Question

Consider steady one-dimensional heat conduction in a plane wall with variable heat generation and constant thermal conductivity. The nodal network of the medium consists of nodes \(0,1,2\), and 3 with a uniform nodal spacing of \(\Delta x\). The temperature at the left boundary (node 0 ) is specified. Using the energy balance approach, obtain the finite difference formulation of boundary node 3 at the right boundary for the case of combined convection and radiation with an emissivity of \(\varepsilon\), convection coefficient of \(h\), ambient temperature of \(T_{\infty}\), and surrounding temperature of $T_{\text {sarr }}$. Also, obtain the finite difference formulation for the rate of heat transfer at the left boundary.

Short answer

Answer: The finite difference formulation for node 3 is \(RT1 - 2RT3 + h(T_{\infty}-T3) + \varepsilon\sigma(T_{\text{sarr}}^4 - T3^4) = 0\), and the finite difference formulation for the rate of heat transfer at node 0 is \(q_0 = k\frac{T1 - T0}{\Delta x}\).

Step-by-step solution

Analyze heat conduction equation

The heat conduction equation for a plane wall with variable heat generation and constant thermal conductivity (\(k\)) is given by: \(\begin{equation} \frac{d^2T}{dx^2}=-\frac{q_g}{k} \end{equation}\) Where \(q_g\) is the heat generation per unit volume, \(T\) is the temperature, and \(x\) is the distance along the wall.

Apply energy balance for a node

For any boundary node \(i\) in the nodal network, the energy balance equation can be written as: \(\begin{equation} \frac{k}{\Delta x^2}(T_{i-1}-2T_i+T_{i+1}) = -q_g \end{equation}\)

Apply boundary conditions for node 3

At node 3, the plane wall is subjected to combined convection and radiation. Hence, the energy balance for node 3 is: \(\begin{equation} \frac{k}{\Delta x^2}(T1 - 2T3) + h(T_{\infty}-T3) + \varepsilon\sigma(T_{\text{sarr}}^4 - T3^4) = 0 \end{equation}\) Here, \(\sigma\) is the Stefan-Boltzmann constant. To simplify this equation, we can set \(R = \frac{k}{\Delta x^2}\), and replace \(T2\) with \(T0 - 2\Delta x\) assuming that node 0 is fixed at \(T0\).

Derive finite difference formulation for node 3

Using the equation obtained in Step 3, we can rewrite the energy balance for node 3 as: \(\begin{equation} RT1 - 2RT3 + h(T_{\infty}-T3) + \varepsilon\sigma(T_{\text{sarr}}^4 - T3^4) = 0 \end{equation}\) This is the finite difference formulation for node 3.

Derive finite difference formulation for rate of heat transfer at node 0

For the left boundary node 0, as the temperature is specified, the rate of heat transfer can be obtained using the energy balance equation with \(q_{0} = k\frac{T_{1}-T_{0}}{\Delta x}\): \(\begin{equation} q_0 = k\frac{T1 - T0}{\Delta x} \end{equation}\) This is the finite difference formulation for the rate of heat transfer at the left boundary node 0.