An astronaut performing an extra-vehicular activity (space walk) shaded from the Sun is wearing a spacesuit that can be approximated as perfectly white ( e = 0 ) except for a 5 cm × 8 cm patch in the form of the astronaut's national flag. The patch has emissivity 0.300. The spacesuit under the patch is 0.500 cm thick, with a thermal conductivity k 0.0600 W/m ℃, and its inner surface is at a temperature of 20.0 ℃. What is the temperature of the patch, and what is the rate of heat loss through it? Assume the patch is so thin that its outer surface is at the same temperature as the outer surface of the spacesuit under it. Also assume the temperature of outer space is 0 K. You will get an equation that is very hand to solve in closed form, so can solve it numerically with a graphing calculator, with software, or even by trial and error with a calculator.
An astronaut performing an extra-vehicular activity (space walk) shaded from the Sun is wearing a spacesuit that can be approximated as perfectly white ( e = 0 ) except for a 5 cm × 8 cm patch in the form of the astronaut's national flag. The patch has emissivity 0.300. The spacesuit under the patch is 0.500 cm thick, with a thermal conductivity k 0.0600 W/m ℃, and its inner surface is at a temperature of 20.0 ℃. What is the temperature of the patch, and what is the rate of heat loss through it? Assume the patch is so thin that its outer surface is at the same temperature as the outer surface of the spacesuit under it. Also assume the temperature of outer space is 0 K. You will get an equation that is very hand to solve in closed form, so can solve it numerically with a graphing calculator, with software, or even by trial and error with a calculator.
An astronaut performing an extra-vehicular activity (space walk) shaded from the Sun is wearing a spacesuit that can be approximated as perfectly white (
e
=
0
) except for a 5 cm × 8 cm patch in the form of the astronaut's national flag. The patch has emissivity 0.300. The spacesuit under the patch is 0.500 cm thick, with a thermal conductivity k 0.0600 W/m ℃, and its inner surface is at a temperature of 20.0 ℃. What is the temperature of the patch, and what is the rate of heat loss through it? Assume the patch is so thin that its outer surface is at the same temperature as the outer surface of the spacesuit under it. Also assume the temperature of outer space is 0 K. You will get an equation that is very hand to solve in closed form, so can solve it numerically with a graphing calculator, with software, or even by trial and error with a calculator.
Study of body parts and their functions. In this combined field of study, anatomy refers to studying the body structure of organisms, whereas physiology refers to their function.
Consider a 6-in * 8-in epoxy glass laminate (k = 0.10 Btu/h·ft·°F) whose thickness is 0.05 in. In order to reduce the thermal resistance across its thickness, cylindrical copper fillings (k = 223 Btu/h·ft·°F) of 0.02 in diameter are to be planted throughout the board, with a center-to-center distance of 0.06 in. Determine the new value of the thermal resistance of the epoxy board for heat conduction across its thickness as a result of this modification.
A person is standing outdoors in the shade where the temperature is 21 °C. (a) What is the radiant energy absorbed per second by his head when it is covered with hair? The surface area of the hair (assumed to be flat) is 130 cm2 and its emissivity is 0.90. (b) What would be the radiant energy absorbed per second by the same person if he were bald and the emissivity of his head were 0.54?
Consider steady heat transfer between two large parallel plates at constant temperatures of T1 = 290 K and T2 = 150 K that are L = 2 cm apart. Assuming the surfaces to be black (emissivity « = 1), determine the rate of heat transfer between the plates per unit surface area assuming the gap between the plates is (a) filled with atmospheric air, (b) evacuated, (c) filled with fiberglass insulation, and (d) filled with superinsulation having an apparent thermal conductivity of 0.00015 W/m·K.
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