂x

∂N

∂x+∂N

∂y

∂N

∂y

)

dx dy

Fe := ρ

∫∫

Ωe

N dx dy


Contents


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References


The Galerkin residual condition

∫∫

Ωe

N

(

∂Uxe∂x

+∂Uye∂y

)

dx dy = 0.

The matrix form of the Galerkin residual condition foreach element can now be expressed

∫∫

Ωe

N∂N

∂xdx dy ·Uxe +

∫∫

Ωe

N∂N

∂ydx dy ·Uye = 0

orKuxe ·Uxe +Kuye ·Uye = 0.


Contents


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Kuxe :=

∫∫

Ωe

N∂N

∂xdx dy

Kuye :=

∫∫

Ωe

N∂N

∂ydx dy

Element equations

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Kpxe · pe +Kxye ·Uxe = gxFe,

Kpye · pe +Kxye ·Uye = gyFe,

Kuxe ·Uxe +Kuye ·Uye = 0

K6×6xye 06×6 K6×3

pxe

06×6 K6×6xye K6×3

pye

K3×6uxe K3×6

uye 03×3

·

U6×1xe

U6×1ye

p3×1e

=

gxF6×1e

gyF6×1e

03×1

K15×15e · u15×1

e = f15×1e

Finite volume method

Contents


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The method

Contents


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Equations are investigated in integral form incomparison with FDM. FVM is more flexible thanFDM in terms of spatial discretisation.Let us consider differential form of the generaltransport equation

∂(ρf)

∂t+∇ · (ρfU) = ∇ · (Γ∇f) + Sf . (203)

To obtain the integral form of this equation one needsGauss’s (divergence) theorem. Two dimensionalversion has the following form

∫∫

Ωi

∇ ·w dΩ =

∮

∂Ω+i

w · dL (204)

where dΩ ≡ dx dy and dL ≡ n dL ≡ ı dy − dx.

Integral form

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Integrating over the two dimensional domain (finite‘volume’) Ωi and utilising Gauss’s theorem results in

d

dt

∫∫

Ωi

ρf dΩ+

∮

∂Ω+i

ρfU · dL =

∮

∂Ω+i

Γ∇f · dL+∫∫

Ωi

Sf dΩ.

For the sake of simplicity it is further assumed that wedeal with an incompressible case ρ = const.First and last integral in the above equation suggestthe following definition of an average fi value of fover Ωi

fi :=1

|Ωi|

∫∫

Ωi

f dΩ. (205)

The average value fi is typically located at the centreof the volume Ωi.

Spatial discretisation

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The next step would be the spatial discretisation overthe volume Ωi boundary ∂Ωi. The line integralrepresents the total flux out of volume Ωi and isreplaced by a sum

∮

∂Ω+i

w · dL ≈∑

k

wk ·∆Lk. (206)

Boundary ∂Ωi consists of lines indexed by subscript k.There are at least three lines (triangle). The vector wis either fU or Γ∇f . Because that vector w istypically not constant along each line it has to beapproximated by a single value wk at the centre ofeach line.

Time discretisation

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The last step would be time discretisation. Amongmany possibilities the simplest is the first orderforward finite difference approximation

dfidt≈ fn+1

i − fni∆t

. (207)

Time step of this approximation is denoted here as ∆t.Finally, one gets the following discretised version oftransport equation (i.e. Finite Volume Scheme)

ρfn+1i − fni

∆t|Ωi|+ ρ

∑

k

(fU)k ·∆Lk =

∑

k

(Γ∇f)k ·∆Lk + Sfi|Ωi|. (208)

|Ωi| stands for the area of control volume Ωi.

1D FVM diffusion problem

Contents


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Finite volumemethod

Monte Carlo method


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References


xi−1 xi xi+1

xi−12

xi+12

∆xi

∆xi−1 ∆xi+1

x0x1 x2 x3 = xN

xN+1

∆x1 ∆xN

∆x12

∆xN2


Contents


Turbulencemodelling



Finite volumemethod

Monte Carlo method


Other methods

References


One dimensional and steady state diffusion equationof a f quantity arises from the general transportequation (convection-diffusion equation)

∂(ρf)

∂t+∇ · (ρUf) = ∇ · (Γ∇f) + Sf . (209)

If the diffusion coefficient Γ is constant then theabove equation simplifies to

∇ · (Γ∇f) + Sf = 0 (210)

or more precisely

d

dx

(

Γdf

dx

)

+ Sf = 0. (211)


Contents


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Finite volumemethod

Monte Carlo method


Other methods

References


The integral form of one dimensional diffusionequation takes the following form

∫ xi+1

2

xi− 1

2

d

dx

(

Γdf

dx

)

dx+

∫ xi+1

2

xi− 1

2

Sf dx = 0. (212)

There is no need to take advantage of Gauss’stheorem. This is because the first term can beintegrated directly

(

Γdf

dx

)

i+ 12

−(

Γdf

dx

)

i− 12

+ Sfi∆xi = 0 (213)

where ∆xi := xi+ 12− xi− 1

2.

