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Consider a 6-in \(\times 8\)-in epoxy glass laminate \((k=\) \(0.10 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft} \cdot{ }^{\circ} \mathrm{F}\) ) whose thickness is \(0.05 \mathrm{in}\). In order to reduce the thermal resistance across its thickness, cylindrical copper fillings \(\left(k=223 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft} \cdot{ }^{\circ} \mathrm{F}\right)\) of \(0.02\) in diameter are to be planted throughout the board, with a center-to-center distance of \(0.06\) in. Determine the new value of the thermal resistance of the epoxy board for heat conduction across its thickness as a result of this modification.

Short Answer

Expert verified
The new thermal resistance of the modified epoxy board is approximately 0.012740 h·°F/Btu.

Step by step solution

01

Calculate the area of the epoxy board

To find the new value of the thermal resistance, we first need to calculate the area of the epoxy board. The board has dimensions of 6-in × 8-in, so its area can be found by multiplying these two values: \(A_{board} = 6 \, in \times 8 \, in = 48 \, in^2\)
02

Calculate the area of one copper filling

Next, we need to find the area of one of the copper fillings. We are given the diameter, so we can find the radius and use the formula for the area of a circle: \(A_{filling} = \pi r^2\) Where \(r = \frac{diameter}{2} = \frac{0.02 \, in}{2} = 0.01 \, in\). So, the area of one copper filling is: \(A_{filling} = \pi (0.01 \, in)^2 = 0.000314 \, in^2\)
03

Calculate the total number of fillings in the epoxy board

In order to find the new thermal resistance, we need to know how many fillings are inserted into the epoxy board. The fillings are spaced with a center-to-center distance of 0.06 in. We can find the number of fillings in the horizontal and vertical directions, and then multiply these together to find the total number of fillings: \(N_{x} = \frac{6 \, in}{0.06 \, in} = 100\) \(N_{y} = \frac{8 \, in}{0.06 \, in} = 133.3\) Since we cannot have a fraction of a filling, we will round down. \(N_{total} = N_{x} \times N_{y} = 100 \times 133 = 13,300\)
04

Calculate the total area of copper fillings

Now, we need to find the total area of the copper fillings. We can multiply the area of one filling by the total number of fillings: \(A_{total\_fillings} = A_{filling} \times N_{total} = 0.000314 \, in^2 \times 13,300 = 4.1734 \, in^2\)
05

Calculate the effective area of the modified epoxy board

In order to find the new thermal resistance, we need to calculate the effective area of the modified epoxy board. We can subtract the total area of the copper fillings from the area of the epoxy board: \(A_{effective} = A_{board} - A_{total\_fillings} = 48 \, in^2 - 4.1734 \, in^2 = 43.8266 \, in^2\)
06

Calculate the new thermal resistance of the modified epoxy board

Finally, we can find the new thermal resistance using the formula: \(R_{new} = \frac{L}{kA}\) Where \(L\) is the thickness of the epoxy board (0.05 in), \(k\) is the thermal conductivity of the epoxy (0.10 Btu/h·ft·°F), converted to Btu/h·in·°F by dividing by 12, and \(A\) is the effective area of the modified epoxy board (43.8266 in²): \(R_{new} = \frac{0.05 \, in}{(0.10 \, Btu/h \cdot ft \cdot°F / 12) \cdot 43.8266 \, in^2} = 0.012740 \, h \cdot °F/Btu\) The new thermal resistance of the modified epoxy board is approximately 0.012740 h·°F/Btu.

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Key Concepts

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

Epoxy Glass Laminate
Epoxy glass laminate is a composite material made by layering thin sheets of glass fibers, which are bonded together using an epoxy resin. It is known for its excellent electrical insulation properties and is widely used in various engineering applications.

This laminate is often utilized in the construction of printed circuit boards (PCBs) and other electrical components because of its good mechanical strength and resistance to environmental damage. The primary reason for its use in thermal applications is its ability to withstand high temperatures and its low thermal conductivity, typically around 0.10 Btu/h·ft·°F, which makes it an effective insulator against heat flow.

However, this same property can present challenges when fast heat dissipation is needed, requiring modifications such as adding conductive paths to adjust its thermal resistance.
Copper Fillings
Copper fillings are small cylindrical inserts made of copper and are used to enhance thermal and electrical conductivity in materials where they are embedded.

In this context, copper fillings are strategically placed within the epoxy glass laminate to reduce thermal resistance and improve heat conduction across the thickness of the board. These fillings have a diameter of 0.02 inches and contribute significantly to heat transfer because copper has a high thermal conductivity of 223 Btu/h·ft·°F.

