Chapter 1: Problem 115
What is asymmetric thermal radiation? How does it cause thermal discomfort in the occupants of a room?
Chapter 1: Problem 115
What is asymmetric thermal radiation? How does it cause thermal discomfort in the occupants of a room?
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Get started for freeA \(4-\mathrm{m} \times 5-\mathrm{m} \times 6-\mathrm{m}\) room is to be heated by one ton \((1000 \mathrm{~kg})\) of liquid water contained in a tank placed in the room. The room is losing heat to the outside at an average rate of $10,000 \mathrm{~kJ} / \mathrm{h}\(. The room is initially at \)20^{\circ} \mathrm{C}$ and \(100 \mathrm{kPa}\) and is maintained at an average temperature of \(20^{\circ} \mathrm{C}\) at all times. If the hot water is to meet the heating requirements of this room for a \(24-\mathrm{h}\) period, determine the minimum temperature of the water when it is first brought into the room. Assume constant specific heats for both air and water at room temperature. Answer: 77.4 \(\mathrm{C}\)
Two surfaces, one highly polished and the other heavily oxidized, are found to be emitting the same amount of energy per unit area. The highly polished surface has an emissivity of \(0.1\) at \(1070^{\circ} \mathrm{C}\), while the emissivity of the heavily oxidized surface is \(0.78\). Determine the temperature of the heavily oxidized surface.
Consider a house in Atlanta, Georgia, that is maintained at $22^{\circ} \mathrm{C}\( and has a total of \)20 \mathrm{~m}^{2}$ of window area. The windows are double-door type with wood frames and metal spacers and have a \(U\)-factor of \(2.5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (see Prob. 1-120 for the definition of \(U\)-factor). The winter average temperature of Atlanta is \(11.3^{\circ} \mathrm{C}\). Determine the average rate of heat loss through the windows in winter.
An engineer who is working on the heat transfer analysis of a house in English units needs the convection heat transfer coefficient on the outer surface of the house. But the only value he can find from his handbooks is $14 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$, which is in SI units. The engineer does not have a direct conversion factor between the two unit systems for the convection heat transfer coefficient. Using the conversion factors between \(\mathrm{W}\) and \(\mathrm{Btu} / \mathrm{h}, \mathrm{m}\) and \(\mathrm{ft}\), and \({ }^{\circ} \mathrm{C}\) and \({ }^{\circ} \mathrm{F}\), express the given convection heat transfer coefficient in Btu/h \(\mathrm{ft}^{2}{ }^{\circ} \mathrm{F}\). Answer: \(2.47\) Btu/h $\cdot \mathrm{ft}^{2}{ }^{\circ} \mathrm{F}$
What is the value of the engineering software packages in \((a)\) engineering education and \((b)\) engineering practice?
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