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Hacking cookbook


 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 2905 \begin{figure}[h!] 15 \centerline{\includegraphics[width=4.in]{figures/onedheatdiff001}} 16 gross 2952 \caption{Example 1: Temperature differential along a single interface between two granite blocks.} 17 gross 2905 \label{fig:onedgbmodel} 18 \end{figure} 19 ahallam 2801 20 gross 2949 \section{Example 1: One Dimensional Heat Diffusion in Granite} 21 gross 2905 \label{Sec:1DHDv00} 22 gross 2878 23 gross 2905 The first model consists of two blocks of isotropic material, for instance granite, sitting next to each other. 24 artak 2957 Initial temperature in \textit{Block 1} is \verb|T1| and in \textit{Block 2} is \verb|T2|. 25 gross 2905 We assume that the system is insulated. 26 What would happen to the temperature distribution in each block over time? 27 artak 2958 Intuition tells us that heat will be transported from the hotter block to the cooler one until both 28 gross 2905 blocks have the same temperature. 29 30 ahallam 2495 \subsection{1D Heat Diffusion Equation} 31 artak 2957 We can model the heat distribution of this problem over time using one dimensional heat diffusion equation\footnote{A detailed discussion on how the heat diffusion equation is derived can be found at \url{http://online.redwoods.edu/instruct/darnold/DEProj/sp02/AbeRichards/paper.pdf}}; 32 ahallam 2494 which is defined as: 33 ahallam 2401 \begin{equation} 34 \rho c\hackscore p \frac{\partial T}{\partial t} - \kappa \frac{\partial^{2} T}{\partial x^{2}} = q\hackscore H 35 \label{eqn:hd} 36 \end{equation} 37 gross 2861 where $\rho$ is the material density, $c\hackscore p$ is the specific heat and $\kappa$ is the thermal 38 conductivity\footnote{A list of some common thermal conductivities is available from Wikipedia \url{http://en.wikipedia.org/wiki/List_of_thermal_conductivities}}. Here we assume that these material 39 parameters are \textbf{constant}. 40 The heat source is defined by the right hand side of \refEq{eqn:hd} as $q\hackscore{H}$; this can take the form of a constant or a function of time and space. For example $q\hackscore{H} = q\hackscore{0}e^{-\gamma t}$ where we have the output of our heat source decaying with time. There are also two partial derivatives in \refEq{eqn:hd}; $\frac{\partial T}{\partial t}$ describes the change in temperature with time while $\frac{\partial ^2 T}{\partial x^2}$ is the spatial change of temperature. As there is only a single spatial dimension to our problem, our temperature solution $T$ is only dependent on the time $t$ and our position along the iron bar $x$. 41 ahallam 2401 42 gross 2861 \subsection{PDEs and the General Form} 43 artak 2958 Potentially, it is possible to solve PDE \refEq{eqn:hd} analytically and obtain an exact solution to our problem. However, it is not always possible or practical to solve a problem this way. Alternatively, computers can be used to solve these kinds of problems. To do this, a numerical approach is required to discretise 44 the PDE \refEq{eqn:hd} in time and space, then we left with a finite number of equations for a finite number of spatial points and time steps in the model. While discretisation introduces approximations and a degree of error, a sufficiently sampled model is generally accurate enough to satisfy the requirements of the modeller. 45 gross 2477 46 artak 2958 Firstly, we discretise the PDE \refEq{eqn:hd} in time. This leaves us with a steady linear PDE which involves spatial derivatives only and needs to be solved in each time step to progress in time. \esc can help us here. 47 gross 2861 48 artak 2958 For time discretization we use the Backwards Euler approximation scheme\footnote{see \url{http://en.wikipedia.org/wiki/Euler_method}}. It is based on the 49 gross 2861 approximation 50 \begin{equation} 51 \frac{\partial T(t)}{\partial t} \approx \frac{T(t)-T(t-h)}{h} 52 \label{eqn:beuler} 53 \end{equation} 54 for $\frac{\partial T}{\partial t}$ at time $t$ 55 where $h$ is the time step size. This can also be written as; 56 \begin{equation} 57 \frac{\partial T}{\partial t}(t^{(n)}) \approx \frac{T^{(n)} - T^{(n-1)}}{h} 58 \label{eqn:Tbeuler} 59 \end{equation} 