Project: Formaldehyde from methanol and air

1
CHAPTER # 1
INTRODUCTION
Formaldehyde, also called Methanal (formulated HCHO), an organic compound, the simplest of the
aldehydes. Pure formaldehyde is a colorless, flammable gas with a strong pungent odor. Upon
condensation, the gas converts to various other forms of formaldehyde (with different chemical
formulas) that are of more practical value. Formaldehyde is an important precursor to many other
materials and chemical compounds. Formaldehyde was first reported in 1859 by the Russian
chemist Aleksandr Butlerov (1828–86) and was conclusively identified in 1869 by August Wilhelm
von Hofmann. It used in large amounts in a variety of chemical manufacturing processes.
The vast number of applications of this material including a building block for other organic
compounds, photographing washing, woodworking, cabinet-making industries, glues, adhesives,
paints, explosives, disinfecting agents, tissue preservation and drug testing.
1.1 Physical properties
1. Description: At room temperature, it is colorless gas with a pungent odour (Reusset al.,
2003)
2. Boiling-point: –19.1 °C
3. Melting-point: –92 °C
4. Density: 0.815 at –20 °C
5. Molecular weight 30.03 kg/kgmol
6. Ignition temperature 430 °C
1.2 Chemical properties
1. Solubility: Soluble in water, ethanol and chloroform; miscible with acetone, benzene
and diethyl ether
2. Stability: Commercial formaldehyde–alcohol solutions are stable; the gas is stable in
the absence of water; incompatible with oxidizers, alkalis, acids, phenols and Urea.
3. Reactivity: Reacts explosively with peroxide, nitrogen oxide and performic acid; can
react with hydrogen chloride or other inorganic chlorides to form chloro-methyl ether
4. Octanol/water partition coefficient (P): log P = 0.35

2
1.3 Applications
i. Formaldehyde is used in the production of following
ii. Formaldehyde is a powerful disinfectant and antiseptic
iii. It is used for preserving anatomical specimens
iv. Formaldehyde with ammonia to give hexamethylenetetramine, a urinary antiseptic,
and is also used to make RDX.
v. Formaldehyde for the production of the polio vaccine, and the tetanus and diphtheria
toxoids
vi. In the production of resins with urea, phenol and melamine
vii. In the manufacture of particle-board, plywood, furniture
viii. For the production of curable moulding materials
ix. As raw materials for surface coating
x. Controlled-release nitrogen fertilizers
xi. Used in the textile, leather, rubber and cement industry
xii. Uses are as binders for foundry sand, stonewool and glasswool mats in insulating
materials
xiii. Abrasive paper and brake linings
xiv. As an intermediate in the synthesis of other industrial chemical compounds, such as
1,4-butanediol, trimethylolpropane and neopentyl glycol
xv. Formaldehyde itself is used to preserve and disinfect, for example, human and
veterinary drugs and biological materials
xvi. Formaldehyde is used as an antimicrobial agent in many cosmetics products
xvii. Formaldehyde is also the basis for products that are used to manufacture dyes,
tanning agents, precursors of dispersion and plastics, extraction agents, crop
protection agents, animal feeds, perfumes, vitamins, flavourings and drugs
xviii. Formaldehyde is also used directly to inhibit corrosion, in mirror finishing and
electroplating

3
Literature Review
1.4 Manufacturing processes
1. It is also stated that it is difficult to obtain formaldehyde free from other aldehydes and by-
products. However, in spite of the above difficulties, improvements have been effected by
the use of special catalysts and better methods of control.
2. Industrially formaldehyde is produce from methanol, however there are ways of producing
formaldehyde from alternative raw materials such as methane or propane. Production from
alternative raw materials, such as methane or propane, are not as profitable and therefore not
used in industrial scale
1.5 Methods
It can be manufactured from methanol via two different routes
1. Dehydrogenation or oxidative dehydrogenation of Methanol in the presence of silver
catalyst
2. Oxidation in the presence of Fe containing MoO3 catalyst
1.5.1 Silver process
There are two kinds of silver processes the methanol ballast process and the BASF process, which
are quite similar, but is described under separate sub headings. As the name indicate both processes
are applying a silver gauze or crystals as the catalyst. Both processes are run at close to atmospheric
pressures and under adiabatic conditions

4
Figure: Process flow diagram for silver catalyst
1.5.2 Metallic oxide
The oxide process was first used by Formox. The process is run at atmospheric pressure and
temperatures between 300 – 400 °C, with proper temperature controls This process uses a
methanol-air feed that are below the lower level of the flame and explosion mixture of 6%. To be
able to have higher methanol concentration, the tail gas is recycled and mixed together with the
ingoing stream of air to reduce the oxygen content to 10 mole.
Figure: Process flow diagram for Metal oxide catalyst

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CHAPTER # 2
CAPACITY & PROCESS SELECTION
2.1 Production plants in Pakistan
Industries Capacity
Super Chemicals, Karachi 100000 Mtons/yr
Wah Noble Chemical 30000 Mtons/yr
Dynea 408000 Mtons/yr
Total 538000 Mtons/yr
In 2017 Total Demand is 572,000 MTPY
So, our plant capacity will be 30,600 MTPY
2.2 Process selection
We choose silver catalytic process due to following reason. Operating cost as well as the
investment cost for the oxide process is greater than for the silver process. In metallic oxide
process, excess air means larger investment and energy consumption as compared with silver
process. And silver process gives stable production
2.3 Economic Analysis
Catalyst Silver Iron molybdenum oxide
Typical technology BASF Perstorp
Battery limits investment
(106
US$)
5.3 7.9
Methanol 0.437 0.425
Fuel (106
kg) (-) 1.4 -
Process water 0.5 1.0
Catalyst and Chemical
(US$)
1.4 2.5
Labor
(operation per shift)
2 2
IFP Petrochemical Processes synthesis gas derivatives and major hydro-carbons by
A.Chauvel

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Table:2.1 Comparison of Formaldehyde CM and DM processes
Silver catalyzed CM process Fe/Mo catalyzed DM
process
Incomplete Conversion Complete Conversion Complete Conversion
Feed
Composition
Methanol rich (40-45%
MeOH, 20-25% air, 30-40%
steam)
Methanol rich (25-
27% MeOH, 46-54%
air, 20-35% steam)
Methanol lean with
recycle or steam (9%
MeOH)
Recycle Yes (MeOH) None Partial recycle of tail
gas
Distillation
(required to
separate
unreacted
methanol)
Yes No No
Catalyst life 2-6 months 2-6 months 18-24 months
Temperature
range
550-650o
C 680-700o
C 275-300o
C wall, 240-
270o
C as inlet. 350-
390o
C gas spot
Pressure
(atm)
1-1.5 1-1.5 1-1.5
Methanol
conversion
(%)
70-80 99-99.5 99-99.5
Yield (%) 90-92 85-90 93-96
Reactor Adiabatic Adiabatic Multi-tubular non-
adiabatic
2.4 Advantages of Silver catalyst process
No other process gives you as much formaldehyde per Euro or US dollar. Dynea Silver has the
lowest operational cost (methanol, KWh, catalyst, cooling water) and highest steam export of the
two formaldehyde processes.
2.4.1 Safe and clean production process
I. No hot oil; using only water/steam cooling reduces fire risk.
II. No oxygen in the absorber improves fire safety as well as product quality.
III. The spent catalyst can be removed easily, quickly and cleanly in a few hours and re-
catalysation takes less than 24 hours.
IV. Fast cooling in waste heat boiler (Very fast cooling time prevents decomposition of
product).
V. Selective formaldehyde absorption, recycling of methanol and water
VI. Improved yield and product quality

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2.4.2 Product quality
I. High formaldehyde concentration, up to 57% by weight
II. Low formic acid concentration in final product, 80ppm
III. Low methanol concentration in final product, down to 0.5%
2.5 Disadvantages of Metal oxide process over Silver process
1. More operating cost
2. More investment cost
3. Less flexibility
4. Size of equipment is large
5. Large energy consumption

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CHAPTER # 3
PROCESS DESCRIPTION
3.1 Process description
1. For the production of Formaldehyde, we need air and Methanol as raw materials. The
Methanol feed is first pass through the vaporizer that converts it into the vapor form. And
then it will mix with the air comes from compressor in a mixer.
2. After mixing these two streams the reaction components pass through heat exchanger. That
raise their temperature at which reaction takes place. The reaction takes place at very high
temperature of 685 0
C.
The following reactions take place
I. CH3OH HCHO + H2
II. H2 + ½ O2 H2O
III. CH3OH + ½ O2 HCHO +H2O
3. Reaction products then contain Formaldehyde that stream will pass through the heat
exchanger. It will cool the gases to the 150 0
C and enter into the absorption column.
The bulk of the water, Formaldehyde, Methanol is condensed in lower section of absorber.
Nearly complete removal of remaining Formaldehyde and Methanol is occurring in the top
of the tower by counter current contact of clean water.
4. Absorber bottom product goes to distillation where remaining Methanol is recovered by
recycling.
The base stream from the distillation, an aqueous solution of formaldehyde, contain upto
55% of Formaldehyde obtained.
5. Add process water in the product from distillation to make the 37% solution Formaldehyde
which is called Formalin

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CHAPTER # 4
MATERIAL BALANCE
4.1 Material balance
Material balance is defined as the mass conserved; the mass entering in a process is
equal to mass exiting from the process. It is based on the law of conservation of mass which states
that mass is neither be created nor be destroyed.
4.2 Law conservation of mass
It states that:
[Rate of mass going into the system]-[Rate of mass going out of the system] +[Rate of mass
generation within the system]-[Rate of mass consumption within the system]=[Rate of mass
accumulation in the system]
Across mixer 1
Stream 1
Component Mole(kgmole) Mole % Weight(kg) Weight %
CH3OH 65.41 100 2093.12 100
Total 65.41 100 2093.12 100
1
2
3

11
Stream 2
Components Mole(kgmole) Mole % Weight(kg) Weight %
CH3OH 0.59 5.59 18.82 9.52
H2O 9.94 94.41 178.87 90.48
Total 10.53 100 197.69 100
Stream 3
CH3OH 66 86.91 2112 92.2
H2O 9.94 13.09 178.92 7.8
Total 75.94 100 2290.92 100
Across mixer 2
Stream 4 from compressor
O2 24.58 21 786.56 23.30
N2 92.48 79 2589.53 76.70
Total 117.06 100 3376.09 100
Stream 5 from mixer 1
CH3OH 66 86.91 2112 92.2
H2O 9.94 13.09 178.92 7.8
Total 75.94 100 2290.92 100
4
5
6

12
Stream 6 outlet of mixer 2
O2 24.58 12.74 786.7 13.88
N2 92.48 47.92 2589.5 45.69
CH3OH 66 34.2 2112 37.27
H2O 9.94 5.14 178.92 3.16
Total 193 100 5666.2 100
Across reactor
Reactions
CH3OH CH2O + H2
H2 +
1
2
O2 H2O
CH3OH +
1
2
O2 CH2O + H2O
Basis: 66 kgmoles
Conversion of methanol: 99 %
Water produced = 885.03kg
6
7

13
Stream 6 from mixer 2
O2 24.58 12.74 786.7 13.88
N2 92.48 47.92 2589.5 45.69
CH3OH 66 34.2 2112 37.27
H2O 9.94 5.14 178.92 3.16
Total 193 100 5667.12 100
Stream 7 Outlet of reactor
CH3OH 0.66 0.282 21.12 0.373
CH2O 65.34 27.95 1960.2 34.589
H2O 59.11 25.82 1063.98 18.77
N2 92.48 39.56 2589.44 45.69
H2 16.17 6.92 32.34 0.571
Total 223.76 100 5667.08 100
Across absorber
7
9
8
10
00
00
0

14
Stream 7 from reactor
CH3OH 0.66 0.282 21.12 0.373
CH2O 65.34 27.95 1960.2 34.589
H2O 59.11 25.82 1063.98 18.77
N2 92.48 39.56 2589.44 45.69
H2 16.17 6.92 32.34 0.571
Total 223.76 100 5667.08 100
Stream 8 (Make up water)
Use Water to Formaldehyde ratio = 1.25
Makeup Water added = 1565.2 kg
Component Mole(kgmole) Mole % Weight(kg) Weight %
H2O 86.957 100 1565.2 100
Stream 10: Off gasses
H2 16.17 10.37 32.34 0.93
N2 92.48 59.28 2589.44 74.37
CH2O 0.65 0.42 19.5 0.56
CH3OH 0.0066 0.004 0.2112 0.006
H20 46.69 29.93 840.42 24.14
Total 155.99 100 3939.10 100
Stream 9: absorption product
H2O 99.37 60.33 1788.68 47.69
CH2O 64.69 39.27 1940.59 51.74
CH3OH 0.6534 0.4 20.91 0.557
Total 164.71 100 3750.189 100

