# Annotation of /trunk/doc/cookbook/example01.tex

Revision 3370 - (hide annotations)
Sun Nov 21 23:22:25 2010 UTC (8 years, 9 months ago) by ahallam
File MIME type: application/x-tex
File size: 37422 byte(s)
Rearranged figures for release.

 1 ahallam 2401 2 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 3 % 4 jfenwick 2881 % Copyright (c) 2003-2010 by University of Queensland 5 ahallam 2401 % Earth Systems Science Computational Center (ESSCC) 6 7 % 8 % Primary Business: Queensland, Australia 9 % Licensed under the Open Software License version 3.0 10 11 % 12 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 13 14 gross 2949 \section{Example 1: One Dimensional Heat Diffusion in Granite} 15 gross 2905 \label{Sec:1DHDv00} 16 gross 2878 17 ahallam 2979 The first model consists of two blocks of isotropic material, for instance 18 ahallam 3370 granite, sitting next to each other (\autoref{fig:onedgbmodel}). 19 ahallam 2979 Initial temperature in \textit{Block 1} is \verb|T1| and in \textit{Block 2} is 20 \verb|T2|. 21 gross 2905 We assume that the system is insulated. 22 What would happen to the temperature distribution in each block over time? 23 ahallam 2979 Intuition tells us that heat will be transported from the hotter block to the 24 cooler one until both 25 gross 2905 blocks have the same temperature. 26 27 ahallam 3370 \begin{figure}[ht] 28 \centerline{\includegraphics[width=4.in]{figures/onedheatdiff001}} 29 \caption{Example 1: Temperature differential along a single interface between 30 two granite blocks.} 31 \label{fig:onedgbmodel} 32 \end{figure} 33 34 ahallam 2495 \subsection{1D Heat Diffusion Equation} 35 caltinay 2982 We can model the heat distribution of this problem over time using the one 36 ahallam 2979 dimensional heat diffusion equation\footnote{A detailed discussion on how the 37 heat diffusion equation is derived can be found at 38 \url{ 39 http://online.redwoods.edu/instruct/darnold/DEProj/sp02/AbeRichards/paper.pdf}}; 40 ahallam 2494 which is defined as: 41 ahallam 2401 \begin{equation} 42 jfenwick 3308 \rho c_p \frac{\partial T}{\partial t} - \kappa \frac{\partial^{2} 43 T}{\partial x^{2}} = q_H 44 ahallam 2401 \label{eqn:hd} 45 \end{equation} 46 jfenwick 3308 where $\rho$ is the material density, $c_p$ is the specific heat and 47 ahallam 2979 $\kappa$ is the thermal 48 conductivity\footnote{A list of some common thermal conductivities is available 49 from Wikipedia 50 \url{http://en.wikipedia.org/wiki/List_of_thermal_conductivities}}. Here we 51 assume that these material 52 gross 2861 parameters are \textbf{constant}. 53 ahallam 2979 The heat source is defined by the right hand side of \refEq{eqn:hd} as 54 jfenwick 3308 $q_{H}$; this can take the form of a constant or a function of time and 55 space. For example $q_{H} = q_{0}e^{-\gamma t}$ where we have 56 ahallam 2979 the output of our heat source decaying with time. There are also two partial 57 derivatives in \refEq{eqn:hd}; $\frac{\partial T}{\partial t}$ describes the 58 change in temperature with time while $\frac{\partial ^2 T}{\partial x^2}$ is 59 the spatial change of temperature. As there is only a single spatial dimension 60 to our problem, our temperature solution $T$ is only dependent on the time $t$ 61 caltinay 2982 and our signed distance from the block-block interface $x$. 62 ahallam 2401 63 gross 2861 \subsection{PDEs and the General Form} 64 ahallam 2979 It is possible to solve PDE \refEq{eqn:hd} analytically and obtain an exact 65 solution to our problem. However, it is not always practical to solve the 66 problem this way. Alternatively, computers can be used to find the solution. To 67 do this, a numerical approach is required to discretise 68 the PDE \refEq{eqn:hd} across time and space, this reduces the problem to a 69 finite number of equations for a finite number of spatial points and time steps. 70 These parameters together define the model. While discretisation introduces 71 approximations and a degree of error, a sufficiently sampled model is generally 72 accurate enough to satisfy the accuracy requirements for the final solution. 