(A) STATEMENT OF THE PROBLEM

Design a plant to manufacture 10 Tonnes/day of FORMALIN

Formaldehyde is manufactured by vapor phase catalytic oxidation of Methanol. Methanol

is diluted with water and is evaporated in the evaporator. The Methanol water vapor

mixture is superheated and mixed with air. Then the mixture is admitted to the Packed bed

catalytic reactor containing silver grains as catalyst. The reaction temperature is

maintained at 650C. Assume a residence time of 10 to 20 seconds. The reaction is first

order with respect to methanol and half order with respect to oxygen. Heat recovery is

made from the outgoing product gases using a waste heat boiler. The product gases are

sent to two absorption columns in series wherein water and dilute Formalin are used for

absorption. Finally the formaldehyde is purified in a distillation column and the methanol

is recycled back as a top product. Assume 90% conversion of methanol. Assume any

missing data suitably if required.

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, 4th edition, 1994

Volume. 11(Page 929 – 947).

5. Industrial & Engg Chemistry by S.J.Green & Raymond.E.Vener,

Volume 47 (Page-103-108) 1955

6. “Introduction to Chemical Engg Thermodynamics “ by Smith & Vannes,

6th edition. (appendix –C)

7. Perry ‘s “Chemical engineer’s handbook” 6th Edition (Chapter 3 & 18)

8. “Plant Design and Economics for Chemical Engineers”

by Peter & Timmerhaus, 4th edition (Chapter – 6).

9. “Process Design of Equipment” Volume 2. by S.D. Dawande, (Page 49-50).

10

.

“Process Equipment Design” by M.V. Joshi 3rd edition. Chapter-9 & page 129

11

.

“Process heat transfer” by Donald Q.Kern (Page 114, 147-148, 836,838)

12

.

“Reaction kinetics for chemical engineers” by Stanley Walas (Page 193) Chapter 8

13

.

(Source http://www.niir.org/projects/tag/z,.,451_0_32/formaldehyde/index.html.)

2

NOMENCLATURE

A Heat transfer surface

a ,a t , asFlow area in general, for tube side, and for shell side, respectively

B Baffle spacing

Cp Specific Heat

C ' Clearance between tubes.

D Diameter

DeEquivalent diameter

d p ,d tDiameter of particle, and diameter of tube

F Feed flow rate

f Friction factor

G ,Gs ,GtMass Velocity in general, for shell side, and for tube side, respectively

H Enthalpy

h ,hi , hoHeat transfer coefficient in general, for inside fluid, and for outside

fluid, respectively

hioValue of hi when referred to outside diameter

ID Inner diameter

J Joules

jH Heat transfer factor, dimensionless

K Temperature measurement, Kelvin

k Thermal conductivity

L Length

M Mass flow

mS ,mwMass flow rate of steam, and water respectively

N Newton

3

N tNumber of tubes

NRE Reynolds number

OD Outer diameter

P Pressure

P Pressure drop

PT Tube pitch

Q Heat flow

RdTotal dirt factor

T Temperature

U ,UC ,U dOverall coefficient of heat transfer, clean coefficient, design coefficient

V Velocity

T Temperature Difference

X Vapor pressure of water

Latent heat of vaporization or condensation

φ Viscosity ratio (μ/ μw )

Viscosity

μwViscosity at the tube wall

Density

Voidage

4

PERFORMANCE DATA OF MAIN EQUIPMENT

REACTOR

Volume of the reactor 0.156 m3

Diameter of the reactor 0.83m

No. of tubes 80

Length of each tube 1 m

ID of tube 50 mm

OD of tube 55 mm

Heat transfer area 6.80 m2

Pressure drop 26.7 kg/ m2

Material of construction

(a) Shell Carbon Steel

(b) Tube SS 410

Catalyst detail

Name: Silver

Diameter 3.5 mm

Void fraction 0.36

5

FRACTIONATION COLUMN

No. of trays 17

No. of plate in enriching section 12

No. of plate in stripping section 5

Column height 6.8 m

Column diameter 0.2 m.

Feed entry 8th tray From top

Operating pressure 1 atm.

Type of plates Sieve tray

Tray spacing 400mm

Material of construction S.S 316

6

SUMMARY OF MECHANICAL DESIGN

Shell side

Shell thickness 10 mm

Inlet nozzle 30 mm.

Head thickness 10 mm

Baffle thickness 6.5 mm

No. of tie rods 6

Dia. of tie rods 12.5 mm

Flange thickness 61 mm

Gasket diameter 835 mm.

Gasket width 24 mm

Tube side

Thickness of tube 5 mm

Thickness of tube sheet 14 mm

Ring gasket width 22 mm

Minimum pitch circle diameter 940 mm

No. of bolts 20

Size of bolt M30

Gasket Flat metal jacketed Asbestos filled

Inlet nozzle dia. 85 mm

Outlet nozzle dia. 93.6 mm

7

INTRODUCTION

Formaldehyde, H2C=O, is a reactive molecule, the first of the series of aliphatic aldehydes

and one of the most important industrial chemicals. Formaldehyde is a colorless gas at

ordinary temperature. Commercially formaldehyde is manufactured in the form of a water

solution usually containing 37% by weight of dissolved formaldehyde, this solution is

called formalin. In 1983 formaldehyde ranked 26th in production among Unites states

chemical products, with an output of 5.40 billion lb of equivalent 37 wt% aqueous

solution. Annual worldwide production capacity now exceeds 15 106 tons

(calculated as 37% solution).Because of its relatively low cost, high purity, and variety of

chemical reactions, formaldehyde has become one of the world’s most important industrial

and research chemicals. Products from formaldehyde are used extensively in the

automobile, construction, paper and textile industry.

HISTORY

Formaldehyde’s public image has always been associated with the funeral homes, doctor

offices and biology classes as an embalming fluid, a disinfectant and a preservative .In

1859, Russian scientist Alexander Mikhailovich Butlerlov discovered Formaldehyde,

accidently as he investigated the structure of organic compounds. Nine years later,

German scientist August Wilhelm Hofmann found a reliable way to make it. Hofmann

Passed a mixture of methanol and air over a heated platinum spiral and then identified

formaldehyde as the product. This method led to the major way in which the

8

Formaldehyde is manufactured today, the oxidation of methanol with air using a metal

catalyst primarily of silver or molybdenum oxide. In 1905 , Dr.Leo Baekeland in Yonkers,

New york made a major breakthrough in the technology of polymer later named Bakelite

after him. The ingredients were Phenol and Formaldehyde, by the 1920’s the growth of

this resin strained wood alcohol (Methanol) producing capacity, but the revolutionary

development of methane reforming route to methanol relieved the situation. Despite the

radical shift in methanol technology, the process for formaldehyde based on methanol

feedstock has remained virtually unchanged even to today, despite volume growth making

it one of the top 25 commodity chemicals.

LITERATURE SURVEY

PYSICAL PROPERTIES

Formaldehyde monomer

Pure anhydrous formaldehyde is a colorless gas at ordinary temperature and at a molecular

weight of 30.26 is sightly heavier than air. It condenses on cooling to -19C and freezes to

a crystalline solid at -118C. The gas is characterized by a pungent odor and is judged

moderately irritating to the eyes, nose and throat by 20% of the population exposed to

concentrations in the 1.5 to 3.0 ppm range. Dry formaldehyde gas is stable and shows no

polymerization tendency at temperature as high as 100C. However, small amounts of

water or other impurities can cause rapid polymerization to poly(oxymethylenes).

Anhydrous formaldehyde gas is readily soluble in polar solvents such as water, methanol,

and n-propanol. It is only moderately soluble in nonpolar solvents such as ethyl ether,

chloroform, and toluene. In water and methanol, its heat of solution is approximately

15kcal/g-mol.  A summary of physical properties of monomeric formaldehyde is given in

Table 1.1

9

Formaldehyde solutions

Formaldehyde is produced and distributed as a water solution, the “standard” strength

being 37 wt%, this being a typical concentration and also the basis for making production

comparisons. Representative commercially available solutions are shown in Table 1.2

Table 1.1 Properties of Monomeric Formaldehyde

Property Value

Density, g/cm3

at -80C 0.9151 at -20C 0.8153Boiling point at 101.3 kPa C -19Melting point, C -118Vapor pressure, Antoine constants, Pa A 9.21876B 959.43C 243.392

Heat of vaporization, ΔH v at 19C, kJ/mol 23.3

Heat of formation, ΔH f at 25C, kJ/mol -115.9

Std free energy, ΔG f at 25C, kJ/mol -109.9Heat capacity, Cp, J/mol.K 35.4

Entropy,So

,J/(mol.k) 218.8Heat of combustion, kJ/mol 563.5Heat of solution at 23C kJ/mol in water 62 in methanol 62.8 in 1-propanol 59.5 in 1-butanol 62.4Critical constants Temperature, C 137.2-141.2 Pressure, MPa 6.784-6.637Flammability in air Lower/upper limits, mol 7.0/73 Ignition temperature, C 430

Varying amounts of methanol are included in the solutions as stabilizers and because of

the expenses of removal of methanol for uses where it is not harmful.

Formaldehyde is highly soluble in water, but in the liquid it reacts readily with water to

10

form the hydrate, methylene glycol, which itself then tends to polymerize to

poly(oxymethylene glycols). Also, some hemiformals and a small amount of formic acid

are produced. At chemical equilibrium the amount of unhydrated formaldehyde is small,

approximately 0.1% at 60C.

In methanol-formaldehyde-water solutions, increasing the concentration of either

methanol or formaldehyde reduces the volatility of the other. The flash point varies with

composition, decreasing from 83 to 60C as the formaldehyde and methanol concentration

increase.

Formaldehyde solutions exists as a mixture of oligomers, HO(CH2O)nH. Methanol

stabilizes aqueous formaldehyde solutions by decreasing the average value of n. Hence

methanolic solutions can be stored at relatively low temperatures without precipitation of

polymer.

Table 1.2 Typical Analyses and Physical properties of Formaldehyde solutions

USP Grades Low-Methanol GradesFormaldehyde (wt%) 37.1 37.1 37.1 44.1 50.3Methanol (wt%) 7.0 11.0 0.9-1.3 0.9-1.3 0.9-1.3Acidity as formic (wt%) 0.012 0.012 0.012 0.02 0.018 Iron as Fe (ppm) 0.3 0.3 0.3 0.3 0.3Turbidity, Hellige 1.5 1.5 2 2 2 Color, APHA 5 5 5 5 5Density (g/cm3), 18C 1.10 1.09 1.11 1.12 1.13Boiling point (C) 97.2 96.7 99.0 99.1 99.5Viscosity (cP), 25C 2.5 2.6 2.0 1.7 1.6Specific heat (cal/g.C) 0.8 0.8 0.8 0.7 0.7Flash point (C) 69 60 83 80 79

Chemical properties

Formaldehyde is noted for its reactivity and its versatility as a chemical intermediate. It is

used in the form of anhydrous monomer solutions, polymers, and derivatives.

Anhydrous, monomeric formaldehyde is not, available commercially. The pure, dry gas

is relatively stable at 80-100C but, slowly polymerizes at lower temperatures. Traces of

11

polar impurities such as acids, alkalies, and water greatly accelerate the polymerization.

Formaldehyde in water solution hydrates to methylene glycol;

Which in turns polymerizes to poly(methylene glycols), HO(CH2O)nH, also called

polyoxymethylenes. From these polymers a specific product, paraformaldehyde (or

“parafrom”), is obtained commercially. Paraformaldehyde is the name given to

polyoxymethanlenes with n values from 8 to 100. It is produced by the vacuum distillation

of concentrated formaldehyde solutions, and is available commercially in powder,

granular, or flakes forms. It has the characteristic pungent odor of formaldehyde, and melts

in the range of 120 to 170C. It is flammable, with a flash point of about 93C. A typical

formaldehyde solution may be obtained by dissolving paraformaldehyde in water.

Paraformaldehyde may be heated together with a strong acid to produce trioxane, the

cyclic trimer of formaldehyde.

This is a colorless crystalline material, melting at 62-64C, boiling without decomposition

at 115C, and having a flash point of 45C. Concentrations of trioxane between 3.6 and

28.7 vol% in air are explosive.

Formaldehyde may be reduced to methanol over a number of metal and metal oxide

catalysts. It may be condensed with itself in an aldol-type reaction to yield lower hydroxyl

aldehydes, hydroxy ketones, and other hydroxyl compounds. Formaldehyde and

acetaldehyde may be reacted in the presence of sodium hydroxide to form pentaerythritol

and sodium formate;

12

Acetylene may be reacted with formaldehyde to form 2-butyne-1,4-idol which, when

hydrogenated, yields 1,4-butanediol;

Formaldehyde and aniline may be condensed to form diphenylmethane diamine:

Reaction of this product with phosgene yields methlenebis (4-phenyl iso-cyanate), or

“MDI,” one of the important types of commercial isocyanates.

Liquid phase condensation of formaldehyde with propylene, catalyzed by BF3 or H2SO4,

gives butadiene.

Hydrogen cyanide reacts with aqueous formaldehyde in the presence of bases to

produce glyconitrile:

HCHO + HCN HOCH2—C=N

This extremely toxic material is an intermediate in the synthesis of nitrilotriacetic acid

(NTA), EDTA, and glycine.

Reaction of formaldehyde, methanol, acetaldehyde, and ammonia over a silica alumina

catalyst at 500C gives pyridine and 3-picoline. This forms the basis of commercial

processes for making pyridines from various aldehydes.

Formaldehyde reacts with syn gas (CO,H2) to produce added value products. Ethylene

glycol (EG).

Alternative choice of manufacture

Currently, the only production technologies for formaldehyde of commercial

significance are based on the partial oxidation and dehydrogenation of methanol using

silver catalyst, or partial oxidation of methanol using metal oxide-based catalyst.

13

Development of New Processes. There has been significant research activity to

develop new processes for producing formaldehyde. Even though this work has been

extensive, no commercial units are known to exist based on the technologies discussed in

the following.

One possible route is to make formaldehyde directly from methane by partial oxidation.

This process has been extensively studied. The incentive for such a process is reduction of

raw material costs by avoiding the capital and expense of producing the methanol from

methane.

Another possible route for producing formaldehyde is by dehydrogenation of methanol

which would produce anhydrous or highly concentrated formaldehyde solutions. For some

formaldehyde users, minimization of the water in the feed reduces energy costs, effluent

generation, and losses while providing more desirable reaction conditions.