15
Across distillation column
Stream 9: Outlet of absorption
H2O 99.37 60.33 1788.68 47.69
CH2O 64.69 39.27 1940.59 51.74
CH3OH 0.6534 0.4 20.91 0.557
Total 164.71 100 3750.189 100
Stream 10: Distillate
CH3OH 0.59 5.58 18.82 9.52
H2O 9.94 94.41 178.87 90.48
Total 10.525 100 197.69 100
Stream 11: Bottom of the column
H2O 89.43 58 1609.814 45.31
CH3OH 0.065 0.042 2.09 0.058
CH2O 64.68 41.95 1940.59 54.63
Total 154.186 100 3552.5 100
9
10
11

16
Overall material balance
OVERALL MATERIAL IN AND OUT
IN OUT
7034.41 7034.2
Reactor Absorber
Distillation
column
Air
CH3OH, H2O
CH20
OFF GASESWater

17
CHAPTER # 5
ENERGY BALANCE
5.1 Introduction
Energy balance is defined as the energy conserved; the energy entering in a process is equal to
energy exiting from the process. It is based on the law of conservation of
energy that energy is neither be created nor be destroyed but it changes from one form to another.
5.2 Law of conservation of energy
It states that:
[Rate of energy going into the system]-[Rate of energy going out of the system] +[Rate of energy
generation within the system]-[Rate of energy consumption within the system]=[Rate of energy
accumulation in the system]
5.3 Reference condition
Condition for the calculations in energy balance is:
Temperature = 298 K
Constants Value for the Calculations of Heat Capacity of Gaseous Components.
Components A 103
B 106
C
CH3OH 13.431 -51.28 131.13
H2O 8.712 1.25 -0.18
CH2O 44.222 0.3986 -1.5358×10-3
O2 3.639 0.506 0
N2 3.280 0.593 0
H2 3.249 0.422 0
CH3OH(g) 2.211 12.216 -3.450
H2O(g) 3.470 1.450 0
CH2O(g) 2.264 7.022 -1.877
Cp = R[A+[B/2×(T0) ×(ῑ+1)] +[C/3×(T0
2)×(ῑ2+ῑ+1)]+[D/(ῑ×T0
2)]]
Where:
R = 8.314 J mol-1
K-1
ῑ = (T-T0)/T0
T0 = 298 K

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A, B, C are constants
Across M-1
Stream 1
Q1 = 0 (reference temp)
Stream 2
QCH3OH = nCp×dT
= 0.59×84.192×(321-298) = 1142.81 KJ
QH2O = nCp×dT
= 9.94×75.50×(321-298) = 17260.81 KJ
Q2 = 1142.81 + 17260.81 = 18403.29 KJ
Stream 3
Here T will find by using Cp relation, so by using it
Methanol
Cp = 3.608×10-4
T2
– 0.1055 T + 80.3988
QCH3OH = nCp×dT
= 0.0238T3
– 14.058T2
+ 7381.29T – 1581283.44
Water
Cp = -4.98×10-7
T2
+ 5.04×10-3
T + 73.936
QH2O = -4.95×10-6
T3
+ 0.052T2
+ 719.95T – 219007.32
Now
QT = 0.0238T3
– 14T2
+ 8101.24T – 1800290.76
Qin = Qout
18403.29 + 0 = 0.0238T3
– 14T2
+ 8101.24T – 1818694.05
M- 2
1
2
3

19
Solving
T = 308 K = 350
C
Across Vaporizer
Inlet Temperature = 309 K
Outlet Temperature = 349 K
ΔHv of Methanol = 36900 KJ/kgmoles
ΔHv of Water = 49687.2 KJ/Kgmoles
Overall Balance
Qin = Qout
Qin = QH2O + QCH4O
Qin = 18403.29 KJ
Now
Qout = QH2O + QCH4O
Methanol
Cp = 50.59 KJ/(kgmole.K)
QCH4O = nCp×dT + nʎ
= 66×50.59×(407-298) + 66×36900 = 2799344.46 KJ
Water
Cp = 33.95 KJ/(Kgmoles.K)
QH2O = nCp×dT + nʎ
= 9.94×33.95×(407-298) + 9.94×(49687.2) = 530674.235 KJ
Qout = 2799344.46 + 530674.235 = 3330018.695 KJ
V-1
1 2

20
Heat required = Qout – Qin
= 3330018.695 - 18403.29 = 3311615 KJ
Across M-2
Inlet temperature of stream 1 = 407 K = 1340
C
Inlet temperature of stream 2 = 407 K = 1340
C
Outlet Temperature = 398 K = 1250
C
Stream 1 (Air)
QO2 = nCp×dT
= 24.58×30.1813×(407-298) = 80862.34 KJ
QN2 = nCp×dT
= 92.48×29.2819×(407-298) = 295170.922 KJ
QT = QO2 + QN2
= 80862.34 + 295170.922 = 376033.2647 KJ
Stream 2 (CH4O + H2O)
= 66×50.59×(407-298) + 66×36900 = 2799344.46 KJ
= 9.94×33.95×(407-298) + 9.94×(49687.2) = 530674.235 KJ
QT = 2799344.46 + 530674.235 = 3330018.695 KJ
Stream 3 (H2O, CH4O, O2, N2)
QCH3OH = nCp×dT
= 66×50.2288×(398-298) = 331510.608 KJ
QH2O = nCp×dT
M- 2
1
2
3

21
= 9.94×33.921×(398-298) = 33717.38 KJ
QN2 = nCp×dT
= 92.48×29.2657×(398-298) = 270649.1936 KJ
QO2 = nCp×dT
= 24.58×30.1272×(398-298) = 74052.65 KJ
QT = Q1 + Q2 + Q3 + Q4
= 331510.608 + 33717.38 + 270649.1936 + 74052.65
= 709929.8361 KJ
Heat loss = Qin – Qout
= 376033.2647 + 3330018.695 - 709929.8361 = 2996122.124 KJ
Across exchanger at reactor inlet
Inlet Temperature T1 = 1250
C
Outlet Temperature T2 = 3850
C
Qin = 709929.8361 KJ
Qout = Q1 + Q2 + Q3 + Q4
QCH3OH = nCp×dT
= 66×60.09×(658-298) = 2617520.4 KJ
QH2O = nCp×dT
= 9.94×77.04×(658-298) = 275679.94 KJ
QN2 = nCp×dT
= 92.48×29.64×(658-298) = 986798.59 KJ
QO2 = nCp×dT
= 24.58×32.26×(658-298) = 285462.288 KJ
E-1
1 2

22
Qout = 2617520.4 + 275679.94 + 986798.59 + 285462.288
= 4165461.218 KJ
Heat loss = 4165461.218 - 709929.8361 = 3455531.382 KJ
Across exchanger at reactor inlet
C
C
Qin = 4165461.218 KJ
Qout = Q1 + Q2 + Q3 + Q4
QCH3OH = nCp×dT
= 66×69.81×(958-298) = 3040923.6 KJ
QH2O = nCp×dT
= 9.94×36.42×(958-298) = 238929.76 KJ
QN2 = nCp×dT
= 92.48×30.366×(958-298) = 1853443.469 KJ
QO2 = nCp×dT
= 24.58×32.896×(958-298) = 533665.22 KJ
Qout = 3040923.6 +238929.76 + 1853443.469 + 533665.22
= 5666962.049 KJ
Heat loss = 5666962.049 - 4165461.218 = 1501500.83KJ
E-2
1 2

23
Across reactor
C
Outlet temperature T2 = 6850
C
Qin = 5666962.049 KJ
Qout = Q1 + Q2 + Q3 + Q4 + Q5
QCH2O = nCp×dT
= 65.34×48.765×(958-298) = 2102961.36 KJ
QH2O = nCp×dT
= 59.11×36.42×(958-298) = 1420838.89 KJ
QH2 = nCp×dT
= 16.17×29.215×(958-298) = 311788.323 KJ
QCH3OH = nCp×dT
= 0.66×69.81×(958-298) = 30409.236 KJ
QN2 = nCp*dT
= 92.48×30.3666×(958-298) = 1853443.469 KJ
Qout = 5719441.27 KJ
Calculation of Heat of rex.
Heat of Reaction = ΔHf (Product) - ΔHf (Reactants)
ΔHf of Rex 1 = 92090 KJ
ΔHf of Rex 2 = -286000 KJ
R-1
1
2

24
ΔHf of Rex 3 = -193910 KJ
CH3OH CH2O + H2
2H2 + O2 2H2O
2CH3OH + O2 2CH2O + 2H2O
Heat of Rex 1 = 130779.2 KJ
Heat of Rex 2 = -302541.411 KJ
Heat of Rex 3 = -183093.92 KJ
Total Heat of Reaction = -354856.13 KJ
Calculation for Mass flow rate required to remove Qrej
Qin + Qrex = Qout + Qrej
QRej = 960588.769 KJ/hr
As
QRej = nCp×dT
960588.769 = n×(33.95×(407-298))
N = 259.58 Kmoles/hr
Mass flow rate = m = 259.58×18 = 4672.46 kg/hr
Across exchanger after reactor
C
C
Qin = 5719441.27 KJ
Qout = QCH2O + QH2O + QH2 + QCH3OH + QN2
QCH2O = nCp×dT
= 65.34×37.82×(423-298) = 308903.0175 KJ
E-3
1 2

25
QH2O = nCp×dT
= 59.11×34.78×(423-298) = 256965.21 KJ
QH2 = nCp×dT
= 16.17×28.28×(423-298) = 57154.886 KJ
QCH3OH = nCp×dT
= 0.66×68.33×(423-298) = 5636.97 KJ
QN2 = nCp×dT
= 92.48×29.04×(423-298) = 335702.4 KJ
Qout = 964362.48 KJ
Heat loss = 5719441.27 - 964362.48 = 4755078.79 KJ
Across exchanger E-4
C
C
Qin = 964362.48 KJ
Qout = QCH2O + QH2O + QH2 + QCH3OH + QN2
QCH2O = nCp×dT
= 65.34×35.93×(363-298) = 105639.098 KJ
QH2O = nCp×dT
= 59.11×32.713×(363-298) = 87014.94 KJ
QH2 = nCp×dT
= 16.17×28.14×(363-298) = 20473.59 KJ
QCH3OH = nCp×dT
= 0.66×47.98×(363-298) = 1425.07 KJ
E-4
1 2

26
QN2 = nCp×dT
= 92.48×28.85×(363-298) = 120062.16 KJ
Qout = 334614.85 KJ
Heat loss = 964362.48 - 334614.85 = 629748 KJ
Across absorber
Stream 1 Temperature = 900
C = 363 K
C = 298 K
C = 359 K
C = 344 K
Stream 1
Q1 = 334614.85 KJ
Stream 2 (water)
QH2O = nCp×dT
= 86.96×72.43×(298-298) = 0 KJ
Total Qin = 470167.1937 KJ
Stream 4
QH2O = nCp×dT + nʎ ʎ = 43.5 KJ/mole
= 46.69×32.65×(344-298) + 46.69×43500 = 2085894.43 KJ
QCH2O = nCp×dT
= 0.65×35.711×(344-298) = 835.63 KJ
A-1
1
2
3
4

27
QN2 = nCp×dT
= 92.48×28.83×(344-298) = 95976.57 KJ
QH2 = nCp×dT
= 16.17×28.10×(344-298) = 16358.58 KJ
= 0.0066×46.62×(344-298) + 0.0066×(38300) = 263.856 KJ
Q4 = 2199329.067 KJ
Stream 3
QcH2O = nCp×dT
= 64.69×127.24×(359-298) = 337477.37 KJ
QCH3OH = nCp×dT
= 0.6534×86.62×(359-298) = 2320.52 KJ
QH2O = nCp×dT
= 99.37×75.58×(359-298) = 307925.768 KJ
Q3 = 647723.6682 KJ
Qin = Qout
Q1 + Q2 = Q3 + Q4
334614.85 + 0 = 647723.6682 + 2199329.067
334614.85 = 2847052.735
Across distillation
D-1
1
2
3

28
C = 359 K
C = 373 K
C = 353 K
Stream 1
Q1 = 647723.6682 KJ
Stream 2
QCH3OH = nCp×dT
= 0.065×88.979×(373-298) = 329.66 KJ
QH2O = nCp×dT
= 89.43×75.664×(373-298) = 385697.99 KJ
QCH2O = nCp×dT
= 64.68×88.84×(373-298) = 327531.758 KJ
Q2 = 713559.415 KJ
Stream 3
QCH3OH = nCp×dT + nʎ
= 0.59×84.19×(353-298) + 0.59×38300 = 23739.45 KJ
= 9.94×75.504×(353-298) + 9.94×43500 = 449651.724 KJ
Q3 = 473391.174 KJ
Qin = Qout
Q1 = Q2 + Q3
647723.6682 = 713559.415 + 473391.174
647723.6682 = 1186950.59 KJ
Across E-4
1 2
E-7

29
Inlet temperature T1 = 373 K
Outlet temperature T2 = 303 K
Stream 1
QCH3OH = nCp×dT
= 0.065×89.79×(373-298) = 1023.606 KJ
QH2O = nCp×dT
= 89.43×74.219×(373-298) =574753.42 KJ
QCH2O = nCp×dT
= 64.68×6296.99 = 617167.98 KJ
Qin = 1192945.016 KJ
Stream 2
QCH3OH = nCp×dT
= 0.065×88.979×(303-298) = 28.918 KJ
QH2O = nCp×dT
= 89.43×75.664×(303-298) = 33833.15 KJ
QCH2O = nCp×dT
= 64.68×88.84×(303-298) = 28730.856 KJ
Qout = 62592.92 KJ
Heat loss = 1192945.016 - 62592.92 = 1130352.096 KJ

30
CHAPTER # 6
EQUIPMENT DESIGN
6.1 Reactor Design
6.1.1 Reactor
Reactor design uses information, knowledge, and experience from a variety of areas-
thermodynamics, chemical kinetics, fluid mechanics, heat transfer, mass transfer, and economics.
Chemical reaction engineering is the synthesis of all these factors with the aim of properly
designing a chemical reactor. Design of the reactor is no routine matter, and many alternatives
can be proposed for a process. In searching for the optimum, it is not just the cost of the reactor
that must be minimized. One design may have low reactor cost, but the materials leaving the unit
may be such that their treatment requires a much higher cost than alternative designs. Hence, the
economics of the overall process must be considered.
The chemical reactor is at the heart of the plant. In size and appearance, it may often seem to be
one of the least impressive items of equipment, but its demands and performance are usually the
most important factors in the design of the whole plant.
6.1.2 Types of Reactor
1. Batch Reactor
2. Continuous Reactor
3. Semi-Batch Reactor
Batch Reactor
The batch reactor is simply a container to hold the contents while they react. All that has to be
determined is the extent of reaction at various times, and this can be followed in a number of
ways, for example:
1. By following the concentration of a given component.