60 where the upper index $n$ denotes the n\textsuperscript{th} time step. So one has 61 \begin{equation} 62 \begin{array}{rcl} 63 t^{(n)} & = & t^{(n-1)}+h \\ 64 T^{(n)} & = & T(t^{(n-1)}) \\ 65 \end{array} 66 \label{eqn:Neuler} 67 \end{equation} 68 Substituting \refEq{eqn:Tbeuler} into \refEq{eqn:hd} we get; 69 \begin{equation} 70 \frac{\rho c\hackscore p}{h} (T^{(n)} - T^{(n-1)}) - \kappa \frac{\partial^{2} T^{(n)}}{\partial x^{2}} = q\hackscore H 71 \label{eqn:hddisc} 72 \end{equation} 73 artak 2958 Notice that we evaluate the spatial derivative term at current time $t^{(n)}$ - therefore the name \textbf{backward Euler} scheme. Alternatively, one can evaluate the spatial derivative term at the previous time $t^{(n-1)}$. This 74 approach is called the \textbf{forward Euler} scheme. This scheme can provide some computational advantages, which 75 are not discussed here. However, this scheme has a major disadvantage, namely depending on the 76 material parameter as well as the discretization of the spatial derivative term, the time step size $h$ needs to be chosen sufficiently small to achieve a stable temperature when progressing in time. The term \textit{stable} means 77 that the approximation of the temperature will not grow beyond its initial bounds and become non-physical. 78 The backward Euler scheme, which we use here, is unconditionally stable meaning that under the assumption of 79 physically correct problem set-up the temperature approximation remains physical for all time steps. 80 gross 2861 The user needs to keep in mind that the discretization error introduced by \refEq{eqn:beuler} 81 artak 2958 is sufficiently small, thus a good approximation of the true temperature is computed. It is 82 gross 2861 therefore crucial that the user remains critical about his/her results and for instance compares 83 the results for different time and spatial step sizes. 84 85 To get the temperature $T^{(n)}$ at time $t^{(n)}$ we need to solve the linear 86 artak 2958 differential equation \refEq{eqn:hddisc} which only includes spatial derivatives. To solve this problem 87 we want to use \esc. 88 gross 2861 89 artak 2958 In \esc any given PDE described by general form. For the purpose of this introduction we illustrate a simpler version of the general form for full linear PDEs which is available in the \esc user's guide. A simplified form that suits our heat diffusion problem\footnote{The form in the \esc users guide which uses the Einstein convention is written as 90 gross 2477 $-(A\hackscore{jl} u\hackscore{,l})\hackscore{,j}+D u =Y$} 91 ahallam 2606 is described by; 92 gross 2477 \begin{equation}\label{eqn:commonform nabla} 93 jfenwick 2657 -\nabla\cdot(A\cdot\nabla u) + Du = f 94 ahallam 2411 \end{equation} 95 gross 2861 where $A$, $D$ and $f$ are known values and $u$ is the unknown solution. The symbol $\nabla$ which is called the \textit{Nabla operator} or \textit{del operator} represents 96 ahallam 2495 the spatial derivative of its subject - in this case $u$. Lets assume for a moment that we deal with a one-dimensional problem then ; 97 gross 2477 \begin{equation} 98 \nabla = \frac{\partial}{\partial x} 99 \end{equation} 100 gross 2861 and we can write \refEq{eqn:commonform nabla} as; 101 ahallam 2411 \begin{equation}\label{eqn:commonform} 102 gross 2477 -A\frac{\partial^{2}u}{\partial x^{2}} + Du = f 103 ahallam 2411 \end{equation} 104 gross 2861 if $A$ is constant. To match this simplified general form to our problem \refEq{eqn:hddisc} 105 we rearrange \refEq{eqn:hddisc}; 106 ahallam 2411 \begin{equation} 107 ahallam 2645 \frac{\rho c\hackscore p}{h} T^{(n)} - \kappa \frac{\partial^2 T^{(n)}}{\partial x^2} = q\hackscore H + \frac{\rho c\hackscore p}{h} T^{(n-1)} 108 ahallam 2494 \label{eqn:hdgenf} 109 \end{equation} 110 ahallam 2775 The PDE is now in a form that satisfies \refEq{eqn:commonform nabla} which is required for \esc to solve our PDE. This can be done by generating a solution for successive increments in the time nodes $t^{(n)}$ where 111 ahallam 2495 $t^{(0)}=0$ and $t^{(n)}=t^{(n-1)}+h$ where $h>0$ is the step size and assumed to be constant. 