Heat Exchanger Design Guide
Heat Exchanger Design Guide
|   |
|   |
=
     |    |
\   .
  \   .
              (1.5) 
You may consider,   1.0
w
  =   
Select the outside tube (shell side) dirt factor (
  do
R ) and inside tube (tube side) dirt factor 
(
  di
R ) ([3] page 845 Table 12).  
Calculate  overall  heat  transfer  coefficient  (
, o cal
U )  based  on  the  outside  tube  area  (you 
may  neglect  the  tube-wall  resistance)  including  dirt  factors: 
1
0 0 0 0
,
1 1
2
  i
o cal   do   di
o   i   w   i   i   i
A   d   d   A   A
U   R   R
h   A   k   A   h   A
   ( |   |   |   | 
=   +   +   +   +
   (    |      |
\   .   \   .    
        (1.6) 
Step #11.   If 
, ,
,
0 30%
o cal   o assm
o assm
U   U
U
=
A
A
=
t T
  t T
t
t
r
h
c
 
API of hot fluid=76; Therefore K
c 
= 1; F
c 
= 0.455 
(The  caloric  temperature  factor,  F
c
  with  API  as  a  function  Kc  is  available  in  reference 
[3] (page 827). 
 
NPTEL  Chemical Engineering  Chemical Engineering Design - II 
 
Joint initiative of IITs and IISc  Funded by MHRD                                                   Page 18 of 41 
Caloric temperature of the hot fluid, 
2 1 2
( )
hc   C
T   T   F   T   T =   +     
      =120+0.455(160-120) 
      =138.2F 
Caloric temperature of the cold fluid, 
1 2 1
( )
cc   C
T   t   F   t   t =   +     
      =75+0.455(120-75) 
       =95.475F 
II.   Fluid properties at caloric temperature     
Viscosity: 
76API gasoline, 
g
=0.2cp (0.484 lb.ft
-1
.h
-1
) 
46API kerosene, 
k
 =1.6 cp (3.872 lb.ft
-1
.h
-1
)                        
Density: 
g
=685 kg.m
-3 
(42.7 lb.ft
-3
) 
k
=800 kg.m
-3
 (49.8 lb.ft
-3
) 
Thermal conductivity: 
k
g
=0.075 Btu h
-1
ft
-1 
F
-1
 
k
k
=0.083 Btu h
-1
ft
-1 
F
-1 
Specific heat capacity: 
C
g 
= 0.57 Btu lb
-1
ft
-1 
C
k
 = 0.48 Btu lb
-1
ft
-1 
Specific gravity: 
S
g 
= 0.685 
S
k 
= 0.80  
III.  Energy balance 
Assume no heat loss to the surrounding.  
) ( ) (
2 1
.
1 2
.
T T C m t t C m Q Q
  g g k k k g
   =  = = =3240000 Btu/h 
1500000.48(120-75)= 
.
g
m 0.57(160-120) 
.
g
m = 142105 lb h
-1 
 
 
 
NPTEL  Chemical Engineering  Chemical Engineering Design - II 
 
Joint initiative of IITs and IISc  Funded by MHRD                                                   Page 19 of 41 
IV.  Calculation of heat transfer area and tube numbers 
Iteration #1: 
The first iteration is started assuming 1 shell pass and 2 tube pass shell and tube 
exchanger with following dimensions and considerations. 
-  Fixed tube plate   
-  1 OD tubes (d
o
) (14 BWG) on 1  square pitch (P
T
)  
-  Outer diameter of tube= 1 
-  Tube length (L
t
) =16  
-  Tube ID (d
i
) = 0.834 
-  Fluid  arrangement:  Kerosene  is  placed  in  tube  side  because  it  has  the  higher 
fouling tendency 
The log mean temperature correction factor (F
T
) for 1-2 shell and tube exchanger: 
 
 
                                 = 0.802 
where,  889 . 0
75 120
120 160
1 2
2 1
=
=
t t
  T T
R ;  529 . 0
75 160
75 120
1 1
1 2
=
=
t T
  t t
S  
                                           
(   )   (   )
2 1 2 1
2 1
2 1
ln
T   T   t   t
LMTD
T   T
t   t
      
=
|   | 
   |
\   .
 
(   )   (   ) 160 120 120 75
160 120
ln
120 75
      
=
 |   |
   |
\   .
 