1D FVM diffusion problem - Dirichlet BC

Contents


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Monte Carlo method


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References


The average value of Sf at the centre of finite volumecan be approximated by means of trapezoidal rule bymeans of known values of Sf at the boundary of finitevolume

Sfi =Sfi− 1

2+ Sfi+ 1

2

2. (214)

Let us consider ODE

y′′(x) + 20x = 0 (215)

subjected to the Dirichlet boundary conditionsy(0) = y(1) = 0. The specific solution is

y(x) := −10

3(x3 − x). (216)


Contents


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Monte Carlo method


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References


In other word, the diffusion coefficient Γ := 1 andsource term Sf := 20x. If so, then discrete version ofone dimensional diffusion equation is now

dfi+ 12

dx−

dfi− 12

dx+ Sfi∆xi = 0. (217)

Derivatives or diffusive fluxes at the boundary of finitevolume are approximated by means of the secondorder scheme as

dfi+ 12

dx≈ fi+1 − fixi+1 − xi

, (218a)

dfi− 12

dx≈ fi − fi−1xi − xi−1

. (218b)


Contents


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Finite volumemethod

Monte Carlo method


Other methods

References


The specific form of a finite volume scheme is now

fi+1 − fixi+1 − xi

− fi − fi−1xi − xi−1

+ Sfi∆xi = 0. (219)

It can also be rewritten to give fi as a function ofsurrounding variables

fi =∆xi+1fi−1 +∆xi−1fi+1 + Sfi∆xi−1∆xi+1∆xi

∆xi−1 +∆xi+1

where ∆xi−1 := xi − xi−1 and ∆xi+1 := xi+1 − xi.For ∆xi−1 = ∆xi+1 = ∆xi = h (i.e. uniform mesh)the finite volume scheme is reduced to a finitedifference scheme

fi =fi−1 + fi+1 + Sfih

2

2. (220)

1D FVM pseudocode

Contents


Turbulencemodelling



Finite volumemethod

Monte Carlo method


Other methods

References


Data: Read volumes data and BCsCreate nodes and ghost nodes;n← 1;repeat

R← 0;for i← 2 to imax − 1 do

fn+1i ← ∆xi+1f

ni−1+∆xi−1f

ni+1+Sfi∆xi−1∆xi+1∆xi

∆xi−1+∆xi+1;

R← max(

|fn+1i − fni |, R

)

;

Update ghost nodes;n← n+ 1;


Results for nonuniform and uniform grids

Contents


Turbulencemodelling



Finite volumemethod

Monte Carlo method


Other methods

References


0.2 0.5 0.72 0.920

0.5

1

1.5

x

y

0.13 0.38 0.63 0.880

0.5

1

1.5

x

y

1D FVM diffusion problem - Neumann BC

Contents


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Finite volumemethod

Monte Carlo method


Other methods

References


Let us consider the same ordinary differential equation

y′′(x) + 20x = 0 (221)

subjected to both the Dirichlet y(0) = 0 andNeumann y′(1) = 0 boundary conditions. The specificsolution is now

y(x) := −10x(

x2

3− 1

)

. (222)

Results for nonuniform and uniform grids

Contents


Turbulencemodelling



Finite volumemethod

Monte Carlo method


Other methods

References


0.2 0.5 0.72 0.920

2

4

6

x

y

0.13 0.38 0.63 0.880

2

4

6

x

y


Contents


Turbulencemodelling



Finite volumemethod

Monte Carlo method


Other methods

References


Two dimensional and steady state diffusion equationof a f quantity arises, as previously, from the generaltransport equation (convection-diffusion equation)

∇ · (Γ∇f) + Sf = 0 (223)

If the diffusion coefficient is constant Γ := 1 and thesource term Sf := a then the above equationsimplifies to

∇ · ∇f + a = 0 (224)

or ∇2f = −a which is Poisson equation.


