Fluid-Structure Interaction with Vortex methods and the ... · 2 • Engineering applications •...

CSElab Computational Science & Engineering Laboratory http://www.cse-lab.ethz.ch

Fluid-Structure Interaction with Vortex methods and the Finite Element Method

Siddhartha Verma Gengyun Li, Petros Koumoutsakos

October 5, 2016

http://www.cse-lab.ethz.ch

2

• Engineering applications • Wind turbines, aircraft frames, buildings,

bridges, bio-inspired devices…

Fluid-Structure Interaction

• Simulations difficult with body fitted grids • Complex shapes, deforming geometries, moving

meshes - large overhead

• Vortex methods + Brinkman penalisation • Fast & accurate, parellelizes easily • Multiple deforming geometries with relative ease

Credit: inhabit.com

Credit: Boeing

Credit: thrombus-vph.eu

Credit: wyss.harvard.edu

• Biomedical applications • Artificial heart valves, stents, aneurysms…

• Vortex methods + Brinkman penalization • Fluid velocity penalized to match

at appropriate grid nodes • Solid influences fluid

• Finite Element Method • Compliant solids deform due to flow-induced

forces on the solid boundary • Fluid influences solid via surface-forces

Objective: Combine vortex methods and FEM

3

Vortex Methods(Fluid mechanics)

D!

Dt= ⌫r2! + �r⇥ (� (us � u))

fext

us

us

fext

Finite Element Method(Solid Mechanics)Mv +Cv +Kv = f

ext

• Weak coupling (partitioned approach) • Staggered stepping of fluid- and solid-solvers

4

Fluid Mechanics (Vortex Methods)

Background• Enforcing Boundary Condition (BC) on

complex, deforming geometries

• Arbitrary Lagrangian Eulerian (ALE) • Body Fitted grid : Difficulty with Multiple bodies

& with Deforming geometries • Expensive : frequent re-meshing

5

Source: https://www.cfd.tu-berlin.de/research/acoustic/AIAA_PAPER_2004-2926/graphics/gitter.png

Source: http://euler.yonsei.ac.kr/2011/09/30/immersed-boundary-method-for-complex-flows/

• Immersed Boundary (IB) method • Simple Cartesian grids • BCs enforced via Dirac-delta forcing

http://euler.yonsei.ac.kr/2011/09/30/immersed-boundary-method-for-complex-flows/

Vortex methods & Brinkman penalization • Remeshed vortex methods

• Solve vorticity form of incompressible Navier-Stokes

6

Koumoutsakos & Leonard, JFM (1995)@!

@t+ u ·r! = ! ·ru+ ⌫r2! + �r⇥ (� (us � u))

Hejlesen et al., JCP (2015)

• Single grid handles both Fluid and Solid

• Fluid:Solid:

� = 0� = 1

DiffusionAdvection

Penalization

0 in 2D

• Brinkman penalization • Relatively easy to implement,

computationally cheap

• Accounts for fluid-solid interaction • Penalty term enforces no-slip BC

Angot et al., Numerische Mathematik (1999)

@u

@t+ u ·ru = �rP

⇢+ ⌫r2u+ �� (us � u)

Curl

Vortex methods & Brinkman penalization • Remeshed vortex methods

• Solve vorticity form of incompressible Navier-Stokes

7

Koumoutsakos & Leonard, JFM (1995)@!

@t+ u ·r! = ! ·ru+ ⌫r2! + �r⇥ (� (us � u))

DiffusionAdvection

Penalization

0 in 2D

@u

@t+ u ·ru = �rP

⇢+ ⌫r2u+ �� (us � u)

Curl

ADVANTAGES: 1) Do advection only where vorticity is non-zero (think uniform flow, where FV/FD

have to do computations in non-important regions) 2) Time step not limited by CFL (i.e., stability issues, but rather limited by accuracy concerns) 3) SL schemes tend to exhibit lower numerical diffusion

Implementation

8

@!

@t+ u ·r! = ! ·ru+ ⌫r2! + �r⇥ (� (us � u))

!? = !n +�t {�r⇥ (� (us � u))}!?? = !? +�t

�r2!?

D!??

Dt= 0 ) !n+1 = !??

