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Derive the nodal finite-difference equations for the following configurations. (a) Node \((m, n)\) on a diagonal boundary subjected to convection with a fluid at \(T_{\infty}\) and a heat transfer coefficient \(h\). Assume \(\Delta x=\Delta y\). (b) Node \((m, n)\) at the tip of a cutting tool with the upper surface exposed to a constant heat flux \(q_{o}^{\prime \prime}\), and the diagonal surface exposed to a convection cooling process with the fluid at \(T_{\infty}\) and a heat transfer coefficient \(h\). Assume \(\Delta x=\Delta y\).

Short Answer

Expert verified
The nodal finite-difference equations for the given configurations are: (a) \(T_{m+1, n} + T_{m-1, n} + T_{m, n+1} + T_{m, n-1} - 4T_{m, n} = 0\) (b) \(T_{m+1, n} + T_{m-1, n} + T_{m, n+1} + T_{m, n-1} - 4T_{m, n} = q_{o}^{\prime \prime} \Delta x^2\)

Step by step solution

01

Configuration (a) Analysis

For the first configuration, we have a node (m, n) on a diagonal boundary subjected to convection. We must consider the boundary conditions and a finite-difference equation that describes the thermal conduction in convection for a 2D grid.
02

Identify boundary conditions

On the diagonal boundary, we have the heat transfer in the x-direction and the y-direction due to convection: \(q_x = -k \frac{\partial T}{\partial x} = h(T_{\infty} - T_{m, n})\) \(q_y = -k \frac{\partial T}{\partial y} = h(T_{\infty} - T_{m, n})\)
03

Apply finite-difference approximation

Using central difference approximation, we get a finite-difference equation: \(\frac{T_{m+1, n} - 2T_{m, n} + T_{m-1, n}}{\Delta x^2} + \frac{T_{m, n+1} - 2T_{m, n} + T_{m, n-1}}{\Delta y^2}\) = 0 Since Δx = Δy, we can simplify this equation \(\frac{T_{m+1, n} - 2T_{m, n} + T_{m-1, n} + T_{m, n+1} - 2T_{m, n} + T_{m, n-1}}{\Delta x^2} = 0\)
04

Derive nodal finite-difference equation

Rearrange the equation to get the nodal finite-difference equation for configuration (a): \(T_{m+1, n} + T_{m-1, n} + T_{m, n+1} + T_{m, n-1} - 4T_{m, n} = 0\)
05

Configuration (b) Analysis

For the second configuration, we have a node (m, n) at the tip of a cutting tool with the upper surface exposed to constant heat flux and the diagonal surface exposed to a convection cooling process.
06

Identify boundary conditions

On the upper surface, we have a constant heat flux \(q_{o}^{\prime \prime}\), and on the diagonal surface, we have the heat transfer due to convection: \(q_x = -k \frac{\partial T}{\partial x} = h(T_{\infty} - T_{m, n})\) \(q_y = q_{o}^{\prime \prime}\)
07

Apply finite-difference approximation

Using central difference approximation, we have: \(\frac{T_{m+1, n} - 2T_{m, n} + T_{m-1, n}}{\Delta x^2} + \frac{T_{m, n+1} - 2T_{m, n} + T_{m, n-1}}{\Delta y^2}\) = \(q_{o}^{\prime \prime}\) Since Δx = Δy, we can simplify this equation \(\frac{T_{m+1, n} - 2T_{m, n} + T_{m-1, n} + T_{m, n+1} - 2T_{m, n} + T_{m, n-1}}{\Delta x^2} = q_{o}^{\prime \prime}\)
08

Derive nodal finite-difference equation

Rearrange the equation to get the nodal finite-difference equation for configuration (b): \(T_{m+1, n} + T_{m-1, n} + T_{m, n+1} + T_{m, n-1} - 4T_{m, n} = q_{o}^{\prime \prime} \Delta x^2\) Now we have derived the nodal finite-difference equations for both configurations.

