Starting with an energy balance on a ring-shaped volume element, derive the two-dimensional steady heat conduction equation in cylindrical coordinates for \(T(r, z)\) for the case of constant thermal conductivity and no heat generation.

Short Answer

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Question: Derive the 2D steady heat conduction equation in cylindrical coordinates for a temperature function, T(r, z), with constant thermal conductivity and no heat generation. Answer: The 2D steady heat conduction equation in cylindrical coordinates is given by: $$\frac{\partial^2 T}{\partial r^2}+\frac{1}{r}\frac{\partial T}{\partial r}+\frac{\partial^2 T}{\partial z^2}=0.$$

Step by step solution

01

Visualize the ring-shaped volume element in cylindrical coordinates

Consider a ring-shaped volume element in cylindrical coordinates \((r, \theta, z)\) with inner radius \(r\), outer radius \(r + \Delta r\), height \(\Delta z\), and thickness \(\Delta \theta\). The volume of the element is given by \(V = \Delta V = (r\Delta\theta)(\Delta r)(\Delta z)\).
02

Set up the energy balance equation

For steady-state heat conduction with no heat generation, the energy balance equation states that the amount of heat entering the volume element is equal to the amount of heat leaving it. We will consider heat transfer in the radial and axial directions only, since the problem is two-dimensional. Thus, we can write the energy balance as: $$q_r(r) - q_r(r+\Delta r) + q_z(z) - q_z(z+\Delta z) = 0.$$
03

Express heat flux in terms of temperature and thermal conductivity

The heat flux in the radial and axial directions can be expressed using Fourier's law of heat conduction: $$q_r = -k\frac{\partial T}{\partial r}\text{ and }q_z = -k\frac{\partial T}{\partial z},$$ where \(k\) is the constant thermal conductivity.
04

Substitute the expressions for heat flux and evaluate the limits

Substitute the expressions for \(q_r\) and \(q_z\) into the energy balance equation and evaluate the result in the limit as \(\Delta r\) and \(\Delta z\) approach zero: $$(-k\frac{\partial T}{\partial r})_{r} + (-k\frac{\partial T}{\partial r})_{r+\Delta r}+(-k\frac{\partial T}{\partial z})_{z} + (-k\frac{\partial T}{\partial z})_{z+\Delta z}=0.$$ Divide the entire equation by \(k\Delta r\Delta z\) and simplify, $$\frac{1}{\Delta r} \left(\frac{\partial T}{\partial r}|_{r+\Delta r} - \frac{\partial T}{\partial r}|_{r}\right) + \frac{1}{\Delta z} \left(\frac{\partial T}{\partial z}|_{z+\Delta z} - \frac{\partial T}{\partial z}|_{z}\right) =0.$$
05

Recognize the limits as partial derivatives and obtain the 2D steady-state heat conduction equation in cylindrical coordinates

As \(\Delta r \to 0\) and \(\Delta z \to 0\), the expression on the left-hand side of the equation represents the sum of two second-order partial derivatives: $$\frac{\partial^2 T}{\partial r^2}+\frac{1}{r}\frac{\partial T}{\partial r}+\frac{\partial^2 T}{\partial z^2}=0.$$

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Most popular questions from this chapter

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