Consider electronic components mounted on the inner surface of a cylindrical tube. The components are cooled...
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Consider electronic components mounted on the inner surface of a cylindrical tube. The components are cooled by air flowing through the hollow center of the tube, as well as by convection between the outer surface of the tube and the surrounding air. The air at the tube inlet is at 27°C and 100 kPa, and the volumetric flow rate at the inlet is 0.0094 m³/s. When the air exits the tube, its temperature has increased to 42°C and its pressure is effectively the same as the inlet pressure. The so-called "thermal conductance" between the outside of the tube and the surrounding air, which is at 27°C, is hA = 4W/K, where h is the convection coefficient and A is the outer area of the tube. The electronic components consume power at a rate of 0.25 kW. a) If the inner diameter of the tube is 0.22 m, determine the average temperature of the outer surface of the tube. (Hint: how is the average temperature of the outer tube surface related to the rate of convection heat loss between the tube and the surrounding air?) b) Determine the rate of entropy production two ways: i. ii. By considering a control volume that only include the tube and its contents. By considering a control volume that extends beyond the outer tube wall to include the surroundings. c) Explain the difference between your results for parts (b-i) and (b-ii) above. Note that neither of these answers is "wrong", but that they do tell us different things. Consider electronic components mounted on the inner surface of a cylindrical tube. The components are cooled by air flowing through the hollow center of the tube, as well as by convection between the outer surface of the tube and the surrounding air. The air at the tube inlet is at 27°C and 100 kPa, and the volumetric flow rate at the inlet is 0.0094 m³/s. When the air exits the tube, its temperature has increased to 42°C and its pressure is effectively the same as the inlet pressure. The so-called "thermal conductance" between the outside of the tube and the surrounding air, which is at 27°C, is hA = 4W/K, where h is the convection coefficient and A is the outer area of the tube. The electronic components consume power at a rate of 0.25 kW. a) If the inner diameter of the tube is 0.22 m, determine the average temperature of the outer surface of the tube. (Hint: how is the average temperature of the outer tube surface related to the rate of convection heat loss between the tube and the surrounding air?) b) Determine the rate of entropy production two ways: i. ii. By considering a control volume that only include the tube and its contents. By considering a control volume that extends beyond the outer tube wall to include the surroundings. c) Explain the difference between your results for parts (b-i) and (b-ii) above. Note that neither of these answers is "wrong", but that they do tell us different things.
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Answer rating: 100% (QA)
a To determine the average temperature of the outer surface of the tube we can use the energy balance equation for the tube and its contents The rate of convection heat loss between the tube and the s... View the full answer
Related Book For
Fundamentals Of Momentum Heat And Mass Transfer
ISBN: 9781118947463
6th Edition
Authors: James Welty, Gregory L. Rorrer, David G. Foster
Posted Date:
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