Table of contents:
# Topic
N. of Pages
1 Literature survey
1
2 Hysys process description
26
3 Equipment design
72
4 Environmental, safety and economics analysis
427
Kuwait University
College of Engineering & Petroleum
CHEMICAL ENGINEERING DEPARTMENT
Literature survey
Production of Aniline
[by vapor phase catalytic reduction of nitrobenzene]
Abdulaziz Alshomer
208113942
YousifAlmuneefi
AbdulhameedAlawadhi
Hasanbu-Taleb
Hussainbu-Taleb
208112671
208115609
207216952
207217011
Instructed by:
Dr.Faheem
9/23/2012
1
Table of Content:
#
Topic
Pages
1
List of Tables
3
2
List of Figures
5
3
Introduction
5
4
History
6
5
World Production and Consumption
6
Uses
7
Feedstock and product Description
10-20
8
Process Flow Sheet
21-29
9
Comparison flow sheets
30
10 Basic economic analysis
31
11 Conclusion
32
12 Recommended flow sheet
32
13 References
33
7-9
9
2
List of Tables:
#
Topic
Pages
Table (1-1)
Aniline Global supply and demand
7
Table (1-2)
Aniline supply and demand in U.S
7
Table (1-3)
Aniline supply and demand in china
8
Table (1-4)
Aniline’s china consumption capacity
9
Table (1-5)
Aniline Uses
9
Table (1-6)
Physical and chemical properties of Benzen
10
Table (1-7)
Physical and chemical properties of Nitrobenze
11
Table (1-8)
Physical and chemical properties of ammonia
12
Table (1-9)
Physical and chemical properties of phenol
12
Table (1-10)
Physical and chemical properties of Hydrogen
13
Table (1-11)
Physical and chemical properties of water
14
Table (1-12)
Physical and chemical properties of Nitric Acid
15
Table (1-13)
Physical and chemical properties of sulfuric acid
16
Table (1-14)
Physical and chemical properties of iron
17
Table (1-15)
Physical and chemical properties of Hydrochloric
Acid
18
Table (1-16)
Physical and chemical properties of ferric chloride
19
Table (1-17)
Physical properties of aniline
20
Table (1-18)
Name and description unit
26
Table (1-19)
Feedstock condition
26
Table (1-20)
Product condition
26
Table (1-21)
Name of stream
27
Table (1-22)
Alternative #1
30
Table (1-23)
Alternative #3
30
Table (1-24)
Comparison between the alternatives
30
Table (1-25)
Cost of materials
31
3
List of Figures:
#
Topic
Pages
Figure (1-1)
Structure of Aniline
5
Figure (1-2)
World consumption of Aniline-2010
8
Figure (1-3)
Process flow sheet of the production of aniline by
Ammonolysis of Phenol
21
Figure (1-4)
Process flow sheet of the production of aniline by
liquid phase reduction of nitrobenzene
23
Figure (1-5)
Process flow sheet of the production of aniline by
vapor phase reduction of nitro benzene
25
4
Introduction:
Aniline, phenylamine or aminobenzene is a colorless organic compound
(aromatic amine) which has the formula C6H5NH2. It contains two groups
(phenyl group and amino group) linked together.
Aniline is toxic by inhalation of the vapor and by skin absorption.It is
flammable liquid that is slightly soluble in water and soluble in alcohol and
ether.
Aniline is a primary amine in which one of hydrogen atoms in the ammonia
molecule is exchanged by the phenyl group. The simplest way to write the
structure of aniline is:
Figure(1-1) Structure of Aniline
It is a main material that is used in chemical industries; also it has a fish
rotten like smell.
History:
Aniline was first isolated from the destructive distillation of indigo in 1826
by crystalline. In 1834, Friedrich Rungeisolated a substance from coal tar
which produced a beautiful blue color on treatment with chloride of lime,
which he named kyanol or cyanol.
In 1841, C. J. Fritzsche showed that by treating indigo with caustic potash it
producesoil, which he named aniline.
5
About the same time N. N. Zinin found that on reducing nitrobenzene, a base
was formed which he named benzidam.
August Wilhelm von Hofmann investigated these variously prepared
substances, and proved them to be identical (1855), and since then they took
their place as one body, under the name aniline or phenylamine.
The first industrial-scale use was in the manufacture of mauveine, a purple
dye discovered in 1856 by William Henry Perkin.
One of aniline derivatives called (p-Toluidine), can be used in qualitative
analysis to prepare carboxylic acid derivatives.
Developments in medicine
In the late 19th century, aniline emerged as an analgesic drug, its cardiacsuppressive side effects countered with caffeine.
World production& Consumption of Aniline:
* World production:
MDI [methylene diphenyldisocyanate] has been the driving force behind the
recovery of the aniline business since 1982 when the industry had acapacity
utilization rate of less than 50%. By 1996‚ capacity utilization
hadapproached 95% in some regions.
Thecapacity ofworld production ofanilinein 1999 was found in these
regions: Western Europe - 47%‚ North America -30% and Asia / Pacific 19%.
million
ton/year
1.5
2.2
2.9
2.97
4
Year
1988
1996
1996
2000
2010
Table (1-1) Aniline Global supply and demand1
6
Historicalproduction in the United States is summarized in Table 1-1.
Millions of Pounds
Type
1995
1996
1997
1998
1999
2000
2001
Capacity
1380
1420
1535
1745
1745
2310
2310
Total prod.
1388
1395
1339
1545
1588
1866
1975
Imports
-
24
60
42
26
11
12
Exports
67
41
19
43
38
58
65
Demand
1321
1378
1380
1544
1574
1819
1922
Table (1-2) Aniline supply and demand in U.S2
Year
ton/year
1996
142,700
2000
200,000
2004
435,000
2005
620,000
Table (1-3) Aniline supply and demand in china3
World consumption of Aniline:
western europe
china
united states
japan
Rep.of korea
Central/Eastern Europe
India
Central/South africa
other
Figure (1-2) World consumption of Aniline-2010
7
World consumption of aniline grew at an average annual rate of 3% during
2006–2010, the result of a growing global economy during 2001–2008,
declines during the economic recession in 2009 and the recovery in 2010,
and growth due to increased MDI capacity.
Strong Asian demand for all applications of MDI boosted world demand
during 2006–2010. World consumption of aniline is forecast to grow at an
average annual rate of 3.8% during 2010–2015.
Continuing rapid demand growth in some regions, particularly in China,
Other Asia and Europe, mainly the result of continued expansion of
nitrobenzene/aniline/MDI units, will balance out moderate growth in
markets such as the Americas.
Chinas consumption structure of aniline is different from developed
countries, primarily used in rubber processing additives, dyes and organic
pigments, pharmaceuticals and organic intermediates production.
Year
consumption(kt)
1993
100
1999
148.4
2000
177.9
2004
387
2010
1100
Table (1-4)Aniline’s china consumption capacity4
Uses
Aniline is mainly used as feed stock for the polyurethane industry. The
largest application of aniline is for the preparation of methylene
diphenyldisocyanate(MDI).Other uses include rubber processing chemicals
(9%), herbicides (2%), and dyes and pigments (2%).Many drugs can be
prepared from aniline such as paracetamol(acetaminophen)and used in the
dye industry as a precursor to indigo.
8
Application
%
Isocyanate
85%
Rubber Chemicals
9%
Agricultural
Chemicals
–
Pesticides
3%
Dyes & Pigments
2%
Specialty Fibers
1%
Miscellaneous
1%
Table (1-5)Aniline Uses5
Feed stock & product description:
Benzene:
Molecular formula
C6H6
Molar mass
78.11 g mol−1
Appearance
Colorless liquid
Density
0.8765(20) g/cm3
Melting point
5.5 °C, 278.7 K
Boiling point
80.1 °C, 353.3 K
Solubility in water
1.8 g/L (15 °C)
Lambda(λ)max
255 nm
Viscosity
0.652 cP at 20 °C
Dipole moment
0D
Table (1-6)Physical and chemical properties of Benzene6
Nitrobenzene:
Molecular formula
Molar mass
C6H5NO2
123.06 g/mol
Appearance
Density
Melting point
yellowish liquid
1.199 g/cm3
5.7 °C
Boiling point
210.9 °C
Solubility in water
0.19 g/100 ml at 20 °C
Table(1-7) Physical and chemical properties of Nitrobenzen7
9
Ammonia:
Molecular formula
NH3
Molar mass
17.031 g/mol
Appearance
Colourless gas with strong pungent odour
Density
0.86 kg/m3 (1.013 bar at boiling point)
0.73 kg/m3 (1.013 bar at 15 °C)
681.9 kg/m3 at −33.3 °C (liquid)
817 kg/m3 at −80 °C (transparent solid)
Melting point
−77.73 °C, 195 K, -108 °F
Boiling point
−33.34 °C, 240 K, -28 °F
Solubility in water
47% (0 °C)
31% (25 °C)
28% (50 °C)
Acidity (pKa)
32.5 (−33 °C), 10.5 (DMSO)
Basicity (pKb)
4.75
Structure
Molecular shape
Trigonal pyramid
Dipole moment
1.42 D
Thermochemistry
Std
enthalpy
formation ΔfHo298
Standard
entropySo298
of
molar
−46 kJ·mol−1
193 J·mol−1·K−1
Table(1-8) Physical and chemical properties of ammonia8
10
Phenol:
Molecular formula
C6H6O
Molar mass
94.11 g mol−1
Appearance
transparent crystalline solid
Density
1.07 g/cm3
Melting point
40.5 °C, 314 K, 105 °F
Boiling point
181.7 °C, 455 K, 359 °F
Solubility in water
8.3 g/100 mL (20 °C)
Acidity (pKa)
9.95 (in water)
29.1 (in acetonitrile)
Dipole moment
1.7 D
Table (1-9)Physical and chemical properties of phenol9
Hydrogen:
Molecular
H2
Phase
gas
Density
(0 °C, 101.325 kPa)
0.08988 g/L
Liquid density at (m.p)
0.07 (0.0763 solid) g·cm−3
Liquid density at (b.p)
0.07099 g·cm−3
Melting point
14.01 K, -259.14 °C, -434.45 °F
Boiling point
20.28 K, -252.87 °C, -423.17 °F
Triple point
13.8033 K (-259°C), 7.042 kPa
Critical point
32.97 K, 1.293 MPa
Heat of fusion
(H2) 0.117 kJ·mol−1
Heat of vaporization
(H2) 0.904 kJ·mol−1
Table (1-10) Physical and chemical properties of Hydrogen10
11
Water:
Molecular formula
H2O
Molar mass
18.01528(33) g/mol
Appearance
white solid or almost colorless, transparent, with a
slight hint of blue, crystalline solid or liquid
Density
1000 kg/m3, liquid
917 kg/m3, solid
Melting point
0 °C, 32 °F, (273.15 K)
Boiling point
99.98 °C, 211.97 °F (373.13 K)
Acidity (pKa)
15.74
~35–36
Basicity (pKb)
15.74
Refractive index (nD)
1.3330
Viscosity
0.001 Pa s at 20 °C
(4 °C)
(62.4
lb/cu.
ft)
Table (1-11) Physical and chemical properties of water11
Nitric Acid:
Molecular formula
HNO3
Molar mass
63.01 g mol−1
Appearance
Colorless liquid
Density
1.5129 g cm−3
Melting point
-42 °C, 231 K, -44 °F
Boiling point
83 °C, 356 K, 181 °F (68% solution boils at
121 °C)
Solubility in water
Completely miscible
Acidity (pKa)
-1.4
Refractive index (nD)
1.397 (16.5 °C)
Dipole moment
2.17 ± 0.02 D
Thermo chemistry
Std enthalpy of formation ΔfHo298
−207 kJ·mol−1
Standard molar entropySo298
146 J·mol−1·K−1
Table(1-12) Physical and chemical properties of Nitric Acid12
12
Sulfuric Acid:
Molecular formula
H2SO4
Molar mass
98.079 g/mol
Appearance
Clear, colorless, odorless liquid
Density
1.84 g/cm3, liquid
Melting point
10 °C, 283 K, 50 °F
Boiling point
337 °C, 610 K, 639 °F
Solubility in
water
Miscible
Acidity (pKa)
−3, 1.99
Viscosity
26.7 cP (20 °C)
Thermo chemistry
Std
enthalpy
formation ΔfHo298
Standard
entropySo298
of
molar
−814 kJ·mol−1
157 J·mol−1·K−1
Table (1-13) Physical and chemical properties of sulfuric acid13
Iron:
Molecular formula
Fe
Density
7.874 g·cm−3
Liquid density at m.p. 6.98 g·cm−3
Melting point
1811 K, 1538 °C, 2800 °F
Boiling point
3134 K, 2862 °C, 5182 °F
Heat of fusion
13.81 kJ·mol−1
Heat of vaporization
340 kJ·mol−1
Molar heat capacity
25.10 J·mol−1·K−1
Table (1-14) Physical and chemical properties of iron14
13
Hydrochloric acid:
Molecular formula
HCl
Molar mass
36.46 g mol−1
Appearance
Colorless gas
Odor
Pungent
Density
1.490 g L−1
Melting point
-114.22 °C, 159 K, -174 °F
Boiling point
-85.05 °C, 188 K, -121 °F
Vapor pressure
4352 kPa (at 21.1 °C)
Acidity (pKa)
-7.0
Basicity (pKb)
21.0
Refractive index (nD)
1.0004456 (gas)
Table (1-15) Physical and chemical properties of Hydrochloric Acid15
Ferric chloride:
Molecular formula
FeCl3
Molar mass
162.2 g/mol (anhydrous)
270.3 g/mol (hexahydrate)
Appearance
green-black by reflected light; purplered by transmitted light
hexahydrate: yellow solid
aq. solutions: brown
Odor
slight HCl
Density
2.898 g/cm3 (anhydrous)
1.82 g/cm3 (hexahydrate)
Melting point
306 °C (anhydrous)
14
37 °C (hexahydrate)
Boiling point
315 °C (anhydrous, decomp)
280 °C (hexahydrate, decomp) (partial
decomposition to FeCl2+ Cl2)
Solubility in water
74.4 g/100 mL (0 °C)
92 g/100 mL (hexahydrate, 20 °C)
Solubility in acetone
63 g/100 ml (18 °C)
Methanol
highly soluble
Ethanol
83 g/100 ml
Diethyl ether
highly soluble
Viscosity
40% solution: 12 cP
Table (1-16) Physical and chemical properties of ferric chloride16
Aniline:
Molecular formula
C6H5NH2
Molar mass
93.13 g/mol
Appearance
colorless to yellow liquid
Density
1.0217 g/mL, liquid
Melting point
-6.3 °C, 267 K, 21 °F
Boiling point
184.13 °C, 457 K, 363 °F
Solubility in water
3.6 g/100 mL at 20°C
Acidity (pKa)
4.7
Basicity (pKb)
9.3
Thermochemistry
Std enthalpy of formation ΔfHo298
-3394 kJ/mol
DMSO:dimethylsulfoxide
Table (1-17) Physical properties of aniline17
15
Process Flow Sheets:
Process 1;Ammonolysis of Phenol
24
14
17
V-100
P-102
21
10
K-100
13
16
CRV-100
9
Amonia
1
MIX-100
phenol
20
19
8
7
2
29
E-100
E-101
27
12
T-100
18
30
E-103
V-101
34
Aniline
27
28
26
11
P-104
T-101 25
31
24
E-104
35
E-102
28
T-102
32
29
P-103
36
33
E-105
37
Figure (1-3) Process flow sheet of the production of aniline by Ammonolysis of Phenol.
Process description:The process is divided into three sections: the feed preparation section, the
reactor section, and the purification section. In the feed preparation section,
The ammonia feed (stream 1) consists of 203 lb-mol/hr liquid ammonia at
90F. The phenol feed (stream 2) supplies 165.8 lb-mol/hr liquid phenol at
110F and atmospheric pressure. The two feed streams are pumped to
increase the pressure before they are mixed with their respective recycle
streams (stream 16 for ammonia and stream 31 for phenol) by using mixer
(MIX-102).(Stream 7) is heated in a heat exchanger (E-100) with the reactor
effluent (stream 10). The heat exchanger effluent (Stream 8) is heated to the
required reactor temperature for the reactor inlet (stream 9). The reactor
16
section includes the adiabatic reactor (CRV-100) that includes a bed Packed
with a silica-alumina catalyst. In the reactor, three reactions occur:
Phenol + NH3 ==> Aniline + H2O
2 Phenol + NH3 ==> Diphenylamine + 2 H2O
2 NH3<==> 3 H2 + N2
The conversion of phenol in the reactor is 95% with a 99% selectivity to
aniline as shown in the First reaction. The second reaction forms a byproduct
(diphenylamine), while the third reaction is the decomposition of ammonia.
The reaction set is exothermic, so the stream leaving the reactor (stream 10)
is hotter than stream 9. The cooling of the reactor effluent begins with the
heat exchanger (E-100) which will be cooled from (stream 10) to (stream
11). After that (stream 11) is sent through a cooler (E-102).
The purification section consists of the distillation columns to separate the
chemicals into Products and non-products. The absorption column (T-100)
separates the gases and the liquids. As a result, all of the hydrogen and
nitrogen go to stream13. Moreover, all of the phenol, aniline and
diphenylamine go to stream 18. From the absorption column, stream 13 goes
to a splitter to split it into stream 14, which is the ammonia recycle stream
that will pass through a compressor (K-100) to increase the pressure. On the
other hand, the splitter also sends small amount of stream 13 to the gaseous
purge (stream 24). The bottoms stream (stream 18) is one of the feeds to the
next column (T-101). The distillate (stream19) is cooled by the unit (E-103).
Stream 20 is then sent to a separator (V-100) to separate the water and the
phenol product. Then, the phenol (stream 21) is recycled to the column (T101) after pressurize it by using (P-102). The aqueous product (stream 24)
from V-100 will be treated. The bottoms stream (stream 25) is the feed to
the next column (T-102). The main component in the distillate (stream 26)
is aniline which should be pumped by (P-104). The resulting stream (stream
17
27) is cooled by (E-104) to produce aniline (Stream 28). Then, (stream 29)
should be pumped by (P-103) to get the suitable condition (stream31) for
mixing by (MIX-102). The bottoms product (stream 32) is cooled b (E-105)
to purchase diphenylamine in (stream 33).
Process 2;Liquidphase reduction of nitrobenzene with metal in mineral
acids.
Sulfuric
acid
13
3
Nitric
acid
2
9
MIX-100
8
Aniline
12
Benzene
1
14
CRV-101
5
4
15
T-101
V-100
6
CRV-100
nitrobenzene
10
11
V-101
7
T-100
RCY-100
Figure (1-4) Process flow sheet of the production of aniline by liquid phase reduction
of nitrobenzene.
Process description :
First of all, nitric acid and sulfuric acid is mixed together by using an acid
mixer(mix-100). the mixed acid and benzene are fed in a nitrate
reactor(CRV-100) and the reaction will occur as the following chemical
equation :C6H6 +HNO3(H2SO4)  C6H5NO2 + H2O(H2SO4)
ΔH=-113KJ
Then the reactor effluent will enter the separator (V-100) which will separate
it into two streams which are crude nitrobenzene and the reactor effluent
acid that will be recovered to the reactor feed. The crude nitrobenzene could
18
be washed by using a dilute alkali (V-101) such as water and sodium
carbonate and distillated by still distillation column (T-100) to produce a
pure nitrobenzene. The crude nitrobenzene will enter a reducer [H2] reactor
(CRV-101) with an iron boring catalyst and hydrochloric acid to produce
aniline and water which will occur as the following chemical equation:4C6H5NO2 + 9Fe + 4H2O  4C6H5NH2 + 3FeO4
And the effluent nitrobenzene will be recovered to the reducer reactor. The
aqueous aniline could be heated by steam to get crude aniline which can be
distilled by using still distillation column (T-101) to produce pure aniline.
Catalyst Reaction : HCl
Yeild : 98%
Process 3;Process catalytic reduction of nitrobenzene in a fluidized bed
reactor.
8
5
E-102
3
E-101
Water
10
Hydrogen
4
6
V-100
Nitrobenzene
9
7
T-100
4
2CRV-100
1
12
P
T-101
11
Aniline
K-100
E-100
Figure (1-5) Process flow sheet of the production of aniline by vapor phase reduction
of nitro benzene.
19
Name of Unit
E-100
CRV-100
E-101
K-100
V-100
T-100
E-102
T-101
Description
Nitrobenzene vaporizer
Reactor
Product Condenser
Hydrogen Recycle Compressor
Aniline water decanter
Crude aniline distillation
Condenser
Aniline product distillation
Table (1-18) Name and description unit
Species
Flowrate ( million lb/yr) Compostion Temperature
(C°)
Nitrobenzene 0.1
25
Pressure (atm)
Hydrogen
1
-
0.9
25
1
Table (1-19) Feedstock condition
Species Flowrate (million lb/yr) Compostion
Temperature (C°) Pressure (bar)
Aniline 200
0.995
-
1
Water
-
-
1
-
Table(1-20) Product condition
Name of stream
S1
S2
S3
S4
S5
S6
S7
S8
S9
S10
S11
S12
Description
Nitrobenzene
Hydrogen Feed
Reactor Product Gases
Condensed Materials
Non-Condensable gas
Crude Aniline
Aqueous Phase
Overheads
Bottom Streams
Water
Aniline product
Recycled bottom
Table (1-21) Name of stream
20
Process description:
The main process for aniline production is the nitrobenzene hydrogenation
reactions. Feed preparation section:
The liquid nitrobenzene feed (contains less than 10ppm – thiophene) (S1) is
vaporized
up by going into vaporizer (E-100) to reach the required
temperature for the fluidized bed reactor after the mixing point between the
nitrobenzene feed (S1) and the hydrogen feed (S2)
(which has been
compressed by Recycle Compressor) (K-100).Hydrogen to nitrobenzene
ratio is 9:1.
Reaction section:
It includes the fluidized bed reactor (CRV-100) with 10-20% copper by
weight on silica catalyst [which is made by spray-drying a silicic acid matrix
(20 to 150 micrometer) with a cuprammonium compound and activated in
position with hydrogen at 250 C] at 270 C and 1.5 atm, the equation for this
process is shown below:
C6H5NO2 + 3H2 --------> C6H5NH2 + 2H2O
The reaction is highly exothermic with enthalpy (-443 KJ/mol) and
approximately 65% of the heat of reaction is removed by circulating a cool
fluid (generally water or low pressure steam) through tubes suspended in the
fluidized bed. The nitrobenzene vapor-hydrogen mixture (300 percent excess
hydrogen) reaction takes place on the surface porous at the bottom of the
fluidized bed reactor. The upper part of the reactor is large enough to allow
the most of the catalyst to fall back into the main catalyst bed. any catalyst
which escapes from the reactor is removed from the product by stainless
steel filters.The conversion of the nitrobenzene in the reactor is 99.7% and
the selectivity to aniline is 99%.
21
Purification section:
The reactor (RCV-100) product gas mixture (aniline , hydrogen and water)
(S3) will enter the condenser (E-101)and the leaving gas stream (3.5%
water, 0.5% aniline and the balance hydrogen) (S5) which has been recycled
to the compressor (K-100), but a small part is vented to avoid the buildup of
gaseous impurities which exist in hydrogen feed. Moreover the aqueous and
organic phases stream (S4) is separated in a decanter (V-100) which
separates the crude aniline (S6) from the aqueous phase solution (S7). The
organic phase (crude aniline) is consist of aniline up to 0.5% nitrobenzene,
and 5% water is purified by two stage distillation column. After that in the
crude still column (T-100)
(stripping) ,aniline and water are removed
overhead while higher boiling organic impurities, such as nitro-benzene
remain in the still bottoms. The overhead product (S9) from the first column
is purified in a finishing still (T-101), Water (S10) is withdrawn from the top
of the column while aniline (S11) is withdrawn in a side stream near the
bottom of the column. The bottom (S12) is recycled to the crude still (T100).
Waste treatment:
The best method of treating the aqueous waste resulting from the following
units (nitrobenzene distillation column overhead(stream8), nitrobenzene
wash water(stream7) and the aniline recovery column purge(stream12)), is
the biological treatment due to its inexpensive cost and it's high efficiency
indicator which could of the toxic nature of the waste which may affect the
environment , moreover the physical method that used as stream stripping
and liquid-liquid extraction.
22
Catalyst Regeneration.
the catalyst can be regenerated with air periodically.At 250-350°C and
subsequent H2 treatment.
Comparison flow sheets:
Alternative 1
Mole
M.wt
Ib
ton
Ib/Ib of aniline
$/Ib
Gross Profit =
phenol
1
94
94
0.047
1.01
0.745
-0.428
ammonia
1
17
17
0.0085
0.182
0.289
aniline
1
93
93
0.0465
1
0.38
Water
1
18
18
0.009
0.193
-
Table(1-22) Alternative #1
Alternative 2:
-There is no enough information to calculate the gross profit for the process.
Alternative 3
Mole
M.wt
Ib
Ib/Ib of aniline
$/Ib
Gross Profit =
nitrobenzene
1
123
123
1.322
0.33
0.076
hydrogen
3
2
6
0.064
0..32
aniline
1
93
93
1
0.38
Water
2
18
36
0.387
-
Table (1-23) Alternative #3
Alternative
1
2
3
No. of equipment
17
9
8
Catalyst
Silica-alumina
Iron borings
Copper
on
silica
Table(1-24) Comparison between the alternatives
23
Raw Material
Phenol + Ammonia
Benzene + Nitric acid
Nitrobenzene + Hydrogen
Basic economic analysis:
lb
Prices,
US$
460.00745.00
385.81771.62
0.37-0.39
300.00351.00
tonne
85.43
tonne
lb
ton
93.7
0.33-0.34
1450
57.0085.00
87
94
73.1
94
50.0062.00
25.0030.00
67
215.00225.00
1472.311507.69
1015.381192.31
0.32
Description
Weight
Ammonia
US Gulf, spot c.f.r. Tampa
tonne
Ammonia
Aniline
ton
lb
Ferric chloride
Hydrochloric
acid
Hydrochloric
acid
Nitrobenzene
Sulphuric acid
US Gulf, spot f.o.b New Orleans
tanks, f.o.b
technical grade, 100% basis, tanks, f.o.b.
works
22 deg. Be, US Gulf dom. ex-works US
NE
22 deg. Be, US Gulf dom. ex-works
USG
tanks, f.o.b.
virgin 100%, tanks, works, East Coast
Sulphuric acid
Sulphuric acid
Sulphuric acid
Sulphuric acid
Sulphuric acid
virgin 100%, tanks, works, Southwest
virgin 100%, tanks, works, Midwest
virgin 100%, tanks, works, Southeast
virgin 100%, tanks, works, West Coast
smelter 100% tanks, works, Gulf Coast
ton
ton
ton
ton
ton
Sulphuric acid
smelter 100%, tanks, works
ton
Sulphuric acid
Sulphuric acid
Nitric acid
smelter, tanks, intro. Southeast
ton
smelter, 93% tanks, dlvd., Northwest
ton
40 deg. Be, 42 deg. Be. tanks, c.l.,
works, 100% basis
ton
Phenol
-
ton
Benzene
Hydrogen gas
-
ton
Ib
Table (1-25) Cost of materials
24
Conclusion:
 We figure out that there are lots of processes to produce aniline.
 The production of aniline is takes an active part in America and china.
 Locally aniline production is not exist.
 At the beginning, aniline was an intermediate substance for (MDS)
production, according to this reason it was produced.
Recommendation:
The process 3 (vapor phase catalytic reduction of nitrobenzene ) is the best
alternative for the production of Aniline due to its inexpensive raw materials
and its highly profit comparing it to the other two alternatives.NitroBenzene
is the classical feedstock for Aniline manufacture. Recently less
chlorobenzene and Phenol are being used in aniline manufacturing processes
in several countries.
25
References:
Web sites;
http://www.springerlink.com
http://www.mpri.lsu.edu
http://www.xakaili.com
http://www.thefreedictionary.com
http://www.chemicalbook.com
http://price.alibaba.com
http://www.icis.com
http://en.wikipedia.org
Books;
McGraw-Hill, Shreve’s Chemical Industries, George T.Austin,1984
26
College of Engineering and Petroleum
Chemical Engineering Department
Plant Design (ChE 491)
Hysys Report
Production of Aniline
Group Members:
Abdulaziz Alshomar
208113942
Yosef Almneffi
208112671
Hassan Butaleb
207216952
Hussain Butaleb
207217011
Abdulhameed Alawadhi
208115609
Presented by:
Prof. Mohamed A. Fahim
Eng. Yusuf Ismail Ali
27
TABLE OF CONTENTS :
Page
Abstract
2
Table of contents
3
List of Tables
4
Introduction
5
Hysys simulation program
6-11
Process Description
7-44
Process Equipments
45-48
Material Balance
49
28
ABSTRACT :
We will used the simulation software hysys to simulate Aniline production
from Nitrobenzene by completed material balance, and we will mention in
detail about all equipment we used. Data and tables will be mentioned in this
report and with comparison between the HYSIS data and measured data.
29
LIST OF TABLES :
Contents
Page
Table-1 : the change of conversion for each reaction in the 21
first reactor
Table-2 : Summary of columns
45
Table-3: Summary of reactors
46
Table-4:Summary of Heat Exchanger,Heaters and coolers
46
Table-5:Summary of Compressors
46
Table-4:Summary of Heat Exchanger,Heaters and coolers
46
Table-6: Summary of separator
47
Table-7: Summary of energy streams
47
Table-8: Mass flow for inlet and outlet streams
48
30
Introduction:
Aniline is a fuel that is produced from Nitrobenzene and Hydrogen and it is
used as feed stock for the polyurethane industry, medical drugs , herbicides ,
Rubber, dyes and preparation of methylene diphenyldisocyanate (MDI)
Reaction Rate:
It includes the fluidized bed reactor (CRV-100) with 10-20% copper by
weight on silica catalyst [which is made by spray-drying a silicic acid matrix
(20 to 150 micrometer) with a cuprammonium compound and activated in
position with hydrogen at 250 C] at 270 C and 1.5 atm, the equation for this
process is shown below:
C6H5NO2 + 3H2 --------> C6H5NH2 + 2H2O
The reaction is highly exothermic with enthalpy (-443 KJ/mol) and
approximately 65% of the heat of reaction is removed by circulating a cool
fluid (generally water or low pressure steam) through tubes suspended in the
fluidized bed. The nitrobenzene vapor-hydrogen mixture (300 percent excess
hydrogen) reaction takes place on the surface porous at the bottom of the
fluidized bed reactor. The upper part of the reactor is large enough to allow
the most of the catalyst to fall back into the main catalyst bed. any catalyst
which escapes from the reactor is removed from the product by stainless
steel filters.The conversion of the nitrobenzene in the reactor is 99.7% and
the selectivity to aniline is 99%.
31
HYSYS Simulation Description:
Objectives :
- To show how to use HYSYS simulation.
- Design Chemical plant using HYSYS simulation.
-Compare between the measured data and HYSIS data.
Introduction:
HYSIS Simulation is a very helpful program for engineering to support them
to simulate Chemical engineering design in many different ways. HYSYS is
a market-leading process modeling system, used by the world’s leading oil
and gas organizations to improve every step of the petrochemical industry.
Platform for Engineering Optimization
HYSYS is a process modeling tool for conceptual design, optimization,
business planning, asset management, and performance monitoring for oil &
gas production, gas processing, petroleum refining, and air separation
industries. It provides these services via a wide variety of internal features
and layered applications. Some of the key features include:
o
Comprehensive toolset for modeling all hydrocarbon processes
o
Dynamic evaluation of process models
o
State of the art assay management
o
Best in class physical properties engine
o
Integration with Aspen PIMS and Aspen Petroleum Scheduler software
o
Value-added layered application
32
Getting Started:
With windows, the installation process creates a shortcut to HYSYS :
Click the icon to start HYSYS
Or
1-click on the Start menu
2-Move from Programs to Hyprotech to HYSYS
3- Select HYSYS .
Fig 2.1:HYSYS starting window
Setting your Session Preferences :
1- To start a new simulation case , do one of the following :
- From the File menu, select New Case.
- Click the New Case icon and the simulation basis manager will appear
 Creating a fluid Package
The next step is to create a fluid Package .As a minimum , a Fluid Package
contains the Components and property method (for example , an Equation
of State )HYSYS will use in its calculations for a particular flow sheet .
33
Depending on what is required in a specific flow sheet , a fluid Package may
also contain other information such as reactions and interaction parameters .
1- On the Simulation Basis Manager view , click the fluid Pkgs tab.
2- Click the Add button , and the property view for your Fluid Package
appears
Fig 2.2 : fluid Package
We used UNIQUAC as a fluid package for the whole Design and PRSV for
The second Distillation. We chose UNIQUAC because the activity
coefficients can be used to predict simple phase equilibrium (vapor–liquid,
liquid–liquid, solid–liquid), or to estimate other physical properties and also
for multicomponent chemical mixtures.
Selecting Component:
After we choose our fluid package the next step is to show you how I choose
my component from feed stock to the products:
1-From Simulation Basis Manager window press view on component List.
In the match space we can find our components by three ways:
 Sim Name : The name appearing within the simulation .
34
 Full Name :IUPAC name (or similar ), and synonyms for many
components
 Formula : The chemical formula of the component .
Fig 2.3 : Selecting Components
Selecting Reaction:
From Simulation Basis Manager windows, we go to the bottom tools bar and
we press on Reaction. First we put our reaction data (stoichiometry,
components in my reaction and conversion percentage) in "Add Rxn". Then
we go "Add Set" and put our reaction there. Finally we go to "Add to FP"
and we add our reaction set to Fluid Package.
The main reaction :
C6H5NO2 + 3H2 --------> C6H5NH2 + 2H2O
35
Fig 2.4: Selecting Reaction
Entering the simulation Environment :
-To leave the Basis environment and enter the Simulation environment, click
the enter Simulation Environment button on the Simulation Basis Manager
view.
Process Description:
Stream (1) which is contain nitrobenzene , water , Diphenylamine and
benzene. It's enter as a liquid phase in heat exchanger (E-101) at condition at
30C and 14.7 psig to outlet stream (1*) with 177C and 6 psig and vapor
phase. It is heated by stream (5**) which leave from the second heat
exchanger at 184.4C and leave the (E-101) with Stream (5h2) with 162.9C.
36
Table 1: Mass flow of compressor inlet feed:
Components
Mass flow( Ib/hr )
Nitrobenzene
15507
Water
15.422
Diphenylamine
19.958
Benzene
0.793
37
Stream (2) is entered with hydrogen and gas impurities at 30C and 14.7 psig
with vapor phase and it enter the mixer with the vapor phase recycled stream
(8rec) with temperature 116.89C and pressure 45psig.
With Composition:
38
And the outlet stream is (2**) in vapor phase with temperature 114C and 14.7
And vapor phase with composition:
39
And the worksheet for this mixer ( mix-101)
Then stream (2**) will enter a heat exchanger (E-102) and leave with
114.1C and 14.7 psig and it is heated by stream (5*) which outlet from
(MIX-102) with temperature 311.6C and its duty to raise the temperature in
stream (2***) to its required temperature 249C. The effluent from (E-102) is
stream (5**) is going to unit (E-103) to heat the Feed stream (1).
40
Then Stream (2***) will enter (TEE-101) and leave the unit with two
streams. This streams include stream (3) and stream (fired h2).
41
The compositions of this unit is:
After that we mix stream (1*) with stream (3) in (C-101) unit to give stream (4) that had
hydrogen and nitrobenzene composition which is the feed for reactor.
42
We now Compress stream (4) to be suitable for reaction condition in (K-100) unit.
43
We did not use heater first in stream (4) because compressor process increases the
temperature while the main purpose for it is increasing pressure. Stream (4*) is entered to
heater to raise the temperature for reaction condition in (E-101) unit.
44
The main target for unit (E-108) is to cool the reactor by circulating tubes
around the reactor using branch feed water.
45
Now Stream (fired H2) with temperature 249C will enter a furnace (F-101) to increase its
temperature and leave as stream (fired 3) with temperature 315C. This stream will enter a
mixer (MIX-102) with the product stream (5) from the reactor.
46
Reaction Section :
* The reactor is conversion reaction ( fluidized bed reactor ).
* This reactor is neither isobaric nor adiabatic.
* The reaction is highly exothermic.
* The feed enter as liquid vapor mixture and exit in a vapor phase.
47
There are two outlet streams from the reactor:
- stream (L) with zero flow rate because the reactions are in the vapor phase ,
since the conversion reactor gives us two streams : liquid and vapor.
- stream 5 with a temperature = 275 C and pressure =35 psia , and it has a
composition mass flow as in the two figures :
48
Now we have Unit (MIX-102) which is the inlet is a mixer for the product stream (5)
from the reactor and stream (fired h2) and the outlet is stream (5*) that entered the heat
exchanger (E-102).
49
Now we will talk about the Hydrogen Section which is start at the outlet from heat
exchanger (E-103) by stream (5h2). This stream is entered a cooler (E-104) and the outlet
is stream (5h3), we put an energy to change also the pressure.
The stream (5h3) will enter a separator (V-101) and the outlet will be stream (6) and that
goes to purification section which will have aniline and water inside it and stream(7)
which will have inside it hydrogen and small amount of water.
50
51
After that stream (7) will enter a (Tee-100) and the outlet are stream (9) as a purge
because we need to vent a few amount of the recycled hydrogen and the impurities to
avoid the build up for the hydrogen feed, and to get rid of the organic phases impurities.
And stream (8).
52
Now stream (8) will enter a Compressor (K-101) to increase its Pressure and the outlet is
stream (8*).
53
After that stream (8*) will be recycled a(RCY-1) and the outlet is Stream (8rec). This
outlet stream will enter a Mixer ( MIX-101) with the hydrogen feed stream (2).
Purification Section :
Stream (6) will enter a mixer (MIX-104) with other two streams (23.3) and
(14.2) and the outlet is stream (6.1).
54
Now Stream (6.1) will enter a three phase separator (V-201) and the outlet streams are
stream (vapor) , stream (15) and stream (10).
55
56
Stream (10) will enter a (MIX-103) with stream (17.4) and the outlet stream (11***) will
enter a heat exchanger (E-201) and leave after heating at stream (11**). The cooling
stream for heat exchanger is stream (13) and the outlet is stream (13.1) as a waste water.
57
After that Stream (11**) will enter a valve (VLV-100) to reduce its pressure and exit as
stream (11*) then it enter a Heater (E-202) to increase the exit temperature stream (11).
58
Stream (11) and stream (12) which is pure water will enter a Stripper (C-201) and the
outlet will be stream (14) and stream (13) which will enter the Heat exchanger (E-201) as
a heating steam.
-number of stage =10
- Pressure of the condenser and the reboiler = 15 psia
59
Then Stream (14) will enter a compressor (K-102) and the outlet stream is (14.0) then it
enters a Cooler (E-203) and leave as stream (14.1) which will recycled (RCY-2) to stream
(14.2) that entered the Mixer (MIX-104).
60
Now Stream (15) will enter a separator (V-202) and the outlet will be stream (15.1) and
stream (15.2).
61
Then stream (15.2) will enter a heat exchanger (E-204) and leave with increase in
temperature by stream (15.3) then it will enter a distillation Column (C-202) and the
outlet streams are stream (16) and stream (17).
This Distillation Column (C-202) have:
-number of stage = 30
- Pressure of the condenser and the reboiler = 5.6 psia
-Reflux Ratio for condenser = 3
-Bottom product Rate (Aniline) = 271.2 lb/hr
62
Stream (17) will enter A compressor (K-203) and leave as stream (17.0) then it is cooled
in (E-205) and leave as stream (17.1).
63
64
Then Stream (17.1) will enter a three phase separator (V-204) and leave as stream (Steam
injection) , stream (18) and stream (17.3). Stream (17.3) will be recycled (RCY-3) to
enter the mixer (MIX-103) by stream (17.4).
65
Now stream (16) will enter a heater (E-212) and the outlet stream (16.) will enter the
second Distillation Column (C-203) and the leaving stream (20) and stream (19) wich is
contain Nitrobenzene.
Information for The second Distillation Column:
-number of stage = 50
- Pressure of the condenser and the reboiler = 5.6 psia
-Reflux Ratio for condenser = 50
-Distillate Rate (Aniline) = 256.4 lb/hr
66
After that stream (20) will enter a Compressor (K-103) and leave as stream (20.1) then it
enters a cooler (E-208) at stream (20.).
67
Then stream (20.) will enter a separator (V-203) and the effluent are stream (20.4) which
we will talk about it later as the Aniline product, and stream (21). Then stream(21) will
enter a mixer (MIX-105) with stream(22) and the leaving stream is (21.1)
68
Then Stream (21.1) will enter a (TEE-102) and the outlet are stream (24) and stream (23),
then stream (23) will enter a valve (VLV-101) to reduce the pressure of the exit stream
(23.1). Then stream (23.1) will enter a cooler (E-211) and the leaving stream (23.2) is
recycled (RCY-4) to enter the mixer (MIX-104) as stream (23.3).
69
70
Now the final Stream (20.4) which will Enter Heat exchanger (E-204) to heat the feeding
stream (15.2) will leave the heat exchanger as stream (20.5) which will enter a cooler (E107) and leave as stream (Product) with aniline composition 98%.
71
72
Process Equipment:
1- Reactors:
Table-3: Summary of reactors
Reactor
Temperature(C)
Pressure(psia)
R-101
275
35
2- Heat Exchangers , Heater, Furnace and coolers:
Unit
ΔT (C)
E-103 (H.E)
147
E-101 (Cooler)
179.3
E-104 (Cooler)
122.9
E-102 (H.E)
134.9
E-108 (Cooler)
118
F-101 (Furnace)
66
E-201(H.E)
50
E-202 (Heater)
9.01
E-203 (Cooler)
121
E-204 (H.E)
78
E-209 (Cooler)
6.81
E-205 (Cooler)
207.1
E-212 (Heater)
4.6
E-208 (Cooler)
111.9
E-211 (Cooler)
99.4
73
3- Stripper & Distillation Columns:
Column name
Number of stage
Top
temperature
(C)
Bottom
temperature
(C)
Top
pressure (psia)
Bottom
pressure
(psia)
C-201
10
100.1
100.6
15
15
C-202
C-203
30
50
78.21
150.7
150.4
175.4
5.6
5.6
5.6
5.6
4- Compressors & valves :
Unit
ΔP(psia)
K-100
K-101
K-102
K-201
K-202
V-100
V-101
34
20
10
19.4
34.3
10
15
5- Separators:
Unit
V-101
V-201
V-202
V-204
V-203
Top flowrate (lb/hr)
82380
0.03132
28870
-3
5.897*10
0.1979
578.2
0
6- Recycles :
Unit
RCY-1
RCY-2
RCY-3
RCY-4
Flowrate (lb/hr)
82010
5026
2426
360
74
Bottom flowrate (lb/hr)
35660
12170
28870
2426
24030
Energy Consumption :
Energy Stream
Heat flow Q(KJ/hr)
Q1
5.57*106
Q2
8.89*106
Qfired
2.38*107
Q4
6.282*107
Q5
2.7*107
Q7
4.281*105
Q8
3.261*106
Q9
2.577*105
Q10
2.666*105
Q11
5.592*106
Q12
5.456*106
Q13
9.073*105
Q14
4.218*105
Q15
1.345*105
Overall Material Balance :
The inlet streams are stream 1 + stream 2 + stream 22 + stream 12 = 34272 +
1977 + 400 + 4697 = 41346
The outlet stream are stream 9 + stream 13.1 + stream 18 + stream 19 +
stream 24 + product = 370.72 + 14264 + 578.15953 + 1833.77320 + 40 +
24030.99236 = 41117.64509
75
Mass flow(lb/hr)
Stream 1 Stream 2 Stream 12
Stream 22
Stream 9
nitroBZ
34187
-
-
-
Tr
H2O
34000
-
4697
400
92.785
H2
-
1829
-
-
230.57
Aniline
-
-
-
-
14.014
Methane
-
29.6
-
-
6.5894
Diphenyl amine
44
-
-
-
Tr
Benzene
1.75
-
-
-
0.2233
N2
-
29.6
-
-
6.5962
CO
-
29.6
-
-
6.5952
CO2
-
29.6
-
-
6.5677
O2
-
29.6
-
-
6.5960
Cyclohexane
1.75
-
-
-
Tr
Niacin
1.75
-
-
-
Tr
Cyclohexylamine
1.75
-
-
-
Tr
Mass flow(lb/hr)
Stream 13.1
Stream 18
Stream 19
Stream 24
Stream
product
nitroBZ
18.246
0.69812
1789.8
-
649.34
H2O
14244
33.374
Tr
40
Tr
H2
Tr
Tr
Tr
-
Tr
Aniline
0.22658
541.22
Tr
-
23382
Methane
Tr
Tr
Tr
-
Tr
Diphenyl amine
1.6164
Tr
42.384
-
Tr
Benzene
Tr
1.4128
Tr
-
Tr
N2
Tr
Tr
Tr
-
Tr
CO
Tr
Tr
Tr
-
Tr
CO2
Tr
Tr
Tr
-
Tr
O2
Tr
Tr
Tr
-
Tr
Cyclohexane
Tr
0.22713
Tr
-
Tr
Niacin
Tr
Tr
1.6118
-
Tr
Cyclohexylamine
Tr
1.1802
Tr
-
Tr
76
Student name :Hassan Ahmed Butaleb
ID number : 207216952
Equipment name
Equipment number
Distillation column
C-202
Cooler
E-208
Reboiler
E-206
Reboiler
E-207
Separator
V-203
Compressor
K-103
Tank
T-203
Tank
T-202
Distillation column
Distillation is a process in which a liquid or vapor mixture of two or more
substances is separated into its component fractions of desired purity, by the
application and removal of heat. Distillation is based on the fact that the
vapor of a boiling mixture will be richer in the components that have lower
boiling points. The composition difference of its products depends dissimilar
of the volatilities or vapor pressure of the components of the liquid mixture.
If such a dissimilar does not exist, then such a separation is not possible. The
distillation column operates as follows; the feed in liquid state is introduced
continuously at a midpoint in the column. The midpoint divides the column
into enriching and stripping sections. By mean of a reboiler located in the
bottom of the column, the liquid feed is vaporized. The vapor ascends within
77
the column and contacts with the descending liquid. The plate or the packing
in the column allows for a thorough contact of the vapor and liquid. The
reminder of the bottom liquid is withdrawn as bottoms. Vapor reaching the
top of the column is condensed to liquid in the overhead condenser. A
portion of the condensed liquid is returned to the column, known as a reflux
liquid. The remainder of the condensate becomes the distillate products. The
reboiler vapor and reflux liquid lead to high purity products and provides a
much greater recovery of the feed. Vapor and liquid phases on a given tray
approach thermal equilibrium to an extent dependent upon the efficiency of
the contacting trays.
Types of distillation columns:
There are many types of distillation columns, each designed to perform
specific types of separations, and each design differs in terms of complexity.
One way of classifying distillation column type is to look at how they are
operated. Thus we have:
78
1- Batch Columns
In batch operation, the feed to the column is introduced batch-wise. That is,
the column is charged with a 'batch' and then the distillation process is
carried out. When the desired task is achieved, a next batch of feed is
introduced.
2- Continuous Columns
In contrast, continuous columns process a continuous feed stream. No
interruptions occur unless there is a problem with the column or surrounding
process units. They are capable of handling high throughputs and are the
most common of the two types. We shall concentrate only on this class of
columns.
3- Column Internals
An important aspect of division in distillations is how a distillation column
is designed from the inside. Either trays or packing are used to allow the
desirable separation. Various types of trays or plates are used for distillation
such as:
- Bubble cap trays
A bubble cap tray has riser or chimney fitted over each hole, and a cap that
covers the riser. The cap is mounted so that there is a space between riser
and cap to allow the passage of vapor. Vapor rises through the chimney and
is directed downward by the cap, finally discharging through slots in the cap,
and finally bubbling through the liquid on the tray.
79
Valve trays
In valve trays, perforations are covered by liftable caps. Vapor flows lift the
caps, thus self-creating a flow area for the passage of vapor. The lifting cap
directs the vapor to flow horizontally into the liquid, thus providing better
mixing than is possible in sieve trays.
Sieve trays
Sieve trays are simply metal plates with holes in them. Vapor passes straight
upward through the liquid on the plate. The arrangement, number and size of
the holes are design parameters. Because of their efficiency, wide operating
range, ease of maintenance and cost factors, sieve and valve trays have
replaced the once highly thought of bubble cap trays in many applications.
80
Packings
There is a clear trend to improve separations by supplementing the use of
trays by additions of packings. Packings are passive devices that are
designed to increase the interfacial area for vapor-liquid contact. The
following picture show different types of packings.
These strangely shaped pieces are supposed to impart good vapor-liquid
contact when a particular type is placed together in numbers, without
causing excessive pressure-drop across a packed section. This is important
because a high pressure drop would mean that more energy is required to
drive the vapor up the distillation column.
Packings Versus Trays Columns
Packed Columns
Trayed Columns
81
82
Advantages of Using Carbon Steel:
Steel is counted to be carbon steel when no minimum content is specified or
required for chromium, cobalt, columbium [niobium], molybdenum, nickel,
titanium, tungsten, vanadium or zirconium, or any other element to be added
to achieve a desired alloying influence; when the specified minimum for
copper does not exceed 0.40 percent; or when the maximum content
specified for any of the following elements does not exceed the percentages
noted: manganese 1.65, silicon 0.60, copper 0.60.As a matter of fact carbon
steel has been used widely in construction by the virtue of its suitable
properties, and mostly because of it is highly corrosion resistance and it
moderate cost
The common types of carbon steels types are:
a) Plain-Carbon
b) Low Carbon Steel
c) Medium Carbon Steel
d) High Carbon Steels
Problems while operating a distillation column:
Foaming:
Foaming refers to the expansion of liquid due to passage of vapour or gas.
Although it provides high interfacial liquid-vapour contact, excessive
foaming often leads to liquid buildup on trays. In some cases, foaming may
be so bad that the foam mixes with liquid on the tray above. Whether
foaming will occur depends primarily on physical properties of the liquid
mixtures, but is sometimes due to tray designs and condition. Whatever the
cause, separation efficiency is always reduced.
83
Entrainment:
Entrainment refers to the liquid carried by vapor up to the tray above and is
again caused by high vapor flow rates. It is detrimental because tray
efficiency is reduced: lower volatile material is carried to a plate holding
liquid of higher volatility. It could also contaminate high purity distillate.
Excessive entrainment can lead to flooding.
Weeping:
This phenomenon is caused by low vapor flow. The pressure exerted by the
vapor is insufficient to hold up the liquid on the tray. Therefore, liquid starts
to leak through perforations. Excessive weeping will lead to dumping. That
is the liquid on all trays will crash (dump) through to the base of the column
(via a domino effect) and the column will have to be re-started. Weeping is
indicated by a sharp pressure drop in the column and reduced separation
efficiency.
Flooding:
Flooding is brought about by excessive vapor flow, causing liquid to be
entrained in the vapor up the column. The increased pressure from excessive
vapor also backs up the liquid in the down comer, causing an increase in
liquid holdup on the plate above. Depending on the degree of flooding, the
maximum capacity of the column may be severely reduced. Flooding is
detected by sharp increases in column differential pressure and significant
decrease in separation efficiency.
84
Assumptions:
1. Tray column.
2. Sieve plate.
3. Material of the distillation is carbon steel.
4. Plate spacing = 0.3 m
5. Efficiency = 95%
6. Flooding % = 85%
7. Weir height = 45 mm
8. Hole diameter = 4 mm
9. Plat thickness = 5 mm
10.Down comer area 12% of total
Nomenclatures
Symbol
FLv
Lw
Vw
ρv
ρL
uf
u`v
Ac
Dc
Ad
An
Aa
Ah
Aap
Ap
how
u`h
hd
hr
hap
hdc
T
P
S
Ri
Ej
Cc
Definition
Liquid vapor flow factor
Liquid mass flow rate (kg/s)
vapor mass flow rate (kg/s)
Vapor density (kg/m 3)
Liquid density (kg/m 3)
flooding vapor velocity (m/s)
flooding at maximum flow rate (kg/s)
Total column cross sectional area (m2)
Column diameter (m)
cross sectional area of down comer (m2)
Net area (m2)
Active area (m2)
Hole area (m2)
Clearance area (m2)
Perforated area (m2)
Weir crest (mm) liquid
Min. vapor velocity (m/s)
Dry plate drop (mm)
Residual head (mm)
Out let weir height (mm)
Head loss in downcomer (mm)
thickness of cylindrical shell (in)
maximum allowable internal pressure (psi)
maximum allowable working stress (psi)
: inside radius of shell (in)
efficiency of joint expressed as fraction
allowance for corrosion (in)
85
Design Procedures:
1. Specify the properties of outlets streams: (flow rate, density and surface
tension) for both vapor and liquid from hysys.
2. Calculate minimum number of trays.
3. Calculate the maximum liquid and vapor outlet flow rate.
4. Choose tray spacing and then determine K1 and K2 using figure (1) from
Appendix A.
5. Calculate correction factor for Bottom K1 and Top K1.
6. Design for X% flooding at maximum flow rate for top and bottom part of
distillation.
7. Calculate the maximum flow rates of liquid.
8. Calculate Net area required.
9. Take down comer area as %Y of the total column cross sectional area.
10. Calculate the column diameter.
11. Calculate the column height using the actual number of stage.
12. Calculate column area, down comer area, active area, net area, hole area
and weir length.
13. Calculate the actual min vapor velocity.
14. Calculate Back-up in down comer.
15. Check residence time.
16. Check entrainment.
17. Calculate number of holes.
18) Calculate area of condenser and re-boiler.
19. Calculate Thickness of the distillation.
86
Distillation Column sample Calculation (T-101)
T-101 column properties
Top
Bottom
Unit
Vapor rate (Vn)
42
330.07
kmol/hr
Mass Density for Vapor ρv
0.2864
1.0337
kg/m3
Molecular Weight (M.Wt)
21.674
113.2
Liquid rate (Ln)
188.58
126
kmol/hr
Mass Density for Liquid ρL
973.36
919.4
kg/m3
Molecular Weight (M.Wt)
76.597
95.685
Surface Tension
0.04293
0.029715
N/m
Number of Stages:
Number of stages
25
Efficiency
0.72
Actual number of stages
35
Column diameter:
Liquid vapor flow factor
Top
Bottom
unit
Mass Density for Vapor ρv
0.2864
0.2
kg/m3
Mass Density for Liquid ρL
973.36
919.4
kg/m3
Surface tension
0.042
0.0297
N/m
87
Bottom = FLV = (L/V)*(ρv/ ρL)0.5 = 0.0056302
Top =
FLV = (L/V)*(ρv/ ρL)0.5 = 0.0770187
Take plate spacing as 0.5 m
Figure 3-8 Flooding velocity for sieve plates
From the figure above:
Base K1 = 0.103
Top K1 = 0.097
Correction for surface tensions
Base K1 = 0.1114876
Top K1 = 0.1130103
Flooding velocity:
Base = uf = K1((ρL- ρv)/ ρv)0.5 = 7.5581652 (m/s)
Top = uf = uf = K1((ρL- ρv)/ ρv)0.5 = 6.5872516 (m/s)
Design for 85% flooding at maximum flow rate
Base uv = uf*0.85 = 6.4244405 (m/s)
Top = uv = uf*0.85 = 5.5991639 (m/s)
88
Maximum volumetric flow rate
Bottom = Vmax= Vn*M.Wt/ρv*3600 = 2.2519255 (m3/s)
Top = Vmax= Vn*M.Wt/ρv*3600 = 0.0441451 (m3/s)
Net area required:
Bottom = A=Vmax/uv = 0.4083822 (m2)
Top = A=Vmax/uv = 0.0078842 (m2)
Taking downcomer area as 12 per cent of total.
Column cross-sectional area
Base =
= 3.2407 /(1 – 0.12 ) = 0.458975 (m2)
Top =
= 2.5517 /( 1 – 0.12 ) = 0.0089594 (m2)
Coloumn diameter:
Bottom = D = (Anet *4/π)0.5 = 0.76444 (m)
Top = D = (Anet *4/π)0.5 = 0.1068071 (m)
Use same diameter above and below feed
D = 0.76 (m) = 2.5 (ft)
Column Height:
Total height = H=(Number of stage * Plate spacing*2)+Column Diameter
= 18.1 (m) = 60 (ft)
Maximum
volumetric
liquid
rate
= 0.003642 (m3/s)
89
=
(
LN*M.Wt)/(ρL*3600)
Figure 3-9 Selection of liquid flow arrangment
From the figure above: Double pass plate is used
Provisional plate design:
Column diameter = Dc = 0.76 (m)
column area = (3.14/4)*(Dc^2) = 0.4589705 (m2)
Downcomer area Ad = 0.12 * 0.3983353
= 0.0550765 (m2)
Net area = An = Ac – Ad = 0.40389 (m2)
Active area = Aa = Ac - 2*Ad = 0.3488176 (m2)
Hole area = Ah = 10% of Aa = 0.0348818 (m2)
Figure 3-10 Relation between downcomer are and weir length
90
= 12
Lw/Dc = 0.76
Weir Length = lw = 0.58097 (m)
Take weir height = hw = 45 (m)
Hole diameter (dh) = 4 (mm)
Plate thickness = 5 (mm)
Check weeping:
Maximum liquid rate
Lw = (Ln*Mwt)/3600 = 3.348975 (kg/s)
Turndown percentage = 0.70
Minimum liquid rate = Lwd *0.7 = 2.3442825 (kg/s)
Maximum = how =750*(Lw/(ρLlw))2/3 = 1.18365 (mm liquid )
Minimum = how =750*(Lw/(ρLlw))2/3 = 20.104388 (mm liquid)
At minimum rate = hw + how = 65.104388 (mm liquid)
Figure 3-11 Weep point correlation (Eduljee, 1959)
91
From the figure above:
K2 = 30.6
Minimum vapor velocity through hole:
uh (min) = (K2-0.90(25.4-dh))/ρv0.5 = 25.357011 (m/s)
Actual minimum vapor velocity = Turn dowd * Max volumetric flow rate
(bot) / Ah
= 52.070245 (m/s)
Plate pressure drop:
Dry plate drop Maximum vapor velocity through holes (uh) = Bottom
Vmax/Hole
uh = 74.386064 (m/s)
Figure 3-12 Discharge coefficient, sieve plates (Liebson et al. 1957)
92
From the figure above:
Plate thickness / hole dia. = 1.25
Ah
x100  0.1
Ap
Co  0.86

