DISSERTATION

1
CHAPTER ONE
1.0 INTRODUCTION
The production of formaldehyde is a straightforward process. Methanol and air are combusted
within a reactor in the presence of a silver catalyst. The product is a mixture of formaldehyde
and methanol in water, which is then run through an absorber to remove inert gases and a
distillation column to recycle residual methanol. The final product contains approximately
thirty-seven weight percentage formaldehyde in water with four weight percentage methanol
added as a stabilizer. The formalin solution may then be stored or used immediately in another
application (Mann, 1969).
Formaldehyde (CH2O) is known as the first series of aliphatic aldehydes. The occurrence of
formaldehyde is abundant in air and is also a by-product of several biological processes. The
average person produces 1.5 ounces of formaldehyde per day as part of normal human
metabolism. Plants andanimals produce formaldehyde as their by-products. For example,
brussels sprouts and cabbage emit formaldehyde when they are cooked. Formaldehyde can be
produced by oxidation of methanol with air in the presence of catalyst. Formaldehyde may be
produced at a relatively low cost, high purity, and from a variety of chemical reactions, making
formaldehyde one of the most produced industrial chemicals in the world (Betsy, 2007
Formaldehyde is a key chemical component in many manufacturing processes. It is relatively
simple to produce, although careful handling, transportation and storage are required. In this
report, analyses on the chemical itself, reactions, safety, plant design, troubleshooting and
economics are performed. Finally, some conclusions and suggestions are presented as well
(Mann, 1969).

2
Formaldehyde industries have grown since 1972, from a yearly global production volume of 7
million metric tons up to 24 million metric tons in recent years. In addition, commercial uses
of formaldehyde have widespread industrial applications; which shows how important the
chemical is in our everyday lives (Mann, 1969).
Formaldehyde has a colourless and distinctive pungent smell even can be detected in low
concentrations. It is a highly flammable gas, with a flashpoint of 50°C. The heat of combustion
is 134.l kcal/mol or 4.47kcal/g. Formaldehyde issoluble in a variety of solvents and miscible
in water. Formaldehyde usuallysold as 37 weight percentage solution in water known as
formalin (Arlington, 2005).
Because of its unique properties, formaldehyde has been used in allkinds of products such as
vaccines, medicines, fertilizers, carpets, plastics, clothing, glues, x-rays, and plywood. Most
formaldehyde products find uses as adhesives and wood coatings to provide weather-
resistance. Formaldehyde is an important ingredient in production of formaldehyde-based
material. The formaldehyde-based resins are used in production of gluesfor household
furnishing. The largest use of formaldehyde is in the manufacturingof amino and phenolic
resins. The phenolic moulding resins are used inappliances, electrical control, telephone and
wiring devices. In the automotive and building industries, formaldehyde-based acetyl resins
are used in theelectrical system, transmission, engine block, door panels, and brake shoes
(Gerberich& Seaman, 2004).

3
1.1 GENERAL OBJECTIVE
The main objective is to design a plant for the production of formalin from methanol
1.2 SPECIFIC OBJECTIVE
In particular the specific objectives are to design a plant to produce 100 tonnes of formalin
annually, 37% weight of formalin and to make the process feasible and very economical
through the following assessments:
 Development of process blocks diagram and flow sheet
 Undertaking of material and energy balances
 Specifications and design of equipment
 Material and fabrication selection
 Economic analysis of the project
 Plant operation and control
 Profitability analysis of the project
 Plant location
 Cash flow for industrial operation
 Health and safety hazards, and
 Environmental protection

4
CHAPTER TWO
LITERATURE REVIEW
2.0 EARLY HISTORY OF FORMALDEHYDE PRODUCTION
Formaldehyde was first prepared by Butlerov in 1859 as the product of an attempted synthesis
of methylene glycol [CH2(OH)2]. The preparation was carried out by hydrolyzing methylene
acetate previously obtained by the reaction of methylene iodide with silver acetate. Butlerov
noticed the characteristic odour of the formaldehyde solution thus produced, but was unable to
isolate the unstable glycol which decomposes to give formaldehyde and water. Butlerov also
prepared a solid polymer of formaldehyde by reacting methylene iodide and silver oxalate. He
showed that this compound was a polymer of oxymethylene, (CH2O), but failed to realize that
it depolymerized on vaporization. He also obtained the new polymer by the reaction of
methylene iodide and silver oxide, which gave additional evidence of its structure. He showed
that it formed a crystalline product with ammonia (hexamethylenetetramine) and even stated
that its reactions were such as one might expect from the unknown "formyl aldehyde". In 1868,
A. W. Hofmann prepared formaldehyde by passing a mixture of methanol vapours and air over
a heated platinum spiral, and definitely identified it (Betsy, 2007).
2.1 SOURCES OF FORMALDEHYDE
Wood as a natural material contains some amount of formaldehyde (Meyer and Boehme, 1997;
Que and Furuno, 2007) which can be released especially during thermal treatment (Schäfer and
Rofael, 2000). However, emission levels depend on numerous factors like species, moisture
content, outside temperature or storing time (Martinez and Belanche, 2000).
(Irle et al., 2008) found correlation between moisture content in material and formaldehyde
release. It was shown that moisture content change from 0.0% to 4.0% resulted in 6-fold

5
increase in formaldehyde emission, which suggests that release is governed by physical
processes rather than chemical ones. Moreover, it was reported that both polysaccharides and
lignin, too, were a source of formaldehyde. A path of formaldehyde release includes
transformation of polysaccharides to hexoses, oxymethylfurfural and its subsequent
imbalances to furfural and formaldehyde (Schäfer and Rofael, 2000). Also it was shown that,
lignin might undergo reactions releasing formaldehyde, especially in acidic environment
(Freudenberg and Harder, 1927).
2.2 PHYSICAL AND CHEMICAL PROPERTIES OF FORMALDEHYDE
2.2.1 PHYSICAL PROPERTIES
Formaldehyde is a colourless gas at ambient temperature that has a pungent, suffocating odour.
At ordinary temperatures formaldehyde gas is readily soluble in water, alcohols and other polar
solvents. It has following physical properties:
Boiling point at 101.3 kPa = -19.2o
C
Melting point = -118o
C
Density at –80o
C = 0.9151g/cm3
At –20o
C = 0 .8153 g/cm3
Vapor density relative to air = 1.04
Critical temperature = 137.2 – 141.2 (o
C)
Critical pressure = 6.784 – 6.637 MPa
Cubic expansion coefficient = 2.83 x 10–3
K-1
(Ullmann - Encyclopaedia Of Chemical Technology, Vol. A11, 1997.)
2.2.2THERMAL PROPERTIES
Heat of formation at 25o
C = -109.6 kJ/mol

6
Heat of combustion at 25o
C = 561.5 kJ/mol
Heat of vaporization at –19.2o
C = 23.32 kJ/mol
Specific heat capacity at 25o
C = 35.425 J/mol K
Heat of solution at 23o
C
In water = 62 kJ/mol
In methanol= 62.8 kJ/mol
In 1-propanal = 59.5 kJ/mol
In 1-butanol = 62.4 kJ/mol
Entropy at 25o
C= 218.8 + 0.4 kJ/mol K
2.2.3 CHEMICAL PROPERTIES
Formaldehyde is one of the most reactive organic compounds known. The various chemical
properties are as follows:
Decomposition
At 150o
C formaldehyde undergoes heterogeneous decomposition to form methanol and CO2
mainly. Above 350o
C it tends to decompose in to CO and H2.
Polymerization
Gaseous formaldehyde polymerizes slowly at temperatures below 100o
C, polymerization
accelerated by traces of polar impurities such as acids, alkalis or water. In water solution
formaldehyde hydrates to methylene glycol.
H

7
H2C=O + H2O HO - C - OH
H
This in turn polymerizes to polymethylene glycols, HO (CH2O)nH, also called polyoxy -
methylenes.(Ullmann - Encyclopaedia Of Chemical Technology, Vol. A11, 1997.)
Reduction and Oxidation
Formaldehyde is readily reduced to methanol with hydrogen over many metal and metal oxide
catalysts. It is oxidized to formic acid or CO2 and H2O. In the presence of strong alkalis or
when heated in the presence of acids formaldehyde undergoes cannizzaro reaction with
formation of methanol and formic acid. In presence of aluminium or magnesium methylate,
paraformaldehyde reacts to form methyl formate (Tishchenko reaction)
2HCHO HCOOCH3
Addition reactions
The formation of sparingly water-soluble formaldehyde bisulphite is an important addition
reaction. Hydrocyanic acid reacts with formaldehyde to give glyconitrile.
HCHO + HCN HOCH2 - CN
Formaldehyde undergoes acid catalysed(Prins reaction) in which it forms α-Hydroxy-
methylated adducts with olefins. Acetylene undergoes a Reppe addition reaction with
formaldehyde to form 2- butyne-1, 4 - diol. 14
2 HCHO + HCCH HOCH2CCH2OH

8
Strong alkalis or calcium hydroxide convert formaldehyde to a mixture of sugars in particular
hexoses, by a multiple aldol condensation, which probably involves a glycolaldehyde
intermediate. Acetaldehyde, for example reacts with formaldehyde to give pentaerythritol,
C(CH2OH)4
Condensation reactions
Important condensation reactions are the reaction of formaldehyde with amino groups to give
Schiff’s bases, as well as the Mannich reaction.
CH3COCH3 + (CH3) 2NH.HCl + HCHO CH3COCH2CH2N(CH3) 2.HCl + H2O
Formaldehyde reacts with ammonia to give hexamethylene teteramine and with ammonium
chloride to give monomethylamine, dimethylamine, or trimethylamine and formic acid,
depending upon reaction conditions.
Aromatic compounds such as benzene, aniline, and toluidine combine with formaldehyde to
produce the corresponding diphenyl methanes. In the presence of hydrochloric acid and
formaldehyde, benzene is chloromethylated to form benzyl chloride. Formaldehyde reacts with
hydroxylamine, hydrazines, or semicardazide to produce formaldehyde oxime, the corresponding
hydrazones, and semicarbazone, respectively (Ullmann - Encyclopaedia Of Chemical
Technology, Vol. A11, 1997).
Resin formation

9
Formaldehyde condenses with urea, melamine, urethanes, cyanamide, aromatic sulfonamides
and amines, and phenols to give wide range of resins (Ullmann - Encyclopaedia Of Chemical
Technology, Vol. A11, 1997).
.
2.3 COMMERCIAL USES OF FORMALDEHYDE
 Formaldehyde resins are one of the major applications of formaldehyde. Some of the
derivatives are given below (Gerberich et al., 1980).
 Urea-formaldehyde resins are produced by the controlled reaction of urea and
formaldehyde. Their major uses are as adhesives for particleboard, fibreboard and
plywood. They are also used for compression moulded plastic parts, as wet-strength
additives for paper treating, and as bonders for glass fibre roofing materials (Gerberich
et al., 1980)
 Phenol formaldehyde is produced by the condensation of phenol with formaldehyde.
The use of these resins is as an adhesive in waterproof plywood. These resins are also
used for binding glass fibre insulation (WHO, 1989)
 Acetylenic chemical uses of formaldehyde involve the reaction with acetylene to form
butynediol, which in turn can be converted to butanediol, butyrolactone and
pyrolidones. Their major applications are as specialty solvent and extractive distillation
agents (Gerberich et al., 1980).
 Polyacetyl resins are produced from the anionic polymerization of formaldehyde. These
resins are used in plumbing materials and automobile components (IARC, 1995).

10
 Pentaerythritol is formed by the reaction of formaldehyde, acetaldehyde and sodium
hydroxide. Its largest use is in the manufacture of alkyd resins for paints and other
protective coatings (WHO, 1989).
.
 Hexamethylene tetramine is formed by the reaction between formaldehyde and
ammonia. It is used as a partial replacement for phosphates as a detergent builder and
as a chelating agent (Gerberich et al., 1980)
.
 Urea-formaldehyde concentrates are used as controlled release nitrogen fertilizers
(Gerberich et al., 1980)
 Melamine resins are thermosetting resins produced from melamine and formaldehyde
and are primarily used for surface coatings. The direct use of formaldehyde is to impart
wrinkle resistance in fabrics (WHO, 1989).
2.4 LIQUID FORMALDEHYDE
Liquid formaldehyde polymerizes rapidly even at temperatures of -80°C unless extremely pure.
However, even the purest samples show signs of polymerization after about four hours. When
the liquid is heated in a sealed tube, polymerization attains almost explosive violence.
Satisfactory agents for inhibiting polymerization have not yet been discovered, although
(Spence, 1933) states that quinol increases stability to a slight extent.
According to (Kekule, 1893), the density of the pure liquid is 0.9151 at -80°C and 0.8153 at -
20°C. The mean coefficient of expansion calculated from these values is 0.00283, a figure
similar to the expansion coefficients of liquid sulphur dioxide and liquid ammonia. At low
temperatures liquid formaldehyde is miscible in all proportions with a wide variety of non-
polar organic solvents, such as toluene, ether, chloroform, and ethyl acetate.

11
According to (Sapgir, 1929), solutions of liquid formaldehyde in acetaldehyde obey the laws
set down for ideal solutions.
Although the solutions obtained with the above-mentioned solvents are somewhat more stable
in most cases than the pure aldehyde, they also precipitate polymer on storage. Liquid
formaldehyde is only slightly miscible with petroleum ether Polar solvents, such as alcohols,
amines, or acids, either act as polymerization catalysts or react to form methyl or methylene
derivatives (Walker, 1933).
2.5 MANUFACTURING PROCESSES
Most of the world’s commercial formaldehyde is manufactured from methanol and air either
by a process using a silver catalyst or one using a metal oxide catalyst. But there exist other
process forms as well (Walker, 1933).
2.5.1 METHANOL PROCESS
Formaldehyde is manufactured principally from methanol; limited amounts are also produced
by the oxidation of natural gas and the lower petroleum hydrocarbons. Although other methods
of preparation involving the hydrogenation of carbon oxides, the pyrofytic decomposition of
formates, and many others have been studied, and do not appear to have achieved commercial
benefits. In connection with the reduction of carbon oxides, (Newton and Dodge, 1937)
determined the equilibrium constant for such reaction, and yield values calculated from this
constant indicated that the reaction would appear to be unfavourable as a means of
formaldehyde synthesis.
Thus, even if a high reaction rate could be attained at a low temperature and if a high pressure
were used, the equilibrium conversion to formaldehyde would be only about 0.8 per cent, while
at high temperatures or lower pressures the yields would be still lower.

