A Project Report on
PRODUCTION OF ACETALDEHYDE
Submitted in partial fulfillment of the requirements for the award of the degree of
BACHELOR OF TECHNOLOGY IN
CHEMICAL ENGINEERING
Submitted By:
Table of Contents
Acknowledgement ................................................................ Ошибка! Закладка не определена.
Table of Contents .............................................................................................................................. i
List of Figures .................................................................................................................................. iv
List of Tables .................................................................................................................................... v
List of Symbols and abbreviations used .......................................................................................... vi
Abstract ......................................................................................................................................... viii
Chapter 1 .......................................................................................................................................... 1
1
Introduction .............................................................................................................................. 1
1.1
Overview ............................................................................................................................ 1
1.2
Occurrence ......................................................................................................................... 1
1.3
Environmental Aspect ........................................................................................................ 2
Chapter 2 .......................................................................................................................................... 3
2
Market Survey .......................................................................................................................... 3
2.1
Global Consumption and demand of Acetaldehyde: ......................................................... 3
2.2
Demand for acetaldehyde in Western Europe and China .................................................. 5
2.3 Demand for acetaldehyde in Japan ......................................................................................... 7
2.4
Indian market for acetaldehyde .......................................................................................... 8
2.5
Present and future condition of demand for acetaldehyde ................................................. 8
Chapter 3 ........................................................................................................................................ 11
3
Literature Survey .................................................................................................................... 11
Chapter 4 ........................................................................................................................................ 12
4
Physical Properties and Uses .................................................................................................. 12
i
4.1 Properties .............................................................................................................................. 12
4.2 Uses ...................................................................................................................................... 19
Chapter 5 ........................................................................................................................................ 20
5
Different Manufacturing Processes ........................................................................................ 20
5.1
Oxidation of Ethylene ...................................................................................................... 20
5.1.1
Two stage process developed by Wacker Chemie ................................................... 21
5.1.2
One stage process developed by Farbwerke Hoechst ............................................... 21
5.2
By Oxidation of ethyl alcohol .......................................................................................... 22
5.3
By Dehydrogenation of ethyl alcohol .............................................................................. 22
5.4
Preparation method of acetaldehyde from acetylene ....................................................... 23
5.5
From synthesis Gas .......................................................................................................... 24
5.6
From acetic acid ............................................................................................................... 24
5.7
From Saturated Hydrocarbons: ........................................................................................ 25
6
Selection of Process with Justification ................................................................................... 26
7
Description of the Process ...................................................................................................... 27
8
Material and Energy Balance ................................................................................................. 29
Chapter 9 ........................................................................................................................................ 45
9
DESIGN OF EQUIPMENT ................................................................................................... 45
9.1
DESIGN OF DISTILLATION COLUMN: ..................................................................... 45
9.2
Design of Heat exchanger ................................................................................................ 88
9.3
Design of Reactor ............................................................................................................ 99
9.4
Design of Condenser ...................................................................................................... 105
Chapter 10 .................................................................................................................................... 109
10 COST ESTIMATION .......................................................................................................... 109
10.1
Estimation of Total Capital Investment: .................................................................... 109
ii
10.2
Estimation of Total Product cost: ............................................................................... 112
11 Profitability Analysis ............................................................................................................ 116
11.1
Gross Earnings/Income: ............................................................................................. 116
11.2
Rate of Return: ........................................................................................................... 116
11.3
Payback Period ........................................................................................................... 116
11.4
Break-Even Analysis .................................................................................................. 117
12 Chapter ................................................................................................................................. 118
PLANT LOCATION AND LAYOUT ......................................................................................... 118
Chapter 13 .................................................................................................................................... 126
13 Pollution and Safety in Process Industries ........................................................................... 126
References .................................................................................................................................... 129
iii
List of Figures
Figure 2.1 The following Pie chart shows the Global Consumption in the year 2012 ..................... 4
Figure 2.2 China's Demand for Acetaldehyde Since 2002 ............................................................... 5
Figure 2.3 Western Europe's Demand for Acetaldehyde Since 2003 .............................................. 6
Figure 2.4 Graph of Japan's Annual import of Acetaldehyde .......................................................... 7
Figure 2.5 India's Acetaldehyde Trade Statistics ( demand and supply) .......................................... 8
Figure 2.6 Chart Showing Major Importers of Acetaldehyde ........................................................ 10
Figure 7.1 Flowsheet ...................................................................................................................... 28
Figure 12.1 Plant Layout .............................................................................................................. 125
iv
List of Tables
Table 2-1 Consumption of Acetaldehyde in Kg ............................................................................... 9
Table 2-2 Some Leading Producers of Acetaldehyde and their Production Capacities ................. 10
Table 4-1 Physical Properties of Acetaldehyde .............................................................................. 13
8-1 Values of constants for calculating Cp ..................................................................................... 31
8-2 Values of CP for different temperature range (kcal/Kg.C) ...................................................... 31
8-3 Mass Balance Table for Vaporizer ........................................................................................... 33
8-4 Energy Balance Table for Vaporizer ........................................................................................ 33
8-5 Mass Balance Table for Reactor .............................................................................................. 34
8-6 Energy Balance Table for Reactor ........................................................................................... 35
8-7 Mass Balance Table for Heat Exchanger ................................................................................. 36
8-8 Energy Balance Table for Heat Exchanger .............................................................................. 36
8-9 Mass Balance Table for Condenser 1 ....................................................................................... 38
8-10 Energy Balance for Condenser 1 ............................................................................................ 38
8-11 Mass Balance Table for Condenser 2 ..................................................................................... 40
8-12 Energy Balance for Condenser 2 ............................................................................................ 40
8-13 Mass Balance for Preheater .................................................................................................... 41
8-14 Energy Balance Table for Preheater....................................................................................... 41
9-1 T-x-y data for ethanol-acetaldehyde ........................................................................................ 46
9-2 List of Parameters used in calculation...................................................................................... 49
9-9-3 Conditions of fluidization ................................................................................................... 103
v
List of Symbols and abbreviations used
Msteam
Mass flow rate of steam.
∆Hsteam
Enthalpy of steam.
E
Mass flow rate of ethanol.
A
Mass flow rate of acetaldehyde.
H
Mass flow rate of hydrogen.
Cp
Specific heat capacity.
λ
Latent heat of vaporization.
F
Molar flow rate of Feed, kmol/hr.
D
Molar flow rate of Distillate, kmol/hr.
W
Molar flow rate of Residue, kmol/hr.
XF
Mole fraction of Acetaldehyde in liquid/Feed.
YD
Mole fraction of Acetaldehyde in Distillate.
XW
Mole fraction of Acetaldehyde in Residue.
MF
Average Molecular weight of Feed, kg/kmol
MD
Average Molecular weight of Distillate, kg/kmol
MW
Average Molecular weight of Residue, kg/kmol
Rm
Minimum Reflux ratio
R
Actual Reflux ratio
L
Molar flow rate of Liquid in the Enriching Section, kmol/hr.
G
Molar flow rate of Vapor in the Enriching Section, kmol/hr.
L
Molar flow rate of Liquid in Stripping Section, kmol/hr.
G
Molar flow rate of Vapor in Stripping Section, kmol/hr.
Q
Thermal condition of Feed
ρL
Density of Liquid, kg/m3.
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.
vi
TL
Temperature of Liquid, K.
TV
Temperature of Vapor, K
vii
Abstract
The present work illustrates the production of acetaldehyde by different production methods, market
survey, literature survey and selection of an appropriate method. The method of production we have
chosen is dehydrogenation of ethyl alcohol. In this process, hydrogen is taken out as a by-product
which can be used elsewhere or which can be used to generate heat. In dehydrogenation process more
conversion-taking place compared to other processes. The mass balance is being performed for the
various units involved in the production of acetaldehyde by dehydrogenation of ethyl alcohol method.
We have also done energy balance for each unit to find out heat utility and amount of auxiliary fluids
required for the heat supply. In design section we have designed a distillation column, heat exchanger,
condenser and packed bed reactor which matches the design specifications required for given plant.
Cost estimation is also done along with profitability studies to show the investment required, rate of
return and payback period. The last sections consists of plant layout, pollution control and safety
measures which are very much essential for and process industry.
viii
Chapter 1
1
Introduction
1.1 Overview
Acetaldehyde, CH3CHO is an important intermediate in industrial organic synthesis with IUPAC
name ethanal. It is one of the most important aldehydes, occurring widely in nature and
being produced on a large scale in industry . In acetaldehyde carbon atom shares a double
bond with an oxygen atom, a single bond with a hydrogen atom, and a single bond with another
atom of carbon. The double bond between carbon and oxygen is characteristic of all aldehydes
and is known as the carbonyl group.
1.2 Occurrence
Acetaldehyde is used in the production of perfumes, polyester resins, and basic
dyes. Acetaldehyde is also used as a fruit and fish preservative, as a flavouring agent, and as a
denaturant for alcohol, in fuel compositions, for hardening gelatin, and as a solvent in the rubber,
tanning, and paper industries. Acetaldehyde is a normal intermediate product in the respiration of
higher plants. It occurs in traces in all ripe fruits that have a tart taste before ripening; the
aldehyde content of the volatiles has been suggested as a chemical index of ripening during cold
storage of apples. Acetaldehyde is an intermediate product of alcoholic fermentation but it is
reduced almost immediately to ethanol. It may form in wine and other alcoholic beverages after
exposure to air, and imparts an unpleasant taste; the aldehyde ordinarily reacts to form diethyl
acetal and ethyl acetate. Acetaldehyde is an intermediate product in the decomposition of sugars
in the body and, hence, occurs in traces in blood. Acetaldehyde is a product of most hydrocarbon
oxidations.
1
1.3 Environmental Aspect
Acetaldehyde is toxic when applied externally for prolonged periods, an irritant, and a probable
carcinogen. It is an air pollutant resulting from combustion, such as automotive exhaust
and tobacco smoke. Acetaldehyde is an important intermediate in the production of acetic acid,
acetic anhydride, ethyl acetate, peracetic acid, pentaerythritol, chloral, glyoxal, alkyl amines, and
pyridines. Acetaldehyde was first used extensively during World War I as an intermediate for
making acetone from acetic acid.
2
Chapter 2
2
Market Survey
2.1 Global Consumption and demand of Acetaldehyde:
The global market for acetaldehyde has been trending downward for the past twenty years as a
result of the commercialization of more efficient technologies to produce those products formerly
based on acetaldehyde. For example, the production of plasticizer alcohols has totally switched
from N-butyraldehyde based on acetaldehyde to the oxonation of propylene, while acetic acid is
now made predominantly by the lower-cost methanol carbonylation process.
With acetic acid manufacturing processes migrating from acetaldehyde based production
techniques towards carbonylation-of-methanol, the world acetaldehyde market are projected to
witness a steady deterioration in consumption. But acetic acid facilities based on acetaldehyde
continue to operate in Asia and South America, although these will eventually be phased out in
favour of methanol carbonylation. In addition to these structural changes, acetaldehyde demand
has also declined in the last few years because of mature end-use markets and the effects of the
economic downturn on these acetaldehyde-derived products.
3
Figure 2.1 The following Pie chart shows the Global Consumption in the year 2012[23]
Sales
1st Qtr
2nd Qtr
3rd Qtr
4th Qtr
.
Production of pyridine and pyridine bases, pentaerythritol, and acetate esters accounted for 16%,
16%, and 14%, respectively, of 2012 global acetaldehyde consumption. Pyridine and pyridine
bases are important raw materials in the production of agricultural chemicals. Pentaerythritol and
acetate esters (mainly ethyl acetate, but also some isobutyl acetate) are both used heavily in
surface coatings. The other uses of acetaldehyde accounted for the remaining global consumption
of acetaldehyde in 2012. This category includes 1, 3-butylene glycol, crotonaldehyde and glyoxal,
along with some smaller-volume derivatives.
As the chart shows, China is the largest consumer of acetaldehyde in the world, accounting for
almost half of global consumption in 2012. Chinese consumption is heavily weighted toward the
production of acetic acid. However, the growth of acetaldehyde in this end use will be limited in
the future because any new acetic acid plants will be based on the methanol carbonylation
process. Other uses such as pyridines and pentaerythritol will grow faster than acetic acid, but the
volumes are not large enough to offset the decline in acetic acid. Consequently, overall Chinese
acetaldehyde consumption will grow only slightly at 1.6% per year through 2018.
4
2.2 Demand for acetaldehyde in Western Europe and China
Western Europe is the second-largest consumer of acetaldehyde worldwide after China,
accounting for 20% of world consumption in 2012. As with China, the Western European
acetaldehyde market is expected to increase only very slightly at 1% per year during 2012 to
2018.
China’s acetaldehyde imports decreased gradually from 2003 and there is a sudden increase in the
value in the year 2010
Figure 2.2 China's Demand for Acetaldehyde Since 2002[23]
Year
5
Figure 2.3 Western Europe's Demand for Acetaldehyde Since 2003[23]
We can see France and Germany are the leading importers of acetaldehyde which shows that they
have a huge demand since 2003.
The US acetaldehyde market will also rise only minimally, at 1–2% per year during 2012–2018.
Acetaldehyde is not consumed for acetic acid and the rest of the markets will experience GDPlike growth in the next five years.
6
2.3 Demand for acetaldehyde in Japan
In contrast, Japan could very well be the brightest spot for acetaldehyde consumption in the next
five years and this hinges upon the on-purpose production of butadiene from acetaldehyde. The
supply of butadiene has been volatile in Japan and the rest of Asia because of the limited
availability of naphtha feedstock. Typically, butadiene and other C4hydrocarbons are coproduced
when naphtha is used as a feedstock for ethylene manufacture. However, the increased production
of natural gas from shale gas, particularly in the United States, has caused many ethylene crackers
to switch feedstock from naphtha to ethane, which yields lower volumes of co product butadiene
and other C4hydrocarbons. This has spurred the revival of on-purpose production for butadiene
and Japan certainly has enough sources of acetaldehyde to support such a project. This new end
use should provide a much needed boost to an otherwise flat acetaldehyde market. If Showa
Denko starts up its butadiene plant as scheduled, acetaldehyde consumption in Japan should
exhibit close to 4% per year growth through 2018.
.
Figure 2.4 Graph of Japan's Annual import of Acetaldehyde[23]
Year
7
2.4 Indian market for acetaldehyde
Indian consumption is anticipated to have moderate growth rate of nearly 3% for the next several
years and demand for acetaldehyde use in acetic acid production will grow 3-4% annually.
Figure 2.5 India's Acetaldehyde Trade Statistics ( demand and supply)[23]
2.5 Present and future condition of demand for acetaldehyde
Overall, the global market for acetaldehyde has grown 2-3% annually during 2009-2014.
However, some of this growth is actually a recovery from the significant decline experienced in
2009(for example, China’s use in acetic acid market). Major regions including Western Europe,
United States of America will have low growth than that of global consumption’s rate. However,
the demand for acetaldehyde worldwide has continued to decrease primarily as a result of less
consumption for acetic acid manufacture, as the industry continues to move toward more efficient
methanol carbonylation process. Its production in North America and Western Europe has
8
decreased but its manufacture in Asia continues and new acetaldehyde capacity has been installed
in china, but this is the only area where new capacity has been added.
Traditionally, acetaldehyde was mainly used as a precursor to acetic acid. In terms of
condensation reactions, acetaldehyde is an important precursor to pyridine derivatives,
pentaerythritol, and crotonaldehyde. Urea and acetaldehyde combine to give a useful resin. Acetic
anhydride reacts with acetaldehyde to give ethylidene diacetate, a precursor to vinyl acetate,
which is used to produce polyvinyl acetate.
China is the largest consumer of acetaldehyde in the world, accounting for almost half of global
consumption in 2012. Major use has been the production of acetic acid. Other uses such
as pyridines and pentaerythritol are expected to grow faster than acetic acid, but the volumes are
not large enough to offset the decline in acetic acid. As a consequence, overall acetaldehyde
consumption in China may grow slightly at 1.6% per year through 2018. Western Europe is the
second-largest consumer of acetaldehyde worldwide, accounting for 20% of world consumption
in 2012. As with China, the Western European acetaldehyde market is expected to increase only
very slightly at 1% per year during 2012–2018. However, Japan could emerge as a potential
consumer for acetaldehyde in next five years due to newfound use in commercial production
of butadiene. The supply of butadiene has been volatile in Japan and the rest of Asia. This should
provide the much needed boost to the flat market, as of 2013
Table 2-1 Consumption of Acetaldehyde in Kg[23]
9
Table 2-2 Some Leading Producers of Acetaldehyde and their Production Capacities[5]
COMPANY
Production ( in 103 t)
Celanese Chemicals Europe, Germany
120
Eastman chemical company
225
ECROS,SA, Spain
90
Japan aldehyde company Ltd, Japan
69
Jilin chemical industrial company, China
180
Kyowa Yuka Company Ltd, Japan
61
Wacker chemie AG, Germany
65
Figure 2.6 Pie Chart Showing Major Importers of Acetaldehyde[5]
10
Chapter 3
3
Literature Survey
Acetaldehyde was first prepared by Scheele in 1774, by the action of manganese dioxide and
sulphuric acid on ethanol[1]. Liebig established the structure of acetaldehyde in 1835 when he
prepared a pure sample by oxidizing ethyl alcohol with chromic acid [2]. Liebig named the
compound “aldehyde” from the Latin words translated as al (cohol) dehyd (rogenated).
Kutscherow observed the formation of acetaldehyde by the addition of water to acetylene in
1881[3]. Commercial processes for the production of acetaldehyde include: the oxidation or
dehydrogenation of ethanol, the addition of water to acetylene, partial oxidation of hydrocarbons,
and the direct oxidation of ethylene. It is estimated that in 1976, 29 companies with more than
82% of the world’s 2.3 megaton per year plant capacity use the Wacker Hoechst processes for the
direct oxidation of ethylene[5].
11
Chapter 4
4
Physical Properties and Uses
4.1 Properties
Chemical name: Acetaldehyde
Othername:Ethanal
Chemical formula CH3CHO
Molecular mass 44.05 g/mol
Appearance: Colourless liquid and Etherial odour
Structure:
12
Physical properties
Acetaldehyde is a colorless, mobile liquid having a pungent suffocating odour that is somewhat
fruity and pleasant in dilute concentrations. Some physical properties of acetaldehyde are given in
Table 4.1.
The freezing points of aqueous solutions of acetaldehyde are as follows: 4.8 wt %, -2.50C; 13.5
wt %, - 7.80 C, and 31.0 wt %, - 23.00 C. Acetaldehyde is miscible in all proportions with water
and most common organic solvents: acetone, benzene, ethyl alcohol, ethyl ether, gasoline,
paraldehyde, toluene, xylenes, turpentine, and acetic acid.
Table 4-1 Physical Properties of Acetaldehyde[5]
Properties
Values
Formula weight g/mol
44.053
Melting point °C
-123.5
Boiling point at 101.3 kPa (1 atm), °C
20.16
Specific Density
0.8045
Coefficient of expansion per °C (0-30 °C)
0.00169
Refractive index
1.33113
Vapor density (air = 1 )
1.52
Absolute viscosity at 15 °CmPa.s b
0.02456
Specific heat at 0 °C,J/(g.K)
2.18
At 25 °C
1.41
α = Cp / Cv at 30 °C and 101.3 kPa
1.145
13
Latent heat of fusion, kJ/mol
3.24
Latent heat of vaporization, kJ/mol
25.71
Heat of solution in water at 0 °C, kJ/mol
- 8.20
At 25 °C
- 6.82
Heat of combustion of liquid at constant
11867.9
pressure, kJ/mol
Heat of formation at 273 K, kJ/mol
-165.48
Free energy of formation at 273 K, kJ/mol
-136.40
Critical temperature, °C
181.5
Critical pressure, MΡa (atm)
6.40(63.2)
Dipole moment, C-m (debyes )
9.04 x 10 – 30
(2.69)
Ionization potential, Ev
10.50
Dissociation constant at 0 °C, K
0.7 x 10 – 14
Flash point, closed cup, °C
-38
Ignition temperature in air, °C
165
14
Chemical Properties
Acetaldehyde is a highly reactive compound exhibiting the general reactions of aldehydes;
under suitable conditions, the oxygen or any hydrogen can be replaced. Acetaldehyde
undergoes numerous condensation, addition, and polymerization reactions.
