/*! 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 29 Consider a plane composite wall ... [FREE SOLUTION] | 91影视

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Chapter 3: Problem 29

Consider a plane composite wall that is composed of two materials of thermal conductivities \(k_{\mathrm{A}}=0.1 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(k_{\mathrm{B}}=0.04 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and thicknesses \(L_{\mathrm{A}}=10 \mathrm{~mm}\) and \(L_{\mathrm{B}}=20 \mathrm{~mm}\). The contact resistance at the interface between the two materials is known to be \(0.30 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\). Material A adjoins a fluid at \(200^{\circ} \mathrm{C}\) for which \(h=10\) \(\mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and material \(\mathrm{B}\) adjoins a fluid at \(40^{\circ} \mathrm{C}\) for which \(h=20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) What is the rate of heat transfer through a wall that is \(2 \mathrm{~m}\) high by \(2.5 \mathrm{~m}\) wide? (b) Sketch the temperature distribution.

Short Answer

Expert verified
The overall heat transfer rate through the composite wall is 711.11 W, with a total thermal resistance of 0.225 K/W. The temperature distribution shows a continuous decrease from the hot fluid (200掳C) through material A, contact resistance, and material B to the cold fluid (40掳C).

Step by step solution

01

Determine the areas and interface resistance

Firstly, calculate the area of the entire wall (\(A_W\)) and the area of the contact (\(A_F\)) using the provided dimensions of the wall. Then, determine the interface resistance (R_contact) using the given contact resistance value. \[A_W = 2 \text{ m} 脳 2.5 \text{ m} = 5 \text{ m}^2\] \[A_F = 2 \text{ m} 脳 2 \text{ m} = 4 \text{ m}^2\] \[R_{contact} = \frac{0.30 \text{ m}^2 \cdot \text{K} / \text{W}}{A_F} = 0.075 \text{ K} / \text{W}\]
02

Calculate the convection and conduction resistances

Calculate the convection resistances on both fluid boundaries (R_conv_A and R_conv_B) using the given convective heat transfer coefficients: \[R_{conv_A} = \frac{1}{h_A A_W} = \frac{1}{(10 \text{ W/m}^2 \text{K})(5 \text{ m}^2)} = 0.02 \text{ K/W}\] \[R_{conv_B} = \frac{1}{h_B A_W} = \frac{1}{(20 \text{ W/m}^2 \text{K})(5 \text{ m}^2)} = 0.01 \text{ K/W}\] Next, calculate the conduction resistances for both materials (R_cond_A and R_cond_B): \[R_{cond_A}= \frac{L_A}{k_A A_W} = \frac{0.01 \text{ m}}{(0.1 \text{ W/m K})(5 \text{ m}^2)} = 0.02 \text{ K/W}\] \[R_{cond_B}= \frac{L_B}{k_B A_W} = \frac{0.02 \text{ m}}{(0.04 \text{ W/m K})(5 \text{ m}^2)} = 0.1 \text{ K/W}\]
03

Calculate the total thermal resistance and heat transfer

Add up all the resistances to find the total thermal resistance (R_total): \[R_{total} = R_{conv_A} + R_{cond_A} + R_{contact} + R_{cond_B} + R_{conv_B} = 0.02 + 0.02 + 0.075 + 0.1 + 0.01 = 0.225 \, \text{K/W}\] Now, calculate the overall heat transfer rate (Q) using the temperature difference between the hot and cold fluids and the total thermal resistance: \[Q = \frac{T_{hot} - T_{cold}}{R_{total}} = \frac{200 - 40}{0.225} = \frac{160}{0.225} \approx 711.11 \, \text{W}\]
04

Sketch the temperature distribution

To sketch the temperature distribution, plot the temperatures at the interfaces between convection and conduction, and conduction and contact resistance. Remember that heat transfer is continuous, meaning that the temperatures have to show a continuous decrease from the hot fluid to the cold fluid along the wall. | | fluid at 200掳C | ----> 0 Convection resistance | Material A| | ----> 0 Conduction resistance | | Interface (contact resistance) | Material B| | ----> 0 Conduction resistance | | fluid at 40掳C V The heat transfer rate through the wall is found to be 711.11 W, and this graph provides a visual representation of the temperature distribution across the composite wall.