12
Moreover, unless the hydrogenation catalyst used were highly selective, any formaldehyde
formed would tend to be completely hydrogenated to methanol (Newton and Dodge, 1937).
2.5.2 SILVER AND COPPER CATALYSTS PROCESS
The first recorded production of formaldehyde was carried out by Butlerov, 1859 when he
hydrolysed methylene acetate and, in 1868 von Hofmann performed the synthesis using heated
platinum spirals to oxidize methanol vapours using air (von Hofmann, 1868). But the first
industrial catalyst to be used for formaldehyde production was a copper gauze catalyst
developed by (Loew, 1886) and (Trillat, 1889). Blank in 1910 was first to use a silver catalyst,
which replaced copper due to higher yields and higher resistance to poisoning.In 1931, the
main catalyst used industrially today was developed by Adkins and Peterson, who used an iron
molybdenum oxide catalyst in the selective oxidation of methanol to formaldehyde (Baekeland,
1909).
Various forms of the silver-catalysed processes are adopted commercially nowadays, described
in many different patents (Halbritter et al., 1974; Wachs and Wang, 1999).Two main variations
are used industrially. The first process referred to as methanol ballast process, in which only
air and pure methanol are fed is used in numerous companies. Features of these processes are
an incomplete conversion of methanol and its distillative recovery. BASF uses another
important process referred to as water ballast process feeding extra water with the mixture,
thereby achieving a complete conversion of methanol (Halbritteret al., 1974; Sperber, 1969).
However, water addition is limited by the requirement of the strength of the final products and
of the additional water needed for tail-gas scrubbing. Normally, the commercial production is
carried out with H2O to CH3OH molar ratio of 40 to 60 or 0.67 (Reuss et al., 2001). Due to its
large heat capacity, water vapour may remove a great deal of reaction heat, thereby both
preventing detrimental overheating as well as sintering of the catalyst. Moreover, because
addition of water vapour helps to burn away the coke on the catalyst surface, it is reported for

13
the industrial production that the life time of silver catalyst in the water ballast process is
significantly longer than that in the methanol ballast process (Weissermel et al., 1998)
2.5.4 SILVER AND IRON MOLYBDATE CATALYSTS PROCESS
Currently there are two main catalysts used industrially producing high yields of formaldehyde
from methanol, which are silver and iron molybdate. Most companies adopt the silver catalyst
during the production of formaldehyde (Quain et al., 2003). Iron molybdate however, is the
preferred choice by the majority of industries due to its clear economic advantages. The
oxidation of methanol to formaldehyde over a silver catalyst operates at temperatures around
650o
C to 680 °C, with approximately99 % conversion of methanol and 90 % selectivity to
formaldehyde (Waterhouse et al., 2004; Lefferts et al., 1986). Iron molybdatebeing economically
more viable to purchase, it also operates at much lower reaction temperatures; less than 400
°C, achieving approximately 99 % conversion and more than 95 % formaldehyde selectivity.
There are however problems associated with the use of both iron molybdate and silver catalysts.
Thus apart from the high reaction temperature needed to achieve high yields of formaldehyde
when employing the silver catalyst, it is also due to contamination with methanol impurities
(Bowker et al., 2002). Iron molybdate catalysts produce an advantage over silver in this aspect,
as it is not known to be easily contaminated, but does deactivate over a period of time due to
molybdenum volatilization, and so needs to be replaced at least yearly (Andersson et al., 2006;
Soares et al., 2004).
2.5.5 SODIUM CATALYST PROCESS
Amongst the two main catalysts used for methanol oxidation to formaldehyde, many other
catalytic systems have attempted to record high yields of formaldehyde. (Ruf et al., 2001)
studied thecatalytic activity of evaporated sodium catalysts which produced 45 % methanol

14
conversion and40 % formaldehyde selectivity, whereas recorded 10 % conversion and 99 %
selectivity using a mixed Ag-SiO2-MgO-Al2O3 catalysts, which are evidently less active than
either silver or iron molybdate catalysts (Ren et al., 2003).
2.5.6 MOLYBDENUM CATALYSTS ON SILICA SUPPORT PROCESS
Relevantly, the most frequently studied catalysts are molybdenum and vanadium based.
Although molybdenum trioxide as a bulk catalyst has been examined to be active for speeding
up reaction, it is mainly studied to be supported on silica. Work studied by Cheng, 1996
revealed that at a reaction temperature of 300°C, bulk MoO3 produces 50 % conversion of
methanol, but when supported on silica (15 wt.%), conversion increased to 95 %, clearly
indicating the benefit of supporting the active material. The disadvantage of supporting MoO3
however, was noticedby the decrease in formaldehyde selectivity, which dropped to 67 %
selectivity (supported) compared to 79 % selectivity achieved with bulk MoO3 recorded that
although the silica support is in fact inert in the oxidation of methanol alone; it is suggested to
have a significant role when applied as a support for MoO3. At 250 °C, it was revealed that
adsorption of methanol resulted in the formation of methoxide species on the silica support,
which are mobile, and can migrate to molybdenum centres to be oxidized to formaldehyde.
(Serman et al., 2004)
2.5.7 VANADIUM CATALYSTS ON SILICA SUPPORT PROCESS
As well as supported molybdenum oxides, vanadium oxides have also been extensively used
as active catalysts for the selective oxidation to formaldehyde. Tatibouet and Germain, 1980
discovered that bulk V2O5 as a catalyst produces appreciable selectivity to formaldehyde with
97 % selectivity, but this is only viable at low methanol conversions. As with MoO3 the most
frequently reported V2O5 catalysts are supported on a range of metal oxide supports such as
TiO2 and ZrO2, among others, with the activity of the vanadium strongly dependent on the

15
support used. Niskala et al., 2010 have examined that the use of 3 wt.% V2O5 on SiO2 and 3
wt.% V2O5 on the V2O5/SiO2 catalyst achieves 75 % selectivity at a conversion of 82 %, but
the use of a mixed oxide support of SiO2/TiO2 and a lower reaction temperature of 410 °C, 96
% formaldehyde selectivity is produced at 91 % conversion. Taking the above approaches, it
shows clearly that the supported molybdenum phosphate and promoted vanadium phosphate
catalysts provide a greater step ahead in the selective oxidation of methanol to formaldehyde.
(Serman et al., 2004)
2.5.8 AEROBIC DEGRADATION PROCESS OF FORMALDEHYDE
Mostly formaldehyde is used in industrial processing of wood and in the production of paper,
leather, resins and glue (Kirk et al., 1980). It occurs in waste waters of various origins and due
to its biocidal action it is toxic to many microorganisms (Omil et al., 1999). It has been shown
that formaldehyde is the product of biodegradation of C1-compounds, like methane and
methanol, and of nitrogen compounds, where it is a major intermediate of degradation (Oren
et al., 1992). Previous works done shows that, the following microorganisms are capable to
degrade formaldehyde: Pseudomonas spp. generum (Kato et al., 1983), Halomonas spp.
(Azachi et al., 1995) and various strains of methylotropha (Attwood et al., 1984);
Debariomycesspp. and Trichosporonspp. yeast genera (Kato et al., 1982), Hansenula spp. (van
Dijken et al., 1975), Candida spp. (Pilat and Prokop., 1976) and Gliocladium spp. fungi
(Sakaguchi et al., 1975). Formic acid is the basic intermediate in formaldehyde biodegradation
(Oren et al., 1992) and (Attwood et al., 1984). Pseudomonas spp. Degrade formaldehyde with
formaldehyde dismutase enzyme (Adroer et al., 1990) while yeast, likeHansenula spp. and
Candida spp. genera, achieves it using the enzymes formaldehyde and formate dehydrogenase
(Pilat and Prokop, 1976; Sakaguchi et al., 1975). The reactions are:

16
2CH2O pseudomonas putida CH3OH + HCOOH
CH2O candida boidinii HCOOH formate dehydrogenase CO2
Not many microorganisms can degrade formaldehyde, which is attributed to its toxic effect on
various parts of bacterial cells such as spores, cell wall and compounds with amino-group. It
has been shown that formaldehyde undergoes biodegradation in synthetic media. There are,
however, scarce reports on its degradation in wastewaters. This can be explained by the fact
that, in addition to formaldehyde, waste waters contain other substances that may affect
formaldehyde degradation. For example, waste waters from the production of melamine resins
do not contain only formaldehyde (70–2700 mg/L), but also methanol (up to 300 mg/L),
butanol (up to 200 mg/L) and melamine (700–5000 mg/L) (Cantó et al., 1998); waste waters
from the production of urea formaldehyde resins contain formaldehyde (200–4000 mg/L), urea
(100–800 mg/L) and ammonium (up to 400 mg/L) (Garrido et al., 2001).
2.6 RECENT MANUFACTURING PROCESSES
2.6.1 PRODUCTION
Formaldehyde has been produced commercially since 1889 by the catalytic oxidation of
methanol. Various specific methods or processes listed above were used in the past, but only
two are widely used currently: the silver catalyst process and the metal oxide catalyst process
(Bizzari, 2000; Reuss et al., 2003; Gerberich & Seaman, 2004).
2.6.2 SILVER CATALYST PROCESS
The silver catalyst process is conducted in one of two ways:

17
(i) Partial oxidation and dehydrogenation with air in the presence of silver crystals,
steam and excess methanol at 680 to 720 °C and at atmospheric pressure (also called
the BASF process with methanol conversion, 97 to 98%) (Reuss et al., 2003).
(ii) Partial oxidation and dehydrogenation with air in the presence of crystalline silver
or silver gauze, steam and excess methanol at 600 to 650 °C (primary conversion
of methanol, 77–87%) thus; the conversion is completed by distilling the product
and recycling the unreacted methanol. Carbon monoxide, carbon dioxide, methyl
formate and formic acid are by-products (Bizzari, 2000; Reuss et al., 2003;
Gerberich & Seaman, 2004).
2.6.3 METAL OXIDE PROCESS
In the metal oxide (Formox) process, methanol is oxidized with excess air in the presence of a
modified iron–molybdenum–vanadium oxide catalyst at 250 to 400 °C and atmospheric
pressure (methanol conversion, 98–99%). By-products are carbon monoxide, dimethyl ether
and small amounts of carbon dioxide and formic acid (Bizzari, 2000; Reuss et al., 2003;
Gerberich and Seaman, 2004).
2.6.4 SELECTION OF PROCESS
It is estimated that nearly 70% of commercial formaldehyde is produced by metal oxide
process. This process has a very low reaction temperature, which permits high catalyst
selectivity, and the very simple method of steam generation. The conversion is around 95-98%
per pass, which is greater than silver oxide process and this project therefore selected this
process type (Gerberich and Seaman, 2004).

18
CHAPTER THREE
3.0 PROCESS DESCRIPTION
Air is compressed to the air compressor in order to raise it pressure. Fresh methanol feed is
mixed with recycled methanol. Pump is used to raise the pressure of the recycled methanol.
The following reactions take place in the reactor.

19
CH3OH + 1
2⁄ O2→ HCOH + H2O 𝜟HR = 37 Kcal/kgmol
HCOH + 1
2⁄ O2→ CO + H2O 𝜟HR=51 Kcal/kgmol
Heat released by the exothermic reaction is removed by vaporization of a high boiling heat
transfer fluid on the outside of the tube. The feed then enters a heat exchanger where the
methanol is vaporized. The reactor is used to ensure the conversion of the methanol. The feed
then goes to the fluid bed reactor where some amount of heat is then taken out by making high
pressure stream from the boiler feed water. Before the feed enters into the reactor, the pressure
is reduced before it goes into the formalin absorber this can either be tray or packed type. The
formalin absorber is set in order to absorb some percentage of formaldehyde that enters. Within
the formalin absorber unit, gas is taken off and heat is supplied before entering the formalin
distillation column. The formalin absorber recovers some percentage weight solution of
formaldehyde in water. A lot of methanol is then removed in the distillate. The distillate is
recycled back to the inlet of the fresh methanol. Formaldehyde concentration in the product is
adjusted by controlling the amount of water added to the top of the absorber. Formic acid is
removed by ion exchange. Deionized water is added to achieve maximum weight percentage
of formaldehyde in water. About 37% weight solution of formaldehyde is in water. The
formalin storage tank must be maintained at a specific temperature (Mann and Hahn, 1969)

21
CHAPTER FOUR
MATERIAL AND ENERGY BALANCE
4.0 MATERIAL BALANCE
See Appendix A for Material Balance Calculations:
4.1.0 Balance on Vaporizer:
Off- gas stream
Component Mole %
Methanol 7.5277 66.89
Water 3.7258 33.11
Total mole (kmol) = 11.2535
Temperature (o
C) = 150
Pressure (kPa) = 101
Output
Component Mole %
Methanol 368.8578 99.98
Water 0.0759 0.02
Temperature (o
C) = 150
Methanol (feed)
Component Mole %
Methanol 372.4253 99
Water 3.8019 1
Total mole (kgmol) = 380.1872
Temperature (o
C) = 25
VAPORIZER

22
4.1.1 Balance on Compressor:
Output
Component Mole %
Oxygen 190.0936 21
Nitrogen 715.1140 79
Temperature (o
C) = 150
Air (feed)
Component Mole %
Oxygen 190.0936 21
Temperature (o
C) = 25
Compressor

23
4.1.2 Balance around mixing point (2):
Output
Component Mole %
Methanol 376.3100 29.35
Formaldehyde 0.2563 0.02
Water 3.7276 0.006
Oxygen 190.0936 14.83
Nitrogen 715.1140 55.79
Temperature (o
C) = 150
Output
Component Mole %
Methanol 368.8578 99.98
Water 0.0759 0.02
Temperature (o
C) = 150
Air (feed)
Component Mole %
Oxygen 190.0936 21
MIXING POINT (2)

24
4.1.3 Balance on Reactor:
Temperature (o
C) = 150
Mixer input
Component Mole %
Methanol 29.35 29.35
Water 0.006 0.006
Oxygen 14.83 14.83
Nitrogen 55.79 55.79
Total mole (kmol) = 100
Temperature (o
C) = 150
REACTOR
(87.4% conversion)

25
Output
Component Mole %
Water 26.6579 23.42
Oxygen 2.0040 1.76
Temperature (o
C) = 343
Off-gas (output)
Component Mole %
Water 24.9685 30.06
Oxygen 2.0040 2.41

26
4.1.4 Balance on Absorber:
99
Temperature (o
C) = 109.12
Pressure (kPa) = 101Fresh water (feed)
Component Mole %
Water 50.1209 100
Temperature (o
C) = 30
Reactor (output)
Component Mole %
Water 26.6579 23.42
Oxygen 2.0040 1.76
Temperature (o
C) = 343
Absorber Bottom Product(output)
Component Mole %
Water 51.8103 64.06
Total mole (kmol) =80.8776
Temperature (o
C) = 109.12
Distillate (output)
Component Mole %
Methanol 3.5905 69.92
Water 1.0362 20.18
ABSORBER
(99%absorption)

27
4.1.5 Balance on Distillation Column:
Absorber Bottom Product
Component Mole %
Water 51.8103 64.06
Total moles (kmol) = 80.8776
Temperature (o
C) = 109.12
Temperature (o
C) = 68.3
Pressure (kPa) = 121.2
Distillation Bottom (output)
Component Mole %
Water 50.7739 67.03
Temperature (o
C) = 110
Distillation Column

28
4.1.6 Balance on mixing point (3):
Distillation Bottom Product
Component Mole %
Water 50.7739 67.03
Temperature (o
C) = 110
Output
Component Mole %
Water 81.2071 76.48
Temperature (o
C) = 109
Deionised water (feed)
Component Mole %
Water 30.4366 100
Temperature (o
C) = 30
MIXING POINT (3)

