(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 650 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.
1
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, 4t edition, 1994 Volume. 11(Page 929 – 929 – 947). 947). 5. Industrial & Engg Chemistry by S.J.Green & Raymond.E.Vener Raymond.E.Vener, Volume 47 (Page-103-108) 1955 6. ―Introduction to Chemical Engg Engg Thermodynamics Thermodynamics ― by Smith & Vannes, 6th edition. (appendix – (appendix – C) C) 7. Perry ‘s ―Chemical engineer’s handbook‖ 6t Edition (Chapter 3 & 18) 8. ―Plant Design and Economics for Chemical Engineers‖ by Peter & Timmerhaus, 4th edition (Chapter – 6). – 6). 9. ―Process Design of Equipment‖ Volume 2. by S.D. Dawande, (Page 49-50). 10. ―Process Equipment Design‖ by M.V. Joshi 3r 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/ta http://www.niir.org/projects/tag/z,.,451_0_32 g/z,.,451_0_32/formaldehyde /formaldehyde/index.html.) /index.html.)
2
NOMENCLATURE A
Heat transfer surface
a, a t , a s
Flow area in general, for tube side, and for shell side, respectively
B
Baffle spacing
Cp
Specific Heat
C '
Clearance between tubes.
D
Diameter
De
Equivalent diameter
d p , d t
Diameter of particle, and diameter of tube
F
Feed flow rate
f
Friction factor
G , G s , Gt
Mass Velocity in general, for shell side, and for tube side, respectively
H
Enthalpy
h, hi , ho
Heat transfer coefficient in general, for inside fluid, and for outside fluid, respectively
hio
Value of hi when referred to outside diameter
ID
Inner diameter
J
Joules
j H
Heat transfer factor, dimensionless
K
Temperature measurement, Kelvin
k
Thermal conductivity
L
Length
M
Mass flow
mS , m w
Mass flow rate of steam, and water respectively
N N t
Newton Number of tubes
NRE
Reynolds number
OD
Outer diameter
P
Pressure
P
Pressure drop
3
P T
Tube pitch
Q
Heat flow
Rd
Total dirt factor
T
Temperature
U , U C , U d
Overall coefficient of heat transfer, clean coefficient, design coefficient
V
Velocity
T
Temperature Difference
X
Vapor pressure of water
Latent heat of vaporization or condensation
Viscosity ratio ( / w )
Viscosity
w
Viscosity at the tube wall
Density
Voidage
4
PERFORMANCE DATA OF MAIN EQUIPMENT REACTOR Volume of the reactor
0.156 m
Diameter of the reactor
0.83m
No. of tubes
80
Length of each tube
1m
ID of tube
50 mm
OD of tube
55 mm
Heat transfer area
6.80 m
Pressure drop
26.7 kg/ m
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
8t 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
7
INTRODUCTION Formaldehyde, H 2C=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 26 th 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 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 wa y 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 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, Yonkers, New york made a major breakthrough in the technology of polymer later named Bakelite after him. The ingredients were Phenol and Formaldehyde, by t he 1920’s the growth of this resin strained wood alcohol (Methanol) producing capacity, but the revolutionary
8
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 -19 C and freezes to a crystalline solid at -118 -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 100 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 physical properties of monomeric formaldehyde is given in Table 1.1
Formaldehyde solutions
Formaldehyde is produced and distributed as a water solution, the ―standard‖ strength being 37 wt%, this being a typical concentration concentrati on and also the basis for making production comparisons. Representative commercially available solutions are shown in Table 1.2
9
Table 1.1 Properties of Monomeric Formaldehyde
Property
Value
Density, g/cm3 at -80 -80C at -20 -20C Boiling point at 101.3 kPa C Melting point, C Vapor pressure, Antoine constants, Pa A B C Heat of vaporization, H v at 19 19C, kJ/mol
0.9151 0.8153 -19 -118 9.21876 959.43 243.392 23.3
Heat of formation, H f at 25 25C, kJ/mol Std free energy,
-115.9
G f at 25 25C, kJ/mol
-109.9
Heat capacity, Cp, J/mol.K Entropy, S o ,J/(mol.k) Heat of combustion, kJ/mol Heat of solution at 23 23 C kJ/mol in water in methanol in 1-propanol in 1-butanol Critical constants Temperature, C Pressure, MPa Flammability in air Lower/upper limits, mol Ignition temperature, C
35.4 218.8 563.5 62 62.8 59.5 62.4 137.2-141.2 6.784-6.637 7.0/73 430
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 re adily with water to form the hydrate, methylene glycol, which itself then tends to polymerize to poly(oxymethylene glycols). 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 60 60 C. In methanol-formaldehyde-water solutions, increasing the concentration of either
10
methanol or formaldehyde reduces the volatility of the other. The flash point varies with composition, decreasing from 83 to 60 60 C 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
Formaldehyde (wt%) Methanol (wt%) Acidity as formic (wt%) Iron as Fe (ppm) Turbidity, Hellige Color, APHA 3 Density (g/cm ), 18 18C Boiling point ( ( C) Viscosity (cP), 25 25C Specific heat (cal/g (cal/g.C) Flash point ( (C)
USP Grades 37.1 37.1 7.0 11.0 0.012 0.012 0.3 0.3 1.5 1.5 5 5 1.10 1.09 97.2 96.7 2.5 2.6 0.8 0.8 69 60
37.1 0.9-1.3 0.012 0.3 2 5 1.11 99.0 2.0 0.8 83
Low-Methanol Grades 44.1 50.3 0.9-1.3 0.9-1.3 0.02 0.018 0.3 0.3 2 2 5 5 1.12 1.13 99.1 99.5 1.7 1.6 0.7 0.7 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 commerciall y. The pure, dry gas is relatively stable at 80-100 80-100C but, slowly polymerizes at lower temperatures. Traces of polar impurities such as acids, alkalies, and water greatly accelerate the polymerization. Formaldehyde in water solution hydrates to methylene glycol;
11
Which in turns polymerizes to poly(methylene glycols), HO(CH 2O)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 va cuum 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 170 C. It is flammable, with a flash point of about 93 93 C. A typical formaldehyde solution may be obtained by dissolving paraformaldeh yde 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 62-64 62-64C, boiling without decomposition at 115 115C, and having a flash point of 45 C. 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;
Acetylene may be reacted with formaldehyde to form 2-butyne-1,4-idol which, when hydrogenated, yields 1,4-butanediol;
12
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 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 500 500C gives pyridine and 3-picoline. This forms the basis of commercial processes for making pyridines from various aldehydes. 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. 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.
13
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 b y 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. 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)
14
by comparison to platinum-cobalt color standards (D1209). (D1209). Table 1.3 Formaldehyde Formaldehyde specifications
Property
Methanol inhibited grades
Formaldehyde, wt%
37
37
Methanol, wt% (max)
6-8
Acidity, wt% (max)
0.02
0.02
Iron, ppm (max)
0.5
Color, APHA (max)
10
37
low methanol uninhibited grades 44
50
56
1.5
1.5-2.0
2.0
0.03
0.05
0.04
0.05 0.5-1.0
0.5
0.5
0.75
10
10
10
10
12-15 1.0-1.8 0.02
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 65 65 C. 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. par aformaldehyde. 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 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
15
about three months is as follows: 35 35C with less than 1% methanol; 21 21 C with 7% methanol; 7 7C with 10% methanol; and 6 6 C 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 variet y 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 ureaformaldehyde 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 insulati ng 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
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
16
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 s aturated 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
diisocyanate (MDI). Its principal end use is rigid urethane foams; other end uses include elastic fibers and elastomers. elastom ers. 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
17
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 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 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 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. Note: The requirement of formaldehyde for certain applications like Pol yacetal, MDI,
1,4-Butanediol and Neopentylglycol does not exist in India.
