/*! 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 82 A plate \((k=10 \mathrm{~W} / \m... [FREE SOLUTION] | 91影视

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Chapter 4: Problem 82

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.

Short Answer

Expert verified
In this problem, we analyze the heat transfer in a rib-stiffened plate using a finite-difference method. For a 5x3 grid, we discretize the heat equation, apply boundary conditions, and solve for the unknown nodal temperatures. To estimate the heat rate from the base, we use Fourier's law and compare these results with the one-dimensional approximation using the fin equation. This procedure can be extended for other tasks with different grid configurations and rib lengths.

Step by step solution

01

Establish the governing equations for the problem

For the given rectangular rib, the heat equation is given by: \[\frac{\partial^{2} T}{\partial x^{2}}+\frac{\partial^{2} T}{\partial y^{2}}=0\]
02

Set up the finite-difference grid

We will use a 5x3 grid with a spacing of 螖x = 螖y = 2 mm. Number the nodal points, and assign index values for each node. Also, note the boundary conditions for each node.
03

Discretize the governing equation

Use finite-difference equations to replace the derivatives in the governing equation: \[\frac{T_{i+1,j}-2T_{i,j}+T_{i-1,j}}{(\Delta x)^{2}} + \frac{T_{i,j+1}-2T_{i,j}+T_{i,j-1}}{(\Delta y)^{2}} = 0\]
04

Apply boundary conditions and solve for the nodal temperatures

For each of the boundary nodes, apply the given boundary conditions, and use the finite-difference equation to solve for the unknown nodal temperatures.
05

Estimate the heat rate from the base

Use the Fourier's law to estimate the heat rate from the base of the plate: \[q = -k\frac{\partial T}{\partial n}\] Calculate this value using the finite-difference method at the base nodes.
06

Calculate the one-dimensional approximation

Assume that heat transfer in the rib is one-dimensional, and use the fin equation to calculate the temperature distribution and heat rate from the base. Compare these results with the finite-difference solution obtained in Step 5. Following this procedure will yield the results for task 1 of the exercise. Similar processes can be followed for tasks 2 and 3 by adjusting the grid spacing and varying the length of the rib, respectively.

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Key Concepts

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

Two-Dimensional Conduction
Two-dimensional conduction refers to heat transfer through a material where the temperature varies in two different directions鈥攖ypically length and width. In the context of the plate with ribs, this means we are considering the temperature distribution through both the length \(L\) and width \(w\) of the ribs. When we address this in practical problems, we often use mathematical models like the heat equation:
  • \[ \frac{\partial^{2} T}{\partial x^{2}} + \frac{\partial^{2} T}{\partial y^{2}} = 0 \]
This equation allows us to understand how heat is distributed across the surface of the rib. To solve this two-dimensional equation, we often employ numerical methods, such as the finite-difference method, which discretizes the area into a grid and approximates the differential equations at each point.
This transforms the continuous problem into a solvable set of algebraic equations, making it feasible to compute the temperatures at various points even for complex geometrical shapes and boundary conditions found in engineering tasks like the given exercise.
Heat Transfer Analysis
Heat transfer analysis involves studying how heat moves from one part of a system to another, primarily through conduction, convection, or radiation. For the ribbed plate, conduction and forced convection are the primary mechanisms considered. Using the given material properties and convection coefficients, we can estimate the heat distribution and flow through the ribbed structure.
The finite-difference method is a powerful tool in this analysis as it segments the area into a grid, calculating temperatures at nodal points. By applying the heat equation:
  • The discrete form becomes:
    \[ \frac{T_{i+1,j} - 2T_{i,j} + T_{i-1,j}}{(\Delta x)^{2}} + \frac{T_{i,j+1} - 2T_{i,j} + T_{i,j-1}}{(\Delta y)^{2}} = 0 \]
This conversion lets us systematically solve for temperature distributions, which are critical for determining the efficiency and effectiveness of the heat transfer in the structure. Analyzing these results allows engineers to make informed decisions regarding design optimizations and heat management strategies.
Boundary Conditions in Heat Transfer
Boundary conditions are essential in defining how heat transfer occurs at the surface of the material. They specify the temperature or heat flux constraints at the boundaries and significantly influence the heat distribution within the material. In our ribbed plate example, boundary conditions include the base temperature \(T_{b}\) and the ambient temperature with convection effects.
For accurate heat transfer analysis, these conditions must be precisely integrated into the finite-difference method. They are used to set up equations that describe scenarios like heat conduction through the base or convection from the rib surfaces:
  • Conduction boundary: Typically fixed temperature conditions or insulated surfaces.
  • Convection boundary: Often involves a convection coefficient \(h\) and ambient temperature \(T_{\infty}\).
By solving the nodal temperatures with these boundary conditions, we predict the heat rate from the base and how much heat the rib loses to the surroundings. Properly defined boundary conditions are crucial for creating realistic simulations that guide engineering design and thermal management.

