Consider transient heat conduction in a plane wall with variable heat generation and constant thermal conductivity. The nodal network of the medium consists of nodes \(0,1,2,3,4\), and 5 with a uniform nodal spacing of \(\Delta x\). The wall is initially at a specified temperature. Using the energy balance approach, obtain the explicit finite difference formulation of the boundary nodes for the case of insulation at the left boundary (node 0 ) and radiation at the right boundary (node 5) with an emissivity of \(\varepsilon\) and surrounding temperature of \(T_{\text {surr }}\).

We will first write the heat balance equation for an internal node, say node i. Considering the heat generation within the volume and neglecting the heat conduction in the y and z directions, the energy balance equation can be written as: \(Q_{in} - Q_{out} = \dot{Q}_{gen} \Delta V\) For the finite difference method, we can write the heat flux using Taylor's series expansion: \(Q_{in} = k \frac{T_{i-1} - T_{i}}{\Delta x}\) and \(Q_{out} = k \frac{T_{i} - T_{i+1}}{\Delta x}\) Then, the energy balance equation becomes: \(k \frac{T_{i-1} - T_{i}}{\Delta x} - k \frac{T_{i} - T_{i+1}}{\Delta x} = \dot{Q}_{gen} \Delta x \Delta y \Delta z\)

Now, we will write the energy balance equation for the boundary nodes. Node 0 (insulated boundary condition): As there is no heat flux through the left boundary (insulated), we need to consider only the heat flux between node 0 and node 1: \(Q_{0 \to 1} = k \frac{T_{1} - T_{0}}{\Delta x}\) Node 5 (radiative boundary condition): The heat flux from node 5 to the surroundings is given by: \(Q_{5 \to \text{surr}} = \varepsilon \sigma\left((T_5)^4 - (T_{\text {surr }})^4\right)\) The heat flux between node 5 to node 4 is given by: \(Q_{5 \to 4} = k \frac{T_{4} - T_{5}}{\Delta x}\) So, the energy balance at node 5 becomes: \(k \frac{T_{4} - T_{5}}{\Delta x} = \varepsilon \sigma \left( (T_5)^4 - (T_{\text {surr }})^4 \right)\)

Now, we need to discretize the energy balance equation for the internal nodes to get the finite difference equations. For internal nodes, the finite difference equation can be derived from the energy balance equation derived in step 1: \(\frac{k}{\Delta x}(T_{i-1} - 2T_i + T_{i+1}) = \dot{Q}_{gen} \Delta x \Delta y \Delta z\) For boundary nodes, we can write the finite difference equations based on the energy balance equation derived in step 2: Node 0 (insulated boundary condition): \(T_{0} - T_{1} = 0\) Node 5 (radiative boundary condition): \(T_{4} - T_{5} = \frac{\varepsilon \sigma \Delta x}{k} (T_5)^4 - \frac{\varepsilon \sigma \Delta x}{k} (T_{\text {surr }})^4\) Now, we have obtained the explicit finite difference equations for the boundary nodes as well as the internal nodes. These equations can be used to solve the transient heat conduction problem in the plane wall with variable heat generation and given boundary conditions.