NPTEL  Chemical Engineering  Chemical Engineering Design - II 
 
Joint initiative of IITs and IISc  Funded by MHRD                                                   Page 20 of 41 
= 42.75 F 
Determining the heat transfer area (A): 
The value of overall heat transfer coefficient (U
o,assm
) of  45 Btu h
-1
ft
-2 
F
-1
 is assumed to 
initiate  the  design  calculation  for  the  kerosene  and  gasoline    heat  exchanger.  The 
approximate range of overall heat transfer coefficient depending on the hot and cold fluid 
can be found out from text books ([3] page 845).  
assm   T
Q
A
U   LMTD   F
=
              (1.1)
 
(   )
.
1 2
g
  g
assm   T
m   C   T   T
U   LMTD   F
=
   
 
(   ) 142105 0.57 160 120
45 42.75 0.802
      
=
   
 
= 2100 ft
2
  
Calculating no. of tubes (n
t
): 
t
o   t
A
n
d  L t
=                     (1.2) 
2100
1
16
12
t
n
t
=
|   |
   
   |
\   .
= 502 
n
  t 
=  518  is  taken  corresponding  to  the  closest  standard  shell  ID  of  35  for  fixed  tube 
sheet,    1-shell  and  2-tube  pass  exchanger  with  1    tube  OD  on  1    square  pitch.  You 
may refer to standard heat transfer books ([3] page 841-842) for the selection of suitable 
shell ID.  
Check for fluid velocity: 
 
.
4 ( / )
Re
  k
  p   t
i
m   n   n
d t   
=                   (1.4) 
2
4 (150000)
518
Re
0.834
3.872
12
t
   
=
   
 
= 2740.2<10
4
 
NPTEL  Chemical Engineering  Chemical Engineering Design - II 
 
Joint initiative of IITs and IISc  Funded by MHRD                                                   Page 21 of 41 
As  Re<<10
4
,  the  design  parameters  and  considerations  needs  to  be  revised  to  meet 
the Reynolds number criteria subject to allowable pressure drop in the tube side of 
the heat exchanger.   
Iteration #2: 
Assumptions: 
-  Fixed tube plate type 
-  1 OD tubes (14 BWG) on 1 square pitch (P
T
) 
-  Tube length (L
t
) = 24 (the tube length is increased from 16) 
-  1 shell pass-6 tube pass (tube passes is increased to 6 from 2) 
-
  Tube ID=0.834 
 
-
  Flow area per tube=0.546 inch
2 
No. of tubes: 
t
o   t
A
n
d  L t
=                   (1.2) 
2100
1
24
12
t
n
t
=
|   |
   
   |
\   .
=335 
n
  t 
=  368  is  taken  corresponding  to  the  closest  standard  shell  ID  of  31  for  fixed  tube 
sheet,    1-shell  and  6-tube  pass  exchanger  with  1    tube  OD  on  1    square  pitch.  The 
tube-counts are available in heat transfer text book ([3] Table 9 & 10 page 841-843). 
Fluid velocity: 
.
4 ( / )
Re
  k
  p   t
i
m   n   n
d t   
=                   (1.4) 
6
4 (150000)
368
Re
0.834
3.872
12
t
   
=
   
 
= 11571.4>10
4 
corresponding to n
p
=6.  
Re
  k
i   k
u
d
 
=                     (1.3) 
11571.4 3.872
0.834
49.8
12
 
NPTEL  Chemical Engineering  Chemical Engineering Design - II 
 
Joint initiative of IITs and IISc  Funded by MHRD                                                   Page 22 of 41 
= 12945.15 ft/h (3.59 ft/s) 
= 1.04 m/s (so the design velocity is within the acceptable range). 
V.  Determination of heat transfer co-efficient 
Tube side heat transfer co-efficient (h
i
): 
1
0.14
3
i   i   k   k
H
k   w
h d   C
j
k   k
   
   
|   |   |   |
=
     |      |
\   .   \   .
               (1.5) 
j
H
=42 for the tube side fluid at Re=11571.4 ([3] page 834)
 
(Lets consider 
t
w
   |
\   .
=  for square pitch 
=0.082 ft 
2
1 1
4 0.86
2 2 4
For triangular pitch, 
1
2
T   T   o
e
o
P   P   d
D
d
t
t
   ( |   |
   
   |    (
\   .
   ( =
   (
   (
   
 
Shell side cross flow area,
  S
s
T
CBD
a
P
=  (please refer to Figure 1.6). 
C= Tube clearance 
=P
T 
- d
o
 
=1 -1=0.25 
NPTEL  Chemical Engineering  Chemical Engineering Design - II 
 
Joint initiative of IITs and IISc  Funded by MHRD                                                   Page 23 of 41 
0.25 15.5 31
12 12 12
1.25
12
s
a
|   ||   ||   |
   |   |   |
\   .\   .\   .
=
|   |
   |
\   .
=0.675 ft
2
 