U 
hd  51 h 
 Co 
hr 
12.5 x10 3
L
2
 V

 L

  0.8300073 mmliquid

 13.595823 mmliquid
Total plate pressure drop
hb  hw  hdc  ht  how  60.606733 mmliquid
Down comer liquid back-up:
Downcomer pressure loss Take
hap  hw 10  45 10  35mm
Area under apron

Aap  weirlengthxhap  0.0203342 m2
2
As this is less than Ad  0.055076 m use
Aap
in the next equation for hdc
2
 max .liquid rate 

  0.05326779
hdc  166
mm



xA
L
ap


Back-up in downcommer
hb  hw  hdc  ht  how  106.869011mm  0.106969011(m)
Residence Time
tr 
hb xAd x L
 1.616 s
Lwd
93

Check Entrainment
UV 
volumetric flowrate
 6.424 m / s
An
Percent Flooding 
UV
x100%  85%
Uf
FLV ( Bottom)  0.0056302
Figure 3-13 Entrainment correlation for sieve plates (Fair, 1961)
94
From the figure above:
ψ = 0.02 , well below 0.1
Perforated area:
Figure 3-14 Relation between angle subtended by chord, chord height and chord
length
From the figure above:
at
lw
 0.76
Dc
  95
Angle subtended by the edge of the plate = 81

Mean length, unperforated edge strips = 1.010026 m
Area of unperforated edge strips= 0.046806 m2
Mean length of calming zone,approx = 0.503513 m
Area of calming zones = 0.050351 m2
Total area for perforations, Ap = 0.205578 m 2
Ah / Ap  0.14726 m 2

95
Figure 3-15 Relation between hole area and pitch
From the figure above:
l p / d h  2.25 satisfactory within 2 to 4
Number of holes:
2
2
Area of one hole = d h  1.257 *10^ 5m
Number of holes = Aa/0.00001 = 2409 hole
Area of condenser
Inlet temperature T1
134.7
Co
Outlet temperature T2
78.2
Co
Mean overall heat transfer coefficient U
1000
W/m2.Co
Heat flow Q
2.81 *106
KW
96
AC 
Q
 49.6m 2
UT
Area of reboiler
Inlet temperature T1
150.2
Co
Outlet temperature T2
150.4
Co
Mean overall heat transfer coefficient U
1200
W/m2.Co
Heat flow Q
2.81 * 106
KW
Ab 
Q
 15575.0 m 2
UT
Thickness Calculations:
Internal raduis of shell before allowance corrosion is added ri
14.018967
in
Maximum allowable internal pressure P
25
psi
Working stress for carbon steel S
13700
psi
Efficincy of joients EJ
0.85
Allowance for corrosin Cc
0.125


Pri
  CC  0.155 in  3.9mm
t  
 SEj  0.6 P 
97
in
Equipment Name
Drying Column
Objective
Separate Aniline
Equipment Number
C-202
Designer
Hassan Ahmed Butaleb
Type
Continuous Tray Distillation Column
Location
After Aniline Stripper (C-100)
Material of Construction
Carbon steel
Insulation
Mineral wool
Operating Condition
Key Components
Light
Aniline
Heavy
Operating Temperature (oC)
113
Operating Pressure (kpa)
172.37
Feed Flow Rate (kg/h)
12539
Diameter (m)
0.71
Height (m)
19.7
Thickness (mm)
3.9
98
Hydrogen
Compressor
A gas compressor is a mechanical device that increases the pressure of a gas
by reducing its volume. Compression of a gas naturally increases its
temperature. Compressors are similar to pumps: both increase the pressure
on a fluid and both can transport the fluid through a pipe. As gases are
compressible, the compressor also reduces the volume of a gas. Liquids are
relatively incompressible, so the main action of a pump is to transport
liquids.
Types of compressors:The main types of gas compressors are discussed below in the table
Compressor
Positive Displacement
Dynamic
Rotary
Centrifugal
Reciprocating
Axial
Centrifugal compressors:
Centrifugal compressors use a vane rotating disk or impeller in a shaped
housing to force the gas to the rim of the impeller, increasing the velocity of
the gas. A diffuser section converts the velocity energy to pressure energy.
They are primarily used for continuous, stationary service in industries such
as oil refineries, chemical and petrochemical plants and natural gas
processing plants. Their application can be from 100 hp (75 kW) to
thousands of horsepower. With multiple staging, they can achieve extremely
high output pressures greater than 10,000 psi (69 MPa). Many large snowmaking operations use this type of compressor. They are also used in
internal combustion engines as superchargers and turbochargers. Centrifugal
compressors are used in small gas turbine engines or as the final
compression stage of medium sized gas turbines.
99
General components of a Centrifugal Pump
Axial Compressors:
Axial-flow compressors are dynamic rotating compressors that use arrays of
fan-like aerofoil to progressively compress the working fluid. They are used
where there is a requirement for a high flow rate or a compact design.
Stationary stator vanes, located downstream of each rotor, redirect the flow
onto the next set of rotor blades. The area of the gas passage diminishes
through the compressor to maintain a roughly constant axial Mach number.
Axial-flow compressors are normally used in high flow applications, such as
medium to large gas turbine engines. They are almost always multi-staged.
Beyond about 4:1 design pressure ratio, variable geometry is often used to
improve operation. Axial compressors can have high efficiencies; around
90% polytrophic at their design conditions. However, they are relatively
expensive, requiring a large number of components, tight tolerances and
high quality materials. Axial-flow compressors can be found in medium to
large gas turbine engines, in natural gas pumping stations, and within certain
chemical plants.
Axial compressors are widely used in gas turbines, such as jet engines, high
speed ship engines, and small scale power stations. They are also used in
industrial applications such as large volume air separation plants, blast
100
furnace air, fluid catalytic cracking air, and propane dehydrogenation. Axial
compressors, known as superchargers, have also been used to boost the
power of automotive reciprocating engines by compressing the intake air,
though these are very rare. A good example of an axial supercharger is the
aftermarket Latham type built between (1955-65), which were used on hot
rods and air cooled Volkswagens at that time, but these didn't catch on.
Axial-flow compressor
Reciprocating compressor:
A reciprocating compressor or piston compressor is a positive-displacement
compressor that uses pistons driven by a crankshaft to deliver gases at high
pressure. The intake gas enters the suction manifold, then flows into the
compression cylinder where it gets compressed by a piston driven in a
reciprocating motion via a crankshaft, and is then discharged. We can
categorize reciprocating compressors into many types and for many
applications. Primarily, it is used in a great many industries, including oil
refineries, gas pipelines, chemical plants, natural gas processing plants and
refrigeration plants. One specialty application is the blowing of plastic
bottles made of Polyethylene Terephthalate
101
Reciprocating compressor
Rotary screw compressors:
Rotary screw compressors use two meshed rotating positive-displacement
helical screws to force the gas into a smaller space. These are usually used
for continuous operation in commercial and industrial applications and may
be either stationary or portable. Their application can be from 3 hp (2.24
kW) to over 500 hp (375 kW) and from low pressure to very high pressure
(>1200 psi or 8.3 MPa).
Rotary vane compressors:
Rotary vane compressors consist of a rotor with a number of blades inserted
in radial slots in the rotor. The rotor is mounted offset in a larger housing
102
which can be circular or a more complex shape. As the rotor turns, blades
slide in and out of the slots keeping contact with the outer wall of the
housing. Thus, a series of decreasing volumes is created by the rotating
blades. Rotary Vane compressors are, with piston compressors one of the
oldest of compressor technologies. With suitable port connections, the
devices may be either a compressor or a vacuum pump. They can be either
stationary or portable, can be single or multi-staged, and can be driven by
electric motors or internal combustion engines. Dry vane machines are used
at relatively low pressures for bulk material movement whilst oil-injected
machines have the necessary volumetric efficiency to achieve pressures up
to about 13 bar in a single stage. A rotary vane compressor is well suited to
electric motor drive and is significantly quieter in operation than the
equivalent piston compressor.
Rotary vane compressors
103
Applications:
Gas compressors are used in various applications where either higher
pressures or lower volumes of gas are needed:

In pipeline transport of purified natural gas to move the gas from the
production site to the consumer.

In petroleum refineries, natural gas processing plants, petrochemical
and chemical plants, and similar large industrial plants for
compressing intermediate and end product gases.

In storing purified or manufactured gases in a small volume, high
pressure cylinders for medical, welding and other uses.

In many various industrial, manufacturing and building processes to
power all types of pneumatic tools.

In pressurized aircraft to provide a breathable atmosphere of higher
than ambient pressure.

In some types of jet engines (such as turbojets and turbofans) to
provide the air required for combustion of the engine fuel. The power
to drive the combustion air compressor comes from the jet's own
turbines,

In SCUBA diving, hyperbaric oxygen therapy and other life support
devices to store breathing gas in a small volume such as in diving
cylinders.

In submarines, to store air for later use in displacing water from
buoyancy chambers, for adjustment of depth.