By planting these fillings with a regular spacing, we ensure they effectively create parallel heat paths within the laminate, thus aiding in better heat distribution. This alteration dramatically lowers the laminate's overall thermal resistance, allowing for quicker heat dissipation without compromising the original structure's integrity.
Heat Conduction
Heat conduction is the process through which heat energy is transferred within a body due to temperature gradients. The principle is based on the fact that energy tends to flow from regions of higher temperature to regions of lower temperature.

In materials like epoxy glass laminate, heat conduction becomes a critical factor when the goal is to manage temperatures, such as when used in electronic devices. The insertion of copper fillings into the laminate facilitates efficient heat conduction, reducing the heat buildup in specific areas.

This is particularly useful for applications where maintaining a stable temperature is vital to the proper functioning of the device. The goal is often to find an optimal balance where the laminate can both protect against external heat sources and conduct internal heat away smoothly.
Thermal Conductivity
Thermal conductivity is a measure of a material's ability to conduct heat. It is expressed as the amount of heat that passes per unit time through a unit area of the material, with the given temperature gradient.

Materials with high thermal conductivity are excellent at transferring heat, which is why copper, with a thermal conductivity of 223 Btu/h·ft·°F, serves as an efficient medium for facilitating heat flow.

In contrast, materials like epoxy glass laminate with a low thermal conductivity (0.10 Btu/h·ft·°F) are good insulators, limiting heat transfer. The design challenge often comes in configuring such materials to balance their insulating properties while improving their ability to dissipate heat efficiently. This is crucial in contexts like electronic devices, where overheating can cause damage or failure. Therefore, understanding the thermal conductivity of each component helps in designing solutions that meet specific thermal management needs.

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Most popular questions from this chapter

Hot water at an average temperature of \(70^{\circ} \mathrm{C}\) is flowing through a \(15-\mathrm{m}\) section of a cast iron pipe \((k=52 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) whose inner and outer diameters are \(4 \mathrm{~cm}\) and \(4.6 \mathrm{~cm}\), respectively. The outer surface of the pipe, whose emissivity is \(0.7\), is exposed to the cold air at \(10^{\circ} \mathrm{C}\) in the basement, with a heat transfer coefficient of \(15 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The heat transfer coefficient at the inner surface of the pipe is \(120 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Taking the walls of the basement to be at \(10^{\circ} \mathrm{C}\) also, determine the rate of heat loss from the hot water. Also, determine the average velocity of the water in the pipe if the temperature of the water drops by \(3^{\circ} \mathrm{C}\) as it passes through the basement.

One wall of a refrigerated warehouse is \(10.0\)-m-high and \(5.0\)-m-wide. The wall is made of three layers: \(1.0\)-cm-thick aluminum \((k=200 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}), 8.0\)-cm-thick fibreglass \((k=\) \(0.038 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\), and \(3.0-\mathrm{cm}\) thick gypsum board \((k=\) \(0.48 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). The warehouse inside and outside temperatures are \(-10^{\circ} \mathrm{C}\) and \(20^{\circ} \mathrm{C}\), respectively, and the average value of both inside and outside heat transfer coefficients is \(40 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) Calculate the rate of heat transfer across the warehouse wall in steady operation. (b) Suppose that 400 metal bolts \((k=43 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\), each \(2.0 \mathrm{~cm}\) in diameter and \(12.0 \mathrm{~cm}\) long, are used to fasten (i.e., hold together) the three wall layers. Calculate the rate of heat transfer for the "bolted" wall. (c) What is the percent change in the rate of heat transfer across the wall due to metal bolts?

Two plate fins of constant rectangular cross section are identical, except that the thickness of one of them is twice the thickness of the other. For which fin is the \((a)\) fin effectiveness and \((b)\) fin efficiency higher? Explain.

Heat is generated steadily in a 3-cm-diameter spherical ball. The ball is exposed to ambient air at \(26^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(7.5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The ball is to be covered with a material of thermal conductivity \(0.15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). The thickness of the covering material that will maximize heat generation within the ball while maintaining ball surface temperature constant is (a) \(0.5 \mathrm{~cm}\) (b) \(1.0 \mathrm{~cm}\) (c) \(1.5 \mathrm{~cm}\) (d) \(2.0 \mathrm{~cm}\) (e) \(2.5 \mathrm{~cm}\)

Consider an insulated pipe exposed to the atmosphere. Will the critical radius of insulation be greater on calm days or on windy days? Why?

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