Formaldehyde Project Report by Abhishek
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Transcript of Formaldehyde Project Report by Abhishek
(A) STATEMENT OF THE PROBLEM
Design a plant to manufacture 10 Tonnes/day of FORMALIN
Formaldehyde is manufactured by vapor phase catalytic oxidation of Methanol. Methanol
is diluted with water and is evaporated in the evaporator. The Methanol water vapor
mixture is superheated and mixed with air. Then the mixture is admitted to the Packed bed
catalytic reactor containing silver grains as catalyst. The reaction temperature is
maintained at 650C. Assume a residence time of 10 to 20 seconds. The reaction is first
order with respect to methanol and half order with respect to oxygen. Heat recovery is
made from the outgoing product gases using a waste heat boiler. The product gases are
sent to two absorption columns in series wherein water and dilute Formalin are used for
absorption. Finally the formaldehyde is purified in a distillation column and the methanol
is recycled back as a top product. Assume 90% conversion of methanol. Assume any
missing data suitably if required.
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BIBLIOGRAPHY
1. “Chemical Engg. Kinetics” by J.M.Smith. 1st edition. (page 290)
2. “Chemical Process industries” by G.T.Austin, Shreves, Page 389-392
3. “Encyclopedia of Chemical processing and design” by John.J. Mcketta Volume 23
(Page351-370)
4. “Encyclopedia of Chemical Technology” by Kirk & Othmer, 4th edition, 1994
Volume. 11(Page 929 – 947).
5. Industrial & Engg Chemistry by S.J.Green & Raymond.E.Vener,
Volume 47 (Page-103-108) 1955
6. “Introduction to Chemical Engg Thermodynamics “ by Smith & Vannes,
6th edition. (appendix –C)
7. Perry ‘s “Chemical engineer’s handbook” 6th Edition (Chapter 3 & 18)
8. “Plant Design and Economics for Chemical Engineers”
by Peter & Timmerhaus, 4th edition (Chapter – 6).
9. “Process Design of Equipment” Volume 2. by S.D. Dawande, (Page 49-50).
10
.
“Process Equipment Design” by M.V. Joshi 3rd edition. Chapter-9 & page 129
11
.
“Process heat transfer” by Donald Q.Kern (Page 114, 147-148, 836,838)
12
.
“Reaction kinetics for chemical engineers” by Stanley Walas (Page 193) Chapter 8
13
.
(Source http://www.niir.org/projects/tag/z,.,451_0_32/formaldehyde/index.html.)
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NOMENCLATURE
A Heat transfer surface
a ,a t , asFlow area in general, for tube side, and for shell side, respectively
B Baffle spacing
Cp Specific Heat
C ' Clearance between tubes.
D Diameter
DeEquivalent diameter
d p ,d tDiameter of particle, and diameter of tube
F Feed flow rate
f Friction factor
G ,Gs ,GtMass Velocity in general, for shell side, and for tube side, respectively
H Enthalpy
h ,hi , hoHeat transfer coefficient in general, for inside fluid, and for outside
fluid, respectively
hioValue of hi when referred to outside diameter
ID Inner diameter
J Joules
jH Heat transfer factor, dimensionless
K Temperature measurement, Kelvin
k Thermal conductivity
L Length
M Mass flow
mS ,mwMass flow rate of steam, and water respectively
N Newton
3
N tNumber of tubes
NRE Reynolds number
OD Outer diameter
P Pressure
P Pressure drop
PT Tube pitch
Q Heat flow
RdTotal dirt factor
T Temperature
U ,UC ,U dOverall coefficient of heat transfer, clean coefficient, design coefficient
V Velocity
T Temperature Difference
X Vapor pressure of water
Latent heat of vaporization or condensation
φ Viscosity ratio (μ/ μw )
Viscosity
μwViscosity at the tube wall
Density
Voidage
4
PERFORMANCE DATA OF MAIN EQUIPMENT
REACTOR
Volume of the reactor 0.156 m3
Diameter of the reactor 0.83m
No. of tubes 80
Length of each tube 1 m
ID of tube 50 mm
OD of tube 55 mm
Heat transfer area 6.80 m2
Pressure drop 26.7 kg/ m2
Material of construction
(a) Shell Carbon Steel
(b) Tube SS 410
Catalyst detail
Name: Silver
Diameter 3.5 mm
Void fraction 0.36
5
FRACTIONATION COLUMN
No. of trays 17
No. of plate in enriching section 12
No. of plate in stripping section 5
Column height 6.8 m
Column diameter 0.2 m.
Feed entry 8th tray From top
Operating pressure 1 atm.
Type of plates Sieve tray
Tray spacing 400mm
Material of construction S.S 316
6
SUMMARY OF MECHANICAL DESIGN
Shell side
Shell thickness 10 mm
Inlet nozzle 30 mm.
Head thickness 10 mm
Baffle thickness 6.5 mm
No. of tie rods 6
Dia. of tie rods 12.5 mm
Flange thickness 61 mm
Gasket diameter 835 mm.
Gasket width 24 mm
Tube side
Thickness of tube 5 mm
Thickness of tube sheet 14 mm
Ring gasket width 22 mm
Minimum pitch circle diameter 940 mm
No. of bolts 20
Size of bolt M30
Gasket Flat metal jacketed Asbestos filled
Inlet nozzle dia. 85 mm
Outlet nozzle dia. 93.6 mm
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INTRODUCTION
Formaldehyde, H2C=O, is a reactive molecule, the first of the series of aliphatic aldehydes
and one of the most important industrial chemicals. Formaldehyde is a colorless gas at
ordinary temperature. Commercially formaldehyde is manufactured in the form of a water
solution usually containing 37% by weight of dissolved formaldehyde, this solution is
called formalin. In 1983 formaldehyde ranked 26th in production among Unites states
chemical products, with an output of 5.40 billion lb of equivalent 37 wt% aqueous
solution. Annual worldwide production capacity now exceeds 15 106 tons
(calculated as 37% solution).Because of its relatively low cost, high purity, and variety of
chemical reactions, formaldehyde has become one of the world’s most important industrial
and research chemicals. Products from formaldehyde are used extensively in the
automobile, construction, paper and textile industry.
HISTORY
Formaldehyde’s public image has always been associated with the funeral homes, doctor
offices and biology classes as an embalming fluid, a disinfectant and a preservative .In
1859, Russian scientist Alexander Mikhailovich Butlerlov discovered Formaldehyde,
accidently as he investigated the structure of organic compounds. Nine years later,
German scientist August Wilhelm Hofmann found a reliable way to make it. Hofmann
Passed a mixture of methanol and air over a heated platinum spiral and then identified
formaldehyde as the product. This method led to the major way in which the
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Formaldehyde is manufactured today, the oxidation of methanol with air using a metal
catalyst primarily of silver or molybdenum oxide. In 1905 , Dr.Leo Baekeland in Yonkers,
New york made a major breakthrough in the technology of polymer later named Bakelite
after him. The ingredients were Phenol and Formaldehyde, by the 1920’s the growth of
this resin strained wood alcohol (Methanol) producing capacity, but the revolutionary
development of methane reforming route to methanol relieved the situation. Despite the
radical shift in methanol technology, the process for formaldehyde based on methanol
feedstock has remained virtually unchanged even to today, despite volume growth making
it one of the top 25 commodity chemicals.
LITERATURE SURVEY
PYSICAL PROPERTIES
Formaldehyde monomer
Pure anhydrous formaldehyde is a colorless gas at ordinary temperature and at a molecular
weight of 30.26 is sightly heavier than air. It condenses on cooling to -19C and freezes to
a crystalline solid at -118C. The gas is characterized by a pungent odor and is judged
moderately irritating to the eyes, nose and throat by 20% of the population exposed to
concentrations in the 1.5 to 3.0 ppm range. Dry formaldehyde gas is stable and shows no
polymerization tendency at temperature as high as 100C. However, small amounts of
water or other impurities can cause rapid polymerization to poly(oxymethylenes).
Anhydrous formaldehyde gas is readily soluble in polar solvents such as water, methanol,
and n-propanol. It is only moderately soluble in nonpolar solvents such as ethyl ether,
chloroform, and toluene. In water and methanol, its heat of solution is approximately
15kcal/g-mol. A summary of physical properties of monomeric formaldehyde is given in
Table 1.1
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Formaldehyde solutions
Formaldehyde is produced and distributed as a water solution, the “standard” strength
being 37 wt%, this being a typical concentration and also the basis for making production
comparisons. Representative commercially available solutions are shown in Table 1.2
Table 1.1 Properties of Monomeric Formaldehyde
Property Value
Density, g/cm3
at -80C 0.9151 at -20C 0.8153Boiling point at 101.3 kPa C -19Melting point, C -118Vapor pressure, Antoine constants, Pa A 9.21876B 959.43C 243.392
Heat of vaporization, ΔH v at 19C, kJ/mol 23.3
Heat of formation, ΔH f at 25C, kJ/mol -115.9
Std free energy, ΔG f at 25C, kJ/mol -109.9Heat capacity, Cp, J/mol.K 35.4
Entropy,So
,J/(mol.k) 218.8Heat of combustion, kJ/mol 563.5Heat of solution at 23C kJ/mol in water 62 in methanol 62.8 in 1-propanol 59.5 in 1-butanol 62.4Critical constants Temperature, C 137.2-141.2 Pressure, MPa 6.784-6.637Flammability in air Lower/upper limits, mol 7.0/73 Ignition temperature, C 430
Varying amounts of methanol are included in the solutions as stabilizers and because of
the expenses of removal of methanol for uses where it is not harmful.
Formaldehyde is highly soluble in water, but in the liquid it reacts readily with water to
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form the hydrate, methylene glycol, which itself then tends to polymerize to
poly(oxymethylene glycols). Also, some hemiformals and a small amount of formic acid
are produced. At chemical equilibrium the amount of unhydrated formaldehyde is small,
approximately 0.1% at 60C.
In methanol-formaldehyde-water solutions, increasing the concentration of either
methanol or formaldehyde reduces the volatility of the other. The flash point varies with
composition, decreasing from 83 to 60C as the formaldehyde and methanol concentration
increase.
Formaldehyde solutions exists as a mixture of oligomers, HO(CH2O)nH. Methanol
stabilizes aqueous formaldehyde solutions by decreasing the average value of n. Hence
methanolic solutions can be stored at relatively low temperatures without precipitation of
polymer.
Table 1.2 Typical Analyses and Physical properties of Formaldehyde solutions
USP Grades Low-Methanol GradesFormaldehyde (wt%) 37.1 37.1 37.1 44.1 50.3Methanol (wt%) 7.0 11.0 0.9-1.3 0.9-1.3 0.9-1.3Acidity as formic (wt%) 0.012 0.012 0.012 0.02 0.018 Iron as Fe (ppm) 0.3 0.3 0.3 0.3 0.3Turbidity, Hellige 1.5 1.5 2 2 2 Color, APHA 5 5 5 5 5Density (g/cm3), 18C 1.10 1.09 1.11 1.12 1.13Boiling point (C) 97.2 96.7 99.0 99.1 99.5Viscosity (cP), 25C 2.5 2.6 2.0 1.7 1.6Specific heat (cal/g.C) 0.8 0.8 0.8 0.7 0.7Flash point (C) 69 60 83 80 79
Chemical properties
Formaldehyde is noted for its reactivity and its versatility as a chemical intermediate. It is
used in the form of anhydrous monomer solutions, polymers, and derivatives.
Anhydrous, monomeric formaldehyde is not, available commercially. The pure, dry gas
is relatively stable at 80-100C but, slowly polymerizes at lower temperatures. Traces of
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polar impurities such as acids, alkalies, and water greatly accelerate the polymerization.
Formaldehyde in water solution hydrates to methylene glycol;
Which in turns polymerizes to poly(methylene glycols), HO(CH2O)nH, also called
polyoxymethylenes. From these polymers a specific product, paraformaldehyde (or
“parafrom”), is obtained commercially. Paraformaldehyde is the name given to
polyoxymethanlenes with n values from 8 to 100. It is produced by the vacuum distillation
of concentrated formaldehyde solutions, and is available commercially in powder,
granular, or flakes forms. It has the characteristic pungent odor of formaldehyde, and melts
in the range of 120 to 170C. It is flammable, with a flash point of about 93C. A typical
formaldehyde solution may be obtained by dissolving paraformaldehyde in water.
Paraformaldehyde may be heated together with a strong acid to produce trioxane, the
cyclic trimer of formaldehyde.
This is a colorless crystalline material, melting at 62-64C, boiling without decomposition
at 115C, and having a flash point of 45C. Concentrations of trioxane between 3.6 and
28.7 vol% in air are explosive.
Formaldehyde may be reduced to methanol over a number of metal and metal oxide
catalysts. It may be condensed with itself in an aldol-type reaction to yield lower hydroxyl
aldehydes, hydroxy ketones, and other hydroxyl compounds. Formaldehyde and
acetaldehyde may be reacted in the presence of sodium hydroxide to form pentaerythritol
and sodium formate;
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Acetylene may be reacted with formaldehyde to form 2-butyne-1,4-idol which, when
hydrogenated, yields 1,4-butanediol;
Formaldehyde and aniline may be condensed to form diphenylmethane diamine:
Reaction of this product with phosgene yields methlenebis (4-phenyl iso-cyanate), or
“MDI,” one of the important types of commercial isocyanates.
Liquid phase condensation of formaldehyde with propylene, catalyzed by BF3 or H2SO4,
gives butadiene.
Hydrogen cyanide reacts with aqueous formaldehyde in the presence of bases to
produce glyconitrile:
HCHO + HCN HOCH2—C=N
This extremely toxic material is an intermediate in the synthesis of nitrilotriacetic acid
(NTA), EDTA, and glycine.
Reaction of formaldehyde, methanol, acetaldehyde, and ammonia over a silica alumina
catalyst at 500C gives pyridine and 3-picoline. This forms the basis of commercial
processes for making pyridines from various aldehydes.
Formaldehyde reacts with syn gas (CO,H2) to produce added value products. Ethylene
glycol (EG).
Alternative choice of manufacture
Currently, the only production technologies for formaldehyde of commercial
significance are based on the partial oxidation and dehydrogenation of methanol using
silver catalyst, or partial oxidation of methanol using metal oxide-based catalyst.
13
Development of New Processes. There has been significant research activity to
develop new processes for producing formaldehyde. Even though this work has been
extensive, no commercial units are known to exist based on the technologies discussed in
the following.
