Section
Hussain Butaleb
207217011
Cooler (E-101)
Heat Exchanger (E-103)
Heat Exchanger (E-102)
Steam Drum (V-102)
Vessel (V-101)
Storage Tank (T-101)
Hydro Cyclone (T-102)
Stripper (C-201)
Distillation (C-101)
Heat Exchangers and Cooler
A heat exchanger is a device designed to transfer heat from one fluid stream to another
without bringing the fluids into direct contact. Heat exchange equipment comes in a wide
variety of forms, with an equal variety of functions.
They are widely used in chemical plants, petroleum refineries, natural gas processing,
refrigeration, power plants, air condition and space heating. Heat exchangers can have
different size and shape depending on the application; it can be made of various materials and
use various fluids for heat transfer.
Types of Heat Exchanger
The heat exchangers can be classified according to:

Boilers and steam generators.

Condensers.

Radiators.

Evaporators.

Cooling towers
Flow arrangements:

Co current of parallel flow.

Countercurrent flow.

Cross flow (single or multiple pass).
Shell and tube heat exchanger:
Shell and tube heat exchangers consist of a series of tubes. One set of these tubes contains the
fluid that must be either heated or cooled. The second fluid runs over the tubes that are being
heated or cooled so that it can either provide the heat or absorb the heat required. A set of
tubes is called the tube bundle. There are several thermal design features that are to be taken
into account when designing the tubes in the shell and tube heat exchangers.
In addition to heating up or cooling down fluids in just a single phase, heat exchangers can be
used either to heat a liquid to evaporate (or boil) it or used as condensers to cool a vapor and
condense it to a liquid. Distillation set-ups typically use condensers to condense distillate
vapors back into liquid. To conserve energy and cooling capacity in chemical and other
plants, regenerative heat exchangers can be used to transfer heat from one stream that needs
to be cooled to another stream that needs to be heated. This term can also refer to heat
exchangers that contain a material within their structure that has a change of phase. This
change of phase effectively acts as a buffer because it occurs at a constant temperature but
still allows for the heat exchanger to accept additional heat.
The transfer of thermal energy between fluids is one of the most important and frequently
used processes in engineering. The transfer of heat is usually accomplished by means of a
device known as a heat exchanger. The basic design of a heat exchanger normally has two
fluids of different temperature separated of some conducting medium. The common design
has one fluid flowing through metal tubes and the other fluid flowing around the tubes. On
either side of the tube, heat is transferred by convection. Heat transferred through the tube
wall by conduction. Single phase exchangers are usually of the tube and shell type.
Design procedure of shell and tube heat exchanger
Calculation procedure:
1) Define fluid flow rates, temperature.
2) Collect together the fluid physical properties required: density, viscosity, thermal
conductivity.
3) Finding the heat load for the process stream.
4) Decide on the type of exchanger to be used.
5) Find the outlet temperature of water flow.
6) Calculate the log mean temperature difference.
7) Find out the temperature correction factor (Ft) using two dimensionless
temperature ratio (R & S), choose the number of shell's and tube passes.
8) Find out the true temperature difference.
9) Assume the overall heat transfer coefficient, Uo .
10) Calculate the provisional area.
11) Choosing tube outside & inside diameter, also tube length, then calculate the area
of one tube.
12) Calculate the number of tubes which is equal to the provisional area over the area
of one tube.
13) Choose a triangle pitch for tube layout and get the constants K1 and n1 for two
tube passes.
14) Calculate the bundle diameter.
15) Using split-ring floating head type, finds out the bundle diametrical clearance, and
then calculates the sell diameter.
16) To calculate the tube side coefficient, find out the following:

Tube cross sectional area.

Tubes per pass.

Total flow area.

Tube mass velocity.

Tube linear velocity.

Reynolds number.

Prantl number.
17) Find the heat transfer factor (jh) and calculate hi.
18) To calculate the shell side coefficient, find out the following:

Choose baffle spacing.

Tube pitch.

Cross flow area.

Mass velocity.

Equivalent diameter.

Reynolds number.

Prandtl number.
Choose 25% baffle cut, and find the heat transfer factor (jh) and calculate hs.
19) The overall heat transfer coefficient must found.
20) The pressure drop can be calculated for both tube-side and shell-side.
21) Calculate thickness of the shell.
Nomenclature
Symbol
Definition
Qh
Heat load transfer in the hot side, KW.
m
Mass flow rate in Kg/s.
T
Temperature difference of the inlet and outlet.
TLM
Log means Temperature.
T1
Inlet shell side fluid temperature (oC).
T2
Outlet shell side fluid temperature (oC).
t1
Inlet tube side temperature (oC).
t2
Outlet tube temperature (oC).
Tm
True temperature difference.
R
Dimensionless temperature ratio.
S
Dimensionless temperature ratio.
Ft
Log mean temperature difference correction factor.
A
Heat transfer area
Nt
Number of tubes in a tube bundle.
Db
Bundle diameter (mm).
d0
Tube outside diameter.
K1
Constant.
n1
Constant.
Ds
Shell diameter.
Ac
Tube cross-sectional area.
di
Tube inside diameter.
At
Total flow area.
Um
Tube mass velocity.
Ut
Tube linear velocity.
 ref
Density.
hi
Film heat-transfer coefficient inside a tube.
Re
Reynolds number

Fluid viscosity at the bulk fluid temperature, Ns/m2.
Pr
Prandtl number.
Cp
Heat capacity.
kf
Thermal conductivity of stream.
lB
Baffle spacing.
pt
Tube pitch
Gs
Mass velocity.
As
Cross-flow area between tubes.
de
Equivalent diameter.
U0
The overall heat transfer coefficient.
hod
Fouling coefficient on outside of tube.
hid
Fouling coefficient on inside of tube.
Pt
Tube- side pressure drop (N/m²) (pa).
Np
Number of tube -side passes
ut
Tube-side fluid viscosity.
L
Length of one tube.
jf
Friction factor.
w
Fluid viscosity at the wall.
Ps
Shell-side pressure drop.
P
Maximum allowable internal pressure (psig).
ri
Internal radius of shell before allowance corrosion is added (in).
Ej
Efficiency of joints.
S
Working stress (psi).
Cc
Allowance for corrosion (in)
Design for cooler (E-101)
Q  mC p T  8.35E  05 W
Where:
Qh = Heat load transfer in the hot side, KW.
m  Mass flow rate in Kg/s.
T  Temperature difference of the inlet and outlet.
Tlm 
(T1  t 2 )  (T2  t1 )
 181.0804C
(T1  t 2 )
ln
(T2  t1 )
Where:
TLM  Log means Temperature.
T1 
Inlet shell side fluid temperature (oC).
T2 
Outlet shell side fluid temperature (oC).
t1 
Inlet tube side temperature (oC).
t 2  Outlet tube temperature (oC).
R
S
(T1  T2 )
 1.073653
(t 2  t1 )
(t 2  t1 )
=0.9314
(T1  t1 )
Tm  Ft Tlm  181.0804 C
Where:
Tm  True temperature difference.
Ft  Temperature correction factor=1
A
Q
 27955.21m2
UTm
Where:
A  Provisional area in m2.
Q  Heat load in W.
Tm  True temperature difference.
A  DL  0.15072 m2
Where:
A  Area of one tube, m2.
N t  Provisional area/Area of one tube.
1
N
Db  d 0 ( t ) n1  1285.214 mm
K1
Where:
Db  Bundle diameter (mm).
d 0  Outside diameter (mm).
N t  Number of tubes.
K1 & n1 are constant.
Ds  Db  Clearance = 1337.214 mm
Where:
Ds  Sell diameter.
Db  Bundle diameter (mm).
Clearance = 52, split ring floating head.
Ac 

4
(d i ) 2 = 1.32665 mm2
Where:
Ac  Tube cross-sectional area.
di  Tube inner diameter.
Tubes N t
= 23184.72

Pass
2
Where:
N t  Number of tubes.
At  Ac
Tubes
= 0.030758 m2
Pass
Where:
At  Total flow area.
Um 
m
= 180.7389 m/s
At
Where:
U m  Tube mass velocity.
At  Total flow area.
m  Mass flow rate in Kg/s.
Ut 
Where:
Um
 ref
= 0.353491 m/s
U t  Tube linear velocity.
 ref  Density.
Re 
U t d i
 309.9418

