Lecture 12
18.086
533
6.7 Fluid Flow and Navier-Stokes
No-slip along a
u = 0 (from viscosity) and v = 0 (no crossing)
horizontal boundary
(8)
Navier-Stokes
/NAVIER-STOKES
Flow
problems
FLUID
FLOW
Along
a sloping
boundary,AND
the velocity
vector is separated into normal and tangen-
6.7
tial components. The inflow and no-slip conditions st ill prescribe both components,
and d/dx in (7) changes to d l d n for outflow (like a free end). Our examples will
incompressible
fluid is governed
by horizontal
the Navier-Stokes
equashow other possibilities,
staying with
and vertical boundaries.
• Tricky equation, tricky numerics…
form brings
the importance
ofuthe
Reynolds
number then
Re.div u = 0 would
If weout
prescribe
the outward flow
- n on
the whole boundary,
ds =Navier-Stokes
0. Please
note equation:
the
u n and duldn = Vu n,
nce-free
vectoruu= nand
the
pressure
is difference
a scalarbetween
p:
•require
Incompressible
a normal component and a normal derivative.
du
1
-+(u-V)u=-Vp+-nu+
dt
Re
f
divu=V=u=O
momentum balance
(1)
The Reynolds Number
continuity/incompr. (2)
equation
To reach the Reynolds number Re, the Navier-Stokes equations have been made
Physical with
conservation
laws
have normalized
dimensions.is The
key
•dimensionless.
Inthe
this momentum
dimensionless
form,
the
onlydensity
physical
parameter
the
Law for
mass
to Reynolds
1. to the physics
is number
the relative
importance of inertial forces and viscous forces.
Re
ften absent. Four terms deserve immediate comments:
-
inertial forces
(a velocity U ) (a length L)
pplies toReynoldsnumber
each component ofR eu=. viscous
Viscosity
produces
dissipation.
forces
(kinematic
viscosity v)
u
nd
%
(9)
Pressure 1
p is Flow
a Lagrange
multiplier,
adjusted
such
that the
incompressibility
Example
Here
L
would
be
the width
of the channel, and
in
a
long
channel
= 0condition
(no time
derivative
in
this
equation)
comes
from
fulfilled.
U could beisthe
inflow velocity. The number v is a ratio p l p of the material constant p
conservation
of mass : const ant density.
(the dynamic viscosity) t o the density p.
•
Poisson for p U* =
of a drag region
u"+'+ grad(pn+' -
Poisson is the equation we know best (also th
a divergence-free part Un+' and a curl-free
are orthogonal complements, with correct bo
p on a finite difference grid is usually called
Example: 2D flow in a lid-driven cavity
change to a backwards facing step
endent geometries
selected streamlines
with an• advection-di↵usion
equation
Consider 2D box, with
no-slip boundary
conditions at three boundaries:
domain ⌦ =Navier-Stokes
[0, lx ]⇥[0, ly ]. TheEquations
four domain boundaries are denoted North,
pressible
d East.
The domain is fixed in time, and we consider no-slip boundary
• In 2D, we can write the NS equations in
e incompressible
Navier-Stokes equations in two space dimensions
ch
wall, component
i.e.
form as
1
+ uyy
(1)
=pux N=(x)(u )x (uv)y + Re (uxxv(x,
ly))Figure
= 0 6.20: Lid-driven cavity
u(x, lyu)t +
flow u =(4)
(u,
on a rectangular
domain
⌦=
[0, lx ]⇥[0,
four
boundaries
a
on a rectangular
domain
⌦ =l[0,
]⇥[0,
ly ].domain
The four
domain bo
y ]. lThe
xLecture
1
2
u(x, 0)
=puy S=(x)
v(x,
0)
=
0
(5)
vt +
(uv)
(v
)
+
(v
+
v
)
(2)
x South,
y West,
xx East.
yy The
South, West,
and
East. and
The
domain
is domain
fixed in istime,
we consider
fixedand
in time,
and we
Re
conditions conditions
on each wall,
on i.e.
eachv(0,
wall,y)
i.e.= vW (y)
u(0, y)
=
0
(6)
ux + vy = 0
(3) on an actua
Those steps need to be executed
2
u(lx , y) = 0
, y)
=different
(y)points, asv(x,
in Figure
uvNE(x)
lThe
ly )at=
uu(x,
ly ) =6.20.