Contents


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Monte Carlo method


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References


The discrete version of the diffusion equation(simplified version of general transport equation) is

∑

k

(Γ∇f)k ·∆Lk + Sfij|Ωij | = 0. (225)

For a structural and Cartesian mesh (next slide) thenormal ∆Lk vectors are

∆LAB = |AB |ı = ∆yi ı, (226a)

∆LBC = |BC| = ∆xi , (226b)

∆LCD = |CD| (−ı) = −∆yi ı, (226c)

∆LDA = |DA| (−) = −∆xi . (226d)


Contents


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References


fi−1j−1

fij−1

fi+1j−1

fi−1j fij fi+1j

fi−1j+1 fij+1 fi+1j+1

A

BC

D∆~LAB

∆~LBC

∆~LCD

∆~LDA


Contents


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Finite volumemethod

Monte Carlo method


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References


The discrete version of two dimensional diffusionequation is now

Γ∂fi+ 1

2j

∂x∆yi + Γ

∂fij+ 12

∂y∆xi

− Γ∂fi− 1

2j

∂x∆yi − Γ

∂fij− 12

∂y∆xi + Sfij|Ωij | = 0

(227)where the area of volume Ωij is |Ωij| = ∆xi∆yi andthe diffusion coefficient is constant Γ := 1. If so, then

∂fi+ 12j

∂x∆yi +

∂fij+ 12

∂y∆xi −

∂fi− 12j

∂x∆yi

−∂fij− 1

2

∂y∆xi + Sfij∆xi∆yi = 0. (228)


Contents


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Finite volumemethod

Monte Carlo method


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References


The average value of Sf at the centre of finite volumeis approximated by means of known values of Sf atthe boundary of finite volume

Sfij =1

4

(

Sfi− 12j− 1

2+ Sfi+ 1

2j− 1

2Sfi− 1

2j+ 1

2+ Sfi+ 1

2j+ 1

2

)

.

Derivatives at the boundary of finite volume areapproximated by means of the second order scheme as

∂fi+ 12j

∂x≈ fi+1j − fijxi+1j − xij

=fi+1j − fij∆xi+1

, (229a)

∂fij+ 12

∂y≈ fij+1 − fijyij+1 − yij

=fij+1 − fij∆yj+1

, (229b)


Contents


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References


∂fi− 12j

∂x≈ fij − fi−1jxij − xi−1j

=fij − fi−1j∆xi−1

, (230a)

∂fij− 12

∂y≈ fij − fij−1yij − yij−1

=fij − fij−1∆yj−1

. (230b)

The specific form of a finite volume scheme is now

Sfij∆xi∆yi+fi−1j∆yi∆xi−1

+fi+1j∆yi∆xi+1

+fij−1∆xi∆yj−1

+fij+1∆xi∆yj+1

−

fij

(

∆yi∆xi−1

+∆yi∆xi+1

+∆xi∆yj−1

+∆xi∆yj+1

)

= 0.


Contents


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Monte Carlo method


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References


It can also be rewritten to give fij as a function ofsurrounding variables

fij =

fi−1j∆yi∆xi−1

+fi+1j∆yi∆xi+1

+fij−1∆xi∆yj−1

+fij+1∆xi∆yj+1

+ Sfij∆xi∆yi∆yi

∆xi−1+ ∆yi

∆xi+1+ ∆xi

∆yj−1+ ∆xi

∆yj+1

.

For∆xi = ∆yi = ∆xi−1 = ∆xi+1 = ∆yi−1 = ∆yi+1 = h(i.e. uniform mesh) the finite volume scheme reducedto a finite difference scheme for Poisson equation

fij =fi−1j + fi+1j + fij−1 + fij+1 + Sfijh

2

4. (231)

2D FVM pseudocode

Contents


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Finite volumemethod

Monte Carlo method


Other methods

References


Data: Read volumes data and BCs

Create nodes and ghost nodes;

n← 1;

repeat

R← 0;

for i← 2 to imax − 1 do

for j ← 2 to jmax − 1 do

fn+1ij ←fni−1j∆yi

∆xi−1+

fni+1j∆yi

∆xi+1+

fnij−1∆xi

∆yj−1+

fnij+1∆xi

∆yj+1+Sfij∆xi∆yi

∆yi∆xi−1

+∆yi

∆xi+1+

∆xi∆yi−1

+∆xi

∆yi+1

;

R← max

(

|fn+1ij − fnij |, R

)

;

Update ghost nodes;

n← n+ 1;


