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first version of testing for transport solver.
1
2 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
3 %
4 % Copyright (c) 2003-2008 by University of Queensland
5 % Earth Systems Science Computational Center (ESSCC)
6 % http://www.uq.edu.au/esscc
7 %
8 % Primary Business: Queensland, Australia
9 % Licensed under the Open Software License version 3.0
10 % http://www.opensource.org/licenses/osl-3.0.php
11 %
12 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
13
14
15 \chapter{Models}
16
17 The following sections give a breif overview of the model classes and their corresponding methods.
18
19 \section{Stokes Cartesian (Saddle Point Problem)}
20
21 \subsection{Description}
22
23 Saddle point type problems emerge in a number of applications throughout physics and engineering. Finite element discretisation of the Navier-Stokes (momentum) equations for incompressible flow leads to equations of a saddle point type, which can be formulated as a solution of the following operator problem for $u \in V$ and $p \in Q$ with suitable Hilbert spaces $V$ and $Q$:
24
25 \begin{equation}
26 \left[ \begin{array}{cc}
27 A & B \\
28 b^{*} & 0 \\
29 \end{array} \right]
30 \left[ \begin{array}{c}
31 u \\
32 p \\
33 \end{array} \right]
34 =\left[ \begin{array}{c}
35 f \\
36 g \\
37 \end{array} \right]
38 \label{SADDLEPOINT}
39 \end{equation}
40
41 where $A$ is coercive, self-adjoint linear operator in $V$, $B$ is a linear operator from $Q$ into $V$ and $B^{*}$ is the adjoint operator of $B$. $f$ and $g$ are given elements from $V$ and $Q$ respectivitly. For more details on the mathematics see references \cite{AAMIRBERKYAN2008,MBENZI2005}.
42
43 The Uzawa scheme scheme is used to solve the momentum equation with the secondary condition of incompressibility \cite{GROSS2006,AAMIRBERKYAN2008}.
44
45 \begin{classdesc}{StokesProblemCartesian}{domain,debug}
46 opens the stokes equations on the \Domain domain. Setting debug=True switches the debug mode to on.
47 \end{classdesc}
48
49 example usage:
50
51 solution=StokesProblemCartesian(mesh) \\
52 solution.setTolerance(TOL) \\
53 solution.initialize(fixed\_u\_mask=b\_c,eta=eta,f=Y) \\
54 velocity,pressure=solution.solve(velocity,pressure,max\_iter=max\_iter,solver=solver) \\
55
56 % \subsection{Benchmark Problem}
57 %
58 % Convection problem
59
60
61 \section{Temperature Cartesian}
62
63 \begin{equation}
64 \rho c\hackscore{p} \left (\frac{\partial T}{\partial t} + \vec{v} \cdot \nabla T \right ) = k \nabla^{2}T
65 \label{HEAT EQUATION}
66 \end{equation}
67
68 where $\vec{v}$ is the velocity vector, $T$ is the temperature, $\rho$ is the density, $\eta$ is the viscosity, $c\hackscore{p}$ is the specific heat at constant pressure and $k$ is the thermal conductivity.
69
70 \subsection{Description}
71
72 \subsection{Method}
73
74 \begin{classdesc}{TemperatureCartesian}{dom,theta=THETA,useSUPG=SUPG}
75 \end{classdesc}
76
77 \subsection{Benchmark Problem}
78
79
80 \section{Level Set Method}
81
82 \subsection{Description}
83
84 \subsection{Method}
85
86 Advection and Reinitialisation
87
88 \begin{classdesc}{LevelSet}{mesh, func\_new, reinit\_max, reinit\_each, tolerance, smooth}
89 \end{classdesc}
90
91 %example usage:
92
93 %levelset = LevelSet(mesh, func\_new, reinit\_max, reinit\_each, tolerance, smooth)
94
95 \begin{methoddesc}[LevelSet]{update\_parameter}{parameter}
96 Update the parameter.
97 \end{methoddesc}
98
99 \begin{methoddesc}[LevelSet]{update\_phi}{paramter}{velocity}{dt}{t\_step}
100 Update level set function; advection and reinitialization
101 \end{methoddesc}
102
103 \subsection{Benchmark Problem}
104
105 Rayleigh-Taylor instability problem
106
107
108 % \section{Drucker Prager Model}
109
110 \section{Isotropic Kelvin Material \label{IKM}}
111 As proposed by Kelvin~\ref{KELVN} material strain $D\hackscore{ij}=v\hackscore{i,j}+v\hackscore{j,i}$ can be decomposed into
112 an elastic part $D\hackscore{ij}^{el}$ and visco-plastic part $D\hackscore{ij}^{vp}$:
113 \begin{equation}\label{IKM-EQU-2}
114 D\hackscore{ij}=D\hackscore{ij}^{el}+D\hackscore{ij}^{vp}
115 \end{equation}
116 with the elastic strain given as
117 \begin{equation}\label{IKM-EQU-3}
118 D\hackscore{ij}'^{el}=\frac{1}{2 \mu} \sigma\hackscore{ij}'
119 \end{equation}
120 where $\sigma'\hackscore{ij}$ is the deviatoric stress (Notice that $\sigma'\hackscore{ii}=0$).