29
4.1.7 Balance on mixing point (1);
Output
Component Mole %
Methanol 379.9756 98.61
Water 4.8381 1.25
Temperature (o
C) =46.87
Recycle Stream (Distillate)
Component Mole %
Methanol 3.5905 69.92
Water 1.0362 20.18
Temperature (o
C) =68.3
Fresh Feed Stream
Component Mole %
Methanol 376.3853 99
Water 3.8019 1
Temperature (o
C) = 25
MIXING POINT (1)

30
4.2 ENERGY BALANCE
4.2.0 Energy balance on vaporizer
V
I P
Q
I = Fresh methanol input
Q = Heat input
V = Amount of methanol and water removed
P = Product outlet
Table1. Energy balance on vaporizer
INPUT OUTPUT
Components KJ/hr Components KJ/hr
Methanol
Water
Q
509.159
14.7848
14560.6801
Methanol
Water
14931.9074
152.7104
Total 15084.6178 15084.6178
VAPORIZER [E-101]

31
4.2.1 Energy balance on Reactor:
J P
Q (steam)
J = Vaporized mixture input
Q = Heat input
P = Product outlet
Table 2 Energy balance on Reactor
INPUT OUTPUT
Methanol
Water
Formaldehyde
Oxygen
Nitrogen
Q
2351.7555
0.3280
1.2176
714.0206
2613.3969
3815138.769
Methanol
Water
Formaldehyde
Oxygen
Nitrogen
67.4239
296.7119
3819910.26
19.72
525.3721
Total 3820819.488 3820819.488
REACTOR

32
4.2.2 Energy balance on Reactor Coolant:
K P
Q
K = Reactor product input
Q = Heat input
P = Product outlet
Table 3 Energy balance on Reactor Coolant
INPUT OUTPUT
Methanol
Water
Formaldehyde
Oxygen
Nitrogen
Q
67.4239
296.7119
343.06
19.72
525.3721
-586.1204
Methanol
Water
Formaldehyde
Oxygen
Nitrogen
33.6352
160.1711
175.4263
10.6217
286.3132
Total 666.1675 666.1675
REACTOR COOLANT [E-103]

33
4.2.3 Energy balance on Heater:
L P
Q
L = Absorber product input
Q = Heat input
P = Product outlet
Table 4 Energy balance on Heater
INPUT OUTPUT
Methanol
Water
Formaldehyde
Q
14.8971
147.8147
79.2634
16.3414
Methanol
Water
Formaldehyde
16.3983
161.8127
80.1056
Total 258.3166 258.3166
HEATER

34
4.2.4 Energy balance on Condenser:
D P
Q
D = Distillate input
Q = Heat input
P = Product outlet
Table 5 Energy balance on Condenser
INPUT OUTPUT
Methanol
Water
Formaldehyde
Q
177.8084
48.5530
2213.8225
-1657.8213
Methanol
Water
Formaldehyde
11.7950
3.2715
767.2961
Total 782.3626 782.3626
CONDENSER

35
CHAPTER FIVE
DESIGN OF EQUIPMENT
5.1.0 CHEMICAL DESIGN OF A DISTILLATION UNIT
Criteria for selection:
This section represents an equipment design and sizing for the distillation unit of the term’s
project on the production of formaldehyde from methanol. The basis for this equipment sizing
is the previously obtained process data for the simulation of the project, which proved to be
reliable and accurate. Preliminary calculations are to be presented first to serve as a baseline of
all the calculations that follows. These calculations include a mass balance of the distillation
unit, average physical properties of the components and relative volatilities. The minimum
reflux ratio of the column is obtained through underwood’s equations. The diameter of the
column is sized in the rectifying section and the stripping section. The minimum tray number
is obtained through Fenske’s relation along with their correlated efficiencies (top and bottom).
The layout of the sieve trays and their hydrodynamic effects are then obtained in a detailed
fashion for the top and bottom sections.
Preliminary Calculations
This first section of the design is set to present the initial calculations needed in the design and
sizing of the distillation column. These calculations include material balance, physical
properties of the system and the relative volatilities of the participating components.
5.1.1 Material Balance
This initial mass balance around the distillation column gives an indication of the accuracy of
the simulated parameters that are to be used in the upcoming calculations on a kmol/hr.
Design parameters
 Minimum reflux ratio
 Minimum number of stages

36
 Actual number of stages
 Column height
 Column diameter
 Column volume
 Cross sectional area
 Number of holes
Basis;
Let n = mole
Distillate = D
Bottom product = B
Mass fraction = x
Mole fraction = y
Methanol = M
Formaldehyde = F
Water = W
Molecular weight = Mw
Assumptions:
Light component: = methanol (M)
Heavy component: = water (W)
Non-heavy component: = formaldehyde (F)
Constant Molal Overflow (CMO)

37
Fig.2. material balance around the distillation column
Table 6.composition of component at the top or rectification section
Component Mol(n)
fraction (y)
Mol
(n) = y×nT
Molecular
weight (Mw)
Mass (m) =
n × Mw
Mass
fraction
(X)= (m/mT)
Methanol 0.6992 3.5905 32.042 115.0468 0.7724
Formaldehyde 0.0989 0.5081 30.026 15.2562 0.1024
Water 0.2018 1.0362 18.000 18.6515 0.1212

38
Table 7: composition of component at the bottom or stripping section
Component Mol (n)
fraction = y
Mole
n = y× ntot
Molecular
weight (Mw)
Mass (m) =
n × Mw
Mass
fraction
(X) = m/mT
Methanol 0.0009 0.0682 32.042 2.3487 0.0014
Formaldehyde 0.3287 24.8956 30.026 747.5153 0.4493
Water 0.6703 50.7739 18.000 913.9330 0.5493
Formaldehyde to water ratio:
𝑋 𝑓
𝑋 𝑤
=
0.4493
0.5493
= 0.818
About 44.93% of formaldehyde
The physical parameters to be included are the molecular weight and average density on basis
of mole fractions of the components in both the rectifying and stripping section. Molecular
weight
Mass fraction of formaldehyde (xf)
Mass fraction of methanol (xm)
Mass fraction of water (xw)
Molecular weight of formaldehyde (MwF)
Molecular weight of methanol (MwM)
Molecular weight of water (MwW)
Density of vapour phase (ρv)
Density of liquid phase (ρL)
Temperature ( T)

39
specific heat in solution = 62KJ/mol
Rectifying Section (MwT): = xm× MwM+ xF×MwF+xw×MwW
0.6992 × 32.042 + 0.0989 × 30.026 +0.2018 × 18.000 = 29.0058g/mol
Stripping Section (MwB):= xm×MwM + xF × MwF + xw× MwM
0.0009 × 32.042 + 0.3287 × 30.026 + 0.6703 × 18 = 21.9638g/mol
Average Density at the rectification section:
ρv=
P𝑀 𝑊𝑇
𝑅𝑇
=
101000×0.0290058
8.314×341.45
= 1.031975kg/m3
Rectifying Section:
(ρL)= ∑(specific heat in solution ) × (water density ) × (mol fraction)
1×62.4×0.0097+ 0.815×62.4×0.0296+0.791×62.4×0.9536 = 49.179kg/m3
Average density at Stripping Section:
ρv =
P𝑀 𝑊𝐵
RT
=
101000×0.0219638
8.314×375.15
= 0.71124kg/m3
(ρL) =∑(specific heat in solution)×(water density) ×(mole fraction)
= 1×62.4×0.6703+0.815×62.4×0.3287+0.791×62.4×0.0009 = 54.6375
5.1.2 Relative Volatility (α)
The volatility of each component is to be calculated for the rectifying and stripping sections
and their average relative to a reference component with is methanol in our case.
Relative volatility of formaldehyde (𝛼F)
Relative volatility of methanol (𝛼M)
Relative volatility of water (αW)

40
Rectification section
𝛼F =
𝑦 𝐹/𝑥 𝐹
𝑦 𝑀/𝑥 𝑀
=
0.0296/0.1024
0.9536/0.77236
= 0.2342
𝛼M =
𝑦 𝑀/𝑥 𝑀
𝑦 𝑀/𝑥 𝑀
=
0.9536/0.77236
0.9536/0.77236
= 1
𝛼w=
𝑦 𝑊/𝑥 𝑊
𝑦 𝑀/𝑥 𝑀
=
0.0097/0.1212
0.9536/0.77236
= 0.0647
Stripping section
𝛼F =
𝑦 𝐹/𝑥 𝐹
𝑦 𝑀/𝑥 𝑀
=
0.32899/0.4493
0.0030/0.0014
= 0.0304
𝛼M =
𝑦 𝑀/𝑥 𝑀
𝑦 𝑀/𝑥 𝑀
=
0.0030/0.0014
0.0030/0.0014
= 1
𝛼w=
𝑦 𝑊/𝑥 𝑊
𝑦 𝑀/𝑥 𝑀
=
0.6703/0.5493
0.0030/0.0014
= 0.5695
Average volatility (for FENSKE’s equation)
𝛼F = √ 𝛼 𝐹𝑡𝑜𝑝 × 𝛼 𝐹𝑏𝑜𝑡 = √ 0.2342 × 0.40497 = 0.30797
𝛼M = √ 𝛼 𝑀𝑡𝑜𝑝 × 𝛼 𝑀𝑏𝑜𝑡 = √1 × 1 = 1
𝛼w = √0.0647 + 0.008241 = 0.192
5.1.3 Minimum reflux ratio
The application is done by utilizing underwood’s shortcut method. To facilitate the underwood’s
approach, we use the following assumptions:
- Constant molar Overflow (CMO)
- Non keys are undistributed with (DxF) = 0 kmol/hr.
- Constant Relative volatilities

41
- Since liquid fraction q=0.9963, saturated liquid feed is assumed.
L = liquid
V = vapour
Using underwood’s second equation (at q≈1):
Where q is quality of saturated feed
∆Vfeed = Vmin - Ṽmin = F (1 – q) = 0
80 (1 – 1) = 0
∆Vfeed= ∑
𝛼𝑖×𝐹×𝑥 𝑖
𝛼𝑖−𝜑
𝑐
𝑖=1 = 0
Where 𝜑 is the root of equilibrium that lies between three volatilities
xiis mass fraction of pieces
F is amount of feed to the column
∆VfeedVolume of feed
∆Vfeed=
𝛼 𝐹×𝐹×𝑥 𝐹
𝛼 𝐹−𝜑
+
𝛼 𝑀×𝐹×𝑥 𝑀
𝛼 𝑀−𝜑
+
𝛼 𝑊×𝐹×𝑥 𝑊
𝛼 𝑊−𝜑
∆Vfeed =
0.30797×80.8776×0.3141
0.30797−𝜑
+
1×80.8776×0.0453
1−𝜑
+
0.5695×80.8776×0.6406
0.5695−𝜑
=0
∆Vfeed =
73.9873
1.87747−3𝜑
= 0
Solving for 𝜑
1.57797-3 𝜑 =0 = 𝜑= 0.6258
Using underwoodfirst equation to find minimum vapour (Vmin):
Vmin = ∑
𝛼𝑖𝐷×𝑥𝑇
𝛼𝑖−𝜑
𝑐
𝑖=1
Minimum vapour volume (vmin) =
𝛼 𝑀×𝐷×𝑥 𝑀
𝛼 𝑀−𝜑
+
𝛼 𝑊×𝐷×𝑥 𝑊
𝑥 𝑊−Ҩ
Vmin=
1×5.1348×0.77236
1−0.6258
+
0.0647×5.1348×0.1212
0.0647−0.6258
= 10.527kmol/hr

42
From the material balance around the condenser
Minimum liquid volume ( Lmin) = Vmin– D
D is the distillate
Lmin = 10.527 – 5.1348 = 5.392kmol/hr
Minimum refluxes
External refluxes ratio:
𝐿 𝑚𝑖𝑛
𝐷
=
5.392
5.1348
= 1.05
Internal refluxes ratio:
𝐿 𝑚𝑖𝑛
𝑉 𝑚𝑖𝑛
=
5.392
10.527
= 0.512
5.1.4 Actual reflux ratios
A conventional multiplier is used to allocate the actual refluxes. According to Wankat(1987),
this multiplier is ranging 1.05 to 1.5. The chosen factor is 1.145 for an economic conservative
design.
The actual external ratio: (
𝐿 𝑚𝑖𝑛
𝐷
) =
5.392
5.1348
×1.145 = 1.20
The actual internal ratio: (
𝐿 𝑚𝑖𝑛
V
) =
5.392
10.527
×1.145 = 0.586
5.1.5 Tray specifications
1. Minimum Number of Stages
An indication of the minimum allowable number of stages is determined using Fenske’s
rigorous solution (1932).
N min = in[
(
𝑦 𝑀
𝑦 𝑤
)𝑇𝑜𝑝/(
𝑥 𝑀
𝑥 𝑤
)𝑏𝑜𝑡
𝑖𝑛(
1
𝛼𝑤
)
= 𝑖𝑛
(
0.6992
0.2018
)/(
0.0014
0.5493
)
𝑖𝑛(
1
0.27
)
= 6.948 stages
5.1.6 Total Number of Stages (theoretical)
The calculation of the theoretical number of stages of the distillation column is presented here
through two distinct approaches: Gilliland correlation (1940) and Molokanov correlation
(1972) as follows:

43
First Approach: GILLILAND CORRELATION
This correlation gives the theoretical number of stages with an accuracy of ±7% in the
following sequence
(
𝐿
𝐷
)min = 1.20
𝐿
𝐷
= 1.20×1.145 = 1.374
Using the Gilliland chart
Abscissa =
(
𝐿
𝐷
)− (
𝐿
𝐷
) 𝑚𝑖𝑛
(
𝐿
𝐷
+ 1)
=
1.374−1.20
1.20+1
= 0.079 the corresponding ordinate = 0.54
𝑁−𝑁 𝑚𝑖𝑛
𝑁
=
𝑁−6.9483
𝑁
=0.54 = 15.105
Solving for N (theoretical) = 15 stages
Second Approach: MOLOKANOV CORRELATION
This method is a refined modern version of the Gilliland correlation that is more accurate and
compatible with our system. It is dependent upon two parameters X and Y as follows:
X =
𝑅−𝑅 𝑚𝑖𝑛
𝑅+1
=
1.374−1.20
1.374+1.00
= 0.0733
Y = 1- exp[(
1+54.4𝑥
11+1117.2𝑥
)(
𝑥−1
𝑥0.5)]
Y = 1- exp [(
1+54.4(0.0733)
11+1117.2(0.0733)
)(
0.0733−1
0.07330.5 )] = 0.168
N =
𝑁𝑚𝑖𝑛+𝑌
1−𝑌
=
6.9483+0.168
1−0.168
= 8.55 stages
This correlation is to be used since it provides more accuracy
5.1.7 Tray efficiencies
The efficiency of the trays is to be determined using O’Connell Correlation which is estimated
the efficiency as a function of the product of the feed liquid viscosity and the volatility of the
key components in the following manner:

