Aheat exchanger contains 400 tubes with inner diameter of 23 mm and outer diameter of 25 mm. the length of each tube is 3.7 m. the corrected log mean temperature difference is 23°c, while the inner surface convection heat transfer coefficient is 3410 w/m2 ·k and the outer surface convection heat transfer coefficient is 6820 w/m2 ·k. if the thermal resistance of the tubes is negligible, determine the heat transfer rate.

0.39J/s

Explanation:

Length of 1 tube = 3.7m

Length of 400 tubes in the heat exchanger = 3.7×400 = 1480m, Temperature difference = 23°C = 296K

Outer surface

Rate of heat transfer = (heat transfer coefficient×area×log mean temperature difference)/length

diameter (d) = 25mm = 0.025m, area = πd^2/4 = (3.142×0.025^2)/4 = 0.000491m^2

Rate of heat transfer = (6820×0.000491×296)/1480 = 0.67J/s

Inner surface

diameter = 23mm = 0.023m, area = (3.142×0.023^2)/4 = 0.000416m^2

Rate of heat transfer = (3410×0.000416×296)/1480 = 0.28J/s

Heat transfer rate between the two surfaces = 0.67 - 0.28 = 0.39J/s

energy can cross the boundaries of a closed system in the form of heat and mass

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