18
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 650 650 C in a steam superheater using low pressure steam. Air is drawn via, a filter filt er 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 CH3OH
HCHO + H 2O HCHO + H 2
+38 kcal/g.mol (exothermic) -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 650 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 650 650C is brought down to 120 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
19
solution is evolved due 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 25 25 C from 120 120 C 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 65 C and bottom reboiler temperature of 93 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 2 nd 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/ 4389.50kgs/ Day
2. AIR
-
4735.2kgs/Day
3. CATALYST
-
90.72kgs/day
20
Recycle Methanol
mixer
Off gas
Methanol super heater
Process Water
Steam FIXED BED Coolant TUBULAR CATALYTI C REACTOR Temp -
Air Superheater Steam
ABSORBER 1
Water 650 C Pressure – 1.05 atm
Condensate
Fractionation Column
ABSORBER 2
EVAPORATOR Coolant
Air Preheater
Fresh Methanol
Coolant
Reboiler
Steam Water
Filter
Formalin (37% Formaldehyde)
Atmospheric air
Air Blower Blower
Fig 2.1 Manufacture of Formaldehyde by Vapor-phase catalytic oxidation of methanol
21
MATERIAL BALANCE
Formaldehyde at atmospheric condition is a gas, commercially formaldehyde is diss olved 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 – 3700 = 6300kgs. Formaldehyde capacity per hour, 3700 24
= 154.167kgs
Assuming wastage of 0.1% through absorber
0.1 100
154.167 0.15kgs.
Total capacity of formaldehyde per hour, 154.167 + 0.154 = 154.32kgs. 154.32kgs.
CH 3 OH
1
O2
-----(3.1) +38kcal/gmole (exothermic) HCHO H 2 O -----(3.1)
MATERIAL BALANCE
Formaldehyde at atmospheric condition is a gas, commercially formaldehyde is diss olved 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 – 3700 = 6300kgs. Formaldehyde capacity per hour, 3700 24
= 154.167kgs
Assuming wastage of 0.1% through absorber
0.1 100
154.167 0.15kgs.
Total capacity of formaldehyde per hour, 154.167 + 0.154 = 154.32kgs. 154.32kgs.
CH 3 OH
1 2
O2
-----(3.1) +38kcal/gmole (exothermic) HCHO H 2 O -----(3.1)
HCHO O CH 3 OH HCH
H 2 -----(3.2)
-20.3kcal/g -2 0.3kcal/gmole mole (endothermic)
For silver process, about half of methanol goes to each rea ction.
John Mcketta, Vol 23, Page 358
Basis: 1 hour of operation of the plant
Formaldehyde production
154.32 30
5.144 kgmoles (MW of HCHO = 30)
Methanol required for the process = 5.144kgmole. Conversion of methanol is 90% Therefore Methanol fed =
5.144 0.90
5.715 kgmole
50% Methanol for each reaction, therefore for react ion 3.1 & reaction 3.2 Methanol fed =
5.715 2
2.8575kgmoles
22
Oxygen required =
Air required =
2.8575 2
1.42875 0.21
1.42875 kgmoles
6.8035kgmoles
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% =
5.715 715 (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 Methanol Water
Entering kgmoles 5.7155 5.7155
Leaving kgmoles 5.7155 5.7155
kg 182.896 102.879
kg 182.896 102.879
Material Balance around Methanol-water vapor Super heater
Here material vaporized is heated to 650 650 C. Here there is no 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 Air
Entering kgmole kg 6.8035 197.30
Leaving kgmole kg 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 change in material balance. Therefore the material balance is same as above. Material Balance around the Mixer
23
Here methanol-water vapor and air are mixed before to the reactor. Component Methanol Water-vapor Air
kgmole 6.8035 5.7155 5.7155
Entering kg 197.30 182.896 102.879
kgmole 6.8035 5.7155 5.7155
Leaving kg 197.30 182.896 102.879
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 0.9 2
Excess oxygen = oxygen supplied = 1.4287
1.2858 kgmoles
oxygen reacted
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.
CH3OH=5.7155kgmoles H2O =5.7155kgmoles O2=1.42875kgmoles N2=5.3748kgmoles
HCHO=5.144kgmoles CH3OH=0.5715kgmoles H2O = 8.233kgmoles 8.233kgmoles O2 = 0.14287 kgmoles N2=5.3748kgmoles H2 = 2.5717kgmoles 2.5717kgmoles
Reactor
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 mater ial 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 (O 2 ) + 5.3748 (N 2 ) + 5.14 (HCHO) + 2.5715 (H 2) + 0.5715 (CH 3 OH ) = 13.8047kgmoles = G Now the temperature of vent gases = 25 25C Vapor pressure of water at 25 25 C = 23.7 mmHg = X Pressure in absorber = 760 mmHg = P Moles of water present in vent gas = G
= 13.8042
23 .7 (760
23 .7)
X P X
0.444 kgmoles = 8kgs.
25
Amount of water leaving with formalin solution 37% (by wt) =
154.32
417.08 kg
0.37
Amount of formalin solution – solution – Amount Amount of HCHO in formalin solution = 417.08 – 417.08 – 154.32 154.32
=
262.76 kgs.
Amount of water present in gas entering the absorber = 149.1705 kgs Amount of water added to absorber =
pres ent in entering gas Amount of water leaving Amount of water present – 262.76 + 8 – 8 – 149.1705 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
H2
2.5717
5.1435
Total
22.0921
481.9882
Net total input = 22.0921 + 6.755 6.755 = 28.8472 28.8472 kgmoles.
26
Net total input = 481.9882 + 121.59 = 603.5782 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. kgmoles. Net output (kgs) = 435.368 + 168.2097 = 603.5777 kgs.
Material Balance around Distillation column
27
HCHO = 5.144kgmole Water = 14.6kgmoles Methanol = 0.5715kgmoles FEED= 20.3155
Fraction ation Column
Methanol =0.5658kgmole HCHO = 0.005144kgmol e DISTILLATE = 0.57094
Methanol =0.005715k gmole HCHO = 5.1389kgmo le Water = 14.6kgmole RESIDUE = 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 25 25C to its B.P 64.7 64.7 C and then vaporizes at the same temperature. Quantity of heat required = mC P T m To calculate for Methanol using Kistya kowsky equation (considering methanol to be a non-polar liquid)
b T b
b T b
8.75 4.571 571 log10T b
( T b = 337.9 337.9K)
8.75 4.571 log(337 337 .9) 20.34
b = 20.34 337.9 = 6860 cal/gmole = 28722.82J/gmole Cp of Methanol between 25 25 C and 64.7 64.7 C Cp a
b
2
(T2
T1 )
c
T 3
2
2
T 1T 2 T 1 2
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
337.7
2
298 337 .7 298 298 2
Cp = 11.205 cal/gmole.K = 46.91J/gmole.K q1 = heat required to raise the temperature of methanol from 298 K to 337.9 337.9 K
q1
mC P dT = 5.7155 103 gmole 46.91J/gmole.K 46.91J/gmole.K 39.7K = 10.6453 106 J.