112 artak 2958 In the following the upper index ${(n)}$ refers to a value at time $t^{(n)}$. Finally, by comparing \refEq{eqn:hdgenf} with \refEq{eqn:commonform} one can see that; 113 gross 2862 \begin{equation}\label{ESCRIPT SET} 114 gross 2861 u=T^{(n)}; 115 ahallam 2494 A = \kappa; D = \frac{\rho c \hackscore{p}}{h}; f = q \hackscore{H} + \frac{\rho c\hackscore p}{h} T^{(n-1)} 116 \end{equation} 117 118 ahallam 2495 \subsection{Boundary Conditions} 119 gross 2878 \label{SEC BOUNDARY COND} 120 gross 2862 With the PDE sufficiently modified, consideration must now be given to the boundary conditions of our model. Typically there are two main types of boundary conditions known as \textbf{Neumann} and \textbf{Dirichlet} boundary conditions\footnote{More information on Boundary Conditions is available at Wikipedia \url{http://en.wikipedia.org/wiki/Boundary_conditions}}, respectively. 121 artak 2958 A \textbf{Dirichlet boundary condition} is conceptually simpler and is used to prescribe a known value to the unknown solution (in our example the temperature) on parts of the boundary or on the entire boundary of the region of interest. 122 gross 2905 We discuss Dirichlet boundary condition in our second example presented in Section~\ref{Sec:1DHDv0}. 123 ahallam 2495 124 gross 2905 We make the model assumption that the system is insulated so we need 125 to add an appropriate boundary condition to prevent 126 any loss or inflow of energy at boundary of our domain. Mathematically this is expressed by prescribing 127 the heat flux $\kappa \frac{\partial T}{\partial x}$ to zero. In our simplified one dimensional model this is expressed 128 gross 2862 in the form; 129 ahallam 2494 \begin{equation} 130 gross 2862 \kappa \frac{\partial T}{\partial x} = 0 131 ahallam 2494 \end{equation} 132 gross 2862 or in a more general case as 133 \begin{equation}\label{NEUMAN 1} 134 \kappa \nabla T \cdot n = 0 135 \end{equation} 136 where $n$ is the outer normal field \index{outer normal field} at the surface of the domain. 137 gross 2905 The $\cdot$ (dot) refers to the dot product of the vectors $\nabla T$ and $n$. In fact, the term $\nabla T \cdot n$ is the normal derivative of 138 artak 2958 the temperature $T$. Other notations used here are\footnote{The \esc notation for the normal 139 gross 2862 derivative is $T\hackscore{,i} n\hackscore i$.}; 140 ahallam 2645 \begin{equation} 141 gross 2862 \nabla T \cdot n = \frac{\partial T}{\partial n} \; . 142 ahallam 2645 \end{equation} 143 gross 2862 A condition of the type \refEq{NEUMAN 1} defines a \textbf{Neuman boundary condition} for the PDE. 144 ahallam 2494 145 gross 2905 The PDE \refEq{eqn:hdgenf} 146 artak 2958 and the Neuman boundary condition~\ref{eqn:hdgenf} (potentially together with the Dirichlet boundary conditions) define a \textbf{boundary value problem}. 147 It is the nature of a boundary value problem to allow making statements about the solution in the 148 gross 2862 interior of the domain from information known on the boundary only. In most cases 149 artak 2958 we use the term partial differential equation but in fact it is a boundary value problem. 150 gross 2862 It is important to keep in mind that boundary conditions need to be complete and consistent in the sense that 151 at any point on the boundary either a Dirichlet or a Neuman boundary condition must be set. 152 153 gross 2878 Conveniently, \esc makes default assumption on the boundary conditions which the user may modify where appropriate. 154 gross 2862 For a problem of the form in~\refEq{eqn:commonform nabla} the default condition\footnote{In the form of the \esc users guide which is using the Einstein convention is written as 155 $n\hackscore{j}A\hackscore{jl} u\hackscore{,l}=0$.} is; 156 \begin{equation}\label{NEUMAN 2} 157 gross 2948 -n\cdot A \cdot\nabla u = 0 158 gross 2862 \end{equation} 159 which is used everywhere on the boundary. Again $n$ denotes the outer normal field. 160 Notice that the coefficient $A$ is the same as in the \esc PDE~\ref{eqn:commonform nabla}. 161 With the settings for the coefficients we have already identified in \refEq{ESCRIPT SET} this 162 condition translates into 163 gross 2867 \begin{equation}\label{NEUMAN 2b} 164 gross 2862 \kappa \frac{\partial T}{\partial x} = 0 165 \end{equation} 166 artak 2958 for the boundary of the domain. This is identical to the Neuman boundary condition we want to set. \esc will take care of this condition for us. We discuss the Dirichlet boundary condition later. 