/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 39 Derive the nodal finite-differen... [FREE SOLUTION] | 91Ó°ÊÓ

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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 \(T_{\mathrm{w}}\) and a heat transfer coefficient \(h\). Assume \(\Delta x=\Delta y\). (b) Node \(m_{4} n\) at the tip of a cutting tool with the upper surface exposed to a constant heat flux \(q_{=}^{\prime \prime}\) and the diagonal surface exposed to a convection cooling process with the fluid at \(T_{w}\) and a heat transfer cocfficient h. Asume \(\Delta \mathrm{r}=\Delta \mathrm{y}\).

Short Answer

Expert verified
Convert boundary conditions into finite-difference equations accounting for convection and heat flux effects.

Step by step solution

01

Understand the Problem

For both configurations, we need to derive finite-difference equations. This involves approximating the heat transfer processes at specific nodes using finite differences, considering convection and boundary conditions.
02

Apply Finite-Difference Approximation for Part (a)

For node \((m, n)\) on a diagonal boundary exposed to convection, we use the finite-difference approximation for heat conduction in both the x and y directions. Let \(T_{m,n}\) be the temperature at node \((m, n)\), and adjacent nodes have temperatures like \(T_{m+1,n}\), \(T_{m-1,n}\), \(T_{m,n+1}\), \(T_{m,n-1}\). The general conduction term becomes:\[\frac{T_{m+1,n} - 2T_{m,n} + T_{m-1,n}}{\Delta x^2} + \frac{T_{m,n+1} - 2T_{m,n} + T_{m,n-1}}{\Delta y^2} = 0\]Convert for convection boundary:\[k(T_{m,n} - T_{w}) = -k\left(\frac{T_{m+1,n} - T_{m-1,n}}{2\Delta x} + \frac{T_{m,n+1} - T_{m,n-1}}{2\Delta y}\right) + h(T_{m,n} - T_w)\]
03

Apply Finite-Difference Approximation for Part (b)

For node \(m_4, n\) at the tip of a cutting tool, consider two boundary conditions: a constant heat flux \(q''\) on the upper surface and convection on the diagonal surface. Using finite differences, the heat flux condition is:\[-k \frac{T_{m+1,n} - T_{m-1,n}}{2\Delta x} = q''\]For convection on the diagonal:\[k(T_{m_4,n} - T_w) = -k\left(\frac{T_{m+1,n} - T_{m-1,n}}{2\Delta x} + \frac{T_{m,n+1} - T_{m,n-1}}{2\Delta y}\right) + h(T_{m_4,n} - T_w)\]
04

Combine Boundary Conditions for Both Nodes

For both nodes, combine the conduction equations with the boundary conditions, adjusting for equal discretization sizes like \(\Delta x = \Delta y\) for (a) and \(\Delta r = \Delta y\) for (b). Simplification results in nodal equations incorporating both convection and any specified heat fluxes.

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Key Concepts

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

Heat Transfer
The study of heat transfer involves understanding how thermal energy moves within and between materials. When conducted through objects, this process can occur in three primary modes: conduction, convection, and radiation.
- **Conduction** is the transfer of heat through a material without the movement of the material itself. It typically occurs in solids where molecules vibrate and pass kinetic energy to nearby molecules. - **Convection** involves bulk movement in fluids (liquids or gases) where the warmer areas of a fluid rise and the cooler areas sink. We'll discuss this more in the convection section. - **Radiation** is the transfer of energy through electromagnetic waves and does not require a medium, meaning it can even occur in a vacuum.
In the context of the finite-difference method for analyzing thermal problems, understanding conduction and convection is critical as they influence the nodal equations. These equations help predict how heat will be distributed across the grid of nodes, allowing for accurate simulations of heat transfer processes.
Convection
Convection is a mode of heat transfer that occurs in fluids (liquids and gases) due to the motion of the fluid itself. In our study, this concept is crucial when exploring boundary conditions where a solid surface is in contact with a fluid.
**The role of convection in heat transfer:**- In convection, heat is transferred from the surface of an object to a fluid. The warmer fluid then moves away, creating space for cooler fluid to come in contact with the surface, and the cycle repeats. - The effectiveness of convection depends primarily on the heat transfer coefficient, denoted as \( h \). This coefficient indicates how well heat is transferred between the solid surface and the fluid. A higher \( h \) value means more effective heating or cooling by convection.
**Understanding convection in the problem:**- In the finite-difference equations, boundary nodes subjected to convection require modifications to account for this process. These equations are often expressed using the convective heat transfer term \( h(T_{surface} - T_{ambient}) \), where \( T_{surface} \) is the temperature at the boundary and \( T_{ambient} \) is the fluid temperature.
Boundary Conditions
Boundary conditions are vital in solving partial differential equations (PDEs) involved in heat transfer and finite-difference methods. They dictate the behavior of a system at its borders and are essential for obtaining unique, physical solutions.
**Types of boundary conditions:** - **Dirichlet Boundary Condition:** Specifies the temperature at a surface; the temperature at the boundary is determined externally or is known before analysis. - **Neumann Boundary Condition:** Specifies the heat flux at a boundary; it can include a constant heat flux or variations due to convection. - **Robin's Condition:** Combines both Dirichlet and Neumann conditions, typically in scenarios involving convection where both temperature and heat flux are involved.
In the exercise presented, we're dealing with boundary conditions involving convection and specific heat flux. These conditions need to be incorporated into the nodal finite-difference equations to ensure the correct modeling of thermal interactions at the boundaries.
Nodal Analysis
Nodal analysis in the context of heat transfer refers to the process of evaluating temperature distribution across nodes in a grid using mathematical models like the finite-difference method. This technique is crucial for breaking down complex heat transfer areas into manageable individual nodes.
**Key steps in nodal analysis:** - Set up a grid system over the region of interest. Each intersection, or node, will have its unique equation reflecting interactions with neighboring nodes. - Use finite-difference approximations to convert differential equations governing heat transfer into algebraic equations. - Incorporate boundary conditions to modify the nodal equations accordingly.
**Applying nodal analysis:** In this specific exercise, we apply nodal analysis to derive finite-difference equations for nodes near boundaries affected by convection. The equations consider both the conduction within the material and the convection heat transfer occurring at the surface, delivering a comprehensive analysis of temperature distribution.