q2
m = 5.7155 103 gmole 28722.82J/gmole = 164.1653 106J
Total heat added = q 2
q2
= 10.6453 106 J + 1641653 106 J = 174.810
30
6
10 J
Heat required for water to vaporize = mC P T m = 5.7155 103 4.18 103 (100-25) + 5.7155 2185 103 9
= 1.8043
10 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 ) =
q
=
174 .810 810 106 1.8043 10 9 2724000
mS = 726kg at 3bar Heat balance of Pre-heater for Methanol vapors
Here gaseous vapors of methanol and water are heated from 64.7 C to 650 650 C (337.9 (337.9K to 923.2 923.2K) Cp a
b
2
(T2
Cp 4.55
T1 )
c
T 3
2.186 10 2 2
2
2
T 1T 2 T 12
(923
337 .7 )
- 0.291 10 -5 3
923
2
337 337 .7 923 337 337 .7 2
= 4.55 + 13.78 – 13.78 – 1.235 1.235 = 17.099 cal/gmole.K = 71.5935J/gmole.K Heat supplied to superheat for Methanol
mC P T
= 5.7155 103 71.5935 (923.2 – (923.2 – 337.9) 337.9) = 239.500
6
10 J
Cp of water = 36.845J/gmole Heat supplied to super heater from 100 100 C to 650 650 C
mC P T
= 5.7155 103gmole gmole 36.845J/gmoleK 36.845J/gmoleK (550) K = 11.5825 Since heat required to preheat = heat given away by steam Amount of steam required assuming steam available at 3 bar
31
7
10 J
Enthalpy of steam = 2724 kJ/kg Mass of steam required ( mS ) =
=
q(methanol water )
( steam)
106 11.5825 107
239.5005
2724000
mS = 130 kgs . Heat balance around Air Pre-heater
Here atmospheric air at 25 25C is heated to 650 650 C (reaction temp) Q
mC P T
= 6.8035 103
(29.1917 35.388) 2
(650 -25) = 137.30
3
10 KJ
Heat required by air = Heat given away awa y 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
10 6
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 650 650C and passes through tubular bed of catalyst (silver grains). The reaction to form formaldehyde is exothermic withheat of reaction (38 – (38 – 20.3) 20.3) kcal/mole i.e, 17.7 kcal/mole Moles of product formed from reaction are = HCHO + H 2O + 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
3
10 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
3
10
= 1.254 106 KJ Since the reaction is exothermic and reaction reacti on temperature is to be maintained at 650 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 35 35C is used as cooling agent. The temperature is expected to reach 100 C, Cp of water = 35.756 J/gmoleK Mass of water required, ( m w ) =
q Cp T
=
=
762 762 .034 034 10 5 J
5
35.756J/gm ole.K (100 - 25)K
2.8415 10 5 18 1000
= 2.8415 10
= 5114.7 kgs .
Energy balance of reactants through heat exchanger Pre-heater
The mixture from outlet of reactor at 650 650 C are brought down to 100 C, by exchanging the heat to the air that is going to the reactor . Products issuing coming out in kgmoles are; ar e; HCHO (5.144) + H 2O (8.2872) + H2 (2.5717) + N2 (5.3748) + O 2 (0.1428) + CH 3OH (0.5715) Heat for cooling the reactants from 650 650 C to 100 100 C 5.3748 103 29.459 550
870.849 10 5J
0.1428 103 30.4353 550
23.9156 105J
0.5715 103 71.593 550
225.036 105 J
qWater ( mC P ) Water T
8.2872 103 36.845 550
1679.418 105 J
(mC P ) HCHO T q Hydroge n ( mC P ) Hydrogen T
5.144 103 53.9578 550
1526.574 105 J
2.5717 103 29.8365 550
422.026 105 J
q Nitrogen Nitrogen
(mC P ) Nitrogen Nitrogen T
qOxygen ( mC P ) OxygenT
q Methanol
(mC P ) Methanol T
q HCHO
Q q Nitrogen Nitrogen qOxygen qmethanol qWater q HCHO q Hydrogen Heat required by air = 137.29
474.7818 106 J
3
10 KJ
33
Heat that will be removed = 474.7818 103 – 137.29 137.29 103 = 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 – 149.1705 – 8 8 = 41.7 kgs. = 7.843 Kgmole. K gmole. Now the heat given out by entering gas to reach the outlet temperature from 120 120C to 25 25C q Nitrogen Nitrogen
5.3748 103 95 29.1708
14.894 106J
qOxygen
0.1428 103 95 29.5183
400.64 103J
qmethanol
0.5715 103 95 46.8944
2.546 106J
qWater
8.2872 103 95 35.1708
27.6895 106J
q HCHO
5.144 103 95 45.2405
221.081 221.081 106J
q Hy droge n
2.57175 103 95 28.4716
6.956 6.956 106J 74.595l 106J
Q total
Heat is removed by using a coolant Heat Balance around Distillation column
Distillate
Methanol =18.10kg/hr HCHO = 0.1543kg/hr Feed
Fractionation column
Methanol = 18.106 kg/hr HCHO = 154.32kg/hr Water = 262.76kg/hr
Residue
34
HCHO = 154.167kg/hr 154.167kg/hr Water = 262.76kg/hr Methanol = 0.182kg/hr
Condenser
Here methanol is cooled from 64.7 64.7 C to 40 40C Heat in = (m ) Methanol ( m ) HCHO
= (18.106 1098 .5 10 3 ) (0.1543 1001 10 3 ) = 20.09 106 J. Heat out mC P T
=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 6
Heat removed = 19.40 10 J.
Water is used to cool the product from 64.7 64.7 C to 40 40 C Q
mC P T m
187 10 3 (40 25) m 2230 10 3 19.40 106 J = m 4.187 mw = 8.46 kgs/hr. Reboiler Heat in bottom of reboiler = (mC P ) Methanol T (mC P ) HCHO T (mC P ) water T
0.183 2508 (64 .7 25) 154 154 .16 0.8 4.18 10 3 (64 .7 25 ) 262 .8 4187
6
= 64.08 10 J
Overall Heat balance around Reboiler
Heat in feed stream
+
Heat added in reboiler
Heat at top product produc t
=
+
Heat loaded in bottom produce
0 + Heat added in reboiler = 20.09 106 J + 64.08 106 J 6
Heat added in reboiler = 84.17 10 J
35
(64.7 25 )
If we use steam of 3 bar, T = 133.5 133.5 C, = 2163 KJ/kg. m S 84.17 10 6 J
mS
84 .17 10 6 2163 10 3
38 .91k gs. (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 condition Void 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 m 3/hr Taking space velocity of 8000 hr -1 Void volume of reactor =
408.454 m 3 / hr 8000hr
-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 particle Diameter of tube
=
d p dt
(Diameter of tube is taken as 50mm)
For, smooth uniform catalyst, porosity
Therefore, actual volume of reactor =
is 0.36
0.051m 0.36
3
= 0.14167 m3
Taking 10% extra volume = 0.14167 1.1 = 0.15584 m 3 0.156 m3 2) To find tubes required
Since tubes are 50mm in diameter the cross section area of each tube is
37
=
4
(50 10-3) 2 = 1.9625 10-3 m2
Taking length of tubes as 1m (standard tubes) Number of tubes, ( N t ) = = 79.49
Volume of reactor Volume of each tube
=
0.156m 1.9625
3
10 -3 m 2 1m
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 m 2 Using 20% excess area = 1.2 0.45 m2 = 0.54m2
Shell diameter required =
0.54m 2 4
d 2 = 0.83m (using A= ) 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 kg/hr 482 kg/hr c) Gas inlet temperature = 650 650C d) Gas outlet temperature = 650 650 C e) Water flow rate = 248.15 kg/hr
38
f) Water inlet temperature = 25C 25C g) Water outlet temperature = 100 100 C Average properties properties of fluids (tube side)