Decomposition Reaction of Acetaldehyde
Acetaldehyde decomposes at temperatures above 400°C, forming principally methane and
carbon monoxide. The activation energy of the pyrolysis reaction is 97.7 kJ/mol
(408.8kcal/mol).
The Hydrate and Enol Form of Acetaldehyde
In aqueous solutions, acetaldehyde exists in equilibrium with the hydrate, CH3CH(OH)2. The
degree of hydration can be computed from an equation derived by Bell and Clunie. The mean
heat of hydration is –21.34 kJ/mol (89.29kcal/mol); hydration has been attributed to hyper
conjugation. The enol form, vinyl alcohol (CH2 = CHOH) exists in equilibrium with
acetaldehyde to the extent of approximately one molecule per 30,000. Acetaldehyde enol has
been acetylated with ketene to form vinyl acetate.
Reduction Reaction of Acetaldehyde
Acetaldehyde is readily reduced to ethanol. Suitable catalysts for vapor-phase hydrogenation
are supported nickel and copper oxide.
15
Oxidation Reactions of Acetaldehyde
Acetaldehyde is readily oxidized with oxygen or air to acetic acid, acetic anhydride, and
peracetic acid. The principal product isolated depends on reaction conditions. Acetic acid is
produced commercially by the liquid – phase oxidation of acetaldehyde at 65°C with cobalt or
manganese acetate dissolved in acetic acid as a catalyst. Liquid – phase oxidation of
acetaldehyde in the presence of mixed acetates of copper and cobalt yields acetic anhydride.
Peroxyacetic acid or a perester is believed to be the precursor of acetic acid and acetic
anhydride. There are two commercial processes for the production of peracetic acid. Low
temperature oxidation of acetaldehyde in the presence of metal salts, ultraviolet irradiation, or
ozone yields acetaldehyde monoperacetate, which can be decomposed to peracetic acid and
acetaldehyde. Peracetic acid can also be formed directly by liquid phase oxidation at 5 - 50°C
with a cobalt salt catalyst. The nitric acid oxidation of acetaldehyde yields glyoxal. Oxidations
of para xylene to terephthalic acid and of ethanol to acetic acid are activated by acetaldehyde.
Polymerization Reactions of Acetaldehyde
Paraldehyde, 2, 4, 6- trimethyl–1,3,5–trioxan, a cyclic trimer of acetaldehyde is formed when a
mineral acid, such as sulfuric, phosphoric, or hydrochloric acid, is added to acetaldehyde.
Paraldehyde can also be formed continuously by feeding acetaldehyde as a liquid at 15 - 20°C
over an acid ion – exchange resin. Depolymerization of paraldehyde occurs in the presence of
acid catalysts.
Metaldehyde, a cyclic tetramer of acetaldehyde, is formed at temperatures below 0°C in the
presence of dry hydrogen chloride or pyridine hydrogen bromide. The metaldehyde crystallizes
from solution and is separated from the paraldehyde by filtration. Metaldehyde melts in a
sealed tube at 246.2°C and sublimes at 115 °C with partial depolymerization.
More recently, it has been shown that white, nontacky, and highly elastic polymer can be
formed by cationic polymerization with BF3 in liquid ethylene. At temperatures below
-75°C with anionic initiators, such as metal alkyls in a hydrocarbon solvent, a crystalline,
isotactic polymer is obtained.
16
This polymer also has an acetal structure [poly (oxymethylene) structure]. Molecular weights
in the range of 800,000 – 3,000,000 have been reported.
Reactions with aldehydes and ketones
The base catalyzed condensation of acetaldehyde leads to the dimmer, acetaldol, which can be
hydrogenated to form 1,3 butandiol or dehydrated to form crotonaldehyde. Crotonaldehyde can
also be made directly by the vapor-phase condensation of acetaldehyde over a catalyst.
Crotonaldehyde was formerly an important intermediate in the production of butyraldehyde,
butanol, and 2-ethylhexanol.
Acetaldehyde forms aldols with other carbonyl compounds containing active hydrogen atoms.
Kinetic studies of the aldol condensation of acetaldehyde and deuterated acetaldehydes have
shown that only the hydrogen atoms bound to the carbon adjacent to the –CHO group takes
part in the condensation reactions and hydrogen exchange. A hexyl alcohol, 2-ethyl-1 butanol,
is produced, industrially by the condensation of acetaldehyde and butaraldehyde in dilute
caustic solution followed by hydrogenation of the acrolein intermediate. condensation of
acetaldehyde in the presence of dimethylamine hydrochloride yields polyenals which can be
hydrogenated to a mixture of alcohols containing from 4 to 22 carbon atoms.
The base catalyzed reaction of acetaldehyde with excess formaldehyde is the commercial route
to pentaerythritol. The aldol condensation of three moles of form aldehyde with one mole of
acetaldehyde is followed by a crossed cannizzaro reaction between pentaerythrose, the
intermediate product, and formaldehyde to give pentaerythritol. The process proceeds to
completion without isolation of the intermediate. Pantaerythrose has been made by condensing
acetaldehyde and formaldehyde at 450 C using magnesium oxide as a catalyst. The vapor-phase
reaction of acetaldehyde and formaldehyde at 450C over a catalyst composed of lanthanum
oxide on silica gel gives acrolein. Ethyl acetate is produced commercially by the Tischenko
condensation of acetaldehyde with an aluminum ethoxide catalyst. The Tischenko reaction of
acetaldehyde with isobutyraldehyde yields a mixture of ethyl acetate, isobutyl acetate, and
isobutyl isobutyrate.
17
Reactions with Ammonia and Amines
Acetaldehyde readily adds ammonia to form acetaldehyde ammonia. Diethyl amine is obtained
when acetaldehyde is added to a saturated aqueous or alcoholic solution of ammonia and the
mixture is heated to 50-750C in the presence of a nickel catalyst and hydrogen at 1.2 MPa
(12atm). Pyridine and pyridine derivates are made from paraldehyde and aqueous ammonia in
the presence of a catalyst at elevated temperatures; acetaldehyde may also be used by the yields
of pyridine are generally lower than when paraldehyde is the staring material.
Reactions with Alcohols and Phenols
Alcohols add readily to acetaldehyde in the presence of a trace of mineral acid to form acetals;
eg, ethanol and acetaldehyde form diethyl acetal. Similarly, cyclic acetals are formed by the
reactions with glycols and other polyhydroxy compounds; eg, the reaction of ethylene glycol
and acetaldehyde gives 2 – methyl – 1,3 – dioxolane.
Mercaptals, CH3CH(SR)2, are formed in a like manner by the addition of mercaptans. The
formation of acetals by a noncatalytic vapor – phase reactions of acetaldehyde and various
alcohols at 3500C has been reported. Butadiene can be made by the reaction of acetaldehyde
and ethyl alcohol at temperature s above 3000C over a tantala – silica catalyst. Aldol and
crotonaldehyde are believed to be intermediates. Butyl acetate has been prepared by the
catalytic reaction of acetaldehyde with butanol at 3000C.
Reactions with Halogens and Halogen compounds
Halogens readily replace the hydrogen atoms of the methyl group. eg, chlorine reacts with
acetaldehyde or paraldehyde at room temperature to give chloroacetaldehyde; increasing the
temperature to 700-800C gives dichloroacetaldehyde; and at a temperature of 80-900C chloral
is formed.
18
4.2 Uses
The manufacturers use about 95% of the acetaldehyde produced internally as an intermediate
for the production of other organic chemicals. Acetic acid and acetic anhydride are the
derivatives of acetaldehyde followed by n-butanol and 2-ethylhexanol. Twenty percent of the
acetaldehyde is consumed in variety of other products, the most important being
pentaerythritol, trimethylolpropane, pyridines, peraceticacid, crotonaldehyde, chloral, lactic
acid. Acetaldehyde is used in the production of perfumes, polyester resins, and basic
dyes. Acetaldehyde is also used as a fruit and fish preservative, as a flavoring agent, and as a
denaturant for alcohol, in fuel compositions, for hardening gelatin, and as a solvent in the
rubber, tanning, and paper industries.
19
Chapter 5
5
Different Manufacturing Processes
The economics of the various processes for the manufacture of acetaldehyde are strongly
dependent on the price of the feedstock used.
5.1 Oxidation of Ethylene [1]
Reactions:
C2H4+PdCl2 + H2O
CH3CHO +Pd +2HCl
Pd + 2CuCl2
PdCl2 +2CuCl
2CuCl+1/2 O2 + 2HCl
2CuCl2 + H2O
The catalyst is an aqueous solution of PdCl2 and CuCl2. In 1894, F.C. Phillips observed the
reaction of ethylene with an aqueous palladium chloride solution to form acetaldehyde. The net
result is a process in which ethylene is oxidized continuously through a series of oxidation –
reduction reactions.
Overall Reaction:
C2H4 + ½ O2
CH3CHO
There are two processes for the production of acetaldehyde by the oxidation of Ethylene.
20
5.1.1 Two stage process developed by Wacker Chemie[5]
In the two – stage process ethylene and oxygen (air) react in the liquid phase in two stages. In
the first stage ethylene is almost completely converted to acetaldehyde in one pass in a tubular
plug-flow reactor made of titanium. The reaction is conducted at 125-1300C and 1.13 Mpa (150
psig) palladium and cupric chloride catalysts. Acetaldehyde produced in the first reactor is
removed from the reaction loop by adiabatic flashing in a tower. The flash step also removes
the heat of reaction. The catalyst solution is recycled from the flash – tower base to the second
stage (or oxidation) reactor where the cuprous salt is oxidized to the cupric state with air. The
high pressure off – gas from the oxidation reactor, mostly nitrogen, is separated from the liquid
– catalyst solution and scrubbed to remove acetaldehyde before venting. The flasher overhead
is fed to a distillation system where water is removed for recycle to the reactor system and
organic impurities, including chlorinated aldehydes, are separated from the purified
acetaldehyde product.
5.1.2 One stage process developed by Farbwerke Hoechst[5]
In the one-stage process ethylene, oxygen, and recycle gas are directed to a vertical reactor for
contact with the catalyst solution under slight pressure. The water evaporated during the
reaction absorbs the heat evolved, and make – up water is fed as necessary to maintain the
catalytic solution concentration. The gases are water –scrubbed and the resulting acetaldehyde
solution is fed to a distillation column and thus acetaldehyde is obtained.
21
5.2 By Oxidation of ethyl alcohol [1,2]
Reaction:
CH3CH2OH + ½ O2
CH3CHO + H2O, ΔH = 242 kJ/mol (57.84 kcal / mol)
Temperature: 4800C
Catalyst: silver
Passing alcohol vapours and preheated air over a silver catalyst at 4800C carries out the
oxidation. With a multi tubular reactor, conversions of 74-82% per pass can be obtained while
generating steam which can be used elsewhere in the process.
5.3 By Dehydrogenation of ethyl alcohol[1]
Reaction:
C2H5OH → CH3CHO + H2
Catalyst: Cu -Co-Cr2O3
Temperature: 280 – 350 0C.
Process description
The raw material i.e., ethanol is vaporized and the vapours so generated are heated in a heat
exchanger to the reaction temperature by hot product stream. The product stream is cooled to –
100C and in doing it; all unreacted ethanol and acetaldehyde are condensed. The outgoing
gaseous stream, containing hydrogen mainly, is scrubbed with dilute alcohol (alcohol + water)
to remove uncondensed products and the undissolved gas. The remaining pure hydrogen (98%)
is burnt in stack. Ethanol is vaporized in vaporizer and heated to the reactor temperature in heat
exchanger. The heated vapours are passed through the converter. The product stream is first
cooled in heat exchanger and then in condensers using water and liquid ammonia. This
condenses most of the unreacted ethanol and the acetaldehyde formed in reactor.
22
The escaping gas, which is almost pure hydrogen, is scrubbed by ethanol to remove all the
traces of the product. The liquid stream consisting of mainly ethanol and acetaldehyde is
distilled in distillation column to get acetaldehyde.
5.4 Preparation method of acetaldehyde from acetylene [2,3]
Reaction:
H2C = CH2 + H2O
CH3CHO
Temperature: 70-900C
Pressure: 103.4 KPa
Process description
Fresh catalyst is fed to the reactor periodically; the catalyst may be added in the mercurous
form. The excess acetylene sweeps out the dissolved acetaldehyde which is condensed by water
and refrigerated brine and scrubbed with water; the crude acetaldehyde is purified by
distillation and the unreacted acetylene is recycled. The catalytic mercuric ion is reduced to
catalytically inactive mercurous sulfate and metallic mercury; this sludge, consisting of reduced
catalyst and tars, is drained from the reactor at intervals and resulfated. Adding ferric or other
salts to the reaction solution can reduce the rate of catalyst depletion. In one variation,
acetylene is completely hydrated with water in a single operation at 68-730C using the
mercuric iron salt catalyst. The acetaldehyde is partially removed by vacuum distillation and
the mother liquor recycled to the reactor. The aldehyde vapors are cooled to about 350C,
compressed to 253 KPa and condensed. It is claimed that this combination of vacuum and
pressure operations substantially reduces heating and refrigeration costs.
23
5.5 From synthesis Gas[2]
Reaction:
CO + H2 → CH3CHO + other products
Catalyst: Rhodium
Temperature: 3000C
The process is a single step comprises passing synthesis gas over 5% rhodium on SiO2 at
3000C and 2.0 Mpa (20 atm). The principal co products are acetaldehyde, 24% acetic acid,
20%; and ethanol, 16%.
5.6 From acetic acid[4]
Reaction:
CH3 COOH + H2 → CH3 CHO + H2 O ΔG300° C. = +0.8 kcal/mole
Catalyst: Iron oxide containing 2.5-90 wt% Pd
Temperature: 250-400 0C
Procedure:
Hydrogen and acetic acid are fed to a reactor in hydrogen to acetic acid ratio of 2:1 to 25:1. The
hydrogenation of acetic acid produces a partially gaseous product, and acetaldehyde is
absorbed from the partially gaseous product with a solvent containing acetic acid. The gas
remaining after the absorption step contains hydrogen, and this gas is recycled for the
hydrogenation of acetic acid. The absorbed acetaldehyde is distilled. After acetaldehyde is
isolated from unreacted acetic acid and the other products via distillation, the unreacted acetic
acid is separated from the other products using azeotropic distillation. Water is contained in the
other products, and the azeotrope is an azeotrope of ethyl acetate and water. The unreacted
24
acetic acid is separated in a column, and the column is controlled to contain an ethyl acetate
rich azeotrope of ethyl acetate and water.
5.7 From Saturated Hydrocarbons:
Acetaldehyde is formed as a co product in the vapour – phase oxidation of saturated
hydrocarbons, such as butane or mixtures containing butane, with air or in higher yield oxygen.
Oxidation of butane yields acetaldehyde, formaldehyde, methanol, acetone, and mixed solvents
as major products; other aldehydes, alcohols, ketones, glycols, acetals, epoxides, and organic
acids are formed in smaller concentrations. It has almost no chance to be used as a major
process.
25
Chapter 6
6
Selection of Process with Justification
Here, ethyl alcohol dehydrogenation is selected for the production of acetaldehyde. Because, in
this process, hydrogen is taken out as a by-product which can be used elsewhere or which can
be used to generate heat. In dehydrogenation process more conversion-taking place compared
to other processes. The dehydrogenation catalyst has a life of several years but requires
periodic reactivation. In dehydrogenation process, number of products is less, so separation of
acetaldehyde from other product is not a difficult problem.
26
Chapter 7
7
Description of the Process
Production of Acetaldehyde by Dehydrogenation of Ethyl alcohol
Reaction:
C2H5OH → CH3CHO + H2
Catalyst
: Cu -Co-Cr2O3
Temperature: 280 – 350 0C.
Process description:
The raw material i.e., ethanol is vaporized and the vapours so generated are heated in a
heat exchanger to the reaction temperature by hot product stream. The product stream is
cooled to –100C and in doing it; all unreacted ethanol and acetaldehyde are condensed.
The outgoing gaseous stream, containing hydrogen mainly, is scrubbed with dilute
alcohol (alcohol + water) to remove uncondensed products and the undissolved gas. The
remaining pure hydrogen (98%) is burnt in stack. Ethanol is vaporized in vaporizer and
heated to the reactor temperature in heat exchanger. The heated vapours are passed
through the converter. The product stream is first cooled in heat exchanger and then in
condensers using water and liquid ammonia. This condenses most of the unreacted
ethanol and the acetaldehyde formed in reactor. The escaping gas, which is almost pure
hydrogen, is scrubbed by ethanol to remove all the traces of the product. The liquid
stream consisting of mainly ethanol and acetaldehyde is distilled in distillation column to
get acetaldehyde.
27
Figure 7.1 Flowsheet
28
Chapter 8
8
Material and Energy Balance
8.1 Reaction
C2H5OH → CH3CHO + H2
Catalyst
: Cu -Co-Cr2O3
Temperature: 280 – 3500 C.
Process
The raw material i.e., ethanol is vaporized and the vapors, so generated, are heated in a
heat exchanger to the reaction temperature by hot product stream. The product stream is
cooled to –100 C and in doing it, all unreacted ethanol and acetaldehyde are condensed.
The outgoing gaseous stream, containing hydrogen mainly, is scrubbed with dilute
alcohol (alcohol + water) to remove uncondensed products and the undissolved gas. The
remaining pure hydrogen (98%) is burnt in stack.
The material and energy balance in a plant design is necessary because this fixes the
relative flow rates of different flow streams and temperatures in the flow sheet.
29
Notations used
Msteam = Mass flow rate of steam.
∆Hsteam = enthalpy of steam.
E = Mass flow rate of ethanol.
A = Mass flow rate of acetaldehyde.
H = Mass flow rate of hydrogen.
Cp = specific heat capacity.
λ = Latent heat of vaporization.
Assumptions :
Let us assume an annual production of acetaldehyde 54000 tons per year.
It is assumed that the plant works for 300 days in a year.
Amount of acetaldehyde produced per day =180 tons per day
Basis: One hour of operation.
Amount of acetaldehyde to be produced = 180 TPD = 7500kg/h.
Molecular weight of ethanol = 46 kg/kmol.
Molecular weight of acetaldehyde = 44 kg/kmol.
Molecular weight of hydrogen = 2 kg/kmol.
Therefore, amount of acetaldehyde to be produced = 170.45kmol/h.
Let conversion be 90%.
Taking into account the losses let, the acetaldehyde produced to be some extra.
Let acetaldehyde to be produced = 8000 kg/h.
Amount of ethanol required for 100% conversion = 8363.63 kg/h.
Therefore, ethanol required for 90% conversion = 9292.92 kg/h.
Temperature dependency of Cp with temperature is given as
30
𝑇
∫
𝑇0
𝐶𝑝
𝑇 2 − 𝑇02
𝑇 3 − 𝑇03
𝑇 4 − 𝑇04
𝑇 5 − 𝑇05
𝑑𝑇 = 𝐴(𝑇 − 𝑇0 ) + 𝐵
+𝐶
+𝐷
+𝐸
𝑅
2
3
4
5
8-1 Values of constants for calculating Cp[22]
Component
A
B*103
C*106
D*10-5
Ethanol
3.518
20.001
-6.002
0
Hydrogen
3.249
0.422
0
0.083
Acetaldehyde
1.693
17.978
-6.158
0
Water
3.470
1.450
0
0.121
8-2 Values of CP for different temperature range (kcal/Kg.C)
Component
-25
40 oC
to 25 to
78.4 oC
78.4
to 100
100 oC
to 200
200 oC
to 40
310 oC
232.36
to 310
o
C
o
C
Ethanol
0.616
0.587
0.4382
0.471
0.539
0.5415
0.549
Acetaldehyde 0.347
-----
------
-----
------
0.417
0.528
Hydrogen
3.399
-----
-----
-----
------
2.485
2.485
Water
---
------
-----
-----
------
------
----
31
8.2 Vaporizer
As shown in the figure,
Ethanol liquid inlet temperature = Ti=25oC.
