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1 ahallam 2401
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14 ahallam 2975
15     \section{Example 2: One Dimensional Heat Diffusion in an Iron Rod}
16     \sslist{example02.py}
17     \label{Sec:1DHDv0}
18 ahallam 2606
19 ahallam 2979 Our second example is of a cold iron bar at a constant temperature of
20 jfenwick 3308 $T_{ref}=20^{\circ} C$, see \reffig{fig:onedhdmodel}. The bar is
21 ahallam 2979 perfectly insulated on all sides with a heating element at one end keeping the
22 jfenwick 3308 temperature at a constant level $T_0=100^{\circ} C$. As heat is
23 caltinay 2982 applied energy will disperse along the bar via conduction. With time the bar
24 ahallam 2979 will reach a constant temperature equivalent to that of the heat source.
25 gross 2905
26 ahallam 3370 \begin{figure}[ht]
27     \centerline{\includegraphics[width=4.in]{figures/onedheatdiff002}}
28     \caption{Example 2: One dimensional model of an Iron bar}
29     \label{fig:onedhdmodel}
30     \end{figure}
31    
32 ahallam 2979 This problem is very similar to the example of temperature diffusion in granite
33 caltinay 2982 blocks presented in the previous Section~\ref{Sec:1DHDv00}. Thus, it is possible
34 ahallam 2979 to modify the script we have already developed for the granite blocks to suit
35     the iron bar problem.
36 caltinay 2982 The obvious differences between the two problems are the dimensions of the
37     domain and different materials involved. This will change the time scale of the
38     model from years to hours. The new settings are
39 ahallam 2801 \begin{python}
40 gross 2905 #Domain related.
41     mx = 1*m #meters - model length
42     my = .1*m #meters - model width
43     ndx = 100 # mesh steps in x direction
44     ndy = 1 # mesh steps in y direction - one dimension means one element
45 ahallam 2606 #PDE related
46 gross 2905 rho = 7874. *kg/m**3 #kg/m^{3} density of iron
47     cp = 449.*J/(kg*K) # J/Kg.K thermal capacity
48     rhocp = rho*cp
49     kappa = 80.*W/m/K # watts/m.Kthermal conductivity
50     qH = 0 * J/(sec*m**3) # J/(sec.m^{3}) no heat source
51     Tref = 20 * Celsius # base temperature of the rod
52     T0 = 100 * Celsius # temperature at heating element
53     tend= 0.5 * day # - time to end simulation
54 ahallam 2801 \end{python}
55 ahallam 2979 We also need to alter the initial value for the temperature. Now we need to set
56 jfenwick 3308 the temperature to $T_{0}$ at the left end of the rod where we have
57     $x_{0}=0$ and
58     $T_{ref}$ elsewhere. Instead of \verb|whereNegative| function we now
59 caltinay 2982 use \verb|whereZero| which returns the value one for those sample points where
60 gross 2905 the argument (almost) equals zero and the value zero elsewhere. The initial
61 caltinay 2982 temperature is set to
62 gross 2905 \begin{python}
63     # ... set initial temperature ....
64     T= T0*whereZero(x[0])+Tref*(1-whereZero(x[0]))
65     \end{python}
66 ahallam 2401
67 gross 2953 \subsection{Dirichlet Boundary Conditions}
68 jfenwick 3308 In the iron rod model we want to keep the initial temperature $T_0$ on
69 ahallam 2979 the left side of the domain constant with time.
70     This implies that when we solve the PDE~\refEq{eqn:hddisc}, the solution must
71 jfenwick 3308 have the value $T_0$ on the left hand side of the domain. As mentioned
72 caltinay 2982 already in Section~\ref{SEC BOUNDARY COND} where we discussed boundary
73     conditions, this kind of scenario can be expressed using a
74     \textbf{Dirichlet boundary condition}. Some people also use the term
75     \textbf{constraint} for the PDE.
76 gross 2905
77 ahallam 2979 To define a Dirichlet boundary condition we need to specify where to apply the
78 caltinay 2982 condition and determine what value the
79 ahallam 2979 solution should have at these locations. In \esc we use $q$ and $r$ to define
80 caltinay 2982 the Dirichlet boundary conditions for a PDE. The solution $u$ of the PDE is set
81     to $r$ for all sample points where $q$ has a positive value.