31
2. By following the change in some physical property of the fluid, such as the electrical
conductivity or refractive index.
3. By following the change in total pressure of a constant-volume system.
4. By following the change in volume of a constant-pressure system.
The experimental batch reactor is usually operated isothermally and at constant volume because
it is easy to interpret the results of such runs. This reactor is a relatively simple device adaptable
to small-scale laboratory set-ups, and it needs but little auxiliary equipment or instrumentation. [1]
Continuous Reactor
The first of the two-ideal steady-state flow reactors is variously known as the
✓ Plug flow
✓ Slug flow
✓ Piston flow
✓ Ideal tubular, and
We refer to it as the plug flow reactor, or PFR, and to this pattern of flow as plug flow. It is
characterized by the fact that the flow of fluid through the reactor is orderly with no element of
fluid overtaking or mixing with any other element ahead or behind. The necessary and sufficient
condition for plug flow is for the residence time in the reactor to be the same for all elements of
fluid. The other ideal steady-state flow reactor is called the mixed reactor, the back-mix reactor,
the ideal stirred tank reactor, the C* (meaning C-star), CSTR, or the CFSTR (constant flow
stirred tank reactor), and, as its names suggest, it is a reactor in which the contents are well
stirred and uniform throughout. Thus, the exit stream from this reactor has the same composition
as the fluid within the reactor. [1]
Homogeneous
In homogeneous reactors only one phase, usually a gas or a liquid, is present. If more than one
reactant is involved, provision must of course be made for mixing them together to form a
homogenous whole. Often, mixing the reactants is the way of starting off the reaction, although
sometimes the reactants are mixed and then brought to the required temperature.

32
Heterogeneous
In heterogeneous reactors two, or possibly three, phases are present, common examples being
gas-liquid, gas-solid, liquid-solid and liquid-liquid systems. In cases where one of the phases is a
solid, it is quite often present as a catalyst; gas-solid catalytic reactors particularly form an
important class of heterogeneous chemical reaction systems. It is worth noting that, in a
heterogeneous reactor, the chemical reaction itself may be truly heterogeneous, but this is not
necessarily so. In a gas-solid catalytic reactor, the reaction takes place on the surface of the solid
and is thus heterogeneous.
6.1.3 Comparison between Reactors
Table: 6.1
Reactor selection
Type of
reactor
Advantages Limitations Area of application
Batch
reactors
1. Small scale
production
2. Suitable for
processes where a
range of different
products involves
3. Not
suitable
for large
batch
sizes
4. Batch
processes are
used in
chemical
(inks, dyes)
and food
industry
Continuous
stirred tank
reactors
1. Flexible device
2. Economically
beneficial to
operate several
CSTR in series
and parallel
3. More
complex
and
expensiv
e than
tubular
reactor
4. All
calculatio
n
required
with
CSTR
assume
perfect
mixing
5. Chemical
industry
involve
liquid/gas
reactions

33
Plug flow
reactors
1. High efficiency
than CSTR for
same volume
2. It may have
several tubes or
pipes in parallel
3. Reagent may be
introduced at
location even
other than inlet
4. Pressure drop is
small
5. Not
economic
al for
small
batches
6. Tubular
reactor is
especially
suited for
very high
temperature
cases
6.1.4 Reactor selection for Formalin production
• It is a continuous reaction, so Batch reactor rejected
• Gaseous phase reaction
• Reaction is exothermic so cooling is required
• Solid catalyst used so mixing is not favorable
CSTR is rejected, and suitable reactor is fixed bed Reactor.
It has four types
1. Packed bed of pellet
2. Slow moving pellet bed
3. Three phase trickle bed reactor
4. Multi tubular fixed bed reactor (MTFBR)
We have selected MTFBR
6.1.5 Catalyst selection
Catalyst is selected on the basis of following aspects
o Conversion
o Catalyst life
o Replacement
o Cost

34
The selection of catalyst is always tradeoff between these factors. The possible catalysts are
Silver, iron-molybdenum in methanol oxidation process. So silver could be recommended
catalyst as its life is large and maintenance is easy too.
Particle Diameter
Size of catalyst is quite important with reference to pressure drop. Smaller the size of catalyst
more will be the pressure drop and vice versa.
Particle Shape
Shape of the catalyst should be such that can bear compressibility and with stand against
crushing and abrasion.
Porosity
Porosity has direct effect on reactivity. More the catalyst porosity less will be the surface area
available for reaction. Hence less will be the reactivity.
Tube diameter
Reducing tube diameter reduce the radial profile. Heat transfer area per unit volume is inversely
proportional to tube diameter and the reaction temperature is affected by change in this area.
Large the diameter of the tube large will be the heat transfer. It is the only source to increase the
capacity of MTFBR but on the other hand large diameter of the tube cause the temperature
control problem.
6.1.6 Design steps:
1. Calculate the Volume of reactor
2. Calculate the Weight of catalyst
3. Calculate the shell height
4. Calculate the shell diameter
5. Calculate the shell side heat transfer coefficient
6. Calculate the tube side heat transfer coefficient
7. Calculate the overall heat transfer coefficient
8. Calculate the area required for heat transfer

35
9. Calculate Area available for heat transfer
10. Calculate shell side pressure drop
11. Calculate tube side pressure drop
6.1.7 Basic information
Flow rate of the feed mixture = 5667.12 kg/hr.
Reactions are
CH3OH CH2O + H2
H2 +
1
2
O2 H2O
CH3OH +
1
2
O2 CH2O + H2O
Conversion = 99%
Catalyst used = Silver
6.1.8 Volume of Reactor:
Space velocity of reactants = s = 4500 per hour
Space time = 𝜏 =
1
𝑠
= 2.22×10-4
hr.
Performance equation for PFR
𝑉
𝐹𝐴0
=
𝜏
𝐶𝐴0
Where
FA0 = 66 Kmoles/hr.
CA0 = 6.09×10-3
Kmole/m3
Volume of reactor = VR = 2.405 m3
6.1.9 Weight of catalyst
VC = VR - ɸ VR

36
Where
VC is Volume of catalyst
VR is volume of reactor
ɸ catalyst porosity
Volume of catalyst VC = 2.405 – (0.3×2.405)
VC = 1.6835 m3
Catalyst dimensions
Catalyst of size 3mm and spherical shape is used
Bulk density of catalyst = 1110 Kg/m3
Weight of catalyst = W = 1868.685 Kg
Catalyst weight per tube = 9.16 Kg
6.1.10 Number of tubes
Volume of a single tube = Vt =
𝜋
4
Di
2
L
= (3.14/4)×(0.01572
)×(6.097)
= 0.01179 m3
Number of tubes =
𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑟𝑒𝑎𝑐𝑡𝑜𝑟
𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑖𝑛𝑔𝑙𝑒 𝑡𝑢𝑏𝑒
= 203.9
By TEMA standards
No. of tubes = 204
6.1.11 Shell Diameter and Height
Shell inner diameter = 0.43 m
Tube length = 20 ft = 6.097 m

37
Assume that shell height is 20% more of tube length
Shell height = 1.2 × L
HR = 7.32 m
6.1.12 Diameter of reactor
Use the Length to diameter ratio
𝐿
𝐷
= 5
Outside diameter of Reactor
DR = 1.46 m
6.1.13 Bundle diameter
Db = d0*(
𝑛
𝑘1
)(1/n1)
Db = 1.108 m
6.1.14 Shell side heat transfer coefficient
ID of shell = 0.43 m
No of passes = 1
Mass velocity
as =
𝐼𝐷 × C B
144𝑝𝑡
= 0.02 m2

38
Gs = Ws/as
= 228491.985 Kg/(m2
.hr)
De = (1.1/d0) (pt
2
– 0.917d0
2
)
= 0.0135 m
µ = 0.053 Kg/(m.hr)
Re = Gs de/µ
= 58307.085
Jh = 0.028
Thermal Conductivity = K = 0.0778 KJ/(hr.K.m)
Cp = 1.892 KJ/(Kg.K)
Pr = 1.28
6.1.15 Heat transfer coefficient
h0 = jh
𝐾
𝑑𝑒
(Re) (Pr)0.33
= 2837 W / (hr.m2
K)
6.1.16 Tube side heat transfer coefficient
Outside diameter of tube OD = 0.019 m
Number of tubes = 204
Tube pitch Pt = triangular 1 inch
BWG = 16
Tube length = 6.097 m
Number of Passes = 1
Viscosity of mixture µ = 0.1359 Kg/(m.hr)

39
Thermal Conductivity of mixture Kf = 0.0716 W/(m.K)
Flow area per tube at = 1.95×10-3
m2
Total flow area a/
t= 0.0397 m2
Mass velocity on tube side Gt = 142553.2012 Kg/(hr. m2
)
Reynold number = Re = 16511.377
Parental number Pr = 0.9266
Heat transfer coefficient equation for catalyst filled tubes
hi (
𝐼𝐷
𝑘
) = 3.5 Re0.7
е-4.6
(
𝐷𝑝
𝐼𝐷
)
hi = 5941 W/ (hr. m2
. K)
hi0 = hi×
𝐼𝐷
𝑂𝐷
= 4902 W/ (hr.K.m2
)
6.1.17 Overall Heat transfer coefficient
1
𝑈𝑑
=
1
ℎ0
+
1
ℎ𝑖0
+ 𝑅𝑑
Dirt factor Rd = 0.0005 (m2
. K)/W
U = 1500 W/ (m2
.K.hr)
6.1.18 Area for heat transfer
Area available for heat transfer = Nt×πDL
= 204×3.14×0.01905×6.097
= 74.4 m2

40
6.1.19 Pressure Drop Calculations
Shell side
Reynold number on shell side Re = 58307.085
Friction factor f = 0.0016 ft2
/in2
Specific gravity = s = 1
No. of crosses = N+1 = 12L/B
= 35
∆Ps =
𝑓𝐺2 𝑠𝐷𝑠(𝑁+1)
5.22×10∧10𝐷𝑒𝑠𝜑𝑡
= 9.79 KPa
Tube Side:
Particle diameter Dp = 3×10-3
m
Density ⍴ = 0.5158 Kg/m3
Viscosity of mixture µ = 0.1359 Kg/(m.hr)
Length of tube L = 6.097 m
Porosity of catalyst particle ɸ = 0.3
∆𝑃
𝐿
=
150 µ𝐺(1 − 𝜀)2
𝑘𝑔⍴𝐷2 𝜀3
+
1.75𝐺2(1 − 𝜀)
𝑘𝑔⍴𝐷𝜀
∆𝑃 = 34.73 KPa

41
Specification sheet
Identification:
Item Reactor
Item No. R-101
No. Required 01
Function: Production of Formalin from Methanol using Silver catalyst
Operation: Continuous
Type: Catalytic Multi-Tubular fixed bed Reactor
Temperature 685 0
C
Pressure 1.5 Atm
Catalyst:
Silver
Spherical shape
3 mm
Length 6.097 m
Diameter 1.4 m
Heat transfer area available: 74.4 m2
Heat transfer Area required: 55 m2
Overall Heat transfer coefficient: 1500 W/ (m2
.K.hr)

