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 \(0,1,2,3,4\), and 5 with a uniform nodal spacing of \(\Delta x\). The temperature at the right boundary (node 5) is specified. Using the energy balance approach, obtain the finite difference formulation of the boundary node 0 on the left boundary for the case of combined convection, radiation, and heat flux at the left boundary with an emissivity of \(\varepsilon\), convection coefficient of \(h\), ambient temperature of \(T_{\infty}\), surrounding temperature of \(T_{\text {surr }}\) and uniform heat flux of \(\dot{q}_{0}\). Also, obtain the finite difference formulation for the rate of heat transfer at the right boundary.

Short answer

Answer: The finite difference formulations for the left and right boundaries are: Left boundary (Node 0): \(T_{0}(k\Delta x^{-1} + h + 4\varepsilon\sigma T_{0}^{3}) - T_{1}(k\Delta x^{-1}) = (hT_{\infty} + \varepsilon\sigma T_{\text{surr}}^{4} + \dot{q}_{0})\) Right boundary (Node 5): \(T_{5}(k\Delta x^{-1}) - T_{4}(k\Delta x^{-1}) = 0\)

Step-by-step solution

Energy balance at the left boundary (Node 0)

At the left boundary, we will have conduction, convection, and radiation heat transfer mechanisms. We can use the energy balance equation, considering all these mechanisms, to calculate the finite difference formulation for the left boundary. The energy balance equation for node 0 can be written as: \(\text{Heat in} - \text{Heat out} = \text{Heat generated}\) In this particular case, we have: 1. Heat in by conduction from Node 1 to Node 0: \(k\frac{T_{1}-T_{0}}{\Delta x}\). 2. Heat out by convection at Node 0: \(h(T_{0}-T_{\infty})\). 3. Heat out by radiation at Node 0: \(\varepsilon\sigma(T_{0}^{4}-T_{\text{surr}}^{4})\). 4. Heat out by heat flux at Node 0: \(\dot{q}_{0}\). 5. Heat generated is assumed to be 0 as per the problem statement. Accordingly, the energy balance equation can be written as: \(k\frac{T_{1}-T_{0}}{\Delta x} - h(T_{0}-T_{\infty}) - \varepsilon\sigma(T_{0}^{4}-T_{\text{surr}}^{4}) = -\dot{q}_{0}\)

Rearrange the equation for the left boundary formulation

We can now rearrange the equation derived in Step 1 for the left boundary finite difference formulation as follows: \(T_{0}(k\Delta x^{-1} + h + 4\varepsilon\sigma T_{0}^{3}) - T_{1}(k\Delta x^{-1}) = (hT_{\infty} + \varepsilon\sigma T_{\text{surr}}^{4} + \dot{q}_{0})\)

Energy balance at the right boundary (Node 5)

At the right boundary, we only have conduction from Node 4 to Node 5. The rate of heat transfer at the right boundary can be calculated using the energy balance equation by considering only heat conduction. The energy balance equation for node 5 can be written as: \(\text{Heat in}-\text{Heat out} = 0\) In this particular case, we have: 1. Heat in by conduction from Node 4 to Node 5: \(k\frac{T_{4}-T_{5}}{\Delta x}\). 2. Heat out by convection and radiation are 0 as per the problem statement. Accordingly, the energy balance equation can be written as: \(k\frac{T_{4}-T_{5}}{\Delta x} = 0\)

Rearrange the equation for the right boundary formulation

We can now rearrange the equation derived in Step 3 for the right boundary finite difference formulation as follows: \(T_{5}(k\Delta x^{-1}) - T_{4}(k\Delta x^{-1}) = 0\) In conclusion, the finite difference formulations for boundary node 0 and the right boundary are as follows: Left boundary (Node 0): \(T_{0}(k\Delta x^{-1} + h + 4\varepsilon\sigma T_{0}^{3}) - T_{1}(k\Delta x^{-1}) = (hT_{\infty} + \varepsilon\sigma T_{\text{surr}}^{4} + \dot{q}_{0})\) Right boundary (Node 5): \(T_{5}(k\Delta x^{-1}) - T_{4}(k\Delta x^{-1}) = 0\)