73 gross 2477 74 ahallam 2979 Firstly, we discretise the PDE \refEq{eqn:hd} in time. This leaves us with a 75 steady linear PDE which involves spatial derivatives only and needs to be solved 76 in each time step to progress in time. \esc can help us here. 77 gross 2861 78 caltinay 2982 For time discretisation we use the Backward Euler approximation 79 ahallam 2979 scheme\footnote{see \url{http://en.wikipedia.org/wiki/Euler_method}}. It is 80 caltinay 2982 based on the approximation 81 gross 2861 \begin{equation} 82 \frac{\partial T(t)}{\partial t} \approx \frac{T(t)-T(t-h)}{h} 83 \label{eqn:beuler} 84 \end{equation} 85 for $\frac{\partial T}{\partial t}$ at time $t$ 86 where $h$ is the time step size. This can also be written as; 87 \begin{equation} 88 \frac{\partial T}{\partial t}(t^{(n)}) \approx \frac{T^{(n)} - T^{(n-1)}}{h} 89 \label{eqn:Tbeuler} 90 \end{equation} 91 ahallam 2979 where the upper index $n$ denotes the n\textsuperscript{th} time step. So one 92 has 93 gross 2861 \begin{equation} 94 \begin{array}{rcl} 95 t^{(n)} & = & t^{(n-1)}+h \\ 96 T^{(n)} & = & T(t^{(n-1)}) \\ 97 \end{array} 98 \label{eqn:Neuler} 99 \end{equation} 100 Substituting \refEq{eqn:Tbeuler} into \refEq{eqn:hd} we get; 101 \begin{equation} 102 jfenwick 3308 \frac{\rho c_p}{h} (T^{(n)} - T^{(n-1)}) - \kappa \frac{\partial^{2} 103 T^{(n)}}{\partial x^{2}} = q_H 104 gross 2861 \label{eqn:hddisc} 105 \end{equation} 106 ahallam 2979 Notice that we evaluate the spatial derivative term at the current time 107 $t^{(n)}$ - therefore the name \textbf{backward Euler} scheme. Alternatively, 108 one can evaluate the spatial derivative term at the previous time $t^{(n-1)}$. 109 caltinay 2982 This approach is called the \textbf{forward Euler} scheme. This scheme can 110 provide some computational advantages, which 111 ahallam 2979 are not discussed here. However, the \textbf{forward Euler} scheme has a major 112 disadvantage. Namely, depending on the 113 material parameters as well as the domain discretization of the spatial 114 derivative term, the time step size $h$ needs to be chosen sufficiently small to 115 caltinay 2982 achieve a stable temperature when progressing in time. Stability is achieved if 116 the temperature does not grow beyond its initial bounds and becomes 117 ahallam 2979 non-physical. 118 The backward Euler scheme, which we use here, is unconditionally stable meaning 119 that under the assumption of a 120 physically correct problem set-up the temperature approximation remains physical 121 for all time steps. 122 caltinay 2982 The user needs to keep in mind that the discretisation error introduced by 123 ahallam 2979 \refEq{eqn:beuler} 124 is sufficiently small, thus a good approximation of the true temperature is 125 caltinay 2982 computed. It is therefore very important that any results are viewed with 126 caution. For example, one may compare the results for different time and 127 spatial step sizes. 128 gross 2861 129 To get the temperature $T^{(n)}$ at time $t^{(n)}$ we need to solve the linear 130 ahallam 2979 differential equation \refEq{eqn:hddisc} which only includes spatial 131 caltinay 2982 derivatives. To solve this problem we want to use \esc. 132 gross 2861 133 ahallam 2979 In \esc any given PDE can be described by the general form. For the purpose of 134 this introduction we illustrate a simpler version of the general form for full 135 linear PDEs which is available in the \esc user's guide. A simplified form that 136 suits our heat diffusion problem\footnote{The form in the \esc users guide which 137 uses the Einstein convention is written as 138 jfenwick 3308 $-(A_{jl} u_{,l})_{,j}+D u =Y$} 139 ahallam 2606 is described by; 140 gross 2477 \begin{equation}\label{eqn:commonform nabla} 141 jfenwick 2657 -\nabla\cdot(A\cdot\nabla u) + Du = f 142 ahallam 2411 \end{equation} 143 ahallam 2979 where $A$, $D$ and $f$ are known values and $u$ is the unknown solution. The 144 symbol $\nabla$ which is called the \textit{Nabla operator} or \textit{del 145 operator} represents 146 the spatial derivative of its subject - in this case $u$. Lets assume for a 147 moment that we deal with a one-dimensional problem then ; 148 gross 2477 \begin{equation} 149 \nabla = \frac{\partial}{\partial x} 150 \end{equation} 151 gross 2861 and we can write \refEq{eqn:commonform nabla} as; 152 ahallam 2411 \begin{equation}\label{eqn:commonform} 153 gross 2477 -A\frac{\partial^{2}u}{\partial x^{2}} + Du = f 154 ahallam 2411 \end{equation} 155 ahallam 2979 if $A$ is constant. To match this simplified general form to our problem 156 \refEq{eqn:hddisc} 157 gross 2861 we rearrange \refEq{eqn:hddisc}; 158 ahallam 2411 \begin{equation} 159 jfenwick 3308 \frac{\rho c_p}{h} T^{(n)} - \kappa \frac{\partial^2 T^{(n)}}{\partial 160 x^2} = q_H + \frac{\rho c_p}{h} T^{(n-1)} 161 ahallam 2494 \label{eqn:hdgenf} 162 \end{equation} 163 ahallam 2979 The PDE is now in a form that satisfies \refEq{eqn:commonform nabla} which is 164 required for \esc to solve our PDE. This can be done by generating a solution 165 for successive increments in the time nodes $t^{(n)}$ where 166 $t^{(0)}=0$ and $t^{(n)}=t^{(n-1)}+h$ where $h>0$ is the step size and assumed 167 to be constant. 