One possible route is to make formaldehyde directly from methane by partial oxidation.
This process has been extensively studied. The incentive for such a process is reduction of
raw material costs by avoiding the capital and expense of producing the methanol from
methane.
Another possible route for producing formaldehyde is by dehydrogenation of methanol
which would produce anhydrous or highly concentrated formaldehyde solutions. For some
formaldehyde users, minimization of the water in the feed reduces energy costs, effluent
generation, and losses while providing more desirable reaction conditions.
A third possible route is to produce formaldehyde from methylal that is produced from
methanol and formaldehyde. The incentive for such a process is twofold. First, a higher
concentrated formaldehyde product of 70% could be made by methylal oxidation as
opposed to methanol oxidation, which makes a 55% product. This higher concentration is
desirable for some formaldehyde users. Secondly, formaldehyde in aqueous recycle
streams from other units could be recovered by reacting with methanol to produce methylal
as opposed to recovery by other more costly means, eg, distillation and evaporation.
Development of this processes is complete.
Specification and Quality control
Formaldehyde is sold in aqueous solutions with concentrations ranging from 25 to 56 wt%
HCHO. Product specifications for typical grades are summarized in Table 1.3.
Formaldehyde is sold as low methanol (uninhibited) and high methanol (inhibited) grades.
Methanol is used to retard paraformaldehyde formation.
14
Procedures for determining the quality of formaldehyde solutions are outlined by
ASTM. Analytical methods relevant to Table 1.3 follow: formaldehyde by the sodium
sulfite method (D2194); methanol by specific gravity (D2380); acidity as formic acid by
titration with sodium hydroxide (D2379); iron by colorimetry (D2087); and color (APHA)
by comparison to platinum-cobalt color standards (D1209).
Table 1.3 Formaldehyde specifications
Property Methanol inhibited grades low methanol uninhibited grades
Formaldehyde, wt% 37 37 37 44 50 56
Methanol, wt% (max) 6-8 12-15 1.0-1.8 1.5 1.5-2.0 2.0
Acidity, wt% (max) 0.02 0.02 0.02 0.03 0.05 0.04
Iron, ppm (max) 0.5 0.05 0.5-1.0 0.5 0.5 0.75
Color, APHA (max) 10 10 10 10 10 10
STORAGE AND TRANSPORTATION
As opposed to gaseous, pure formaldehyde, solutions of formaldehyde are unstable. Both
formic acid (acidity) and paraformaldehyde (solids) concentrations increase with time and
depend on temperature. Formic acid concentration builds at a rate of 1.5-3 ppm/d at 35 C
and 10-20 ppm/d at 65C. Trace metallic impurities such as iron can boost the rate of
formation of formic acid. Although low storage temperature minimizes acidity, it also
increase the tendency to precipitate paraformaldehyde.
Paraformaldehyde solids can be minimized by storing formaldehyde solutions above a
minimum temperature for less than a given time period. The addition of methanol as an
inhibitor or of another chemical as a stabilizer allows storage at lower temperatures and/or
for longer times. Stabilizers for formaldehyde solutions include
hydroxypropylmethylcellulose, methyl- and ethylcellulose, poly(vinyl alcohol)s, or
isophthalobisguanamine at concentrations ranging from 10 to 1000ppm. Inhibited
15
formaldehyde typically contains 5-15 wt% methanol.
Most formaldehyde producers recommend a minimum storage temperature for both
stabilized and unstabilized solutions. The minimum temperature to prevent
paraformaldehyde formation in unstabilised 37% formaldehyde solutions stored for one to
about three months is as follows: 35C with less than 1% methanol; 21C with 7%
methanol; 7C with 10% methanol; and 6C with 12% methanol.
Materials of construction preferred for storage vessels are 304-, 316-, and 347-type
stainless steels or lined carbon steel.
USES
Formaldehyde is a basic chemical building block for the production of a wide range of
chemicals finding a wide variety of end uses such as wood products, plastics, and coatings.
Amino and Phenolic Resins. The largest use of formaldehyde is in the manufacture of
urea-formaldehyde, phenol-formaldehyde, and melamine-formaldehyde resins, accounting
for over one-half (51%) of the total demand. These resins find use as adhesives for binding
wood products that comprise particle board, fiber board, and plywood. Plywood is the
largest market for phenol-formaldehyde resins; particle board is the largest for urea-
formaldehyde resins.
Phenol-formaldehyde resins are used as molding compounds. Their thermal and
electrical properties allow use in electrical, automotive, and kitchen parts. Other uses for
phenol-formaldehyde resins include phenolic foam insulation, foundry mold binders,
decorative and industrial laminates, and binders for insulating materials.
Urea-formaldehyde resins are also used as molding compounds and as wet strength
additives for paper products. Melamine-formaldehyde resins find use in decorative
laminates, thermoset surface coatings, and molding compounds such as dinnerware.
1,4-Butanediol. market for formaldehyde represents 11% of its demand. It is used to
16
produce tetrahydrofuran (THF), which is used for polyurethane elastomer; -
butyrolactone, which is used to make various pyrrolidinone derivatives; poly(butylenes
terephthalate) (PBT), which is an engineering plastic; and polyurethanes.
Polyols. The principal ones include pentaerythritol, trimethylolpropane and neopentyl
glycol. These polyols find use in the alkyd resin and synthetic lubricants markets.
Pentaerythritol is also used to produce rosin/tall oil esters and explosives (pentaerythritol
tetranitrate). Trimethylolpropane is also used in urethane coatings, polyurethane foams,
and multifunctional monomers. Neopentyl glycol finds use in plastics produced from
unsaturated polyester resins and in coatings based in saturated polyesters.
The formaldehyde demands for pentaerythritol, trimethylolpropane, and neopentyl
glycol are about 7, 2, and 1% respectively, of production.
Acetal Resins. These are high performance plastics produced from formaldehyde that
are used for automotive parts, in building products, and in consumer goods. The acetal
resins formaldehyde demand are 9% of production.
Hexamethylenetetramine. Pure hexamethylenetetramine (also called hexamine and
HMTA), the production of hexamethylenetetramine consumes about 6% of the U.S.
formaldehyde supply. Its principle use is as a thermosetting catalyst for phenolic resins.
Other significant uses are for the manufacture of RDX (cyclonite) high explosives, in
molding compounds, and for rubber vulcanization accelerators. It is an unisolated
intermediate in the manufacture of nitrilotriacetic acid.
Slow-Release Fertilizers. Products containing urea-formaldehyde are used to
manufacture slow release fertilizers. These products can be either solids, liquid
concentrates, or liquid solutions. This market consumes almost 6% of the formaldehyde
produced.
Methylenebis(4-phenyl isocyanate). This compound is also known as methyl
17
diisocyanate (MDI). Its principal end use is rigid urethane foams; other end uses include
elastic fibers and elastomers. Total formaldehyde use is 5% of production.
Chelating Agents. The chelating agents produced from formaldehyde include the
aminopolycarboxylic acids, their salts, and organophosphonates. The largest demand for
formaldehyde is for ethylenediaminetetraacetic acid (EDTA); the next largest is for
nitrilotriacetic acid (NTA). Chelating agents find use in industrial and houseland cleaners
and for water treatment. Overall, chelating agents represent a modest demand for
formaldehyde of about 3%.
Formaldehyde-Alcohol Solutions. These solutions are blends of concentrated aqueous
formaldehyde, the alcohol, and the hemiacetal. These solutions are used to produce urea
and melamine resins; the alcohol can act as the resin solvent and as a reactant.
Paraformaldehyde. It is used by resin manufacturers seeking low water content or more
favorable control of reaction rates. It is often used in making phenol-urea-.resorcinol-, and
melamine-formaldehyde resins. It is EPA registered disinfectant, “Steri –dri” sanitizer and
fungicide for barber and beauty and for households, ships, bedding, clothing,
nonfood/non/feed transporting trucks.
Trioxane and Tetraoxane. It is mainly used for the production of acetal resins.
Other Applications. Formaldehyde derivatives, such as dimethyl dihydroxyethylene, are
used in textiles to produce permanent press fabrics. Other formaldehyde derivatives are
used in this industry to produce fire-retardant fabrics, Paraquat made from Pyridine
chemicals, are used for agricultural chemicals (Herbicides). Formaldehyde and
paraformaldehyde have found use as a corrosion inhibitor, hydrogen sulfide scavenger, and
biocide in oil production operations such as drilling, waterflood, and enhanced oil
recovery. Other used for formaldehyde and formaldehyde derivatives include fungicides,
embalming fluids, silage preservatives, and disinfectants.
18
Note: The requirement of formaldehyde for certain applications like Polyacetal, MDI,
1,4-Butanediol and Neopentylglycol does not exist in India.
PROCESS DESCRIPTION
Fresh methanol, which is free from iron carbonyls and sulfur compounds (catalyst poisons)
is combined with recycle methanol and diluted with equal amount of water, which is then
pumped to a evaporator, where methanol is vaporized along with water. Methanol-water
vapor mixture is then superheated to 650C in a steam superheater using low pressure
steam. Air is drawn via, a filter and compressed in a blower for feed to the process.
Filtered air is preheated with outgoing reactor effluent gases and then superheated to 650 C
in a additional superheater. Superheated air and methanol water vapor is mixed in a mixer,
the mixture is then passed into a Fixed bed tubular catalytic reactor which is nothing but a
1-1 shell and tube heat exchanger, the reactor tubes is packed with silver grains as catalyst,
where the below mentioned reactions takes place i,e formaldehyde is produced both by
oxidation and by dehydrogenation of methanol. About half of the methanol goes to each
reaction, hence the combination is net exothermic.
CH3OH + ½ O2 HCHO + H2O +38 kcal/g.mol (exothermic)
CH3OH HCHO + H 2 -20.3 kcal/g.mol (endothermic)
The conversion of methanol is 90% and the reactor pressure is kept slightly above
atmospheric. Methanol-air ratio in reactor is kept above the rich side of the explosive
limits (6.5 to 36.5 vol% in air), i.e, above 36.5%. Since the reactor temperature is to be
maintained at 650C the net exothermic heat of reaction is removed by circulating water
through the shell side of the reactor, which in turn takes away net exothermic heat of
reaction to give low pressure steam. The reactor effluent gases, at 650C is brought down
19
to 120C by preheating the feed air in a air-reactor effluent gases preheater. The reactor
effluent contains oxygen, nitrogen, hydrogen, HCHO, water, and unreacted methanol. The
effluent gases enters the packed bed absorbers (2nos.) in series where dilute formalin and
water are used for absorption, here HCHO is cooled and gets dissolved in water, heat of
solution is evolved due to absorption, assuming heat of solution to be negligible. Some
amount is required to vaporize water which passes via the vent. Heat given out by the
entering gases to reach a temperature of 25C from 120C is removed by using a coolant
which is circulated via tower external circulation. The bottom material from 2nd absorber
contains HCHO-methanol water solution which is sent to a fractionation column. The
fractionation column is sieve plate column containing 17 trays where the feed is introduced
on 8th tray from top. The column overhead temperature is maintained around 65C and
bottom reboiler temperature of 93C, methanol is condensed in the overhead condenser as
top product and recycled back to evaporator. The bottom contains 37% wt of HCHO and
less than 1% of methanol and remaining water. The water content of the bottoms is
controlled by the amount of makeup water added at the top of the absorber. The tail gas
coming out of 2nd absorber contains hydrogen which is used a source of fuel in boiler for
steam generation. The formalin product tapped out of the distillation still is cooled and sent
for storage.
RAW MATERIAL REQUIPMENTS
(For 10 Tones Per Day)
1. METHANOL - 4389.50kgs/ Day
2. AIR - 4735.2kgs/Day
3. CATALYST - 90.72kgs/day
20
mixer
Air Blower
Atmosphericair
Water
Fresh Methanol
Recycle Methanol
ABSORBER 2
Off gas
Steam
Coolant
Coolant
Fractionation Column
Reboiler
Water
Fig 2.1 Manufacture of Formaldehyde by Vapor-phase catalytic oxidation of methanol
ProcessWater
Coolant
Formalin (37% Formaldehyde)
FIXED BED TUBULAR CATALYTIC REACTORTemp -650CPressure – 1.05 atm
Methanol super heater
Filter
EVAPORATOR
Steam
Condensate
AirPreheater
Air Superheater
ABSORBER 1
Steam
21
MATERIAL BALANCE
Formaldehyde at atmospheric condition is a gas, commercially formaldehyde is dissolved
in water known as Formalin (37% wt of formaldehyde)
10 TPD of Formalin
Formaldehyde production per day, 10 0.37 = 3.7tons = 3700kgs.
Weight of water in final product per day, = 10000 – 3700 = 6300kgs.
Formaldehyde capacity per hour,
370024 = 154.167kgs
Assuming wastage of 0.1% through absorber
0 .1100
× 154 .167 = 0 . 15kgs .
Total capacity of formaldehyde per hour, 154.167 + 0.154 = 154.32kgs.
CH3 OH + 12O2→HCHO+H2O
-----(3.1) +38kcal/gmole (exothermic)
CH3 OH →HCHO+H2 -----(3.2) -20.3kcal/gmole (endothermic)
For silver process, about half of methanol goes to each reaction.
Basis: 1 hour of operation of the plant
Formaldehyde production
154 . 3230
= 5 . 144 kgmoles(MW of HCHO = 30)
Methanol required for the process = 5.144kgmole.
Conversion of methanol is 90%
Therefore Methanol fed =
5. 1440 .90
= 5 . 715 kgmole
50% Methanol for each reaction, therefore for reaction 3.1 & reaction 3.2
John Mcketta, Vol 23, Page 358
22
Methanol fed =
5. 7152
= 2 .8575 kgmoles
Oxygen required =
2. 85752
= 1 . 42875 kgmoles
Air required =
1. 428750 .21
= 6 . 8035 kgmoles
Nitrogen sent along with air = 6.8035¿ 0.79 = 5.3747 kgmole.
Since Methanol-air have explosive range of 6.7% to 36.5% (mole%), Methanol oxidation
must be brought outside this range. That is keeping methanol ratio above 36.5%
Methanol mole% =
5 .715(5 .715+6 .8035 )
×100= 45 . 65% > 36 .5%.
Material Balance around Evaporator
Methanol is diluted with water and vaporized in evaporator assuming equal quantities of
water and methanol is mixed.
