 # Contents of /trunk/doc/cookbook/example02.tex

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cookbook review final final 3.1 - artak and tony corrections

 1 2 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 3 % 4 % Copyright (c) 2003-2010 by University of Queensland 5 % 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 15 \section{Example 2: One Dimensional Heat Diffusion in an Iron Rod} 16 \sslist{example02.py} 17 18 \label{Sec:1DHDv0} 19 \begin{figure}[ht] 20 \centerline{\includegraphics[width=4.in]{figures/onedheatdiff002}} 21 \caption{Example 2: One dimensional model of an Iron bar.} 22 \label{fig:onedhdmodel} 23 \end{figure} 24 25 Our second example is of a cold iron bar at a constant temperature of 26 $T\hackscore{ref}=20^{\circ} C$, see \reffig{fig:onedhdmodel}. The bar is 27 perfectly insulated on all sides with a heating element at one end keeping the 28 the temperature at a constant level $T\hackscore0=100^{\circ} C$. As heat is 29 applied; energy will disperse along the bar via conduction. With time the bar 30 will reach a constant temperature equivalent to that of the heat source. 31 32 This problem is very similar to the example of temperature diffusion in granite 33 blocks presented in the previous section~\ref{Sec:1DHDv00}. Thus, it is possible 34 to modify the script we have already developed for the granite blocks to suit 35 the iron bar problem. 36 The obvious difference between the two problems are the dimensions of the domain 37 and different materials involved. This will change the time scale of the model 38 from years to hours. 39 The new settings are; 40 \begin{python} 41 #Domain related. 42 mx = 1*m #meters - model length 43 my = .1*m #meters - model width 44 ndx = 100 # mesh steps in x direction 45 ndy = 1 # mesh steps in y direction - one dimension means one element 46 #PDE related 47 rho = 7874. *kg/m**3 #kg/m^{3} density of iron 48 cp = 449.*J/(kg*K) # J/Kg.K thermal capacity 49 rhocp = rho*cp 50 kappa = 80.*W/m/K # watts/m.Kthermal conductivity 51 qH = 0 * J/(sec*m**3) # J/(sec.m^{3}) no heat source 52 Tref = 20 * Celsius # base temperature of the rod 53 T0 = 100 * Celsius # temperature at heating element 54 tend= 0.5 * day # - time to end simulation 55 \end{python} 56 We also need to alter the initial value for the temperature. Now we need to set 57 the 58 temperature to $T\hackscore{0}$ at the left end of the rod where we have 59 $x\hackscore{0}=0$ and 60 $T\hackscore{ref}$ elsewhere. Instead of \verb|whereNegative| function we use 61 now the 62 \verb|whereZero| which returns the value one for those sample points where 63 the argument (almost) equals zero and the value zero elsewhere. The initial 64 temperature is set to; 65 \begin{python} 66 # ... set initial temperature .... 67 T= T0*whereZero(x)+Tref*(1-whereZero(x)) 68 \end{python} 69 70 \subsection{Dirichlet Boundary Conditions} 71 In the iron rod model we want to keep the initial temperature $T\hackscore0$ on 72 the left side of the domain constant with time. 73 This implies that when we solve the PDE~\refEq{eqn:hddisc}, the solution must 74 have the value $T\hackscore0$ on the left hand 75 side of the domain. As mentioned already in Section~\ref{SEC BOUNDARY COND} 76 where we discussed 77 boundary conditions, this kind of scenario can be expressed using a 78 \textbf{Dirichlet boundary condition}. Some people also 79 use the term \textbf{constraint} for the PDE. 80 81 To define a Dirichlet boundary condition we need to specify where to apply the 82 condition and determine what value the 83 solution should have at these locations. In \esc we use $q$ and $r$ to define 84 the Dirichlet boundary conditions 85 for a PDE. The solution $u$ of the PDE is set to $r$ for all sample points where 86 $q$ has a positive value. 87 Mathematically this is expressed in the form; 88 \begin{equation} 89 u(x) = r(x) \mbox{ for any } x \mbox{ with } q(x) > 0 90 \end{equation} 91 In the case of the iron rod 92 we can set; 93 \begin{python} 94 q=whereZero(x) 95 r=T0 96 \end{python} 97 to prescribe the value $T\hackscore{0}$ for the temperature at the left end of 98 the rod where $x\hackscore{0}=0$. 99 Here we use the \verb|whereZero| function again which we have already used to 100 set the initial value. 101 Notice that $r$ is set to the constant value $T\hackscore{0}$ for all sample 102 points. In fact, 103 values of $r$ are used only where $q$ is positive. Where $q$ is non-positive, 104 $r$ may have any value as these values are not used by the PDE solver. 105 106 To set the Dirichlet boundary conditions for the PDE to be solved in each time 107 step we need 108 to add some statements; 109 \begin{python} 110 mypde=LinearPDE(rod) 111 A=zeros((2,2))) 112 A[0,0]=kappa 113 q=whereZero(x) 114 mypde.setValue(A=A, D=rhocp/h, q=q, r=T0) 115 \end{python} 116 It is important to remark here that if a Dirichlet boundary condition is 117 prescribed on the same location as any Neuman boundary condition, the Neuman 118 boundary condition will be \textbf{overwritten}. This applies to Neuman 119 boundary conditions that \esc sets by default and those defined by the user. 120 121 \begin{figure} 122 \begin{center} 123 \includegraphics[width=4in]{figures/ttrodpyplot150} 124 \caption{Example 2: Total Energy in the Iron Rod over Time (in seconds).} 125 \label{fig:onedheatout1 002} 126 \end{center} 127 \end{figure} 128 129 \begin{figure} 130 \begin{center} 131 \includegraphics[width=4in]{figures/rodpyplot001} 132 \includegraphics[width=4in]{figures/rodpyplot050} 133 \includegraphics[width=4in]{figures/rodpyplot200} 134 \caption{Example 2: Temperature ($T$) distribution in the iron rod at time steps 135 $1$, $50$ and $200$.} 136 \label{fig:onedheatout 002} 137 \end{center} 138 \end{figure} 139 140 Besides some cosmetic modification this all we need to change. The total energy 141 over time is shown in \reffig{fig:onedheatout1 002}. As heat 142 is transfered into the rod by the heater the total energy is growing over time 143 but reaches a plateau 144 when the temperature is constant is the rod, see \reffig{fig:onedheatout 002}. 145 You will notice that the time scale of this model is several order of magnitudes 146 faster than 147 for the granite rock problem due to the different length scale and material 148 parameters. 149 In practice it can take a few models run before the right time scale has been 150 chosen\footnote{An estimate of the 151 time scale for a diffusion problem is given by the formula $\frac{\rho 152 c\hackscore{p} L\hackscore{0}^2}{4 \kappa}$, see 153 \url{http://en.wikipedia.org/wiki/Fick\%27s_laws_of_diffusion}}. 154 155 156 157 158 159 160 \section{For the Reader} 161 \begin{enumerate} 162 \item Move the boundary line between the two granite blocks to another part of 163 the domain. 164 \item Split the domain into multiple granite blocks with varying temperatures. 165 \item Vary the mesh step size. Do you see a difference in the answers? What 166 does happen with the compute time? 167 \item Insert an internal heat source (Hint: The internal heat source is given 168 by $q\hackscore{H}$.) 169 \item Change the boundary condition for iron rod example such that the 170 temperature 171 at the right end is kept at a constant level $T\hackscore{ref}$, which 172 corresponds to the installation of a cooling element (Hint: Modify $q$ and 173 $r$). 174 \end{enumerate} 175