A third possible route is to produce formaldehyde from methylal that is produced from

methanol and formaldehyde. The incentive for such a process is twofold. First, a higher

concentrated formaldehyde product of 70% could be made by methylal oxidation as

opposed to methanol oxidation, which makes a 55% product. This higher concentration is

desirable for some formaldehyde users. Secondly, formaldehyde in aqueous recycle

streams from other units could be recovered by reacting with methanol to produce methylal

as opposed to recovery by other more costly means, eg, distillation and evaporation.

Development of this processes is complete.

Specification and Quality control

Formaldehyde is sold in aqueous solutions with concentrations ranging from 25 to 56 wt%

HCHO. Product specifications for typical grades are summarized in Table 1.3.

Formaldehyde is sold as low methanol (uninhibited) and high methanol (inhibited) grades.

Methanol is used to retard paraformaldehyde formation.

14

Procedures for determining the quality of formaldehyde solutions are outlined by

ASTM. Analytical methods relevant to Table 1.3 follow: formaldehyde by the sodium

sulfite method (D2194); methanol by specific gravity (D2380); acidity as formic acid by

titration with sodium hydroxide (D2379); iron by colorimetry (D2087); and color (APHA)

by comparison to platinum-cobalt color standards (D1209).

Table 1.3 Formaldehyde specifications

Property Methanol inhibited grades low methanol uninhibited grades

Formaldehyde, wt% 37 37 37 44 50 56

Methanol, wt% (max) 6-8 12-15 1.0-1.8 1.5 1.5-2.0 2.0

Acidity, wt% (max) 0.02 0.02 0.02 0.03 0.05 0.04

Iron, ppm (max) 0.5 0.05 0.5-1.0 0.5 0.5 0.75

Color, APHA (max) 10 10 10 10 10 10

STORAGE AND TRANSPORTATION

As opposed to gaseous, pure formaldehyde, solutions of formaldehyde are unstable. Both

formic acid (acidity) and paraformaldehyde (solids) concentrations increase with time and

depend on temperature. Formic acid concentration builds at a rate of 1.5-3 ppm/d at 35 C

and 10-20 ppm/d at 65C. Trace metallic impurities such as iron can boost the rate of

formation of formic acid. Although low storage temperature minimizes acidity, it also

increase the tendency to precipitate paraformaldehyde.

Paraformaldehyde solids can be minimized by storing formaldehyde solutions above a

minimum temperature for less than a given time period. The addition of methanol as an

inhibitor or of another chemical as a stabilizer allows storage at lower temperatures and/or

for longer times. Stabilizers for formaldehyde solutions include

hydroxypropylmethylcellulose, methyl- and ethylcellulose, poly(vinyl alcohol)s, or

isophthalobisguanamine at concentrations ranging from 10 to 1000ppm. Inhibited

15

formaldehyde typically contains 5-15 wt% methanol.

Most formaldehyde producers recommend a minimum storage temperature for both

stabilized and unstabilized solutions. The minimum temperature to prevent

paraformaldehyde formation in unstabilised 37% formaldehyde solutions stored for one to

about three months is as follows: 35C with less than 1% methanol; 21C with 7%

methanol; 7C with 10% methanol; and 6C with 12% methanol.

Materials of construction preferred for storage vessels are 304-, 316-, and 347-type

stainless steels or lined carbon steel.

USES

Formaldehyde is a basic chemical building block for the production of a wide range of

chemicals finding a wide variety of end uses such as wood products, plastics, and coatings.

Amino and Phenolic Resins. The largest use of formaldehyde is in the manufacture of

urea-formaldehyde, phenol-formaldehyde, and melamine-formaldehyde resins, accounting

for over one-half (51%) of the total demand. These resins find use as adhesives for binding

wood products that comprise particle board, fiber board, and plywood. Plywood is the

largest market for phenol-formaldehyde resins; particle board is the largest for urea-

formaldehyde resins.

Phenol-formaldehyde resins are used as molding compounds. Their thermal and

electrical properties allow use in electrical, automotive, and kitchen parts. Other uses for

phenol-formaldehyde resins include phenolic foam insulation, foundry mold binders,

decorative and industrial laminates, and binders for insulating materials.

Urea-formaldehyde resins are also used as molding compounds and as wet strength

additives for paper products. Melamine-formaldehyde resins find use in decorative

laminates, thermoset surface coatings, and molding compounds such as dinnerware.

1,4-Butanediol. market for formaldehyde represents 11% of its demand. It is used to

16

produce tetrahydrofuran (THF), which is used for polyurethane elastomer; -

butyrolactone, which is used to make various pyrrolidinone derivatives; poly(butylenes

terephthalate) (PBT), which is an engineering plastic; and polyurethanes.

Polyols. The principal ones include pentaerythritol, trimethylolpropane and neopentyl

glycol. These polyols find use in the alkyd resin and synthetic lubricants markets.

Pentaerythritol is also used to produce rosin/tall oil esters and explosives (pentaerythritol

tetranitrate). Trimethylolpropane is also used in urethane coatings, polyurethane foams,

and multifunctional monomers. Neopentyl glycol finds use in plastics produced from

unsaturated polyester resins and in coatings based in saturated polyesters.

The formaldehyde demands for pentaerythritol, trimethylolpropane, and neopentyl

glycol are about 7, 2, and 1% respectively, of production.

Acetal Resins. These are high performance plastics produced from formaldehyde that

are used for automotive parts, in building products, and in consumer goods. The acetal

resins formaldehyde demand are 9% of production.

Hexamethylenetetramine. Pure hexamethylenetetramine (also called hexamine and

HMTA), the production of hexamethylenetetramine consumes about 6% of the U.S.

formaldehyde supply. Its principle use is as a thermosetting catalyst for phenolic resins.

Other significant uses are for the manufacture of RDX (cyclonite) high explosives, in

molding compounds, and for rubber vulcanization accelerators. It is an unisolated

intermediate in the manufacture of nitrilotriacetic acid.

Slow-Release Fertilizers. Products containing urea-formaldehyde are used to

manufacture slow release fertilizers. These products can be either solids, liquid

concentrates, or liquid solutions. This market consumes almost 6% of the formaldehyde

produced.

Methylenebis(4-phenyl isocyanate). This compound is also known as methyl

17

diisocyanate (MDI). Its principal end use is rigid urethane foams; other end uses include

elastic fibers and elastomers. Total formaldehyde use is 5% of production.

Chelating Agents. The chelating agents produced from formaldehyde include the

aminopolycarboxylic acids, their salts, and organophosphonates. The largest demand for

formaldehyde is for ethylenediaminetetraacetic acid (EDTA); the next largest is for

nitrilotriacetic acid (NTA). Chelating agents find use in industrial and houseland cleaners

and for water treatment. Overall, chelating agents represent a modest demand for

formaldehyde of about 3%.

Formaldehyde-Alcohol Solutions. These solutions are blends of concentrated aqueous

formaldehyde, the alcohol, and the hemiacetal. These solutions are used to produce urea

and melamine resins; the alcohol can act as the resin solvent and as a reactant.

Paraformaldehyde. It is used by resin manufacturers seeking low water content or more

favorable control of reaction rates. It is often used in making phenol-urea-.resorcinol-, and

melamine-formaldehyde resins. It is EPA registered disinfectant, “Steri –dri” sanitizer and

fungicide for barber and beauty and for households, ships, bedding, clothing,

nonfood/non/feed transporting trucks.

Trioxane and Tetraoxane. It is mainly used for the production of acetal resins.

Other Applications. Formaldehyde derivatives, such as dimethyl dihydroxyethylene, are

used in textiles to produce permanent press fabrics. Other formaldehyde derivatives are

used in this industry to produce fire-retardant fabrics, Paraquat made from Pyridine

chemicals, are used for agricultural chemicals (Herbicides). Formaldehyde and

paraformaldehyde have found use as a corrosion inhibitor, hydrogen sulfide scavenger, and

biocide in oil production operations such as drilling, waterflood, and enhanced oil

recovery. Other used for formaldehyde and formaldehyde derivatives include fungicides,

embalming fluids, silage preservatives, and disinfectants.

18

Note: The requirement of formaldehyde for certain applications like Polyacetal, MDI,

1,4-Butanediol and Neopentylglycol does not exist in India.

PROCESS DESCRIPTION

Fresh methanol, which is free from iron carbonyls and sulfur compounds (catalyst poisons)

is combined with recycle methanol and diluted with equal amount of water, which is then

pumped to a evaporator, where methanol is vaporized along with water. Methanol-water

vapor mixture is then superheated to 650C in a steam superheater using low pressure

steam. Air is drawn via, a filter and compressed in a blower for feed to the process.

Filtered air is preheated with outgoing reactor effluent gases and then superheated to 650 C

in a additional superheater. Superheated air and methanol water vapor is mixed in a mixer,

the mixture is then passed into a Fixed bed tubular catalytic reactor which is nothing but a

1-1 shell and tube heat exchanger, the reactor tubes is packed with silver grains as catalyst,

where the below mentioned reactions takes place i,e formaldehyde is produced both by

oxidation and by dehydrogenation of methanol. About half of the methanol goes to each

reaction, hence the combination is net exothermic.

CH3OH + ½ O2 HCHO + H2O +38 kcal/g.mol (exothermic)

CH3OH HCHO + H 2 -20.3 kcal/g.mol (endothermic)

The conversion of methanol is 90% and the reactor pressure is kept slightly above

atmospheric. Methanol-air ratio in reactor is kept above the rich side of the explosive

limits (6.5 to 36.5 vol% in air), i.e, above 36.5%. Since the reactor temperature is to be

maintained at 650C the net exothermic heat of reaction is removed by circulating water

through the shell side of the reactor, which in turn takes away net exothermic heat of

reaction to give low pressure steam. The reactor effluent gases, at 650C is brought down

19

to 120C by preheating the feed air in a air-reactor effluent gases preheater. The reactor

effluent contains oxygen, nitrogen, hydrogen, HCHO, water, and unreacted methanol. The

effluent gases enters the packed bed absorbers (2nos.) in series where dilute formalin and

water are used for absorption, here HCHO is cooled and gets dissolved in water, heat of

solution is evolved due to absorption, assuming heat of solution to be negligible. Some

amount is required to vaporize water which passes via the vent. Heat given out by the

entering gases to reach a temperature of 25C from 120C is removed by using a coolant

which is circulated via tower external circulation. The bottom material from 2nd absorber

contains HCHO-methanol water solution which is sent to a fractionation column. The

fractionation column is sieve plate column containing 17 trays where the feed is introduced

on 8th tray from top. The column overhead temperature is maintained around 65C and

bottom reboiler temperature of 93C, methanol is condensed in the overhead condenser as

top product and recycled back to evaporator. The bottom contains 37% wt of HCHO and

less than 1% of methanol and remaining water. The water content of the bottoms is

controlled by the amount of makeup water added at the top of the absorber. The tail gas

coming out of 2nd absorber contains hydrogen which is used a source of fuel in boiler for

steam generation. The formalin product tapped out of the distillation still is cooled and sent

for storage.

RAW MATERIAL REQUIPMENTS

(For 10 Tones Per Day)

1. METHANOL - 4389.50kgs/ Day

2. AIR - 4735.2kgs/Day

3. CATALYST - 90.72kgs/day

20

mixer

Air Blower

Atmosphericair

Water

Fresh Methanol

Recycle Methanol

ABSORBER 2

Off gas

Steam

Coolant

Coolant

Fractionation Column

Reboiler

Water

Fig 2.1 Manufacture of Formaldehyde by Vapor-phase catalytic oxidation of methanol

ProcessWater

Coolant

Formalin (37% Formaldehyde)

FIXED BED TUBULAR CATALYTIC REACTORTemp -650CPressure – 1.05 atm

Methanol super heater

Filter

EVAPORATOR

Steam

Condensate

AirPreheater

Air Superheater

ABSORBER 1

Steam

21

MATERIAL BALANCE

Formaldehyde at atmospheric condition is a gas, commercially formaldehyde is dissolved

in water known as Formalin (37% wt of formaldehyde)

10 TPD of Formalin

Formaldehyde production per day, 10 0.37 = 3.7tons = 3700kgs.

Weight of water in final product per day, = 10000 – 3700 = 6300kgs.

Formaldehyde capacity per hour,

370024 = 154.167kgs

Assuming wastage of 0.1% through absorber

0 .1100

× 154 .167 = 0 . 15kgs .

Total capacity of formaldehyde per hour, 154.167 + 0.154 = 154.32kgs.

CH3 OH + 12O2→HCHO+H2O

-----(3.1) +38kcal/gmole (exothermic)

CH3 OH →HCHO+H2 -----(3.2) -20.3kcal/gmole (endothermic)

For silver process, about half of methanol goes to each reaction.

Basis: 1 hour of operation of the plant

Formaldehyde production

154 . 3230

= 5 . 144 kgmoles(MW of HCHO = 30)

Methanol required for the process = 5.144kgmole.

Conversion of methanol is 90%

Therefore Methanol fed =

5. 1440 .90

= 5 . 715 kgmole

50% Methanol for each reaction, therefore for reaction 3.1 & reaction 3.2

John Mcketta, Vol 23, Page 358

22

Methanol fed =

5. 7152

= 2 .8575 kgmoles

Oxygen required =

2. 85752

= 1 . 42875 kgmoles

Air required =

1. 428750 .21

= 6 . 8035 kgmoles

Nitrogen sent along with air = 6.8035¿ 0.79 = 5.3747 kgmole.

Since Methanol-air have explosive range of 6.7% to 36.5% (mole%), Methanol oxidation

must be brought outside this range. That is keeping methanol ratio above 36.5%

Methanol mole% =

5 .715(5 .715+6 .8035 )

×100= 45 . 65% > 36 .5%.

Material Balance around Evaporator

Methanol is diluted with water and vaporized in evaporator assuming equal quantities of

water and methanol is mixed.

Component Entering Leavingkgmoles kg kgmoles kg

Methanol 5.7155 182.896 5.7155 182.896Water 5.7155 102.879 5.7155 102.879

Material Balance around Methanol-water vapor Super heater

Here material vaporized is heated to 650C. Here there is no loss of reactant and the

material balance is same as above.

Material Balance for Air blower and Air filters

Here air from atmosphere is sucked from the blower and passed to air filter and to the

process. Here is no chemical change involved.

