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15
16 \section{Example 2: One Dimensional Heat Diffusion in an Iron Rod}
17 \sslist{example02.py}
18 \label{Sec:1DHDv0}
19
20 Our second example is of a cold iron bar at a constant temperature of
21 $T_{ref}=20^{\circ} C$, see \reffig{fig:onedhdmodel}. The bar is
22 perfectly insulated on all sides with a heating element at one end keeping the
23 temperature at a constant level $T_0=100^{\circ} C$. As heat is
24 applied energy will disperse along the bar via conduction. With time the bar
25 will reach a constant temperature equivalent to that of the heat source.
26
27 \begin{figure}[ht]
28 \centerline{\includegraphics[width=4.in]{figures/onedheatdiff002}}
29 \caption{Example 2: One dimensional model of an Iron bar}
30 \label{fig:onedhdmodel}
31 \end{figure}
32
33 This problem is very similar to the example of temperature diffusion in granite
34 blocks presented in the previous Section~\ref{Sec:1DHDv00}. Thus, it is possible
35 to modify the script we have already developed for the granite blocks to suit
36 the iron bar problem.
37 The obvious differences between the two problems are the dimensions of the
38 domain and different materials involved. This will change the time scale of the
39 model from years to hours. 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 temperature to $T_{0}$ at the left end of the rod where we have
58 $x_{0}=0$ and
59 $T_{ref}$ elsewhere. Instead of \verb|whereNegative| function we now
60 use \verb|whereZero| which returns the value one for those sample points where
61 the argument (almost) equals zero and the value zero elsewhere. The initial
62 temperature is set to
63 \begin{python}
64 # ... set initial temperature ....
65 T= T0*whereZero(x[0])+Tref*(1-whereZero(x[0]))
66 \end{python}
67
68 \subsection{Dirichlet Boundary Conditions}
69 In the iron rod model we want to keep the initial temperature $T_0$ on
70 the left side of the domain constant with time.
71 This implies that when we solve the PDE~\refEq{eqn:hddisc}, the solution must
72 have the value $T_0$ on the left hand side of the domain. As mentioned
73 already in Section~\ref{SEC BOUNDARY COND} where we discussed boundary
74 conditions, this kind of scenario can be expressed using a
75 \textbf{Dirichlet boundary condition}. Some people also use the term
76 \textbf{constraint} for the PDE.
77
78 To define a Dirichlet boundary condition we need to specify where to apply the
79 condition and determine what value the
80 solution should have at these locations. In \esc we use $q$ and $r$ to define
81 the Dirichlet boundary conditions for a PDE. The solution $u$ of the PDE is set
82 to $r$ for all sample points where $q$ has a positive value.
83 Mathematically this is expressed in the form;
84 \begin{equation}
85 u(x) = r(x) \mbox{ for any } x \mbox{ with } q(x) > 0
86 \end{equation}
87 In the case of the iron rod we can set
88 \begin{python}
89 q=whereZero(x[0])
90 r=T0
91 \end{python}
92 to prescribe the value $T_{0}$ for the temperature at the left end of
93 the rod where $x_{0}=0$.
94 Here we use the \verb|whereZero| function again which we have already used to
95 set the initial value.
96 Notice that $r$ is set to the constant value $T_{0}$ for all sample
97 points. In fact, values of $r$ are used only where $q$ is positive. Where $q$
98 is non-positive, $r$ may have any value as these values are not used by the PDE
99 solver.
100
101 To set the Dirichlet boundary conditions for the PDE to be solved in each time
102 step we need to add some statements;
103 \begin{python}
104 mypde=LinearPDE(rod)
105 A=zeros((2,2)))
106 A[0,0]=kappa
107 q=whereZero(x[0])
108 mypde.setValue(A=A, D=rhocp/h, q=q, r=T0)
109 \end{python}
110 It is important to remark here that if a Dirichlet boundary condition is
111 prescribed on the same location as any Neumann boundary condition, the Neumann
112 boundary condition will be \textbf{overwritten}. This applies to Neumann
113 boundary conditions that \esc sets by default and those defined by the user.
114
115 Besides some cosmetic modification this is all we need to change. The total
116 energy over time is shown in \reffig{fig:onedheatout1 002}. As heat
117 is transferred into the rod by the heater the total energy is growing over time
118 but reaches a plateau when the temperature is constant in the rod, see
119 \reffig{fig:onedheatout 002}.
120 You will notice that the time scale of this model is several order of
121 magnitudes faster than for the granite rock problem due to the different length
122 scale and material parameters.
123 In practice it can take a few model runs before the right time scale has been
124 chosen\footnote{An estimate of the
125 time scale for a diffusion problem is given by the formula $\frac{\rho
126 c_{p} L_{0}^2}{4 \kappa}$, see
127 \url{http://en.wikipedia.org/wiki/Fick\%27s_laws_of_diffusion}}.
128
129 \begin{figure}[ht]
130 \begin{center}
131 \includegraphics[width=4in]{figures/ttrodpyplot150}
132 \caption{Example 2: Total Energy in the Iron Rod over Time (in seconds)}
133 \label{fig:onedheatout1 002}
134 \end{center}
135 \end{figure}
136
137 \begin{figure}[ht]
138 \begin{center}
139 \includegraphics[width=4in]{figures/rodpyplot001}
140 \includegraphics[width=4in]{figures/rodpyplot050}
141 \includegraphics[width=4in]{figures/rodpyplot200}
142 \caption{Example 2: Temperature ($T$) distribution in the iron rod at time steps
143 $1$, $50$ and $200$}
144 \label{fig:onedheatout 002}
145 \end{center}
146 \end{figure}
147
148 \section{For the Reader}
149 \begin{enumerate}
150 \item Move the boundary line between the two granite blocks to another part of
151 the domain.
152 \item Split the domain into multiple granite blocks with varying temperatures.
153 \item Vary the mesh step size. Do you see a difference in the answers? What
154 does happen with the compute time?
155 \item Insert an internal heat source (Hint: The internal heat source is given
156 by $q_{H}$.)
157 \item Change the boundary condition for the iron rod example such that the
158 temperature
159 at the right end is kept at a constant level $T_{ref}$, which
160 corresponds to the installation of a cooling element (Hint: Modify $q$ and
161 $r$).
162 \end{enumerate}
163

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