Heat Exchanger :

Summary Here we shall learn more about geometry input and triangulation files, as well as read and write#triangulationfiles,aswellasreadandwrite operations.

The problem Let \(\{C_{i}\}_{1,2} \), be 2 thermal conductors within an enclosure \(C_0\). The first one is held at a constant temperature \({u} _{1} \) the other one has a given thermal conductivity \(\kappa_2\) 5 times larger than the one of \(C_0\). We assume that the border of enclosure \(C_0\) is held at temperature \(20^\circ C\) and that we have waited long enough for thermal equilibrium.

In order to know \({u} (x) \) at any point \(x\) of the domain \(\Omega\), we must solve $$ \nabla\cdot(\kappa\nabla{u}) =0    \hbox{ in } \Omega, \quad {u}_{|\Gamma} =g $$ where \(\Omega\) is the interior of \(C_0\) minus the conductors \(C_1\) and \(\Gamma\) is the boundary of \(\Omega\), that is \(C_0\cup C_1 \) Here \(g\) is any function of \(x\) equal to \({u}_i\) on \(C_i\). The second equation is a reduced form for: $$ {u} ={u} _{i} \hbox{ on }    C_{i}, \quad i=0,1. $$ The variational formulation for this problem is in the subspace \(H^1_0(\Omega) \subset H^1(\Omega)\) of functions which have zero traces on \(\Gamma\). $$ u-g\in H^1_0(\Omega) : \int_\Omega\nabla u\nabla v =0   \forall v\in H^1_0(\Omega) $$ Let us assume that \(C_0\) is a circle of radius 5 centered at the origin,

 \(C_i\) are

rectangles, \(C_1\) being at the constant temperature \(u_1=60^\circ C\).

Example[heatex.edp]

 // file heatex.edp
 int C1=99, C2=98; // could be anything such that $\ne 0$ and $ C1 \ne C2$
 border C0(t=0,2*pi){x=5*cos(t); y=5*sin(t);}

 border C11(t=0,1){ x=1+t;  y=3;      label=C1;}
 border C12(t=0,1){ x=2;    y=3-6*t;  label=C1;}
 border C13(t=0,1){ x=2-t;  y=-3;     label=C1;}
 border C14(t=0,1){ x=1;    y=-3+6*t; label=C1;}

 border C21(t=0,1){ x=-2+t; y=3;      label=C2;}
 border C22(t=0,1){ x=-1;   y=3-6*t;  label=C2;}
 border C23(t=0,1){ x=-1-t; y=-3;     label=C2;}
 border C24(t=0,1){ x=-2;   y=-3+6*t; label=C2;}

 plot(    C0(50)            // to see the border of the domain
         + C11(5)+C12(20)+C13(5)+C14(20)
         + C21(-5)+C22(-20)+C23(-5)+C24(-20),
         wait=true, ps="heatexb.eps");

 mesh Th=buildmesh(    C0(50)
                     + C11(5)+C12(20)+C13(5)+C14(20)
                     + C21(-5)+C22(-20)+C23(-5)+C24(-20));
 plot(Th,wait=1);

 fespace Vh(Th,P1); Vh u,v;
 Vh kappa=1+2*(x<-1)*(x>-2)*(y<3)*(y>-3);
 solve a(u,v)= int2d(Th)(kappa*(dx(u)*dx(v)+dy(u)*dy(v)))
                 +on(C0,u=20)+on(C1,u=60);
 plot(u,wait=true, value=true, fill=true, ps="heatex.eps");

Note the following:

     and C2 is oriented counterclockwise.
    This is why C1 is viewed as a hole by \tt buildmesh.
    same logical unit to input the boundary conditions in a readable way we
    assigned a label on the boundaries.  As said earlier, borders
    have an internal number corresponding to their order in the program (check it by
    adding a \tt cout<<C22; above). This is essential to understand how a mesh can be
    output to a file and re-read (see below).
    boundary. It is not possible to change the (uniform) distribution of vertices but a
    piece of boundary can always be cut in two or more parts, for instance C12 could be
    replaced by C121+C122:
// border C12(t=0,1){ x=2; y=3-6*t; label=C1;}
     border C121(t=0,0.7){ x=2;    y=3-6*t;  label=C1;}
     border C122(t=0.7,1){ x=2;    y=3-6*t;  label=C1;}
     ... buildmesh(.../* C12(20) */ + C121(12)+C122(8)+...);

[htbp]

   

[figheatex] The heat exchanger

Exercise Use the symmetry of the problem with respect to the axes; triangulate only one half of the domain, and set Dirichlet conditions on the vertical axis, and Neumann conditions on the horizontal axis. \bigskip

Writing and reading triangulation files Suppose that at the end of the previous program we added the line

 savemesh(Th,"condensor.msh");

and then later on we write a similar program but we wish to read the mesh from that file. Then this is how the condenser should be computed:

 mesh Sh=readmesh("condensor.msh");
 fespace Wh(Sh,P1); Wh us,vs;
 solve b(us,vs)= int2d(Sh)(dx(us)*dx(vs)+dy(us)*dy(vs))
                 +on(1,us=0)+on(99,us=1)+on(98,us=-1);
 plot(us);

Note that the names of the boundaries are lost but either their internal number (in the case of C0) or their label number (for C1 and C2) are kept.