168 In the following the upper index ${(n)}$ refers to a value at time $t^{(n)}$. 169 Finally, by comparing \refEq{eqn:hdgenf} with \refEq{eqn:commonform} one can see 170 that; 171 gross 2862 \begin{equation}\label{ESCRIPT SET} 172 gross 2861 u=T^{(n)}; 173 jfenwick 3308 A = \kappa; D = \frac{\rho c _{p}}{h}; f = q _{H} + \frac{\rho 174 c_p}{h} T^{(n-1)} 175 ahallam 2494 \end{equation} 176 177 ahallam 2495 \subsection{Boundary Conditions} 178 gross 2878 \label{SEC BOUNDARY COND} 179 ahallam 2979 With the PDE sufficiently modified, consideration must now be given to the 180 boundary conditions of our model. Typically there are two main types of boundary 181 conditions known as \textbf{Neumann} and \textbf{Dirichlet} boundary 182 conditions\footnote{More information on Boundary Conditions is available at 183 Wikipedia \url{http://en.wikipedia.org/wiki/Boundary_conditions}}, 184 respectively. 185 A \textbf{Dirichlet boundary condition} is conceptually simpler and is used to 186 prescribe a known value to the unknown solution (in our example the temperature) 187 on parts of the boundary or on the entire boundary of the region of interest. 188 caltinay 2982 We discuss the Dirichlet boundary condition in our second example presented in 189 ahallam 2979 Section~\ref{Sec:1DHDv0}. 190 ahallam 2495 191 ahallam 2979 However, for this example we have made the model assumption that the system is 192 caltinay 2982 insulated, so we need to add an appropriate boundary condition to prevent 193 ahallam 2979 any loss or inflow of energy at the boundary of our domain. Mathematically this 194 is expressed by prescribing 195 the heat flux $\kappa \frac{\partial T}{\partial x}$ to zero. In our simplified 196 one dimensional model this is expressed 197 gross 2862 in the form; 198 ahallam 2494 \begin{equation} 199 gross 2862 \kappa \frac{\partial T}{\partial x} = 0 200 ahallam 2494 \end{equation} 201 gross 2862 or in a more general case as 202 \begin{equation}\label{NEUMAN 1} 203 \kappa \nabla T \cdot n = 0 204 \end{equation} 205 ahallam 2979 where $n$ is the outer normal field \index{outer normal field} at the surface 206 of the domain. 207 caltinay 2982 The $\cdot$ (dot) refers to the dot product of the vectors $\nabla T$ and $n$. 208 ahallam 2979 In fact, the term $\nabla T \cdot n$ is the normal derivative of 209 the temperature $T$. Other notations used here are\footnote{The \esc notation 210 for the normal 211 jfenwick 3308 derivative is $T_{,i} n_i$.}; 212 ahallam 2645 \begin{equation} 213 gross 2862 \nabla T \cdot n = \frac{\partial T}{\partial n} \; . 214 ahallam 2645 \end{equation} 215 caltinay 2982 A condition of the type \refEq{NEUMAN 1} defines a \textbf{Neumann boundary 216 ahallam 2979 condition} for the PDE. 217 ahallam 2494 218 gross 2905 The PDE \refEq{eqn:hdgenf} 219 caltinay 2982 and the Neumann boundary condition~\ref{eqn:hdgenf} (potentially together with 220 ahallam 2979 the Dirichlet boundary conditions) define a \textbf{boundary value problem}. 221 It is the nature of a boundary value problem to allow making statements about 222 the solution in the 223 interior of the domain from information known on the boundary only. In most 224 caltinay 2982 cases we use the term partial differential equation but in fact it is a 225 boundary value problem. 226 ahallam 2979 It is important to keep in mind that boundary conditions need to be complete and 227 consistent in the sense that 228 caltinay 2982 at any point on the boundary either a Dirichlet or a Neumann boundary condition 229 ahallam 2979 must be set. 230 gross 2862 231 caltinay 2982 Conveniently, \esc makes a default assumption on the boundary conditions which 232 the user may modify where appropriate. 233 ahallam 2979 For a problem of the form in~\refEq{eqn:commonform nabla} the default 234 caltinay 2982 condition\footnote{In the \esc user guide which uses the Einstein convention 235 this is written as 236 jfenwick 3308 $n_{j}A_{jl} u_{,l}=0$.} is; 237 gross 2862 \begin{equation}\label{NEUMAN 2} 238 gross 2948 -n\cdot A \cdot\nabla u = 0 239 gross 2862 \end{equation} 240 ahallam 2979 which is used everywhere on the boundary. Again $n$ denotes the outer normal 241 field. 242 Notice that the coefficient $A$ is the same as in the \esc 243 PDE~\ref{eqn:commonform nabla}. 