0v(x,(7)
u(x, ly ) =v(l
y)
Nx(x)
of the i,j cell. The horizontal velocity U is
• The boundary conditions will
be
(x)(p and U arev(x,
v(x,
u(x, 0) = uu(x,
0) = 0on
S (x)0)
of =
theuScell
staggered
row0)2
2
he Navier-Stokes equations can be
in [2].
momentum
(againv(0,
on the
cellvv(0,
edges
so
y)below
= The
0 the center
y)
u(0,found
y) = 0u(0,
y)equations
=
W (y)
The
three
gridsinertial
for p, U, V and
bring many
conve
ibe the time evolution of the velocity
field
(u,
v)
under
viscous
u(lx , y) = 0u(lx , y) = 0
v(lx , y) = vv(l
x , y)
E (y)
A first question is notationcondition
for the grid va
sure p is a Lagrange multiplier to satisfy the incompressibility
change to a backwards facing step
Easiest numerical scheme: FD with
projection
with an advection-di↵usion equation
ndent geometries
• The incompressibility
equation
is not time-dependent, but needs to be satisfied
pressible
Navier-Stokes
Equations
at every point in time. => Cannot simply integrate NS equation in time…
e incompressible
Navier-Stokes equations in two space dimensions
•
Also, we have advective and diffusive terms (i.e. time step problems!) e.g.:
1
(uxx + uyy )
ut + px = (u )x (uv)y +
Re
1
2
(v )y +
(vxx + vyy )
t + py = (uv)
x
• vNumerical
approach:
Re
ux + vy = 0
2
•
Treat nonlinear terms (advection) explicitly in time first
(1)
(2)
(3)
2
•
Add linear diffusive terms using an implicit scheme
•
Correct the pressure to satisfy incompressibility, update
velocities to be divergence-free
The
termsSolution
are treatedApproach
explicitly. This circumvents the sol
4 nonlinear
Numerical
terms
system, but introduces a CFL condition which limits the time step
The
general
approachthe
of solution
the code of
is described
in Section 6.7 in the book
sSolution
are treated Approach
explicitly.the
This
circumvents
a nonlinear
spacial
resolution.
pressible Navier-Stokes
Equations
Science and Engineering
[4].
uces a CFL condition which limits the time step
by
times
⇤a constant
n
U
U
h of the code is described inWhile
Section
the qbook
Computational
u, 6.7
v, pinand
are the
solutions to then Navier-Stokes
equations,
2
n n
=
((U
)
)
(U
V
)y
x
on.
ering
incompressible
Navier-Stokes
equations in two
space
dimensions
[4].
numerical approximations
by capital
Assume we have the velocity fi
t letters.
th time step
⇤ solutions
⇤and
n the (3) is satisfied. We find the
d q areUthe
to the
Navier-Stokes
equations,
we
denote
atnthe
n
(time
t),
condition
Un
V
V
2
n n
n n
n 2
n
n
=
((U
)
)
(U
V
)
(12)
=
(U
V
)
((V
) )y
st
1
xwetime
y
x step approach:
ations by capital letters. Assume
havestep
the (time
velocity
field
Uby and
V
(n
+
1)
t
+
t)
the
following
three
2
(uxx + uyy )t
(1)
ut t+ px = (u )x (uv)y +
(time t),⇤and condition
(3) is satisfied. We
the solution at the
Refind terms
V
Vn
1.
Treat
nonlinear
n
n 2will detail how to discretize the nonlinear terms.
5 we
time t + t) by the=following
three
step
approach:
(UInn VSection
)The
((V
) 1)terms
(13)
x nonlinear
y
2
are
treated
explicitly.
This circumvents
vt t+ py = (uv)x (v )y +
(vxx + vyy )
(2) the solutio
ear terms
Re
system,
but introduces
a CFL condition which limits the time step by a
2. Implicit
viscosity
lerms
detail
toexplicitly.
discretize
the
nonlinear
terms.
are how
treated
This
circumvents
the
of a nonlinear
the
spacial
resolution.
ux + vy = 0 The viscosity terms solution
(3) treated e
are treated
implicitly. If they were
condition which limits the time step by a constant
times
⇤
n
U
U
yroduces a• CFL
have a time step restriction proportional
to)2 )the (U
spacial
discretiz
n
nand
U*
= ((Uestimate
V n )V*:
olution. Nonlinear advection is solved explicitly in time (Euler) to obtain
x
y
t implicit
s are treated
implicitly.