Nonuniform and uniform volumes and nodes

Contents


Turbulencemodelling



Finite volumemethod

Monte Carlo method


Other methods

References


Results for nonuniform and uniform mesh

Contents


Turbulencemodelling



Finite volumemethod

Monte Carlo method


Other methods

References


02

46

810 0

24

6810

0

3

6

02

46

810 0

24

6810

0

3

6

Results for nonuniform and uniform mesh

Contents


Turbulencemodelling



Finite volumemethod

Monte Carlo method


Other methods

References


Monte Carlo method

Contents


Turbulencemodelling



Finite volumemethod

Monte Carlo method


Other methods

References


Monte Carlo integration

Contents


Turbulencemodelling



Finite volumemethod

Monte Carlo method


Other methods

References


0 0.2 0.4 0.6 0.8 10

0.2

0.4

0.6

0.8

1

x

y

0 0.2 0.4 0.6 0.8 10

0.2

0.4

0.6

0.8

1

xy

f(x) := xx

Monte Carlo integration

Contents


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Finite volumemethod

Monte Carlo method


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References


The simplest Monte Carlo integration is based onsampling uniformly distributed points (U ,U)

∫ 1

0

f(x) dx ≈ 1

n

n∑

i=1

F (U ,U) (232)

where

F (x, y) :=

1; if f(x) ≥ y

0; otherwise. (233)

Estimation of π

Contents


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Finite volumemethod

Monte Carlo method


Other methods

References


−1 −0.5 0 0.5 1

−1

−0.5

0

0.5

1

x

y

Estimation of π

Contents


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Monte Carlo method


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References


The area |D| = π of an unit circle

D :=

(x, y) : x2 + y2 ≤ 1

(234)

is estimated as

π =

∫∫

D

F (x) dx ≈ 4

n

n∑

i=1

F (U ,U) (235)

where

F (x, y) :=

1; if x2 + y2 ≤ 1

0; otherwise. (236)

Wiener process and random walk 1D

Contents


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Finite volumemethod

Monte Carlo method


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References


0 20 40 60 80 1000

0.2

0.4

0.6

0.8

1

Poisson equation

Contents


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References


Poisson equation subjected to the Dirichlet boundarycondition

∇2U(x) = −f(x); ∀x ∈ Ω, (237a)

U(x) = g(x); ∀x ∈ ∂Ω. (237b)

It can be solved as an expected value of random pathsof a stochastic process

U(x) = E

[

g(Wτ ) +12

∫ τ

0

f(Wt) dt

]

(238)

where t is a terminal time of a random walk

τ = inf t :Wt ∈ ∂Ω . (239)

Poisson equation

Contents


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Monte Carlo method


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References


If the Dirichlet boundary condition is g(x) = 0 then

U(x) = 12E

[∫ τ

0

f(Wt) dt

]

. (240)

We can also estimate

∫ τ

0

f(Wt) dt ≈m∑

i=1

fi(Wτ )∆t =m∑

i=1

fi(Wτ )h

V (h).

(241)Let us consider an ordinary differential equationy′′(x) = −20x subjected to boundary conditionsy(0) = y(1) = 0 (i.e. U := y, f(x) := 20x,g(x) := 0).

Monte Carlo ODE pseudocode

Contents


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Monte Carlo method


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References


Data: Read input variablesfor i← 1 to imax do

if not boundary(xi) thenS ← 0;for k ← 1 to n do

I ← 0;α← i;while not boundary(xα) do

α← α + 2⌊U(0, 1) + 12⌋ − 1;

I ← I + 20f(xα);

S ← S + I;

yi ← h2

2Sn;

Results – ODE

Contents


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Finite volumemethod

Monte Carlo method


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References


n = 10

0 0.2 0.4 0.6 0.8 10

0.5

1

1.5

x

y

Results – ODE

Contents


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Finite volumemethod

Monte Carlo method


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References


n = 100

0 0.2 0.4 0.6 0.8 10

0.5

1

1.5

x

y

Results – ODE

Contents


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Monte Carlo method


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References


n = 1000

0 0.2 0.4 0.6 0.8 10

0.5

1

1.5

x

y

Monte Carlo Poisson pseudocode

Contents


Turbulencemodelling



Finite volumemethod

Monte Carlo method


Other methods

References


Data: Read input variables and BCsfor i← 1 to imax do


if not boundary(Uij) thenτn ← 0;for k ← 1 to n do

S ← 0;α← i; β ← j;while not boundary(Uαβ) do

α← α+ 2⌊U(0, 1) + 12⌋ − 1;

