the immersed interface method for flow around …...the immersed interface method for flow around...

The Immersed Interface Method for Flow AroundNon-smooth Boundaries

Yang Liu Sheng Xu

Department of MathematicsSouthern Methodist University

International Workshop on Fluid-Structure Interaction ProblemsJune 1 2016

Yang Liu, Sheng Xu (SMU) Immersed Interface Method 1 / 22

Outline

1 Motivations

2 Introduction

3 Jump Conditions

4 Pressure Solver

5 Results

6 Conclusions

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Motivations

Figure: Flow paststationary square cylinderat Re=200

Figure: Flow past twostationary square cylindersat Re=200

Figure: Flow around threehovering rectangleflappers

Development of the Immersed Interface Method for flows withstationary/moving smooth/non-smooth complex objects

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Formulation for the Immersed Interface MethodThe Naiver-Stokes equation for incompressible viscous flow past rigid objects is

∂~u∂t +5 · (~u~u) = −5 p + 1

Re ∆~u + ~q +∫

Γ~f (~X , t)δ(~x − ~X )dl (1)

5 ·~u = 0 (2)

where singular force ~F =∫

Γ~f (~X , t)δ(~x − ~X )dl

x

y

B

Ω+

Ω−

Γ

Γ+

Γ−

~n

~τ

~Xb

References: Peskin, JCP, 72; Xu, JCP, 08

Body force ~q = θ · (~X − ~xc) (nonzero onlyinside the boundary) to enforce the rigidmotion of the fluid enclosed by theboundary

Singular force ~F to represent the effect ofthe object boundary

Jump conditions induced by the singularforce and the discontinuous body force

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Questions of Jump Conditions

What jump conditions we need?How to derive these jump conditions?How to use these jump conditions?

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Overview of the Immersed Interface Method

z

g(z)

z0 z1 z2 z3 zm−1 zm zm+1

+

-

+

-

+

-

+

-

+ -+

-

h h

b b b| |

zi−1 zi zi+1ξ η

Figure: Examples for generalized Taylor expansion and finite difference scheme

General Taylor expansion with jump conditions:

g(

z−m+1

)=

∞∑n=0

g(n)(z+0 )

n!(zm+1 − z0)n +

m∑l=1

∞∑n=0

[g(n)(zl )

]n!

(zm+1 − zl )n, (3)

Cartesian grid finite difference schemes incorporated with jump conditions

dg(

z−i)

dz=

g(

z−i+1

)− g(

z+i−1

)2h

+ O(

h2)−

12h

(2∑

n=0

[gn(ξ)]n!

(zi−1 − ξ)n +

2∑n=0

[gn(η)]n!

(zi+1 − η)n

)(4)

Reference: LeVeque & Li, SINUM, 94

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Overview of the Immersed Interface Method

~n

~τ ~νA

B

C

D

E

F

M1 M2

b b

Cartesian Jump Conditions[∂~u∂x

],[∂~u∂y

],[∂2~u∂x2

],[∂2~u∂y2

],[∂2~u∂x∂y

][∂p∂x

],[∂p∂y

],[∂2p∂x2

],[∂2p∂y2

],[∂2p∂x∂y

]Computing Cartesian Jump Conditionsfrom Principal Jump Conditions usingTriangular Mesh Representation of aBoundary in 3DReference: Xu & Pearson, JCP, 2015Principal Jump Conditions[~u],

[∂~u∂n

], [∆~u], [p],

[∂p∂n

], [∆p]

Implementation: MAC grid,FFT/Helmholtz Poisson Solver, RK4Time Marching

Yang Liu, Sheng Xu (SMU) Immersed Interface Method 7 / 22

Principle Jump Conditions for ~u

δx

δy

δn

δn

Γ

IV III

I II

S2

S1

S0

Ω+

Ω−

τ

n

Principle jump conditions for ~u:

[~u] = 0 (5)[∂~u∂n

]= ∂~u∂n |

+−θ · ~τ (6)

[4~u] = ∂2~u∂n2 |

++κ[∂~u∂n

](7)

θ is angular velocity of prescribedmotionκ is the curvature of the boundary∂~u∂n |

+ and ∂2~u∂n2 |+ using one-sided finite

difference method∂~u∂n|+=

−3~u(S0) + 4~u(S1)− ~u(S2)2δn

+ O(δn2) (8)