42
6.2 Design of heat exchanger
Heat exchanger is a device that is used to transfer heat between two fluids at different temperature.
6.2.1 Types
The principle types of heat Exchanger used in Chemical and allied industries are as follows:
1. Double Pipe Heat Exchanger
2. Shell and Tube Heat Exchanger
3. Plate and Frame Heat Exchanger
4. Plate and Fin Type Heat Exchanger
5. Spiral Type Heat Exchanger
6. A Cooled: Cooler and Condenser
6.2.2 Selection criteria
Selection process includes a No. of factors all of these are related to the heat transfer
application.
1. Thermal Requirements
2. Material Compatibility
3. Operational Maintains
4. Environmental, Health & Safety Consideration
5. Availability
6. Cost
6.2.3 Shell & tube heat exchanger
Shell and tube heat exchangers represent the most widely used vehicle for the transfer of heat in
industrial process applications. They are frequently selected for such duties as:
• Process liquid or gas cooling
• Process or refrigerant vapor or steam condensing
• 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

43
• Many other industrial applications
Shell and tube heat exchangers have the ability to transfer large amounts of heat in relatively low
cost, serviceable designs. They can provide large amounts of effective tube surface while
minimizing the requirements of floor space, liquid volume and weight. Shell and tube exchangers
are available in a wide range of sizes. They have been used in industry for over 150 years, so the
thermal technologies and manufacturing methods are well defined and applied by modern
competitive manufacturers. Tube surfaces from standard to exotic metals with plain or enhanced
surface characteristics are widely available. They can help provide the least costly mechanical
design for the flows, liquids and temperatures involved.
6.2.4 Fluid Stream Allocations
There are a number of practical guidelines which can lead to the optimum design of a given heat
exchanger. Remembering that the primary duty is to perform its thermal duty with the lowest cost
yet provide excellent in-service reliability, the selection of fluid stream allocations should be of
primary concern to the designer. There are many trade-offs in fluid allocation in heat transfer
coefficients, available pressure drop, fouling tendencies and operating pressure.
1. The higher-pressure fluid normally flows through the tube side. With their small diameter
and nominal wall thicknesses, they are easily able to accept high pressures and avoids more
expensive, larger diameter components to be designed for high pressure. If it is necessary
to put the higher-pressure stream in the shell, it should be placed in a smaller diameter and
longer shell.
2. Place corrosive fluids in the tubes, other items being equal. Corrosion is resisted by using
special alloys and it is much less expensive than using special alloy shell materials. Other
tube side materials can be clad with corrosion resistant materials or epoxy coated.
3. Flow the higher fouling fluids through the tubes. Tubes are easier to clean using common
mechanical methods.
6.2.5 Tubes
Tubing that is generally used in TEMA sizes is made from low carbon steel, copper, Admiralty,
Copper-Nickel, stainless steel, Hastalloy, Inconel, titanium and a few others. It is common to use
tubing from 5/8 to 1-1/2 in these designs. Tubes are either generally drawn and seamless or

44
welded. High quality ERW (electro-resistance welded) tubes exhibit superior grain structure at
the weld. Extruded tube with low fins and interior rifling is specified for certain applications.
6.2.6 Tube sheets
Tube sheets are usually made from a round flat piece of metal with holes drilled for the tube ends
in a precise location and pattern relative to one another. Tube sheet materials range as tube
materials. Tubes are attached to the tube sheet by pneumatic or hydraulic pressure or by roller
expansion. Tube holes can be drilled and reamed and can be machined with one or more grooves.
This greatly increases the strength of the tube joint.
The tube sheet is in contact with both fluids and so must have corrosion resistance allowances
and have metallurgical and electrochemical properties appropriate for the fluids and velocities.
6.2.7 Baffles
Baffles serve two important functions. They support the tubes during assembly and operation and
help prevent vibration from flow induced eddies and direct the shell side fluid back and forth
across the tube bundle to provide effective velocity and heat transfer rates. The diameter of the
baffle must be slightly less than the shell inside diameter to allow assembly, but must be close
enough to avoid the substantial performance penalty caused by fluid bypass around the baffles.
6.2.8 Standard design steps
a. Define the duty; heat transfer rate and temperature
b. Collection of fluid physical properties
c. Assume the value of heat transfer coefficient
d. Calculate the mean temperature difference
e. Calculate the area required
f. Decide the heat exchanger layout
g. Calculate the pressure drop

45
6.2.9 Stream conditions
Hot fluid (Mixture of 0.059% CH3OH, 54.626% CH2O, 45.315%H2O)
Inlet Temperature T1 =100o
C
Outlet Temperature T2 =30o
C
Mass flow rate mh =3552.5 kg/hr
Cold fluid (Water)
Inlet Temperature t1 =25o
C
Outlet Temperature t2 =40o
C
Mass flow rate mc =7547.14 kg/hr
6.2.10 Physical properties
Hot fluid
Specific heat Cp = 3.03 x 103
J/kgK
Thermal conductivity k = 0.304W/mK
Density 𝝆 = 447.9kg/m3
Viscosity μ = 0.00024kg/ms
Cold fluid
Specific Heat Cp =4.194 kJ/kg K
Thermal conductivity k = 0.624W/m K
Density 𝝆 = 994.6kg/m3
Viscosity μ = 0.75 x 10-3
kg/ms
6.2.11 Heat balance
Mixture: Q = 474337.62kJ
Water: Q = mcp∆T
474337.62 = m (4.192) (40-25)
m = 7547.14 kg/hr

46
6.2.12 True temperature difference
Hot fluid Cold fluid Difference
100 Higher temperature 40 60
30 Lower temperature 25 5
70 Difference 15 55
LMTD =
∆t2−∆t1
ln
∆t2
∆t1
=
60−5
𝑙𝑛
60
5
= 17.39 o
C
R =
T1−𝑇2
𝑡2−𝑡1
= 3.47
S =
𝑡2−𝑡1
𝑇1−𝑡1
= 0.62
Ft =
√𝑅2+1ln(
1−𝑆
1−𝑅𝑆
)
(𝑅−1) ln[
2−𝑆(𝑅+1−√ 𝑅2+1)
2−𝑆(𝑅+1+√ 𝑅2+1)
]
= 0.76
∆t = LMTD x Ft
= 13.035o
C
Take U = 640W/m2
K
Area =
𝑄
𝑈∆𝑡
= 31.28m2
Choose 19.05mm o.d, 15.748mm i.d, 3.66m long tube, material: cupro-nickel
Area of one tube = πdol = 3.14 x 19.05 x 10-3
x 3.66
= 0.1955m2
Number of tubes = 31.28/0.1955 = 160
Tube pitch = 1.25do = 23.8125mm
Bundle diameter = 𝑑𝑜 (
𝑁𝑡
𝐾1
)
1
𝑛1

47
n1 = 2.207, K1 = 0.249
= 19.05 (
160
0.249
)
1
2.207
= 375.59mm
Use a U-tube heat exchanger
Bundle clearance = 11.55mm
Shell diameter = 375.59 + 11.55 = 387.35mm
6.2.13 Shell side coefficient
Mean shell side temperature =
100+30
2
= 65o
C
Baffle spacing = 0.5 (Ds)
= 193.675mm
No. of baffles = L/B = 18
Area for cross flow =
(p 𝑡 − do)(Ds)(Bs)
𝑝 𝑡
= 7693.04mm2
Mass velocity = w/as
=
3552.5
3600×0.00769304
= 128.27kg/sm2
Velocity = Gs /𝝆 = 0.3m/s
Shell equivalent diameter = de =
1.1(𝑝𝑡2−0.917𝑑𝑜2)
𝑑𝑜
=
1.1(23.81252
− 0.917(19.05)2
)
19.05
= 13.526mm
Reynolds number = Gsde/µ
=
128.27×13.526×10−3
0.00024
= 13526

48
Prandtl number = cp µ/k
=
3.03×103×0.00024
0.304
= 2.392
jH = 7 x 10-3
ho = j 𝐻RePr0.33 𝑘
𝑑𝑒
(
𝜇
𝜇 𝑤
)
0.14
Assume µ/µw= 1
ho = 1516.6W/m2
K
6.2.14 Estimate wall temperature
Mean temperature difference across all resistances = 55-32.5 = 22.5o
C
Mean temperature across mixture film =
𝑈
ℎ𝑜
×∆𝑇
= 640 x 22.5/1516.6 = 9.495 o
C
Mean wall temperature = 55-9.495 = 45.51 o
C
Viscosity at wall = µw = 0.000251kg/ms
(
𝜇
𝜇 𝑤
)
0.14
= 0.985 It shows that correction for low viscosity fluid is not significant.
6.2.15 Tube side coefficient
Mean water temperature =
40+25
2
= 32.5o
C
Tube side area = πdi2
/4
= 3.14 x 15.7482
/4 = 194.68mm2
Tubes per pass = 160/2 = 80
Total flow area = 80 x 194.68 x 10-6
= 0.0156m2
Water mass velocity = 2.096/0.0156 = 134.39kg/s

49
Density of water = 994.6 kg/m3
Water linear velocity = 134.39/994.6 = 0.14m/s
Re =
𝜌𝑢𝑑𝑖
𝜇
=
994.6×0.14×15.748×10−3
0.00075
= 10157
ℎ𝑖 =
4200(1.35+0.02𝑡)(𝑢 𝑡)0.8
𝑑𝑖
0.2
= 2800W/m2
K
6.2.17 Overall Coefficient
Take fouling factors for
Mixture of gases 5000W/m2
K, Water 6000W/m2
K
Also take thermal conductivity of Cupro-Nickel = 50W/mK
1/Uo =
1
ℎ𝑜
+
1
ℎ 𝑑0
+
𝑑0 𝑙𝑛(
𝑑𝑜
𝑑𝑖
)
2𝑘 𝑤
+
𝑑 𝑜
𝑑𝑖ℎ 𝑖
+
𝑑 𝑜
𝑑𝑖ℎ 𝑑𝑖
Uo = 643.16W/m2
K
This value is approximately equal to assumed value.
6.2.18 Pressure Drop
Shell side pressure drop
For Res = 10157
jf = 0.049
∆Ps = 8𝑗 𝑓 (
𝐷𝑠
𝑑 𝑒
) (
𝐿
𝐵
) (
𝜌𝑢2
2
) (
𝜇
𝜇 𝑤
)
−0.14
= 7243Pa
Tube side pressure drop
For Ret = 7229
jf = 0.0048

50
∆Pt = 𝑁𝑝 (8𝑗 𝑓 (
𝐿
𝑑𝑖
) (
𝜇
𝜇 𝑤
)
−0.14
+ 2.5) (
𝜌𝑢2
2
)
= 8615Pa
Specification sheet:
Type Shell and Tube Heat Exchanger
Heat transfer area 31.28m
2
No. of tubes 160
Inner diameter of tubes 15.748mm
Outer diameter of tubes 19.05mm
Length of tubes 3.66m
BWG 16
Tube material of construction Cupro-nickel alloy
Inner diameter of shell 387.35mm
No. of baffles 18
Shell material of construction Carbon Steel

51
6.3 Design of absorber
6.3.1 Absorption
The removal of one or more component from the mixture of gases by using a suitable solvent is
second major operation of Chemical Engineering that based on mass transfer.
In gas absorption, a soluble vapor is more or less absorbed in the solvent from its mixture with
inert gas. The purpose of such gas absorption operations may be any of the following;
a) For Separation of component having the economic value.
b) As a stage in the preparation of some compound.
c) For removing of undesired component (pollution).
6.3.2 Types
1) Physical absorption
2) Chemical absorption.
Physical Absorption
In physical absorption mass transfer take place, purely by diffusion and physical absorption is
governed by the physical equilibria.
Chemical Absorption
In this type of absorption as soon as a particular component comes in contact with the absorbing
liquid a chemical reaction take place.
6.3.3 Types of absorber
There are two major types of absorbers which are used for absorption purposes:
➢ Packed column
➢ Plate column
6.3.4 Comparison between packed and plate column
1. Contact
The packed column provides continuous contact between vapor and liquid phases

52
while the plate column brings the two phases into contact on stage wise basis.
2. Scale
For column, less diameter. It is more usual to employ packed towers because of high
fabrication cost of small trays. But if the column is very large then the liquid distribution
is problem and large volume of packing and its weight is problem.
3. Pressure drop
Pressure drop in packed column is less than the plate/tray column. because the
packing open area approaches the tower cross-sectional area, while the tray’s open area is
only 8 to 15 percent of the tower cross-sectional area. If there are large No. of Plates in
the tower, this pressure drop may be quite high and the use of packed column could affect
considerable saving.
4. Liquid holdup
Packings have lower liquid holdup than do trays. This is often advantageous for
reducing polymerization, degradation, or the inventory of hazardous materials
5. Size and cost
The practical range of packing material is wider. Ceramic and plastic packing are
cheap and effective as compared to trays.
From the above consideration, packed column is selected as the absorber, because in our case the
diameter of the column is less. It is easy to operate.
6.3.5 Packing
The packing is the most important component of the system. The packing provides sufficient
area for intimate contact between phases. The efficiency of the packing with respect to both HTU
and flow capacity determines to a significance extent the overall size of the tower. The
economics of the installation is therefore tied up with packing choice.
Packings are generally divided into two classes
1. Random/dumped packing are discrete pieces of packing of specific geometric shape that
are dumped or randomly packed into shell of column.
2. Structured/arranged packing are crimped layers of corrugated sheets or wire mesh.

