/*! 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 177 A nanolaminated material is fabr... [FREE SOLUTION] | 91影视

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Chapter 3: Problem 177

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.

Short Answer

Expert verified
The effective thermal conductivity of the nanolaminated material considering the theoretical thermal conductivities is calculated using the formula \(k_{\mathrm{effective}}=\frac{\delta}{R_{\mathrm{total}}}\), where the total thermal resistance, \(R_{\mathrm{total}} = R_{\mathrm{A}} + R_{\mathrm{T}} + R_{t,i}^{\prime \prime}\). After determining the effective thermal conductivity, compare it with the bulk thermal conductivities of aluminum oxide and tungsten. Repeat the process for bulk thermal conductivities to obtain the effective thermal conductivity in that case.

Step by step solution

01

Calculate the individual thermal resistances of the aluminum oxide and tungsten layers

In order to find the total thermal resistance within the nanolaminated material, we first need to calculate the thermal resistance of each layer. The formula for thermal resistance is given by: \[R = \frac{\delta}{k}\] where \(R\) = thermal resistance of the layer, \(\delta\) = thickness of the layer, \(k\) = thermal conductivity of the layer. Now we can calculate the thermal resistance of each layer: For the aluminum oxide layer, \[R_{\mathrm{A}}=\frac{\delta}{k_{\mathrm{A}}}\] For the tungsten layer, \[R_{\mathrm{T}}=\frac{\delta}{k_{\mathrm{T}}}\]
02

Calculate the total thermal resistance of the nanolaminated material

Considering we have alternating layers of aluminum oxide and tungsten with thermal interface resistance between them, the total thermal resistance will be the sum of the individual thermal resistances of aluminum oxide and tungsten layers as well as the interface thermal resistances: \[R_{\mathrm{total}} = R_{\mathrm{A}} + R_{\mathrm{T}} + R_{t,i}^{\prime \prime}\]
03

Calculate the effective thermal conductivity of the nanolaminated material

Now that we have found the total thermal resistance of the nanolaminated material, we can determine the effective thermal conductivity (k_effective) using the thickness of either layer. In this case, we'll use the thickness of the aluminum oxide layer: \[k_{\mathrm{effective}}=\frac{\delta}{R_{\mathrm{total}}}\]
04

Repeat the procedure for the case with bulk thermal conductivities

In part (b) of the exercise, we are asked to determine the effective thermal conductivity when considering the bulk thermal conductivities of the aluminum oxide and tungsten layers. We'll follow the same process as in Steps 1 to 3 but replace the theoretical thermal conductivities with the bulk values, found in Tables A.1 and A.2. Calculate \(R_{\mathrm{A}}\), \(R_{\mathrm{T}}\), \(R_{\mathrm{total}}\), and ultimately the effective thermal conductivity (\(k_{\mathrm{effective}}\)). Now we have determined the effective thermal conductivities considering both the theoretical and bulk thermal conductivities of the aluminum oxide and tungsten layers.

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Key Concepts

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

Thermal Resistance
When heat travels through materials, it encounters a kind of friction, which is quantified as thermal resistance. Just as electrical resistance determines how easily electricity flows through a circuit, thermal resistance measures how easily heat conducts through a material. It is commonly denoted by the symbol 'R'.

The mathematical expression for thermal resistance is \[R = \frac{\Delta x}{k}\] where \(\Delta x\) is the thickness of the material and 'k' is the thermal conductivity. The unit of thermal resistance is squared meters Kelvin per Watt (\(\mathrm{m}^2\cdot \mathrm{K} / \mathrm{W}\)). In a layered system such as a nanolaminated material, each layer's thermal resistance contributes to the overall resistance to heat flow. This is of particular interest in designing materials with tailored thermal properties for applications in electronics and insulation.
Nanolaminated Materials
Nanolaminated materials consist of alternating layers of different substances, often on a nanometer scale. These complex structures exhibit unique properties that are not found in bulk materials. Due to their extremely thin layers, they can have different thermal, electrical, and mechanical properties, making them ideal for a broad range of applications, including protective coatings and electronic devices.

In the provided exercise, the material is made of tungsten and aluminum oxide layers, with each layer being only 0.5 nm thick. Such precise control over layer thickness is crucial for tailoring the material's properties. For instance, the effective thermal conductivity of a nanolaminated material can be significantly different from that of its constituent layers when in bulk form, leading to potentially advantageous characteristics.
Atomic Layer Deposition
Atomic layer deposition (ALD) is a process for creating thin films with atomic-level precision. It uses sequential, self-limiting chemical reactions to build up material one atomic layer at a time, which makes it possible to achieve extremely uniform and conformal coatings. ALD is particularly valuable for producing nanolaminated materials where each layer's thickness must be controlled very accurately.

