Fluid Mechanics Tutorial for Biomedical Engineering

Mirjam Abegglen

Contents1 Introduction 1

2 Fluid Mechanics in 1D 22.1 Conservation of mass . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22.2 Equation of motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

3 Fluid Mechanics in 3D 53.1 Conservation of mass . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53.2 Conservation of momentum: The Navier-Stokes Equation . . . . . . . . . . . . . . 6

3.2.1 Local acceleration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73.2.2 Convective acceleration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73.2.3 Volume forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73.2.4 Pressure term . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73.2.5 Friction term . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

4 Numerical Solution: The Lattice Boltzmann Method 84.1 Discrete vs Continuous . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84.2 The density distribution function . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84.3 Macroscopic quantities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94.4 Streaming and Collision . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94.5 The equilibrium distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104.6 Incorporating Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104.7 Boundary Conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104.8 Implementation Guidance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

5 Exercise 13

6 Solution Exercise 14

7 Matlab Code 167.1 LBM-poiseuille . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 167.2 streaming . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 187.3 constantvelocityBC . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 197.4 rhoandugenerator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 197.5 uequgenerator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 197.6 fequgenerator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 207.7 collision . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

1 IntroductionFluid Flow in general plays a huge role in several biomedical processes. One example is interstitialfluid flow in the bone. In previous studies it has been shown, that interstitial fluid flow (mechan-ically) activates the osteocytes to regulate bone remodeling [1] [2]. For this case of microscalefluid flow (but also for other biomedical flows) in comparably complex structures, continuum ap-proaches to model fluid flow might not be valid anymore. As a consequence Lattice BoltzmannMethod (LBM) has been suggested to give good results for micro- and nano-flow simulations [3].Moreover LBM has already been applied to microscale fluid flow analysis in canaliculi (bone) [4].

1

This script should give you a broad idea of the basic principles (equations) in fluid mechanics,how to fit them to your desired set up and how to solve them finally with different methods(analytically and numerically).

2 Fluid Mechanics in 1DThe sections Fluid Mechanics in 1D and 3D are based on the book "Strömungsmechanik: Grundlagen-Grundgleichungen-Lösungsmethoden-Softwarebeispiele" [5].

There are two basic princples that determine the motion of a fluid in one dimension, conser-vation of mass and conservation of momentum. Note that for simplicity reasons we look atincompressible fluids in the following sections. Incompressible means, that the density does notdepend on the pressure. If you exert a force on a volume element it will not shrink and thereforethe density stays constant.

First lets have a look on kinematics. Imagine a defined line where the variable s indicates theposition of your "fluid particle". We can describe the motion of the particle with c(s) representingthe velocity on that line at location s. This is the case in a stationary flow (time independent). Ifthe flow is in stationary the velocity will depend also on t ⇒ c(s, t).

Even though we describe the motion of the fluid one dimensionally, we can extend this byintroducing several lines to show flow in a tube as you might see in figure 1

Figure 1: flow tube

For the following explanations we also need the quantities p(s) representing the pressure at acertain location s on the streaming line. Additionally for the the tube setting shown in figure 1 weneed the surface A(s) that also depends on the one dimensional coordinate s.

2.1 Conservation of massIn general we derive:

m = ρ · V (1)

m = ρ · V (2)

V = c ·A (3)

m = c · ρ ·A = const. (4)

Where m represents the mass, ρ density, and V volume.Conservation of mass applied to the tube set up shown in figure 1 implies that the inflow equals

the outflow if a flow occurs, this can be expressed as follows.

c1 · ρ1 ·A1 = c2 · ρ2 ·A2 (5)

Note that in case of an incompressible fluid the density is constant (ρ1 = ρ2).

2

2.2 Equation of motionThe equation of motion also known as bernoullis principle can be derived through the sum of allacting forces. We still take the set up from figure 1. We know that motion can be describedthrough:

m · −→a =∑−→

F (6)

with a=acceleration, F=force.Since we are in the one dimensional case, we can go on with scalar quantities: a(s) is obtained bycalculation of the total differential of c(s, t) (we first look at the general case of location and timedependency and simplify later).

a =dc

dt=∂c

∂t+ c · ∂c

∂s(7)

Note that dt represents the total differential and ∂t and ∂s represent the partial differentials.If you don’t remember the difference from your math class, it may be worth it to look it up again.

The motion that is computed from equation 6 is dependent on the forces that acts on thefluid. In the set up that is described in figure 2 we consider the gravitational force and a pressuredifference ∆p, where the pressure p(s) needs to be multiplied with the surface A(s) to obtain theforce.

Figure 2: set up for the equation of motion

Rewrite equation 6 with equation 7 and the gravitational force, as well as the pressure termp(s).

ρ · dA · ds · (∂c∂t

+ c · ∂c∂s

) = p · dA− (p+∂p

∂s· ds) · dA− ρ · g · dA · ds · cos(φ) (8)

Note that the mass m has been replaced by ρ · dA · ds since dA · ds represents a volume.With cos(φ) = dz

ds and with division by ρ · dA · ds we can simplify equation 8 to:

dc

dt=∂c

∂t+ c · ∂c

∂s= −1

ρ· ∂p∂s− g · dz

ds(9)

Next we consider our flow to be stationary, which means that the velocity s does not dependon the time but only on the coordinate s. We can therefore simplify equation 9 by crossing all theterms that involve t, meaning all the differentials ∂t.

c · dcds

=d

ds·∫c · dc =

d

ds· (1

2· c2) = −1

ρ· dpds− g · dz

ds(10)

(Now that our velocity is only dependent on s we can note dcds as a total differential.)

