@@ -103,14 +103,39 @@ <h2 id="mathematicalformulation"><a name="Mathematical formulation"></a>Mathemat
103103 < img src ="../assets/HeatConduction1DWall.png " width ="100 " />
104104
105105 < p >
106- The above schematic illustrates the problem domain and outlines the associated boundary conditions. In
107- particular, a convection boundary condition, implemented as Robin type, is applied at the inner side of
108- the wall (\(x = 0\)). The latter is expressed as \(\frac{dT}{dx}|_{x=0}=-{\frac{h}{k}}(T-T_{in})\),
109- where \(h\) is the heat transfer coefficient, \(k\) the thermal conductivity and \(T_{in}\) is the
110- internal temperature. We assume here that \({\frac{h}{k}}\) = 1 m< sup > -1</ sup > and \(T_{in}\) = 25
111- °C. At the outer side of the wall (\(x = W\), where \(W\) = 0.15 m, is the width of the wall), we
112- apply a constant temperature (\(T_0\) = 5 °C) boundary condition, implemented as Dirichlet type in
113- the finite element code.
106+ The above schematic illustrates the problem domain and outlines the associated boundary conditions. Our
107+ objective is to determine the temperature distribution within the wall itself. In this problem, a
108+ convection boundary condition, implemented as Robin type, is applied at the inner side of the wall (\(x
109+ = 0\)). The latter is expressed as \(\frac{dT}{dx}|_{x=0}=-{\frac{h}{k}}(T-T_{in})\), where \(h\) is the
110+ heat transfer coefficient, \(k\) the thermal conductivity, and \(T_{in}\) is the internal temperature of
111+ the house (room temperature). We assume here that \({\frac{h}{k}}\) = 1 m< sup > -1</ sup > and \(T_{in}\) =
112+ 25 °C. At the outer side of the wall (\(x = W\), where \(W\) = 0.15 m, is the width of the wall), we
113+ apply a constant temperature (\(T_0\) = 5 °C) boundary condition, representing the outdoor
114+ temperature, implemented as Dirichlet type in the finite element code.
115+ </ p >
116+
117+ < p >
118+ This problem has an analytical solution. Since the governing equation is \(\frac{d^2T}{dx^2} = 0\),
119+ the general solution is a linear function \(T(x) = Ax + B\), where \(A\) and \(B\) are constants
120+ determined by the boundary conditions:
121+ </ p >
122+ < ol >
123+ < li >
124+ At \(x=0\): \(\frac{dT}{dx} = -\frac{h}{k}(T-T_{in})\)
125+ </ li >
126+ < li >
127+ At \(x=W\): \(T(W) = T_0\)
128+ </ li >
129+ </ ol >
130+ < p >
131+ Solving these equations for our specific values (\(\frac{h}{k} = 1\) m< sup > -1</ sup > , \(T_{in} = 25\) °C,
132+ \(T_0 = 5\) °C, and \(W = 0.15\) m), we get the analytical solution:
133+ </ p >
134+ < p >
135+ \(T(x) \approx 23.53x + 1.47\)
136+ </ p >
137+ < p >
138+ The numerical solution using finite elements should closely match this analytical result.
114139 </ p >
115140
116141 < h2 id ="solvingwithfeascript "> < a name ="Solving with FEAScript "> </ a > Solving with FEAScript</ h2 >
@@ -228,8 +253,8 @@ <h2 id="results"><a name="Results"></a>Results</h2>
228253 } ) ;
229254
230255 // Define boundary conditions
231- // model.addBoundaryCondition("0", ["convection", 1, 25]);
232- model . addBoundaryCondition ( "0" , [ "constantTemp" , 25 ] ) ; // Temporary BC for testing
256+ model . addBoundaryCondition ( "0" , [ "convection" , 1 , 25 ] ) ;
257+ // model.addBoundaryCondition("0", ["constantTemp", 25]); // Temporary BC for testing
233258 model . addBoundaryCondition ( "1" , [ "constantTemp" , 5 ] ) ;
234259
235260 // Set solver method
@@ -276,8 +301,8 @@ <h2 id="results"><a name="Results"></a>Results</h2>
276301 } ) ;
277302
278303 // Define boundary conditions
279- // model.addBoundaryCondition("0", ["convection", 1, 25]);
280- model . addBoundaryCondition ( "0" , [ "constantTemp" , 25 ] ) ; // Temporary BC for testing
304+ model . addBoundaryCondition ( "0" , [ "convection" , 1 , 25 ] ) ;
305+ // model.addBoundaryCondition("0", ["constantTemp", 25]); // Temporary BC for testing
281306 model . addBoundaryCondition ( "1" , [ "constantTemp" , 5 ] ) ;
282307
283308 // Set solver method
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