Where:
Re  Reynolds number.
  Fluid viscosity at the bulk fluid temperature, Ns/m2.
Pr 
Cp
kf
= 5.293297
Where:
Pr  Prandtl number.
C p  Heat capacity.
k f  Thermal conductivity of stream.
hi 
k f j h Re(Pr) 0.33
di
= 502.75 W/m2C
Where:
hi  Inside coefficient (W/m2 oC).
j h  Tube side heat transfer factor.
k f  Thermal conductivity of stream.
Pr  Prandtl number.
lB 
Where:
Ds
= 222.869 mm
5
l B  Baffle spacing.
Ds  Shell diameter.
pt  1.25d 0 = 5mm
Where:
pt  Tube pitch.
d 0  Outside diameter (mm).
As 
( p t  d 0 ) Ds l B
= 0.059605 m2
pt
Where:
As  Cross-flow area.
pt  Tube pitch.
d 0  Outside diameter (mm).
Ds  Shell diameter.
Gs 
m
= 27.61555 kg/m2s
As
Where:
Gs  Mass velocity.
As  Cross-flow area.
m  Mass flow rate in Kg/s.
de 
1.1 2
2
( pt  0.917 d 0 ) = 2.8402 mm
d0
Where:
d e  Equivalent diameter (mm).
d 0  Outside diameter (mm).
pt  Tube pitch.
Re 
Gs d e

= 52.28913
Where:
Re  Reynolds number.
d e  Equivalent diameter (mm).
Gs  Mass velocity.
  Fluid viscosity at the bulk fluid temperature, Ns/m2.
Choose 25% baffle cut jh = 0.42
hi 
k f j h Re(Pr) 0.33
di
= 5793.259 W/m2C
Where:
hi  Inside coefficient (W/m2 oC).
j h  Tube side heat transfer factor.
k f  Thermal conductivity of stream.
Pr  Prandtl number.
1
1
1



U 0 h0 hod
d 0 ln(
d0
)
di
2k w
Where:
U 0  The overall heat transfer coefficient.
hod  Outside coefficient (fouling factor).

d0 1 d0 1

0.007249
d i hid d i hi =
hid  Inside coefficient (fouling factor).
Uo = 137.9562
 L
Pt  N p 8 j f 
  d i
  


  w 
m
 u 2
 2.5 t
 2
= 76.12681 kpa
Where:
Pt  Tube- side pressure drop (N/m²) (pa).
N p  Number of tube -side passes.
u t  Tube-side velocity, m/s.
L  Length of one tube.
j f  Friction factor.
 w  Fluid viscosity at the wall.
  Fluid viscosity at the bulk fluid temperature, Ns/m2.
D
Ps  8 j f  s
 de
 L  u s 2
 
 l B  2
  



 w
0.14
= 32.54424 kpa
Where:
Ps  Shell-side pressure drop (N/m²) (pa).
j f  Friction factor.
L  Length of tube.
For carbon steel
t
Where:
t  Shell thickness (in).
Pri
 Cc = 5.476371 mm
SEj  0.6 P
P  Maximum allowable internal pressure (psig) = 40 psi
ri  Internal radius of shell before allowance corrosion is added (in) = 26.32306 inch
E j  Efficiency of joints = 0.85
S  Working stress (psi) = 13700 psi
Cc  Allowance for corrosion (in) = 0.125 inch
Specification sheet for heat exchanger ( E-101)
Equipment Name
Heat exchanger
Objective
To cool the feed of the reactor which is
contain hydrogen and nitro benzene
Equipment Number
E-101
Designer
Hussain Butaleb
Type
Shell and tube heat exchanger
Location
After heat exchanger (E-103)
Utility
cooling water
Material of Construction
Carbon steel
Operating Condition
Shell Side
Inlet temperature (oC)
5
Outlet temperature (oC)
170
Tube Side
Inlet temperature (oC)
357.3
Outlet temperature (oC)
178
Number of passes
8
Thickness (mm)
5.476371
Number of Tube Per Pass
23184.72
Number of Tubes
185477.7
Tube bundle Diameter (m)
1337.214
LMTD (oC)
181.0804
U (W/C.m2)
137.9562
Heat Exchanger Area (m2)
27.95521
Heat Exchangers:
Nomenclature
Symbol
Definition
T1
Inlet shell side fluid temperature (°C)
T2
Outlet shell side fluid temperature(°C)
t1
Inlet tube side fluid temperature (°C)
t2
Outlet tube side fluid temperature (°C)
µ
Fluid viscosity (m N s /m2)
kf
Thermal conductivity ( W/ m °C)
Cp
Mass heat capacity (kJ / Kg °C)
Р
Density of the fluid (Kg/ m3)
Q
Heat load (Kw)
∆Tlm
Log mean temperature difference (°C)
A
Area (m2)
U
Overall heat transfer coefficient (W/m2. °C)
do
Tube outside diameter (mm)
di
Tube inner diameter (mm)
Lt
Tube length
Re
Reynolds number
Pr
Prandtl number
Gs
Mass velocity (m/s)
lb
Baffle spacing (m)
T
Shell Thickness
∆Pt
Tube side pressure drop (N/m2)
Np
Number of tube side passes
Ej
Efficiency of joints
S
Working stress (psi)
Cc
Allowance for corrosion (in)
ri
Internal radius of shell
Calculation procedure
a. Define the duty: heat transfer rate, fluid flow rates, temperature.
b. Collect together the fluid physical properties required: density, viscosity,
c. Thermal conductivity.
d. Select a trail value for the overall coefficient, U.
e. Calculate the mean temperature difference, ΔTm.
f.
Calculate the area required from Q=UAΔTm.
g. Calculate the bundle and shell diameter
h. Calculate the individual coefficients.
i.
Calculate the overall coefficient and compare with the trail value.
j.
Calculate the exchanger pressure drop.
k. Calculate thickness of the shell.
l.
Find the price of the heat exchanger based on the heat transfer area and the material
of construction
Detailed calculation procedure
1- Heat load
Q = (m Cp ΔT) hot = (m Cp ΔT) cold, (kW)
2-Tube side flow
, (Kg/hr) mcold 
Qhot
C p Tcold
3- Log mean temperature
Tlm 
T2  T1
 T 
LN  2 
 T1 
, (°C)
T1  T1  t 2
T2  T2  t1
Where,
T1: is inlet shell side fluid temperature (°C)
T2: is outlet shell side fluid temperature (°C)
t1: is inlet tube side temperature (°C)
t2: is outlet tube side temperature (°C)
3-Calculate the mean (true) temperature ∆Tm
ΔTm= Ft * ΔTlm
For more than one tube passes
 (1  S ) 

( R 2  1) LN 
(1  RS ) 

Ft 
 2  S ( R  1  ( R 2  1) 

( R  1) LN 
 2  S ( R  1 ( R 2  1) 


R
(T1  T2 )
(t 2  t1 )
S
(t 2  t1 )
(T1  t1 )
Where,
Ft: is the temperature correction factor
R: is the shell side flow *specific heat / tube side flow*specific heat,
(Dimensionless).
S: is temperature efficiency of the heat exchanger, (dimensionless)
4- Provisional Area
A
Q
UTm
, (m2)
Where,
Area of one tube = Lt * do *π , (mm2)
Outer diameter (do), (mm)
Length of tube (Lt), (mm)
Number of tubes = provisional area / area of one tube
5- Bundle diameter
N 
, (mm) Db  d o  t 
 K1 
Where,
Db: bundle diameter, (mm)
Nt: number of tubes
K1, n1: constants.
1 / n1
6- Shell diameter
Ds = Db + (Bundle diameter clearance) , (mm)
Using split-ring floating head type (bundle).
From figure (A.12) we get bundle diameter clearance.
7-Tube side Coefficient
Cold stream mean temperature=
Tube cross sectional area =