• Godunov splitting 1. Penalization 2. Diffusion 3. Advection

• Wavelet-based adaptive grid • Cost-effective compared to uniform grids

• Each thread receives a small data block (up to 60% CPU utilisation)

• Open source code: available on github • https://github.com/cselab/MRAG-I2D

Rossinelli et al., JCP (2015)

https://github.com/cselab/MRAG-I2D

• Force computations require the stress tensor σ • Drag = net force opposing velocity

• Compute surface traction, dot product with velocity vector • +ve => Thrust; -ve => Drag

Surface forces

Fsurf =

Z� · nds

FD/T =Fsurf · u

|u|

• Stress tensor requires two things • Pressure • Strain-rate tensor

Z�Pn · dS

Z2µD · ndS

D =1

2

�ru+ruT

�

• Vorticity form of Navier Stokes • Advantage: no need to deal with pressure • Disadvantage: pressure term unavailable!

9

@u

@t+ u ·ru = �rP

⇢+ ⌫r2u+ �� (us � u)

Cur

l

@!

@t+ u ·r! = ! ·ru+ ⌫r2! + �r⇥ (� (us � u))

Computing pressure• For Pressure : divg. of the N-S equans

10

r ·✓@u

@t+ u ·ru = �rP

⇢+ ⌫r2u+ �� (us � u)

◆

r · u = 0

ruT : ru = �r2P

⇢+

1

⇢2r⇢ ·rP +r · (�� (us � u))

r2P = ⇢��ruT : ru+r · (�� (us � u))

�

0

• Poisson’s equation: solved using free-space Green’s function r2

G = �(x� x0)

rG = 0 on dS

) G =1

2⇡ln|r|

P = G ⇤RHS

• Strain-rate tensor (for frictional forces)

D =1

2

�ru+ruT

�

STRESS that Pressure solve is not done on the entire domain, but a very small subset consisting of the solid surface points.

The Fast Multipole Method (FMM)• Extremely fast solutions for the

Poisson equation (‘N-body’ problem) • Speed-up: 10,000X and more

11

O

z

R1

R2

zc1zc2

Group 1

Group 2

• Speed-up results from combination of • Tree-code: groups together sources into clusters • Truncated series expansions

• Impulsively started flow around a stationary cylinder

Flow around a cylinder

12

0

0.5

1

1.5

2

0 1 2 3 4 5 6 7

Cd

T

CdPCd

νRe = 550

0

0.5

1

1.5

2

0 1 2 3 4 5

Cd

T

CdPCd

νRe = 1000Vorticity

Pressure

-3

-2

-1

0

1

2

0 0.25 0.5 0.75 1

Cp

θ/π

T = 1T = 3T = 5

Surface-pressure (Re = 550)

Koumoutsakos & Leonard, JFM (1995)Li et al., JFM (2004)

• Spatial distribution of surface-pressure & drag time-evolution agree with reference data

Surface vorticity distribution

13

Re = 1000

Re = 550

• Body-undulations displace fluid • Reaction forces propels swimmer forward

Self-propelled swimmer (Re=4000)

14

-0.01

0

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0.08

0 1 2 3 4 5 6 7 8 9 10

ucm

T

uv

Surface-force vectors

• Compare unsteady acceleration computed using surface-forces to penalisation accel.

• Good agreement: Surface-force computation accurate for complex, temporally deforming bodies

Validation for self-propelled swimmer

15

-0.02

0

0.02

0.04

0.06

0.08

0.1

0 1 2 3 4 5 6 7 8 9 10

ax

T

apenalaSF

-0.04

-0.02

0

0.02

0.04

0 1 2 3 4 5 6 7 8 9 10

ay

T

Horizontal acceleration

Vertical acceleration-0.02

0

0.02

0.04

0.06

0.08

0.1

0 1 2 3 4 5

ax

T

10242

20482

40962

81922

• Low sensitivity to grid-resolution

16

Solid Mechanics (now ‘u’ = displacement!)

1.Mesh the object (Delaunay triangulation) • Equi-angular triangles (few ‘skinny’ triangles) • No vertex lies inside the circumcircle of other triangles • Returns list of element connectivity

2.Assign degrees of freedom • Each node has 2DoFs: u, v

(displacements in x and y)

FEM Steps : Pre-processing

17

Credit: http://www.cescg.org/

• u : Displacement • M : Mass matrix - inertia • K : Stiffness matrix - elasticity

• Governing equation

Mu+Cu+Ku = fext

http://www.cescg.org/

FEM Steps : Generate system matrices

18

3.Generate local ‘mass’ and ‘stiffness’ matrices for each CST element

Source: mae.uta.edu/~lawrenceme5310/course_materials/... : Concepts and applications of FEA (Robert Cook et al.)