244 With the settings for the coefficients we have already identified in 245 \refEq{ESCRIPT SET} this 246 gross 2862 condition translates into 247 gross 2867 \begin{equation}\label{NEUMAN 2b} 248 gross 2862 \kappa \frac{\partial T}{\partial x} = 0 249 \end{equation} 250 caltinay 2982 for the boundary of the domain. This is identical to the Neumann boundary 251 ahallam 2979 condition we want to set. \esc will take care of this condition for us. We 252 discuss the Dirichlet boundary condition later. 253 gross 2862 254 gross 2870 \subsection{Outline of the Implementation} 255 \label{sec:outline} 256 ahallam 2979 To solve the heat diffusion equation (\refEq{eqn:hd}) we write a simple \pyt 257 script. At this point we assume that you have some basic understanding of the 258 \pyt programming language. If not, there are some pointers and links available 259 in Section \ref{sec:escpybas}. The script (discussed in \refSec{sec:key}) has 260 four major steps. Firstly, we need to define the domain where we want to 261 calculate the temperature. For our problem this is the joint blocks of granite 262 which has a rectangular shape. Secondly, we need to define the PDE to solve in 263 each time step to get the updated temperature. Thirdly, we need to define the 264 coefficients of the PDE and finally we need to solve the PDE. The last two steps 265 need to be repeated until the final time marker has been reached. The work flow 266 caltinay 2982 is described in \reffig{fig:wf}. 267 ahallam 2975 % \begin{enumerate} 268 % \item create domain 269 % \item create PDE 270 % \item while end time not reached: 271 % \begin{enumerate} 272 % \item set PDE coefficients 273 % \item solve PDE 274 % \item update time marker 275 % \end{enumerate} 276 % \item end of calculation 277 % \end{enumerate} 278 279 \begin{figure}[h!] 280 \centering 281 \includegraphics[width=1in]{figures/workflow.png} 282 caltinay 2982 \caption{Workflow for developing an \esc model and solution} 283 ahallam 2975 \label{fig:wf} 284 \end{figure} 285 286 ahallam 2979 In the terminology of \pyt, the domain and PDE are represented by 287 \textbf{objects}. The nice feature of an object is that it is defined by its 288 usage and features 289 rather than its actual representation. So we will create a domain object to 290 describe the geometry of the two 291 granite blocks. Then we define PDEs and spatially distributed values such as the 292 temperature 293 on this domain. Similarly, to define a PDE object we use the fact that one needs 294 only to define the coefficients of the PDE and solve the PDE. The PDE object has 295 advanced features, but these are not required in simple cases. 296 gross 2870 297 298 ahallam 3029 \begin{figure}[htp] 299 gross 2870 \centering 300 \includegraphics[width=6in]{figures/functionspace.pdf} 301 ahallam 2975 \caption{\esc domain construction overview} 302 gross 2870 \label{fig:fs} 303 \end{figure} 304 305 \subsection{The Domain Constructor in \esc} 306 \label{ss:domcon} 307 ahallam 2979 Whilst it is not strictly relevant or necessary, a better understanding of 308 caltinay 2982 how values are spatially distributed (\textit{e.g.} Temperature) and how PDE 309 ahallam 2979 coefficients are interpreted in \esc can be helpful. 310 gross 2870 311 ahallam 2979 There are various ways to construct domain objects. The simplest form is a 312 rectangular shaped region with a length and height. There is 313 a ready to use function for this named \verb rectangle(). Besides the spatial 314 dimensions this function requires to specify the number of 315 elements or cells to be used along the length and height, see \reffig{fig:fs}. 316 Any spatially distributed value 317 and the PDE is represented in discrete form using this element 318 representation\footnote{We use the finite element method (FEM), see 319 \url{http://en.wikipedia.org/wiki/Finite_element_method} for details.}. 320 Therefore we will have access to an approximation of the true PDE solution 321 only. 322 caltinay 2982 The quality of the approximation depends - besides other factors - mainly on the 323 ahallam 2979 number of elements being used. In fact, the 324 approximation becomes better when more elements are used. However, computational 325 cost grows with the number of 326 elements being used. It is therefore important that you find the right balance 327 between the demand in accuracy and acceptable resource usage. 328 gross 2870 329 ahallam 2979 In general, one can think about a domain object as a composition of nodes and 330 elements. 