Component Entering Leavingkgmoles kg kgmoles kg
Methanol 5.7155 182.896 5.7155 182.896Water 5.7155 102.879 5.7155 102.879
Material Balance around Methanol-water vapor Super heater
Here material vaporized is heated to 650C. Here there is no loss of reactant and the
material balance is same as above.
Material Balance for Air blower and Air filters
Here air from atmosphere is sucked from the blower and passed to air filter and to the
process. Here is no chemical change involved.
Component Entering Leavingkgmole kg kgmole kg
Air 6.8035 197.30 6.8035 197.30
Material Balance of Air Pre-heater
Here air is heated from atmospheric condition to the reaction temperature. Since there is no
23
change in material balance. Therefore the material balance is same as above.
Material Balance around the MixerHere methanol-water vapor and air are mixed before to the reactor.
Component Entering Leaving kgmole kg kgmole kg
Methanol 6.8035 197.30 6.8035 197.30Water-vapor 5.7155 182.896 5.7155 182.896Air 5.7155 102.879 5.7155 102.879
Material Balance around the Reactor
Conversion is 90% based on Methanol
Moles of formaldehyde formed = 5.144 kgmole.
Moles of Methanol reacted = 4.144 kgmole.
Moles of Methanol unreacted = 0.5715 kgmole.
Moles of oxygen reacted =
2. 8575×0 .92
= 1 . 2858 kgmoles
Excess oxygen = oxygen supplied oxygen reacted
= 1.4287 1.2858 = 0.14287 kgmole.
Moles of hydrogen formed = 2.8575 0.9 = 2.5717 kgmole.
Moles of water formed = 2.5717 kgmole.
Total moles of water = Moles of water vaporized in evaporator + Moles of water formed
= 5.7155 + 2.5717
= 8.28725 kgmole.
Reactor
CH3OH=5.7155kgmolesH2O =5.7155kgmolesO2=1.42875kgmolesN2=5.3748kgmoles
HCHO=5.144kgmolesCH3OH=0.5715kgmolesH2O = 8.233kgmolesO2 = 0.14287 kgmolesN2=5.3748kgmolesH2 = 2.5717kgmoles
24
Reactor Mass entering = Reactor Mass leaving
Component Entering Leaving
kgmole kg kgmole kg
CH3 OH 5.7155 182.896 0.5715 18.288
O2 1.4287 45.72 0.14287 4.5718
N2 5.3748 150.4944 5.3748 150.4944
H2O 5.7155 102.879 8.2872 149.1705
HCHO 0 0 5.144 154.32
H2 0 0 2.5717 5.1435
Total 18.2345 481.9894 22.0921 481.9882
Quenching : Waste heat boiler comes under this unit. The material goes through without
any change.
Material Balance around Absorber.
Here we assume that all the formaldehyde gets absorbed in the absorber. So the gases
leaving contains N2, O2, H2 & water vapor(traces) Assuming no, N2, O2 & H2 are absorbed.
Therefore kgmoles of vent gas on dry basis = 0.14287 (O2 ) + 5.3748 (N2 ) + 5.14
(HCHO) + 2.5715 (H2) + 0.5715 (CH3 OH )
= 13.8047kgmoles = G
Now the temperature of vent gases = 25C
Vapor pressure of water at 25C = 23.7 mmHg = X
Pressure in absorber = 760 mmHg = P
25
Moles of water present in vent gas = G
XP−X
= 13.8042
23 .7(760−23. 7 )
= 0 . 444 kgmoles = 8kgs.
Amount of water leaving with formalin solution 37% (by wt)
=
154 . 320 .37
= 417 .08 kg
Amount of formalin solution – Amount of HCHO in formalin solution
= 417.08 – 154.32 = 262.76 kgs.
Amount of water present in gas entering the absorber = 149.1705 kgs
Amount of water added to absorber =
[ Amount of water leaving ¿ ] ¿¿
¿¿–
[ Amount of water present in entering gas ¿ ] ¿¿
¿¿
262.76 + 8 – 149.1705 =121.5895 kgs.
Overall Material Balance around Absorber
1) Liquid stream entering
Component kgmole kg
Water 6.755 121.5895
2) Gases stream entering
Component kgmoles kgs
HCHO 5.144 154.32
H2O 8.2872 149.1705
N2 5.3748 150.4944
O2 0.1428 4.5718
CH3 OH 0.5715 18.288
26
H2 2.5717 5.1435
Total 22.0921 481.9882
Net total input = 22.0921 + 6.755 = 28.8472 kgmoles.
Net total input = 481.9882 + 121.59 = 603.5782 kgs.
Material output (Vent gas Leaving)
Component kgmoles kg
H2O 0.444 8.0
N2 5.3748 150.4944
O2 0.1428 4.5718
H2 2.5717 5.4135
Total 8.5334 168.2097
Material Output (Solution leaving)
Component kgmoles kg
HCHO 5.144 154.32
H2O 14.6 262.76
CH3 OH 0.5715 18.288
Total 20.3155 435.368
Net output (kgmoles) = 20.315 + 8.5334 = 28.8489 kgmoles.
Net output (kgs) = 435.368 + 168.2097 = 603.5777 kgs.
Material Balance around Distillation column
27
Fractionation ColumnHCHO = 5.144kgmoleWater = 14.6kgmolesMethanol = 0.5715kgmolesFEED= 20.3155
Methanol =0.5658kgmoleHCHO = 0.005144kgmoleDISTILLATE = 0.57094
Methanol =0.005715kgmoleHCHO = 5.1389kgmoleWater = 14.6kgmoleRESIDUE = 19.745
1) Feed Entering
Component kgmole kg
HCHO 5.144 154.32
H2O 14.6 262.76
CH3 OH 0.5715 18.288
Total 20.3155 435.368
28
2) Distillate leaving
Component kgmoles kg
CH3 OH 0.5658 18.105
HCHO 0.005144 0.154
Total 0.56631 18.259
3) Residue
Component kgmoles kg
HCHO 5.1389 154.167
CH3 OH 0.005715 0.183
H2O 14.6 262.76
Total 19.7446 417.11
Total mass entering (Feed) = Total mass leaving (Distillate + Residue)
435.368kgs = 435.369 (18.259 + 417.11)
29
ENERGY BALANCE
Heat Balance around Evaporator
Here raw material methanol and water is vaporized. Methanol gets heated from 25C to its
B.P 64.7C and then vaporizes at the same temperature.
Quantity of heat required = mCP ΔT+mλ
To calculate for Methanol using Kistya kowsky equation (considering methanol to be a
non-polar liquid)
λbTb
=8 .75+4 . 571 log 10T b (T b = 337.9K)
λbTb
=8 .75+4 . 571 log(337 . 9)=20 .34
b = 20.34 337.9 = 6860 cal/gmole = 28722.82J/gmole
Cp of Methanol between 25C and 64.7C
Cp=a+ b
2(T2+ T1 )+
c3 (T 2
2+T 1T 2+T12)
a = 4.55; b = 2.186 10-2; c = -0.291 10-5
Cp=4 .55+2 .186×10−2
2(337 .7 +298 )+-0 . 291×10-5
3(337 .72+298×337 .7+2982)
Cp = 11.205 cal/gmole.K = 46.91J/gmole.K
q1 = heat required to raise the temperature of methanol from 298K to 337.9K
q1=mCP dT = 5.7155 103 gmole 46.91J/gmole.K 39.7K
= 10.6453 106 J.
q2=mλ= 5.7155 103 gmole 28722.82J/gmole
= 164.1653 106J
Total heat added = q2+q2
= 10.6453 106 J + 1641653 106 J = 174.810 106 J
30
Heat required for water to vaporize = mCP ΔT+mλ
= 5.7155 103 4.18 103 (100-25) + 5.7155 2185 103
= 1.8043 109 J
Heat balance:
Heat required to vaporize = Heat given away by steam
Amount of steam required assuming steam available at 20psia
Enthalpy of steam = 2724 kJ/kg
Mass of steam required, (mS ) =
qλ =
174 .810×106+1 . 8043×109
2724000
mS = 726kg at 3bar
Heat balance of Pre-heater for Methanol vapors
Here gaseous vapors of methanol and water are heated from 64.7C to 650C (337.9K
to 923.2K)
Cp=a+ b2(T2+ T1 )+
c3 (T 2
2+T 1T 2+T12)
Cp=4 .55+2 .186×10−2
2(923 +337 .7 )+-0 . 291×10-5
3(9232+337 .7×923+337 .72 )
= 4.55 + 13.78 – 1.235 = 17.099 cal/gmole.K = 71.5935J/gmole.K
Heat supplied to superheat for Methanol =mCP ΔT
= 5.7155 103 71.5935 (923.2 – 337.9)
= 239.500 106 J
Cp of water = 36.845J/gmole
Heat supplied to super heater from 100C to 650C =mCP ΔT
= 5.7155 103gmole 36.845J/gmoleK (550) K = 11.5825 107J
Since heat required to preheat = heat given away by steam
Amount of steam required assuming steam available at 3 bar
31
Enthalpy of steam = 2724 kJ/kg
Mass of steam required (mS ) =
q(methanol+water )λ (steam )
=
239 .5005× 106+ 11. 5825 × 107
2724000
mS = 130 kgs.
Heat balance around Air Pre-heater
Here atmospheric air at 25C is heated to 650C (reaction temp)
Q =mCP ΔT
= 6.8035 103
(29 .1917+35 .388 )2 (650 -25) = 137.30 103 KJ
Heat required by air = Heat given away by steam
Amount of steam required assuming steam available at 3 bar
Enthalpy of steam = 2724 kJ/kg
Mass of steam required, (mS ) =
137 .292×106
2724000 = 51Kgs.
Heat balance around Mixer
Heat entering through inlet stream = Heat leaving out through outlet
Energy/Heat balance around the Reactor
The raw material Methanol + air + Water vapor enters the reactor at 650C and passes
through tubular bed of catalyst (silver grains). The reaction to form formaldehyde is
exothermic withheat of reaction (38 – 20.3) kcal/mole i.e, 17.7 kcal/mole
Moles of product formed from reaction are = HCHO + H2O + H2
= 5.144 + 2.5715 + 2.5715 = 10.2871 kgmole
Heat of reaction = 10.287 103 gmole 17.7 103 cal/gmole = 1.82 108 cal.
= 762.034 103 KJ
32
Heat leaving the reactor = Heat entering with reactant + Heat of reaction
= 239.500 103 (CH3 OH) +11.5825 104J (H2O) +137.30 103 KJ (Air) + 762.034 103
= 1.254 106 KJ
Since the reaction is exothermic and reaction temperature is to be maintained at 650C all
the heat that is formed by reaction is to taken out by a arrangement of cooling system.
Heat taken out by cooler = 762.034 103 KJ
Water at 35C is used as cooling agent. The temperature is expected to reach 100C,
Cp of water = 35.756 J/gmoleK
Mass of water required, (mw ) =
qCp ΔT =
762 .034× 105 J35 .756J/gmole . K×(100-25 )K = 2.8415 105
=
2.8415× 105×181000 = 5114.7 kgs.
Energy balance of reactants through heat exchanger Pre-heater
The mixture from outlet of reactor at 650C are brought down to 100C, by exchanging
the heat to the air that is going to the reactor.
Products issuing coming out in kgmoles are; HCHO (5.144) + H2O (8.2872) + H2 (2.5717)
+ N2 (5.3748) + O2 (0.1428) + CH3OH (0.5715)
Heat for cooling the reactants from 650C to 100C
qNitrogen=(mCP)Nitrogen ΔT 5.3748 103 29.459 550 870.849 10 5J
qOxygen=(mCP)Oxygen ΔT 0.1428 103 30.4353 550
23.9156 105J
qMethanol=(mCP )Methanol ΔT 0.5715 103 71.593 550 225.036 105 J
qWater=(mCP)Water ΔT 8.2872 103 36.845 550 1679.418 105 J
qHCHO=(mCP)HCHO ΔT 5.144 103 53.9578 550 1526.574 105 J
qHydrogen=(mCP )Hydrogen ΔT 2.5717 103 29.8365 550
422.026 105 J
33
ΣQ=qNitrogen+qOxygen+qmethanol
+qWater+qHCHO+qHydrogen
474.7818 106 J
Heat required by air = 137.29 103 KJ
Heat that will be removed = 474.7818 103 –137.29 103 = 337.5 103 KJ
Energy balance around the Absorber
Here HCHO is cooled and gets dissolved in water entering at the top. Here the heat of
reaction is evolved due to absorption. Some amount of heat is required to vaporize 8kg of
water which passes through vent. Assuming heat of reaction is negligible. Some water
enters as vapors, in the product inlet stream, this amount is 149.1705 Kgs.
Considering the same water is passing out, the water to be condensed in the absorber
= 149.1705 – 8 = 41.7 kgs. = 7.843 Kgmole.
Now the heat given out by entering gas to reach the outlet temperature from 120C to 25C
qNitrogen5.3748 103 95 29.1708 14.894 106J
qOxygen0.1428 103 95 29.5183 400.64 103J
qmethanol0.5715 103 95 46.8944 2.546 106J
qWater8.2872 103 95 35.1708 27.6895 106J
qHCHO5.144 103 95 45.2405 221.081 106J
qHydrogen2.57175 103 95 28.4716
6.956 106J
Q total 74.595l 106J
Heat is removed by using a coolant
Heat Balance around Distillation column
Distillate Methanol =18.10kg/hr HCHO = 0.1543kg/hr Feed
Methanol = 18.106 kg/hr
Fractionation column
34
HCHO = 154.32kg/hrWater = 262.76kg/hr HCHO = 154.167kg/hr Water = 262.76kg/hr Residue Methanol = 0.182kg/hr Condenser
Here methanol is cooled from 64.7C to 40C
Heat in = (mλ)Methanol+(mλ)HCHO
= (18 .106×1098 .5×103 )+(0 . 1543×1001×103 )
= 20.09 106 J.
Heat out=mCP ΔT
=18.106 2508 (40-25) = 6822800J.
Overall heat balance in condenser
Heat in = Heat out of condenser + Heat removed
20.09 106 J. = 6822800J. + Heat removed
Heat removed = 19.40 106 J.
Water is used to cool the product from 64.7C to 40C
Q=mCP ΔT+mλ
19.40 106 J = m×4 .187×103×(40−25)+m×2230×103
mw = 8.46 kgs/hr.
