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Chapter 3: Problem 62

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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Most popular questions from this chapter

A \(2.2\)-mm-diameter and 10-m-long electric wire is tightly wrapped with a \(1-m m\)-thick plastic cover whose thermal conductivity is \(k=0.15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Electrical measurements indicate that a current of \(13 \mathrm{~A}\) passes through the wire and there is a voltage drop of \(8 \mathrm{~V}\) along the wire. If the insulated wire is exposed to a medium at \(T_{\infty}=30^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(h=24 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the temperature at the interface of the wire and the plastic cover in steady operation. Also determine if doubling the thickness of the plastic cover will increase or decrease this interface temperature.

Consider a wall that consists of two layers, \(A\) and \(B\), with the following values: \(k_{A}=0.8 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, L_{A}=8 \mathrm{~cm}, k_{B}=\) \(0.2 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, L_{B}=5 \mathrm{~cm}\). If the temperature drop across the wall is \(18^{\circ} \mathrm{C}\), the rate of heat transfer through the wall per unit area of the wall is (a) \(180 \mathrm{~W} / \mathrm{m}^{2}\) (b) \(153 \mathrm{~W} / \mathrm{m}^{2}\) (c) \(89.6 \mathrm{~W} / \mathrm{m}^{2}\) (d) \(72 \mathrm{~W} / \mathrm{m}^{2}\) (e) \(51.4 \mathrm{~W} / \mathrm{m}^{2}\)

Consider two metal plates pressed against each other. Other things being equal, which of the measures below will cause the thermal contact resistance to increase? (a) Cleaning the surfaces to make them shinier. (b) Pressing the plates against each other with a greater force. (c) Filling the gap with a conducting fluid. (d) Using softer metals. (e) Coating the contact surfaces with a thin layer of soft metal such as tin.

The unit thermal resistances ( \(R\)-values) of both 40-mm and 90-mm vertical air spaces are given in Table 3-9 to be \(0.22 \mathrm{~m}^{2} \cdot \mathrm{C} / \mathrm{W}\), which implies that more than doubling the thickness of air space in a wall has no effect on heat transfer through the wall. Do you think this is a typing error? Explain.

Two finned surfaces are identical, except that the convection heat transfer coefficient of one of them is twice that of the other. For which finned surface is the \((a)\) fin effectiveness and \((b)\) fin efficiency higher? Explain.
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