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Cartesian Jump Conditions for ~u

First order Cartesian jump conditions:As [~u] = 0 and

[∂~u∂τ

]= 0, we can build the matrix form below:[

τx τynx ny

][[∂~u∂x][

∂~u∂y

]] =[

0[∂~u∂n]] (9)

Second order Cartesian jump conditions:

~n

~τ ~νA

B

C

D

E

F

M1 M2

b b

1 0 1τx τy 00 τx τy

[∂2~u∂x2

][∂2~u∂x∂y

][∂2~u∂y2

] =

[4~u]∂∂τ

[∂~u∂x]

∂∂τ

[∂~u∂y

] (10)

∂

∂τ

[∂~u∂x

]A

= 1| AB |

([∂~u∂x

]B−[∂~u∂x

]A

)(11)

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Principle Jump Conditions for pTaking jump conditions of Navier-Stokes equation

∂~u∂t +5 · (~u~u) = −5 p + 1

Re ∆~u + ~q +∫

Γ~f (~X , t)δ(~x − ~X )dl (12)

[~u] = 0[D~uDt]

=[∂~u∂t +5 · (~u~u)

]= 0[

~F]

= 0

=⇒[5p] = 1

Re [4~u] + [~q][∂p∂n

]= [5p] · ~n

Taking divergence of (12)4p = sp +5 ·

(~q + ~F

)(13a)

D = 5 · ~u (13b)

sp = −(∂D∂t

+5 · (2~uD)−1

Re4 D

)+2(∂u∂x

∂v∂y−∂u∂y

∂v∂x

)(13c)

=⇒ [4p] = 2[∂u∂x

∂v∂y

]− 2

[∂u∂y

∂v∂x

]

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Principle Jump Conditions for p

[∂p∂τ

]A

= [5p]A · ~τA (14)∫ B

A

[∂p∂τ

]dτ = [p]B − [p]A ≈

| AB |6

([∂p∂τ

]A

+ 4[∂p∂τ

]M1

+[∂p∂τ

]B

)(15a)∫ F

A

[∂p∂ν

]dν = [p]F − [p]A ≈

| AF |6

([∂p∂ν

]A

+ 4[∂p∂ν

]M2

+[∂p∂ν

]F

)(15b)

~n

~τ ~νA

B

C

D

E

F

M1 M2

b b

[p]B + [p]F − 2 [p]A = rhsA, (16)Toeplitz matrix problem: It’s singular matrix

−2 1 . . . 11 −2 . . . 0...

.... . .

...0 . . . −2 11 . . . 1 −2

[p]A[p]B

...[p]F

=

rhsArhsB

...rhsF

(17)

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Cartesian Jump Conditions for p

As [5p] = 1Re [4~u] + [~q], first order Cartesian jump conditions of p:[

∂p∂x

]= 1

Re [4u] + [qx ] (18a)

[∂p∂y

]= 1

Re [4v ] + [qy ] (18b)

Similar as ~u, second order Cartesian jump conditions of p:

1 0 1τx τy 00 τx τy

[∂2p∂x2

][∂2p∂x∂y

][∂2p∂y2

] =

[4p]∂∂τ

[∂p∂x

]∂∂τ

[∂p∂y

] (19)

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Poisson Solver

Recall that

4p = −(∂D∂t +5 · (2~uD)− 1

Re 4 D)

+ 2(∂u∂x

∂v∂y −

∂u∂y

∂v∂x

)+5 ·

(~q + ~F

)(20)

where D = 5 · ~uAfter discretization,

(4p)i,j = pi−1,j − 2pi,j + pi+1,jδx2 + pi,j−1 − 2pi,j + pi,j+1

δy2 + ci,j (21)

Lp + c = f (22)

Principle jump conditions are known, c is independent of p. FFT solver can beused to solve Lp = f − c.

