Question

Starting with an energy balance on a cylindrical shell volume element, derive the steady one-dimensional heat conduction equation for a long cylinder with constant thermal conductivity in which heat is generated at a rate of \(\dot{e}_{\text {gen }}\).

Short answer

Question: Derive the steady one-dimensional heat conduction equation for a long cylinder with constant thermal conductivity and a heat generation rate within the cylinder. Answer: The steady one-dimensional heat conduction equation for a long cylinder with constant thermal conductivity and heat generation rate \(\dot{e}_{\text {gen }}\) is given by: (d/dr)(k dQ/dr) + \(\dot{e}_{\text {gen }}\)(d(r^2)/dr) = 0

Step-by-step solution

Create the differential cylindrical element

Let's create a cylindrical element with an inner radius r, an outer radius (r+dr), and a length L. The volume of the cylindrical shell ΔV is given by the difference in the volume of the two cylinders: ΔV = π[(r + dr)^2 - r^2]L

Calculate the heat generated in the element

If there is a heat generating rate \(\dot{e}_{\text {gen }}\) per unit volume, the total heat generated in the cylindrical shell ΔV is given by: ΔQ_gen = \(\dot{e}_{\text {gen }}\)ΔV

Calculate the heat conducted across the element

Using Fourier's law of heat conduction, the rate of heat flow across the inner and outer boundaries of the cylindrical element is given by: q_in = -k(2πrL) (dQ/dr)|_r q_out = -k(2π(r+dr)L) (dQ/dr)|_(r+dr) where k is the constant thermal conductivity.

Perform an energy balance on the cylindrical element

Since this is a steady-state system, there is no accumulation of heat within the element. Therefore, the energy balance equation is: Heat conducted into the element + Heat generated in the element = Heat conducted out of the element Which can be rewritten as: q_in + ΔQ_gen = q_out

Substitute equations to get the heat conduction equation

Now we substitute the expressions for q_in, q_out, and ΔQ_gen from steps 2 and 3 into the energy balance equation: -k(2πrL) (dQ/dr)|_r + \(\dot{e}_{\text {gen }}\)π[(r + dr)^2 - r^2]L = -k(2π(r+dr)L) (dQ/dr)|_(r+dr)

Simplify the equation and remove constants

Now we rearrange and simplify the equation: -k(2πL) [(dQ/dr)|_(r+dr) - (dQ/dr)|_r] = \(\dot{e}_{\text {gen }}\)π[(r + dr)^2 - r^2]L Next, we divide both sides by 2πLΔr to remove the constants and take the limit as dr approaches 0: (1/dr)[-k (dQ/dr)|_(r+dr) + k (dQ/dr)|_r] = (1/dr)[\(\dot{e}_{\text {gen }}\)[(r + dr)^2 - r^2]] In the limit as dr approaches 0: (-d/dr)(k dQ/dr) = \(\dot{e}_{\text {gen }}\) d(r^2)/dr

Finalize the heat conduction equation

Rearrange the equation to obtain the one-dimensional heat conduction equation for a long cylinder with constant thermal conductivity and heat generation rate \(\dot{e}_{\text {gen }}\): (d/dr)(k dQ/dr) + \(\dot{e}_{\text {gen }}\)(d(r^2)/dr) = 0 The derived equation represents the heat conduction equation for a long cylinder with constant thermal conductivity and heat generation.