Mass velocity,
.
142105
0.675
g
S
s
m
G
a
=   =  
=210526 lb. h
-1
.ft
-2
 
Re
  e   S
g
D G
=  
(   ) 0.082 210526
0.484
=  
=35668 
Now for the shell side, 
1/ 3
0.14
g   g
o   e
H
g   g   w
C
h D
j
k   k
  
|   |
  |   |
=
     |
     |
   |
  \   .
\   .
        (1.5) 
j
H
=110 for the shell side fluid at Re=35668 with 25% cut segmental baffles ([3] page 
838)
       
1/ 3
0
(0.082) 0.57 0.484
110
(0.075) 0.075
h
  
 |   |
=
     |
\   .
      
(
  s
w
   ( |   |   |   | 
=   +   +   +   +
   (    |      |
\   .   \   .    
        (1.6) 
 
Let select, Admirality brass as tube material with thermal conductivity,  k
w
=70 Btu h
-1 
ft
-1 
F
-1
.  
1
2 2 2
, 2 2 2
1 0.834
1 (1) (1) 1 (1)
12 12
0.0005 0.001
155.3 (0.834) 2 70 (0.834) 141.3 (0.834)
o cal
U
  t   t   t
t   t   t
( | |
   (    |
  |   |
=   +   +   +   +   
   (    |      |
  \   .
   (    |
   (
\   .    
 
, o cal
U   = 53.5 Btu h
-1 
ft
-2 
F
-1 
 
Now, 
, ,
,
53.5 45
100
45
18.9% 30%
o cal   o assm
o assm
U   U
U
  
=   
=   <
 
Therefore, the calculated overall heat transfer co-efficient is well within the design 
criteria.  
VI.  Pressure drop calculation 
VI.1.  Tube side pressure drop: 
Friction factor  0.00028 144 0.04032 f =      =  ft
2
/ft
2
 for Re=11571.4 ([3] page 836]  
t
a = (no. of tubes)(flow area per tube)/(no. of passes) 
2
368 0.546
ft
6 144
 
=0.232 ft
2
 
Tube side mass velocity: 
150000
0.232
k
t
t
m
G
a
=   =
 
=646552 lb. h
-1
.ft
-2 
 
NPTEL  Chemical Engineering  Chemical Engineering Design - II 
 
Joint initiative of IITs and IISc  Funded by MHRD                                                   Page 25 of 41 
Frictional pressure drop:
2
12
7.5 10
t   t   p
t
i   k   t
fG  L n
P
d S |
A   =
   
  
2
12
0.04032 646552 24 6
0.834
7.5 10 0.8 1
12
      
=
         
 
=5.81 psi 
Return loss 
rt
P A : (due to change in flow direction of the tube side fluid) 
(   )
2
13
1.334 10 2 1.5
  t
tr   p
k
G
p   n
S
A   =        
(   )
 (   )
(   )
2
13
646552
1.334 10 2 6 1.5
0.8
=         
=0.73 psi 
Total tube side drop neglecting nozzle loss: 
T   t   tr
P   P   P A   = A  + A                   (1.8) 
=5.81+0.73 
=6.54 psi<10 psi  
Therefore the tube side pressure drop is within the maximum allowable pressure 
drop of 10 psi. 
VI.2.  Shell side pressure drop calculation 
Tube clearance, C=0.25 
Spacing, B=15.5 
0.675
s
a = ft
2 
 
Mass velocity,
  S
G = 210526 lb. h
-1
.ft
-2
 
Re=35668 
No of baffles, 
tube length 24
=  18.6 19
baffle spacing 15.5/12
b
n =   =   ~  
Friction factor,  0.0017 144 0.2448 f =      =  ft
2
/ft
2
 with 25% cut segmental baffles ([3] 
page 839) 
 
 
NPTEL  Chemical Engineering  Chemical Engineering Design - II 
 
Joint initiative of IITs and IISc  Funded by MHRD                                                   Page 26 of 41 
Shell side frictional pressure drop 
s
P A : 
(   )
2
12
1
7.5 10
s   S   b
s
e   k   k
fG  D   n
P
D S |
+
A   =
   
   
(   )
2
12
31
0.2376 210526 19 1
12
7.5 10 0.082 0.685 1
      +   
=
         
 
=1.4 psi <7 psi 
 
0
rs
P A   =  (in case of single shell pass flow) 
Total shell side drop neglecting nozzle loss: 
S   s   sr
P   P   P A   = A   + A =1.4 psi                (1.9) 
Therefore the shell side pressure drop is within the maximum allowable pressure 
drop of 7 psi. 
VII.  Over surface and over design 
Over surface =
, C   o cal
C
U   U
U
  
