Question

Consider steady heat conduction in a plane wall with variable heat generation and constant thermal conductivity. The nodal network of the medium consists of nodes \(0,1,2,3\), and 4 with a uniform nodal spacing of \(\Delta x\). Using the energy balance approach, obtain the finite difference formulation of the boundary nodes for the case of uniform heat flux \(\dot{q}_{0}\) at the left boundary (node 0 ) and convection at the right boundary (node 4) with a convection coefficient of \(h\) and an ambient temperature of \(T_{\infty}\).

Short answer

#Question# Using the energy balance approach, obtain the finite difference formulation for the boundary nodes of a nodal network that consists of nodes 0 through 4, evenly spaced by a distance of Δx. The left boundary (node 0) has a uniform heat flux \(q_0\), and the right boundary (node 4) has convection with a convection coefficient h, and an ambient temperature \(T_\infty\). #Answer# For node 0: \(T_{0} = T_{1} + \frac{\dot{q}_{0}\Delta x}{k}\) For node 4: \(T_{4} = \frac{h\Delta x T_{\infty} + k T_{3}}{h\Delta x + k}\)

Step-by-step solution

Analyze boundary node 0 conditions

At node 0, there's a uniform heat flux \(\dot{q}_{0}\). Using Fourier's law of heat conduction, we have: \(\dot{q}_{0} = -k\frac{dT}{dx}|_{x=0}\), where \(k\) is the thermal conductivity.

Apply a finite difference approximation to the temperature gradient

The spatial derivative of the temperature at node 0 can be approximated using a forward finite difference method: \(\frac{dT}{dx}|_{x=0} \approx \frac{T_{1} - T_{0}}{\Delta x}\)

Calculate the temperature at node 0

Using the results from Step 1 and Step 2, we can now write an equation for the temperature at node 0: \(\dot{q}_{0} = -k\frac{T_{1} - T_{0}}{\Delta x}\) Now, isolate \(T_{0}\): \(T_{0} = T_{1} + \frac{\dot{q}_{0}\Delta x}{k}\)

Analyze boundary node 4 conditions

At node 4, we have convection with convection coefficient \(h\) and an ambient temperature of \(T_{\infty}\). Using Newton's law of cooling, we have: \(\dot{q}_{conv} = h(T_{4} - T_{\infty})\)

Apply a finite difference approximation to the energy balance equation

At node 4, the energy balance equation can be given as: \(\dot{q}_{conv} = -k\frac{dT}{dx}|_{x=4\Delta x}\) Using a backward finite difference approximation for the spatial derivative at node 4: \(\frac{dT}{dx}|_{x=4\Delta x} \approx \frac{T_{4} - T_{3}}{\Delta x}\)

Calculate the temperature at node 4

Using the results from Step 4 and Step 5, we can now write an equation for the temperature at node 4: \(h(T_{4} - T_{\infty}) = -k\frac{T_{4} - T_{3}}{\Delta x}\) Solving for \(T_{4}\), we get: \(T_{4} = \frac{h\Delta x T_{\infty} + k T_{3}}{h\Delta x + k}\) We've now obtained the finite difference formulation for the boundary nodes: - For node 0: \(T_{0} = T_{1} + \frac{\dot{q}_{0}\Delta x}{k}\) - For node 4: \(T_{4} = \frac{h\Delta x T_{\infty} + k T_{3}}{h\Delta x + k}\)