Question

Heat is generated in a long wire of radius \(r_{e}\) at a constant rate of \(\dot{e}_{\text {gen }}\) per unit volume. The wire is covered with a plastic insulation layer. Express the heat flux boundary condition at the interface in terms of the heat generated.

Short answer

To express the heat flux boundary condition at the interface in terms of the heat generated, we first calculate the total heat generated in the wire as \(Q_{gen} = \dot{e}_{\text{gen}} \times \pi r_e^2 l\). Then, we apply Fourier's law of heat conduction, \(q_r = -k_r \frac{dT}{dr}\), and calculate the heat flux at the interface as \(q_r(r_e) = -k_r \frac{dT}{dr}\bigg|_{r=r_e}\). Finally, we substitute the heat flux per unit area in terms of heat generated and obtain the heat flux boundary condition: \(q_r(r_e) = -k_r \frac{dT}{dr}\bigg|_{r=r_e} =\dot{e}_{\text{gen}} r_e\).

Step-by-step solution

Calculate the total heat generated in the wire

Since heat is generated at a constant rate, the total heat generated in the wire (\(Q_{gen}\)) can be calculated by multiplying the volume of the wire by the heat generated per unit volume. The volume of the wire can be calculated using the formula for the volume of a cylinder: \(V = \pi r_e^2 l\), where \(l\) is the length of the wire. Hence, the total heat generated is given by: \(Q_{gen} = \dot{e}_{\text{gen}} \times V = \dot{e}_{\text{gen}} \times \pi r_e^2 l\).

Apply Fourier's law of heat conduction

Fourier's law of heat conduction states that heat flux (q) is proportional to the temperature gradient, i.e., \(q = -k \nabla T\), where \(k\) is the thermal conductivity, and \(\nabla T\) is the temperature gradient. In this situation, the temperature gradient will be along the radial direction, so the heat flux \(q\) can be written as: \(q_r = -k_r \frac{dT}{dr}\), where \(k_r\) is the radial thermal conductivity, and \(\frac{dT}{dr}\) is the temperature gradient in the radial direction.

Calculate the heat flux boundary condition

Since we want to find the heat flux boundary condition at the interface, we need to determine the heat flux at \(r = r_e\). We can write the heat flux at the interface as follows: \(q_r(r_e) = -k_r \frac{dT}{dr}\bigg|_{r=r_e}\). Now, using the total heat generated \(Q_{gen}\) determined in Step 1, we can calculate the heat flux per unit area at the interface: \(\frac{Q_{gen}}{2 \pi r_e l} = \dot{e}_{\text{gen}} r_e\).

Express the heat flux boundary condition in terms of heat generated

Finally, we can substitute the heat flux per unit area calculated in Step 3 into the expression for the heat flux at the interface: \(q_r(r_e) = -k_r \frac{dT}{dr}\bigg|_{r=r_e} =\dot{e}_{\text{gen}} r_e\). That's the heat flux boundary condition at the interface in terms of the heat generated.