If
they
were
treated
explicitly,
we would
have
no
such
limitation
for
the
treatment. The price t
⇤
n
⇤
n
2 n n
U
U
V
V
n 2
n n
n 2
= ((U
)
)
(U
V
)
(12)
estriction proportional
to
the
spacial
discretization
squared.
We
=
(U
V
)
((V
) )y
systems
in
each
time
step.
x to be solved
y
x
and
t
t
⇤
ation for Vthe
implicit
treatment.
The price to pay⇤⇤is two
⇤ linear
Vn
1 the nonlinear
U
n n In Section
n 25 we will detail U
how
to
discretize
⇤⇤
⇤⇤ terms.
=
(U
V
)
((V
)
)
(13)
x
y
=V**: (Uxx + Uyy )
d in each
timet step.
• Implicit
diffusion is is added to obtain estimate U**
and
t
Re
2.
Implicit
viscosity
⇤⇤ toUdiscretize
⇤
⇤⇤
⇤
e will detailUhow
nonlinear terms.
1 theThe
V
V
1If they
⇤⇤ viscosity
⇤⇤
⇤⇤ were⇤⇤
terms
are
treated
implicitly.
treated expli
=
(Uxx + Uyy )
(14)
=
(V
and
xx + Vyy )
t
Re have a time step restriction proportional
t
Re
osity
to the spacial discretizatio
V ⇤⇤ implicitly.
V⇤
1 If they
erms are treated
treated
explicitly,
we would
have
no ⇤⇤
such
limitation
for the
implicit treatment. The price to pa
⇤⇤ were
3.=Pressure
(V
+correction
Vyy
)be solved in each
(15)
xx
ep restriction
proportional
to
the
spacial
discretization
squared.
We
systems
to
time
step.
• The current
estimate
U** and V** does not yet satisfy incompressibility,
t
Re
⇤⇤ , V ⇤⇤ ) by the gra
We
correct
the
intermediate
velocity
field
(U
imitation for
the implicit
treatment.
The
price
to pay is two
⇤⇤ linear
⇤
which
is corrected
in
the
last
step
1
U
U
⇤⇤
⇤⇤
n+1
ion in each time step. P
=
(Uxx
+ Uyy
)
to enforce incompressibility.
solved
t
Re
⇤⇤
⇤⇤
⇤⇤
⇤ ⇤⇤ 1
ermediate velocity
) by the gradientVof
a Vpressure
n+1
U ⇤⇤ U ⇤field1(U ⇤⇤, V ⇤⇤
U
=
(Uxx + Uyy )
(14)U=
(V ⇤⇤ +n+1
V ⇤⇤ )
Nonlinear and diffusive terms
xx
yy
to be solved insquared.
each timeWe
step.
striction proportional to the systems
spacial discretization
⇤⇤
tion for the implicit treatment. The price to pay is twoU linear
1
U⇤
⇤⇤
⇤⇤
=
(U
+
U
xx
yy )
d
in
each
time
step.
pressible Navier-Stokes Equations ⇤⇤ t ⇤ Re
V
V
1
1
U ⇤⇤ U ⇤
⇤⇤
⇤⇤
⇤⇤
⇤⇤
(Vxx
+ Vyy
)
=
(Uxx + Uyy )
(14) =
t
Re
t
Re
e incompressible
Navier-Stokes
equations in two space dimensions
V ⇤⇤ V ⇤
1
⇤⇤Pressure
⇤⇤
=
(V3.xx
+ Vyy
) correction
(15)
1
t
Re 2 We correct the intermediate velocity field (U ⇤⇤ , V ⇤⇤ ) by the gradie
(uxx + uyy )
(1)
ut + px = (u )x (uv)y +
Re incompressibility.
P n+1 to enforce
on
1
⇤⇤ , V ⇤⇤ ) by
2 the gradient
rmediate velocity
field
(U
vt + py = (uv)x (v )y +
(vxx +ofvayy pressure
) U n+1 U ⇤⇤
n+1 (2)
• In the last step, U** and V** are
= (P term:
)x
Recorrected by a pressure gradient
ompressibility.
t
uUxn+1
+ vy U=⇤⇤0
(3)
V n+1 V ⇤⇤
n+1
= (P
)x
(16)
= (P n+1 )y
and
t
t
n+1
n+1 , since it is only given implicitly. It is ob
Vn+1
V ⇤⇤
n+1 2
The
pressure
is
denoted
n+1 P
= (P needed
)y
(17)
• P
is tthe pressure
to make U divergence-free.
a linear system. In vector notation the correction equations read as
• , The
ted P n+1
sincecorrection
it is only given
implicitly.