44
Top side efficiency
Viscosity (µ) = 0.1329
Relative volatility (αF, top) = 0.2342
Eo = 0.52782 – 0.27511×log(αF×µ) + 0.044923×[log(αµ)]2
Eo = 0.52782 – 0.27511×log (0.03113) + 0.044923[ log (0.03113)]2
Eo = 75%
Bottom side efficiency
Viscosity (μ, simulated) = 0.1329
Relative volatility(αF,bot)=0.0304
Eo = 0.52782 – 0.27511×log(αµ) + 0.044923×[log(α𝐹 ×µ)]2
= 0.52782 – 0.27511×log (0.00402) + 0.044923[log(0.00402)]2
Eo=84%
Efficiency (E0) =
𝑛𝑢𝑚𝑏𝑒𝑟𝑜𝑓𝑡ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙𝑠𝑡𝑎𝑔𝑒𝑠
𝑛𝑢𝑚𝑏𝑒𝑟𝑜𝑓𝑎𝑐𝑡𝑢𝑎𝑙𝑠𝑡𝑎𝑔𝑒𝑠
For the rectified section:
The actual number of stages =
8.55
0.75
= 11stages
For the stripping section
The actual number of stages =
8.55
0.84
= 10stages
The average of the rectified section + the stripping section = the actual stages of the column
The column actual stages =
10.962+9.488
2
= 10.58 ~10.6 stages
Number of stages at the rectified section (ND)
Number of stages at the striping section (NB
𝑁 𝐷
𝑁 𝐵
= 0.95
N = NB + ND…………… (1)

45
10.6 = NB + ND…………. (2)
By solving eq1 & 2
ND = 5.1 stages above the feed
NB = 5.44 stages below the feed
5.1.8 Column height
In our design, 0.4572m was chosen for spacing to provide a reasonable space to ease the
accessibility for manuals between the plates for maintenance. According to
Turton’sDistillation Column Design Heuristics(1955), a safety factor of 10% is to be added to
the number of stages. The column height is determined as follows:
1 stage of partial condenser is to be added to the total height
Total Actual number of stages = 11 + 1
Safety Factor = 12× (0.1) = 1.2 stages
The actual height of column = number of stages X tray spacing
= 12 X 0.4572 = 5.486m
Total construction stages = 1.2 +12 + 1 = 114.2 ~ 14 stages
The column height = tray spaces ×(number of stages + safety factor)
0.4572×(14+1.2) =6.95m ~ 7m
Column height is 7m

46
5.1.9 The volume of the distillation column
For the rectification section
Total mole of the component = 5.134kmol/hr
Average molecular weight of the component = 29.0058g/mol
Density of component = 49.179kg/m3
Mole =
𝑚𝑎𝑠𝑠
𝑚𝑜𝑙𝑎𝑟 𝑤𝑒𝑖𝑔ℎ𝑡
Mass flow rate = moles X average molecular weight of the component
Mass flow rate = 5.1348
𝑘𝑚𝑜𝑙
ℎ𝑟
X 29.0058
𝑔
𝑚𝑜𝑙
X
1𝑘𝑔
1000𝑔
X
1000𝑚𝑜𝑙
1 𝑘𝑚𝑜𝑙
= 148.939kg/hr
Density (ρ) =
𝑚𝑎𝑠𝑠 𝑓𝑙𝑜𝑤 𝑟𝑎𝑡𝑒
𝑣𝑜𝑙𝑢𝑚𝑒 𝑓𝑙𝑜𝑤 𝑟𝑎𝑡𝑒
Volume flow rate =
𝑚𝑎𝑠𝑠
𝑑𝑒𝑛𝑠𝑖𝑡𝑦
Volume flow rate =
148.939𝑘𝑔/ℎ𝑟
49.179𝑘𝑔/𝑚3 = 3.029m3
/hr
Assuming a hold up time of 0.5hr
Volume = volume flow rate X hold up time
V = 3.028m3
/hr X 0.5hr
V = 1.5145m3

47
Diameter at the rectifying section (DT)
Volume (V) =
𝜋𝐷2 𝐻
4
Therefore DT = √
4𝑉
𝜋𝐻
DT = √
4×1.5145
𝜋×7
= 0.53m
For the stripping section
Total mole of the component = 75.7429kmol/hr
Average molecular weight of the component = 21.9638g/mol
Density of component = 54.6375kg/m3
Mole =
𝑚𝑎𝑠𝑠
𝑚𝑜𝑙𝑎𝑟 𝑤𝑒𝑖𝑔ℎ𝑡
Mass flow rate = moles X average molecular weight of the component
Mass flow rate = 75.7429
𝑘𝑚𝑜𝑙
ℎ𝑟
X 21.9638
𝑔
𝑚𝑜𝑙
X
1𝑘𝑔
1000𝑔
X
1000𝑚𝑜𝑙
1 𝑘𝑚𝑜𝑙
= 1663.60kg/hr
Density (ρ) =
𝑣𝑜𝑙𝑢𝑚𝑒 𝑓𝑙𝑜𝑤 𝑟𝑎𝑡𝑒
Volume flow rate =
𝑑𝑒𝑛𝑠𝑖𝑡𝑦
Volume flow rate =
1663.60𝑘𝑔/ℎ𝑟
54.6375𝑘𝑔/𝑚3 = 30.4479m3
/hr
Assuming a hold up time of 0.5hr
Volume = volume flow rate X hold up time

48
V = 30.4479m3
/hr X 0.5hr
V = 15.234m3
Volume (V) =
𝜋𝐷2 𝐻
4
Therefore DB = √
4𝑉
𝜋𝐻
DB = √
4×15.234
𝜋×7
= 1.66m
Total volume of the column
= Volume of the rectification section + volume of the stripping section
= 15.234 + 1.5145
= 16.7485m3
5.1.10 Determination of diameter of the distillation column (D)
Volume (V) =
𝜋𝐷2 𝐻
4
Therefore D = √
4𝑉
𝜋𝐻
D = √
4 𝑋 16.7485𝑚3
𝜋 𝑋 7𝑚
= 1.7m
Active area top (AT) =
𝜋𝐷2
4
AaT =
𝜋×(0.52)2
4
= 0.196m2
Active area of bottom (AaB) = 1.978m2

49
Assuming hole diameter = 5×10-3
m
Area of hole (Ah) =
𝜋×(5×10−3)2
4
= 3.93×10-3
m2
Assuming hole covers 10% of active area per tray (R.K.SINNIOT):
Area covered by hole
Area of hole at the top (AhT) = AaT×0.1 = 0.019m2
Area of hole at the top (AhB) = (AaB)×0.1 = 0.100m2
Number of hole per tray:
𝑎𝑟𝑒𝑎 𝑐𝑜𝑣𝑒𝑟𝑒𝑑 𝑏𝑦 ℎ𝑜𝑙𝑒
𝑎𝑟𝑒𝑎 𝑜𝑓 ℎ𝑜𝑙𝑒
Rectified section (top) =
0.019
0.00393
= 5holes
Stripping section (bottom) =
0.1978
0.00393
= 25holes
Total number of holes above the feed stage = 5×5.1 = 26holes
Total number of holes below the feed stage = 5.44×25 = 136holes
Total number of hole = 26 + 136 = 162holes
Tray layout
This section is a detailed representation of the design layout calculations for the sieve plates in
the top section. The decided type of tray is a single pass sieve plate counter-flow tray with a
straight segmental vertical downcomer and a weir. The use of single pass tray is due to the
relatively small diameter of the column and its liquid load. Also to avoid the propagation of
mal-distribution of the liquid, this could lead to a major decrease in the efficiency of the tray
and the capacity of the column if a multiple-pass tray was used. The decision to use a segmental
straight down-comer is due to its simple geometry, low cost. Also because it utilizes most of

50
the column area for the large downflow in our system and the ease at which it’s operated and
maintained.
The sequence of the tray layout design is applied as follows:
Tray Dimensions
Diameter of column 1.7m
Entrainment at a flooding of 75%
Fig. 3 entrainment chart (Colson and Richardson, volume 6)
Flooding Point = 0.03993 →read fromchart: fractional entrainment (ψ) = 0.07
External mass balance
V =
𝐷
1−
𝐿
𝑉
=
5.1348
1−0586
= 12.4
L = internal ratio ×Volume = 0.586×12.40 = 7.2681 kgmol/hr

51
Entrained liquid (Le)
(Le) =
𝛹×𝐿
1−𝛹
=
0.07×7.2681
1−0.07
= 0.5471kgmol/hr
AMOUNT ENTRAINED ON TOP
L + Le = 7.2681+0.5471 = 7.815kgmol/hr
L is liquid
Le is entering liquid
5.1.11 Column cross-sectional area (A)
A=
𝜋×(𝐷𝑖𝑎𝑡𝑜𝑝)2
4
=
𝜋×(1.587)2
4
= 1.978m2
~ 2m2
downcomer area (Ad)
Ad = (1- ŋ) A= (1- 0.9) ×1.978m2
= 0.2m2
As for the fraction of cross-sectional area that is available for vapor flow η, Wankat(1987)
= η is 0.90
The
𝐿𝑤𝑒𝑖𝑟
𝐷𝑖𝑎
ratio is provided by Wankat (1987) as 0.726
Weir length Lw = Dia × 0.726 = 1.684m ×0.726 = 1.152m
Active area of the tray(Aa)
Aa = A(1-2(1-ŋ)
= 1.978m2
× (1-2(1- 0.90)
= 1.58m2
5.1.12 Total area of the holes
Assuming hole covers 10% of active area per tray (R.K.SINNIOT)
A hole = A active = 1.58× 0.1 = 0.158m2
Chosen tray is a std. 14 gauge tray with thickness (T tray) = 0.078 in, 0.0019812m with a
common hole diameter do= 3/16 inch, 0.0762/0.4064m for normal operation and clean service.
Pitch Std. spacing between the holes of 3.8do = 0.1725 inches. A 2.5 in,0.0635mspace between

52
the edge holes and the column wall is chosen, and a space of 4 in,0.016m between the edge
hole and the tray weir
Table 8: design items specification
Design item (Distillation unit) Specification
Material of construction Stainless steel
Type of flow Gas-liquid counter-flow
Tray type SS sieve trays
Tray spacing 0.4572m
Tray thickness 0.0019812m
Minimum number of stages 6.48
Number of tray 14 stages plus reboiler
Number of tray pass Single
column diameter 1.7m
Top tray efficiency 75%
Bottom tray efficiency 84%
Column height 8.5m
Volume of the column 16.75m3
Down-comer type Vertical straight segment
Down-comer area 0.20m2
Active area of tray 1.581m2
Weir length 1.152m
Cross sectional area of the column 1.978m2

53
5.2 CHEMICAL ENGINEERING DESIGN OF A PLUG FLOW REACTOR
5.2.0 Criterion selected:
In this section, designing a plug flow reactor for multi reaction and non- isothermal condition
was done. This reactor is supported with a heat exchange to remove the heat generated
from the exothermic reaction. in this designing section, mole balances were considered to be
in the form of the final mole which is the remaining at the end of the reaction period.
Since the reaction is parallel, taking in mind the reaction rates is too important by
combining all these rates for each material. Evaluating the concentration of each material
were done in which all the pressure and temperature effect was considered. Here, one
assumption was used which is the ideality of the gas introduced to the reactor. By the
end of this step, combination all previous steps can be done to reduce the number of
equations. Using the Ergun equation, pressure drop across the reactor was evaluated. In energy
balance, to increase the accuracy of the results, we used the integrated heat capacity instead of
assuming it constant.
The design parameters to be calculated;
Residence time
Length of reactor
Diameter of reactor
Pressure drop in the shell
Heat transfer area
Number of tubes
Bundle diameter

54
Volume of reactor
Area of reactor
Superficial velocity
Weight of silver catalyst
Volume of silver catalyst
Log Mean Temperature Difference
Mean Temperature Difference
Tube spacing
Tube pitch
Baffle spacing
5.2.1 Reactor Design:
Reactor conversion = 87.4%
Average temperature of reactor = 246.5o
C
Average pressure of reactor = 182.7kPa
Reaction rate constant for main and side reactions = K1 and K2
Density of silver catalyst used = 960kg/m3
Oxidation reaction of methanol is given as;
CH3OH + 1
2⁄ O2 → HCOH + H2O ………………………………… main reaction
HCOH + 1
2⁄ O2 → CO + H2O …………………………………….. side reaction

55
Table 9 Reactor feed input compositions
5.2.2 Calculating of the residence time in the reactor:
Basic performance equation for a Plug Flow Reactor is:
1
𝐹 𝐴0
( 𝑉) = ∫
𝑑 𝑋𝐴
−𝑟 𝐴
𝑋 𝐴𝐹
0
𝑜𝑟 𝝉 = 𝐶𝐴𝑂 ∫
𝑑 𝑋𝐴
−𝑟 𝐴
𝑋 𝐴𝐹
0
…………….. (1)
The reaction rate (-r), expression for methanol-formaldehyde system is:
−𝑟 =
𝐾1 𝑃 𝑚
1+𝐾2 𝑃 𝑚
………………… (2)
But, T = 246.50
C = 519.5K
Where Log10 K1 = 10.79 -
5640
𝑇
= 10.79 −
5640
519.5
= 0.0666
K1 = antilog (0.0666) = 1.1657
Also Log10 K2 = 11.43 -
3810
𝑇
= 11.43 −
3810
519.5
= 4.0960
K2 =antilog (4.0960) = 12474.5
From equation (2),
PmV = nRT
Where; Pm = pressure of methanol within reactor
V = volume of components entering
Compositions of feed into the reactor
Component Mole (kmol/hr) %
Methanol 376.3100 29.35
Water 3.7276 0.006
Oxygen 190.0936 14.83
Nitrogen 715.1140 55.79
Total 1281.5929 100

56
T = temperature within the reactor
R = gas rate constant
Pm = CART ……………………….. (3)
Methanol reaction rate, (−rm ) =
K1CA RT
1 + K2CART
=
RT K1CA
RT(1
RT⁄ + K2 CA)
………………… (4)
Putting –rm into equation (1):
τ = CAO ∫
(1
RT⁄ + K2CA) dXA
K1CA
XA
0
………………. (5)
Where; τ = residence time
CAO = concentration of methanol initially
XA = conversion rate
ԐA = extent of reaction
Now concentration of methanol at time (t) (CA):
CA =
CAO(1−XA)
(1+ εAXA)
…………………………. (6)
But from the main reaction equation:
CH3OH + 0.5O2 CH2O + H2O
Extent of reaction, (ԐA) =
⅀products stoich−⅀reactants stoich
⅀reactants stoich
=
2− 1.5
1.5
= 0.333

