Chapter 5: Problem 17
How can a node on an insulated boundary be treated as an interior node in the finite difference formulation of a plane wall? Explain.
Chapter 5: Problem 17
How can a node on an insulated boundary be treated as an interior node in the finite difference formulation of a plane wall? Explain.
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Get started for freeConsider a large uranium plate of thickness \(5 \mathrm{~cm}\) and thermal conductivity \(k=28 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) in which heat is generated uniformly at a constant rate of \(\dot{e}=6 \times 10^{5} \mathrm{~W} / \mathrm{m}^{3}\). One side of the plate is insulated while the other side is subjected to convection to an environment at \(30^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(h=60 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Considering six equally spaced nodes with a nodal spacing of \(1 \mathrm{~cm},(a)\) obtain the finite difference formulation of this problem and \((b)\) determine the nodal temperatures under steady conditions by solving those equations.
Consider a large plane wall of thickness \(L=0.3 \mathrm{~m}\), thermal conductivity \(k=2.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and surface area \(A=24 \mathrm{~m}^{2}\). The left side of the wall is subjected to a heat flux of \(\dot{q}_{0}=350 \mathrm{~W} / \mathrm{m}^{2}\) while the temperature at that surface is measured to be \(T_{0}=60^{\circ} \mathrm{C}\). Assuming steady one-dimensional heat transfer and taking the nodal spacing to be \(6 \mathrm{~cm},(a)\) obtain the finite difference formulation for the six nodes and (b) determine the temperature of the other surface of the wall by solving those equations.
A hot brass plate is having its upper surface cooled by impinging jet of air at temperature of \(15^{\circ} \mathrm{C}\) and convection heat transfer coefficient of \(220 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The \(10-\mathrm{cm}-\) thick brass plate \(\left(\rho=8530 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=380 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, k=\right.\) \(110 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and \(\left.\alpha=33.9 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\right)\) had a uniform initial temperature of \(650^{\circ} \mathrm{C}\), and the lower surface of the plate is insulated. Using a uniform nodal spacing of \(\Delta x=\) \(2.5 \mathrm{~cm}\) determine \((a)\) the explicit finite difference equations, (b) the maximum allowable value of the time step, \((c)\) the temperature at the center plane of the brass plate after 1 minute of cooling, and \((d)\) compare the result in \((c)\) with the approximate analytical solution from Chapter 4 .
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 \(0,1,2,3\), and 4 with a uniform nodal spacing of \(\Delta x\). The wall is initially at a specified temperature. The temperature at the right boundary (node 4) is specified. Using the energy balance approach, obtain the explicit finite difference formulation of the boundary node 0 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 \(\dot{q}_{0}\) toward the wall. Also, obtain the finite difference formulation for the total amount of heat transfer at the right boundary for the first 20 time steps.
Consider a large plane wall of thickness \(L=0.4 \mathrm{~m}\), thermal conductivity \(k=2.3 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and surface area \(A=20 \mathrm{~m}^{2}\). The left side of the wall is maintained at a constant temperature of \(95^{\circ} \mathrm{C}\), while the right side loses heat by convection to the surrounding air at \(T_{\infty}=15^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(h=18 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Assuming steady one-dimensional heat transfer and taking the nodal spacing to be \(10 \mathrm{~cm},(a)\) obtain the finite difference formulation for all nodes, \((b)\) determine the nodal temperatures by solving those equations, and (c) evaluate the rate of heat transfer through the wall.
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