Component Entering Leavingkgmole kg kgmole kg

Air 6.8035 197.30 6.8035 197.30

Material Balance of Air Pre-heater

Here air is heated from atmospheric condition to the reaction temperature. Since there is no

23

change in material balance. Therefore the material balance is same as above.

Material Balance around the MixerHere methanol-water vapor and air are mixed before to the reactor.

Component Entering Leaving kgmole kg kgmole kg

Methanol 6.8035 197.30 6.8035 197.30Water-vapor 5.7155 182.896 5.7155 182.896Air 5.7155 102.879 5.7155 102.879

Material Balance around the Reactor

Conversion is 90% based on Methanol

Moles of formaldehyde formed = 5.144 kgmole.

Moles of Methanol reacted = 4.144 kgmole.

Moles of Methanol unreacted = 0.5715 kgmole.

Moles of oxygen reacted =

2. 8575×0 .92

= 1 . 2858 kgmoles

Excess oxygen = oxygen supplied oxygen reacted

= 1.4287 1.2858 = 0.14287 kgmole.

Moles of hydrogen formed = 2.8575 0.9 = 2.5717 kgmole.

Moles of water formed = 2.5717 kgmole.

Total moles of water = Moles of water vaporized in evaporator + Moles of water formed

= 5.7155 + 2.5717

= 8.28725 kgmole.

Reactor

CH3OH=5.7155kgmolesH2O =5.7155kgmolesO2=1.42875kgmolesN2=5.3748kgmoles

HCHO=5.144kgmolesCH3OH=0.5715kgmolesH2O = 8.233kgmolesO2 = 0.14287 kgmolesN2=5.3748kgmolesH2 = 2.5717kgmoles

24

Reactor Mass entering = Reactor Mass leaving

Component Entering Leaving

kgmole kg kgmole kg

CH3 OH 5.7155 182.896 0.5715 18.288

O2 1.4287 45.72 0.14287 4.5718

N2 5.3748 150.4944 5.3748 150.4944

H2O 5.7155 102.879 8.2872 149.1705

HCHO 0 0 5.144 154.32

H2 0 0 2.5717 5.1435

Total 18.2345 481.9894 22.0921 481.9882

Quenching : Waste heat boiler comes under this unit. The material goes through without

any change.

Material Balance around Absorber.

Here we assume that all the formaldehyde gets absorbed in the absorber. So the gases

leaving contains N2, O2, H2 & water vapor(traces) Assuming no, N2, O2 & H2 are absorbed.

Therefore kgmoles of vent gas on dry basis = 0.14287 (O2 ) + 5.3748 (N2 ) + 5.14

(HCHO) + 2.5715 (H2) + 0.5715 (CH3 OH )

= 13.8047kgmoles = G

Now the temperature of vent gases = 25C

Vapor pressure of water at 25C = 23.7 mmHg = X

Pressure in absorber = 760 mmHg = P

25

Moles of water present in vent gas = G

XP−X

= 13.8042

23 .7(760−23. 7 )

= 0 . 444 kgmoles = 8kgs.

Amount of water leaving with formalin solution 37% (by wt)

=

154 . 320 .37

= 417 .08 kg

Amount of formalin solution – Amount of HCHO in formalin solution

= 417.08 – 154.32 = 262.76 kgs.

Amount of water present in gas entering the absorber = 149.1705 kgs

Amount of water added to absorber =

[ Amount of water leaving ¿ ] ¿¿

¿¿–

[ Amount of water present in entering gas ¿ ] ¿¿

¿¿

262.76 + 8 – 149.1705 =121.5895 kgs.

Overall Material Balance around Absorber

1) Liquid stream entering

Component kgmole kg

Water 6.755 121.5895

2) Gases stream entering

Component kgmoles kgs

HCHO 5.144 154.32

H2O 8.2872 149.1705

N2 5.3748 150.4944

O2 0.1428 4.5718

CH3 OH 0.5715 18.288

26

H2 2.5717 5.1435

Total 22.0921 481.9882

Net total input = 22.0921 + 6.755 = 28.8472 kgmoles.

Net total input = 481.9882 + 121.59 = 603.5782 kgs.

Material output (Vent gas Leaving)

Component kgmoles kg

H2O 0.444 8.0

N2 5.3748 150.4944

O2 0.1428 4.5718

H2 2.5717 5.4135

Total 8.5334 168.2097

Material Output (Solution leaving)

Component kgmoles kg

HCHO 5.144 154.32

H2O 14.6 262.76

CH3 OH 0.5715 18.288

Total 20.3155 435.368

Net output (kgmoles) = 20.315 + 8.5334 = 28.8489 kgmoles.

Net output (kgs) = 435.368 + 168.2097 = 603.5777 kgs.

Material Balance around Distillation column

27

Fractionation ColumnHCHO = 5.144kgmoleWater = 14.6kgmolesMethanol = 0.5715kgmolesFEED= 20.3155

Methanol =0.5658kgmoleHCHO = 0.005144kgmoleDISTILLATE = 0.57094

Methanol =0.005715kgmoleHCHO = 5.1389kgmoleWater = 14.6kgmoleRESIDUE = 19.745

1) Feed Entering

Component kgmole kg

HCHO 5.144 154.32

H2O 14.6 262.76

CH3 OH 0.5715 18.288

Total 20.3155 435.368

28

2) Distillate leaving

Component kgmoles kg

CH3 OH 0.5658 18.105

HCHO 0.005144 0.154

Total 0.56631 18.259

3) Residue

Component kgmoles kg

HCHO 5.1389 154.167

CH3 OH 0.005715 0.183

H2O 14.6 262.76

Total 19.7446 417.11

Total mass entering (Feed) = Total mass leaving (Distillate + Residue)

435.368kgs = 435.369 (18.259 + 417.11)

29

ENERGY BALANCE

Heat Balance around Evaporator

Here raw material methanol and water is vaporized. Methanol gets heated from 25C to its

B.P 64.7C and then vaporizes at the same temperature.

Quantity of heat required = mCP ΔT+mλ

To calculate for Methanol using Kistya kowsky equation (considering methanol to be a

non-polar liquid)

λbTb

=8 .75+4 . 571 log 10T b (T b = 337.9K)

λbTb

=8 .75+4 . 571 log(337 . 9)=20 .34

b = 20.34 337.9 = 6860 cal/gmole = 28722.82J/gmole

Cp of Methanol between 25C and 64.7C

Cp=a+ b

2(T2+ T1 )+

c3 (T 2

2+T 1T 2+T12)

a = 4.55; b = 2.186 10-2; c = -0.291 10-5

Cp=4 .55+2 .186×10−2

2(337 .7 +298 )+-0 . 291×10-5

3(337 .72+298×337 .7+2982)

Cp = 11.205 cal/gmole.K = 46.91J/gmole.K

q1 = heat required to raise the temperature of methanol from 298K to 337.9K

q1=mCP dT = 5.7155 103 gmole 46.91J/gmole.K 39.7K

= 10.6453 106 J.

q2=mλ= 5.7155 103 gmole 28722.82J/gmole

= 164.1653 106J

Total heat added = q2+q2

= 10.6453 106 J + 1641653 106 J = 174.810 106 J

30

Heat required for water to vaporize = mCP ΔT+mλ

= 5.7155 103 4.18 103 (100-25) + 5.7155 2185 103

= 1.8043 109 J

Heat balance:

Heat required to vaporize = Heat given away by steam

Amount of steam required assuming steam available at 20psia

Enthalpy of steam = 2724 kJ/kg

Mass of steam required, (mS ) =

qλ =

174 .810×106+1 . 8043×109

2724000

mS = 726kg at 3bar

Heat balance of Pre-heater for Methanol vapors

Here gaseous vapors of methanol and water are heated from 64.7C to 650C (337.9K

to 923.2K)

Cp=a+ b2(T2+ T1 )+

c3 (T 2

2+T 1T 2+T12)

Cp=4 .55+2 .186×10−2

2(923 +337 .7 )+-0 . 291×10-5

3(9232+337 .7×923+337 .72 )

= 4.55 + 13.78 – 1.235 = 17.099 cal/gmole.K = 71.5935J/gmole.K

Heat supplied to superheat for Methanol =mCP ΔT

= 5.7155 103 71.5935 (923.2 – 337.9)

= 239.500 106 J

Cp of water = 36.845J/gmole

Heat supplied to super heater from 100C to 650C =mCP ΔT

= 5.7155 103gmole 36.845J/gmoleK (550) K = 11.5825 107J

Since heat required to preheat = heat given away by steam

Amount of steam required assuming steam available at 3 bar

31

Enthalpy of steam = 2724 kJ/kg

Mass of steam required (mS ) =

q(methanol+water )λ (steam )

=

239 .5005× 106+ 11. 5825 × 107

2724000

mS = 130 kgs.

Heat balance around Air Pre-heater

Here atmospheric air at 25C is heated to 650C (reaction temp)

Q =mCP ΔT

= 6.8035 103

(29 .1917+35 .388 )2 (650 -25) = 137.30 103 KJ

Heat required by air = Heat given away by steam

Amount of steam required assuming steam available at 3 bar

Enthalpy of steam = 2724 kJ/kg

Mass of steam required, (mS ) =

137 .292×106

2724000 = 51Kgs.

Heat balance around Mixer

Heat entering through inlet stream = Heat leaving out through outlet

Energy/Heat balance around the Reactor

The raw material Methanol + air + Water vapor enters the reactor at 650C and passes

through tubular bed of catalyst (silver grains). The reaction to form formaldehyde is

exothermic withheat of reaction (38 – 20.3) kcal/mole i.e, 17.7 kcal/mole

Moles of product formed from reaction are = HCHO + H2O + H2

= 5.144 + 2.5715 + 2.5715 = 10.2871 kgmole

Heat of reaction = 10.287 103 gmole 17.7 103 cal/gmole = 1.82 108 cal.

= 762.034 103 KJ

32

Heat leaving the reactor = Heat entering with reactant + Heat of reaction

= 239.500 103 (CH3 OH) +11.5825 104J (H2O) +137.30 103 KJ (Air) + 762.034 103

= 1.254 106 KJ

Since the reaction is exothermic and reaction temperature is to be maintained at 650C all

the heat that is formed by reaction is to taken out by a arrangement of cooling system.

Heat taken out by cooler = 762.034 103 KJ

Water at 35C is used as cooling agent. The temperature is expected to reach 100C,

Cp of water = 35.756 J/gmoleK

Mass of water required, (mw ) =

qCp ΔT =

762 .034× 105 J35 .756J/gmole . K×(100-25 )K = 2.8415 105

=

2.8415× 105×181000 = 5114.7 kgs.

Energy balance of reactants through heat exchanger Pre-heater

The mixture from outlet of reactor at 650C are brought down to 100C, by exchanging

the heat to the air that is going to the reactor.

Products issuing coming out in kgmoles are; HCHO (5.144) + H2O (8.2872) + H2 (2.5717)

+ N2 (5.3748) + O2 (0.1428) + CH3OH (0.5715)

Heat for cooling the reactants from 650C to 100C

qNitrogen=(mCP)Nitrogen ΔT 5.3748 103 29.459 550 870.849 10 5J

qOxygen=(mCP)Oxygen ΔT 0.1428 103 30.4353 550

23.9156 105J

qMethanol=(mCP )Methanol ΔT 0.5715 103 71.593 550 225.036 105 J

qWater=(mCP)Water ΔT 8.2872 103 36.845 550 1679.418 105 J

qHCHO=(mCP)HCHO ΔT 5.144 103 53.9578 550 1526.574 105 J

qHydrogen=(mCP )Hydrogen ΔT 2.5717 103 29.8365 550

422.026 105 J

33

ΣQ=qNitrogen+qOxygen+qmethanol

+qWater+qHCHO+qHydrogen

474.7818 106 J

Heat required by air = 137.29 103 KJ

Heat that will be removed = 474.7818 103 –137.29 103 = 337.5 103 KJ

Energy balance around the Absorber

Here HCHO is cooled and gets dissolved in water entering at the top. Here the heat of

reaction is evolved due to absorption. Some amount of heat is required to vaporize 8kg of

water which passes through vent. Assuming heat of reaction is negligible. Some water

enters as vapors, in the product inlet stream, this amount is 149.1705 Kgs.

Considering the same water is passing out, the water to be condensed in the absorber

= 149.1705 – 8 = 41.7 kgs. = 7.843 Kgmole.

Now the heat given out by entering gas to reach the outlet temperature from 120C to 25C

qNitrogen5.3748 103 95 29.1708 14.894 106J

qOxygen0.1428 103 95 29.5183 400.64 103J

qmethanol0.5715 103 95 46.8944 2.546 106J

qWater8.2872 103 95 35.1708 27.6895 106J

qHCHO5.144 103 95 45.2405 221.081 106J

qHydrogen2.57175 103 95 28.4716

6.956 106J

Q total 74.595l 106J

Heat is removed by using a coolant

Heat Balance around Distillation column

Distillate Methanol =18.10kg/hr HCHO = 0.1543kg/hr Feed

Methanol = 18.106 kg/hr

Fractionation column

34

HCHO = 154.32kg/hrWater = 262.76kg/hr HCHO = 154.167kg/hr Water = 262.76kg/hr Residue Methanol = 0.182kg/hr Condenser

Here methanol is cooled from 64.7C to 40C

Heat in = (mλ)Methanol+(mλ)HCHO

= (18 .106×1098 .5×103 )+(0 . 1543×1001×103 )

= 20.09 106 J.

Heat out=mCP ΔT

=18.106 2508 (40-25) = 6822800J.

Overall heat balance in condenser

Heat in = Heat out of condenser + Heat removed

20.09 106 J. = 6822800J. + Heat removed

Heat removed = 19.40 106 J.

Water is used to cool the product from 64.7C to 40C

Q=mCP ΔT+mλ

19.40 106 J = m×4 .187×103×(40−25)+m×2230×103

mw = 8.46 kgs/hr.

Reboiler

Heat in bottom of reboiler = (mCP)Methanol ΔT+(mCP )HCHO ΔT+(mCP)water ΔT

0 .183×2508×(64 . 7−25)+154 .16×0 . 8×4 .18×103×(64 .7−25 )+262.8×4187×(64 .7−25)

= 64.08 106 J

Overall Heat balance around Reboiler

35

[ Heat in ¿ ]¿¿

¿¿ +

[ Heat added in ¿ ]¿¿

¿¿ =

[ Heat at top ¿ ]¿¿

¿¿ +

[ Heat loaded in ¿ ]¿¿

¿¿

0 + Heat added in reboiler = 20.09 106 J + 64.08 106 J

Heat added in reboiler = 84.17 106 J

If we use steam of 3 bar,

T = 133.5C, λ = 2163 KJ/kg.

mS λ=84 .17×106J

mS=84 .17×106

2163×103=38.91kgs .