4
t 2  t1
, (°C)
2
2
d i , (mm2)
Tubes per pass = no. of tubes / number of passes
Total flow area = tubes per pass * cross sectional area, (m2)
Mass velocity = mass flow rate / total flow area, (kg /sec.m2)
Linear velocity (ų) = mass velocity / density, (m/s)
Reynolds number (Re) =ρ ų di / μ
Prandtl number (Pr) = Cp μ / κ
(hi di / κ) = jh Re Pr0.33 * (μ/μwall)0.14
Using Fig. to find jh
8-Shell side Coefficient
Baffle spacing (Lb) = 0.2 * Ds, (mm)
Tube pitch (pt) = 1.25 * do, (mm)
Cross flow area (As) = (pt - do)* Ds* Lb / pt , (m2)
Mass velocity (Gs) = mass flow rate / cross flow area, (kg/s.m2)
Equivalent diameter for triangular arrangement (de) =1.1*(pt2-0.917do2) /do, (mm)
Mean shell side temperature = (Thi +Tho)/2, (°C)
Reynolds number (Re) = Gs de / μ
Prandtl number (Pr) = Cp μ / κ
And from fig find jh.
hs = K * jh *Re *Pr (1/3) / de , W/m2.°C
Overall heat transfer coefficient
d 
d o LN  o 
 1  1
1
 di   do

  

2K w
di
 U o  ho hod
 1

 hid
 do
 
 di
1
 
 hi 
,(W/m2.°C)
9- Pressure drop
Tube side

 u  2
 L / di 
  2.5 
, (KPa) Pt  N p 8 j f 
 M /Mw 

 2 
Where,
ΔPt: tube side pressure drop (N/m2= pa)
Np : number of tube side passes
u : tube side velocity (m/s)
L: length of one tube, (m)
Use the fig.(A.14)
Shell side
Linear velocity = Gs /р
D
p s  8 j f  s
 do
 L  u 2
 
 lb  2
 M

 M w



0.14
Where,
L: tube length, (m)
lb: baffle spacing(m)
Use fig.(12.30) to get jf.
10-Shell thickness
t
t: shell thickness (in)
P : internal pressure (psig)
ri: internal radius of shell (in)
EJ: efficiency of joints
S : working stress (psi)
Cc: allowance for corrosion (in)
Pri
 Cc
SE j  0.6 P
Sample Calculation:
Heat exchanger (E-103)
Shell side
Prameter
Unit
Inlet
Outlet
Mean
Tempreture Ti
C
30
177
103.5
k
W/m.C
1.37E-01
1.49E-02
0.075769
Mass Density ρ
kg/m3
1191.8
1.35E+00
596.5761
Viscosity μ
mPa.s
2.40E+00
7.54E-03
1.203118
Specfic Heat Cp
KJ/Kg.K
142.9
165.67
154.285
Heat Of Vaporization
KJ/kg
Mass Flow Rate
kg/s
Thermal Conductivty
1.92E+02
5.448889
Tube side
Prameter
Unit
Temperture ti
C
Thermal Conductvity
Inlet
Outlet
Mean
184.81
163.87
174.34
W/m.C
2.06E-01
0.19891
0.202635
Mass Density
kg/m3
0.2473
0.28409
0.265695
Viscosity
mPa.s
1.20E-02
1.14E-02
0.011703
Specfic Heat Cp
KJ/Kg.K
30.727
30.585
30.656
Mass Flow Rate
kg/s
0.249167
Q = (m Cp ΔT) hot = 1.05E+03 KW
T1
C
30
T2
C
177
t1
C
184.81
t2
C
163.87
Tlm 
T2  T1
 T 
LN  2 
 T1 
T lm= -44.36445101 °C
Using one shell pass and one tube passes
R
(T1  T2 )
(t 2  t1 )
R= 7.020057307
S
(t 2  t1 )
(T1  t1 )
S= 0.13526258
Using fig to find Ft
Ft= 1
= -44.364451 °C Tm  Ft * Tlm
assume U=160 W/m2°C
Provisional area
m2 = 147.3956 A 
Q
UTm
Choose,
Assume Outler diameter (do)
50 mm
Assume inside diameter (di)
4 mm
Assume Length of tubes (L)
10 m
Area of one tube = Lt * do *π = 0.019635 m2
Number of tubes Nt = provisional area / area of one tube= 7506.796 tube
As the shell – side fluid is relatively clean use 1.25 triangular pitch.
N 
Bundle diameter Db= d o  t 
 K1 
1 / n1
K1= 0.319
N1= 2.142
Db= 5494.096 mm
Use a split – ring floating head type.
Bundle diametrical clearance = 77 mm
Shell diameter, Ds = Db + bundle diametrical clearance
Ds= 5571.096 mm
Tube – side coefficient
  2  
2
d i =   * (26 E  3) = 1.26E-05 m2
4
4
Tube cross sectional area = 
Tubes per pass =
Nt
= 7506.796
1
Total flow area(area/pass) = tube per pass * cross sectional area
= 0.094333 m2
Linear velocity (ut) = mass velocity/density = 9.941288 m/s
The coefficient can be calculated from the following equation
  
hi d i

 j h Re Pr 0.33 
Kf
 w 
902.7943 Re 
ud i


1.770509 Pr 
Cp

Kf
From figure
jh= 3.00E-03
hs = 90.33455W/m2.C
Assume that the viscosity of the fluid is the same as at the wall

1
w
Shell - side coefficient
Choose baffle spacing Lb=
Ds
 1114.219 mm
5
Tube pitch (pt)=1.25*do= 62.5 mm
 ( pt  d o ) * Ds * Lb 
  1.241484 m 2
pt


Cross-flow area As= 


 1.1  2
 pt  0.917d o 2  0.035503 mm
 do 
Equivalent diameter de = 
Re = 129.5142
Pr = 486.7794
Choose 25% baffle cut.
From figure
jh= 7.00E-02
hs = 285.1998 W/m2.C
Overall heat transfer coefficient
d
d o LN  o
 1  1  1 
 di

     
 
2k w
 U o   ho   hod 
Uo= 143.8143 (W/m2 °C) Acceptable
Pressure drop:


   d o
d
 i
 1

 hid
 do
 
 di
1

 hi



Tube side
From figure for Re = 902.7943
jf= 3.00E-03
Neglecting the viscosity correction term

 u 2
 L / di 
  2.5
pt  N p 8 j f 
  / w 

 2



pt  1477.507 pa
Shell side
From figure for Re = 129.5142
jf= 7.00E-02
Neglecting the viscosity correction term
D
Pt  8 j f  s
 de
 Lt

 lb
 u 2

 2
 

  w
Pt  12.20782 pa
Shell thickness
P= 5.2752 psig
ri = 109.667 in
S= 13,700 psi
EJ =0.85
Cc = 0.125 in
t = 0.174693 in = 4.437197 mm



0.14
Heat Exchanger (E-103) specification sheet
E-103
Equipment Name
To heat the nitro benzene feed by the reactor
Objective
(R-101) product
Hussain Butaleb
Designer
Type
Shell and tube heat exchanger
Location
After heat exchanger (E-102)
Carbon steel
Material of Construction
Quartz wool – Glass wool
Insulation
Operating Condition
Shell Side
Inlet temperature (oC)
30
Outlet temperature (oC)
77
Tube Side
Inlet temperature (oC)
Number of passes
184.81
Outlet temperature (oC)
163.87
1
Number of Tubes
7506.796
Tube bundle Diameter (mm)
5571.096
Thickness (mm)
4.437197
Q total (W)
1.05E+06
LMTD (oC)
-44.3645
U (W/m2 oC)
143.8143
Heat Exchanger Area (m2)
147.3956
Sample Calculation:
Heat exchanger (E-102)
Shell side
Prameter
Unit
Inlet
Outlet
Mean
Tempreture Ti
C
311.64
184.81
248.225
k
W/m.C
2.47E-01
2.06E-01
0.226535
Mass Density ρ
kg/m3
0.21182
2.47E-01
0.22956
Viscosity μ
mPa.s
1.48E-02
1.20E-02
0.013364
Specfic Heat Cp
KJ/Kg.K
31.52
30.727
31.1235
Heat Of Vaporization
KJ/kg
Mass Flow Rate
kg/s
Thermal Conductivty
2.60E+02
5.449167
Tube side
Prameter
Unit
Temperture ti
C
Thermal Conductvity
Inlet
Outlet
Mean
114.06
249
181.53
W/m.C
1.94E-01
0.24214
0.218045
Mass Density
kg/m3
9.54E-02
7.07E-02
0.083072
Viscosity
mPa.s
1.04E-02
1.33E-02
0.011838
Specfic Heat Cp
KJ/Kg.K
29.045
29.525
29.285
Mass Flow Rate
kg/s
4.318056
Q = (m Cp ΔT) hot = 1.42E+03 KW
T1
C
311.64
T2
C
184.81
t1
C
114.06
t2
C
249
Tlm 
T2  T1
 T 
LN  2 
 T1 
T lm= 66.61273864 °C
Using one shell pass and one tube passes
R
(T1  T2 )
(t 2  t1 )
R= 0.939899214
S
(t 2  t1 )
(T1  t1 )
S= 0.682963863
Using fig to find Ft
Ft=0.85
= 56.62082784 °C Tm  Ft * Tlm
Assume U= 135 W/m2°C
Provisional area
= 185.5697 m2 A 
Q
UTm
Choose,
Assume Outler diameter (do)
25 mm
Assume inside diameter (di)
22 mm
Assume Length of tubes (L)
1 m
Area of one tube = Lt * do *π = 0.000491 m2
Number of tubes Nt = provisional area / area of one tube= 378039.5 tube
As the shell – side fluid is relatively clean use 1.25 triangular pitch.
N 
Bundle diameter Db= d o  t 
 K1 
K1= 0.319
1 / n1
N1= 2.148
Db= 16810.04 mm
Use a split – ring floating head type.
Bundle diametrical clearance = 78 mm
Shell diameter, Ds = Db + bundle diametrical clearance
Ds= 16888.04 mm
Tube – side coefficient
  2  
2
d i =   * (26 E  3) = 0.00038 m2
4
4
 
