Chapter 1: Problem 31
Define thermal conductivity, and explain its significance in heat transfer.
Chapter 1: Problem 31
Define thermal conductivity, and explain its significance in heat transfer.
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Heat treatment is common in processing of semiconductor material. A 200 -mm- diameter silicon wafer with thickness of \(725 \mu \mathrm{m}\) is being heat treated in a vacuum chamber by infrared heat. The surrounding walls of the chamber have a uniform temperature of \(310 \mathrm{~K}\). The infrared heater provides an incident radiation flux of \(200 \mathrm{~kW} / \mathrm{m}^{2}\) on the upper surface of the wafer, and the emissivity and absorptivity of the wafer surface are 0.70. Using a pyrometer, the lower surface temperature of the wafer is measured to be \(1000 \mathrm{~K}\). Assuming there is no radiation exchange between the lower surface of the wafer and the surroundings, determine the upper surface temperature of the wafer. (Note: A pyrometer is a noncontacting device that intercepts and measures thermal radiation. This device can be used to determine the temperature of an object's surface.)
The rate of heat loss through a unit surface area of a window per unit temperature difference between the indoors and the outdoors is called the \(U\)-factor. The value of the \(U\)-factor ranges from about $1.25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\( (or \)0.22 \mathrm{Btw} / \mathrm{h} \cdot \mathrm{ft}^{2} \cdot{ }^{\circ} \mathrm{F}$ ) for low-e coated, argon-filled, quadruple-pane windows to \(6.25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (or $1.1 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2}{ }^{\circ} \mathrm{F}$ ) for a single-pane window with aluminum frames. Determine the range for the rate of heat loss through a $1.2-\mathrm{m} \times 1.8-\mathrm{m}\( window of a house that is maintained at \)20^{\circ} \mathrm{C}\( when the outdoor air temperature is \)-8^{\circ} \mathrm{C}$.
Solar radiation is incident on a \(5-\mathrm{m}^{2}\) solar absorber plate surface at a rate of \(800 \mathrm{~W} / \mathrm{m}^{2}\). Ninety-three percent of the solar radiation is absorbed by the absorber plate, while the remaining 7 percent is reflected away. The solar absorber plate has a surface temperature of \(40^{\circ} \mathrm{C}\) with an emissivity of \(0.9\) that experiences radiation exchange with the surrounding temperature of $-5^{\circ} \mathrm{C}$. In addition, convective heat transfer occurs between the absorber plate surface and the ambient air of \(20^{\circ} \mathrm{C}\) with a convection heat transfer coefficient of \(7 \mathrm{~W} / \mathrm{m}^{2}\). \(\mathrm{K}\). Determine the efficiency of the solar absorber, which is defined as the ratio of the usable heat collected by the absorber to the incident solar radiation on the absorber.
A 1000-W iron is left on an ironing board with its base exposed to the air at \(20^{\circ} \mathrm{C}\). The convection heat transfer coefficient between the base surface and the surrounding air is $35 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\(. If the base has an emissivity of \)0.6$ and a surface area of \(0.02 \mathrm{~m}^{2}\), determine the temperature of the base of the iron. Answer: \(674^{\circ} \mathrm{C}\)
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