53
Packing should
1) Be chemically inert to the fluids in the tower.
2) Provide for large interfacial area between gas and liquid
3) Ensure low gas pressure drop
4) Provide good contact between liquid and gas.
5) Be reasonable in cost.
6) Permit passage of large volumes of gas and liquid through small tower cross-sections
Thus, most packing is made of cheap, inert, fairly light materials such as clay, porcelain, or
graphite. Thin-walled metal rings of steel or aluminum are some limes used.
Common Packings are
a) Berl Saddle.
b) Intalox Saddle.
c) Raschingrings.
d) Lessing rings.
e) Cross-partition rings.
f) Single spiral ring.
g) Double - Spiral ring.
h) Triple - Spiral ring.

54
6.3.6 Designing steps for absorption column
1. Selection of column.
2. Selection of packing and material
3. Determining the no. of transfer units (NOG)
4. Calculating the diameter of column
5. Determining the height of packing
6. Determining the height of the colum
7. Determining the pressure drop.
8. Select and design the column internal features: packing support, liquid distributer and re-
distributer.
Source: “absorption and stripping” by p Chattopadhyay

55
Selection of column
➢ The liquid holdup is lower in packed columns.
➢ Pressure drop is lower in packing as compared to plates.
➢ Low cost of packing compared to plates.
So, packed column is selected
6.3.7 Type of packing
Intalox saddles have been selected because;
➢ It provides a larger contact area per unit volume.
➢ It has an open structure and high bed porosity.
➢ Also provides high flooding limits and low pressure drop.
➢ Material of packing is ceramic because it resists corrosion.

56
6.3.8 Size of packing
➢ Packing size is taken as 38mm.
6.3.9 Design of Absorber
Material Balance
Gm (y1 – y2) = Lm (x1 – x2)
Gm=flow rate of gas entering (Kgmoles/hr)
Lm = flow rate of solvent entering (Kgmoles/ hr)
Y1=Mole fraction of HCHO in entering streams
Y2= Mole fraction of HCHO in leaving streams
X1= Mole fraction of HCHO in leaving solvent stream
X2=Mole fraction of HCHO in entering solvent stream
233.82(Y-.2919) =86.95(X-0.3917)
X=0.661(Y-0.0002) ……………… (1)
Y=1.513X +0.0002 ……………… (2)
Equations (1) and (2) are the operating line equations.
Equation for Equilibrium Curve
LET
Y1=Mole fraction of HCHO in entering gas stream=0.2919
Y2= Mole fraction of HCHO in leaving gas stream=0.004473
Y1/Y2 = 0.2919/0.004473 = 65.26
AS
Gm (y1 – y2) = Lm (x1 – x2)
y1 – y2 = (Lm/Gm) (x1 – x2)
The above equation is in the form y = mX + 0
The NOG using Y1/Y2 &mGm/Lm. Where ‘m ‘is slope of equilibrium line.
Colburn has suggested that the economic range for mGm/Lm lies from 0.7 to 0.8. For our system.

57
m=0.2604
Gm/Lm=233.8234/86.956 = 2.6888
m Gm/Lm= 0.70
From graph
Area Under the curve= NOG=10
6.3.10 Calculation of Diameter of Column
Flow rate of entering gases =G =5488.226 Kg/hr
Flow rate of entering solvent=L= 1563.57 Kg/hr
Temperature of entering gas=Tg=90o
C =363K
Temperature of entering Solvent=TL=25o
C =298K
Pressure of entering gases=P= 1.4 atm
Average molecular weight of entering gases=22.01 Kg
Density of gas mixture=ρg = PM /RTg
= (1.4×22.01) / (0.08205×343)
=1.0937 Kg/m3
Density of liquid solvent at 25o
C=ρL=1000 Kg/m3
Viscosity of liquid solvent at 25o
C = µL =1/1000 Ns/m2
Viscosity of Gaseous mixture at 70o
C = µg = 1.5*10^-5 Ns/m2
Now
Abscissa of fig 11.44
= 0.02022
For pressure drop 20 mm of H2O /m of packing
K4 = 1
At flooding
K4’=6
% flooding = (K4/ K4’) 0.5
×100
=41.40%
Packing factor for 1.5-inch ceramic Intalox - saddles =Fp=170/m
G* = [k4× ρg × (ρL-ρg) / 13.1×Fp× (µL /ρL) 0.1] ½
L
g


G
L

58
G*= [1.2×1.0936× (1000-1.0936) /13.1×22× (1×10-3
/1000)0.1
]1/2
G*=1.53 Kg/m2
-sec.
Flow rate of gas entering =G =5488.226 /3600
=1.524 Kg/sec.
Area =A= G / G* =0.996 m2
Diameter of column=D= 4[A]½
[3.14] ½
Diameter of column= 1.13 m
Round off D’=1.2 m
then column area =A’=1.13 m2
Packing size to column D ratio = D’/38/1000 = 31.57
% flooding at selected diameter = 41.4(A/A’) = 36.47 %
6.3.11 Calculation of Height of Transfer Units
Equation for calculation of effective interfacial area is given as.
Where
aw = effective interfacial area of packing per unit volume m2
/m3
Lw= liquid mass velocity kg/m2
s
a = actual area of packing per unit volume m2
/m3
σc = critical surface tension for particular packing material
σL = liquid surface tension N/m
a = 194 m2
/m3
Lw = 0.3844 kg/m2
s
σc = 61 x 10-3
N/m
σL = 72 x 10-3
N/m
µL=1 cP
ρL =1000Kg /m3





































 2.0205.0
2
21.075.0
45.1exp1
a
L
g
aL
a
L
a
a
LL
w
L
w
L
w
l
cw



59
aw = 44.74 m2/m3
6.3.12 Calculation of liquid film mass transfer coefficient
KL = liquid film coefficient m/s
dp = packing size =38 x 10-3
m
DL = diffusivity of liquid = 1.7 x 10-9
m2
/sec
Then, by substituting the values,
KL = 4.197 x 10-5
m/s
6.3.13 Calculation of gas film mass transfer coefficient
Where KG = Gas film coefficient, kmol/m2
s.bar
VW= Gas mass velocity = 1.347 Kg m2
/sec
K5= 5.23
Dg =Diffusivity of gas = 1.5 x 10-5
m2
/sec
Then, by substituting the values,
KG =1.835 x 10-3
kmol/m2
s.bar



































 


2.0
3
205.0
2
21.0
3
75.0
19410721000
3844.
8.91000
1943844.
10194
3844.
70
61
45.1exp1
194
wa
  4.0
2
1
3
2
3
1
0051.0 p
LL
L
Lw
w
L
L ad
Da
L
g
K L

























  2
3
1
7.0
5

















 p
gg
g
g
w
g
gG
ad
Da
V
K
aD
RTK



P
w
a
G
K
m
G
G
H 

60
Where,
HG = Gas-film transfer unit height
Gm = 1.347/22 = 0.06127 Kmol/m2
.sec
Then,
HG = 0.06127/ (1.835 x 10-3
× 44.74 ×1.4)
= 0.53 m
And
HL= Liquid-film transfer unit height
Lm= .3844/18 = 0.021355 Kmol/m2
.sec
Ct = Concentration of solvent = 1000/18 = 55.5 Kmol/m3
Then,
HL = 0.021355 / (4.197 × 10-5
× 44.74 ×55.5)
= 0.204 m
6.3.14 Calculation of height of transfer units
As,
HG = 0.53 m
HL = 0.204 m
So,
Height of transfer units=HOG = 0.53 + 0.7 × 0.204
HOG = 0.67 m (From Coulson & Richardson,range is 0.6 to 1m, topic 11.14.3)
LH
mL
mmG
GHoGH 
t
C
w
a
L
K
m
L
L
H 

61
6.3.15 Calculation of height of tower
Total height of packing =Z= NOG × HOG
Z = 10 × 0.67 = 6.7 m
Z=7 m (round off)
Allowances for liquid distribution = 1m
Allowances for liquid re-distribution =1m
Total height of tower = 7 + 1 + 1
Zt = 9 m
Total height of tower= 9 m
6.3.16 Calculation of wetting rate
If very low liquid rates have to be used the packing wetting rate should be checked to make it
sure it is above the minimum recommended by packing manufacturer Wetting rate is defined by
following relation.
Wetting rate = Liquid volumetric flow rate per unit cross-sectional area
Specific area of packing per unit volume
Liquid volumetric flow rate/Unit cross-sectional area =5488.226/ (3600×1000×1.13)
=1.349×10-3
m3
/m2
-sec
Specific area of packing = 194 m2
/m3
Wetting rate =6.954×10-6 m3sec-1/m2.
6.3.19 Calculation of pressure drop at flooding
From McCabe & smith 5th edition, Eq.22.1,
Pressure drop at flooding is given by relation.
ΔPflooding=0.115Fp 0.7
Where
ΔPflooding =Pressure drop at flooding.
Fp =Packing factor for 3-inch ceramic Intalox saddles = 52
ΔPflooding=0.115(52)0.7
=1.8in.H2O/ft of packing

62
6.3.20 Calculation of Total Pressure Drop
= 0.03308
Here, Gx = L (lb/sec.ft2
)
Gy= G (lb/sec.ft2
)
y
 =
g
 (lb/ft3
)
x
 =
L
 (lb/ft3
)
Also,
G2
×Fp×µL
0.1
/ ρg (ρL - ρg)gc = 0.02889
ΔP = 0.25 in.H2O/ft of packing
ΔP = 20.833 mmH2O/m of packing (Recommended pressure drop for absorber is 15 to 50
mmH2O/m of packing, topic 11.14.4, Coulson & Richardson)
Total Pressure Drop = 20.833x 7
= 145.83 mmH2O/m of packing
6.3.21 Calculation of number of streams for liquid distribution at top of the
packing
Number of liquid distribution streams at the top of the packing
Ns= (D/6)2
D = Diameter of the absorption column in inches = 1.2m = 47.23 inch
Putting values in above equations, we get, Ns = 61.96
 yx
y


y
x
G
G

63
Specifications
Identification:
Item: Packed Absorption Column
Item No: A-104
No. required: 01
Function: To absorb formaldehyde gas in water.
Operation: Continuous
Design Data:
No. of transfer units = 10
Height of transfer units = 0.6763 m
Height of packing section = 7 m
Total height of column = 9 m
Diameter = 1.2 m
Pressure drop = 20.833 mmH2O /m of packing
Internals:
Size and type = 38 mm Intalox saddles
Material of packing: Ceramic
Packing arrangement: Dumped
Type of packing support: Simple grid & perforated support

64
6.4 Design of distillation column
6.4.1 Types of distillation columns
There are basically two types of distillation columns used in industries.
➢ Batch columns
➢ Continuous columns
Batch columns
In batch distillation, the more volatile component is evaporated from the still which therefore
becomes progressively richer in the less volatile constituent. Distillation is continued, either until
the residue of the still contains a material with an acceptably low content of the volatile material,
or until the distillate is no longer sufficiently pure in respect of volatile content. 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. Most distillation processes
operate in a continuous fashion, but there is a growing interest in batch distillation, particularly
in the food, pharmaceutical, and biotechnology industries. The advantage of this separation
process is that the distillation unit can be used repeatedly, after cleaning, to separate a variety of
products. The unit generally is quite simple, but because concentration are continuously
changing, the process becomes more difficult to control.
Continuous distillation
In contrast to batch columns, a continuous feed is given to the column. 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 more common used. I will put light only on this type of
distillation column.
6.4.2 Choice between plate and packed column
The choice between use of tray column or a packed column for a given mass transfer operation
should, theoretically, be based on a detail cost analysis for the two types of contactors. However,
the decision can be made on the basis of a qualitative analysis of relative advantages and
disadvantages, eliminating the need for a detailed cost comparison.

