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Conduction within relatively complex geometries can sometimes be evaluated using the finite-difference methods of this text that are applied to subdomains and patched together. Consider the two-dimensional domain formed by rectangular and cylindrical subdomains patched at the common, dashed control surface. Note that, along the dashed control surface, temperatures in the two subdomains are identical and local conduction heat fluxes to the cylindrical subdomain are identical to local conduction heat fluxes from the rectangular subdomain. Calculate the heat transfer per unit depth into the page, \(q^{\prime}\), using \(\Delta x=\Delta y=\Delta r=10 \mathrm{~mm}\) and \(\Delta \phi=\pi / 8\). The base of the rectangular subdomain is held at \(T_{h}=20^{\circ} \mathrm{C}\), while the vertical surface of the cylindrical subdomain and the surface at outer radius \(r_{o}\) are at \(T_{c}=0^{\circ} \mathrm{C}\). The remaining surfaces are adiabatic, and the thermal conductivity is \(k=10 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\).

Short Answer

Expert verified
To calculate the heat transfer per unit depth, \(q^{\prime}\), we used the finite difference method on both the rectangular and cylindrical subdomains, created grids of nodal points, and applied boundary conditions. We then patched the subdomains together by equating their heat fluxes along the common boundary. Finally, we computed the local conduction heat fluxes in the x, y, r, and \(\phi\) directions and summed them to obtain the total heat transfer per unit depth, \(q^{\prime}\).

Step by step solution

01

Apply the finite difference method to the rectangular subdomain

We will create a grid of nodal points, spacing them \(\Delta x=10 \mathrm{~mm}\) apart horizontally and \(\Delta y=10 \mathrm{~mm}\) apart vertically. Then, we will apply the finite difference method to each nodal point. For an 2-D node in a rectangular grid, the energy balance can be expressed as: \[k \frac{T_{i-1,j} - 2T_{i,j} + T_{i+1,j}}{(\Delta x)^2} + k \frac{T_{i,j-1} - 2T_{i,j} + T_{i,j+1}}{(\Delta y)^2} = 0\] where \(T_{i,j}\) represents the temperature at \(x_i\) and \(y_j\). For this subdomain, the base is held at \(T_h=20\), and the remaining surfaces are adiabatic.
02

Apply the finite difference method to the cylindrical subdomain

In the cylindrical subdomain, we will create a polar grid of nodal points, with the radial spacing as \(\Delta r=10 \mathrm{~mm}\) and angular spacing \(\Delta \phi=\pi / 8\). Then, we will apply the finite difference method to each nodal point. For a 2-D node in a cylindrical grid, the energy balance can be expressed as: \[k \frac{T_{i-1,j} - 2T_{i,j} + T_{i+1,j}}{(\Delta r)^2} + \frac{k}{r_i} \frac{T_{i,j-1} - 2T_{i,j} + T_{i,j+1}}{(\Delta \phi)^2} = 0\] where \(T_{i,j}\) represents the temperature at \(r_i\) and \(\phi_j\). For this subdomain, \(T_c=0\) at the vertical and outer radial surfaces, with the remaining surfaces having the same heat flux as in the rectangular subdomain.
03

Patching the subdomains together

Along the dashed control surface, the heat transfer across the subdomains must be equal. We need to equate the heat flux from the rectangular subdomain at the interface to the heat flux into the cylindrical subdomain at the corresponding radial and angular nodes. By matching the heat fluxes, we will obtain a total heat transfer per unit depth. For the last row of the rectangular subdomain (adiabatic) and the first radial node of the cylindrical subdomain, use the boundary conditions given above to find the remaining temperatures.
04

Calculate the heat transfer per unit depth

At the common boundary, the heat flux is the same between the two subdomains. To calculate the heat transfer per unit depth (\(q^{\prime}\)), we compute the local conduction heat flux in the x, y, r, and \(\phi\) directions, then sum them up to obtain the total heat transfer per unit depth. Ensure that the boundary equations have been appropriately applied to the subdomain temperatures before computing the heat flux. Finally, use the numerical grid to compute the heat transfer and sum the results. The heat transfer per unit depth (\(q^{\prime}\)) will be the result of this calculation.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Conduction in Complex Geometries
Understanding heat transfer in complex geometries can be quite challenging, especially when the structures do not conform to simple shapes. In such scenarios, we cannot rely solely on analytical methods because they are primarily suited to simple, symmetrical geometries. Instead, we turn to numerical techniques like the finite difference method (FDM).