 
Tube cross sectional area = 
Tubes per pass = 378039.5
Total flow area(area/pass) = tube per pass * cross sectional area
= 143.7052 m2
Linear velocity (ut) = mass velocity/density = 0.361711 m/s
The coefficient can be calculated from the following equation
  
hi d i

 j h Re Pr 0.33 
Kf
 w 
55.84426 Re 
ud i


1.589860751 Pr 
Cp

Kf
Assume that the viscosity of the fluid is the same as at the wall

1
w
From figure
jh= 2.50E-02

hi  8.806477 (W / m 2 C )
Shell - side coefficient
Choose baffle spacing Lb=
Ds
 3377.608 mm
5
Tube pitch (pt)=1.25*do= 31.25 mm
 ( pt  d o ) * Ds * Lb 
  11.40824 m 2
pt


Cross-flow area As= 


 1.1  2
 pt  0.917d o 2  0.017751 mm
 do 
Equivalent diameter de = 
Re = 634.4598
Pr = 1.727612687
Choose 25% baffle cut.
From figure
jh= 2.00E-02
hs = 197.2752 W/m2.C
Overall heat transfer coefficient
d
d o LN  o
 1  1  1 
 di

     
 
2k w
 U o   ho   hod 


   d o
d
 i
Uo= 152.4811 (W/m2 °C) Acceptable
Pressure drop:
Tube side
From figure
jf= 2.50E-02
Neglecting the viscosity correction term
 1

 hid
 do
 
 di
1

 hi




 u 2
 L / di 
  2.5
pt  N p 8 j f 
  / w 

 2



pt  0.104043 pa
Shell side
From figure for Re = 634.4598
jf= 2.00E-02
Neglecting the viscosity correction term
D
Pt  8 j f  s
 de
 Lt

 lb
 u 2

 2
 

  w
Pt  40.31546 pa
Shell thickness
P= 14.7 psig
ri = 332.4411 in
S= 13,700 psi
EJ =0.85
Cc = 0.125 in
t = 0.544973 in = 13.84232 mm



0.14
Heat Exchanger (E-102) specification sheet
E-102
Equipment Name
to heat hydrogen feed from the reactor (R-101)
Objective
product
Designer
Hussain Butaleb
Type
Shell and tube
Location
After reactor (R-101)
Material of Construction
Carbon steel
Insulation
Quartz wool – Glass wool
Operating Condition
Shell Side
Inlet temperature (oC)
311.64
Outlet temperature (oC)
184.81
Inlet temperature (oC)
114.06
Outlet temperature (oC)
249
Number of passes
1
Number of Tubes
378039.5
Tube bundle Diameter (m)
16888.04
Thickness (mm)
13.84232
Q total (W)
1.42E+03
LMTD (oC)
56.62083
U (W/m2 oC)
152.4811
Heat Exchanger Area (m2)
185.5697
Tube Side
Steam Drum (V-102)
A steam drum is a standard feature of a water-tube boiler. It is a reservoir of water/steam at
the top end of the water tubes. The drum stores the steam generated in the water tubes and
acts as a phase-separator for the steam/water mixture. The difference in densities between hot
and cold water helps in the accumulation of the "hotter"-water/and saturated-steam into the
steam-drum.
Since the drum serves at high pressures and temperatures, it is expensive to manufacture and
there is considerable economic incentive to keep it as small as possible.
Two Phase Separator Design:
A vapor-liquid separator is a device in which a liquid and vapor mixture is fed and the liquid
is fall by gravity to the bottom of the vessel while the vapor travels upward to the top of the
vessel. These separators are used after flashing a hot liquid across a valve (flash drum) .The
most reason for using vapor-liquid separators is to recover valuable products.
Design Procedure:
1. State assumptions.
2. Calculate the settling velocity in m/s using the following equation:
U t  0.07
( L  V )
V
,
Where ρL is the liquid density (kg/m3) and ρv is the gas density (kg/m3).
3. Calculate the actual settling velocity in m/s by:
Ua=0.15Ut.
4. Calculate the volumetric flow rates for both the vapor ( Vv) and liquid (VL) in
m3/susing:
VV 
mV
VL 
mL
V
And
L
,
Where mv, L is the mass flow rate for vapor and liquid in kg/h respectively.
5. Get the cross-sectional area in terms of Dv (minimum vessel diameter, m):
Ac 
 Dv 2f v
4
,
Where fv is the fraction of the total cross-sectional area occupied by vapor which
equals 0.5.
6. Get an expression for the vapor residence time for the droplets to settle to liquid
surface in terms of Dv:
tr 
hv
,
Ua
Where hv is the liquid level (m), hv=0.5Dv.
7. Get an expression for the actual residence time in terms of Dv:
tra 
Lv
,
Uv
Where Lv is the vessel length (m), Lv=4Dv.
And Uv is the vapor velocity (m/s),
Uv 
Vv
.
Ac
8. Find the value of Dv by equalizing tr and tra. (Solve tr-tra=0). Then find the length of
the separator Lv=4Dv.
9. Calculate the thickness of the separator using the following equation:
t
Pri
Co ,
SE j  0.6P
Where P is the operating pressure in psig (P=405 psig), ri is the radius of the vessel (ri=Dv/2
(in)), S is the stress value of carbon steel (S=13700 psia), Ej is the joint efficiency (Ej=0.85 for
spot examined welding), and Cc is the corrosion allowance (Cc=1/8 in). These values were
obtained from the metals table (in the Appendix B).
Assumptions:
The horizontal separator is assumed not to have a demister pad.
Detailed Calculation:
Operating conditions:
T= 208 oC
P= 264 psig.
Design Consideration:
rL=
1220
kg/m3
r V=
8.197
kg/m3
mv= 20243.31 kg/h
mL=
24400
kg/h
Calculating the settling velocity:
( L  V )
U t  0.07
V
= 0.85111 m/s
Calculating the actual velocity:
Ua = 0.15Ut = 0.15* = 0.12767 m/s
Calculating both vapor and liquid volumetric flow rates:
o VV 
VL 
mL
L
mV
= 0.686 m3/s
V
= 0.00556 m3/s
Expressing the cross-sectional area in terms of Dv:
o
Ac 
 Dv 2f v
4