65
Which are:
➢ Because of liquid dispersion difficulties in packed columns, the design of tray column is
considerably more reliable.
➢ Tray columns can be designed to handle wide ranges liquid rates without flooding.
➢ If the operation involves liquids that contain dispersed solids, use of a tray column is
preferred because the plates are more accessible for cleaning.
➢ For non-foaming systems the plate column is preferred.
➢ If periodic cleaning is required, man holes will be provided for cleaning. In packed
columns packing must be removed before cleaning.
➢ For large column heights, weight of the packed column is more than plate column.
➢ Design information for plate column is more readily available and more reliable than that
for packed column.
➢ Inter stage cooling can be provided to remove heat of reaction or solution in plate
column.
➢ When temperature change is involved, packing may be damaged.
➢ Random-packed columns generally are not designed with diameters larger than 1.0 m,
and diameters of commercial tray column are seldom less than 0.67m.
As my system is non-foaming and diameter calculated is larger than 1.0m so I am going to use
tray column.
Also, as average temperature calculated for my distillation column is higher that is
approximately equal to 103 ºC. So, I prefer Tray column.
6.4.3 Plate contactors
Cross flow plate is the most commonly used plate contactor in distillation. In which liquid flows,
downward and vapors flow upward. The liquid moves from plate to plate via down comer. A
certain level of liquid is maintained on the plates by weir. Other types of plate are used which
have no down comer (non-cross flow) the liquid showering down the column through large
opening in the plates (called shower plates). Used when low pressure drop is required.
Three basic types of cross flow trays used are
➢ Sieve Plate (Perforated Plate)

66
➢ Bubble Cap Plates
➢ Valve plates (floating cap plates)
I prefer sieve plate because:
➢ Their fundamentals are well established, entailing low risk.
➢ The trays are low in cost relative to many other types of trays.
➢ They can easily handle wide variations in flow rates.
➢ They are lighter in weight. It is easier and cheaper to install.
➢ Pressure drop is low as compared to bubble cap trays.
➢ Peak efficiency is generally high.
➢ Maintenance cost is reduced due to the ease of cleaning.
6.4.4 Labeled diagram
6.4.5 Factors affecting selection of trays
➢ Relative Cost of plate will depend upon material of construction used.
➢ For mild steel, the ratio of cost between plates is
Sieve plate : valve plate : bubble-cap plate
Man Way
Plate
support ring
Down comer
And weir
Calming
zone
Major Beam

67
3.0 : 1.5 : 1.0
➢ There is little difference in Capacity Rating of the three types (the column diameter
required for a given flow rate).
Sieve tray > valve tray > bubble-cap tray
➢ Operating Range means the range of liquid and vapour flow rates which must be above
the weeping conditions and below the flooding conditions. Operating range flexibility
comparison is.
Bubble cape tray > Valve tray > Sieve tray
➢ Sieve plate depends on the vapours flow through the holes to hold the liquid on the plate,
and cannot operate at very low vapour flow rates. But with good design, sieve plate gives
satisfactory operating range.
➢ The Plate pressure drop will depends on the detailed design of plate but, in general, sieve
plate gives the lowest pressure drop, followed by valves, with bubble-caps giving the
highest.
6.4.6 Operation of typical distillation column
The operation of typical distillation column may by followed by figure. The column consists of a
cylindrical structure divided into sections by a series of perforated trays which permit the upward
flow of vapor. The liquid reflux flows across each tray, over a weir and down a down comer to
the tray below. The vapor rising from the top tray passes to condenser and then through an
accumulator or reflux drum and a reflux divider, where part is withdrawn as the overhead
product D and the remainder is returned to the top tray as reflux R.
In the bottom, there is reboiler which is used to give heat to the system. Liquid from the bottom
of distillation column is fed to the reboiler which vaporizes the incoming liquid. These vapors in
turn move towards the bottom plate interact with the liquid over that plate. Due to which partial
condensation of vapors occur. Also, partial vaporization of liquid occurs too. That is less volatile
component condensed first and more volatile component vaporizes first. This phenomenon
occurs on each plate. Causing enrichment on each plate

68
6.4.7 Factors affecting distillation column operation
1. Vapor flow conditions
Adverse vapor flow conditions can cause:
➢ Foaming
➢ Entrainment
➢ Weeping/dumping
➢ Flooding
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 build-up 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.
Entrainment
Entrainment refers to the liquid carried by vapour up to the tray above and is again caused by
high vapour 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/Dumping
This phenomenon is caused by low vapour flow. The pressure exerted by the vapour 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.

69
Flooding
Flooding is brought about by excessive vapour flow, causing liquid to be entrained in the vapour
up the column. The increased pressure from excessive vapour also backs up the liquid in the
down comer, causing an increase in liquid hold-up 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.
2. Reflux conditions
Minimum trays are required under total reflux conditions, i.e. there is no withdrawal of distillate.
On the other hand, as reflux is decreased, more and more trays are required.
3. Feed conditions
The state of the feed mixture and feed composition affects the operating lines and hence the
number of stages required for separation. It also affects the location of feed tray.
4. State of trays
Remember that the actual number of trays required for a particular separation duty is determined
by the efficiency of the plate. Thus, any factors that cause a decrease in tray efficiency will also
change the performance of the column. Tray efficiencies are affected by fouling, wear and tear
and corrosion, and the rates at which these occur depends on the properties of the liquids being
processed. Thus appropriate materials should be specified for tray construction.
5. Column diameter
Vapor flow velocity is dependent on column diameter. Weeping determines the minimum vapor
flow required while flooding determines the maximum vapor flow allowed, hence column
capacity. Thus, if the column diameter is not sized properly, the column will not perform well.

70
6.4.8 Preliminary Calculations
Material balance
Components Feed
(kmole/hr)
Feed
(%)
Top
((kmole/hr)
Top (%) Bottom
(kmole/hr)
Bottom
(%)
Water 99.37 60.33 9.94 94.41 89.43 58
Formaldehyde 64.69 39.27 0 0 64.68 41.95
Methanol 0.6534 0.4 0.59 5.58 0.065 0.042
Total 164.71 100 10.525 100 154.186 100
F = D + W
6.4.9 Selection of Light and Heavy Key Components
Methanol = Light Key (L.K.)
Formaldehyde =Heavy Key (H.K.)
Water = Heavy Non Key (H.N.K)
6.4.10 Nature of Feed
Feed is entering in the column as a saturated liquid at T=100O
C & P=141 kPa
6.4.11Standard Design Steps of Distillation Column
• Calculation of minimum number of Plates.
• Calculation of minimum Reflux Ratio Rm.
• Calculation of actual reflux ratio.
• Calculation of theoretical number of stages.
• Calculation of diameter of the column.
• Calculation of weeping point.

71
• Calculation of entrainment.
• Calculation of pressure drop.
• Calculation of the height of the column
6.4.12 Calculation of Minimum number of Plates
The minimum No. of Stages 𝑁 𝑚𝑖𝑛can be finding from Fenske Equation which is,
Nmin= 11
6.4.13 Calculation of minimum reflux ratio
By using Underwood Equation;
∑
𝛼𝑖 𝑥 𝐹 𝑖
𝛼𝑖−𝛳
𝑛
𝑖=1 = 1 − q
As feed is saturated liquid, so q=1
By iterations in excel we find, θ = 1.49
Now using;
∑
𝛼𝑖 𝑥 𝐷 𝑖
𝛼𝑖−𝛳
𝑛
𝑖=1 = 𝑅 𝑚𝑖𝑛 + 1
We find, Rmin= 3.82
6.4.14 Calculation of actual reflux ratio
We take actual reflux 1.2-1.5 times Rmin
R = 5.348 ( 1.4 Times)
6.4.15 Calculation of theoretical number of stages
Using Gilliland Correlation
𝑋0 = (
𝑅 − 𝑅 𝑚𝑖𝑛
𝑅 + 1
) = 0.240
Using Gilliland Graph, we have
𝑌0 = 0.427

72
𝑌0 = (
𝑁 − 𝑁 𝑚𝑖𝑛
𝑁 + 1
)
Ne = 20 Plates
6.4.16 Location of Feed Plate
Using Kirkbride Equation
𝑁 𝑅
𝑁𝑆
= ([
𝑋 𝐻𝐾
𝑋 𝐿𝐾
]
𝐹
[
𝐵
𝐷
] [
(𝑋 𝐿𝐾) 𝐵
(𝑋 𝐻𝐾) 𝐷
]
2
)
0.206
𝑁 𝑅
𝑁𝑆
= 3.49
And 𝑁 𝑅 + 𝑁𝑆 = 20( Excluding Reboiler)
So, NS = 5 NR = 15
Feed Plate =6th
plate from bottom, including reboiler.
6.4.17 Flow rates inside the column
For Rectifying Section
Ln = D × R =56.154Kgmol/hr
Vn = Ln + D = 66.679Kgmol/hr
For stripping section
Lm = Ln + F =56.154+164.71Kgmol/hr
Vm = Lm – W = 66.7Kgmol/hr
Let assume the100 mmH2O pressure drop per plate.
6.4.18 Densities at Top
ρ of the gas = 1.0294 kg/m3
ρ of liquid = 971 kg/m3
6.4.19 Densities at the bottom
ρ of the gas = 1.037 kg/m3

73
ρ of liquid = 970 kg/m3
Let assume the100 mmH2O pressure drop per plate.
Total column pressure drop = 100 x 10-3
x970x9.8x20
= 19012 Pa
Top pressure 1.1 atm = 111.4 x 103 Pa
6.4.20 Estimated bottom pressure
= Top Pressure + Total column pressure drop
= 19012+111.4x103
= 130412 Pa
= 1.3 atm
6.4.21 Diameter of the column
Column Diameter = 𝐷 = √
4𝐴
𝜋
6.4.22 Flooding velocity
𝐹𝐿𝑉=
𝐿 𝑤
𝑉 𝑤
√
𝜌 𝑣
𝜌 𝐿
𝐹𝐿𝑉bottom =
220.86
66.68
√
1.037
970
= 0.0602 𝐹𝐿𝑉top=
56.154
66.7
√
1.0294
971
= 0.01467
Take plate spacing as 0.5 m
From Figure 11.27
base K1= 3 x 10-2
top K1 = 2.7 x 10-2
6.4.23 Correction for surface tensions
baseK1 =[Ϭ/0.02]0.2
x K1 = [0.06/0.02]0.02
x 3 x 10-2
= 3.1 x 10-2
topK1 =[Ϭ/0.02]0.2
x K1 = [0.08/0.02]0.02
x 2.7 x 10-2
= 3 x 10-2
6.4.24 flooding velocity
Base uf = K1 √
𝜌 𝑙−𝜌 𝑣
𝜌 𝑣
= 3.1 X 10-2
X √
970.21−0.3181
0.3181
= 1.65
Top uf=K1 √
𝜌 𝑙−𝜌 𝑣
𝜌 𝑣
= 3 X 10-2
X √
971.4−0.294
0.294
= 1.72
Maximum volumetric flow-rate =
𝐹𝑙𝑜𝑤 𝑟𝑎𝑡𝑒 𝑥 (𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑚𝑎𝑠𝑠)𝑎𝑣𝑔
𝜌 𝑣 𝑥 3600
Base =
66.8𝑥35
1.037 𝑥 3600
= 0.6262 m3
/s

74
top =
66.8𝑥23
1.0294 𝑥 3600
= 0.4145 m3
/s
Net area required = A =
Maximum volumetric flow rate
𝐹𝑙𝑜𝑜𝑑𝑖𝑛𝑔 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦
For design we take 80% of the flooding velocity so
At top : 1.72 × 0.80 = 1.376
At bottom: 1.65 × 0.80 = 1.32
bottom =
0.6262
1.32
= 0.4744 top =
0.4145
1.376
= 0.3012
As first trial take downcomer area as 15 per cent of total.
Column cross-sectioned area = A/0.85
base= 0.4843/0.85 = 0.5698 m2
top = 0.3012/0.85 = 0.3543 m2
Column diameter = D = √
4𝐴
𝜋
base=√
4 𝑥 0.5698
3.14
= 0.85𝑚 top =√
4 𝑥 0.3543
3.14
= 0.70𝑚
Use same diameter above and below feed, reducing the perforated area for plates above the
feed.So both diameters = 1 m
6.4.25 Provisional plate design
For provision, we use these values of diameter and area.
Column diameter = 𝐷𝑐 = 1 m
Column area = 𝐴 𝐶 = 0.6268 m2
Down-comer area = 𝐴 𝑑 = 0.10 x 0.6268 = 0.06268 m2
Net area = 𝐴 𝑛 =𝐴 𝐶– 𝐴 𝑑 = 0.6268 – 0.06268 = 0.5641 m2
Active area =𝐴 𝑎 =𝐴 𝐶 - 2𝐴 𝑑 = 0.6268– 2(0.06268) = 0.5014 m2
Hole area =𝐴ℎ(take 10 per cent 𝐴 𝑎 as first trial) = 0.05014 m2
Take weir height 50 mm
Hole diameter 5 mm
Plate thickness 5 mm

75
6.4.26 Weir Length
As we have,
𝐴 𝑑
𝐴 𝐶
𝑥 100 =
0.06268
0.6268
x 100 = 10%
From fig. 11.31,
𝑙 𝑤
𝐷𝑐
= 0.7
𝑙 𝑤 = 𝐷𝑐 𝑥 0.73 = 1 𝑥 0.73
𝑙 𝑤 = 0.73 𝑚
6.4.27 Weir Liquid Crest
Let, maximum liquid rate =
𝑙𝑖𝑞𝑢𝑖𝑑 𝑓𝑙𝑜𝑤 𝑟𝑎𝑡𝑒 𝑥 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑚𝑎𝑠𝑠
3600
=
220.86𝑥23
3600
= 1.411 m3
/s
Using equation
ℎ 𝑜𝑤 = 750 [
𝐿 𝑤
𝜌𝑙 𝑥 𝑙 𝑤
]
= 750 [
1.4
970𝑥0.73
]
ℎ 𝑜𝑤 = 11.81 𝑚𝑚𝑙𝑖𝑞
For minimum crest, its 70% is
ℎ 𝑜𝑤(min) = 0.70 𝑥 ℎ 𝑜𝑤 = 0.70 𝑥 11.81
= 8mmliq
Hence,
Min. Liquid Crest = ℎ 𝑜𝑤 + ℎ 𝑜𝑤(min)= 50 + 8= 58 mmliq