β ← β + 2⌊U(0, 1) + 12⌋ − 1;

S ← S + 1;τn ← τn + S;

Uij ← −a h2

2τnn;

Results – Poisson equation

Contents


Turbulencemodelling



Finite volumemethod

Monte Carlo method


Other methods

References


24

68

102

4

6

810

0

3

6

n = 10


Contents


Turbulencemodelling



Finite volumemethod

Monte Carlo method


Other methods

References


24

68

102

4

6

810

0

3

6

n = 50


Contents


Turbulencemodelling



Finite volumemethod

Monte Carlo method


Other methods

References


24

68

102

4

6

810

0

3

6

n = 100


Contents


Turbulencemodelling



Finite volumemethod

Monte Carlo method


Other methods

References


24

68

102

4

6

810

0

3

6

n = 500


Contents


Turbulencemodelling



Finite volumemethod

Monte Carlo method


Other methods

References


24

68

102

4

6

810

0

3

6

n = 5000

Lattice Boltzmann method

Contents


Turbulencemodelling



Finite volumemethod

Monte Carlo method


Other methods

References


Definitions and ideas

Contents


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References


The kinetic theory of gases treats gas as a largenumber of small molecules. They are in constant (andrandom) motion and constantly collide with oneanother.Knowing the position and velocity of each particle atsome instant in time it would be possible to know theexact dynamical state of the whole system. Themotion of particles could then be described by meansof classical mechanics. This would allow for predictionof all future states of the system.Due to the large number of molecules a statisticaltreatment is possible and necessary.

Assumptions

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The gas is compose of small molecules whichmeans that the average distance separatingparticles is large in comparison with their size.

Molecules are in constant and random motion The large number of molecules make it possible to

apply statistical treatment Molecules have the same mass and spherical

shape Molecules constantly and elastically collide The only interaction is due to collision (no other

forces on one another)

Phase space and the distribution function

Contents


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For N molecules we can think of phase space in whichthe coordinates consist of the position xi, velocityvectors (vi) and the time t. For a three dimensionalcase we have 6N dimensional phase space (threecoordinates + three velocities times N molecules).The system can be described by a probabilitydistribution function f that depends on 6N variablesplus time t.For a single molecule this reduces to 6 dimensionalphase space (x1, x2, x3, v1, v2, v3). This can betreated as a statistical approach in which a system isrepresented by an ensemble of many copies.

Interpretation of the distribution function

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References


The elementary volume and dV and product dv ofelementary velocities are defined as

dV :=D∏

i=1

dxi, dv :=D∏

i=1

dvi (242)

where D means the physical dimension size. Thedistribution f that depends on r,v, t represents theprobability of finding a particular molecule mass witha given position and velocity per unit phase space.

∫

RD

∫

RD

f(r,v, t) dV dv (243)

The above integrate represents the total mass ofmolecules.

Continuous Boltzmann equation

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If no collisions occur then the probability of finding aparticular molecule mass with a given position andvelocity at (r,v, t) equals the probability at(r+ dr,v + dv, t + dt)

f(r+ dr,v+ dv, t+ dt) dV dv−f(r,v, t) dV dv = 0.

If, however, collisions take place then

f(r+ dr,v+ dv, t+ dt) dV dv−f(r,v, t) dV dv =

Ω(f) dV dv dt

where Ω is so called collision operator. It takes underconsideration collisions during dt interval.

Continuous Boltzmann equation

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References


We can now expand the left hand side of the previousequation by means of Taylor’s theorem

f(r+ dr,v + dv, t + dt) ≈

f(r,v, t) + dr · ∇f + dv · ∇vf +∂f

∂tdt. (244)

The two above equations give the Boltzmann equation

∂f

∂t+ v · ∇f +

F

m· ∇vf = Ω(f) (245)

where v = drdt

and mdvdt

= F.

Collision operator

Contents


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References


The simplification of the complicated collisionoperator Ω is needed. It should, however, fulfil at leasttwo conditions:

conservation of collision invariants ϕ

∫

RD

ϕΩdv = 0 (246)

where collision invariants are: 1 (obvious), v and12‖v‖2.

tendency to the Maxwell-Boltzmann distribution(relaxation to local equilibrium)

BGK approximation

Contents


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The BGK (Bhatnagar-Gross-Krook) approximation isthe most popular simplification of the collisionoperator

Ω =1

τ(f eq − f) . (247)

It expresses relaxation to local equilibrium f eq withthe relaxation time τ . Both conditions are fulfilled.The Boltzmann equations without external forces F isnow

∂f

∂t+ v · ∇f =

1

τ(f eq − f) . (248)

Now the equation is linear! More precisely it is a linearpartial differential equation.