121 If the material is composed by materials $q$ the visco-plastic strain can be decomposed as
122 \begin{equation}\label{IKM-EQU-4}
123 D\hackscore{ij}'^{vp}=\sum\hackscore{q} D\hackscore{ij}'^{q}
124 \end{equation}
125 where $D\hackscore{ij}^{q}$ is the strain in material $q$ given as
126 \begin{equation}\label{IKM-EQU-5}
127 D\hackscore{ij}'^{q}=\frac{1}{2 \eta^{q}} \sigma'\hackscore{ij}
128 \end{equation}
129 where $\eta^{q}$ is the viscosity of material $q$. We assume the following
130 betwee the the strain in material $q$
131 \begin{equation}\label{IKM-EQU-5b}
132 \eta^{q}=\eta^{q}\hackscore{N} \left(\frac{\tau}{\tau\hackscore{t}^q}\right)^{\frac{1}{n^{q}}-1}
133 \mbox{ with } \tau=\sqrt{\frac{1}{2}\sigma'\hackscore{ij} \sigma'\hackscore{ij}}
134 \end{equation}
135 for a given power law coefficients $n^{q}$ and transition stresses $\tau\hackscore{t}^q$, see~\ref{POERLAW}.
136 Notice that $n^{q}=1$ gives a constant viscosity.
137 After inserting equation~\ref{IKM-EQU-5} into equation \ref{IKM-EQU-4} one gets:
138 \begin{equation}\label{IKM-EQU-6}
139 D\hackscore{ij}'^{vp}=\frac{1}{2 \eta^{vp}} \sigma'\hackscore{ij} \mbox{ with } \frac{1}{\eta^{vp}} \sum\hackscore{q} \frac{1}{\eta^{q}} \;.
140 \end{equation}
141 With
142 \begin{equation}\label{IKM-EQU-8}
143 \dot{\gamma}=\sqrt{2 D\hackscore{ij} D\hackscore{ij}}
144 \end{equation}
145 one gets
146 \begin{equation}\label{IKM-EQU-8b}
147 \tau = \eta^{vp} \dot{\gamma}^{vp} \;.
148 \end{equation}
149 With the Drucker-Prager cohesion factor $\tau\hackscore{Y}$, Drucker-Prager friction $\beta$ and total pressure $p$ we want to achieve
150 \begin{equation}\label{IKM-EQU-8c}
151 \tau \le \tau\hackscore{Y} + \beta \; p
152 \end{equation}
153 which leads to the condition
154 \begin{equation}\label{IKM-EQU-8d}
155 \eta^{vp} \le \frac{\tau\hackscore{Y} + \beta \; p}{ \dot{\gamma}^{vp}} \; .
156 \end{equation}
157 Therefore we modify the definition of $\eta^{vp}$ to the form
158 \begin{equation}\label{IKM-EQU-6b}
159 \frac{1}{\eta^{vp}}=\max(\sum\hackscore{q} \frac{1}{\eta^{q}}, \frac{\dot{\gamma}^{vp}} {\tau\hackscore{Y} + \beta \; p})
160 \end{equation}
161 Now we can combine equations~\ref{IKM-EQU-2}, \ref{IKM-EQU-3} and~\ref{IKM-EQU-6b} to get
162 \begin{equation}\label{IKM-EQU-10}
163 D\hackscore{ij}'=\frac{1}{2 \eta\hackscore{eff}} \sigma\hackscore{ij}' \mbox{ with }
164 \frac{1}{\eta\hackscore{eff}}=\frac{1}{\mu}+\frac{1}{\eta^{vp}}
165 \end{equation}
166 The deviatoric stress needs to fullfill the equilibrion equation
167 \begin{equation}\label{IKM-EQU-1}
168 -\sigma'\hackscore{ij,j}+p\hackscore{,i}=F\hackscore{i}
169 \end{equation}
170 where $F\hackscore{j}$ is a given external fource. We assume an incompressible media:
171 \begin{equation}\label{IKM-EQU-2}
172 -v\hackscore{i,i}=0
173 \end{equation}
174 After inserting~\ref{IKM-EQU-10} into~\ref{IKM-EQU-1} we get
175 \begin{equation}\label{IKM-EQU-1ib}
176 -\left(\eta\hackscore{eff} (v\hackscore{i,j}+ v\hackscore{i,j})\right)\hackscore{,j}+p\hackscore{,i}=F\hackscore{i}
177 \end{equation}
178

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