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Most popular questions from this chapter

A heat sink for cooling computer chips is fabricated from copper \(\left(k_{3}=400 \mathrm{~W} / \mathrm{m}-\mathrm{K}\right)\), with machined microchannels passing a cooling fluid for which \(T_{v}=25^{\circ} \mathrm{C}\) 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 dimensicns 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 \(q\) ' to the coolant per unit channel length \((W / m)\) for a maximum allowable chip termerature \(T_{\text {cmat }}=75^{\circ} \mathrm{C}\). Estimate the corresponding thermal resistance between the chip surface and the fluid, \(R_{\text {U- }-}(\mathrm{m} \cdot \mathrm{K} / \mathrm{W})\). What is the maximum allowable heat dissipation for a chip that measures \(10 \mathrm{~mm} \times 10 \mathrm{~mm}\) cn a side? (b) The grid spacing wed 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?

Consider heat transfer in a one-dimensional (radial) cylindrical coordinate system under steady-state conditions with volumetric heat generation. (a) Derive the finite-differenee equation for any interior node \(m\). (b) Derive the finite-difference equation for the node \(n\) located at the extemal boundary subjected to the convection process \(\left(T_{w}, h\right)\).

Small diameter electrical heating clements dissipating 50 Whan (length norunal to the sketch) are used to heat a ceramic plate of thermal conductivity 2 Whm \(+\mathrm{K}\). The upper surface of the plate is cxposed to ambient air at \(30^{\circ} \mathrm{C}\). with a convection coefficient of \(100 \mathrm{~W} / \mathrm{m}^{3} \cdot \mathrm{K}\), while the lower surface is well insulated. (a) Using the Gauss-Seidel method with a grid spacing of \(\mathrm{At}=6 \mathrm{~mm}\) and \(\Delta \mathrm{y}=2 \mathrm{~mm}\), 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) Calculute the heat loss by convection from the plate to the ftuid. Compare this value with the clement dissipation rate, (d) What advantage, if any, is there in not making \(\Delta r=\Delta y\) for thik situation? (c) With \(\Delta x=\Delta y=2 \mathrm{~mm}\), calculate the icmperatare field within the plate and the rate of heal transfer from the plate. Under no circumstances may the temperature at any location in the plate excecd \(400^{\circ} \mathrm{C}\). Would this limit be excecded if the air flow were terminated and heat transfer to the air was by natural convection with \(h=10 \mathrm{~W} / \mathrm{m}^{2}+\mathrm{K}\) ?

A plate \((k=10 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) is stiffened by a series of longitudinal ribs having a rectangulur cross section with Iength \(L=8 \mathrm{~mm}\) and width \(w=4 \mathrm{~mm}\). The base of the plate is maintained at a uniform temperature \(T_{b}=\) \(4^{\circ} \mathrm{C}\), while the rib surfaces are exposed to air at a temperature of \(T_{z}=25^{\circ} \mathrm{C}\) and a convection coefficient of \(h=600 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) Using a finite-difference method with \(\Delta t=\Delta v=\) \(2 \mathrm{~mm}\) 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, therchy approximating the behavior of a fin. (b) The grid spacing used in the foregoing finitedifference solution is coarse, resulting in poor precision for cxtimates of temperatures and the heat rate. Investigate the effect of grid refinement by reducing the nodul spacing to \(\Delta \mathrm{r}=\Delta y=1 \mathrm{~mm}\) (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 \(1.5 \leq L / \leq 10\), keeping the length \(L\) constant. Compare your results with those determined by approximating the rib as a fin.

Hot water at \(85^{\circ} \mathrm{C}\) flows through a thin-walled copper tube of \(30 \mathrm{~mm}\) diameter. The tube is enclosed by an eccentric cylindrical shell that is maintained at \(35^{\circ} \mathrm{C}\) and has a diameter of \(120 \mathrm{~mm}\). The eccentricity, defined as the separation between the centers of the tube and shell, is \(20 \mathrm{~mm}\). The space between the tube and shell is filled with an insulating material having a thermal conductivity of \(0.05 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Cakulate the heat loss per unit length of the tube und compare the result with the heat loss for a concentric artangement.

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