Ethanol leaves as superheated steam at 1000C = To
Heating fluid is assumed to be saturated steam here and to provide sufficient temperature
gradient; it is taken to be at about 3 atmospheric pressure. At this pressure it condenses at
133.890 C and because process streams are normally available at this pressure.
Condensing temperature of water = 133.890 C.
From steam table enthalpy of steam at this temperature = ∆Hsteam =514.9kcal/kg. [6]
Boiling point of ethanol = Tb = 78.40 C. [6]
Latent heat of vaporization of ethanol= λ ethanol = 200.6 kcal/kg [6]
From heat balance we have,
Msteam = E*[Cpi*(Tb-Ti) + λ ethanol + Cpo*(To-Tb)]
Msteam = 9292.92* [0.5876 * (78.4 – 25) + 200.6 + 0.4382 * (100 – 78.4)]/514.9
= 4357.56 kg.
32
8-3 Mass Balance Table for Vaporizer
Component
Input Feed Stream
Output Feed Stream
(Kg)
(Kg)
Ethanol
9292.92
9292.92
Water
4357.56
4357.56
Total
13650.48
13650.48
For Water
𝑇
Heat in = (𝑚 ∫𝑇 𝐷 𝐶𝑃 𝑑𝑇)
𝑜
𝑊𝑎𝑡𝑒𝑟
+ m ∗ λ𝑤𝑎𝑡𝑒𝑟 = 2714324.124kcal
Where To=298.15 K and TD=406.15K
406.15
Heat out=(𝑚 ∫298.15 𝐶𝑃 𝑑𝑇)
𝑊𝑎𝑡𝑒𝑟
=2714324.124 kcal
For ethanol (Tref= 298.15 K)
Heat in= 0 kcal
Heat out=470616.48kcal
8-4 Energy Balance Table for Vaporizer
Component
Stream In kcal
Stream Out kcal
Ethanol
0
2243710.112
Water
2714324.124
470616.48
Total
2714324.124
2714326.592
8.3 Reactor
33
The reaction in the reactor:
C2H5OH → CH3CHO + H2
Optimum reaction temperature = 3100C.
Conversion = 90%.
From material balance we have,
Amount of acetaldehyde produced = .90 * 44* 9292.92 / 46 =8000 kg.
=181.81kmol
Amount of hydrogen produced = .90 * 2* 9292.92 / 46 = 363.63 kg.
Amount of ethanol unreacted = 9292.92 – (8000 + 363.63)
= 929.28kg =20.20 kmol
8-5 Mass Balance Table for Reactor
Component
Input Feed stream (Kg)
Output feed stream (kg)
Ethanol
9292.92
929.28
Acetaldehyde
0
8000
Hydrogen
0
363.64
Dowtherm
32447.6042
32447.6042
Total
41740.5252
41740.5252
It is decided to use saturated vapours of dowtherm, at 3200 C, for cooling purposes
λ dowtherm = 56.5 kcal / kg.
Heat of reaction = ∆Hrxn = 332.64 kcal / kg. [6]
Assuming ethanol vapors enter the reactor at 2000 C.
From heat balance we can found amount of dowtherm required = Md.
Specific heat capacity of ethanol = 0.549 kcal / kg 0C.
Energy in by ethanol =929.28*[0.5876*(98.2-25)+200.6+0.549*(310-98.2)]
=334439.0624 kcal/kg
Energy in by hydrogen=363.63*2.485*(310-25)=257531.8568 kcal/kg
34
Energy in by acetaldehyde
=8000*0.528*(310-25)
=1203840 kcal/kg
Energy required for the reaction
=0.90*9292.92*332.64
=2782077.218 kcal/kg
Q= Amount of heat supplied by dowtherm
Q={Energy out from the reactor+ Heat of reaction }-{Energy entering into the reactor}
= 1833289.639 kcal
Md *λdowtherm=Q
Md=1833289.639 /56.5=32447.6042Kg
8-6 Energy Balance Table for Reactor
Component
Stream In (kcal)
Stream Out(kcal)
Ethanol
2744598.5
334439.0624
Hydrogen
0
257531.8568
Acetaldehyde
0
1203840
Heat Exchanger
This is used only for heat recovery. Since it is assumed vapor is heated up to 200 0 C by
the product stream of the reactor at 3100 C.
Let outlet temperature = T0 C.
Cp of ethanol at 3100 C = 0.549 kcal / kg 0C.
35
Cp of acetaldehyde at 3100 C = 0.528 kcal / kg 0C.
Cp of hydrogen at 3100 C = 2.485 kcal / kg 0C.
8-7 Mass Balance Table for Heat Exchanger
Component
Ethanol
Acetaldehyde
Water
Total
Shell in (kg)
929.28
8000
363.63
9292.91
Tube in (kg)
9292.92
9292.92
Shell out (Kg)
929.28
8000
363.63
9292.91
Tube out (kg)
9292.92
9292.92
From heat balance we can find the outlet temperature.
E*Cp,ethanol*(200–100)=E*Cp,ethanol*(310–T) +H*Cp,hydrogen*(310–T)+ A*Cp,acet*(310-T)
9292.92*0.471*(200-100)=929.92*0.549*(310-T)+363.63*2.485*(310T)+8000*0.528*(310-T)
Therefore, T = 232.360 C.
8-8 Energy Balance Table for Heat Exchanger
Component
Shell side(kcal)
Tube side(kcal)
Ethanol
437696
39637.24
Hydrogen
-
70157.099
Acetaldehyde
-
327951.36
Total
437696
437745.699
8.4 Condenser C1
In condenser 1 it is decided to use cooling water at 300 C. the outlet temperature of
cooling water is not allowed to go above 500C, because above this temperature, there is a
problem of vaporization. Normally the approach temp difference is about 100C. Since the
product can at best be cooled to 400C, at this temperature the product stream would be a
two-phase mixture and the mixture composition can be found out from VLE data.
We make an approximate that; the information given at 699 mmHg is taken.
36
At 4000C, ethanol in vapor phase = 4.1 mol%.[6]
Ethanol in liquid phase = 55 mol%.[6]
Let, ml = moles of liquid consisting of ethanol and acetaldehyde.
mv = moles of vapor consisting of ethanol and acetaldehyde.
Therefore from mole balance we have
0.55 * ml + 0.041 * mv = 20.20
0.45 * ml + 0.959 * mv = 181.81
On solving above two equations we get, ml = 23.413 kmol.
mv = 178.596 kmol.
Vapor phase composition,
Acetaldehyde = 171.27 kmol = 7536.03 kg.
Ethanol = 7.32 kmol = 336.83 kg.
Liquid phase composition,
Acetaldehyde = 10.535 kmol = 463.57 kg.
Ethanol = 12.977 kmol = 592.348 kg
Heat Balance
At 232.690C,
Cp, hydrogen = 2.485 kcal / kg 0C.
Cp, acetaldehyde = 0.417 kcal / kg 0C.
Cp, ethanol = 0.5415 kcal / kg 0C.
λ acetaldehyde = 139.5 kcal / kg.
λ ethanol = 200.6 kcal / kg.
Heat given out by hydrogen = 363.636 * 2.485 * (232.69 – 40) = 173.823 * 103 kcal.
Heat given out by acetaldehyde = 8000* 0.417 * (232.69 – 40) + 463.57 * 139.5
= 706.38 * 103 kcal.
Heat given out by ethanol = 929.92* 0.5415 * (232.69 – 40) + 592.348 *200.6
= 215.621 * 103 kcal.
Total heat given out = 1095.82 * 103 kcal.
Let, Mw = mass flow rate of cooling water
Cp of water = 1 kcal / kg 0C.
37
Therefore, Mw = 1095.8 * 103 / (1 * (5030)). = 54.79 * 103 kg.
8-9 Mass Balance Table for Condenser 1
Component
In (Kg)
Vapor
Phase
Out (Kg)
Liquid phase out (Kg)
Ethanol
929.28
336.83
592.348
Acetaldehyde
8000
7536.03
463.57
Hydrogen
363.64
363.636
--------
Water
54791.29
----------
54791.29
Total
64084.21
64083.704
8-10 Energy Balance for Condenser 1
Component
Tube side (kcal)
Shell side (kcal)
Ethanol
215621
------
Hydrogen
173823
------
Acetaldehyde
706380
------
Water
--------
1095800
Total
1095824
1095800
Condenser C2
In condenser c2, it is desired to condense all ethanol and acetaldehyde. If the working
38
pressure is 1 atm. From the equilibrium data it is seen that for temperatures below 300C,
there is going to be no ethanol in vapor phase and acetaldehyde would exert its vapor
pressure at that temperature. If it is desired to achieve about 97% recovery of
acetaldehyde, the outlet temperature of the product stream should be about –250C.This is
because at –22.60C, its vapor pressure is 100 mmHg and the vapor phase will consists of
13.15 mol%. In view of this, the cooling fluid chosen is saturated NH3 at 1 atm. At which
it boils at –33.60C[6].
Heat balance:
At 400C,
Cp, hydrogen = 3.399 kcal / kg 0C.
Cp, acetaldehyde = 0.347 kcal / kg 0C.
Cp, ethanol = 0.616 kcal / kg 0C.
λ acetaldehyde = 139.5 kcal / kg.
λ ethanol = 200.6 kcal / kg.
λ ammonia = 590 kcal / kg.
Heat given out by hydrogen = 363.64 * 3.399 * (40 + 25) = 80.339 * 103 kcal.
Heat given out by acetaldehyde = 7536.03* 0.347 * (40+ 25) + 7309.95 * 139.5
= 1189.182 * 103 kcal.
Heat given out by ethanol = 336.83* 0.616 * (40+25) + 336.83 * 200.6
39
= 81.054 * 103 kcal.
Total heat given out = 1350.575 * 103 kcal.
Let Mammonia = mass flow rate of ammonia.
Therefore, Mammonia = 1350.575 * 103 / 590.= 2289.11 kg.
8-11 Mass Balance Table for Condenser 2
Component
Feed In (Kg)
Vapor Phase Out (Kg)
Liquid Phase out (Kg)
Ethanol
336.83
--------
336.83
Acetaldehyde
7536.03
226.10
7309.95
Hydrogen
363.64
363.64
----------
NH3
2289.11
2289.11
----------
Total
10525.61
10525.63
8-12 Energy Balance for Condenser 2
Component
Tube side (kcal)
Shell side (kcal)
Ethanol
81.054*103
------
Hydrogen
80.339*103
------
Acetaldehyde
1189.182*103
------
Ammonia
--------
1350574.9
Total
1350.575*103
1350.5749*103
Preheater
The preheater to the distillation column is necessary because the feed plate will be
completely chilled if the feed is not heated. The water stream from condenser c1 is
40
available at 500C and is used in the preheater. If the maximum approach temperature
difference is 100C, the product stream can at best be heated to 400C. the distillation
column pressure is chosen to be 1158 mmHg so that pure acetaldehyde is obtained as
liquid product at 400C. In view of this, the stream coming out of the preheater is liquid.
Heat balance
At 40 oC
Cp, acetaldehyde = 0.347 kcal / kg 0C.
Cp, ethanol = 0.616 kcal / kg 0C.
Let, Mw = mass flow rate of cooling water.
Mw * (50-30) = 929.178 * (40 + 14.54872)* 0.616 + 7773.52*(40+14.54872)*0.347
Therefore, Mw = 8918.1296 kg
8-13 Mass Balance for Preheater
Component
Input (Kg)
Out(Kg)
Ethanol
929.178
929.178
Acetaldehyde
7773.52
7773.52
Water
10626.80
10626.80
Total
19329.498
19329.498
8-14 Energy Balance Table for Preheater
component
Tube side (kcal)
Shell side (kcal)
Ethanol
31222.249
0
Acetaldehyde
147140.3414
0
Water
0
178362.59
Total
178362.5904
178362.59
8.7 Distillation column
In distillation column acetaldehyde condenses at 400C. since vapor pressure data’s of
pure gas is not available, it is estimated using Antoine’s equation.
41
ln P = A + B/T
Where, A and B are constants, they can be determined from boiling point data
at, Pressures 760 mmHg and 400 mmHg.
At 760 mmHg T = 20.20C= 293.20K. [6]
400 mmHg T = 4.90C = 277.90K. [6]
Therefore, ln 760 = A + B/293.2
ln 760 = A + B/277.9
On solving above two equations we get,
A = 18.29 and B = -3418.2[6]
Therefore, ln P = 18.29 – 3418.2/T
Therefore at 400C, P = 1586.41 mmHg.
Assume 99% acetaldehyde recovery in overhead product and ethanol recovery
as 10%.
In overhead:
Acetaldehyde = 7695.78 kg.
Ethanol = 92.917 kg.
Total D = 7788.69 kg.
In bottom:
Acetaldehyde = 77.73 kg. Ethanol = 836.26 kg. Total W = 913.99 kg.
42
F=D+W
xd = .989
F * xf = D * xd + W * x w
Therefore, xw = F * xf – D * xd
W
Therefore, xw = .0815
Assume reflux ratio = 0.3
Therefore L / D = 0.3
L = 0.3 * 7788.69 = 2336.60 kg.
Vapor going to the condenser = L + D = 2336.60 + 7788.69
=10125.30 kg.
Of this 98.9% is acetylaldehyde
Therefore, vapor composition going to the condenser:
Acetaldehyde = 10013.92 kg, Ethanol = 111.37 kg.
Heat load to condenser
Let L/D=0.3
L=2336.60 kg
Vapor going to condenser = L+D =10125.30 kg
Out of this 98.9% is acetaldehyde
Acetaldehyde=10013.92 kg
Ethanol= 111.37 kg.
Heat load to the condenser = Methanol*λethanol + Macetaldehyde*λacetaldehyde
=111.37*200.6+10013.92*139.5 =1419282.662 Kcal
Reboiler load
Let “m” be the amount of liquid vaporized.
Let “L” be liquid going into the reboiler.
Let L/W = 10
43
L= 9139.9 kg
Therefore, m= L-W
m=9139.9-913.99=8225.91 kg
we know xw = .0815, so
Acetaldehyde=8225.91/45.976*.0815=14.581 kg
Ethanol=8211.32 kg
Heat load to the reboiler = Methanol*λethanol + Macetaldehyde*λacetaldehyde
=8211.32*200.6+14.581*139.5
=1649226.492 kcal.
44
Chapter 9
9
DESIGN OF EQUIPMENT
9.1 DESIGN OF DISTILLATION COLUMN:
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.
XF = mole fraction of Acetaldehyde in liquid/Feed.
YD = mole fraction of Acetaldehyde in Distillate.
XW = mole fraction of Acetaldehyde in Residue.
MF = Average Molecular weight of Feed, kg/kmol
MD = Average Molecular weight of Distillate, kg/kmol
MW = Average Molecular weight of Residue, kg/kmol
Rm = Minimum Reflux ratio
R = Actual Reflux ratio
L = Molar flow rate of Liquid in the Enriching Section, kmol/hr.
G = Molar flow rate of Vapor in the Enriching Section, kmol/hr.
L = Molar flow rate of Liquid in Stripping Section, kmol/hr.
G = Molar flow rate of Vapor in Stripping Section, kmol/hr.
q = Thermal condition of Feed
ρL = Density of Liquid, kg/m3.
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.
45
TL = Temperature of Liquid, 0K.
TV = Temperature of Vapor, 0K
T – x- y data:
9-1 T-x-y data for ethanol-acetaldehyde[6]
T 0C
98.5
89.9
80
71
60.5
50
39
X
0.000
0.069
0.164
0.286
0.445
0.664
1.000
Y
0.000
0.317
0.578
0.761
0.879
0.954
1.000
Preliminary calculations:
F = 196.867 kmol/hr,
XF = 0.897,
MF = 44.123 kg/kmol.
D = 177.574 kmol/hr,
XD = 0.989,
MD = 44.022 kg/kmol.
W = 19.293 kmol/hr,
XW = 0.082,
MW = 45.976 kg/kmol.
Distillation column temperature = 40 0C.
Distillation column pressure = 2.08 atm. = 1586.41 mm Hg.
Basis: One-hour operation.
46
From graph
Figure 8: Calculating number of stages using Mc Cabe Thiele Method[9]
Number of stages = 5 (including the reboiler).
Reboiler is the last tray.
Feed tray = 3
Number of trays in Enriching Section = 3
Number of trays in Stripping Section = 2
Minimum Reflux ratio = ( xD-yF)/(yF-xF)=(.989-.976)/(.99-.897)=0.096643
Actual Reflux ratio
=1.5*.096643=.145
Now, we know that,
47
R = Lo/ D
=> Lo = R*D
i.e., Lo= .145*177.574
i.e., Lo =25.748 kmol/hr.
Therefore, Lo = 25.748 kmol/hr.
L= Liquid flow rate on the Top tray = 25.748 kmol/hr.
Since feed is Liquid, entering at bubble point,
q= (HV-HF) / (HV-HL) = 1
Now,
Slope of q-line = q/ (q-1)
= 1/ (1-1) = 1/0 = ∞
Now we know that,
(L̅ -L)/F = q = 1
(̅L - L) = F
L̅ = F + L
i.e., L̅ = 25.748 + 196.867
i.e., L̅ = 222.615 kmol/hr.
Therefore, liquid flow rate in the Stripping Section = 222.615 kmol/hr.
Also, we know that,
G̅ = [(q-1) ×F] + G
i.e., G̅ = [(1-1) ×F] + G
i.e., G̅ = [0×F] +G
i.e., G̅ = 0 +G
G̅ = G
Now, we know that,
G=L+D
i.e., G = Lo +D
i.e., G= 25.748 + 177.574
i.e., G= 203.322 kmol/hr.
Thus, the flow rate of Vapor in the Enriching Section = 203.322 kmol/hr.
Since G̅ =G
48
G̅ = G = 203.322 kmol/hr.
Therefore, the flow rate of Vapor in the Stripping Section = 203.322kmol/hr.
List of parameters used in calculation:
SECTION
ENRICHING SECTION
STRIPING SECTION
PROPERTY
TOP
BOTTOM
TOP
BOTTOM
X
.983
.719
.719
.102
Y
.999
.966
.966
.415
Liquid
L(kmol/hr)
25.748
25.748
222.615
222.615
Vapour
G(kmol/hr)
203.322
203.322
203.322
203.322
Tliquid 0C
41.93
51.56
51.56
85.70
Mavg Liquid(kg/kmol)
44.02
44.56
44.56
45.79
Mavg Vapour(kg/kmol)
44.01
44.07
44.07
44.17
Liquid,
L kg/hr.
1133.427
1147.331
9919.724
10193.541
Vapor,
G kg/hr
8948.201
8960.401
8960.401
8980.733
Density
ρ1(kg/m3)
784.69
774.50
774.50
747.87
Density
ρg(kg/m3)
3.4376
3.425
3.425
3.391
.008
.008
.073
.074
(L/G)( ρ1/ ρg).5
9-2 List of Parameters used in calculation
Design Specification[6]:
a) Design of Enriching Section:
Tray Hydraulics,
The design of a sieve plate tower is described below. The equations and correlations are
borrowed from the 6th and 7th editions of Perry’s Chemical Engineers’ Handbook.
1. Tray Spacing, (ts) :
Let ts = 18” = 457 mm. (range 0.15 – 1.0 m).
49
2. Hole Diameter, (dh):
Let dh = 5 mm. (range 2.5 – 12 mm).
3. Hole Pitch (lp):
Let lp = 3* dh (range 2.5 to 4.0 times dh).
i.e., lp = 3*5 = 15 mm.
4. Tray thickness (tT):
Let tT = 0.6* dh (range 0.4 to 0.7 times dh).
i.e., tT = 0.6*5 = 3 mm.