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Helmholtz Solver

If principle jump conditions are unknown, then [p] = p |+ −p |−.

p |+= 3p1 − 3p2 + p3 (23a)

p |−= −d2xc

dt2 x −d2yc

dt2 y +12

(dθdt

)2 ((x − xc )2 + (y − yc )2)+ pc (23b)

[p] is a linear function about p, then c = Cp + c0,

(L + C) p = f − c0 (24)

Introduce the pseudo-time∂p∂t = Lp +

(Cp + c0

)− f (25)

Use implicit method for Lp term and let γ = 1∆t

γ(pn+1 − pn) = Lpn+1 +

(Cpn + c0

)− f (26)

Finally we have (L− γI) pn+1 = −(

(C + γI) pn + c0 − f)

Yang Liu, Sheng Xu (SMU) Immersed Interface Method 14 / 22

Results

Test of Accuracy: Circular Couette FlowFFT solverHelmholtz solver

Code Validation:Flow Past a Stationary Circular CylinderFlow Past a Stationary Square Cylinder

Test of Efficiency: Flow Around Hovering Flappers

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Test of Accuracy: Circular Couette Flow(FFT)

lx

lyr1

r2

x

y

B

Π1

Π2

n ||eu||∞ order ||ev ||∞ order ||ep||∞ order30 3.90× 10−2 - 4.01× 10−2 - 1.71× 10−2 -60 7.28× 10−3 2.4215 7.28× 10−3 2.4615 2.98× 10−3 2.5215

120 1.82× 10−3 2.0008 1.83× 10−3 1.9952 2.45× 10−3 0.2805240 4.53× 10−4 2.0066 4.55× 10−4 2.0068 1.09× 10−3 1.1697

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Test of Accuracy: Circular Couette Flow(Helmholtz)

Figure: no. of iterations vs. time

n ||eu ||∞ order ||ev ||∞ order ||ep ||∞ order30 3.14× 10−2 - 3.22× 10−2 - 3.66× 10−2 -60 6.00× 10−3 2.3958 6.16× 10−3 2.3877 1.51× 10−2 1.3087

120 1.68× 10−3 1.8318 1.69× 10−3 1.8634 8.18× 10−3 0.7893240 6.99× 10−4 1.285 6.92× 10−4 1.3012 9.80× 10−3 -0.2592

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Code Validation: Circular Cylinder

Figure: Stream function at Re=20 & 40

Re = 20 Re = 40 Re = 100 Re = 200L Cd L Cd Cd St Cd St

Tritton - 2.22 - 1.48 Braza 1.36 - 1.40 -Dennis 0.94 2.05 2.13 1.52 Russell 1.43 0.172 1.45 0.201

Fornberg 0.91 2.00 2.24 1.50 Le 1.37 0.169 1.34 0.200Xu 0.92 2.23 2.21 1.66 Xu 1.32 0.171 1.42 0.202

Linnick 0.93 2.16 2.23 1.61 Linnick 1.38 0.169 1.37 0.200Present 0.98 2.06 2.4 1.56 Present 1.39 0.169 1.41 0.200

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Code Validation: Square Cylinder

Figure: Left: Stream function at Re=40; Right: Vorticity filed at Re=100

B L/D Cd Re = 100Re = 40 Re = 5 Re = 40 Cd St

Paliwal 0.067 2.7000 4.8140 1.8990 Sohankar 1.4770 0.1460Sen 0.067 2.7348 5.2641 1.8565 Robichaux 1.5300 0.1540

Present 0.067 2.77 5.1334 1.7664 Sharma 1.4936 0.1488Present 0.0625 2.79 5.0616 1.7518 Singh 1.5100 0.1470Dhiman 0.050 2.8220 4.8400 1.7670 Sahu 1.4878 0.1486

Sen 0.050 2.8065 4.9535 1.7871 Sen 1.5287 0.1452Present 0.050 2.86 4.8759 1.7154 Present 1.4941 0.1479

(At Re=100, blockage ratio B = 0.5)

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Test of Efficiency: Flow Around Hovering Flappers

no. of plates 1 2 3 4Rounded plate 1 1.0509 1.1534 1.1977

Rectangle 1.0195 1.0790 1.1526 1.4100(The computational time corresponding to the unit value is 0.636 hours.)

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Conclusions

Conclusions:

The Helmholtz solver is not efficient as FFT, but shows good results.

The method can simulate flows with multiple moving non-smooth complexobjects.

The method is second order accurate in the infinity norm for the velocity.

The method is stable at all the Reynolds numbers(Re = 5 ∼ 1000) in ourtests.

The method is efficient to handle multiple moving objects. The extra cost tohandle an additional object is proportional to the number of the verticesused to represent the objects

Current Work:

The ongoing work focuses on the parallelization of the method fordistributed-memory parallel computing with MPI.

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Acknowledgement

We want to thank the support of this work from NSF through the grant NSFDMS #1320317

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