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   This problem shows a complex heatflow application.

   A rod laser is glued inside a cylinder of copper.

   Manufacturing errors allow the rod to move inside the glue, leaving a

   non-uniform glue layer around the rod.  The glue is an insulator, and

   traps heat in the rod.  The copper cylinder is cooled only on a 60-degree

   portion of its outer surface.

   The laser rod has a temperature-dependent conductivity.

   We wish to find the temperature distribution in the laser rod.

   The heat flow equation is

      div(K*grad(Temp)) + Source = 0.

   We will model a cross-section of the cylinder. While this is a cylindrical

   structure, in cross-section there is no implied rotation out of the

   cartesian plane, so the equations are cartesian.

                               -- Submitted by Luis Zapata, LLNL



title "Nd:YAG Rod - End pumped. 200 W/cm3 volume source. 0.005in uropol"



   temp   { declare "temp" to be the system variable }



   k = 3               { declare the conductivity parameter for later use }

   krod=39.8/(300+temp){ Nonlinear conductivity in the rod.(W/cm/K) }

   Rod=0.2             { cm Rod radius }

   Qheat=200           { W/cc, heat source in the rod }

   kuropol=.0019   { Uropol conductivity }

   Qu=0           { Volumetric source in the Uropol }

   Ur=0.005       { Uropol annulus thickness in r dim }

   kcopper=3.0     { Copper conductivity }

   Rcu=0.5         { Copper convection surface radius }

   tcoolant=0.     { Edge coolant temperature }

   ASE=0.         { ASE heat/area to apply to edge, heat bar or mount }


initial values

   temp = 50       { estimate solution for quicker convergence }

equations           { define the heatflow equation }

   temp : div(k*grad(temp)) + source = 0;


  region 1       { the outer boundary defines the copper region }

       k = kcopper

      start (0,-Rcu)

      natural(temp) = -2 * temp       {convection boundary}

          arc(center=0,0) angle 60

      natural(temp) = 0               {insulated boundary}

          arc(center=0,0) angle 300

          arc(center=0,0) to close

  region 2       { next, overlay the Uropol in a central cylinder }

       k = kuropol

      start (0,-Rod-Ur) arc(center=0,0) angle 360


  region 3       { next, overlay the rod on a shifted center }

       k = krod

       Source = Qheat

      start (0,-Rod-Ur/2) arc(center=0,-Ur/2) angle 360


  grid(x,y) zoom(-8*Ur, -(Rod+8*Ur),16*Ur,16*Ur)




  contour (temp)

  contour(temp) zoom(-(Rod+Ur),-(Rod+Ur),2*(Rod+Ur),2*(Rod+Ur))

  contour(temp) zoom(-(Rod+Ur)/4,-(Rod+Ur),(Rod+Ur)/2,(Rod+Ur)/2)

  vector(-k*dx(temp),-k*dy(temp)) as "heat flow"