In turbochargers and superchargers to increase the performance of
internal combustion engines by increasing mass flow.
104

In rail and heavy road transport to provide compressed air for
operation of rail vehicle brakes or road vehicle brakes and various
other systems.
Nomenclature:
Symbol
Nomenclature
K
the heat capacity ratio
P1
the inlet pressure in psi
P2
the discharge pressure in psi
m
n
the mass flow rate in kg/s
polytropic exponent
T1
The inlet temperature in R
T2
The discharge temperature in R
ZC
is the average compressibility factor.
Hp
horse power, Hp
M
molar flow rate, lbmol/s
W
work done, Btu/lbmol
Ep
effeciency of the compressor
Cp
heat capacity, Btu/lb F
105
Design Procedure:
1. Calculate the compression factor using the following equation:
 n 


P1  T1  n 1 
,
 
P2  T 2 
Where P1,2 is the pressure of inlet and outlet respectively (psia),
And T1,2 is the temperature of the inlet and outlet respectively (R).
2. Calculate the work done in Btu/lbmol by:
W 
nR (T 1 T 2 )
,
1 n
Where R is the ratio of the specific heat capacities (Cp/Cv).
3. Calculate the horse power, Hp using the following equation:
Hp=W*M,
Where M is the molar flow rate in lbmol/s.
4. Calculate the efficiency of the compressor using the following
equation:
Ep 
n
K
n 1 ,
K 1
Where K 
MwC p
MwC p  1.986
and Mw is the molecular weight of the gas
in the stream and Cp is the specific heat capacity (Btu/lboF).
106
Calculation:
1. ln ( P1 / P2 ) = ln ( 14.7 / 35 ) = -0.867500568
ln ( T1 / T2 ) = ln ( 536.67 / 735.21) = -0.314772792
n/(n-1) = ln(P1/P2) / ln(T1/T2) = 0
n = 1.569489731
2. w = ( 1.569489731 * 1.3986 * ( 536.67 - 735.21 ))/( 1 - 1.569489731 )
= 765.2690028 ( Btu / lbmol )
3. Hp = 765.2690028 * 0.048611111 = 50.0347
4. k = ( 28.85 * 6.9675 ) / ( 28.85 * 6.9675 – 1.986 ) = 1.009978577
Ep = (1.569489731 / ( 1.569489731 – 1 )) * (1.009978577-1)/(
1.009978577) *100
Ep = 2.722883313 %
Specification sheet for Air compressor(K-103)
Equipment Name
Compressor
Objective
To increase the pressure
Equipment Number
K-103
Designer
Hassan Ahmed Butaleb
Type
Centrifugal Compressor
Material of Construction
Carbon steel
Insulation
Quartz wool
Operating Condition
Inlet Temperature (R)
536.67
Outlet Temperature (R)
735.21
Inlet Pressure (psia)
14.7
Outlet Pressure (psia)
35
Efficiency (%)
2.722883313 %
Power (Hp)
50.0347
107
Heat Exchanger:
Introduction:
A heat exchanger is a device built for efficient heat transfer from one
medium to another, whether the media are separated by a solid wall so that
they never mix, or the media are in direct contact. They are widely used in
space heating, refrigeration, air conditioning, power plants, chemical plants,
petrochemical plants, petroleum refineries, and natural gas processing. One
common example of a heat exchanger is the radiator in a car, in which the
heat source, being a hot engine-cooling fluid, water, transfers heat to air
flowing through the radiator. Heat exchanger has a barrier which separates
the fluids and permits heat to flow from the hotter to the colder stream
without mixing of the streams. The shell and tube heat exchanger provides
comparatively large ratios of heat transfer are to volume and weight. It
provides this surface in a form which is relatively easy to construct in a wide
range of sizes and which is mechanically rugged enough to withstand normal
fabrication stress and normal operating conditions. There are many
modifications of the basic configuration which can be used to solve special
problems. The shell and tube heat exchanger can be reasonably easily
cleaned. Finally good design methods exist, and the expertise and the shop
facilities for successful design and construction of shell and tube exchangers
are available thought the world.
108
Types of heat exchangers:
Plat heat exchanger
Another type of heat exchanger is the plate heat exchanger. One is composed
of multiple, thin, slightly-separated plates that have very large surface areas
and fluid flow passages for heat transfer. This stacked-plate arrangement can
be more effective, in a given space, than the shell and tube heat exchanger.
Advances in gasket and brazing technology have made the plate-type heat
exchanger increasingly practical. In HVAC applications, large heat
exchangers of this type are called plate-and-frame; when used in open loops,
these heat exchangers are normally of the casketed type to allow periodic
disassembly, cleaning, and inspection. There are many types of permanentlybonded plate heat exchangers, such as dip-brazed and vacuum-brazed plate
varieties, and they are often specified for closed-loop applications such as
refrigeration. Plate heat exchangers also differ in the types of plates that are
used, and in the configurations of those plates. Some plates may be stamped
with "chevron" or other patterns, where others may have machined fins
and/or grooves.
109
Plate heat exchanger
Phase-change 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 it or used as
condensers to cool a vapor and condense it to a liquid. In chemical plants
and refineries, reboilers used to heat incoming feed for distillation towers are
often heat exchangers. Distillation set-ups typically use condensers to
condense distillate vapors back into liquid.
Power plants which have steam-driven turbines commonly use heat
exchangers to boil water into steam. Heat exchangers or similar units for
producing steam from water are often called boilers or steam generators
110
Spiral heat exchangers
A spiral heat exchanger may refer to a coiled tube configuration, more
generally, the term refers to a pair of flat surfaces that are coiled to form the
two channels in a counter-flow arrangement. Each of the two channels has
one long curved path. A pair of fluid ports is connected tangentially to the
outer arms of the spiral, and axial ports are common, but optional.
The main advantage is highly efficient use of space. This attribute is often
leveraged and partially reallocated to gain other improvements in
performance, according to well known tradeoffs in heat exchanger design.
(A notable tradeoff is capital cost vs. operating cost.) A compact spiral heat
exchanger may be used to have a smaller footprint and thus lower all-around
capital costs, or an over-sized may be used to have less pressure drop, less
pumping energy, higher thermal efficiency, and lower energy costs.
spiral heat exchanger
Shell and tube heat exchanger
The tubes are the basic component of the shell and tube exchanger. They
provide heat transfer surface between one fluid flowing inside the tubes and
the other fluid flowing across the outside of the tubes. They are made of
copper and steel alloys. Other alloys of nickel, titanium, or aluminum may
also be used. The shell is simply the container for the shell side fluid and the
nozzles are the inlet and exit ports. The shell normally has a circular cross
section and is commonly made by roiling a metal plate of appropriate
111
dimensions into a cylinder into a cylinder and welding the longitudinal joint.
In large exchangers, the shell is made out of low carbon steel wherever
possible for reasons of economy, through other alloys can be and is used
when corrosion or high temperature strength demands must be met.
Baffles:
Baffles serve two functions:
- They support the tubes in the proper position during assembly and
operation and prevent vibration of the tubes caused by flow-induced
eddies.
- They guide the shell side flow back and forth across the tube field,
increasing the velocity and the heat transfer coefficients.
The most common baffle shape is the single segment. The segment sheared
off must be less than half of the diameter in order to insure that adjacent
baffles overlap at least one full tube row. For liquid flows on shell side, a
baffle cut of 20 to 25 percent of the diameter is common; for low pressure
gas flow, 40 to 50 percent is more common in order to minimize pressure
drop. For many high velocity gas flow s, the single segment baffle
configuration results in an undesirably high shell side pressure drop. One
way to retain the structural advantages of the segment baffle and reduce the
pressure drop is to use the double segment baffle. It halves the local velocity
and therefore reduces the pressure drop by a factor of 4 from a comparable
size single segment unit.
Applications
Shell and tube heat exchangers are frequently selected for such applications
as:
 Process liquid or gas cooling.
 Process or refrigerant vapor or steam condensing.
112
 Process liquid, steam or refrigerant evaporation.
 Process heat removal and preheating of feed water.
 Thermal energy conservation efforts, heat recovery.
 Compressor, turbine and engine cooling, oil and jacket water.
 Hydraulic and lube oil cooling.
Selection of heat exchanger type:
The selection process normally includes a number of factors, all of which are
related to the heat transfer application. These factors include, but not limited
to, the following items:
1. Thermal and hydraulic requirements.
2. Material compatibility.
3. Operational maintenance.
4. Environmental, health, and safety considerations and regulations.
5. Cost.
Design parameter
The critical design factors for a heat exchanger application are: flow rate,
temperature, pressure drop, heat needed to be transferred.
Selection of tube material
To be able to transfer heat well, the tube material should have good thermal
conductivity. Because heat is transferred from a hot to a cold side through
the tubes, there is a temperature difference through the width of the tubes.
Because of the tendency of the tube material to thermally expand differently
at various temperatures, thermal stresses occur during operation. This is in
addition to any stress from high pressures from the fluids themselves. The
tube material also should be compatible with both the shell and tube side
fluids for long periods under the operating conditions (temperatures,
pressures, pH, etc.) to minimize deterioration such as corrosion. All of these
113
requirements call for careful selection of strong, thermally-conductive,
corrosion-resistant, high quality tube materials, typically metals. Poor choice
of tube material could result in a leak through a tube between the shell and
tube sides causing fluid cross-contamination and possibly loss of pressure.
Carbon Steel
In our heat exchangers the material of construction used is carbon steel.
Advantage: Low cost, easy to fabricate, abundant, most common material
and resists most alkaline environments well.
Disadvantage: Very poor resistance to acids and stronger alkaline streams
and more brittle than other materials, especially at low temperatures.
Glass wool
Glass wool offers the best mix advantages for insulation:
1. Thermal Resistance.
2. Acoustical Absorption.
3. Fire resistance.
4. Lightness.
5. Environmentally friendly.
6. Easy Insulation.
Assumptions
1-
For cooling the fluid, Chilled water has been used for many reasons,
mostly because of its suitable temperature range in addition to its
treatment.
2-
Assuming value for overall heat transfer coefficient based on table
12.1, which is close to the calculated value.
3-
The type of heat exchanger is shell and tube, while the material of
construction is carbon steel.
114
Nomenclature
Table 3.1 : Nomenclature of Heat exchanger
Symbol
T1
T2
t1
t2
µ
kf
Cp
Р
Q
∆Tlm
A
U
do
di
Lt
Re
Pr
Gs
lb
T
∆Pt
Np
Ej
S
Cc
ri
Definition
Inlet shell side fluid temperature (°C)
Outlet shell side fluid temperature(°C)
Inlet tube side fluid temperature (°C)
Outlet tube side fluid temperature (°C)
Fluid viscosity (m N s /m2)
Thermal conductivity ( W/ m °C)
Mass heat capacity (kJ / Kg °C)
Density of the fluid (Kg/ m3)
Heat load (Kw)
Log mean temperature difference (°C)
Area (m2)
Overall heat transfer coefficient (W/m2. °C)
Tube outside diameter (mm)
Tube inner diameter (mm)
Tube length
Reynolds number
Prandtl number
Mass velocity (m/s)
Baffle spacing (m)
Shell Thickness
Tube side pressure drop (N/m2)
Number of tube side passes
Efficiency of joints
Working stress (psi)
Allowance for corrosion (in)
Internal radius of shell
115
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
Sample calculation:
Q = (m Cp ΔT) = 1741.510264 KW
mcold = (Qh)/(Cp*∆T)= 0.27777 Kg/s
000
Tlm 
T2  T1
 T 
LN  2 
 T1 
T lm=( 144 – 25 ) - ( 225 - 130.7 ) / LN(( 144 – 25 )/( 225 - 130.7 ))
= 106.6578 °C
116
Using one shell pass and two tube passes
R
(T1  T2 )
(t 2  t1 )
R= (25 – 130.7)/ (144 – 225.9) = 1.2905983
S
(t 2  t1 )
(T1  t1 )
S= (144 – 225.9)/ (25 – 225.9)
= 0.4076655
Using fig. (A.11) to find Ft
Ft= 0.98
Tm  Ft * Tlm
= 104.52464 C
From table in appendix assume U= 25 W/m2°C
Provisional area
A
Q
= 604.46982 m2
UTm
Choose
Do= 32 mm
Di= 28 mm
117
Assume,
Lt= 2.1 m
Area of one tube = Lt * do *π = 0.211115 m2
Number of tubes Nt = provisional area / area of one tube= 2863
As the shell – side fluid is relatively clean use 1.25 triangular pitch.
Using Table (A.4) in appendix
N 
Bundle diameter Db= d o  t 
 K1 
1 / n1
K1= 0.249
N1= 2.207
Db= 50* 
2.285 

 0.17 
1 / 2.207
= 2313 mm
Use a split – ring floating head type.
From figure (A.12) in appendix
Bundle diametrical clearance = 67 mm
Shell diameter, Ds = Db + bundle diametrical clearance
Ds= 2313 + 67 = 2280 mm
Tube – side coefficient
  2  
Tube cross sectional area =  d i =   * (28E  3) 2 = 615.75216 mm2
4
4
Tubes per pass= N t = 3157 = 789
4
4
Total flow area = tube per pass * cross sectional area * 10^-6
= 0.440759 m2
118
Mass velocity=mass flow rate/total flow area
=
11510
0.4859529
= 2.61 * 10^4 Kg/m2.s
Linear velocity (ut) =mass velocity/density
=
2.37 *10^4
461.707
= 56.559744 m/s
The coefficient can be calculated from the following equation
 
hi d i
 j h Re Pr 0.33 
Kf
 w
Re 
Pr 



ud i
 1.46 * 10^6

Cp
 7.65 * 10^2
Kf
L 2.1 *1000

 75
di
28
From figure (A.13) in appendix
jh= 2 * 10^-3

hi  1.16*10^5(W / m 2 C )
from previous equation
Shell - side coefficient
Choose baffle spacing Lb= Ds
5

2380
 456.0684 mm
5
Tube pitch (pt)=1.25*do= 1.25 * 32 = 40 mm
Cross-flow area As=
 ( pt  d o ) * Ds * Lb

pt

 (40 E  3  32) * 2380 * 476.0865
 
 0.20798m 2
40

Mass velocity Gs = mass flow rate / cross flow area
119
=
1000 / 3600
 1.211281 Kg / m 2 s
1 / 0.2268
Equivalent diameter de =


 1.1  2
1.1
  pt  0.917d o 2    (40) 2  0.917 * (32) 2  22.7mm
 32 
 do 


Mean shell side temperature =
Re 
Gs d e
 60

pr 
Cp
 84.01
Kf
(T1  T2 ) (130.7  25)

 77.85 C
2
2
Choose 15 per cent baffle cut.
From figure (A.15) in appendix
jh= 6 * 10^-2
hs= 224.7233 W/m^2 C
Overall heat transfer coefficient
Thermal conductivity of cupro-nickel alloys kw = 45 W/m °C
Take the fouling coefficient from Table in appendix
Taking fouling coefficients
From table in appendix
Outside coefficient (fouling factor) (hod) = 5000
Inside coefficient (fouling factor) (hid) =5000
d
d o LN  o
 1  1  1 
 di
      
 
2k w
 U o   ho   hod 


   d o  1
 d  h
 i  id
120
 do
 
 di
1

 hi



Uo= 202.60019 (W/m2 °C) Acceptable
Pressure drop:
Tube side
From figure (A.14) in appendix and for Re = 1.46 * 10^6
jf= 2 * 10^-3
Neglecting the viscosity correction term
  L / di 
 u 2 


pt  N p 8 j f 
  2.5 2 

   / w 

pt  1585.3416 psia
Shell side
Linear velocity= Gs

 00.0024015
From figure (A.16) in appendix and for Re = 60
jf= 0.35
Neglecting the viscosity correction term
 D  L
Pt  8 j f  s  t
 d e  lb
 u 2   

 
2
  w 

0.14
Pt  0.0018819 Kpa
Shell thickness
P= 40 psi
ri = 44.88864805 in
S= 13706.66 psi
EJ =0.85
Cc = 0.125 in
121
t
Pri
 Cc  0.279 in  7.1 mm
SE j  0.6P
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)
Heat transfer area = 6506.46 ft2
Specification Sheet
Equipment Name
Cooler
Objective
To cool the aniline product
Equipment Number
E-208
Designer
Hassan Ahmed Butaleb
Type
Shell and tube heat exchanger
Location
After the distillation column (C-203)
Utility
Chilled water
Material of Construction
Carbon steel
Insulation
Quartz wool – Glass wool
Operating Condition
Shell Side
Inlet temperature (oC)
25
Outlet temperature (oC)
130.7
Inlet temperature (oC)
225.9
Outlet temperature (oC)
144
Number of Tube Rows
5
Number of Tubes
2863
Tube bundle Diameter (m)
2.2133
Shell Diameter (m)
2.28034
Q total (Kw)
1579.549
LMTD (oC)
106.6
U (W/m2 oC)
25
Heat Exchanger Area (m2)
603.8697
Tube Side
122
Reboiler (E-206)
Sample calculation:
Tlm 
T2  T1
 T 
LN  2 
 T1 
T lm=( 150.4 – 95 ) - ( 150.2 – 75.02 ) / LN(( 150.4 – 95 )/( 150.2 – 75.02 ))
hnb = 0.00122 (Tw – Ts)0.24 (pw – ps)0.75 ((kl0.79 Cpl0.45 ρL0.49)/( σ0.5 μL0.29
ρv0.24 L0.24)
L : latent heat
g : gravitational acceleration
σ : Surface tension
hnb = -0.672 W / m2 co
qc = 0.131 * L (σ g (ρL - ρv ) ρv2 )0.25
qc = 6.84 * 10^5
Rel = (ρL 2*10^-3 Di) / (μL * 10^-3) = 5.51 * 10^4
jh From fig 12.29 = 0.003
Pr = (Cpl μL) / (kl) = 12.9
hfc = (kl * jh * Rel ) * ( Pr )^0.33 / (Di*10^-3) = 1811.7626 W / m2 co
Xtt = 1 / ((( x / 1-x)0.9 * (ρL / ρv)0.5 = 0.8398513
fc from fig 12.52 = 2.5
hfc* = fc * hfc = 4529.4064 W / m2 co
Heat load = ( L * U ) / ( 3600 Mw.t ) = 3.2317 kW
Area of reboiler = ( Heat load * 10^3 ) / ( Q ) = 32.3177 m2
Q : Heat flux
Area of one tube = 3.14 * Do * Lo = 0.2299646 m2
Lo : Tube length
123
Number of tubes = Area of reboiler / Area of one tube = 141
Reboiler (E-207)
Sample calculation:
Tlm 
T2  T1
 T 
LN  2 
 T1 
T lm=( 175.4 – 95 ) - ( 174.9 – 75.02 ) / LN(( 175.4 – 95 )/( 174.9 – 75.02 ))
hnb = 0.00122 (Tw – Ts)0.24 (pw – ps)0.75 ((kl0.79 Cpl0.45 ρL0.49)/( σ0.5 μL0.29
ρv0.24 L0.24)
L : latent heat
g : gravitational acceleration
σ : Surface tension
hnb = - 7.9 *10^-2 W / m2 co
qc = 0.131 * L (σ g (ρL - ρv ) ρv2 )0.25
qc = 8.67 * 10^3
Rel = (ρL 2*10^-3 Di) / (μL * 10^-3) = 8.56 * 10^3
jh From fig 12.29 = 0.007
Pr = (Cpl μL) / (kl) = 0.792
hfc = (kl * jh * Rel ) * ( Pr )^0.33 / (Di*10^-3) = 32.641 W / m2 co
Xtt = 1 / ((( x / 1-x)0.9 * (ρL / ρv)0.5 = 0.0032204
fc from fig 12.52 = 1
hfc* = fc * hfc = 32.6417 W / m2 co
Heat load = ( L * U ) / ( 3600 Mw.t ) = 2.08925 kW
Area of reboiler = ( Heat load * 10^3 ) / ( Q ) = 69.64186 m2
Q : Heat flux
124
Area of one tube = 3.14 * Do * Lo = 0.2299646 m2
Lo : Tube length
Number of tubes = Area of reboiler / Area of one tube = 303
Flash Separator
Separator is a device that separate liquid from gasses. There are two types of
separator: two phase separator and three phase separator. The choice of
separator depends on the process. In our process we use two phase separator.
Separators are mechanical devices for removing and collecting liquids from
natural gas. A properly designed separator will also provide for the release
of entrained gases from the accumulated hydrocarbon liquids. Gravity causes
the liquid to settle to the bottom of the vessel, where it is withdrawn. The
vapor travels upward at a design velocity which minimizes the entrainment
of any liquid droplets in the vapor as it exits the top of the vessel.
125
The feed to a vapor-liquid separator may also be a liquid that is being
partially or totally flashed into a vapor and liquid as it enters the separator.
A vapor-liquid separator may also be referred to as a flash drum, knock-out
drum, knock-out pot, compressor suction drum or compressor inlet drum.
When used to remove suspended water droplets from streams of air, a vaporliquid separator is often called a demister. Vapor-liquid separators are very
widely used in a great many industries and applications, such as:
1. Oil refineries
2. Natural gas processing plants
3. Petrochemical and chemical plants
4. Refrigeration systems
5. Air conditioning
6. Compressor systems for air or other gases
7. Gas pipelines
8. Steam condensate flash drums
Vertical Separator
Vertical separator is a device with its cylindrical axes perpendicular to the
ground that is used to separate liquids from gases. There are two types of
separator depending on the process, two phases and three Phase's separator.
Most refinery processes include one or more separators with different types.
Vertical Separators perform four distinct functions – inlet control, vapor
demisting, liquid separation, and liquid outlet control.
Some of Features of vertical separators:
1. Designed for high liquid loading applications.
2. High efficiency over wide flow range.
3. No required maintenance.
126
Horizontal
Separator
Horizontal separator is a device with its cylindrical axes perpendicular to the
ground that is used to separate liquids from gases. There are two types of
separator depending on the process, two phases and three Phase's separator.
Most refinery processes include one or more separators with different types.
Horizontal Separators perform four distinct functions – inlet control, vapor
demisting, liquid separation, and liquid outlet control.
Some of Features of Horizontal separators:
1. Designed for high liquid loading applications.
2. High efficiency over wide flow range.
3. No required maintenance.
127
Material of Construction:
We can use stainless steel but carbon steel is good and cheaper.
Insulation:
Material of insulation depends on the operating temperatures, since
temperature in the separator is not high so from figure 3 in appendix, we can
see that the possible materials that cover the temperature are glass fiber and
mineral wool. And we choose mineral wool as insulation.
Design Procedure
1. Settling velocity
Ut = 0.07 [(ρL – ρv ) / ρv ]0.5 = 0.773209297 (m/s)
2. Volumetric flow rate
Vv = Mv / (3600 * ρv ) = 0.000223714 (m3/s)
Lv = ML / (3600 * ρL ) = 0.00348878 (m3/s)
3. Volume held in vessel
VHV = 10 * 60 * Lv = 2.093267716 (m3)
4. Minimum vessel diameter
Dv = [(4 * Vv ) / (pi * Us )]0.5 = 2.2 (m) = 0.755837249 (in)
5. Liquid depth
128
Hv = VHV / [(pi / 4) * (Dv )2 ] = 7234.847249 (m)
ri = Dv / 2 = 1.103 (m) = 0.377918625 (in)
Thickness = Cc + [(P * ri ) / (S * Ej – 0.6 * P)] = 1.024816852 (in) = 0.02603
(m)
h = [3 * (Dv / 2)] + Dv + Hv + 0.4 = 7235.295244 (m)
6.area of vessel = 2*pi*(dv/2)*ht = 436.1629606 (m2)
7. Metal
Vm = VDo - VDv = 11.35349638 (m3)
Wm = Vm * Density of the steel = 87421.9221 (kg)
Specification sheet for separator V-203
Equipment Name
Separator
Objective
To separate Aniline from the other gases
Equipment Number
V-203
Designer
Hassan Ahmed Butaleb
Type
Vertical
Location
After C-203
Material of Construction
Carbon Steel
Insulation
Glass wall and quartz
Operating Condition
Operating Temperature (oC) 144
Operating Pressure (psi)
40
Storage Tank
A storage tank is a container, usually for holding liquids, sometimes for
compressed gases (gas tank). The term can be used for reservoirs (artificial
129
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-202)
1- Assume:
 Cylindrical tank
 Square shape from inside (H=0.2D)
 Daily storage (time hold-up=0.399 hr)
2-Volume of the liquid:
Volume of the liquid = Total mass flow rate in* time hold-up
Where time hold-up is the time where the liquid is hold inside the tank
Volume of the liquid = 0.399 * 7.11678 = 2.8396 m3
3- The total volume:
The total volume = volume of liquid/0.3 = 9.46532 m3
The actual diameter=(5*the total volume/π)^(1/3) = 2.47016 m
5- The actual high:
The actual high = 0.2* diameter = 0.49403 m
6-Area of the storage tank:
Area of the storage tank =V/H = 19.1593 m2
130
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) = 1.23508
t(m) = 0.00583
8- Vapor pressure:
Vapor pressure = Log10 P* = A- (B/C+T) = 760 mmhg
Specification sheet of Tank T-202
Equipment Name
T-202
Objective
To store Aniline and Water
Designer
Hassan Ahmed Butaleb
Location
After three phase separator
Material of Construction
Carbon steel
Insulation
Quartz Wool
Operating Condition
Operating
(oC)
Temperature
40
Diameter (m)
2.47016
Operating Pressure (Pisa)
25
Height (m)
0.49403
Feed Flow Rate (mole/s)
7.11678
Thickness (m)
0.00583
Vapor Pressure (atm)
1
Total Area (m2)
19.1593
131
Design and Calculation For Tank (V-202)
1- Assume:
 Cylindrical tank
 Square shape from inside (D=0.2H)
 Daily storage (time hold-up= 0.6 hr)
2-Volume of the liquid:
Volume of the liquid = Total mass flow rate in* time hold-up
Where time hold-up is the time where the liquid is hold inside the tank
Volume of the liquid = 0.6 * 12.3125 = 7.38747 m3
3- The total volume:
The total volume = volume of liquid/0.3 = 24.6249 m3
4-The actual diameter:
The actual diameter=(5*the total volume/π)^(1/3) = 3.39733 m
5- The actual high:
The actual high = 0.2* diameter = 0.67947 m
6-Area of the storage tank:
Area of the storage tank =V/H = 36.2415 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
132
Cc : allowance for corrosion (m ) = 0.003175
ri(m) = 1.69867
t(m) = 0.00682
8- Vapor pressure:
Vapor pressure = Log10 P* = A- (B/C+T) = 760 mmhg
Specification sheet of Tank T-101
Equipment Name
V-202
Objective
To store Aniline and Water
Designer
Hassan Ahmed Butaleb
Location
After three phase separator
Material of Construction
Carbon steel
Insulation
Quartz Wool
Operating Condition
Operating
(oC)
Temperature
40
Diameter (m)
3.39733
Operating Pressure (Pisa)
25
Height (m)
0.67947
Feed Flow Rate (mole/s)
24.6249
Thickness (m)
0.00682
Vapor Pressure (atm)
1
Total Area (m2)
36.2415
133
Valves
A valve is a device that regulates, directs or controls the flow of a fluid
(gases, liquids, fluidized solids, or slurries) by opening, closing, or partially
obstructing various passageways. Valves are technically pipe fittings, but are
usually discussed as a separate category. In an open valve, fluid flows in a
direction from higher pressure to lower pressure. Valves are used in a variety
of contexts, including industrial, military, commercial, residential, and
transport. The industries in which the majority of valves are used are oil and
gas, power generation, mining, water reticulation, sewage and chemical
manufacturing. In nature, veins acting as valves are controlling the blood
circulation; heart valves control the flow of blood in the chambers of the
heart and maintain the correct pumping action. Valves play a vital role in
industrial applications ranging from transportation of drinking water to
control of ignition in a rocket engine. Valves may be operated manually,
either by a handle, lever or pedal. Valves may also be automatic, driven by
changes in pressure, temperature, or flow. These changes may act upon
a diaphragm or a piston which in turn activates the valve, examples of this
type of valve found commonly are safety valves fitted to hot water systems
or boilers.
Valve control
Control valves are imperative elements in any system where fluid flow must
be monitored and manipulated. Selection of the proper valve involves a
thorough knowledge of the process for which it will be used. Involved in
selecting the proper valve is not only which type of valve to use, but the
material of which it is made and the size it must be to perform its designated
task. The basic valve is used to permit or restrain the flow of fluid and/or
adjust the pressure in a system. A complete control valve is made of the
134
valve itself, an actuator, and, if necessary, a valve control device. The
actuator is what provides the required force to cause the closing part of the
valve to move. Valve control devices keep the valves in the proper operating
conditions; they can ensure appropriate position, interpret signals, and
manipulate responses. When implementing a valve into a process, one must
consider the possible adverse occurrences in the system. This can include
noise due to the movement of the valve, which can ultimately produce shock
waves and damage the construction of the system. Cavitations and flashing,
which involve the rapid expansion and collapse of vapor bubbles inside the
pipe, can also damage the system and may corrode the valve material and
reduce the fluid flow. No matter which avenue you take, the following
criteria should be considered to assure you select the right valve, the first
time:
• Process Parameters:
• Flow
• Pressure
• Temperature
• Chemical Compatibility:
• Media
• Concentration
• % of Solids
• Specific Gravity (sg)
• Process Requirements:
• On/Off versus control service
• Allowable leakage rate
• Cleanliness
• Emissions Control
• Available space and structural considerations
135
There are many types of valves like :
Globe Valves
A globe valve is a type of valve used for regulating flow in a pipeline,
consisting of a movable disk-type element and a stationary ring seat in a
generally spherical body. The valve can have a stem or a cage. The fluid's
flow characteristics can be controlled by the design of the plug being used in
the valve. A seal is used to stop leakage through the valve. Globe valves are
designed to be easily maintained. They usually have a top that can be easily
removed, exposing the plug and seal. Globe valves are good for on, off, and
accurate throttling purposes but especially for situations when noise and
cavitations are factors. A common example would be the valves that control
the hot and cold water for a kitchen or bathroom sink.
Butterfly Valves
Butterfly valves consist of a disc attached to a shaft with bearings used to
facilitate rotation. The characteristics of the flow can be controlled by
changing the design of the disk being use. Butterfly valves are good for
situations with straight flow and where a small pressure drop is desired.
There are also high performance butterfly valves.
136
Ball Valves
A ball valve is a valve with a spherical disc, the part of the valve which
controls the flow through it. The sphere has a hole, or port, through the
middle so that when the port is in line with both ends of the valve, flow will
occur. When the valve is closed, the hole is perpendicular to the ends of the
valve, and flow is blocked. Ball valves are good for on/off situations. A
common use for a ball valve is the emergency shut off for a sink.
The Gate Valve
This is the most common type of valve in use in industry and is used to start
or stop the flow of fluids. It gives a positive shut-off when closed and is
often used as a 'Block Valve' for isolating systems. The gate valve must be
either fully closed or fully open and never used to control flow, as the fluid
velocity will erode the valve internals - gate and body seats.
137
Section B
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.
138
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
139
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.
140
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 .
141
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 K 1 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.
142
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.
143
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
Qh
Definition
m
T
Mass flow rate in Kg/s.
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
S
Dimensionless temperature ratio.
Dimensionless temperature ratio.
Ft
Log mean temperature difference correction factor.
A
Heat transfer area
Nt
Number of tubes in a tube bundle.
Heat load transfer in the hot side, KW.
144
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.
Prandtl number.

Pr
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.
145
jf
Friction factor.
w
Fluid viscosity at the wall.
Ps
Shell-side pressure drop.
P
Maximum allowable internal pressure (psig).
Internal radius of shell before allowance corrosion is added
(in).
ri
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).
146
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).
147
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

Pass
2
= 23184.72
Where:
N t  Number of tubes.
At  Ac
Tubes
Pass
= 0.030758 m2
Where:
At  Total flow area.
Um 
m
= 180.7389 m/s
At
Where:
148
U m  Tube mass velocity.
At  Total flow area.
m  Mass flow rate in Kg/s.
Ut 
Um
 ref
= 0.353491 m/s
Where:
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).
149
j h  Tube side heat transfer factor.
k f  Thermal conductivity of stream.
Pr  Prandtl number.
lB 
Ds
5
= 222.869 mm
Where:
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:
150
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.
151
k f  Thermal conductivity of stream.
Pr  Prandtl number.
1
1
1



U 0 h0 hod
d 0 ln(
d0
)
di

2k w
d0 1
d 1
 0
d i hid d i hi
0.007249
=
Where:
U 0  The overall heat transfer coefficient.
hod  Outside coefficient (fouling factor).
hid  Inside coefficient (fouling factor).
Uo = 137.9562

L
Pt  N p 8 j f 

 di
  



 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
152
Where:
Ps  Shell-side pressure drop (N/m²) (pa).
j f  Friction factor.
L  Length of tube.
For carbon steel
t
Pri
 Cc = 5.476371 mm
SEj  0.6 P
Where:
t  Shell thickness (in).
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
Objective
Equipment Number
Designer
Type
Location
Utility
Material of Construction
Operating Condition
Heat exchanger
To cool the feed of the reactor which is contain hydrogen and
nitro benzene
E-101
Hussain Butaleb
Shell and tube heat exchanger
After heat exchanger (E-103)
cooling water
Carbon steel
Shell Side
153
Inlet temperature (oC)
Tube Side
Inlet temperature (oC)
Number of passes
Number of Tube Per Pass
Tube bundle Diameter (m)
U (W/C.m2)
5
Outlet temperature (oC)
170
357.3
8
23184.72
1337.214
137.9562
Outlet temperature (oC)
Thickness (mm)
Number of Tubes
LMTD (oC)
Heat Exchanger Area (m2)
178
5.476371
185477.7
181.0804
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)
154
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
m. Define the duty: heat transfer rate, fluid flow rates, temperature.
n. Collect together the fluid physical properties required: density,
viscosity,
o. Thermal conductivity.
p. Select a trail value for the overall coefficient, U.
q. Calculate the mean temperature difference, ΔTm.
r. Calculate the area required from Q=UAΔTm.
s. Calculate the bundle and shell diameter
t. Calculate the individual coefficients.
u. Calculate the overall coefficient and compare with the trail value.
v. Calculate the exchanger pressure drop.
w. Calculate thickness of the shell.
x. Find the price of the heat exchanger based on the heat transfer area
and the material of construction
Detailed calculation procedure
1- Heat load
155
Q = (m Cp ΔT) hot = (m Cp ΔT) cold, (kW)
2-Tube side flow
mcold 
Qhot
, (Kg/hr)
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 )
156
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 
Db  d o  t 
 K1 
1 / n1
, (mm)
Where,
Db: bundle diameter, (mm)
Nt: number of tubes
K1, n1: constants.
6- Shell diameter
Ds = Db + (Bundle diameter clearance) , (mm)
Using split-ring floating head type (bundle).
157
From figure (A.12) we get bundle diameter clearance.
7-Tube side Coefficient
Cold stream mean temperature= t 2  t1 , (°C)
2
Tube cross sectional area = 
4
di
2
, (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*(pt 2-0.917do2) /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.
158
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
,(W/m2.°C)
9- Pressure drop
Tube side
  L / di 
 u  2
  2.5  , (KPa)
Pt  N p 8 j f 
M
/
M
w 
 
 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
Where,
L: tube length, (m)
lb: baffle spacing(m)
159
 M

 M w



0.14
1
 
 hi 
Use fig.(12.30) to get jf.
10-Shell thickness
t
Pri
 Cc
SE j  0.6 P
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)
Sample Calculation:
Heat exchanger (E-103)
Shell side
Prameter
Unit
Inlet
Outlet
Mean
Tempreture Ti
C
30
177
103.5
Thermal Conductivty 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
Specfic Heat Cp
KJ/Kg.K 142.9
Heat Of Vaporization
KJ/kg
1.92E+02
Mass Flow Rate
kg/s
5.448889
165.67
1.203118
154.285
Tube side
Prameter
Unit
Inlet
160
Outlet
Mean
Temperture ti
C
184.81
163.87
174.34
Thermal Conductvity
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
161
Tm  Ft * Tlm =
-44.364451 °C
assume U=160 W/m2°C
Provisional area
A
Q
= 147.3956 m2
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
162
Tube – side coefficient
  2  
Tube cross sectional area =  d i =   * (26 E  3) 2 = 1.26E-05 m2
4
4
Tubes per pass = N t = 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
Re 
Pr 



ud i
 902.7943

Cp
 1.770509
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
5
 1114.219 mm
Tube pitch (pt)=1.25*do= 62.5 mm
163
Cross-flow area As=  ( pt  d o ) * Ds * Lb   1.241484 m 2


pt
Equivalent diameter de =  1.1 pt 2  0.917d o 2   0.035503 mm
 do 
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 


   d o  1
 d  h
 i  id
Uo= 143.8143 (W/m2 °C) Acceptable
Pressure drop:
Tube side
From figure for Re = 902.7943
jf= 3.00E-03
Neglecting the viscosity correction term
  L / di 
 u 2 
  2.5

pt  N p 8 j f 

/

2


w




pt  1477.507 pa
Shell side
From figure for Re = 129.5142
164
 do
 
 di
1

 hi



jf= 7.00E-02
Neglecting the viscosity correction term
 D  L
Pt  8 j f  s  t
 d e  lb
 u 2   

