The Boundary value problem when an electric current produces a rate of heat generation linear in temperature

In this section, Qgen in equation (3.1) is developed as a function of temperature to obtain a second order linear ODE with specified boundary conditions.

Now suppose the rod of radius R is of finite length L. On the end surfaces, z = 0 and z = L, let the temperatures be maintained at the constant values T0 and TL, respectively. As before, assume that the thermal conductivity k is constant, and that (dT / dr) = 0 at r = R (the cylindrical surface is insulated). Let a constant current I (amperes) pass through the rod and assume that the electrical resistivity, , of the rod is a linear function of temperature:
(3.5)
units: ohms . m
The power density based on Ohm's law is
(3.6)
units: ohms . m
where Ac = cross-sectional area of the rod (m2) and is given in (3.5).
Substituting from (3.5) and (3.6) into (3.1), we get the differential equation
 
 
which can be simplified to
(3.7)
 
where    
(3.8)
 

Note that a, b (and therefore A, B) are to be understood as prescribed positive constants, which depend on the material making up the rod.

We now have to solve the boundary value problem consisting of the second order ODE (3.7) with the boundary conditions
(3.9)
 

to obtain the temperature distribution in the rod.


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