 * Dv 2 *0.5
4
= 0.393 Dv2 m2
Expressing the residence time in terms of Dv:
o
tr 
hv
= 3.916 Dv s
Ua
Expressing the actual residence time in terms of Dv:
Uv 
Vv
= 1.746Dv-2 m/s
Ac
Finding the value of Dv:
Lv
= 2.291 Dv3 s
Uv
o
tra 
o
tr-tra= Dv - Dv3 = 0
Dv= 1.3074 m
Lv=4Dv= 4*1.3074 = 5.2296 m
Calculating the thickness of the vessel:
Ej =0.85
S = 13700 psi
Cc = 0.125 in
ri = Dv/2 = 0.6537
o
t
Pri
Co
SE j  0.6P
= 0.7165 in = 0.0182 m
Time hold up = 14.5 hr
Volume = 9.946999 m3
Steam Drum Specification Sheet V-102
Steam Drum
Equipment Name
Separate the steam (upward) from the water
Objective
liquid (downward)
Equipment Number
V-102
Designer
Hussain Butaleb
Type
Horizontal Separator
Location
After heater (E-105)
Material of Construction
Carbon steel
Insulation
Glass wool
Operating Condition
Operating Temperature (oC)
208
Operating Pressure (psig)
264
Liquid Density (kg/m3)
1220
Gas Density (kg/m3)
8.197
Gas Flow rate (kg/h)
20243.31
Liquid Flow rate (kg/h)
24400
1.3074
Height (m)
5.2296
Design Considerations
Dimensions
Diameter (m)
Vessel (V-101)
A vapor-liquid separator is a device in which a liquid and vapor mixture is fed and the liquid
is fall by gravity to the bottom of the vessel while the vapor travels upward to the top of the
vessel. These separators are used after flashing a hot liquid across a valve (flash drum) .The
most reason for using vapor-liquid separators is to recover valuable products.
Vertical separators have the advantage of lower space requirement and easy to install control
systems.
Material of insulation depends on the operating temperatures, since temperature in the
separator is not high so the possible materials that cover the temperature are glass
fiber and mineral wool.
Design Procedures and Equations:
Properties
3342
Vapor flow rate (Mv)
Kg/h
Liquid flow rate (ML)
Kg/h
44000
Vapor density (ρv)
Kg/m3
0.20298
Liquid density (ρL)
Kg/m3
1014.4
Inlet Pressure (P)
psi
25
Max allowable working stress (S)
psi
13700
Efficiency expressed as a fraction (Ej)
0.85
Allowance for corrosion (Cc)
Steel density
0.125
Kg/m3
7700
1- To estimate the settling velocity of the liquid droplets:
1
ut = 0.07 X [
(ρl − ρv ) 2
ρv
] = 4.948033 m/s
ut = settling velocity, m/s
ρv = vapor density, kg/m3
ρl = liquid density, kg/m3
2- Volumetric flow rate:
Vv =
lv =
kg
)
s
kg
Vapor density ( 3)
m
= 4.573521 m3/s
kg
)
s
kg
liquid density ( 3)
m
= 0.012049 m3/s
Vapor flow rate (
liquid flow rate (
3- Volume held in vessel
VHV = 10 * 60 * Lv = 7.229232 (m3)
4- Minimum vessel diameter
Dv = [(4 * Vv ) / (pi * Us )]0.5 = 1.085111 (m) =42.72091 (in)
Dv = minimum vessel diameter, m
Vv = gas, or vapor volumetric flow rate, m3/s
5- Liquid depth
Hv = VHV / [(pi / 4) * (Dv )2 ]= 7.821212(m)
ri = Dv / 2= 3.99962 (m) = 0.542556 (in)
Thickness = Cc + [(P * ri ) / (S * Ej – 0.6 * P)]= 0.170917 (in) =0.004341 (m)
h = [3 * (Dv / 2)] + Dv + Hv + 0.4 = 10.93399 (m)
CC = allowance for corrosion, in.
Ej = efficiency of joints expressed as a fraction.
S = maximum allowable working stress, psi.
rj = inside radius of the shell, before corrosion allowance is added, in.
6- Area of vessel :
Area of vessel = 2*pi*(dv/2)*ht= 37.25483 (m2)
7- Metal
Volume of metal Vm = area of vessel*thickness = 0.161734 (m3)
Weight of metal Wm= Vm*Steel density= 1245.35 (kg)
Specification sheet for separator V-101
Separator
Equipment Name
To separate hydrogen (vapor) from the
Objective
aniline and water (liquid)
Equipment Number
V-101
Designer
Hussain Butaleb
Type
Vertical
Location
After cooler E-104
Material of Construction
Carbon Steel
Insulation
Glass wall and quartz
Operating Condition
Operating Temperature (oC)
40
Operating Pressure (psi)
25
Dimensions
Diameter (m)
1.085111
Height (m)
10.93399
Storage Tank (T-101)
A storage tank is a container, usually for holding liquids, sometimes for compressed gases
(gas tank). The term can be used for reservoirs (artificial lakes and ponds), and for
manufactured containers. Storage tanks are available in many shapes: vertical and horizontal
cylindrical; open top and closed top; flat bottom, cone bottom, slope bottom and dish bottom.
Large tanks tend to be vertical cylindrical, or to have rounded corners transition from vertical
side wall to bottom profile, to easier withstand hydraulic hydrostatically induced pressure of
contained liquid. Most container tanks for handling liquids during transportation are designed
to handle varying degrees of pressure.
The top space in the tank for the vapor pressure of the component will be 12 % of the tank
volume.
Design and Calculation:
For Tank (T-101)
1- Assume:

Cylindrical tank

Square shape from inside (D=0.2H)