Maxwell-Boltzmann distribution

Contents


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References


It is the basic law of the kinetic theory of gases. TheMaxwell-Boltzmann distribution is used for moleculesbeing not far from thermodynamic equilibrium. Othereffects like quantum effects and relativistic speeds areneglected. The distribution is

f eq = ρ (2πRT )−D2 e−

‖v−U‖2

2RT =

ρ(√

2πcs

)−D

e−

‖c‖2

2c2s . (249)

This distribution is valid for freely moving moleculeswithout interacting with one another. The exceptionsare only elastic collisions.

From Boltzmann eq. to conservation eqs

Contents


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References


It is possible to derive the conservation laws from theBoltzmann equation. Firstly, from the interpretationof distribution function f it arises the definition ofmacroscopic density

ρ(r, t) =

∫

RD

f(r,v, t) dv. (250)

The average value of a quantity ϕ is defined as

〈ϕ〉 =

∫

RD

ϕf dv

∫

RD

f dv=

1

ρ

∫

RD

ϕf dv. (251)

The integration is carried out over velocity space.

Averaged Boltzmann equation

Contents


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Multiplying the Boltzmann equation by ϕ and thenintegrating over velocity space results in

∫

RD

ϕ∂f

∂tdv +

∫

RD

ϕv · ∇f dv +∫

RD

ϕF

m· ∇vf dv =

∫

RD

ϕΩ(f) dv. (252)

Taking advantage of the average definition theaveraged Boltzmann equation may now be rewrittenas

∂

∂t(ρ〈ϕ〉) +∇ · (ρ〈ϕv〉)− ρf · 〈∇vϕ〉 = 0. (253)


Contents


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Substituting ϕ := 1 into the averaged Boltzmannequation we have

∂ρ

∂t+∇ · (ρ〈v〉) = 0. (254)

Comparing the above equation with the massconservation equation

∂ρ

∂t+∇ · (ρU) = 0 (255)

it becomes obvious that the macroscopic velocity U

must be

U = 〈v〉 = 1

ρ

∫

RD

v f dv. (256)


Contents


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References


Substituting ϕ← v into the averaged Boltzmannequation we have

∂

∂t(ρU) +∇ · (ρ〈vv〉)− ρf = 0. (257)

Introducing the microscopic velocity c in the meanvelocity frame

c(r,v, t) := v −U(r, t) (258)

it is possible to define the stress tensor

σ := −ρ〈cc〉 = −∫

RD

ccf dv. (259)

Now, the averaged Boltzmann equations becomes themacroscopic momentum conservation equation

∂

∂t(ρU) +∇ · (ρUU) = ρf +∇ · σ. (260)

Energy conservation equation

Contents


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Substituting ϕ := 12‖v‖2 ≡ 1

2v · v into the averaged

Boltzmann equation we have

e :=1

2〈‖c‖2〉 = 1

2ρ

∫

RD

‖c‖2f dv. (261)

Introducing the heat vector q definition

q :=1

2ρ〈c‖c‖2〉 = 1

2

∫

RD

c‖c‖2f dv (262)

we have the macroscopic energy conservation equation

∂

∂t

(

ρ

(

e+1

2‖U‖2

))

+∇·(

ρ

(

e+1

2‖U‖2

)

U

)

=

ρf ·U+∇ · (σ ·U− q) . (263)

Equation of state

Contents


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From the definition of the stress tensor σ := −ρ〈cc〉and the internal energy e := 2−1〈c · c〉 we have

tr〈cc〉 = 2e. (264)

Additionally, by means of the stress tensor we havepressure definition p := −D−1 trσ. Combining theseresults in

2ρe = pD. (265)

The above equation together with the equipartition ofenergy for mono-atomic gases gives the equation ofstate

p = ρRT = ρc2s. (266)

Velocity space discretisation

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The velocity space v is discretised into a finite set ofQ velocities vn where Q = |vn|. Discretedistributions are defined by means of the discretisedvelocity space

fn(r, t) = Wn f(r,vn, t), (267a)

f eqn (r, t) = Wn feq(r,vn, t). (267b)

Wn are the weights of the Gaussian quadrature rule.The density may now be approximated as

∫

RD

f(r,v, t) dv ≈∑

n

Wnf(r,vn, t) =∑

n

fn(r, t).