Next we multiply by ds and bring everything to the left side:

d(c2

2) +

1

ρ· dp+ g · dz = 0 (11)

Finally we want to integrate this over the streaming line from s1 with the quantities c1, p1 andz1 to s2 with c2, p2 and z2.

The Integral writes as follows.

3

1

2(c22 − c21) +

∫ p2

p1

1

ρ+

∫ z2

z1

g · dz = 0 (12)

In the next step we solve the integrals that are left. Note that since we look at a incompressiblefluid ρ does not depend on the pressure p and can therefore be treated as a constant.

1

2(c22 − c21) + (

p2

ρ− p1

ρ+ g · (z2 − z1) = 0 (13)

In a last step we put all the terms from state 1 at s1 to the left and all the terms from stat 2at s2 to the right. Since we know, that the derivation equals zero (from equation 11, we can taketheses states as constant.

c222

+p2

ρ+ g · z2 =

c212

+p1

ρ+ g · z1 = const. (14)

This is the general form of Bernoullis Equation for incompressible inviscid (frictionless) andstationary flow in one dimension. These three terms, incompressible inviscid and stationary arereally critical. In future if you want to apply any equation to your setting you should exactly nowwhat requirement your flow has, e.g. it makes a huge difference if you consider a flow to be inviscidor not. But more on that later.

Since you know now how to derive Bernoullis Equation, you can also obtain the equation ofmotion for instationary flow by just involving the partial differential ∂c

∂t .

From textbooks you might also know another form of Bernoullis Equation. We also discussthis "version" here for the intuitive understanding of the equation.

Figure 3: bernoullis equation

The conditions stay the same, we have incompressible (ρ = const.), inviscid and stationary flow.Moreover we have the three parameters P (pressure), v (velocity) and h (height, before expressedas z). As you see in figure 3 we multiplied equation 14 with ρ to obtain the energy per unitvolume. Imagine now, that the energy is conserved if you follow one and the same streaming line.If you now lower the pressure P , either v or h or both are increased to conserve the total energyper unit volume. Make sure that with Bernoullis Equation, even though the set up in figure 3 isin 2D you always follow one streaming line (red) because Bernoullis Equation describes motionin 1D.

Usually you solve flow problems with the a combination of conservation of mass and conservationof momentum.

4

3 Fluid Mechanics in 3DIn 3D flow problems, the principles stay the same. We still use conservation of mass and conserva-tion of momentum (also conservation of energy, not discussed in this script) but of course thingsget a little bit more complicated. Since we already discussed the derivation of the equations in 1D,and since the derivation in 3D is more complex, we will not go in to much detail for the derivationof the equations in 3D. Important is that you understand the idea behind the derivation (whereto start?) and intuitively understand the different terms of the final equations, so you can use theequations in the right way.

3.1 Conservation of massIn the following sections we consider the fluid to be homogeneous, so we can treat it as a continuum.

We define the velocity as:

−→u (x, y, z, t) =

uvw

(15)

Of course we can optionally simplify this to a stationary flow by leaving out the dependencyon the time t.

In the 1D set up we looked at the tube as a whole, now we break down the complexity of thesystem and introduce a volume element (figure 4) with the volume:

V = dx · dy · dz (16)

Figure 4: voluem element

The mass is defined as:

m = ρ · V = ρ · dx · dy · dz (17)

Before we used the approach of mass conservation in the whole system (tube), now since we’rejust looking at a volume element we compute the time dependent change of mass in this volumeelement.

time dependent change of mass in the volume element=∑

mass inflow -∑

mass outflow

Temporal change of mass m:

m =d(ρ · dx · dy · dz)

dt=dρ

dt· dx · dy · dz (18)

What is done next is the computation of the massflow through every surface of the volumeelement. If we do this properly we get the equation for mass conservation in its most general form:

∂p

∂t+∂(ρ · u)

∂x+∂(ρ · v)

∂y+∂(ρ · w)

∂z= 0 (19)

For an incompressible fluid (ρ = const.) this simplifies to:

∂u

∂x+∂v

∂y+∂w

∂z= 0 (20)

5

since ∂u∂t = 0 and the rest of the term can be divided by ρ.

Equations 19 and 20 can be written in vector form:

∂p

∂t+5(ρ · −→v ) = 0 (21)

5−→v = 0 (22)

Remember the nabla operator 5 from your math class:

5 =

∂∂x∂∂x∂∂x

(23)

For complete understanding of derivation I refer to the book I mentioned in the beginning [6].