167 gross 2862 168 gross 2870 \subsection{Outline of the Implementation} 169 \label{sec:outline} 170 artak 2958 To solve the heat diffusion equation (equation \refEq{eqn:hd}) we write a simple \pyt script. At this point we assume that you have some basic understanding of the \pyt programming language. If not, there are some pointers and links available in Section \ref{sec:escpybas}. The script we discuss later in details have four major steps. Firstly, we need to define the domain where we want to 171 calculate the temperature. For our problem this is the joint blocks of granite which has a rectangular shape. Secondly, we need to define the PDE to solve in each time step to get the updated temperature. Thirdly, we need to define the coefficients of the PDE and finally we need to solve the PDE. The last two steps need to be repeated until the final time marker has been reached. The work flow is as follows: 172 gross 2878 \begin{enumerate} 173 \item create domain 174 \item create PDE 175 \item while end time not reached: 176 \begin{enumerate} 177 \item set PDE coefficients 178 \item solve PDE 179 \item update time marker 180 \end{enumerate} 181 \item end of calculation 182 \end{enumerate} 183 artak 2959 In the terminology of \pyt, the domain and PDE are represented by \textbf{objects}. The nice feature of an object is that it is defined by its usage and features 184 gross 2905 rather than its actual representation. So we will create a domain object to describe the geometry of the two 185 artak 2959 granite blocks. Then we define PDEs and spatially distributed values such as the temperature 186 on this domain. Similarly, to define a PDE object we use the fact that one needs only to define the coefficients of the PDE and solve the PDE. The PDE object has advanced features, but these are not required in simple cases. 187 gross 2870 188 189 \begin{figure}[t] 190 \centering 191 \includegraphics[width=6in]{figures/functionspace.pdf} 192 \label{fig:fs} 193 \caption{\esc domain construction overview} 194 \end{figure} 195 196 \subsection{The Domain Constructor in \esc} 197 \label{ss:domcon} 198 It is helpful to have a better understanding how spatially distributed value such as the temperature or PDE coefficients are interpreted in \esc. Again 199 from the user's point of view the representation of these spatially distributed values is not relevant. 200 201 There are various ways to construct domain objects. The simplest form is as rectangular shaped region with a length and height. There is 202 a ready to use function call for this. Besides the spatial dimensions the function call will require you to specify the number 203 gross 2905 elements or cells to be used along the length and height, see \reffig{fig:fs}. Any spatially distributed value 204 and the PDE is represented in discrete form using this element representation\footnote{We will use the finite element method (FEM), see \url{http://en.wikipedia.org/wiki/Finite_element_method} for details.}. Therefore we will have access to an approximation of the true PDE solution only. 205 gross 2870 The quality of the approximation depends - besides other factors- mainly on the number of elements being used. In fact, the 206 approximation becomes better the more elements are used. However, computational costs and compute time grow with the number of 207 gross 2878 elements being used. It therefore important that you find the right balance between the demand in accuracy and acceptable resource usage. 208 gross 2870 209 In general, one can thinks about a domain object as a composition of nodes and elements. 210 gross 2905 As shown in \reffig{fig:fs}, an element is defined by the nodes used to describe its vertices. 211 gross 2870 To represent spatial distributed values the user can use 212 the values at the nodes, at the elements in the interior of the domain or at elements located at the surface of the domain. 