76
6.4.28 Weeping check
At minimum Liquid rate = ℎ 𝑜𝑤 + ℎ 𝑜𝑤(min) = 58 mm
From Figure 11.30, K2= 30.3
Now, minimum design vapour velocity is given by:
𝑈 𝑚𝑖𝑛 =
𝐾2 − 0.90(25.4 − 𝑑ℎ)
(𝜌 𝑣)1/2
=
30.3 − 0.90(25.4 − 5)
(0.3186)1/2
= 6.78 m/s
And
𝐴𝑐𝑡𝑢𝑎𝑙 𝑚𝑖𝑛𝑖𝑚𝑢𝑚 𝑣𝑎𝑝𝑜𝑢𝑟 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 =
min𝑖𝑚𝑢𝑚 𝑣𝑎𝑝𝑜𝑢𝑟 𝑟𝑎𝑡𝑒
𝐴ℎ
=
0.6262x0.70
0.05014
= 8.70 m/s (at 10% hole area)
Which is greater than minimum design vapour velocity. So minimum operating rate will be well
above weep point.
6.4.29 Entrainment-check
Actual velocity (based on net area) = Un =
𝑀𝑎𝑥. 𝑣𝑜𝑙𝑢𝑚𝑒𝑡𝑟𝑖𝑐 𝑓𝑙𝑜𝑤𝑟𝑎𝑡𝑒
𝑁𝑒𝑡 𝑎𝑟𝑒𝑎
=
0.6362
0.5641
= 1.13
Percentage flooding =
𝐴𝑐𝑡𝑢𝑎𝑙 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 (𝑏𝑎𝑠𝑒𝑑 𝑜𝑛 𝑛𝑒𝑡 𝑎𝑟𝑒𝑎)
𝐹𝑙𝑜𝑜𝑑𝑖𝑛𝑔 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦
=
1.13
1.65
= 70 %
Also 𝐹𝑙𝑣(𝑏𝑜𝑡𝑡𝑜𝑚) = 0.0602
Ψ = 0.03 (Which is below 0.1). Hence, no Entrainment there.
[As a rough guide the upper limit of Ψ can be taken as 0.1; below this figure the effect on
efficiency will be small. The optimum design value may be above this figure, see Fair (1963).]

77
6.4.30 Perforated area
From Figure, at
𝑙 𝑤
𝐷𝑐
= 0.73
ϴc = 94o
Angle subtended by the edge of the plate = 180-94 = 86o
Mean length, unperforated edge strips = (0.5014 - 50 x 10-3
)x
86𝜋
180
= 0.67 m
Area of unperforated edge strips = As = 50 x 10-3
x 0.67 = 0.034 m2
Mean length of calming zone, approx.
= weir length + width of unperforated strip
= 0.73 + 50 x 10-3
= 0.78 m
Area of calming zones = Acz = 2(0.78 x 50 x 10-3
) = 0.078 m2
Total area for perforations, AP =Aa – As - Acz
= 0.5014-0.034-0.078 = 0.3894 m2
Now, Ah/Ap = 0.05014/0.3894 = 0.129
𝑙 𝑝
𝑑ℎ
=2.75, satisfactory, within 2.5 to 4.0.
6.4.31 Number of holes
Area of one hole = 1.964 x 10-5
m2
Number of holes =
0.05014
1.964 𝑥 10−5
= 2550 holes
6.4.32 Hole Pitch
𝑙 𝑝
𝑑ℎ
= 2.75
So
𝑙 𝑝 = 2.65 x 𝑑ℎ = 2.75 x 5mm = 0.01375m
6.4.33 Plate pressure drop
Total plate pressure drop =
Dry plate drop + Residual head + ℎ 𝑤 + ℎ 𝑜𝑤(min)
Dry plate drop
Velocity through holes = Uh =
volumetric flowrate
𝐻𝑜𝑙𝑒 𝐴𝑟𝑒𝑎

78
=
0.6220
0.05014
= 12.4 𝑚/𝑠
At
Ah
Ap
= 0.12% and
plate thickness
𝐻𝑜𝑙𝑒 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟
=
5 𝑚𝑚
5 𝑚𝑚
= 1
From figure CO = 0.86
So,
ℎ 𝑑 = 51 [
𝑈ℎ
𝐶 𝑜
]
2
𝜌 𝑣
𝜌𝑙
= 51 [
12.4
0.86
]
2 0.3186
970
= 3.5 mmliq
6.4.34 Residual head
hr =
12.5 𝑥 103
𝜌 𝑣
= 13 mmliq
total plate pressure drops per plate = ℎ 𝑡= 3.5+13+(50+8) = 74 mm liq
= 0.074 m liq.
Less than 100mmliq (assumed) so acceptable.
Pressure drop per plate in terms of Pascal = ρgL = (𝜌 𝐿) ×g×(ℎ 𝑡)
ΔP = 970 x 9.8 x 0.074
= 0.7034 KPa
6.4.35 Total Pressure drop in the column
ΔPtot. = No. Of plates × ΔP
= 20x703 = 14060 Pa = 14 kPa
6.4.36 Down-comer liquid back-up
Back-up = hb =ℎ 𝑑𝑐 + ht+ ℎ 𝑤 + ℎ 𝑜𝑤(min)
6.4.37 Down-comer pressure loss (𝒉 𝒅𝒄)
Take hap =hw - 10 = 50 -10 = 40 mm.
Area under apron, Aap= hap (lw) = 0.73 x 40 x 10-3
= 0.0292 mm2
ℎ 𝑑𝑐 = 166 [
𝐿 𝑤
𝜌𝑙 𝑥 Aap
]
2
= 0.1102 mm
Down-comer Back-up = (50 + 8) + 74 + 0.1102 = 132mmliq
= 0.132mliq

79
0.13<
1
2
(plate spacing + weir height)
0.13<
1
2
(500 + 50) 0.118< 0.275
so plate spacing is acceptable
6.4.38 Check residence time
Residence time is given by
tr=
𝐴 𝑑ℎ 𝑏 𝜌 𝐿
𝐿 𝑤
Ad = down-comer area
=
0.13 𝑥 0.06268 𝑥 970
1.4
= 5.6 sechb=down-comer backup
= 5.6 s >3 s, satisfactory.
6.4.39 Height of the column
Height (z) of the distillation column can be find by the given below formula,
z = NAZt+ Ls + 1.22 m
Ls =
4 𝑉 𝐵 𝑡 𝑠
𝜋𝐷2
VB = Volumetric flow rate of bottom
=
4 x 0.6262 x 5
3.14 𝑥 (1)2
= 4 𝑚 (where t’s= surge time = 5-10 min)
z = (20 x 0.5) + 4 + 1.22 z = 15 m

80
Identification
Item Distillation Column
Item no T-1
Type Sieve Tray
Operation Continuous
No of items 1
Design Specification
Column
No of trays 20
Column Pressure 1.3 atm
Height 15m
Diameter 1m
Reflux Ratio 5.348
Tray Spacing 0.50 m
Material of Construction Stainless Steel
Tray
Pressure Drop per tray 0.70 kPa
Residence time 5.6 sec
Plate thickness 5mm
No of holes 2550
Weir Height 50mm
Weir Length 0.73m
Active Area 0.50142
Hole Diameter 5mm

81
CHAPTER # 7
Process Instrumentation and control
7.1 Instrumentation
Instruments are provided to monitor the key process variables during plant operation. They may
be incorporated in automatic control loops or used for the manual monitoring of the process
operation. They may also be part of an automatic computer data logging system. Instruments
monitoring critical process variables will be fitted with automatic alarms to alert the operators to
critical and hazardous situations. It is desirable that the process variable to be monitored be
measured directly. However, this is often impractical and some dependent variable that is easier
to measure, monitored in its place.
For example, in the control of distillation columns the continuous on-line, analysis of the over-
head product is desirable but it is difficult and expensive to achieve reliably, so temperature is
often monitored as an indication of composition.
7.2 Instrumentation and Control Objective
There might be lot of control objectives depending upon the situation, condition and equipment.
Some of common and major objectives are given bellow:
7.2.1 Safe plant operation
To keep the process variables within known safe operating limits.
To detect dangerous situations as they develop and to provide alarms and automatics hut-
down systems.
To provide inter locks and alarms to prevent dangerous operating procedures.
7.2.2 Production Rate
To achieve the design product output.
To increase production rate by fast and accurate control action
7.2.3 Product Quality
To maintain the product composition within the specified quality standards

82
7.2.4 Cost
To operate at the lowest production cost, commensurate with the other objectives.
These are not separate objectives and must be considered together. The order in which they
are listed is not meant to imply the precedence of any objective over another, other than that
of putting safety first.
In a typical chemical processing plant these objectives are achieved by a combination of
automatic control, manual monitoring and laboratory analysis.
7.3 Components of Control System
7.3.1 Process
Any operation or series of operations that produces a desired final result is a process. In
this discussion, the process is the production of formaldehyde using methanol.
7.3.2 Measuring Means
Of all the parts of the control system the measuring element is perhaps the most
important. If measurements are not made properly the remainder of the system cannot
operate satisfactorily.
I. Pressure measurements
II. Temperature measurements
III. Flow Rate measurements
IV. Level measurements
7.3.3 Controller
The controller is the mechanism that responds to any error indicated by the error
detecting mechanism. The output of the controller is some predetermined function of the
error. In the controller, there is also an error-detecting mechanism which compares the
measured variables with the desired value of the measured variable, the difference being
the error.
7.3.4 Final Control Element
The final control element receives the signal from the controller and by some
predetermined relationships changes the energy input to the process.
Classification of Controller

83
The four basic modes of control are:
i) On-off Control
ii) Integral Control
iii) Proportional Control
iv) Derivative Control
Mode Advantage Disadvantage
On-off Simple, inexpensive Constant cycling
Proportional Time lag is not included Almost has an offset
Integral Eliminates offset Adds time lag to the system
Derivative Speeds up the response Responds to noise
7.4 Different types of controllers
1. Alarms and Safety Trips and Interlocks
2. Alarms are used to alert operators of serious and potentially hazardous, deviations in
process conditions. Key instruments are fitted with switches and relays to operate audible
and visual alarms on the control panels.

84
7.5 Basic components of an automatic trip system
1. A sensor to monitor the control variable and provide an output signal when a preset valve
is exceeded (the instrument).
2. A link to transfer the signal to the actuator usually consisting of a system of pneumatic or
electric relays.
3. An actuator to carry out the required action; close or open a valve, switch off a motor.
4. A safety trip can be incorporated in control loop. In this system, the high-temperature
alarm operates a solenoid valve, releasing the air on the pneumatic activator closing the
valve on high temperature.
5. Interlocks where it is necessary to follow the fixed sequence of operations for example,
during a plant start-up and shut-down, or in batch operations-inter-locks are included to
prevent operators departed from the required sequence. They may be incorporated in the
control system design, as pneumatic and electric relays or may be mechanical interlocks.
7.6 Classification of Controllers
7.6.1 Flow controllers
Flow-indicator-controllers are used to control the amount of liquid. Also, all manually set
streams require some flow indication or some easy means for occasional sample
measurement. They are also used to control feed rate into a process unit. Orifice plates are by
far the most common type off low rate sensor. Normally, orifice plates are designed to give
pressure drops in the range of 20 to200inch of water. Venture tubes and turbine meters are
also used.
7.6.2 Temperature Controller
Thermocouples are the most commonly used temperature sensing devices. The two dissimilar
wires produce a millivolt emf that varies with the "hot-junction" temperature. Iron constrict
ant thermocouples are commonly used over the 0 to 1300°F temperature range.
7.6.3 Pressure Controller
Bourdon tubes, bellows and diaphragms are used to sense pressure and differential pressure.
For example, in a mechanical system the process pressure force is balanced by the movement
of a spring. The spring position can be related to process pressure.

85
7.6.4 Level Controller
Liquid levels are detected in a variety of ways. The three most common are:
1. Following the position of a float, that is lighter them the fluid.
2. Measuring the apparent weight of a heavy cylinder as it buoyed up more or less by
the liquid (these are called displacement meters).
3. Measuring the difference between static pressure of two fixed elevation, one on the
vapor which is above the liquid and the other under the liquid surface. The
differential pressure between the two-level taps is directly related to the liquid level in
the vessel.
7.6.5 Transmitter
The transmitter is the interface between the process and its control system. The job of the
transmitter is to convert the sensor signal (milli volts, mechanical movement, pressure
differential, etc.) into a control signal 3 to 15 psig air-pressure signal, 1 to 5 or 10 to 50
milliampere electrical signal, etc.
7.6.6 Control Valves
The interface with the process at the other end of the control loop is made by the final control
element is an automatic control valve which throttles the flow of a stem that opens or closes
an orifice opening as the stem is raised or lowered. The stem is attached to a diaphragm that
is driven by changing air-pressure above the diaphragm. The force of the air pressure is
opposed by a spring.