3.2 Conservation of momentum: The Navier-Stokes EquationWe still derive the equation for conservation of momentum by looking at the small volume elementdx · dy · dz. You find the approach for the derivation of the Navier-Stokes equation in the boxbelow:

temporal change of the momentum in the volume element ∆p∆t =∑

entering momentum flow -∑leaving momentum flow +∑forces that act on the volume element (shear forces, normal forces) +∑forces that act on the mass of the volume element (gravitational force, magnetic force,

electric force) →−→k

With temporal we mean change of momentum over time.I think the term conservation of momentum in this case is a little bit misleading, since what you

see in the box is a formulation of a equation of forces. This can be explained by the relationshipof force (

−→F = m · −→a ) and momentum (−→p = m · −→v ).

d−→pdt

=−→F (24)

Meaning that the force equals the momentum flow. After setting up this equation of forces,and several steps of refinement we obtain the navier stokes equation in its most general (vector)form:

ρ · (∂−→v∂t

+ (−→v · 5)−→v ) =−→k −5p+ µ · 4−→v (25)

with:

5 p =

∂p∂x∂p∂y∂p∂z

(26)

−→v · 5 = u · ∂∂x

+ v · ∂∂y

+ w · ∂∂z

(27)

−→5 ·−→5 = 4 (28)

4−→v =∂2−→v∂x2

+∂2−→v∂y2

+∂2−→v∂z2

(29)

6

DiscussionThe Navier Stokes Equation written like in equation 25 contains three scalar nonlinear par-

tial differential equations. (As an exercise: try to "extract" those three differential equations).If you’re just solving for the velocity components u, v, w you are perfectly served with the Navier-Stokes equation. However sometimes you have more unknowns, in that case you can get anotherdifferential equation out of the conservation of mass (equation 19 or 21) Since these partial differ-ential equations are non-linear its hard to get an analytical solutions, alternatively you can solvethis equation numerically. We’ll discuss one numerical approach later in this script. Nevertheless,in some circumstances the Navier Stokes equation can be simplified depending on what fluid andflow properties you assume. In some cases, as we will find out, the non linear terms can be canceledout, such as equation can be solved analytically.

Unfortunately its hard to recognize the terms out of the box (eg. we don’t recognize shear forcein the final form anymore). Anyway we are still able to understand intuitively what the differentterms of the Navier-Stokes Equation (equ. 25) state. This is important for the proper applicationof the equation depending on your set up of interest (eg. what are the properties of the fluid, ofthe flow etc.)

∂−→v∂t (−→v · 5) · −→v

−→k −5 p µ ∗ 4−→v

local acceleration convective acceleration force on volume element pressure friction

Table 1: meaning of the different terms

3.2.1 Local acceleration

This term tells you if at a given fixed point in space the velocity changes over time (your fluid isaccelerated). This is only the case in an instationary flow.

3.2.2 Convective acceleration

Convective acceleration is due to changes in the velocity field. This is for example the case if youhave constriction of the tube at a certain point (as shown in figure 3) or any arbitrary complexstructure your fluid is flowing through. This kind of acceleration also takes place when you have astationary flow. The acceleration is not time dependent but depends on the structure. Or explainedin other words, if you follow a streaming line in the set up of figure 3 you (as a fluid particle) willexperience an acceleration between point with P1 and point with P2 (point of constriction).

3.2.3 Volume forces

Volume forces (−→k ) could be gravitational force, magnetic force or electric force. These forces act

on the mass of the volume element.

3.2.4 Pressure term

This term represents the pressure difference from one point to another, that acts as a driving force.

3.2.5 Friction term

With this term we model friction. The friction in turn depends on the viscosity µ of the fluid andthe change of the velocity expressed by the laplace operator 4 (equ. 28).

7

4 Numerical Solution: The Lattice Boltzmann MethodThis section has been developed according to the book "Lattice Boltzmann Modeling" [6] and thepaper "Lattice Boltzmann method for fluid simulation" [7].

In the previous section it was already mentioned, that the analytical solving of the Navier-Stokes equations is mostly impossible (it depends on the setting). In this section we discuss onepossible numerical solution approach: the Lattice Boltzmann Method (LBM).

To be precise LBM is a discrete numerical solving approach for the Boltzmann-Equation. TheBoltzmann-Equation models a fluid with fluid-particles, where the particles are described with adensity distribution function f . Since the Boltzmann-Equation looks at particle that "fly" around(statistical part) and collide with each other (mechanical part), the movement is described bystatistical and mechanical means. In comparison, Navier-Stokes equation models the fluid as acontinuum, this is why no statistical view is required. This makes up the "big" difference betweenthe Boltzmann-Equation and the Navier-Stokes Equation. However, they are still closely relatedto each other. One can even get from Boltzmann Equation to Navier-Stokes. This is why solvingof the Boltzmann Equation with the Lattice Boltzmann Method gives also appropriate solutionsfor the Navier-Stokes equation.

Since the derivation of the Boltzmann-Equation is highly complex, we will not focus on thatbut try to understand the method itself.

4.1 Discrete vs ContinuousContinuous means, that for example the velocity v of a fluid is defined in any arbitrary point inspace. In contrast if we discrete the problem we restrict the points in space where this quantityv is known to a set of well known points. (Analogy: A Signal measured from a sensor is usuallycontinous, whereas the digital representation of the signal is discrete)

In our example we restrict the places where a fluid particle can reside to the lattice nodes of agrid as shown in figure 5

Figure 5: Lattice with density distributions

4.2 The density distribution functionIn 2D we describe every single fluid particle sitting on a lattice node with 9 density distributionfunctions f0, ...f8 that point in direction of e0, e1, ...e8 (shown in figure 5), this makes up the D2Q9model. These density distributions represent again a discretization of the density into a distinctset of directions.