5. Ratio of hole area to perforated area (Ah/Ap):
Now, for a triangular pitch, we know that,
Ratio of hole area to perforated area (Ah/Ap) = ½ (π/4*dh2)/ [(√3/4) *lp2]
i.e., (Ah/Ap) = 0.90* (dh/lp)2
i.e., (Ah/Ap) = 0.90* (5/15)2
i.e., (Ah/Ap) = 0.1
Thus,
(Ah/Ap) = 0.1
6. Plate Diameter (Dc):
The plate diameter is calculated based on entrainment flooding considerations
L/G {ρg/ρl}0.5 = 0.008 ---------- (maximum value)
Now for,
L/G {ρg/ρl}0.5 = 0.008 and for a tray spacing of 500 mm.
We have,
From the flooding curve, ---------- (fig.18.10, page 18.7, 6th edition Perry.)
Flooding parameter, Csb, flood = 0.29 ft/s .
Now,
Unf = Csb, flood * (σ / 20) 0.2 [(ρl - ρg) / ρg]0.5
---- {eqn. 18.2, page 18.6, 6th edition Perry.}
Where,
Unf = gas velocity through the net area at flood, m/s (ft/s)
Csb, flood = capacity parameter, m/s (ft/s, as in fig.18.10)
σ = liquid surface tension, mN/m (dyne/cm.)
50
ρl = liquid density, kg/m3 (lb/ft3)
ρg = gas density, kg/m3 (lb/ft3)
Now, we have,
σ = 19.325 mN/m = 19.325 dyne/cm.
ρl = 774.50kg/m3.
ρg = 3.425 kg/m3.
Therefore,
Unf = 0.29*(19.325/20)0.2*[(774.50-3.4250)/ 3.4250]0.5
i.e.,Unf = 4.321 ft/s = 1.317 m/s.
Let,
Actual velocity, Un= 0.8*Unf
i.e., Un = 0.8∗1.317
i.e., Un = 1.054 m/s
It is desired to design with volumetric flow rate maximum (therefore, actual is less than
the maximum).
Volumetric flow rate of Vapor at the bottom of the Enriching Section
= qo = 8960.401/ (3600*3.4250) = 0.727 m3/s.
Now,
Net area available for gas flow (An)
Net area = (Column cross sectional area) - (Down comer area.)
An = Ac - Ad
Thus,
Net Active area, An = qo/ Un = 0.727/ 1.054 = 0.690 m2.
Let Lw / Dc = 0.77 (range 0.6 to 0.85 times Dc ).
Where, Lw = weir length, m
Dc = Column diameter, m
Now,
θc = 2*sin-1(Lw / Dc) = 2*sin-1 (0.77) = 100.70
Now,
Ac=(π /4)*Dc2= 0.785*Dc2 , m2
Ad = [(π/4) * Dc2 * (θc/3600)] - [(Lw/2) * (Dc/2) *cos (θc/2)]
i.e., Ad = [0.785*Dc2 *(100.70/3600)]-[(1/4)* (Lw / Dc) * Dc2 * cos(100.70)]
51
i.e., Ad = (0.2196* Dc2) - (0.1288* Dc2)
i.e., Ad = 0.0968*Dc2, m2
Since An = Ac -Ad
0.690 = (0.785*Dc2) - (0.0968* Dc2)
i.e., 0.6882* Dc2 = 0.690
⇒ Dc2 = 0.690/ 0.6882 = 1.003
⇒ Dc = √ 1.003
Dc = 1.001 m
Since Lw / Dc = 0.77,
⇒ Lw = 0.77* Dc = 0.77*1.001 = 0.771 m.
Therefore, Lw = 0.771 m.
Now,
Ac = 0.785*1.003 = 0.787 m2
Ad = 0.0968*Dc2 = 0.0968*1.003 = 0.097 m2
Aa = Ac –2* Ad
i.e., Aa = 0.787- 2*0.097 ⇒ Aa= 0.593 m2
7. Perforated plate area (Ap):
Now,
Lw / Dc = 0.771/ 1.001 = 0.7701
θc = 100.73 0
α = 180 0- θc
⇒ α = 79.27 0
Now,
Acz = 2* Lw* (thickness of distribution)
Where, Acz = area of calming zone, m2 (5 to 20% of Ac )
Acz = 2*0.771* (30×10-3) = 0.046 m2 -------- (which is 6.76% of Ac)
Also,
Awz= 2*{(π/4)*Dc2*(θc/3600)- (π/4)*(Dc-30*10-3)2*(θc/3600)}
Where, Awz = area of waste periphery, m2 (range 2 to 5% of Ac)
i.e., Awz = 2*{(π/4)*1.0012*(100.730/3600)- (π/4)*(1.001-30*10-3)2*(100.730/3600)}
i.e., Awz = 0.026 m2
--------- (which is 3.8% of Ac)
52
Now,
Ap = Ac - (2*Ad) - Acz - Awz
i.e., Ap = 0.787- (2*0.097) - 0.046 - 0.026
Thus, Ap = 0.521 m2.
8. Total Hole Area (Ah):
Since,
Ah / Ap = 0.1
⇒ Ah = 0.1* Ap
i.e., Ah = 0.1*0.521
⇒ Ah = 0.0521 m2
Thus, Total Hole Area = 0.0521 m2
Now we know that,
Ah = nh*(π/4)*dh2
Where, nh = number of holes.
⇒ nh = (4*Ah)/ (π*dh2) =2654(Aprox)
Therefore, Number of holes = 2654.
9. Weir Height (hw):
Let hw = 50 mm.
10. Weeping Check
The static pressure below the tray should be capable enough to hold the liquid
above the tray so that no liquid sweeps through the holes.
All the pressure drops calculated in this section are represented as mm head of
liquid on the plate. This serves as a common basis for evaluating the pressure
drops.
Notations used and their units:
hd = Pressure drop through the dry plate, mm of liquid on the plate
uh = Vapor velocity based on the hole area, m/s
how = Height of liquid over weir, mm of liquid on the plate
hσ = Pressure drop due to bubble formation, mm of liquid
53
hds= Dynamic seal of liquid, mm of liquid
hl = Pressure drop due to foaming, mm of liquid
hf = Pressure drop due to foaming, actual, mm of liquid
Df = Average flow length of the liquid, m
Rh = Hydraulic radius of liquid flow, m
uf = Velocity of foam, m/s
(NRe) = Reynolds number of flow
f = Friction factor
hhg = Hydraulic gradient, mm of liquid
hda = Loss under down comer apron, mm of liquid
Ads = Area under the down comer apron, m2
c = Down comer clearance, m
hdc = Down comer backup, mm of liquid
Calculations:
Head loss through dry hole
hd = head loss across the dry hole
hd = k1 + [k2* (ρg/ρl) *Uh2] --------- (eqn. 18.6, page 18.9, 6th edition Perry)
Where, Uh =gas velocity through hole area
k1, k2 are constants
For sieve plates,
k1 = 0 and
k2 = 50.8 / (Cv)2
Where, Cv=discharge coefficient, taken from fig 18.14, page 18.9 6th edition Perry.
Now,
(Ah /Aa) = 0.0521/ 0.593 = 0.088
Also, tT/dh = 3/5 = 0.60
Thus for (Ah/Aa) = 0.088 and tT/dh = 0.60
We have from fig. edition 18.14, page 18.9 6th Perry.
Cv = 0.74
54
⇒ k2 = 50.8 / 0.742 = 92.77
Volumetric flow rate of Vapor at the top of the Enriching Section
=qt = 8948.201/ (3.4376*3600) = 0.723 m3/s -------- (minimum at top)
Volumetric flow rate of Vapor at the bottom of the Enriching Section
= qo = 8960.401/(3.425*3600) = 0.727 m3/s. ---- (maximum at bottom)
Velocity through the hole area (Uh):
Now,
Velocity through the hole area at the top = Uh, top = qt /Ah
= 0.723/0.0521= 13.877 m/s
Also, Velocity through the hole area at the bottom= Uh, bottom = qo /Ah
= 0.727/0.0521 = 13.954 m/s
Now,
hd, top = k2 [ρg/ρl] (Uh,top)2
= 92.77∗(3.4376/784.69) ∗13.8772
⇒ hd, top = 78.263 mm clear liquid. -------- (minimum at top)
Also,
hd, bottom = k2 [ρg/ρl] (Uh, bottom)2
= 92.77∗(3.425/774.50)×13.9542
⇒ hd, bottom = 79.881 mm clear liquid ----- (maximum at bottom)
Head Loss Due to Bubble Formation
hσ = 409 [σ / ( ρL∗dh) ]
where σ =surface tension, mN/m (dyne/cm) = 19.325 dyne/cm.
dh =Hole diameter, mm
ρl = density of liquid in the section, kg/m3
= 784.69 kg/m3
hσ = 409 [ 19.325/(784.69 *5)]
hσ= 2.014 mm clear liquid
Height of Liquid Crest over Weir:
how = 664∗Fw [(q/Lw)2/3]
55
q = liquid flow rate at top, m3/s
= 1133.427/60* (784.69)
= 0.024 m3/min.
Thus, q’ = 6.339 gal/min.
Lw = weir length = 0.771 m = 2.530 ft
Now,
q’/Lw2.5 = 6.339/ (2.530)2.5 = 0.623
Now for q’/Lw2.5 = 0.623 and Lw /Dc =0.77
We have from fig.18.16, page 18.11, 6th edition Perry
Fw= correction factor =1.03
Thus,
how = 1.03∗664∗ [0.024/60*0.771] 2/3
⇒ how = 4.416 mm clear liquid.
Now,
(hd + hσ) = 78.263 + 2.014 = 80.227 mm ------ Design value
(hw + how) = 50 + 4.416 = 54.416 mm
For, Ah/Aa = 0.088 and (hw + how) = 54.416 mm
The minimum value of (hd + hσ) required is calculated from a graph given in
Perry,
plotted against Ah/Aa.
i.e., we have from fig. 18.11, page 18.7, 6th edition Perry
(hd + hσ)min = 13.0 mm ------- Theoretical value.
The minimum value as found is 13.0 mm.
Since the design value is greater than the minimum value, there is no
problem of
weeping.
Down comer Flooding:
hds =hw + how + (hhg /2) ------- (eqn 18.10, page 18.10, 6th edition Perry)
Where,
hw = weir height, mm
hds = static slot seal (weir height minus height of top of slot above plate floor,
56
height equivalent clear liquid, mm)
how = height of crest over weir, equivalent clear liquid, mm
hhg = hydraulic gradient across the plate, height of equivalent clear liquid, mm.
Hydraulic gradient, hhg
Let
hhg = 0.5 mm.
hds = hw + how + hhg/2
= 50 + 4.416 + 0.5/2 = 54.666 mm.
Now, Fga = Ua ∗ρg0.5
Where Fga = gas-phase kinetic energy factor,
Ua = superficial gas velocity, m/s (ft/s),
ρg = gas density, kg/m3 (lb/ft3)
Here Ua is calculated at the bottom of the section.
Thus, Ua = (Gb/ρg)/ Aa = 8960.401/(3600*3.425 * 0.593) = 1.225 m/s
Thus, Ua = 4.019 ft/s
ρg = 3.4250 kg/m3 = 0.209 lb/ft3
Therefore, Fga = 4.019∗(0.209) 0.5
Fga = 1.837
Now for Fga = 1.837, we have from fig. 18.15, page 18.10 6th edition Perry
Aeration factor = β = 0.6
Relative Froth Density = φt = 0.2
Now hl’= β∗hds ---- (eqn. 18.8, page 18.10, 6th edition Perry)
Where, hl’= pressure drop through the aerated mass over and around the disperser,
mm liquid,
⇒ hl’= 0.6∗54.666 = 32.800 mm.
Now,
hf = hl’/φt ------- (eqn. 18.9, page 18.10, 6th edition Perry)
⇒ hf = 32.8/ 0.2 = 164.0 mm.
Average width of liquid flow path, Df = (Dc + Lw)/2
= (1.001 + 0.771)/2 = 0.886 m.
Hydraulic radius of aerated mass Rh = hf * Df /(2*hf + 1000*Df) (from eq. 18.23, page
57
18.12 6th edition Perry)
Rh = 164*.886/(2*164 + 1000*0.886)
= 0.120 m.
Velocity of aerated mass, Uf = 1000*q/ (hl’ * Df )
Volumetric flow rate(liquid at bottom), q = 4.167*10-4 m3/s.
Uf = 1000* 4.167*10-4 / (32.8* 0.886)
= 0.014 m/s.
Reynolds modulus NRe = Rh * Uf * ρl / μliq
= 0.12 * 0.014 * 774.5 /(1.03 * 10-3)
= 1263.26
hhg = 1000* f* Uf2 *Lf/(g * Rh)
f = 0.6 for hw = 1.97” and NRe = 1871.373
Lf = 2 * Dc/2 cos(θc/ 2) = 1.569 m
hhg = 1000* 0.6 *0.0142*1.569/(9.81* 0.12)
= 0.157 mm.
Head loss over down comer apron:
hda = 165.2 {q/ Ada}2 ----- (eqn. 18.19, page 18.10, 6th edition Perry)
Where, hda = head loss under the down comer apron, as millimeters of liquid,
q = liquid flow rate calculated at the bottom of section, m3/s
And Ada = minimum area of flow under the down comer apron, m2
Now,
q = 4.167∗10-4 m3/s
Take clearance, C = 1” = 25.4 mm
hap = hds - C = 54.666 - 25.4 = 29.266 mm
Ada = Lw * hap = 0.771∗29.266∗10-3 = 0.023 m2
hda = 165.2[(4.167* 10-4)/ (0.023)] 2
hda = 0.054 mm
Now,
ht = hd + hl`
58
Here hd and hl’ are calculated at bottom of the enriching section.
Now we have,
hd, bottom = 79.881 mm
hl, bottom = 32.8 mm
ht = hd + hl`
= 79.881+32.8
ht = 112.681 mm
Down comer Backup:
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.
hdc = 112.681 +50 +4.416 + 0.054 + 0.157
hdc = 167.308 mm.
Let φdc = average relative froth density (ratio of froth density to liquid density)
=0.5
h`dc = hdc / φdc = 167.308/ 0.5
h`dc = 334.616 mm.
which is less than the tray spacing, ts= 457 mm.
Hence no flooding in the enriching section and hence the design calculations are
acceptable.
b). Design of Stripping Section:
Tray Hydraulics,
The design of a sieve plate tower is described below. The equations and correlations are
borrowed from the 6th and 7th editions of Perry’s Chemical Engineers’ Handbook.
59
1. Tray Spacing, (ts) :
Let ts = 18” = 457 mm. (range 0.15 – 1.0 m).
2. Hole Diameter, (dh):
Let dh = 5 mm. (range 2.5 – 12 mm).
3. Hole Pitch (lp):
Let lp = 3* dh (range 2.5 to 4.0 times dh).
i.e., lp = 3*5 = 15 mm.
4. Tray thickness (tT):
Let tT = 0.6* dh (range 0.4 to 0.7 times dh).
i.e., tT = 0.6*5 = 3 mm.
5. Ratio of hole area to perforated area (Ah/Ap):
Now, for a triangular pitch, we know that,
Ratio of hole area to perforated area (Ah/Ap) = ½ (π/4*dh2)/ [(√3/4) *lp2]
i.e., (Ah/Ap) = 0.90* (dh/lp)2
i.e., (Ah/Ap) = 0.90* (5/15)2
i.e., (Ah/Ap) = 0.1
Thus,
(Ah/Ap) = 0.1
6. Plate Diameter (Dc):
The plate diameter is calculated based on entrainment flooding considerations
L/G {ρg/ρl} 0.5 = 0.0705 ---------- (maximum value)
Now for,
L/G {ρg/ρl} 0.5 = 0.0705 and for a tray spacing of 457 mm.
We have,
From the flooding curve, ---------- (fig.18.10, page 18.7, 6th edition Perry.)
Flooding parameter, Csb, flood = 0.27 ft/s .
Now,
Unf = Csb, flood * (σ / 20) 0.2 [(ρl - ρg) / ρg]0.5
---- {eqn. 18.2, page 18.6, 6th edition Perry.}
Where,
Unf = gas velocity through the net area at flood, m/s (ft/s)
60
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)
Now, we have,
σ = 18.330 mN/m = 18.330 dyne/cm.
ρl = 747.87kg/m3.
ρg = 3.391 kg/m3.
Therefore,
Unf = 0.27*(18.33/20)0.2*[(747.87-3.391)/ 3.391]0.5
i.e.,Unf = 3.931 ft/s
Let,
Actual velocity, Un= 0.8*Unf
i.e., Un = 0.8∗3.931 ft/s
i.e., Un = 3.145 ft/s = .959 m/s
It is desired to design with volumetric flow rate maximum (therefore, actual is less than
the maximum).
Volumetric flow rate of Vapor at the bottom of the Stripping Section
= qo = 8980.733/ (3600*3.391) = 0.736 m3/s.
Now,
Net area available for gas flow (An)
Net area = (Column cross sectional area) - (Down comer area.)
An = Ac - Ad
Thus,
Net Active area, An = qo/ Un = 0.736/.959 = 0.767 m2.
Let Lw / Dc = 0.77 (range 0.6 to 0.85 times Dc ).
Where, Lw = weir length, m
Dc = Column diameter, m
Now,
θc = 2*sin-1(Lw / Dc) = 2*sin-1 (0.77) = 100.70
Now,
Ac=(π /4)*Dc2= 0.785*Dc2 , m2
61
Ad = [(π/4) * Dc2 * (θc/3600)] - [(Lw/2) * (Dc/2) *cos (θc/2)]
i.e., Ad = [0.785*Dc2 *(100.70/3600)]-[(1/4)* (Lw / Dc) * Dc2 * cos(100.70)]
i.e., Ad = (0.2196* Dc2) - (0.1288* Dc2)
i.e., Ad = 0.0968*Dc2, m2
Since An = Ac -Ad
0.767 = (0.785*Dc2) - (0.0968* Dc2)
i.e., 0.6882* Dc2 = 0.767
⇒ Dc2 = 0.767/ 0.6882 = 1.115
⇒ Dc = √ 1.115
Dc = 1.056 m
Since Lw / Dc = 0.77,
⇒ Lw = 0.77* Dc = 0.77*1.056 = 0.813 m.
Therefore, Lw = 0.813 m.
Now,
Ac = 0.785*1.0562 = 0.875 m2
Ad = 0.0968*Dc2 = 0.0968*1.0562= 0.108 m2
Aa = Ac –2* Ad
i.e., Aa = 0.875- 2*0.108 ⇒ Aa= 0.659 m2
7. Perforated plate area (Ap):
Now,
Lw / Dc = 0.813/ 1.056 = 0.77
θc = 100.73 0
α = 180 0- θc
⇒ α = 79.27 0
Now,
Acz = 2* Lw* (thickness of distribution)
Where, Acz = area of calming zone, m2 (5 to 20% of Ac )
Acz = 2*0.813* (30×10-3) = 0.043 m2 -------- (which is 6.76% of Ac)
Also,
Awz= 2*{(π/4)*Dc2*(θc/3600)- (π/4)*(Dc-30*10-3)2*(θc/3600)}
Where, Awz = area of waste periphery, m2 (range 2 to 5% of Ac)
62
i.e., Awz = 2*{(π/4)*1.0562*(100.730/3600)- (π/4)*(1.056-30*10-3)2*(100.730/3600)}
i.e., Awz = 0.027 m2
--------- (which is 3.8% of Ac)
Now,
Ap = Ac - (2*Ad) - Acz - Awz
i.e., Ap = 0.813- (2*.108) - 0.043 - 0.027
Thus, Ap = 0.590 m2.
8. Total Hole Area (Ah):
Since,
Ah / Ap = 0.1
⇒ Ah = 0.1* Ap
i.e., Ah = 0.1*0.590
⇒ Ah = 0.0590 m2
Thus, Total Hole Area = 0.059 m2
Now we know that,
Ah = nh*(π/4)*dh2
Where, nh = number of holes.
⇒ nh = (4*Ah)/ (π*dh2) =3005(Aprox)
Therefore, Number of holes = 3005.
9. Weir Height (hw):
Let hw = 50 mm.
10. Weeping Check
All the pressure drops calculated in this section are represented as mm head of
liquid on the plate. This serves as a common basis for evaluating the pressure
drops.
