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

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A node on an insulated boundary can be treated as an interior node in the finite difference formulation of a plane wall by using the concept of ghost nodes and considering the temperature gradient across the boundary as zero. The ghost node is an imaginary node located outside the boundary of the problem domain, ensuring that the desired insulated boundary condition is met. By replacing the equations of the insulated boundary node with those of the interior nodes and keeping the temperature gradient zero at the boundary, the node on the insulated boundary can effectively be treated as an interior node, allowing for efficient problem-solving while maintaining the insulated boundary condition.

Step by step solution

01

Understanding Insulated Boundary and Interior Nodes

An insulated boundary is a type of boundary condition where there is no heat transfer through the boundary. In other words, the rate of heat transfer across the boundary is zero. This is different from other types of boundary conditions like fixed temperature or convective boundary conditions, where heat transfer occurs. An interior node is a point within the domain of the problem, not located on the boundaries. In the case of a plane wall, these are points inside the wall, away from the wall boundaries.
02

Finite Difference Formulation

Finite difference formulation is a numerical method used to solve partial differential equations (PDEs), such as heat conduction equation. The method involves dividing the problem domain into a grid of discrete points (nodes) and equations that describe the behavior of the PDE at each node. The finite difference equations approximate the behavior of the continuous PDE over the nodes.
03

Applying Insulated Boundary Condition in Finite Difference Formulation

In a finite difference formulation for heat conduction in a plane wall, when an insulated boundary is present, it means that the heat transfer at the boundary is zero. Mathematically, this translates to the derivative of temperature with respect to the spatial coordinate(s) being zero (i.e., no temperature gradient at the boundary). To incorporate an insulated boundary in the finite difference formulation, we can use a "ghost node" strategy. A ghost node is an imaginary node located outside the boundary of the problem domain, and its purpose is to ensure that the desired boundary condition is met.
04

Treating Node on Insulated Boundary as an Interior Node

When a node is located on an insulated boundary, it can be treated as an interior node by incorporating the ghost node concept. Since the insulated boundary condition states that the temperature gradient across the boundary is zero, the temperature of the ghost node should be the same as the temperature at the insulated boundary node. By creating a ghost node outside the boundary and replacing the equations of the insulated boundary node with those of the interior nodes, while considering the condition that the temperature gradient is zero at the boundary, the node on the insulated boundary can effectively be treated as an interior node in the finite difference formulation. In conclusion, a node on an insulated boundary can be treated as an interior node in the finite difference formulation of a plane wall by using the concept of ghost nodes and considering the temperature gradient across the boundary as zero. This allows us to solve the heat conduction problem efficiently while maintaining the insulated boundary condition.

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Most popular questions from this chapter

Consider 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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