Hot fluid stream gas mixture from Pre-heater Inlet temperature = 650 650C, Outlet temperature = 650 650 C Average pressure = Atmospheric = 1.033 Kg/cm 2 (absolute) Gas flow rate = 482 Kg/hr Average temperature = 650 650C Volume at average temperature at NTP the volume is 408.5m 3/hr =
923 408.454 273
= 1380.1 m 3
Average density of gas = 1kg/ m 3 Average viscosity of gas = 0.126kg/m.hr Average thermal conductivity, k = 203 w/m.K Average properties properties of fluids (shell side)
Stream is sent through the shell Cold fluid stream = water Inlet temperature = 25 25C, Outlet temperature = 100 100 C Average pressure = 1.033 Kg/cm2 (absolute) Water flow rate = 248.15 m 3 /hr Average density = 1000 kg/m 3 Average Cp = 4.187 KJ/kg.K Average viscosity = 1Kg/m.hr, Average thermal conductivity, K = 2445.208 w/m.K Log mean temperature, Tm =
t 2 t 1 t2 ln t 1
Hot fluid; Inlet = 650 650 C, Outlet = 650 650 C
39
Cold fluid; Inlet = 25C 25C , Outlet = 100 100 C 650 – 100 100 = 550 550 C t 1 = 650 –
t 2 = 650 – 650 – 25 25 = 625 625C Tm =
550 550 625 ln 625 550 550
625 625
= 586 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 m 2 Mass flow rate, (G t ) =
Mass flow rate of gases (G) (G) tube side flowarea (a t )
=
482k g / hr 0.1568m
2
= 3073.98 kg /m 2.hr
For packed tubes with air flowing in inner side Heat transfer coefficient is given by
h p
k d .G 3.5 P t d t
d P .Gt 3.8 50
0.7
=
0.7
e
dp 4.6 dt
3.8 10 3073.98 0.126
(Page 290 eqn. 10-10 Chemical engineering st kinetics by J.M Smith 1 edition) 0.7
= 23.82
- 4.6
e
= 0.7049
h p = 3.5
hid = hp
203 50 10
ID OD
-3
23.82 0.7049 = 238.60 KJ/hr.m 2.K
= 238.60
50 2 55 = 216.90 KJ/hr.m .K
Shell side Heat transfer coefficient Cold fluid:
Flow area, (a S ) '
shell shell dia( I . D) baffle spacing spacing ( B) clearance(C ' ) pitch( P t )
40
, m2
a S
G ' s
'
=
0.83m 20 10 3 m 0.15 m 3
75 10 m
= 0.0332 m 2
mass flowrate of water (m) (m) Shell side flow area(a ' s )
5114.7kg/h r 2 = 154057.23 kg/hr.m 2 0.0332m
Reynolds No. N Re=
De = 4
De G s
(0.5 Pt 0.86 Pt 0.5 D
2
4
0.5 Do
)
57.04 mm
NRe = 57.04 10-3 154057.23 = 8787.42 Assuming there is no considerable change in Viscosity
ho
j H
k Cp De
0.33
w
0.14
(Eq 6.15b, page111, Process heat transfer by D.kern)
When NRe = 8787.42 j H = 50 (page 838 figure 28 Process heat transfer by D.Kern) Cp. k
=
4187 1 2445.208
ho = 50
0.33
= 1.194
2445.208 57.04 10
-3
1.195
= 2561.14 KJl/hr.m 2.K
Clean overall Heat transfer coefficient = U C
U C =
hid .ho hid ho
=
216.90 2561.14 216.90 2561.14
= 199.97
Let total dirt factor Rd = 1.43 10 5 hr.m2.K/KJ
Design overall heat transfer coefficient, U d
1 = = 1 U C Rd
=194.40KJ/hr.m2.K
41
1 1 5 1.43 10 199.97
Area required, (A) =
q U .t d ln
=
762 .034 034 10 3 KJ 194 .40 KJ / m K 586 K 2
= 6.7m2
5) Pressure drop calculation Tube side pressure drop
P L
0.4G 2 a Gt
=
g 2
0.1
Page 193, eqn, 24, Reaction kinetics for chemical engineers by Stanley Walas
a
Where, Gt = mass velocity(kg/hrm2),
= free space or porosity
g = gravity constant(m/s2), = viscosity of gases(kg/ms) a = specific surface of bed m2/m3 = 6
a=6
(1 - 0.36) 3.8 10
-3
(1(1- ) d p
= 1010.5 m2/m3
3073.98 P 0.4 (3073.98) 2 1010.5 = L 1010.5 0.126 1.27 108 1!(0.36) 3 2
P = 26.7 1 = 26.7 Kg/cm
= 26.7
(L = 1mts)
Pressure drop on shell side
NRe=
De G S
57 .04 10 3 154057 .23 1
= 8787.4
Therefore f = 2.7 10-5 (page 836,figure 26, Process heat transfer by D.Kern) Number of cross, N + 1 =
L B
Where, L = tube length (m), B = baffle space (m) =
1 0.15
= 6.67
N + 1 = 7 fG fGs D( N 1) 2
Ps =
2 g . De De s
; s s =
w
=1
(Page 147 eqn. 7.44 Process heat transfer by D.Kern)
42
Where, f = friction factor dimensionless
=
2.7 10 5 (1.54 10 5 ) 2 0.83 7 2 1.27 10 8 57.04 10 3 1
Ps = 2.57
-4
2
10 Kg/ m
43
MECHANICAL DESIGN OF REACTOR As temperature in reactor is 650 650C, we consider working stress at same temperature Taking factor of safety = 3 Therefore working stress for carbon steel = 200Kg/cm 2 Design pressure = 1.2 times of 1Kg/cm 2 (20% extra)
Tube side 1) Thickness of tube
Minimum thickness is given by, t =
pD
2 f
C
P = average operating pressure = 1Kg/cm 2 D = diameter of tubes = 50mm C = corrosion allowance = 3mm f = working stress = 200Kg/cm 2 t=
1.2 50 2 200
3
= 4.15mm = 5mm
2) Tube sheet thickness
t FGc
0.25 P
P=
f
1.013 105 1.2 3 2
(10 )
0.1026 N / mm2
Where, F = 1 for most Heat exchanger except U type 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 0.1026 100 100
= 14mm
3) Channel and Channel cover
Thickness of channel portion
44
t c
kp
Gc
f
Where, k = 0.3 for ring type gasket, Gc = mean gasket dia for cover = 835mm f = permissible stress at design temperature 95N/mm2
t h
0.3 0.126 126
835
95
25 .7mm 26 mm
4) Flange diameter (between tube sheet and channel) Gc = 835mm, Ring gasket width = 22mm bo
W m1
W
8
22 8
2.75mm ,
y a
126 .60 N / mm 2 , m = 5.5, b =
Gby 835 835 2.75 121 .6
= 912814.5N
W m 2
2 bGmP
W m 2
2 2.75 835 5.5 0.1026 8352 0.1026
4
G 2 P
4
= 64292.7N Wm 1
Am1
Am1
912814.5
Am 2
64292.5
Am
2
permissibl permissibl e stress for bolt material(1 40.6N/mm 2 )
140.6
140.6
N b
6520mm 2
457.3mm 2
D 2 4
No. of bolts, N b
835 10 2.5
34 bolts
45
bo
4 A m2
Diameter of bolt =
Nb
4 457.3
34
= 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
K=
0.3
P Kf
C hG
1 1.5 1.5 Wm h G
= Radial distance from gas load reaction
HG
to bolt circle = BG
2 H
= Total hydrostatic end force = /4 G2 P =
4
8352 0.1026
1
K= 0 .3
1.5 912814.5
940 835 835 835 56155.2 2
= 56155.2 N
= 0.54mm 835 tf = 835
1.5 0.54 95
38mm
6) Nozzle diameter of the reactor
a) Reactant inlet Nozzle diameter (Tube side)
mo
481.99 18.24 22.4 1.18 10
Av
3
= 1.18 10-4Kg/cm3 = 11.8Kg/m 3
, Where v = 2m/sec for gases, mo = 481.88kg/hr (from material balance)
46
Cross sectional area (A) =
mo
.v
481.99
=
11.8 2 3600
= 56.571cm 2
Therefore r =
56.7
0.5
4.25cm
Diameter = 2r = 8.5cm. b) Product outlet Nozzle diameter (Tube side) =
481.99 22.093 22.4 1000
Cross sectional area =
= 9.74 10-4Kg/cm3
mo
.v
=
481.99 9.74 2 3600
= 68.73cm2 Therefore r =
68.73
0.5
4.68cm
Diameter = 2r = 9.36cm
b) Shell side 1) Thickness of shell
pD
Minimum thickness of shell is given by, t =
2 fe
C
(Page 48, eqn 1.123 Process equipment design by D.Dawande
P = average operating pressure = 1Kg/cm 2 D = diameter of shell = 3.8m f = working stress = 200Kg/cm 2 C = corrosion allowance = 3mm e = weld joint efficiency = 0.6 t=