331 As shown in \reffig{fig:fs}, an element is defined by the nodes that are used to 332 describe its vertices. 333 caltinay 2982 To represent spatially distributed values the user can use 334 ahallam 2979 the values at the nodes, at the elements in the interior of the domain or at the 335 caltinay 2982 elements located on the surface of the domain. 336 ahallam 2979 The different approach used to represent values is called \textbf{function 337 space} and is attached to all objects 338 caltinay 2982 in \esc representing a spatially distributed value such as the solution of 339 a PDE. The three function spaces we use at the moment are; 340 gross 2870 \begin{enumerate} 341 \item the nodes, called by \verb|ContinuousFunction(domain)| ; 342 \item the elements/cells, called by \verb|Function(domain)| ; and 343 caltinay 2982 \item the boundary, called by \verb|FunctionOnBoundary(domain)|. 344 gross 2870 \end{enumerate} 345 ahallam 2979 A function space object such as \verb|ContinuousFunction(domain)| has the method 346 \verb|getX| attached to it. This method returns the 347 location of the so-called \textbf{sample points} used to represent values of the 348 particular function space. So the 349 call \verb|ContinuousFunction(domain).getX()| will return the coordinates of the 350 nodes used to describe the domain while 351 caltinay 2982 \verb|Function(domain).getX()| returns the coordinates of numerical 352 integration points within elements, see \reffig{fig:fs}. 353 gross 2870 354 ahallam 2979 This distinction between different representations of spatially distributed 355 values 356 is important in order to be able to vary the degrees of smoothness in a PDE 357 problem. 358 The coefficients of a PDE do not need to be continuous, thus this qualifies as a 359 \verb|Function()| type. 360 On the other hand a temperature distribution must be continuous and needs to be 361 represented with a \verb|ContinuousFunction()| function space. 362 caltinay 2982 An influx may only be defined at the boundary and is therefore a 363 \verb|FunctionOnBoundary()| object. 364 \esc allows certain transformations of the function spaces. A 365 \verb|ContinuousFunction()| can be transformed into a 366 \verb|FunctionOnBoundary()| or \verb|Function()|. On the other hand there is 367 not enough information in a \verb|FunctionOnBoundary()| to transform it to a 368 \verb|ContinuousFunction()|. 369 ahallam 2979 These transformations, which are called \textbf{interpolation} are invoked 370 automatically by \esc if needed. 371 gross 2870 372 artak 2963 Later in this introduction we discuss how 373 ahallam 2979 to define specific areas of geometry with different materials which are 374 caltinay 2982 represented by different material coefficients such as the 375 ahallam 2979 thermal conductivities $\kappa$. A very powerful technique to define these types 376 of PDE 377 coefficients is tagging. Blocks of materials and boundaries can be named and 378 values can be defined on subregions based on their names. 379 This is a method for simplifying PDE coefficient and flux definitions. It makes 380 scripting much easier and we will discuss this technique in 381 Section~\ref{STEADY-STATE HEAT REFRACTION}. 382 gross 2870 383 384 \subsection{A Clarification for the 1D Case} 385 gross 2931 \label{SEC: 1D CLARIFICATION} 386 ahallam 2979 It is necessary for clarification that we revisit our general PDE from 387 caltinay 2982 \refeq{eqn:commonform nabla} for a two dimensional domain. \esc is inherently 388 designed to solve problems that are multi-dimensional and so 389 ahallam 2979 \refEq{eqn:commonform nabla} needs to be read as a higher dimensional problem. 390 In the case of two spatial dimensions the \textit{Nabla operator} has in fact 391 two components $\nabla = (\frac{\partial}{\partial x}, \frac{\partial}{\partial 392 y})$. Assuming the coefficient $A$ is constant, the \refEq{eqn:commonform nabla} 393 takes the following form; 394 gross 2477 \begin{equation}\label{eqn:commonform2D} 395 jfenwick 3308 -A_{00}\frac{\partial^{2}u}{\partial x^{2}} 396 -A_{01}\frac{\partial^{2}u}{\partial x\partial y} 397 -A_{10}\frac{\partial^{2}u}{\partial y\partial x} 398 -A_{11}\frac{\partial^{2}u}{\partial y^{2}} 399 gross 2477 + Du = f 400 \end{equation} 401 ahallam 2606 Notice that for the higher dimensional case $A$ becomes a matrix. It is also 402 ahallam 2495 important to notice that the usage of the Nabla operator creates 403 a compact formulation which is also independent from the spatial dimension. 