Electronic Properties of Materials: Conduction - Problems and Solutions

Back to Conduction Problems and Solutions

Consider a coaxial cable that has a copper core conductor and polyethylene dielectric with the following properties: Core conductor resistivity of 19 n-ohm m, core radius of 4 mm, dielectric thickness of 3.5 mm, dielectric thermal conductivity of 0.3 W (m-K)-1. The outside temperature is 25oC. The cable is carrying a current of 500A. What is the temperature of the inner conductor?       


We need to apply the Fourier's law of heat conduction. Consider a small element of thickness dr at a distance r from the center of the coaxial cable of length L as shown in the figure below. The surface area of this element will be A = 2{pi}rL. Now apply the Fourier's law:

Q{prime} = - A{kappa}{{dT}/{dr}}
Or, Q{prime} = - 2{pi}rL{kappa}{{dT}/{dr}}
Or, Q{prime}dr = - 2{pi}rL{kappa}{dT}
Or, Q{prime}int{r_i}{r_o}{{dr/r}} = - 2{pi}L{kappa}int{T_i}{T_o}{dT}

Solving this integration gives a general equation for a coaxial cable:
Q{prime} = - ({T_o}-{T_i})[{{{2{pi}{kappa}L}/{ln{{r_o}/{r_i}}}}}] = -{Delta}T [{{{2{pi}{kappa}L}/{ln{{r_o}/{r_i}}}}}] = {{-{Delta}T} /{theta}}
Where, thermal resistance {theta} = {ln{{r_o}/{r_i}}}/{2{pi}{kappa}L}


The outside temperature, To, is 25 + 273 = 298 K. We have been asked to determine the inside temperature, Ti, for which we can use the general equation derived above. Now we need to determine the heat flow per unit time (Q').
Q{prime} = {I^2}R = {I^2}{{{rho}L}/{A}}
doubleright Q{prime} = 94.498L Watts
and from the thermal resistance formula we get {theta} = {{0.333}/{L}} KW-1.
Now using the equation {Delta}T = - Q{prime}{theta}, we find {Delta}T = - 31.468 K.
Therefore, the inner temperature (Ti) = 31.468 + 298 = 329.468 K or 56.468oC.