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How does the thermal resistance network associated with a single-layer plane wall differ from the one associated with a five-layer composite wall?

Short Answer

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The primary difference between the thermal resistance network of a single-layer plane wall and a five-layer composite wall is the complexity of their networks. A single-layer wall has a straightforward one-dimensional conduction described by a single-value thermal resistance (\(R_{total} = \frac{L}{kA}\)). On the other hand, a five-layer composite wall has a more complex thermal resistance network, where the total thermal resistance is the sum of the individual contributions from each layer (\(R_{total} = R_1+R_2+R_3+R_4+R_5\)). Each layer in a composite wall will have its own heat transfer rate, which is dictated by the properties and thickness of the material. By understanding these differences in thermal resistance networks, we can better analyze and design wall assemblies to meet specific heat transfer and insulation requirements.

Step by step solution

01

Understand thermal resistance and its application in heat transfer

Thermal resistance is a property of a material that quantifies its ability to resist the flow of heat. In the context of building walls, it enables us to calculate the heat transfer rate between two environments separated by the wall. When multiple layers are combined to form a wall, each layer contributes its own thermal resistance, which is added linearly. It is important to note the difference between one-layer walls and composite walls, as we will see in the following steps.
02

Describe the thermal resistance network of a single-layer plane wall

A single-layer plane wall consists of one layer of material with a certain thickness. Heat transfer through this wall can be modeled by a simple one-dimensional conduction. In this case, the total thermal resistance of the wall, \(R_{total}\), is given by: \(R_{total} = \frac{L}{kA}\) where: - \(L\) is the thickness of the wall, - \(k\) is the thermal conductivity of the material, - \(A\) is the area through which heat transfer occurs.
03

Describe the thermal resistance network of a five-layer composite wall

A five-layer composite wall consists of several layers of different materials, each contributing to the wall's overall thermal resistance. In this case, the total thermal resistance of the wall can be obtained by summing the individual thermal resistances: \(R_{total} = R_1+R_2+R_3+R_4+R_5\) where: - \(R_i\) is the thermal resistance contribution of the \(i^{th}\) layer, - \(R_i = \frac{L_i}{k_iA}\) (with \(L_i\) being the thickness, and \(k_i\) the thermal conductivity of the \(i^{th}\) layer).
04

Compare the thermal resistance network between a single-layer and a five-layer composite wall

In summary, the primary difference between a single-layer plane wall and a five-layer composite wall is the number of layers and thus the complexity of their thermal resistance networks: - A single-layer plane wall has a straightforward one-dimensional conduction described by a single-value thermal resistance (\(R_{total} = \frac{L}{kA}\)). - A five-layer composite wall has a more complex thermal resistance network, where the total thermal resistance is the sum of the individual contributions from each layer (\(R_{total} = R_1+R_2+R_3+R_4+R_5\)). Each layer will have its own heat transfer rate, which will be dictated by the material's properties and thickness. By understanding the differences in thermal resistance networks, we can better analyze and design different wall assemblies for specific heat transfer and insulation requirements.

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Most popular questions from this chapter

Consider two finned surfaces that are identical except that the fins on the first surface are formed by casting or extrusion, whereas they are attached to the second surface afterwards by welding or tight fitting. For which case do you think the fins will provide greater enhancement in heat transfer? Explain.

A spherical vessel, \(3.0 \mathrm{~m}\) in diameter (and negligible wall thickness), is used for storing a fluid at a temperature of $0^{\circ} \mathrm{C}\(. The vessel is covered with a \)5.0$-cm-thick layer of an insulation \((k=0.20 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). The surrounding air is at \(22^{\circ} \mathrm{C}\). The inside and outside heat transfer coefficients are 40 and 10 $\mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}\(, respectively. Calculate \)(a)$ all thermal resistances, in \(\mathrm{K} / \mathrm{W},(b)\) the steady rate of heat transfer, and \((c)\) the temperature difference across the insulation layer.

A \(0.4-\mathrm{cm}\)-thick, 12 -cm-high, and \(18-\mathrm{cm}\)-long circuit board houses 80 closely spaced logic chips on one side, each dissipating $0.04 \mathrm{~W}$. The board is impregnated with copper fillings and has an effective thermal conductivity of $30 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}$. All the heat generated in the chips is conducted across the circuit board and is dissipated from the back side of the board to a medium at \(40^{\circ} \mathrm{C}\), with a heat transfer coefficient of $52 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$. (a) Determine the temperatures on the two sides of the circuit board. (b) Now a \(0.2-\mathrm{cm}\)-thick, \(12-\mathrm{cm}\)-high, and \(18-\mathrm{cm}\)-long aluminum plate $(k=237 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\( with \)8642 \cdot \mathrm{cm}-$ long aluminum pin fins of diameter \(0.25 \mathrm{~cm}\) is attached to the back side of the circuit board with a \(0.02-\mathrm{cm}\)-thick epoxy adhesive \((k=1.8 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). Determine the new temperatures on the two sides of the circuit board.

A hot surface at \(100^{\circ} \mathrm{C}\) is to be cooled by attaching 3 -cm- long, \(0.25\)-cm-diameter aluminum pin fins \((k=237\) $\mathrm{W} / \mathrm{m} \cdot \mathrm{K})\( to it, with a center-to-center distance of \)0.6 \mathrm{~cm}\(. The temperature of the surrounding medium is \)30^{\circ} \mathrm{C}\(, and the heat transfer coefficient on the surfaces is \)35 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$. Determine the rate of heat transfer from the surface for a \(1-\mathrm{m} \times 1-\mathrm{m}\) section of the plate. Also determine the overall effectiveness of the fins.

Circular cooling fins of diameter \(D=1 \mathrm{~mm}\) and length $L=30 \mathrm{~mm}\(, made of copper \)(k=380 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})$, are used to enhance heat transfer from a surface that is maintained at temperature \(T_{s 1}=132^{\circ} \mathrm{C}\). Each rod has one end attached to this surface \((x=0)\), while the opposite end \((x=L)\) is joined to a second surface, which is maintained at \(T_{s 2}=0^{\circ} \mathrm{C}\). The air flowing between the surfaces and the rods is also at \(T_{\infty}=0^{\circ} \mathrm{C}\). and the convection coefficient is $h=100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$. (a) Express the function \(\theta(x)=T(x)-T_{\infty}\) along a fin, and calculate the temperature at \(x=L / 2\). (b) Determine the rate of heat transferred from the hot surface through each fin and the fin effectiveness. Is the use of fins justified? Why? (c) What is the total rate of heat transfer from a \(10-\mathrm{cm}\) by \(10-\mathrm{cm}\) section of the wall, which has 625 uniformly distributed fins? Assume the same convection coefficient for the fin and for the unfinned wall surface.

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