404 artak 2963 To make the general PDE \refEq{eqn:commonform2D} one dimensional as 405 gross 2861 shown in \refEq{eqn:commonform} we need to set 406 ahallam 2606 \begin{equation} 407 jfenwick 3308 A_{00}=A; A_{01}=A_{10}=A_{11}=0 408 gross 2477 \end{equation} 409 410 gross 2867 411 ahallam 2495 \subsection{Developing a PDE Solution Script} 412 ahallam 2801 \label{sec:key} 413 gross 2949 \sslist{example01a.py} 414 ahallam 2979 We write a simple \pyt script which uses the \modescript, \modfinley and \modmpl 415 modules. 416 By developing a script for \esc, the heat diffusion equation can be solved at 417 successive time steps for a predefined period using our general form 418 \refEq{eqn:hdgenf}. Firstly it is necessary to import all the 419 libraries\footnote{The libraries contain predefined scripts that are required to 420 solve certain problems, these can be simple like sine and cosine functions or 421 more complicated like those from our \esc library.} 422 ahallam 2495 that we will require. 423 ahallam 2775 \begin{python} 424 ahallam 2495 from esys.escript import * 425 ahallam 2606 # This defines the LinearPDE module as LinearPDE 426 from esys.escript.linearPDEs import LinearPDE 427 # This imports the rectangle domain function from finley. 428 from esys.finley import Rectangle 429 # A useful unit handling package which will make sure all our units 430 # match up in the equations under SI. 431 from esys.escript.unitsSI import * 432 ahallam 2775 \end{python} 433 ahallam 2979 It is generally a good idea to import all of the \modescript library, although 434 if the functions and classes required are known they can be specified 435 individually. The function \verb|LinearPDE| has been imported explicitly for 436 ease of use later in the script. \verb|Rectangle| is going to be our type of 437 caltinay 2982 domain. The module \verb|unitsSI| provides support for SI unit definitions with 438 ahallam 2979 our variables. 439 gross 2477 440 ahallam 2979 Once our library dependencies have been established, defining the problem 441 specific variables is the next step. In general the number of variables needed 442 will vary between problems. These variables belong to two categories. They are 443 either directly related to the PDE and can be used as inputs into the \esc 444 solver, or they are script variables used to control internal functions and 445 iterations in our problem. For this PDE there are a number of constants which 446 need values. Firstly, the domain upon which we wish to solve our problem needs 447 to be defined. There are different types of domains in \modescript which we 448 demonstrate in later tutorials but for our granite blocks, we simply use a 449 rectangular domain. 450 ahallam 2401 451 ahallam 2979 Using a rectangular domain simplifies our granite blocks (which would in reality 452 be a \textit{3D} object) into a single dimension. The granite blocks will have a 453 lengthways cross section that looks like a rectangle. As a result we do not 454 need to model the volume of the block due to symmetry. There are four arguments 455 we must consider when we decide to create a rectangular domain, the domain 456 \textit{length}, \textit{width} and \textit{step size} in each direction. When 457 defining the size of our problem it will help us determine appropriate values 458 for our model arguments. If we make our dimensions large but our step sizes very 459 small we increase the accuracy of our solution. Unfortunately we also increase 460 the number of calculations that must be solved per time step. This means more 461 computational time is required to produce a solution. In this \textit{1D} 462 problem, the bar is defined as being 1 metre long. An appropriate step size 463 caltinay 2982 \verb|ndx| would be 1 to 10\% of the length. Our \verb|ndy| needs only be 1, 464 this is because our problem stipulates no partial derivatives in the $y$ 465 direction. 466 ahallam 2979 Thus the temperature does not vary with $y$. Hence, the model parameters can be 467 caltinay 2982 defined as follows; note we have used the \verb|unitsSI| convention to make sure 468 ahallam 2979 all our input units are converted to SI. 469 ahallam 2775 \begin{python} 470 gross 2905 mx = 500.*m #meters - model length 471 my = 100.*m #meters - model width 472 ndx = 50 # mesh steps in x direction 473 ndy = 1 # mesh steps in y direction 474 boundloc = mx/2 # location of boundary between the two blocks 475 ahallam 2775 \end{python} 476 ahallam 2979 The material constants and the temperature variables must also be defined. For 477 the granite in the model they are defined as: 478 ahallam 2775 \begin{python} 479 ahallam 2495 #PDE related 480 gross 2905 rho = 2750. *kg/m**3 #kg/m^{3} density of iron 481 cp = 790.