# thermal conductivity, pde toolbox

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juan ugalde on 12 Oct 2018
Commented: Ravi Kumar on 22 Oct 2018
Dear all,
I have a question concerning the pde tool box. Indeed, I would like to model the thermal behaviour of a cyclindrical battery. For said purpose, I would like to specify the thermal conductivities in the x and y directions. How can I specify said parameters using the following code?
specifyCoefficients(Thermal, 'm',0,...% 'd',(para.rho)/para.cp,...%rho*cp 'c',???,...%k 'a',0,...%0 'f',Heat)%source

Ravi Kumar on 15 Oct 2018
Hi Juan,
I recommend using the new ThermalModel workflow.
In the ThermalModel based workflow you can use the function thermalProperties to define thermal conductivity, mass density, and specific heat, all of which can also be function spatial coordinates (x,y) and the temperature itself (nonlinear).
For example if your thermal conductivity is linear function of x & y, then you can define the thermal conductivity as a function:
function k = kFunc(region,state)
k = 1 + 0.1*region.x + 0.01*region.y;
end
You can then specify kFunc in thermalProperties function as in the example below:
model = createpde('thermal','transient');
importGeometry(model,'Block.stl');
thermalProperties(model,'ThermalConductivity',@kFunc,'MassDensity',1,'SpecificHeat',1)
generateMesh(model);
thermalBC(model,'Face',1,'Temperature',1)
thermalIC(model,0);
R = solve(model,[0,0.01])
pdeplot3D(model,'ColorMapData',R.Temperature(:,2))
Regards, Ravi
juan ugalde on 16 Oct 2018
Dear Ravi,
If the radial thermal conductivity (x) is equal to 0.15(W/m.K) and the axial(y) is equal to 30, is the following code correct?
function k = kFunc(region,state)
k = 0.15*region.x + 30*region.y;
end
Moreover, can I specify a heat source, f, such as
function f = fFunc(region,state)
f = 2*state.t;
end
where the heat source increases as time goes by?
Thanks again.
Juan.

Ravi Kumar on 16 Oct 2018
Hi Juan,
Looks like you have orthotropic thermal conductivity, not spatially varying conductivity. in that case, you can just use the matrix form of thermal conductivity, k = [0.15, 0 ; 0, 30].
thermalProperties(model,'ThermalConductivity',k,'MassDensity',1,'SpecificHeat',1)
Note if the problem is in 3-D space, then you need to specify a 3x3 matrix of conductivity. You don't need a function for the case you describe now.
You can specify heat source as a function of time using the function fFunc. Note that you can specify such heat source as internalHeatSource, if it is a domain heat source. If it is a boundary heat input, then you need to use 'HeatFlux' option of thermalBC function.
Regards,
Ravi
##### 2 CommentsShowHide 1 older comment
Ravi Kumar on 22 Oct 2018
Hi Juan,
I think you might have figured out answer for your question by now. Solver works by discretization of the domain. So, solver needs value of heat source at several points on the boundary. The coordinates points at which heat source should be provided by the user are passed to the function as location.x, location.y, etc. So what you have done is correct.
Regards,
Ravi