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Key Concepts

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

Convection Boundary Conditions
Understanding convection boundary conditions is crucial when analyzing heat transfer problems in engineering and physics. Convection is a mode of heat transfer that involves the movement of fluid, which can carry heat away from or towards a surface. In the given exercise, a surface (a diagonal boundary) is exposed to a fluid at a temperature, denoted as \(T_{\infty}\), with a known heat transfer coefficient \(h\).

The heat transfer from the surface to the fluid is quantified by Newton's law of cooling, which states that the heat transfer rate per unit area, or the heat flux \(q\), is proportional to the temperature difference between the surface and the fluid:
\(q = h(T_{\infty} - T_{m, n})\).

This relation becomes a boundary condition when solving heat transfer problems. In the finite-difference method, the boundary conditions are used to relate the nodal temperature values to the surrounding fluid temperature, enabling the construction of discrete equations that can be solved numerically for the temperature distribution in the domain.
Heat Flux
Heat flux, represented by \(q\), is the rate of heat energy transfer through a given surface per unit area. It is an important concept in thermal engineering, dictating how heat is distributed and dissipated in materials and systems. In the context of the exercise, a constant heat flux condition, \(q_{o}^{\prime \prime}\), is applied to the upper surface of a cutting tool.

When a surface is subject to a constant heat flux, this implies that a specified amount of heat energy is continuously added to or removed from the surface independent of the temperature difference. This condition affects the thermal gradient and, subsequently, the temperature distribution in the neighboring material. Establishing the heat flux boundary condition in finite-difference equations enables the prediction of temperature fields in systems where heat application or dissipation is controlled, such as in the case of the cutting tool.
Finite-Difference Approximation
The finite-difference approximation is a numerical method for solving differential equations by approximating derivatives with differences. In thermal analysis, it is used to discretize the heat conduction equation, transforming it into algebraic equations that relate temperatures at discrete points, or nodes.

In the exercise, the central difference approximation was employed, replacing the partial derivatives in the heat equation with differences between nodal temperatures, like so:
\(\frac{\partial T}{\partial x} ≈ \frac{T_{m+1, n} - T_{m-1, n}}{2\Delta x}\).
By doing this for all relevant directions, and under the condition that \(\Delta x = \Delta y\), we simplify the expression to a discrete form. This approach converts a continuous problem into a set of linear equations solvable using various numerical methods, providing an estimate of the temperature distribution within the studied domain.

The accuracy of finite-difference approximations depends on the grid resolution and the careful implementation of boundary and initial conditions. These approximations enable engineers and scientists to predict complex thermal behaviors in various applications, from aerospace components to electronic devices.

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Most popular questions from this chapter

An electrical heater \(100 \mathrm{~mm}\) long and \(5 \mathrm{~mm}\) in diameter is inserted into a hole drilled normal to the surface of a large block of material having a thermal conductivity of \(5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Estimate the temperature reached by the heater when dissipating \(50 \mathrm{~W}\) with the surface of the block at a temperature of \(25^{\circ} \mathrm{C}\).

A long bar of rectangular cross section is \(60 \mathrm{~mm} \times\) \(90 \mathrm{~mm}\) on a side and has a thermal conductivity of \(1 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). One surface is exposed to a convection process with air at \(100^{\circ} \mathrm{C}\) and a convection coeffient of \(100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), while the remaining surfaces are maintained at \(50^{\circ} \mathrm{C}\). (a) Using a grid spacing of \(30 \mathrm{~mm}\) and the Gauss-Seidel iteration method, determine the nodal temperatures and the heat rate per unit length normal to the page into the bar from the air. (b) Determine the effect of grid spacing on the temperature field and heat rate. Specifically, consider a grid spacing of \(15 \mathrm{~mm}\). For this grid, explore the effect of changes in \(h\) on the temperature field and the isotherms.