57
Also total molar flowrate components entering (Fao) is:
Fao = 1281.5929kmol/hr
And PVo = nRT, but n = Fao
Volumetric flowrate of components, (Vo) =
𝐹𝑎𝑜 𝑅𝑇
𝑃
=
1281.5929 x 8.314 x 519.5
182.7
=
30297.52m3 x 1hr
hr x 60min
= 504.96m3
/min
But mole flowrate of methanol, (Fm) =
376.3100kmol x 1hr
hr x 60min
= 6.27kmol/min
Thus concentration of methanol initially, (CAO) =
Fm
Vo
=
6.27𝑘𝑚𝑜𝑙 𝑥 1 𝑚𝑖𝑛
min 𝑥504.96𝑚3
= 0.0124kmol/m3
From equation (5), we have:
τ = CAO ∫
(1
RT⁄ + K2CA)dXA
K1CA
XAF
0
But
1
𝑅𝑇
=
1
8.314 𝑥 519.5
= 0.0002
Initial rate constant, K1 = 1.1657, final rate constant, K2 = 12474.5 conversion rate of methanol
to formaldehyde, XAF = 0.874
Thus: τ = CAO ∫
(0.0002+12474.5CA)dXA
1.1657CA
0.874
0

58
Since CA =
CAO(1− XA)
(1+0.333XA)
,
Equation (5) becomes:
τ = CAO ∫
(1+0.333 XA)
CAO (1− XA)
(0.0002+13967CA)dXA
1
0.874
0
= CAO ∫
(1+0.333 XA)
CAO (1− XA)
[0.0002((1+0.333XA)+12474.5CAo(1− XA)]dXA
(1+0.333XA)
0.874
0
Putting CAO = 0.0124 into the above relation:
τ = 0.8578 ∫
[0.0002((1+0.333XA)+12474.5 x 0.0124(1− XA)]dXA
(1− XA)
0.874
0
Solving integrally:
τ = ∫
[0.0002((1+0.333XA)dXA
(1− XA)
0.874
0
+ 154.7 ∫
(1− 𝑋 𝐴)2 𝑑 𝑋𝐴
(1− XA)
0.874
0
+ ∫ 𝑋𝐴
0.874
0
= 0.8578 (0.0002[∫
1 𝑑 𝑋𝐴
(1− XA)
0.874
0
+ 0.333∫
1𝑑 𝑋𝐴
(1− XA)
0.874
0
] + 154.7(0.874))
= 0.8578 (0.0002[In(1 − XA) ∫ +
0.874
0
0.333∫
(1−1+ 𝑋 𝐴)𝑑 𝑋𝐴
(1− XA)
0.874
0
] + 135.2)
= 0.8578(0.0002[-In(1–0.874)–In (1)] + 0.333[∫
1 𝑑 𝑋𝐴
(1− XA)
0.874
0
− 0.333∫
(1+ 𝑋 𝐴)𝑑 𝑋𝐴
(1− XA)
0.874
0
]
+ 135.2)
= 0.8578 (0.0002(2.07) + 0.333[-In (1 – XA) ∫ −
0.874
0
XA ∫ ]
0.874
0
+ 135.2)
= 0.8578 [0.0004 + 0.333(2.07 – 0.874) + 135.2]
= 0.8578 (0.0004 + 0.3983 + 135.2)
= 116.32hrs
Thus with the help of catalyst used: τ =
116.32kmol x kg.hr x m3
m3 x kmol x 950kg
=
0.1224ℎ𝑟 𝑥 60𝑚𝑖𝑛
1ℎ𝑟
= 7.35 minutes

59
Hence the residence time achieved is 7.35 minutes
Also τ =
CAO W
FAO
, making weight of catalyst, (W):
It implies that: (W) = τ
𝐹 𝐴𝑂
𝐶 𝐴𝑂
=
7.35 𝑥 6.27
0.0124
= 3716.5𝑘𝑔
And volume of catalyst used (VC) =
Weight of catalyst
density of catalyst
=
3716.5kg
950kg/m3
= 4m3
Catalyst Dimensions:
Diameter of catalyst from rules of thumbs ranges from 2mm - 5mm
Hence average diameter of silver catalyst = 3mm
Spherical form silver catalyst is to be used with a voidage (e) = 60% = 0.6
Density of silver catalyst used = 950kg/m3
5.2.3 Tube Dimensions:
Assumptions:
Based on standard tube dimensions commonly used provided by Tubular Heat Exchanger
Association (TEMA) with 14 British Wire Gauge specified for tubes;
Outer diameter of tube (d) = 1in = 0.0254m
Length of tube chosen (Lt) = 16ft = 4.87m
Inner diameter of tube = 0.834in = 0.0212m
Wall thickness of tube = 0.083in = 0.0021m
5.2.4 Volume of tube (Vt):
Volume of tube, (Vt) =
π x d2 x 𝐿 𝑡
4
=
3.142 x (0.0245)2 x 4.87
4
= 0.0025m3

60
Pitch of tube (Pt):
For square pitches, due to easy cleaning and spacing;
Pitch = 1.25 x OD
= 1.25 x 0.0254m = 0.0317m
Tube spacing (Ts):
Spacing of tubes, Ts = pitch – outer diameter of tube
= 0.0317 – 0.0254
= 0.0063m
5.2.5 Log Mean Temperature Difference (LMTD):
LMTD =
(𝑇ℎ𝑖−𝑇𝑐𝑜)− (𝑇ℎ𝑜−𝑇 𝑐𝑖)
ln(
𝑇ℎ𝑖−𝑇 𝑐𝑜
𝑇ℎ𝑜−𝑇 𝑐𝑖
)
Assumptions; inlet and outlet temperature of cold stream is 145o
C
Where; Thi = temperature of hot fluid inlet = 150o
C
Tho = temperature of hot fluid outlet = 343o
C
Tci = temperature of cold stream inlet = 145o
C
Tco = temperature of cold stream outlet = 145o
C
Also in our design system, saturated water is used to cool the reactor. Thus water enters the
reactor and leave at same temperature of 145o
C but in steam phase.
Therefore, LMTD =
( 𝑇ℎ𝑖−𝑇𝑐𝑜)− (𝑇ℎ𝑜−𝑇 𝑐𝑖)
ln(
𝑇ℎ𝑖−𝑇 𝑐𝑜
𝑇ℎ𝑜−𝑇 𝑐𝑖
)
=
(343−145)− (150−145)
ln(
343− 145
150−145
)
= 52.46o
C

61
Assumption;
Temperature correction factor (ft) = 1.0;
Hence mean temperature difference (𝜟Tm) = ft  LMTD = 1.0  52.46 = 52.46C = 325.46K
5.2.6 Heat transfer area (A)
From rule of thumbs heat transfer coefficient (U) for liquid (methanol) in tube and steam
in shell; (1000 – 3500)W/m2
.K, chosen UAV = 2250W/m2
.K
Since methanol has the maximum composition among the components the liquid was
assumed to be methanol.
Heat input, (Q) = ṁCpT
Heat capacity (Cp) at 150o
C, [Cpmx = CP (CH3OH + H2O + N2 + O2 + CH2O)]
= 1920 + 4320 + 1039 + 910 + 847
= 9036J/kg.K
Mass flow rate of liquid, (ṁmix) = ṁ (CH3OH + H2O + N2 + O2 + CH2O)
= Molar mass of mixture x Mole flow rate
= 140kg/kmol x 1281.5929kmol/hr
= 179423kg/hr
Therefore heat, Q =
179423kg x 9036J x 193K
hr x kg.K
= 86917883.89W
But heat, Q = UA𝜟Tm

62
A =
𝑄
UΔ𝑇 𝑚
=
86917883.89𝑊
2250𝑊/𝑚2.𝐾 × 325.46𝐾
= 118.7m2
Hence heat transfer area = 118.7m2
5.2.7 Number of tubes (Nt):
Nt =
𝐴
 𝑥 (𝑑𝑜𝐿)
Where; A = heat transfer area
do = outer diameter of tube
Lt = length of tube
Nt = number of tubes
Nt =
118.7𝑚2
𝜋 ×0.0254𝑚 ×4.87𝑚
= 305tubes
Assumption:
Due to large number of tubes, three reactors will be used instead of one.
Therefore, number of tubes in one reactor =
305
3
= 102 tubes
Reactor dimensions:
Assumptions:
Let, length of reactor (LR) = 1.5 x length of tube
= 1.5 x 4.87
= 7.3m

63
Bundle diameter (Bd):
(Bd) = number of tubes x tube outer diameter
= 102 x 0.0254m
= 2.6m
Using a bundle diameter clearance of 20cm for effective heat transfer and cleaning of tubes,
we have;
Shell diameter, (Ds) = bundle diameter + bundle diameter clearance
= bundle diameter + 0.2m
= 2.6 – 0.2
= 2.8m
5.2.8 Area of reactor (Ar):
Reactor area, (Ar) =
π xD2
4
Where; D = shell diameter
(Ar) =
𝜋 𝑥 (2.8)2
4
= 6.16m2
5.2.9 Volume of reactor (Vr):
Reactor volume, (Vr) =
π xD2x 𝐿 𝑅
4
Where; D = shell diameter
LR = length of reactor
(Vr) =
𝜋 𝑥 (2.8)2 𝑥 7.3
4
= 44.95m3
5.2.1.0 Baffle spacing (Bs):
Spacing of baffles, (Bs) =
1
5
x diameter of reactor shell
= 0.2 x 2.8 = 0.56m

64
5.2.1.1 Pressure drop in the reactor:
Components molar mass, Mc = MW(O2) + MW(CH3OH) + MW(H2O) + MW(N2) + MW(CH2O)
= (32) + (32) + (18) + (28) + (30)
= 140kg/kmol
Also, PV = nRT =
m
M
RT
PM =
𝑚
𝑉
𝑅𝑇 , but
𝑚
𝑉
= ρ (density)
= RT
Density of components entering, (ρ) =
PM
RT
=
182.6kPa x 140kg x kmol.K
kmol x 8.314 x 519.5K
= 5.92kg/m3
5.2.1.2 Superficial velocity of components (uc):
Thus, (uc) =
1281.5929kmol x 140kg
hr x kmol x 3600s
= 49.84kg/s
=
49.84kg x m3
s x 5.92kg x 6.6m2
= 1.4m/s
Pressure drop (𝜟P) calculation:
Given that: Superficial velocity of fluid (uc) = 1.4m/s
Viscosity of fluid, (𝝁) = 0.54cP =0.00054N.s/m2
Length of tube, (Lt) = 4.87m
Outer diameter of tube, (d) = 0.0254m
Voidage of spherical silver catalyst, (e) = 0.6 (Scott and Kilgour, 1969).
And using the Ergun equation:
𝛥𝑃
𝐿 𝑡
=
150(1−𝑒)2(𝑈 𝐶 𝑥 𝜇)
𝑒3 𝑑2
+
1.75(1−𝑒)((𝑈 𝐶)2 𝑥 ₱)
𝑒3 𝑑
……… (8)
Putting values of the parameters in the above equation:
𝛥𝑃 = (130.2 + 1480) x 4.87
= 7843.77Pa = 7.84kPa

65
Reactor vessel dimensions:
Diameter of reactor, (DR) = 2.8m
Length of reactor, (LR) = 7.3m
Design pressure = 182.7kPa = 1.83bar = 0.183N/mm2
Taking 15% above operating pressure = (1.83) x 1.15
= 2.10bar
= 0.21N/mm2
Hence design pressure (P1) = 0.21 + 0.183
= 0.4N/mm2
For carbon steel, allowable stress (f) = 70N/mm2
Thus, cylindrical section allowance =
P1 𝑥 𝐷 𝑅
2f−P1
=
0.4 x 2.8 x 103
(2 x 70)−0.4
= 8mm
Adding corrosion allowance of 30% for plant life span:
Thickness of the vessel will be = 8 + (0.3 x 10)
= 11mm

66
5.3 CHEMICAL DESIGN OF A DISTILLATION UNIT VAPORIZER
CH3OH= 0.6689
H2O= 0.3311
380.1872kmol/hr 376.3853kmol/hr
CH3OH= 0.99 CH3OH= 0.9998
H2O= 0.01 H2O= 0.0002
Table 10 Compositions of liquid entering the vaporizer
5.3.1 Shell-side flowrate
Thi = 185C and Tho = 160C
Where; Thi = inlet temperature of hot fluid (steam)
Tho = outlet temperature of hot fluid (steam)
But, Tci = 46.87C and Tco = 150C
Where; Tci = inlet temperature of cold feed (methanol & water)
Tco = outlet temperature of cold feed (methanol & water)
For any heat exchanger design, assume three known temperature and find the fourth one or
four temperature values and find one of the shell or tube side flow rate using the heat duty
equation ),()( ,inhouthhhinoucc TTcpmTcTccpmq  where subscripts c and h refer to
Component Mole flowrate Mole fraction
CH3OH 376.3100 0.9997
H2O 0.0753 0.0002
VAPORIZER

67
cold and hot streams. Then obtain the heat duty, q. (Coulson and Richardson, Chemical
Engineering, Volume 6)
The following key parameters are determined
I. Shell side flowrate
II. Long Mean Temperature Difference
III. Heat exchanger surface area
IV. Number of tubes
V. Bundle diameter
VI. Fanning factor at shell and tube sides
VII. Overall heat transfer at shell and tube sides
VIII. Prandtle number at shell and tube sides
IX. Reynolds’s number at shell and tube sides
X. Area of maximum velocity (m) in annulus using Colburn correlation:
XI. Shell diameter
XII. Pressure drop at the shell and tube side
Heat supplied, (Q) = ṁCpT (theoretical heat of fluid)
Heat capacity of liquid mixture, Cpmx = (2510 + 4190) = 6700J/kg. K
And mass flow rate of liquid mixture, (ṁmx) = ṁ (H2O) + ṁ (CH30H)
= 0.0753
𝑘𝑚𝑜𝑙
ℎ𝑟
 18
𝑘𝑔
𝑘𝑚𝑜𝑙
+ 376.3100
𝑘𝑚𝑜𝑙
ℎ𝑟
 32
𝑘𝑔
𝑘𝑚𝑜𝑙
= 12043.2754kg/hr
Q = 12043.2754
𝑘𝑔
ℎ𝑟
 6700
𝐽
𝑘𝑔
 376.13K = 30349909.08
𝑘𝐽
ℎ𝑟

1ℎ𝑟
3600𝑠
= 8430530.3W
For steam side (shell-side): T = 185 – 160 = 25C = = 298K

68
Heat supplied, Q = ṁsCpT,
where; Q = heat supplied,
ṁ = mass flowrate of steam
Cp = specific heat capacity
T = change in temperature of fluid
30349909.08
𝑘𝐽
ℎ𝑟
= ṁs  4190
𝐽
𝑘𝑔.𝐾
 298K
ṁs = 30349909080
𝐽
ℎ𝑟

1𝑘𝑔
4190𝐽  298
=
30349909080𝑘𝑔
1248620ℎ𝑟
= 24306.762kg/hr
5.3.2 Log Mean Temperature difference (LMTD):
(LMTD) =
(𝑇ℎ𝑖−𝑇𝑐 𝑜)− (𝑇ℎ 𝑜−𝑇𝑐𝑖)
ln(
𝑇ℎ 𝑖−𝑇𝑐 𝑜
𝑇ℎ 𝑜−𝑇𝑐 𝑖
)
Where; Thi = inlet temperature of hot fluid (steam)
Tho = outlet temperature of hot fluid (steam)
Tci = inlet temperature of cold feed (methanol & water)
Tco = outlet temperature of cold feed (methanol & water)
(LMTD) =
(185−150)− (160−46.87)
ln(
185−150
160−46.87
)
=
78.13
1.1732
= 66.6o
C
For steam, it is assumed that temperature correction factor is ft = 1.0;
Hence mean temperature difference (𝜟Tm) = ft  LMTD = 1.0  66.6 = 66.6C = 339.6K