 
 2   w 
0.14
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
Heat Exchanger (E-103) specification sheet
Equipment Name
Objective
Designer
Type
Location
Material of Construction
Insulation
Operating Condition
Shell Side
Inlet temperature (oC)
Tube Side
Inlet temperature (oC)
E-103
To heat the nitro benzene feed by the reactor (R-101) product
Hussain Butaleb
Shell and tube heat exchanger
After heat exchanger (E-102)
Carbon steel
Quartz wool – Glass wool
30
Outlet temperature (oC)
77
184.81
Outlet temperature (oC)
163.87
Number of passes
Tube bundle Diameter (mm)
1
5571.096
Number of Tubes
Thickness (mm)
7506.796
4.437197
Q total (W)
1.05E+06
LMTD (oC)
-44.3645
U (W/m2 oC)
143.8143
Heat Exchanger Area (m2)
147.3956
165
Sample Calculation:
Heat exchanger (E-102)
Shell side
Prameter
Unit
Inlet
Outlet
Mean
Tempreture Ti
C
311.64
184.81
248.225
Thermal Conductivty 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
2.60E+02
Mass Flow Rate
kg/s
5.449167
Prameter
Unit
Inlet
Outlet
Mean
Temperture ti
C
114.06
249
181.53
Thermal Conductvity
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
Tube side
4.318056
166
Q = (m Cp ΔT) hot = 1.42E+03 KW
Tlm 
T1
C
311.64
T2
C
184.81
t1
C
114.06
t2
C
249
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
Tm  Ft * Tlm =
56.62082784 °C
Assume U= 135 W/m2°C
Provisional area
A
Q
= 185.5697 m2
UTm
167
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 
1 / n1
K1= 0.319
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  
Tube cross sectional area =  d i =   * (26 E  3) 2 = 0.00038 m2
4
4
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
168
The coefficient can be calculated from the following equation
 
hi d i
 j h Re Pr 0.33 
Kf
 w
Re 
Pr 



ud i
 55.84426

Cp
 1.589860751
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
5
 3377.608 mm
Tube pitch (pt)=1.25*do= 31.25 mm
Cross-flow area As=  ( pt  d o ) * Ds * Lb   11.40824 m 2

pt

Equivalent diameter de =  1.1  pt 2  0.917d o 2   0.017751 mm
 do 
Re = 634.4598
Pr = 1.727612687
Choose 25% baffle cut.
From figure
169
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  1
 d  h
 i  id
Uo= 152.4811 (W/m2 °C) Acceptable
Pressure drop:
Tube side
From figure
jf= 2.50E-02
Neglecting the viscosity correction term
  L / di 
 u 2 
  2.5

pt  N p 8 j f 

/

2


w




pt  0.104043 pa
Shell side
From figure for Re = 634.4598
jf= 2.00E-02
Neglecting the viscosity correction term
 D  L
Pt  8 j f  s  t
 d e  lb
 u 2   

 
2
  w 

0.14
Pt  40.31546 pa
Shell thickness
170
 do
 
 di
1

 hi



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
Heat Exchanger (E-102) specification sheet
Equipment Name
E-102
Objective
to heat hydrogen feed from the reactor (R-101) 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)
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
184.81
Tube Side
171
172
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 ( V v) and liquid
(VL) in m3/susing:
173
mV
VV 
V
And
mL
VL 
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 2 f 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:
174
t
Pri
Co ,
SE j  0.6P
Where P is the operating pressure in psig (P=405 psig), r i is the radius of the
vessel (ri=Dv/2 (in)), S is the stress value of carbon steel (S=13700 psia), E j
is the joint efficiency (Ej=0.85 for spot examined welding), and C c 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
rV =
8.197
kg/m3
mv=
20243.31 kg/h
mL=
24400
kg/h
Calculating the settling velocity:
U t  0.07
( L  V )
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:
175
o
VL 
VV 
= 0.686 m3/s
mV
V
= 0.00556 m3/s
mL
L
Expressing the cross-sectional area in terms of Dv:
o
Ac 
 Dv 2 f v
4

 * Dv 2 *0.5
4
= 0.393 Dv2 m2
Expressing the residence time in terms of Dv:
o
tr 
hv
Ua
= 3.916 Dv s
Expressing the actual residence time in terms of Dv:
Uv 
Vv
Ac
= 1.746Dv-2 m/s
Finding the value of Dv:
o
tra 
Lv
Uv
= 2.291 Dv3 s
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
176
Steam Drum Specification Sheet V-102
Equipment Name
Steam Drum
Objective
Separate the steam (upward) from the water 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.
177
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
Vapor flow rate (Mv)
Kg/h
3342
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)
0.125
Steel density
Kg/m3
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
178
7700
2- Volumetric flow rate:
Vv =
lv =
kg
)
s
kg
Vapor density ( 3 )
m
Vapor flow rate (
kg
)
s
kg
liquid density ( 3 )
m
liquid flow rate (
= 4.573521 m3/s
= 0.012049 m3/s
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)
179
Specification sheet for separator V-101
Equipment Name
Objective
Separator
To separate hydrogen (vapor) from the 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
1.085111
Height (m)
10.93399
Dimensions
Diameter (m)
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
180
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.
181
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)
127.0822199 Thickness (m)
0.01178
Vapor Pressure (atm)
1
583.2946
Total Area (m2)
182
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 …………
183
The typical proportions of the hydrocyclone are shown in figure ………
Design equations:
𝑑50
𝐷𝑐 3 𝜇
= 4.5 [ 1.2
]
𝐿 (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
184
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
𝑔
𝑐𝑚3
185
𝜌𝑎𝑖𝑟 = 1.2344
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
186
Specification sheet for hydro cyclone (T-102)
Equipment Name
T-102
Objective
To separate the air from the solid waste 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.
187
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
188
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.
189
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,
190
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.
191
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
192
options include heavy/light valve combination, flush designs and nonrotating 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
193
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.
194
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.
195
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)
196
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)
197
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.
198
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:
199
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 
LxMwt
L
Where:
L=liquid flow rate.
Choose plat pass from figure
200
Column area:
AC 

4
D2
Where:
AC= column area.
Net area:
An  AC  Ad
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 ;
201
Weir length:
weir length  0.75D
Wier crest:
2
max how
 max liquid rate  3

 750
  L xweir length 
2
 min liquid rate  3

min how  750

xweir
length
 L

Where:
how= weir crest.
Vapor velocity:
U h (min) 
K 2  0.9(25.4  hole diameter)
 0.5
Where:
Uh= vapor velocity.
Choose Turndown percentage = 70%
From figure get K2
202
Actual min. vapor velocity:
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 
12.5 x10 3
L
Where:
203
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


2
hb  hw  hdc  ht  how
Where:
hb= back-up in downcomer.
Downcomer residence time:
tr 
hb xAdx L
Lwd
Where:
tr= downcomer residence time.
Lwd= minimum liquid flow rate.
UV 
volumetric flow rate
An
204
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) = 2.91E+03 kmol/hr
ρv
= 3.28769 kg/m3
MW = 107.2743
Liquid rate (L) = 2.42E+03 kmol/hr
ρL
= 1050.493 kg/m3
MW = 122.5036
Surface Tension = 4.74E+01 N/m
Bottom properties
Vapor rate (Vm) = 2.90E+03 kmol/hr
ρv
= 3.906938 kg/m3
205
MW = 1.23E+02
Liquid rate (Lm) = 2.91E+03 kmol/hr
ρL
= 981.5581 kg/m3
MW = 123.0347
Surface Tension = 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:
FLV
L

V
0.5
 v 
  = 0.046591
 L 
For bottom:
FLV
L

V
0.5
 v 
  = 0.063308
 L 
Where:
FLV= liquid vapor flow rate.
L= liquid flow rate.
V= vapor flow rate.
Take tray spacing 0.75 m
Find K1 from figure:
206
Bottom K1 =1.00E-01
Top K1 = 1.5E-01
Flooding velocity:
For top:
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
V
= 3.26E+01 m3/s
For bottom:
Vmax 
MwtV
V
= 1.84E+01 m3/s
Where:
207
MwtV= vapor molecular weight.
Xw= 0.990221
Xd= 0.724112
Volatility= 0.45
Nmin= 4.574458
Net area required:
For top:
Anet 
Vmax
= 3.440657 m2
UV
For bottom:
Anet 
Vmax
= 3.437593 m2
UV
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
D2
= 3.440759 m2
Net area:
208
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
 =
 750
  L xweir length 
118.5807 mm liquid
2
 min liquid rate  3
 = 93.48587 mm liquid
min how  750

xweir
length
 L

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
Ah
Where:
Ah= hole area.
209
= 49.19689 m/s
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:
210
tr= downcomer residence time.
Lwd= minimum liquid flow rate.
UV 
volumetric flow rate
An
= 6.070
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 (for carbon steel)
Cc = 1.25E-01 in
211
Specification sheet of distillation (C-101)
Equipment Name
Distillation
Objective
To mix hydrogen gas with nitrobenzene gas
Equipment Number
C-101
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:
212
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.
213
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:
214
215
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
216
Design for pressure drop 125mm H2O/m packing
From Figure:
K4 = 2.2
At flooding K4 = 4
Percentage Flooding = 74.16198487 %
L (kg/h)
7101
x1
0.018787
x2
0.006161
y1
0
y2
0.99975
y2'
0.000003
M
0.0625
Vmin
1434.874335
V actual = Vmin *1.5
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
217
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
218
Specification sheet of stripper (C-201)
Equipment Name
Stripper
Objective
To separate water and aniline (vapor) from waste
water (liquid)
Equipment Number
C-201
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
Student name :Abdulhameedyousef al-awadhi
219
ID number : 208115609
Equipment name
Equipment number
Distillation column
C-203
Cooler
V-103
Cooler
E-104
Heat exchanger
E-201
Heater
E-202
Compressor
K-101
Tank
T-103
Tank
T-104
Tank
V-104
Distillation Columns
Introduction :Distillation is the most common form of separation technology used in
petroleum refineries, petrochemical and chemical plants and natural gas
processing plants. As a matter of fact the separation of liquid flows depends
on difference in volatility between the components of the feed, whereas the
compound with higher relative volatilities will be separated first. It is
possible to classify distillation according to many things:
 Firstly…
( to the types of distillation which can be generally divided into two
types):1) Continuous distillation: This receives a feed, and parts it into two or
more products continuously. In this specific distillation the principle is
220
that at any point the amount entering the still and the amount coming
out from the still must be equal each other. It is widely used with large
volume products like benzene, plastic, jet fuel and monomers.
2) Batch distillation: receives on lot (or batch) at a time of feed and
divide it into products by selectively removing the more volatile
fractions over time. That's why batch distillation used for smaller
volume products and in plants that produce many of various products
but use the same still for these multiple products (in different batches).
 Secondly…
( to equipment typewhich can be either one of the following type ):1) Trays (or Plates): This forces a rising vapor to bubble through a pool
of descending liquid.
2) Packing: creates a surface for liquid in order to extend on. For the
sake of mass-transfer between the liquid and vapor; the thin liquid
film must have a great surface area.
Fig1 : structured of packing
221
Types of plates:
1) Sieve plates: This is the simplest type of cross- flow plate. The vapor
passes through perforation in the plate and the liquid is retained on the
plate by the vapor flow. At low flow rates liquid will weep through
the holes reducing the plate efficiency. The perforations are usually
small holes.
Fig2 : Sieve plate
2) Bubble cap plates : The vapor passes up through short pipes called
risers, covered by a cap. The bubble cap plate is traditional, oldest,
type of cross flow plate.
Fig3 : shows typical bubble-cap trays in a distillation tower.
222
3) Valve plates:Valve plates are essentially sieve plates with large
diameter holes covered by movable flaps which lift as the vapor flow
increases. As the area for vapor flow varies with the flow rate, valve
plates can operate efficiently at lower flow rates than sieve plates.
 Thirdly …
( to process type):1) Refining
2) petrochemical
3) Chemical
4) gas treating.
 Fourthly …
(to distillation process configuration):1) Distillation
2) absorption
3) stripping
4) azeotropic
5) extractive
6) complex.
Main Components of Distillation Columns : A vertical shell in which the separation of liquid components is
accomplished.
 Column internals, for example trays and/or packings.
 A reboiler for supplying the required vaporization for the distillation
process.
 A condenser , its main function is to cool and condense the vapor
coming out from the top of the column
 A reflux drum, in order to recycle the liquid reflex this reflux drum is
used to hold the condensed vapor from the top of the column.
223
Fig4 :A typical distillation unit with a single feed and two product streams.
Fig5 : the vapor moves up the column, and as it exits the top of the unit, it is cooled
by a condenser.
Fig6. Shows the heat is supplied to the reboiler to generate vapor.
224
Fig7 : Distillation column
225
1.1.4.2. Calculation Procedure
1. Specify the properties of outlets streams: (flow rate, density and
surface tension) for both vapor and liquid from HYSYS.
2. Calculate the maximum liquid and vapor outlet flow rate.
3. Choose tray spacing and then determine K1 and K2 .
4. Calculate correction factor for Bottom K1 and Top K1.
5. Design for X% flooding at maximum flow rate for top and
bottom part of distillation.
6. Calculate the maximum flow rates of liquid.
7. Calculate Net area required.
8. Take down comer area as %Y of the total column Cross sectional
area.
9. Calculate the column diameter.
10.Calculate the column height using the actual number of stage.
11.Calculate column area, down comer area, active area, net area,
hole area and weir length.
12.Calculate the actual min vapor velocity.
13.Calculate Back-up in down comer.
14.Check residence time.
15.Check entrainment.
16.Calculate number of holes.
17.Calculate area of condenser and reboiler.
18.Calculate Thickness of the distillation.
19.Calculate the feed point location.
226
20.Calculate cost.
symbol
Nomenclature
V
Vapor flow rate (Kmol/hr)
L
Liquid flow rate (Kmol/hr)
ρ
Density (Kg/m3 )
M.Wt
Molecular weight
FLV
Liquid vapor flow factor
K1
Constant
uf
Flooding vapor velocity (m/s)
uv
Designed vapor velocity (m/s)
Dc
Column diameter
Ac
Column area
Ad
Downcomer area
An
Net area
Aa
Active area
Ah
Hole area
Iw
Weir length
hw
Weir hight
dh
Hole diameter
Lwd
Maximum liquid rate
how
Weir crest mm liquid
uh
Minimum vapor velocity (m/s)
Co
Orifice coefficient
hd
Dry plate drop mm liquid
hr
Residual head mm liquid
ht
Total plate drop mm liquid
hap
Height of the bottom edge of the apron above the plate (mm)
Aap
Clearance area under the downcomer (m2 )
hdc
Head loss in the downcomer mm
hb
Downcomer back-up mm
tr
Residence time (sec)
227
Hc
Column hight
Q
Heat flow (W)
ri
Inner diameter (inch)
S
Maximum allowable working stress (psi)
Ej
Efficiency of joints expressed as fraction
Cc
Allowance for corrosion m
distillation column (T-101)
Assumptions1) Tray column.
2) Sieve plate.
3) Material of the distillation is carbon steel.
4) Plate spacing= 0.6 m
5) Efficiency = 51%
6) Flooding % = 85%
7) Weir height = 45 mm
8) Hole diameter = 4 mm
9) Plate thickness =5 mm
10) downcomer area 12% of total
228
Distillation Column sample Calculation (C-203)
Table1: (C-203) column properties:
Top
Bottom
Unit
Vapor rate (Vn)
5930.5
3500
kmol/hr
Mass Density for Vapor ρv
1.027
9.2962
kg/m3
Molecular Weight (M.Wt)
93.74
92.45
Liquid rate (Ln)
124
5672.2
kmol/hr
Mass Density for Liquid ρL
909.57
1038.3
kg/m3
Molecular Weight (M.Wt)
93.74
123.06
Surface Tension
0.028495
0.45945
N/m
Number of Stages:
Table 3-2 Actual and Theoretical number of stage
Number of stages
50
Efficiency
80%
Actual number of stages
56
Column diameter:
Liquid vapor flow factor:
Top
Bottom
unit
Mass Density for Vapor ρv
1.027
9.2962
kg/m3
Mass Density for Liquid ρL
909.57
1038.3
kg/m3
Surface tension
0.028495
4.5945
N/m
229
Bottom = FLV = (L/V)*(ρv/ ρL)0.5 =0.15334683
Top =
FLV = (L/V)*(ρv/ ρL)0.5 = 0.0329432
Take plate spacing as 0.6 m
Fig1: Flooding velocity for sieve plates
From the figure above:-
Base K1 = 0.009
Top K1 = 0.01
Correction for surface tensions
Base K1 = 0.1.68E-01
Top K1 = 1.07E-01
Flooding velocity:
Base = uf = K1((ρL- ρv)/ ρv)0.5 = 3.192532 (m/s)
Top = uf = uf = K1((ρL- ρv)/ ρv)0.5 =1.772311 (m/s)
Design for 85% flooding at maximum flow rate
Base uv = uf*0.85 = 1.506464 (m/s)
Top = uv = uf*0.85 = 2.713652 (m/s)
230
Maximum volumetric flow rate
Bottom = Vmax= Vn*M.Wt/ρv*3600 = 9.668676 (m3/s)
Top = Vmax= Vn*M.Wt/ρv*3600 =7.51819 (m3/s)
Net area required:
Bottom = A=Vmax/uv = 6.418125 (m2)
Top = A=Vmax/uv = 2.770506 (m2)
“Taking downcomer area as 12 per cent of total”
Column cross-sectional area
= 3.2407 /(1 – 0.12 ) = 7.293324 (m2)
Base =
Top =
= 2.5517 /( 1 – 0.12 ) = 3.148303 (m2)
Coloumn diameter:
Bottom = D = (Anet *4/π)0.5 = 3.047318 (m)
Top = D = (Anet *4/π)0.5 = 1.002135 (m)
Use same diameter above and below feed
D = 3.04731816 (m) = 10 (ft)
Column Height:
Total height = H=(Number of stage * Plate spacing)+Clolumn Diameter
= 36.6473182 (m) = 117 (ft)
Maximum volumetric liquid rate = ( LN*M.Wt)/(ρL*3600) = 0.186742
(m3/s)
231
Fig2: Selection of liquid flow arrangment
From the figure above: single pass plate is used
Provisional plate design:
Column diameter = Dc = 3.04731816 (m)
column area = (3.14/4)*(Dc^2) = 7.29194769 (m2)
Down comer area Ad = 0.87503372 (m2)
Net area = An = Ac – Ad = 6.41691397 (m2)
Active area = Aa = Ac - 2*Ad = 5.54188025 (m2)
Hole area = Ah = 10% of Aa= 0.55418802 (m2)
232
Fig3: Relation between down comer are and weir length
From the figure above:
= 12
Lw/Dc = 0.76
Weir Length = lw= 2.3159618 (m)
Take weir height = hw = 50 (mm)
Hole diameter (dh) = 5 (mm)
Plate thickness = 5 (mm)
Check weeping:
Maximum liquid rate Lw = (Ln*Mwt)/3600 = 193.8947 (kg/s)
Turndown percentage = 0.70
Minimum liquid rate = Lwd *0.8 = 135.7263 (kg/s)
Maximum = how =750*(Lw/(ρLlw))2/3 = 139.980991 (mm liquid )
Minimum = how =750*(Lw/(ρLlw))2/3 = 110.357306 (mm liquid)
At minimum rate = hw + how = 160.3573 (mm liquid)
233
Fig4: Weep point correlation (Eduljee, 1959)
From the figure above:
K2 = 31.05
Minimum vapor velocity through hole:
uh (min) = (K2-0.90(25.4-dh))/ρv0.5= 4.162065 (m/s)
Actual minimum vapor velocity = Minimum vapor rate/Ah = 12.21259 (m/s)
So minimum operating rate will be well above weep point.
Plate pressure drop:
Dry plate drop
Maximum vapor velocity through holes (uh) = Bottom Vmax/Hole area
Ah=17.44656 (m/s)
234
Figure 3-12 Discharge coefficient, sieve plates (Liebson et al. 1957)
From the figure above:
Plate thickness / hole dia. = 1
Ah/Ap = Ah/Aa = 0.1
Co = 0.8
VaporVelocityMAX 
VolemetricFlowRate
Ah
 PlateThickness 


Holediamet er 

Re ad
 Find Co Fig .11.34 
 Ah Ah




 AP Aa

2
u  
hd  51 h  v
 Co   L
12.5 103
hr 
L
ht  hd  hr  hw  howMAX 
235
hd = 196.9763 mm
Residual head (hr) = 12.0389098 mm
Total plate drop (ht) = 398.9962 mm
Downcomer Liquid Back-Up:
hap  hw  10
Aap  hap lw 
 L

hdc  166 wd ; Am  MIN Ad ; Aap 
  L  Am 
hb  ht  hdc  hw  howMAX 
Find :
PlateSpacing  hw
2
hap = 40 mm
Aap = 0.09263847 m2
Since Aap is less than Ad use equation 11.92:
hdc = 0.674546 mm
hb = 589.6517 mm
Check Residence Time:
Ah 
t r  d bc L
Lwd
tr = 6.969157 s
Check Entrainment:
uv 
VolumetricflowRate
An
Flooding % 
uv
u fMAX
Find  f F LV Fig .11.29
uv = 1.506749 m/s
Percent flooding = 85.0160381 %
236
From Figure 11.29 :
Number of Holes:
AreaOfOneHole  0.25Dh
2
Ah
AreaOfOneHole
# holes 
Area of one hole = 1.9635 * 10^-5 m2
Total number of holes = 28224.56
hole on one plate = 504.01
Thickness:
rj 
t
D
2
Pr j
SE j  0.6 P
 Cc
rj
59.98661
inch
P
105
psi
Ej
0.85
Cc
0.125
inch
S
13700
psi
t
0.668826
inch
t
16.98818
mm
237
Area of Condenser:
∆T=
0.2
Co
U=
1000
W/m2.Co
Q=
5957000
W
Area =
29785
m2
320603.1
ft2
∆T=
0.5
Co
U=
900
W/m2.Co
Q=
4354000
W
Area =
9675.556
m2
104146.8
ft2
Area of Reboiler:
238
Equipment Name
Distillation column
Objective
To separate anilne from other residue
Equipment Number
C-203
Designer
Abdulhameed al-awadhi
Type
Tray column
Location
aniline recovery section
Material of Construction
Carbon steel
Insulation
foam wool
Total Cost ($)
1.85E+05
Key Components
Light
aniline
Heavy
Residues
Height (m)
10.687
Dimensions
Diameter (m)
Bottom 1.067 (m)
Top 1.0043 (m)
Number of Trays
50
Reflux Ratio
0.98
Tray Spacing
0.74
Type of tray
Sieve tray
Number of Holes
28224.56
Cost( $)
Vessel($)
5.53E+04
Trays($)
3900
Condenser Unit ($)
55400
Reboiler ($)
66400
Heat Exchanger
239
Introduction:In general a heat exchanger is a device made for efficient heat transfer from
one fluid to another across a solid surface. It has been used a lot in
engineering plants whereas exchanger analysis and design include both
convection and conduction. If the exchanger is un-insulated and its external
surfaces are very hot the radiative transfer between the exchanger and the
environment cannot be eliminated.
Application:1. petrochemical processing
2. petroleum refining
3. Waste water treatment.
4. Refrigeration systems.
5. Wine-brewery industry.
6. Petroleum industry.
Flow arrangement:-
Fig1: Countercurrent (A) and parallel (B) flows.
Heat exchangers may be classified according to their flow arrangement. In
parallel flow heat exchangers, the two fluids enter the exchanger at the same
end, and travel in parallel to one another to the other side. In counter flow
heat exchangers the fluids enter the exchanger from opposite ends. The
240
counter current design is most efficient, in that it can transfer the most heat.
In a cross flow heat exchanger, the fluids travel roughly perpendicular to one
another through the exchanger.
For efficiency, heat exchangers are designed to maximize the surface area of
the wall between the two fluids, while minimizing resistance to fluid flow
through the exchanger. The exchanger's performance can also be affected by
the addition of fins or corrugations in one or both directions, which increase
surface area and may channel fluid flow or induce turbulence.
The driving temperature across the heat transfer surface varies with position,
but an appropriate mean temperature can be defined. In most simple systems
this is the log mean temperature difference (LMTD). Sometimes direct
knowledge of the LMTD is not available and the NTU method is used.
1- Parallel-flow Heat Exchanger:
Fig2: Temperature distribution along tube axis(Parallel flow).
Figure above shows a fluid flowing through a pipe and exchanges heat with
another fluid through an annulus surrounding the pipe. In a parallel flow heat
exchanger fluids flow in the same direction. If the specific heat capacity of
fluids are constant, it can be shown that:
241
where,
dQ/dt= Rate of heat transfer between two fluids.
U= Overall heat transfer coefficient.
A= Area of the tube.
T= Logarithmic mean temperature difference defined by:
2- Counter-flow Heat Exchanger:
Fig3: Temperature distribution along tube axis(Counter –flow).
Figure above shows a fluid flowing through a pipe and exchanges heat with
another fluid through an annulus surrounding the pipe. In a counter-flow
heat exchanger fluids flow in the opposite direction. If the specific heat
capacity of fluids are constant, it can be shown that:
242
where,
dQ/dt= Rate of heat transfer between two fluids
U= Overall heat transfer coefficient
A= Area of the tube
T= Logarithmic mean temperature difference defined by:
3- Cross-flow Heat Exchanger:
Fig4: cross-flow heat exchanger.
In a cross-flow heat exchanger the direction of fluids are perpendicular to
each other. The required surface area, Across for this heat exchanger is
usually calculated by using tables. It is between the required surface area for
counter flow, Acounter and parallel-flow, Aparallel i.e.
Acounter< Across <Aparallel
243
Fig5: Temperature profile of concurrent and counter current flow.
Types of heat exchangers:
-Shell and Tube heat exchanger.
-Plate heat exchanger.
-Regenerative heat exchanger.
-Adiabatic Wheel heat exchanger.
-Fluid heat exchangers.
-Dynamic Scraped surface heat exchanger.
-Spiral Heat Exchangers.
244
Fig5: Types of Heat Exchanger.
shell and tube heat exchanger:
A shell and tube heat exchanger is a class of heat exchanger designs. It is the
most common type of heat exchanger in oil refineries and other large
chemical processes, and is suited for higher-pressure applications. As its
name implies, this type of heat exchanger consists of a shell (a large pressure
vessel) with a bundle of tubes inside it. One fluid runs through the tubes, and
another fluid flows over the tubes (through the shell) to transfer heat
between the two fluids. The set of tubes is called a tube bundle, and may be
composed by several types of tubes: plain, longitudinally finned, etc.
245
Fig6: shell & tube heat exchanger
The function of heat exchanger depends on the fact that the heat is
transferred whenever a temperature difference occurs. In a heat exchanger,
there is a cold stream and a hot stream. The two streams are separated by a
solid wall. The wall must be thin and conductive in order for heat exchange
to occur. Yet the wall must be strong enough to withstand any pressure by
the fluid.
Shell and tube heat exchanger design
There can be many variations on the shell and tube design. Typically, the
ends of each tube are connected to plenums (sometimes called water boxes)
through holes in tube sheets. The tubes may be straight or bent in the shape
of a U, called U-tubes.
246
Fig7: U-tube heat exchanger.
In nuclear power plants called pressurized water reactors, large heat
exchangers called steam generators are two-phase, shell-and-tube heat
exchangers which typically have U-tubes. They are used to boil water
recycled from a surface condenser into steam to drive the turbine to produce
power. Most shell-and-tube heat exchangers are either 1, 2, or 4 pass designs
on the tube side. This refers to the number of times the fluid in the tubes
passes through the fluid in the shell. In a single pass heat exchanger, the
fluid goes in one end of each tube and out the other.
247
Fig8: 1-pass straight-tube heat exchanger.
Surface condensers in power plants are often 1-pass straight-tube heat
exchangers. Two and four pass designs are common because the fluid can
enter and exit on the same side. This makes construction much simpler.
Fig8: 2-pass straight-tube heat exchanger
248
There are often baffles directing flow through the shell side so the fluid does
not take a short cut through the shell side leaving ineffective low flow
volumes.
Counter current heat exchangers are most efficient because they allow the
highest log mean temperature difference between the hot and cold streams.
Many companies however do not use single pass heat exchangers because
they can break easily in addition to being more expensive to build. Often
multiple heat exchangers can be used to simulate the counter current flow of
a single large exchanger.
Selection of tube material
To be able to transfer heat well, the tube material should have good thermal
conductivity. Because heat is transferred from a hot to a cold side through
the tubes, there is a temperature difference through the width of the tubes.
Because of the tendency of the tube material to thermally expand differently
at various temperatures, thermal stresses occur during operation. This is in
addition to any stress from high pressures from the fluids themselves. The
tube material also should be compatible with both the shell and tube side
fluids for long periods under the operating conditions (temperatures,
pressures, pH, etc.) to minimize deterioration such as corrosion. All of these
requirements call for careful selection of strong, thermally-conductive,
corrosion-resistant, high quality tube materials, typically metals. Poor choice
of tube material could result in a leak through a tube between the shell and
tube sides causing fluid cross-contamination and possibly loss of pressure.
Any component or the entire unit can be made of materials such as carbon
steel, stainless steel, nickel, nickel alloys or other special alloys.
249
Advantages:
 connections that come in standardized sizes for easy assembly and
feature additional thread and surface protection for clean installation.
 U-bend tubes expanded into a tubesheet which allow for tube
expansions and contractions due to thermal fluctuations.
 gaskets that are made of high quality compressed fibres which lends to
reusability.
 A standard cast-iron or steel head for heavy duty services (also
available as a spare part).
 saddle attaches which make for quick and easy mounting.
 punched baffles with minimal clearances between tubes guaranteeing
correct fluid flow and minimized bypass.
 A welded shell protected with high quality paint for corrosion
resistance.
Plate heat exchanger:
A plate heat exchanger is an type of heat exchanger that uses metal plates to
transfer heat between two fluids. This has a major advantage over a
conventional heat exchanger in that the fluids are exposed to a much larger
surface area because the fluids spread out over the plates. This facilitates the
transfer of heat, and greatly increases the speed of the temperature change. It
is not as common to see plate heat exchangers due to the fact that they need
well-sealed gaskets to prevent the fluids from escaping, although modern
manufacturing processes have made them feasible.
The concept behind a heat exchanger is the use of pipes or other containment
vessels to heat or cool one fluid by transferring heat between it and another
fluid. In most cases, the exchanger consists of a coiled pipe containing one
250
fluid that passes through a chamber containing another fluid. The walls of
the pipe are usually made of metal, or another substance with a high thermal
conductivity, to facilitate the interchange, whereas the outer casing of the
larger chamber is made of a plastic or coated with thermal insulation, to
discourage heat from escaping from the exchanger.
Design of plate and frame heat exchangers
Fig9: Schematic conceptual diagram of a plate and frame heat exchanger.
The plate heat exchanger is a specialized design well suited to transferring
heat between medium- and low-pressure liquids. Welded, semi-welded and
brazed heat exchangers are used for heat exchange between high-pressure
fluids or where a more compact product is required. In place of a pipe
passing through a chamber, there are instead two alternating chambers,
usually thin in depth, separated at their largest surface by a corrugated metal
plate.
The plates used in a plate and frame heat exchanger are obtained by one
piece pressing of metal plates. Stainless steel is commonly used metal for the
plates because it is able to withstand high temperatures, resistance to rusting
while also being a strong material. The plates are often spaced by rubber
sealing gaskets which are cemented into a section around the edge of the
plates. The plates are pressed to form troughs at right angles to the direction
251
of flow of the liquid which runs through the channels in the heat exchanger.
These troughs are arranged so that they interlink with the other plates which
forms the channel with gaps of 1.3-1.5 mm between the plates.
The plates produce an extremely large surface area, which allows for the
fastest possible transfer. Making each chamber thin ensures that the majority
of the volume of the liquid contacts the plate, again aiding exchange. The
troughs also create and maintain a turbulent flow in the liquid to maximize
heat transfer inthe exchanger. A high degree of turbulence can be obtained at
low flow rates and high heat transfer coefficient can then be achieved.
A plate heat exchanger consists of a series of thin, corrugated plates which
are mentioned above. These plates are gasketed, welded or brazed together
depending on the application of the heat exchanger. The plates are
compressed together in a rigid frame to form an arrangement of parallel flow
channels with alternating hot and cold fluids.
As compared to shell and tube heat exchangers, the temperature approach in
a plate heat exchangers may be as low as 1 °C whereas shell and tube heat
exchangers require an approach of gives 5°C or more.
252
Advantages :
 Plates are attractive when material costs are high.
 Plate Heat Exchanger are easier to maintain.
 Plate Heat Exchanger are more flexible , it is easy to add extra plates .
 Plate Heat Exchanger are more suitable for highly viscous materials.
 the temperature correction factor ,Ft, will normally be higher with
plate heat exchangers , as the flow is closer to true counter – current
flow.
 Fouling tends to be significantly less in plate heat exchangers .
 Expansion and reduction of the heat transfer area is possible in a plate
heat exchanger.
 It is easily dismantled for inspection and cleaning.
 the plates are also easily replaceable due the fact that plates can be
removed and replaced individually.
Disadvantage:
 The main weakness of the plate and frame heat exchanger is the
necessity for the long gaskets which holds the plates together,
Although these gaskets are seen as a weakness towards this type of
heat exchanger, it has been successfully run at high temperatures and
pressures.
 A plate is not a good shape to resist pressure and plate heat
exchangers are not suitable for pressure greater than about 30 bar.
 The maximum operating temperature is limited to about 250°C , due
to the performance of the available gasket materials .
253
Selection: High/ Low pressure limits.
 Thermal Performance.
 Temperature ranges.
 Product Mix (liquid/liquid, particulates or high-solids liquid).
 Pressure Drops across the exchanger.
 Fluid flow capacity.
 Cleanability, maintenance and repair.
 Materials required for construction.
 Ability and ease of future expansion.
254
Nomenclature :Symbol
Definition
Q
Heat load (kw)
Cp
Heat capacity of water(kj/kg oC)
t1
Inlet hot stream temperature (˚C)
t2
Outlet stream temperature (˚C)
T1
Inlet cold stream temperature (˚C)
T2
Outlet cold temperature (˚C)
ΔTlm
Log mean Temperature (˚C)
Ft
Temperature correction factor
ΔTm
True temperature difference (˚C)
R
is the shell side flow × specific heat / tube side
flow×specificheat
S
is temperature efficiency of the heat exchanger
A
Provisional Area (m2)
do
Outer diameter (mm)
Lt
Length of tube(mm)
Nt
Number of tubes
Db
bundle diameter (mm)
Ds
Shell diameter (mm)
di
Tube inside diameter (mm)
Ų
Linear velocity (m/s)
Ρ
Density (kg/ m3)
Re
Reynolds number
255
Pr
Prendelt number
jh
Heat transfer function
µ
Viscosity
Lb
Baffle spacing (mm)
Pt
Tube pitch (mm)
As
Cross flow area (m2)
Gs
Mass velocity (kg/s.m2)
De
Equivalent diameter for triangular arrangement (mm)
hs
shell side coefficient (w/m2. °C)
Uo
Overall heat transfer coefficient (w/m2. °C)
ΔP
Tube side pressure drop (N/m^2). (pa)
Np
number of tube side passes
Υ
tube side velocity (m/s)
L
length of one tube, (m)
T
shell thickness (in)
P
internal pressure (psi)
ri
internal radius of shell (in)
Ej
efficiency of joint expressed as fraction
Cc
allowance for corrosion (in)
S
maximum allowable working stress (psi)
256
Assumptions :1-Assume a counter current flow heat exchanger because it provides more
effective heat transfer.
2-The value of the overall heat transfer coefficient was assumed in the shell
and tube heat exchanger design.
3-Assume the outer, the inner diameter and the length of the tube on both the
Shell and tube heat exchanger design.
Procedure for calculation of Air Coolers
1- calculate the heat duty :
Q= m Cp ΔT
Where,
Q: heat duty
m: mass flow rate
Cp: heat capacity
ΔT : Temperature difference .
2-Overall Heat Transfer Coefficient
Assume from Book (chemical engineering design) coulson& Richardson
4th Edition Table 12.1
3-Outlet Air Temperature
 T  T2