Complex geometries often involve combinations of different shapes, such as rectangles adjacent to cylinders. The finite difference approach deals with this by dividing the entire domain into smaller, manageable subdomains. Each subdomain can be analyzed separately, and the results can be 'patched' together at the boundaries. This patching is done keeping in mind that the temperature must be continuous across subdomain interfaces, and heat flux must be conserved. Solving heat conduction in complex geometries also involves applying relevant boundary conditions, which will be discussed in their respective section.

By discretizing the domain into a grid or mesh of nodes, the differential equations governing heat transfer are transformed into algebraic equations that can be iteratively solved using computational resources. This approach is popular in engineering and allows for the calculation of temperature distribution and heat flux within objects that would otherwise be impractical, if not impossible, to analyze through direct analytical methods. The example exercise illustrates how FDM can handle an irregular domain by breaking it down into a combination of rectangular and cylindrical subdomains.
Two-dimensional Heat Conduction
Two-dimensional heat conduction is particularly significant in situations where heat transfer is not uniform or axial symmetry is not present. This necessitates considering heat flow in two distinct directions, which is inherently more complex than one-dimensional conduction.

In the finite difference method, a 2D domain is gridded into nodes where temperatures are calculated based on the thermal balance. This involves applying the discretization of the Fourier heat conduction equation in both the x and y directions, or r and \(\phi\) in polar coordinates. The provided exercise demonstrates the discretization process using equal spacing in both dimensions, which simplifies the calculation but does not reduce its accuracy if the mesh is fine enough.

Energy Balance for Nodes

For an inner node within a rectangular domain, the heat balance is a sum of the heat conduction fluxes in and out of the node. Similarly, in cylindrical coordinates, consideration for the radial position is crucial because the area associated with heat transfer changes with radius. Consequently, the discretization formula varies slightly with the inclusion of radial terms. The result of this numerical approach is a set of linear algebraic equations, which must all be satisfied simultaneously to determine the temperature distribution across the entire domain.
Numerical Heat Transfer Analysis
Numerical heat transfer analysis is an indispensable tool for engineers and scientists working with systems where analytical solutions are unattainable or overly complex. It uses computational methods to simulate and predict heat transfer processes within various materials and under diverse boundary conditions. The finite difference method, illustrated in the exercise, is one such numerical technique.

Numerical analysis allows for the accommodation of irregular geometries, varying material properties, and convective or radiative boundary conditions. It also simplifies the inclusion of internal heat generation sources or non-linear temperature-dependent properties. The marriage of numerical methods with high-speed computing has enabled complex heat transfer problems to be divided into tiny, solvable pieces, just like the example of handling the rectangular and cylindrical subdomains.

Once the correct formulation of the discrete nodal equations is done, and all the boundary conditions are set, iterative solvers or direct matrix solvers can be employed to find the nodal temperatures. The accurate prediction of these temperatures is crucial for assessing the thermal states of materials and can influence the design and safety of engineering systems.
Boundary Conditions in Heat Transfer
Boundary conditions play a pivotal role in any heat transfer analysis because they define how a material or system interacts with its surroundings. They are the key to ensuring that a numerical model is both mathematically well-posed and physically realistic. Different types of boundary conditions include temperature-based (Dirichlet), heat flux-based (Neumann), and convective (Robin).

The exercise provided focuses on Dirichlet and adiabatic (a type of Neumann) boundary conditions. It's essential to apply boundary conditions appropriately as they directly influence the temperature distribution and therefore the heat flow within the material. For instance, in the exercise, one part of the boundary has a fixed temperature, while adiabatic boundaries are assumed where there is no heat transfer across. This assumption simplifies the calculations at those boundaries, as the temperature gradient normal to the surface is zero.

Implementing Boundary Conditions

During the finite difference analysis, special attention should be given to the nodes at or near the boundaries. The temperature at a node affected by a Dirichlet condition is known, so it's fixed in the numerical model. For Neumann conditions, the heat flux at the boundary must be computed and used in the nodal equations. These correct implementations lead to an accurate and reliable prediction of temperature fields and thereby heat transfer within the entirety of the domain.

