Page 41 - Design and Operation of Heat Exchangers and their Networks

28 Design and operation of heat exchangers and their networks

For thermally and hydrodynamically developing laminar flow, the
asymptote of Gnielinski (2010b, 2013b) can be used to calculate the mean
Nusselt number:
h i 3
3 3 1=3
Nu ¼ Nu +1:615fd i =d o Þ RePrd=LÞ
ð
ð
T ∞
" # 3
1=6
2 1=2
+ ð RePrd=LÞ (2.43)
1 + 22Pr
2.1.1.8 Heat transfer in curved tubes
Curved tubes such as helically coiled tubes are widely used in industries as
heat exchangers. Compared with the fluid flow and heat transfer in straight
tubes, a higher heat transfer coefficient arises due to the centrifugal
force generated by curvature of the tubes. A secondary flow is induced
by the centrifugal force and enhances the heat transfer rate. Naphon and
Wongwises (2006) presented a review of the work done on the character-
istics of single-phase and two-phase heat transfer and flow in curved
tubes. The pressure drop correlations for flow through helically coiled tubes
were summarized by Ali (2001). A systematic review of heat transfer and
pressure drop correlations for laminar flow in curved tubes including the
correlations for critical Reynolds number was made by Ghobadi and
Muzychka (2016).
Based on the experimental data of air, water, and oil, Schmidt (1967)
proposed the following correlations for single-phase flow and heat transfer
in curved tubes in the laminar, transition, and turbulent regions. In his
experiments, the fluids were heated by steam condensation on the tube out-
side; therefore, the boundary condition can be considered as the constant
wall temperature. The ratio of the tube radius r to the radius of the tube
curvature r c of the tested curved tubes covered the range of 0.12–0.2.
For laminar flow (100<Re<Re cr ),

0:194
0:9 0:5+ 0:2903 r=r c Þ 1=3
ð
Nu ¼ 3:65 + 0:08 1 + 0:8 r=r c Þ Re Pr (2.44)
ð
4
In the transition region (Re cr <Re<2.2 10 ),
h i
1=3 0:8 0:22 r=r c Þ 0:1 1=3
ð
ð
ð
Nu ¼ 0:023 1 + 14:81 + r=r c Þ r=r c Þ Re Pr (2.45)
4 5
For turbulent flow (2 10 <Re<1.5 10 ),
0:8 0:8 1=3
Nu ¼ 0:023 1 + 3:61 r=r c Þ r=r c Þ Re Pr (2.46)
ð
ð