Chapter 5: Problem 134
Can the global (accumulated) discretization error be less than the local error during a step? Explain.
Chapter 5: Problem 134
Can the global (accumulated) discretization error be less than the local error during a step? Explain.
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Get started for freeA nonmetal plate \((k=0.05 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) is attached on the upper surface of an ASME SB-96 copper-silicon plate $(k=36 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})$. The nonmetal plate and the ASME SB-96 plate have thicknesses of \(20 \mathrm{~mm}\) and \(30 \mathrm{~mm}\), respectively. The bottom surface of the ASME SB-96 plate (surface 1) is subjected to a uniform heat flux of \(150 \mathrm{~W} / \mathrm{m}^{2}\). The top nonmetal plate surface (surface 2) is exposed to convection at an air temperature of \(15^{\circ} \mathrm{C}\) and a convection heat transfer coefficient of \(5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Copper- silicon alloys are not always suitable for applications where they are exposed to certain median and high temperatures. Therefore, the ASME Boiler and Pressure Vessel Code (ASME BPVC.IV-2015, HF-300) limits equipment constructed with ASME SB-96 plate to be operated at a temperature not exceeding \(93^{\circ} \mathrm{C}\). Using the finite difference method with a uniform nodal spacing of \(\Delta x=5 \mathrm{~mm}\) along the the temperature distribution as a function of \(x\). Would the use of the ASME SB- 96 plate under these conditions be in compliance with the ASME Boiler and Pressure Vessel Code? What is the lowest value of the convection heat transfer coefficient for the air so that the ASME SB-96 plate is below $93^{\circ} \mathrm{C}$ ?
Consider transient heat conduction in a plane wall whose left surface (node 0 ) is maintained at \(80^{\circ} \mathrm{C}\) while the right surface (node 6) is subjected to a solar heat flux of \(600 \mathrm{~W} / \mathrm{m}^{2}\). The wall is initially at a uniform temperature of \(50^{\circ} \mathrm{C}\). Express the explicit finite difference formulation of the boundary nodes 0 and 6 for the case of no heat generation. Also, obtain the finite difference formulation for the total amount of heat transfer at the left boundary during the first three time steps.
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 min have elapsed using \(\Delta t=10 \mathrm{~s}\) and a uniform mesh size of \(\Delta x=\Delta y=10 \mathrm{~cm}\).
Consider a \(2-\mathrm{cm} \times 4-\mathrm{cm}\) ceramic strip $(k=3 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\(, \)\rho=1600 \mathrm{~kg} / \mathrm{m}^{3}\(, and \)\left.c_{p}=800 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)$ embedded in very high conductivity material as shown in Fig. P5-131. The two sides of the ceramic strip are maintained at a constant temperature of \(300^{\circ} \mathrm{C}\). The bottom surface of the strip is insulated, while the top surface is exposed to a convective environment with \(h=200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and ambient temperature of \(50^{\circ} \mathrm{C}\). Initially at \(t=0\), the ceramic strip is at a uniform temperature of \(300^{\circ} \mathrm{C}\). Using the implicit finite difference formulation and a time step of \(2 \mathrm{~s}\), determine the nodal temperatures after \(12 \mathrm{~s}\) for a uniform mesh size of $1 \mathrm{~cm}$.
Consider steady one-dimensional heat conduction in a composite plane wall consisting of two layers \(A\) and \(B\) in perfect contact at the interface. The wall involves no heat generation. The nodal network of the medium consists of nodes 0,1 (at the interface), and 2 with a uniform nodal spacing of $\Delta x$. Using the energy balance approach, obtain the finite difference formulation of this problem for the case of insulation at the left boundary (node 0 ) and radiation at the right boundary (node 2 ) with an emissivity of \(\varepsilon\) and surrounding temperature of \(T_{\text {sarr }}\).
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