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Chapter 3: Problem 217

The walls of a food storage facility are made of a 2 -cm-thick layer of wood \((k=0.1 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) in contact with a 5 -cm- thick layer of polyurethane foam \((k=0.03 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). If the temperature of the surface of the wood is \(-10^{\circ} \mathrm{C}\) and the temperature of the surface of the polyurethane foam is \(20^{\circ} \mathrm{C}\), the temperature of the surface where the two layers are in contact is (a) \(-7^{\circ} \mathrm{C}\) (b) \(-2^{\circ} \mathrm{C}\) (c) \(3^{\circ} \mathrm{C}\) (d) \(8^{\circ} \mathrm{C}\) (e) \(11^{\circ} \mathrm{C}\)

Short Answer

Expert verified
a) 4°C b) 6°C c) 7°C d) 8°C Answer: d) 8°C

Step by step solution

01

Understand the heat flow through the layers

The heat flow through the layers can be represented as a thermal resistance network, where the heat flow (Q) is the same through both layers, and the temperature at the surface contact can be found by comparing the thermal resistances.
02

Calculate the thermal resistance of each layer

The thermal resistance (R) of a layer can be calculated using the following formula: \(R = \frac{L}{kA}\), where L is the thickness, k is the thermal conductivity, and A is the area of heat transfer (which is the same for both layers). For the wood layer, we have: \(R_{wood} = \frac{L_{wood}}{k_{wood}A} = \frac{0.02\,\text{m}}{0.1\,\text{W/m}⋅\text{K}⋅A}\). For the polyurethane foam layer, we have: \(R_{foam} = \frac{L_{foam}}{k_{foam}A} = \frac{0.05\,\text{m}}{0.03\,\text{W/m}⋅\text{K}⋅A}\).
03

Use the heat flow equation to find the contact point temperature

Since the heat flow (Q) is the same through both layers, we can set up the following equation based on Fourier's Law: \(Q = \frac{\Delta T_{wood}}{R_{wood}} = \frac{\Delta T_{foam}}{R_{foam}}\). Substituting the values from Step 2 into this equation, we get: \(\frac{(-10 - T_{contact})}{\frac{0.02\,\text{m}}{0.1\,\text{W/m}⋅\text{K}⋅A}} = \frac{(T_{contact}-20)}{\frac{0.05\,\text{m}}{0.03\,\text{W/m}⋅\text{K}⋅A}}\).
04

Solve for the contact point temperature

To find the temperature at the contact point (T_contact), we can first simplify the equation from Step 3 and then solve for T_contact: \((-10 - T_{contact})(0.03)(0.05) = (T_{contact}-20)(0.1)(0.02)\). Simplifying, we get: \(-0.015 - 0.0015T_{contact} = 0.002T_{contact} - 0.04\). Combining like terms and moving variables to one side gives: \(0.002T_{contact} + 0.0015T_{contact} = -0.015 + 0.04\), or \(0.0035T_{contact} = 0.025\). Finally, we find the contact point temperature by dividing both sides by 0.0035: \(T_{contact} = \frac{0.025}{0.0035} = 7.14^{\circ}\text{C}\). Considering the options provided, the closest answer is (d) \(8^{\circ}\text{C}\).

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Most popular questions from this chapter

A 0.3-cm-thick, 12-cm-high, and 18-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 \(40 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) Determine the temperatures on the two sides of the circuit board. (b) Now a \(0.2\)-cm-thick, 12-cm-high, and 18-cmlong aluminum plate \((k=237 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) with 864 2-cm-long aluminum pin fins of diameter \(0.25 \mathrm{~cm}\) is attached to the back side of the circuit board with a \(0.02\)-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 1.4-m-diameter spherical steel tank filled with iced water at \(0^{\circ} \mathrm{C}\) is buried underground at a location where the thermal conductivity of the soil is \(k=0.55 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). The distance between the tank center and the ground surface is \(2.4 \mathrm{~m}\). For ground surface temperature of \(18^{\circ} \mathrm{C}\), determine the rate of heat transfer to the iced water in the tank. What would your answer be if the soil temperature were \(18^{\circ} \mathrm{C}\) and the ground surface were insulated?

What is the reason for the widespread use of fins on surfaces?

Two flow passages with different cross-sectional shapes, one circular another square, are each centered in a square solid bar of the same dimension and thermal conductivity. Both configurations have the same length, \(T_{1}\), and \(T_{2}\). Determine which configuration has the higher rate of heat transfer through the square solid bar for \((a) a=1.2 b\) and \((b) a=2 b\).

A 0.6-m-diameter, 1.9-m-long cylindrical tank containing liquefied natural gas (LNG) at \(-160^{\circ} \mathrm{C}\) is placed at the center of a 1.9-m-long \(1.4-\mathrm{m} \times 1.4-\mathrm{m}\) square solid bar made of an insulating material with \(k=0.0002 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). If the outer surface temperature of the bar is \(12^{\circ} \mathrm{C}\), determine the rate of heat transfer to the tank. Also, determine the LNG temperature after one month. Take the density and the specific heat of LNG to be \(425 \mathrm{~kg} / \mathrm{m}^{3}\) and \(3.475 \mathrm{~kJ} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}\), respectively.
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