The density distribution can be viewed as frequency of occurrence in the given direction.

8

Figure 6: density distribution function

4.3 Macroscopic quantitiesThe density distribution function f represents a microscopic quantity, but what we are interestedin, is the macroscopic density of the fluid at a certain point in space and of course the velocity ofthe fluid.

With this set of density distribution functions we can easily obtain ρ trough the sum of thesedensity distributions f0, f1...f8:

ρ(−→x , t) =

8∑a=0

fa(−→x , t) (30)

a = 0, 1, 2....8 indicate the direction of the density distribution and −→x indicates a distinct latticenode.

Moreover the velocity is defined as follows:

−→u (−→x , t) =1

ρ

8∑a=0

fa(−→x , t) · −→ea (31)

4.4 Streaming and CollisionThe movement of the particle is modeled by two computationally seperated events: the streaming(red) and the collision (blue).

fa(−→x +−→ea∆t, t+ ∆t)− fa(−→x , t) = − [fa(−→x , t)− feqa (−→x , t)]τ

(32)

In the streaming step, the density distributions are moved to the next lattice node along theirdirection. (figure 4.4)

Figure 7: streaming

The collision step is not this intuitive, since it does not represent a mechanical collision likewe know it from physics but it is modeled as a relaxation of the density distribution to a local

9

equilibrium feqa (−→x , t). Where τ represent the relaxation time towards the local equilibrium. τ is afluid specific parameter, and can be related to physical properties like diffusivity D and kinematicviscosity ϑ.

4.5 The equilibrium distributionAs discussed in the previous section, collision involves a relaxation to the equilibrium distribution.There are different definitions of that term, probably mostly empirically derived. We are using theBhatnagar-Gross-Krook (BGK) collision where feqa is defined as follows:

feqa (−→x ) = wa · ρ(−→x ) ·

1 + 3 ·−→ea ·−→ueq

c2+

9

2·

(−→ea · −→ueq)2

c4− 3

2·−→ueq

2

c2

(33)

With the weights wa:

direction a weight wa

a=0 4/9a=1,2,3,4 1/9a=5,6,7,8 1/36

Table 2: weights for each direction

4.6 Incorporating ForcesIf we take the tube set up from sections in the beginning (figure 3) of this script, the driving forceis made up from a pressure gradient. If we now want to solve a flow problem in the tube withthe Lattice Boltzmann Model, we need a way to incorporate any external force (could be pressuredifference or gravitational force, or even electric/magnetic force). This is achieved through the ueqterm, that in turn is needed for the computation of the equilibrium distribution feq.

−−→uequ = −→u +

τ−−→Fext

ρ(34)

Note that −→u was computed in 31.

4.7 Boundary ConditionsWe know boundary conditions from the analytical solution of differential equations, where we usedthe boundary conditions (e.g. v(0) = 0) to find the values for the integration constants C.

Here boundary conditions work differently, but you can produce the same effect. First lets showa detailed set up of our flow problem.We’d like to show a 2D flow in a tube, where virtually everything that flows out on the right side(see in figure 8), just flows in on the left side again (periodic boundary conditions at the left andright wall). Moreover we use constant velocity boundary conditions on the north and south wall.

Figure 8: set up for the flow

10

Caption for figure 8

• solid node → 1

• fluid node → 0

• boundary layer with constant velocity boundary conditions → 2

• boundary layer with constant density boundary conditions → 3

• boundary layer with bounce back boundary conditions → 4

• boundary layer with periodic boundary conditions → 5

If you want to apply two different boundary conditions at one node (like at the edges in figure8) we just add the number for the individual boundary conditions.In the code the set up is implemented with a matrix solid-node, where every node gets a number(you find them in the caption) indicating the boundary condition or if it is a fluid node or a solidnode.

Generally (meaning for fluid node → 0)we solve equation 32 for f of the next time step, usingstreaming and collision. This is repeated through many iterations to get propagation in time.If you think of a boundary node, for example at the north boundary wall, streaming might bedifficult, because there will be no streaming from a solid node into a fluid, which in case of thenorth boundary wall leads to unknown density distributions f7, f4, f8 (figure 9). We need theboundary condition to compute the unknown density distributions after streaming.

Figure 9: north boundary wall

In our example we set the constant velocity boundary condition:

−→u0 =

(0v0

)(35)

We have four unknowns: f4, f7, f8 and ρ. Using equations 30 and 31 we obtain three equations,one out of 30 and two out of 31.

From equation 30:

ρ = f0 + f1 + f2 + f3 + f4 + f5 + f6 + f7 + f8 (36)

From equation 31:

0 = f1 − f3 + f5 − f6 − f7 + f8 (37)

and

ρ ∗ v0 = f2 − f4 + f5 + f6 − f7 − f8 (38)

Since we have four unknowns, these equations are not sufficient. We obtain one additionalequations by assuming bounce back for the density distributions normal to the surface we look at,in this case f2, f4:

11

f2 − feq2 = f4 − feq4 (39)

Exercise

Try to obtain the four unknowns f4, f7, f8 and ρ out of the four equations written above.

