Page 70 - Process Equipment and Plant Design Principles and Practices by Subhabrata Ray Gargi Das

66 Chapter 3 Double pipe heat exchanger

0:8 0:33
o
Nu o ¼ 0:027:ðRe o Þ :ðPr o Þ :f ¼ 78:2158
2
h i ¼ Nu t :ðk w=D i Þ¼ 7911:7W=ðm .K
2
h o ¼ Nu o :ðk c=ðD o D i ÞÞ ¼ 5184:2W=ðm .K
h o D io
¼ 0:7886 > 0:75, and hence no need of finned tubes.
h i D o
h i T w;avg þ h o ðD o =D i ÞT c;avg +
¼ 48:2592 C
T w ¼
h i þ h o ðD o =D i Þ
Estimating f and f e
t
o
o 0:14 4 0:14
m @39 C 6:65 10
w
f i;new ¼ ¼ ¼ 1:0225
o
m @48:2592 C 5:6741 10 4
w
m @40 C 5:8 10
o 0:14 4 0:14
c
f o;new ¼ ¼ ¼ 0:9712
o
m @48:2592 C 7:1437 10 4
w
h o D io
Based f and f , h i ¼ 5035:2 and ¼ 0:75. This is a marginal case when one need not
t;new o;new
h i D o
go for finned tubes and the same is opted for ease of fabrication.
h i T w;avg þ h o ðD o =D i ÞT c;avg
¼ 47:9941 C

T w;new ¼
h i þ h o ðD o =D i Þ
Recalculated f and f does not show significant change and the iteration is stopped.
t;new o;new
U c ¼ 1=ð1 = h o þ D io = ðh i :D i Þþ D io :lnðD io = D i Þ = ð2:k wall ÞÞ ¼ 1700:3
U D ¼ 1=f1 = U c þ R c þ R w :ðD io = D i Þg ¼ 865:98
2
Total heat transfer area based on outer surface of inner tube, A ¼ Q/(U D .LMTD) ¼ 5.8752 m .
Minimum length of tube ¼ A/(pDio) ¼ 35.63 m.
We adopt a standard tube length of 6 m and provide three hairpins that make the total tube length to
be 36 m. N HP ¼ 3, L total ¼ 36 m.
Pressure drop.
Pressure drop in straight pipe length.
Re t ¼ 96,497, is turbulent flow and Eq. 3.16b is applicable.
0:42
¼ 0:0055
f i ¼ 0:0035 þ 0:246=ðRe t Þ
2
4f i G L total
i
2
DH f;i ¼ ¼ 3:8759 m
2gr D i
i
Re o ¼ 12,895, is turbulent flow and Eq. 3.16b is applicable.
0:42
¼ 0:0081
f o ¼ 0:0035 þ 0:246=ðRe o Þ