Page 222 - Design and Operation of Heat Exchangers and their Networks
P. 222
Optimal design of heat exchangers 211
The shell-side hydraulic diameter in window section is expressed by
Eq. (5.76):
4A sw 4 0:01287
d hw ¼ ¼
N tw πd o + d s θ ds =2 18:53 π 0:019 + 0:336 2:131=2
¼ 0:03517 m
(3) Calculation of tube-side heat transfer coefficient
The tube-side Reynolds number is given by
4 _m t 4 18:1
Re t ¼ ¼ ¼ 32,401
N t,p πd i μ 52 π 0:0166 8:2398 10 4
t
4
Because Re t 10 , we can use the Dittus-Boelter correlation (Dittus and
Boelter, 1930) to calculate the tube-side Nusselt number:
Nu t ¼ 0:023Re 0:8 Pr n t
t
with n¼0.4 for heating,
Nu t ¼ 0:023 32,401 0:8 5:3429 0:4 ¼ 182:51
and then obtain the tube-side heat transfer coefficient as
2
α t ¼ Nu t λ t =d i ¼ 182:51 0:61814=0:0166 ¼ 6796:2W=m K
(4) Calculation of shell-side heat transfer coefficient
Since s l <d o , we use Eq. (5.42) to calculate the heat transfer coefficient for
ideal crossflow. We calculate at first the tube bundle Reynolds number as
follows:
Because s l <s l,min , we have
_ m s s t =2 36:3 0:03536=2
G s,max ¼ ¼
ð
ð
d s l bc s d d o Þ 0:336 0:279 0:025 0:019Þ
2
¼ 1140:9kg=m s
Re tb ¼ G s,max d o =μ ¼ 727:25
s
The Hagen number for laminar flow is calculated with Eq. (5.51):
0:5 2
ð s l =d o Þ 0:6 +0:75
Hg lam ¼ 140Re tb 1:6
∗
ð
ð
s =d o ½ 4 s l =d o Þ s t =d o Þ=π 1
t
0:5 2
ð 0:01768=0:019Þ 0:6 +0:75
¼ 140 727:25 1:6
ð
ð 0:025=0:019Þ ½ 4 0:01768=0:019Þ 0:03536=0:019Þ=π 1
ð
¼ 48,114