Solution

ρ =f0 + f1 + f3 + 2(f2 + f5 + f6)

1 + v0(40)

f7 = f5 +1

2(f1 − f3)− 1

6ρv0 (41)

f4 = f2 −2

3ρv0 (42)

f8 = f6 −1

2(f1 − f3)− 1

6ρv0 (43)

The south boundary wall is solved in an analogous way. Try to find the equations and unknownsfor the south boundary wall. Of course you can take this boundary condition for every wall (north,east, south, west).

In the caption for figure 9 you can see that there are several other boundary conditions thatare not discussed in this script.

4.8 Implementation Guidance1. Initialize ρ,−→u ,fa and fequa .

2. Streaming step: move fa to the next lattice node. Use two different storage arrays fa andftempa (temporary storage). Stream fa to the next lattice node and store it in ftempa.Apply the boundary conditions for density distributions that are missing in ftempa now.

3. Compute macroscopic ρ and −→u with ftempa

4. Collision step: Calculate the updated distribution function using

fa = ftempa −1

τ(ftempa − fequa ) (44)

5. Repeat step 2 to 5

12

5 Exercise1. We want to investigate the steady state flow of an incompressible viscid fluid in a 2D tube

(see figure 10). Simplify Navier-Stokes equation to obtain the partial differential equationfor this set up.

2. Solve the PDE you obtained in exercise 2. analytically. Given the boundary conditionsv(0) = 0, v(H) = 0, P (0, y, t) = P0 and P (L, y, t) = P1. (Hint: you can set ∂p

∂x = P1−P0

L ).When you obtained the solution, check the units to verify your solution.

3. Open the .m files out of the LBM.zip folder in Matlab. Implement your analytical solution inthe LBM-poiseuille.m file below the LBM method and visualize your solution in the plot thatis already implemented. Use the following parameters (parameters are already implemented):

• P0 = 0

• P1 = −0.05

• µ = 4.2mPa · s (possible parameter for the viscosity of blood)

• L = 40

• H = 32

Discussion: How does the velocity profile look like? What is the peak velocity?

4. In the next step we want to empirically produce the same result with LBM like in theanalytical case. Try different parameters for the numbers of iterations iter and the parameterτ to obtain the same velocity profile. Important: we are looking for the stationary flow profile.In this case it means, if you increase the number of iteration at your solution, there shouldnot be a further gain of velocity.Discussion:

• What could the parameters iter and τ stand for intuitively?

Figure 10: steady state poiseuille flow

13

6 Solution Exercise1.

dp

dx= µ · ∂

2u

∂2y(45)

Explanation:

• stationary flow: the term ∂v∂t can be cancels out (no dependency on time)

• simple structure: since we have no constriction or something comparable, there will beknow convective acceleration. Or in other words if you follow one streaming line, theparticle has the same velocity at every location. → (−→v · 5) · −→v can be canceled out.

• There were no external forces mentioned in the exercise description→−→k can be canceled

out.What is left are the terms − 5 p (standing for the pressure difference) and µ · 4−→v(representing the friction (viscid flow)) leading to:

5−→p = µ · 4−→v (46)

Since we’re only looking at a one dimensional flow problem we can bring this vector intoscalar form using:

5 p =

∂p∂x∂p∂y∂p∂z

(47)

and

4−→v =∂2−→v∂x2

+∂2−→v∂y2

+∂2−→v∂z2

(48)

we get ∂p∂x∂p∂y∂p∂z

= µ

(∂2−→v∂x2

+∂2−→v∂y2

+∂2−→v∂z2

)(49)

Where on the left side p only depends on x and on the right side v only depends on y. ∂p∂x00

= µ∂2−→v∂y2

(50)

Remember:

−→v =

uvw

(51)

Where in this case v = 0 and w = 0. So we can further simplify: ∂p∂x00

= µ

∂2u∂2y

00

(52)

Which then finally gives us the "scalar" partial differential equation that is writtenabove.

14

2. We first set∂p

∂x=P1 − P0

L(53)

so we obtainP1 − P0

L= µ

∂2u

∂2y(54)

We solve this differential equation by integrating twice and applying the boundary conditions.

The integration gives us:

u(y) =P1 − P0

2µL· y2 + C1 · y + C2 (55)

From boundary conditions v(0) = 0 we can easily set the constant C2 equal to zero.

From boundary condition v(H) = 0 we get:

C1 = −P1 − P0

2µL·H (56)

Which in the end gives us the solution:

u(y) =P1 − P0

2µL· (y2 −H · y) (57)

3. You find the solution in the LBM-poiseuille-solution.m file.

Figure 11: velocity profile of poiseuille flow

4. Choice of parameters:

• iter around 1400 to have approximately steady state flow.

• τ = 2.425

Role of the parameters:

Since you probably used different values for iter, you might have found out, that with in-creasing value for iter the velocity increases until it reaches the steady state flow. iter inthat sense relates to the propagation in time.

The parameter τ is rather related to the viscosity, and influences the shape of the velocityprofile as well as the peak velocity given a fixed external force. If you go from τ = 2 to τ = 5you might see, that the peak velocity is reduced remarkably.

How do we find these parameters: Start for example with τ = 2, find the value for iter to getthe same peak velocity as in the analytical solution. If you reached the same peak velocitycheck either if the shape is the same, if true, increase the value for iter to check if the fluidis still accelerating. If it is the steady state flow for τ = 1 will be at a higher peak velocitylike the analytical solution: no match! Go on with a higher τ and repete this procedure.