82 gross 2905 Mathematically this is expressed in the form;
83     \begin{equation}
84     u(x) = r(x) \mbox{ for any } x \mbox{ with } q(x) > 0
85     \end{equation}
86 caltinay 2982 In the case of the iron rod we can set
87 ahallam 2801 \begin{python}
88 gross 2905 q=whereZero(x[0])
89     r=T0
90 ahallam 2801 \end{python}
91 jfenwick 3308 to prescribe the value $T_{0}$ for the temperature at the left end of
92     the rod where $x_{0}=0$.
93 ahallam 2979 Here we use the \verb|whereZero| function again which we have already used to
94     set the initial value.
95 jfenwick 3308 Notice that $r$ is set to the constant value $T_{0}$ for all sample
96 caltinay 2982 points. In fact, values of $r$ are used only where $q$ is positive. Where $q$
97     is non-positive, $r$ may have any value as these values are not used by the PDE
98     solver.
99 ahallam 2401
100 ahallam 2979 To set the Dirichlet boundary conditions for the PDE to be solved in each time
101 caltinay 2982 step we need to add some statements;
102 ahallam 2801 \begin{python}
103 gross 2905 mypde=LinearPDE(rod)
104     A=zeros((2,2)))
105     A[0,0]=kappa
106     q=whereZero(x[0])
107     mypde.setValue(A=A, D=rhocp/h, q=q, r=T0)
108 ahallam 2801 \end{python}
109 ahallam 2979 It is important to remark here that if a Dirichlet boundary condition is
110 caltinay 2982 prescribed on the same location as any Neumann boundary condition, the Neumann
111     boundary condition will be \textbf{overwritten}. This applies to Neumann
112 ahallam 2979 boundary conditions that \esc sets by default and those defined by the user.
113 ahallam 2401
114 ahallam 3370 Besides some cosmetic modification this is all we need to change. The total
115     energy over time is shown in \reffig{fig:onedheatout1 002}. As heat
116     is transferred into the rod by the heater the total energy is growing over time
117     but reaches a plateau when the temperature is constant in the rod, see
118     \reffig{fig:onedheatout 002}.
119     You will notice that the time scale of this model is several order of
120     magnitudes faster than for the granite rock problem due to the different length
121     scale and material parameters.
122     In practice it can take a few model runs before the right time scale has been
123     chosen\footnote{An estimate of the
124     time scale for a diffusion problem is given by the formula $\frac{\rho
125     c_{p} L_{0}^2}{4 \kappa}$, see
126     \url{http://en.wikipedia.org/wiki/Fick\%27s_laws_of_diffusion}}.
127    
128     \begin{figure}[ht]
129 gross 2905 \begin{center}
130     \includegraphics[width=4in]{figures/ttrodpyplot150}
131 caltinay 2982 \caption{Example 2: Total Energy in the Iron Rod over Time (in seconds)}
132 gross 2905 \label{fig:onedheatout1 002}
133     \end{center}
134     \end{figure}
135    
136 ahallam 3370 \begin{figure}[ht]
137 gross 2905 \begin{center}
138     \includegraphics[width=4in]{figures/rodpyplot001}
139     \includegraphics[width=4in]{figures/rodpyplot050}
140     \includegraphics[width=4in]{figures/rodpyplot200}
141 ahallam 2979 \caption{Example 2: Temperature ($T$) distribution in the iron rod at time steps
142 caltinay 2982 $1$, $50$ and $200$}
143 gross 2905 \label{fig:onedheatout 002}
144     \end{center}
145     \end{figure}
146    
147     \section{For the Reader}
148 ahallam 2401 \begin{enumerate}
149 ahallam 2979 \item Move the boundary line between the two granite blocks to another part of
150     the domain.
151 gross 2905 \item Split the domain into multiple granite blocks with varying temperatures.
152 ahallam 2979 \item Vary the mesh step size. Do you see a difference in the answers? What
153     does happen with the compute time?
154     \item Insert an internal heat source (Hint: The internal heat source is given
155 jfenwick 3308 by $q_{H}$.)
156 caltinay 2982 \item Change the boundary condition for the iron rod example such that the
157 ahallam 2979 temperature
158 jfenwick 3308 at the right end is kept at a constant level $T_{ref}$, which
159 ahallam 2979 corresponds to the installation of a cooling element (Hint: Modify $q$ and
160     $r$).
161 ahallam 2401 \end{enumerate}
162    

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