1.2 83 2 200 0.6
0.3 = 0.415 +0.3 = 0.715cms =
1cm
2) Flange thickness (shell side)
Gasket diameter (G) =
Shell I.D I. D Shell O.D 2
47
830 840 2
835mm
Gasket width (N) = 24mm bo
N
2
24 2
b 2.5 12
12mm
8.64mm
Where, b = effective gasket seating se ating width b0 = basic gasket seating width Gasket factor (m) = 3.75 (flat metal jacket asbestos filled)
(Table 5.4 page 129
Seating stress(y) = 53.4 N/mm 2
―Process equipment design
Gby 0.835 10 3 8.64 53 .40
W m1
by M.V.Joshi)
= 1209681.73N
W m 2
2 bGmP
W m 2
2 8.64 0.835 10 3 3.75 1.2
4
G 2 P
4
0.835 10
3 2
1.2
= 860664.89N Flange Thickness, tf = G
K=
0.3
P Kf
C
1 1.5 1.5 Wm h G
hG
= Radial distance from gas load reaction
HG
to bolt circle = BG
2 H
= Total hydrostatic end force = /4 G2 P =
0.835 10
3 2
4
1
K= 0 .3
1.5 860664.89
940 835 835 656785.95 2
= 656785.95N
48
1 .2
=2.36 835 tf = 835
1 .5
61mm
2.36 95
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 =
t=
0.885 1.2 840
0.885 885 PL fe fe 0.1 P
C
3 = 7.44 + 3 = 10.44mm
200 0.6 0.1 1.2
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 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
(Page 51, table1.10 Process equipment design by S.D Dawande Vol 2 (Page 51, table1.8, Process equipment design by S.D Dawande Vol 2)
6) Nozzle diameter of the reactor
a) Diameter of cold water nozzle inlet (Shell side) mo
Av
, Where v = 2m/sec for gases, mo = 5114.7kg/hr (from material balance)
Cross sectional area =
Therefore r =
7.10
mo
.v
=
5114.7 1000 2 3600
= 7.10cm2
0.5
1.5cm
Diameter = 1.5 2 = 3cm
49
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 (2 (2cold water inlet dia) is taken.
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
X
0
0.06
0.126 0.199
Y
0
0.2655
0.44
83.7
79.1
76.9
74.4
72.1
69.3
67.1
0.279
0.369
0.4704
0.5844
0.7142
0.864
1.0
0.554 0.6414 0.7105
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/hr xD = 0.99
Methanol =0.5658kgmole HCHO = 0.005144kgmole
F= 20.3155kgmol/hr ZF = 0.02813
HCHO = 5.144kgmole Water = 14.6kgmoles Methanol = 0.5715kgmoles
Fraction ation Column
W = 19.745kgmol/hr xw = 0.000289
Methanol =0.005715kgmole HCHO = 5.1389kgmole
51
64.3
Glossary of notations used:
F = molar molar flow rate of Feed, kmol/hr, kmol/hr, D = molar molar flow rate of of Distillate, kmol/hr. W = molar flow rate of Residue, kmol/hr. z F = mole fraction of methanol in liquid/Feed. x D , = 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 ρL = Density of Liquid, kg/m3, 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:
z F
x D
xW
0.5715 0.5715 14.6 5.144 0.5658 0.5658 5.144 10 3
= 0.02813
= 0.99
5.715 10 3 5.1389
14 .6 5.715 10 3
= 0.000289
Marking z F , x D , xW on the x-y graph and assuming liquid enters as saturated (q = 1) From graph (Figure 1) x D R m
1
0.99 0.15
= 0.15
Rm 1
52
Rm = 6.6-1 = 5.6 Assuming reflux ratio of 1.5 times of Rm R = Rm
Now
1.5 = 5.6
x D R 1
0.99 8.4 1
1.5 = 8.4
= 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
=
8.4 0.57094
=
4.796kgmole
G = L + D = RD + D = (R+1) D = (8.4 +1) +1) 0.57094 = 5.367kgmole
(q =1 for saturated liquid) =
L+qF
=
4.796 +1 20.3155
=
25.115kgmole
G
=
G + (q-1) F
G
=
G (Since q = 1 )
=
5.367kgmole
L
(q = 1)
Table 4.2 List of Parameters Parameters used in calculation
Consider 4 points in the column, top and bottom of enriching section and top and bottom of stripping section.
53
Enriching Section Top
Stripping Section 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/kmole
31.98
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
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
0.034
0.0303
0.159
0.162
L / G G L
0 .5
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, ( t S ) = 400mm 2) Hole diameter, ( d L ) = 5mm 3) Hole pitch (triangular),( (triangular),( L P ) = 3 d L = 15mm
54
4) Tray thickness, ( t r ) = 0.6 d L = 3mm 5) Plate diameter ( DC ) (Plate hydraulic table)
L / G G L
0.5
= 0.162 (maximum value)
For t S = 16‖, 16‖, C Sb flood = 0.23
We have U nf C Sb flood 20
0.2
L G G
0.5
{eqn. 18.2, page 18.6, 6th edition Perry.}
Where, U nf = gas velocity through the net area at flood, fl ood, m/s
C Sb flood flood = capacity parameter, m/s (ft/s, as in fig.18.10)
σ = liquid surface tension, mN/m (dyne/cm.) L = liquid density, kg/m3 (lb/ft3) G = gas density, kg/m3 (lb/ft3)
Mixture
Methanol Water 2
1
4
P L G ---( eqn. 3-151, page 3-288, table 3-343, 6th edition Perry.}
(at low pressure, where L >> G , so neglect G ) 1
1
4
Water
1 P L = 4 Water 51 18
Water = 64.56 dyne/cm 1
792 0.792 , Me thanol = 19.86 dyne/cm 32
4 Methanol 85.3
Mixture
Methanol Water 2
=
19.86 64.45 2
= 42.16dynes/cm
55
0.2
Flooding velocity U nf
42.16 1000.9 1.2 = 0.23 20 1 . 2
0.5
= 7.706 ft/sec = 2.348m/sec
Let us take U n = 80% of U nf 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
Volumetric flow rate of vapor (at top of stripping section) =
168.9 3600 1.86
An =
0.0252 1.86
0.0252m3 / sec
0.01356m2
Ad = down comer area can be taken 10-12% of Ac (let us take Ad = 11% Ac ) Ac =
Dc 2 4
= 0.785 DC 2
Ad = 0.7850.11 DC 2 = 0.0864 DC 2 DC 2 = 0.0194m2 = 0.14m
Taking DC = 0.2m Ac = 0.7850.22 = 0.0314m2 Ad = 0.0864 Dc 2 = 0.08640.22 = 3.45610-3m2
Active area; Aa = Ac - 2 Ad = 0.0314-23.45610-3 = 0.0245m 2 6) Perforated area, ( Ap )
Let Lw / DC = 0.75 ( Lw = weir length)
56
Where, Lw = weir length, m DC = Column diameter, m
Lw = 0.750.2 = 0.15m
Lw 97.18 = 2 Sin 1 (0.75) = 97.18 Dc Dc
2 Sin 1
= 180 180 – 97.18 – 97.18 = 82.82 82.82
Periphery waste = 50mm = 50 10-3 m Area of periphery waste AWZ
2 2 Dc2 Dc 0.02 4 360 4 360
82.82 2 82.82 2 0.2 2 0.2 0.02 360 4 360 4
= 0.01186m 2
2 -3 Acz = area of calming zone, m = 2 LW 50 10
= 2 0.15 20 10 -3 = 6 10 -3 m2 Ap Ac 2 Ad Acz Awz -3 0.0314 – 2 2 3.456 10-3 – 6 – 6 10 – 0.0107 – 0.0107 Ap = 0.0314 –
= 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
2 nh d h 4
7.788 10 4 Number of holes, N h = = 40 3 5 10 4
57
8) Weir height,( hw ):
Let us take hw = 50mm 9) Weeping check
Head loss across the dry hole is
G 2 U h L
hd K 1 K 2
---- (eqn. 18.6, page 18.9, 6 th edition Perry)
Where, U h = gas velocity through hole area K 1 , K 2 are constants
Volumetric flow rate of vapor =
U h
0.0252
168.9 3600 1.86
= 0.0252 m 3/sec
= 32.37 m/sec
7.788 10 4
For sieve plate K 1 = 0 ; K 2 = 50.8/Cv 2 Where, Cv =discharge coefficient, taken from fig 18.14, page 18.9 6 th edition Perry.
