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Chapter 5: Problem 21

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 \(\Delta x\). Using the energy balance approach, obtain the 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 \(T_{\text {surr }}\)

Short Answer

Expert verified
Answer: The finite difference formulations for the boundary nodes are: For node 0 (insulation): \(T_1 = T_0 + \frac{Δx^2}{k}q_0\) For node 5 (radiation): \(T_5=\frac{εσΔx^2T^4_5+(kΔx^2+εσΔx^2T^4_{\text{surr}})T_4-q_5Δx^2}{kΔx^2+εσΔx^2T^4_5}\)

Step by step solution

01

Energy balance for insulation at node 0

Considering the left boundary node 0, which has insulation, no heat is allowed to transfer in or out of the boundary. The energy balance equation for the insulated boundary node 0 can be written as follows: \([\frac{k}{Δx^2}(T_1 - T_0)] - q_0 = 0\) Where: - \(k\) is the constant thermal conductivity - \(Δx\) is the nodal spacing - \(T_0\) and \(T_1\) are temperatures at nodes 0 and 1, respectively - \(q_0\) is the heat generation at node 0
02

Energy balance for radiation at node 5

Considering the right boundary node 5, which has radiation, the energy balance equation can be written as follows: \([\frac{k}{Δx^2}(T_4 - T_5)] - q_5 = εσ(T^4_5 - T^4_{\text{surr}})\) Where: - \(T_4\) and \(T_5\) are temperatures at nodes 4 and 5, respectively - \(q_5\) is the heat generation at node 5 - \(ε\) is the emissivity - \(σ\) is the Stefan-Boltzmann constant - \(T_{\text{surr}}\) is the surrounding temperature
03

Finite difference formulation for boundary nodes

Now, we can write the finite difference formulations for the boundary nodes by rearranging the energy balance equations from Steps 1 and 2: For node 0 (insulation): \(T_1 = T_0 + \frac{Δx^2}{k}q_0\) For node 5 (radiation): \(T_5=\frac{εσΔx^2T^4_5+(kΔx^2+εσΔx^2T^4_{\text{surr}})T_4-q_5Δx^2}{kΔx^2+εσΔx^2T^4_5}\) The finite difference formulation for the boundary nodes in the given heat conduction problem with insulation at node 0 and radiation at node 5 are derived as shown in the steps above. These formulations can be used to solve for the temperature at the boundary nodes given the material properties, heat generation, and surrounding temperature.

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Most popular questions from this chapter

A plane wall with surface temperature of \(350^{\circ} \mathrm{C}\) is attached with straight rectangular fins \((k=235 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). The fins are exposed to an ambient air condition of \(25^{\circ} \mathrm{C}\) and the convection heat transfer coefficient is \(154 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Each fin has a length of \(50 \mathrm{~mm}\), a base of \(5 \mathrm{~mm}\) thick, and a width of \(100 \mathrm{~mm}\). For a single fin, using a uniform nodal spacing of \(10 \mathrm{~mm}\), determine \((a)\) the finite difference equations, \((b)\) the nodal temperatures by solving the finite difference equations, and \((c)\) the heat transfer rate and compare the result with analytical solution.

Quench hardening is a mechanical process in which the ferrous metals or alloys are first heated and then quickly cooled down to improve their physical properties and avoid phase transformation. Consider a \(40 \mathrm{~cm} \times 20 \mathrm{~cm}\) block of copper alloy \(\left(k=120 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \alpha=3.91 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\right)\) being heated uniformly until it reaches a temperature of \(800^{\circ} \mathrm{C}\). It is then suddenly immersed into the water bath maintained at \(15^{\circ} \mathrm{C}\) with \(h=100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) for quenching process. However, the upper surface of the metal is not submerged in the water and is exposed to air at \(15^{\circ} \mathrm{C}\) with a convective heat transfer coefficient of \(10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Using an explicit finite difference formulation, calculate the temperature distribution in the copper alloy block after \(10 \mathrm{~min}\) have elapsed using \(\Delta t=10 \mathrm{~s}\) and a uniform mesh size of \(\Delta x=\Delta y=10 \mathrm{~cm}\).

Can the global (accumulated) discretization error be less than the local error during a step? Explain.

Consider transient one-dimensional heat conduction in a plane wall that is to be solved by the explicit method. If both sides of the wall are at specified temperatures, express the stability criterion for this problem in its simplest form.

Consider a medium in which the finite difference formulation of a general interior node is given in its simplest form as $$ T_{\text {node }}=\left(T_{\text {left }}+T_{\text {top }}+T_{\text {right }}+T_{\text {bottom }}\right) / 4 $$ (a) Is heat transfer in this medium steady or transient? (b) Is heat transfer one-, two-, or three-dimensional? (c) Is there heat generation in the medium? (d) Is the nodal spacing constant or variable? (e) Is the thermal conductivity of the medium constant or variable?
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