4.Assemble local element matrices into a global system matrix

• Each node may belong to multiple elements

Ke = teAe

�B0DB

�

M e =

ZZZ⇢NTNdV

Node 1

234

1

34

Ele 1

23

1

Ele 2

Discretization

1

234

Assembly

http://mae.uta.edu/~lawrenceme5310/course_materials/%5D

• u : Displacement • M : Mass matrix - inertia • K : Stiffness matrix - elasticity

FEM Steps : Constrain system & Solve5.Apply constraints to fix object in space

• Remove rows and columns corresponding to known displacements

2

6664

u2

u3

u4...

3

7775+

2

6664

m22 m23 m24 . . .m32 m33 m34 . . .m42 m43 m44 . . .

......

3

7775

2

6664

u2

u3

u4...

3

7775=

2

6664

f2f3f4...

3

7775�

2

6664

k21k31k41...

3

7775u1 �

2

6664

m21

m31

m41...

3

7775u1

2

666664

k11 k12 k13 k14 . . .k21 k22 k23 k24 . . .k31 k32 k33 k34 . . .k41 k42 k43 k44 . . .

......

3

777775

19

RHS knownUnknown

un+1 = 2

�un+1 � un

�

�t� un

un+1 = 4

�un+1 � un

�

�t2� 4

un

�t� un

Plug into matrix system, solve for un+16.Advance in time using implicit

integration scheme (Newmark’s method)

Mu+Cu+Ku = fext

• Vortex methods + Brinkman penalization • Fluid velocity penalized to match

at appropriate grid nodes • Solid influences fluid

• Finite Element Method • Compliant solids deform due to flow-induced

forces on the solid boundary

Objective: Combine vortex methods and FEM

20

Vortex Methods(Fluid mechanics)

D!

Dt= ⌫r2! + �r⇥ (� (us � u))

fext

us

us

Finite Element Method(Solid Mechanics)Mv +Cv +Kv = f

ext

B. Patzák, OOFEM project home pagehttp://www.oofem.org, 2000

• OOFEM: Open source FEM solver

Coupling the Fluids solver with FEM

21

MultipolesVelocity

Projection

Penalization

Diffusion

Advection

RK2+LTS

Explicit Euler

r2 n = �!n

un = r⇥ n

@!n�

@t= ⌫r2!n

�

@!n�

@t+ u

n� ·r!n

� = 0

!n+1 = !n+1�

x

n+1i = x

ni + u

t,ni �tn

✓n+1i = ✓n + ✓

n

i �tn

RK2+LTS / particles

x

n+1i = x

ni + u

t,ni �tn, ✓n+1

i = ✓ni + ✓

n

i �tn

⇢n = ⇢f +NX

i=1

(⇢ni � ⇢f )�ni

u

t,ni =

1

mi

Z

⌦⇢n�n

i un dx

✓n

i = J

�1i

Z

⌦⇢n�n

i (x� x

ni )⇥ u

n dx

u

r,ni = ✓

n

i ⇥ (x� x

ni )

!� = !n +r⇥ (u� � un)u� =un + ��tuS

1 + ��t

Surface forces r2P = ⇢��ruT : ru+r · (�� (us � u))

�

D =1

2

�ru+ruT

�

Vortex methods + Brinkmann penalization

0.44

0.46

0.48

0.5

0.52

0.54

0.56

0.1 0.15 0.2 0.25 0.3 0.35 0.4

y

x

Coupling the Fluids solver with FEM

22

MultipolesVelocity

Projection

Penalization

Diffusion

Advection

RK2+LTS

Explicit Euler

r2 n = �!n

un = r⇥ n

@!n�

@t= ⌫r2!n

�

@!n�

@t+ u

n� ·r!n

� = 0

!n+1 = !n+1�

x

n+1i = x

ni + u

t,ni �tn

✓n+1i = ✓n + ✓

n

i �tn

RK2+LTS / particles

x

n+1i = x

ni + u

t,ni �tn, ✓n+1

i = ✓ni + ✓

n

i �tn

⇢n = ⇢f +NX

i=1

(⇢ni � ⇢f )�ni

u

t,ni =

1

mi

Z

⌦⇢n�n

i un dx

✓n

i = J

�1i

Z

⌦⇢n�n

i (x� x

ni )⇥ u

n dx

u

r,ni = ✓

n

i ⇥ (x� x

ni )