69
5.3.3 Heat transfer area (A):
From rule of thumbs overall heat transfer coefficient (U) for tubular evaporation ranges from
(600 – 1700)W/m2
.K, hence average heat transfer coefficient used = 1150W/m2
.K
Q = UA Δ𝑇 𝑚 but, A =
𝑄
UΔ𝑇 𝑚
Where; U = overall heat transfer coefficient
A = heat transfer area
Δ𝑇 𝑚 = Mean Temperature Difference
A =
8430530.3𝑊
1150𝑊/𝑚2.𝐾 × 339.6𝐾
Therefore heat transfer area is 21.52m2
5.3.4 Number of tubes (Nt):
Nt =
𝐴
(𝑑𝑜𝐿)
Where; A = heat transfer area
do = outer diameter of tube
Lt = length of tube
Hence, (Nt) =
21.58𝑚2
𝜋 ×0.0234𝑚 ×1.7𝑚
= 173tubes
Tube pitch (Pt):
Pt = 1.25 x outer diameter of tube
= 1.25  0.0234m = 0.0292m
Bundle diameter (Bd) = do(
𝑁𝑡
𝐾𝑖
)
1
𝑛𝑖

70
where; Nt = number of tubes
But for square pitch with two passes, Ki and ni are given by 0.156 and 2.291 respectively.
Bd = 0.0234 x (
173
0.156
)1/2.291
= 0.4992m = 0.50m
Shell side calculations
For fixed tube and U-tube heat exchanger with bundle diameter 0.50m from graphs of Bundle
Diameter clearance against Bundle Diameter is, BDC = 13mm.
Shell Diameter = Bundle Diameter + Bundle Diameter Clearance
= 0.50m + 0.013 = 0.513m
Baffle spacing =
1
5
 Ds = 0.2  0.513m = 0.10m;
Where; Ds = shell diameter
5.3.5 Cross flow-area (Ac):
Ac =
(𝑃𝑡−𝑑𝑜)𝐷𝑠𝐵𝑠
𝑝𝑡
Where; Pt = tube pitch
Bs = baffle spacing
Ds = shell diameter
Hence, (Ac) =
(0.0292−0.0234) × 0.513 × 0.10
0.0292
= 0.0102m2
;
Shell-side mass velocity (Gs):
(Gs) =
ṁ 𝑠
𝐴𝑠
Where; Gs = shell-side mass velocity

71
ṁs = mass flowrate of steam
As = cross flow area
Hence, (Gs) =
24306.762𝑘𝑔/ℎ𝑟
0.0102𝑚2
= 2385418.601
𝑘𝑔
𝑚2ℎ𝑟

1ℎ𝑟
3600𝑠
= 662.62kg/m2
.s
Shell equivalent diameter (De) for a square pitch arrangement:
De =
1.27
𝑑𝑜
(Pt2
– 0.785do2
)
Where; De = equivalent diameter
Pt = tube pitch
do = outer diameter of shell
Hence, (De) =
1.27
0.0234
[(0.0292)2
– 0.785(0.0234)2
] = 0.023m;
5.3.6 Shell-side Reynold’s number (Re):
Re =
𝐺𝑠 𝑥 𝐷𝑒
µ
=
662.62𝑘𝑔/𝑚2.𝑠 × 0.023𝑚
0.0002968𝑘𝑔/𝑚.𝑠
= 53138.98
Where; Gs = shell-side mass velocity
De = equivalent diameter
µ = viscosity of fluid (water)
Prandtle number at shell-side (Pr):
Pr =
µ 𝑥 𝐶𝑝
𝐾 𝑤
=
0.0002868𝑘𝑔/𝑚2.𝑠 ×4190𝐽/𝑘𝑔°.𝐶
0.58𝑊/𝑚2.°𝐶
= 2.0179
Where; µ = viscosity of fluid (steam)
Cp = heat capacity of steam
K = heat conductivity of steam (water)

72
5.3.7 Area of maximum velocity (m) in annulus using Colburn correlation:
m =
𝐷22− 𝐷12
4 ln(
𝐷2
𝐷1
)
=
[0.02342− (0.0122)2]
4 ln(
0.0234
0.0122
)
= 0.00015m2
Where: m = maximum velocity outside the tubes
D2 = outer tube diameter
D1 = inner tube diameter
Fanning factor at shell side (f2) = 0.24(1.82logNRe – 1.64)-2
= 0.25[1.82log (53138.98) – 1.64]-2
= 0.0052
But f2 = 2 x JH2
JH2 = f2/2 = 0.0052/2 = 0.0026
Where; JH2 = friction factor outside the tube
f2 = fanning factor at shell side
Fanning factor at tube side, f1 =
𝐹2 𝐷2( 𝑚−𝐷1
2
)
𝐷1(𝐷2
2−  𝑚)
Where; f1 = fanning factor at tube side
f2 = fanning factor at shell side
D2 = outer tube diameter
D1 = inner tube diameter
f1 =
0.0026 ×0.0234[0.00015−(0.0122)2]
0.0122[(0.0234)2− 0.0015]
= 0.00014

73
But JH1 = f1/2 = 0.000014/2 = 0.000007
where; f1 = fanning factor at tube side
Jhi = friction factor inside the tube
5.3.8 Shell-side heat transfer coefficient (hs)
Nusselt Number, Nu = hsDe/kf = Jhi x Re x Rr1/3
x (
µ
µ𝑤
)1/4
But T across methanol f1/m= 70 of average total temperature = 0.71(150 – 46.87) = 72.2C
T across steam f1/m = 10 of average total temperature = 0.1(150 – 16.87) = 10.3C
Therefore wall temperature, Ts = 46.87 + 72.2 = 119C
But viscosity of CH3OH at 119C = 0.179  10-3
kg/m.s
µ/µw = 0.237/0.179 = 1.32
Nu = hsDe/kf = Jhi x Re x Rr1/3
x (
µ
µ𝑤
)1/4
where; Kf = conductivity factor
µ/µw = relative viscosity of methanol to water
hs = shell side heat transfer coefficient
Jhi = friction factor inside the tube
Hence, (hs) =
0.026 × 53138.98 × (2.0119)0.33× (1.32)1/4×0.58
0.023
= 47449.32W/m2
.C

74
5.3.9 Pressure drop in the shell (Ps):
Ps = 8 x JH2 x (
𝐷𝑠
𝐷𝑒
) x (
𝐿
𝐵 𝑆
) 𝑥
ℎ𝑠2
2
𝑥 (
𝑈
𝑈𝑤
)–0.14
= 105.5kg/m.s2
Where; Ds = shell diameter
L = length of tube
Bs = baffle spacing
µ/µW = relative viscosity of methanol to water
5.3.1 Number of tubes per pass (Ntpp):
Ntpp = Nt/Number of passes = 173/2 = 86tubes per pass
Tube-side mass velocity (Gmt) =
𝑇𝑢𝑏𝑒 𝑠𝑖𝑑𝑒 𝑓𝑙𝑜𝑤𝑟𝑎𝑡𝑒 (
𝑘𝑔
𝑠
)
𝑁𝑡𝑝𝑝 ×
𝜋𝑑𝑖2
4
Where; ṅtotal = total moles
Mwt = total molecular weight
Gmt = tube-side mass velocity
Ntpp = number of tubes per pass
di = inner diameter of tube
But tube side mass flowrate = ṅtotal × Mwt
= 380.1872 kmol/hr × (18+32) kg/mol
= 19009.36kg/hr × 1hr/3600s = 5.2808kg/s

75
Gmt =
5.2808𝑘𝑔/𝑠
86 ×
𝜋 × 0.01222
4
= 525.24kg/m2
.s
Tube-side velocity (Vt) =
𝐺 𝑚𝑡
𝜌𝑖
But 𝜌i = ym × 𝜌m + yw × 𝜌w
Where 𝜌i = total density of fluid
m = methanol
w = water
y = mole fractions of components
𝜌w = density of water
𝜌m = density of methanol
Vt = velocity at tube side
At average temperature of 98˚C: 𝜌m = 721.36kg/m3
𝜌w = 960.2kg/m3
And mole fractions of each component are;
ym = 0.99
yw = 0.01
𝜌1= 0.99(721.36) + 0.01(960.2) = 723.75kg/m3
Hence, (Vt ) =
525.24𝑘𝑔/𝑚2.𝑠
723.75𝑘𝑔/𝑚3
= 0.7257m/s;
Prandtle number (tube-side), Pr =
µ x Cp
𝐾 𝑚
Since the composition of methanol is high, we chose it as in the tube fluid at an average temperature
of 98˚C. Thus;

76
µm (viscosity of methanol) = µ 𝐶𝐻3 𝑂𝐻 = µ = 0.237×10-3
kg/m.s
Km (conductivity of methanol) = 𝐾𝐶𝐻3 𝑂𝐻 = 0.1987W/m.K
Cpm (heat capacity of methanol) = 𝐶𝑝 𝐶𝐻3 𝑂𝐻 = 2510J/kg.K
Pr =
0.237×10−3 𝑘𝑔/𝑚𝑠×2510𝐽/𝑘𝑔.𝐾
0.1987𝑊/𝑚..𝐾
= 2.994
Also, Reynolds Number, Re =
𝜌𝑖 x di x Vt
µi
Where; Vt = tube side velocity
ρi = density of
Re (Reynold’s number) =
723.75𝑘𝑔/𝑚 𝑒×0.0122𝑚×0.7257𝑚/𝑠
0.0237×10−3 𝑘𝑔/𝑚𝑠
= 27036.92
5.3.2 Tube side film coefficient for sensible heat transfer in turbulent flow (hi):
hi = 0.23 𝑥 (
𝐾
𝐷1
)(Re)0.8
(Pr)0.33
(
µ
µ 𝑤
)
0.14
where; K = conductivity of methanol
D1 = inner diameter of tube
Re = Reynolds number
Pr = Prandtle number
𝝁 = viscosity of methanol used
𝝁w = viscosity of water
hi = 0.23 𝑥 (
0.1987
0.0122
) 𝑥 (27036.92)0.8
x (2.994)0.33
𝑥 (
µ
µ 𝑤
)
0.14
= 19641.52W/m2
.K
5.3.3 Overall heat transfer coefficient for inner tube (Ui);

77
Ui =
1
1
hs
+
1
hdi
+ di
ln(
do
di
)
2K
+
di
dohs
Where; do = outer diameter of tube
K = conductivity of material of construction used
hdi = fouling factor inside tube
U1 =
1
1
19641.52
+
1
5000
+
0.0122 𝑙𝑛(0.234/0.0122)
2(45)
+
0.0122
0.0034(47449.32)
= 1887W/m2
.K
5.3.4 Overall heat transfer coefficient for outer tube (Uo);
U0 =
1
1
ℎ𝑠
+
1
ℎ𝑑𝑜
+ 𝑑𝑜
ln(
𝑑𝑜
𝑑𝑖
)
2𝐾𝑚
+
𝑑𝑜
𝑑𝑖ℎ𝑠
K = conductivity of material of construction used
hdo = fouling factor outside tube
=
1
1
19641.52
+
1
5000
+
0.0234 𝑙𝑛(0.234/0.0122)
2(45)
+
0.0234
0.0034(47449.32)
=
1
0.0001+0.0002+0.000169+0.000404
= 1016W/m2
.K

78
5.4 CHEMICAL ENGINEERING DESIGN ON ABSORBER
THE ABSORBER
This section is concerned with the design of the absorption column. The absorber’s functions
are to absorb any formaldehyde vapour from the reactor product stream and remove un-reacted
gases. The basis for this equipment sizing is the previously obtained process data from the
project, which proved to be reliable and accurate.
O = 50.1209kmol/hr P = 83.0556kmol/hr
H2O = 1.0 CH3OH = 0.0004
CH2O = 0.0031
H2O = 0.3006
O2 = 0.0241
N2 = 0.6717
N = 113.802kmol/hr Q = 80.8776kmol/hr
CH3OH = 0.0325 CH3OH = 0.0453
CH2O = 0.2255 CH2O = 0.3141
H2O = 0.2342 H2O = 0.6406
O2 = 0.0176
N2 = 0.671
ABSORBER

79
Plant capacity = 100 tons/yr = 11.4 kg/hr
yformalin = Mole fraction of formalin entering the absorber
ymethanol = Mole fraction of methanol in the off gas
ywater= Mole fraction of water added into the absorber (top of column)
y2 =
𝑆𝑜𝑙𝑢𝑡𝑒 𝑚𝑜𝑙𝑒 𝑓𝑟𝑎𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑓𝑜𝑟𝑚𝑎𝑙𝑖𝑛
𝑆𝑜𝑙𝑢𝑡𝑒 𝑓𝑟𝑒𝑒 𝑚𝑜𝑙𝑒 𝑓𝑟𝑎𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑓𝑜𝑟𝑚𝑎𝑙𝑖𝑛
mx2
= mole fraction solute in equilibrium with incoming solvent (Zero for pure solvent)
Ae= Absorption factor = Molar liquid to gas ratio =
𝐿 𝑀
𝑀𝐺 𝑀
Mw (formalin) = Molecular weight of formalin
Mw (water) = Molecular weight of water
Mw (Methanol) = Molecular weight of methanol
Mw (Oxygen) = Molecular weight of oxygen
Mw (Nitrogen) = Molecular weight of nitrogen
Mwt= Sum total of all molecular weight of individual components
R = Gas constant = 8.314 kpa.m3
/ mol.k
T = Temperature at inlet stream
Uvd= Actual velocity
nA = Net plate area

80
P = Atmospheric pressure
Eoc= overall column efficiency, fractional
Nt = Theoretical number of plates
Na = Actual number of plates
Kv= vapor-liquid equilibrium coefficient
ht= total pressure drop, mm liquid
Lw = length of weir, mm
𝜌L= Density of water, kg/m3
Aa= Active area
Ad = Area occupied by downcomer
Ah = Hole area
At = Total area
𝜌G =Density of mixture entering the absorber from the reactor
T = Temperature of mixture entering absorber, K
Cv = Coefficient of discharge
how= height of crest over weir equivalent clear liquid, mm
Q= Volumetric flow rate, m3
/s
Qt = Total Flow rate into the tower
Qtwater = Flowrate of water into the tower