t 2  t1  0.0009U  1
 t1 
 2

Th  T1  t 2
Tc  T2  t1
lMTD 
Th  Tc
ln Th / Tc
257

5-Calculate correction on mean temperature difference
S

t 2  t1
T1  t1
R

T1  T2
t 2  t1
From figure 12.19 take Ft
6- Calculate Area
Δ Tm=lMTD*Ft
Q=UAΔTm
A=Q/(UΔTm)
7- # tube
Nt=A/A assume
8- Calculate horse power requirement
hydraulic Hp=ACFM*density of air *diff head/(33000)
diff head=total dela @fan ins H2O*5.193/Density
Bhp=hydraulic hp/η
9- Cost (from matche.com)
258
Cooler Design: V-103
Fluid properties:
Shell Side ( Cooling water stream )
Flow rate
3.97E+01
Kg/h
Inlet Temperature ,T1
135
o
C
Outlet Temperature ,T2
266
o
C
Average Heat Cpavg
75.8675
KJ/kgoC
Average Mass Density, ρavg
1039.8
kg/m3
Average Viscosity of stream, µavg
0.542
mNs/m2
Average Thermal conductivity, Kf
0.498
W/moC
Calculation of Heat load:
Qh = mh *Cp * ( T2-T1)
where:
Qh = heat load in the hot side (KW)
mh = mass flowrate of hot fluid (Kg/h)
Cp = heat capacity of hot fluid (kJ/kgoC)
T1 = inlet temperature (oC)
T2 =outlet temperatue (oC)
Heat load = -1090.4 KW> 1000 KW
The type of the heat exchanger is shell and tube heat exchanger
Tube side ( gas stream)
Flowrate
71.01
Kg/h
Average Heat Capacity, Cp
76.49
kJ/kgoC
Average Mass Density, ρ
502.97
kg/m3
Average Viscosity of stream, µ
0.159
mNs/m2
259
Average Thermal conductivity of stream, Kf
0.317
W/moC
inlet Temperature , T1
310
o
C
outlet Temperature, t2
295
o
C
Log mean Temperature calculation:
∆Tlm =(T1-t2)-(T2-t1) / ln((T1-t2)/(T2-t1))
where:
∆Tlm = log mean temperature differace
T1 = inlet shell side fluid temoerature (oC)
T2 = outlet shell side fluid temerature (oC)
t1
= inlet tube side temoerature (oC)
t2 = outlet tube side temerature (oC)
∆Tlm = -89.854 oC
Temperature correction Fctor calculation:
Using two shell pass and four or multiple of Two tube passes Np= 2
R = (T1-T2) / (t2-t1)
R = 8.7333333
S = (t2-t1) / (T1-t1
S = 0.0857143
Ft = 0.99
From figure 12.20
∆Tm = Ft * ∆Tlm
where:
∆Tm = true temperature difference
Ft
= the temperature correction factor
260
∆Tlm = log mean temperature differace
∆Tm = -88.95539 oC
Assuming U = 100 W/m2 oC ( From table 12.1)
A= Q / U * ∆Tm
where:
A = provisional area (m2)
Q = heat load (kW)
U = overall heat transfer coefficient (W/m2 oC)
Provisional area = 97.37333 m2
Choosing
Tube outside diameter(do) = 32 mm 1.25984
Tube inner diameter(di) = 28 mm 1.10236
Tube length(L) = 3.4 m 133.858
Take tube materail is cupro- nickel
Area of one tube = L* do *π
Area of one tube = 0.3418053 m2
Number of tubes = provisinal area / area of one tube
Number of tubes = 285
Using 1.25 triangular pitch
K1 = 0.249 Use table 12.4
n1 = 2.207
Db = (do)*( Nt / K1)^ (1/n1)
where;
Db =bundle diameter (mm)
261
do = outer diameter (mm)
Nt : number of tubes
K1& n1 are constant
Bundle diameter (Db) = 778 mm
0.777948 m
Using split ring floating head type
Bundle diametrical clearance = 64 mm from fig 12.10
Ds = Db + Bundle diametrical clearance
Shell diameter(Ds) = 842 mm
0.841948 m
Tube side coefficient
Method 1
Mean Tube temperature=(t1+t2)/2 = 302.5 oC
Tube cross-sectional area = p/4 *di2= 615.75216 mm2
Tube per pass=(Nt/NP) = 142
Total flow area = tubes per pass * cross sectional area
Total flow area = 0.0877076 m2
mass velocity = mass flow rate / total flow area
Tube mass velocity = 2.25E-01 kg/s.m2
linear velosity (ut ) = mass velosity / density
Tube linear velocity (ut) = 4.47E-04 m/s
hi = (4200* (1.35+0.02t)* ut0.8)/(di0.2)
where:
hi =inside coefficient (W/m2 oC)
262
t =mean temperature (oC)
ut =linear velocity (m/s)
di =tube inside diameter (mm)
hi = 33.373333 W/m2 oC
Method 2
Renolds number (Re) = ρ* ut*di/ µ
Re = 3.96E+01
Prandtl number (Pr) = Cp µ / kf
Pr = 3.84E+01
L/di = 1.21E+02
From Figuer 12.23 Tube -side heat transfer factor:
jh = 2.00E-03
where jh is the heat transfer factor
assume that the viscisity of the fluid is the same as at the wall
(µ/µwall) = 1
(hi di / kf) = jh Re Pr0.33 * (µ/µwall)0.14
hi = 2.99E+00 W/m2 oC
Using hi from method 1 as it has low value
hi = 3 W/m2 oC
Shell-side coefficient
Choose baffle spacing (Lb)= (Ds/5) = 168.3896 mm
Tube pitch (pt) =1.25 * do= 40 mm
Cross flow area (As) =((pt - do)* Ds* Lb)/pt
Cross flow area As = 0.0283551 m2
263
Mass velocity (Gs) = mass flow rate / cross flow area
Mass velocity (Gs) = 0.3884277 kg/s.m2
Equivalent diameter de =(1.1/do)(pt2-0.917do2)
de = 2.27E+01 mm
Mean Shell side temperature =(T1+T2)/2
Mean Shell side temperature = 200.5 oC
Renolds number (Re) = (Gs de)/ m
Re = 2.E+01
Prandtl number (Pr) = Cp µ / kf
Pr = 82.66
Choose 15% baffle cut
jh = 8.50E-03 from fig 12.29
Without the viscosity correction term, (µ/µw) = 1
hs = kf* jh *Re *Pr^(1/3) / d
hs = 13.19995 W/m2 oC
Overall Heat Transfer Coefficient
Thermal conductivity of cupro-nickel alloy= 45 W/moC
Taking fouling coefficients from table 12.2
Outside coefficient(fouling factor)=hod 4500 W/m2 oC
Inside coefficient(fouling factor) =hid 5000 W/m2 oC
1/Uo =(1/ho) + (1/hod) + ( do(ln(do/di))/2kw) + (do/di) * (1/hid) + (do/di) *
(1/hi)
1/Uo = 4.59E-01
Uo= 2.18E+00 W/m2 oC
Close to initial value assumed
264
Pressure Drop
Tube side
Re = 3.96E+01
jf = 7.00E-03 from fig 12.24
where jf is the friction factor
ΔPt= Np [ 8jf (L/di)(µ/µw)^(-m) +2.5 ] ρut²/2
where , ΔPt = tube side pressur drop (N/m²)(pa)
Np= number of tube side passes
ut= tube side velosity (m/s)
L = length of one tube
Neglecting the viscosity correction term, (µ/µw) = 1
∆pt= 0.0009352 N/m2
9.352E-07 kPa
1.356E-07 psi
Shell side (Acceptable)
Linear velocity =Gs /ρ= 0.0003736 m/s
Re = 2.E+01
jf = 5.00E-02 from fig 12.30
ΔPs = 8jf (Ds/de)(L/Lb)( ρus^2/2)(µ/µw)^(-0.14)
where :
L : tube length
Lb : baffle spacing
∆Ps = 0.0217125 N/m2
2.171E-05 kPa
265
3.149E-06 psi
Thickness Calculations:
t =(Pri/(SEJ-0.6P))+Cc
where:
t = shell thichness (in)
P = Maximum allowable internal pressure (psig)
ri= internal raduis of shell before allowance corrosion is added (in)
EJ = efficincy of joients
S = working stress (psi)
Cc = allowance for corrosin (in)
ri = 16.57378856 in
P = 25 psi
S = 13700 psi (for carbon steel)
EJ =0.85
Cc = 0.125 in
t = 0.161 in
t = 4.1 mm
Cost Calculations:
Heat transfer area = 1048.12 ft2
Cost = $784,600 (www.matche.com)
266
Heat Exchanger Design: E-201
Fluid properties:
Shell Side ( Cooling water stream )
Flow rate
6.82E+03
Kg/h
Inlet Temperature ,T1
100.6
o
C
Outlet Temperature ,T2
50.22
o
C
Average Heat Capacity, Cpavg
75.8675
KJ/kgoC
Average Mass Density, ρavg
968.02
kg/m3
Average Viscosity of stream, µavg
0.410
mNs/m2
Average Thermal conductivity, Kf
0.662
W/moC
Calculation of Heat load:
Qh = mh *Cp * ( T2-T1)
where
Qh = heat load in the hot side (KW)
mh = mass flowrate of hot fluid (Kg/h)
Cp = heat capacity of hot fluid (kJ/kgoC)
T1 = inlet temperature (oC)
T2 =outlet temperatue (oC)
Heat load = -7237.8 KW > 1000
The type of the heat exchanger is shell and tube heat exchanger
267
Tube side ( gas stream)
Flowrate
7101
Kg/h
Average Heat Capacity, Cp
76.49
kJ/kgoC
Average Mass Density, ρ
969.79
kg/m3
Average Viscosity of stream, µ
0.522
mNs/m2
Average Thermal conductivity of stream, Kf
0.649
W/moC
inlet Temperature , T1
40
o
C
outlet Temperature, t2
90
o
C
Log mean Temperature calculation:
∆Tlm =(T1-t2)-(T2-t1) / ln((T1-t2)/(T2-t1))
where:
∆Tlm = log mean temperature differace
T1 = inlet shell side fluid temoerature (oC)
T2 = outlet shell side fluid temerature (oC)
t1
= inlet tube side temoerature (oC)
t2 = outlet tube side temerature (oC)
∆Tlm = -10.409 oC
Temperature correction Fctor calculation:
Using two shell pass and four or multiple of Two tube passes Np=
R = (T1-T2) / (t2-t1)
R = 1.0076
S = (t2-t1) / (T1-t1)
S = 0.8250825
268
Ft = 0.99 From figure 12.20
∆Tm = Ft * ∆Tlm
where:
∆Tm = true temperature difference
Ft
= the temperature correction factor
∆Tlm = log mean temperature differace
∆Tm = -10.30476 oC
Assuming U = 1200 W/m2 oC ( From table 12.1)
A= Q / U * ∆Tm
where:
A = provisional area (m2)
Q = heat load (kW)
U = overall heat transfer coefficient (W/m2 oC)
Provisional area = 62.152868 m2
Choosing
Tube outside diameter(do) = 32 mm
Tube inner diameter(di) = 28 mm
Tube length(L) = 3.4 m
Take tube materail is cupro- nickel
Area of one tube = L* do *π
Area of one tube = 0.3418053 m2
Number of tubes = provisinal area / area of one tube
Number of tubes =182
269
Using 1.25 triangular pitch
K1 = 0.249 Use table 12.4
n1 = 2.207
Db = (do)*( Nt / K1)^ (1/n1)
where;
Db =bundle diameter (mm)
do = outer diameter (mm)
Nt : number of tubes
K1& n1 are constant
Bundle diameter (Db) = 635 mm
0.6347532 m
Using split ring floating head type
Bundle diametrical clearance = 64 mm
Ds = Db + Bundle diametrical clearance
Shell diameter(Ds) = 699 mm
0.6987532 m
Shell length(m)= 4.0987532 m
Tube side coefficient
Method 1
Mean Tube temperature=(t1+t2)/2 = 65 oC
Tube cross-sectional area = p/4 *di2= 615.75216 mm2
Tube per pass=(Nt/NP) = 91
Total flow area = tubes per pass * cross sectional area
270
Total flow area = 0.0559833 m2
mass velocity = mass flow rate / total flow area
Tube mass velocity = 3.52E+01 kg/s.m2
linear velosity (ut ) = mass velosity / density
Tube linear velocity (ut) = 3.63E-02 m/s
hi = (4200* (1.35+0.02t)* ut0.8)/(di0.2)
where:
hi =inside coefficient (W/m2 oC)
t =mean temperature (oC)
ut =linear velocity (m/s)
di =tube inside diameter (mm)
hi = 402.98448 W/m2 oC
Method 2
Renolds number (Re) = ρ* ut*di/ µ
Re = 1.89E+03
Prandtl number (Pr) = Cp µ / kf
Pr = 6.15E+01
L/di = 1.21E+02
From Figuer 12.23 Tube -side heat transfer factor:
jh = 2.00E-03
where jh is the heat transfer factor
assume that the viscisity of the fluid is the same as at the wall
µ/µwall) = 1
271
(hi di / kf) = jh Re Pr0.33 * (µ/µwall)0.14
hi = 3.41E+02 W/m2 oC
Using hi from method 1 as it has low value
hi = 341 W/m2 oC
Shell-side coefficient
Choose baffle spacing (Lb)= (Ds/5) = 139.75065 mm
Tube pitch (pt) =1.25 * do= 40 mm
Cross flow area (As) =((pt - do)* Ds* Lb)/pt
Cross flow area As = 0.0195302 m2
Mass velocity (Gs) = mass flow rate / cross flow area
Mass velocity (Gs) = 96.957887 kg/s.m2
Equivalent diameter de =(1.1/do)(pt2-0.917do2)
de = 2.27E+01 mm
Mean Shell side temperature =(T1+T2)/2
Mean Shell side temperature = 75.41 oC
Renolds number (Re) = (Gs de)/ m
Re = 5.E+03
Prandtl number (Pr) = Cp µ / kf
Pr = 46.94
Choose 15% baffle cut
jh = 8.50E-03
Without the viscosity correction term, (µ/µw) = 1
hs = kf* jh *Re *Pr^(1/3) / de
272
hs = 4805.0743 W/m2 oC
Overall Heat Transfer Coefficient
Thermal conductivity of cupro-nickel alloy= 45 W/moC
Taking fouling coefficients from table 12.2
Outside coefficient(fouling factor)=hod 4500 W/m2 oC
Inside coefficient(fouling factor) =hid 5000 W/m2 oC
1/Uo =(1/ho) + (1/hod) + ( do(ln(do/di))/2kw) + (do/di) * (1/hid) + (do/di) *
(1/hi)
1/Uo = 3.96E-03
Uo= 2.53E+02 W/m2 oC
Close to initial value assumed
Pressure Drop
Tube side
Re = 1.89E+03
jf = 7.00E-03
where jf is the friction factor
ΔPt= Np [ 8jf (L/di)(µ/µw)^(-m) +2.5 ] ρut²/2
where
ΔPt = tube side pressur drop (N/m²)(pa)
Np= number of tube side passes
ut= tube side velosity (m/s)
L = length of one tube
Neglecting the viscosity correction term, (µ/µw) = 1
273
∆pt= 11.904812 N/m2
0.0119048 kPa
0.0017268 psi
Shell side
(Acceptable)
Linear velocity =Gs /ρ= 0.100161 m/s
Re = 5.E+03
jf = 5.00E-02 from fig 12.30
ΔPs = 8jf (Ds/de)(L/Lb)( ρus^2/2)(µ/µw)^(-0.14)
where
L : tube length
Lb : baffle spacing
∆Ps = 1453.1885 N/m2
1.4531885 kPa
0.2107812 psi
Thickness Calculations:
(Acceptable)
t =(Pri/(SEJ-0.6P))+Cc
where:
t = shell thichness (in)
P = Maximum allowable internal pressure (psig)
ri= internal raduis of shell before allowance corrosion is added (in)
EJ = efficincy of joients
S = working stress (psi)
274
Cc = allowance for corrosin (in)
ri = 13.75499235 in
P = 25 psi
S = 13700 psi (for carbon steel)
EJ = 0.85
Cc = 0.125 in
t = 0.155 in
t = 3.9 mm
Cost Calculations:
Heat transfer area = 669.01 ft2
Cost = $604,300 www.matche. com
275
Cooler Design: E-104
Fluid properties:
Shell Side ( Cooling water stream )
Flow rate
2.42E+04
Kg/h
Inlet Temperature ,T1
25
o
C
Outlet Temperature ,T2
115.6
o
C
Average Heat Capacity, Cpavg
58.105
KJ/kgoC
Average Mass Density, ρavg
504.13
kg/m3
Average Viscosity of stream, µavg
0.452
mNs/m2
Average Thermal conductivity, Kf
0.318
W/moC
Calculation of Heat load:
Qh = mh *Cp * ( T2-T1)
where:
Qh = heat load in the hot side (KW)
mh = mass flowrate of hot fluid (Kg/h)
Cp = heat capacity of hot fluid (kJ/kgoC)
T1 = inlet temperature (oC)
T2 =outlet temperatue (oC)
Heat load = 35402.5 KW > 1000 KW
The type of the heat exchanger is shell and tube heat exchanger
276
Tube side ( gas stream)
Flowrate
19617
Kg/h
Average Heat Capacity, Cp
30.8455
kJ/kgoC
Average Mass Density, ρ
0.28745
kg/m3
Average Viscosity of stream, µ
0.011
mNs/m2
Average Thermal conductivity of stream, Kf
0.199
W/moC
inlet Temperature , T1
162.9
o
C
outlet Temperature, t2
40
o
C
Log mean Temperature calculation:
∆Tlm =(T1-t2)-(T2-t1) / ln((T1-t2)/(T2-t1))
where:
∆Tlm = log mean temperature differace
T1 = inlet shell side fluid temoerature (oC)
T2 = outlet shell side fluid temerature (oC)
t1
= inlet tube side temoerature (oC)
t2 = outlet tube side temerature (oC)
∆Tlm = 28.1246 oC
Temperature correction Fctor calculation:
Using two shell pass and four or multiple of Two tube passes Np=
R = (T1-T2) / (t2-t1)
R = 0.7371847
S = (t2-t1) / (T1-t1)
S = 0.8912255
277
Ft = 0.98 From figure 12.20
∆Tm = Ft * ∆Tlm
where:
∆Tm = true temperature difference
Ft
= the temperature correction factor
∆Tlm = log mean temperature differace
∆Tm = 27.562124 oC
Assuming U = 2400 W/m2 oC ( From table 12.1)
A= Q / U * ∆Tm
where:
A = provisional area (m2)
Q = heat load (kW)
U = overall heat transfer coefficient (W/m2 oC)
Provisional area = 406.5701 m2
Choosing
Tube outside diameter(do) = 19 mm
Tube inner diameter(di) = 15 mm
Tube length(L) = 6 m
Take tube materail is cupro- nickel
Area of one tube = L* do *π
Area of one tube = 0.3581416 m2
Number of tubes = provisinal area / area of one tube
Number of tubes = 1135
278
Using 1.25 triangular pitch
K1 = 0.249 Use table 12.4
n1 = 2.207
Db = (do)*( Nt / K1)^ (1/n1)
where;
Db =bundle diameter (mm)
do = outer diameter (mm)
Nt : number of tubes
K1& n1 are constant
Bundle diameter (Db) = 864 mm
0.8641846 m
Using split ring floating head type
Bundle diametrical clearance = 78 mm
Ds = Db + Bundle diametrical clearance
Shell diameter(Ds) = 942 mm
0.9421846 m
Shell length(m)= 6.9421846 m
Tube side coefficient
Method 1
Mean Tube temperature=(t1+t2)/2 = 101.45 oC
Tube cross-sectional area = p/4 *di2= 176.71459 mm2
Tube per pass=(Nt/NP) = 568
Total flow area = tubes per pass * cross sectional area
Total flow area = 0.1003051 m2
279
mass velocity = mass flow rate / total flow area
Tube mass velocity = 5.43E+01 kg/s.m2
linear velosity (ut ) = mass velosity / density
Tube linear velocity (ut) = 1.89E+02 m/s
hi = (4200* (1.35+0.02t)* ut0.8)/(di0.2)
where:
hi =inside coefficient (W/m2 oC)
t =mean temperature (oC)
ut =linear velocity (m/s)
di =tube inside diameter (mm)
hi = 546984.42 W/m2 oC
Method 2
Renolds number (Re) = ρ* ut*di/ µ
Re = 7.12E+04
Prandtl number (Pr) = Cp µ / kf
Pr = 1.78E+00
L/di = 4.00E+02
From Figuer 12.23 Tube -side heat transfer factor
jh = 3.50E-03
where jh is the heat transfer factor
assume that the viscisity of the fluid is the same as at the wall
(µ/µwall) = 1
(hi di / kf) = jh Re Pr0.33 * (µ/µwall)0.14
280
hi = 3.99E+03 W/m2 oC
Using hi from method 1 as it has low value
hi = 3993 W/m2 oC
Shell-side coefficient
Choose baffle spacing (Lb)= (Ds/5) = 188.43691 mm
Tube pitch (pt) =1.25 * do= 23.75 mm
Cross flow area (As) =((pt - do)* Ds* Lb)/pt
Cross flow area As = 0.0355085 m2
Mass velocity (Gs) = mass flow rate / cross flow area
Mass velocity (Gs) = 189.39143 kg/s.m2
Equivalent diameter de =(1.1/do)(pt2-0.917do2)
de = 1.35E+01 mm
Mean Shell side temperature =(T1+T2)/2
Mean Shell side temperature = 70.3 oC
Renolds number (Re) = (Gs de)/ m
Re = 6.E+03
Prandtl number (Pr) = Cp µ / kf
Pr = 82.41
Choose 15% baffle cut
jh = 1.00E-02
Without the viscosity correction term, (µ/µw) = 1
hs = kf* jh *Re *Pr^(1/3) / de
hs = 5810.9256 W/m2 oC
281
Overall Heat Transfer Coefficient
Thermal conductivity of cupro-nickel alloy= 50 W/moC
Taking fouling coefficients from table 12.2
Outside coefficient(fouling factor)=hod 5000 W/m2 oC
Inside coefficient(fouling factor) =hid 5000 W/m2 oC
1/Uo =(1/ho) + (1/hod) + ( do(ln(do/di))/2kw) + (do/di) * (1/hid) + (do/di) *
(1/hi)
1/Uo = 8.98E-04
Uo= 1.11E+03 W/m2 oC
Close to initial value assumed
Pressure Drop
Tube side
Re = 7.12E+04
jf = 3.50E-03
where jf is the friction factor
ΔPt= Np [ 8jf (L/di)(µ/µw)^(-m) +2.5 ] ρut²/2
Where
ΔPt = tube side pressur drop (N/m²)(pa)
Np= number of tube side passes
ut= tube side velosity (m/s)
L = length of one tube
Neglecting the viscosity correction term, (µ/µw) = 1
∆pt= 140660.51 N/m2
140.66051 kPa
20.402436 psi
282
Shell side
(Acceptable)
Linear velocity =Gs /ρ= 0.3756797 m/s
Re = 6.E+03
jf = 6.00E-02 from fig 12.30
ΔPs = 8jf (Ds/de)(L/Lb)( ρus^2/2)(µ/µw)^(-0.14)
where :
L : tube length
Lb : baffle spacing
∆Ps = 37972.401 N/m2
37.972401 kPa
5.5077965 psi
Thickness Calculations:
(Acceptable)
t =(Pri/(SEJ-0.6P))+Cc
where:
t = shell thichness (in)
P = Maximum allowable internal pressure (psig)
ri= internal raduis of shell before allowance corrosion is added (in)
EJ = efficincy of joients
S = working stress (psi)
Cc = allowance for corrosin (in)
ri = 18.54695041 in
P = 25 psi
283
S = 13706.66 psi (for carbon steel)
EJ = 0.85
Cc = 0.125 in
t = 0.165 in
t = 4.2 mm
Cost Calculations:
Heat transfer area = 4376.28 ft2
Cost = $784,60 www.matche.com
284
Heater Design: E-202
Fluid properties:
Shell Side ( Cooling water stream )
Flow rate
3.27E+04
Kg/h
Inlet Temperature ,T1
105
o
C
Outlet Temperature ,T2
100
o
C
Average Heat Capacity, Cpavg
76.1155
KJ/kgoC
Average Mass Density, ρavg
1002.85
kg/m3
Average Viscosity of stream, µavg
0.782
mNs/m2
Average Thermal conductivity, Kf
0.620
W/moC
Calculation of Heat load:
Qh = mh *Cp * ( T2-T1)
where:
Qh = heat load in the hot side (KW)
mh = mass flowrate of hot fluid (Kg/h)
Cp = heat capacity of hot fluid (kJ/kgoC)
T1 = inlet temperature (oC)
T2 =outlet temperatue (oC)
Heat load = -3455.4 KW > 1000
The type of the heat exchanger is shell and tube heat exchanger
285
Tube side ( gas stream)
Flowrate
7101
Kg/h
Average Heat Capacity, Cp
76.2595
kJ/kgoC
Average Mass Density, ρ
637.25
kg/m3
Average Viscosity of stream, µ
0.712
mNs/m2
Average Thermal conductivity of stream, Kf
0.596
W/moC
inlet Temperature , T1
89.992
o
C
outlet Temperature, t2
99
o
C
Log mean Temperature calculation:
∆Tlm =(T1-t2)-(T2-t1) / ln((T1-t2)/(T2-t1))
where:
∆Tlm = log mean temperature differace
T1 = inlet shell side fluid temoerature (oC)
T2 = outlet shell side fluid temerature (oC)
t1
= inlet tube side temoerature (oC)
t2 = outlet tube side temerature (oC)
∆Tlm = -7.8339 oC
Temperature correction Fctor calculation:
Using two shell pass and four or multiple of Two tube passes Np=
R = (T1-T2) / (t2-t1)
R = 0.5550622
286
S = (t2-t1) / (T1-t1)
S = 0.6002132
Ft = 0.98 From figure 12.20
∆Tm = Ft * ∆Tlm
where:
∆Tm = true temperature difference
Ft
= the temperature correction factor
∆Tlm = log mean temperature differace
∆Tm = -7.677181 oC
Assuming U = 1000 W/m2 oC (From table 12.1)
A= Q / U * ∆Tm
where:
A = provisional area (m2)
Q = heat load (kW)
U = overall heat transfer coefficient (W/m2 oC)
Provisional area = 26.527979 m2
Choosing
Tube outside diameter(do) = 21 mm
Tube inner diameter(di) = 17 mm
Tube length(L) = 5 m
Take tube materail is cupro- nickel
Area of one tube = L* do *π
Area of one tube = 0.3298672 m2
Number of tubes = provisinal area / area of one tube
287
Number of tubes = 80
Using 1.25 triangular pitch
K1 = 0.249 Use table 12.4
n1 = 2.207
Db = (do)*( Nt / K1)^ (1/n1)
where;
Db =bundle diameter (mm)
do = outer diameter (mm)
Nt : number of tubes
K1& n1 are constant
Bundle diameter (Db) = 288 mm
0.2878271 m
Using split ring floating head type
Bundle diametrical clearance = 75 mm
Ds = Db + Bundle diametrical clearance
Shell diameter(Ds) = 363 mm
0.3628271 m
Shell length(m)= 5.3628271 m
Tube side coefficient
Method 1
Mean Tube temperature=(t1+t2)/2 = 94.496 oC
Tube cross-sectional area = p/4 *di2= 226.98007 mm2
Tube per pass=(Nt/NP) = 40
Total flow area = tubes per pass * cross sectional area
288
Total flow area = 0.0091269 m2
mass velocity = mass flow rate / total flow area
Tube mass velocity = 2.16E+02 kg/s.m2
linear velosity (ut ) = mass velosity / density
Tube linear velocity (ut) = 3.39E-01 m/s
hi = (4200* (1.35+0.02t)* ut0.8)/(di0.2)
where:
hi =inside coefficient (W/m2 oC)
t =mean temperature (oC)
ut =linear velocity (m/s)
di =tube inside diameter (mm)
hi = 3250.8831 W/m2 oC
Method 2
Renolds number (Re) = ρ* ut*di/ µ
Re = 5.16E+03
Prandtl number (Pr) = Cp µ / kf
Pr = 9.11E+01
L/di = 2.94E+02
From Figuer 12.23 Tube -side heat transfer factor:
jh = 1.70E-02
where jh is the heat transfer factor
assume that the viscisity of the fluid is the same as at the wall
(µ/µwall) = 1
(hi di / kf) = jh Re Pr0.33 * (µ/µwall)0.14
289
hi = 1.36E+04 W/m2 oC
Using hi from method 1 as it has low value
hi = 3251 W/m2 oC
Shell-side coefficient
Choose baffle spacing (Lb)= (Ds/5) = 72.56542 mm
Tube pitch (pt) =1.25 * do= 26.25 mm
Cross flow area (As) =((pt - do)* Ds* Lb)/pt
Cross flow area As = 0.0052657 m2
Mass velocity (Gs) = mass flow rate / cross flow area
Mass velocity (Gs) = 1724.2485 kg/s.m2
Equivalent diameter de =(1.1/do)(pt2-0.917do2)
de = 1.49E+01 mm
Mean Shell side temperature =(T1+T2)/2
Mean Shell side temperature = 102.5 oC
Renolds number (Re) = (Gs de)/ m
Re = 3.E+04
Prandtl number (Pr) = Cp µ / kf
Pr = 96.07
Choose 15% baffle cut
jh = 1.00E-02
Without the viscosity correction term, (µ/µw) = 1
hs = kf* jh *Re *Pr^(1/3) / de
hs = 62566.755 W/m2 oC
290
Overall Heat Transfer Coefficient
Thermal conductivity of cupro-nickel alloy= 50 W/moC
Taking fouling coefficients from table 12.2
Outside coefficient(fouling factor)=hod 3500 W/m2 oC
Inside coefficient(fouling factor) =hid 5000 W/m2 oC
1/Uo =(1/ho) + (1/hod) + ( do(ln(do/di))/2kw) + (do/di) * (1/hid) + (do/di) *
(1/hi)
1/Uo = 8.84E-04
Uo= 1.13E+03 W/m2 oC
Close to initial value assumed
Pressure Drop
Tube side
Re = 5.16E+03
jf = 3.00E-02
where jf is the friction factor
ΔPt= Np [ 8jf (L/di)(µ/µw)^(-m) +2.5 ] ρut²/2
where ,
ΔPt = tube side pressur drop (N/m²)(pa)
Np= number of tube side passes
ut= tube side velosity (m/s)
L = length of one tube
Neglecting the viscosity correction term, (µ/µw) = 1
∆pt= 5357.0565 N/m2
5.3570565 kPa
0.7770269 psi
291
Shell side
(Acceptable)
Linear velocity =Gs /ρ= 1.7193483 m/s
Re = 3.E+04
jf = 6.00E-02 from fig 12.30
ΔPs = 8jf (Ds/de)(L/Lb)( ρus^2/2)(µ/µw)^(-0.14)
where :
L : tube length
Lb : baffle spacing
∆Ps = 1192907.4 N/m2
1192.9074 kPa
173.02807 psi
Thickness Calculations:
(Acceptable)
t =(Pri/(SEJ-0.6P))+Cc
where:
t = shell thichness (in)
P = Maximum allowable internal pressure (psig)
ri= internal raduis of shell before allowance corrosion is added (in)
EJ = efficincy of joients
S = working stress (psi)
Cc = allowance for corrosin (in
ri = 7.142269604 in
P = 15 psi
292
S = 13706.66 psi (for carbon steel)
EJ = 0.85
Cc = 0.125 in
t = 0.134 in
t = 3.4 mm
Cost Calculations:
Heat transfer area = 285.54 ft2
Cost = $784,600 www.matche.co
293
Compressor
Introduction :A gas compressor is a mechanical device that increases the pressure of a gas
by reducing its volume. Compressors are similar to pumps: both increase the
pressure on a fluid and both can transport the fluid through a pipe. As gases
are compressible, the compressor also reduces the volume of a gas. Liquids
are relatively incompressible, so the main action of a pump is to transport
liquids.
Types of Compressors
Fig1: types of compressors
1- Centrifugal compressors
Centrifugal compressors use a rotating disk or impeller in a shaped housing
to force the gas to the rim of the impeller, increasing the velocity of the gas.
A diffuser (divergent duct) section converts the velocity energy to pressure
energy. They are primarily used for continuous, stationary service in
industries such as oil refineries, chemical and petrochemical plants and
natural gas processing plants.[1][2][3] Their application can be from 100 hp (75
294
kW) to thousands of horsepower. With multiple staging, they can achieve
extremely high output pressures greater than 10,000 psi (69 MPa).
Many large snow-making operations (like ski resorts) use this type of
compressor. They are also used in internal combustion engines as
superchargers and turbochargers. Centrifugal compressors are used in small
gas turbineengines or as the final compression stage of medium sized gas
turbines.
Fig2: A single stage centrifugal compresso
2- Axial-flow compressors
Axial-flow compressors are dynamic rotating compressors that use arrays of
fan-like aero foils to progressively compress the working fluid. They are
used where there is a requirement for a high flows or a compact design.
The arrays of aero foils are set in rows, usually as pairs: one rotating and one
stationary. The rotating aero foils, also known as blades or rotors, decelerate
and pressurize the fluid. The stationary aero foils, also known as a stators or
vanes, turn and decelerate the fluid; preparing and redirecting the flow for
the rotor blades of the next stage.[1] Axial compressors are almost always
multi-staged, with the cross-sectional area of the gas passage diminishing
295
along the compressor to maintain an optimum axial Mach number. Beyond
about 5 stages or a 4:1 design pressure ratio, variable geometry is normally
used to improve operation.
Axial compressors can have high efficiencies; around 90% polytrophic at
their design conditions. However, they are relatively expensive, requiring a
large number of components, tight tolerances and high quality materials.
Axial-flow compressors can be found in medium to large gas turbine
engines, in natural gas pumping stations, and within certain chemical plants
Fig3: an axial compressor
3- Reciprocating compressors
Reciprocating compressors use pistons driven by a crankshaft. They can be
either stationary or portable, can be single or multi-staged, and can be driven
by electric motors or internal combustion engines.
Small reciprocating compressors from 5 to 30 horsepower (hp) are
commonly seen in automotive applications and are typically for intermittent
duty. Larger reciprocating compressors up to 1000 hp are still commonly
found in large industrial applications, but their numbers are declining as they
are replaced by various other types of compressors. Discharge pressures can
range from low pressure to very high pressure (>5000 psi or 35 MPa). In
certain applications, such as air compression, multi-stage double-acting
compressors are said to be the most efficient compressors available, and are
typically larger, noisier, and more costly than comparable rotary units.
296
Fig4: A reciprocating compressor
4- Rotary screw compressors
Rotary screw compressors use two meshed rotating positive-displacement
helical screws to force the gas into a smaller space.[1][7][8] These are usually
used for continuous operation in commercial and industrial applications and
may be either stationary or portable. Their application can be from 3 hp
(2.24 kW) to over 500 hp (375 kW) and from low pressure to very high
pressure (>1200 psi or 8.3 MPa). They are commonly seen with roadside
repair crews powering air-tools. This type is also used for many automobile
engine superchargers because it is easily matched to the induction capacity
of a piston Engine.
Fig5: a rotary screw compressor
297
4- scroll compressor
A scroll compressor, also known as scroll pump and scroll vacuum pump,
uses two interleaved spiral-like vanes to pump or compressfluids such as
liquids and gases. The vane geometry may be involute, archimedean spiral,
or hybrid curves.[9][10][11] They operate more smoothly, quietly, and reliably
than other types of compressors in the lower volume range.
Often, one of the scrolls is fixed, while the other orbits eccentrically without
rotating, thereby trapping and pumping or compressing pockets of fluid or
gas between the scrolls.
Fig6: Mechanism of a scroll pump
Applications:Gas compressors are used in various applications where either higher
pressures or lower volumes of gas are needed:

In pipeline transport of purified natural gas to move the gas from the
production site to the consumer.
298

In petroleum refineries, natural gas processing plants, petrochemical
and chemical plants, and similar large industrial plants for
compressing intermediate and end product gases.