!� = !n +r⇥ (u� � un)u� =un + ��tuS

1 + ��t

Surface forces r2P = ⇢��ruT : ru+r · (�� (us � u))

�

D =1

2

�ru+ruT

�

Vortex methods + Brinkmann penalization

OOFEM Solve

DeformablemeshForce

distribution

Deformed configuration

FSI: Dynamic tests• Cantilever beam (fixed at lower end)

• Uniform inflow U∞ = 0.1 Re = 2000

23

Material: Biological Tissue equivalent

0.4

0.45

0.5

0.55

0.6

0.65

0.36 0.38 0.4 0.42 0.44

• Beam deformation • abrupt initially: impulsive force • then exhibits ‘wobbling’

• Flow induced force largest in the centre (high pressure due to flow stagnation)

• Beam rigid up until 1s, then allowed to deform

Stiffer beams

24

Stiff rubber Lead

Vortex induced vibrations• Solid cylinder with attached tail

• Tail rigid until steady state attained U∞ = 2.0 Re = 200

• Exhibits large-amplitude oscillations

25

Lee & You (Phys. Fluids, 2013)

• Still to do • Quantitative comparisons with

FSI benchmarksTurek et al. (2010),

Fluid Structure Interaction II, LNCS

Summary• Method for determining surface

pressure and shear forces • Using Vortex methods and

Brinkman penalisation

26

• Coupled with the Finite Element Method • Flow-induced deformation: cantilever beams • Vortex-induced oscillation of flexible tail

r2P = ⇢��ruT : ru+r · (�� (us � u))

�D =

1

2

�ru+ruT

�

@!

@t+ u ·r! = ! ·ru+ ⌫r2! + �r⇥ (� (us � u))

• Validated using simulations of rigid and deforming objects

Backup

27

Application: ‘Smart’ follower learning to maximise efficiency

• ‘Siphoning’ energy from the leader’s wake

• Follower autonomously learns optimal manoeuvres for maximising efficiency • Reinforcement Learning

28

R⌘ =Tu

Tu+max(Pdef , 0)

• Overall, 28% gain in average efficiency • Energetic savings of 64%

• Relatively long occurrences of η=1 • Follower effectively hitching a ‘free ride’

29

• Variation of all three parameters

Efficiency, Thrust power & Def power

⌘ =PThrust

PThrust + PDef

t=0 to 2 t=8 to 10

t=0 to 10

Spatial & temporal evolution

30

31

• Max η - neck and tail no drag, tail tip barely moving, fore section low curvature

• Min η - tail tip max velocity, neck substantial drag

• Max Power - majority of the body high vel

Spatial & temporal evolutionMax η

Max P(Thrust) & P(Def) Min η

Def. power along the body• Majority of effort at tail

• Head: suction peak results in large thrust, with almost no effort • Rest of body: thrust closely follows deformation power

32

Pressure

x x

Thrust generation along the body• Majority of thrust from mid-body & tail

• Head: drag caused by stagnation pressure. Thrust at bulge (suction peak) • ‘Neck’ region : no thrust, negligible drag • Tail end: Mostly drag!

33

Pressure

x

Pressure FMM

34

then the logarithm is equal to:

log(z � z

j

) = log r + log(ei✓) = log r + i✓ = log |z � z

j

|+ i✓.

We found that Re(log(z � z

j

)) = log |z � z

j

|. Thus the potential can be computed as the realpart of the complex number:

�(z) =mX

j=1

m

j

log(z � z

j

).