The clean overall heat fransfer co-efficient:  
o io
o io
C
  h   h
U
h   h
=
+
 
i
io i
o
d
h   h
d
=    =141.30.834=117.8 Btu h 
-1
 ft
-2
 F
-1
 
C
U  =66.98 Btu h 
-1
 ft
-2
 F
-1
 
% Over surface =
66.98 53.5
100
66.98
    
=20% (acceptable) 
Over design: 
  -
% Overdesign = 100
reqd
reqd
A   A
A
                 (1.7) 
The design area of heat transfer in the exchanger (
  t
n =318): 
o   t   t
A   d Ln t = =
12
1
24368=2312 ft
2
  
 
NPTEL  Chemical Engineering  Chemical Engineering Design - II 
 
Joint initiative of IITs and IISc  Funded by MHRD                                                   Page 27 of 41 
The required heat transfer area (where, 
t
n =335): 
reqd   o   t   t
A   d Ln t = = 
12
1
24335=2105 ft
2
 
% Overdesign =9.8% which is within the acceptable limit.  
Refer module # 2 for the mechanical design of shell and tube heat exchanger.  
Lecture 5: Shell and Tube Exchanger for Two Phase Heat Transfer 
 
2.    PROCESS DESIGN OF SHELL AND TUBE 
EXCHANGER FOR TWO PHASE HEAT TRANSFER 
2.1.   Condenser 
The change from liquid phase to vapor phase is called vaporization and the reverse phase 
transfer is condensation. The change from liquid to vapor or vapor to liquid occurs at one 
temperature (called saturation or equilibrium temperature) for a pure fluid compound at a 
given pressure. The industrial practice of vaporization and condensation occurs at almost 
constant pressure; therefore the phase change occurs isothermally.  
Condensation occurs by two different physical mechanisms i.e. drop-wise condensation 
and film condensation. 
The  nature  of  the  condensation  depends  upon  whether  the  condensate  (liquid  formed 
from vapor) wets or does not wet the solid surface. If the condensate wets the surface and 
flows  on  the  surface  in  the  form  of  a  film,  it  is  called  film  condensation.  When  the 
condensate does not wet the solid surface and the condensate is accumulated in the form 
of  droplets,  is  drop-wise  condensation.  Heat  transfer  coefficient  is  about  4  to  8  times 
higher  for  drop  wise  condensation.  The  condensate  forms  a  liquid  film  on  the  bare-
surface  in  case  of  film  condensation.  The  heat  transfer  coefficient  is  lower  for  film 
condensation due to the resistance of this liquid film. 
Dropwise  condensation  occurs  usually  on  new,  clean  and  polished  surfaces.  The  heat 
exchanger  used  for  condensation  is  called  condenser.  In  industrial  condensers,  film 
condensation normally occurs.   
 
 
NPTEL  Chemical Engineering  Chemical Engineering Design - II 
 
Joint initiative of IITs and IISc  Funded by MHRD                                                   Page 28 of 41 
2.1.1. Types of condensers 
There two general types of condensers: 
i.  Vertical condenser  
Downflow vertical condenser: The vapor enters at the top of condenser and flows 
down  inside  tubes.  The  condensate  drains  from  the  tubes  by  gravity  and  vapor 
induced shear (Figure 1.7). 
Upflow  vertical  condenser:  In  case  of  upflow  condenser,  the  vapor  enters  at  the 
bottom and flows upwards inside the tubes. The condensate drains down the tubes 
by gravity only. 
ii.  Horizontal  condenser:  The  condensation  may  occur  inside  or  outside  the 
horizontal  tubes  (Figure  1.8).  Condensation  in  the  tube-side  is  common  in  air-
cooled  condensers.  The  main  disadvantage  of  this  type  of  condenser  is  that  the 
liquid  tends  to  build  up  in  the  tubes.  Therefore  the  effective  heat  transfer  co-
efficient is reduced significantly.  
 
NPTEL  Chemical Engineering  Chemical Engineering Design - II 
 
Joint initiative of IITs and IISc  Funded by MHRD                                                   Page 29 of 41 
 
Figure 1.7. Downflow vertical condenser with condensation inside tube [5]. 
 
