LaserXC
Previous  Top  Next

{ LASERXC.PDE }

{ **************************************************************

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

************************************************************** }

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

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

definitions
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}
source=0

Initial values
temp = 50 { estimate solution for quicker convergence }

equations { define the heatflow equation }
div(k*grad(temp)) + source = 0;

boundaries
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

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

plots
grid(x,y)
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"
surface(temp)

end