Sample Problems > Applications > Lasers > laser

{ LASER_HEATFLOW.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, LLNL

}

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 }

temp : 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

laser_heatflow

laser_heatflow