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How can a node on an insulated boundary be treated as an interior node in the finite difference formulation of a plane wall? Explain.

Short Answer

Expert verified
Answer: A node on an insulated boundary can be treated as an interior node by modifying the finite difference equation at the boundary. We first use a forward difference approximation to represent the insulated boundary condition. Following this, we replace the term in the second derivative approximation equation with the result from the forward difference approximation. This results in an equation with the same format as the equation for interior nodes, allowing us to treat the insulated boundary node as an interior node in the finite difference formulation.

Step by step solution

01

1. Recap of Finite Difference Method for Plane Wall Heat Conduction

The Finite Difference Method (FDM) is a numerical technique used to solve partial differential equations by discretizing the domain into discrete nodes. In the context of heat conduction in a plane wall, we use FDM to solve the 1-dimensional steady state heat conduction equation. The governing equation for 1D steady state heat conduction is: \(\frac{d^2T}{dx^2} = 0\) where T is the temperature and x is the spatial coordinate.
02

2. Discretization of the Governing Equation

To apply the finite difference formulation, we need to discretize the governing equation. For a node i, we can approximate the second derivative using the central difference approximation, which results in: \(\frac{T_{i-1} - 2T_i + T_{i+1}}{(\Delta x)^2} = 0\) where \(\Delta x\) is the distance between the nodes, and \(T_{i-1}\), \(T_i\), \(T_{i+1}\) are the temperatures of nodes i-1, i, and i+1, respectively.
03

3. Insulated Boundary Condition

An insulated boundary condition means that there is no heat flux across the boundary. Mathematically, this means the derivative of temperature with respect to x is zero at the boundary: \(\frac{dT}{dx} = 0\) at the insulated boundary.
04

4. Treatment of Insulated Boundary as an Interior Node

To treat a node on an insulated boundary as an interior node, we need to modify the finite difference equation at the boundary. We can use a forward difference approximation to represent the insulated boundary condition: \(\frac{T_{i+1} - T_i}{\Delta x} = 0\) Solving for \(T_{i+1}\), we get: \(T_{i+1} = T_i\) Now, we can replace \(T_{i+1}\) in the second derivative approximation equation with \(T_i\): \(\frac{T_{i-1} - 2T_i + T_i}{(\Delta x)^2} = 0\) which simplifies to: \(\frac{T_{i-1} - T_i}{(\Delta x)^2} = 0\) Now the equation for the boundary node has the same format as the equation for the interior nodes. This allows us to treat the insulated boundary node as an interior node in the finite difference formulation of a plane wall.

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Most popular questions from this chapter

Consider a large uranium plate of thickness \(5 \mathrm{~cm}\) and thermal conductivity \(k=34 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) 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 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 \(T_{\text {sarr }}\).

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}, \mathrm{~K}\(. Each fin has a length of \)50 \mathrm{~mm}$, a base \(5 \mathrm{~mm}\) 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 the analytical solution.

A nonmetal plate is connected to a stainless steel plate by long ASTM A437 B4B stainless steel bolts \(9.5 \mathrm{~mm}\) in diameter. The portion of the bolts exposed to convection heat transfer with a cryogenic fluid is \(5 \mathrm{~cm}\) long. The fluid temperature for convection is at \(-50^{\circ} \mathrm{C}\) with a convection heat transfer coefficient of $100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\(. The thermal conductivity of the bolts is known to be \)23.9 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}$. Both the nonmetal and stainless steel plates maintain a uniform temperature of \(0^{\circ} \mathrm{C}\). According to the ASME Code for Process Piping (ASME B31.3-2014, Table A-2M), the minimum temperature suitable for ASTM A437 B4B stainless steel bolts is \(-30^{\circ} \mathrm{C}\). Using the finite difference method with a uniform nodal spacing of \(\Delta x=5 \mathrm{~mm}\) along the bolt, determine the temperature at each node. Compare the numerical results with the analytical solution. Plot the temperature distribution along the bolt. Would any part of the ASTM A437 B4B bolts be lower than the minimum suitable temperature of \(-30^{\circ} \mathrm{C}\) ?

Consider a medium in which the finite difference formulation of a general interior node is given in its simplest form as $$ T_{\text {left }}+T_{\text {top }}+T_{\text {right }}+T_{\text {boetom }}-4 T_{\text {node }}+\frac{\dot{e}_{\text {node }} P^{2}}{k}=0 $$ (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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