Daily storage (time hold-up=12 hr)
2-Volume of the liquid:
Volume of the liquid = Total mass flow rate in* time hold-up = 190.8 m3
Where time hold-up is the time where the liquid is hold inside the tank =12 hours.
3- The total volume:
The total volume=volume of liquid/0.12 = 1590 m3
4-The actual diameter:
The actual diameter=(5*the total volume/π)^(1/3) = 13.62948 m
5- The actual high:
The actual high = 0.2* diameter = 2.725895 m
6-Area of the storage tank:
Area of the storage tank =V/H = 583.2946 m2
7- Thickness:
The best materiel is construction is carbon steel.
t = (P r i / S E – 0.6P) + Cc
P : internal pressure (kpa gage)
r i : internal radius of shell (m)
E : efficiency of joints = 0.85
S : working stress (kpa) = 94500
Cc : allowance for corrosion (m ) = 0.003175
ri(m) = 6.814738
t(m) = 0.01178
8- Vapor pressure:
Vapor pressure = Log10 P* = A- (B/C+T) = 760 mmhg
Specification sheet of Tank T-101
Equipment Name
T-101
Objective
NitroBenzene feed tank
Designer
Hussain Butaleb
Location
The first equipment
Material of Construction
Carbon steel
Insulation
Quartz Wool
Operating Condition
Operating Temperature (oC)
30
Diameter (m)
13.62948
Operating Pressure (Pisa)
14.7
Height (m)
2.725895
Feed Flow Rate (kmole/h)
Vapor Pressure (atm)
127.0822199 Thickness (m)
1
Total Area (m2)
0.01178
583.2946
Hydro cyclone (T-102)
Hydrocyclones, also known as liquid cyclones, are an important device for the separation of
solid-liquid suspensions. The principle employed is centrifugal sedimentation, i.e., the
particles in the suspension are subjected to centrifugal forces, which cause their separation
from the fluid. Like centrifuges, which make use of the same principle, hydrocyclones do not
have moving parts, require a low installation and maintenance investment and are simple to
operate. Hence, these devices are widely utilized in mineral, chemical, petrochemical, textile
and metallurgical industries.
The hydrocyclone separates solid and liquid or liquid and liquid by the difference in density
between the fluid and the material to be separated in this equipment. Due to the fluid acquires
a spiraling motion caused by the tangent feeding, the material of larger density is thrown
against the wall of the hydrocyclone and dragged to the underflow while the one of smaller
density proceeds for the overflow, forming a free vortex (outer vortex) and a forced vortex
(inner vortex) in agreement with Figure …………
The typical proportions of the hydrocyclone are shown in figure ………
Design equations:
𝑑50 = 4.5 [
𝐷𝑐 3 𝜇
]
𝐿1.2 (𝜌𝑠 − 𝜌𝐿 )
Where
𝑑50 = 𝑡ℎ𝑒 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟 𝑓𝑜𝑟 𝑤ℎ𝑖𝑐ℎ 𝑡ℎ𝑒 𝑐𝑦𝑐𝑙𝑜𝑛𝑒 𝑖𝑠 50% 𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡, 𝜇𝑚
𝐷𝑐 = 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟 𝑜𝑓 𝑡ℎ𝑒 𝑐𝑦𝑐𝑙𝑜𝑛𝑒 𝑐ℎ𝑎𝑚𝑝𝑒𝑟, 𝑐𝑚
𝜇 = 𝑣𝑖𝑠𝑐𝑜𝑠𝑖𝑡𝑦, 𝑐𝑒𝑛𝑡𝑖𝑝𝑜𝑖𝑠𝑒 (𝑚𝑁
𝑠
)
𝑚2
𝐿 = 𝑓𝑒𝑒𝑑 𝑓𝑙𝑜𝑤 𝑟𝑎𝑡𝑒 𝑙/𝑚𝑖𝑛
𝜌𝐿 = 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑓 𝑡ℎ𝑒 𝑙𝑖𝑞𝑢𝑖𝑑, 𝑔/𝑐𝑚3
𝜌𝑠 = 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑜𝑙𝑖𝑑, 𝑔/𝑐𝑚3
Thickness:
The best materiel is construction is carbon steel.
t = (P r i / S E – 0.6P) + Cc
P : internal pressure (kpa gage)
r i : internal radius of shell (m)
E : efficiency of joints
S : working stress (kpa)
Cc : allowance for corrosion (m
ri(m)
t(m)
Area (m2) = 3.14*r2*h
Volume (m3) = area*h /3
Tank (T-102)
T-102 is a simple solid separator used to separate solids from the mixture. Hydrocyclone was
chosen because it is a simple separator that can be used over the particle size range from 4 to
500 𝜇𝑚.
The efficiency of the hydrocyclone is assumed to be 95% and according to equation ……
Where
𝜇 = 7 𝑐𝑃
𝐿 = 295.18
𝐿
𝑚𝑖𝑛
𝜌𝑠𝑜𝑙𝑖𝑑 = 1.601808
𝜌𝑎𝑖𝑟 = 1.2344
𝑔
𝑐𝑚3
𝑔
𝑐𝑚3
𝐷𝑐 3 (7)
50 = 4.5 [
]
(295.18)1.2 (1.601808 − 1.2344)
The chamber diameter of the hydrocyclone Dc = 8.127753 cm
Height = 1.6 m
The best materiel is construction is carbon steel.
t = (P r i / S E – 0.6P) + Cc
P : internal pressure (kpa gage)
r i : internal radius of shell (m)
E : efficiency of joints = 0.85
S : working stress (kpa) = 94500
Cc : allowance for corrosion (m ) = 0.003175
ri(m) = 0.004064
t(m) = 0.003189
Area = 82.97183 m2
Volume = 44.25164 m3
Specification sheet for hydro cyclone (T-102)
T-102
Equipment Name
To separate the air from the solid waste
Objective
catalyst
Designer
Hussain Butaleb
Location
After the reactor (R-101)
Material of Construction
Carbon steel
Operating Condition
Volumetric flow rate (Air)
161.887
Chamber Diameter Dc (cm)
8.127753
Volumetric flow rate (Solid)
133.2925
Thickness (cm)
3.189
Distillation Column and Stripper
Distillation is probably the most widely used separation process in chemical and allied
industries. The separation of liquid mixtures by distillation depends on differences in
volatility between the components. The greater the relative volatilities, the easier the
separation. The basic equipments required for continuous distillation are a reflex column, a
condenser, and a reboiler. Vapor flows up the column and liquid counter-currently down the
column. The vapor and liquid are brought into contact on plates or packing. Part of the
condensate from the condenser is reflex back to the top of the column, while part of the
bottom liquid is vaporized in the reboiler and returned to provide the vapor flow.
In the section below the feed , the more volatile component and stripped from the liquid and
this is known as stropping section .Above the feed, the concentration of the more volatile
components is increased and this is called enrichment ,or more commonly, the recifying
section.
Stripping works on the basis of mass transfer. The idea is to make the conditions favorable for
the component, A, in the liquid phase to transfer to the vapor phase. This involves a gas-liquid
interface that A must cross. The total amount of A that has moved across this boundary can be
defined as the flux of A, NA.
Stripping is mainly conducted in trayed towers (plate columns) and packed columns, and less
often in spray towers, bubble columns, and centrifugal contactors.
The Different Types of Distillation
There are several methods of distillation depending on the procedure and the instrument
setup. Each distillation type is used for purification of compounds having different properties.
Following are the common types.
Simple Distillation
Simple distillation is practiced for a mixture in which the boiling point of the components
differ by at least 70° C. It is also followed for the mixtures contaminated with nonvolatile
particles (solid or oil) and those that are nearly pure with less than 10 percent contamination.
Double distillation is the process of repeating distillation on the collected liquid in order to
enhance the purity of the separated compounds.
Fractional Distillation
Those mixtures, in which the volatility of the components is nearly similar or differs by 25° C
(at 1 atmosphere pressure), cannot be separated by simple distillation. In such cases,
fractional distillation is used whereby the constituents are separated by a fractionating
column. In the fractionating column, the plates are arranged and the compound with the least
boiling point are collected at the top while those with higher boiling point are present at the
bottom. A series of compounds are separated simultaneously one after another. Fractional
distillation is used for the alcohol purification and gasoline purification in petroleum refining
industries.
Steam Distillation
Steam distillation is used for the purification of mixtures, in which the components are
temperature or heat sensitive; for example, organic compounds. In the instrument setup, steam
is introduced by heating water, which allows the compounds to boil at a lower temperature.
This way, the temperature sensitive compounds are separated before decomposition. The
vapors are collected and condensed in the same way as other distillation types. The resultant
liquid consists of two phases, water and compound, which is then purified by using simple
distillation. Steam distillation is practiced for the large-scale separation of essential oils and
perfumes.
Vacuum Distillation
Vacuum distillation is a special method of separating compounds at pressure lower than the
standard atmospheric pressure. Under this condition, the compounds boil below their normal
boiling temperature. Hence, vacuum distillation is best suited for separation of compounds
with higher boiling points (more than 200°C), which tend to decompose at their boiling
temperature. Vacuum distillation can be conducted without heating the mixture, as usually
followed in other distillation types. For the separation of some aromatic compounds, vacuum
distillation is used along with steam distillation.
Short Path Distillation
Thermal sensitive compounds can also be separated by following short path distillation. In
this technique, the separated compounds are condensed immediately without traveling the
condenser. The condenser is configured in a vertical manner between the heating flask and the
collecting flask. Similar to vacuum type, the pressure is maintained below the atmospheric
pressure. Short path distillation is used for the separation of organic compounds with high
molecular weight, especially in the pharmaceutical industries.
Another method of classifying distillation is based on the column type used in the process.
There are two types of distillation columns namely, batch and continuous.
Columns for the distillation process can be of the following types:
1. The 'PACKED' Tower.
2. The 'TRAY' Tower.
1. THE PACKED TOWER
As its name implies, the packed tower is a vertical, steel column which contains 'Beds' of
packing material which are used to bring the rising vapours into intimate contact with falling
liquid within the tower. The heat added to the mixture before entering the tower partially
vaporises the mixture and the vapours rise up the tower and begin to cool.
The liquid falls towards the bottom of the tower. At the tower bottom, in general, more heat is
added to the liquid by a 'Reboiler' which may be steam heated or a fuel fired furnace type.
The addition of heat here causes more vapours to rise up the column. As the two phases of the
mixture - falling liquid and rising vapour - come together, light components are stripped out
of the liquid and enter the gas phase while heavy components in the vapour are condensed
into the liquid phase.
In this way, as the vapour rises and gradually cools, it becomes lighter and, as the liquid falls,
it becomes hotter and heavier.
With this type of distillation column there is generally only a top and bottom product. The
quality of the products depends upon the height of the tower, the number of contacting
devices, the tower temperature and pressure and their control, and the velocity of the rising
vapours.
The type of packing materials used, also plays a part in the separation process.
Distillation Packing types:
2. THE TRAY TYPE TOWER
This is also a tall, cylindrical column. Inside, a series of trays are placed, one above the other.
The trays are used to bring the rising vapour and falling liquid into intimate contact. Tray
towers do the same job as packed towers but they are very much more efficient in the
separation process than packed towers and, they are also more costly. There are various types
of tray in use and the type selected depends upon the degree of product purity required, the
type of fluids, fluid velocity and other process parameters of the system.
Distillation Trays:
Sieve or Perforated Trays
Sieve trays are made from a flat perforated plate which allows the passage of vapor through
the liquid. They are the most economical tray option when low turndown is required. They
have better anti-fouling characteristics and lower pressure drop than valve or bubble cap
trays. Perforations are typically 1/2” diameter, but Ambani Metals can provide designs with
smaller hole size.
Valve Trays
Ambani Metals valve trays have better turndown and slightly higher efficiency than sieve
trays. Ambani Metals offers different valve selections including fixed valves, floating valves
and combination valves. Valve trays cost more than sieve trays, but are more economical than
bubble cap trays.
One Piece Valve
This is the most commonly used valve. This design features integral legs for tray decks up to
1/4” thickness. Anti-stick dimples are standard. Other options include heavy/light valve
combination, flush designs and non-rotating tabs in the tray deck.
3 Piece Valve
This valve consists of a light weight orifice plate, a valve, and a restraining cage. This design
is recommended for higher turndowns.
Fixed Valve
This valve is integral with the tray deck. This is the preferred option for fouling conditions.
However, it provides lower turndown and less efficiency than floating valves.
One Piece Rectangular Valve
<
Ambani Metals offers rectangular valves, caged valves, and also venturi type openings to
provide lower pressure drop.
Bubble Cap Trays
Bubble cap trays are best suited for applications with low liquid flows and/or high turndown
ratios. In terms of capacity, however, they are slightly lower than valve or sieve trays. They
are also the most expensive tray option.
Dual Flow Trays
Dual flow trays are sieve trays that do not have downcomers. The term dual flow comes from
the countercurrent flow of the vapor and liquid through the perforations. Typical perforation
sizes range between 1/2” and 1” in diameter. Dual flow trays best suit systems containing a
moderate to high solids content or polymerizable compounds. High open area dual flow trays
have a higher capacity and lower pressure drop than comparably spaced fractionation trays.
However, their primary drawback is their narrow operating range. Most often, they are
efficient when used in smaller tower diameters. Dual flow trays are also sensitive to levelness
and may be subject to gross liquid and vapor flow partitioning through the deck if not level.
Baffle Trays
Because of their open design, baffle trays are used in applications requiring high capacity,
fouling resistance and low pressure drop. Vapor-liquid contacting takes place when the vapor
passes through a curtain of liquid falling between trays, or through rivulets of liquid flowing
through tray deck perforations. Tray decks may be level or slightly inclined and typically
occupy 40-60% of the tower cross-sectional area. “Disk and donut” trays have circular baffles
and are a popular variant of this deign.
Baffle trays are well suited for heattransfer applications including heavy oil refining and
petrochemical oil refining and petrochemical heat transfer services with high solids or
petroleum coke content.
High Strength Trays
For applications with potentially damaging uplift surges, Ambani Metals can equip trays with
special heavy duty features. These include special fasteners, increased tray thickness or
additional support beams. Depending upon specific operating criteria, design adjustments can
be made to take into account parameters such as corrosion, temperature, vibration and
pressure surges. Contact your Ambani Metals separation specialist for design assistance.
Design and equations for distillation:
Nomenclature
Symbol
Definition
FLv
Liquid vapor flow factor
Lw
Liquid mass flow rate (kg/s)
Vw
vapor mass flow rate (kg/s)
ρv
Vapor density (kg/m 3)
ρL
Liquid density (kg/m 3)
uf
flooding vapor velocity (m/s)
u`v
flooding at maximum flow rate (kg/s)
Ac
Total column cross sectional area (m2)
Dc
Column diameter (m)
Ad
cross sectional area of down comer (m2)
An
Net area (m2)
Aa
Active area (m2)
Ah
Hole area (m2)
Aap
Clearance area (m2)
Ap
Perforated area (m2)
how
Weir crest (mm) liquid
u`h
Min. vapor velocity (m/s)
hd
Dry plate drop (mm)
hr
Residual head (mm)
hap
Out let weir height (mm)
hdc
Head loss in downcomer (mm)
T
thickness of cylindrical shell (in)
P
maximum allowable internal pressure (psi)
S
maximum allowable working stress (psi)
Ri
: inside radius of shell (in)
Ej
efficiency of joint expressed as fraction
Cc
allowance for corrosion (in)
Design procedures:
1. Collect, or estimate, the system physical properties.
2. Select trial plate spacing.
3. Calculate Nmin.
4. Calculate the height of the column.
5. Calculate the column diameter based on flooding consideration.
6. Make a trial plate layout: downcomer area, active area, hole area, hole size, weir
height.
7. Calculate the weeping rate.
8. Calculate the plate pressure drop.
9. Calculate downcomer back-up.
10. Calculate residence time.
11. Calculate heat transfer area for the condenser and reboiler.
12. Calculate thickness.
13. Calculate cost.
Equations:
Lquid vapor flow rate:
For top and bottom:
FLV
L