213 The different approach used to represent values is called \textbf{function space} and is attached to all objects 214 in \esc representing a spatial distributed value such as the solution of a PDE. The three 215 function spaces we will use at the moment are; 216 \begin{enumerate} 217 \item the nodes, called by \verb|ContinuousFunction(domain)| ; 218 \item the elements/cells, called by \verb|Function(domain)| ; and 219 \item the boundary, called by \verb|FunctionOnBoundary(domain)| . 220 \end{enumerate} 221 A function space object such as \verb|ContinuousFunction(domain)| has the method \verb|getX| attached to it. This method returns the 222 location of the so-called \textbf{sample points} used to represent values with the particular function space attached to it. So the 223 call \verb|ContinuousFunction(domain).getX()| will return the coordinates of the nodes used to describe the domain while 224 the \verb|Function(domain).getX()| returns the coordinates of numerical integration points within elements, see 225 gross 2905 \reffig{fig:fs}. 226 gross 2870 227 gross 2878 This distinction between different representations of spatial distributed values 228 gross 2870 is important in order to be able to vary the degrees of smoothness in a PDE problem. 229 The coefficients of a PDE need not be continuous thus this qualifies as a \verb|Function()| type. 230 On the other hand a temperature distribution must be continuous and needs to be represented with a \verb|ContinuousFunction()| function space. 231 An influx may only be defined at the boundary and is therefore a \verb FunctionOnBoundary() object. 232 \esc allows certain transformations of the function spaces. A \verb ContinuousFunction() can be transformed into a \verb|FunctionOnBoundary()| 233 or \verb|Function()|. On the other hand there is not enough information in a \verb FunctionOnBoundary() to transform it to a \verb ContinuousFunction() . 234 gross 2878 These transformations, which are called \textbf{interpolation} are invoked automatically by \esc if needed. 235 gross 2870 236 Later in this introduction we will discuss how 237 to define specific areas of geometry with different materials which are represented by different material coefficients such the 238 gross 2878 thermal conductivities $kappa$. A very powerful technique to define these types of PDE 239 gross 2870 coefficients is tagging. Blocks of materials and boundaries can be named and values can be defined on subregions based on their names. 240 gross 2878 This is simplifying PDE coefficient and flux definitions. It makes for much easier scripting. We will discuss this technique in Section~\ref{STEADY-STATE HEAT REFRACTION}. 241 gross 2870 242 243 \subsection{A Clarification for the 1D Case} 244 gross 2931 \label{SEC: 1D CLARIFICATION} 245 gross 2861 It is necessary for clarification that we revisit the general PDE from \refeq{eqn:commonform nabla} under the light of a two dimensional domain. \esc is inherently designed to solve problems that are greater than one dimension and so \refEq{eqn:commonform nabla} needs to be read as a higher dimensional problem. In the case of two spatial dimensions the \textit{Nabla operator} has in fact two components $\nabla = (\frac{\partial}{\partial x}, \frac{\partial}{\partial y})$. In full, \refEq{eqn:commonform nabla} assuming a constant coefficient $A$, takes the form; 246 gross 2477 \begin{equation}\label{eqn:commonform2D} 247 -A\hackscore{00}\frac{\partial^{2}u}{\partial x^{2}} 248 -A\hackscore{01}\frac{\partial^{2}u}{\partial x\partial y} 249 -A\hackscore{10}\frac{\partial^{2}u}{\partial y\partial x} 250 -A\hackscore{11}\frac{\partial^{2}u}{\partial y^{2}} 251 + Du = f 252 \end{equation} 253 ahallam 2606 Notice that for the higher dimensional case $A$ becomes a matrix. It is also 254 ahallam 2495 important to notice that the usage of the Nabla operator creates 255 a compact formulation which is also independent from the spatial dimension. 256 gross 2861 So to make the general PDE \refEq{eqn:commonform2D} one dimensional as 257 shown in \refEq{eqn:commonform} we need to set 258 ahallam 2606 \begin{equation} 259 ahallam 2494 A\hackscore{00}=A; A\hackscore{01}=A\hackscore{10}=A\hackscore{11}=0 260 gross 2477 \end{equation} 261 262 gross 2867 263 ahallam 2495 \subsection{Developing a PDE Solution Script} 264 ahallam 2801 \label{sec:key} 265 gross 2949 \sslist{example01a.py} 266 gross 2870 We will write a simple \pyt script which uses the \modescript, \modfinley and \modmpl modules. 