The exercise mentions the use of ALD to fabricate alternating layers of tungsten and aluminum oxide, showcasing ALD's ability to create complex, multilayer structures with precise interfaces. These well-defined interfaces can contribute additional thermal resistance, which is accounted for in thermal analyses of the material. ALD's precision is essential in industries where the thermal management of devices is critical, such as in semiconductor manufacturing.
Bulk Thermal Conductivity
Bulk thermal conductivity represents the innate ability of a material in its bulk form鈥攎eaning not on the nanoscale or as part of a composite鈥攖o conduct heat. It is an intrinsic property determined by the material's structure and the motion of atoms and electrons within it. High bulk thermal conductivity materials, like metals, are effective at transferring heat, while low conductivity materials, such as ceramics, are better insulators.

In our exercise, the bulk thermal conductivities of aluminum oxide and tungsten are given for comparison to the effective thermal conductivity of the nanolaminated material. Frequently, the effective thermal conductivity of nanolaminated materials will differ from the bulk values due to size effects and interface thermal resistances. By comparing the effective thermal conductivity to the bulk values, one can evaluate how the nanoscale lamination influences the material's overall thermal behavior, which is crucial for optimizing thermal management in practical applications.

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Most popular questions from this chapter

Measurements show that steady-state conduction through a plane wall without heat generation produced a convex temperature distribution such that the midpoint temperature was \(\Delta T_{o}\) higher than expected for a linear temperature distribution. Assuming that the thermal conductivity has a linear dependence on temperature, \(k=k_{o}(1+\alpha T)\), where \(\alpha\) is a constant, develop a relationship to evaluate \(\alpha\) in terms of \(\Delta T_{o}, T_{1}\), and \(T_{2}\).

An electrical current of 700 A flows through a stainless steel cable having a diameter of \(5 \mathrm{~mm}\) and an electrical resistance of \(6 \times 10^{-4} \mathrm{\Omega} / \mathrm{m}\) (i.e., per meter of cable length). The cable is in an environment having a temperature of \(30^{\circ} \mathrm{C}\), and the total coefficient associated with convection and radiation between the cable and the environment is approximately \(25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) If the cable is bare, what is its surface temperature? (b) If a very thin coating of electrical insulation is applied to the cable, with a contact resistance of \(0.02 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\), what are the insulation and cable surface temperatures? (c) There is some concern about the ability of the insulation to withstand elevated temperatures. What thickness of this insulation \((k=0.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) will yield the lowest value of the maximum insulation temperature? What is the value of the maximum temperature when this thickness is used?

A bakelite coating is to be used with a 10 -mm-diameter conducting rod, whose surface is maintained at \(200^{\circ} \mathrm{C}\) by passage of an electrical current. The rod is in a fluid at \(25^{\circ} \mathrm{C}\), and the convection coefficient is \(140 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). What is the critical radius associated with the coating? What is the heat transfer rate per unit length for the bare rod and for the rod with a coating of bakelite that corresponds to the critical radius? How much bakelite should be added to reduce the heat transfer associated with the bare rod by \(25 \%\) ?

A scheme for concurrently heating separate water and air streams involves passing them through and over an array of tubes, respectively, while the tube wall is heated electrically. To enhance gas-side heat transfer, annular fins of rectangular profile are attached to the outer tube surface. Attachment is facilitated with a dielectric adhesive that electrically isolates the fins from the current-carrying tube wall. (a) Assuming uniform volumetric heat generation within the tube wall, obtain expressions for the heat rate per unit tube length \((\mathrm{W} / \mathrm{m})\) at the inner \(\left(r_{i}\right)\) and outer \(\left(r_{o}\right)\) surfaces of the wall. Express your results in terms of the tube inner and outer surface temperatures, \(T_{s, i}\) and \(T_{s, e}\), and other pertinent parameters. (b) Obtain expressions that could be used to determine \(T_{s, i}\) and \(T_{s, o}\) in terms of parameters associated with the water- and air-side conditions. (c) Consider conditions for which the water and air are at \(T_{\infty, i}=T_{\infty, o}=300 \mathrm{~K}\), with corresponding convection coefficients of \(h_{i}=2000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(h_{o}=100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Heat is uniformly dissipated in a stainless steel tube \(\left(k_{w}=15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right)\), having inner and outer radii of \(r_{i}=25 \mathrm{~mm}\) and \(r_{o}=30\) \(\mathrm{mm}\), and aluminum fins \(\left(t=\delta=2 \mathrm{~mm}, r_{t}=55\right.\) \(\mathrm{mm}\) ) are attached to the outer surface, with \(R_{t, c}^{\prime \prime}=\) \(10^{-4} \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\). Determine the heat rates and temperatures at the inner and outer surfaces as a function of the rate of volumetric heating \(\dot{q}\). The upper limit to \(\dot{q}\) will be determined by the constraints that \(T_{s, i}\) not exceed the boiling point of water \(\left(100^{\circ} \mathrm{C}\right)\) and \(T_{s, o}\) not exceed the decomposition temperature of the adhesive \(\left(250^{\circ} \mathrm{C}\right)\).

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 \(S\).
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