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
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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
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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)
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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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