267 gross 2905 By developing a script for \esc, the heat diffusion equation can be solved at successive time steps for a predefined period using our general form \refEq{eqn:hdgenf}. Firstly it is necessary to import all the libraries\footnote{The libraries contain predefined scripts that are required to solve certain problems, these can be simple like sine and cosine functions or more complicated like those from our \esc library.} 268 ahallam 2495 that we will require. 269 ahallam 2775 \begin{python} 270 ahallam 2495 from esys.escript import * 271 ahallam 2606 # This defines the LinearPDE module as LinearPDE 272 from esys.escript.linearPDEs import LinearPDE 273 # This imports the rectangle domain function from finley. 274 from esys.finley import Rectangle 275 # A useful unit handling package which will make sure all our units 276 # match up in the equations under SI. 277 from esys.escript.unitsSI import * 278 ahallam 2775 \end{python} 279 gross 2878 It is generally a good idea to import all of the \modescript library, although if the functions and classes required are known they can be specified individually. The function \verb|LinearPDE| has been imported explicitly for ease of use later in the script. \verb|Rectangle| is going to be our type of model. The module \verb unitsSI provides support for SI unit definitions with our variables. 280 gross 2477 281 gross 2905 Once our library dependencies have been established, defining the problem specific variables is the next step. In general the number of variables needed will vary between problems. These variables belong to two categories. They are either directly related to the PDE and can be used as inputs into the \esc solver, or they are script variables used to control internal functions and iterations in our problem. For this PDE there are a number of constants which will need values. Firstly, the model upon which we wish to solve our problem needs to be defined. There are many different types of models in \modescript which we will demonstrate in later tutorials but for our granite blocks, we will simply use a rectangular model. 282 ahallam 2401 283 gross 2905 Using a rectangular model simplifies our granite blocks which would in reality be a \textit{3D} object, into a single dimension. The granite blocks will have a lengthways cross section that looks like a rectangle. As a result we do not need to model the volume of the block. There are four arguments we must consider when we decide to create a rectangular model, the model \textit{length}, \textit{width} and \textit{step size} in each direction. When defining the size of our problem it will help us determine appropriate values for our model arguments. If we make our dimensions large but our step sizes very small we will to a point, increase the accuracy of our solution. Unfortunately we also increase the number of calculations that must be solved per time step. This means more computational time is required to produce a solution. In this \textit{1D} problem, the bar is defined as being 1 metre long. An appropriate step size \verb|ndx| would be 1 to 10\% of the length. Our \verb|ndy| need only be 1, this is because our problem stipulates no partial derivatives in the $y$ direction. Thus the temperature does not vary with $y$. Hence, the model parameters can be defined as follows; note we have used the \verb unitsSI convention to make sure all our input units are converted to SI. 284 ahallam 2775 \begin{python} 285 gross 2905 mx = 500.*m #meters - model length 286 my = 100.*m #meters - model width 287 ndx = 50 # mesh steps in x direction 288 ndy = 1 # mesh steps in y direction 289 boundloc = mx/2 # location of boundary between the two blocks 290 ahallam 2775 \end{python} 291 gross 2905 The material constants and the temperature variables must also be defined. For the granite in the model they are defined as: 292 ahallam 2775 \begin{python} 293 ahallam 2495 #PDE related 294 gross 2905 rho = 2750. *kg/m**3 #kg/m^{3} density of iron 295 cp = 790.