a. Explanation Given: Length of the layer of insulation L i = 3 in . Length of the layer of siding L s = 0.75 in . The inner temperature of the insulation T i = 67 ° F . The outer temperature of the siding T s = − 8 ° F . The thermal conductivity of the insulation layer, k i = 0.029 BTU-hr − 1 - o R − 1 . The thermal conductivity of the sliding layer, k s = 0.058 BTU-hr − 1 - o R − 1 . The heat transfer rate is the rate by which an amount of heat flows from one place to another place. The rate of heat transfer is calculated with the following formula. Q = − k A T o − T i L Here, Q is the rate of heat transfer, k is thermal conductivity A is the area, L is the length of the wall. T o is the temperature at the outer surface, and T i is the inner surface temperature. Calculation: There are 12 inches in one foot. Thus, divide the length of the layer of insulation and layer of sliding by 12 to convert them into feet. L i = 3 in 12 L i = 0.25 ft L s = 0 .75 in 12 L s = 0 b. To determine The rate of heat transfer through the wall of surface area.

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Author: MORAN

Publisher: WILEY

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Chapter 2, Problem 2.37P

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The temperature at the interface of the two layers.

To determine

The rate of heat transfer through the wall of surface area.

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Chapter 2, Problem 2.36P

Chapter 2, Problem 2.38P

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What is the weight of a 5-kg substance in N, kN, kg·m/s², kgf, Ibm-ft/s², and lbf? The weight of a 5-kg substance in N is 49.05 N. The weight of a 5-kg substance in kN is KN. The weight of a 5-kg substance in kg·m/s² is 49.05 kg-m/s². The weight of a 5-kg substance in kgf is 5.0 kgf. The weight of a 5-kg substance in Ibm-ft/s² is 11.02 lbm-ft/s². The weight of a 5-kg substance in lbf is 11.023 lbf.

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The temperature at the interface of the two layers.

Solution Summary: The author explains the heat transfer rate, which is the rate by which an amount of heat flows from one place to another.

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The temperature at the interface of the two layers.

The rate of heat transfer through the wall of surface area.

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Chapter 2 Solutions