(amount of steam required)

36

REACTOR DESIGN (PROCESS)

From Kinetic consideration

1) To find volume of reactor

Space velocity is 8000-10000 hr-1

Space velocity =

Volumetric feed rate at standard conditionVoid volume of reactor

Therefore, Void volume of reactor =

Volumetric feed rate Space velocity

Volumetric feed rate at standard condition = 18.23455 kgmole/hr

= 18.23455 22.4 m3/kgmole

= 408.454 m3/hr

Taking space velocity of 8000 hr-1

Void volume of reactor =

408 . 454 m3 /hr8000hr-1

= 0.051 m3

Porosity of packing;

Specific area of catalyst 10m2/gm with cylindrical dimension of 3mm high to 3.8mm dia of

packed tubes.

Diameter of particleDiameter of tube =

d p

d t (Diameter of tube is taken as 50mm)

For, smooth uniform catalyst, porosity is 0.36

Therefore, actual volume of reactor =

0 .051m3

0 .36 = 0.14167 m3

Taking 10% extra volume = 0.14167 1.1

37

= 0.15584 m3 0.156 m3

2) To find tubes required

Since tubes are 50mm in diameter the cross section area of each tube is

=

π4 (50 10-3) 2 = 1.9625 10-3 m2

Taking length of tubes as 1m (standard tubes)

Number of tubes, (N t ) =

Volume of reactor Volume of each tube =

0 .156m3

1. 9625 × 10-3m2×1m

= 79.49 80 tubes.

3) Steel tubing data and inner diameter of shell

Stainless steel pipes are provided for process condition O.D = 55mm, I.D = 50mm

Thickness =5mm.

Surface area/m inside = 19.6 10-4 m2

Surface area/m outside = 23.7 10-4 m2

Tube arrangement: Tubes are laid out in Triangular pitch of 1.5D i.e, 75mm, triangular

pitch is chosen for efficient heat transfer.

Minimum required area = (pitch ) 2 No. of tubes(N t )= (0.075)2 m 80m = 0.45 m2

Using 20% excess area = 1.2 0.45 m2 = 0.54m2

Shell diameter required = √ 0.54m2×4π = 0.83m (using A=

{πd2

4 })

From heat transfer consideration, reactor is 1-1 heat exchanger in which the tubes are

packed with catalyst bed.

Specification of reactor

a) Heat duty of reactor = 762.034 103 KJ.

b) Gas flow rate = 481.9894 kg/hr 482 kg/hr

c) Gas inlet temperature = 650C

38

d) Gas outlet temperature = 650C

e) Water flow rate = 248.15 kg/hr

f) Water inlet temperature = 25C

g) Water outlet temperature = 100C

Average properties of fluids (tube side)

Hot fluid stream gas mixture from Pre-heater

Inlet temperature = 650C, Outlet temperature = 650C

Average pressure = Atmospheric = 1.033 Kg/cm2 (absolute)

Gas flow rate = 482 Kg/hr

Average temperature = 650C

Volume at average temperature at NTP the volume is 408.5m3/hr

=

923 × 408 . 454273 = 1380.1 m3

Average density of gas = 1kg/ m3

Average viscosity of gas = 0.126kg/m.hr

Average thermal conductivity, k = 203 w/m.K

Average properties of fluids (shell side)

Stream is sent through the shell

Cold fluid stream = water

Inlet temperature = 25C, Outlet temperature = 100C

Average pressure = 1.033 Kg/cm2 (absolute)

Water flow rate = 248.15 m3 /hr

Average density = 1000 kg/m3 Average Cp = 4.187 KJ/kg.K

Average viscosity = 1Kg/m.hr, Average thermal conductivity, K = 2445.208 w/m.K

39

Log mean temperature, Tm =

Δt 2−Δt 1

[ ln Δ t2Δt 1 ]

Hot fluid; Inlet = 650C, Outlet = 650C

Cold fluid; Inlet = 25C, Outlet = 100C

Δt1 = 650 – 100 = 550C

Δt2 = 650 – 25 = 625C

Tm =

625−550

[ ln625550 ]

= 586C

4) Heat transfer coefficient

Hot fluid (Tube side)

Flow area (a t ) = Number of tubes (N t ) cross sectional area of each tube

= 80 19.6 10-4 = 0.1568 m2

Mass flow rate, (Gt ) =

Mass flow rate of gases (G )tube side flowarea (at ) =

482kg /hr0 .1568m2

= 3073.98 kg /m2.hr

For packed tubes with air flowing in inner side Heat transfer coefficient is given by

h p=3. 5[ kd t ] [d P .Gt

μ ]0 .7

e−4 .6[ dpdt ]

[ dP .Gt

μ ]0.7

= [ 3 . 8×10×3073 .98

0 .126 ]0.7

= 23.82

e-4 . 6×( 3. 8

50 ) = 0.7049

h p = 3.5

203

50 × 10-3 23.82 0.7049 = 238.60 KJ/hr.m2.K

hid = hp [IDOD ] = 238.60

[5055 ]

= 216.90 KJ/hr.m2.K

(Page 290 eqn. 10-10 Chemical engineering kinetics by J.M Smith 1st edition)

40

Shell side Heat transfer coefficient

Cold fluid:

Flow area, (a

'S )=

shell dia ( I . D)×baffle spacing(B )×clearance(C ' )pitch(Pt )

,m2

a'S=0 .83m×20×10−3 m×0 .15m

75×10−3m = 0.0332 m2

G's=

mass flowrate of water (m)

Shell side flow area( a's )

= [5114.7kg/hr

0 .0332m2 ] = 154057.23 kg/hr.m2

Reynolds No. NRe=

DeGs'

μ

De = 4

(0 .5Pt×0 .86Pt−0 .5×π×D2

4)

0 .5×π Do=

57.04 mm

NRe = 57.04 10-3 154057.23 = 8787.42

Assuming there is no considerable change in Viscosity

ho= jHkD e

[Cpμ ]0.33 [ μ

μw ]0.14

When NRe = 8787.42 jH = 50 (page 838 figure 28 Process heat transfer by D.Kern)

Cp . μk =

[4187×12445. 208 ]

0.33

= 1.194

ho = 50×2445. 208

57 .04 × 10-3×1 .195

= 2561.14 KJl/hr.m2.K

Clean overall Heat transfer coefficient = UC

UC = [ hid .hohid+ho ]

=

216 . 90×2561 .14216 . 90+2561 .14 = 199.97

(Eq 6.15b, page111, Process heat transfer by D.kern)

41

Let total dirt factor Rd = 1 .43×10−5 hr.m2.K/KJ

Design overall heat transfer coefficient, Ud =[ 1

1UC

+Rd ]=

[ 11

199 . 97+1 . 43×10−5 ]

=194.40KJ/hr.m2.K

Area required, (A) = [ qU d . Δt ln ]

=

762 .034×103 KJ194 . 40 KJ /m2 K×586 K = 6.7m2

5) Pressure drop calculation

Tube side pressure drop

ΔPL =

0 .4G2agρ∈2 [ Gt

aμ ]−0 .1

Where, Gt = mass velocity(kg/hrm2), ∈ = free space or porosity

g = gravity constant(m/s2), μ = viscosity of gases(kg/ms)

a = specific surface of bed m2/m3 = 6(1-∈)d p

a = 6

(1-0 .36 )3. 8 × 10 -3

= 1010.5 m2/m3

ΔPL =

0 .4×(3073 .98)2×1010 .5

1. 27×108×1×! (0 .36 )3×3073 . 98

1010 . 5×0. 126 = 26.7

P = 26.7 1 = 26.7 Kg/cm2 (L = 1mts)

Pressure drop on shell side

NRe=

DeGS'

μ=

57 .04×10−3×154057 .231 = 8787.4

Therefore f = 2.7 10-5 (page 836,figure 26, Process heat transfer by D.Kern)

Number of cross, N + 1 =

LB

Page 193, eqn, 24, Reaction kinetics for chemical engineers by Stanley Walas

42

Where, L = tube length (m), B = baffle space (m)

=

10 .15 = 6.67

N + 1 = 7

Ps =

fGs2 D(N+1)2g . ρ Deφs ; φ s =

μμw = 1

Where, f = friction factor dimensionless

=

2. 7×10−5×(1 .54×105 )2×0 . 83×7

2×1.27×108×57 . 04×10−3×1

Ps = 2.57 10-4 Kg/ m2

(Page 147 eqn. 7.44 Process heat transfer by D.Kern)

43

MECHANICAL DESIGN OF REACTOR

As temperature in reactor is 650C, we consider working stress at same temperature

Taking factor of safety = 3

Therefore working stress for carbon steel = 200Kg/cm2

Design pressure = 1.2 times of 1Kg/cm2 (20% extra)

Tube side

1) Thickness of tube

Minimum thickness is given by, t =

pD2 f

+C

P = average operating pressure = 1Kg/cm2

D = diameter of tubes = 50mm

C = corrosion allowance = 3mm

f = working stress = 200Kg/cm2

t =

1. 2×502×200

+3 = 4.15mm = 5mm

2) Tube sheet thickness

t=FGc√ 0 . 25Pf P =

1. 013×105×1. 2(103 )2

=0 .1026 N /mm2

Where, F = 1 for most Heat exchanger except U type

44

Gc = mean diameter of gasket

F = allowable stress at appropriate temperature = 100N/mm2; SS IS grade-10

t = effective thickness of tube sheet

t=1×835√ 0 .25×0 .1026100 = 14mm

3) Channel and Channel cover

Thickness of channel portion

t c=G c√ kpf

Where, k = 0.3 for ring type gasket, Gc = mean gasket dia for cover = 835mm

f = permissible stress at design temperature 95N/mm2

th=835√ 0 . 3×0 . 12695

=25 .7 mm≈26mm

4) Flange diameter (between tube sheet and channel)

Gc = 835mm, Ring gasket width = 22mm

bo=W8

=228

=2 .75mm, ya=126 .60 N /mm2

, m = 5.5, b = bo

W m1=π Gby=π×835×2. 75×121 . 6

= 912814.5N

W m2=2 π bGmP+ π4G2P

W m2=2×π×2 .75×835×5 .5×0. 1026+ π4

(835 )2×0 . 1026

= 64292.7N

Am1=W m1

permissible stress for bolt material(140 .6N/mm2 )

45

Am1=912814 . 5140 . 6

=6520mm2

Am2=64292 .5140 . 6

=457 .3mm2

Am2= N b¿

πD 2

4

No. of bolts, Nb=

83510×2.5

= 34 bolts

Diameter of bolt = √ 4× Am2

Nb ×π =√4 × 457 . 3

34 ×π = 5mm

Minimum pitch circle dia = outside of gasket + 2 ¿ dia of bolts + 22mm

= 835 + 2 x 12 +22 = 881mm

Take, B = 940mm

5) Flange Thickness, tf = G√ P

Kf+ C

K =

1

[0 .3+1 .5 Wm hG

HG ]hG = Radial distance from gas load reaction

to bolt circle

= B−G2

H = Total hydrostatic end force = /4 G2 P

= π4

(835 )2×0 . 1026

K =

1

0 .3+ 1 .5 ×912814 . 5835×56155 . 2 [940−835

2 ] = 56155.2 N

= 0.54mm

46

tf = 835×√ 1 . 5

0 . 54×95= 38mm

6) Nozzle diameter of the reactor

a) Reactant inlet Nozzle diameter (Tube side)

ρ=481 . 99

18. 24×22 . 4×1. 18×10−3 = 1.18 10-4Kg/cm3 = 11.8Kg/m3

mo=ρ Av , Where v = 2m/sec for gases, mo

= 481.88kg/hr (from material balance)

Cross sectional area (A) =

mo

ρ . v =

481 . 9911. 8×2×3600

= 56.571cm2

Therefore r = (56 . 7

π )0 .5

=4 .25 cm

Diameter = 2r = 8.5cm.

b) Product outlet Nozzle diameter (Tube side)

=

481 . 9922 .093×22 . 4×1000 = 9.74 10-4Kg/cm3

Cross sectional area =

mo

ρ . v =

481 . 999 .74×2×3600

= 68.73cm2

Therefore r = (68 . 73

π )0.5

=4 .68cm

Diameter = 2r = 9.36cm

b) Shell side

1) Thickness of shell

Minimum thickness of shell is given by, t =

pD2 fe

+C

P = average operating pressure = 1Kg/cm2

(Page 48, eqn 1.123 Process equipment design by D.Dawande)

47

D = diameter of shell = 3.8m

f = working stress = 200Kg/cm2

C = corrosion allowance = 3mm

e = weld joint efficiency = 0.6

t =

1. 2×832×200×0 .6

+0 . 3= 0.415 +0.3 = 0.715cms = 1cm

2) Flange thickness (shell side)

Gasket diameter (G) =

Shell I . D + Shell O. D2

=830+8402

=835mm

Gasket width (N) = 24mm

bo=N2

=242

=12mm

b=2 .5√12=8 .64mm

Where, b = effective gasket seating width

b0 = basic gasket seating width

Gasket factor (m) = 3.75 (flat metal jacket asbestos filled)

Seating stress(y) = 53.4 N/mm2

W m1=π Gby=π×0 .835×103×8 .64×53 .40

= 1209681.73N

W m2=2 π bGmP+ π4G2P

W m2=2×π×8 . 64×0. 835×103×3 .75×1. 2+ π

4(0 . 835×103)2×1 .2

= 860664.89N

Flange Thickness, tf = G√ P

Kf+ C

(Table 5.4 page 129

“Process equipment design

by M.V.Joshi)

48

K =

1

[0 .3+1 .5 Wm hG

HG ]hG = Radial distance from gas load reaction

to bolt circle

= B−G2

H = Total hydrostatic end force = /4 G2 P

= π4

(0 .835×103)2×1 .2

K =

1

0 .3+ 1 .5 ×860664 .89835×656785 .95 [940−835

2 ] = 656785.95N

=2.36

tf = 835×√ 1 .5

2 .36×95= 61mm

3) Head of reactor

I.D of shell = 830mm

O.D of shell = 840

For normal pressure, torispherical heads are used

Here crown radius, L = O.D of shell = 840mm

Knuckle radius = 0.6 Crown radius = 504mm

Thickness of head is given by t =

0 .885 PLfe−0 .1 P

+C

t =

0 . 885×1 . 2×840200×0 .6−0 . 1×1 .2

+3 = 7.44 + 3 = 10.44mm

Therefore thickness of head is taken as 1cms

4) Baffle

For baffle spacing of 0.15m and inside shell

diameter of 0.83m the baffle thickness is 6.5mm

(Page 51, table1.10 Process equipment design by S.D Dawande Vol 2)

49

5) Size and number of tie rods and spacers

For shell diameter of 0.83m, number of tie rods = 6

Minimum diameter of tie rods = 12.5mm

6) Nozzle diameter of the reactor

a) Diameter of cold water nozzle inlet (Shell side)

mo=ρ Av , Where v = 2m/sec for gases, mo

= 5114.7kg/hr (from material balance)

Cross sectional area =

mo

ρ . v =

5114 .71000×2×3600 = 7.10cm2

Therefore r = ( 7 .10

π )0. 5

=1 .5cm

Diameter = 1.5 2 = 3cm

b) Diameter of hot water nozzle outlet (Shell side)

Since the water is hot in this case and expanded the diameter has to be more. So diameter

of 6cm (2cold water inlet dia) is taken.