*J/(kg*K) # J/Kg.K thermal capacity 296 gross 2878 rhocp = rho*cp 297 gross 2905 kappa = 2.2*W/m/K # watts/m.Kthermal conductivity 298 gross 2878 qH=0 * J/(sec*m**3) # J/(sec.m^{3}) no heat source 299 gross 2905 T1=20 * Celsius # initial temperature at Block 1 300 T2=2273. * Celsius # base temperature at Block 2 301 ahallam 2775 \end{python} 302 ahallam 2495 Finally, to control our script we will have to specify our timing controls and where we would like to save the output from the solver. This is simple enough: 303 ahallam 2775 \begin{python} 304 gross 2878 t=0 * day #our start time, usually zero 305 tend=1. * day # - time to end simulation 306 ahallam 2495 outputs = 200 # number of time steps required. 307 h=(tend-t)/outputs #size of time step 308 ahallam 2606 #user warning statement 309 print "Expected Number of time outputs is: ", (tend-t)/h 310 i=0 #loop counter 311 ahallam 2775 \end{python} 312 gross 2905 Now that we know our inputs we will build a domain using the \verb Rectangle() function from \verb finley . The four arguments allow us to define our domain \verb model as: 313 ahallam 2775 \begin{python} 314 ahallam 2606 #generate domain using rectangle 315 gross 2905 blocks = Rectangle(l0=mx,l1=my,n0=ndx, n1=ndy) 316 ahallam 2775 \end{python} 317 gross 2905 \verb blocks now describes a domain in the manner of Section \ref{ss:domcon}. T 318 ahallam 2658 319 ahallam 2775 With a domain and all our required variables established, it is now possible to set up our PDE so that it can be solved by \esc. The first step is to define the type of PDE that we are trying to solve in each time step. In this example it is a single linear PDE\footnote{in comparison to a system of PDEs which will be discussed later.}. We also need to state the values of our general form variables. 320 \begin{python} 321 gross 2905 mypde=LinearPDE(blocks) 322 gross 2878 A=zeros((2,2))) 323 A[0,0]=kappa 324 gross 2905 mypde.setValue(A=A, D=rhocp/h) 325 ahallam 2775 \end{python} 326 gross 2878 In a many cases it may be possible to decrease the computational time of the solver if the PDE is symmetric. 327 Symmetry of a PDE is defined by; 328 ahallam 2495 \begin{equation}\label{eqn:symm} 329 A\hackscore{jl}=A\hackscore{lj} 330 \end{equation} 331 gross 2878 Symmetry is only dependent on the $A$ coefficient in the general form and the other coefficients $D$ as well as the right hand side $Y$ may take any value. From the above definition we can see that our PDE is symmetric. The \verb LinearPDE class provides the method \method{checkSymmetry} to check if the given PDE is symmetric. As our PDE is symmetrical we will enable symmetry via; 332 ahallam 2775 \begin{python} 333 ahallam 2495 myPDE.setSymmetryOn() 334 ahallam 2775 \end{python} 335 gross 2905 Next we need to establish the initial temperature distribution \verb|T|. We need to 336 assign the value \verb|T1| to all sample points left to the contact interface at $x\hackscore{0}=\frac{mx}{2}$ 337 and the value \verb|T2| right to the contact interface. \esc 338 provides the \verb|whereNegative| function to construct this. In fact, 339 \verb|whereNegative| returns the value $1$ at those sample points where the argument 340 has a negative value. Otherwise zero is returned. If \verb|x| are the $x\hackscore{0}$ 341 coordinates of the sample points used to represent the temperature distribution 342 then \verb|x[0]-boundloc| gives us a negative value for 343 all sample points left to the interface and non-negative value to 344 the right of the interface. So with; 345 gross 2878 \begin{python} 346 # ... set initial temperature .... 347 gross 2905 T= T1*whereNegative(x[0]-boundloc)+T2*(1-whereNegative(x[0]-boundloc)) 348 gross 2878 \end{python} 349 gross 2905 we get the desired temperature distribution. To get the actual sample points \verb|x| we use 350 the \verb|getX()| method of the function space \verb|Solution(blocks)| 351 which is used to represent the solution of a PDE; 352 gross 2878 \begin{python} 353 gross 2905 x=Solution(blocks).getX() 354 \end{python} 355 As \verb|x| are the sample points for the function space \verb|Solution(blocks)| 356 the initial temperature \verb|T| is using these sample points for representation. 