In refrigeration and air conditioner equipment to move heat from one
place to another in refrigerant cycles: see Vapor-compression
refrigeration.

In gas turbine systems to compress the intake combustion air

In storing purified or manufactured gases in a small volume, high
pressure cylinders for medical, welding and other uses.

In many various industrial, manufacturing and building processes to
power all types of pneumatic tools.

as a medium for transferring energy, such as to power pneumatic
equipment.

In pressurized aircraft to provide a breathable atmosphere of higher
than ambient pressure.

In some types of jet engines (such as turbojets and turbofans) to
provide the air required for combustion of the engine fuel. The power
to drive the combustion air compressor comes from the jet's own
turbines.
Design Procedure:1. Calculate the compression factor using the following equation:
 n 


P1  T1  n 1 
,
 
P2  T 2 
Where P1,2 is the pressure of inlet and outlet respectively (psia),
And T1,2 is the temperature of the inlet and outlet respectively (R).
299
2. Calculate the work done in Btu/lbmol by:
W 
nR (T 1 T 2 )
,
1 n
Where R is the ratio of the specific heat capacities (Cp/Cv).
3. Calculate the horse power, Hp using the following equation:
Hp=W*M,
Where M is the molar flow rate in lbmol/s.
4. Calculate the efficiency of the compressor using the following
equation:
Ep 
n
K
n 1 ,
K 1
Where K 
MwC p
MwC p  1.986
and Mw is the molecular weight of the gas
in the stream and Cp is the specific heat capacity (Btu/lboF).
5. Calculate the cost of the compressor from www.matche.com
300
Calculation :Table1 : calculation of compressor for “k101”
Variable
Value
Molar flow rate (lbmol/s) =
0.504166667
P1 (psia) =
14.7
P2 (psia) =
45
T1 (R) =
563.67
T2 (R) =
702.054
Ln(P1/P2) =
-1.118814996
Ln(T1/T2) =
-0.21954135
Ln(P1/P2)/ln(T1/T2) =
5.096147011
N=
1.244131863
Ln(P1/P2)/ln(T1/T2)-n/(n-1) =
0
Mw =
3.066
Z=
1
R (Cv/Cp) =
1.4053
Cp (Btu/lb˚F) =
6.88593
Rc =
3.06122449
W (Btu/lbmol) =
991.0529848
Hp (Hp) =
706.3446498
K=
1.103836289
Ep (%) =
47.93872045
Cost ($) =
186,000
301
Equipment Name
Compressor
Objective
To increase the pressure
Equipment Number
K-101
Designer
Abdulhameed al-awadhi
Type
Centrifugal Compressor
Material of Construction
Carbon steel
Insulation
Quartz wool
Cost
$186,000
Operating Condition
Inlet Temperature (◦C)
563.67
Outlet Temperature (◦C)
702.054
Inlet Pressure (psia)
14.7
Outlet Pressure (psia)
45
Efficiency (%)
47.93872045
Power (Hp)
706.3446498
Storage tank
Introduction :Tanks are basically was made to hold, transport or store fluid and solid.
Gases are stored at high pressures where this process requirement and to
reduce the storage volume .for some gases the volume can be further
reduced by liquefying the gas by pressure or refrigeration .cylindrical and
spherical vessels are used. Liquids are usually stored in bulk in vertical
cylindrical steel tanks; Fixed and floating-roof tanks are used. In a floatingroof tank a movable piston floats on the surface of the liquid and is sealed to
the tank walls. Floating-roof tanks are used to eliminate evaporation losses
and, for liquids, to obviate the need for inert gas blanketing to prevent an
302
explosive mixture forming above the liquid, as would be the situation with a
fixed-roof tank. Horizontal cylindrical tanks and rectangular tanks are used
also used for storing liquids, usually for relatively small quantities. Storage
of solid is usually more expensive than the movement of liquid and gases,
which can easily pumped down a pipeline. The design more flexible and
moderate the international conditions.
Storage tanks can be divided as Aboveground Storage Tanks (ASTs) and
Underground Storage Tanks (USTs). While the ASTs have more capacity,
they require frequent maintenance operations and inspection as compared to
USTs that are of lower capacity.
These tanks are usually made of Carbon Steel/Stainless Steel or Aluminum.
Where carbon steel is used, the concentration of carbon should be less than
0.3%. The material has to be decided carefully depending upon the costing
and material that is to be stored in the tank. The construction basically
involves welding the steel plates together to form the structure and proceeds
in three stages:Stage 1: A concrete base is constructed upon which the tank ring walls
reside. The first tank ring is erected, which is approximately 8 to 10 feet
303
high. The only means of entry/exit is a manhole cut into one of the steel
plates that comprises the ring.
Stage 2: A second ring is erected on top of the first ring and welded into
place. With this ring in place, the tank walls are 16 to 20 feet high. A "door
sheet" is then cut out, creating an opening approximately 8 feet wide by 10
feet high. This opening provides access for personnel, equipment and
material.
Airflow
is
typically
not
a
problem
at
this
stage.
Stage 3: The top is erected and forced air ventilation is used to assist in air
flow.
Usually the roofs of ASTs are designed to be rounded so as to prevent it
from buckling under wind loads. Now days, most tanks are built in with a
floating fixed roof inside the rounded one, which rises and sinks according
to the fluid level inside the container. This acts as a safety point as it
decreases the vapor space above the liquid level
Another safety requirement to prevent leakage is to build a dike around the
container that holds the spilled liquid and prevents it from leaking out. The
leakage is caused as a result of corrosive action, which may result in pits,
cracks, crevices etc. Therefore, it is necessary to regularly inspect the tanks,
keep records for tracking, plan out inspections and assess results for
appropriate measures to be taken. Besides, it is also necessary to inspect the
newly constructed tanks for welding.
304
Design of storage tank
In the design of the storage there are some main principles to be considered
to make the design more flexible and moderate the international conditions.
In the case of this project the tanks which will be used is cylindrical in
shape. This shape is the proper one for the liquid phase in our case. Also the
top space in the tank for the vapor pressure of the component will be 12 % of
the tank volume.
The steps in the design will begin in finding the volume of the liquid part in
the tank and then determining the height and diameter of this part. After that
the total volume will be determined . finally all the dimensions could be
determined from this volume.
Volume of the liquid= Total mass flow rate in* time hold-up
Time hold-up: the time where the liquid is hold inside the tank
Volume of cylinder = π R2 H … (1)
Get the volume of the liquid (assume H=0.2D)
To determine vapor pressure Antoine equation is used:
Log10 P* = A- (B/C+T) … (2)
T: the temperature in ˚C
The values of A, B, and C is taken from table Antoine equation constant
At this vapor pressure the top space in the tank can be determined.
the totel volume=volume of liquid/0.12 … (3)
Diameter3 = (5*total volume)/П
Height = 0.2diameter
Area of the tank = total volume/ height … (4)
305
Thickness
t = (P r i / S E - 0.6P) + Cc …(5)
Assumption :
 Cylindrical tank
 Square shape from inside (H =0.2 D)
Sample calculation:Thickness
t = (P r i / S E – 0.6P) + Cc
P : internal pressure (kpa gage)
r i : internal raduis of shell (m)
E : efficincy of joients=0.85
S : working stress (kpa)=94500
Cc : allowance for corrosin (m)=0.003175
Tank (T-104)
Storage no.
Flow rate(m3/hr)
time hold(hr)
volume of liquid(m3)
T_104
35.37066246
0.5
17.68533123
Storage no.
Total Volume(m3)
D^3
ActD(tank)m
ActH(tank)m
Area(m2)
T_104
147.3777603
234.6779622
6.168185643
1.233637129
119.46605
Storage no.
P(kpa)
ri(m)
E(m)
S(kpa)
Cc(m)
t(m)
T_104
101.3
37.47089627
0.85
94500
0.003175
0.050466
306
Tank (T-103)
Storage no.
Flow rate(m3/hr)
time hold(hr)
volume of liquid(m3)
V_103
380.0460732
0.1
38.00460732
Storage no.
Total
Volume(m3)
D^3
ActD(tank)m
ActH(tank)m Area(m2)
T_103
316.705061
504.3074219
7.959732142
1.591946428 198.94203
Storage no.
P(kpa)
ri(m)
E(m)
S(kpa)
Cc(m)
t(m)
T_103
101.3
3.979866071
0.85
94500
0.003175
0.008198
Tank (V-104)
Storage no.
Flow rate(m3/hr)
time hold(hr)
volume of liquid(m3)
V_104
54.569
0.1
5.4569
Storage Total
no.
Volume(m3) D^3
V_104
ActD(tank)m ActH(tank)m Area(m2)
45.47416667 72.41109342 4.168070299 0.83361406
54.550623
Storage no.
P(kpa)
ri(m)
E(m)
S(kpa)
Cc(m)
t(m)
V_104
101.3
2.084035149
0.85
94500
0.003175
0.005805
307
Equipment Name
Start up tank
Objective
To store nitrobenzene feed
Equipment Number
T_104
Designer
Abdulhameed al-awadhi
Type
Fixed Head
Location
after stream 6
Material of Construction
Carbon steel
Insulation
Quartz Wool
Cost
265500$
Operating Condition
Operating Temperature (oC)
100
Diameter (m)
22.34
Operating Pressure (kpa)
2742
Height (m)
4.47
Feed Flow Rate (kgmole/s)
377
Thickness (m)
0.187
Vapor Pressure (atm)
3077.68
Total Area (m)
119.46605
308
Equipment Name
Catalyst hopper
Objective
To store catalyt feed
Equipment Number
T_103
Designer
Abdulhameed al-awadhi
Type
Fixed Head
Location
Before K-102
Material of Construction
Carbon steel
Insulation
Quartz Wool
Cost
148000$
Operating Condition
Operating Temperature (oC)
200
Diameter (m)
20. 4
Operating Pressure (kpa)
101.3
Height (m)
4. 76
Feed Flow Rate (kgmole/s)
10.53
Thickness (m)
0.167
Vapor Pressure (atm)
4162. 8
Total Area (m)
198.94203
309
Equipment Name
Discharge drum
Objective
To store nitrobenzene & hydrogen
Equipment Number
V_104
Designer
Abdulhameed al-awadhi
Type
Fixed Head
Location
After hydrogen feed
Material of Construction
Carbon steel
Insulation
Quartz Wool
Cost
172500$
Operating Condition
Operating Temperature (oC)
25
Diameter (m)
20.82
Operating Pressure (kpa)
101.3
Height (m)
5.07
Feed Flow Rate (kgmole/s)
835
Thickness (m)
0.204
Vapor Pressure (atm)
4002. 7
Total Area (m)
54.550623
310
Yousef A.Almunifi
208112671
-Heat exchanger(E-203)
-Heat Exchanger (E-204)
-Heat Exchanger(E-205)
-Furnace(F-101)
-Compressor(k-102)
-Tank T-201 A&B
-V-105
-V-204
311
Heat Exchangers
What are heat exchangers?
Heat exchangers are devices used to transfer heat energy from one fluid to
another. Typical heat exchangers experienced by us in our daily lives include
condensers and evaporators used in air conditioning units and refrigerators.
Boilers and condensers in thermal power plants are examples of large
industrial heat exchangers. There are heat exchangers in our automobiles in
the form of radiators and oil coolers. Heat exchangers are also abundant in
chemical and process industries.
There is a wide variety of heat exchangers for diverse kinds of uses, hence
the construction also would differ widely. However, in spite of the variety,
most heat exchangers can be classified into some common types based on
some fundamental design concepts. We will consider only the more common
types here for discussing some analysis and design methodologies.
Heat Transfer Considerations
The energy flow between hot and cold streams, with hot stream in the bigger
diameter tube, is as shown in Figure 1. Heat transfer mode is by convection
on the inside as well as outside of the inner tube and by conduction across
the tube. Since the heat transfer occurs across the smaller tube, it is this
internal surface which controls the heat transfer process. By convention, it is
the outer surface, termed Ao, of this central tube which is referred to in
describing heat exchanger area. Applying the principles of thermal
resistance,
312
End view of a tubular heat exchanger
If we define overall the heat transfer coefficient, Uc, as:
Substituting the value of the thermal resistance R yields:
Standard convective correlations are available in text books and handbooks
for the convective coefficients, ho and hi. The thermal conductivity, k,
corresponds to that for the material of the internal tube. To evaluate the
thermal resistances, geometrical quantities (areas and radii) are determined
from the internal tube dimensions available.
Fouling
Material deposits on the surfaces of the heat exchanger tubes may add more
thermal resistances to heat transfer. Such deposits, which are detrimental to
the heat exchange process, are known as fouling. Fouling can be caused by a
variety of reasons and may significantly affect heat exchanger performance.
313
With the addition of fouling resistance, the overall heat transfer coefficient,
Uc, may be modified as:
where R” is the fouling resistance.
Fouling can be caused by the following sources:
1) Scaling is the most common form of fouling and is associated with
inverse solubility salts. Examples of such salts are CaCO3, CaSO4,
Ca3(PO4)2, CaSiO3, Ca(OH)2, Mg(OH)2, MgSiO3, Na2SO4, LiSO4, and
Li2CO3.
2) Corrosion fouling is caused by chemical reaction of some fluid
constituents with the heat exchanger tube material.
3) Chemical reaction fouling involves chemical reactions in the process
stream which results in deposition of material on the heat exchanger tubes.
This commonly occurs in food processing industries.
4) Freezing fouling is occurs when a portion of the hot stream is cooled to
near the freezing point for one of its components. This commonly occurs in
refineries where paraffin frequently solidifies from petroleum products at
various stages in the refining process. , obstructing both flow and heat
transfer.
5) Biological fouling is common where untreated water from natural
resources such as rivers and lakes is used as a coolant. Biological microorganisms such as algae or other microbes can grow inside the heat
exchanger and hinder heat transfer.
6) Particulate fouling results from the presence of microscale sized particles
in solution. When such particles accumulate on a heat exchanger surface
314
they sometimes fuse and harden. Like scale these deposits are difficult to
remove.
With fouling, the expression for overall heat transfer coefficient becomes:
Basic Heat Exchanger Flow Arrangements
Two basic flow arrangements are as shown in Figure 2. Parallel and counter
flow provide alternative arrangements for certain specialized applications. In
parallel flow both the hot and cold streams enter the heat exchanger at the
same end and travel to the opposite end in parallel streams. Energy is
transferred along the length from the hot to the cold fluid so the outlet
temperatures asymptotically approach each other. In a counter flow
arrangement, the two streams enter at opposite ends of the heat exchanger
and flow in parallel but opposite directions. Temperatures within the two
streams tend to approach one another in a nearly linearly fashion resulting in
a much more uniform heating pattern. Shown below the heat exchangers are
representations of the axial temperature profiles for each. Parallel flow
results in rapid initial rates of heat exchange near the entrance, but heat
transfer rates rapidly decrease as the temperatures of the two streams
approach one another. This leads to higher energy loss during heat exchange.
Counter flow provides for relatively uniform temperature differences and,
consequently, lead toward relatively uniform heat rates throughout the length
of the unit.
315
Basic Flow Arrangements for Tubular Heat Exchangers.
Log Mean Temperature Differences
Heat flows between the hot and cold streams due to the temperature
difference across the tube acting as a driving force. As seen in the Figure
7.3, the temperature difference will vary along the length of the HX, and this
must be taken into account in the analysis.
Temperature Differences Between Hot and Cold Process Streams
316
From the heat exchanger equations shown earlier, it can be shown that the
integrated average temperature difference for either parallel or counter flow
may be written as:
The effective temperature difference calculated from this equation is known
as the log mean temperature difference, frequently abbreviated as LMTD,
based on the type of mathematical average that it describes. While the
equation applies to either parallel or counter flow, it can be shown that
∆Ɵeff will always be greater in the counter flow arrangement.
Another interesting observation from the above Figure is that counter flow is
more appropriate for maximum energy recovery. In a number of industrial
applications there will be considerable energy available within a hot waste
stream which may be recovered before the stream is discharged. This is done
by recovering energy into a fresh cold stream. Note in the Figures shown
above that the hot stream may be cooled to t1 for counter flow, but may only
be cooled to t2 for parallel flow. Counter flow allows for a greater degree of
energy recovery. Similar arguments may be made to show the advantage of
counter flow for energy recovery from refrigerated cold streams.
Applications for Counter and Parallel Flows
We have seen two advantages for counter flow, (a) larger effective LMTD
and (b) greater potential energy recovery. The advantage of the larger
LMTD, as seen from the heat exchanger equation, is that a larger LMTD
permits a smaller heat exchanger area, Ao, for a given heat transfer, Q. This
would normally be expected to result in smaller, less expensive equipment
for a given application.
317
Sometimes, however, parallel flows are desirable (a) where the high initial
heating rate may be used to advantage and (b) where it is required the
temperatures developed at the tube walls are moderate. In heating very
viscous fluids, parallel flow provides for rapid initial heating and consequent
decrease in fluid viscosity and reduction in pumping requirement. In
applications where moderation of tube wall temperatures is required, parallel
flow results in cooler walls. This is especially beneficial in cases where the
tubes are sensitive to fouling effects which are aggravated by high
temperature.
Multipass Flow Arrangements
In order to increase the surface area for convection relative to the fluid
volume, it is common to design for multiple tubes within a single heat
exchanger. With multiple tubes it is possible to arrange to flow so that one
region will be in parallel and another portion in counter flow. An
arrangement where the tube side fluid passes through once in parallel and
once in counter flow is shown in the Figure 4. Normal terminology would
refer to this arrangement as a 1-2 pass heat exchanger, indicating that the
shell side fluid passes through the unit once, the tube side twice. By
convention the number of shell side passes is always listed first.
318
Multipass flow arrangement
The primary reason for using multipass designs is to increase the
average tube side fluid velocity in a given arrangement. In a two pass
arrangement the fluid flows through only half the tubes and any one
point, so that the Reynold’s number is effectively doubled. Increasing
the Reynolds’s number results in increased turbulence, increased
Nusselt numbers and, finally, in increased convection coefficients. Even
though the parallel portion of the flow results in a lower effective ∆T,
the increase in overall heat transfer coefficient will frequently
compensate so that the overall heat exchanger size will be smaller for a
specific service. The improvement achievable with multipass heat
exchangers is substantialy large. Accordingly, it is a more accepted
practice in modern industries compared to conventional true parallel or
counter flow designs.
The LMTD formulas developed earlier are no longer adequate for
multipass heat exchangers. Normal practice is to calculate the LMTD for
counter flow, LMTDcf, and to apply a correction factor, FT, such that
The correction factors, FT, can be found theoretically and presented in
analytical form. The equation given below has been shown to be
accurate for any arrangement having 2, 4, 6,.....,2n tube passes per shell
pass to within 2%.
where the capacity ratio, R, is defined as:
The effectiveness may be given by the equation:
Provided that R>1. In the case that R=1, the effectiveness is given by:
Effectiveness-NTU Method:
Quite often, heat exchanger analysts are faced with the situation that
only the inlet temperatures are known and the heat transfer
characteristics (UA value) are known, but the outlet temperatures have to
be calculated. Clearly, LMTH method will not be applicable here. In this
regard, an alternative method known as the ε-NTU method is used.
320
321
NTUmax can be obtained from figures in textbooks/handbooks
First, however, we must determine which fluid has Cmin.
Theory of Heat Exchanger:
 Q  mC p T
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 )
(T  t )
ln 1 2
(T2  t1 )
322
Where:
- TLM
 Log
means Temperature.
- T1  Inlet shell side fluid temperature (oC).
- T2
- t1


Outlet shell side fluid temperature (oC).
Inlet tube side temperature (oC).
- t 2  Outlet tube temperature (oC).
 R
(T1  T2 )
(t 2  t1 )
 S
(t 2  t1 )
(T1  t1 )
 Tm  Ft Tlm
Where:
- Tm  True temperature difference.
- Ft  Temperature correction factor.
 A
Q
UTm
Where:
- A  Provisional area in m2.
- Q  Heat load in W.
- Tm  True temperature difference.

A  DL
323
Where:
- A  Area of one tube, m2.

N t  Provisional area/Area of one tube.
Where:
- N t  Number of tubes.
1

N
Db  d 0 ( t ) n1
K1
Where:
- Db  Bundle diameter (mm).
- d 0  Outside diameter (mm).
- N t  Number of tubes.
-K1 & n1 are constant.

Ds  Db  Clearance
Where:
- Ds  Sell diameter.
- Db  Bundle diameter (mm).

Ac 

4
(d i ) 2
Where:
324
- Ac  Tube cross-sectional area.
- di  Tube inner diameter.

N
Tubes
 t
Pass
4
Where:
- N t  Number of tubes.

At  Ac
Tubes
Pass
Where:
- At  Total flow area.

Um 
m
At
Where:
- U m  Tube mass velocity.
- At  Total flow area.
- m  Mass flow rate in Kg/s.

Ut 
Um
 ref
Where:
- U t  Tube linear velocity.
-  ref  Density.
325

hi 
(4200 * (1.35  0.02t ) * U t
di
0.8
)
0.2
Where:
- hi  Inside coefficient (W/m2 oC).
- U t  Tube linear velocity.
- t  Mean temperature (oC).

Re 
U t d i

Where:
- Re  Reynolds number.
-   Fluid viscosity at the bulk fluid temperature, Ns/m2.

Pr 
Cp
kf
Where:
- Pr  Prandtl number.
- C p  Heat capacity.
- k f  Thermal conductivity of stream.

hi 
k f j h Re(Pr) 0.33
di
Where:
- hi  Inside coefficient (W/m2 oC).
326
- j h  Tube side heat transfer factor.
- k f  Thermal conductivity of stream.
- Pr  Prandtl number.

lB 
Ds
5
Where:
- l B  Baffle spacing.
- Ds  Shell diameter.

pt  1.25d 0
Where:
- pt  Tube pitch.
- d 0  Outside diameter (mm).

As 
( p t  d 0 ) Ds l B
pt
Where:
- As  Cross-flow area.
- pt  Tube pitch.
- d 0  Outside diameter (mm).

Gs 
m
As
Where:
327
- Ds  Shell diameter.
- Gs  Mass velocity.
- As  Cross-flow area.
- m  Mass flow rate in Kg/s.

de 
1.1 2
2
( pt  0.917 d 0 )
d0
Where:
- d e  Equivalent diameter (mm).
- d 0  Outside diameter (mm).
- pt  Tube pitch.

Re 
Gs d e

Where:
- Re  Reynolds number.
- d e  Equivalent diameter (mm).
- Gs  Mass velocity.
-   Fluid viscosity at the bulk fluid temperature, Ns/m2.

1
1
1



U 0 h0 hod
d 0 ln(
d0
)
di
2k w

d0 1
d 1
 0
d i hid d i hi
Where:
- U 0  The overall heat transfer coefficient.
328
- hod  Outside coefficient (fouling factor).
- hid  Inside coefficient (fouling factor).


L
Pt  N p 8 j f 

 di
  



 w 
m
 u 2
 2.5 t
 2
Where:
- Pt  Tube- side pressure drop (N/m²) (pa).
- N p  Number of tube -side passes.
- ut  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.

 Ds  L  u s 2   
 
Ps  8 j f   
 d e  l B  2   w 
0.14
Where:
- Ps  Shell-side pressure drop (N/m²) (pa).
- j f  Friction factor.
- L  Length of tube.