It isisobtained by 1solving
in vector
notation
1 n
n+1
n+1
U
U
=
rP
vector notation
the correction
equations read as
n+1
⇤⇤
t
t
~
~
U
U
n+1
rPn+1
1 n+1
1 n = Applying
the divergence to both sides yields the linear system
U
U
=
rP
(18)
t
t
t
1
n
n+1n+1
P
=
r
·
U
Applying
the divergence,
we obtain an equation for P :
ence to •both
sides yields
the linear system
t
1 1Hence,
n+1
⇤⇤ pressure correction step is
the
n+1
n· U
~
4P
=
r
P
=
r·U
(19)
t t
correction step is
4
• This means we need to solve a Poisson-problem!
Projection step
Poisson problem
•
Pressure correction steps:
(a) Compute F n = r
r ·· U
Un⇤⇤
(b) Solve Poisson equation
P n+1 =
1
n
F
t
(c) Compute Gn+1 = rP n+1
n+1
n+1
(d) Update velocity field UU
==
UnU⇤⇤ tGn+1
tGn+1
(b): question,
Pij is a matrix
withboundary
nx∙ny components
(nx,are
ny :appropriate
number of spatial
(resolution)
The
which
conditions
for thecells
Poisson
equation
for the
pressure
, is complicated.
A standard
approach is to prescribe homogeF can
also beP written
as a matrix with
nx∙ny components
neous Neumann boundary conditions for P wherever no-slip boundary conditions are
ΔP can now be expressed using finite differences, resulting in a matrix
prescribed for the velocity field. For the lid driven cavity problem this means that
L. We then need to solve
homogeneous Neumann boundary conditions are prescribed everywhere. This implies
1 nP is only defined up to a constant, which is fine, since
n+1
in particular
that
the
pressure
LP
=
F
only the gradient of P enters
t the momentum equation.
=> Invert L matrix! Need a way to invert large matrices (see later)
In addition to the solution steps, we have the visualization step, in which the stream function
Qn is computed. Similarly to the pressure is is obtained by the following steps
n
n
n
Consider to have nx ⇥ ny cells. Figure 1 shows a staggered grid with nx = 5 and ny = 3.
When speaking of the fields P , U and V (and Q), care has to be taken about interior and
boundary points. Any point truly inside the domain is an interior point, while points on or
outside boundaries are boundary points. Dark markers in Figure 1 stand for interior points,
while light markers represent boundary points. The fields have the following sizes:
Spatial discretization
field quantity
pressure P
velocity component U
velocity component V
stream function Q
interior resolution
nx ⇥ ny
(nx 1) ⇥ ny
nx ⇥ (ny 1)
(nx 1) ⇥ (ny 1)
resolution with boundary points
(nx + 2) ⇥ (ny + 2)
(nx + 1) ⇥ (ny + 2)
(nx + 2) ⇥ (ny + 1)
(nx + 1) ⇥ (ny + 1)
The values at boundary points are no unknown variables. For Dirichlet boundary conditions
they are prescribed, and for Neumann boundry conditions they can be expressed in term
Staggered
of interior points. However, boundary points of U and V are used for
the finitegrid
di↵erence
approximation of the nonlinear advection terms. Note that the boundary
points of
in cells
the four
nx,ny: Number
corner are never used.
interior U
boundary U
5
interior V
Uij Pij
boundary V
U P
j
Vij
x pressure
i
• time
and velocity
memoryfield
efficiency
for
large
computations
the
centers,
and UxPand
canVbe
approximated
1. The divergence
of the
F cell
=r
·U
computes
y . Both
3. The
gradient
of added
the pressure
values then live in the cell centers,
i.e. they
can be
directly.G = rP requires t
The code qualifies as a basis for modifications
and extensions. The followi
1
P
.
Again,
a
look
at
Figure
1
indicates
that
t
2. For the have
pressure
Poisson
equation
P
=
F
,
both
sides
are
defined
in
y
t
already been applied to the code by MIT students and other users:
position of U ,as
while
the latter
live at the position
the cell centers, and
P can be approximated
described
above.
update
the velocity
• external
forces G = are
3. The gradient
of the pressure
rP needed
requirestothe
computation
ofU PPxfield.
and
V
Py . Again, a look at Figure
1 indicates
that the former then live at
the
– Stream
function
• ,inflow
and
outflow
boundaries
• position
Second of
derivatives:
Centered
differences
U
while the
latter
live
at the positions
of V . grid
Exactly
where
Similarly,
theas
staggered
works
fine both
for the stream
usual.
