MATLAB: Specify boundary conditions for a thermal model (PDE)

boundary conditionspdethermal model

Hello everyone,
I need to simulate the heat transfer on a surface by using the PDE Toolbox. I have defined the geometry, the mesh, thermal properties of the material, the internal heat source, but I don't know how to define the following boundary conditions:
  • Surface boundary condition (z=0): where S is the heat flux proportional to the incidence angle (i), sigma is the Stefan-Boltzmann constant, epsilon is the emissivity, k the thermal conductivity.
  • Bottom boundary condition (z=D): where Q is the heat production of the body
Does anyone have any ideas?
Thanks for any help!
Pamela
thermalmodel = createpde('thermal','transient');
%% Geometry
gm = multicuboid(1,1,[0.1 1.9],'ZOffset',[0 0.1]);
thermalmodel.Geometry = gm;
gm2 = rotate(gm,180,[0 0 0],[0 1 0]);
thermalmodel.Geometry = gm2;
%pdegplot(thermalmodel,'CellLabels','on','FaceAlpha',0.5)
%% Generate a mesh for the geometry
msh = generateMesh(thermalmodel,'Hmax', 5);
%% Specify thermal properties of the material
%--------------------------------------------------------------------------
%DENSITY (rho)
rhob = 1000;
rho = @(location,state)rhob*((location.z + 0.122)/(location.z + 0.18));
%SPECIFIC HEAT (cp)
ca=1;
cb=3;
cc=5;
cd=0.4;
ce=9;
cp = @(location,state)ca+cb*state.u+cc*(state.u)^2+cd*(state.u)^3+ce*(state.u)^4;
%THERMAL CONDUCTIVITY (k)
kb = 0.8;
ks = 0.7;
rhos = 2100;
sigma = 8.67e-08;
epsilon = 0.95;
l = 5.5e-05;
beta = 4*sigma*epsilon*l;
kc = @(location,state)kb-(kb-ks)*((rhob-(rho(location,state))))/(rhob-rhos);
k = @(location,state)beta*(state.u)^3+(kc(location,state));
thermalProperties(thermalmodel,'ThermalConductivity',k, 'MassDensity',rho,'SpecificHeat',cp)
%% Internal heat source [W m^-2]
Q = 0.020
internalHeatSource(thermalmodel,Q)

Best Answer

You may refer the applyBoundaryCondition function and the solvepde function to see if it fits your requirement. You may also refer this example to use the functions mentioned.
You may refer to a similar solution here.