NPTEL  Chemical Engineering  Chemical Engineering Design - II 
 
Joint initiative of IITs and IISc  Funded by MHRD                                                   Page 30 of 41 
 
Figure 1.8. Horizontal condenser with condensation outside horizontal tubes [5]. 
2.1.2. Condenser design  
The  design  of  condenser  is  similar  to  a  typical  shell  and  tube  exchangers.  A  condenser 
must  have  a  vent  for  removal  of  non-condensable  gas.  The  non-condensable  gas 
decreases  the  heat  transfer  rate.    Condenser  usually  use  a  wider  baffle  spacing  of 
s
B   D = (ID  of  shell)  as  the  allowable  pressure  drop  in  shell  side  vapor  is  usually  less. 
Vertical  cut-segmental  baffles  are  generally  used  in  condensers  for  side-to-side  vapor 
flow  and  not  for  top  to  bottom.  An  opening  at  the  bottom  of  the  baffles  is  provided  to 
allow draining of condensate.  
2.1.2.1. Mean temperature difference 
The condensation occurs almost at a fixed temperature (isothermally) at constant pressure 
for  a  pure  saturated  vapor  compound.  The  logarithmic  mean  temperature  difference  can 
be  used  for  condenser  design.  No  correction  factor  for  multiple  pass  condensers  is 
required. The logarithmic mean temperature difference: 
1 2 2 1
1 1
2 2
( ) ( ) ( )
( ) ( )
ln ln
( ) ( )
sat   sat
sat   sat
sat   sat
T   t   T   t   t   t
LMTD
T   t   T   t
T   t   T   t
         
=   =
   
   
           (1.10)
 
Where, T
sat
= Saturation vapor temperature 
t
1
 = Coolant inlet temperature 
t
2
 = Coolant outlet temperature   
 
 
 
NPTEL  Chemical Engineering  Chemical Engineering Design - II 
 
Joint initiative of IITs and IISc  Funded by MHRD                                                   Page 31 of 41 
2.1.2.2. Calculation of heat transfer co-efficient during condensation  
Calculation of tube side heat transfer co-efficient (h
i
): The calculation of heat transfer 
co-efficient for the cold fluid (coolant) can be performed similarly as discussed in design 
of shell and tube heat exchanger (heat transfer without phase change).  Here it is assumed 
that  the  coolant  flows  the  in  tube  side  and  the  condensing  saturated  vapor  flows  in  the 
shell side.  If the condensation occurs in the tube side, follow the procedure discussed in 
next section for shell side calculation. 
Calculation  of  shell-side  heat  transfer  coefficient  (condensing  film  heat  transfer 
coefficient)  (h
o
):  The  Kern  method  is  discussed  here  to  calculate  the  individual  heat 
transfer co-efficient of the condensing fluid by trial and error calculation.   
i.  Assume, 
( ) o  assm
h   in  the  range  from  100  to  300  BTU.h
-1
.ft
-2
.F
-1
.  The  film 
coefficient  of  condensing  hydrocarbons  generally  varies  in  this  range.  Air-free 
condensing steam has a coefficient of 1500 BTU.h
-1
.ft
-2
.F
-1
. 
ii.  Calculate the tube wall temperature (
  w
T ): 
( )
( )
( )
( )
o   v   C  avg
w   C  avg
io   o
h   T   T
T   T
h   h
=   +
+
                (1.11)
 
or 
( )
( )
o   v   cc
w   cc
io   o
h   T   T
T   T
h   h
=   +
+
                  (1.12)
 
Where,   
i
io   i
io
d
h   h
d
=     (
  i
d  tube ID and 
o
d  tube OD) 
( ) C  avg
T = Average temperature of the cold fluid  
cc
T =Caloric temperature of the cold fluid 
iii.   Calculate condensate film temperature, 
( )
2
w   v
f
  T   T
T
  +
=        (1.13) 
v
T =Condensation temperature (For pure fluid compound 
v
T is the saturation temperature. 
Average  of  condensation  over  a  temperature  range  also  can  be  used  for  non-isothermal 
condensation).  
 