*J/(kg*K) # J/Kg.K thermal capacity 482 gross 2878 rhocp = rho*cp 483 gross 2905 kappa = 2.2*W/m/K # watts/m.Kthermal conductivity 484 gross 2878 qH=0 * J/(sec*m**3) # J/(sec.m^{3}) no heat source 485 gross 2905 T1=20 * Celsius # initial temperature at Block 1 486 T2=2273. * Celsius # base temperature at Block 2 487 ahallam 2775 \end{python} 488 ahallam 2979 Finally, to control our script we will have to specify our timing controls and 489 where we would like to save the output from the solver. This is simple enough: 490 ahallam 2775 \begin{python} 491 gross 2878 t=0 * day #our start time, usually zero 492 tend=1. * day # - time to end simulation 493 ahallam 2495 outputs = 200 # number of time steps required. 494 h=(tend-t)/outputs #size of time step 495 ahallam 2606 #user warning statement 496 print "Expected Number of time outputs is: ", (tend-t)/h 497 i=0 #loop counter 498 ahallam 2775 \end{python} 499 ahallam 2979 Now that we know our inputs we will build a domain using the 500 caltinay 2982 \verb|Rectangle()| function from \FINLEY. The four arguments allow us to 501 define our domain \verb|model| as: 502 ahallam 2775 \begin{python} 503 ahallam 2606 #generate domain using rectangle 504 gross 2905 blocks = Rectangle(l0=mx,l1=my,n0=ndx, n1=ndy) 505 ahallam 2775 \end{python} 506 caltinay 2982 \verb|blocks| now describes a domain in the manner of Section \ref{ss:domcon}. 507 ahallam 2658 508 caltinay 2982 With a domain and all the required variables established, it is now possible to 509 ahallam 2979 set up our PDE so that it can be solved by \esc. The first step is to define the 510 type of PDE that we are trying to solve in each time step. In this example it is 511 caltinay 2982 a single linear PDE\footnote{in contrast to a system of PDEs which we discuss 512 ahallam 2979 later.}. We also need to state the values of our general form variables. 513 ahallam 2775 \begin{python} 514 gross 2905 mypde=LinearPDE(blocks) 515 gross 2878 A=zeros((2,2))) 516 A[0,0]=kappa 517 gross 2905 mypde.setValue(A=A, D=rhocp/h) 518 ahallam 2775 \end{python} 519 caltinay 2982 In many cases it may be possible to decrease the computational time of the 520 ahallam 2979 solver if the PDE is symmetric. 521 gross 2878 Symmetry of a PDE is defined by; 522 ahallam 2495 \begin{equation}\label{eqn:symm} 523 jfenwick 3308 A_{jl}=A_{lj} 524 ahallam 2495 \end{equation} 525 ahallam 2979 Symmetry is only dependent on the $A$ coefficient in the general form and the 526 other coefficients $D$ as well as the right hand side $Y$. From the above 527 caltinay 2982 definition we can see that our PDE is symmetric. The \verb|LinearPDE| class 528 ahallam 2979 provides the method \method{checkSymmetry} to check if the given PDE is 529 symmetric. As our PDE is symmetrical we enable symmetry via; 530 ahallam 2775 \begin{python} 531 caltinay 2982 myPDE.setSymmetryOn() 532 ahallam 2775 \end{python} 533 ahallam 2979 Next we need to establish the initial temperature distribution \verb|T|. We need 534 to 535 assign the value \verb|T1| to all sample points left to the contact interface at 536 jfenwick 3308 $x_{0}=\frac{mx}{2}$ 537 gross 2905 and the value \verb|T2| right to the contact interface. \esc 538 caltinay 2982 provides the \verb|whereNegative| function to construct this. More 539 specifically, \verb|whereNegative| returns the value $1$ at those sample points 540 where the argument has a negative value. Otherwise zero is returned. 541 jfenwick 3308 If \verb|x| are the $x_{0}$ 542 gross 2905 coordinates of the sample points used to represent the temperature distribution 543 then \verb|x[0]-boundloc| gives us a negative value for 544 all sample points left to the interface and non-negative value to 545 the right of the interface. So with; 546 gross 2878 \begin{python} 547 # ... set initial temperature .... 548 gross 2905 T= T1*whereNegative(x[0]-boundloc)+T2*(1-whereNegative(x[0]-boundloc)) 549 gross 2878 \end{python} 550 ahallam 2979 we get the desired temperature distribution. To get the actual sample points 551 caltinay 2982 \verb|x| we use the \verb|getX()| method of the function space 552 \verb|Solution(blocks)| which is used to represent the solution of a PDE; 553 gross 2878 \begin{python} 554 gross 2905 x=Solution(blocks).getX() 555 \end{python} 556 ahallam 2979 As \verb|x| are the sample points for the function space 557 \verb|Solution(blocks)| 558 the initial temperature \verb|T| is using these sample points for 559 representation. 