(Page 51, table1.8, Process equipment design by S.D Dawande Vol 2)

50

PROCESS DESIGN FOR DISTILLATION COLUMN

Equilibrium data of methanol in formalin solution

Table 4.1 T-x-y data

T 93 92.5 87.4 83.7 79.1 76.9 74.4 72.1 69.3 67.1 64.3

X 0 0.06 0.126 0.19

9

0.279 0.369 0.4704 0.5844 0.7142 0.864 1.0

Y 0 0.2655 0.44 0.55

4

0.6414 0.710

5

0.775 0.839 0.884 0.9445 1.0

Source: Vapor-liquid equilbria of formaldehyde- methanol-water

by S.J. Green & Raymond.E.Vener, Industrial & engineering chemistry, 1955

Page103 to 108, Volume 47 no 1,

D = 0.5709kgmol/hrxD = 0.99

Methanol =0.5658kgmoleHCHO =

51

Fractionation Column

F = 20.3155kgmol/hr ZF = 0.02813

HCHO = 5.144kgmoleWater = 14.6kgmolesMethanol = 0.5715kgmoles

W = 19.745kgmol/hrxw = 0.000289Methanol =0.005715kgmoleHCHO = 5.1389kgmoleWater = 14.6kgmoleRESIDUE = 19.745

Glossary of notations used:

F = molar flow rate of Feed, kmol/hr, D = molar flow rate of Distillate, kmol/hr.

W = molar flow rate of Residue, kmol/hr.

zF = mole fraction of methanol in liquid/Feed.

xD , = mole fraction of methanol in Distillate, xW = mole fraction of methanol in Residue.

Rm = Minimum Reflux ratio, R = Actual Reflux ratio

L = Molar flow rate of Liquid in the Enriching Section, kmol/hr.

G = Molar flow rate of Vapor in the Enriching Section, kmol/hr.

L = Molar flow rate of Liquid in Stripping Section, kmol/hr.

G = Molar flow rate of Vapor in Stripping Section, kmol/hr.

q = Thermal condition of Feed

D = 0.5709kgmol/hrxD = 0.99

Methanol =0.5658kgmoleHCHO =

52

ρL = Density of Liquid, kg/m3, ρV = Density of Vapor, kg/m3.

qL = Volumetric flow rate of Liquid, m3/s, qV = Volumetric flow rate of Vapor, m3/s

μL = Viscosity of Liquid, cP

Preliminary calculations:

zF=0 .5715

0 .5715+14 . 6+5 . 144 = 0.02813

xD= 0 .5658

0. 5658+5 .144×10−3 = 0.99

xW= 5 .715×10−3

5 .1389+14 .6+5. 715×10−3 = 0.000289

Marking zF , xD , xW on the x-y graph and assuming liquid enters as saturated (q = 1)

From graph (Figure 1)

xD

Rm+1 = 0.15

0 .990 .15

=Rm+1

Rm= 6.6-1 = 5.6

Assuming reflux ratio of 1.5 times of Rm

R = Rm 1.5 = 5.6 1.5 = 8.4

Now

xD

R+1= 0 . 99

8 .4+1 = 0.105

Plotting the operating line and feed, the number of theoretical plates = 11 (from graph 1)

Number of trays in enriching section = 7

Number of trays in stripping section = 4 (including re-boiler)

Therefore total number of trays = 10 (without re-boiler)

Flow streams

L = Lo = R.D

53

= 8.4 0.57094

= 4.796kgmole

G = L + D = RD + D = (R+1) D = (8.4 +1)0.57094 = 5.367kgmole

(q =1 for saturated liquid)

L = L+qF (q = 1)

= 4.796 +1 20.3155

= 25.115kgmole

G = G + (q-1) F

G = G (Since q = 1 )

= 5.367kgmole

Table 4.2

List of Parameters used in calculation

Consider 4 points in the column, top and bottom of enriching section and top and bottom

of stripping section.

Enriching Section Stripping Section

Top Bottom Top Bottom

Liquid(kgmole/hr) L = 4.796 4.796 25.115 25.115

Vapor(kgmole/hr) G = 5.367 5.367 5.367 5.367

X 0.99 0.02813 0.02813 0.00028

Y 0.99 0.9719 0.9719 0.00028

M liquid kg/kmole31.98 21.43 21.43 21.13

M vap kg/kmole

31.98 31.47 31.47 21.13

Liquid (kg/hr) 153.38 102.78 538.21 530.68

54

Vapor (kg/hr) 171.64 168.9 168.9 113.40

TC liquid 64 91 91 93

TC Vapor 65 92 92 93

L (kg/m3) 793.98 995.08 995.08 1000.9

G (kg/m3) 1.147 1.15 1.15 1.2

L/G( ρG

ρL)0 .5 0.034 0.0303 0.159 0.162

a) Design of Enriching section

Plate hydraulics,

The design of a sieve plate tower is described below. The equations and correlations

are borrowed from the 6th edition of Perry’s Chemical Engineers’ Handbook.

1) Plate spacing, (tS ) = 400mm

2) Hole diameter, (d L ) = 5mm

3) Hole pitch (triangular),(LP) = 3d L = 15mm

4) Tray thickness, (t r ) = 0.6d L = 3mm

5) Plate diameter (DC ) (Plate hydraulic table)

L/G( ρG

ρL)0 .5

= 0.162 (maximum value)

For tS = 16”,CSb flood = 0.23

We have

Unf =CSb flood ( σ20 )

0 . 2

( ρL− ρG

ρG)0 .5

{eqn. 18.2, page 18.6, 6th edition Perry.}

Where,

55

Unf = gas velocity through the net area at flood, m/s

CSb flood = capacity parameter, m/s (ft/s, as in fig.18.10)

σ = liquid surface tension, mN/m (dyne/cm.)

ρL = liquid density, kg/m3 (lb/ft3)

ρG = gas density, kg/m3 (lb/ft3)

σMixture= σ Methanol −σWater

2

σ14=[ P ] ( ρL−ρG ) ---( eqn. 3-151, page 3-288, table 3-343, 6th edition Perry.}

(at low pressure, where ρL>> ρG , so neglect ρG )

σ14Water=[P ] ( ρL ) =

σ14Water=[ 51 ]( 1

18 )

σWater = 64.56 dyne/cm

σ14Methanol=[ 85 .3 ]( 0 . 792

32 ), σMethanol = 19.86 dyne/cm

σMixture= σ Methanol −σWater

2 =

19 . 86−64 . 452 = 42.16dynes/cm

Flooding velocity Unf = 0.23(42. 1620 )

0 .2

(1000 . 9−1 . 21 . 2 )

0.5

= 7.706 ft/sec = 2.348m/sec

Let us take Un = 80% of Unf

0.80 2.348 = 1.86m/sec

Now,

Net area available for gas flow, (An )

Net area = (Column cross sectional area) - (Down comer area.)

An = Ac - Ad

56

Volumetric flow rate of vapor (at top of stripping section)

=

168 . 93600×1. 86

= 0. 0252m3 /sec

An =

0 .02521.86

= 0 . 01356m2

Ad = down comer area can be taken 10-12% of Ac (let us take Ad= 11%Ac )

Ac = ( π4 )Dc2

= 0.785DC 2

Ad = 0.7850.11DC 2 = 0.0864DC 2

DC 2 = 0.0194m2 = 0.14m

Taking DC = 0.2m

Ac = 0.7850.22 = 0.0314m2

Ad = 0.0864 Dc2 = 0.08640.22 = 3.45610-3m2

Active area; Aa = Ac - 2 Ad

= 0.0314-23.45610-3 = 0.0245m2

6) Perforated area, (Ap )

Let Lw /DC = 0.75 (Lw= weir length)

Where, Lw = weir length, m

DC = Column diameter, m

Lw = 0.750.2 = 0.15m

θ=2×Sin−1( LwDc ) = 2 Sin

−1(0.75) = 97.18

σ = 180 – 97.18 = 82.82

Periphery waste = 50mm = 5010-3 m

57

Area of periphery waste AWZ = 2 {π4 Dc2 ( α

360 )− π4

(Dc−0 .02 )2( α360 )}

= 2 {π4 0 . 22(82. 82

360 )−π4

(0 .2−0 . 02 )2(82 . 82360 )}

= 0.01186m2

Acz = area of calming zone, m2 = 2 LW 50 10-3

= 2 0.15 20 10 -3 = 6 10 -3 m2

Ap=Ac−2 Ad−Acz−Awz

Ap = 0.0314 – 2 3.456 10-3 – 6 10-3 – 0.0107

= 7.788 10 -3 m2

7) Total Hole area,(Ah ):

Since,

Ah

Ap = 0.1

Ah = 0.1 7.788 10 -3

= 7.788 10 -4 m2

Now we know that,

Ah=nh( π4 )dh2

Number of holes, Nh =

7 . 788×10−4

π4×(5×10−3 )

= 40

8) Weir height,(hw ):

Let us takehw= 50mm

9) Weeping check

Head loss across the dry hole is

58

hd=K1+K 2( ρG

ρL)Uh

2

---- (eqn. 18.6, page 18.9, 6th edition Perry)

Where, Uh = gas velocity through hole area

K1 , K2 are constants

Volumetric flow rate of vapor =

168 . 93600×1. 86 = 0.0252 m3/sec

Uh=

0 .0252

7 .788×10−4 = 32.37 m/sec

For sieve plateK1 = 0 ; K2 = 50.8/Cv2

Where, Cv =discharge coefficient, taken from fig 18.14, page 18.9 6th edition Perry.

Holeareaactivearea

=Ah

Aa

=7 .788×10−4

0 . 0245 = 0.032

Tray thicknessHole area

=t Rd L

=35 = 0.6

Thus for (Ah/Aa) = 0.032 and

tRdL = 0.60

We have from fig. edition 18.14, page 18.9 6th Perry. CV (discharge coefficient) = 0.7

K2=50 .8

(0.7 )2=103 .67

hd=103 .67×( 1 . 151000 .2 )×(32 . 37 )2

= 124.9 mm of liquid

Height of liquid crest over weir

hOW=44300 FW ( q '

LW)0 .704

Thus, q = liquid flow rate, m3

/s; Lw = weir length

59

Liquid load q '=538 .82

( 995.08×3600 )=1 .504×10−4m3 /sec

hOW=44300( 1 .504×10−4

150 )0 .704

=2 .65mm

Head loss due to bubble formation (hσ ) is given by (page 18-17, equation 18-2a )

hσ=409σ

(ρL×dh)=(409×42. 16

995 . 08×5 )= 3.47mm liquid

Where,

σ =surface tension, mN/m (dyne/cm) = 42.16 dyne/cm.

dh = diameter of perforation, (mm) = 5mm

ρL = density of liquid in the section,(kg/m3) = 995.08 kg/m3

hd+ hσ = 124.9 + 3.47 = 128.37mm

hW + hOW = 50 + 2.65 = 52.65mm

Ah

Aa

=0 .032

From figure 18-11, 6th edition Perry, hd+ hσ = 10mm

Since design value of hd + hσ is well above, the value obtained from graph, no weeping

occurs.

8) Check for down comer flooding

Down comer back up is given by

hdc=ht+hw+how+hda+hhg --- (eqn 18.3, page 18.7, 6th edition Perry)

a) hydraulic gradient across the plate hhg for stable operation hd>2.5hhg

For sieve plates hhg is generally small or negligible, let us take hhg = 0.5mm of liquid

60

b) Total pressure drop across the plate, (ht );

ht=hd+hl'

Now, hl'=β×hdS ---- (eqn. 18.5, page 18.9, 6th edition Perry)

Where,hl'

= pressure drop through the aerated liquid (mm)

β =Aeration factor, hds = Calculated height of clear liquid over the dispersers, (mm)

hds=hw+how+hhg

2 ----(eqn. 18.10, page 18.10, 6th edition Perry)

Where, how = height of crest over weir equivalent clear liquid, (mm)

hhg = hydraulic gradient across plate, height of clear liquid column, (mm)

hds=50+2 .65+ 0 . 5

2 = 52.9mm

To find β

Now, Fga=U a (ρg )12

Where

Fga = gas-phase kinetic energy factor,Ua = superficial gas velocity, m/s (ft/s),ρg = gas density, kg/m3

(lb/ft3)

Ua =

168 . 93600×1. 15×0. 0245 = 1.665m/s

Fga = 5.46¿ (1 .15×10−3×62 .4 )0. 5= 1.463

From (figure 18-15, page 18-10, 6th edition Perry) β = 0.59

hl'

= 0.59 52.9 = 31.211

ht=hd+hl'

= 124.9 + 31.211 = 156.11 mm

61

c) Loss under down comer, (hda)

hda=165 .2( qAda )

2

----- (eqn. 18.19, page 18.10, 6th edition Perry)

Where, hda = head loss due to liquid flow under down comer, apron mm liquid,

q = liquid flow rate, m3/s

Ada = minimum area of flow under the down comer apron, m2

hap=hds−c '

(Take clearance, c'= 1” = 25.4mm)

hap=52. 9−25 . 4 = 27.5mm liquid

Ada=Lw×hap=0. 15×27 .5×10−3 = 41.25 10-3 mm2

hda=165 .2( 1 .504×10−4

41 .25×10−3 )2

= 2.196 10-3 mm

Now, hdc=ht+hw+how+hda+hhg ---- (eqn 18.3, page 18.7, 6th edition Perry)

ht = total pressure drop across the plate (mm liquid) = hd + hl`

hdc = height in down comer, mm liquid,

hw = height of weir at the plate outlet, mm liquid,

how =height of crest over the weir, mm liquid,

hda = head loss due to liquid flow under the down comer apron, mm liquid,

hhg = liquid gradient across the plate, mm liquid.

