Transcript 2-4 Shell and Tube Heat exchanger

2-4 Shell and Tube
Heat exchanger
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Outline
2-4 Shell and tube heat exchanger
 Why we use it ?
 Problem 8.1

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Problem Statement

33,114 lb/hr of n-butyl alcohol at 210 0F is to be
cooled to 105 0F using water from 95 to 115 0F.
Available for the purpose is a 19¼ in. ID, twopass shell exchanger with 204 tubes ¾. OD , 16
BWG, 16’0’’ long on 1-in .square pitch arranged
for four passes. Vertically cut baffles are spaced
7 in. apart. Pressure drops of 10psi are
allowable.

What is the Dirt factor ?
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SOLUTION
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Data Available
Shell Side Data





Inside Shell Diameter
Number of Passes
Baffle spacing
Baffle type
Allowable Pressure Drop
=
=
=
=
=
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19¼ in
2
7 in
Vertically Cut
10psi
Data Available
Tube Side Data







Outside Diameter of Tubes
BWG
Length of tubes
Tubes Pitch
Number of tubes
Number of tube passes
Allowable Pressure Drop
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=
=
=
=
=
=
=
¾ in
16
16’0’’
1 in. Square
204
4
10psi
Location of Fluids
Tube Side Fluid
 As water has more scaling tendency than
n-butyl alcohol that is why it is taken in
tube side
Shell Side Fluid
 n-butyl alcohol certainly
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Data Available
Hot Fluid (n-butyl alcohol)



Inlet temperature (T1)
Outlet temperature (T2)
Mass Flow rate (mh)
=
=
=
210 0F
105 0F
33114 lb/hr
Cold Fluid (Water)


Inlet temperature (t1)
Outlet temperature (t2)
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=
=
95 0F
115 0F
Diagram
mh = 33114 lb/hr

(n-butyl alcohol)
210 0F
105 0F

(Water)
115 0F
95 0F
T1
1
Temperature Profile
Tx
t2
2
4
T2
3
2
t1
1
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L
Step #1
Heat Duty
Qh
 mh
 Cph
=
=
=
mhCph(T1 - T2)
(1)
33,114 lb/hr
0.69 Btu/lboF (from fig.2)
Qh
=
=
33114*(0.69)*(210-105) Btu/hr
2399109.3 Btu/hr


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Step # 1 contd.
Mass flow rate of water
 As
Qh =
Qc
 mc
=
Qh / {Cpw*(t2 – t1)}

=
2399109.3 / {1*(115 - 95)}
=
119955.46 lb/hr
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Step # 2
LMTD Calculation

(n-butyl alcohol)

(Water)
LMTD =

210 0F
115 0F
(T1-t2) – (T2-t1)
105 0F
95 0F
=
ln(T1-t2)/(T2-t1)
(210 – 115 ) – (105 - 95)
ln(210 – 115 )/(105 - 95)
=
37.75 0F
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True temperature Difference

Δt
=
FT * LMTD

R
=
S
=
=
FT
 Δt
=
=
=
T1 – T2 =
210 - 105
t2 – t1
115 – 95
5.25
t2 – t1
=
115 - 95
T1 – t1
210 – 95
0.174
0.95
(from fig 19)
0.95 * 37.75 =
35.860F


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Step # 3
Tc and tc
 These liquids are not viscous and the
viscosity correction will be negligible

(μ/μw)s =
(μ/μw)t
=

Average temperatures can be used
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1
Step # 4a
Shell Side Calculations
 Hot Fluid (n-butyl alcohol)
 Flow area (as)
=
I.D*C*B
n*PT*144

as
=
(19.25)*(.25)*(7)
(2)*(1)*144

=
0.117 ft2
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Step # 5a
Mass velocity
 Gs
=
W/as
=
33114
0.117

=
283025.6 lb/hr.ft2
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Step # 6a
Reynold Number Res
 Res
=
De * Gs / μ
 De
=
4*(PT2 – (3.14/4)*do2)
3.14 * do
=
4 * (12 – (3.14/4)*0.752)
3.14 * 0.75
=
0.95/12
=
0.0789ft
from figure 14
μ
=
1cp * 2.42
=
2.42
 Re
=
9356

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Step # 7a
 jH
=
jH Factor
from figure 28
54
Step # 8a
ho
=
k
 ho
=
=


=
jH * (k / De) * (C μ / k)1/3
from Table 4
0.096 Btu/ft.0F
54*(0.096 / 0.0789)*(0.69*2.42/0.096)1/3
170 Btu / hr.ft2.0F
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Step # 4b
Tube Side Calculations
 Tubes flow area
from Table 10
 at
=
0.302 in2 / tube
=
204 * (0.302) / (144 * 4)
=
0.1069 ft2
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Step # 5b

Mass velocity Gt

Gt
=
=
=
w/at
119955.46
0.1069
1122127.78 lb / hr ft2
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Tube Side Velocity

V
=
=
Gt / p
1122127.78
62.5 *3600
4.987 fps
=
OR
=
1.52 ms-1
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Step # 6b
Reynold Number Ret
 Ret
=
di * Gt / μ
from figure 17
μ
=
0.7 * 2.42 =
from table 10
 di
=
0.620 in =
 Ret
=
34180.5

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1.694 lb / ft hr
0.0516 ft
Step # 7b
Tube side heat transfer coefficient hi
from Figure 25
 hi
=
1240 Btu / hr ft2 0F
 hio
=
1240 * ID / OD
=
1240 * 0.620 / 0.75
=
1025 Btu / hr ft2 0F

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Step # 8
Clean Overall Coefficient Uc
 Uc
=
hio * ho
hio + ho

=
145.8 Btu / hr ft2 0F
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Step # 9
Design Overall Coefficient UD
from Fourier Equation
 UD
=
Q/A. Δt
From Table 10
 a’’
=
0.1963 *ft2/ lin. Ft
A
=
204 * 0.1963 * 16
=
640.72 ft2
 UD
=
2399109.3 / 640.72 * 35.86
=
104.47 Btu / hr . Ft2 .0F

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Step # 10

Rd
= Uc-Ud
Uc*Ud
= 145.8 - 104.47
145.8 * 104.47
= .0027 hr ft2 0F/Btu
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Step # 11a

Pressure drop: (on shell side
For Res= 9356

(from fig.29)
f=0.0035 ft2/in.2
 No of crosses, N+1=12L/B
N+1=(12 × 16)/7
N+1=27.42 ( Say,28)
 Ds=19.25 in./12
Ds=1.604 ft
s=?
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
Step # 11a
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Step # 11a
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Step # 11a

∆Ps = f×Gs2×Ds×(N+1)
5.22×1010×De×s×Φs
∆Ps =0.0035× 283025.6 2×1.604×28
5.22×1010× 0.0789ft ×?×1
∆Ps =7.0psi (allowable=10psi
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Step # 11b
Pressure drop: (on tube side)
 Ret = 34180.5(from fig.26)
f=0.0002ft2/in.2
∆Pt=(f×Gt2×L×n)/(5.22×1010×Ds×Φt)
∆Pt= 4 psi
Gt=973500,v2/2g=0.13 (from fig.)
∆Pr=(4×n×v2)/(2g×s)
∆Pr=3.2 psi
∆PT=∆Pt+∆Pr=7.2psi(allowable=10psi)

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Step # 11b
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