=> Consistent
withvelocity
staggered
grid
are
needed
to update the
field.
live
on the cell corners, so the Poisson right hand side
• addition of a drag
region
Figure 1: Staggered grid with bo
P
2P
+
P
P
2P
+
P
yield
instabilities,
as
shown
in
many
textbooks
on
numerical
anali 1,j stream
i,j function
i+1,j Q lives
i,j 1 at the
i,jcell corners.
i,j+1
– Stream
function
4Pij = Pxx,ij + Pyy,ij ⇡
+
2
5.1 Both
Approximating
derivatives
xbackwards
x2 Vx and
gered Similarly,
grid comes
Assume,
wefine
not
in step
the
• play.
geometry
toare
afor
facing
theinto
staggered
gridchange
works
theinterested
stream
function.
Uy
– Nonlinear terms (central
di↵erencing)
• Second derivatives
sitionlive
of• Uon
, but
instead
we
want
value
in the
middle
cell
corners,
so
the the
Poisson
right
hand
side between
F = Vx points
Uy and
thus
the
First
derivatives:
“Centered”
differences
around
half-grid
(good
i,j the
Finite
di↵erences
can approximate second deriva
The nonlinear terms are the stencil.
onlyAtplace
where the disc
•
time
dependent
geometries
an interior point U we approximate t
function
Q livesnumerical
at the cellinstabilities):
corners.
d Ui,jstream
. Then
the approximation
accuracy,
avoids
Why staggered grid?
i,j
grid does not work directly. For
instance, theU product
2U + U
U = (U ) + (U ) ⇡
h
– Nonlinear terms
(central
di↵erencing)
• coupling
with
an
advection-di↵usion
equation
Ui+1,j
U
since
U
and
V
live
in
di↵erent
positions.
The
solutio
i,j
(U
)
⇡
(22) on
1
Here the
one or staggered
two of the 2
neighboring points migh
The nonlinear
x i+ ,j terms are the only place where the discretization
backwards: For updating U ,formula
we holds
need
(Ucomponent
)x and
2
for the
V , and(U
for V
the
hx
grid does not work directly. For instance, the product U V is not the
directly
defined,
stream function
Q. If the unknown quantity
step isNavier-Stokes
comparably slow, we
wish
toapproximation
use thecansame
cen
then
the above
be represented
2
Incompressible
Equations
•
approximation
toV Ulive
middle
between
the
two
points.
In
derivatives
at
center
of cells =>
ideal
choice
pressure
correction
sinceFirst
U and
inthe
di↵erent
positions.
The
solution
isfortothe
take
thefromarguments
applied
the
left. Thus, solving the Poisson
x in“live”
2
as before.
This requires U tosolving
be implicitly
defined
theterms
cellin Ucen
for thein
viscosity
and
2
(which
lives
at
cell
centers
as
well):
backwards:
For updating
, we needof(U
inaseach
time
his position
happens
to be theUposition
Pi,j).x and (U V )y . If the flow
be solved,
detailed in
Section 7.
inuse
thethe
cellsame
corners.
We equations
obtain
these
quantities
by int
We consider
the
incompressible
Navier-Stokes
in
two
space
dimens
step is comparably
slow,
we
wish
to
centered
staggered
derivatives
•
First
derivatives
1
n+1
~ ⇤⇤
A first derivative in a grid point can be approxim
4P
=
r·U
values.
rection
2
as before. This requires
U to be defined in the cell centers, and U V 1to the defined
t
U
U
✓
◆
2 staggered
(U ) ⇡
correction,
the
approximation
of
first
derivatives
on
a
(uinterpolating
)x (uv)y +two (u
Only
terms
in u
yyi,j) + Ui+1,j2h
in •the
cellproblem:
corners.Nonlinear
We obtain
these
quantities
neighboring
xx + uU
t + px = by
2 Re
(U
)i+ 1 ,j =
perfectly:
live
at
cell
center,
but
are
needed
at
edges!
6
values.