V
 v

 L



0.5
Where:
FLV= liquid vapor flow rate.
L= liquid flow rate.
V= vapor flow rate.
Find K1 from figure:
Flooding velocity:
 (   V ) 

U f  K 1  L
V


0.5
Where:
Uf= flooding velocity.
K1= constant.
Actual velocity:
U V  Percentage Flooding x U f
Where:
Uv= actual velocity.
Maximum volumetric flow rate:
Vmax 
MwtV
V
Where:
Vmax= maximum volumetric flow rate.
MwtV= vapor molecular weight.
Nmin:
Nmin = (log(xd/1-xd)*(xw/1-xw)) /(log volatility)
Net area required:
Anet 
Vmax
UV
Anet= net area required.
Diameter:
4

D   Anet 


0.5
Where:
D = diameter.
Max volumetric liquid rate:
max volumetric liquid rate 
Where:
L=liquid flow rate.
Choose plat pass from figure
Column area:
AC 

4
D2
Where:
AC= column area.
Net area:
An  AC  Ad
LxMwt
L
Where:
An= net area.
Active area:
Aa  Ac  2 Ad
Aa= active area.
Hole area:
Ah  0.1xAa
Ah= hole area.
From figure get lw/Dc ;
Weir length:
weir length  0.75D
Wier crest:
 max liquid rate 

max how  750
  L xweir length 
2
3
2
 min liquid rate  3

min how  750
  L xweir length 
Where:
how= weir crest.
Vapor velocity:
U h (min) 
K 2  0.9(25.4  hole diameter)
Where:
Uh= vapor velocity.
Choose Turndown percentage = 70%
From figure get K2
Actual min. vapor velocity:
 0.5
actual min . vapor velocity 
min . vapor rate
Ah
Where:
Ah= hole area.
From figure get Co
Pressure drop through dry plate:
U
hd  51 h
 Co



2
 V

 L
Where:
hd= pressure drop through dry plate.
Co= orifice coefficient.
Residual head:
hr 
Where:
12.5 x10 3
L



hr= residual head.
Height of bottom edge of apron above plate:
hap  hw  10
Where:
hap= height of bottom edge of apron above plate.
hw= weir height.
Area under apron Aap :
area under apron Aap  weir lengthxhap
Back-up in downcomer:
 max . liquid rate 

hdc  166



xA
L
ap


hb  hw  hdc  ht  how
Where:
hb= back-up in downcomer.
Downcomer residence time:
tr 
Where:
tr= downcomer residence time.
Lwd= minimum liquid flow rate.
hb xAdx L
Lwd
2
UV 
volumetric flow rate
An
Thickness:
Pxri


  CC
t  
 SxEj  0.6 xP 
Where:
t= thickness.
P= pressure.
r= radius.
S= working stress.
Ej= efficiency of joints.
CC= allowance for corrosion.
Design for distillation:
Properties:
Top properties
Vapor rate (V) =
ρv
=
MW =
Liquid rate (L) =
ρL
=
MW =
Surface Tension =
2.91E+03 kmol/hr
3.28769 kg/m3
107.2743
2.42E+03 kmol/hr
1050.493 kg/m3
122.5036
4.74E+01 N/m
Bottom properties
Vapor rate (Vm) =
ρv
=
MW =
Liquid rate (Lm) =
ρL
=
MW =
Surface Tension =
2.90E+03 kmol/hr
3.906938 kg/m3
1.23E+02
2.91E+03 kmol/hr
981.5581 kg/m3
123.0347
3.77E+01 N/m
Calculation:
Number of Stages
(No. of stages)min =
5
Efficiency = 75% (Assumed)
Actual stages =
6
Liquid vapor flow rate:
For top:
 v

 L



0.5
FLV
L

V
 v

 L



0.5
FLV
L

V
= 0.046591
For bottom:
Where:
FLV= liquid vapor flow rate.
L= liquid flow rate.
V= vapor flow rate.
Take tray spacing 0.75 m
Find K1 from figure:
Bottom K1 =1.00E-01
Top K1 = 1.5E-01
Flooding velocity:
For top:
= 0.063308
0.5
 (   V ) 
 = 12.66702 m/s
U f  K 1  L