Now for the multiple expansion, consider the case in which |z| is larger than any |zj

|. We maywrite the previous as:

�(z) =mX

j=1

m

j

log(z � z

j

) =mX

j=1

m

j

hlog z + log

⇣1� z

j

z

⌘i= M log z +

mX

j=1

m

j

1X

l=1

⇣z

j

z

⌘l

l

=

= M log z +1X

l=1

mX

j=1

m

j

⇣z

j

z

⌘l

l

= M log z +1X

l=1

2

4z

�l

mX

j=1

m

j

z

l

j

l

3

5 = M log z +1X

l=1

↵

l

z

l

where ↵

l

=mPj=1

m

j

z

l

j

l

. We can now introduce the first theorem for the expansion. Having to do

with a Quadtree, we would use it to compute the FMM expansions of the leaves, truncating theinfinite sum to a certain order p.

Lemma 1. Suppose that m charges of strengths {qi

, i = 1, . . . ,m} are located at points {zi

, i =1, . . . ,m} with |z

i

| < r. Then for any z with |z| > r the potential �(z) induced by the charges isgiven by:

�(z) = a0 log(z) +1X

k=1

a

k

z

k

where:

a0 =mX

i=1

q

i

and a

k

=mX

i=1

�q

i

z

k

i

k

for k � 1

The gain in speed is achieved by using a grid which is as coarse as possible. In order to dothis we should be able to “unify” the expansions of neighboring leaves. This is possible by usingthe next lemma, which allows us to translate the center of a certain expansion. In this way, forexample, we could compute the expansions for 4 leaves, center them onto their common parentnode and sum the expansions, so that we have the expansion for the entire parent node.

Lemma 2. Suppose that:

�(z) = a0 log(z � z0) +1X

k=1

a

k

(z � z0)k

is a multipole expansion of the potential due to a set of m charges of strength q1, q2, . . . qm, all ofwhich are located inside the circle D of radius R with center at z0. Then for z outside the circleD1 of radius (R+ |z0|) and center at the origin,

�(z) = a0 log(z) +1X

l=1

b

l

z

l

3

• Poisson equation: solved using free-space Green’s function r2

G = �(x� x0)

rG = 0 on dS

) G =1

2⇡ln|r|

P = G ⇤RHS


log(z � z

j


j

|+ i✓.


j

)) = log |z � z

j


�(z) =mX

j=1

m

j

log(z � z

j

).



�(z) =mX

j=1

m

j

log(z � z

j

) =mX

j=1

m

j

hlog z + log

⇣1� z

j

z

⌘i= M log z +

mX

j=1

m

j

1X

l=1

⇣z

j

z

⌘l

l

=

= M log z +1X

l=1

mX

j=1

m

j

⇣z

j

z

⌘l

l

= M log z +1X

l=1

2

4z

�l

mX

j=1

m

j

z

l

j

l

3

5 = M log z +1X

l=1

↵

l

z

l

where ↵

l

=mPj=1

m

j

z

l

j

l





, i =1, . . . ,m} with |z

i


�(z) = a0 log(z) +1X

k=1

a

k

z

k

where:

a0 =mX

i=1

q

i

and a

k

=mX

i=1

�q

i

z

k

i

k

for k � 1



�(z) = a0 log(z � z0) +1X

k=1

a

k

(z � z0)k


�(z) = a0 log(z) +1X

l=1

b

l

z

l

3


log(z � z

j


j

|+ i✓.


j

)) = log |z � z

j


�(z) =mX

j=1

m

j

log(z � z

j

).



�(z) =mX

j=1

m

j

log(z � z

j

) =mX

j=1

m

j

hlog z + log

⇣1� z

j

z

⌘i= M log z +

mX

j=1

m

j

1X

l=1

⇣z

j

z

⌘l

l

=

= M log z +1X

l=1

mX

j=1

m

j

⇣z

j

z

⌘l

l

= M log z +1X

l=1

2

4z

�l

mX

j=1

m

j

z

l

j

l

3

5 = M log z +1X

l=1

↵

l

z

l

where ↵

l

=mPj=1

m

j

z

l

j

l





, i =1, . . . ,m} with |z

i


�(z) = a0 log(z) +1X

k=1

a

k

z

k

where:

a0 =mX

i=1

q

i

and a

k

=mX

i=1

�q

i

z

k

i

k

for k � 1



�(z) = a0 log(z � z0) +1X

k=1

a

k

(z � z0)k


�(z) = a0 log(z) +1X

l=1

b

l

z

l

3

• The algorithm Speedy Q-Learning was implemented and tried, but it did not convergefaster than normal Q-Learning. The update of the Q-table is the following:

Q

k+1(x, a) = Q

k

(x, a)+↵

k

(Tk

Q

k�1(x, a)�Q

k

(x, a))+(1�↵

k

)(Tk

Q

k

(x, a)�Tk

Q

k�1(x, a))

where we have:Tk

Q(x, a) = r(x, a) + �MQ(yk

).