2
1
2
2
✓
◆
v
+
p
=
(uv)
(v
)y +
(vxx + vyy )
t
y Ux and V
x2y . Both
ence of the velocity field F = r · U computes
Re
Ui,j + Ui,j+1
Ui,j + Ui+1,j
2
= added directly.
(23)
(U )can
U
1 =
=>cell
Interpolate,
e.g.they
and
n live in the
centers, i.e.
i+ 21 ,jbe
i,j+
ux + vy =2 0
2
2
1
essure Poisson equation
P =
sides
are defined in
Vi,j + Vi+1,j
Ui,j +
Ui,j+1
t F , both
i,j
xx i,j
yy i,j
i 1,j
x i,j
i,j
2
x
i
i+1,j
x
between them. At the points that lie on the boundary the value is dire
as U at the west and east boundary, and V at the north and south bou
north and south boundary and for V at west and east boundary, one
between two data points. The idea is the same as above: average the
For instance, the north boundary lies between points with velocity U
below respectively above the boundary be Ui,j and Ui,j+1 , and let the
value be UN . Then the boundary condition is
Some notes
•
•
•
Ui,j+1accuracy (see code
Centered differences can be “upwinded”Ui,j
for+better
= UN () Ui,j + Ui,j+1 = 2UN
2
documentation)
Analogously at the south boundary, and for V at the west and east bo
Boundary conditions: Need a bit of care due
@P to the staggered grid…
The normal derivative @n at the boundary is defined using two p
Boundary conditions
for pressure???
This isFor
a delicate
for no- presc
boundary
in their middle.
instance,issue.
at theNormally,
north boundary,
slip velocity BC, one
employs
Neumann
BCs foryields
pressure, i.e.
Neumann
boundary
conditions
@P
Pi,j+1 Pi,j
=0
thus, e.g. at the top boundary:
= 0 () Pi,j+1 = Pi,j
@~n @⌦
hy
The stream
function
is defined on cell boundary points. Thus all its
=> “Mirror” pressure value
in boundary
cells
on the domain boundary. We prescibe homogeneous Dirichlet bounda
can be prescribed directly.
6
U P
Solving the Linear Systems
V
The pressure correction and the implicit discretization of the viscosity
systems to be solved in every time step. Additionally, the comput
function requires another system to be solved whenever the data is p
Figure 1: Staggered grid with boundary cells
Demo
•
•
CSE website has a demo implementation (and documentation)!
Have a look at the code (you won’t yet understand how the Poisson problem is solved)
Lattice-Boltzmann method
•
Poisson problem for FD (or, for that matter, finite volume etc.) discretization:
Matrix inversion=>very costly in 3D, not easy to parallelize (non-local)
Boltzmann’s equation (one for each fluid):
•
An alternative that is much easier to implement numerically is the LatticeBoltzmann method (LBM)
•
Starting point of LBM is kinetic gas theory:
probability distribution function for particle at position x
with speed v
number density of molecules inside “volume” (x,x+dx) and (v,v+dv) at
t
f is the Probability Distributiontime
Function
of the particles in (x,v,t) space
dN = f (x, v, t)dxdv
•
Z
• 7 dimensional
space (3 space, 3 velocity, 1 time)!! Can handle, if clever!
• F:t)forces
over
the interfaces
density
⇢(x,
= betweenffluids
(x, v,
t)dv
velocity
• Collision term:ZThis is where VISCOSITY originates
~u(~x, t)⇢(~x, t) =
~v f (x, v, t)dv mean velocity at (x,t)
velocity
• In an equilibrium:
f is the isotropic Maxwell-Boltzmann distribution
Gaussian type of
PDF)quantities
etc. (aforsymmetric
other macroscopic
fluid
Probability function (1D)
Phase space (v,x) for 1 space dimension X
PDF shown as contours:
Question: How does the probability in a “volume”
change over time? => Since its a probability: We
expect a conservation law for f!!!
V
t
3 possibilities to change f(x,v,t) inside a small volume:
t+Δt
•
Particle can move: x => x+v dt (v:
current particle speed)
•
Particle can accelerate from v => v+F/m dt
(F: external force, m: particle mass)
•
Particles can collide: (x1,v1,x2,v2) => (x’1,v’1,x’2,v’2)
X
We can express this (in 3D) as:
Integration over V gives hydrodynamic quantities at chosen X
3
3
3
~
t)d xd v = f (~x, ~v , t)d xd3 v + Collisions
f (~x + ~v t, ~v + F /m t, t +
By Taylor-expansion of the LHS, one can derive the famous Boltzmann equation:
@f
@f
@f
~
+ ~v ·
+F ·
@t
@~x
@~v
✓
@f
@t
◆
=0
coll.