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Key Concepts

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

Thermal Conductivity
Thermal conductivity, denoted as k, is a measure of a material's ability to conduct heat. It quantifies how easily heat can flow through a material due to a temperature gradient. In the exercise, two different materials, A and B, are given distinct thermal conductivity values, kA = 0.1 W/m·K and kB = 0.04 W/m·K. Higher k values indicate better heat conduction. When thermal conductivities are mixed in a composite wall, understanding each material鈥檚 k helps us assess how they affect overall heat transfer.
Convection Resistance
Convection resistance, represented by Rconv, relates to the opposition a fluid provides to heat flow due to convection. The inverse of the convective heat transfer coefficient, h, and the surface area A, the formula is Rconv = 1 / (h·A). In the exercise, the wall experiences convection on two sides, with coefficients hA = 10 W/m2·K and hB = 20 W/m2·K. The resistance to heat transfer by convection is lower where h is higher, as seen with Rconv_B being half the value of Rconv_A.
Conduction Resistance
Conduction resistance is a measure of how strongly a material opposes the flow of heat through its thickness. Represented by Rcond, it is determined by the thickness of the material L, its thermal conductivity k, and the surface area A, following the equation Rcond = L / (k·A). Materials A and B exhibit different conduction resistances in the exercise, reflecting their distinct thicknesses and thermal conductivities. Conduction resistance is critical for understanding how heat travels across solid materials and the impact of varying material properties on overall heat flow.
Contact Resistance
Contact resistance arises when two materials converge and there is a thermal barrier at their interface. This resistance is due to surface roughness and imperfections that restrict the flow of heat. In the given exercise, the contact resistance at the interface between materials A and B is 0.30 m2·K/W, assuming perfect contact over the interface area. The actual resistance to heat flow, Rcontact, is obtained by dividing the contact resistance per unit area by the actual surface area of contact. Contact resistance often acts as a bottleneck in heat transfer across composite structures.
Heat Transfer Rate
The heat transfer rate, denoted by Q, describes the amount of heat flowing through a material per unit time. It is calculated using the overall temperature difference and the total thermal resistance of the heat path. In the exercise, the formula Q = (Thot - Tcold) / Rtotal is used, where Thot and Tcold are the temperatures of the hot and cold fluids, respectively. The heat transfer rate is essential in engineering applications as it defines the efficiency of thermal systems and plays a crucial role in thermal management and control.
Temperature Distribution
Temperature distribution refers to how temperature varies across a system from one location to another. In the context of a composite wall, it is vital to understand how temperature reduces from the hotter side to the cooler side and how various resistances affect this gradient. Conduction, convection, and contact resistance each cause a 'drop' in temperature akin to voltage drop in electrical circuits. The solution's sketch visualizes the continuous decrease in temperature across different media in the system. Understanding temperature distribution allows for predictions and enhancements of material performance and energy efficiency in thermal systems.

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Most popular questions from this chapter

The fin array of Problem \(3.142\) is commonly found in compact heat exchangers, whose function is to provide a large surface area per unit volume in transferring heat from one fluid to another. Consider conditions for which the second fluid maintains equivalent temperatures at the parallel plates, \(T_{o}=T_{L}\), thereby establishing symmetry about the midplane of the fin array. The heat exchanger is \(1 \mathrm{~m}\) long in the direction of the flow of air (first fluid) and \(1 \mathrm{~m}\) wide in a direction normal to both the airflow and the fin surfaces. The length of the fin passages between adjoining parallel plates is \(L=8 \mathrm{~mm}\), whereas the fin thermal conductivity and convection coefficient are \(k=200 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) (aluminum) and \(h=150 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), respectively. (a) If the fin thickness and pitch are \(t=1 \mathrm{~mm}\) and \(S=4 \mathrm{~mm}\), respectively, what is the value of the thermal resistance \(R_{t, o}\) for a one-half section of the fin array? (b) Subject to the constraints that the fin thickness and pitch may not be less than \(0.5\) and \(3 \mathrm{~mm}\), respectively, assess the effect of changes in \(t\) and \(S\).