Discrete Boltzmann equation

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The discrete distribution function fn satisfies thediscrete Boltzmann equation (with BGKapproximation)

∂fn∂t

+ vn · ∇fn =1

τ(f eqn − fn) . (268)

The fluid density, velocity and internal energy are nowcalculated from the discrete distribution function:

Quantity Continuous discreteρ

∫

RD f dv∑

n fnρU

∫

RD v f dv∑

n vnfnρe 1

2

∫

RD ‖v −U‖2f dv 12

∑

n ‖vn −U‖2fn

Quadratures

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To recover correct form of the Navier-Stokesequations the discrete velocity set has to be chosen sothe following quadratures hold exactly

∀0≤m≤3

∫

RD

f eqm∏

i=0

v dv =∑

n

f0n

m∏

i=0

vn. (269)

The above may be reduced to

I :=

∫

RD

e−

‖v‖2

2c2s ψ(v) dv ≈∑

n

Wne−

‖vn‖2

2c2s ψ(vn) (270)

where

Wn = e‖vn‖2

2c2s

(√2πcs

)D

wn (271)and

wn = π−D2

D∏

i=1

ωi. (272)

Space discretisation

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The space discretisation follows the velocity spacediscretisation. This means it is discretised into alattices (D1Q3, D2Q9, D3Q27 discussed further).From the quadratures it arises speed of the model

c =√3cs. (273)

The speed c is used for space discretisation in thefollowing manner ∆xi = c∆t where ∆t representstime step (time space discretisation).One may also introduce dimensionless lattice velocities

en :=vn

c. (274)

Lattice Boltzmann equation

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Introducing the substantial derivative symboldndt

:= ∂∂t+ vn · ∇ makes it possible to rewrite the

discretised Boltzmann equation

dnfndt

=1

τ(f eqn − fn) . (275)

The substantial derivative is approximated by means of

dnfn(r, t)

dt=fn(r+ vn∆t, t+∆t)− fn(r, t)

∆t+O (∆t) .

From the two above one gets the Lattice Boltzmannequation

fn(r+vn∆t, t+∆t)−fn(r, t) =1

τ

(

f0n(r, t)− fn(r, t)

)

.

where dimensionless collision time is τ := τ∆t.

Equations

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Continuous Boltzmann equation∂f

∂t+ v · ∇f = Ω(f) (276)

Continuous Boltzmann equation with BGKapproximation

∂f

∂t+ v · ∇f =

1

τ(f eq − f) (277)

Discrete Boltzmann equation

∂fn∂t

+ vn · ∇fn =1

τ(f eqn − fn) (278)

Lattice Boltzmann equation

fn(r+vn∆t, t+∆t)−fn(r, t) =1

τ

(


)

(279)

Typical lattices

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D1Q3 lattice

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wn =

23, n = 0;

16, n ∈ 1, 2.

D2Q9 lattice

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wn =

49, n = 0;

19, n ∈ 1, . . . , 4;

136, n ∈ 5, . . . , 8.

D3Q27 lattice

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wn =

827, n = 0;

227, n ∈ 1, . . . , 6;

154, n ∈ 15, . . . , 26;

1216

, n ∈ 7, . . . , 14;

D3Q19 lattice

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wn =

13, n = 0;

218, n ∈ 1, . . . , 6;

136, n ∈ 7, . . . , 18;

D3Q15 lattice

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wn =

29, n = 0;

19, n ∈ 1, . . . , 6;

172, n ∈ 7, . . . , 14;

Expansion of the Maxwell-Boltzmann

distribution

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The Maxwell-Boltzmann distribution can berearranged

f eq = ρ(√

2πcs

)−D

e−

‖v‖2

2c2s e−

(

‖U‖2

2c2s−v·U

c2s

)

. (280)

Now f eq can be expanded into a Taylor series in termsof the fluid velocity

f0 := ρ(√

2πcs

)−D

e−

‖v‖2

2c2s

(

1 +v ·Uc2s

+(v ·U)2

2c4s− ‖U‖

2

2c2s

)

.

This is valid for low Mach numbers

f eq = f0 +O(‖U‖3

c3s

)

= f0 +O(

Ma3)

. (281)

Discrete Maxwell-Boltzmann distribution

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The discrete equilibrium distribution is defined bymeans of the discretised velocity space

f0n(r, t) = Wnf

0(r,vn, t) (282)

where weights are

Wn = e‖vn‖2

2c2s

(√2πcs

)D

wn. (283)

Together with the Taylor expansion for low Machnumbers we have

f0n := wnρ

(

1 +vn ·Uc2s

+(vn ·U)2

2c4s− ‖U‖

2

2c2s

)

.

