Reconsider the heat exchanger of Problem 6.38 with an inline tube arrangement, where the centerlines of the tubes are spaced 7.5 cm apart, both in the longitudinal and transverse directions. Compare the results with those for the staggered tube arrangement and comment upon the difference. Also, which of the two arrangements is expected to have a larger pressure drop?
GIVEN
• Air flow through a tube bank
• Tube spacing (S) = 7.5 cm = 0.075 m
• Air temperature (Ta) = 60°C
• Air velocity (Us) = 1 m/s
• Tube outside diameter (D) = 6 cm = 0.06 m
• Tube wall temperature (Tw) = 117°C FIND
• The average heat transfer coefficient ( ch )
• Steady state
SKETCH
Therefore, the minimum free flow area is between adjacent tubes in a row, and the maximum air
velocity for inline arrangement is
The Reynolds number is
Applying
Adjusting the Nusselt number for the first two rows
Where f is calculated for inline tubes
For (ST/D-1)(SL/D-1)=1/16= 0.0625 and ReD=15464 from the figure we get x=6 For SL/D= 7.5/6=1.25 and ReD=15464 f/x=0.52
The change in the geometry from Problem 6.38 lead to a 50% increase in the heat transfer coefficient
and 240 % increase in pressure drop. Thus inline tube arrangement is expected to have higher pressure
drop.
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