Reboiler
Heat in bottom of reboiler = (mCP)Methanol ΔT+(mCP )HCHO ΔT+(mCP)water ΔT
0 .183×2508×(64 . 7−25)+154 .16×0 . 8×4 .18×103×(64 .7−25 )+262.8×4187×(64 .7−25)
= 64.08 106 J
Overall Heat balance around Reboiler
35
[ Heat in ¿ ]¿¿
¿¿ +
[ Heat added in ¿ ]¿¿
¿¿ =
[ Heat at top ¿ ]¿¿
¿¿ +
[ Heat loaded in ¿ ]¿¿
¿¿
0 + Heat added in reboiler = 20.09 106 J + 64.08 106 J
Heat added in reboiler = 84.17 106 J
If we use steam of 3 bar,
T = 133.5C, λ = 2163 KJ/kg.
mS λ=84 .17×106J
mS=84 .17×106
2163×103=38.91kgs .
(amount of steam required)
36
REACTOR DESIGN (PROCESS)
From Kinetic consideration
1) To find volume of reactor
Space velocity is 8000-10000 hr-1
Space velocity =
Volumetric feed rate at standard conditionVoid volume of reactor
Therefore, Void volume of reactor =
Volumetric feed rate Space velocity
Volumetric feed rate at standard condition = 18.23455 kgmole/hr
= 18.23455 22.4 m3/kgmole
= 408.454 m3/hr
Taking space velocity of 8000 hr-1
Void volume of reactor =
408 . 454 m3 /hr8000hr-1
= 0.051 m3
Porosity of packing;
Specific area of catalyst 10m2/gm with cylindrical dimension of 3mm high to 3.8mm dia of
packed tubes.
Diameter of particleDiameter of tube =
d p
d t (Diameter of tube is taken as 50mm)
For, smooth uniform catalyst, porosity is 0.36
Therefore, actual volume of reactor =
0 .051m3
0 .36 = 0.14167 m3
Taking 10% extra volume = 0.14167 1.1
37
= 0.15584 m3 0.156 m3
2) To find tubes required
Since tubes are 50mm in diameter the cross section area of each tube is
=
π4 (50 10-3) 2 = 1.9625 10-3 m2
Taking length of tubes as 1m (standard tubes)
Number of tubes, (N t ) =
Volume of reactor Volume of each tube =
0 .156m3
1. 9625 × 10-3m2×1m
= 79.49 80 tubes.
3) Steel tubing data and inner diameter of shell
Stainless steel pipes are provided for process condition O.D = 55mm, I.D = 50mm
Thickness =5mm.
Surface area/m inside = 19.6 10-4 m2
Surface area/m outside = 23.7 10-4 m2
Tube arrangement: Tubes are laid out in Triangular pitch of 1.5D i.e, 75mm, triangular
pitch is chosen for efficient heat transfer.
Minimum required area = (pitch ) 2 No. of tubes(N t )= (0.075)2 m 80m = 0.45 m2
Using 20% excess area = 1.2 0.45 m2 = 0.54m2
Shell diameter required = √ 0.54m2×4π = 0.83m (using A=
{πd2
4 })
From heat transfer consideration, reactor is 1-1 heat exchanger in which the tubes are
packed with catalyst bed.
Specification of reactor
a) Heat duty of reactor = 762.034 103 KJ.
b) Gas flow rate = 481.9894 kg/hr 482 kg/hr
c) Gas inlet temperature = 650C
38
d) Gas outlet temperature = 650C
e) Water flow rate = 248.15 kg/hr
f) Water inlet temperature = 25C
g) Water outlet temperature = 100C
Average properties of fluids (tube side)
Hot fluid stream gas mixture from Pre-heater
Inlet temperature = 650C, Outlet temperature = 650C
Average pressure = Atmospheric = 1.033 Kg/cm2 (absolute)
Gas flow rate = 482 Kg/hr
Average temperature = 650C
Volume at average temperature at NTP the volume is 408.5m3/hr
=
923 × 408 . 454273 = 1380.1 m3
Average density of gas = 1kg/ m3
Average viscosity of gas = 0.126kg/m.hr
Average thermal conductivity, k = 203 w/m.K
Average properties of fluids (shell side)
Stream is sent through the shell
Cold fluid stream = water
Inlet temperature = 25C, Outlet temperature = 100C
Average pressure = 1.033 Kg/cm2 (absolute)
Water flow rate = 248.15 m3 /hr
Average density = 1000 kg/m3 Average Cp = 4.187 KJ/kg.K
Average viscosity = 1Kg/m.hr, Average thermal conductivity, K = 2445.208 w/m.K
39
Log mean temperature, Tm =
Δt 2−Δt 1
[ ln Δ t2Δt 1 ]
Hot fluid; Inlet = 650C, Outlet = 650C
Cold fluid; Inlet = 25C, Outlet = 100C
Δt1 = 650 – 100 = 550C
Δt2 = 650 – 25 = 625C
Tm =
625−550
[ ln625550 ]
= 586C
4) Heat transfer coefficient
Hot fluid (Tube side)
Flow area (a t ) = Number of tubes (N t ) cross sectional area of each tube
= 80 19.6 10-4 = 0.1568 m2
Mass flow rate, (Gt ) =
Mass flow rate of gases (G )tube side flowarea (at ) =
482kg /hr0 .1568m2
= 3073.98 kg /m2.hr
For packed tubes with air flowing in inner side Heat transfer coefficient is given by
h p=3. 5[ kd t ] [d P .Gt
μ ]0 .7
e−4 .6[ dpdt ]
[ dP .Gt
μ ]0.7
= [ 3 . 8×10×3073 .98
0 .126 ]0.7
= 23.82
e-4 . 6×( 3. 8
50 ) = 0.7049
h p = 3.5
203
50 × 10-3 23.82 0.7049 = 238.60 KJ/hr.m2.K
hid = hp [IDOD ] = 238.60
[5055 ]
= 216.90 KJ/hr.m2.K
(Page 290 eqn. 10-10 Chemical engineering kinetics by J.M Smith 1st edition)
40
Shell side Heat transfer coefficient
Cold fluid:
Flow area, (a
'S )=
shell dia ( I . D)×baffle spacing(B )×clearance(C ' )pitch(Pt )
,m2
a'S=0 .83m×20×10−3 m×0 .15m
75×10−3m = 0.0332 m2
G's=
mass flowrate of water (m)
Shell side flow area( a's )
= [5114.7kg/hr
0 .0332m2 ] = 154057.23 kg/hr.m2
Reynolds No. NRe=
DeGs'
μ
De = 4
(0 .5Pt×0 .86Pt−0 .5×π×D2
4)
0 .5×π Do=
57.04 mm
NRe = 57.04 10-3 154057.23 = 8787.42
Assuming there is no considerable change in Viscosity
ho= jHkD e
[Cpμ ]0.33 [ μ
μw ]0.14
When NRe = 8787.42 jH = 50 (page 838 figure 28 Process heat transfer by D.Kern)
Cp . μk =
[4187×12445. 208 ]
0.33
= 1.194
ho = 50×2445. 208
57 .04 × 10-3×1 .195
= 2561.14 KJl/hr.m2.K
Clean overall Heat transfer coefficient = UC
UC = [ hid .hohid+ho ]
=
216 . 90×2561 .14216 . 90+2561 .14 = 199.97
(Eq 6.15b, page111, Process heat transfer by D.kern)
41
Let total dirt factor Rd = 1 .43×10−5 hr.m2.K/KJ
Design overall heat transfer coefficient, Ud =[ 1
1UC
+Rd ]=
[ 11
199 . 97+1 . 43×10−5 ]
=194.40KJ/hr.m2.K
Area required, (A) = [ qU d . Δt ln ]
=
762 .034×103 KJ194 . 40 KJ /m2 K×586 K = 6.7m2
5) Pressure drop calculation
Tube side pressure drop
ΔPL =
0 .4G2agρ∈2 [ Gt
aμ ]−0 .1
Where, Gt = mass velocity(kg/hrm2), ∈ = free space or porosity
g = gravity constant(m/s2), μ = viscosity of gases(kg/ms)
a = specific surface of bed m2/m3 = 6(1-∈)d p
a = 6
(1-0 .36 )3. 8 × 10 -3
= 1010.5 m2/m3
ΔPL =
0 .4×(3073 .98)2×1010 .5
1. 27×108×1×! (0 .36 )3×3073 . 98
1010 . 5×0. 126 = 26.7
P = 26.7 1 = 26.7 Kg/cm2 (L = 1mts)
Pressure drop on shell side
NRe=
DeGS'
μ=
57 .04×10−3×154057 .231 = 8787.4
Therefore f = 2.7 10-5 (page 836,figure 26, Process heat transfer by D.Kern)
Number of cross, N + 1 =
LB
Page 193, eqn, 24, Reaction kinetics for chemical engineers by Stanley Walas
42
Where, L = tube length (m), B = baffle space (m)
=
10 .15 = 6.67
N + 1 = 7
Ps =
fGs2 D(N+1)2g . ρ Deφs ; φ s =
μμw = 1
Where, f = friction factor dimensionless
=
2. 7×10−5×(1 .54×105 )2×0 . 83×7
2×1.27×108×57 . 04×10−3×1
Ps = 2.57 10-4 Kg/ m2
(Page 147 eqn. 7.44 Process heat transfer by D.Kern)
43
MECHANICAL DESIGN OF REACTOR
As temperature in reactor is 650C, we consider working stress at same temperature
Taking factor of safety = 3
Therefore working stress for carbon steel = 200Kg/cm2
Design pressure = 1.2 times of 1Kg/cm2 (20% extra)
Tube side
1) Thickness of tube
Minimum thickness is given by, t =
pD2 f
+C
P = average operating pressure = 1Kg/cm2
D = diameter of tubes = 50mm
C = corrosion allowance = 3mm
f = working stress = 200Kg/cm2
t =
1. 2×502×200
+3 = 4.15mm = 5mm
2) Tube sheet thickness
t=FGc√ 0 . 25Pf P =
1. 013×105×1. 2(103 )2
=0 .1026 N /mm2
Where, F = 1 for most Heat exchanger except U type
44
Gc = mean diameter of gasket
F = allowable stress at appropriate temperature = 100N/mm2; SS IS grade-10
t = effective thickness of tube sheet
t=1×835√ 0 .25×0 .1026100 = 14mm
3) Channel and Channel cover
Thickness of channel portion
t c=G c√ kpf
Where, k = 0.3 for ring type gasket, Gc = mean gasket dia for cover = 835mm
f = permissible stress at design temperature 95N/mm2
th=835√ 0 . 3×0 . 12695
=25 .7 mm≈26mm
4) Flange diameter (between tube sheet and channel)
Gc = 835mm, Ring gasket width = 22mm
bo=W8
=228
=2 .75mm, ya=126 .60 N /mm2
, m = 5.5, b = bo
W m1=π Gby=π×835×2. 75×121 . 6
= 912814.5N
W m2=2 π bGmP+ π4G2P
W m2=2×π×2 .75×835×5 .5×0. 1026+ π4
(835 )2×0 . 1026
= 64292.7N
Am1=W m1
permissible stress for bolt material(140 .6N/mm2 )
45
Am1=912814 . 5140 . 6
=6520mm2
Am2=64292 .5140 . 6
=457 .3mm2
Am2= N b¿
πD 2
4
No. of bolts, Nb=
83510×2.5
= 34 bolts
Diameter of bolt = √ 4× Am2
Nb ×π =√4 × 457 . 3
34 ×π = 5mm
Minimum pitch circle dia = outside of gasket + 2 ¿ dia of bolts + 22mm
= 835 + 2 x 12 +22 = 881mm
Take, B = 940mm
5) Flange Thickness, tf = G√ P
Kf+ C
K =
1
[0 .3+1 .5 Wm hG
HG ]hG = Radial distance from gas load reaction
to bolt circle
= B−G2
H = Total hydrostatic end force = /4 G2 P
= π4
(835 )2×0 . 1026
K =
1
0 .3+ 1 .5 ×912814 . 5835×56155 . 2 [940−835
2 ] = 56155.2 N
= 0.54mm
46
tf = 835×√ 1 . 5
0 . 54×95= 38mm
6) Nozzle diameter of the reactor
a) Reactant inlet Nozzle diameter (Tube side)
ρ=481 . 99
18. 24×22 . 4×1. 18×10−3 = 1.18 10-4Kg/cm3 = 11.8Kg/m3
mo=ρ Av , Where v = 2m/sec for gases, mo
= 481.88kg/hr (from material balance)
Cross sectional area (A) =
mo
ρ . v =
481 . 9911. 8×2×3600
= 56.571cm2
Therefore r = (56 . 7
π )0 .5
=4 .25 cm
Diameter = 2r = 8.5cm.
b) Product outlet Nozzle diameter (Tube side)
=
481 . 9922 .093×22 . 4×1000 = 9.74 10-4Kg/cm3
Cross sectional area =
mo
ρ . v =
481 . 999 .74×2×3600
= 68.73cm2
Therefore r = (68 . 73
π )0.5
=4 .68cm
Diameter = 2r = 9.36cm
b) Shell side
1) Thickness of shell
Minimum thickness of shell is given by, t =
pD2 fe
+C
P = average operating pressure = 1Kg/cm2
(Page 48, eqn 1.123 Process equipment design by D.Dawande)
47
D = diameter of shell = 3.8m
f = working stress = 200Kg/cm2
C = corrosion allowance = 3mm
e = weld joint efficiency = 0.6
t =
1. 2×832×200×0 .6
+0 . 3= 0.415 +0.3 = 0.715cms = 1cm
2) Flange thickness (shell side)
Gasket diameter (G) =
Shell I . D + Shell O. D2
=830+8402
=835mm
Gasket width (N) = 24mm
bo=N2
=242
=12mm
b=2 .5√12=8 .64mm
Where, b = effective gasket seating width
b0 = basic gasket seating width
Gasket factor (m) = 3.75 (flat metal jacket asbestos filled)
Seating stress(y) = 53.4 N/mm2
W m1=π Gby=π×0 .835×103×8 .64×53 .40
= 1209681.73N
W m2=2 π bGmP+ π4G2P
W m2=2×π×8 . 64×0. 835×103×3 .75×1. 2+ π
4(0 . 835×103)2×1 .2
= 860664.89N
Flange Thickness, tf = G√ P
Kf+ C
(Table 5.4 page 129
“Process equipment design
by M.V.Joshi)
48
K =
1
[0 .3+1 .5 Wm hG
HG ]hG = Radial distance from gas load reaction
to bolt circle
= B−G2
H = Total hydrostatic end force = /4 G2 P
= π4
(0 .835×103)2×1 .2
K =
1
0 .3+ 1 .5 ×860664 .89835×656785 .95 [940−835
2 ] = 656785.95N
=2.36
tf = 835×√ 1 .5
2 .36×95= 61mm
3) Head of reactor
I.D of shell = 830mm
O.D of shell = 840
For normal pressure, torispherical heads are used
Here crown radius, L = O.D of shell = 840mm
Knuckle radius = 0.6 Crown radius = 504mm
Thickness of head is given by t =
0 .885 PLfe−0 .1 P
+C
t =
0 . 885×1 . 2×840200×0 .6−0 . 1×1 .2
+3 = 7.44 + 3 = 10.44mm
Therefore thickness of head is taken as 1cms
4) Baffle
For baffle spacing of 0.15m and inside shell
diameter of 0.83m the baffle thickness is 6.5mm
(Page 51, table1.10 Process equipment design by S.D Dawande Vol 2)
49
5) Size and number of tie rods and spacers
For shell diameter of 0.83m, number of tie rods = 6
Minimum diameter of tie rods = 12.5mm
6) Nozzle diameter of the reactor
a) Diameter of cold water nozzle inlet (Shell side)
mo=ρ Av , Where v = 2m/sec for gases, mo
= 5114.7kg/hr (from material balance)
Cross sectional area =
mo
ρ . v =
5114 .71000×2×3600 = 7.10cm2
Therefore r = ( 7 .10
π )0. 5
=1 .5cm
Diameter = 1.5 2 = 3cm
b) Diameter of hot water nozzle outlet (Shell side)
Since the water is hot in this case and expanded the diameter has to be more. So diameter
of 6cm (2cold water inlet dia) is taken.