= 156.11 + 50 + 2.65 + 2.196 + 2.196 10-3 + 0.5

= 209.2 mm of clear liquid

Actual back up with aeration; hdc' =

hdc

φdc

62

For system with low gas velocity, low liquid viscosity and low foamability we can take

φdc=0 .6

hdc' =209.2

0 .6=348. 72mm

(Which is less than the tray spacing)

tS = 400mm

Since tS >hdc'

(no down comer flooding will occur)

Column efficiency

The efficiency calculation are based on average condition prevailing in each section

Enriching section

Average molar liquid rate = 4.796 kgmole/hr

Average mass liquid rate =

153 .38+102. 782

=128. 08 kg /hr

Average molar vapor rate = 5.367 kgmole/hr

Average mass vapor rate =

171 .64+168 . 92

=170 .27 kg /hr

Average density of liquid =

793 . 98+995 . 082

=894 .53 kg/m3

Average density of vapor =

1. 147+1 .152

=1 .1485kg/m3

Average liquid temperature =

64+912

=77 . 5oC

Average vapor temperature =

65+922

=78 .5oC

Viscosity of formaldehyde at 77.50C = 0.06cp

Viscosity of Methanol at 77.50C = 0.07cp

Average viscosity of liquid is calculated using Knedall-maserol equation

63

μ13L=x1μ

131+ x1 μ

132

x1=0 . 99+0 . 02813

2=0 . 509

x2=1−x1=0.491

μmLiquid=[0 .509×(0 .07 )13 +0 .06× (0.491 )

13 ]3

= 0.0162cP

Viscosity of formalin at 78.50C = 0.05cP

Viscosity of methanol at 78.50C = 0.062cP

y1=0. 99+0. 9719

2=0 . 98

y2=1− y1=1−0. 98=0 .02

μmvapor=

∑ [ y iμ i M

i

12 ]

∑ y i M

i

12

μmvapor=0.98×0 .062×32

12+0.02×0.05×30

12

0.98×3212×0 .12×30

12

= 0.054cP

Liquid phase diffusivity

Wilke chang equation status

DL=[7 . 4×10−8×(φ×MB )0 .5×T ]

ηBVA

0. 6

Where, DL = mutual diffusion coefficient of solute A at very low conc, in solvent B,

φ = association factor of solvent B, MB = mol wt of solvent B

ηB = viscosity of solvent B, cP

T = 273+77.5 = 350.5K,

V A = solute molal volume (methanol)

64

= 16.5 1 + 1.98 4 + 5.48 = 29.9

DL=[7 . 4×10−8×(2 .6×30 )0 . 5×350. 5 ]

0 .7×(29 . 9 )0 . 6 = 4.26 10-5 cm2 /sec

Vapor phase density

Fullers equation

Dg=

10−3T 1. 75( 1M A

+ 1MB )

0 .5

P[∑V A

13 +∑V

B

13 ]

2

T = 78.5+ 273 = 351.5K

M A = 32; MB = 30; P = 1atmosphere

∑V A = 29.9; ∑V B= HCHO = 1 16.5 + 1.98 2 + 5.48 = 25.94

Dg=10−3×3511.75×( 1

32+ 1

30 )0 .5

1×[29 . 913+25. 94

13 ]

2

= 0.19cm2 /sec

N Scg=μg

ρg×Dg

= 0 . 054×10−3

1. 20×0.19×10−4 = 2.3

Stripping section

Average molar liquid flow rate = 25.115 kgmole/hr

Average mass liquid flow rate =

538 .21+530 .682

=534 . 45kg /hr

Average molar vapor flow rate = 5.367kg/hr

Average mass vapor flow rate =

168 . 9+113. 42

=141 .15 kg/hr

ρav ( Liquid )=

995 .08+1000 . 92

=997 .94 kg /m3

65

ρav (Vapor )=

1. 15+1 .22

=1 .175kg /m3

Tav (Liquid )=

91+932

=920C

Tav (Vapor )=

92+932

=92. 50C

μMethanol at 91.50C = 0.034cP μHCHO at 91.50C = 0.02cP

x1=0 . 02813+0 . 00028

2=0 . 0142

x2=1−x1=1−0 .0142=0 .9858

μm=[0 .0142×(0 .034 )13+0 .9858×(0 .02 )

13 ]3

= 0.01865cP

μ of Methanol at 92.50C = 0.032cP, μ of HCHO at 92.50C = 0.019cP

y1=0. 9719+0 . 00028

2=0 . 486

y2=1− y1=1−0 . 486=0 . 514

μmvapor=0.486×0 .032×32

13 +0 .514×0.019×30

13

0.486×3213×0 .514×30

13

= 0.0254cP

Diffusivity of liquid phase

DL=[7 . 4×10−8×(2 .6×30 )0 . 5×364 .5 ]

0 .60× (29 . 9 )0 . 6 = 2.382 10-4

= 5.169 10-5 cm2 /sec

Vapor phase diffusivity

Dg=10−3×365 .51 . 75×( 1

32+ 1

30 )0. 5

1×[29 .913+25 . 94

13 ]

2

= 0.202 cm2 /sec

66

N Scg=0 .025×10−3

1 .15×0 . 202×10−4=1 . 08

Table 4.2

Condition Enriching section Stripping section

Liquid flow rate kgmole/hr 4.796 25.115

Liquid flow rate kgmole/hr 128.08 534.45

ρLKg /m3 894.53 997.94

T L0C 77.5 92

μLcP0.0162 0.019

DLcm2 /sec 4.26 10-5 5.169 10-5

Vapor flow rate kgmoles/hr 5.367 5.367

Vapor flow rate kgmoles/hr 170.27 141.15

ρV Kg /m3 1.1485 1.175

T V0C 78.5 92.5

μV cP0.054 0.025

Dgcm2 /sec 0.189 0.202

N Scg2.3 1.08

Enriching section

a) Point efficiency, (Eog );

N g=[0 .776+0 .00457×hW−0 .238U a ρ

g0. 5+0 .0712W ]

NScg0 .5

Where,

(eqn. 18.36, page 18.15, 6th edition Perry)

67

W = Liquid flow rate, m3/ (s.m) of width of flow path on the plate,

hW = weir height = 50.00 mm

Ua = Gas velocity through active area, m/s

DL = liquid phase diffusion coefficient, = 4 .26×10−9cm2 /sec

Ua=170 .273600×1 .1485×0 .0245

=1 .68m /sec

q=128 . 08894 . 53×60

=2. 39×10−3m3 /sec

D f=( DC+LW )

2=0. 2+0 . 15

2=0 .175

W= qD f

=2 .39×10−3

60×0 . 175=2. 276×10−4 m3 /sec.m

N Scg= Schmidt number (dimensionless), N Scg=

μg

ρg×Dg

= 0 . 054×10−3

1. 20×0.19×10−4= 2.3

N g=0 . 776+0 .00457×50−0 . 238×1 . 68×1 .14850. 5+0. 0712×2.276×10−6

(2. 3 )0 . 5=0.39

Number of liquid transfer unit is given by N l=k LaθL

Where,

k L= Liquid phase transfer coefficient, m/s

a = effective interfacial area for mass transfer m2

/m3 froth or spray on the plate,

θL = residence time of liquid in the froth or spray zone, s

θL=hL Aa

1000×q ---- (eqn. 18.38, page 18.16, 6th edition Perry)

θL=31 . 211×0. 0245

1000×(2 .39×10−3 /60 )=19. 196 sec

68

K La=(3 .875×108×DL)0.5

(0 . 4×Ua×ρg0 .5+0 . 17)

K La=( 3.875×108×4 .26×10−9 )0 .5 (0 .4×1.68×1.14850 .5+0 .17 )=1 .1438 sec−1

N l=1. 1438×19.196=21. 96

Nog=1

1N g

+λN l ----- (eqn. 18.34, page 18.15, 6th edition Perry)

Where,

N l = Liquid phase transfer units,

N g = Gas phase transfer units,

m = slope of Equilibrium Curve,

Gm = Gas flow rate, mol/s (from table 4.2)

Lm= Liquid flow rate, mol/s (from table 4.2)

Nog = overall transfer units

λ=mGm

Lm = Stripping factor,

(from graph 4.1 for enriching section straight line slope is 1.4/1.6 = 0.875)

Nog=1

10 .39

+0 . 97921 . 96

=0 . 39

Eog=1−e−Nog=1−e−0. 39=0 .33 -------- (eqn. 18.33, page 18.15, 6th edition Perry)

b) Calculation of Murphree plate efficiency, (Emv );

Now, N Pe=

ZL2

DE×θL

979.0796.4

367.5875.0

m

m

L

Gm

69

ZL = length of liquid travel, m

ZL=2×[ DCCos( θ2 )2 ]

=

2×[ 0 . 2Cos(97 .182 )

2 ] = 0.1323

θL = 19.196sec

DE=6 . 675×10−3Va

1.44+0 . 922×10−4×hl−0 . 00562

=6 .675×10−3×1.681 . 44+0 . 922×10−4×31. 211−0 . 00562

= 0.01135m3 /sec

Where, DE = Eddy diffusion coefficient, m2/s

N Pc=0 .13232

0.01135×19 .196=0 . 0804

λEog=0 . 979×0. 33=0 .323

Now, for λEog = 0.323 and N Pe = 0.0804 value, (we have from fig.18.29a, page 18.18, 6th

edition Perry)

Emv

Eog

=1 .8

(Murphree plate efficiency)Emv = 1.8 x 0.33 = 0.60

c) Overall column efficiency (EOC );

EOC=N t

Na

=log [1+Ea ( λ−1 ) ]log λ ----- (eqn. 18.46, page 18.17, 6th edition Perry)

Where,

Eα

Emv

= 1

[1+Emv( ψ1−ψ )]

----(eqn. 18.27, page 18.13, 6th edition Perry)

ψ= eL+e ----(eqn. 18.26, page 18.13, 6th edition Perry)

(eqn. 18.45, page 18.17, 6th edition Perry)

70

e = absolute entrainment of liquid

L = liquid down flow rate without entrainment.

Emv = Murphee Vapor efficiency,

Eα = Murphee Vapor efficiency, corrected for recycle effect of liquid entrainment.

ψ = fractional entrainment, moles/mole gross down flow

Taking ( LV ) .( ρG

ρL)0 .5

=0.162 at 80% flooding, (we have from fig.18.22, page 18.14, 6th

edition Perry) ψ = 0.011

Ea

0 .6=1

[1+0 .6 (0 .0111−0. 011 )]

Ea=0. 596

(Actual trays)Nactual=

N T

EOC

=

Ideal trays Overall efficiency

Actual number of tray in enriching section =

70 .596 = 12 trays

Stripping section

a) Calculation of point efficiency, (Eog )

Ua=141 . 153600×1 .175×0 .0245

=1 .362m /sec

q '=534 . 45997 . 94×60

=8 .925×10−3 m3 /sec

D f=0 .175m

w=8 . 925×10−3

60×0 .175=8 .5×10−4m3 /sec .m

N Scg= 1.08

71

N g=0 . 776+0 .00457×50−0 . 238×1 . 362×1. 1750 . 5+0 . 712×8 . 5×10−4

(1. 08 )0 .5=0.63

θL=31 . 211×0. 0245

(1000×8 .925×10−3

60 )=5 .14

k L×a=( 3. 875×108×5 .169×10−9 )0 .5 (0 . 4×1. 362×1 .1750.5+0.17 )

= 1 .076sec−1

N l=1. 076×5 .14=5 . 54

Nog=1

1N g

+λN l

= 11

0.63+

1 .0685 .54

=0 .562

m=1 .5

0 .3=5

; m

Gm

Lm

=5× 5 .36725. 115

=1 . 068

Eog=1−e−Nog=1−e−0. 562=0 . 43

b) Calculation of Murphree plate efficiency, (Emv );

N Pe=Z

l2

DE×θL

DE=6 . 675×10−3×1. 3621.44+0 . 922×10−4×31 .211−0 . 00562

= 7.6810-3m3 /sec

ZL=2[ DC cosθC

22 ]=0 .1323m

θL=5 . 14

N Pe=0 .13232

7 . 68×10−3×5. 14=0 . 444

λEog=1 .068×0 .43=0 .46

Emv

Eog

=1 .6

72

Emv = 1.6 0.43 = 0.69

c) Overall column efficiency, (EOC )

ψ=0.011 (refer enriching section EOC )

Ea

0 .69=1

[1+0 .69(0 .0111−0 .011 )]

Ea=0.6852

(Actual trays)Nactual=

N T

EOC

=

Ideal trays Overall efficiency

Actual Number of trays in stripping section =

30 .6852

=5 trays

Height of enriching section = 12trays 400mm = 4800 = 4.8 mts

Height of stripping section = 5 400 = 2mts

Total height of column = 4.8 + 2 = 6.8mts.

73

INSTRUMENTATION AND PROCESS CONTROL OF REACTOR

1. FI 001 & FI 002 are flow indicator of Methanol & air respectively.

2. FV 001 & FV 002 are Methanol/ Air flow control value respectively.

74

3. TI 001, PI 001 are Temperature & pressure indicator of reactor Inlet

4. TI 002 PI 002 are Temperature & pressure indicator of Reactor outlet

5. TC 002 is Temperature controller, which is located in the Outlet of reactor,

TC 002 is the cascade controller (Master control) which controls FV 003

Water flow rate (Slave controller)

6. R/C is Ratio controller, which controls FV 001 & FV 002

7. PDI 001 is the Pressure Difference Indicator, reads the pressure between

PI 001 – PI 002

8. Safety interlocks :

SDV 01 & 02 (Control valve) to trip for below abnormalities,

a) PSH is pressure high switch, if pressure reaches more than 1.5atm

b) Also a TSH (temperature switch high) is provided at reactor outlet if

temperature reaches more than set range will trip the same SDV’s

c) If Methanol pump or Air compressor fails (or u/s) upset of plants

9. Local Pressure gauges and Temperature gauges are provided for reactor inlet

and outlet, and also a rupture disc is mounted on the head of reactor.