Of course you can also write a function, that finds the right set of parameters.

15

7 Matlab Code

7.1 LBM-poiseuille

1 %% Author : Mirjam Abegglen2 %% Sing l e Component , S i ng l e Phase La t t i c e Boltzmann f o r P o i s e u i l l e Flow3

4 d i g i t s (6 ) %% g iv e s the output p r e c i s i o n5 nx=4;6 ny=32;7 H=32; %% Height in cm8 L=40; %% Length in cm9 dy=H/ny ;

10 dt=1; %% timestep −−> here 1 f o r s imp l i c i t y11 c=dy/dt %% l a t t i c e speed −−> here 1 f o r s imp l i c i t y12 q=9; %% number o f d i r e c t i o n s f o r the dens i ty d i s t r i b u t i o n13 tau =2.425; %% RELAXTION FACTOR14 v0=0; %% ve l o c i t y f o r constant v e l o c i t y boundary cond i t i on15 F=H∗0 . 0 5 ; %% dr i v i ng f o r c e16 i t e r =1400; %% NUMBER OF ITERATIONS17

18 %% i n i t i a l i z e sol id_node ( g i v e s the s e t up o f the l a t t i c e )19 sol id_node=ze ro s (ny , nx ) ; %% sol id_node==1 means i s s o l i d ( wa l l ) ,

sol id_node==0 means i s f l u i d20 f o r x=1:nx21 sol id_node (1 , x )=1;22 sol id_node (ny , x )=1;23 end24

25 f o r x=1:nx26 sol id_node (2 , x )=2; %% sol id_node==2 means constant v e l o c i t y

boundary cond i t i on s27 sol id_node (ny−1,x )=2;28 end29

30

31 %% i n i t i a l i z e the dens i ty d i s t r i b u t i o n func t i on f32 f=ze ro s (ny , nx , q ) ;33 f o r x=1:nx34 f o r y=1:ny35 i f so l id_node (y , x )==0 | sol id_node (y , x )==236 f (y , x , : ) =10/9;%10/9 ;37 f (y , x , 3 ) =0.1 ;%0 . 1 ;38 f (y , x , 8 ) =0.5 ;39 end40 end41 end42 f ;43

44 %% i n i t i a l i s e the d i r e c t i o n array e45 e =[0 ,0 ;1 ,0 ;0 ,1 ;−1 ,0 ;0 ,−1;1 ,1 ;−1 ,1 ;−1 ,−1;1 ,−1] ;46 e=e .∗ c ;47 x=ze ro s (q , 1 ) ;48 y=ze ro s (q , 1 ) ;49

50 f i g u r e51 qu iver (x , y , e ( : , 1 ) , e ( : , 2 ) ) %% v i s u a l i z a t i o n o f the d i r e c t i o n s e52 t i t l e ( ’ d i r e c t i o n s e ’ )

16

53

54 %% computing i n i t i a l rho and i n i t i a l u55

56 [ rho , u_x ,u_y ] = rhoandugenerator ( ny , nx , f , q , e , sol id_node ) ;57

58 %% Sta r t i ng with the I t e r a t i o n59 f o r time=1: i t e r60 %% Streaming61 [ ftemp ] = streaming ( f , ny , nx , q , sol id_node ) ;62

63 %% Constant v e l o c i t y Boundary Condit ions64

65 [ ftemp ] = constantve loc i tyBC ( ftemp , ny , nx , v0 ) ;66

67 %% update rho and u us ing ftemp68

69 [ rho , u_x ,u_y ] = rhoandugenerator ( ny , nx , ftemp , q , e , sol id_node ) ;70

71 %% Computing u_equ72

73 [ u_equ_x , u_equ_y ] = uequgenerator ( u_x , u_y , tau ,F , rho , ny , nx ,sol id_node ) ;

74

75 %% Computing f_equ76

77 [ f_equ ] = fequgene ra to r ( u_equ_x , u_equ_y , e , rho , q , ny , nx , sol id_node, c ) ;

78

79 %% Co l l i s i o n80

81 [ f ] = c o l l i s i o n ( f_equ , ftemp , ny , nx , q , sol id_node , tau ) ;82

83

84 end85

86 % ca l c u l a t e the peak v e l o c i t y87 di sp ( ’ Peak v e l o c i t y LBM’ )88 max(u_x)89

90 %% plo t o f the v e l o c i t y u_x91 u_x( 2 : ny−1 ,2) ;92 l ength (u_x( 2 : ny−1 ,2) ) ;93 y=(0:dy :H−3∗dy ) ;94 l ength (y ) ;95 p lo t (u_x( 2 : ny−1 ,2) , y , ’−−or ’ )96 t i t l e ( ’ v e l o c i t y p r o f i l e ’ )97 hold on98

99

100 %% Ana ly t i c a l So lu t i on to P o i s e u i l l e Flow101 %% Underlying p a r t i a l d i f f e r e n t i a l equat ion mu∗d^2u/d^2y=dp/dx102

103 %%parameters104 P1=−0.05;105 P0=0;106 L=40;107 H=ny−3;108 v i sk =4.2∗10^−3;

17

109

110 %% ana l y t i c a l s o l u t i o n111 v=0.5∗(P1−P0) /( v i sk ∗L) ∗( y.^2−H∗y ) ;112