(Page 51, table1.8, Process equipment design by S.D Dawande Vol 2)
50
PROCESS DESIGN FOR DISTILLATION COLUMN
Equilibrium data of methanol in formalin solution
Table 4.1 T-x-y data
T 93 92.5 87.4 83.7 79.1 76.9 74.4 72.1 69.3 67.1 64.3
X 0 0.06 0.126 0.19
9
0.279 0.369 0.4704 0.5844 0.7142 0.864 1.0
Y 0 0.2655 0.44 0.55
4
0.6414 0.710
5
0.775 0.839 0.884 0.9445 1.0
Source: Vapor-liquid equilbria of formaldehyde- methanol-water
by S.J. Green & Raymond.E.Vener, Industrial & engineering chemistry, 1955
Page103 to 108, Volume 47 no 1,
D = 0.5709kgmol/hrxD = 0.99
Methanol =0.5658kgmoleHCHO =
51
Fractionation Column
F = 20.3155kgmol/hr ZF = 0.02813
HCHO = 5.144kgmoleWater = 14.6kgmolesMethanol = 0.5715kgmoles
W = 19.745kgmol/hrxw = 0.000289Methanol =0.005715kgmoleHCHO = 5.1389kgmoleWater = 14.6kgmoleRESIDUE = 19.745
Glossary of notations used:
F = molar flow rate of Feed, kmol/hr, D = molar flow rate of Distillate, kmol/hr.
W = molar flow rate of Residue, kmol/hr.
zF = mole fraction of methanol in liquid/Feed.
xD , = mole fraction of methanol in Distillate, xW = mole fraction of methanol in Residue.
Rm = Minimum Reflux ratio, R = Actual Reflux ratio
L = Molar flow rate of Liquid in the Enriching Section, kmol/hr.
G = Molar flow rate of Vapor in the Enriching Section, kmol/hr.
L = Molar flow rate of Liquid in Stripping Section, kmol/hr.
G = Molar flow rate of Vapor in Stripping Section, kmol/hr.
q = Thermal condition of Feed
D = 0.5709kgmol/hrxD = 0.99
Methanol =0.5658kgmoleHCHO =
52
ρL = Density of Liquid, kg/m3, ρV = Density of Vapor, kg/m3.
qL = Volumetric flow rate of Liquid, m3/s, qV = Volumetric flow rate of Vapor, m3/s
μL = Viscosity of Liquid, cP
Preliminary calculations:
zF=0 .5715
0 .5715+14 . 6+5 . 144 = 0.02813
xD= 0 .5658
0. 5658+5 .144×10−3 = 0.99
xW= 5 .715×10−3
5 .1389+14 .6+5. 715×10−3 = 0.000289
Marking zF , xD , xW on the x-y graph and assuming liquid enters as saturated (q = 1)
From graph (Figure 1)
xD
Rm+1 = 0.15
0 .990 .15
=Rm+1
Rm= 6.6-1 = 5.6
Assuming reflux ratio of 1.5 times of Rm
R = Rm 1.5 = 5.6 1.5 = 8.4
Now
xD
R+1= 0 . 99
8 .4+1 = 0.105
Plotting the operating line and feed, the number of theoretical plates = 11 (from graph 1)
Number of trays in enriching section = 7
Number of trays in stripping section = 4 (including re-boiler)
Therefore total number of trays = 10 (without re-boiler)
Flow streams
L = Lo = R.D
53
= 8.4 0.57094
= 4.796kgmole
G = L + D = RD + D = (R+1) D = (8.4 +1)0.57094 = 5.367kgmole
(q =1 for saturated liquid)
L = L+qF (q = 1)
= 4.796 +1 20.3155
= 25.115kgmole
G = G + (q-1) F
G = G (Since q = 1 )
= 5.367kgmole
Table 4.2
List of Parameters used in calculation
Consider 4 points in the column, top and bottom of enriching section and top and bottom
of stripping section.
Enriching Section Stripping Section
Top Bottom Top Bottom
Liquid(kgmole/hr) L = 4.796 4.796 25.115 25.115
Vapor(kgmole/hr) G = 5.367 5.367 5.367 5.367
X 0.99 0.02813 0.02813 0.00028
Y 0.99 0.9719 0.9719 0.00028
M liquid kg/kmole31.98 21.43 21.43 21.13
M vap kg/kmole
31.98 31.47 31.47 21.13
Liquid (kg/hr) 153.38 102.78 538.21 530.68
54
Vapor (kg/hr) 171.64 168.9 168.9 113.40
TC liquid 64 91 91 93
TC Vapor 65 92 92 93
L (kg/m3) 793.98 995.08 995.08 1000.9
G (kg/m3) 1.147 1.15 1.15 1.2
L/G( ρG
ρL)0 .5 0.034 0.0303 0.159 0.162
a) Design of Enriching section
Plate hydraulics,
The design of a sieve plate tower is described below. The equations and correlations
are borrowed from the 6th edition of Perry’s Chemical Engineers’ Handbook.
1) Plate spacing, (tS ) = 400mm
2) Hole diameter, (d L ) = 5mm
3) Hole pitch (triangular),(LP) = 3d L = 15mm
4) Tray thickness, (t r ) = 0.6d L = 3mm
5) Plate diameter (DC ) (Plate hydraulic table)
L/G( ρG
ρL)0 .5
= 0.162 (maximum value)
For tS = 16”,CSb flood = 0.23
We have
Unf =CSb flood ( σ20 )
0 . 2
( ρL− ρG
ρG)0 .5
{eqn. 18.2, page 18.6, 6th edition Perry.}
Where,
55
Unf = gas velocity through the net area at flood, m/s
CSb flood = capacity parameter, m/s (ft/s, as in fig.18.10)
σ = liquid surface tension, mN/m (dyne/cm.)
ρL = liquid density, kg/m3 (lb/ft3)
ρG = gas density, kg/m3 (lb/ft3)
σMixture= σ Methanol −σWater
2
σ14=[ P ] ( ρL−ρG ) ---( eqn. 3-151, page 3-288, table 3-343, 6th edition Perry.}
(at low pressure, where ρL>> ρG , so neglect ρG )
σ14Water=[P ] ( ρL ) =
σ14Water=[ 51 ]( 1
18 )
σWater = 64.56 dyne/cm
σ14Methanol=[ 85 .3 ]( 0 . 792
32 ), σMethanol = 19.86 dyne/cm
σMixture= σ Methanol −σWater
2 =
19 . 86−64 . 452 = 42.16dynes/cm
Flooding velocity Unf = 0.23(42. 1620 )
0 .2
(1000 . 9−1 . 21 . 2 )
0.5
= 7.706 ft/sec = 2.348m/sec
Let us take Un = 80% of Unf
0.80 2.348 = 1.86m/sec
Now,
Net area available for gas flow, (An )
Net area = (Column cross sectional area) - (Down comer area.)
An = Ac - Ad
56
Volumetric flow rate of vapor (at top of stripping section)
=
168 . 93600×1. 86
= 0. 0252m3 /sec
An =
0 .02521.86
= 0 . 01356m2
Ad = down comer area can be taken 10-12% of Ac (let us take Ad= 11%Ac )
Ac = ( π4 )Dc2
= 0.785DC 2
Ad = 0.7850.11DC 2 = 0.0864DC 2
DC 2 = 0.0194m2 = 0.14m
Taking DC = 0.2m
Ac = 0.7850.22 = 0.0314m2
Ad = 0.0864 Dc2 = 0.08640.22 = 3.45610-3m2
Active area; Aa = Ac - 2 Ad
= 0.0314-23.45610-3 = 0.0245m2
6) Perforated area, (Ap )
Let Lw /DC = 0.75 (Lw= weir length)
Where, Lw = weir length, m
DC = Column diameter, m
Lw = 0.750.2 = 0.15m
θ=2×Sin−1( LwDc ) = 2 Sin
−1(0.75) = 97.18
σ = 180 – 97.18 = 82.82
Periphery waste = 50mm = 5010-3 m
57
Area of periphery waste AWZ = 2 {π4 Dc2 ( α
360 )− π4
(Dc−0 .02 )2( α360 )}
= 2 {π4 0 . 22(82. 82
360 )−π4
(0 .2−0 . 02 )2(82 . 82360 )}
= 0.01186m2
Acz = area of calming zone, m2 = 2 LW 50 10-3
= 2 0.15 20 10 -3 = 6 10 -3 m2
Ap=Ac−2 Ad−Acz−Awz
Ap = 0.0314 – 2 3.456 10-3 – 6 10-3 – 0.0107
= 7.788 10 -3 m2
7) Total Hole area,(Ah ):
Since,
Ah
Ap = 0.1
Ah = 0.1 7.788 10 -3
= 7.788 10 -4 m2
Now we know that,
Ah=nh( π4 )dh2
Number of holes, Nh =
7 . 788×10−4
π4×(5×10−3 )
= 40
8) Weir height,(hw ):
Let us takehw= 50mm
9) Weeping check
Head loss across the dry hole is
58
hd=K1+K 2( ρG
ρL)Uh
2
---- (eqn. 18.6, page 18.9, 6th edition Perry)
Where, Uh = gas velocity through hole area
K1 , K2 are constants
Volumetric flow rate of vapor =
168 . 93600×1. 86 = 0.0252 m3/sec
Uh=
0 .0252
7 .788×10−4 = 32.37 m/sec
For sieve plateK1 = 0 ; K2 = 50.8/Cv2
Where, Cv =discharge coefficient, taken from fig 18.14, page 18.9 6th edition Perry.
Holeareaactivearea
=Ah
Aa
=7 .788×10−4
0 . 0245 = 0.032
Tray thicknessHole area
=t Rd L
=35 = 0.6
Thus for (Ah/Aa) = 0.032 and
tRdL = 0.60
We have from fig. edition 18.14, page 18.9 6th Perry. CV (discharge coefficient) = 0.7
K2=50 .8
(0.7 )2=103 .67
hd=103 .67×( 1 . 151000 .2 )×(32 . 37 )2
= 124.9 mm of liquid
Height of liquid crest over weir
hOW=44300 FW ( q '
LW)0 .704
Thus, q = liquid flow rate, m3
/s; Lw = weir length
59
Liquid load q '=538 .82
( 995.08×3600 )=1 .504×10−4m3 /sec
hOW=44300( 1 .504×10−4
150 )0 .704
=2 .65mm
Head loss due to bubble formation (hσ ) is given by (page 18-17, equation 18-2a )
hσ=409σ
(ρL×dh)=(409×42. 16
995 . 08×5 )= 3.47mm liquid
Where,
σ =surface tension, mN/m (dyne/cm) = 42.16 dyne/cm.
dh = diameter of perforation, (mm) = 5mm
ρL = density of liquid in the section,(kg/m3) = 995.08 kg/m3
hd+ hσ = 124.9 + 3.47 = 128.37mm
hW + hOW = 50 + 2.65 = 52.65mm
Ah
Aa
=0 .032
From figure 18-11, 6th edition Perry, hd+ hσ = 10mm
Since design value of hd + hσ is well above, the value obtained from graph, no weeping
occurs.
8) Check for down comer flooding
Down comer back up is given by
hdc=ht+hw+how+hda+hhg --- (eqn 18.3, page 18.7, 6th edition Perry)
a) hydraulic gradient across the plate hhg for stable operation hd>2.5hhg
For sieve plates hhg is generally small or negligible, let us take hhg = 0.5mm of liquid
60
b) Total pressure drop across the plate, (ht );
ht=hd+hl'
Now, hl'=β×hdS ---- (eqn. 18.5, page 18.9, 6th edition Perry)
Where,hl'
= pressure drop through the aerated liquid (mm)
β =Aeration factor, hds = Calculated height of clear liquid over the dispersers, (mm)
hds=hw+how+hhg
2 ----(eqn. 18.10, page 18.10, 6th edition Perry)
Where, how = height of crest over weir equivalent clear liquid, (mm)
hhg = hydraulic gradient across plate, height of clear liquid column, (mm)
hds=50+2 .65+ 0 . 5
2 = 52.9mm
To find β
Now, Fga=U a (ρg )12
Where
Fga = gas-phase kinetic energy factor,Ua = superficial gas velocity, m/s (ft/s),ρg = gas density, kg/m3
(lb/ft3)
Ua =
168 . 93600×1. 15×0. 0245 = 1.665m/s
Fga = 5.46¿ (1 .15×10−3×62 .4 )0. 5= 1.463
From (figure 18-15, page 18-10, 6th edition Perry) β = 0.59
hl'
= 0.59 52.9 = 31.211
ht=hd+hl'
= 124.9 + 31.211 = 156.11 mm
61
c) Loss under down comer, (hda)
hda=165 .2( qAda )
2
----- (eqn. 18.19, page 18.10, 6th edition Perry)
Where, hda = head loss due to liquid flow under down comer, apron mm liquid,
q = liquid flow rate, m3/s
Ada = minimum area of flow under the down comer apron, m2
hap=hds−c '
(Take clearance, c'= 1” = 25.4mm)
hap=52. 9−25 . 4 = 27.5mm liquid
Ada=Lw×hap=0. 15×27 .5×10−3 = 41.25 10-3 mm2
hda=165 .2( 1 .504×10−4
41 .25×10−3 )2
= 2.196 10-3 mm
Now, hdc=ht+hw+how+hda+hhg ---- (eqn 18.3, page 18.7, 6th edition Perry)
ht = total pressure drop across the plate (mm liquid) = hd + hl`
hdc = height in down comer, mm liquid,
hw = height of weir at the plate outlet, mm liquid,
how =height of crest over the weir, mm liquid,
hda = head loss due to liquid flow under the down comer apron, mm liquid,
hhg = liquid gradient across the plate, mm liquid.