NPTEL  Chemical Engineering  Chemical Engineering Design - II 
 
Joint initiative of IITs and IISc  Funded by MHRD                                                   Page 32 of 41 
iv.  Calculate all thermophysical property of the condensing fluid at film temperature 
(
  f
T ). 
v.   Recalculate, 
( ) o  cal
h from j
H
 factor. 
Now again set, 
( ) ( ) o  assm   o  cal
h   h = and continue the calculation till
( ) ( ) o  assm   o  cal
h   h ~ . 
vi.  Calculate the overall heat transfer-coefficient (
  d
U ) including the dirt factors.  
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
NPTEL  Chemical Engineering  Chemical Engineering Design - II 
 
Joint initiative of IITs and IISc  Funded by MHRD                                                   Page 33 of 41 
Lecture 6: Condenser and Reboiler Design 
2.1.2.3. Pressure drop calculation 
i.  Tube side pressure drop 
In case of tube side condensation: 
For  condensation  in  the  tube  side  by  taking  one-half  of  the  conventional  pressure  drop 
relation can be used.  
2
12
1
2 7.5 10
t   t   p
t
i   t   t
fG  L n
P
d S |
|   |
A   =
     |
   |
   
\   .
, psi              (1.14) 
Where, 
f = friction factor 
t
G =mass velocity [lb. h
-1
.ft
-2
] 
t
L =Tube length [ft] 
p
n =Number of tube passes 
i
d =Tube ID [ft] 
t
S =Specific gravity of the tube side fluid 
t
| =Viscosity correction factor 
(
  t
w
 
NPTEL  Chemical Engineering  Chemical Engineering Design - II 
 
Joint initiative of IITs and IISc  Funded by MHRD                                                   Page 35 of 41 
Practice problem: 
Design a horizontal condenser for the condensation of 45,000 lb/h of almost pure normal 
propyl alcohol available at 15 psig. At this pressure, the boiling point of n-propyl alcohol 
is 244F. Water available in the temperature range of 95 to 120F can be as the coolant. 
The  maximum  pressure  drop  of  2  psi  and  10  psi  is  permissible  for  the  vapor  phase  and 
water respectively.  
2.2.  Reboilers 
2.2.1. Classification of reboilers 
There are three major types of reboilers: 
i.   Thermosyphon natural circulation reboiler: The boiling occurs inside the tubes 
in  vertical  thermosyphon  reboiler  and  inside  shell  in  horizontal  thermosyphon 
reboiler  (Figure  1.10).  In  vertical  thermosyphon  reboiler,  the  liquid  circulation 
occurs due to density difference between vapor-liquid mixture (two phase) in the 
exchanger from the reboiler and the liquid through the downcomer to the reboiler. 
Advantages:  most economical because no pump is required. 
Limitations:  not  suitable  for  heavily  viscous  fluid;  high  construction  cost  for  the 
installation  of  the  column  base  at  suitable  elevation  to  get  thermosyphon  effect; 
not  suitable  for  low  temperature  difference  processes  due  to  boiling  point 
elevation imposed by static head.  
 
 
Figure 1.10. Thermosyphon reboiler [5]. (a) Horizontal thermosyphon reboiler. (b) Vertical thermosyphon reboiler 
 
 
NPTEL  Chemical Engineering  Chemical Engineering Design - II 
 
Joint initiative of IITs and IISc  Funded by MHRD                                                   Page 36 of 41 
ii.  Forced  circulation  reboiler:  The  liquid  is  fed  by  means  of  a  pump.  Forced 
circulation  reboilers  with  vertical  or  horizontal  tubes  boiling  may  be  designed. 
Forced circulation reboilers are similar to vertical thermosiphon reboilers, except 
the  pump  is  used  for  the  circulation  of  the  liquid  and  the  hot  liquid  flows  inside 
column. To calculate the heat transfer coefficient it is generally assumed that, heat 
is  transferred  only  by  forced  convection.  The  usual  method  of  shell  and  tube 
exchanger design can be used.  
Advantage: suitable for viscous and highly fouling fluids. 
Disadvantage: high pumping and maintenance cost; pump is required to circulate 
the boiling liquid through the tubes and back into the column. 
iii.  Kettle reboiler: The tube bundle is immerged in a pool of liquid at the base of the 
column  in  an  oversize  shell  (Figure  1.11).  Kettle  reboiler  is  also  called  a 
submerged bundle reboiler. The height of the tube bundle is usually 40-60% of 
the shell ID. The submergence of the tube bundle is assured by an overflow weir 
at height of typically 5-15 cm from the upper surface of topmost tubes.  
Advantage: suitable for vacuum operation and high vaporization rate up to about 
80% of the feed. 
Limitations: low heat transfer rate than other types as there is no liquid circulation 
(low velocity); not appropriate for fouling fluids; kettle reboiler is not suitable for 
heat sensitive materials as it has higher residence time. 
The  bundle  diameter  D
b
,  can  be  obtained  from  the  empirical  equation  ([2]  page 
647-649 ): 
1
1/
1
n
t
b   o
  n
D   d
K
|   |
=
     |
\   .
                (1.16) 
where, 
b
D =  bundle  diameter  [mm],
  t
n   =  number  of  tubes, 
o
d    =  tube  outside 
diameter [mm]. The values of the constants 
1
K and 
1
n are in Table 1.5. 
 