81
Lw= weir length, m
𝛽 = aeration factor, dimensionless
hds= calculated height of clear liquid over the dispersers, mm(dynamic seal)
ûf= flooding velocity through risers (bubble caps) or perforations(sieve plate), m/s.
K1= Drop through the slots.For sieve plates, K1 = 0
K2= Dropthrough the riser, reversal, and annular areas.
ht= total pressure drop, mm liquid
hd= pressure drop across the dispersion unit (dry hole forsieve plates; dry valve for valve plates;
dry cap, riser, andslot drop for bubble caps, mm liquid
hL= pressure drop through aerated mass over and aroundthe disperser, mm liquid
Vc= Volume of column
HC = Height of column
5.4.1 Theoretical number of plates required
At bottom of tower yformalin = 0.2255
y2= 𝑆𝑜𝑙𝑢𝑡𝑒 𝑚𝑜𝑙𝑒 𝑓𝑟𝑎𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑓𝑜𝑟𝑚𝑎𝑙𝑖𝑛
𝑆𝑜𝑙𝑢𝑡𝑒 𝑓𝑟𝑒𝑒 𝑚𝑜𝑙𝑒 𝑓𝑟𝑎𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑓𝑜𝑟𝑚𝑎𝑙𝑖𝑛
= 0.2255
1−0.2255
= 0.291
Qt = 113.802 kg mol / hr = 0.0316 kg mol / s
ymethanol = 0.0004
By assuming that the operating and equilibrium curves are straight lines and that heat effects
are negligible, Souders and Brown developed the equation:

82
𝑦2 − 𝑦 𝑚𝑒𝑡ℎ𝑎𝑛𝑜𝑙
𝑦2 − 𝑚𝑥2
=
𝐴 𝑒
𝑁𝑡+1
− 𝐴 𝑒
𝐴 𝑒
𝑁𝑡+1
− 1
Absorption factor Ae=
(𝑦 𝑤𝑎𝑡𝑒𝑟)(𝑄𝑡𝑤𝑎𝑡𝑒𝑟)
(𝑦 𝑓𝑜𝑟𝑚𝑎𝑙𝑖𝑛)(𝑄𝑡)
( 𝑦 𝑤𝑎𝑡𝑒𝑟)(50.1209)
( 𝑦 𝑓𝑜𝑟𝑚𝑎𝑙𝑖𝑛)(113.802)
=
(1)(50.1209)
(0.2255)(113.802)
= 1.948
0.291 − 0.0004
0.291 − 0
=
1.948 𝑁 𝑡+1 − 1.948
1.948 𝑁 𝑡+1 − 1
Nt = 8.7
Hence the theoretical number of plates needed is 8.7
5.4.2 Column diameter
Assumptions;
1. Hole diameter of 10 mm (Glitsch, 1970)
2. Tray thickness of 2 mm (Smith, 1963)
3. Tray spacing of 0.6 m (24 in) (Douglas, 1988).
4. Downcomer occupies 15% of the columns area
ρL = 1000 kg / m 3
Mw (formalin) = 30g / mol
Mw (water) = 18g / mol
Mw (Methanol) = 32g / mol

83
Mw (Oxygen) = 32g / mol
Mw (Nitrogen) = 28g / mol
T = 438.15K
P = 101.3 kpa
Mwt = 0.0325(32) + 0.2255(32) + 0.2342(30) + 0.0176(18) + 0.4902(28) = 29.32 g /g mol
ρG =
(𝑃) (𝑀𝑤𝑡)
(𝑅)(𝑇)
=
(101.3) (29.32)
(8.314)(438.15)
= 0.815kg / m 3
Figure 4- Rough estimate values (±25%) for Kv (J. R. Fair 1961)
Assuming a tray spacing of 0.6m (24 in), from figure ------,
The vapour-liquid equilibrium coefficient Kv = 0.3 ft (0.09 m)
The flooding velocity, ûf =Kv√
𝜌 𝐿 −𝜌 𝐺
𝜌 𝐺
= 0.09 √
1000 −0.815
0.815
= 3.15m / s

84
To limit the absorption column fromflooding, we choose a velocity of 80 percent of flooding
velocity (Douglas, 1988).
Design for 80%of flooding velocity: Uvd = 0.80 x 0.03m / s = 2.52 m / s
The volumetric flow rate in tower: Q =
(0.0316 k mol/s) ( 29.32 g/mol )
(0.815 𝑘𝑔 / 𝑚3)
= 1.14 m3
/ s
Hence the net plate area, An =
𝑄
𝑈 𝑣𝑑
=
0.00163 m3 / s
0.024 m / s
= 0.452 m 2
If the downcomer occupies 15% of the area, then the column x-sectional area is:
=
𝐴 𝑛
1−0.15
=
0.07 𝑚2
1−0.15
= 0.53 m 2
The column diameter =√
(4)(Qt) (Mwt)
(𝜋)(𝜌 𝐺)(Uvd) (1−0.15m2)
= √
(4)(0.0316 k mol/s) (29.32g/gmol)
(3.142)(0.815𝑘𝑔/𝑚3)(2.52 m/s) (1−0.15m2)
= 0.82 m
Active area Aa =
𝜋𝐷2
4
=
𝜋0.822
4
= 0.53 m 2
Area occupied by downcomer = 𝐴 𝑑 = 0.15 𝑥 Aa= 0.15 x 0.53 = 0.0795 m 2
Total area At = Aa + 2Ad = 0.53 + 2(0.0795) = 0.689 m 2
Assuming;
Hole diameter = 10 mm

85
Tray thickness = 2 mm
𝑇𝑟𝑎𝑦 𝑡ℎ𝑖𝑐𝑘𝑛𝑒𝑠𝑠
𝐻𝑜𝑙𝑒 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟
=
2
10
= 0.2
Hole area Ah = Aa x 0.2= 0.53 x 0.2 = 0.106 m 2
𝐻𝑜𝑙𝑒 𝑎𝑟𝑒𝑎
𝐴𝑐𝑡𝑖𝑣𝑒 𝑎𝑟𝑒𝑎
=
0.106
0.53
= 0.2
5.4.3 Pressure Drop
Methods for estimating fluid-dynamic behaviour of crossflow plates are analogous, whether
the plates be bubble-cap,sieve, or valve. The total pressure drop across a plate is defined by the
general equation:
ht= hd+ hL
Fig 5- Discharge coefficients for gas flow sieve trays. [Liebson, Kelley and Bullington, Pet.
Refiner, (1957)](From M & S Fig.18.27)

86
It is convenient and consistent to relate all of these pressure-dropterms to height of equivalent
clear liquid (deaerated basis) on the plate, in either millimeters or inches of liquid.Pressure drop
across the disperser is calculated by variations of thestandard orifice equation:
From Fig -5, Cv = 0.78
hd= K1 + K2(
𝜌 𝐺
𝜌 𝐿
)û𝑓
2
K2 =
50.8
𝐶 𝑣
2
K2=
50.8
0.782
= 83.5
hd= 83.5 (
0.815
1000
)3.152
= 0.675 mm of liquid
hL= 𝛽hds
Hole F-factor = ρG (ûf
1/2
) = 0.815 x 3.15 1/2
= 1.45
Q = 1.14 x 0.006 = 0.00684 m 3
/ s

87
Fig.6 – Aeration factor 𝛽 for pressure drop calculation sieve plates. [Bolles and Fair1982]
From Fig.6- 𝛽 = 0.7
Length of weir LW= 0.6 to 0.85 of the column diameter (Douglas, 1991)
= 0.82m (0.725) = 0.6 m = 600 mm
how= 664(
𝑄
𝐿 𝑊
)2/3
how= 664(
1.14
600
)2/3
= 10.2 mm
hds = how + hw = 10.2 + 600 = 610 mm
hL= 𝛽hds
hL= 0.7(610) = 427 mm
ht= hd+ hL

88
ht= 427 + 0.675= 427.675 mm liquid
5.4.4 Plate efficiency
The efficiency of a plate for mass transferdepends upon three sets of design parameters:
1. The system—composition and properties
2. Flow conditions—rates of throughput
3. Geometry—plate type and dimensions
The ultimate concern is with overall column efficiency or the ratio of theoretical plates to actual
plates required in making the separation:
Eoc= Nt/Na
Na = Nt/Eoc
The methods adopted in predicting the plate efficiency are the empirical methods. Two
empirical correlations which have found wide use are the one of Drickamer and Bradford
[Trans. Am. Inst. Chem. Eng., 39, 319 (1943)] and a modification of it by O’Connell [Trans.
Am. Inst. Chem. Eng., 42, 741 (1946)]. A semi-theoretical method which gives overall
efficiency is that of Bakowski [Br. Chem. Eng., 8, 384, 472 (1963); 14, 945 (1969)]. It is based
on the assumption that the mass-transfer rate for a component moving to the vapour phase is
proportional to the concentration of the component in the liquid and to its vapour pressure.
Also, the interfacial area is assumed proportional to liquid depth, and surface renewal rate is
assumed proportional to gas velocity. The resulting general equation is;
Eoc=
1
1.15+3.7(104)
𝐾𝑣 𝑀 𝑤𝑡
𝐿 𝑤 𝜌 𝐿 𝑇

89
Eoc=
1
1.15+3.7(105)
0.09 × 29.32
600 × 1000 × 438.15
= 0.87 = 87%
Nt= 8.7
Actual number of plates Na = Nt/Eoc
= 8.7/0.87
= 10 plates
5.4.5 COLUMN HEIGHT
The column height is determined as follows:
The column height = tray spaces × (number of stages)
According to Turton (1955), a safety factor of 10% is to be added to the final design height.
The column height Hc = 0.6 x 10
= 6.0 m
After adding a safety factor of 0.1,
The column height Hc becomes = 6.0 + 0.10
= 6.10 m
Volume of the column Vc =
𝜋
4
× D 2
×Hc =
𝜋
4
× (0.82) 2
× 6.1 = 3.2 m 3

90
Table 11. Specification sheet
Material of construction Steel
Tower type Packed tower
Type of packing Stacked
Mode of flow Cross – flow
Column diameter 0.82 m
Column Height 6.1 m
Tray spacing 24 in = 0.6 m
Actual Number of trays 10
Inlet Temperature 438.15
Outlet temperature 382.27 K
Flooding velocity 3.15 m / s
Gas flowrate 1.14 m 3
/ s
Volume of column 3.2 m 3
Overall column efficiency 87%
Maximum allowable pressure drop 427.675 mm liquid

91
5.5 CHEMICAL ENGINEERING CENTRIFUGAL PUMP DESIGN
Given that;
Fluid = (CH2O, H2O and CH3OH)
Vapour pressure (CH2O) at 45˚C = 1317.22N/m2
Vapour pressure (CH3OH) at 45˚C = 47500N/m2
Vapour pressure (H2O) at 45˚C = 9.85N/m2
Total vapour pressure (VP) = 48827N/m2
Average density of fluid = 857kg/m3
Acceleration due to gravity (g) = 9.81m/s2
Velocity of fluid from rules of thumbs ranges from 1 – 3m/s, hence average velocity = 2m/s
Ps = pressure in the fluid before the impeller (P1)
Mass flow rate of fluid (ethanol, formalin and water) through pump = 12276.6076kg/hr
Average density of fluid (ethanol,formalin and water) = 857kg/m3
5.5.1 Volumetric flowrate of fluid = mass flowrate/average density
=
12276.6076𝑘𝑔/ℎ𝑟
857𝑘𝑔/𝑚3 = 144.33m3
/hr
= 14.33m3
/hr × 264.17Gal/1m3
× 1hr/60min
= 63GPM
5.5.2 Head (h) developed by the pump can be calculated using the relation below;
h =
(𝑃2− 𝑃1)
𝜌𝑔
+
𝑉2
2𝑔
where; P2 = outlet pressure of centrifugal pump = 303975N/m2
(3atm)

92
P1 = inlet pressure of centrifugal pump = 101325N/m2
(1atm)
𝜌 = density of fluid = 857kg/m3
g = acceleration due to gravity
v = velocity of fluid = (1 to 3) m/s from the rule of thumbs
h =
(303975−101325)𝑘𝑔/𝑚.𝑠2
857𝑘𝑔/𝑚3+9.81𝑚/𝑠2
+
(2𝑚/𝑠)2
2(9.8)𝑚/𝑠
= 24.10m + 19.62m = 43.72m = 143ft
5.5.3 Efficiency of the centrifugal pump (Ep):
(Ep) = 80 – 0.0285(h) + 3.78×10-4
(h)(GPM) – 2.38×10-7
(h)(GPM)2
+ 5.39×10-4
(h)2
–
6.39×10-7
(h)2
(GPM) + 4×10-10
(h)2
(GPM)2
= 80 – 0.0285(143) + 3.78×10-4
(143)(63) – 2.38×10-7
(143)(63)2
+ 5.39×10-4
(143)2
–
6.39×10-7
(143)2
(63) + 4×10-10
(143)2
(63)2
= 90 – 0.85
= 89%
5.5.4 Power of centrifugal pump (P) = (GPM)∆P/(1715)(Efficiency)
=
63𝑔𝑎𝑙/min × (303975−101325)𝑁/𝑚2
1715 ×0.89
= 8364.36gal/min × N/m2
× 1m3
/264.17gal × 1min/60s
= 0.53Nm/s
= 0.53J/s × 1.341 ×
10−3ℎ𝑝
1𝐽/𝑠
= 0.0007hp
5.5.5 Centrifugal Pump specific speed (Ns) = (RPM)(GPM)0.5
/(h)0.75
But speed of impeller used = 2500rpm
Ns =
(2500)(63)0.5
(143)0.75
= 479.85rpm

93
5.5.6 Impeller Diameter (D2) of centrifugal pump =
(1840)(𝑘𝑣)(ℎ)0.5
𝑅𝑃𝑀
Fig.7 Impeller Eye Diameter/outside diameter ratio (Colson and Richardson, volume 6)
From graph head constant (Kv) and specific speed (Ns) above for 5&6 vanes,
Kv = 0.99
D2 =
(1840 ×0.99)×(143)0.5
2800
= 8.71inch

94
5.5.7 Eye diameter of centrifugal pump (from chart between D1/D2 against Ns):
Fig 9 chart between D1/D2 against Ns
From the chart above, with the calculated specific speed,
D1/D2 = 0.333
D1 = 0.333 × 8.71inch = 2.87inch
Impeller eye area of centrifugal pump:
Eye area =
𝜋
4
×D1
2
=
𝜋
4
×(2.87)2
= 6.5in2
Suction eye velocity of centrifugal pump:
SEV = 0.321×GPM/Eye area =
0.321×63gal/min
6.5𝑚2
= 3.1gal/min × 1/in2
× 1min/60s × (12in)2
/1ft2
× 1ft3
/7.4805gal = 0.99ft/s

95
5.5.8 Peripheral velocity (Ut) of centrifugal pump impeller:
Ut = Eye diameter × (speed of pump/229)
= 2.87 × (250/229) = 31.3inch/min = 2.61ft/min = 0.04ft/s
Specific weight (𝛾) of fluid (ethanol and water) = 𝜌×g = 𝛾 = 857kg/m3 × 9.81m/s2
Suction head (Hs):
Hs = Ps/𝛾 + V2/2g
Vapour head= Pv/𝛾
5.5.9 Net Positive Suction Head of centrifugal pump,(NPSH):
NPSH= suction head – vapour head
= (Ps/𝛾 + Vs2
/2g) – (Pv/𝛾)
=
101325𝑁/𝑚2
8407.17𝑘𝑔/𝑚2.𝑠2
+
(2𝑚/𝑠)2
2×9.81𝑚/𝑠2
−
4827.0𝑁/𝑚2
8407.17𝑘𝑔/𝑚2.𝑠2
=
12𝑘𝑔/𝑚.𝑠2
𝑘𝑔/𝑚2.𝑠2
+
0.2𝑚2/𝑠2
𝑚/𝑠2
−
5.80𝑘𝑔/𝑚.𝑠2
𝑘𝑔/𝑚2.𝑠2
= 6.4m

96
5.6 CHEMICAL ENGINEERING DESIGN OF AIR HEATER
Air heater is a general term used to describe any device designed to heat air before another
process with the primary objective of increasing the thermal efficiency of the process. Air
heaters are commonly used in industrial settings to heat up air before the process of combustion,
or to be used to dry out or heat up other substances. An air heater works on the same principle
as a heat exchanger to heat the air passing through (Lienhard, 2004).
Heat exchangers are devices that facilitate the exchange of heat between two fluids that are at
different temperatures. The principal types of heat exchangers used in the chemical process
and allied industries are:
1. Double-pipe exchanger-The simplest type, used for cooling and heating.
2. Shell and tube exchangers-Used for all applications.
3. Plate and frame exchangers-Used for cooling and heating
4. Plate-fin exchangers.
5. Spiral heat exchangers etc.
The use of air heaters is becoming more and more important these days as they enhance the
drying process, increase the thermal efficiency ratio and thus lead to economic advantages due
to primary energy and time being saved (Lienhard, 2004).
Selection of heat exchangers
The selection or design of a heat exchanger depends on several factors such as the heat transfer
rate, cost, pressure drop, size, weight, construction type, materials and operating environment.
There are other considerations in the selection of heat exchangers that may or may not be
important, depending on the application.

