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HX Design

This document summarizes the steps to design a shell and tube heat exchanger. Key parameters and calculations are presented to determine the log mean temperature difference (LMTD), heat transfer coefficients, required heat transfer area, and overall heat transfer coefficients. The design achieves a 4.73% excess area which is within the acceptable range of 0-20%.

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56 views5 pages

HX Design

This document summarizes the steps to design a shell and tube heat exchanger. Key parameters and calculations are presented to determine the log mean temperature difference (LMTD), heat transfer coefficients, required heat transfer area, and overall heat transfer coefficients. The design achieves a 4.73% excess area which is within the acceptable range of 0-20%.

Uploaded by

Awais
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as DOCX, PDF, TXT or read online on Scribd
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Designing Shell and Tube Heat Exchanger

So,
Hot
 t1 = 85 F
 t2 = 100 0F
 T1 = 200 0F
 T2 = 115 0F
Calculate LMTD:

( T 1−t 2 )−( T 2−t 1 )


LMTD=
T 1−t 2
ln ( T 2−t 1 )
( 100 ) −( 30 )
LMTD=
100
( )
ln
30
LMTD=58.31 F
To calculate;
T 1−T 2
R= =5.66
t 2−t 1
t 2−t 1
S= =0.1304
T 1−T 2

From Fig. 18 (DQ Kern)


F t=0.96

∆ t=LMTD × F t

∆ t=55.97 F
From table 8 (D.Q Kern)
Assume UD = 70 Btu/hr.ft2.F
Q = UD A ∆t
So,
1 Hr . ft 2. F
11100530.1× × ⇒A
55.97 F 70 Btu

A=2833 ft 2
From fig. 17 D.Q. Kern
Kc = 0.35
Fc = 0.4
So now,
Tc = T2 + FC (T1 - T2)
Tc = 115 + 0.4 (200-115)
Tc = 149 F
tc = t1 + FC (t2 – t1)
tc = 85 + 0.4 (100-85)
tc = 91 0F
Calculations for Tube Side:
From the table 10 D.Q Kern
0.041 ft oD
Length BNG = 20 ft
di = ID ft = 0.0358
a” surface per = 0.1309 linear ft / ft2
Flow area per tube = 2.57 ft2
Total number of tubes
A
Nt=
¿¿
Total area of tubes
N t × Flow area per tube
a ( t )=
No of passes
1082× 0.01
a ( t )=
2

a ( t ) ⟹ 0.54 ft 2
Mass flow rate in tube side
m
G (t)=
a (t )
70000 lb /hr
G (t)=
0.541 ft 2
lb 2
G=129390.01 . ft
hr
Calculations for Reynold’s number
DeG
Ra =
μ
To calculate equivalent Dia (De)

( oD2−di 2)
De =
di

D e ⟹ 0.01 0 9 ft 2

μ = 2.41 lb/ft.hr

0.0109 ×129390.01
Re = =585.20
2.41
20
L/ D= =487
0.041
From fig. 24 (D.Q Kern)
j H =2∗10−2
−1
h D Cpμ μ −0.14
j H= i
k ( )( )
k
3
μw
−1
k C μ h
j ( ) .( )p 3 i
H =
d k ∅

Cp = 0.5 Btu / lb0F


μ = 4.83 lb/ft.hr
k = 0.1 Btu / hr.ft2.
hi / ∅ = 93.24 Btu/hr

hi = 94.18 Btu/hr
Calculations for shell
( ID ×CB )
Flow area=
( Pt × 144 )
ID ⟹ 0.1962 ft
Baffle space = 0.404 ft
Pt = 0.051 ft
C = Pt – OD
C = 15/16 – 1/2 = 0.0030 ft2
Flow area = 0.16 ft2
Mass flow rate
Mass Velocity=
Flow area
70000 lb /hr
Mass Velocity=
0.16 ft 2
lb
Mass Velocity=437500
h−ft 2

D .G
Re =
μ
μ = 2.41 lb/ft hr
De = 0.0825 ft
Re = 14976
From Fig. 28 jH = 2 * 10 ^ -3
1
ho De C p μ μ −0.14
j H=
k k( )( ) 3
μw

k = 0.36 Btu/hr.ft2
Cp = 1 Btu / lbF
−1
K Cpμ
h0 = j H ( )( )
D
.
k
3
∅

h0 Btu
=1.59 .F
∅ hr
T w =t c +1 ( h 0 /∅ ) / [ hi /∅ +h0 /∅ ] × ( T c −t c )

T w =91.97 F
0.4
μ
( )
μW
=0.99

h0 =2261× 0.26=587.6 Btu /d


Calculations for Uc:
(hi . ho )
U c=
(hi +ho )

( 151.253)
U c=
( 95.786)
Btu 2
U c =1.579 . ft ℃
hr
Calculations for UD:
Q
U D=
A∆T
A=N t ×a ×

A=2832 ft 2
1958950
U D=
2832 ×55.97
U D=66

As it is less than selected value


U D=6 6

Required Area:

Q= U.A.∆ T
A = 2704
%Excess Area: (2832-2704)/2704 * 100 = 4.73%
Area is acceptable (0 – 20)%

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