Here (MQ)(x) = maxa2A Q(x, a) and ↵

k

is the decaying learning rate, going as 1/n wheren is the number of the trial. If you really use ↵

k

= 1/k the algorithm does not seem toconverge, while you substitute ↵

k

with a small but constant learning rate (e.g. 0.01) youcan observe a slow convergence.

The problems now are:

• Convergence is slow

• Is the reward ok? (pressure?)

• Learning: the Q table might not be the best choice, it is not flexible and cannot be refined,you need to restart everything. The next approach might be to emulate the Q functionvalue with a neural network (Batch Programming).

• E�ciency: once the fish had learned how to follow a point, it has to learn how to followthat e�ciently, and the di↵erence in e�ciency between a good and a bad case might bequantitatively small (reward?).

Fast Multiple Method [5] [6]

We need to solve the potential problem of 2-dimensional point charges. Every point chargelocated in x

0

2 R2 gives rise to the potential:

�

x0(x, y) = � log(||x� x

0

||),

and away from the charge the potential is harmonic:

4� = 0.

First of all, substitute (x, y) with the complex number z = x+ iy, so that we have:

�

x0(x) = Re(� log(z � z0)).

Consider we have m particles located at z1, z2, . . . , zm, they will generate a potential:

�(x, y) =mX

j=1

m

j

log r =mX

j=1

m

j

log

q(x� x

j

)2 + (y � y

j

)2�.

We can write r as:

r =q(x� x

j

)2 + (y � y

j

)2 = |z � z

j

|

which in polar coordinates becomes:

z � z

j

= re

i✓ = r cos ✓ + ir sin ✓

2

So solution is the real part of

Pressure FMM

35


log(z � z

j


j

|+ i✓.


j

)) = log |z � z

j


�(z) =mX

j=1

m

j

log(z � z

j

).



�(z) =mX

j=1

m

j

log(z � z

j

) =mX

j=1

m

j

hlog z + log

⇣1� z

j

z

⌘i= M log z +

mX

j=1

m

j

1X

l=1

⇣z

j

z

⌘l

l

=

= M log z +1X

l=1

mX

j=1

m

j

⇣z

j

z

⌘l

l

= M log z +1X

l=1

2

4z

�l

mX

j=1

m

j

z

l

j

l

3

5 = M log z +1X

l=1

↵

l

z

l

where ↵

l

=mPj=1

m

j

z

l

j

l





, i =1, . . . ,m} with |z

i


�(z) = a0 log(z) +1X

k=1

a

k

z

k

where:

a0 =mX

i=1

q

i

and a

k

=mX

i=1

�q

i

z

k

i

k

for k � 1



�(z) = a0 log(z � z0) +1X

k=1

a

k

(z � z0)k


�(z) = a0 log(z) +1X

l=1

b

l

z

l

3


log(z � z

j


j

|+ i✓.


j

)) = log |z � z

j


�(z) =mX

j=1

m

j

log(z � z

j

).



�(z) =mX

j=1

m

j

log(z � z

j

) =mX

j=1

m

j

hlog z + log

⇣1� z

j

z

⌘i= M log z +

mX

j=1

m

j

1X

l=1

⇣z

j

z

⌘l

l

=

= M log z +1X

l=1

mX

j=1

m

j

⇣z

j

z

⌘l

l

= M log z +1X

l=1

2

4z

�l

mX

j=1

m

j

z

l

j

l

3

5 = M log z +1X

l=1

↵

l

z

l

where ↵

l

=mPj=1

m

j

z

l

j

l





, i =1, . . . ,m} with |z

i


�(z) = a0 log(z) +1X

k=1

a

k

z

k

where:

a0 =mX

i=1

q

i

and a

k

=mX

i=1

�q

i

z

k

i

k

for k � 1



�(z) = a0 log(z � z0) +1X

k=1

a

k

(z � z0)k


�(z) = a0 log(z) +1X

l=1

b

l

z

l

3


log(z � z

j


j

|+ i✓.


j

)) = log |z � z

j


�(z) =mX

j=1

m

j

log(z � z

j

).