113 %% output the peak v e l o c i t y114 di sp ( ’ Peak v e l o c i t y a n a l y t i c a l ’ )115 max(v )116

117 %% plo t the a n a l y t i c a l s o l u t i o n118 p lo t (v , y , ’−g ’ )119 h = legend ( ’LBM’ , ’ a n a l y t i c a l ’ , 2) ;120 x l ab e l ( ’ v e l o c i t y u(y ) ’ )121 y l ab e l ( ’ he ight y ’ )122 s e t (h , ’ I n t e r p r e t e r ’ , ’ none ’ )

7.2 streaming

1 f unc t i on [ ftemp ] = streaming ( f , ny , nx , q , sol id_node )2 ftemp=ze ro s (ny , nx , q ) ; %% i n i t i a l i z e temporary s to rage array3 f o r x=1:nx %%streaming f o r every l a t t i c e po int where sol id_node

==04 f o r y=1:ny %%without the f i r s t and l a s t colum5 i f so l id_node (y , x )==0 | sol id_node (y , x )==26 i f x<nx %% i f statements needed f o r p e r i o d i c boundary

cond i t i on7 x_p=x+1; %% de f i n i n g x p o s i t i v ( go 1 in p o s i t i v e x

d i r e c t i o n )8 e l s e9 x_p=1; %% de f i n i n g x negat iv ( go 1 in negat ive x

d i r e c t i o n )10 end11 i f x>112 x_n=x−1;13 e l s e14 x_n=nx ;15 end16 y_p=y+1; %% y po s i t i v e ( go 1 in p o s i t i v e y d i r e c t i o n )17 y_n=y−1; %% y negat ive ( go 1 in negat ive y d i r e c t i o n )18 ftemp (y , x , 1 )=f (y , x , 1 ) ; %% streaming to the next l a t t i c e

node19 ftemp (y , x_p , 2 )=f (y , x , 2 ) ;20 ftemp (y_n , x , 3 )=f (y , x , 3 ) ;21 ftemp (y , x_n , 4 )=f (y , x , 4 ) ;22 ftemp (y_p , x , 5 )=f (y , x , 5 ) ;23 ftemp (y_n , x_p , 6 )=f (y , x , 6 ) ;24 ftemp (y_n , x_n , 7 )=f (y , x , 7 ) ;25 ftemp (y_p , x_n , 8 )=f (y , x , 8 ) ;26 ftemp (y_p , x_p , 9 )=f (y , x , 9 ) ;27 end28 end29 end30

31 ftemp ( 1 , : , : ) =0; %% r ed e f i n e row one and ny 0 ( s i n c e they are s o l i dnodes )

32 ftemp (ny , : , : ) =0;33

34 end

18

7.3 constantvelocityBC

1 f unc t i on [ ftemp ] = constantve loc i tyBC ( ftemp , ny , nx , v0 )2 rhozw=ze ro s (ny , nx ) ; %% de f i n e temporary s to rage array f o r rho3

4 f o r x=1:nx5 % North Boundary6 rhozw (2 , x )=(ftemp (2 , x , 1 )+ftemp (2 , x , 2 )+ftemp (2 , x , 4 ) +2∗( ftemp (2 , x

, 3 )+ftemp (2 , x , 6 )+ftemp (2 , x , 7 ) ) ) /(1+v0 ) ;7 ftemp (2 , x , 5 )=ftemp (2 , x , 3 )−2/3∗rhozw (1 , x ) ∗v0 ;8 ftemp (2 , x , 8 )=ftemp (2 , x , 6 ) +1/2∗( ftemp (2 , x , 2 )−ftemp (2 , x , 4 ) )−1/6∗

rhozw (2 , x ) ∗v0 ;9 ftemp (2 , x , 9 )=ftemp (2 , x , 7 ) +1/2∗( ftemp (2 , x , 4 )−ftemp (2 , x , 2 ) )−1/6∗

rhozw (2 , x ) ∗v0 ;10 % South Boundary11 rhozw (ny−1,x )=(ftemp (ny−1,x , 1 )+ftemp (ny−1,x , 2 )+ftemp (ny−1,x , 4 )

+2∗( ftemp (ny−1,x , 5 )+ftemp (ny−1,x , 8 )+ftemp (ny−1,x , 9 ) ) ) /(1+v0) ;

12 ftemp (ny−1,x , 3 )=ftemp (ny−1,x , 5 )−2/3∗rhozw (ny−1,x ) ∗v0 ;13 ftemp (ny−1,x , 6 )=ftemp (ny−1,x , 8 ) −1/2∗( ftemp (ny−1,x , 2 )−ftemp (ny

−1,x , 4 ) )+1/6∗rhozw (ny−1,x ) ∗v0 ;14 ftemp (ny−1,x , 7 )=ftemp (ny−1,x , 9 ) +1/2∗( ftemp (ny−1,x , 2 )−ftemp (ny

−1,x , 4 ) )+1/6∗rhozw (ny−1,x ) ∗v0 ;15 end16

17 end

7.4 rhoandugenerator

1 f unc t i on [ rho , u_x ,u_y ] = rhoandugenerator ( ny , nx , f , q , e , sol id_node )2 rho=ze ro s (ny , nx ) ;3 rho ( : , : )=sum( f , 3 ) ;4 u_x=ze ro s (ny , nx ) ;5 u_y=ze ro s (ny , nx ) ;6