357 Although \esc is trying to be forgiving with the choice of sample points and to convert 358 where necessary the adjustment of the function space is not always possible. So it is 359 advisable to make a careful choice on the function space used. 360 361 Finally we will initialise an iteration loop to solve our PDE for all the time steps we specified in the variable section. As the right hand side of the general form is dependent on the previous values for temperature \verb T across the bar this must be updated in the loop. Our output at each time step is \verb T the heat distribution and \verb totT the total heat in the system. 362 \begin{python} 363 gross 2878 while t < tend: 364 i+=1 #increment the counter 365 t+=h #increment the current time 366 mypde.setValue(Y=qH+rhocp/h*T) # set variable PDE coefficients 367 T=mypde.getSolution() #get the PDE solution 368 totE = integrate(rhocp*T) #get the total heat (energy) in the system 369 \end{python} 370 The last statement in this script calculates the total energy in the system as volume integral 371 gross 2905 of $\rho c\hackscore{p} T$ over the block. As the blocks are insulated no energy should be get lost or added. 372 The total energy should stay constant for the example discussed here. 373 ahallam 2401 374 gross 2905 \subsection{Running the Script} 375 The script presented so for is available under 376 gross 2949 \verb|example01a.py|. You can edit this file with your favourite text editor. 377 jfenwick 2911 On most operating systems\footnote{The you can use \texttt{run-escript} launcher is not supported under {\it MS Windows} yet.} you can use the \program{run-escript} command 378 gross 2905 to launch {\it escript} scripts. For the example script use; 379 \begin{verbatim} 380 gross 2949 run-escript example01a.py 381 gross 2905 \end{verbatim} 382 The program will print a progress report. Alternatively, you can use 383 the python interpreter directly; 384 \begin{verbatim} 385 gross 2949 python example01a.py 386 gross 2905 \end{verbatim} 387 if the system is configured correctly (Please talk to your system administrator). 388 389 \begin{figure} 390 \begin{center} 391 \includegraphics[width=4in]{figures/ttblockspyplot150} 392 gross 2952 \caption{Example 1b: Total Energy in the Blocks over Time (in seconds).} 393 gross 2905 \label{fig:onedheatout1} 394 \end{center} 395 \end{figure} 396 397 gross 2878 \subsection{Plotting the Total Energy} 398 gross 2949 \sslist{example01b.py} 399 gross 2878 400 gross 2905 \esc does not include its own plotting capabilities. However, it is possible to use a variety of free \pyt packages for visualisation. 401 gross 2878 Two types will be demonstrated in this cookbook; \mpl\footnote{\url{http://matplotlib.sourceforge.net/}} and \verb VTK \footnote{\url{http://www.vtk.org/}} visualisation. 402 The \mpl package is a component of SciPy\footnote{\url{http://www.scipy.org}} and is good for basic graphs and plots. 403 For more complex visualisation tasks in particular when it comes to two and three dimensional problems it is recommended to us more advanced tools for instance \mayavi \footnote{\url{http://code.enthought.com/projects/mayavi/}} 404 gross 2905 which bases on the \verb|VTK| toolkit. We will discuss the usage of \verb|VTK| based 405 gross 2878 visualization in Chapter~\ref{Sec:2DHD} where will discuss a two dimensional PDE. 406 407 For our simple problem we have two plotting tasks: Firstly we are interested in showing the 408 gross 2905 behaviour of the total energy over time and secondly in how the temperature distribution within the block is 409 gross 2878 developing over time. Lets start with the first task. 410 411 The trick is to create a record of the time marks and the corresponding total energies observed. 412 \pyt provides the concept of lists for this. Before 413 the time loop is opened we create empty lists for the time marks \verb|t_list| and the total energies \verb|E_list|. 414 After the new temperature as been calculated by solving the PDE we append the new time marker and total energy 415 to the corresponding list using the \verb|append| method. With these modifications the script looks as follows: 416 ahallam 2775 \begin{python} 417 gross 2878 t_list=[] 418 E_list=[] 419 # ... start iteration: 420 while t