Notations used and their units:
hd = Pressure drop through the dry plate, mm of liquid on the plate
uh = Vapor velocity based on the hole area, m/s
how = Height of liquid over weir, mm of liquid on the plate
hσ = Pressure drop due to bubble formation, mm of liquid
63
hds= Dynamic seal of liquid, mm of liquid
hl = Pressure drop due to foaming, mm of liquid
hf = Pressure drop due to foaming, actual, mm of liquid
Df = Average flow length of the liquid, m
Rh = Hydraulic radius of liquid flow, m
uf = Velocity of foam, m/s
(NRe) = Reynolds number of flow
f = Friction factor
hhg = Hydraulic gradient, mm of liquid
hda = Loss under down comer apron, mm of liquid
Ads = Area under the down comer apron, m2
c = Down comer clearance, m
hdc = Down comer backup, mm of liquid
Calculations:
Head loss through dry hole
hd = head loss across the dry hole
hd = k1 + [k2* (ρg/ρl) *Uh2] --------- (eqn. 18.6, page 18.9, 6th edition Perry)
Where, Uh =gas velocity through hole area
k1, k2 are constants
For sieve plates,
k1 = 0 and
k2 = 50.8 / (Cv)2
Where, Cv=discharge coefficient, taken from fig 18.14, page 18.9 6th edition Perry.
Now,
(Ah /Aa) = 0.059/ 0.659 = 0.090
Also, tT/dh = 3/5 = 0.60
Thus for (Ah/Aa) = 0.09 and tT/dh = 0.60
We have from fig. edition 18.14, page 18.9 6th Perry.
Cv = 0.75
64
⇒ k2 = 50.8 / 0.752 = 90.311
Volumetric flow rate of Vapor at the top of the Stripping Section
=qt = 8960.401/ (3.425*3600) = 0.727 m3/s -------- (minimum at top)
Volumetric flow rate of Vapor at the bottom of the Stripping Section
= qo = 8980.733/(3.391*3600) = 0.736 m3/s. ---- (maximum at bottom)
Velocity through the hole area (Uh):
Now,
Velocity through the hole area at the top = Uh, top = qt /Ah
= 0.727/0.059= 12.322 m/s
Also, Velocity through the hole area at the bottom= Uh, bottom = qo /Ah
= 0.736/0.059 = 12.475 m/s
Now,
hd, top = k2 [ρg/ρl] (Uh,top)2
= 90.311∗(3.425/774.50) ∗12.4752
⇒ hd, top = 62.153 mm clear liquid. -------- (minimum at top)
Also,
hd, bottom = k2 [ρg/ρl] (Uh, bottom)2
= 90.311∗(3.391/747.87)×12.4752
⇒ hd, bottom = 63.727 mm clear liquid ----- (maximum at bottom)
Head Loss Due to Bubble Formation
hσ = 409 [σ / ( ρL∗dh) ]
where σ =surface tension, mN/m (dyne/cm) = 18.330 dyne/cm.
dh =Hole diameter, mm
ρl = density of liquid in the section, kg/m3
= 774.50 kg/m3
hσ = 409 [ 18.33/(774.50 *5)]
hσ= 1.936 mm clear liquid
Height of Liquid Crest over Weir:
how = 664∗Fw [(q/Lw)2/3]
65
q = liquid flow rate at top, m3/s
= 9919.724/60* (774.50)
= 0.213 m3/min.
Thus, q’ = 56.269 gal/min.
Lw = weir length = 0.813 m = 2.667 ft
Now,
q’/Lw2.5 = 56.269/ (2.667)2.5 = 4.844
Now for q’/Lw2.5 = 4.844 and Lw /Dc =0.77
We have from fig.18.16, page 18.11, 6th edition Perry
Fw= correction factor =1.035
Thus,
how = 1.035∗664∗ [0.213/60*0.813] 2/3
⇒ how = 18.360 mm clear liquid.
Now,
(hd + hσ) = 62.153 + 1.936 = 64.089 mm ------ Design value
(hw + how) = 50 + 18.36 = 68.360 mm
For, Ah/Aa = 0.09 and (hw + how) = 68.360 mm
The minimum value of (hd + hσ) required is calculated from a graph given in
Perry,
plotted against Ah/Aa.
i.e., we have from fig. 18.11, page 18.7, 6th edition Perry
(hd + hσ)min = 17.0 mm ------- Theoretical value.
The minimum value as found is 17.0 mm.
Since the design value is greater than the minimum value, there is no
problem of
weeping.
Down comer Flooding:
hds =hw + how + (hhg /2) ------- (eqn 18.10, page 18.10, 6th edition Perry)
Where,
hw = weir height, mm
66
hds = static slot seal (weir height minus height of top of slot above plate
floor,
height equivalent clear liquid, mm)
how = height of crest over weir, equivalent clear liquid, mm
hhg = hydraulic gradient across the plate, height of equivalent clear liquid,
mm.
Hydraulic gradient, hhg
Let
hhg = 0.5 mm.
hds = hw + how + hhg/2
= 50 + 18.36 + 0.5/2 = 68.610 mm.
Now, Fga = Ua ∗ρg0.5
Where Fga = gas-phase kinetic energy factor,
Ua = superficial gas velocity, m/s (ft/s),
ρg = gas density, kg/m3 (lb/ft3)
Here Ua is calculated at the bottom of the section.
Thus, Ua = (Gb/ρg)/ Aa = 8980.733/(3600*3.391 * 0.659) = 1.116 m/s
Thus, Ua = 3.661 ft/s
ρg = 3.391 kg/m3 = 0.207 lb/ft3
Therefore, Fga = 3.661∗(0.207) 0.5
Fga = 1.666
Now for Fga = 1.666, we have from fig. 18.15, page 18.10 6th edition Perry
Aeration factor = β = 0.61
Relative Froth Density = φt = 0.21
Now hl’= β∗hds ---- (eqn. 18.8, page 18.10, 6th edition Perry)
Where, hl’= pressure drop through the aerated mass over and around the disperser,
mm liquid,
⇒ hl’= 0.61∗68.610 = 41.852 mm.
Now,
hf = hl’/φt ------- (eqn. 18.9, page 18.10, 6th edition Perry)
⇒ hf = 41.852/ 0.21 = 199.30 mm.
67
Average width of liquid flow path, Df = (Dc + Lw)/2
= (1.056 + 0.813)/2 = 0.935 m.
Hydraulic radius of aerated mass Rh = hf * Df /(2*hf + 1000*Df) (from eq. 18.23,
page 18.12 6th edition Perry)
Rh = 199.3*.935/(2*199.3 + 1000*0.935)
= 0.140 m.
Velocity of aerated mass, Uf = 1000*q/ (hl’ * Df )
Volumetric flow rate(liquid at bottom), q = 40*10-4 m3/s.
Uf = 1000* 40*10-4 / (41.852* 0.935)
= 0.102 m/s.
Reynolds modulus NRe = Rh * Uf * ρl / μliq
= 0.14 * 0.102 * 747.87 /(.924 * 10-3)
= 11557.99
hhg = 1000* f* Uf2 *Lf/(g * Rh)
f = 0.18 for hw = 1.97” and NRe = 11557.9
Lf = 2 * Dc/2 cos(θc/ 2) = .668 m
hhg = 1000* 0.18 *0.1022*.668/(9.81* 0.14)
= 0.911 mm.
Head loss over down comer apron:
hda = 165.2 {q/ Ada}2 ----- (eqn. 18.19, page 18.10, 6th edition Perry)
Where, hda = head loss under the down comer apron, as millimeters of liquid,
q = liquid flow rate calculated at the bottom of section, m3/s
And Ada = minimum area of flow under the down comer apron, m2
Now,
q = 40∗10-4 m3/s
Take clearance, C = 1” = 25.4 mm
hap = hds - C = 68.610 - 25.4 = 43.21 mm
Ada = Lw * hap = 0.813∗43.21∗10-3 = 0.035 m2
hda = 165.2[(40* 10-4)/ (0.035)] 2
hda = 2.158 mm
68
Now,
ht = hd + hl`
Here hd and hl’ are calculated at bottom of the stripping section.
Now we have,
hd, bottom = 63.727 mm
hl, bottom = 41.852 mm
ht = hd + hl`
= 63.727+41.852
ht = 105.579 mm
Down comer Backup:
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.
hdc = 105.579 +50 +18.36 + 2.158 + 0.911
hdc = 177.008 mm.
Let φdc = average relative froth density (ratio of froth density to liquid density)
=0.5
h`dc = hdc / φdc = 177.008/ 0.5
h`dc = 354.016 mm.
which is less than the tray spacing, ts= 457 mm.
Hence no flooding in the enriching section and hence the design calculations are
acceptable.
69
Formulas used in calculation of properties:
1 VISCOSITY:
(i). Average Liquid Viscosity:
(μliq)1/3 = [x1× (μ1)1/3] + [x2 × (μ2)1/3]
2 DIFFUSIVITIES:
(i). Liquid Phase Diffusivity:
For the case of Organic solutes diffusing in Organic solvents
DAB = (1.173*10-13*(Ɵ*M)0.5*T)/[ ƞB × (VA)0.6] –(Richardson – coulson vol.6)
Where,
Ɵ=constant
M = molecular weight.
T = absolute temperature, 0K,
ƞB = viscosity of solvent B, cP,
VA =molar volume of solute A at its normal boiling temperature, cm3/g-mol.
DAB =mutual diffusivity coefficient of solute A at very low concentration in
solvent B, cm2/s
(ii). Gas Phase Diffusivity:
DAB = 1.013*10-7×T1.75× [(MA+MB)/ (MA×MB)]1/2}/{P×[(∑VA)1/3+ (∑VB)1/3]2
------ (Richardson – coulson vol.6 ).
Where P = Pressure in atmospheres,
T = Temperature in 0K
DAB = Diffusivity, cm2/s
∑VA and ∑VB = summation of atomic diffusion volumes for
components A and B respectively.
MA and MB = Molecular weights of components A and B respectively.
3. SURFACE TENSION:
σ =[Pch *(ρ1-ρg)/M]4 ×10-12 ----- (eqn. 8.23, page 293, Coulson and Richardson
vol.6)
70
Where,
σ =Surface tension, dyne/cm
Pch =Sugden’s Parachor,
ρ1 = liquid density, kg/m3
ρg = density of saturated vapor, kg/m3
M = Molecular weight
σ , ρ1, ρg are evaluated at system temperature.
σ mix =∑ (xi*σi) where i=1,2,3,……n.
4. LIQUID DENSITY:
ρ = Pc/ ( R * Tc * Zc[ 1 + ( 1 – Tr)2/7] ) (Coulson and Richardson vol.6)
Where,
Pc = critical pressure = M/(0.34 + (∑ ∆P)2 )
M = Molecular weight.
Tc = Critical temperature = Tb / (0.567 + ∑ ∆T – (∑ ∆T)2 )
Tb = Normal boiling temperature 0K.
Zc = Pc * Vc / (R * Tc)
Vc = critical volume
R = universal gas constant.
5. GAS DENSITY:
ρ = P * M /( R * T )
P = pressure
M = Molecular weight.
R = universal gas constant.
T = temperature.
Enriching section:
Column efficiency ( AIChe method )
1. Point Efficiency, (Eog):
Eog = 1-e-Nog = 1-exp (-Nog) ----- (eqn. 18.33, page 18.15, 6th edition Perry)
71
Where Nog = Overall transfer units
Nog = 1/ [(1/Ng)+(λ/ N1)] ---- (eqn. e 18.34, page 18.15, 6th edition Perry)
Where Nl = Liquid phase transfer units,
Ng = Gas phase transfer units,
λ=(m*Gm)/ Lm = Stripping factor,
m = slope of Equilibrium Curve,
Gm = Gas flow rate, mol/s
Lm = Liquid flow rate, mol/s
Ng= (0.776 + (0.0045*hw) - (0.238*Ua*ρg0.5) + (105*W))/ (NSc, g)0.5
----- (eqn. 18., page 18., 6th edition Perry)--- *
Where,
hw = weir height = 50.00 mm
Ua = Gas velocity through active area, m/s
= 1.225 m/s.
Ua = 1.225 m/s
Df = (Lw + Dc)/2 = (0.771 + 1.001)/2 = 0.886 m
q = 416.7 * 10 -6 m3/s
W = Liquid flow rate, m3/ (s.m) of width of flow path on the plate,
= q/Df = 416.7*10-6/0.886 = 270.316*10-6 m3/ (s.m)
NSc, g = Schmidt number =μg/(ρg*Dg) = 9.4*10-6/(3.425*4.433*10-6) = 0.619
Dg = Diffusivity = 4.433 * 10-6 m2/s.
Now,
Number of gas phase transfer units,
Ng=(0.776+(0.0045*50)-(0.238*1.225*3.4250.5)+(105*270.316*10-6))/ (0.619)0.5
Ng = 0.623
Also,
Number of liquid phase transfer units,
Nl = kl* a*θl ----- (eqn 18.36a, page 18.15, 6th edition Perry)
Where kl = Liquid phase transfer coefficient kmol/ (m2 kmol/m3) or m/s
72
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, s
θl = (hl*Aa)/ (1000*q) ---- (eqn. 18.38, page 18.16, 6th edition Perry)
Now, q = liquid flow rate, m3/s
q =416.7*10 -6 m3/s
hl = hl’ = 32.8 mm
Aa = 0.593 m2
θl = 32.8*0.593/ (1000*416.7*10-6) = 46.677 s
kl *a = (3.875*108*DL)0.5* ((0.40*Ua*ρg0.5) + 0.17)
--- (eqn. 18.40a, page 18.16, 6th edition Perry)
DL= liquid phase diffusion coefficient, m2/s
kl *a = (3.875*108*2.002*10-9)0.5* ((0.40*1.225*3.4250.5) + 0.17)
kl *a = 0.948 m/s
Nl = kl* a* θl
i.e.,
Nl = 0.948*46.677
= 44.250
λm = mm * Gm/Lm
λE = mE * GE/LE =.125*203.322/25.748
table for enriching section)
⇒ λ = 0.987
Nog = 1/ [(1/Ng)+( λ /Nl)]
= 1/ [(1/.623) + (0.987/44.25)]
Nog = .614
Eog = 1-e-Nog = 1-exp (-Nog)
= 1-e-.614 = 1-exp (-.614)
Eog = 0.459
Point Efficiency = Eog = 0.459
2 Murphree Plate Efficiency (Emv):
Now, Peclet number =NPe = Zl2 / (DE* ql)
73
=.987(slope calculated from property
Zl = length of liquid travel, m
DE = (6.675 * 10 –3* (Ua) 1.44) + (0.922 * 10 –4* hl) - 0.00562
---- (eqn. 18.45, page 18.17, 6th edition Perry)
Where DE = Eddy diffusion coefficient, m2/s
DE = (6.675 * 10 –3* (1.225) 1.44) + (0.922 * 10 –4* 32.8) - 0.00562
DE = 0.006 m2/s
Also,
Zl = Dc *cos(Ɵc/2) = 1.001* cos (100.73 0/2) = 0.639 m
NPe = Zl2 / (DE* θl)
= 0.639 2/ (0.006 * 44.25)
NPe = 1.809
λ*Eog = 0.987 * 0.459 = 0.453
Now, for λ*Eog = 0.453 and NPe = 1.809
We have from fig.18.29a, page 18.18, 6th edition Perry
Emv/ Eog = 1.17
Emv = 1.17* Eog = 1.17*0.459 = 0.537
Murphee Plate Efficiency = Emv = 0.537
3 Overall Efficiency ( EOC):
Overall Efficiency = EOC = log [1 + Eα ( λ - 1)]
log λ
----- (eqn. 18.46, page 18.17, 6th edition Perry)
Where, Eα /Emv= 1/ (1 + EMV [ψ/ (1- ψ)])
----- (eqn. 18.27, page 18.13, 6th edition Perry)
Emv = Murphee Vapor efficiency,
Eα = Murphee Vapor efficiency, corrected for recycle effect of liquid
entrainment.
(L/G)*{ρg/ρl}0.5 = 0.004
Thus, for (L/G)*{ρg/ρl}0.5 = 0.008 and at 80 % of the flooding value,
We have from fig.18.22, page 18.14, 6th edition Perry
ψ = fractional entrainment, moles/mole gross down flow = 0.035
⇒ Eα / Emv = 1 / (1 + Emv [ψ/ (1- ψ)]
74
⇒ Eα = Emv / ( 1 + Emv [ψ/ (1- ψ)] )
= 0.537/ (1+0.537[0.035/ (1-0.035)])
⇒ Eα = 0.527
Overall Efficiency = EOC = log [1 + Eα ( λ - 1)]/log λ
EOC = log [1+ 0.527(0.987-1)]/ log 0.987
Overall Efficiency = EOC = 0.525
Actual trays = Nact = NT/EOC = (ideal trays)/ (overall efficiency)
Where NT = Theoretical plates,
Nact = actual trays
Nact = 3/0.5767 = 5.714=6(aprox) _
Thus, Actual trays in the Enriching Section = 6
Thus 6th tray is the feed tray.
Total Height of Enriching section = 6*ts = 6*457 = 2742 mm = 2.742 m
Striping section:
Column efficiency ( AIChe method )
1. Point Efficiency, (Eog):
Eog = 1-e-Nog = 1-exp (-Nog) ----- (eqn. 18.33, page 18.15, 6th edition Perry)
Where Nog = Overall transfer units
Nog = 1/ [(1/Ng)+(λ/ N1)] ---- (eqn. 18.34, page 18.15, 6th edition Perry)
Where Nl = Liquid phase transfer units,
Ng = Gas phase transfer units,
λ=(m*Gm)/ Lm = Stripping factor,
m = slope of Equilibrium Curve,
Gm = Gas flow rate, mol/s
Lm = Liquid flow rate, mol/s
Ng= (0.776 + (0.0045*hw) - (0.238*Ua*ρg0.5) + (105*W))/ (NSc, g)0.5
----- (eqn. 18., page 18., 6th edition Perry)--- *
Where,
hw = weir height = 50.00 mm
Ua = Gas velocity through active area, m/s
75
= 1.116 m/s.
Ua = 1.116 m/s
Df = (Lw + Dc)/2 = (0.813 + 1.056)/2 = 0.935 m
q = 3550 * 10 -6 m3/s
W = Liquid flow rate, m3/ (s.m) of width of flow path on the plate,
= q/Df = 3550*10-6/0.935 = 3796.79*10-6 m3/ (s.m)
NSc, g = Schmidt number =μg/(ρg*Dg) = 9.4*10-6/(3.361*4.433*10-6) = 0.631
Dg = Diffusivity = 4.433 * 10-6 m2/s.