t
Pri
 Cc
SEj  0.6 P
329
Where:
- t  Shell thickness (in).
- P  Maximum allowable internal pressure (psig).
- ri  Internal radius of shell before allowance corrosion is added
(in).
- E j  Efficiency of joints.
- S  Working stress (psi).
- Cc  Allowance for corrosion (in)
330
Sample Calculation:
(E-203):
331
332
333
334
335
(E-204):
336
337
338
339
340
(E-205):
341
342
343
344
345
Furnace(F-101):
An industrial furnace or direct fired heater, is an equipment used to
provide heat for a process or can serve as reactor which provides heats
of reaction. Furnace designs vary as to its function, heating duty, type of
fuel and method of introducing combustion air. However, most process
furnaces have some common features.
Fuel flows into the burner and is burnt with air provided from an air
blower. There can be more than one burner in a particular furnace which
can be arranged in cells which heat a particular set of tubes. Burners can
also be floor mounted, wall mounted or roof mounted depending on
design. The flames heat up the tubes, which in turn heat the fluid inside
in the first part of the furnace known as the radiant section or firebox. In
this chamber where combustion takes place, the heat is transferred
mainly by radiation to tubes around the fire in the chamber. The heating
fluid passes through the tubes and is thus heated to the desired
temperature. The gases from the combustion are known as flue gas.
After the flue gas leaves the firebox, most furnace designs include a
convection section where more heat is recovered before venting to the
atmosphere through the flue gas stack. (HTF=Heat Transfer Fluid.
Industries commonly use their furnaces to heat a secondary fluid with
special additives like anti-rust and high heat transfer efficiency. This
heated fluid is then circulated round the whole plant to heat exchangers
to be used wherever heat is needed instead of directly heating the
product line as the product or material may be volatile or prone to
cracking at the furnace temperature.)
346
Radiant section
The radiant section is where the tubes receive almost all its heat by
radiation from the flame. In a vertical, cylindrical furnace, the tubes are
vertical. Tubes can be vertical or horizontal, placed along the refractory
wall, in the middle, or arranged in cells. Studs are used to hold the
insulation together and on the wall of the furnace. They are placed about
1 ft (300 mm) apart in this picture of the inside of a furnace. The tubes
which are reddish brown from corrosion, are carbon steel tubes and run
the height of the radiant section. The tubes are a distance away from the
insulation so radiation can be reflected to the back of the tubes to
maintain a uniform tube wall temperature. Tube guides at the top,
middle and bottom hold the tubes in place.
347
Convection section
The convection section is located above the radiant section where it is
cooler to recover additional heat. Heat transfer takes place by convection
here, and the tubes are finned to increase heat transfer. The first two tube
rows in the bottom of the convection section and at the top of the radiant
section is an area of bare tubes (without fins) and are known as the
shield section, so named because they are still exposed to plenty of
radiation from the firebox and they also act to shield the convection
section tubes, which are normally of less resistant material from the high
temperatures in the firebox. The area of the radiant section just before
flue gas enters the shield section and into the convection section called
the bridge zone. Crossover is the term used to describe the tube that
connects from the convection section outlet to the radiant section inlet.
348
The crossover piping is normally located outside so that the temperature
can be monitored and the efficiency of the convection section can be
calculated. The sightglass at the top allows personnel to see the flame
shape and pattern from above and visually inspect if flame impingement
is occurring. Flame impingement happens when the flame touches the
tubes and causes small isolated spots of very high temperature.
Burner
The burner in the vertical, cylindrical furnace as above, is located in the
floor and fires upward. Some furnaces have side fired burners, such as in
train locomotives. The burner tile is made of high temperature refractory
and is where the flame is contained in.
Air registers located below the burner and at the outlet of the air blower
are devices with movable flaps or vanes that control the shape and
349
pattern of the flame, whether it spreads out or even swirls around.
Flames should not spread out too much, as this will cause flame
impingement. Air registers can be classified as primary, secondary and
if applicable, tertiary, depending on when their air is introduced. The
primary air register supplies primary air, which is the first to be
introduced in the burner. Secondary air is added to supplement primary
air. Burners may include a premixer to mix the air and fuel for better
combustion before introducing into the burner. Some burners even use
steam as premix to preheat the air and create better mixing of the fuel
and heated air. The floor of the furnace is mostly made of a different
material from that of the wall, typically hard cast able refractory to allow
technicians to walk on its floor during maintenance.
350
Direct radiation in the radiant section of a direct fired heater can be
described by the equation shown below.
qr
4
cpF(Tg
- Tw4)
Where,
qr = Radiant heat transfer, Btu/hr
s = Stefan-Boltzman constant, 0.173E-8 Btu/ft2-hr-R4
a = Relative effectiveness factor of the tube bank
Acp = Cold plane area of the tube bank, ft2
F = Exchange factor
Tg = Effective gas temperature in firebox, °R
Tw = Average tube wall temperature, °R
Relative Effectiveness Factor, a:
Because the tube bank does not absorb all the heat radiated to the cold
plane, an absorption effectiveness factor, a, can be used to correct the
cold plane area, depending on the arrangement of the tubes. The relative
effectiveness factor can be described by the following curves.
For a single row in front of a refractory wall, use Total One Row. For
two rows in front of a refractory wall, use Total Two Rows. For double
sided firing, use Direct One Row.
351
Cold Plane Area, Acp:
The normal heat-absorbing surfaces in a fired heater consist of a number
of parallel tubes. In the case of a fired heater design where the tubes are
fired from one side only, the tubes are normally positioned in front of a
refractory wall. Part of the radiation from the hot gas strikes the tubes
directly, while the rest passes through and is radiated back into the
chamber, where part is absorbed by the tubes. In the case of tubes fired
from both sides, as when the tubes are positioned in the center of the
chamber, the tubes absorb direct radiation from both sides. Expressing
the tube area as an equivalent plane area simplifies this calculation. The
calculated cold plane area is the area of a plane through the tube center
lines, whether they are in a curved plane, such as in a cylindrical pattern
or in a row side-by-side. For most tube panels, the width would be equal
to the center/center spacing of the tubes times the number of tubes.
352
The length is the length of tube exposed to the radiation. In the case of
tubes penetrating a tube sheet it is the length between tube sheets. But
for tubes with the return bends inside the firebox, the length may be
taken as the distance from the centerline of the return on one end to the
centerline of the return on the other end.
For a firebox with the tubes down the center, or other pattern which
results in the tubes being fired from both sides, the cold plane area
would be twice the projected area.
Exchange Factor, F:
Because the flue gas in the firebox is a poor radiator, the equation must
be corrected using an exchange factor which is dependent on the
emissivity of the gas and the ratio of refractory area to cold plane area.
Since the radiant heat is reflected back into the firebox, by the
refractory, a heater having a larger ratio of refractory surface relative to
the tube surface will absorb more heat. Since the tubes themselves are
not perfect absorbers, the curves are based on a tube-surface absorptivity
of 0.9. This is a value considered typical for oxidized metal surfaces.
353
The overall radiant exchange factor, F, can be taken from the curve
below.
Where,
Aw/aAcp :
The equivalent cold plane area,aAcp, is the product of the effectiveness
factor and the cold plane area as described above. The A w can be
described as follows,
Aw = Ar -
cp
and,
Aw = Effective refractory area, ft2
Ar = Total refractory area, ft2
354
cp
= Equivalent cold plane area, ft2
The total refractory area, Ar, is simply the total of the refractory area
exposed to the radiant section of the heater.
Flue Gas Emissivity :
The gas emissivity can be described by the curve.The tube wall
temperature has only a minor effect. Therefore, the emissivity can be
correlated as a function of PL product and the gas temperature, T g.
Variations in tube wall temperatures between 600 and 1200°F cause less
than 1% deviation from these curves.
And,
PL = Product of the Partial Pressure of the carbon dioxide and water
times the Beam Length, in atm-ft.
355
Where,
Partial Pressure Of CO2 & H2O :
The only constituents normally in the flue gas that contribute
significantly to the radiant emission are the carbon dioxide and the
water, the sum of these are all that are considered. The Partial pressure
of a gas component in atom's is the mole volume fraction percent of that
component.
For Box Type Heaters
Dimension Ratio
Mean Beam Length
1-1-1 to 1-1-3
2/3(Furnace Volume)1/3
1-2-1 to 1-2-4
1-1-4 to 1-1-inf
1 x Smallest Dimension
1-2-5 to 1-2-inf
1.3 x Smallest Dimension
1-3-3 to 1-inf-inf
1.8 x Smallest Dimension
For Vertical Cylindrical Heaters
Length/Diameter < 2
(((L/D)-1)*0.33 + 0.67)*D
Length/Diameter >= 2
Diameter
Mean Beam Length:
In computing the mean beam length, placement of the tubes must be
taken into account. If the firebox is a rectangular shape with the tubes
down the center, the beam length would be based on half the box. Beam
lengths for other configurations, such as a cylindrical heater with an
octagonal tube or cross tube layout, must be calculated with
consideration for those cavities.
356
The mean beam length for heaters can be accounted for according to
Wimpress in Hydrocarbon Processing.
Effective gas temperature in firebox, Tg
For a radiant section that is considered "well mixed", this temperature is
assumed to be equal to the temperature leaving the radiant section, i.e.,
the bridge wall temperature. For most applications, this is an acceptable
assumption. But in a high temperature heater with a tall narrow firebox
and wall firing, the Tg controlling radiant transfer may be 200 to 300 °F
higher than the exit temperature. Tall, bottom fired cylindrical heaters
are somewhere in between normal and this extreme. For those cases
where this is true, either an adjustment such as using another
temperature in this equation while using the exit temperature for the heat
balance or dividing the radiant into zones for the balance calculations,
should be considered.
Average tube wall temperature, Tw
Tube wall temperature depends on the temperature of the process fluid
and its transfer coefficient inside the tube, the thermal resistance of the
tube wall, the heat flux, and the fouling. The calculation of this
temperature will be treated in another section of this guide.
The average tube wall temperature as used herein may be one of either
the average temperature of the front 180° face of the tube, or the overall
average for the full circumference. Some engineers follow one method
while the others go the other way. Either way, the overall difference
between methods is relatively small.
357
Convective Heat Transfer In The Radiant Section
Even though most of the heat exchanged in the radiant section is from
radiant heat transfer, the convective heat transfer cannot be ignored. The
heat exchanged by convection can be described with the following
equation:
qc = hcAt(Tg - Tw)
Where,
qc = Convection heat transfer, Btu/hr
hc = Film heat transfer coefficient, Btu/hr-ft2- °R
At = Area of the tubes in bank, ft2
Tg = Effective gas temperature in firebox, °R
Tw = Average tube wall temperature, °R
Film heat transfer coefficient, hc
This value cannot be calculated precisely, and is usually selected by
experience or rule of thumb. The arrangement of the tubes as well as the
firebox design contributes to this factor. For horizontal tube, cabin type
heater, which is normally small in size, this coefficient might = 1.5,
where on large box heaters with multiple tube cells, it may be as high as
2.8. Vertical heaters with an L/D less than 2 would normally be
designed with hc = 2, where for an L/D greater than 2.0, you could use
3.0.
358
Total Radiant Heat Absorption In The Radiant Section
The total heat absorbed by the radiant section tubes, now can be
expressed by the following equation.
q R = qr + qc
Where,
qR = Total heat transfered to radiant tubes, Btu/hr
qr = Radiant heat transfer, Btu/hr
qc = Convective heat transfer, Btu/hr
Heat Balance In The Radiant Section
The procedures we have reviewed above gives us a method to either
compute the heat absorbed, or we can calculate the temperature we
would need to transfer a specific amount of heat into our process coil.
For us to make a heat balance, we will need to determine the firing rate
necessary to maintain that temperature. This is accomplished by a heat
balance around the fire box.
There are three primary sources of heat input to the radiant section, the
burner release, qrls, the sensible heat of the combustion air, q air, and the
sensible heat of the fuel and any atomizing medium, qother. Heat is taken
out of the radiant section by the two heat transfer methods previously
explained, qR and qS, and by losses through the casing, qloss, and sensible
heat of the exiting flue gas, qout.
We can now set up the heat balance equation as follows:
Where,
359
qrls + qair + qother = qR + qS + qloss + qout
qrls = Heat released by burners, Btu/hr
qair = Heat in the combustion air, Btu/hr
qother = Heat in other items, Btu/hr
qR = Heat absorbed by radiant tubes, Btu/hr
qS = Radiant heat to shield tubes, Btu/hr
qloss = Heat loss through setting, Btu/hr
qout = Heat in gas leaving radiant section, Btu/hr
qrls = Heat release by burners, Btu/hr
The burner release can be easily calculated for a gas when we know the
composition of the fuel and the heating values of the various
components. For liquid fuels, the heating values are obtained by a
calorimeter test.
From these values and using the standard combustion equation, we can
determine the composition of the flue gas. As an example, the
combustion of methane could be stated :
CH4 + 2O2 → CO2 + 2H2O
Of course for fuel gases containing many more components and burning
in air rather than pure oxygen, the equation gets more complicated.
Therefore, a task that in itself is quite simple, becomes a burden to do by
hand, but can be easily accomplished by a simple computer program.
The heating values normally used in fired heater design are the (LHV),
lower heating values.
To try some calculations, click the button below to open another
window to do some fuel combustion calculations:
360
qair = Heat in the combustion air, Btu/hr
The heat available in the combustion air, such as from preheated air, or
using Gas Turbine Exhaust, etc., is taken as the heat content above 60
°F, since that is the design datum temperature for fired heaters. For the
purpose of this discussion, radiant heat transfer, we are not going to take
this into account, i.e., we will consider the air at 60 °F.
qother = Heat in other items, Btu/hr
The heat available in other items would include such things as the fuel
when it is above 60 °F, atomizing air or steam, etc. These must be taken
into account in heater design; however, for the purposes of these
discussions, we are not going to include them.
qloss = Heat loss through setting, Btu/hr
These losses, referred to as Setting Loss or Radiation Loss are usually
not calculated during heater rating calculations. They are normally
accounted for by allowances, such as a percent of burner release or a
percent of heat absorbed. Either way, the loss amounts to a rule of
thumb. The actual losses may be calculated for the various surfaces and
these methods are described elsewhere.
qout = Sensible heat in flue gas leaving radiant section, Btu/hr
From the flue gas composition, we can calculate the overall enthalpy of
the flue gas, at a specific temperature, by adding the proportion each of
th components contribute to the total. These enthalpies can be obtained
from the following curves:
361
Since the flue gas composition remains consistent for a given fuel and
excess air, the first thing we need to do is perform the combustion
calculation. This is what we did above to compute the burner release.
The enthalpy for a given temperature can now be calculated by
obtaining the enthalpy for each component and adding them together.
362
Convection Section Design
In the convection section, heat is transferred by both radiation and
convection. The convection transfer coefficients for fin and stud tubes
are explored here as well as bare tube transfer. The short beam radiation
is
treated
separately
from
the
convection
transfer
below.
This section of Fired Heater Design is divided into five main areas:
-Convection Transfer, Bare Tubes.
-Convection Transfer, Fin Tubes.
-Convection Transfer, Stud Tubes.
-Short Beam, Reflective Radiation.
-Convection Section Design.
Convection Transfer, Bare Tubes
Overall Heat Transfer Coefficient, Uo:
Uo = 1/Rto
Where,
Uo = Overall heat transfer coefficient, Btu/hr-ft2-F
Rto = Total outside thermal resistance, hr-ft2-F/Bt
And,
Rto = Ro + Rwo + Rio
Ro = Outside thermal resistance, hr-ft2-F/Btu
Rwo = Tube wall thermal resistance, hr-ft2-F/Btu
Rio = Inside thermal resistance, hr-ft2-F/Btu
363
And the resistances are computed as,
Ro = 1/he
Rwo = (tw/12*kw)(Ao/Aw)
Rio = ((1/hi)+Rfi)(Ao/Ai)
he = Effective outside heat transfer coefficient, Btu/hr-ft2-F
hi = Inside film heat transfer coefficient, Btu/hr-ft2-F
tw = Tubewall thickness, in
kw = Tube wall thermal conductivity, Btu/hr-ft-F
Ao = Outside tube surface area, ft2/ft
Aw = Mean area of tube wall, ft2/ft
Ai = Inside tube surface area, ft2/ft
Rfi = Inside fouling resistance, hr-ft2-F/Btu
Where,
Inside film heat transfer coefficient, hi:
The inside heat transfer coefficient calculation procedure is covered in
detail, elsewhere in this course.
Effective outside heat transfer coefficient, he
he = 1/(1/(hc+hr)+Rfo)
Where,
hc = Outside heat transfer coefficient, Btu/hr-ft2-F
hr = Outside radiation heat transfer coefficient, Btu/hr-ft2-F
Rfo = Outside fouling resistance, hr-ft2-F/Btu
364
Outside film heat transfer coefficient, hc:
The bare tube heat transfer film coefficient, h c, can be described by the
following equations.
For a staggered tube arrangement,
hc = 0.33*kb(12/do)((cp
b)/kb)
1/3
((do/12)(Gn
b)))
0.6
b)))
0.6
And for an inline tube arrangement,
hc = 0.26*kb(12/do)((cp
b)/kb)
1/3
((do/12)(Gn
Where,
hc = Convection heat transfer coefficient, Btu/hr-ft2-F
do = Tube outside diameter, in
kb = Gas thermal conductivity, Btu/hr-ft-F
cp = Gas heat capacity, Btu/lb-F
mb = Gas dynamic viscosity, lb/hr-ft
Gn = Mass velocity of gas, lb/hr-ft2
Convection Transfer, Fin Tubes
You will notice that the heat transfer equations for the fin tubes are
basically the same as for the bare tubes untill you reach the he factor,
where a new concept is introduced to account for the fin or extended
surface. The procedure presented herein are taken from the Escoa
manual which can be downloaded in full from the internet.
-Overall Heat Transfer Coefficient, Uo:
Uo = 1/Rto
365
Where,
Uo = Overall heat transfer coefficient, Btu/hr-ft2-F
Rto = Total outside thermal resistance, hr-ft2-F/Btu
And,
Rto = Ro + Rwo + Rio
Ro = Outside thermal resistance, hr-ft2-F/Btu
Rwo = Tube wall thermal resistance, hr-ft2-F/Btu
Rio = Inside thermal resistance, hr-ft2-F/Btu
And the resistances are computed as,
Ro = 1/he
Rwo = (tw/12*kw)(Ao/Aw)
Rio = ((1/hi)+Rfi)(Ao/Ai)
Where,
he = Effective outside heat transfer coefficient, Btu/hr-ft2-F
hi = Inside film heat transfer coefficient, Btu/hr-ft2-F
tw = Tubewall thickness, in
kw = Tube wall thermal conductivity, Btu/hr-ft-F
Ao = Total outside surface area, ft2/ft
Inside film heat transfer coefficient, hi:
The inside heat transfer coefficient calculation procedure is covered in
detail, elsewhere in this course.
366
Effective outside heat transfer coefficient, he:
he = ho(E*Afo+Apo)/Ao
Where,
ho = Average outside heat transfer coefficient, Btu/hr-ft2-F
E = Fin efficiency
Ao = Total outside surface area, ft2/ft
Afo = Fin outside surface area, ft2/ft
Apo = Outside tube surface area, ft2/ft
Average outside heat transfer coefficient, ho:
ho = 1/(1/(hc+hr)+Rfo)
Where,
hc = Outside heat transfer coefficient, Btu/hr-ft2-F
hr = Outside radiation heat transfer coefficient, Btu/hr-ft2-F
Rfo = Outside fouling resistance, hr-ft2-F/Btu
Outside film heat transfer coefficient, hc:
hc = j*Gn*cp(kb/(cp*mb))0.67
Where,
j = Colburn heat transfer factor
Gn = Mass velocity based on net free area, lb/hr-ft2
cp = Heat capacity, Btu/lb-F
kb = Gas thermal conductivity, Btu/hr-ft-F
367
mb = Gas dynamic viscosity, lb/hr-ft
Colburn heat transfer factor, j:
j = C1*C3*C5(df/do)0.5((Tb+460)/(Ts+460))0.25
Where,
C1 = Reynolds number correction
C3 = Geometry correction
C5 = Non-equilateral & row correction
df = Outside diameter of fin, in
do = Outside diameter of tube, in
Tb = Average gas temperature, F
Ts = Average fin temperature, F
Reynolds number correction, C1:
C1 = 0.25*Re-0.35
Where,
Re = Reynolds number
Geometry correction, C3:
For segmented fin tubes arranged in, a staggered pattern,
C3 = 0.55+0.45*e(-0.35*lf/Sf)
an inline pattern,
C3 = 0.35+0.50*e(-0.35*lf/Sf)
For solid fin tubes arranged in, a staggered pattern,
C3 = 0.35+0.65*e(-0.25*lf/Sf)
368
an inline pattern,
C3 = 0.20+0.65*e(-0.25*lf/Sf)
Where,
lf = Fin height, in
sf = Fin spacing, in
Non-equilateral & row correction, C5:
For fin tubes arranged in, a staggered pattern,
C5 = 0.7+(0.70-0.8*e(-0.15*Nr^2))*e(-1.0*Pl/Pt)
an inline pattern,
C5 = 1.1-(0.75-1.5*e(-0.70*Nr))*e(-2.0*Pl/Pt)
Where,
Nr = Number of tube rows
Pl = Longitudinal tube pitch, in
Pt = Transverse tube pitch, in
Mass Velocity, Gn:
Gn = Wg/An
Where,
Wg = Mass gas flow, lb/hr
An = Net free area, ft2
Net Free Area, An:
An = Ad - Ac * Le * Nt
369
Where,
Ad = Cross sectional area of box, ft2
Ac = Fin tube cross sectional area/ft, ft2/ft
Le = Effective tube length, ft
Nt = Number tubes wide
And,
Ad = Nt * Le * Pt / 12
Ac = (do + 2 * lf * tf * nf) / 12
tf = fin thickness, in
nf = number of fins, fins/in
Surface Area Calculations:
For the prime tube,
Apo = Pi * do (1- nf * tf) / 12
And for solid fins,
Ao = Pi*do(1-nf* tf)/12+Pi*nf(2*lf(do+lf)+tf(do+2*lf))/12
And for segmented fins,
Ao = Pi*do(1-nf* tf)/12+0.4*Pi*nf(do+0.2)/12+Pi*nf (do+0.2)((2*lf0.4)(wn+tf)+ws*tf)/(12*ws)
And then,
Afo = Ao - Apo
Where,
370
ws = Width of fin segment, in
Fin Efficiency, E:
For segmented fins,
E = x * (0.9 + 0.1 * x)
And for solid fins,
E = y * (0.45 * ln(df / do) * (y - 1) + 1)
Where,
y = x * (0.7 + 0.3 * x)
And,
x = tanh(m * B) / (m * B)
Where,
B = lf + (tf /2)
For segmented fins,
m = (ho (tf + ws) / (6 * kf * tf * ws))0.5
And for solid fins,
m = (ho / (6 * kf * tf))0.5
Fin Tip Temperature, Ts:
The average fin tip temperature is calculated as follows,
Ts = Tg + (Tw - Tg) * 1/((e1.4142mB+e-1.4142mB)/2)
Maximum Fin Tip Temperature, Tfm:
The maximum fin tip temperature is calculated as follows,
371
Tsm = Twm
gm
- Twm)
Where,
Tsm = Maximum Fin Tip Temperature, F
Tgm = Maximum Gas Temperature, F
Twm = Maximum Tube Wall Temperature, F
372
Sample calculation:
373
374
375
Compressor(k-102)
A gas compressor is a mechanical device that increases the pressure of a
gas by reducing its volume. Compressors are similar to pumps: both
increase the pressure on a fluid and both can transport the fluid through
a pipe. As gases are compressible, the compressor also reduces the
volume of a gas. Liquids are relatively incompressible; while some can
be compressed, the main action of a pump is to pressurize and transport
liquids.
Types of compressors
376
The main types of gas compressors are illustrated and discussed below
Centrifugal compressors
Centrifugal compressors use a rotating disk or impeller in a shaped
housing to force the gas to the rim of the impeller, increasing the
velocity of the gas. A diffuser (divergent duct) section converts the
velocity energy to pressure energy. They are primarily used for
continuous, stationary service in industries such as oil refineries,
chemical and petrochemical plants and natural gas processing plants.
Their application can be from 100 horsepower (75 kW) to thousands of
horsepower. With multiple staging, they can achieve extremely high
output pressures greater than 10,000 psi (69 MPa)
Axial-flow compressors
Axial-flow compressors are dynamic rotating compressors that use
arrays of fan-like airfoils to progressively compress the working fluid.
They are used where there is a requirement for a high flow rate or a
compact design.The arrays of airfoils are set in rows, usually as pairs:
one rotating and one stationary. The rotating airfoils, also known as
blades or rotors, accelerate the fluid. The stationary airfoils, also known
377
as stators or vanes, decelerate and redirect the flow direction of the
fluid, preparing it for the rotor blades of the next stage. Axial
compressors are almost always multi-staged, with the cross-sectional
area of the gas passage diminishing along the compressor to maintain an
optimum axial Mach number. Beyond about 5 stages or a 4:1 design
pressure ratio, variable geometry is normally used to improve
operation.Axial compressors can have high efficiencies; around 90%
polytropic at their design conditions. However, they are relatively
expensive, requiring a large number of components, tight tolerances and
high quality materials. Axial-flow compressors can be found in medium
to large gas turbine engines, in natural gas pumping stations, and within
certain chemical plants.
Reciprocating compressors
Reciprocating compressors use pistons driven by a crankshaft. They can
be either stationary or portable, can be single or multi-staged, and can be
driven by electric motors or internal combustion engines. Small
reciprocating compressors from 5 to 30 horsepower (hp) are commonly
seen in automotive applications and are typically for intermittent duty.
Larger reciprocating compressors well over 1,000 hp (750 kW) are
commonly found in large industrial and petroleum applications.
Discharge pressures can range from low pressure to very high pressure
(>18000 psi or 180 MPa). In certain applications, such as air
compression, multi-stage double-acting compressors are said to be the
most efficient compressors available, and are typically larger, and more
costly than comparable rotary units. Another type of reciprocating
compressor is the swash plate compressor, which uses pistons which are
378
moved by a swash plate mounted on a shaft - see Axial Piston Pump.
Household, home workshop, and smaller job site compressors are
typically reciprocating compressors 1½ hp or less with an attached
receiver tank.
Rotary compressors
There are many types of rotary compressor one of them is the rotary
screw compressors. The rotary screw compressors use two meshed
rotating positive-displacement helical screws to force the gas into a
smaller space. These are usually used for continuous operation in
commercial and industrial applications and may be either stationary or
portable. Their application can be from 3 horsepower (2.2 kW) to over
1,200 horsepower (890 kW) and from low pressure to moderately high
pressure (>1,200 psi or 8.3 MPa).Rotary screw compressors are
commercially produced in Oil Flooded, Water Flooded and Dry type.
Another type of rotary compressor is rotary vane compressor. Rotary
vane compressors consist of a rotor with a number of blades inserted in
radial slots in the rotor. The rotor is mounted offset in a larger housing
which can be circular or a more complex shape. As the rotor turns,
blades slide in and out of the slots keeping contact with the outer wall of
the housing. Thus, a series of decreasing volumes is created by the
rotating blades. Rotary Vane compressors are, with piston compressors
one of the oldest of compressor technologies. With suitable port
connections, the devices may be either a compressor or a vacuum pump.
They can be either stationary or portable, can be single or multi-staged,
and can be driven by electric motors or internal combustion engines.
379
Dry vane machines are used at relatively low pressures (e.g., 2 bar or
200 kPa; 29 psi) for bulk material movement while oil-injected
machines have the necessary volumetric efficiency to achieve pressures
up to about 13 bar (1,300 kPa; 190 psi) in a single stage. A rotary vane
compressor is well suited to electric motor drive and is significantly
quieter in operation than the equivalent piston compressor. Rotary vane
compressors can have mechanical efficiencies of about 90%
Theory used in Compressor calculation
 n 


P1  T1  n 1 
 
P2  T 2 
Where
P1 : inlet pressure, psia
P2 : outlet pressure, psia
T1 : inlet temperature, R
T2 : outlet temperature, R
n : compression factor
W 
nR (T 1 T 2 )
1 n
Where
W : work done, Btu/lbmol
R : Cp/Cv
Hp=W*M
Where
380
Hp : horse power, Hp
M : molar flow rate, lbmol/s
Ep 
n
n 1
K
K 1
Where
Ep : efficiency of the compressor
K 
MwC p
MwC p  1.986
Where
Cp: heat capacity, Btu/lb F0
Mw: molecular weight of the gas
381
382
Tank T-201A&B,V-105:
Tanks are basically was made to hold, transport or store fluid and solid.
Gases are stored at high pressures where this process requirement and to
reduce the storage volume .for some gases the volume can be further
reduced by liquefying the gas by pressure or refrigeration .cylindrical
and spherical vessels are used. Liquids are usually stored in bulk in
vertical cylindrical steel tanks; Fixed and floating-roof tanks are used. In
a floating-roof tank a movable piston floats on the surface of the liquid
and is sealed to the tank walls. Floating-roof tanks are used to eliminate
evaporation losses and, for liquids, to obviate the need for inert gas
blanketing to prevent an explosive mixture forming above the liquid, as
would be the situation with a fixed-roof tank. Horizontal cylindrical
tanks and rectangular tanks are used also used for storing liquids,
usually for relatively small quantities. Storage of solid is usually more
expensive than the movement of liquid and gases, which can easily
383
pumped down a pipeline. The design more flexible and moderate the
international conditions.
Design of storage tank
In the design of the storage there are some main principles to be
considered to make
The design more flexible and more moderate the international
conditions.
In the case of this project the tanks which will be used is cylindrical in
shape. This shape is the proper one for the liquid phase in our case. Also
the top space in the tank for the vapor pressure of the component will be
12 % of the tank volume.
The steps in the design will begin in finding the volume of the liquid
part in the tank and then determining the height and diameter of this
part. After that the total volume will be determined. Finally all the
dimensions could be determined from this volume.
Volume of the liquid= Total mass flow rate in* time hold-up
Time hold-up: the time where the liquid is hold inside the tank
Volume of cylinder = π R2 H … (1)
Get the volume of the liquid (assume H=5D)
To determine vapor pressure Antoine equation is used:
Log10 P* = A- (B/C+T) … (2)
T: the temperature in ˚C
The values of A, B, and C is taken from table antoin equation constant
384
At this vapor pressure the top space in the tank can be determined.
Total Volume = free volume + liquid V … (3)
Diameter3 = (5*total volume)/П
Height = 0.2diameter
Area of the tank = total volume/ height … (4)
Thickness
t = (P r I / S E - 0.6P) + Cc …(5)
Sample calculation(Tank T-201A&B):
385
Sample calculation(V-105):
386
THREE- PHASE SEPARATOR(V-201):
Introduction:Horizontal three- phase separator used to separate water, gas and oils
from the inlet . We do this design to calculate the diameter and length
for the horizontal three- phase separator and determine the cost.
Horizontal three- phase separator
387
Procedure design :-
The diameter and length of the horizontal three-phase separator can
be determined by the following steps:
 Step #1:Determine maximum allowable oil pad thickness Ho,max .
Ho,max = [ ( 1.28 x 10-3 ) (to ) (γo – γw ) ( dm2) ] / µo
 Step # 2:Determine the ratio of Aw/A.
Aw/A = ( 0.5)[ (Qw) (tw) / [ (Qo ) (to) + (Qw) (tw) ]]
 Step # 3 :From Fig.(11) in appendix (A) determine the ratio of Ho/D at Aw/A
 Step # 4 :Determine the maximum diameter of the vessel associated with
the maximum oil pad height.
Dmax = [ (Ho,max ) ] / [ (Ho) (D) ]
 Step # 5 :determine the gas and oil densities.
388
ρg = [ (2.7) (γg) (P) ] / [ (T) (Z) ]
ρl = [ (ρw) (γl) ]
 Step # 6 :Determine the gas capacity constraint.
DL = (420) [ (T) (Z) (Qg) / (P) ] [(ρg) (Cd) /(dm) (ρo - ρg) ]0.5
 Step # 7 :Check the liquid capacity (retention time) constraint.
D2L = (1.429) [ (Qo) (to) + (Qw) (tw) ]
 Step # 8 :Assume diameter smaller than determined maximum diameter and
determine the corresponding effective length from equation used in
step # 7 then determine seam-to-seam length by
Ls= (4L) / 3
 Step # 9 :Select a reasonable diameter and length by recommended with
slenderness ratio in the rang of 3 to 5 , where slenderness ratio
(SR):
SR=L/(D/12)
389
 Step # 10:Calculate thickness (t) of three phase separator by using
design equation for cylindrical shells:
t = (P ri) / (S Ej - 0.6 P) + Cc
 Step # 11:Calculate the cost from internet.
do = di + 2t
Volume of cylinder using (do) = Л (do/2)2 L
Volume of cylinder using (di) = Л (di/2)2 L
Volume of metal = Volume of cylinder using (do) - Volume of cylinder using (di)
390
Sample
calculation(v-201):
391
Abdulaziz Alshomar
208113942
Section #1
Reactor (R-101)
Flash Tank (V-204)
Cooler Desgin (E-209)
Cooler Desgin (E-211)
Heater Desgin (E-212)
Reactor
The reactor is the heart of a chemical process. It is the only place in the
process where raw materials are converted into products, and reactor
design is a vital step in the overall design of the process.
The design of an industrial chemical reactor must satisfy the following
requirements:
1. The chemical factors: The kinetics of the reaction. The design must
provide sufficient residence time for the desired reaction to proceed to
the required degree of conversion.
2. The mass transfer factors: With heterogeneous reactions the reaction
rate may be controlled by the rates of diffusion of the reacting species,
rather than the chemical kinetics.
3. The heat transfer factors: The removal, or addition, of the heat of
reaction.
392
4. The safety factors: The confinement of hazardous reactants and
products, and the control of the reaction and the process conditions.
The need to satisfy these interrelated and often contradictory factors
makes reactor design a complex and difficult task. However, in many
instances one of the factors will predominate and will determine the
choice of reactor type and the design method.
The following characteristics are normally used to classify reactor
designs:
1. Mode of operation: batch or continuous.
2. Phases present: homogeneous or heterogeneous.
3. Reactor geometry: flow pattern and manner of contacting the phases:
i. Stirred tank reactor;
ii. Tubular reactor;
iii. Packed bed, fixed and moving;
iv. Fluidized bed.
Fluidized-Bed Reactors
The essential feature of a fluidized-bed reactor is that the solids are held
in suspension by the upward flow of the reacting fluid; this promotes
high mass and heat transfer rates and good mixing. Heat transfer
coefficients in the order of 200W/m28C to jackets and internal coils are
typically obtained. The solids may be a catalyst, a reactant in fluidized
combustion processes, or an inert powder added to promote heat
transfer.
393
Though the principal advantage of a fluidized bed over a fixed bed is the
higher heat transfer rate, fluidized beds are also useful where it is
necessary to transport large quantities of solids as part of the reaction
processes, such as where catalysts are transferred to another vessel for
regeneration.
Fluidization can be used only with relatively small-sized particles, <300
mm with gases.
A great deal of research and development work has been done on
fluidized-bed reactors in recent years, but the design and scale-up of
large diameter reactors is still an uncertain process and design methods
are largely empirical.
The principles of fluidization processes are described in Richardson et
al. (2002).
The design of fluidized bed reactors is discussed by Rase (1977).
Reactor R-101:
HYSYS data
FA0 (Kgmol/hr)
T0 (C)
T (C)
PT0 (psia)
PT (psia)
X
yA0
yB0
Table (3-1) Hysys reactor worksheet
394
126.07
451
548
40
35
0.997
0.2534
0.7056
C6H5NO2 + 3H2 - C6H5NH2 + 2H2O
The initial conversions and the initial reaction rates at different reaction
conditions were obtained by extrapolating the conversion curves to time
t=0 min. It could be
shown that hydrogenation of nitrobenzene on the fresh catalyst follows a
Langmuir Hinshelwood mechanism considering the surface reaction of
the adsorbed nitrobenzene molecule and one adsorbed hydrogen atom as
the rate determining step (Amon et al., 1999a):
. Parameter
Value of estimation
Standard deviation
k0 (Kmol/Kgcat.s)
1.86x10-4
1.27x10-5
EA (KJ/mol)
10
0.9
KNB (Kpa-1)
1.51x10-2
2.98x10-3
KH2 (Kpa-0.5)
0.14
0.02
Table ( 3-2 ) Kinetic constants of the initial reaction rate
rA = kKNBKH2PNBPH20.5/(1+ KNBPNB + KH2PH20.5)2
PA0  y A0 PT 0  10.136 psia
  y A0
= 0.534*(1+2-3-1) = -0.534
A 
y A0
=1
y A0
B 
y B0
 0.7056 / 0.2534  2.78453
y A0
PA  PA0
 A  x  T0  P 
   = PNB
1  x  T  P0 
 0.04682907 psia =
0.3227 Kpa
395
PB  PA0
 B  x  T0  P  = P
H2
  
1  x  T  P0 
 27.9 psia =
192.263 Kpa
Kinetic Data:
k = 1.642x10-7 Kmol/Kgcat.s
-rA = kKNBKH2PNBPH20.5/(1+ KNBPNB + KH2PH20.5)2
= (1.642x10-7*1.51x10-2*0.14*0.3227*192.2630.5) / (1+ 1.51x102*
0.3227 + 0.14*192.2630.5)2 = 1.79x10-10 Kmol/Kgcat.s
= 1.79x10-7 mol/Kgcat.s
= 6.45X10-4 mol/Kgcat.hr
dx  rA

dW FA0
W
FA0 ( x)
 194827 Kg
 rA
Wcopper = 15% of the total weight = 29224 Kg
Wsilica = 1548.211376 – 29224 = 165603 Kg
  ( Bulk )  100
lb / ft 3
= 45.359237 Kg/ft3
  = 0.3
 V (column) 
W
3
 6136 ft
(1   ) * 
V
 D   
 