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Most popular questions from this chapter

A long constantan wire of \(1-\mathrm{mm}\) diameter is butt welded to the surface of a large copper block, forming a thermocouple junction. The wire behaves as a fin, permitting heat to flow from the surface, thereby depressing the sensing junction temperature \(T_{j}\) below that of the block \(T_{\sigma}\). Copper block, \(T_{o}\) (a) If the wire is in air at \(25^{\circ} \mathrm{C}\) with a convection coefficient of \(10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), estimate the measurement error \(\left(T_{j}-T_{o}\right)\) for the thermocouple when the block is at \(125^{\circ} \mathrm{C}\). (b) For convection coefficients of 5,10 , and \(25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), plot the measurement error as a function of the thermal conductivity of the block material over the range 15 to \(400 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Under what circumstances is it advantageous to use smaller diameter wire?

The elemental unit of an air heater consists of a long circular rod of diameter \(D\), which is encapsulated by a finned sleeve and in which thermal energy is generated by ohmic heating. The \(N\) fins of thickness \(t\) and length \(L\) are integrally fabricated with the square sleeve of width \(w\). Under steady-state operating conditions, the rate of thermal energy generation corresponds to the rate of heat transfer to airflow over the sleeve. (a) Under conditions for which a uniform surface temperature \(T_{s}\) is maintained around the circumference of the heater and the temperature \(T_{\infty}\) and convection coefficient \(h\) of the airflow are known, obtain an expression for the rate of heat transfer per unit length to the air. Evaluate the heat rate for \(T_{s}=300^{\circ} \mathrm{C}, D=20 \mathrm{~mm}\), an aluminum sleeve \(\left(k_{s}=240 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right), w=40 \mathrm{~mm}\), \(N=16, t=4 \mathrm{~mm}, L=20 \mathrm{~mm}, T_{\infty}=50^{\circ} \mathrm{C}\), and \(h=500 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (b) For the foregoing heat rate and a copper heater of thermal conductivity \(k_{h}=400 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), what is the required volumetric heat generation within the heater and its corresponding centerline temperature? (c) With all other quantities unchanged, explore the effect of variations in the fin parameters \((N, L, t)\) on the heat rate, subject to the constraint that the fin thickness and the spacing between fins cannot be less than \(2 \mathrm{~mm}\).

Consider an aluminum heat sink \((k=240 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\), such as that shown schematically in Problem 4.28. The inner and outer widths of the square channel are \(w=20 \mathrm{~mm}\) and \(W=40 \mathrm{~mm}\), respectively, and an outer surface temperature of \(T_{s}=50^{\circ} \mathrm{C}\) is maintained by the array of electronic chips. In this case, it is not the inner surface temperature that is known, but conditions \(\left(T_{\infty}, h\right)\) associated with coolant flow through the channel, and we wish to determine the rate of heat transfer to the coolant per unit length of channel. For this purpose, consider a symmetrical section of the channel and a two-dimensional grid with \(\Delta x=\Delta y=5 \mathrm{~mm}\). (a) For \(T_{\infty}=20^{\circ} \mathrm{C}\) and \(h=5000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the unknown temperatures, \(T_{1}, \ldots, T_{7}\), and the rate of heat transfer per unit length of channel, \(q^{\prime}\). (b) Assess the effect of variations in \(h\) on the unknown temperatures and the heat rate.

A small device is used to measure the surface temperature of an object. A thermocouple bead of diameter \(D=120 \mu \mathrm{m}\) is positioned a distance \(z=100 \mu \mathrm{m}\) from the surface of interest. The two thermocouple wires, each of diameter \(d=25 \mu \mathrm{m}\) and length \(L=300 \mu \mathrm{m}\), are held by a large manipulator that is at a temperature of \(T_{m}=23^{\circ} \mathrm{C}\). If the thermocouple registers a temperature of \(T_{\mathrm{tc}}=29^{\circ} \mathrm{C}\), what is the surface temperature? The thermal conductivities of the chromel and alumel thermocouple wires are \(k_{\mathrm{Ch}_{h}}=19 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(k_{\mathrm{Al}}=29 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), respectively. You may neglect radiation and convection effects.

Hot water is transported from a cogeneration power station to commercial and industrial users through steel pipes of diameter \(D=150 \mathrm{~mm}\), with each pipe centered in concrete \((k=1.4 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) of square cross section \((w=300 \mathrm{~mm})\). The outer surfaces of the concrete are exposed to ambient air for which \(T_{\infty}=0^{\circ} \mathrm{C}\) and \(h=\) \(25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) If the inlet temperature of water flowing through the pipe is \(T_{i}=90^{\circ} \mathrm{C}\), what is the heat loss per unit length of pipe in proximity to the inlet? The temperature of the pipe \(T_{1}\) may be assumed to be that of the inlet water. (b) If the difference between the inlet and outlet temperatures of water flowing through a 100 -m-long pipe is not to exceed \(5^{\circ} \mathrm{C}\), estimate the minimum allowable flow rate \(\dot{m}\). A value of \(c=4207 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\) may be used for the specific heat of the water.
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