Now,
Number of gas phase transfer units,
Ng=(0.776+(0.0045*50)-(0.238*1.116*3.3610.5)+(104.6*3796.79*10-6))/ (0.631)0.5
Ng = 1.147
Also,
Number of liquid phase transfer units,
Nl = kl* a*θl ----- (eqn 18.36a, page 18.15, 6th edition Perry)
Where kl = Liquid phase transfer coefficient kmol/ (m2 kmol/m3) or 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, s
θl = (hl*Aa)/ (1000*q) ---- (eqn. 18.38, page 18.16, 6th edition Perry)
Now, q = liquid flow rate, m3/s
q =3796.79*10 -6 m3/s
hl = hl’ = 41.852 mm
Aa = 0.659 m2
θl = 41.852 *0.659/ (1000*3796.79*10-6) = 7.266 s
kl *a = (3.875*108*DL)0.5* ((0.40*Ua*ρg0.5) + 0.17)
--- (eqn. 18.40a, page 18.16, 6th edition Perry)
DL= liquid phase diffusion coefficient, m2/s
kl *a = (3.875*108*2.002*10-9)0.5* ((0.40*1.116*3.3610.5) + 0.17)
kl *a = 0.891 m/s
Nl = kl* a* θl
76
i.e.,
Nl = 0.891*7.266
= 6.474
λm = mm * Gm/Lm
λE = mS * GS/LS =.893*203.322/222.615
=.816(slope calculated from property
table for enriching section)
⇒ λ = 0.816
Nog = 1/ [(1/Ng)+( λ /Nl)]
= 1/ [(1/1.147) + (0.816/7.266)]
Nog = 1.016
Eog = 1-e-Nog = 1-exp (-Nog)
= 1-e-1.016 = 1-exp (-1.016)
Eog = 0.638
Point Efficiency = Eog = 0.638
2 Murphree Plate Efficiency (Emv):
Now, Peclet number =NPe = Zl2 / (DE* ql)
Zl = length of liquid travel, m
DE = (6.675 * 10 –3* (Ua) 1.44) + (0.922 * 10 –4* hl) - 0.00562
---- (eqn. 18.45, page 18.17, 6th edition Perry)
Where DE = Eddy diffusion coefficient, m2/s
DE = (6.675 * 10 –3* (1.116) 1.44) + (0.922 * 10 –4* 41.852) - 0.00562
DE = 0.0061 m2/s
Also,
Zl = Dc *cos(Ɵc/2) = 1.056* cos (100.73 0/2) = 0.674 m
NPe = Zl2 / (DE* θl)
= 0.674 2/ (0.0061 * 7.266)
NPe = 10.25
λ*Eog = 0.816 * 0.638 = 0.521
Now, for λ*Eog = 0.521 and NPe = 10.25
We have from fig.18.29a, page 18.18, 6th edition Perry
77
Emv/ Eog = 1.45
Emv = 1.45* Eog = 1.45*0.638 = 0.925
Murphee Plate Efficiency = Emv = 0.925
3 Overall Efficiency ( EOC):
Overall Efficiency = EOC = log [1 + Eα ( λ - 1)]
log λ
----- (eqn. 18.46, page 18.17, 6th edition Perry)
Where, Eα /Emv= 1/ (1 + EMV [ψ/ (1- ψ)])
----- (eqn. 18.27, page 18.13, 6th edition Perry)
Emv = Murphee Vapor efficiency,
Eα = Murphee Vapor efficiency, corrected for recycle effect of liquid
entrainment.
(L/G)*{ρg/ρl}0.5 = 0.074
Thus, for (L/G)*{ρg/ρl}0.5 = 0.008 and at 80 % of the flooding value,
We have from fig.18.22, page 18.14, 6th edition Perry
ψ = fractional entrainment, moles/mole gross down flow = 0.04
⇒ Eα / Emv = 1 / (1 + Emv [ψ/ (1- ψ)]
⇒ Eα = Emv / ( 1 + Emv [ψ/ (1- ψ)] )
= 0.925/ (1+0.925[0.04/ (1-0.04)])
⇒ Eα = 0.891
Overall Efficiency = EOC = log [1 + Eα ( λ - 1)]
log λ
EOC = log [1+ 0.891(0.816-1)]/ log 0.816
Overall Efficiency = EOC = 0.881
Actual trays = Nact = NT/EOC = (ideal trays)/ (overall efficiency)
Where NT = Theoretical plates,
Nact = actual trays
Nact = 2/0.881 = 2.27=3(aprox) _
Thus, Actual trays in the Stripping Section = 3
78
Total Height of Stripping section = 3*ts = 3*457 = 1371 mm = 1.371 m
SUMMARY OF THE DISTILLATION COLUMN:
A) Enriching section
Tray spacing = 457 mm
Column diameter = 1.001 m
Weir length = 0.771 m
Weir height = 50 mm
Hole diameter = 5 mm
Hole pitch = 15 mm, triangular
Tray thickness = 3 mm
Number of holes = 2654
Flooding % = 80%
B) Stripping section
Tray spacing = 457 mm
Column diameter = 1.056 m
Weir length = 0.813 m
Weir height = 50 mm
Hole diameter = 5 mm
Hole pitch = 15 mm, triangular
Tray thickness = 3 mm
Number of holes = 3005,
Flooding % = 80%
5.2 MECHANICAL DESIGN OF DISTILLATION COLUMN:
a) Shell:
Diameter of the tower =Di = 1060 mm =1.06 m
Working/Operating Pressure = 2.087 atmosphere = 2.1558 kg/cm2
Design pressure = 1.1*Operating Pressure = 1.1*2.1558 = 2.37138 kg/cm2
Working temperature = 95 0C = 368 0K
Design temperature = 104.5 0C = 377.5 0K
Shell material - IS: 2002-1962 Carbon steel (specific gravity 7.7)
79
Permissible tensile stress (ft) = 95 MN/m2 = 970 kg/cm2
Insulation material - asbestos
Insulation thickness = 2”= 50.8 mm
Density of insulation = 2700 kg/m3
Top disengaging space = 0.3 m
Bottom separator space = 0.4 m
Weir height = 50 mm
Down comer clearance = 1” = 25.4 mm
b) Head - torispherical dished head:
Material - IS: 2002-1962 Carbon steel
Allowable tensile stress = 95 MN/m2 = 970 kg/cm2
c) Support skirt:
Height of support = 1000 mm = 1.0 m
Material - Carbon Steel
d) Trays-sieve type:
Number of trays = 9
Hole Diameter = 5 mm
Number of holes:
Enriching section = 2654
Stripping section = 3005
Tray spacing:
Enriching section: 18” = 457 mm
Stripping section: 18” = 457 mm
Thickness = 3 mm
e) Support for tray:
Purlins - Channels and Angles
Material - Carbon Steel
Permissible Stress = 127.5 MN/m2 =1299.7 kg/cm2
1. Shell minimum thickness:
Considering the vessel as an internal pressure vessel.
ts = ((P*Di)/ ((2*ft*J)- P)) + C
80
where ts = thickness of shell, mm
P = design pressure, kg/cm2
Di = diameter of shell, mm
ft = permissible/allowable tensile stress, kg/cm2
C = Corrosion allowance, mm
J = Joint factor
Considering double welded butt joint with backing strip
J= 85% = 0.85
Thus, ts = ((2.37138*1060)/ ((2*970*0.85)- 2.37138)) + 3 = 4.527 mm
Taking the thickness of the shell = 6 mm (standard)
2. Head Design- Shallow dished and Torispherical head:
Thickness of head = th = (P*Rc*W)/ (2*f*J)
P =internal design pressure, kg/cm2
Rc = crown radius = diameter of shell, mm
W= stress intensification factor or stress concentration factor for torispherical
head,
W= ¼ * (3 + (Rc/Rk)0.5)
Rk = knuckle radius, which is at least 6% of crown radius, mm
Now, Rc = 1060 mm
Rk = 6% of Rc = 0.06*1060 = 63.6 mm
W= ¼ * (3 + (Rc/Rk)0.5) = ¼ * (3 + (1060/63.6)0.5) = 1.7706 mm
th = (2.37138*1060*1.7706)/ (2*970*0.85) = 2.70 mm
Including corrosion allowance take the thickness of head = 6 mm
Weight of Head:
Diameter = O.D + (O.D/24) + (2*sf) + (2*icr/3) --- (eqn. 5.12 Brownell and Young)
Where O.D. = Outer diameter of the dish, inch
icr = inside cover radius, inch
sf = straight flange length, inch
From table 5.7 and 5.8 of Brownell and Young
sf =1.5”
icr = 2.31”
81
Also, O.D.= 1060 mm = 41.73”
Diameter = 41.73+ (41.73/24) + (2*1.5)+(2/3*2.31)
d = 48.01” = 1219.45mm.
Weight of Head = ((π*d2*t)/4)*(ρ/1728)
=(( π *48.012 *0.2362)/4) * (7700/1728) = 1905.37 lb
= 864.26 kg
3. Shell thickness at different heights
At a distance ‘X’m from the top of the shell the stresses are:
3.1 Axial Tensile Stress due to Pressure:
fap =(P*Di)/ 4(ts -c) = 2.37138*1060/4(6 - 3)= 209.47 kgf/cm2 .
This is the same through out the column height.
3.2 Circumferential stress
2 * fap = 2*209.47 = 418.94 kgf/cm2
3.3 Compressive stress due Dead Loads:
3.3.1 Compressive stress due to Weight of shell up to a distance ‘X’ meter from top.
fds = weight of shell/cross-section of shell
=
(π/4)*(Do2-Di2)*ρs*X = weight of shell per unit height X
(π /4)* (Do2-Di2)
(π*Dm*(ts-c))
Where Do and Di are external and internal diameter of shell.
ρs = density of shell material, kg/m3
Dm = mean diameter of shell,
ts = thickness of shell,
c = corrosion allowance
Now
ρs = 7700 kg/m3 =0.0077 kg/cm3
Fds=ρs* X = (7700*X) kg/m2 = (0.77*X) kg/cm2
The vessel contains manholes, nozzles etc., additional weight may be estimated 20% of
the weight of the shell.
fT,ds = 1.2 * .77*X = 0.924* (X) kg/cm2
3.3.2 Compressive stress due to weight of insulation at a height X meter:
fd(ins)
= π *Dins* tins* ρins *X/( π *Dm* (ts - c) )
= weight of insulation per unit height (X)/( π*Dm* (ts - c))
82
where Dins, tins, ρins are diameter, thickness and density of insulation respectively.
Dm = (Dc+ (Dc+2ts))/2
Assuming asbestos is to be used as insulation material.
ρins = 2700 kg/m3
tins = 2” = 5.08 cm.
Dins =Dc+2ts+2tins = 106+ (2*0.6) + (2*5.08) = 117.36 cm.
Dm = (106+ (106+ (2*0.6)))/2 = 106.60 cm.
fd(ins) = π *117.36* 5.08*2700*X
( π *106.6* (0.6 - 0.3))
= 50334.89 *X kg/m2
= 5.033489*X kg/cm3
3.3.3 Stress due to the weight of the liquid and tray in the column up to a height X
meter.
fd, liq. = ∑weight of liquid and tray per unit height X
(π*Dm* (ts - c))
The top chamber height is 0.3 m and it does not contain any liquid or tray. Tray
spacing is 457 mm.
Average liquid density in column = 770.39 kg/m3
Liquid and tray weight for X meter
Fliq-tray = [(X-.03)/0.5+1]*(π*Di2/4) ×ρl
= [(X-.03)/0.5+1]*(π*1.062/4) ×770.39
= [2X + 0.4] * 679.85 kg
fd (liq) = Fliq-tray *10/ (π*Dm* (ts - c))
= [2X + 0.4] * 679.85 *10/ (π*1066* (6 - 3))
= [2X + 0.4] * 0.677
= 1.354*X + 0.271 kg/cm2
3.3.4 Compressive stress due to attachments such as internals, top head, platforms
and ladder up to height X meter.
fd (attch.) =∑weight of attachments per unit height X
( π*Dm* (ts - c))
83
Now total weight up to height X meter = weight of top head + pipes +ladder, etc.,
Taking the weight of pipes, ladder and platforms as 25 kg/m = 0.25 kg/cm
Total weight up to height X meter = (864.26 +25X) kg
fd (attch.) = (864.26 +25X) * 10/ π*1060* (6 - 3) = 0.865 + 0.025X kg/cm2
Total compressive dead weight stress:
fdx = fds + fins +fd (liq) + fd (attch)
= 0.924X + 5.033X + [1.354X+0.271] + [0.865 +0.025X]
fdx = 7.336X + 1.136 kg/cm2
4. Tensile stress due to wind load in self supporting vessels:
fwx = Mw /Z
Where, Mw = bending moment due to wind load = (wind load* distance)/2
= 0.7*Pw*D*X2/2
Z = modulus for the section for the area of shell ≈ π*Dm2* (ts-c)/4
Thus, fwx =1.4*Pw*X2/ π*Dm* (ts-c)
Now
Pw = 25 lb/ft2 --- (from table 9.1 Brownell and Young)
= 37.204 kg/m2
Bending moment due to wind load
Mwx = 0.7*37.204*1.06*X2/2 = 13.8(X2) kg-m
fwx= 1.4*37.204*X2 /(π*1.06* (6-3) *10-3)/10-4 = 0.5213(X2) kg/cm2
5. Stresses due to Seismic load:
fsx = Msx /(π*Dm2* (ts-c)/ 4)
Where, bending moment Msx at a distance X meter is given by
Msx = [C*W*X2/3] * [(3H-X)/H2]
Where, C = seismic coefficient,
W= total weight of column, kg
H = height of column
Total weight of column = W= Cv *π*ρm*Dm*g* (Hv+ (0.8*Dm))*ts*10-3
----- (eqn. 13.75, page 743, Coulson and Richardson 6th volume)
84
Where W = total weight of column, excluding the internal fittings like plates, N
Cv = a factor to account for the weight of nozzles, man ways, internal
supports, etc.
= 1.5 for distillation column with several man ways, and with plate
support rings or equivalent fittings
Hv = height or length between tangent lines (length of cylindrical section)
g = gravitational acceleration = 9.81 m/s2
t = wall thickness
ρm = density of vessel material, kg/m3
Dm = mean diameter of vessel = Di + (t *10-3) = 1.06+ (6 *10-3) = 1.066 m
W=1.5*π*7700*1.066*9.81*(3+(.8*1.066))*6*10-3 =8771.74 N=894.16 kg.
Weight of plates: ------- (Coulson and Richardson 6th volume)
Plate Area = π *1.062/4 = 0.882 m2
Weight of each plate = 1.2*0.882 = 1.059 kN
Weight of 9 plates = 9*0.8328 = 9.531 kN = 971.55 kg.
Total weight of column = 894.16 + 971.55 = 1865.71 kg.
Let, C = seismic coefficient = 0.08
Msx = [0.08*1865.71*X2/3] * [((3*4.113)-X)/4.1132]
= 49.75X2 * [0.729-0.059X] kg-m
fsx = Msx*103/(π*Dm2* (ts-c))/4
=(49.75X2 * [0.729-0.059X ])* 103/(π*106.62* (6-3)/4)
= [1.354X2- 0.109X3], kg/cm2
On the up wind side:
ft,max = (fwx or fsx) + fap -fdx
Since the chances of, stresses due to wind load and seismic load, to occur together is rare
hence it is assumed that the stresses due to wind load and earthquake load will not occur
simultaneously and hence the maximum value of either is therefore accepted and
considered for evaluation of combined stresses.
Thus,
ft,max = 0.5213X2 + 209.47 - [7.336X + 1.136]
i.e., 0.5213X2- 7.336X + 209.47 - 1.136- 824.5 = 0
85
0.5213X2- 7.336X – 616.166 =0
=>
X = 42.12 m
On the down side:
fc,max = (fwx or fsx) - fap +fdx
The column height is 4.113 m, for which the maximum value is
fc,max
= 0.5213(4.113)2 - 209.47 + [7.336(4.113) + 1.136]
= -169.34 kg/cm2
this shows that the stress on the down wind side is tensile.
ft,max = 85% of allowable tensile stress.
ft,max = 970 * 0.85 = 824.5 kg/cm2.
ft,max = 0.5213 (X)2 – 209.47 + [7.336X + 1.136]= 824.5
Therefore, X = 38.03 m.
Hence we see that the design value of the column height is more than 4.113 m, which is
the
actual column height. So we conclude that the design is safe and thus the design
calculations are acceptable.
Hence a thickness of 6 mm is taken throughout the length of shell.
Height of the head = Dc/4 = 1.06/4 = 0.265 m
Skirt support Height = 1.0 m
Total actual height = 4.113 + 1 + 0.265 = 5.378 m
86
87
9.2 Design of Heat exchanger
Heat exchanger used is shell and tube.
The ethanol entering from vaporizer must be heated from 1000C to 2000C using ethanol,
acetaldehyde and hydrogen mixture available at 3100C.
Shell side:
Feed (mh) (ethanol) =9292.92 kg/h=20487.3573 lb/h
Inlet temperature (T1)= 1000C
Outlet temperature (T2)= 2000C
Tube side:
Inlet temperature (t1)= 3100C
Outlet temperature (t2)= 232.360C
1) Heat balance
Qh=mh Cp (T2-T1)
= 1749825.187 btu/h
2) LMTD
T1-t2
=310-200
=110 oC
T2-t1
=232.36-100
=132.36 oC
LMTD
=120.835 oC
FT=LMTD correction factor.
R=0.7764 & S=0.4761
From graph of FT Vs S
FT =0.905
∆t= LMTD×Correction Factor(FT)
LMTD (corrected)
88
∆t = 109.3550 C.
From the table, assume UD=45 btu/hr.ft2 as the process fluids are light organics
categeory.
3) Heat transfer area:
Choose overall heat transfer coefficient= 45 btu/hr.ft2
Q = UA(LMTD)
Area= 1749825.187 / (45*109.355)
A=355.58 ft2
4) Tube selection:
Length of the tube L=16’0’’(3/4 in.OD,16BWG)-Square pitch
From the table, Flow area per tube a’t=0.302in2
Outside area per linear ft a’’=0.1963 ft2
Tube inside diameter D=0.620 in
Number of tubes= 320.026/16*0.1963=113.21
Floating head type: Nearest tube count from tube count table NT= 124
Shell inside diameter =15.25 in
2 tube passes and 1 shell pass
Corrected heat transfer area=124*16*0.1963=389.4592 ft2
Corrected over all heat transfer coefficient (UD)=41.086 btu/hr.ft2
Average properties of fluids
a) Shell side (ethanol)
Cp=0.854058Btu/lb*Fo
k=0.0856794Btu*ft/hr*ft2*F
viscosity=0.321388cP
b) Tube side
Cp=0.58659Btu/lb*Fo
k=0.0611937Btu*ft/hr*ft2*F
89
Viscosity=0.188235cP
5) Flow area, at:
a’t=0.302in2
at =
𝑁𝑡 ∗𝑎′𝑡
144𝑛
=(124*0.302)/144*2
=0.130027ft2
6) Mass velocity Gt
Gt = w/at
=20487.3573/0.13003
=157561.381b/(hr)(ft2)
7) Reynolds number(Ret):
𝑅𝑒𝑡 =
Viscosity μ
𝐷 ∗ 𝐺𝑡
µ
=0.188235cP
=0.188235×2.42
=0.4555287lb/ft hr
D=0.62/12=0.0517ft
Ret =0.0517*157561.38/0.4555287
=17882.350
8) Jh Factor for heat transfer curve jh:
At L/D=16/0.0517=25.8 and Reynolds number=17882.350
Jh =66 (from graph)
9) Inside fluid heat transfer coefficient(hi)
ℎ𝑖 = 𝑗𝐻 ∗
𝐾 (𝐶µ)1/3
∗
𝐷 (𝐾)1/3
hi=127.198 btu/hr.ft2 Fo
10) Value of hi when referred to the tube outside diameter( hio )
ℎ𝑖𝑜 = ℎ𝑖 ∗
𝐼𝐷
𝑂𝐷
90
hio
=127.198*(0.62/0.75)
=105.1507 btu/hr.ft2 oF
Check the pressure drop.If unsatisfactory assume new pass arrangment.
Pressure drop in tube side
11) Obtain friction factor (f)
f=0.00023 ft2/in2 at Ret=17882.350
12) Tube side pressure drop (∆Pt)
𝑓𝐺𝑡2 𝐿𝑛
△ 𝑃𝑡 =
5.22 ∗ 1010 𝐷𝑠
∆Pt=6.97 psi
13) Return pressure drop(∆Pr)
At Gt=157561.38, V2/2g’=0.0033 (from graph)
△ 𝑃𝑟 =
4𝑛 𝑉 2
∗
𝑠 2𝑔′
∆Pr=2.83 psi
Total Pressure drop, △PT = △Pt +△P
∆PT=6.97+2.83
=9.80 psi
Since pressure drop is less than 10 psi therefore we can use one shell and two tube
pass.
Calculation of shell side pressure drop
14) Flow area, as:
Assume baffle spacing B=12in.
Clearance C’ =0.25 in.