 
 
 
 
NPTEL  Chemical Engineering  Chemical Engineering Design - II 
 
Joint initiative of IITs and IISc  Funded by MHRD                                                   Page 37 of 41 
Table 1.5. Constants used to calculate the tube bundle diameter. 
Pitch type  Constants 
Number of tube passes (
  t
n ) 
1  2  4  6  8 
Triangular 
( 1.25
T   o
P   d = ) 
1
K  
0.319  0.249  0.175  0.0743  0.0365 
1
n  
2.142  2.207  2.285  2.499  2.675 
Square 
( 1.25
T   o
P   d = ) 
1
K  
0.215  0.156  0.158  0.0402  0.0331 
1
n  
2.207  2.291  2.263  2.617  2.643 
 
 
Figure 1.11. Kettle type reboiler [1]. 
2.2.2. Design of kettle reboiler  
The  Kern  method  for  designing  of  Kettle  reboiler  for  isothermal  boiling  is  summarized 
below. It is assumed that the degree of sub-cooling and super-heating of the cold fluid is 
negligible  i.e.  vaporization  of  close  boiling  compounds  with  negligible  super-heating  of 
vapors formed.  
i.  Make energy balance and determine the heat duty.  
ii.  Calculate  of  fluid  property  at  the  caloric  temperature  (or  at  arithmetic  mean 
temperature) as already shown.  
iii.  Follow  the  same  guideline  and  design  requirements  for  shell  containing  the 
vaporing liquid.  
iv.  Calculation of heat transfer co-efficient  
NPTEL  Chemical Engineering  Chemical Engineering Design - II 
 
Joint initiative of IITs and IISc  Funded by MHRD                                                   Page 38 of 41 
Calculation  of  individual  heat  transfer  co-efficient  hot  fluid:  The  calculation 
of heat transfer co-efficient of the hot fluid can be performed similarly as in case 
of design of shell and tube heat exchanger for single phase.   
Calculation  of  individual  heat  transfer  coefficient  of  the  boiling  liquid:  The 
Kern method is discussed here to calculate the individual heat transfer co-efficient 
of the boiling liquid by trial and error procedure.   
Kern [2] recommends that the maximum allowable vaporizing film coefficients:   
-  300 Btu/h.ft
2 
F for natural or forced circulation vaporizing organics. 
-  1000 Btu/h.ft
2 
F for natural or forced circulation vaporizing aqueous solution 
of low concentration. 
The maximum allowable heat flux:   
-   20000  Btu/(h)ft
2
)  for  forced  circulation  reboilers  and  12000  Btu/(h)ft
2
)  for 
natural circulation reboilers vaporizing organics. 
 
-  30000  Btu/(h)ft
2
)  for  both  forced  or  natural  circulation  reboilers  vaporizing 
aqueous solution.  
Assume that h
(assm)
 = 300 Btu/h.ft
2 
F for organics or 1000 Btu/h.ft
2 
F for water. 
With this assumed value, calculate the tube wall temperature (
  w
T ): 
( )
( )
( )
( )
io   hc   h  avg
w   h  avg
io   o
h   T   T
T   T
h   h
=   +
+
              (1.17)
 
Where,   
i
io   i
io
d
h   h
d
=     (
  i
d  tube ID and 
o
d  tube OD) 
( ) h  avg
T = Average temperature of the hot fluid  
hc
T =Caloric temperature of the hot fluid  
Now,  re-determine  h
cal
  (latent  heat  transfer)  from  the  Figure  1.12  corresponding 
to( )
w
T   t  . ( t  is the cold fluid boiling temperature). 
Continue the calculation till, 
cal
h
    ~
( ) assm
h .  
If the calculated 
cal
h is greater than the maximum heat transfer co-efficient of 300 
Btu/h.ft
2 
F for organics and 1000 Btu/h.ft
2 
F for water, take 
cal
h   = 300 Btu/h.ft
2 
F for organics and 
cal
h   =1000 Btu/h.ft
2 
F for water. 
Calculate the overall heat transfer-coefficient (
  d
U ) including the dirt factors.   
NPTEL  Chemical Engineering  Chemical Engineering Design - II 
 
Joint initiative of IITs and IISc  Funded by MHRD                                                   Page 39 of 41 
10 100
100
1000
 