V


0.5
 (   V ) 
 = 7.147454 m/s
U f  K 1  L

V


Where:
K1= constant.
Design for 85% flooding at maximum flow rate
Actual velocity:
For top:
U V  Percentage Flooding x U f = 10.76697 m/s
For bottom:
U V  Percentage Flooding x U f = 6.075336 m/s
Maximum volumetric flow rate:
For top:
Vmax 
MwtV
= 3.26E+01 m3/s
Vmax 
MwtV
= 1.84E+01 m3/s
V
For bottom:
Where:
MwtV= vapor molecular weight.
Xw=
0.990221
Xd=
0.724112
Volatility=
0.45
Nmin=
4.574458
V
Net area required:
For top:
Anet 
Vmax
= 3.440657 m2
UV
Anet 
Vmax
= 3.437593 m2
UV
For bottom:
Take downcomer area as 12% of total
Diameter:
4

D   Anet 


0.5
= 2.09 m
Max volumetric liquid rate:
max volumetric liquid rate 
LxMwt
L
= 0.101322 m3/h
Where:
L=liquid flow rate.
Column area:
AC 

4
D 2 = 3.440759 m2
Net area:
An  AC  Ad = 3.027868 m2
Active area:
Aa  Ac  2 Ad = 2.614977 m2
Hole area:
Ah  0.1xAa = 0.261498 m2
Taking
Weir height (hw) =
50 mm
Hole diameter (dh) =
5 mm
Plate thickness =
5 mm
Weir length:
weir length  0.75D = 1.611658 m
Wier crest:
2
max how
 max liquid rate  3
 = 118.5807 mm liquid
 750
  L xweir length 
 min liquid rate 

min how  750
  L xweir length 
2
3
= 93.48587 mm liquid
Vapor velocity:
U h (min) 
K 2  0.9(25.4  hole diameter)
 0.5
= 6.900747 m/s
Actual min. vapor velocity:
actual min . vapor velocity 
min . vapor rate
= 49.19689 m/s
Ah
Where:
Ah= hole area.
Pressure drop through dry plate:
U
hd  51 h
 Co



2
 V

 L

 = 1648.089 mm liquid

Where:
hd= pressure drop through dry plate.
Co= orifice coefficient.
Residual head:
hr 
12.5 x10 3
L
= 12.73485 mm liquid
Height of bottom edge of apron above plate:
hap  hw  10 = 40 mm liquid
Where:
hap= height of bottom edge of apron above plate.
hw= weir height.
Area under apron Aap :
area under apron Aap  weir lengthxhap = 0.064466 m2
Back-up in downcomer:
2
 max . liquid rate 
 = 410.0589 mm
hdc  166



xA
L
ap


hb  hw  hdc  ht  how = 2.408044 m
Downcomer residence time:
tr 
hb xAdx L
= 9.813 s
Lwd
Where:
tr= downcomer residence time.
Lwd= minimum liquid flow rate.
UV 
volumetric flow rate
= 6.070
An
Thickness:
Pxri


  CC = 5.0 mm
t  
 SxEj  0.6 xP 
Where:
For carbon steel
t= thickness.
P= pressure.
r= radius.
S= working stress.
Ej= efficiency of joints.
CC= allowance for corrosion.
ri =
41.20204 in
P=
20 psi
S=
13700 psi
Ej =
0.85
Cc =
1.25E-01 in
(for carbon steel)
Specification sheet of distillation (C-101)
Distillation
Equipment Name
Objective
To mix hydrogen gas with nitrobenzene gas
C-101
Equipment Number
Designer
Hussain Butaleb
Location
After cooler (E-101)
Material of Construction
Carbon steel
Insulation
Mineral wool
Operating Condition
Operating Temperature (oC)
140.9
Operating Pressure (psi)
30
Feed Flow Rate (kgmol/h)
497.520473339072
Diameter (m)
2.09
Height (m)
6.6
Thickness (mm)
5
Stripper (C-2101)
Typical applications:
Stripping is commonly used in industrial applications to remove harmful contaminants from
waste streams. One example is decontaminating soils almost completely.
Steam is also frequently used as a stripping agent for water treatment. Volatile organic
compounds are partially soluble in water and because of environmental considerations and
regulations, must be removed from groundwater, surface water, and wastewater. These
compounds can be present because of industrial, agricultural, and commercial activity.
Packed column:
There are a number of advantages to the packed column design. Improvements can yield
proof of about 190, and the still can be run either continuously or on a batch basis. On a small
scale, packed columns are inexpensive to build and quite easy to operate. However, on a large
scale the design presents problems.
Design of Stripper Columns:
1- Estimate the slope m from equilibrium data of process.
2- Calculate the amount mGm/Lm, and y1/y2 to determine NOG from figure.
Assume 38mm and 1.5 ceramic interlocks saddles.
3- Calculate Flow of liquid vapor:
FLV
Lw

Vw
V
L
4- Design for pressure drop 125mmH2O/m packing and get K4 from the figure.
5- Calculate percentage flooding = (K4 / K4 @flooding )^.5 *100
6- Calculate Vmin and V actual:
Vmin = L(x1-x2)/m(y1-y2)
V actual = Vmin *1.5
7- Calculate gas mass flow rate per unit column cross sectional area.
8- Calculate column area required in m² = gas flow rate /V*w
9- Calculate diameter and round of diameter and approximate area.
10- Estimate packing size to column diameter ratio.
11- Estimating HOG and height of the column using figures and equations:
12- Get HOG = Hg + ( HL * mGm/Lm), and z = HOG * NOG.
Stripper (C-201)
(Packed)
Feed Property
Gas
Liquid
Flow rate (Kg/h)
2.28E+03
6.47E+03
Flow rate (Kgmole/h)
1.17E+02
3.59E+02
Density р (Kg/m3)
0.648029375
947.674435
Viscosity µ (N s/m2)
9.27E-03
2.77E-01
Molecular weight (g/mol)
19.44657143
18.0370529
Surface tension (N/m)
-
58.4970599
m slop of the equilibrium = 0.0625
m Gm/Lm = 2.04E-02
y1/y2 = 32.30769231
From fig:
NOG = 0.5
Column Diameter
Gas Flow rate = 6.33E-01 Kg/s
Liquid Flow rate = 1.797301255 Kg/s
Select 38mm (1 1/2) in ceramic Intalox saddles
From Table:
Fp = 35 m-1
FLV = 0.074209109
Design for pressure drop 125mm H2O/m packing
From Figure:
K4 = 2.2
At flooding K4 = 4
Percentage Flooding = 74.16198487 %
L (kg/h)
x1
x2
y1
y2
y2'
m
7101
0.018787
0.006161
0
0.99975
0.000003
0.0625
Vmin
V actual = Vmin *1.5
1434.874335
2152.311502
V*w = 2.577539424 Kg/m2.s
Column Area required = 2.46E-01 m2
Actual Diameter = 1 m
Column Area = 0.78525 m2
Packing size to column diameter ratio = 26.3157895
Large packing size can be considered
Percentage of flooding for selected diameter = 2.32E+01
Estimate for HOG
DL= 1.8E-09 m2/s
DV= 0.00000152 m2/s
(Sc)v = 9406.149061 μ /ρ*DL
(Sc)L = 162563.2521 μ /ρ*DL
L*w = 2.28882681
From Figure 11.41
K3=
1
From Figure 11.42
ψh=
24
From Figure 11.43
φh=
1.00E-01
HOG can be expected to be around = 5 m
Estimate Z(height) = 2.5 m
Diameter Correction =1 m
f1=f2=f3 =1
HG = 36.4529333 m
HL = 11.93595174 m
HOG = 3.67E+01 m
Z = 18.3 m
Specification sheet of stripper (C-201)
Stripper
Equipment Name
Objective
To separate water and aniline (vapor) from
waste water (liquid)
C-201
Equipment Number
Designer
Hussain Butaleb
Location
After aniline-water tank (T-202)
Material of Construction
Carbon steel
Insulation
Mineral wool
Operating Condition
Operating Temperature (oC)
100.1
Operating Pressure (psi)
15
Feed Flow Rate (kgmol/h)
475.9659
Diameter (m)
1
Height (m)
18.3
HOG
36.7
NOG
0.5