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A 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 $93^{\circ} \mathrm{C}$ ?

Short Answer

Expert verified
Answer: The finite difference method for heat conduction can be used to calculate the temperature distribution in the composite structure. By iterating the temperature calculation to convergence, checking for ASME compliance, and adjusting the convection heat transfer coefficient, the lowest value for compliance with the ASME Boiler and Pressure Vessel Code can be determined.

Step by step solution

01

Calculate temperature at each node

Since we have a uniform nodal spacing of Δx = 5mm, we will discretize the problem into 10 nodes, with 6 nodes representing the ASME SB-96 plate (30mm thickness) and 4 nodes representing the nonmetal plate (20mm thickness). We know the heat flux at the bottom of the ASME SB-96 plate is 150 W/m². The boundary condition at the top of the nonmetal plate is given by: q'' = h(T_surf - T_air) where h is the convection heat transfer coefficient, T_surf is the temperature at the top surface (surface 2), and T_air is the air temperature. Using the general finite difference equation for steady-state two-dimensional heat conduction: k1(T_1 - T_2) / Δx + k2(T_4 - T_2) / Δx = 0 we can calculate the temperature at each node.
02

Iterate temperature calculation to convergence

We can start with an initial guess for the temperature at each node, and iterate the finite difference equation until the difference in temperature between iterations is minimized. At the end of this process, we will obtain the temperature distribution in the composite structure.
03

Check ASME compliance

To check if the temperatures in the ASME SB-96 plate do not exceed 93°C, compare the calculated temperatures at nodes 1 through 6. If all of them are below 93°C, the structure is in compliance with the ASME Boiler and Pressure Vessel Code.
04

Calculate lowest convection coefficient for compliance

To find the lowest convection heat transfer coefficient that ensures the compliance with the ASME Boiler and Pressure Vessel Code, we need to: 1. Start with the given convection coefficient value (h = 5 W/m²K) and calculate the temperature distribution. 2. Increment or decrement the convection coefficient and calculate the temperature distribution again. 3. Repeat this process until the highest temperature in the copper-silicon plate is exactly 93°C. The convection coefficient value found at this point corresponds to the lowest value for which the structure remains in compliance with the ASME Boiler and Pressure Vessel Code.

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Most popular questions from this chapter

A 1-m-long and \(0.1\)-m-thick steel plate of thermal conductivity $35 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}$ is well insulated on both sides, while the top surface is exposed to a uniform heat flux of $5500 \mathrm{~W} / \mathrm{m}^{2}$. The bottom surface is convectively cooled by a fluid at \(10^{\circ} \mathrm{C}\) having a convective heat transfer coefficient of $150 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$. Assuming one-dimensional heat conduction in the lateral direction, find the temperature at the midpoint of the plate. Discretize the plate thickness into four equal parts.

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 \(\Delta x=\Delta y=10 \mathrm{~cm}\).

Starting with an energy balance on the volume element, obtain the steady three-dimensional finite difference equation for a general interior node in rectangular coordinates for \(T(x, y, z)\) for the case of constant thermal conductivity and uniform heat generation.

Consider steady two-dimensional heat transfer in a long solid bar $(k=25 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\( of square cross section ( \)3 \mathrm{~cm} \times 3 \mathrm{~cm}$ ) with the prescribed temperatures at the top, right, bottom, and left surfaces to be $100^{\circ} \mathrm{C}, 200^{\circ} \mathrm{C}, 300^{\circ} \mathrm{C}\(, and \)500^{\circ} \mathrm{C}$, respectively. Heat is generated in the bar uniformly at a rate of $\dot{e}=5 \times 10^{6} \mathrm{~W} / \mathrm{m}^{3}\(. Using a uniform mesh size \)\Delta x=\Delta y=1 \mathrm{~cm}\(, determine \)(a)$ the finite difference equations and \((b)\) the nodal temperatures with the Gauss-Seidel iterative method.

Consider a long concrete dam $(k=0.6 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\(, \)\left.\alpha_{s}=0.7\right)$ of triangular cross section whose exposed surface is subjected to solar heat flux of $\dot{q}_{s}=800 \mathrm{~W} / \mathrm{m}^{2}$ and to convection and radiation to the environment at \(25^{\circ} \mathrm{C}\) with a combined heat transfer coefficient of \(30 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The 2-m-high vertical section of the dam is subjected to convection by water at \(15^{\circ} \mathrm{C}\) with a heat transfer coefficient of $150 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$, and heat transfer through the 2 -m-long base is considered to be negligible. Using the finite difference method with a mesh size of \(\Delta x=\Delta y=1 \mathrm{~m}\) and assuming steady two-dimensional heat transfer, determine the temperature of the top, middle, and bottom of the exposed surface of the dam. Answers: \(21.3^{\circ} \mathrm{C}\). $43.2^{\circ} \mathrm{C}, 43.6^{\circ} \mathrm{C}$

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