Tube pitch PT= Clearance + Tube OD=0.75+0.25=1 in
Flow area,
aa= ID*C’B/144PT
as=15.25*0.25*12/144*1
as=0.3177ft2
15) Mass velocity Gs:
Mass velocity, Gs =W/as
Gs=20487.3573/0.3177
91
16) Reynolds number(Res):
De=0.95/12=0.0792 ft (from the graph)
𝑅𝑒𝑠 =
𝐷 ∗ 𝐺𝑠
µ
Res=(0.0792*64484.797)/(0.321388*2.42)
Res=6566.55
17) Jh Factor for heat transfer curve jh:
At Reynolds number=6566.55
Jh =46 (from graph)
18) Out side fluid heat transfer coefficient(h0):
𝐾 (𝐶µ)1/3
ℎ𝑜 = 𝑗𝐻 ∗ ∗
𝛷
𝐷 (𝐾)1/3 𝑠
ho=98.49 btu/hr.ft2 Fo
Check the pressure drop.If unsatisfactory assume new baffle spacing.
Pressure drop in shell side
19) Obtain friction factor (f)
f=0.0023 ft2/in2 at Res=6566.55(from graph)
20) No. of crosses:
N+1=12*(L/B)
=12*(16/12)
=16
Shell diameter=15.25/12=1.2708
21) Shell side pressure drop
𝑓𝐺𝑠2 𝐷𝑠 (𝑁 + 1)
△ 𝑃𝑡 =
5.22 ∗ 1010 𝐷𝑠
∆Ps=2.13 psi
22) Clean overall coefficient Uc
Uc=
𝒉𝒊𝒐 𝒉𝒐
𝒉𝒊𝒐 +𝒉𝒐
92
=(105.1507*98.49)/( 105.1507+98.49)
=50.855
R d=
𝑈𝐶 −𝑈𝐷
𝑈𝐶 𝑈𝐷
Rd=(50.855-41.086)/( 50.855*41.086)
=4.675*10-3 hr*ft2*oF/Btu
Mechanical design of Heat Exchanger:
(a) Shell side details:
Material: carbon steel
Number of shell passes: one
Diameter of shell: 388 mm
Working pressure: 0.3N/mm2
Design pressure: 0.33N/mm2
Inlet temperature: 1000C
Out let temperature:2000C
Permissible stress for carbon steel: 95N/mm2
(b) Tube side details:
Number tubes: 124
Number of passes : 2
Outside diameter: 19.05mm
Inside diameter: 15.748 mm.
Length: 4.88m
Pitch square:1 inch
Working pressure: 0.3 N/mm2
Design pressure: 0.33N/mm2
Inlet temperature: 3100C
Outlet temperature: 232.360C
93
Shell side:
1.Shell thickness:
ts= PD/(2fJ+P)
= 0.33*388/(2*95*0.85+0.33) = 0.791 mm
Minimum thickness of shell must be=6.0mm
Including corrosion allowance shell thickness is 8mm
2.Head thickness:
Shallow dished and torispherical
ts = PRcW/2fJ
P =internal design pressure,kg/cm2
Rc = crown radius = diameter of shell, mm
W= stress intensification factor or stress concentration factor for torispherical head,
W= ¼ * (3 + (Rc/Rk)0.5)
Rk = knuckle radius, which is at least 6% of crown radius, mm
Now,
Rc = 388 mm
Rk = 6% of Rc = 0.06*388 = 23.28 mm
W= ¼ * (3 + (Rc/Rk)0.5) = ¼ * (3 + (388/23.28)0.5) = 1.7706 mm
ts = PRcW/2fJ
= 0.33*388*1.77/(2*95*0.85)
= 1.403 mm.
Since head thickness is less than shell thickness use same thickness as of shell.
3. Inside depth of head (hi) can be calculated as:
ℎ𝑖 = 𝑅𝑖 − [(𝑅𝑖 −
1/2
𝐷𝑠
𝐷𝑠
) (𝑅𝑖 + ) + 2𝑟𝑖 ]
2
2
Ri= crown radius =(assume this is equal to diameter of shell)
Ri =Ds=388 mm
94
ri=6% of Ds=23.28 mm
hi=51.9128 mm
4. Effective length of heat exchanger:
Leff
=Lt +2* hi
=4.88+2*0.05191
=4.983 m
5. Transverse Baffles:
Baffle spacing =0.8*Dc = 0.8*388 = 310.40 mm
Number of baffles,
Nb+1=L/LS=4.88/0.3104=16
Nb=15
Thickness of baffles, tb=6mm
6. Tie Rods and spacers
For shell diameter, 300-500mm
Diameter of Rod = 9mm
Number of rods=4
7. Flanges
Design pressure=0.33 N/mm2
Flange material IS: 2004-1962,class 2
Bolting steel: 5% Cr-Mo steel
Gasket material: asbestos composition
Shell thickness: 8mm=go
Outside diameter of shell: 404 mm
Allowable stress of flange material: 100MN/m2
Allowable stress of bolting material = 138 MN/m2
8. Determination of gasket width:
dO/di = [(y-Pm)/(y-P(m+1))]0.5
Assume a gasket thickness of 10 mm
95
y = minimum design yield seating stress = 25.5MN/m2
m = gasket factor = 2.75
dO/di = [(25.5-0.33*2.75)/(25.50.33(2.75+1))]0.5
dO/di = 1.00678
Let, di of gasket equal 414 mm
do= 1.00678*di
do= 416.80692 mm
Minimum gasket width = (do-di)/2 = 1.4034 mm =0.0014034 m.
Taking gasket width of N=10 mm = 0.010m
Basic gasket seating width, bo=N/2=5 mm
Diameter of location of gasket load reaction is
G= di + N = 0.414 + 0.01 = 0.424 m
9. Estimation of Bolt Load
Load due to design pressure
H = πG2P/4 = 3.14*0.4242*0.33/4 = 0.0465945MN
Load to keep joint tight under operation
b = 2.5 (b0)0.5 = 5.59 mm.
Hp=π*G*(2b)*m*p = 3.14*0.424*(2*0.00559)*2.75*0.33
= 0.0135077 MN
Total operating load,
Wo= H + Hp = 0.060102294 MN.
Load to seat gasket under bolting condition
Wg = π*G*b*y = 0.1809186 MN.
Wg>Wo, controlling load=0.1809186 MN
Calculation of optimum bolting area:
Am = Ag = Wg/Sg = 0.1809186 /138 = 1.1311*10-3 m2
96
Calculation of optimum bolt size:
Bolt size, M18 X 2
Actual number of bolts =20
Radial clearance from bolt circle to point of connection of hub or nozzle and back
of flange = R = 0.027 m
Bolt circle diameter, C
C =ID + 2(1.415g + R) = 404 +2[11.315+0.027*103] = 480.63 mm
Bolt circle diameter = 0.48063 m.
Calculation of flange outside diameter Let, bolt diameter = 18 mm.
A=C+ bolt diameter +0.02 = 0.48063+0.018+0.02 = 0.51863 m. Check for
gasket width,
AbSG / (πGN) = 1.54*10-4*20*138/(3.14*0.424*10-2) = 39.21<2*y.
Where, SG is the Allowable stress for the gasket material.
Flange moment computation:
For operating condition Wo=W1+W2+W3
W1=∏*B2*P/4 = ∏*0.4042*0.33/4 = 0.0423025 MN
W2 = H-W1= 0.0465945-0.0423025 = 0.0042919 MN. W3= Wo-H
= Hp= 0.0135077 MN.
Mo=Total flange moment Mo=W1a1 + W2a2 + W3a3
a1=(C-B)/2=(0.48063-0.404)/2
a1=0.038315 m
a3=(C-G)/2=(0.48063-0.424)/2
a3=0.028315 m
a2=(a1 + a3)/2= (0.038315 +0.028315)/2=0.033315 m
Mo=0.0423025*0.038315+0.0042919*0.033315+0.0135077*0.028315
2.1462*10-3 MN-m
97
M o=
For bolting condition
Mg=Wa3 W=(Am+Ab)*Sg/2
Ab=20*1.54*10-4 =3.08*10-3 m2 Am= 1.1311*10-03 m2 W=(1.1311*10-3
+3.08*10-3)*138/2
W= 0.29056 MN
Mg= 0.29056 *0.028315 = 8.227773*10-3 MN-m
Mg>Mo , Hence moment under operating condition Mg is controlling,
Mg=M
Calculation of flange thickness
t2 = M CF Y / (B SF),
SF is the allowable stress for the flange material
K =A/B = 0.43963/0.404 = 1.0882
For choosen bolts , A=0.43963
For K = 1.3527, Y = 10 Assuming
CF =1
t2 = 8.228*10-3 *1*10/(0.404*100)
t= 0.04512 m=45.12 mm
Actual bolt spacing BS = π*C/n = (3.14*0.48063)/(20)
= 0.075458m=75.458 mm
Bolt Pitch Correction Factor
CF = [Bs / (2d+t)]0.5 = (0.075458/(2*0.018+0.04512)1/2 = 0.96446
√CF=0.98207
Actual flange thickness = √CF*t = 0.98207*0.04512 = 0.04431099 m
= 44.31099 mm.
Standard flange thickness available is 50 mm
98
Channel and channel Cover
th=Gc√(K*P/f) = 0.424*√(0.3*0.33/95) = 0.013687m =13.687mm
th=16mm including corrosion allowance
Tube sheet thickness
tts=F*G√(0.25*P/f) = 1*0.424√(0.25*0.33/95) = 0.01249m=12.49mm tts=15 mm
including corrosion allowance.
9.3 Design of Reactor
The reactor used in our process is packed bed reactor
Reactor conversion = 90%
Temperature of the reactor = 310°C (583 K)
Catalyst used is cu-co-cr2o3
Rate expression:
-rA = k CE
[15]
Where, k is rate constant
CE is the concentration of ethanol.
Basic performance equation for the Plug Flow Reactor is given by:
99
𝑥𝐴
𝑊
𝑑𝑥𝐴
=∫
𝐹𝐴𝑂
0 −𝑟𝐴
Here FAO is the molar flow rate of ethanol
W is the weight of the catalyst
T = 583 K
From literature, Activation energy E = 17150 cal/mol [19]
Frequency factor A= 8.39*105 m3 kg-1 hr-1
[19]
Production of acetaldehyde
Rate constant k = A e-(E/RT)
= 8.39*105 e-(11510/1.987*583)
= 0.31 m3 kg-1 hr-1
Putting in above equation, we have,
0.9
𝑊
𝑑𝑥𝐴
=∫
𝐹𝐴𝑂
−𝐾𝐶𝐴
0
Now,
𝐶𝐴𝑂 (1 − 𝑥𝐴 )
(1 + Ɛ𝐴 𝑥𝐴 )
𝐶𝐴 =
C2H5OH → CH3CHO + H2
Here,
VXA=1 −VXA=0
ƐA =
VXA=0
=
2−1
1
=1
CAo = FAo /Vo
CAo is the initial concentration of ethanol
Vo is the volumetric flow rate of ethanol
From,
0.9
𝑊 = 𝐹𝐴𝑂 ∫
0
3
-1
𝑑𝑥𝐴 (1 + 𝑥𝐴 )
−𝐾𝐶𝐴𝑂 (1 − 𝑥𝐴 )
-1
Putting K= 0.312 m kg hr
100
Substituting CAo = FAo /Vo
Vo= mass flow rate/density of fluid
Density of fluid (ρfluid) = 521.686 kg/m3
Vo= 9292.92/521.686 m3/hr
Vo= 17.81 m3/hr
Putting it in above equation
Weight of the catalyst required (W) = 211.5 Kg
Therefore, Volume of the catalyst = W / ρcat = 211.5/ 658 = 0.3214 m3
Volume of the catalyst = .3214 m3
Weight of the catalyst = 211.5 kg
Catalyst Dimensions
Diameter of the catalyst = 5mm
Voidage = 55% - 60%
Let the voidage be 60 % = 0.60
Density of catalyst ρ = 573 kg/m3
Volume of the vessel = VB
𝑉𝐵 − 𝑉𝑃
𝑉𝐵
= 0.6
Putting VP = 0.3214 m3, we get
VB = 0.8035 m3
Diameter of the tube = 3 in = 0.0762 m
Length of the tube = 10 ft = 3.048 m
Volume of the tube = 0.01389 m3
No. of tubes = (0.8035 / 0.01389) = 57 tubes
Pitch (square) = 1.25 * O.D = 1.25 *3.50 = 4.37 in = 111.125 mm
Now, Do = Outer diameter = 3.5 in = 0.0889 m
DB = Bundle diameter =
DO (𝑁t / 𝐾1)1/n1
For square pitch, K1 = 0.215, n1 = 2.207
DB = 1.114m
Length of the reactor = 1.2*( Length of tube)
101
= 1.2*3.048
= 3.6576 m
Conditions of Fluidization:
𝑔(ρp − ρ) =
1.75ρu2
ε3 dp
+
150(1 − ε)uµ
ε3 d2p
Average molecular weight = 46 kg/mol
ρ = density of fluid = 521.686 kg/m3
Mass flow rate (G) = 9292.2 kg/hr
Volumetric flow rate (Vo) = G/ ρ = 9292.2/521.686 =17.812 m3/hr
Area = Nt x At = 0.2599 m2
Operating velocity = Vo/area =68.52 m/hr= 0.019 m/s
Where Vom = minimum velocity of fluidization
μ (viscosity of medium) = 0.223 cP
ε = 0.6 (assumed)
ϕs (sphericity of particle) = 1
dp(diameter of the particle) = 5 mm
ρp (density of particle) = 658 kg/m3
ρ (density of fluid) = 521.686 kg/m3
Therefore, Vom = 0.0384m/s
As operating velocity is less than minimum fluidization velocity (0.0384 m/s),
fluidization does not occur.
102
9-9-3 Conditions of fluidization
Temperature
310oC
Volume of the catalyst
.3214 m3
Weight of the catalyst
211.5 kg
Volume of the vessel
0.8035 m3
Porosity of bed
0.6
Number of tubes
57
Pitch
Square
Bundle diameter
1.114 m
Length of the reactor
3.6576 m
Minimum fluidization velocity
0.0384 m/s
Operating Velocity
0.019 m/s
103
104
9.4 Design of Condenser
To calculate the amount of Ammonia required:
Q = mCP ΔT = mλ
m = Q/ λ = 1350.575 * 103 /590 = 2289.11 kg/hr
The Ammonia required = 2289.11 kg/hr
To find the LMTD:
ΔTln=
(40−(−33.86))−((−25)−(−33.86))
40−(−33.86)
)
(−25)−(−33.86)
ln(
LMTD = 30.650C
To Calculate the Heat Transfer Area:
From the DQ Kern table 8 page 840 assume the heat transfer coefficient = 45 W/m2K
Q= U A ΔTln
A = Q/U ΔTln = 1.138 m2
To Calculate the Number of Tubes:
Take the pipe to be a 16 BWG pipe with 0.75”
O.D. = 0.75” , I.D. = 0.62”=0.0157m , Length = 1 m ,
a = 0.0598 m2 /m
Number of tubes Nt = A/(a*L)
Nt = (1.138/(0.0598*1) = 19
105
To find the dimensions of the Shell:
From the table 9 (D.Q. Kern Page 842) triangular pitch 1”,
1-2 pass Heat Exchanger
Nt = 26 , Shell ID = 8 in = 20.32 cm
Corrected Heat transfer Area = 19 x 0.0598 x 1 = 1.1362 m2
Corrected Heat transfer coefficient = Uo = 45.0717 W/m2K
Shell Side FTC:
Properties:
ρ = 757.21 kg/m3
μ=.388x 10-3 N-s/m2
k = 0.171 W/m2K
Cp=1.42 kJ/kg K
Assumption: Shell Dia is equal to tube bundle dia.
Pitch: Equilateral Triangular Pitch is used.
P' = standard pitch = 1 in = 25.4 mm.
pp = pitch parallel to flow = √3/4 P‟ = 21.997 mm
pn = pitch normal to flow = (1 / 2 ) P' = 12.7 mm
Sm = Cross flow area at center of shell = [(P' - Do)Ls]Ds / P'
Do= outer diameter of the tube = 0.75” =0.01905
Ls = baffle spacing = Ds / 2 = 0.1016 m
Nb = number of baffles
Nb + 1 = L / Ls = 9
Nb = 8
Sm = 0.0116 m2
Equivalent Diameter:
106
𝐷𝑒 =
1
√3 2
𝑃 − 𝜋𝐷𝑜2
4 𝑇
2
𝜋𝐷𝑂
4
= 9.33 x 10-3 m
Shell side velocity vs = ms / Sm ρ
= 8236.5 / 3600*0.0116 x 757.21
= 0.26047 m/sec
Reynolds Number NRe= vs De ρ/μ
= 0.26 x 9.33 x 10-3 x 757.21/(.388 x 10-3)
= 4734.12
Prandtl Number NPr = ρCpμ/K
= 757.21 x17.2 x .388 x 10-3 / 0.171
= 29.55
Nusselt Number NNu = jH (NPr)1/3 (From DQ Kern, Fig 28, Page No: 838 )
jH = 30
NNu = 92.7
Heat transfer coefficient ho = NNu K / Do
= 834.75 W/m2 K.
Tube Side FTC:
Properties:
ρ = 605Kg/m3
μ=2.38 x 10-4 N-s/m2
k = 0.02219 W/m2K
Cp = 0.028 KJ/Kg ºC
Flow Area at = π/ 4 x Di2 x(Nt / Np )
= π/ 4*0.01572*(26/2) = 2.5167m2
Velocity vt = ( mt /ρ at) = 0.635 / 605x 2.5167 x 10-3
107
= 0.417 m / sec
Reynolds Number NRe = vt Di ρ/μ
= 0.417 x 605 x 0.0157/ 2.388 x 10-4
= 16586.5
Prandtl Number NPr = ρCp μ /K = 605*0.028*2.388*10-4/0.02219 =0.1823
Nusselt Number NNu = 0.023 ( NRe)0.8 (NPr)1/3 =365.295
Heat transfer coefficient hi = NNu K / Di = 51.63 W/m2K.
Overall clean Heat Transfer Coefficient:
UC =
𝒉𝒊𝒐𝒉𝒐
𝒉𝒊𝒐+𝒉𝒐
Uc = 48.622 W/m2 K
Uc > U0 assumed
Therefore the above value of shell and tube dimension can be accepted
108
Chapter 10
10
COST ESTIMATION
Cost of acetaldehyde plant of capacity 150 TPD in 1971 is Rs. 7*108 [24,7].
Chemical Engineering Plant Cost Index:
Cost index in 1971 = 132 [24]
Cost index in 2014(August) = 578.9
[24]
Thus, Present cost of Plant = (original cost) * (present cost index)/ (past cost index)
= (7*108) * (578.9/132) = Rs. 30.6992*108
i.e., Fixed Capital Cost (FCI) = Rs. 30.6992*108
10.1 Estimation of Total Capital Investment:
1. Direct Costs: material and labour involved in actual installation of complete facility
(70-85% of fixed-capital investment)
A) Equipment + installation + instrumentation + piping + electrical + insulation +
Painting (50-60% of Fixed-capital investment)
1. Purchased equipment cost (PEC): (15-40% of Fixed-capital
investment) Consider purchased equipment cost = 25% of Fixed-capital
investment i.e., PEC = 25% of 30.6992*108 = 0.25*30.6992*108 = Rs.
7.6748*108
2. Installation, including insulation and painting:
(25-55% of purchased equipment cost.)
Consider the Installation cost = 40% of Purchased equipment cost
109
40% of 7.6748*108 = 0.40 *7.6748*108 = Rs.3.0699*108
3. Instrumentation and controls, installed:
(6-30% of Purchased equipment cost.)