Holearea Holearea activearea
Ah Aa
Tray thick ness Hole area
7.788 10 4
t R d L
0.0245
= 0.032
3
= 0.6 5
Thus for (Ah/Aa) = 0.032 and
t R d L
= 0.60
We have from fig. edition editi on 18.14, page 18.9 6 th Perry. C V (discharge coefficient) = 0.7 K 2
hd
50.8
0.7 2
103 .67
1.15 2 103 .67 32 .37 = 124.9 mm of liquid 1000 .2
Height of liquid crest over weir
58
hOW
q 44300 F W LW
0.704
Thus, q = liquid flow rate, m 3 /s; Liquid load q
538 .82
995 .08 3600
1.504 10 4 hOW 44300 150
Lw = weir length
1.504 10 4 m 3 / sec
0.704
2.65mm
Head loss due to bubble formation ( h ) is given by (page 18-17, equation 18-2a ) h
409 409
409 409 42.16 = 3.47mm liquid L d h 995 995 .08 5
Where, σ =surface tension, mN/m (dyne/cm) = 42.16 dyne/cm. d h = diameter of perforation, (mm) = 5mm 3 3 L = density of liquid in the section,(kg/m ) = 995.08 kg/m
hd + h = 124.9 + 3.47 = 128.37mm
hW + hOW = 50 + 2.65 = 52.65mm Ah Aa
032 0.032
From figure 18-11, 6th edition Perry, hd + h = 10mm Since design value of hd + h 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 editi on Perry)
a) hydraulic gradient across the plate hhg for stable operation hd >2.5 hhg
59
For sieve plates hhg is generally small or negligible, let us take hhg = 0.5mm of liquid b) Total pressure drop across the plate, ( ht ); ht hd hl
Now, hl
hdS
---- (eqn. 18.5, page 18.9, 6 th edition Perry)
Where, hl = pressure drop through the aerated liquid (mm) =Aeration factor, hds = Calculated height of clear liquid over the dispersers, (mm) hds
hw how
hhg
2
----(eqn. 18.10, page 18.10, 6 th 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 1
Now, F ga
U a g 2
Where F ga = gas-phase kinetic energy factor, U a = superficial gas velocity, m/s (ft/s), g = gas density, kg/m 3 (lb/ft3) U a =
168.9 3600 1.15 0.0245
= 1.665m/s
F ga = 5.46 1.15 103 62.4
0.5
1.463
From (figure 18-15, page 18-10, 6 th edition Perry) = 0.59 hl = 0.59 52.9 = 31.211
ht hd hl = 124.9 + 31.211 = 156.11 mm
c) Loss under down comer, ( hda ) 60
2
q ----- (eqn. 18.19, page 18.10, 6th edition Perr y) hda 165.2 A da 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, m 2 hap
hds c (Take clearance, c = 1‖ = 25.4mm)
hap
52.9 25.4 = 27.5mm liquid
Ad a
Lw hap 0.15 27.5 103
= 41.25 10-3 mm2
2
hda
1.504 10 4 = 2.196 10-3 mm 165.2 3 41.25 10
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 dc
For system with low gas velocity, low liquid viscosity and low foamability we can take
dc
0.6
61
hdc
209.2 0.6
348.72mm
(Which is less than the tray spacing)
t S = 400mm
(no down comer flooding will occur) Since t S > hdc Column efficiency
The efficiency calculation are based on average av erage condition prevailing in each section Enriching section
Average molar liquid rate = 4.796 kgmole/hr Average mass liquid rate =
153.38 102.78 2
128.08kg / hr
Average molar vapor rate = 5.367 kgmole/hr Average mass vapor rate =
171.64 168.9 2
793.98 995.08
Average density of liquid =
Average density of vapor =
170.27kg / hr
2 1.147 1.15
Average liquid temperature =
Average vapor temperature =
2 64 91 2 65 92 2
894.53kg / m3
1.1485kg / m 3
77.5o C 78.5 o C
Viscosity of formaldehyde at 77.50C = 0.06cp Viscosity of Methanol at 77.5 0C = 0.07cp Average viscosity of liquid is calculated using Knedall-maser ol equation 1
3
1 L
1
x1 1 x1 3 2 3
0.99 0.02813
x1
x2
1 x1 0.491 491
2
0.509
62
m Liquid Liquid
1 1 0.509 0.073 0.06 0.4913
3
= 0.0162cP
Viscosity of formalin at 78.5 0C = 0.05cP Viscosity of methanol at 78.5 0C = 0.062cP
0.99 0.9719
y1
y 2
1 y1 1 0.98 0.02
mvapor
2
0.98
y M 12 i i i
y M i
mvapor
1
i
2
0.98 0.062 32
1
1
2
0.02 0.05 30 2
1
1
0.98 32 2
0.12 30 2
= 0.054cP
Liquid phase diffusivity
Wilke chang equation status
7.4 10 M
0.5
8
D L
B
BV A
T
0.6
Where, D L = mutual diffusion coefficient of solute A at very low conc, in solvent B, = association factor of solvent B, M B = mol wt of solvent B B = viscosity of solvent B, cP
T = 273+77.5 = 350.5K, V A = solute molal volume (methanol)
= 16.5 1 + 1.98 4 + 5.48 = 29.9
7.4 10
2.6 300.5 350.5 -5 2 = 4.26 cm /sec 10 0.6 0.7 29.9
8
D L
Vapor phase density
Fullers equation
63
0.5
1 1 10 3 T 1.75 M M B A D g 2 1 1 P V A 3 V B 3 T = 78.5+ 273 = 351.5K
M A = 32; M B = 30; P = 1atmosphere
V
A
= 29.9;
V = HCHO = 1 16.5 + 1.98 2 + 5.48 = 25.94 B
1 1 10 351 32 30 D g 2 1 1 1 29.9 3 25.94 3 3
N Scg
0.5
1.75
g g D g
= 0.19cm2 /sec
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 =
538.21 530.68 2
534.45kg / hr
Average molar vapor flow rate = 5.367kg/hr Average mass vapor flow rate =
av Liquid Liquid
av Vapor
T av Liquid Liquid
T av Vapor
995.08 1000.9 2 1.15 1.2 2 91 93 2
92 93 2
168.9 113.4 2
141.15k g / hr
997.94kg / m 3
1.175kg / m3
92 0 C 92.50 C
Methanol at 91.50C = 0.034cP HCHO at 91.50C = 0.02cP
64
0.02813 0.00028
x1
x2
1 x1 1 0.0142 0.9858
2
0.0142
3
1 1 m 0.0142 0.034 3 0.9858 0.02 3 = 0.01865cP
of Methanol at 92.50C = 0.032cP, of HCHO at 92.5 0C = 0.019cP 0.9719 0.00028