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Most popular questions from this chapter

A hole of diameter \(D=0.25 \mathrm{~m}\) is drilled through the center of a solid block of square cross section with \(w=1 \mathrm{~m}\) on a side. The hole is drilled along the length, \(l=2 \mathrm{~m}\), of the block, which has a thermal conductivity of \(k=150 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). The four outer surfaces are exposed to ambient air, with \(T_{\infty, 2}=25^{\circ} \mathrm{C}\) and \(h_{2}=4 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), while hot oil flowing through the hole is characterized by \(T_{\infty, 1}=300^{\circ} \mathrm{C}\) and \(h_{1}=50 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine the corresponding heat rate and surface temperatures.

A straight fin of uniform cross section is fabricated from a material of thermal conductivity \(50 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), thickness \(w=6 \mathrm{~mm}\), and length \(L=48 \mathrm{~mm}\), and it is very long in the direction normal to the page. The convection heat transfer coefficient is \(500 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) with an ambient air temperature of \(T_{\infty}=30^{\circ} \mathrm{C}\). The base of the \(\mathrm{f}\) is maintained at \(T_{b}=100^{\circ} \mathrm{C}\), while the fi tip is well insulated. (a) Using a finite-difference method with a space increment of \(4 \mathrm{~mm}\), estimate the temperature distribution within the fin. Is the assumption of onedimensional heat transfer reasonable for this fin? (b) Estimate the fin heat transfer rate per unit length normal to the page. Compare your result with the one-dimensional, analytical solution, Equation \(3.81\). (c) Using the finite-difference mesh of part (a), compute and plot the fin temperature distribution for values of \(h=10,100,500\), and \(1000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine and plot the fin heat transfer rate as a function of \(h\).

Small-diameter electrical heating elements dissipating \(50 \mathrm{~W} / \mathrm{m}\) (length normal to the sketch) are used to heat a ceramic plate of thermal conductivity \(2 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). The upper surface of the plate is exposed to ambient air at \(30^{\circ} \mathrm{C}\) with a convection coeftient of \(100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), while the lower surface is well insulated. (a) Using the Gauss-Seidel method with a grid spacing of \(\Delta x=6 \mathrm{~mm}\) and \(\Delta y=2 \mathrm{~mm}\), obtain the temperature distribution within the plate. (b) Using the calculated nodal temperatures, sketch four isotherms to illustrate the temperature distribution in the plate. (c) Calculate the heat loss by convection from the plate to the fluid. Compare this value with the element dissipation rate. (d) What advantage, if any, is there in not making \(\Delta x=\Delta y\) for this situation? (e) With \(\Delta x=\Delta y=2 \mathrm{~mm}\), calculate the temperature field within the plate and the rate of heat transfer from the plate. Under no circumstances may the temperature at any location in the plate exceed \(400^{\circ} \mathrm{C}\). Would this limit be exceeded if the airfw were terminated and heat transfer to the air were by natural convection with \(h=10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) ?

Consider a one-dimensional fin of uniform crosssectional area, insulated at its tip, \(x=L\). (See Table \(3.4\), case B). The temperature at the base of the fin \(T_{b}\) and of the adjoining fluid \(T_{\infty}\), as well as the heat transfer coefficient \(h\) and the thermal conductivity \(k\), are known. (a) Derive the finite-difference equation for any interior node \(m\). (b) Derive the finite-difference equation for a node \(n\) located at the insulated tip.

The top surface of a plate, including its grooves, is maintained at a uniform temperature of \(T_{1}=200^{\circ} \mathrm{C}\). The lower surface is at \(T_{2}=20^{\circ} \mathrm{C}\), the thermal conductivity is \(15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and the groove spacing is \(0.16 \mathrm{~m}\). (a) Using a finite-difference method with a mesh size of \(\Delta x=\Delta y=40 \mathrm{~mm}\), calculate the unknown nodal temperatures and the heat transfer rate per width of groove spacing \((w)\) and per unit length normal to the page. (b) With a mesh size of \(\Delta x=\Delta y=10 \mathrm{~mm}\), repeat the foregoing calculations, determining the temperature field and the heat rate. Also, consider conditions for which the bottom surface is not at a uniform temperature \(T_{2}\) but is exposed to a fluid at \(T_{\infty}=20^{\circ} \mathrm{C}\). With \(\Delta x=\Delta y=10 \mathrm{~mm}\), determine the temperature field and heat rate for values of \(h=5,200\), and 1000 \(\mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}\), as well as for \(h \rightarrow \infty\).

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