moving mesh-moving boundary method for two-phase … · •the shape functions are 1 at the node...
TRANSCRIPT
MOVING MESH-MOVING BOUNDARY METHOD FOR TWO-PHASE FLOWS
WITH PHASE CHANGE
Tutorial 11-4-2: Two Phase Boiling Computational Modelling Challenges
Gustavo R. ANJOS
San Francisco - CA, InterPACK2015 July 6th, 2015
http://gustavo.rabello.org http://www.gesar.uerj.br
OUTLINE
• Intro to Finite Element Method• Variational method: the weak form;• Function approximations: Galerkin method;• 1D example;• Tasks: 1D and 2D examples;
2
Fundamentals of the Finite Element Method for Heat and Fluid Flow
Authors: Roland W. Lewis, Perumal Nithiarasu, Kankanhally e N.
Seetharamu
Basic
Basic-advancedThe Finite Element Method - Linear Static and Dynamic Finite Element
AnalysisAuthors: Thomas J.R. Hughues
BIBLIOGRAPHY
Find in such that:u ⌦ = [0, 1]⇥ [0, 1]
321
1 2 3 4(0,0) (0,1)
x
1D PROBLEM - STRONG FORM
d
2u
dx
2+ u+ 1 = 0
u(0) = 0 boundaryconditiondu
dx
(1) = 1
h1 = h2 = h3 = 1/3domain:Answer: u2=1.049; u3=1,874; u4=2,386
3
1D PROBLEM - WEAK FORM
Z
⌦w
✓d
2u
dx
2+ u+ 1
◆d⌦ = 0
Z 1
0w
d
2u
dx
2dx+
Z 1
0wudx+
Z 1
0dx = 0
w
du
dx
����1
0
�Z 1
0
du
dx
dw
dx
dx+
Z 1
0wudx+
Z 1
0wdx = 0
w
du
dx
����1
� w
du
dx
����0
�Z 1
0
du
dx
dw
dx
dx+
Z 1
0wudx+
Z 1
0wdx = 0
weight functionmathematical procedure (integration by parts)
Find in such that:u H1
4
•the shape functions are 1 at the node and zero everywhere else;•the sum of all shape function at the element is 1 everywhere, including boundary.
Finite element properties:
Chart:function node, i node, j x
Ni 1 0 between 0 e 1Nj 0 1 between 0 e 1
Ni+Nj 1 1 1
CHOOSING ELEMENT
5
FEM SHAPE FUNCTIONS1D Problem - linear :
T (x) = ↵1 + ↵2x T (x) = ↵1 + ↵2x+ ↵3x2
1D problem - quadratic:
shapefunction
function
shapefunction
derivative
6
GALERKIN METHOD
u =4X
i=1
Niui w =4X
j=1
Njwj
4X
i=j=1
✓Z 1
0
dNi
dx
dNj
dx
dx�Z 1
0NiNjdx
◆ui =
4X
j=1
Z 1
0Njdx+
du
dx
(1)� du
dx
(0)
w(1)du
dx
(1)� w(0)du
dx
(0)�4X
i=j=1
Z 1
0
dNi
dx
uidNj
dx
wjdx+4X
i=j=1
Z 1
0NiuiNjwjdx+
4X
j=1
Z 1
0Njwjdx = 0
w(1)du
dx
(1)� w(0)du
dx
(0)�4X
i=j=1
Z 1
0
dNi
dx
uidNj
dx
wjdx+4X
i=j=1
Z 1
0NiuiNjwjdx+
4X
j=1
Z 1
0Njwjdx = 0
(Kij �Mij)ui = bistiffness matrix
mass matrix(Kij �Mij)ui = bi(Kij �Mij)ui = bi
right hand side.
Approximated functions:
boundarycondition
7(Kij �Mij)ui = bi + b.c.