75

Plant Layout

Good plant layout keeps overall costs, erecting cost, safety, appearance, convenience,

operating and maintenance cost to the minimum. The unit should be planned in minimum

space. Equipments should be positioned in such a manner that the piping cost is minimum.

Feed lines, product lines, utility lines should be planned in such a way that future

expansion is easily possible. Furnaces, fire hazard, explosive units should be isolated from

other process units to avoid major hazards. It is customary that the heat exchange units are

positioned vertically while pipe lines are laid in rectangles.

Safety measures have to be installed for process equipments, raw materials storages,

personnel. Study of all these aspects are part of chemical process plant design.

The key to economic construction and efficient is a carefully planned, functional

arrangement of equipments, piping and buildings. Furthermore, an accessible and

aesthetically pleasing plot plan can make major contributions to safety, employee

satisfaction and sound community relations. The physical layout of the equipments is very

important. A modern process plant installed today shall remain in operation for 20 years or

even more. Any error in layout can prove costly at the later stage.

There cannot be any ideal plot plan because the chemical processes differ in many ways.

The following parameters have to be considered while designing a plot plan.

76

1) Scale and scope of the operation.

2) Available property limitations.

3) Safety considerations.

4) Operating supervision and labour scheduling.

5) Utilities supply.

6) Solid materials handling requirements.

7) Maintenance convenience.

8) Construction economics.

9) Possible future expansion.

Units can be placed in one of the two general forms.

1) Grouped layout

2) Flow line layout

In grouped layout similar equipments such as towers, pumps, heat exchangers etc. are

grouped together. In flow line pattern, the towers, pumps, and heat exchangers are

arranged in the layout as they appear on the flow plan.

The grouping layout is advantageous and economical for large chemical process plants,

while flow line layout is useful for small process plants or large plants having relatively

less pump or exchangers.

For designing a plant layout the following guidelines should be followed:

1) Study the process flow diagram and equipment list and find out the scope of equipment

to be included in the unit area.

2) Integrate as many process operations as possible. This helps to minimize operating staff.

3) Decide the equipment elevation. It is dependent upon the process requirement as well as

pump suction requirements. The process, project and mechanical engineers should work in

co-ordination to achieve the satisfactory operation of the process.

77

4) Make a detailed study of process flow, and operating procedure. The function of each

process equipment should be easily accessible for maintenance.

5) Study the maintenance, shutdown method for each process equipment. The equipment

should be easily accessible for maintenance.

6) Make a detailed study of operating hazards. This helps to device safest arrangement of

equipments.

7) Adequate clearance should be available between two equipments. For example, a

rectangular plan with a central overhead pipe rack permit equipments to be located along

both sides of the pipe way.

8) Locate large field fabricated equipments, such as reactor, or fractionating columns, at

one end of the plot so that the erecting staff can unload, assemble, erect, weld and test

these vessels without interfering or delaying work in the rest of the area.

The process plant should be located on one side of a tank farm while shipping, transport,

loading/unloading facilities on another side. It helps to reduce the piping length between

process units and storage tanks and between unloading points and storage tanks.

Administration and service facilities should be located near the process plant entrance and

to initially installed process units. It reduces the distance between service units,

laboratories, storehouse etc.

Warehouse, salvage yard should be close together. Cooling towers should be located where

water drift from the towers will not cause excessive corrosion of process equipment. They

should be oriented crossways to the wind direction in order to minimize recycling of air

from the discharge of one tower to the section of an adjacent tower.

Hazardous, toxic chemicals storage should be planned close to the unloading of tank car.

This minimizes line washing. Hazardous tanks should be provided with fire protection

78

walls and a clear space of one diameter between any pair of tank. Effluent treatment should

be located near the natural drain facilities.

Pumping arrangement of liquids from the tanks should be decentralized. This minimizes

damage in the event of fire.

In a process plant, there should be sufficient space between the process equipments. Its

avoids congestion after piping valves instrumentation is done on equipments. Pump and

compressor lines should be small, short as far as possible.

Furnace transfer lines should be short as far as possible. Hot lines should be long enough.

It helps flexibility. Valve stems should never be located at face level. It becomes a hazard

to operating personnel. Piping should not be a grade level in operation area. Large vertical

vessels and reactors should be spaced at minimum 3 diameters centre to centre from each

other.

While arranging pipeline network, 30 percent space should be left for future pipelines

requirement. Double deck pipelines are useful.

Process units should be located side by side with a distance of 1.5 times the plant size

between them. A centralized control house should have as many instruments as possible.

In large process plants, plant layout has its own importance. Lot of money can be saved

by having a good plant layout. A process plant designer usually talks about the production

schedule, related performance but not about the plant layout. Production schedule can be

affected by improper plant layout. Operating costs and maintenance costs are influenced by

plant layout which can ultimately affect the profitability of process plant.

79

80

PRODUCT STORAGE

AREA

RAW MATERIAL STORAGE AREA

EFFLUENTTREATMENT

SITE

MAIN GATE

PLANT LAYOUT

FUTURE EXPANSIONSBOILER SECTION

UTILITIES

SUB-STATION

WORKSHOP FIRE & SAFETY DEPT.

EVAPORATERREACTOR ABSORBER ABSORBER FRACTINATION COLUMN

PUMP HOUSE

CENTRAL CONTROL ROOM

CANTEENADMIN BLOCKMATERIALS DEPT.

MARKETING &DISPATCH AREA

81

HEALTH AND SAFETY FACTORS

Hazard identification

SAF-T-DATA(tm) Ratings

Health Rating: 3 - Severe (Poison)

Flammability Rating: 2 – Moderate

Reactivity Rating: 2 – Moderate

Contact Rating: 3 - Severe (Corrosive)

Lab Protective Equip: GOGGLES & SHIELD; LAB COAT & APRON; VENT HOOD;

PROPER GLOVES; CLASS B EXTINGUISHER

Storage Color Code: Red (Flammable)

Potential Health effects

The perception of formaldehyde by odor and eye irritation becomes less sensitive with

time as one adapts to formaldehyde. This can lead to overexposure if a worker is relying on

formaldehyde's warning properties to alert him or her to the potential for exposure.

Inhalation: May cause sore throat, coughing, and shortness of breath. Causes irritation and

sensitization of the respiratory tract. Concentrations of 25 to 30 ppm cause severe

respiratory tract injury leading to pulmonary edema and pneumonitis. May be fatal in high

concentrations.

Ingestion: Can cause severe abdominal pain, violent vomiting, headache, and diarrhea.

Larger doses may produce decreased body temperature, pain in the digestive tract, shallow

respiration, weak irregular pulse, unconsciousness and death. Methanol component affects

the optic nerve and may cause blindness.

Skin Contact: Toxic: May cause irritation to skin with redness, pain, and possibly burns.

Skin absorption may occur with symptoms paralleling those from ingestion. Formaldehyde

is a severe skin irritant and sensitizer. Contact causes white discoloration, smarting,

82

cracking and scaling.

Eye Contact: Vapors cause irritation to the eyes with redness, pain, and blurred vision.

Higher concentrations or splashes may cause irreversible eye damage.

Chronic Exposure: Frequent or prolonged exposure to formaldehyde may cause

hypersensitivity leading to contact dermatitis. Repeated or prolonged skin contact with

formaldehyde may cause an allergic reaction in some people. Vision impairment and

enlargement of liver may occur from methanol component. Formaldehyde is a suspected

carcinogen (positive animal inhalation studies).

Aggravation of Pre-existing Conditions:

Persons with pre-existing skin disorders or eye problems, or impaired liver, kidney or

respiratory function may be more susceptible to the effects of the substance. Previously

exposed persons may have an allergic reaction to future exposures.

First Aid measures

Inhalation: Remove to fresh air. If not breathing, give artificial respiration. If breathing is

difficult, give oxygen. Call a physician.

Ingestion: If swallowed and the victim is conscious, dilute, inactivate, or absorb the

ingested formaldehyde by giving milk, activated charcoal, or water. Any organic material

will inactivate formaldehyde. Keep affected person warm and at rest. Get medical attention

immediately. If vomiting occurs, keep head lower than hips.

Skin Contact: In case of contact, immediately flush skin with plenty of water for at least

15 minutes while removing contaminated clothing and shoe. Get medical attention

immediately.

Eye Contact: Immediately flush eyes with plenty of water for at least 15 minutes, lifting

lower and upper eyelids occasionally. Get medical attention immediately.

Note to Physician: Monitor arterial blood gases and methanol levels after significant

83

ingestion. Hemodyalysis may be effective in formaldehyde removal. Use formic acid in

urine and formaldehyde in blood or expired air as diagnostic tests.

Fire fighting measures

Fire: Flash point: 60C (140F) CC Autoignition temperature: 300C (572F)Flammable limits in air % by volume: lel: 7.0; uel: 73

Flammable liquid and vapor! Gas vaporizes readily from solution and is flammable in air. Explosion: Above flash point, vapor-air mixtures are explosive within flammable limits

noted above. Containers may explode when involved in a fire. Fire Extinguishing Media: Water spray, dry chemical, alcohol foam, or carbon dioxide.

Special Information: In the event of a fire, wear full protective clothing and NIOSH-

approved self-contained breathing apparatus with full facepiece operated in the pressure

demand or other positive pressure mode. Water may be used to flush spills away from

exposures and to dilute spills to non-flammable mixtures.

Exposure controls/Personal protectionAirborne Exposure Limits:

-OSHA Permissible Exposure Limit (PEL):

0.75 ppm (TWA), 2 ppm (STEL), 0.5 ppm (TWA) action level for formaldehyde

200 ppm (TWA) for methanol

-ACGIH Threshold Limit Value (TLV):

0.3 ppm Ceiling formaldehyde, Sensitizer, A2 Suspected Human Carcinogen

200 ppm (TWA) 250 ppm (STEL) skin for methanol

Personal Respirators (NIOSH Approved):

If the exposure limit is exceeded and engineering controls are not feasible, a full facepiece

respirator with a formaldehyde cartridge may be worn up to 50 times the exposure limit or

the maximum use concentration specified by the appropriate regulatory agency or

respirator supplier, whichever is lowest. For emergencies or instances where the exposure

84

levels are not known, use a full-face piece positive-pressure, air-supplied respirator.

WARNING: Air purifying respirators do not protect workers in oxygen-deficient

atmospheres. Irritation also provides warning. For Methanol: If the exposure limit is

exceeded and engineering controls are not feasible, wear a supplied air, full-facepiece

respirator, airlined hood, or full-facepiece self-contained breathing apparatus. Breathing air

quality must meet the requirements of the OSHA respiratory protection standard

(29CFR1910.134). Where respirators are required, you must have a written program

covering the basic requirements in the OSHA respirator standard. These include training,

fit testing, medical approval, cleaning, maintenance, cartridge change schedules, etc.

Skin Protection: Wear impervious protective clothing, including boots, gloves, lab coat,

apron or coveralls, as appropriate, to prevent skin contact.

Eye Protection: Use chemical safety goggles and/or a full face shield where splashing is

possible. Maintain eye wash fountain and quick-drench facilities in work area.

Accidental release measures

Small Spill: Dilute with water and mop up, or absorb with an inert dry material and place

in an appropriate waste disposal container.

Large Spill: Flammable liquid. Keep away from heat. Keep away from sources of ignition.

Stop leak if without risk. Absorb with DRY earth, sand or other non-combustible material.

Do not touch spilled material. Prevent entry into sewers, basements or confined areas; dike

if needed. Be careful that the product is not present at a concentration level above TLV.

Check TLV on the MSDS and with local authorities.

Handling and Storage

Precautions: Keep away from heat. Keep away from sources of ignition. Ground all

equipment containing material. Do not ingest. Do not breathe gas/fumes/ vapor/spray. In

case of insufficient ventilation, wear suitable respiratory equipment. If ingested, seek

85

medical advice immediately and show the container or the label. Avoid contact with

skin and eyes. Keep away from incompatibles such as oxidizing agents, reducing agents,

acids, alkalis, moisture.

Storage: Store in a segregated and approved area. Keep container in a cool, well-ventilated

area. Keep container tightly closed and sealed until ready for use. Avoid all possible

sources of ignition (spark or flame).

Stability and Reactivity

Stability: Stable under ordinary conditions of use and storage. Hazardous Decomposition Products: May form carbon dioxide, carbon monoxide, and

formaldehyde when heated to decomposition. Hazardous Polymerization: Trioxymethylene precipitate can be formed on long standing

at very low temperatures. Nonhazardous polymerization may occur at low temperatures,

forming paraformaldehyde, a white solid.

Incompatibilities: Incompatible with oxidizing agents and alkalis. Reacts explosively with

nitrogen dioxide at ca. 180C (356F). Reacts violently with perchloric acid,

perchloricacid-aniline mixtures, and nitromethane. Reaction with hydrochloric acid may

form bis-chloromethyl ether, an OSHA regulated carcinogen.

Conditions to Avoid: Heat, flames, ignition sources and incompatibles

Environmental Impact and Pollution control

Formaldehyde is present in the environment as a result of natural processes and from

manmade sources. It is formed in large quantities in the troposphere by the oxidation of

hydrocarbons. Minor natural sources include the decomposition of plant residues and the

transformation of various chemicals emitted by foliage.

Formaldehyde is produced industrially in large quantities and used in many

applications. Two other important man-made sources are automotive exhaust from engines

86

without catalytic converters, and residues, emissions, or wastes produced during the

manufacture of formaldehyde or by materials derived from, or treated with it.

When released into the soil, this material is expected to leach into groundwater. When

released into water, this material is expected to readily biodegrade. While formaldehyde is

biodegradable under both aerobic and anaerobic conditions. When released into water, this

material is not expected to evaporate significantly. This material is not expected to

significantly bioaccumulate. When released into the air, this material is expected to be

readily degraded by reaction with photochemically produced hydroxyl radicals. When

released into the air, this material is expected to be readily degraded by photolysis. When

released into the air, this material is expected to be readily removed from the atmosphere

by dry and wet deposition. When released into the air, this material is expected to have a

half-life of less than 1 day.

Formaldehyde is proven to be 100% biodegradable. That means it breaks down to

simpler molecules (like Co2 & H2O) through the natural action of oxygen, sunlight,

bacteria and heat. Studies show that formaldehyde breaks down much more easily than the

active ingredients of most other deodorant products, so it can be easily treated in waste

treatment systems.