560 Although \esc is trying to be forgiving with the choice of sample points and to 561 caltinay 2982 convert 562 ahallam 2979 where necessary the adjustment of the function space is not always possible. So 563 caltinay 2982 it is advisable to make a careful choice on the function space used. 564 gross 2905 565 ahallam 2979 Finally we initialise an iteration loop to solve our PDE for all the time steps 566 we specified in the variable section. As the right hand side of the general form 567 is dependent on the previous values for temperature \verb T across the bar this 568 must be updated in the loop. Our output at each time step is \verb T the heat 569 distribution and \verb totT the total heat in the system. 570 gross 2905 \begin{python} 571 gross 2878 while t < tend: 572 i+=1 #increment the counter 573 t+=h #increment the current time 574 mypde.setValue(Y=qH+rhocp/h*T) # set variable PDE coefficients 575 T=mypde.getSolution() #get the PDE solution 576 totE = integrate(rhocp*T) #get the total heat (energy) in the system 577 \end{python} 578 ahallam 2979 The last statement in this script calculates the total energy in the system as 579 jfenwick 3308 the volume integral of $\rho c_{p} T$ over the block. 580 caltinay 2982 As the blocks are insulated no energy should be lost or added. 581 gross 2905 The total energy should stay constant for the example discussed here. 582 ahallam 2401 583 gross 2905 \subsection{Running the Script} 584 ahallam 2975 The script presented so far is available under 585 gross 2949 \verb|example01a.py|. You can edit this file with your favourite text editor. 586 caltinay 2982 On most operating systems\footnote{The \texttt{run-escript} launcher is not 587 supported under {\it MS Windows} yet.} you can use the 588 ahallam 2979 \program{run-escript} command 589 gross 2905 to launch {\it escript} scripts. For the example script use; 590 \begin{verbatim} 591 gross 2949 run-escript example01a.py 592 gross 2905 \end{verbatim} 593 The program will print a progress report. Alternatively, you can use 594 the python interpreter directly; 595 \begin{verbatim} 596 gross 2949 python example01a.py 597 gross 2905 \end{verbatim} 598 caltinay 2982 if the system is configured correctly (please talk to your system 599 ahallam 2979 administrator). 600 gross 2905 601 gross 2878 \subsection{Plotting the Total Energy} 602 gross 2949 \sslist{example01b.py} 603 gross 2878 604 ahallam 2979 \esc does not include its own plotting capabilities. However, it is possible to 605 use a variety of free \pyt packages for visualisation. 606 Two types will be demonstrated in this cookbook; 607 \mpl\footnote{\url{http://matplotlib.sourceforge.net/}} and 608 caltinay 2982 \verb|VTK|\footnote{\url{http://www.vtk.org/}}. 609 ahallam 2979 The \mpl package is a component of SciPy\footnote{\url{http://www.scipy.org}} 610 and is good for basic graphs and plots. 611 caltinay 2982 For more complex visualisation tasks, in particular two and three dimensional 612 ahallam 2979 problems we recommend the use of more advanced tools. For instance, \mayavi 613 \footnote{\url{http://code.enthought.com/projects/mayavi/}} 614 ahallam 2975 which is based upon the \verb|VTK| toolkit. The usage of \verb|VTK| based 615 caltinay 2982 visualisation is discussed in Chapter~\ref{Sec:2DHD} which focuses on a two 616 ahallam 2979 dimensional PDE. 617 gross 2878 618 ahallam 2979 For our simple granite block problem, we have two plotting tasks. Firstly, we 619 are interested in showing the 620 behaviour of the total energy over time and secondly, how the temperature 621 caltinay 2982 distribution within the block is developing over time. 622 Let us start with the first task. 623 gross 2878 624 caltinay 2982 The idea is to create a record of the time marks and the corresponding total 625 ahallam 2979 energies observed. 626 gross 2878 \pyt provides the concept of lists for this. Before 627 ahallam 2979 the time loop is opened we create empty lists for the time marks \verb|t_list| 628 and the total energies \verb|E_list|. 629 After the new temperature has been calculated by solving the PDE we append the 630 new time marker and the total energy value for that time 631 to the corresponding list using the \verb|append| method. With these 632 modifications our script looks as follows: 633 ahallam 2775 \begin{python} 634 gross 2878 t_list=[] 635 E_list=[] 636 # ... start iteration: 637 while t

 ViewVC Help Powered by ViewVC 1.1.26