1D PROBLEM - LINEAR
3211 2 3 4
(0,0) (1.0)x
(0.3) (0.6)
1element
2element
3element
N1 = �3x+ 1N2 = 3x
N2 = �3x+ 2N3 = 3x� 1
N3 = �3x+ 3
N4 = 3x� 2
⌦e2 = [1/3, 2/3]
⌦e1 = [0, 1/3]
⌦e3 = [2/3, 1]
N1 = �3x+ 1N2 = 3x N3 = 3x� 1N4 = 3x� 2domainand shapefunctions:
8
MATRIX FORM
1element
N1 = �3x+ 1N2 = 3x
⌦e1 = [0, 1/3]
K11 �M11 =
Z 1/3
0
dN1
dx
dN1
dx
dx�Z 1/3
0N1N1dx
K12 �M12 =
Z 1/3
0
dN1
dx
dN2
dx
dx�Z 1/3
0N1N2dx
K21 �M21 =
Z 1/3
0
dN2
dx
dN1
dx
dx�Z 1/3
0N2N1dx
K22 �M22 =
Z 1/3
0
dN2
dx
dN2
dx
dx�Z 1/3
0N2N2dx
b1 =
Z 1/3
0N1dx� du
dx
(0)
b2 =
Z 1/3
0N2dx
9
matrix
vector
Ke1 �Me
1 =
"K11 K12
K21 K22
#�
"M11 M12
M21 M22
#
"b1
b2
#
MATRIX FORM
2element
9
matrix
vector
N2 = �3x+ 2
N3 = 3x� 1
⌦e2 = [1/3, 2/3]
K22 �M22 =
Z 2/3
1/3
dN2
dx
dN2
dx
dx�Z 2/3
1/3N2N2dx
K23 �M23 =
Z 2/3
1/3
dN2
dx
dN3
dx
dx�Z 2/3
1/3N2N3dx
K32 �M32 =
Z 2/3
1/3
dN3
dx
dN2
dx
dx�Z 2/3
1/3N3N2dx
K33 �M33 =
Z 2/3
1/3
dN3
dx
dN3
dx
dx�Z 2/3
1/3N3N3dx
b2 =
Z 2/3
1/3N2dx
b3 =
Z 2/3
1/3N3dx
Ke2 �Me
2 =
"K22 K23
K32 K33
#�
"M22 M23
M32 M33
#
be2 =
"b2
b3
#=
"1/6
1/6
#
MATRIX FORM
3element
9
matrix
vector⌦e
3 = [2/3, 1]
N3 = �3x+ 3
N4 = 3x� 2
Ke3 �Me
3 =
"K33 K34
K43 K44
#�
"M33 M34
M43 M44
#
"b3
b4
#
K33 �M33 =
Z 1
2/3
dN3
dx
dN3
dx
dx�Z 1
2/3N3N3dx
K34 �M34 =
Z 1
2/3
dN3
dx
dN4
dx
dx�Z 1
2/3N3N4dx
K43 �M43 =
Z 1
2/3
dN4
dx
dN3
dx
dx�Z 1
2/3N4N3dx
K44 �M44 =
Z 1
2/3
dN4
dx
dN4
dx
dx�Z 1
2/3N4N4dx
b3 =
Z 1
2/3N3dx
b4 =
Z 1
2/3N4dx
MATRIX FORM
3211 2 3 4
(0,0) (1.0)x
(0.3) (0.6)
1element⌦e
1 = [0, 1/3]
N1 = �3x+ 1N2 = 3x N3 = 3x� 1N4 = 3x� 2domainand shapefunctions:
Ke1 �Me
1 =
"K11 K12
K21 K22
#�
"M11 M12
M21 M22
#
Ke1 �Me
1 =
"3� 2
18 �3� 118
�3� 118 3� 2
18
#=
"5218 � 55
18
� 5518
5218
#
b
e1 =
"b1 + b.c. at x = 0
b2
#=
"1/6
1/6
#
10
MATRIX FORM
3211 2 3 4
(0,0) (1.0)x
(0.3) (0.6)
2element
N1 = �3x+ 1N2 = 3x N3 = 3x� 1N4 = 3x� 2domainand shapefunctions:
be2 =
"b2
b3
#=
"1/6
1/6
#
Ke2 �Me
2 =
"K22 K23
K32 K33
#�
"M22 M23
M32 M33
#
Ke2 �Me
2 =
"3� 2
18 �3� 118
�3� 118 3� 2
18
#=
"5218 � 55
18
� 5518
5218
#⌦e
2 = [1/3, 2/3]
11
MATRIX FORM
3211 2 3 4
(0,0) (1.0)x
(0.3) (0.6)
3element
N1 = �3x+ 1N2 = 3x N3 = 3x� 1N4 = 3x� 2domainand shapefunctions:
⌦e3 = [2/3, 1]
Ke3 �Me
3 =
"3� 2
18 �3� 118
�3� 118 3� 2
18
#=
"5218 � 55
18
� 5518
5218
#Ke
3 �Me3 =
"K33 K34
K43 K44
#�
"M33 M34
M43 M44
#
b
e3 =
"b3
b4 + b.c. at x = 1
#=
"1/6
1/6 + 1
#
12
ASSEMBLING
12
1 2 3 4
1
2
3
4
=
linear system of equations:
(Kij �Mij)ui = bi(Kij �Mij)ui = bi bi + b.c.
(Kij �Mij)ui = bi + b.c.
Find in such that:u ⌦ = [0, 1]⇥ [0, 1]
321
1 2 3 4(0,0) (0,1)
x
1D PROBLEM
d
2u
dx
2+ u+ 1 = 0
u(0) = 0 boundarycondition
h1 = h2 = h3 = 1/3domain:Answer: u2=0.251; u3=0.363; u4=0.328
du
dx
(1) = �u
13
Find in such that:
Answer: u6=0.611; u7=0.889; u10=0.889; u11=1.167
u
boundarycondition
⌦ = [0, 1]⇥ [0, 1]
r2u = 0
6
5
4
32
1
12
11
10
98
7
18
17
16
1514
13
1 2 3 4
5
9
13
8
12
16
6 7
10 11
14 15
u = y2 + 1
u = x
2 + 1
u = y
u = x
Equation:
(0,0) (0,1)
(1,1)(0,1)
x
y
2D PROBLEM
14