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Wed Oct 8 03:03:37 2008 UTC (11 years, 4 months ago) by gross
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first version of testing for transport solver.
1 ksteube 1811
2     %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
3 lgraham 1702 %
4 ksteube 1811 % Copyright (c) 2003-2008 by University of Queensland
5     % Earth Systems Science Computational Center (ESSCC)
6     % http://www.uq.edu.au/esscc
7 lgraham 1702 %
8 ksteube 1811 % 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 lgraham 1702 %
12 ksteube 1811 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
13 lgraham 1702
14 ksteube 1811
15 lgraham 1702 \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 gross 1859 % \subsection{Benchmark Problem}
57     %
58     % Convection problem
59 lgraham 1702
60    
61     \section{Temperature Cartesian}
62    
63     \begin{equation}
64 lgraham 1709 \rho c\hackscore{p} \left (\frac{\partial T}{\partial t} + \vec{v} \cdot \nabla T \right ) = k \nabla^{2}T
65 lgraham 1702 \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 gross 1859 % \section{Drucker Prager Model}
109 lgraham 1702
110 gross 1841 \section{Isotropic Kelvin Material \label{IKM}}
111 gross 1859 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 gross 1841 \begin{equation}\label{IKM-EQU-2}
114 gross 1859 D\hackscore{ij}=D\hackscore{ij}^{el}+D\hackscore{ij}^{vp}
115 gross 1841 \end{equation}
116 gross 1859 with the elastic strain given as
117 gross 1841 \begin{equation}\label{IKM-EQU-3}
118 gross 1859 D\hackscore{ij}'^{el}=\frac{1}{2 \mu} \sigma\hackscore{ij}'
119 gross 1841 \end{equation}
120 gross 1859 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 gross 1841 \begin{equation}\label{IKM-EQU-4}
123 gross 1859 D\hackscore{ij}'^{vp}=\sum\hackscore{q} D\hackscore{ij}'^{q}
124 gross 1841 \end{equation}
125 gross 1859 where $D\hackscore{ij}^{q}$ is the strain in material $q$ given as
126 gross 1841 \begin{equation}\label{IKM-EQU-5}
127 gross 1859 D\hackscore{ij}'^{q}=\frac{1}{2 \eta^{q}} \sigma'\hackscore{ij}
128 gross 1841 \end{equation}
129 gross 1859 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 gross 1841 After inserting equation~\ref{IKM-EQU-5} into equation \ref{IKM-EQU-4} one gets:
138 gross 1859 \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 gross 1841 \end{equation}
141 gross 1859 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 gross 1841 \begin{equation}\label{IKM-EQU-1}
168 gross 1859 -\sigma'\hackscore{ij,j}+p\hackscore{,i}=F\hackscore{i}
169 gross 1841 \end{equation}
170 gross 1859 where $F\hackscore{j}$ is a given external fource. We assume an incompressible media:
171 gross 1841 \begin{equation}\label{IKM-EQU-2}
172 gross 1859 -v\hackscore{i,i}=0
173 gross 1841 \end{equation}
174 gross 1859 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 gross 1841

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