�(z) =mX

j=1

m

j

log(z � z

j

) =mX

j=1

m

j

hlog z + log

⇣1� z

j

z

⌘i= M log z +

mX

j=1

m

j

1X

l=1

⇣z

j

z

⌘l

l

=

= M log z +1X

l=1

mX

j=1

m

j

⇣z

j

z

⌘l

l

= M log z +1X

l=1

2

4z

�l

mX

j=1

m

j

z

l

j

l

3

5 = M log z +1X

l=1

↵

l

z

l

where ↵

l

=mPj=1

m

j

z

l

j

l





, i =1, . . . ,m} with |z

i


�(z) = a0 log(z) +1X

k=1

a

k

z

k

where:

a0 =mX

i=1

q

i

and a

k

=mX

i=1

�q

i

z

k

i

k

for k � 1



�(z) = a0 log(z � z0) +1X

k=1

a

k

(z � z0)k


�(z) = a0 log(z) +1X

l=1

b

l

z

l

3

where:

b

l

= �a0zl

0

l

+lX

k=1

a

k

z

l�k

0

✓l � 1k � 1

◆

with

✓l � 1k � 1

◆the binomial coe�cients.

4

where:

b

l

= �a0zl

0

l

+lX

k=1

a

k

z

l�k

0

✓l � 1k � 1

◆

with

✓l � 1k � 1

◆the binomial coe�cients.

4

FSI Algorithm details

@!

@t+ (u ·r)! = (! ·r)u+ ⌫r2! +

1

⇢2r⇢⇥rp

d(Msx)

dt= F

d(Is!)

dt= ⌧

r · u = 0

u = uS ⌘ ut + ur + udef

fluid flow

fluid2solid solid2fluid

continuum

r · u = 0@!

@t+ (u ·r)! = (! ·r)u+ ⌫r2!

� r⇢

⇢⇥

✓@u

@t+ (u ·r)u� ⌫r2u� g

◆

+ ��(uS � u)

single-fluid model for FSI

u

ti =

1

mi

Z

⌦⇢�iu dx

✓i = J

�1i

Z

⌦⇢�i(x� xi)⇥ u dx

u

ri = ✓i ⇥ (x� xi)

@!n

@t= !n ·run

MultipolesVelocity

Projection

Penalization

Diffusion

Advection

Refinement

Compression

tn

tn+1

RK2+LTS

RK2

Explicit Euler

r2 n = �!n

un = r⇥ n

@!n�

@t= ⌫r2!n

�

@!n�

@t+ u

n� ·r!n

� = 0

!n+1 = !n+1�

x

n+1i = x

ni + u

t,ni �tn

✓n+1i = ✓n + ✓

n

i �tn

R(!nand/or un

)|tr

C(!n+1and/or un+1

)|tc

Stretching

RK2+LTS / particles

x

n+1i = x

ni + u

t,ni �tn, ✓n+1

i = ✓ni + ✓

n

i �tn

Baroclinicity @!n�

@t= �r⇢n

⇢n⇥

✓@un

�

@t+ (un

� ·r)un� � ⌫r2un

� � g

◆Explicit Euler

⇢n = ⇢f +NX

i=1

(⇢ni � ⇢f )�ni

u

t,ni =

1

mi

Z

⌦⇢n�n

i un dx

✓n

i = J

�1i

Z

⌦⇢n�n

i (x� x

ni )⇥ u

n dx

u

r,ni = ✓

n

i ⇥ (x� x

ni )

disc

reti

zati

on

!� = !n +r⇥ (u� � un)u� =un + ��tuS

1 + ��t

Fish surface pressure-force• Time varying force

37

• Thrust magnitude • Almost all segments

contribute to thrust • Either side of head -

thrust (negative pressure)

• Largest drag: tip of head, and tail-end

• Sum : Net Fx and Fy • For validation

against ax/y from velocity

Pressure and thrust• Thrust contribution:

depends on • Pressure sign • Body curvature

38

• Regions close to head contribute to thrust via pressure • Caveat: still need to

examine viscous contribution

Fluid-Structure Interaction with Vortex methods and the ... · 2 • Engineering applications •...

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Transcript of Fluid-Structure Interaction with Vortex methods and the ... · 2 • Engineering applications •...