= 156.11 + 50 + 2.65 + 2.196 + 2.196 10-3 + 0.5
= 209.2 mm of clear liquid
Actual back up with aeration; hdc' =
hdc
φdc
62
For system with low gas velocity, low liquid viscosity and low foamability we can take
φdc=0 .6
hdc' =209.2
0 .6=348. 72mm
(Which is less than the tray spacing)
tS = 400mm
Since tS >hdc'
(no down comer flooding will occur)
Column efficiency
The efficiency calculation are based on average condition prevailing in each section
Enriching section
Average molar liquid rate = 4.796 kgmole/hr
Average mass liquid rate =
153 .38+102. 782
=128. 08 kg /hr
Average molar vapor rate = 5.367 kgmole/hr
Average mass vapor rate =
171 .64+168 . 92
=170 .27 kg /hr
Average density of liquid =
793 . 98+995 . 082
=894 .53 kg/m3
Average density of vapor =
1. 147+1 .152
=1 .1485kg/m3
Average liquid temperature =
64+912
=77 . 5oC
Average vapor temperature =
65+922
=78 .5oC
Viscosity of formaldehyde at 77.50C = 0.06cp
Viscosity of Methanol at 77.50C = 0.07cp
Average viscosity of liquid is calculated using Knedall-maserol equation
63
μ13L=x1μ
131+ x1 μ
132
x1=0 . 99+0 . 02813
2=0 . 509
x2=1−x1=0.491
μmLiquid=[0 .509×(0 .07 )13 +0 .06× (0.491 )
13 ]3
= 0.0162cP
Viscosity of formalin at 78.50C = 0.05cP
Viscosity of methanol at 78.50C = 0.062cP
y1=0. 99+0. 9719
2=0 . 98
y2=1− y1=1−0. 98=0 .02
μmvapor=
∑ [ y iμ i M
i
12 ]
∑ y i M
i
12
μmvapor=0.98×0 .062×32
12+0.02×0.05×30
12
0.98×3212×0 .12×30
12
= 0.054cP
Liquid phase diffusivity
Wilke chang equation status
DL=[7 . 4×10−8×(φ×MB )0 .5×T ]
ηBVA
0. 6
Where, DL = mutual diffusion coefficient of solute A at very low conc, in solvent B,
φ = association factor of solvent B, MB = mol wt of solvent B
ηB = viscosity of solvent B, cP
T = 273+77.5 = 350.5K,
V A = solute molal volume (methanol)
64
= 16.5 1 + 1.98 4 + 5.48 = 29.9
DL=[7 . 4×10−8×(2 .6×30 )0 . 5×350. 5 ]
0 .7×(29 . 9 )0 . 6 = 4.26 10-5 cm2 /sec
Vapor phase density
Fullers equation
Dg=
10−3T 1. 75( 1M A
+ 1MB )
0 .5
P[∑V A
13 +∑V
B
13 ]
2
T = 78.5+ 273 = 351.5K
M A = 32; MB = 30; P = 1atmosphere
∑V A = 29.9; ∑V B= HCHO = 1 16.5 + 1.98 2 + 5.48 = 25.94
Dg=10−3×3511.75×( 1
32+ 1
30 )0 .5
1×[29 . 913+25. 94
13 ]
2
= 0.19cm2 /sec
N Scg=μg
ρg×Dg
= 0 . 054×10−3
1. 20×0.19×10−4 = 2.3
Stripping section
Average molar liquid flow rate = 25.115 kgmole/hr
Average mass liquid flow rate =
538 .21+530 .682
=534 . 45kg /hr
Average molar vapor flow rate = 5.367kg/hr
Average mass vapor flow rate =
168 . 9+113. 42
=141 .15 kg/hr
ρav ( Liquid )=
995 .08+1000 . 92
=997 .94 kg /m3
65
ρav (Vapor )=
1. 15+1 .22
=1 .175kg /m3
Tav (Liquid )=
91+932
=920C
Tav (Vapor )=
92+932
=92. 50C
μMethanol at 91.50C = 0.034cP μHCHO at 91.50C = 0.02cP
x1=0 . 02813+0 . 00028
2=0 . 0142
x2=1−x1=1−0 .0142=0 .9858
μm=[0 .0142×(0 .034 )13+0 .9858×(0 .02 )
13 ]3
= 0.01865cP
μ of Methanol at 92.50C = 0.032cP, μ of HCHO at 92.50C = 0.019cP
y1=0. 9719+0 . 00028
2=0 . 486
y2=1− y1=1−0 . 486=0 . 514
μmvapor=0.486×0 .032×32
13 +0 .514×0.019×30
13
0.486×3213×0 .514×30
13
= 0.0254cP
Diffusivity of liquid phase
DL=[7 . 4×10−8×(2 .6×30 )0 . 5×364 .5 ]
0 .60× (29 . 9 )0 . 6 = 2.382 10-4
= 5.169 10-5 cm2 /sec
Vapor phase diffusivity
Dg=10−3×365 .51 . 75×( 1
32+ 1
30 )0. 5
1×[29 .913+25 . 94
13 ]
2
= 0.202 cm2 /sec
66
N Scg=0 .025×10−3
1 .15×0 . 202×10−4=1 . 08
Table 4.2
Condition Enriching section Stripping section
Liquid flow rate kgmole/hr 4.796 25.115
Liquid flow rate kgmole/hr 128.08 534.45
ρLKg /m3 894.53 997.94
T L0C 77.5 92
μLcP0.0162 0.019
DLcm2 /sec 4.26 10-5 5.169 10-5
Vapor flow rate kgmoles/hr 5.367 5.367
Vapor flow rate kgmoles/hr 170.27 141.15
ρV Kg /m3 1.1485 1.175
T V0C 78.5 92.5
μV cP0.054 0.025
Dgcm2 /sec 0.189 0.202
N Scg2.3 1.08
Enriching section
a) Point efficiency, (Eog );
N g=[0 .776+0 .00457×hW−0 .238U a ρ
g0. 5+0 .0712W ]
NScg0 .5
Where,
(eqn. 18.36, page 18.15, 6th edition Perry)
67
W = Liquid flow rate, m3/ (s.m) of width of flow path on the plate,
hW = weir height = 50.00 mm
Ua = Gas velocity through active area, m/s
DL = liquid phase diffusion coefficient, = 4 .26×10−9cm2 /sec
Ua=170 .273600×1 .1485×0 .0245
=1 .68m /sec
q=128 . 08894 . 53×60
=2. 39×10−3m3 /sec
D f=( DC+LW )
2=0. 2+0 . 15
2=0 .175
W= qD f
=2 .39×10−3
60×0 . 175=2. 276×10−4 m3 /sec.m
N Scg= Schmidt number (dimensionless), N Scg=
μg
ρg×Dg
= 0 . 054×10−3
1. 20×0.19×10−4= 2.3
N g=0 . 776+0 .00457×50−0 . 238×1 . 68×1 .14850. 5+0. 0712×2.276×10−6
(2. 3 )0 . 5=0.39
Number of liquid transfer unit is given by N l=k LaθL
Where,
k L= Liquid phase transfer coefficient, m/s
a = effective interfacial area for mass transfer m2
/m3 froth or spray on the plate,
θL = residence time of liquid in the froth or spray zone, s
θL=hL Aa
1000×q ---- (eqn. 18.38, page 18.16, 6th edition Perry)
θL=31 . 211×0. 0245
1000×(2 .39×10−3 /60 )=19. 196 sec
68
K La=(3 .875×108×DL)0.5
(0 . 4×Ua×ρg0 .5+0 . 17)
K La=( 3.875×108×4 .26×10−9 )0 .5 (0 .4×1.68×1.14850 .5+0 .17 )=1 .1438 sec−1
N l=1. 1438×19.196=21. 96
Nog=1
1N g
+λN l ----- (eqn. 18.34, page 18.15, 6th edition Perry)
Where,
N l = Liquid phase transfer units,
N g = Gas phase transfer units,
m = slope of Equilibrium Curve,
Gm = Gas flow rate, mol/s (from table 4.2)
Lm= Liquid flow rate, mol/s (from table 4.2)
Nog = overall transfer units
λ=mGm
Lm = Stripping factor,
(from graph 4.1 for enriching section straight line slope is 1.4/1.6 = 0.875)
Nog=1
10 .39
+0 . 97921 . 96
=0 . 39
Eog=1−e−Nog=1−e−0. 39=0 .33 -------- (eqn. 18.33, page 18.15, 6th edition Perry)
b) Calculation of Murphree plate efficiency, (Emv );
Now, N Pe=
ZL2
DE×θL
979.0796.4
367.5875.0
m
m
L
Gm
69
ZL = length of liquid travel, m
ZL=2×[ DCCos( θ2 )2 ]
=
2×[ 0 . 2Cos(97 .182 )
2 ] = 0.1323
θL = 19.196sec
DE=6 . 675×10−3Va
1.44+0 . 922×10−4×hl−0 . 00562
=6 .675×10−3×1.681 . 44+0 . 922×10−4×31. 211−0 . 00562
= 0.01135m3 /sec
Where, DE = Eddy diffusion coefficient, m2/s
N Pc=0 .13232
0.01135×19 .196=0 . 0804
λEog=0 . 979×0. 33=0 .323
Now, for λEog = 0.323 and N Pe = 0.0804 value, (we have from fig.18.29a, page 18.18, 6th
edition Perry)
Emv
Eog
=1 .8
(Murphree plate efficiency)Emv = 1.8 x 0.33 = 0.60
c) Overall column efficiency (EOC );
EOC=N t
Na
=log [1+Ea ( λ−1 ) ]log λ ----- (eqn. 18.46, page 18.17, 6th edition Perry)
Where,
Eα
Emv
= 1
[1+Emv( ψ1−ψ )]
----(eqn. 18.27, page 18.13, 6th edition Perry)
ψ= eL+e ----(eqn. 18.26, page 18.13, 6th edition Perry)
(eqn. 18.45, page 18.17, 6th edition Perry)
70
e = absolute entrainment of liquid
L = liquid down flow rate without entrainment.
Emv = Murphee Vapor efficiency,
Eα = Murphee Vapor efficiency, corrected for recycle effect of liquid entrainment.
ψ = fractional entrainment, moles/mole gross down flow
Taking ( LV ) .( ρG
ρL)0 .5
=0.162 at 80% flooding, (we have from fig.18.22, page 18.14, 6th
edition Perry) ψ = 0.011
Ea
0 .6=1
[1+0 .6 (0 .0111−0. 011 )]
Ea=0. 596
(Actual trays)Nactual=
N T
EOC
=
Ideal trays Overall efficiency
Actual number of tray in enriching section =
70 .596 = 12 trays
Stripping section
a) Calculation of point efficiency, (Eog )
Ua=141 . 153600×1 .175×0 .0245
=1 .362m /sec
q '=534 . 45997 . 94×60
=8 .925×10−3 m3 /sec
D f=0 .175m
w=8 . 925×10−3
60×0 .175=8 .5×10−4m3 /sec .m
N Scg= 1.08
71
N g=0 . 776+0 .00457×50−0 . 238×1 . 362×1. 1750 . 5+0 . 712×8 . 5×10−4
(1. 08 )0 .5=0.63
θL=31 . 211×0. 0245
(1000×8 .925×10−3
60 )=5 .14
k L×a=( 3. 875×108×5 .169×10−9 )0 .5 (0 . 4×1. 362×1 .1750.5+0.17 )
= 1 .076sec−1
N l=1. 076×5 .14=5 . 54
Nog=1
1N g
+λN l
= 11
0.63+
1 .0685 .54
=0 .562
m=1 .5
0 .3=5
; m
Gm
Lm
=5× 5 .36725. 115
=1 . 068
Eog=1−e−Nog=1−e−0. 562=0 . 43
b) Calculation of Murphree plate efficiency, (Emv );
N Pe=Z
l2
DE×θL
DE=6 . 675×10−3×1. 3621.44+0 . 922×10−4×31 .211−0 . 00562
= 7.6810-3m3 /sec
ZL=2[ DC cosθC
22 ]=0 .1323m
θL=5 . 14
N Pe=0 .13232
7 . 68×10−3×5. 14=0 . 444
λEog=1 .068×0 .43=0 .46
Emv
Eog
=1 .6
72
Emv = 1.6 0.43 = 0.69
c) Overall column efficiency, (EOC )
ψ=0.011 (refer enriching section EOC )
Ea
0 .69=1
[1+0 .69(0 .0111−0 .011 )]
Ea=0.6852
(Actual trays)Nactual=
N T
EOC
=
Ideal trays Overall efficiency
Actual Number of trays in stripping section =
30 .6852
=5 trays
Height of enriching section = 12trays 400mm = 4800 = 4.8 mts
Height of stripping section = 5 400 = 2mts
Total height of column = 4.8 + 2 = 6.8mts.
73
INSTRUMENTATION AND PROCESS CONTROL OF REACTOR
1. FI 001 & FI 002 are flow indicator of Methanol & air respectively.
2. FV 001 & FV 002 are Methanol/ Air flow control value respectively.
74
3. TI 001, PI 001 are Temperature & pressure indicator of reactor Inlet
4. TI 002 PI 002 are Temperature & pressure indicator of Reactor outlet
5. TC 002 is Temperature controller, which is located in the Outlet of reactor,
TC 002 is the cascade controller (Master control) which controls FV 003
Water flow rate (Slave controller)
6. R/C is Ratio controller, which controls FV 001 & FV 002
7. PDI 001 is the Pressure Difference Indicator, reads the pressure between
PI 001 – PI 002
8. Safety interlocks :
SDV 01 & 02 (Control valve) to trip for below abnormalities,
a) PSH is pressure high switch, if pressure reaches more than 1.5atm
b) Also a TSH (temperature switch high) is provided at reactor outlet if
temperature reaches more than set range will trip the same SDV’s
c) If Methanol pump or Air compressor fails (or u/s) upset of plants
9. Local Pressure gauges and Temperature gauges are provided for reactor inlet
and outlet, and also a rupture disc is mounted on the head of reactor.