86
7.7 Control scheme around Reactor
Controlled variable
➢ Temperature of reactor
Manipulated variable
➢ Flow rate of water
Controller type
➢ Selective control system

87
7.8 Control scheme around absorber
Controlled variable
➢ Level and pressure inside of absorber
➢ Flow rate
Controller type
FC
FC
LC
PC
Absorber

88
7.9 Distillation column
Controlled variable
➢ Pressure in distillation column
➢ Flow rate
Controller type
DISTILATION COLUMN
FC
TT
3
FT
FE
LT
LT
PT PC
LC
FC
TT TC
5
4
LC
7
TC
FE
FT
FC
1
2
6

89
CHAPTER # 8
Plant Layout
8.1 Introduction
The economic construction and efficient operation of a process unit will depend upon how well
the plant and equipment specified on the process flow sheet is laid out and on the profitability of
the project with its scope for future expansion. Plant location and site selection should be made
before the plant layout.
8.2 Plant location and site selection
The location of the plant has a crucial effect on the profitability of the project. The important
factors that are to be considered while selecting a site are:
1. Location, with respect to market area
2. Raw material supply
3. Transport facilities
4. Availability of Labor
5. Availability of utilities
6. Availability of suitable land
7. Environmental impact and effluent disposal
8. Local community considerations
9. Climate
10. Political and strategic considerations
8.2.1 Marketing area
For materials that are produced in bulk quantities, such as cement, mineral acids, and fertilizers
where the cost of product per ton is relatively low and the cost of transport a significant fraction
of the sales price, the plant should be located close to the primary product. This consideration
will be less important for low volume production, high-priced products, such as pharmaceutical.

90
8.2.2 Raw materials
The availability and price of suitable raw materials will often determine the site location. Plants
producing bulk chemicals are best located close to the source of major raw material, where this is
also close to the marketing area. For the production of formaldehyde, the site should be
preferably near a methanol plant.
8.2.3 Transport
Transport of raw materials and products is an important factor to be considered. Transport of
products can be in any of the four modes of transport.
8.2.4 Availability of labor
Labor will be needed for construction of the plant and its operation. Skilled construction workers
will usually be brought in from outside the site area, but there should be an adequate pool of un
skilled labors available locally; and labor suitable for training to operate the plant. Skilled
tradesman will be needed for plant maintenance. Local trade union customs and restrictive
practices will have to be considered when assessing the availability and suitability of the local
labor for recruitment and training.
8.2.5 Environmental impact and effluent disposal
All industrial processes produce waste products, and full consideration must be given to the
difficulties and cost of their disposal. The disposal of toxic and harmful effluents will be covered
by the local regulations and the appropriate authorities must be consulted during the initial
survey to determine the standards that must be met.
8.2.6 Local community consideration
The proposed plant must fit with and be acceptable to the local community. Full consideration
must be given to the safe location of the plant so that it does not impose a significant additional
risk to the community on a new site, the local community must be able to provide adequate
facilities for the plant personnel.

91
8.2.7 Land
Sufficient suitable land must be available for the proposed plant and for future expansion. The
land should ideally be flat, well drained and have suitable loadbearing characteristics full site
evaluation should be made to determine the need for piling or other special foundations.
8.2.8 Climate
Adverse climatic conditions, at a site will increase costs. Abnormally low temperatures will
require the provision of additional insulation and special heating for equipment and pipe runs.
8.2.9 Political and strategic considerations
Capital grants, tax concessions and other inducements are often given by governments to direct
new investment to preferred locations; such as areas of high unemployment. The availability of
such grants can be overriding consideration in the site selection. After considering the location of
the site the plant layout is completed.
It involves placing of equipment so that the following are minimized:
1. Damage to persons and property in case of fire explosion or toxic release
2. Maintenance costs
3. Number of people required to operate the plant.
4. Construction costs
5. Cost of planned expansion.
In plant layout first thing that should be done is to determine the direction of the prevailing wind.
Wind direction will decide the location of the plant. List of items that should be placed upwind
and downwind of the plant is given down.
8.3 Items that should be located upwind of the plant
1. Laboratories
2. Fire station
3. Offices building
4. Canteen and Change house
5. Storehouse
6. Medical facilities
7. Electrical substation

92
8. Water treatment plant
9. Water pumps
10. Workshops
8.4 Items that should be located downwind of the plant
1. Blow down tanks
2. Settling tanks
3. Burning flares
8.5 The various units that should be laid out include
1. Main processing unit
2. Storage for raw materials and products
3. Maintenance workshops
4. Laboratories for process control
5. Fire stations and other emergency services
6. Utilities: steam boilers, compressed air, power generation, refrigeration
7. Effluent disposal plant
8. Offices for general administration
9. Canteens and other amenity buildings, such as medical centers
10. Car parks
8.5.1 Processing area
Processing area also known as plant area is the main part of the plant where the actual production
takes place. There are two ways of laying out the processing area
1. Grouped layout
2. Flow line layout
Grouped layout
Grouped layout places all similar pieces of equipment adjacent. This provides for ease of
operation and switching from one unit to another. This is suitable for all plants.

93
Flow line layout
Flow line layout uses the line system, which locates all the equipment in the order in which it
occurs on the flow sheet. This minimizes the length of transfer lines and therefore reduces the
energy needed to transport materials. This is used mainly for small volume products.
8.5.2 Storage house
The main stage areas should be placed between the loading and unloading facilities and the
process they serve. The amount of space required for storage is determined from how much is to
be stored in what containers. In raw material storage, liquids are stored in small containers or in a
pile on the ground. Automatic storage and retrieving equipment can be substantially cut down
storage.
8.5.3 Laboratories
Quality control laboratories are a necessary part of any plant and must be included in all cost
estimates. Adequate space must be provided in them for performing all tests, and for clearing and
storing laboratory sampling and testing containers.
8.5.4 Transport
The transport of materials and products to and from the plant will be an overriding consideration
in site selection. If practicable, a site should be selected that is close to at least two major forms
of transport: road, rail, waterway or a seaport. Rail transport will be cheaper for long distance
transport of bulk chemicals. Road transport is being increasingly used and is suitable for local
distribution. Road area also used for firefighting equipment and other emergency vehicles and
for maintenance equipment. This means that there should be a road around the perimeter of the
site. No roads should be a dead end. All major traffic should be kept away from the processing
areas. It is wise to locate all loading and unloading facilities, as well as plant offices, personnel
facilities near the main road to minimize traffic congestion within the plant and to reduce danger.
8.5.5 Utilities
The word “Utilities” is now generally used for ancillary services needed in the operation of any
production process. These services will normally be supplied from a central site facility and will
include:

94
1. Electricity
2. Steam for process heating
3. Cooling water
4. Water for general use
5. Inert gas supplies
Electricity
Electrical power will be needed at all the sites. Electrochemical processes that require large
quantities of power need to be located close to a cheap source of power. Transformers will be
used to step down the supply voltage to the voltages used on the purpose.
Steam for process heating
The steam for process heating is usually generated in water tube boilers using the most
economical fuel available. The process temperature can be obtained with low-pressure steam. A
competitively priced fuel must be available on site for steam generation.
Cooling water
Chemical processes invariably require large quantities of water for cooling. The cooling water
required can be taken from a river or lake or from the sea. Water for general use Water is needed
in large quantities for general purpose and the plant must be located near the sources of water of
suitable quality, process water may be drawn from river from wells or purchased from a local
authority.
8.5.6 Offices
The location of this building should be arranged so as to minimize the time spent by personnel in
travelling between buildings. Administration offices in which a relatively large number of people
working should be located well from potentially hazardous process.
8.5.7 Canteen
Canteen should be spacious and large enough for the workers with good and hygienic food.
8.5.8 Fire station
Fire station should be located adjacent to the plant area, so that in case of fire or emergency, the
service can be put into action.

95
8.5.9 Medical facilities
Medical facilities should be provided with at least basic facilities giving first aid to the injured
workers. Provision must be made for the environmentally acceptable disposal of effluent.
8.6 The layout of the plant can be made effective by:
1. Adopting the shortest run of connecting pipe between equipment’s and the least amount
of structural steel work and thereby reducing the cost.
2. Equipment that need frequent operator attention should be located convenient to control
rooms.
3. Locating the vessels that require frequent replacement of packing or catalyst outside the
building.
4. Providing at least two escape routes for operators from each level in process buildings.
5. Convenient location of the equipment so that it can be tied with any future expansion of
the process.

96
CHAPTER # 9
Material of Construction
9.1 Introduction
As chemical process plants turn to higher temperatures and flow rates to boost yields and
throughputs, selection of construction materials takes on added importance. This trend to more
severe operating conditions forces the chemical engineer to search for more dependable, more
corrosion-resistant materials of construction for these process plants, because all these severe
conditions intensify corrosive action.
Fortunately, a broad range of materials is now available for corrosive service. However, this
apparent abundance of materials also complicates the task of choosing the “best” material
because, in many cases, a number of alloys and plastics will have sufficient corrosion resistance
for a particular application. Final choice cannot be based simply on choosing a suitable material
from a corrosion table but must be based on a sound economic analysis of competing materials.
The chemical engineer would hardly expect a metallurgist to handle the design and operation of
a complex chemical plant.
Many factors have to be considered when selecting engineering materials, but for chemical
process plant the overriding consideration is usually the ability to resist corrosion. The material
selected must have sufficient strength and be easily worked. The most economical material that
satisfies both process and mechanical requirements should be selected; this will be the material
that gives the lowest cost over the working life of the plant, allowing for maintenance and
replacement. Other factors, such as product contamination and process safety, must also be
considered.
9.2 Material Properties
The most important characteristics to be considered when selecting a material of construction
are;
1. Mechanical Properties
2. Strength- tensile strength

97
3. Stiffness- elastic modulus (Young’s modulus)
4. Toughness- fracture resistance
5. Hardness- wear resistance
The effect of high and low temperatures on the mechanical properties; corrosion resistance
Any special properties required: such as
1. Thermal conductivity
2. Electrical resistance
3. Magnetic properties ease of fabrication-forming103
4. Welding
5. Casting availability in standard sizes-plates
6. Sections
7. Tubes
8. Cost
9.3 Selection for Corrosion Resistance
In order to select the correct material of construction, the process environment to which the
material will be exposed must be clearly defined. Additional to main corrosive chemicals
present, the following factors must be considered:
1. Temperature (affects corrosion rate and mechanical properties.
2. Pressure
3. Presence of trace impurities-stress corrosion
4. The amount of aeration-differential oxidation cells
5. Stream velocity and agitation-erosion-corrosion
6. Heat transfer rates- differential temperatures
The conditions that may arise during abnormal operation, such as at start-up and shutdown, must
be considered, in addition to normal, steady state operation.
9.4 Commonly used Materials of Construction
The general mechanical properties, corrosion resistance, and typical areas of use of some of the
materials commonly used in the construction of chemical plant are described as under.

98
9.4.1 Iron and steel
Low carbon steel (mild steel) is the most commonly used engineering material. It is cheap; is
available in a wide range of standard forms and sizes; and can be easily worked and welded. It
has good tensile strength and ductility.
The carbon steel and iron are not resistant to corrosion, except in certain specific environments,
such as concentrated sulfuric acid and the caustic alkalis. They are suitable for use with most
organic solvents, except chlorinated solvents; but traces of corrosion products may cause
discoloration. Mild steel is susceptible to stress-corrosion cracking in certain environments.
9.4.2 Stainless steel
The stainless steels are the most frequently used corrosion resistant materials in the
Chemical industry. To impart corrosion resistance the chromium content must be above 12%,and
the higher the chromium content, the more resistant is the alloy to corrosion in
oxidizing104conditions. Nickel is added to improve to improve the corrosion resistance in non-
oxidizing environments.
9.5 Types
A wide range of stainless steels is available, with compositions tailored to give the properties
required for specific applications. They can be divided into three broad classes according to their
microstructure:
1. Ferritic:- 13-20% cr, 0.1% c, with no Nickel
2. Austonitic -18-20% Cr, >7.0% Ni
3. Marionsitic -12-1-% Cr, 0.2-0.4% C, up to 2% N
9.5.1 Monel
Monel, the classic nickel copper alloy with the metals in the ratio 2:1, is probably, after the
stainless steels, the most commonly used alloy for chemical plant. It is easily worked and has got
mechanical properties up to 500 C. it is more expensive than stainless steels but is not
susceptible to stress corrosion cracking in chloride solutions. Monel as good resistance to ductile
mineral acids and can be used in reducing conditions, where the stainless steels would be
unsuitable. It may be used for equipment handling, alkalis, organic acids and salts, and sea water.

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