ENGINEERING FUNDAMENTALS
ENGINEERING FUNDAMENTALS
6th Edition
ISBN: 9781337705011
Author: MOAVENI
Publisher: CENGAGE L
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Chapter 11, Problem 29P
To determine

Calculate the heat loss through the side walls of a refrigerator.

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12. Thermal efficiency using temperature values 13. Thermal efficiency using rc and k values
Air at 25 ° C blows over the hot steel plate whose surface temperature is maintained at 150 ° C. The plates are 50 cm x 40 cm in dimensions and 2.5 cm thick. The convection heat transfer coefficient on the upper surface is 20 W / (m² ° C). The thermal conductivity of steel is 45 W / (m ° C). Calculate the hourly heat loss from the plate surface. a. heat loss per hour = kJ. Answer b. If the reverse side surface temperature is maintained, determine the hourly heat loss incurred = kJ.Answer
A double-pane glass window consists of two pieces of glass, each having athickness of 8 mm, with a thermal conductivity of k = 1.4 W⁄m ⋅ K . The twoglass panes are separated by an air gap of 10 mm, as shown in Figure 11.10 .Assuming the thermal conductivity of air to be k = 0.025 W⁄m ⋅ K , determinethe total R-value for this window. The total thermal resistance of the window is obtained by adding the resistance offered by each pane of glass and the air gap in the following manner:

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ENGINEERING FUNDAMENTALS

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