A hot fluid passes through circular channels of a cast iron platen (A) of thickness \(L_{\mathrm{A}}=30 \mathrm{~mm}\) which is in poor contact with the cover plates (B) of thickness \(L_{\mathrm{B}}=7.5 \mathrm{~mm}\). The channels are of diameter \(D=15 \mathrm{~mm}\) with a centerline spacing of \(L_{o}=60 \mathrm{~mm}\). The thermal conductivities of the materials are \(k_{\mathrm{A}}=20 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(k_{\mathrm{B}}=75 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), while the contact resistance between the two materials is \(R_{t, c}^{r}=2.0 \times 10^{-4} \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\). The hot fluid is at \(T_{i}=150^{\circ} \mathrm{C}\), and the convection coeftient is \(1000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The cover plate is exposed to ambient air at \(T_{\infty}=25^{\circ} \mathrm{C}\) with a convection coeftient of \(200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The shape factor between one channel and the platen top and bottom surfaces is 4.25.

A heat sink for cooling computer chips is fabricated from copper \(\left(k_{s}=400 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right)\), with machined microchannels passing a cooling fluid for which \(T=25^{\circ} \mathrm{C}\) and \(h=30,000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The lower side of the sink experiences no heat removal, and a preliminary heat sink design calls for dimensions of \(a=b=w_{s}=w_{f}=200 \mu \mathrm{m}\). A symmetrical element of the heat path from the chip to the fluid is shown in the inset. (a) Using the symmetrical element with a square nodal network of \(\Delta x=\Delta y=100 \mu \mathrm{m}\), determine the corresponding temperature field and the heat rate \(q^{\prime}\) to the coolant per unit channel length \((\mathrm{W} / \mathrm{m})\) for a maximum allowable chip temperature \(T_{c, \max }=75^{\circ} \mathrm{C}\). Estimate the corresponding thermal resistance between the chip surface and the fluid, \(R_{t, c-f}^{\prime}(\mathrm{m} \cdot \mathrm{K} / \mathrm{W})\). What is the maximum allowable heat dissipation for a chip that measures \(10 \mathrm{~mm} \times 10 \mathrm{~mm}\) on a side? (b) The grid spacing used in the foregoing finite-difference solution is coarse, resulting in poor precision for the temperature distribution and heat removal rate. Investigate the effect of grid spacing by considering spatial increments of 50 and \(25 \mu \mathrm{m}\). (c) Consistent with the requirement that \(a+b=400 \mu \mathrm{m}\), can the heat sink dimensions be altered in a manner that reduces the overall thermal resistance?

A plate \((k=10 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) is stiffened by a series of longitudinal ribs having a rectangular cross section with length \(L=8 \mathrm{~mm}\) and width \(w=4 \mathrm{~mm}\). The base of the plate is maintained at a uniform temperature \(T_{b}=\) \(45^{\circ} \mathrm{C}\), while the rib surfaces are exposed to air at a temperature of \(T_{\infty}=25^{\circ} \mathrm{C}\) and a convection coeffient of \(h=600 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) Using a finite-difference method with \(\Delta x=\Delta y=\) \(2 \mathrm{~mm}\) and a total of \(5 \times 3\) nodal points and regions, estimate the temperature distribution and the heat rate from the base. Compare these results with those obtained by assuming that heat transfer in the rib is one- dimensional, thereby approximating the behavior of a fin. (b) The grid spacing used in the foregoing finitedifference solution is coarse, resulting in poor precision for estimates of temperatures and the heat rate. Investigate the effect of grid refinement by reducing the nodal spacing to \(\Delta x=\Delta y=1 \mathrm{~mm}\) (a \(9 \times 3\) grid) considering symmetry of the center line. (c) Investigate the nature of two-dimensional conduction in the rib and determine a criterion for which the one-dimensional approximation is reasonable. Do so by extending your finite-difference analysis to determine the heat rate from the base as a function of the length of the rib for the range \(1.5 \leq L w \leq 10\), keeping the length \(L\) constant. Compare your results with those determined by approximating the rib as a fin.

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