 
H
e
a
t
 
t
r
a
n
s
f
e
r
 
c
o
-
e
f
f
i
c
i
e
n
t
 
(
h
)
,
 
B
t
u
/
(
h
r
)
(
f
t
2
)
(
o
F
)
(T
w
-t), 
o
F
Temperature difference between tube wall and boiling liquid
Maximum for organics
Maximum for water
S
e
n
s
ib
le
 h
e
a
t tr
a
n
s
fe
r
 b
y
 fo
r
c
e
d
 c
o
n
v
e
c
tio
n
L
a
t
e
n
t
 
h
e
a
t
 
t
r
a
n
s
f
e
r
20
4
500
50
200
Figure 1.12. Natural circulation boiling and sensible heat transfer [3]. 
v.  Decide type of exchanger i.e. fixed tube sheet or U- shell (use U-tube reboiler for 
large  temperature  difference),  tube  size  (diameter,  length,  tube  pitch),  layout, 
effective tube length.  A tube pitch of between 1.5 to 2 times the tubes OD should 
be used to avoid vapor blanketing. 
vi.  Calculate  exchanger  area  (
( )
d
Q
A
U   LMTD
= )  and  number  of  tubes  (
  t
o   t
A
n
d  L t
= ). 
The  number  of  tubes  should  be  calculated  based  on  the  effective  tube  length  for 
U-tube reboilers. The effective tube length is less than physical tube length due to 
U-bend. 
vii.  Calculate  the  heat  flux=
 Q
A
    [Btu/(h.ft
2
)].  This  value  should  be  less  than  the 
maximum  heat  flux  of  20000  Btu/(h)ft
2
)  for  forced  circulation  reboilers 
vaporizing  organics  and  30000  Btu/(h)ft
2
)  for  both  forced  or  natural  circulation 
NPTEL  Chemical Engineering  Chemical Engineering Design - II 
 
Joint initiative of IITs and IISc  Funded by MHRD                                                   Page 40 of 41 
reboilers  vaporizing  aqueous  solution.  Otherwise,  go  to  step  #  v,  repeat  the 
calculation until within the allowable limits.  
viii.  Check for allowable vapor velocity (
  v
u ) ([3] page 749): 
The  maximum  vapor  velocity 
v
u (m/s)  at  the  liquid  surface  should  be  less  than 
that given by the expression below to avoid too much entrainment.
 
 
1/ 2
0.2
  l   v
v
v
u
     
|   | 
<
     |
\   .
                (1.18) 
 
where,
  l
 = liquid density and,
  l
 = vapor density 
If this criterion is not satisfied, go to step # v and revise the calculation.  
ix.  Pressure drop calculation 
Tube side pressure drop (hot fluid): The pressure drop  calculation of the hot fluid 
can be carried out as already presented. 
Shell  side  pressure  drop  (vaporizing  liquid):  There  will  be  negligible  hydrostatic 
head for the  flow of liquid from the column to  reboilers (low circulation  velocity) if 
the liquid level above the tube bundle is not too high.  Therefore, shell side pressure 
drop may be considered negligible. 
x.   Calculate over surface and over design 
xi.  Go for mechanical design 
Design problem: 
Gasoline (65API gravity) flow rate of 60,000 lb/h with a small boiling range at 400F is 
to be vaporized to form 37,050 lb/h vapor at an operating pressure of 200 psig.  Use gas 
oil  (30API  gravity)  in  the  temperature  range  from  600  to  500F  at  120  psig  operating 
pressure as the heating medium. A tube side pressure drop of 10 psi is allowable. Design 
a suitable Kettle reboiler to serve the purpose.  
 
 
 
 
 
 
NPTEL  Chemical Engineering  Chemical Engineering Design - II 
 
Joint initiative of IITs and IISc  Funded by MHRD                                                   Page 41 of 41 
References 
[1].  Indian Standard (IS: 4503-1967): Specification for Shell and Tube Type Heat 
Exchangers, BIS 2007, New Delhi.  
 
[2].   R. K. Sinnott, Coulson & Richardsons Chemical Engineering: Chemical 
Engineering Design (volume 6), Butterworth-Heinemann, 3
rd
 ed. 1999.  
 
[3].   D. Q. Kern, Process Heat Transfer, McGraw-Hill Book Company, Int. ed. 1965. 
 
[4]   Dutta B.K. Heat Transfer-Principles and Applications,   PHI Pvt. Ltd., New 
Delhi, 1
st
 ed. 2006.  
 
[5]   James R. Couper; W. Roy Penney, James R. Fair, Stanley M. Walas, Chemical 
Process Equipment: selection and design, Elsevier Inc., 2
nd
 ed. 2005.