Boltzmann equation
From Boltzmann to NavierStokes
•
It can be shown: Boltzmann equation => Navier Stokes equation
•
=> instead of solving Navier-Stokes (a PDE for average molecule
velocities), we could solve the Boltzmann equation (a PDE for
probabilities f(x,v,t)), and then obtain the average velocities by
integrating f(x,v,t), e.g.
~u(~x, t)⇢(~x, t) =
•
Z
~v f (x, v, t)dv
velocity
Lattice-Boltzmann method: Discretize f(x,v,t), then numerically
integrate Boltzmann equation!
Lattice Boltzmann method
•
Let’s consider the previous equation for the prob. density function f(x,v,t), in absence of a force F:
f (~x + ~v t, ~v , t +
t)d3 xd3 v = f (~x, ~v , t)d3 xd3 v + Collisions
D2Q9
2
5
3
0
1
7
4
8
6
•
Discretize position (xi) and velocities vj on a lattice
•
At each lattice site i, we have a value fj (~
xi , t) per
= f (~xi , ~vj , t)
discrete velocity j. We write
fj (~xi , t) = f (~xi , ~vj , t)
•
One can show: Only very few discrete velocities are
necessary! Simplest scheme (D2Q9): 9 velocities are
sufficient:
~vj=5
v0 = |~v0 | =0
v1..4 =1
p
v5..8 = 2
lattice
lattice
unit
(lu)
spacing
Figure 2: Discrete lattice velocities for the D2Q9 m
Lattice Boltzmann method
D2Q9
•
6
2
5
3
0
1
7
4
8
~vj=5
Macroscopic quantities: Integrals => sums:
⇢(x, t) =
Z
⇢=
f (x, v, t)dv
velocity
Z
9
X
fj (~xi , t)
j=1
8
X
~u(~x, t)⇢(~x, t) =
~v f (x, v, t)dv
~uthe
⇢=
xi , t)~vj
j (~
make sense to stream DFs from walls towards the fluid). For this reason,
two steps f
(streamvelocity
j=0
ng & collision) are usually treated separately in actual numerical implementations.
The streaming step, where the DFs are translated to the neighbouring sites according to
•
LBM algorithm: Velocities “stream” f from site i to neighboring
he respective discrete velocity direction, is illustrated in Fig. (3), in the D2Q9 model for
sites and collide:
lattice
unit (lu)
f The(~xcollision
vstepj (illustrated
t, t +in Fig.t)[4])=consists
fj (~
x , t) + Coll(f (~xi , t))
i +~
implicity. j
of aire-distribution of the DFsj
LBM
owards the local discretized Maxwellian equilibrium DFs, but in such a way that local mass
nd momentum are invariant5 .
f (~x + ~v t, ~v , t +
3
3
3
3
t)d xd v = f (~x, ~v , t)d xd v + Collisions
continuum
Figure 2: Discrete lattice velocities f
The macroscopic variables are defined as function
streaming
streaming
(hereafter DFs) according to:
length of arrow: value of fj
⇢=
X1
fi
i=0
and
1
(mac
Collision term
•
Bhatnagar, Gross and Krook showed: Collision term can be approximated by the
relaxation ansatz (BGK approximation):
Coll(fj (~xi , t)) =
•
fjeq (~xi , t)
fj (~xi , t)
⌧
wj=0
Equilibrium distribution for f:

2
~
v
·
~
u
(~
u
·
~
v
)
j
j
eq
fj (~xi , t) = wj ⇢ 1 + 2 +
c
2c4
•
~u2
2c2
with weights wj=1..4
Side node: This equilibrium distribution is related to a series expansion of the Maxwell-Boltzmann velocity distribution for small velocities
•
𝜏 is the relaxation time scale
•
c is the lattice speed of sound
⇢, ~v
⇢, ~v
collision
wj=5..8
4
=
9
1
=
9
1
=
36
Notes
•
2
8
X
The speed of sound is c =
transported)
j=0
wj ~vj2
1
=
3
(speed by which inform. can be
ion of the collision process on a D2Q9 lattice. Note that the local den•
The collision relaxation time is related to the viscosity of the fluid. For 2D9Q model:
✓ but the
◆ DFs change according to the relaxation-to-localy~
v are conserved,
t
(typically one chooses Δt=1; it’s an 2
⌫=c ⌧
“artificial” time scale anyway)
2
For a given viscosity, the relaxation time is thus fixed:
⌧ = 3⌫ + t/2
6
7
8
e-mentioned
assumption
of
a
low
Mach
number,
and
further
taking
Kn
,
,
t
x
(for 2D9Q scheme, c2=1/3)
•
vers the incompressible Navier-Stokes equations:
•
Note: It can be shown that the continuum limit of the 2D9Q model is the Navier-Stokes
equation
⇢@tu
~ + ⇢~
ur · u
~ =
•
!
