Page 224 - Design and Operation of Heat Exchangers and their Networks

Optimal design of heat exchangers 213

1
1 d o ln d o =d i Þ d o 1
ð
k s ¼ + R f ,s + + R f ,t +
α s 2λ w d i α t

1 0:019 ln 0:019=0:0166Þ 0:019
ð
¼ +0:000176 + +
581:6 2 111 0:0166
1
1
2
0:0000881 + ¼ 459:5W=m K (5.86)
6796
The shell-side heat transfer area is calculated with Eq. (5.87):
A s ¼ N t πd o L ¼ 104 π 0:019 4:3 ¼ 26:69 m 2 (5.87)
Then, we can obtain the number of heat transfer units

k s A s 459:5 26:69
NTU t ¼ ¼ ¼ 0:1691
_ m t c p,t 18:1 4008
The thermal capacity flow rate ratio is expressed as

18:1 4008
_ m t c p,t
R t ¼ ¼ ¼ 0:9624
_ m s c p,s 36:3 2077
We take the ratio of NTU of parallel-flow passes to total NTU as
NTU parallel flow passes
ξ ¼ ¼ 0:5,
NTU t
and calculate the heat exchanger effectiveness as
q ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2
S ¼ 1+ R +2R t 2ξ 1ð Þ
t
(5.88)
q ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2
¼ 1+ 0:9624 +2 0:9624 2 0:5 1ð Þ ¼ 1:388
2
t t,in t t,out
P t ¼ ¼
t t,in t s,in 1+ R t + Scoth NTU t S=2ð Þ
(5.89)
2
¼ ¼ 0:1444
1+ 0:9624 + 1:388 coth 0:1691 1:388=2Þ
ð
Thus, we can obtain the heat transfer rate, the outlet temperatures of
seawater (tube side) and oil (shell side), and the mean oil temperature at
tube outside wall as
_
ð
Q ¼ P t t s,in t t,in Þ _m t c p,t ¼ 0:1444 66 32ð Þ 18:1 4008
5
¼ 3:563 10 W
_ Q 3:563 10 5
t t,out ¼ t t,in + ¼ 32 + ¼ 36:9°C
_ m t c p,t 18:1 4008