= 173.75 m3
1/ 3
 3.81
m
 L  6D  22.86 m
396
Equipment Name
Reactor
Objective
Convert nitrobenzene to aniline
Equipment Number
R-101
Designer
Abdulaziz Alshomer
Type
Fluidized bed vapor catalytic reactor
Material of Construction
Carbon steel
Table (3-3) reactor 101 information
C6H5NO2 + S ↔ C6H5NO2.S
-rNB = kNB(PNBCv - CNB.s/KNB)
-rNB/kNB = 0
CNB.s = KNB PNBCv
0.5H2 + S ↔ H.S
-rH2 = kH2(PH20.5Cv – CH.s/KH2)
-rH2/kH2 = 0
CH.s = KH2PH20.5Cv
C6H5NO2.S + H.S C6H5NH2 + H2O + 2S
-ra = K CNB.s CH.s
Ct = CNB.s + CH.s + Cv
Ct = KNBPNBCv + KH2PH20.5Cv+ Cv
Cv = Ct / (1 + KNB PNB + KH2 PH20.5 )
Surface Reaction rate limiting
-ra = rs = KsCt KNB KH2 PNB PH20.5 / (1 + KNB PNB + KH2 PH20.5 )2
397
Let k = Ks Ct
-rA = kKNBKH2PNBPH20.5/(1+ KNBPNB + KH2PH20.5)2
GAS-LIQUID SEPARATORS
The separation of liquid droplets and mists from gas or vapor streams is
analogous to the separation of solid particles and, with the possible
exception of filtration, the same techniques and equipment can be used.
Where the carryover of some fine droplets can be tolerated, it is often
sufficient to rely on gravity settling in a vertical or horizontal separating
vessel (knockout pot).
Knitted mesh demisting pads are frequently used to improve the
performance of separating vessels where the droplets are likely to be
small, down to 1 mm, and where high separating efficiencies are
required. Proprietary demister pads are available in a wide range of
materials, metals and plastics; thicknesses; and pad densities. For liquid
separators, stainless steel pads around 100mm thick and with a nominal
density of 150 kg/m3 would generally be used. Use of a demister pad
allows a smaller vessel to be used. Separating efficiencies above 99%
can be obtained with low pressure drop.
The design and specification of demister pads for gas-liquid separators
is discussed by Pryce Bailey and Davies (1973).
The design methods for horizontal separators in following sections are
based on a procedure given by Gerunda (1981).
Cyclone separators are also frequently used for gas-liquid separation.
They can be designed using the same methods for gas-solids cyclones.
398
The inlet velocity should be kept below 30 m/s to avoid pickup of liquid
from the cyclone surfaces.
The equation below can be used to estimate the settling velocity of the
liquid droplets, for the design of separating vessels:
If a demister pad is not used, the value of ut obtained from equation
10.10 should be multiplied by a factor of 0.15 to provide a margin of
safety and to allow for flow surges.
Horizontal Separators
Figure (3-1) Horizontal liquid-vapor separator
The layout of a typical horizontal separator is shown in Figure 3-1.
399
A horizontal separator would be selected when a long liquid holdup time
is required.
In the design of a horizontal separator, the vessel diameter cannot be
determined independently of its length, unlike for a vertical separator.
The diameter and length and the liquid level must be chosen to give
sufficient vapor residence time for the liquid droplets to settle out and
for the required liquid holdup time to be met.
The most economical length to diameter ratio will depend on the
operating pressure.As a general guide, the following values can be used:
For preliminary designs, set the liquid height at half the vessel diameter,
where fv is the fraction of the total cross-sectional area occupied by the
vapor.
Flash Tank (V-204)
vapor flow 1362.463 Kg/h
liquid flow 1362.374 Kg/h
flow rate 2724.837 Kg/h
Temp. 40 C
400
liquid density 1004.23 kg/m3
vapor density 0.302391 kg/m3
Ut 4.033341 m/s
Us 0.605001 m/s
vapor volumtric flow rate 1.251566 m3/s
cross sectional area 0.393Dv^2
Vapor velocity 3.184647 Dv^-2
Actual residence time 0.94202 Dv^3
hv/us 0.826445 Dv
Dv 0.936649 m standard dv 1.55 m
liquid volumetric flow rate 0.000377 m3/s
Liquid cross sectional area 0.943459 m^2
Length 4.65 m
Hold Volume 4.387083 m^3
risidence time 11641.67 s 194.0278 min
401
Cooler Design (E-209):Fluid properties:
Shell Side ( gas stream ) shell side (water stream)
Flow rate
7.00E+04
Kg/h
Inlet Temperature ,T1
90
o
Outlet Temperature ,T2
93.92
o
Heat Capacity of inlet stream, Cpin
4.1943
KJ/kg°C
Heat Capacity of outlet stream, Cpout
4.193
KJ/kgoC
Average Heat Capacity, Cpavg
4.19365
KJ/kgoC
Mass Density of inlet stream , ρin
956.22
kg/m3
Mass Density of outlet stream , ρout
953.05
kg/m3
Average Mass Density, ρavg
954.635
kg/m3
Average Viscosity of stream, µavg
0.305
mNs/m2
Average Thermal conductivity, Kf
0.677
W/moC
Calculation of Heat load:
Qh = mh *Cp * (T1-T2)
where:
Qh = heat load in the hot side (KW)
mh = mass flowrate of hot fluid (Kg/h)
Cp = heat capacity of hot fluid (kJ/kgoC)
T1 = inlet temperature (oC)
T2 =outlet temperatue (oC)
Heat load = 319.6493222 KW > 1000 KW
402
C
C
Tube side ( gas stream)
Flowrate
1.15E+04
Kg/hr
Average Heat Capacity, Cp
1.82285
kJ/kgoC
Average Mass Density, ρ
1009
kg/m3
Average Viscosity of stream, µ
2.488
mNs/m2
Average Thermal conductivity of stream, Kf
0.146
W/moC
inlet Temperature , t1
46.816
o
outlet Temperature, t2
40
o
C
C
Log mean Temperature calculation:
∆Tlm =(T1-t2)-(T2-t1) / ln((T1-t2)/(T2-t1))
where:
∆Tlm = log mean temperature differace
T1 = inlet shell side fluid temoerature (oC)
T2 = outlet shell side fluid temerature (oC)
t1
= inlet tube side temoerature (oC)
t2 = outlet tube side temerature (oC)
∆Tlm = 1627.484556 oC
Temperature correction Fctor calculation:
Using two shell pass and four or multiple of four tube passes
R = (T1-T2) / (t2-t1)
R = 0.5751174
S = (t2-t1) / (T1-t1)
S = 0.1578362
Ft = 0.99 From figure 12.20
403
∆Tm = F
t
* ∆Tlm
where:
∆Tm = true temperature difference
Ft
= the temperature correction factor
∆Tlm = log mean temperature differace
∆Tm = 1611.20971 oC
Assuming U = 10 W/m2 oC ( From table 12.1)
Tube outside diameter(do) = 38 mm 1.496063
A= Q / U *
∆Tm
where:
A = provisional area (m2)
Q = heat load
(kW)
U = overall heat transfer coefficient (W/m2 oC)
Provisional area = 19.839089 m2
Tube inner diameter(di) = 26 mm 1.023622
Tube length(L) = 4.88 m 192.1256
Choosing
Take tube materail is cupro- nickel
Area of one tube = L* do *π
404
Area of one tube = 0.5825769 m2
Number of tubes = provisinal area / area of one tube
Number of tubes = 34
Using 1.25 triangular pitch
K1 = 0.175 Use table 12.4
n1 = 2.285
Db = (do)*( Nt / K1)^ (1/n1)
where;
Db =bundle diameter (mm)
do = outer diameter (mm)
Nt : number of tubes
K1 & n1 are constant
Bundle diameter (Db) = 382 mm
0.3815849 m
Using split ring floating head type
Bundle diametrical clearance = 67 mm
Ds = Db + Bundle diametrical clearance
Shell diameter(Ds) = 449 mm
0.4485849 m
Shell length(m)= 5.3285849 m
Tube side coefficient
405
Method 1
Mean Tube temperature=(t1+t2)/2 = 43.408 oC
Tube cross-sectional area = p/4 *di2= 530.92916 mm2
Tube per pass=(Nt/4) = 9
Total flow area = tubes per pass * cross sectional area
Total flow area = 0.0045201 m2
mass velocity = mass flow rate / total flow area
Tube mass velocity = 2.55E+06 kg/s.m2
linear velosity (ut ) = mass velosity / density
Tube linear velocity (ut) = 2.52E+03 m/s
hi = (4200* (1.35+0.02t)* ut0.8)/(di0.2)
where:
hi =inside coefficient (W/m2 oC)
t =mean temperature (oC)
ut =linear velocity (m/s)
di =tube inside diameter (mm)
hi = 2557872 W/m2 oC
Method 2
Renolds number (Re) = ρ* ut *di/ µ
Re = 2.66E+07
Prandtl number (Pr) = Cp µ / kf
Pr = 3.11E+01
406
L/di = 1.88E+02
From Figuer 12.23 Tube -side heat transfer factor:
jh = 2.00E-03
where jh is the heat transfer factor
assume that the viscisity of the fluid is the same as at the wall
(µ/µwall) = 1
(hi di / kf) = jh Re Pr0.33 * (µ/µwall)0.14
hi = 9.28E+05 W/m2 oC
Using hi from method 1 as it has low value
hi = 927661 W/m2 oC
Shell-side coefficient
Choose baffle spacing (Lb)= (Ds/5) = 89.71698 mm
Tube pitch (pt) =1.25 * do= 47.5 mm
Cross flow area (As) =((pt - do)* Ds* Lb)/pt
Cross flow area As = 0.008049 m2
Mass velocity (Gs) = mass flow rate / cross flow area
Mass velocity (Gs) = 2415.718 kg/s.m2
Equivalent diameter de =(1.1/do)(pt2-0.917do2)
de = 2.70E+01 mm
Mean Shell side temperature =(T1+T2)/2
Mean Shell side temperature = 91.96 oC
Renolds number (Re) = (Gs de)/ m
407
Re = 2.E+05
Prandtl number (Pr) = Cp µ / kf
Pr = 1.89
Choose 25% baffle cut
jh = 1.50E-03
Without the viscosity correction term, (µ/µw) = 1
hs = kf * jh *Re *Pr^(1/3) / de
hs = 9950.843 W/m2 oC
Overall Heat Transfer Coefficient
Thermal conductivity of cupro-nickel alloy= 50 W/moC
Taking fouling coefficients from table 12.2
Outside coefficient(fouling factor)=hod 5000 W/m2 oC
Inside coefficient(fouling factor) =hid 5000 W/m2 oC
/Uo =(1/ho) + (1/hod) + ( do(ln(do/di))/2kw) + (do/di) * (1/hid) + (do/di) *
(1/hi)
1/Uo = 0.0007386
Uo = 1353.9435 W/m2 oC
Close to initial value assumed
Pressure Drop
Tube side
Re = 2.66E+07
jf = 2.00E-03
408
where jf is the friction factor
ΔPt = Np [ 8jf (L/di)(µ/µw)^(-m) +2.5 ] ρut²/2
where ,
ΔPt = tube side pressur drop (N/m²)(pa)
Np = number of tube side passes
ut = tube side velosity (m/s)
L = length of one tube
Neglecting the viscosity correction term, (µ/µw) = 1
∆pt= 2.236E+10 N/m2
22361599 kPa
3243490.8 psi
Shell side
(Acceptable)
Linear velocity =Gs /ρ= 2.5305147 m/s
Re = 2.E+05
jf = 1.50E-03 from fig 12.29
ΔPs = 8jf (Ds/de)(L/Lb)( ρus^2/2)(µ/µw)^(-0.14)
where :
L : tube length
Lb : baffle spacing
∆Ps = 331.68331 N/m2
0.3316833 kPa
409
0.0481098 psi
Thickness Calculations:
t =(Pri/(SEJ-0.6P))+Cc
where:
t = shell thichness (in)
P = Maximum allowable internal pressure (psig)
ri = internal raduis of shell before allowance corrosion is added (in)
EJ = efficincy of joients
S = working stress (psi)
Cc = allowance for corrosin (in)
ri = 8.830416549 in
P = 587.8 psi
S = 13706.66 psi (for carbon steel)
EJ = 0.85
Cc = 0.125 in
t = 0.584 in
t = 14.8 mm
Cost Calculations:
Heat transfer area = 213.55 ft2
Cost = $50,100
410
Cooler Design (E-211):Fluid properties:
Shell Side ( gas stream ) shell side (water stream)
Flow rate
5.00E+03 Kg/h
Inlet Temperature ,T1
25
o
C
Outlet Temperature ,T2
26.19
o
C
Heat Capacity of inlet stream, Cpin
4.2025
KJ/kg°C
Heat Capacity of outlet stream, Cpout
4.2022
KJ/kgoC
Average Heat Capacity, Cpavg
4.20235
KJ/kgoC
Mass Density of inlet stream , ρin
1007.3
kg/m3
Mass Density of outlet stream , ρout
1006.5
kg/m3
Average Mass Density, ρavg
1006.9
kg/m3
Average Viscosity of stream, µavg
0.879
mNs/m2
Average Thermal conductivity, Kf
0.612
W/moC
Calculation of Heat load:
Qh = mh *Cp * (T1-T2)
where:
Qh = heat load in the hot side (KW)
mh = mass flowrate of hot fluid (Kg/h)
Cp = heat capacity of hot fluid (kJ/kgoC)
T1 = inlet temperature (oC)
T2 =outlet temperatue (oC)
Heat load = 6.945550694 KW > 1000 KW
411
The type of the heat exchanger is shell and tube heat exchanger
Tube side ( gas stream)
Flowrate
1.74E+02
Kg/hr
Average Heat Capacity, Cp
3.17025
kJ/kgoC
Average Mass Density, ρ
498.44266 kg/m3
Average Viscosity of stream, µ
0.333
mNs/m2
Average Thermal conductivity of stream, Kf
0.330
W/moC
inlet Temperature , t1
139
o
C
outlet Temperature, t2
40
o
C
Log mean Temperature calculation:
∆Tlm =(T1-t2)-(T2-t1) / ln((T1-t2)/(T2-t1))
where:
∆Tlm = log mean temperature differace
T1 = inlet shell side fluid temoerature (oC)
T2 = outlet shell side fluid temerature (oC)
t1
= inlet tube side temoerature (oC)
t2 = outlet tube side temerature (oC)
∆Tlm = 63.34582152 oC
Temperature correction Fctor calculation:
R = (T1-T2) / (t2-t1)
R = 0.0120202
S = (t2-t1) / (T1-t1)
412
S = 0.8684211
Ft = 0.98
∆Tm = Ft * ∆Tlm
where:
∆Tm = true temperature difference
Ft
= the temperature correction factor
∆Tlm = log mean temperature differace
∆Tm = 62.07890509 oC
Assuming U = 10 W/m2 oC ( From table 12.1)
A= Q / U * ∆Tm
where:
A = provisional area (m2)
Q = heat load (kW)
U = overall heat transfer coefficient (W/m2 oC)
Provisional area = 11.188262 m2
Choosing
Tube outside diameter(do) = 38 mm 1.496063
Tube inner diameter(di) = 26 mm 1.023622
Tube length(L) = 4.88 m 192.1256
Take tube materail is cupro- nickel
Area of one tube = L* do *π
Area of one tube = 0.5825769 m2
413
Number of tubes = provisinal area / area of one tube
Number of tubes = 19
Using 1.25 triangular pitch
K1 = 0.175 Use table 12.4
n1 = 2.285
Db = (do)*( Nt / K1)^ (1/n1)
where;
Db =bundle diameter (mm)
do = outer diameter (mm)
Nt : number of tubes
K1 & n1 are constant
Bundle diameter (Db) = 297 mm 0.2969786 m
Bundle diametrical clearance = 67 mm from fig 12.10
Ds = Db + Bundle diametrical clearance
Shell diameter(Ds) = 364 mm
0.3639786 m
Shell length(m)= 5.2439786 m
Tube side coefficient
Method 1
Mean Tube temperature=(t1+t2)/2 = 89.5 oC
Tube cross-sectional area = p/4 *di2= 530.92916 mm2
Tube per pass=(Nt/4) = 5
414
Total flow area = tubes per pass * cross sectional area
Total flow area = 0.0025491 m2
mass velocity = mass flow rate / total flow area
Tube mass velocity = 6.83E+04 kg/s.m2
linear velosity (ut ) = mass velosity / density
Tube linear velocity (ut) = 1.37E+02 m/s
hi = (4200* (1.35+0.02t)* ut0.8)/(di0.2)
where:
hi =inside coefficient (W/m2 oC)
t =mean temperature (oC)
ut =linear velocity (m/s)
di =tube inside diameter (mm)
hi = 351902.83 W/m2 oC
Method 2
Renolds number (Re) = ρ* ut *di/ µ
Re = 5.34E+06
Prandtl number (Pr) = Cp µ / kf
Pr = 3.20E+00
L/di = 1.88E+02
From Figuer 12.23 Tube -side heat transfer factor:
jh = 2.00E-03
415
where jh is the heat transfer factor assume that the viscisity of the fluid
is the same as at the wall
(µ/µwall) = 1
(hi di / kf) = jh Re Pr0.33 * (µ/µwall)0.14
hi = 1.99E+05 W/m2 oC
Using hi from method 1 as it has low value
hi = 198667 W/m2 oC
Shell-side coefficient
Choose baffle spacing (Lb)= (Ds/5) = 72.79571 mm
Tube pitch (pt) =1.25 * do= 47.5 mm
Cross flow area (As) = ((pt - do)* Ds* Lb)/pt
Cross flow area As = 0.005299 m2
Mass velocity (Gs) = mass flow rate / cross flow area
Mass velocity (Gs) = 262.0933 kg/s.m2
Equivalent diameter de =(1.1/do)(pt2-0.917do2)
de = 2.70E+01 mm
Mean Shell side temperature =(T1+T2)/2
Mean Shell side temperature = 25.595 oC
Renolds number (Re) = (Gs de)/ m
Re = 8.E+03
Prandtl number (Pr) = Cp µ / kf
Pr = 6.03
416
Choose 25% baffle cut
jh = 1.50E-03
Without the viscosity correction term, (µ/µw) = 1
hs = kf * jh *Re *Pr^(1/3) / de
hs = 498.4646 W/m2 oC
Overall Heat Transfer Coefficient
Thermal conductivity of cupro-nickel alloy= 50 W/moC
Taking fouling coefficients from table 12.2
Outside coefficient(fouling factor)=hod 5000 W/m2 oC
Inside coefficient(fouling factor) =hid 5000 W/m2 oC
1/Uo =(1/ho) + (1/hod) + ( do(ln(do/di))/2kw) + (do/di) * (1/hid) + (do/di)
* (1/hi)
1/Uo = 0.00265
Uo = 377.35409 W/m2 oC
Close to initial value assumed
Pressure Drop
Tube side
Re = 5.34E+06
jf = 2.00E-03
where jf is the friction factor
ΔPt = Np [ 8jf (L/di)(µ/µw)^(-m) +2.5 ] ρut²/2
where
417
ΔPt = tube side pressur drop (N/m²)(pa)
Np = number of tube side passes
ut = tube side velosity (m/s)
L = length of one tube
Neglecting the viscosity correction term, (µ/µw) = 1
∆pt= 114515262 N/m2
114515.26 kPa
16610.136 psi
Shell side (Acceptable)
Linear velocity =Gs /ρ= 0.2602973 m/s
Re = 8.E+03
jf = 1.50E-03 from fig 12.29
ΔPs = 8jf (Ds/de)(L/Lb)( ρus^2/2)(µ/µw)^(-0.14)
where
L : tube length
Lb : baffle spacing
∆Ps = 3.701639 N/m2
0.0037016 kPa
0.0005369 psi
Thickness Calculations: (Acceptable)
t =(Pri/(SEJ-0.6P))+Cc
where:
418
t = shell thichness (in)
P = Maximum allowable internal pressure (psig)
ri = internal raduis of shell before allowance corrosion is added (in)
EJ = efficincy of joients
S = working stress (psi)
Cc = allowance for corrosin (in)
ri = 7.164935966 in
P = 587.8 psi
S = 13706.66 psi (for carbon steel)
EJ = 0.85
Cc = 0.125 in
t = 0.498 in
t = 12.6 mm
Cost Calculations:
Heat transfer area = 120.43 ft2
Cost = $50,100 ( from www.matche.com)
419
Heater Design (E-212):Fluid properties:
Shell Side ( steam stream )
Flow rate
8.00E+04
Kg/h
Inlet Temperature ,T1
110
o
Outlet Temperature ,T2
75.02
o
Heat Capacity of inlet stream, Cpin
2.2022
KJ/kg°C
Heat Capacity of outlet stream, Cpout
2.2368
KJ/kgoC
Average Heat Capacity, Cpavg
2.2195
KJ/kgoC
Mass Density of inlet stream , ρin
0.8101
kg/m3
Mass Density of outlet stream , ρout
0.27327
kg/m3
Average Mass Density, ρavg
0.541685
kg/m3
Average Viscosity of stream, µavg
0.012
mNs/m2
Average Thermal conductivity, Kf
0.025
W/moC
Calculation of Heat load:
Qh = mh *Cp * (T1-T2)
where:
Qh = heat load in the hot side (KW)
mh = mass flowrate of hot fluid (Kg/h)
Cp = heat capacity of hot fluid (kJ/kgoC)
T1 = inlet temperature (oC)
T2 =outlet temperatue (oC)
Heat load = 1725.291333 KW > 1000 KW
420
C
C
The type of the heat exchanger is shell and tube heat exchanger
Tube side ( gas stream)
Flowrate
1.17E+04
Kg/hr
Average Heat Capacity, Cp
1.8554
kJ/kgoC
Average Mass Density, ρ
460.2192
kg/m3
Average Viscosity of stream, µ
0.467
mNs/m2
Average Thermal conductivity of stream, Kf
0.122
W/moC
inlet Temperature , t1
150.41
o
C
outlet Temperature, t2
155
o
C
Log mean Temperature calculation:
∆Tlm =(T1-t2)-(T2-t1) / ln((T1-t2)/(T2-t1))
where:
∆Tlm = log mean temperature differace
T1 = inlet shell side fluid temoerature (oC)
T2 = outlet shell side fluid temerature (oC)
t1
= inlet tube side temoerature (oC)
t2 = outlet tube side temerature (oC)
∆Tlm = 58.89396204 oC
Temperature correction Fctor calculation:
R = (T1-T2) / (t2-t1)
R = 7.620915
S = (t2-t1) / (T1-t1)
421
S = 0.1135857
Ft = 0.97
∆Tm = Ft * ∆Tlm
where:
∆Tm = true temperature difference
Ft
= the temperature correction factor
∆Tlm = log mean temperature differace
∆Tm = 57.12714318 oC
Assuming U = 850 W/m2 oC ( From table 12.1)
A= Q / U * ∆Tm
where:
A = provisional area (m2)
Q = heat load (kW)
U = overall heat transfer coefficient (W/m2 oC)
Provisional area = 35.530475 m2
Choosing
Tube outside diameter(do) = 50 mm 1.968504 in
Tube inner diameter(di) = 46 mm 1.811024 in
Tube length(L) = 4.88 m 192.1256 in
Take tube materail is cupro- nickel
Area of one tube = L* do *π
Area of one tube = 0.7665486 m2
422
Number of tubes = provisinal area / area of one tube
Number of tubes = 46
Using 1.25 triangular pitch
K1 = 0.175 Use table 12.4
n1 = 2.285
Db = (do)*( Nt / K1)^ (1/n1)
where;
Db =bundle diameter (mm)
do = outer diameter (mm)
Nt : number of tubes
K1 & n1 are constant
Bundle diameter (Db) = 575 mm
0.5746112 m
Using split ring floating head type
Bundle diametrical clearance = 67 Mm
Ds = Db + Bundle diametrical clearance
Shell diameter(Ds) = 642 mm
0.6416112 m
Shell length(m)= 5.5216112 m
Tube side coefficient
Method 1
Mean Tube temperature=(t1+t2)/2 = 152.705 oC
423
Tube cross-sectional area = p/4 *di2= 1661.9025 mm2
Tube per pass=(Nt/4) = 12
Total flow area = tubes per pass * cross sectional area
Total flow area = 0.0192578 m2
mass velocity = mass flow rate / total flow area
Tube mass velocity = 6.09E+05 kg/s.m2
linear velosity (ut ) = mass velosity / density
Tube linear velocity (ut) = 1.32E+03 m/s
hi = (4200* (1.35+0.02t)* ut0.8)/(di0.2)
where:
hi =inside coefficient (W/m2 oC)
t =mean temperature (oC)
ut =linear velocity (m/s)
di =tube inside diameter (mm)
hi = 2703375.7 W/m2 oC
Method 2
Renolds number (Re) = ρ* ut *di/ µ
Re = 6.00E+07
Prandtl number (Pr) = Cp µ / kf
Pr = 7.11E+00
L/di = 1.06E+02
From Figuer 12.23 Tube -side heat transfer factor:
424
jh = 2.00E-03
where jh is the heat transfer factor
assume that the viscisity of the fluid is the same as at the wall
(µ/µwall) = 1
(hi di / kf) = jh Re Pr0.33 * (µ/µwall)0.14
hi = 6.07E+05 W/m2 oC
Using hi from method 1 as it has low value
hi = 607083 W/m2 oC
Shell-side coefficient
Choose baffle spacing (Lb)= (Ds/5) = 128.3222 Mm
Tube pitch (pt) =1.25 * do= 62.5 Mm
Cross flow area (As) =((pt - do)* Ds* Lb)/pt
Cross flow area As = 0.016467 m2
Mass velocity (Gs) = mass flow rate / cross flow area
Mass velocity (Gs) = 1349.534 kg/s.m2
Equivalent diameter de =(1.1/do)(pt2-0.917do2)
de = 3.55E+01 Mm
Mean Shell side temperature =(T1+T2)/2
Mean Shell side temperature = 92.51 oC
Renolds number (Re) = (Gs de)/ m
Re = 4.E+06
Prandtl number (Pr) = Cp µ / kf
425
Pr = 1.09
Choose 25% baffle cut
jh = 1.50E-03
Without the viscosity correction term, (µ/µw) = 1
hs = kf * jh *Re *Pr^(1/3) / de
hs = 4238.841 W/m2 oC
Overall Heat Transfer Coefficient
Thermal conductivity of cupro-nickel alloy= 50 W/moC
Taking fouling coefficients from table 12.2
Outside coefficient(fouling factor)=hod 5000 W/m2 oC
Inside coefficient(fouling factor) =hid 5000 W/m2 oC
1/Uo =(1/ho) + (1/hod) + ( do(ln(do/di))/2kw) + (do/di) * (1/hid) + (do/di)
* (1/hi)
1/Uo = 0.0006968
Uo = 1435.1606 W/m2 oC
Close to initial value assumed
Pressure Drop
Tube side
Re = 6.00E+07
jf = 2.00E-03
where jf is the friction factor
ΔPt = Np [ 8jf (L/di)(µ/µw)^(-m) +2.5 ] ρut²/2
426
where
ΔPt = tube side pressur drop (N/m²)(pa)
Np = number of tube side passes
ut = tube side velosity (m/s)
L = length of one tube
∆pt= 4.774E+09 N/m2
4773795.8 kPa
692426.46 psi
Shell side
(Acceptable)
Linear velocity =Gs /ρ= 2491.3622 m/s
Re = 4.E+06
jf = 1.50E-03 from fig 12.29
ΔPs = 8jf (Ds/de)(L/Lb)( ρus^2/2)(µ/µw)^(-0.14)
where
L : tube length
Lb : baffle spacing
∆Ps = 138644.51 N/m2
138.64451 kPa
20.11002 psi
Thickness Calculations:
(Acceptable)
427
t =(Pri/(SEJ-0.6P))+Cc
where:
t = shell thichness (in)
P = Maximum allowable internal pressure (psig)
ri = internal raduis of shell before allowance corrosion is added (in)
EJ = efficincy of joients
S = working stress (psi)
Cc = allowance for corrosin (in)
ri = 12.63014778 in
P = 587.8 psi
S = 13706.66 psi (for carbon steel)
EJ = 0.85
Cc = 0.125 in
t = 0.782 in
t = 19.9 mm
Cost Calculations:
Heat transfer area = 382.45 ft2
Cost = $50,100 ( from www.matche.com)
428
Cost:
Probable Variation of Key Parameters over Plant Life
Lower Limit
Upper Limit
Base Value
FCIL
-20%
30%
$
22,900,000
Price of Product
-10%
10%
$
82,333,862
Working Capital
-50%
10%
$
13,600,000
Income Tax Rate*
-20%
20%
0%
Interest Rate*
-10%
20%
5%
Raw Material Price
-10%
15%
$ 112,982,285
Salvage Value
-80%
20%
0
* Please note that variations for percentages are a percent of a percent.
For example, a 10% variance on a 12% interst rate would imply a 1.2%
uncertainty
Net Present Value Data
Low NPV
-172.9
High NPV
186.4
Bins
Upper Value
# points/bin
Cumulative
0
-172.9
0
0
1
-137.0
5
5
2
-101.0
22
27
3
-65.1
74
101
4
-29.2
156
257
5
6.7
232
489
6
42.7
235
724
7
78.6
156
880
8
114.5
99
979
9
150.5
19
998
10
186.4
2
1000
429
Cumulative Number of Data Points
1000
750
500
250
0
-200
-150
-100
-50
0
50
100
150
Net Present Value (millions of dollars)
Discounted Cash Flow Rate of Return Data
Low DCFROR
0.00
High DCFROR
0.27
Bins
Upper
#/bin
Cumulative
0
0.00
0
0
1
0.03
63
63
2
0.05
68
131
3
0.08
126
257
4
0.11
140
397
5
0.13
169
566
6
0.16
159
725
7
0.19
108
833
8
0.21
68
901
9
0.24
29
930
10
0.27
3
933
430
200
250
Hazop:
What is HAZOP?
Systematic technique to identify
potential hazard and Operating
problems
• A formal systematic rigorous examination to the process and
engineering facets of a production facility
• A qualitative technique based on “guide-words” to help provoke
thoughts about the way deviations from the intended operating
conditions can lead to hazardous situations or operability problems
• hazop is basically for safety
- Hazards are the main concern
- Operability problems degrade plant performance (product
quality, production rate, profit)
• Considerable engineering insight is required - engineers working
independently could develop different results
• Objective of HAZOP For identifying cause and the consequences of
operations of equipment and associated operator interfaces in the
context of the complete system.
• How and Why HAZOP is Used HAZOP identifies potential hazards ,
failures and operability problems.
• Its use is recommended as a principal method by professional
institutions and legislators on the basis of proven capabilities for
years.
431
• It is most effective as a team effort consists of plant and prices
designers, operating personnel, control and instrumentation engineer.
• It encourages creativity in design concept evaluation.
• Necessary changes to a system for eliminating or reducing the
probability of operating deviations are suggested by the analytical
procedure.
• HAZOP provides a necessary management tool and bonus in so far
that it demonstrates to insurers and inspectors evidence of
comprehensive thoroughness.
Guide Words
No forward flow when there should be
NONE MORE
More of
any parameter than there should be, more flow, more pressure, more
temperature. LESS As above, but "less of" in each instance PART
System composition difference from what it should be.
MORE THAN
More "components" present than there should be for
example, extra phase, impurities.
OTHER
What needs to happen other than normal operation, e.g.
start up, shutdown, maintenance
NONE
e.g., NO FLOW caused by blockage; pump failure; valve
closed or jammed : leak: valve open ;suction vessel empty; delivery side
over - pressurized : vapor lock ; control failure
432
REVERSE e.g., REVERSE FLOW caused by pump failure : NRV
failure or wrongly inserted ; wrong routing; delivery over pressured;
back- siphoning ; pump reversed
MORE OF e.g., MORE FLOW caused by reduced delivery head ;
surging ; suction pressurised ; controller failure ; valve stuck open leak ;
incorrect instrument reading.
MORE OF MORE TEMPERATURE, pressure caused by external
fires; blockage ; shot spots; loss of control ; foaming; gas release;
reaction;explosion; valve closed; loss of level in heater; sun.
LESS OF
e.g., LESS FLOW caused by pump failure; leak; scale in
delivery; partial blockage ; sediments ; poor suction head; process
turndown.
LESS
e.g., low temperature, pressure caused by Heat loss;
vaporization ; ambient conditions; rain ; imbalance of input and output ;
sealing ; blocked vent .
PART OF
Change in composition high or low concentration of
mixture; additional reactions in reactor or other location ; feed change.
MORE THAN
Impurities or extra phase Ingress of contaminants
such as air, water, lube oils; corrosion products; presence of other
process materials due to internal leakage ; failure of isolation ; start-up
features.
OTHER
Activities other than normal operation start-up and
shutdown of plant ; testing and inspection ; sampling ; maintenance;
activating catalyst; removing blockage or scale ; corrosion; process
433
emergency ; safety procedures activated ; failure of power, fuel, steam ,
air, water or inert gas; emissions and lack of compatibility with other
emission and effluents.
Case Study – Shell & Tube Heat Exchanger
Using relevant guide works, perform HAZOP study on shell and tube
heat exchanger
434
Guide Word
Less
Deviation
-Less flow of
cooling water.
-Less pressure
in tubes.
Causes
Pipe
blockage.Pipe
leakage.-Heat tubes
burst.
-Burst pipe.
More
-More cooling
flow.
-More pressure
or tube side.
-Failure of
cooling water
valve.
-Failure of
fluid valve.
-Tube
blockage.
-Failure of
inlet cooling
water valve
to open.
-No cooling
water flow.
None
-Process fluid
contamination.
Contaminatio
n in cooling
water
-Equipment
failure, pipe
leak
-Outlet
temperature
too low.
-low fuel gas.
-low cooling
water.
-Empty
storage.
-Organic acid
present.
-Corrosion of
tube.
-Cooler
freezing
-Decrease
absorption.
-Increase
pollution.
-Increase
impurities in
syngas
product.
-Less cooling
and crack of
tube
-Hardness of
cooling water.
Contaminati
on
-Maintenance
Other
Part of
More than
Consequences
-Temperature
of process
fluid remains
constant.
-Process fluid
temperature
too low.
-Explosion
-Temperature
of process
fluid decrease
(Too low).
-Bursting of
tube.
-Tube failure.
-Process fluid
temperature
does not
decreased.
435
-Process stops
Action
-High temperature
Alarm.
-Installation of
flow meter.
-Flow alarm or
shutdown.
-Alarm low
pressure.
-Low temperature
alarm.
-Flow alarm.
-Install high
pressure alarm.
-Pressure relief
system on tubes
-Install
Temperature
indicator before
and after the
process fluid line.
-Proper
maintenance and
operator alert.
-Ensure all pipes
and fittings are
constructed of the
right material and
are stress relieved.
-Install low flow
alarm in fuel gas.
-Install low level
alarm on cooling
water.
-Proper
Maintenance to
check suitability of
material of
construction.
CONCLUSION
Benefit of hazop
During the design of a new plant, design personnel are under pressure to
keep the project on schedule. This pressure frequently results in errors
and oversights. HAZOP study is an opportunity to correct these before
such changes become too expensive, or impossible to accomplish.
Besides safety hazards, the HAZOP technique is very effective for
identifying plant operability problems, threats to the environment,
product quality, plant throughput and for highlighting critical
maintenance requirements.
Limitation of Hazop
HAZOP is a powerful technique but the extent to which it can uncover
all foreseeable hazards is limited by the knowledge, experience and
deductive skills of the HAZOP team. For these reasons, it is difficult to
assess the ‘quality' of a given HAZOP in any objective or auditable way.
Audits can be carried out to establish that the process has been followed,
but they cannot verify the competence of the team.
436
Environmental :
Benzene
PHYSICAL STATE; APPEARANCE:
COLOURLESS LIQUID , WITH CHARACTERISTIC ODOUR.
PHYSICAL DANGERS:
The vapour is heavier than air and may travel along the ground; distant
ignition possible. As a result of flow, agitation, etc., electrostatic charges
can be generated.
CHEMICAL DANGERS:
Reacts violently with oxidants, nitric acid, sulfuric acid and halogens
causing fire and explosion hazard. Attacks plastic and rubber.
OCCUPATIONAL EXPOSURE LIMITS:
TLV: 0.5 ppm as TWA; 2.5 ppm as STEL; (skin); A1; BEI issued;
(ACGIH 2004).MAC: H;
Carcinogen category: 1; Germ cell mutagen group: 3A; (DFG 2004).
OSHA PEL: 1910.1028 TWA 1 ppm ST 5 ppm
NIOSH REL: Ca TWA 0.1 ppm ST 1 ppm
NIOSH IDLH: Potential occupational carcinogen 500 ppm
ROUTES OF EXPOSURE:
The substance can be absorbed into the body by inhalation , through the
skin and by ingestion .
437
INHALATION RISK:
A harmful contamination of the air can be reached very quickly on
evaporation of this substance at 20°C.
EFFECTS OF SHORT-TERM EXPOSURE:
The substance is irritating to the eyes , the skin and the respiratory tract .
Swallowing the liquid may cause aspiration into the lungs with the risk
of chemical pneumonitis. The substance may cause effects on the central
nervous system , resulting in lowering of consciousness . Exposure far
above the occupational exposure limit value may result in
unconsciousness and death .
EFFECTS OF LONG-TERM OR REPEATED EXPOSURE:
The liquid defats the skin. The substance may have effects on the bone
marrow and immune system , resulting in a decrease of blood cells. This
substance is carcinogenic to humans.
Methane
PHYSICAL STATE; APPEARANCE:
COLOURLESS, COMPRESSED OR LIQUEFIED GAS , WITH NO
ODOUR.
PHYSICAL DANGERS:
The gas is lighter than air.
OCCUPATIONAL EXPOSURE LIMITS:
438
TLV: (aliphatic hydrocarbons gases, Alkane C1-C4) 1000 ppm (as
TWA).
MAC not established.
ROUTES OF EXPOSURE:
The substance can be absorbed into the body by inhalation.
INHALATION RISK:
On loss of containment this gas can cause suffocation by lowering the
oxygen content of the air in confined areas.
EFFECTS OF SHORT-TERM EXPOSURE:
Rapid evaporation of the liquid may cause frostbite.
An Introduction to Indoor Air Quality (IAQ) Carbon Monoxide (CO)
Carbon monoxide is an odorless, colorless and toxic gas. Because it is
impossible to see, taste or smell the toxic fumes, CO can kill you before
you are aware it is in your home. At lower levels of exposure, CO
causes mild effects that are often mistaken for the flu. These symptoms
include headaches, dizziness, disorientation, nausea and fatigue. The
effects of CO exposure can vary greatly from person to person
depending on age, overall health and the concentration and length of
exposure.
Health Effects Associated with Carbon Monoxide
At low concentrations, fatigue in healthy people and chest pain in people
with heart disease. At higher concentrations, impaired vision and
coordination; headaches; dizziness; confusion; nausea. Can cause flulike symptoms that clear up after leaving home. Fatal at very high
439
concentrations.
Acute
effects
are
due
to
the
formation
of
carboxyhemoglobin in the blood, which inhibits oxygen intake. At
moderate concentrations, angina, impaired vision, and reduced brain
function may result. At higher concentrations, CO exposure can be fatal.
Levels in Homes
Average levels in homes without gas stoves vary from 0.5 to 5 parts per
million (ppm). Levels near properly adjusted gas stoves are often 5 to 15
ppm and those near poorly adjusted stoves may be 30 ppm or higher.
Steps to Reduce Exposure to Carbon Monoxide
It is most important to be sure combustion equipment is maintained and
properly adjusted. Vehicular use should be carefully managed adjacent
to buildings and in vocational programs. Additional ventilation can be
used as a temporary measure when high levels of CO are expected for
short periods of time.

Keep gas appliances properly adjusted.

Consider purchasing a vented space heater when replacing an
unvented one.

Use proper fuel in kerosene space heaters.

Install and use an exhaust fan vented to outdoors over gas stoves.

Open flues when fireplaces are in use.

Choose properly sized wood stoves that are certified to meet EPA
emission standards. Make certain that doors on all wood stoves fit
tightly.
440

Have a trained professional inspect, clean, and tune-up central
heating system (furnaces, flues, and chimneys) annually. Repair
any leaks promptly.

Do not idle the car inside garage.
Measurement Methods
Some
relatively
high-cost
infrared
radiation
adsorption
and
electrochemical instruments do exist. Moderately priced real-time
measuring devices are also available. A passive monitor is currently
under development.
Exposure Limits
Occupational Safety and Health Guideline for Carbon Monoxide
[OSHA
PEL]
The
current
Occupational
Safety
and
Health
Administration (OSHA) permissible exposure limit (PEL) for carbon
monoxide is 50 parts per million (ppm) parts of air (55 milligrams per
cubic meter (mg/m(3))) as an 8-hour time-weighted average (TWA)
concentration [29 CFR Table Z-1].
[NIOSH REL] The National Institute for Occupational Safety and
Health (NIOSH) has established a recommended exposure limit (REL)
for carbon monoxide of 35 ppm (40 mg/m(3)) as an 8-hour TWA and
200 ppm (229 mg/m(3)) as a ceiling [NIOSH 1992]. The NIOSH limit is
based on the risk of cardiovascular effects.
[ACGIH TLV] The American Conference of Governmental Industrial
Hygienists (ACGIH) has assigned carbon monoxide a threshold limit
value (TLV) of 25 ppm (29 mg/m(3)) as a TWA for a normal 8-hour
441
workday and a 40-hour workweek [ACGIH 1994, p. 15]. The ACGIH
limit is based on the risk of elevated carboxyhemoglobin levels [ACGIH
1991, p. 229].
442