y1
y 2
1 y1 1 0.486 0.514 514
2
mvapor
0.486
0.486 0.032 32
1
1
3
0.514 0.019 30 3
1
1
0.486 32 3
0.514 30 3
= 0.0254cP
Diffusivity of liquid phase
7.4 10
2.6 300.5 364.5 = 2.382 10-4 0.6 0.60 29.9 8
D L
= 5.169 10-5 cm2 /sec Vapor phase diffusivity
1 1 10 365.5 32 30 D g 2 1 1 1 29.9 3 25.94 3 3
N Scg
0.5
1.75
0.025 025 10 3 1.15 0.202 202 10
4
= 0.202 cm 2 /sec
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
65
L Kg / m 3
894.53
997.94
T L 0 C
77.5
92
L cP
0.0162
0.019
D L cm2 / sec
4.26 10-
5.169 10-
Vapor flow rate kgmoles/hr
5.367
5.367
Vapor flow rate kgmoles/hr
170.27
141.15
V Kg / m 3
1.1485
1.175
T V 0 C
78.5
92.5
V cP
0.054
0.025
D g cm2 / sec
0.189
0.202
N Scg
2.3
1.08
Enriching section a) Point efficiency, ( E og );
N g
0.776 0.00457 hW
(eqn. 18.36, page 18.15, 6th edition Perry)
0.238U a g 0.5 0.0712W 0.5
N Scg
Where, W = Liquid flow rate, m 3 / (s.m) of width of flow path on the plate, hW = weir height = 50.00 mm U a = Gas velocity through active area, m/s D L = liquid phase diffusion coefficient, = 4.26 10 cm / sec 9
U a
170.27 3600 1.1485 0.0245
1.68m / sec
66
2
128.08
q
894.53 60
D f
DC LW 2 q
W
2.39 103 m3 / sec
D f
0.2 0.15 2
2.39 10 3 60 0.175
0.175
2.276 10 4 m3 / sec.m
N Scg = Schmidt number (dimensionless), N Scg
N g
g g D g
0.776 0.00457 50 0.238 1.68 1.14850.5
0.054 10 3 1.20 0.19 10 4
0.0712 2.276 10 6
2.30.5
Number of liquid transfer unit is given by N l
= 2.3
0.39
k L a L
Where, k L = Liquid phase transfer coefficient, m/s
a = effective interfacial area for mass transfer m 2 / m 3 froth or spray on the plate, L = residence time of liquid in the froth or spray zone, s
L
L
h L Aa
1000
q
---- (eqn. 18.38, page 18.16, 6th edition Perr y)
31 .211 0.0245 1000
2.39 10 3 / 60
K L a 3.875 108 D L
19 .196 sec
0.4 U 0.5
K L a 3.875 108 4.26 109 N l
a
g 0.5 0.17
0.4 1.68 1.1485 0.5
0.5
0.17 1.1438sec1
1.1438 19.196 21 .96
N og
1
1 N g
----- (eqn. 18.34, page 18.15, 6th edition Perr y)
N l
Where, N l = Liquid phase transfer units,
67
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) N og = overall transfer units
m
m
Gm Lm Gm Lm
= Stripping factor,
0.875
5.367 367 4.796
0.979
(from graph 4.1 for enriching section straight line slope is 1.4/1.6 = 0.875) N og
1 1 0.39
0.979
0.39
21.96
E og 1 e Nog 1 e 0.39
0.33
-------- (eqn. 18.33, page 18.15, 6th edition editi on Perry)
b) Calculation of Murphree Murphree plate efficiency, ( E mv );
Now, N Pe
Z L
2
D E L
Z L = length of liquid travel, m
Z L
D Cos C 2 2 2
=
97.18 0 . 2 Cos 2 2 2
= 0.1323
L = 19.196sec
D E
6.675103V a1.44 0.922104 hl 0.00562
6.675 10 3 1.681.44 0.922 10 4 31.211 0.00562 = 0.01135m 3 /sec
68
(eqn. 18.45, page 18.17, 6th edition Perry)
Where, D E = Eddy diffusion coefficient, m 2/s N Pc
0.1323 2 0.01135 19 .196
0.0804
Eog 0.979 0.33 0.323 Now, for Eog = 0.323 and N Pe = 0.0804 value, (we have from fig.18.29a, page 18.18, 6th edition Perry)
E mv
1.8
E og
(Murphree plate efficiency) E mv = 1.8 x 0.33 = 0.60 c) Overall column efficiency ( E OC ); E OC
N t N a
Where,
E E mv
e L e
log1 E a 1 log
----- (eqn. 18.46, page 18.17, 6th edition Perry)
1
1 E mv 1
----(eqn. 18.27, page 18.13, 6th edition Perry)
----(eqn. 18.26, page 18.13, 6th edition Perry)
e = absolute entrainment of liquid L = liquid down flow rate without entrainment. E mv = Murphee Vapor efficiency, E = Murphee Vapor efficiency, corrected for recycle effect of liquid entrainment.
= fractional entrainment, moles/mole gross down flow 0.5
L Taking . G =0.162 at 80% flooding, (we have from fig.18.22, page 18.14, 6th V L edition Perry) = 0.011
69
E a
0 .6 E a
1
0.011 1 0 . 6 1 0.011
0.596
(Actual trays) N actual
N T E OC
Ideal trays Overall efficiency
Actual number of tray in enriching section =
7 0.596
= 12 trays
Stripping section a) Calculation of point efficiency, ( E og )
U a
141.15
3600 1.175 0.0245 534.45
q
997.94 60
1.362m / sec
8.925 10 3 m3 / sec
175 m 0.175
D f w
8.925 10 3 60 0.175
8.5 10 4 m 3 / sec .m
N Scg = 1.08
N g
L
0.776 0.00457 50 0.238 1.362 1.1750.5
1.080.5 31.211 0.0245
8.925 10 3 1000 60
5.14
k L a 3.875 108 5.169 109
0.4 1.362 1.175 0.5
0.5
1 = 1.076 sec
N l
0.712 8.5 10 4
1.076 5.14 5.54
70
0.17
0.63
1
N og
1 N g
1 1.068
1 0.63
N l
0.562
m
1.5 0.3
5;
m
5.54
1 e Nog 1 e 0.562 0.43
E og
b) Calculation of Murphree Murphree plate efficiency, ( E mv );
N Pe
Z l
2
D E L
6.675 10 3 1.362 1.44 0.922 922 10 4 31 .211 0.00562
D E
= 7.6810-3m3 /sec
Z L
C D c os C 2 0.1323m 2 2
L
5.14
N Pe
E og E mv E og
0.1323 2 7.68 10 3 5.14
0.444
1.068 0.43 0.46
1.6
E mv = 1.6 0.43 = 0.69 c) Overall column efficiency, ( E OC )
0.011 (refer enriching section E OC ) E a
0.69 E a
1
0.011 1 0.69 1 0.011
0.6852
(Actual trays) N actual
N T E OC
Ideal trays Overall efficiency
71
Gm Lm
5
5.367 367 25.115 115
1.068 068
Actual Number of trays in stripping section =
3 0.6852
5trays
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.
