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\begin{figure}[ht] 
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\centerline{\includegraphics[width=4.in]{figures/onedheatdiff002}} 
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\caption{One dimensional model of an Iron bar.} 
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\label{fig:onedhdmodel} 
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\end{figure} 
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\section{One Dimensional Heat Diffusion in an Iron Rod} 
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\sslist{onedheatdiff002.py} 
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\label{Sec:1DHDv0} 
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Our second example is a cold iron bar at a constant temperature of $T\hackscore{ref}=20^{\circ} C$, see \reffig{fig:onedhdmodel}. The bar is perfectly insulated on all sides with a heating element at one end keeping the the temperature at a constant level $T\hackscore0=100^{\circ} C$. as heat is applied; energy will disperse along the bar via conduction. With time the bar will reach a constant temperature equivalent to that of the heat source. 
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This problem is very similar to the example of temperature diffusion in granit blocks presented in the previous section~\ref{Sec:1DHDv00}. So we will modify the script we have already developed for the granit blocks to adjust 
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it to the iron bar problem. 
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The obvious difference between the two problems are the dimensions of the domain and different materials involved. This will change the time scale of the model from years to hours. 
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The new settings are; 
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\begin{python} 
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#Domain related. 
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mx = 1*m #meters  model length 
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my = .1*m #meters  model width 
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ndx = 100 # mesh steps in x direction 
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ndy = 1 # mesh steps in y direction  one dimension means one element 
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#PDE related 
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rho = 7874. *kg/m**3 #kg/m^{3} density of iron 
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cp = 449.*J/(kg*K) # J/Kg.K thermal capacity 
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rhocp = rho*cp 
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kappa = 80.*W/m/K # watts/m.Kthermal conductivity 
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qH = 0 * J/(sec*m**3) # J/(sec.m^{3}) no heat source 
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Tref = 20 * Celsius # base temperature of the rod 
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T0 = 100 * Celsius # temperature at heating element 
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tend= 0.5 * day #  time to end simulation 
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\end{python} 
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We also need to alter the initial value for the temperature. Now we need to set the 
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temperature to $T\hackscore{0}$ at the left end of the rod where we have $x\hackscore{0}=0$ and 
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$T\hackscore{ref}$ elsewhere. Instead of \verbwhereNegative function we use now the 
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\verbwhereZero which returns the value one for those sample points where 
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the argument (almost) equals zero and the value zero elsewhere. The initial 
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temperature is set to; 
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\begin{python} 
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# ... set initial temperature .... 
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T= T0*whereZero(x[0])+Tref*(1whereZero(x[0])) 
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\end{python} 
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\subsection{Dirchlet Boundary Conditions} 
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In iron rod model we want to keep the initial temperature $T\hackscore0$ on the left side of the domain over time. 
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So when we solve the PDE~\refEq{eqn:hddisc} the solution must have the value $T\hackscore0$ on the left hand 
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side of the domain. As mentioned already in Section~\ref{SEC BOUNDARY COND} where we discussed 
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boundary condition this kind of condition are called a \textbf{Dirichlet boundary condition}. Some people also 
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use the term \textbf{constraint} for the PDE. 
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To define a Dirichlet boundary condition we need to define where to apply the condition and what value the 
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solution should have at these locations. In \esc we use $q$ and $r$ to define the Dirichlet boundary conditions 
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for a PDE. The solution $u$ of the PDE is set to $r$ for all sample points where $q$ has a positive value. 
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Mathematically this is expressed in the form; 
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\begin{equation} 
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u(x) = r(x) \mbox{ for any } x \mbox{ with } q(x) > 0 
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\end{equation} 
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In the case of the iron rod 
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we can set; 
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\begin{python} 
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q=whereZero(x[0]) 
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r=T0 
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\end{python} 
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to prescibe the value $T0$ for the temperature at the left end of the rod where $x\hackscore{0}=0$. 
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Here we use the \verbwhereZero function again which we have alread used to set the initial value. 
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Notice that $r$ is set to the constant value $r$ for all sample points. In fact, 
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values of $r$ are used only where $q$ is positive. Where $q$ is nonpositive, 
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$r$ may have any value as these values are not used by the PDE solver. 
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To set the Dirichlet boundary conditions for the PDE to be solved in each time step we need 
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to add some statements; 
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\begin{python} 
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mypde=LinearPDE(rod) 
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A=zeros((2,2))) 
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A[0,0]=kappa 
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q=whereZero(x[0]) 
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mypde.setValue(A=A, D=rhocp/h, q=q, r=T0) 
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\end{python} 
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It is important to remark here that the Dirichlet condition \textbf{overwrites} any Neuman boundary 
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condition \esc sets by default (or you may set). 
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\begin{figure} 
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\begin{center} 
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\includegraphics[width=4in]{figures/ttrodpyplot150} 
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\caption{Total Energy in the Iron Rod over Time (in seconds).} 
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\label{fig:onedheatout1 002} 
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\end{center} 
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\end{figure} 
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\begin{figure} 
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\begin{center} 
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\includegraphics[width=4in]{figures/rodpyplot001} 
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\includegraphics[width=4in]{figures/rodpyplot050} 
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\includegraphics[width=4in]{figures/rodpyplot200} 
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\caption{Temperature ($T$) distribution in the iron rod at time steps $1$, $50$ and $200$.} 
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\label{fig:onedheatout 002} 
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\end{center} 
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\end{figure} 
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Besides some cosmetic modification this all we need to change. The total energy over time is shown in \reffig{fig:onedheatout1 002}. As heat 
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is transfered into the rod by the heater the total energy is growing over time but reaches a plateau 
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when the temperature is constant is the rod, see \reffig{fig:onedheatout 002}. 
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YOu will notice that the time scale of this model is several order of magnitudes faster than 
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for the granite rock problem due to the different length scale and material parameters. 
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In practice it can take a few models run before the right time scale has been chosen\footnote{An estimate of the 
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time scale for a diffusion problem is given by the formula $\frac{\rho c\hackscore{p} L\hackscore{0}^2}{4 \kappa}$, see 
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\url{http://en.wikipedia.org/wiki/Fick\%27s_laws_of_diffusion}}. 
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\section{For the Reader} 
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\begin{enumerate} 
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\item Move the boundary line between the two granite blocks to another part of the domain. 
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\item Split the domain into multiple granite blocks with varying temperatures. 
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\item Vary the mesh step size. Do you see a difference in the answers? What does happen with the compute time? 
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\item Insert an internal heat source (Hint: The internal heat source is given by $q\hackscore{H}$.) 
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\item Change the boundary condition for iron rod example such that the temperature 
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at the right end is kept at a constant level $T\hackscore{ref}$, which corresponds to the installation of a cooling element (Hint: Modify $q$ and $r$). 
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\end{enumerate} 
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