Consider the flat plate of Problem \(3.112\), but with the heat sinks at different temperatures, \(T(0)=T_{o}\) and \(T(L)=T_{L}\), and with the bottom surface no longer insulated. Convection heat transfer is now allowed to occur between this surface and a fluid at \(T_{\infty}\), with a convection coefficient \(h\). (a) Derive the differential equation that determines the steady-state temperature distribution \(T(x)\) in the plate. (b) Solve the foregoing equation for the temperature distribution, and obtain an expression for the rate of heat transfer from the plate to the heat sinks. (c) For \(q_{o}^{\prime \prime}=20,000 \mathrm{~W} / \mathrm{m}^{2}, T_{o}=100^{\circ} \mathrm{C}, T_{L}=35^{\circ} \mathrm{C}\), \(T_{\infty}=25^{\circ} \mathrm{C}, k=25 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, h=50 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), \(L=100 \mathrm{~mm}, t=5 \mathrm{~mm}\), and a plate width of \(W=\) \(30 \mathrm{~mm}\), plot the temperature distribution and determine the sink heat rates, \(q_{x}(0)\) and \(q_{x}(L)\). On the same graph, plot three additional temperature distributions corresponding to changes in the following parameters, with the remaining parameters unchanged: (i) \(q_{o}^{\prime \prime}=30,000 \mathrm{~W} / \mathrm{m}^{2}\), (ii) \(h=200\) \(\mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and (iii) the value of \(q_{o}^{\prime \prime}\) for which \(q_{x}(0)=0\) when \(h=200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\).

A nanolaminated material is fabricated with an atomic layer deposition process, resulting in a series of stacked, alternating layers of tungsten and aluminum oxide, each layer being \(\delta=0.5 \mathrm{~nm}\) thick. Each tungsten-aluminum oxide interface is associated with a thermal resistance of \(R_{t, i}^{\prime \prime}=3.85 \times 10^{-9} \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\). The theoretical values of the thermal conductivities of the thin aluminum oxide and tungsten layers are \(k_{\mathrm{A}}=1.65 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(k_{\mathrm{T}}=6.10 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), respectively. The properties are evaluated at \(T=300 \mathrm{~K}\). (a) Determine the effective thermal conductivity of the nanolaminated material. Compare the value of the effective thermal conductivity to the bulk thermal conductivities of aluminum oxide and tungsten, given in Tables A.1 and A.2. (b) Determine the effective thermal conductivity of the nanolaminated material assuming that the thermal conductivities of the tungsten and aluminum oxide layers are equal to their bulk values.

A very long rod of \(5-\mathrm{mm}\) diameter and uniform thermal conductivity \(k=25 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) is subjected to a heat treatment process. The center, 30 -mm-long portion of the rod within the induction heating coil experiences uniform volumetric heat generation of \(7.5 \times 10^{6} \mathrm{~W} / \mathrm{m}^{3}\). The unheated portions of the rod, which protrude from the heating coil on either side, experience convection with the ambient air at \(T_{\infty}=20^{\circ} \mathrm{C}\) and \(h=10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Assume that there is no convection from the surface of the rod within the coil. (a) Calculate the steady-state temperature \(T_{o}\) of the rod at the midpoint of the heated portion in the coil. (b) Calculate the temperature of the rod \(T_{b}\) at the edge of the heated portion.

Circular copper rods of diameter \(D=1 \mathrm{~mm}\) and length \(L=25 \mathrm{~mm}\) are used to enhance heat transfer from a surface that is maintained at \(T_{s, 1}=100^{\circ} \mathrm{C}\). One end of the rod is attached to this surface (at \(x=0\) ), while the other end \((x=25 \mathrm{~mm})\) is joined to a second surface, which is maintained at \(T_{s, 2}=0^{\circ} \mathrm{C}\). Air flowing between the surfaces (and over the rods) is also at a temperature of \(T_{\infty}=0^{\circ} \mathrm{C}\), and a convection coefficient of \(h=100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) is maintained. (a) What is the rate of heat transfer by convection from a single copper rod to the air? (b) What is the total rate of heat transfer from a \(1 \mathrm{~m} \times 1 \mathrm{~m}\) section of the surface at \(100^{\circ} \mathrm{C}\), if a bundle of the rods is installed on 4 -mm centers?
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