75
Plant Layout
Good plant layout keeps overall costs, erecting cost, safety, appearance, convenience,
operating and maintenance cost to the minimum. The unit should be planned in minimum
space. Equipments should be positioned in such a manner that the piping cost is minimum.
Feed lines, product lines, utility lines should be planned in such a way that future
expansion is easily possible. Furnaces, fire hazard, explosive units should be isolated from
other process units to avoid major hazards. It is customary that the heat exchange units are
positioned vertically while pipe lines are laid in rectangles.
Safety measures have to be installed for process equipments, raw materials storages,
personnel. Study of all these aspects are part of chemical process plant design.
The key to economic construction and efficient is a carefully planned, functional
arrangement of equipments, piping and buildings. Furthermore, an accessible and
aesthetically pleasing plot plan can make major contributions to safety, employee
satisfaction and sound community relations. The physical layout of the equipments is very
important. A modern process plant installed today shall remain in operation for 20 years or
even more. Any error in layout can prove costly at the later stage.
There cannot be any ideal plot plan because the chemical processes differ in many ways.
The following parameters have to be considered while designing a plot plan.
76
1) Scale and scope of the operation.
2) Available property limitations.
3) Safety considerations.
4) Operating supervision and labour scheduling.
5) Utilities supply.
6) Solid materials handling requirements.
7) Maintenance convenience.
8) Construction economics.
9) Possible future expansion.
Units can be placed in one of the two general forms.
1) Grouped layout
2) Flow line layout
In grouped layout similar equipments such as towers, pumps, heat exchangers etc. are
grouped together. In flow line pattern, the towers, pumps, and heat exchangers are
arranged in the layout as they appear on the flow plan.
The grouping layout is advantageous and economical for large chemical process plants,
while flow line layout is useful for small process plants or large plants having relatively
less pump or exchangers.
For designing a plant layout the following guidelines should be followed:
1) Study the process flow diagram and equipment list and find out the scope of equipment
to be included in the unit area.
2) Integrate as many process operations as possible. This helps to minimize operating staff.
3) Decide the equipment elevation. It is dependent upon the process requirement as well as
pump suction requirements. The process, project and mechanical engineers should work in
co-ordination to achieve the satisfactory operation of the process.
77
4) Make a detailed study of process flow, and operating procedure. The function of each
process equipment should be easily accessible for maintenance.
5) Study the maintenance, shutdown method for each process equipment. The equipment
should be easily accessible for maintenance.
6) Make a detailed study of operating hazards. This helps to device safest arrangement of
equipments.
7) Adequate clearance should be available between two equipments. For example, a
rectangular plan with a central overhead pipe rack permit equipments to be located along
both sides of the pipe way.
8) Locate large field fabricated equipments, such as reactor, or fractionating columns, at
one end of the plot so that the erecting staff can unload, assemble, erect, weld and test
these vessels without interfering or delaying work in the rest of the area.
The process plant should be located on one side of a tank farm while shipping, transport,
loading/unloading facilities on another side. It helps to reduce the piping length between
process units and storage tanks and between unloading points and storage tanks.
Administration and service facilities should be located near the process plant entrance and
to initially installed process units. It reduces the distance between service units,
laboratories, storehouse etc.
Warehouse, salvage yard should be close together. Cooling towers should be located where
water drift from the towers will not cause excessive corrosion of process equipment. They
should be oriented crossways to the wind direction in order to minimize recycling of air
from the discharge of one tower to the section of an adjacent tower.
Hazardous, toxic chemicals storage should be planned close to the unloading of tank car.
This minimizes line washing. Hazardous tanks should be provided with fire protection
78
walls and a clear space of one diameter between any pair of tank. Effluent treatment should
be located near the natural drain facilities.
Pumping arrangement of liquids from the tanks should be decentralized. This minimizes
damage in the event of fire.
In a process plant, there should be sufficient space between the process equipments. Its
avoids congestion after piping valves instrumentation is done on equipments. Pump and
compressor lines should be small, short as far as possible.
Furnace transfer lines should be short as far as possible. Hot lines should be long enough.
It helps flexibility. Valve stems should never be located at face level. It becomes a hazard
to operating personnel. Piping should not be a grade level in operation area. Large vertical
vessels and reactors should be spaced at minimum 3 diameters centre to centre from each
other.
While arranging pipeline network, 30 percent space should be left for future pipelines
requirement. Double deck pipelines are useful.
Process units should be located side by side with a distance of 1.5 times the plant size
between them. A centralized control house should have as many instruments as possible.
In large process plants, plant layout has its own importance. Lot of money can be saved
by having a good plant layout. A process plant designer usually talks about the production
schedule, related performance but not about the plant layout. Production schedule can be
affected by improper plant layout. Operating costs and maintenance costs are influenced by
plant layout which can ultimately affect the profitability of process plant.
79
80
PRODUCT STORAGE
AREA
RAW MATERIAL STORAGE AREA
EFFLUENTTREATMENT
SITE
MAIN GATE
PLANT LAYOUT
FUTURE EXPANSIONSBOILER SECTION
UTILITIES
SUB-STATION
WORKSHOP FIRE & SAFETY DEPT.
EVAPORATERREACTOR ABSORBER ABSORBER FRACTINATION COLUMN
PUMP HOUSE
CENTRAL CONTROL ROOM
CANTEENADMIN BLOCKMATERIALS DEPT.
MARKETING &DISPATCH AREA
81
HEALTH AND SAFETY FACTORS
Hazard identification
SAF-T-DATA(tm) Ratings
Health Rating: 3 - Severe (Poison)
Flammability Rating: 2 – Moderate
Reactivity Rating: 2 – Moderate
Contact Rating: 3 - Severe (Corrosive)
Lab Protective Equip: GOGGLES & SHIELD; LAB COAT & APRON; VENT HOOD;
PROPER GLOVES; CLASS B EXTINGUISHER
Storage Color Code: Red (Flammable)
Potential Health effects
The perception of formaldehyde by odor and eye irritation becomes less sensitive with
time as one adapts to formaldehyde. This can lead to overexposure if a worker is relying on
formaldehyde's warning properties to alert him or her to the potential for exposure.
Inhalation: May cause sore throat, coughing, and shortness of breath. Causes irritation and
sensitization of the respiratory tract. Concentrations of 25 to 30 ppm cause severe
respiratory tract injury leading to pulmonary edema and pneumonitis. May be fatal in high
concentrations.
Ingestion: Can cause severe abdominal pain, violent vomiting, headache, and diarrhea.
Larger doses may produce decreased body temperature, pain in the digestive tract, shallow
respiration, weak irregular pulse, unconsciousness and death. Methanol component affects
the optic nerve and may cause blindness.
Skin Contact: Toxic: May cause irritation to skin with redness, pain, and possibly burns.
Skin absorption may occur with symptoms paralleling those from ingestion. Formaldehyde
is a severe skin irritant and sensitizer. Contact causes white discoloration, smarting,
82
cracking and scaling.
Eye Contact: Vapors cause irritation to the eyes with redness, pain, and blurred vision.
Higher concentrations or splashes may cause irreversible eye damage.
Chronic Exposure: Frequent or prolonged exposure to formaldehyde may cause
hypersensitivity leading to contact dermatitis. Repeated or prolonged skin contact with
formaldehyde may cause an allergic reaction in some people. Vision impairment and
enlargement of liver may occur from methanol component. Formaldehyde is a suspected
carcinogen (positive animal inhalation studies).
Aggravation of Pre-existing Conditions:
Persons with pre-existing skin disorders or eye problems, or impaired liver, kidney or
respiratory function may be more susceptible to the effects of the substance. Previously
exposed persons may have an allergic reaction to future exposures.
First Aid measures
Inhalation: Remove to fresh air. If not breathing, give artificial respiration. If breathing is
difficult, give oxygen. Call a physician.
Ingestion: If swallowed and the victim is conscious, dilute, inactivate, or absorb the
ingested formaldehyde by giving milk, activated charcoal, or water. Any organic material
will inactivate formaldehyde. Keep affected person warm and at rest. Get medical attention
immediately. If vomiting occurs, keep head lower than hips.
Skin Contact: In case of contact, immediately flush skin with plenty of water for at least
15 minutes while removing contaminated clothing and shoe. Get medical attention
immediately.
Eye Contact: Immediately flush eyes with plenty of water for at least 15 minutes, lifting
lower and upper eyelids occasionally. Get medical attention immediately.
Note to Physician: Monitor arterial blood gases and methanol levels after significant
83
ingestion. Hemodyalysis may be effective in formaldehyde removal. Use formic acid in
urine and formaldehyde in blood or expired air as diagnostic tests.
Fire fighting measures
Fire: Flash point: 60C (140F) CC Autoignition temperature: 300C (572F)Flammable limits in air % by volume: lel: 7.0; uel: 73
Flammable liquid and vapor! Gas vaporizes readily from solution and is flammable in air. Explosion: Above flash point, vapor-air mixtures are explosive within flammable limits
noted above. Containers may explode when involved in a fire. Fire Extinguishing Media: Water spray, dry chemical, alcohol foam, or carbon dioxide.
Special Information: In the event of a fire, wear full protective clothing and NIOSH-
approved self-contained breathing apparatus with full facepiece operated in the pressure
demand or other positive pressure mode. Water may be used to flush spills away from
exposures and to dilute spills to non-flammable mixtures.
Exposure controls/Personal protectionAirborne Exposure Limits:
-OSHA Permissible Exposure Limit (PEL):
0.75 ppm (TWA), 2 ppm (STEL), 0.5 ppm (TWA) action level for formaldehyde
200 ppm (TWA) for methanol
-ACGIH Threshold Limit Value (TLV):
0.3 ppm Ceiling formaldehyde, Sensitizer, A2 Suspected Human Carcinogen
200 ppm (TWA) 250 ppm (STEL) skin for methanol
Personal Respirators (NIOSH Approved):
If the exposure limit is exceeded and engineering controls are not feasible, a full facepiece
respirator with a formaldehyde cartridge may be worn up to 50 times the exposure limit or
the maximum use concentration specified by the appropriate regulatory agency or
respirator supplier, whichever is lowest. For emergencies or instances where the exposure
84
levels are not known, use a full-face piece positive-pressure, air-supplied respirator.
WARNING: Air purifying respirators do not protect workers in oxygen-deficient
atmospheres. Irritation also provides warning. For Methanol: If the exposure limit is
exceeded and engineering controls are not feasible, wear a supplied air, full-facepiece
respirator, airlined hood, or full-facepiece self-contained breathing apparatus. Breathing air
quality must meet the requirements of the OSHA respiratory protection standard
(29CFR1910.134). Where respirators are required, you must have a written program
covering the basic requirements in the OSHA respirator standard. These include training,
fit testing, medical approval, cleaning, maintenance, cartridge change schedules, etc.
Skin Protection: Wear impervious protective clothing, including boots, gloves, lab coat,
apron or coveralls, as appropriate, to prevent skin contact.
Eye Protection: Use chemical safety goggles and/or a full face shield where splashing is
possible. Maintain eye wash fountain and quick-drench facilities in work area.
Accidental release measures
Small Spill: Dilute with water and mop up, or absorb with an inert dry material and place
in an appropriate waste disposal container.
Large Spill: Flammable liquid. Keep away from heat. Keep away from sources of ignition.
Stop leak if without risk. Absorb with DRY earth, sand or other non-combustible material.
Do not touch spilled material. Prevent entry into sewers, basements or confined areas; dike
if needed. Be careful that the product is not present at a concentration level above TLV.
Check TLV on the MSDS and with local authorities.
Handling and Storage
Precautions: Keep away from heat. Keep away from sources of ignition. Ground all
equipment containing material. Do not ingest. Do not breathe gas/fumes/ vapor/spray. In
case of insufficient ventilation, wear suitable respiratory equipment. If ingested, seek
85
medical advice immediately and show the container or the label. Avoid contact with
skin and eyes. Keep away from incompatibles such as oxidizing agents, reducing agents,
acids, alkalis, moisture.
Storage: Store in a segregated and approved area. Keep container in a cool, well-ventilated
area. Keep container tightly closed and sealed until ready for use. Avoid all possible
sources of ignition (spark or flame).
Stability and Reactivity
Stability: Stable under ordinary conditions of use and storage. Hazardous Decomposition Products: May form carbon dioxide, carbon monoxide, and
formaldehyde when heated to decomposition. Hazardous Polymerization: Trioxymethylene precipitate can be formed on long standing
at very low temperatures. Nonhazardous polymerization may occur at low temperatures,
forming paraformaldehyde, a white solid.
Incompatibilities: Incompatible with oxidizing agents and alkalis. Reacts explosively with
nitrogen dioxide at ca. 180C (356F). Reacts violently with perchloric acid,
perchloricacid-aniline mixtures, and nitromethane. Reaction with hydrochloric acid may
form bis-chloromethyl ether, an OSHA regulated carcinogen.
Conditions to Avoid: Heat, flames, ignition sources and incompatibles
Environmental Impact and Pollution control
Formaldehyde is present in the environment as a result of natural processes and from
manmade sources. It is formed in large quantities in the troposphere by the oxidation of
hydrocarbons. Minor natural sources include the decomposition of plant residues and the
transformation of various chemicals emitted by foliage.
Formaldehyde is produced industrially in large quantities and used in many
applications. Two other important man-made sources are automotive exhaust from engines
86
without catalytic converters, and residues, emissions, or wastes produced during the
manufacture of formaldehyde or by materials derived from, or treated with it.
When released into the soil, this material is expected to leach into groundwater. When
released into water, this material is expected to readily biodegrade. While formaldehyde is
biodegradable under both aerobic and anaerobic conditions. When released into water, this
material is not expected to evaporate significantly. This material is not expected to
significantly bioaccumulate. When released into the air, this material is expected to be
readily degraded by reaction with photochemically produced hydroxyl radicals. When
released into the air, this material is expected to be readily degraded by photolysis. When
released into the air, this material is expected to be readily removed from the atmosphere
by dry and wet deposition. When released into the air, this material is expected to have a
half-life of less than 1 day.