72
INSTRUMENTATION AND PROCESS CONTROL OF REACTOR
1. FI 001 001 & FI 002 are flow indicator indicator of Methanol & air respectively. 2. FV 001 & FV 002 are Methanol/ Air flow control control value respectively. 3. TI 001, PI 001 are Temperature & pressure indicator indicator of reactor Inlet 4. TI 002 PI 002 002 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 002 002 7. PDI 001 is the Pressure Difference Indicator, reads the pressure between
73
PI 001 – 001 – PI 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 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 gauges and Temperature gauges gauges are provided for reactor inlet and outlet, and also a rupture disc is mounted on the head of reactor.
74
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 i s 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. 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.
75
8) Construction economics. 9) Possible future expansion. Units can be placed in one of the two general form s. 1) Grouped layout 2) Flow line layout la yout 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. 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.
76
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 locat ed 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 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.
77
Furnace transfer lines should be short as far fa r 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 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.
78
PLANT LAYOUT BOILER SECTION
FUTURE EXPANSIONS
MARKETING & DISPATCH AREA
UTILITIES
EVAPOR REACTOR ABSORBER ABSORBER FRACTINATION ATER COLUMN
PRODUCT STORAGE AREA
SUBSTATION
WORKSHOP
MATERIALS DEPT.
PUMP HOUSE
CENTRAL CONTROL ROOM
RAW MATERIAL STORAGE AREA
FIRE & SAFETY DEPT. ADMIN BLOCK
CANTEEN
EFFLUENT TREATMENT SITE
MAIN GATE
79
HEALTH AND SAFETY FACTORS Hazard identification SAF-T-DATA
(tm)
Ratings
Health Rating: 3 - Severe (Poison) Flammability Rating: 2 – 2 – Moderate Moderate Reactivity Rating: 2 – 2 – Moderate 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
HEALTH AND SAFETY FACTORS Hazard identification SAF-T-DATA
(tm)
Ratings
Health Rating: 3 - Severe (Poison) Flammability Rating: 2 – 2 – Moderate Moderate Reactivity Rating: 2 – 2 – Moderate 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 irritati on 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 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, r edness, pain, and possibly burns.
Skin absorption may occur with symptoms s ymptoms paralleling those from ingestion. Formaldehyde is a severe skin irritant and sensitizer. Contact causes white discoloration, smarting,
80
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 s ome 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 eff ects 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 r espiration. If breathing is
difficult, give oxygen. Call a physician. Ingestion: If swallowed and the victim is conscious, dilute, inactivate, or a bsorb 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 le ast 15 minutes, lifting
lower and upper eyelids occasionally. Get medical attention immediately. Note to Physician: Monitor arterial blood gases and methanol levels afte r significant
81
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
60 C (140F) CC Fire: Flash point: 60 Autoignition temperature: 300 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 limit s
noted above. Containers may explode when involved in a fire. Fire Extinguishing Media: Water spray, dry chemical, alcohol foam, or carbon dioxide. Special Information: 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 spill s away from exposures and to dilute spills to non-flammable mixtures. Exposure controls/Personal protection Airborne Exposure Limits:
-OSHA Permissible Exposure Limit (PEL): 0.75 ppm (TWA), 2 ppm (STEL), 0.5 ppm (TWA) action l evel 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 metha nol 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
82
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: Protection: Wear impervious protective clothing, including boots, gloves, l ab 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 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, bas ements 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
83
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 unti l 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. 180 180 C (356F). Reacts violently with perchloric acid, perchloricacid-aniline mixtures, and nitromethane. Reaction with hydrochloric hydrochloric acid may form
bis-chloromethyl
ether,
an
OSHA
regulated
carcinogen.
Conditions to Avoid: Heat, flames, ignition sources and incompatibles Environmental Environmental Impact I mpact 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
84
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 dr y and wet deposition. When released rel eased 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 easil y 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 formaldeh yde 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 System (In Metal Oxide Process) Leakage of formaldehyde through pumps and equipments to be avoided.
85
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 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 f luids.
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 I NGREDIENTS: Description
CAS No.
EINECS Number
Formaldehyde
50-00-0
200-001-8
86
Methanol
67-56-1
200-659-6
Water
7732-18-5
3. HAZARD IDENTIFICATION: 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 for 15 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 p ositive 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
87
Store in tightly closed containers in a dry, cool, well ventilated, flammable liquid storage area.
8. EXPOSURE CONTROLS / PROTECTION: 50-00-0 Formaldehyde Short term value *3 mg/m3 Long term value **1.5mg m3
OES
* 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 Complete 20 oC 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, 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)
88
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 toxicity
The product is expected to be slightly toxic to aquatic life. The LC 50/96 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.
89
COST ESTIMATION (Source http://www.niir.org/projects http://www.niir.org/projects/tag/z,.,451_0_3 /tag/z,.,451_0_32/formaldehyde 2/formaldehyde/index.html.) /index.html.)
Total Capital cost of the plant for 10 tones/ day or 3000 tones/year (for 300 working days plant) = 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
90
(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) 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% (40-50% of PEC) Assuming 40% of PEC = 0.4 1.2623 = 0.505 crores
(h)
Land (1-2% of FCI or 4-8% 4-8% PEC)
91
Assuming 1.5% of FCI =
1 .5 100
4.9
= 0.0735 crores (i)
Direct cost (DC) = 1.2623 + 0.505 + 0.227 + 0.505 + 0.1894 + 0.505 + 0.505 + 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% (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
0.921 4.9
100
= 18.8%
92
Working Capital (10-20% of FCI) Assuming 15% of FCI = 0.15 4.9 = 0.735 crores 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 – (0.4 – 1% 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
93
= 0.09 [0.505 + 0.0735] = 0.052 crores Fixed charges = 0.588 + 0.098 + 0.0245 + 0.052 = 0.7625 crores 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
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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) 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
95
= 0.305 crores Manufacturing Cost = fixed charge + Direct product cost + plant overhead cost = 0.765 + 2.12 + 0.305 = 3.188 crores. General Expenses
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 – (2 – 6%) 6%) of TPC) Assuming 4% of TPC = 0.04 3.8125 = 0.1525 0.1525 crores (c)
R & D ( 2 – 5% – 5% of TPC) Assuming 3% of TPC = 0.03 3.8125 = 0.114 crores
(d) Finances (0 – (0 – 7% 7% of Total 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
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= 3.188 + 0.4238 = 3.612 crores Gross earnings = total income – income – total total production cost for 1 kg of Formalin Selling price is Rs. 25/-
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 – earning – Total Total product cost = 6 – 6 – 3.8125 3.8125 = 2.1875 crores Net annual earning after depreciation = 2.1875 – 2.1875 – 0.588 0.588 = 1.5995 crores Net profit after tax (46% tax rate) = (1 – (1 – 0.46) 0.46) 1.5995 = 0.8638 crores
FCI = 4.59 crores. Payout period =
=
FCI FCI (Net profi profit after tax
depreciation)
4.59 (0.8638
0.588)
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= 3.16 years
Rate of return =
=
Net Net profit profit Fixed capital investment
(0.8638 0.588) 4.59
100
= 31.63%
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