Solid and liquid waste are not generated in the formaldehyde plants. The off-gas is the only possible source of pollution. Most of the domestic units in India are recycling 2/3rd

part of the tail gases and remaining is incinerated (in silver catalyst process) or oxidized to

water and carbon dioxide by Emission Control System (In Metal Oxide Process)

Leakage of formaldehyde through pumps and equipments to be avoided.

87

 

General Information:

CAS Number 50 - 00 - 0

EINECS Number 200 - 001 - 8

Chemical Name Formaldehyde

Chemical Classification Aldehyde

Synonyms Formalin; formic aldehyde; formal; methyl aldehyde; methylene glycol; methylene oxide; methanal; morbicid; oxamethane; oxymethylene; paraform; polyoxymethylene glycol; superlysoform.

Formula CH2O

Molecular weight 30.03

Shipping Name Formaldehyde solution

Codes / Label Flammable liquid, class - 3.1

Hazardous Waste ID No. 5

Hazchem Code 2 SE / 2 T

UN Number 1198

Description Colourless liquid with characteristic pungent odour.

Product uses In the manufacturing of phenolic resins, artificial silk and cellulose esters, dyes, organic chemicals, glass mirrors, explosives; disinfectant for dwellings, ship, storage houses, utensils, clothes etc., as a germicide & fungicide for plants & vegetables, improving fastness of dyes on fabrics, tanning & preserving hides;, mordanting & water proofing fabrics; preserving & coagulating rubber latex; in embalming fluids.

  Packaging In SS tankers, ISO-containers and in HDPE drums / carboys.

      

 MATERIAL SAFETY DATA SHEET

(According to 91/155 EC)  

 1. IDENTIFICATION OF THE SUBSTANCE / PREPARATION AND OF THE COMPANY / UNDERTAKING:

   Product details:  

  Trade name FORMALDEHYDE, 37% SOLN. IN METHANOL/WATER

 

 

 2. COMPOSITION/ INFORMATION ON INGREDIENTS:

 

Description CAS No. EINECS Number

Formaldehyde 50-00-0 200-001-8

Methanol 67-56-1 200-659-6

Water 7732-18-5  

 

 3. HAZARD IDENTIFICATION:

88

 

Hazard description Extremely hazardous product. Harmful if swallowed or inhaled. Causes severe irritation or burns to skin, eyes, upper respiratory tract, gastro-intestinal irritation and burns to mouth and throat Lachrymator at levels from less than 20 ppm upwards.

Chronic Exposure Kidney and lever damage.

Carcinogen city (NTP/IARC/OSHA) Yes

 

 4. FIRST-AID MEASURES:

 

After inhalation Remove to fresh air, restore breathing, get medical attention.

After skin contact

Remove contaminated clothing, flush with plenty of water atleast for15 minutes.

After eye contact Immediately flush opened eye with plenty of running water for atleast 15 minutes.

After swallowing Induce vomiting of conscious patient by giving plenty of water to drink. Consult physician immediately in case of an unconscious victim.

 

 5. FIRE-FIGHTING MEASURES:

 Suitable extinguishing agents

CO2 , Dry chemical powder, water spray and alcohol foam.

 

 6. ACCIDENTAL RELEASE MEASURES:

 

Person-related safety precautions

Proper protective equipment and self contained breathing apparatus with full face piece operated in positive pressure mode.

Measures for cleaning/collecting Shut off sources of ignition, no flares, smoking or flames In area. Stop leakage if possible. Use water spray to Reduce vapour. Take up with sand or other non combustible absorbent and place into container for disposal according to item 13.

   7. HANDLING AND STORAGE:

 

Handling  

Information for safe handling Avoid breathing vapours, avoid contact with eyes, skin and clothing. Decontaminate soiled clothing thoroughly before use.

Information about fire and explosion protection Flammable liquid. Closed containers exposed to heat may explode.

Storage  

Requirements to be met by store rooms and receptacles

Store in tightly closed containers in a dry, cool, well ventilated, flammable liquid storage area.

 

 8. EXPOSURE CONTROLS / PROTECTION:

 

50-00-0 Formaldehyde

OES Short term value *3 mg/m3

Long term value **1.5mgm3

89

* total inhalable vapour, ** permissible exposure

Personal protective equipment

Self contained breathing apparatus and full protective clothing.

Respiratory protection Self contained breathing apparatus.

Protection of hands Gloves of natural rubber.

Eye protection Safety glasses W/face shield.

 

 9. PHYSICAL AND CHEMICAL PROPERTIES:

 

General information:

Form Liquid

Colour Colourless

Odour Pungent odour

Change in condition  

Melting point --

Boiling Point / Boiling range 96oC

Flash point: (CC) 59oC

Flammable limits Upper 73.0%, Lower 7.0%

Auto ignition temperature 423oC

Specific gravity 1.08

Vapour density (Air = 1) 1.0

Vapour Pressure (MMHG) at 20 oC 1.3

Solubility/Miscibility with water at 20 oC

Complete

 

 10. STABILITY & REACTIVITY:

 This product is stable, strong reducing agent, especially in alkaline solution. Keep away from strong bases and acids, oxidizing agents, aniline,phenol, isocyanates, anhydride. Combustible light and air sensitive, polymerizes spontaneously.

 

Dangerous reactions With bases / acids and oxidizing agents.

Dangerous decomposition products

CO, CO2 & vapours of formaldehyde

 11. TOXICOLOGICAL INFORMATION:

 

Acute toxicity:  

LD/LC50 values relevant for classification:

Oral/LD 50 : 100 mg/Kg (rat)SKN LD 50: 270 µl/kg (rabbit)

Primary irritant effect:  

On the skin Irritating effect (severe)

On the eye Irritating effect (severe)

Sensitization Prolonged contact may cause skin sensitization.

 

 12. ECOLOGICAL INFORMATION:  Environmental The product is expected to be slightly toxic to aquatic life. The LC 50/96

90

toxicity hrs values for fish are between 10 and 100 mg/l.

Environmental fate When released into the soil, material is expected to reach into ground water. When released into water, the material is expected to readily bio-degrade and is not expected to bio-accumulate. When released into air, the material is expected to readily degrade by photolysis, readily removed from atmosphere by dry and deposition. Half life of the material is less than one day, when released into air.

 

 13. DISPOSAL CONDITIONS:

 Whatever cannot be recycled, should be absorbed in sand or other non-combustible absorbent, containerized and transferred to appropriate and approved waste disposal facility. Dispose waste, containers and unused contents in accordance with official regulations.

   14. TRANSIT INFORMATION:

 

Shipping name FORMALDEHYDE SOLUTION

UN number 1198 Class: 3, 3.3, 8

Packing Group III Label: Flammable liquid.

 

 15. REGULATORY INFORMATION:

 This product is extremely hazardous. It is listed as an "ACGIH" 'suspected' human carcinogen and a "NTP" anticipated human carcinogen. It may cause (mutagenic) reproductive effects.

    

COST ESTIMATION

(Source http://www.niir.org/projects/tag/z,.,451_0_32/formaldehyde/index.html.)

Total Capital cost of the plant for 10 tones/ day or 3000 tones/year (for 300 working days

plant)

91

= 5.4 crores

Fixed Capital Investment (80-90% of Total Capital Investment)

FCI = 0.85 5.4 = 4.59 crores

Working Capital investment = Direct cost + Indirect cost

Direct Cost

Materials & labours involved in actual installation of complete facility (Assumed 70-85%

of FCI )

Assuming = 77.5% of FCI

= 0.775 4.59

= 3.558 crores

Equipment + Installation + Instrumentation + Piping + Electrical + Insulation + Painting

(50-60% of FCI)

Assuming value is 55% of FCI = 0.55 4.59

= 2.525 crores

(a) Purchased equipment cost ( PEC) (15 to 40% of FCI)

Assuming Value is 27.5% of FCI

= 0.275 4.59

= 1.2623 crores

(b) Installation including insulation & painting (35-45% of PEC)

Assuming 40% of PEC

= 0.4 1.2623

= 0.505 crores.

(c) Instrumentation & control installation ( 6-30% of PEC)

92

Assuming value is 18% of PEC

= 0.18 1.2623

= 0.227 crores.

(d) Piping Installation (10-80% of PEC)

Assuming 40% of PEC

= 0.4 1.2623

= 0.505 crores.

(e) Electrical Installation (10-20% of PEC )

Assuming 15% of PEC

= 0.15% 1.2623

= 0.1894 crores.

(f) Building process & auxillary (10-70% of PEC)

Assuming 40% of PEC

= 0.4 1.2623

= 0.505 crores

(g) Service facility & yard improvement (40-50% of PEC)

Assuming 40% of PEC

= 0.4 1.2623

= 0.505 crores

(h) Land (1-2% of FCI or 4-8% PEC)

Assuming 1.5% of FCI

=

1 .5100 4.9

= 0.0735 crores

(i) Direct cost (DC) = 1.2623 + 0.505 + 0.227 + 0.505 + 0.1894 + 0.505 + 0.505

93

+ 0.0735

= 3.773 crores

Indirect Cost

Expenses which are not directly involved with material & labour of actual installation of

complete facility (15-30% of FCI)

(1) Engg. & supervisors (5-15% of DC)

Assuming 6% of DC

= 0.06 3.773

= 0.227 crores

(2) Construction expenses & contractor’s fees (7-20% of DC)

Assuming 8% of DC

= 0.08 3.773

= 0.3019 crores

3) Contingency ( 5-15% of FCI)

Assuming 8% of FCI

= 0.08 4.9

= 0.392 crores

Indirect Cost = 0.227 + 0.3019 + 0.392

= 0.921 crores

This is equal to

0 .9214 . 9

×100 = 18.8%

Working Capital (10-20% of FCI)

Assuming 15% of FCI

= 0.15 4.9

= 0.735 crores

94

Total Capital investment = FCI + working Capital

= 4.9 + 0.735

= 5.64 crores

Estimation of Total product cost (TPC)

Manufacturing Cost = Direct product cost + fixed charges + plant overhead cost

I. Fixed Charges (10-20% of TPC)

1) Depreciation depends on life period, salvage value, method of calculation about

10% of FCI for machinery & equipment 2-3% for building value.

= 0.1 4.9 + 0.02 4.9

= 0.588 crores

2) Local Taxes (1-4% of FCI)

Assuming 2% of FCI

= 0.02 4.9

= 0.098 crores

3) Insurance (0.4 – 1% of FCI)

Assuming 0.5% of FCI

= 0.005 4.9

= 0.0245 crores

4) Rent (8 - 10% value rented land & building )

Assuming 9% of rented land & building

= 0.09 [0.505 + 0.0735]

= 0.052 crores

Fixed charges = 0.588 + 0.098 + 0.0245 + 0.052

= 0.7625 crores

95

Total product cost (TPC) = 5 to 10 times of fixed charges

Assuming 5 times of fixed charges

= 5 0.7625

= 3.8125 crores

II. Direct Product Cost

1) Raw material (10-50% of TPC )

Assuming 20% of TPC

= 0.2 3.8125

= 0.7625 crores

2) Operating labour cost (10-20% of TPC)

Assuming 10% of TPC

= 0.1 3.8125

= 0.3812 crores

3) Direct supervisory & Electrical labour (10-25% of operating labour)

= 0.15 0.38125

= 0.0572 crores

4) Utilities (10-20% of TPC)

Assuming 15% of TPC

= 0.15 3.8125

= 0.572 crores

5) Maintenance & repairs (2-10% of FCI)

Assuming 5% of FCI

= 0.05 4.9

= 0.245 crores

6) Operating expenses (0.5-1.0% of FCI)

96

Assuming 0.75% of FCI

= 0.0075 4.9

= 0.0368 crores

7) Lab charges (10-20% of operating labour)

Assuming 12% of operating labour

= 0.12 0.38125

= 0.0458 crores

8) Patent & Royalties (0-6% of TPC)

Assuming 0.5% of TPC

= 0.005 3.8125

= 0.019 crores

Direct Product cost = 0.7625 + 0.38125 + 0.0572 + 0.572 +0.245 + 0.0368 +

0.0458 + 0.019

= 2.12 crores

III. Plant overhead Cost (5-15% of TPC)

It involves cost of general plant up keep & overhead payroll, packaging, medical services,

safety & production, restaurant recreation, salvage, lab & storage facilities.

Assuming 8% of TPC

= 0.08 3.8125

= 0.305 crores

Manufacturing Cost = fixed charge + Direct product cost + plant overhead cost

= 0.765 + 2.12 + 0.305

= 3.188 crores.

General Expenses

97

General Expenses = Administration cost + Distribution & Selling cost

(a) Administration cost (2 to 6% of TPC)

It includes cost for executive salaries, clerical wages, legal fees, and communication.

Assuming 2% of TPC

= 0.02 3.8125

= 0.0763 crores

(b

)

Distribution & Selling cost (2 – 6%) of TPC)

Assuming 4% of TPC

= 0.04 3.8125 = 0.1525 crores

(c) R & D ( 2 – 5% of TPC)

Assuming 3% of TPC

= 0.03 3.8125

= 0.114 crores

(d

)

Finances (0 – 7% of Total capital expenses)

Assuming 1.5% of Total capital expenses

= 0.015 5.4

= 0.081 crores

General expenses = 0.0763 + 0.1525 + 0.114 + 0.081

= 0.4238 crores

Total product cost = Manufacturing cost + General expenses

= 3.188 + 0.4238

= 3.612 crores

Gross earnings = total income – total production cost

for 1 kg of Formalin

Selling price is Rs. 25/-

98

Assuming 20% profit for seller

Selling price = Rs. 25 0.80 = Rs. 20

Gross Annual earning = 10 103 300 20

= 6 crores

Net annual income = gross annual earning – Total product cost

= 6 – 3.8125

= 2.1875 crores

Net annual earning after depreciation

= 2.1875 – 0.588

= 1.5995 crores

Net profit after tax (46% tax rate)

= (1 – 0.46) 1.5995

= 0.8638 crores

FCI = 4.59 crores.

Payout period =

FCI(Net profit after tax + depreciation )

=

4 .59(0 . 8638 + 0 . 588 )

= 3.16 years

Rate of return =

Net profitFixed capital investment

=

(0 . 8638 + 0 . 588 )4 .59

×100

= 31.63%

99

100