97
For example, being leak-tight is an important consideration when toxic or expensive fluids are
involved, ease of servicing, low maintenance cost, safety and reliability (Lienhard, 2004).
A shell and tube heat exchange will be used to heat the air. The shell and tube heat exchanger
is by far the most common type of heat transfer equipment used in the chemical and allied
industries. The advantages of this type are as follows:
1. Easily cleaned.
2. Has a good design procedure.
3. Can be constructed from a wide range of materials.
4. Good mechanical layout: has a good shape for pressure operation.
5. The configuration gives a large surface area in a small volume.
CHEMICAL ENGINEERING DESIGN
From our material balance around the air heater:
J = 905.2076kmol/hr K = 905.2076kmol/hr
O2= 0.21 O2= 0.21
N2= 0.79 N2= 0.79
Where,
J is the amount of air input
K is the amount of air output
The following key parameters are to be determined;
1. Heat transfer rate
2. Heat transfer area
AIR HEATER

98
3. Number of tubes
4. Bundle diameter
5. Shell equivalent diameter
6. Tube and shell side heat transfer coefficient
7. Overall heat transfer coefficient
8. Pressure drop at shell side
9. Reynolds number at shell and tube side
10. Fanning factor at shell and tube side
But total mole flow rate of air (ṅair) = 905.2076 kmol/hr
Molecular weight of air used (Mw) = 60kg/kmol
Mass flow rate of air (Mf) = Mw x ṅair
= 60kg/kmol x 905.2076kmol/hr
= 48986.187 kg/hr = 13.6kg/s
Heat transferred (Q) calculation:
Using the design equation Q = ṁCpT
Where, A = Heat transfer area
Q = Heat transfer rate
U = Overall heat transfer coefficient
T = Temperature difference
Using standard dimensions of tubes commonly used provided by Tubular Exchanger
Manufacturing Association (TEMA):
Tube outside diameter (do) = 1inch =0.0254m
Tube inside diameter (di) = 0.02m

99
Tube length (L)= 10ft = 3.048m
Therefore, using the relation below;
The long mean temperature difference, TLM
TLM =
(𝑇 𝑊𝐼− 𝑇 𝐴𝑂)−(𝑇 𝑊𝑂− 𝑇 𝐴𝐼)
𝑙𝑛
(𝑇 𝑊𝐼−𝑇 𝐴𝑂)
(𝑇 𝑊𝑂− 𝑇 𝐴𝐼)
(Sinnott, 2005)
Where,
TWO = Water outlet temperature (C)
TWI= Water inlet temperature (C)
TAO= Air outlet temperature(C)
TAI= Air inlet temperature(C)
Thus, (LMTD) = TLM =
(185−150)−(150−37.3)
𝑙𝑛
(185−150)
(150−37.3)
=70 C
Heat transferred, Q = ṁCpT
Q= 13.6
𝑘𝑔
𝑠
 1013.5
𝐽
𝑘𝑔.𝐾
 385.7K = 5316334.5W
For tube side:
Change in temperature, T = 185-150
= 35C
= 308K
But mass flow rate at shell side (ṁt) = 5316334.5
𝐽
𝑠

1𝑘𝑔.𝐾
4178.9𝐽  308K
= 4.1kg/s
From rules of thumb, with steam in tube and air within shell, heat transfer (U) ranges from (25–
250)W/m2
.K. Therefore average heat transfer coefficient used (Uav) = 180W/m2
.K

100
(Coulson and Richardson, 2005)
Heat transfer area (A)
A =
𝑄
U𝑇 𝐿𝑀
=
5316334.5W
180𝑊/𝑚2.𝐾 × 343𝐾
= 86m2
But Number of Tube =
𝐴
 x 𝑑 𝑜 𝑥 𝐿
=
12.7𝑚2
𝜋 ×0.0254𝑚 ×3.048𝑚
= 57tubes
Tubes pitch (Pt)
Tubes pitch, (Pt) = 1.25do = 1.25  0.0254m = 0.0317m
Area of tubes (At)
(At) =
𝝅 𝒙 𝒅 𝒐
𝟐
𝟒
=
𝝅 𝒙 (𝟎.𝟎𝟐𝟓𝟒) 𝟐
𝟒
= 0.0005m2
Volume of tubes (Vt)
(Vt) =
𝝅 𝒙 𝒅𝒐 𝟐 𝒙 𝑳
𝟒
=
𝝅 𝒙 (𝟎.𝟎𝟐𝟓𝟒) 𝟐 𝒙 3.048
𝟒
= 0.0015m3
Shell dimensions
Assumptions: using an increment of 10% on tube length for clearance between shell and tube;
Length of shell (Ls) = 1.10 x length of tube
= 1.10 x 3.048
= 3.35m
Bundle diameter (Bd)
(Bd) = number of tubes x tube outer diameter
= 86 x 0.0254m
= 2.15m
Using a bundle diameter clearance of 10cm for effective heat transfer and cleaning of tubes,
we have;
Shell diameter, (Ds) = bundle diameter + bundle diameter clearance
= bundle diameter + 0.1m
= 2.15m + 0.1m

101
= 2.25m
Area of shell (As)
Shell area, (As) =
π x𝐷𝑠
2
4
=
𝜋 𝑥 (2.25)2
4
= 3.97m
Volume of shell (Vs)
Shell volume, (Vs) =
π xD2x L
4
=
𝜋 𝑥 (2.25)2 𝑥 3.35
4
= 13.3m3
Baffle spacing (Bs)
Baffle spacing, (Bs) =
1
5
 Ds = 0.2  2.25m = 0.45m
Cross flow-area (Ac)
Cross flow-area, (Ac) =
(𝑃𝑡−𝑑 𝑜)𝐷 𝑆 𝐵 𝑆
𝑝𝑡
Pt is the tube pitch
do is the tube outside diameter
Ds is the shell diameter
Bs is the baffle spacing
=
(0.0317−0.0254)×2.25×0.45
0.0317
= 0.213m2
Shell-side mass velocity (Gs)
(Gs) =
ṁ 𝑠
𝐴𝑐
𝑚̇ is the shell mass velocity
Ac shell side cross flow rate
=
13.6𝑘𝑔/𝑠
0.213𝑚2 = 63kg/m2
.s
Shell equivalent diameter for a square pitch arrangement (De)
De =
1.27
𝑑 𝑜
(Pt2
– 0.785𝑑 𝑜
2
)

102
Where,
De is the shell equivalent diameter
Pt is the tube pitch
=
1.27
0.0254
[(0.0317)2
– 0.785(0.0254)2
] = 0.025m
Shell-side Reynolds number (Re)
Re =
𝐺 𝑆 𝐷 𝑒
µ
Where,
GS is the shell mass velocity
De is the shell equivalent diameter
µ is the viscosity of the steam
For Shell-side Reynolds number (Re)
Re =
𝐺𝑠𝐷𝑒
µ
=
68𝑘𝑔/𝑚2.𝑠 × 0.025𝑚
0.0002968𝑘𝑔/𝑚.𝑠
= 5306
Prandtle (Pr) number at the (shell-side) gives
Pr =
µ𝐶 𝑝
𝐾
Where,
µ is the viscosity of the steam
cp is the specific heat capacity of the steam
k is the thermal conductivity
=
0.0002868𝑘𝑔/𝑚2.𝑠 ×4190𝐽/𝑘𝑔°.𝐶
0.32𝑊/𝑚2.°𝐶
= 3.66
Shell side fanning factor (f2)
Fanning factor at shell side (f2) = 0.25(1.82logNRe – 1.64)-2

103
= 0.25[1.82log (5306) – 1.64]-2
= 0.0052
But f2 = 2JH2
Where JH2 is fanning factor at the shell side
JH2 = f2/2 = 0.0026/2 = 0.00103
Area of maximum velocity (m) in annulus using Colburn correlation
m =
𝑑 𝑜
2
− 𝑑𝑖
2
4 ln(
𝑑 𝑜
𝑑 𝑖
)
=
[0.02542− (0.0212)2]
4 ln(
0.0254
0.0212
)
= 0.0003m2
Tube side fanning factor (f1)
Fanning factor at tube side, (f1) =
𝐹2 𝑑2[(𝑚+(−𝑑1
2
)]
𝑑1(𝑑 𝑜2− 𝑚)
f1 =
0.00212 ×0.0254[0.0003+(−0.0212)2]
0.0212[(0.0254)2− 0.0003]
= 5.5×10-15
But JH1
Where JH1 is the fanning factor at tube side
= f1/2 = 5.5×10-15
/2 = 2.75×10-15
Shell-side heat transfer coefficient (hs)
hs= 0.023(
𝑘
𝐷𝑖
)(Re)0.8
(Pr)1/3
(
µ
µ𝑤
)0.14
(
µ
µ𝒘
) = Viscosity correction factor
(
µ
µ𝒘
) = 1
Thus, hs = 0.023(
0.32
0.0212
)(5306)0.8
(3.66)1/3
(1)0.14
= 507W/m2
.K
Pressure drop (𝜟P) in the shell calculation

104
Density of steam at 94C = 0.949004kg/m3
Ps = 8JH2 x (
𝐷𝑠
𝐷 𝑒
) x (
𝐿
𝐵 𝑠
) x
ρ 𝑠
2
2
𝑥 (
µ
µ 𝑤
)–0.14
= 8 x (0.00106) x (
2.25
0.0197
) x (
3.048
0.45
) 𝑥
0.9006
2
𝑥 (
1
1
)–0.14
= 24kg/m.s2
Number of tubes per pass (Ntpp)
Ntpp = Nt/Number of passes = 281/2 = 140 tubes per pass
Tube-side mass velocity (Gmt)
(Gmt) =
𝑇𝑢𝑏𝑒 𝑠𝑖𝑑𝑒 𝑓𝑙𝑜𝑤𝑟𝑎𝑡𝑒 (
𝑘𝑔
𝑠
)
𝑁𝑡𝑝𝑝 ×
𝜋𝑑𝑖2
4
Gmt =
4.4𝑘𝑔/𝑠
140 ×
𝜋 × (0.0212)2
4
= 89kg/m2
.s
Tube-side velocity (Vts) =
𝐺 𝑚𝑡
ρ 𝑎𝑖𝑟
Hence tube side velocity, (Vts) =
89𝑘𝑔/𝑚2.𝑠
3.895𝑘𝑔/𝑚3 = 22m/s
Prandtle number of steam is given as
Pr = 0.7071
Also, Reynolds Number, Re =
ρ 𝑠
×𝑑𝑖×𝑉𝑡
µ 𝑡
Re =
994𝑘𝑔/𝑚 𝑒×0.0212𝑚×22𝑚/𝑠
0.0007𝑘𝑔/𝑚.𝑠
= 66228
Tube side film coefficient (hi)
hi = 1.86(
𝐾
𝑑𝑖
)(Re)(Pr)0.33
(
𝑑𝑖
𝐿
)
0.33
(
µ
µ 𝑤
)
0.14
but, µ = µ 𝑤
= 1.86 𝑥 (
35.25×10−3
0.0212
) 𝑥 (66228) x (0.7)0.33
x (0.0069)0.33
x (
1
1
)
0.14
= 34189W/m2
.K
Fouling factors within and outside the tubes, hdo=2000W/m.K and hdi= 5000 (Coulson and
Richardson, 2005)

105
The overall heat transfer coefficient based on inside and outside tube flow (Uo) and (Ui) can be
estimated as:
Ui =
1
1
ℎ 𝑖
+
1
ℎ𝑑 𝑖
+
𝑑 𝑖ln(𝑑 𝑜/𝑑 𝑖
)
2𝑘 𝑤
+
𝑑 𝑖
𝑑 𝑜ℎ𝑑 𝑜
+
𝑑 𝑖
𝑑 𝑜ℎ 𝑜
Uo =
1
1
ℎ 𝑜
+
1
ℎ𝑑 𝑜
+
𝑑 𝑜ln(𝑑 𝑜/𝑑 𝑖
)
2𝑘 𝑤
+
𝑑0
𝑑 𝑖ℎ𝑑 𝑖
+
𝑑 𝑜
𝑑 𝑖ℎ 𝑖
Kw = conductivity of material of construction used
ho = shell side heat transfer coefficient
hdi = fouling factor inside tube
hi = tube side heat transfer coefficient
Ui =
1
1
34189
+
1
2000
+
0.0212ln(0.025/0.0212)
2(45)
+
0.0212
0.0254(5000)
+
0.0212
0.0254(507)
= 428 W/m2
.K
Uo =
1
1
507
+
1
5000
+
0.0254ln(0.025/0.0212)
2(45)
+
0.0254
0.0212(2000)
+
0.0212
0.0212(34189)
= 362 W/m2
.K

107
CHAPTER SIX
EQUIPMENT SPECIFICATIONS
6.1 Vaporizer
Function: to heat the liquid mixture (Methanol and Water) into its vapour state
Capacity:
Material of construction: Carbon Steel
Quantity: 1
6.2 Absorber
Function: To absorb formaldehyde from liquid mixture
Capacity: 17.26 m 3
Quantity: 1
6.3 Air heater
Function: to increase the temperature of air to its vapour state
Capacity: 48986kg/hr
Quantity: 1
6.4 Liquid heater (1)
Function: to increase the temperature of feed entering the distillation unit
Capacity: 1809.65kg/hr
Quantity: 1

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