r·u
~ =0,
(9)
rP + ⇢⌫r2 u
~
(10)
In other words: Viscosity in NS has its origin in collisions in the Boltzmann-picture
equation of state:
2
has been proven to be only first-order accurate in time and space [Pan et al., 2006]. A straightFluid
forward improvement is to consider the wall-fluid interface to be situated halfway between the
Solid
wall and fluid lattice nodes [Ziegler, 1993]. This simple translation (which is actually nothing
more than a slightly different post-processing procedure), commonly referred to as half-way
Boundary conditions
bounce-back in the literature, is illustrated in Fig. (5).
has been proven to be only first-order accurate in time and space [Pan et al., 2006]. A straight•
Fluid
forwardboundary
improvement
is to consider the wall-fluid interface to be situated halfway between the
Possible
conditions:
Solid
• wall
and fluid lattice nodes [Ziegler, 1993]. This simple translation (which is actually nothing
No slip
Fluid
before stream (t)
Solid
than a(straight
slightly different
post-processing procedure), commonly referred to as half-way
• more
Periodic
forward)
the literature,
is illustratednormal
in Fig. (5).
• bounce-back
Inlet/outletinvelocities
(prescribe
velocities
•
to boundary)
Pressure difference (a bit more complex)
o be only first-order accurate in time and space [Pan et al., 2006]. A straightFluid
ment is to consider the wall-fluid interface to be situated halfway between the
Fluid
before stream (t)
Solid
Solid
•
No-slip
boundary
conditions:
Addnothing
a layer
of
ice nodes [Ziegler, 1993]. This simple translation
(which is actually
“ghost” sites.
Fluid
after stream
Solid
tly different post-processing procedure), commonly referred to as half-way
e literature, is illustrated in Fig. (5).
Fluid
Fluid
before stream (t)
Solid
Solid
before “stream” at time t
Fluid
Solid
after “stream” at time t:
stream to ghosts
Fluid
after stream
Solid
Fluid
after stream
Solid
Fluid
after bounce-back
Solid
collide in bulk,
bounce back, resulting in
reverse values at
the state at time t+Δt
ghostsFigure
sites5: Illustration of the half-way bounce-back algorit
Fluid
after bounce-back
after
Solid
before stream (t +
[Sukop and Thorne, 2006]).
t)
DEMO
Demo shows flow through a 2D porous medium, inlet: left,
outlet: right, top/bottom: no-slip BC
Parallelization
•
Parallelization is simple: Domain-decomposition
In each iteration
- On each processor:
Stream and collide
- Exchange values at domain
boundaries
- Next iteration
•
Today’s (gaming) graphics cards often have >256 single cores!
Cheap parallelization using
GPUs (real time!)
sailfish.us.edu.pl
LBM: Easy to model
multiphase-flow
•
Multiphase flow (water & oil, e.g.) is easily incorporated: Use two different prop.
densities f (one per phase)
https://www.youtube.com/watch?v=ucqgRo6Fd_c B. Ahrenholz
https://www.youtube.com/watch?v=1tasbL39xds A. I. Kupershtokh, 2013
Disadvantages of LBM
⌫
Kn =
L
r
⇡m
2kB T
•
Only working for small Knudsen numbers Kn<<1.
•
L: char. length scale (i.e domain size), kB: Boltzmann constant, T: temperature
r
kB T
Kinetic gas theory: Speed of sound of a medium is: cs ⇠
m
Taking the problem domain, with N lattice sites spaced by 1 length unit as
characteristic length, and c (lattice speed of sound) as maximal sound speed, we find
•
•
⌫
Kn ⇠
cN
•
Other problems:
•
No consistent thermo-hydrodynamic scheme
•
Similar to Kn condition: Problems for high Mach numbers