Streaming and collision

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The Lattice Boltzmann equation

fn(r+vn∆t, t+∆t) = fn(r, t)+1

τ

(


)

.

The collision step

f tn(r, t+∆t) = fn(r, t) +1

τ

(


)

.

The streaming step

fn(r+ vn∆t, t +∆t) = f tn(r, t+∆t).

Streaming

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LBM pseudocode

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k ← 0;repeat

R← 0;ρ←

∑

n fn;U← 1

ρ

∑

n vnfn;

Calculate residue;

f0n ← wnρ

(

1 + vn·Uc2s

+ (vn·U)2

2c4s− ‖U‖2

2c2s

)

;

f tn(r, t+∆t)← fn(r, t) +1τ(f0n(r, t)− fn(r, t));

Apply Bounceback;fn(r+ vn∆t, t +∆t)← f tn(r, t+∆t);Apply other BCs;k ← k + 1;

until k < kmax and R > Rmin;

Chapman-Enskog expansion

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The Navier-Stokes equations can be recovered fromthe Lattice Boltzmann equation

∂

∂t(ρU) +∇ · (ρUU) = −∇p+ µ∇2U (284)

through the Chapman-Enskog expansion (multi-scaleanalysis). The expansion of the discreteMaxwell-Boltzmann distribution is used

f0n := wnρ

(

1 +vn ·Uc2s

+(vn ·U)2

2c4s− ‖U‖

2

2c2s

)

.

The first RHS term is responsible for ∇p, the secondfor ∇2U, the last two terms are related to ρUU.

Simplifications

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Knowing the structure of the discreteMaxwell-Boltzmann distribution we can now drop thenonlinear terms

f0n := wnρ

(

1 +vn ·Uc2s

)

(285)

to recover Stokes equations

∂

∂t(ρU) = −∇p+ µ∇2U. (286)

For both cases the dynamic viscosity is defined as

µ := ρ

(

τ − 1

2

)

c2s∆t. (287)

Example - input

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1 60 2023 1000000000122222222222222222222222222222222222222222222222224 1000000000122222222222222222222222222222222222222222222222225 1000000000122222222222222222222222222222222222222222222222226 1000000000122222222222222222222222222222222222222222222222227 1000000000122222222222222222222222222222222222222222222222228 1000000000111111111111111111111111111111111111111111111111119 100000000000000000000000000000000000000000000000000000000001

10 10000000000000000000000000000000000000000000000000000000000111 10000000000000000000000000000000000000000000000000000000000112 10000000000000000000000000000000000000000000000000000000000113 10000000000000000000000000000000000000000000000000000000000114 10000000000000000000000000000000000000000000000000000000000115 10000000000000000000000000000000000000000000000000000000000116 10000000000000000000000000000000000000000000000000000000000117 10000000000000000000000000000000000000000000000000000000000118 10000000000000000000000000000000000000000000000000000000000119 10000000000000000000000000000000000000000000000000000000000120 10000000000000000000000000000000000000000000000000000000000121 10000000000000000000000000000000000000000000000000000000000122 1111111111111111111111111111111111111111111111111100000000012324 22526 PT27 0 20 10 2028 0 .42930 VB31 50 0 60 032 0 .01

Advantages over conventional methods

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Possibility of microscopic interactionsincorporation

Easier dealing with complex boundaries Parallelisation of the method (the collision and

streaming processes are local)

Other methods

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Monte Carlo method


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References


Other methods

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Turbulencemodelling



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Boundary element method Smoothed-particle hydrodynamics

References

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List of references

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[1] Ferziger J.H., Peric M., Computational Methods for Fluid Dynamics,Springer-Verlag, Berlin, 2002

[2] Pope S. B., Turbulent Flows, Cambridge University Press, Cambridge,2000

[3] Succi S., The Lattice Boltzmann Equation for Fluid Dynamics and

Beyond, Oxford University Press, Oxford, 2001

[4] Tesch K., Fluid Mechanics, Wyd. PG, 2008, 2013 (in Polish)

[5] Wilcox D. C., Turbulence Modeling for CFD, DCW Industries,California, 1994

[6] Zienkiewicz O. C., Taylor R. L., The Finite Element Method,Butterworth-Heinemann, Oxford, 2000

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