7 f o r i =1:q %% summing up ea∗ f a f o r every d i r e c t i o n a={0 ,1 ,2 ,3 ,4 ,5 ,6 ,7 ,8}8 u_x ( : , : )=u_x ( : , : )+e ( i , 1 ) ∗ f ( : , : , i ) ;9 u_y ( : , : )=u_y ( : , : )+e ( i , 2 ) ∗ f ( : , : , i ) ;

10 end11

12 f o r y=1:ny13 f o r x=1:nx14 i f so l id_node (y , x )==0 | sol id_node (y , x )==2 %% only computes u

f o r f l u i d nodes15 u_x(y , x )=u_x(y , x ) / rho (y , x ) ;16 u_y(y , x )=u_y(y , x ) / rho (y , x ) ;17 end18 end19 end20

21

22

23

24

25 end

7.5 uequgenerator

1 f unc t i on [ u_equ_x , u_equ_y ] = uequgenerator ( u_x , u_y , tau ,F , rho , ny , nx ,sol id_node )

19

2 u_equ_x=ze ro s (ny , nx ) ; %% i n i t i a l i z e3 u_equ_y=ze ro s (ny , nx ) ;4

5 f o r x=1:nx6 f o r y=1:ny7 i f ( sol id_node (y , x )==0 | sol id_node (y , x )==2) %% only

computes u_equ i f node i s a f l u i d node8 u_equ_y(y , x )=u_y(y , x ) ;9 u_equ_x(y , x )=u_x(y , x )+tau∗F/rho (y , x ) ; %% inco rpo ra t e

ex t e rna l Force10 end11 end12

13

14

15 end

7.6 fequgenerator

1 f unc t i on [ f_equ ] = fequgene ra to r ( u_equ_x , u_equ_y , e , rho , q , ny , nx ,sol id_node , c )

2 f_equ=ze ro s (ny , nx , q ) ;3 w=[4/9 ,1/9 ,1/9 ,1/9 ,1/9 ,1/36 ,1/36 ,1/36 ,1/36 ] ; %% weights w_a4

5 f o r x=1:nx6 f o r y=1:ny7 f o r a=1:q8 i f ( sol id_node (y , x )==0 | sol id_node (y , x )==2) %% only

compute f_euq i f node i s a f l u i d node9 f_equ (y , x , a )=w( a ) ∗ rho (y , x ) ∗(1+3∗(( e ( a , 1 ) ∗u_equ_x(y ,

x )+e (a , 2 ) ∗u_equ_y(y , x ) ) ) /( c^2)+9/2∗(( e ( a , 1 ) ∗u_equ_x(y , x )+e (a , 2 ) ∗u_equ_y(y , x ) ) ^2) /( c^4)−3/2∗(u_equ_x(y , x )^2+u_equ_y(y , x ) ^2) /( c^2) ) ;

10 end11 end12 end13 end14

15

16

17

18 end

7.7 collision

1 f unc t i on [ f ] = c o l l i s i o n ( f_equ , ftemp , ny , nx , q , sol id_node , tau )2 f=ze ro s (ny , nx , q ) ;3

4 f o r x=1:nx5 f o r y=1:ny6 f o r a=1:q7 i f ( sol id_node (y , x ) == 0 | sol id_node (y , x )==2)%% only

compute c o l l i s i o n f o r f l u i d nodes8 f (y , x , a )=ftemp (y , x , a )−(ftemp (y , x , a )−f_equ (y , x , a ) ) /

tau ;9 end

10 end11 end12 end

20

13

14

15 end

References[1] Yoshitaka Kameo and Taiji Adachi. Interstitial fluid flow in canaliculi as a mechanical stimulus

for cancellous bone remodeling: in silico validation. Biomechanics and modeling in mechanobi-ology, 13(4):851–860, 2014.

[2] Abhishek Kumar Tiwari and Jitendra Prasad. Computer modelling of bone’s adaptation: therole of normal strain, shear strain and fluid flow. Biomechanics and Modeling in Mechanobiol-ogy, pages 1–16, 2016.

[3] Xiaowen Shan. Lattice boltzmann in micro-and nano-flow simulations. IMA journal of appliedmathematics, 76(5):650–660, 2011.

[4] Hiroshi Kamioka, Yoshitaka Kameo, Yuichi Imai, Astrid D Bakker, Rommel G Bacabac, NaokoYamada, Akio Takaoka, Takashi Yamashiro, Taiji Adachi, and Jenneke Klein-Nulend. Mi-croscale fluid flow analysis in a human osteocyte canaliculus using a realistic high-resolutionimage-based three-dimensional model. Integrative Biology, 4(10):1198–1206, 2012.

[5] Herbert Oertel jr, Martin Böhle, and Thomas Reviol. Strömungsmechanik: Grundlagen-Grundgleichungen-Lösungsmethoden-Softwarebeispiele. Springer-Verlag, 2011.

[6] MC Sukop. Dt thorne, jr. lattice boltzmann modeling lattice boltzmann modeling. 2006.

[7] Yuanxun Bill Bao and Justin Meskas. Lattice boltzmann method for fluid simulations. De-partment of Mathematics, Courant Institute of Mathematical Sciences, New York University,2011.

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