Formaldehyde is proven to be 100% biodegradable. That means it breaks down to
simpler molecules (like Co2 & H2O) through the natural action of oxygen, sunlight,
bacteria and heat. Studies show that formaldehyde breaks down much more easily than the
active ingredients of most other deodorant products, so it can be easily treated in waste
treatment systems.
Solid and liquid waste are not generated in the formaldehyde plants. The off-gas is the only possible source of pollution. Most of the domestic units in India are recycling 2/3rd
part of the tail gases and remaining is incinerated (in silver catalyst process) or oxidized to
water and carbon dioxide by Emission Control System (In Metal Oxide Process)
Leakage of formaldehyde through pumps and equipments to be avoided.
87
General Information:
CAS Number 50 - 00 - 0
EINECS Number 200 - 001 - 8
Chemical Name Formaldehyde
Chemical Classification Aldehyde
Synonyms Formalin; formic aldehyde; formal; methyl aldehyde; methylene glycol; methylene oxide; methanal; morbicid; oxamethane; oxymethylene; paraform; polyoxymethylene glycol; superlysoform.
Formula CH2O
Molecular weight 30.03
Shipping Name Formaldehyde solution
Codes / Label Flammable liquid, class - 3.1
Hazardous Waste ID No. 5
Hazchem Code 2 SE / 2 T
UN Number 1198
Description Colourless liquid with characteristic pungent odour.
Product uses In the manufacturing of phenolic resins, artificial silk and cellulose esters, dyes, organic chemicals, glass mirrors, explosives; disinfectant for dwellings, ship, storage houses, utensils, clothes etc., as a germicide & fungicide for plants & vegetables, improving fastness of dyes on fabrics, tanning & preserving hides;, mordanting & water proofing fabrics; preserving & coagulating rubber latex; in embalming fluids.
Packaging In SS tankers, ISO-containers and in HDPE drums / carboys.
MATERIAL SAFETY DATA SHEET
(According to 91/155 EC)
1. IDENTIFICATION OF THE SUBSTANCE / PREPARATION AND OF THE COMPANY / UNDERTAKING:
Product details:
Trade name FORMALDEHYDE, 37% SOLN. IN METHANOL/WATER
2. COMPOSITION/ INFORMATION ON INGREDIENTS:
Description CAS No. EINECS Number
Formaldehyde 50-00-0 200-001-8
Methanol 67-56-1 200-659-6
Water 7732-18-5
3. HAZARD IDENTIFICATION:
88
Hazard description Extremely hazardous product. Harmful if swallowed or inhaled. Causes severe irritation or burns to skin, eyes, upper respiratory tract, gastro-intestinal irritation and burns to mouth and throat Lachrymator at levels from less than 20 ppm upwards.
Chronic Exposure Kidney and lever damage.
Carcinogen city (NTP/IARC/OSHA) Yes
4. FIRST-AID MEASURES:
After inhalation Remove to fresh air, restore breathing, get medical attention.
After skin contact
Remove contaminated clothing, flush with plenty of water atleast for15 minutes.
After eye contact Immediately flush opened eye with plenty of running water for atleast 15 minutes.
After swallowing Induce vomiting of conscious patient by giving plenty of water to drink. Consult physician immediately in case of an unconscious victim.
5. FIRE-FIGHTING MEASURES:
Suitable extinguishing agents
CO2 , Dry chemical powder, water spray and alcohol foam.
6. ACCIDENTAL RELEASE MEASURES:
Person-related safety precautions
Proper protective equipment and self contained breathing apparatus with full face piece operated in positive pressure mode.
Measures for cleaning/collecting Shut off sources of ignition, no flares, smoking or flames In area. Stop leakage if possible. Use water spray to Reduce vapour. Take up with sand or other non combustible absorbent and place into container for disposal according to item 13.
7. HANDLING AND STORAGE:
Handling
Information for safe handling Avoid breathing vapours, avoid contact with eyes, skin and clothing. Decontaminate soiled clothing thoroughly before use.
Information about fire and explosion protection Flammable liquid. Closed containers exposed to heat may explode.
Storage
Requirements to be met by store rooms and receptacles
Store in tightly closed containers in a dry, cool, well ventilated, flammable liquid storage area.
8. EXPOSURE CONTROLS / PROTECTION:
50-00-0 Formaldehyde
OES Short term value *3 mg/m3
Long term value **1.5mgm3
89
* total inhalable vapour, ** permissible exposure
Personal protective equipment
Self contained breathing apparatus and full protective clothing.
Respiratory protection Self contained breathing apparatus.
Protection of hands Gloves of natural rubber.
Eye protection Safety glasses W/face shield.
9. PHYSICAL AND CHEMICAL PROPERTIES:
General information:
Form Liquid
Colour Colourless
Odour Pungent odour
Change in condition
Melting point --
Boiling Point / Boiling range 96oC
Flash point: (CC) 59oC
Flammable limits Upper 73.0%, Lower 7.0%
Auto ignition temperature 423oC
Specific gravity 1.08
Vapour density (Air = 1) 1.0
Vapour Pressure (MMHG) at 20 oC 1.3
Solubility/Miscibility with water at 20 oC
Complete
10. STABILITY & REACTIVITY:
This product is stable, strong reducing agent, especially in alkaline solution. Keep away from strong bases and acids, oxidizing agents, aniline,phenol, isocyanates, anhydride. Combustible light and air sensitive, polymerizes spontaneously.
Dangerous reactions With bases / acids and oxidizing agents.
Dangerous decomposition products
CO, CO2 & vapours of formaldehyde
11. TOXICOLOGICAL INFORMATION:
Acute toxicity:
LD/LC50 values relevant for classification:
Oral/LD 50 : 100 mg/Kg (rat)SKN LD 50: 270 µl/kg (rabbit)
Primary irritant effect:
On the skin Irritating effect (severe)
On the eye Irritating effect (severe)
Sensitization Prolonged contact may cause skin sensitization.
12. ECOLOGICAL INFORMATION: Environmental The product is expected to be slightly toxic to aquatic life. The LC 50/96
90
toxicity hrs values for fish are between 10 and 100 mg/l.
Environmental fate When released into the soil, material is expected to reach into ground water. When released into water, the material is expected to readily bio-degrade and is not expected to bio-accumulate. When released into air, the material is expected to readily degrade by photolysis, readily removed from atmosphere by dry and deposition. Half life of the material is less than one day, when released into air.
13. DISPOSAL CONDITIONS:
Whatever cannot be recycled, should be absorbed in sand or other non-combustible absorbent, containerized and transferred to appropriate and approved waste disposal facility. Dispose waste, containers and unused contents in accordance with official regulations.
14. TRANSIT INFORMATION:
Shipping name FORMALDEHYDE SOLUTION
UN number 1198 Class: 3, 3.3, 8
Packing Group III Label: Flammable liquid.
15. REGULATORY INFORMATION:
This product is extremely hazardous. It is listed as an "ACGIH" 'suspected' human carcinogen and a "NTP" anticipated human carcinogen. It may cause (mutagenic) reproductive effects.
COST ESTIMATION
(Source http://www.niir.org/projects/tag/z,.,451_0_32/formaldehyde/index.html.)
Total Capital cost of the plant for 10 tones/ day or 3000 tones/year (for 300 working days
plant)
91
= 5.4 crores
Fixed Capital Investment (80-90% of Total Capital Investment)
FCI = 0.85 5.4 = 4.59 crores
Working Capital investment = Direct cost + Indirect cost
Direct Cost
Materials & labours involved in actual installation of complete facility (Assumed 70-85%
of FCI )
Assuming = 77.5% of FCI
= 0.775 4.59
= 3.558 crores
Equipment + Installation + Instrumentation + Piping + Electrical + Insulation + Painting
(50-60% of FCI)
Assuming value is 55% of FCI = 0.55 4.59
= 2.525 crores
(a) Purchased equipment cost ( PEC) (15 to 40% of FCI)
Assuming Value is 27.5% of FCI
= 0.275 4.59
= 1.2623 crores
(b) Installation including insulation & painting (35-45% of PEC)
Assuming 40% of PEC
= 0.4 1.2623
= 0.505 crores.
(c) Instrumentation & control installation ( 6-30% of PEC)
92
Assuming value is 18% of PEC
= 0.18 1.2623
= 0.227 crores.
(d) Piping Installation (10-80% of PEC)
Assuming 40% of PEC
= 0.4 1.2623
= 0.505 crores.
(e) Electrical Installation (10-20% of PEC )
Assuming 15% of PEC
= 0.15% 1.2623
= 0.1894 crores.
(f) Building process & auxillary (10-70% of PEC)
Assuming 40% of PEC
= 0.4 1.2623
= 0.505 crores
(g) Service facility & yard improvement (40-50% of PEC)
Assuming 40% of PEC
= 0.4 1.2623
= 0.505 crores
(h) Land (1-2% of FCI or 4-8% PEC)
Assuming 1.5% of FCI
=
1 .5100 4.9
= 0.0735 crores
(i) Direct cost (DC) = 1.2623 + 0.505 + 0.227 + 0.505 + 0.1894 + 0.505 + 0.505
93
+ 0.0735
= 3.773 crores
Indirect Cost
Expenses which are not directly involved with material & labour of actual installation of
complete facility (15-30% of FCI)
(1) Engg. & supervisors (5-15% of DC)
Assuming 6% of DC
= 0.06 3.773
= 0.227 crores
(2) Construction expenses & contractor’s fees (7-20% of DC)
Assuming 8% of DC
= 0.08 3.773
= 0.3019 crores
3) Contingency ( 5-15% of FCI)
Assuming 8% of FCI
= 0.08 4.9
= 0.392 crores
Indirect Cost = 0.227 + 0.3019 + 0.392
= 0.921 crores
This is equal to
0 .9214 . 9
×100 = 18.8%
Working Capital (10-20% of FCI)
Assuming 15% of FCI
= 0.15 4.9
= 0.735 crores
94
Total Capital investment = FCI + working Capital
= 4.9 + 0.735
= 5.64 crores
Estimation of Total product cost (TPC)
Manufacturing Cost = Direct product cost + fixed charges + plant overhead cost
I. Fixed Charges (10-20% of TPC)
1) Depreciation depends on life period, salvage value, method of calculation about
10% of FCI for machinery & equipment 2-3% for building value.
= 0.1 4.9 + 0.02 4.9
= 0.588 crores
2) Local Taxes (1-4% of FCI)
Assuming 2% of FCI
= 0.02 4.9
= 0.098 crores
3) Insurance (0.4 – 1% of FCI)
Assuming 0.5% of FCI
= 0.005 4.9
= 0.0245 crores
4) Rent (8 - 10% value rented land & building )
Assuming 9% of rented land & building
= 0.09 [0.505 + 0.0735]
= 0.052 crores
Fixed charges = 0.588 + 0.098 + 0.0245 + 0.052
= 0.7625 crores
95
Total product cost (TPC) = 5 to 10 times of fixed charges
Assuming 5 times of fixed charges
= 5 0.7625
= 3.8125 crores
II. Direct Product Cost
1) Raw material (10-50% of TPC )
Assuming 20% of TPC
= 0.2 3.8125
= 0.7625 crores
2) Operating labour cost (10-20% of TPC)
Assuming 10% of TPC
= 0.1 3.8125
= 0.3812 crores
3) Direct supervisory & Electrical labour (10-25% of operating labour)
= 0.15 0.38125
= 0.0572 crores
4) Utilities (10-20% of TPC)
Assuming 15% of TPC
= 0.15 3.8125
= 0.572 crores
5) Maintenance & repairs (2-10% of FCI)
Assuming 5% of FCI
= 0.05 4.9
= 0.245 crores
6) Operating expenses (0.5-1.0% of FCI)
96
Assuming 0.75% of FCI
= 0.0075 4.9
= 0.0368 crores
7) Lab charges (10-20% of operating labour)
Assuming 12% of operating labour
= 0.12 0.38125
= 0.0458 crores
8) Patent & Royalties (0-6% of TPC)
Assuming 0.5% of TPC
= 0.005 3.8125
= 0.019 crores
Direct Product cost = 0.7625 + 0.38125 + 0.0572 + 0.572 +0.245 + 0.0368 +
0.0458 + 0.019
= 2.12 crores
III. Plant overhead Cost (5-15% of TPC)
It involves cost of general plant up keep & overhead payroll, packaging, medical services,
safety & production, restaurant recreation, salvage, lab & storage facilities.
Assuming 8% of TPC
= 0.08 3.8125
= 0.305 crores
Manufacturing Cost = fixed charge + Direct product cost + plant overhead cost
= 0.765 + 2.12 + 0.305
= 3.188 crores.
General Expenses
97
General Expenses = Administration cost + Distribution & Selling cost
(a) Administration cost (2 to 6% of TPC)
It includes cost for executive salaries, clerical wages, legal fees, and communication.
Assuming 2% of TPC
= 0.02 3.8125
= 0.0763 crores
(b
)
Distribution & Selling cost (2 – 6%) of TPC)
Assuming 4% of TPC
= 0.04 3.8125 = 0.1525 crores
(c) R & D ( 2 – 5% of TPC)
Assuming 3% of TPC
= 0.03 3.8125
= 0.114 crores
(d
)
Finances (0 – 7% of Total capital expenses)
Assuming 1.5% of Total capital expenses
= 0.015 5.4
= 0.081 crores
General expenses = 0.0763 + 0.1525 + 0.114 + 0.081
= 0.4238 crores
Total product cost = Manufacturing cost + General expenses
= 3.188 + 0.4238
= 3.612 crores
Gross earnings = total income – total production cost
for 1 kg of Formalin
Selling price is Rs. 25/-
98
Assuming 20% profit for seller
Selling price = Rs. 25 0.80 = Rs. 20
Gross Annual earning = 10 103 300 20
= 6 crores
Net annual income = gross annual earning – Total product cost
= 6 – 3.8125
= 2.1875 crores
Net annual earning after depreciation
= 2.1875 – 0.588
= 1.5995 crores
Net profit after tax (46% tax rate)
= (1 – 0.46) 1.5995
= 0.8638 crores
FCI = 4.59 crores.
Payout period =
FCI(Net profit after tax + depreciation )
=
4 .59(0 . 8638 + 0 . 588 )
= 3.16 years
Rate of return =
Net profitFixed capital investment
=
(0 . 8638 + 0 . 588 )4 .59
×100
= 31.63%
99
100