Consider the installation cost = 15% of Purchased equipment cost
15% of 7.6748x108 = 0.15 *7.6748*108 = Rs. 1.1512*108
4. Piping installed: (10-80% of Purchased equipment cost)
Consider the piping cost = 40% Purchased equipment cost
=40% of 7.6748*108 = 0.40 *7.6748*108
=Rs. 3.0699*108
5. Electrical, installed: (10-40% of Purchased equipment cost)
Consider Electrical cost = 25% of Purchased equipment cost
= 25% of 7.6748*108 = 0.25 *7.6748*108 = Rs.1.9187*108
Hence total cost (PEC + Installation + Instrumentation and controls+
Piping + Electrical) =16.8845*108 Rs= 54.99% of FCI
B. Buildings, process and Auxiliary: (10-70% of Purchased equipment cost)
Consider Buildings, process and auxiliary cost = 40% of PEC
= 40% of 7.6748*108 = 0.40 *7.6748*108 = Rs. 3.0699*108
C. Service facilities and yard improvement: (40-100% of Purchased equipment cost)
Consider the cost of service facilities and yard improvement = 65% of PEC
= 65% of 7.6748*108= 0.65 *7.6748*108 = Rs. 4.9886*108
D. Land: (1-2% of fixed capital investment or 4-8% of Purchased equipment cost)
Consider the cost of land = 5% of PEC = 5% of 7.6748*108 = 0.05 *7.6748*108
= Rs. 0.38374*108
110
Thus, Direct cost = Rs. 25.3267*108 = (82.49% of FCI)
II.Indirect costs: Expenses which are not directly involved with material and labour of
actual installation of complete facility (15-30% of Fixed-capital investment)
A. Engineering and Supervision: (5-30% of direct costs)
Consider the cost of engineering and supervision = 15% of Direct costs
i.e., cost of engineering and supervision = 15% of 25.3267*108
=0.15*25.3267*108 = Rs. 3.7990*108
B. Construction Expense and Contractor’s fee: (6-30% of direct costs)
Consider the construction expense and contractor’s fee = 10% of Direct costs
i.e., construction expense and contractor’s fee = 10% of 25.3267*108
=0.1*25.3267*108 = Rs. 2.5326*108
C. Contingency: (5-15% of Fixed-capital investment)
Consider the contingency cost = 10% of Fixed-capital investment
i.e., Contingency cost = 10% of 30.6992*108 = 0.10 * 30.6992*108
= Rs. 3.0699*108
Thus, Indirect Costs = Rs. 9.4015*108 =30.6 % of FCI
III. Fixed Capital Investment:
Fixed capital investment = Direct costs + Indirect costs
=
(25.3267*108)
+
(9.4015*108)
i.e., Fixed capital investment = Rs. 34.7282*108
IV. Working Capital: (10-20% of Fixed-capital investment)
Consider the Working Capital = 15% of Fixed-capital investment
i.e., Working capital = 15% of 34.7282*108 = 0.15 * 34.7282*108
= Rs. 4.3192*108
V. Total Capital Investment (TCI):
Total capital investment = Fixed capital investment + Working capital
111
= (34.7282*108) + (4.3192*108) i.e.,
Total capital investment = Rs. 39.0474*108
10.2 Estimation of Total Product cost:
I.
Manufacturing Cost = Direct production cost + Fixed charges + Plant
overhead cost.
A. Fixed Charges: (10-20% total product cost)
i) Depreciation: (depends on life period, salvage value and method of calculationabout 10% of FCI for machinery and equipment, and 2-3% for Building Value for
Buildings)
Consider depreciation = 10% of FCI for machinery and equipment, and 3%
for Building Value for Buildings)
i.e., Depreciation = (0.10*30.6992*108) + (0.03*30.6992*108)
= Rs. 3.9909*108
ii. Local Taxes: (1-4% of fixed capital investment)
Consider the local taxes = 4% of fixed capital investment
i.e. Local Taxes = 0.04*30.6992*108 = Rs. 1.2279*108
iii.Insurances: (0.4-1% of fixed capital investment)
Consider the Insurance = 0.6% of fixed capital investment
i.e. Insurance = 0.006*30.6992*108 = Rs. 0.1842*108
iv. Rent: (8-12% fixed capital investment )
Consider rent = 10% of fixed capital investment
= 10% of 30.6992*108
= 0.10* 30.6992*108
Rent = Rs. 3.0699*108
Thus, Fixed Charges = Rs. 8.4729*108
112
B. Direct Production Cost:
Now we have Fixed charges = 10-20% of total product charges –(given)
Consider the Fixed charges = 15% of total product cost
Total product cost = fixed charges/15%
= 8.4729*108/15%
= 8.4729*108/0.15
Therefore, total product cost (TPC) = Rs. 56.486*108
i. Raw Materials: (10-50% of total product cost)
Consider the cost of raw materials = 30% of total product cost
Raw material cost = 30% of 56.4860*108 = 0.30*56.486*108
Raw material cost = Rs. 16.9458*108
ii. Operating Labor (OL): (10-20% of total product cost)
Consider the cost of operating labor = 15% of total product cost
Operating labor cost = 15% of 56.486*108 = 0.15*56.486*108
Operating labor cost = Rs. 8.4729*108
iii.Direct Supervisory and Clerical Labor (DS & CL): (10-25% of OL)
Consider the cost for Direct supervisory and clerical labor = 12% of OL
Direct supervisory and clerical labor cost = 12% of 8.4729*108
= 0.12*8.4729*108
iv.
Direct supervisory and clerical labor cost = Rs. 1.01675*108
Utilities: (10-20% of total product cost)
Consider the cost of Utilities = 15% of total product cost
Utilities cost= 15% of 56.486*108= 0.15*56.486*108
Utilities cost = Rs. 8.4729*108
iv.
Maintenance and repairs (M & R): (2-10% of fixed capital investment) Consider
the maintenance and repair cost = 5% of fixed capital investment i.e. Maintenance
and repair cost = 0.05*30.6992*108 = Rs. 1.5349*108
v.
Operating Supplies: (10-20% of M & R or 0.5-1% of FCI)
113
Consider the cost of Operating supplies = 15% of M & R
Operating supplies cost = 15% of 1.5349*108 = 0.15 *1.5349*108
Operating supplies cost = Rs. 0.2302*108
vii Laboratory Charges: (10-20% of OL)
Consider the Laboratory charges = 15% of OL
Laboratory charges = 15% of 8.4729*108= 0.15*8.4729*108
Laboratory charges = Rs. 1.2709*108
viii.
Patent and Royalties: (2-6% of total product cost)
Consider the cost of Patent and royalties = 5% of total product cost
Patent and Royalties = 5% of 56.486*108 = 0.05*56.486*108
Patent and Royalties cost = Rs. 2.8243*108
Thus, Direct Production Cost = Rs.40.7686*108
C. Plant overhead Costs (50-70% of Operating labor, supervision, and maintenance or 515% of total product cost); includes for the following: general plant upkeep and overhead,
payroll overhead, packaging, medical services, safety and protection, restaurants,
recreation, salvage, laboratories, and storage facilities.
Consider the plant overhead cost = 60% of OL, DS & CL, and M & R
Plant overhead cost = 60% of ((8.4729*108) + (1.0167*108) + (1.5349*108)) Plant
overhead cost = 0.60 * ((8.4729*108) + (1.0167*108) + (1.5349*108)) Plant overhead
cost = Rs. 6.6147*108
Thus, Manufacture cost = Direct production cost + Fixed charges + Plant overhead costs.
Manufacture cost = 40.7686*108 + (8.4729*108) + 6.6147*108
Manufacture cost = Rs. 55.8562*108
II General Expenses = Administrative costs + distribution and selling costs + research
and development costs
A. Administrative costs :( 40-60% of operating labor)
114
Consider the Administrative costs = 50% of operating labor
Administrative costs = 0.5 * 8.4729*108
Administrative costs = Rs. 4.2365*108
B. Distribution and Selling costs: (2-20% of total product cost);
Includes costs for sales offices, salesmen, shipping, and advertising.
Consider the Distribution and selling costs = 10% of total product cost Distribution and
selling costs = 10% of 56.486*108
Distribution and selling costs = 0.1 * 56.486*108
Distribution and Selling costs = Rs. 5.6486*108
C. Research and Development costs: (about 3% of total product cost) Consider the Research
and development costs = 3% of total product cost Research and Development costs = 3% of
56.486*108
Research and development costs = 0.03 * 56.486*108
Research and Development costs = Rs. 1.6946*108
Thus, General Expenses = Rs. 11.5797*108
III. Total Production cost = Manufacture cost + General Expenses
= (55.8562*108) + (11.5797*108)
Total production Cost = Rs. 67.4359*108
115
11
Profitability Analysis
11.1 Gross Earnings/Income:
Selling Price of acetaldehyde per kg = 3$ = 3*62= Rs. 186
Total Income = Selling price * Quantity of product manufactured
= 186 * (150*103 /day) * (300 days/year)
Total Income = Rs. 83.7 * 108
Gross income= Total Income – Total Production Cost
= (83.7*108) – (67.4359*108)
Gross Income = Rs. 16.2641*108
Let the Tax rate be 45% (common) Taxes = 45% of Gross income
= 45% of 16.2641*108 = 0.45*16.2641*108
Taxes = Rs. 7.318845*108
Net Profit = Gross income - Taxes = Gross income* (1- Tax rate)
Net profit =16.2641*108*(1-0.45) = Rs. 8.945255*108
11.2 Rate of Return:
Rate of return = Net profit*100/Total Capital Investment
= 8.945255*108/ (39.0474*108)
Therefore, Rate of Return = 0.229 = 22.9%
11.3 Payback Period
Payback period is the duration of time in which the fixed capital investment made
in the industry at the beginning is recovered through annual net profits (and depreciation
claims). All revenue made after this time gives rise to a net positive cash flow in the
industry.
116
Payback Period = Fixed Capital Investment / (Net Profit + Depreciation)
= 34.7282*108/ ( 8.945255*108+ 3.9909*108) = 2.68 years
11.4 Break-Even Analysis
Let plant operate at ‘x’ fraction of its capacity which is its Break-Even Point (BEP) of
Zero profit.
Revenue at BEP = ₹ 186*150,000*300*x = ₹ 83.7*108* x
Total fixed costs = fixed cost + overhead + general expenses
= ₹ 8.4729*108 + ₹ 6.6147*108 + ₹11.5797*108
= ₹ 26.6673*108
Total direct costs = ₹ 40.7686*108(at full capacity)
Direct costs at BEP = ₹ 40.7686*108 * x
At break-even point:
Revenue at BEP = Total fixed costs + Direct costs at BEP
(83.7*108 * x) = (26.6673*108) + (40.7686*108 * x)
On solving, we get, x = 0.6211 = 62.11%
In terms of quantity, BEP = 0.6211 * 45 KTA = 27.9495 KTA
Plant must operate at a minimum of 62.11% of its capacity to break-even.
117
12
Chapter
PLANT LOCATION AND LAYOUT
The Geographical Location of the Plant can have a crucial effect on the Profitability of a
Project and the Scope for Future Expansion. Considerable care must be exercised in
selecting the plant site and many different factors must be considered. Primarily, the plant
should be located where the minimum cost of production and distribution can be
obtained, but other factors, such as room for expansion and safe living conditions for
expansion and safe living conditions for plant operation as well as the surrounding
community are also important.
A general consensus as to the plant location should be obtained before a design project
reaches the detailed estimate stage, and a firm location should be established upon
completion of the detailed-estimate design. The choice of the final site should first be
based on a complete survey of the advantages and disadvantages of various geographical
areas and ultimately, on the advantages and disadvantages of available real estate. The
Principal factors to be considered in selecting a plant site are:
1. Location, with respect to the Marketing Area.
2. Raw Material.
3. Transport Facilities.
4. Availability Of Labour.
5. Energy availability
6. Water supply
7. Site characteristics.
8. Environmental Impact, And Effluent Disposal.
9. Local Community Considerations.
118
10. Climate.
11. Flood and fire protection.
1. Marketing Area:
The location of markets affects the cost of product distribution and the time
required for shipping. Proximity to the major markets is an important consideration in the
selection of a plant site , because the buyer usually finds it advantageous to purchase from
nearby sources. It should be noted that markets are needed for by-products as well as for
major final products.
2. Raw Materials:
The availability and price of suitable raw materials will often determine the Site
Location. Plants producing bulk chemicals are best located close to the source of the
major raw material, because this permits considerable reduction in transportation and
storage charges. Attention should be given to the purchased price of the raw materials,
distance from the source of supply, freight or transportation expenses, availability and
reliability of supply, purity of the raw materials.
3. Transport:
The transport of materials and products to and from plant will be an overriding
consideration in site selection. A Site should be selected that is close to at least two Major
forms of transport: Road, Rail, Waterway or a Seaport. Road Transport is being
increasingly used and is suitable for Local distribution from a central warehouse. Rail
transport will be cheaper for the Long-Distance transport of bulk chemicals.
Air Transport is convenient and efficient for the movement of personnel and
essential equipment and supplies, and the proximity of the site to a major airport should
be considered.
119
4. Availability of Labour:
Labour will be needed for construction of the plant and its operation. There
should be an adequate pool of skilled Labour available locally. Skilled tradesmen will be
needed for plant maintenance. Consideration should be given to prevailing pay scales,
restrictions on number of hours worked per week, competing industries that can cause
dissatisfaction or high turnover rates among the workers.
5. Energy availability:
Power and heat requirements are high in most industrial plants, and fuel is
ordinarily required to supply these utilities. Consequently, power and fuel can be
combined as one major factor in the choice of a plant site. Electrolytic processes require a
cheap source of electricity, and plants using electrolytic processes are often located near
large hydroelectric installations. If the plant requires large quantities of coal or oil,
location near a source of fuel supply may be essential for economic operation. The local
cost of power can help determine whether power should be purchased or self-generated.


6. Water supply:
The process industries use large quantities of water for cooling, washing ,
steam generation, and as a raw material. The plant, therefore , must be located where
a dependable supply of water is available. A large river or lake is preferable, although
deep wells may be satisfactory if the amount of water required is not too great. The
level of the existing water table can be checked by consulting the state geological
survey, and the information on the constancy of the water table and the year – round
capacity of local rivers or lakes should be obtained. If the water supply shows
seasonal fluctuations, it may be desirable to construct a reservoir or to drill several
standby wells. The temperature, material content, and cost for supply and purification
treatment must also be considered when choosing a water supply.
120
7. Site characteristics:
The characteristics of the land at a proposed plant site should be examined
carefully. The topography of the tract of land and the soil structure must be considered,
since either or both may have a pronounced effect on construction costs. The cost of the
land is important, as well as local building costs and living conditions. Future changes
may make it desirable or necessary to expand plant facilities. Therefore, even though no
immediate expansion is planned, a new plant should be constructed at a location where
additional space is available.
8. Environmental Impact, And Effluent Disposal:
All Industrial processes produce waste products, and full consideration must be
given to the difficulties and Cost of their disposal. In recent years, many legal restrictions
have been placed on the methods for disposing of materials from the process industries.
The site selected for a plant should have adequate capacity and facilities for waste
disposal. In choosing a plant site, the permissible tolerance levels for various methods of
waste disposal should be considered carefully, and attention should be given to potential
requirements for additional waste – treatment facilities.
9. Local Community Considerations:
The proposed plant must fit in with and be acceptable to the local community. Full
consideration must be given to the safe location of the plant so that it does not impose a
significant additional risk to the Community.
10. Climate:
Adverse climatic conditions at site will increase costs. Abnormally Low
Temperatures will require the provision of additional insulation and special heating for
equipment and Piping. Stronger Equipments will be needed at locations subjected to high
wind loads or Earthquakes. Special cooling towers or air – conditioning equipment may
be required if the temperatures are high. Excessive humidity or extremes of hot or cold
weather can effect on economic operation of a plant, and these factors should be
examined when selecting a plant site.
121
11. Flood and Fire protection:
Many industrial plants are located along rivers or large water bodies of water,
and there are risks of flood and hurricane damage. Before selecting a plant site, the
regional history of natural events of this type should be examined and the
consequences of such occurrences considered.
Protection from losses by fire is another important factor in selecting a plant location.
In case of a major fire, assistance from outside fire departments should be available.
Fire hazards in the immediate area surrounding the plant site must not be overlooked.
12. Political and Strategic Considerations:
Capital Grants, Governments often give tax concessions, and other inducements
to direct new investment to preferred locations; such as areas of high Unemployment.
The availability of Such Grants can be the overriding consideration in site selection.
13. Costs:
The cost of construction can be minimized by adopting a layout that gives the
shortest run of connecting pipe between equipment, and at least amount of structural steel
work. However, this will not necessarily be the best arrangement for operation and
maintenance.
14. Process requirements:
An example of the need to take into account process consideration is the need to
elevate the base of columns to provide the necessary net positive suction head to a pump
or the operating head for a thermosyphon reboiler.
122
15. Operations:
Equipment that needs to have frequent attention should be located convenient to
the control room. Valves, sample points, and instruments should be located at convenient
positions and heights. Sufficient working space and headroom must be provided to allow
easy access to equipment.
16. Maintenance:
Heat exchangers need to be sited so that the tube bundles can be easily withdrawn
for cleaning and tube replacement. Vessels that require frequent replacement of catalyst
or packing should be located on the out side of buildings. Equipment that requires
dismantling for maintenance, such as compressors and large pumps, should be places
under cover.
17. Safety:
Blast walls may be needed to isolate potentially hazardous equipment, and confine
the effects of an explosion. At least two escape routes for operators must be provided
from each level in process buildings.
18. Plant expansion:
Equipment should be located so that it can be conveniently tied in with any future
expansion of the process. Space should be left on pipe alleys for future needs, and service
pipes over-sized to allow for future requirements.
19. Modular construction:
In recent years there has been a move to assemble sections of plant at the plant
manufacturer’s site. These modules will include the equipment, structural steel, piping
and instrumentation. The modules are then transported to the plant site, by road or sea.
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The advantages of modular construction are:
 Improved quality control.
 Reduced construction cost.
 Less need for skilled labour on site.
Some of the disadvantages are:
 Higher design costs & more structural steel work.
 More flanged constructions & possible problems with assembly, on site.
PLANT LAYOUT:
The economic construction and efficient operation of a process unit will depend on how
well the plant and equipment specified on the process flow sheet is laid out. Since each
plant differs in many ways and no two plant sites are exactly alike, there is no one ideal
plant layout. However, proper layout in each case will include arrangement of processing
areas, storage areas, handling areas in efficient coordination and with regard to such
factors as:
New site development or addition to previously developed site
Type and quantity of products to be produced
Type of process and product control
Operational convenience and accessibility.
Economic distribution of utilities and services.
Type of buildings and building- code requirements.
Health and safety considerations
Waste disposal requirements
Auxiliary equipment
Space available and space required.
Roads and railroads
Possible future Expansion
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Figure 12.1 Plant Layout
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Chapter 13
13
Pollution and Safety in Process Industries
Health and safety factors:
Exposure Routes: inhalation, ingestion, skin contact and eye contact.
Symptoms: irritation eyes, skin, upper respiratory system, coughing, burning sensation,
headaches, and stupefaction
First aid:
Eye Contact: Check for and remove any contact lenses. In case of contact, immediately
flush eyes with plenty of water for at least 15 minutes. Cold water may be used. Warm
water must be used. Get medical attention immediately.
Skin Contact: In case of contact, immediately flush skin with plenty of water for at least
15 minutes while removing contaminated clothing and shoes. Cover the irritated skin with
an emollient. Wash clothing before reuse. Thoroughly clean shoes before reuse. Get
medical attention immediately.
Serious Skin Contact: Wash with a disinfectant soap and cover the contaminated skin
with an anti-bacterial cream. Seek immediate medical attention.
Inhalation: If inhaled, remove to fresh air. If not breathing, give artificial respiration. If
breathing is difficult, give oxygen. Get medical attention immediately.
Serious Inhalation: Evacuate the victim to a safe area as soon as possible. Loosen tight
clothing such as a collar, tie, belt or waistband. If breathing is difficult, administer
oxygen. If the victim is not breathing, perform mouth-to-mouth resuscitation.
Warning: It may be hazardous to the person providing aid to give mouth-to-mouth
resuscitation when the inhaled material is toxic, infectious or corrosive. Seek immediate
medical attention.
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The main toxic effect of acetaldehyde is irritation of the skin and mucous membrane.
Anesthesia is the dominant effect of acetaldehyde. As an example, paraldehyde
(acetaldehyde trimer) is an extremely safe hypnotic, but its objectionable odour and taste
have restricted its use.
One of the main products of the photooxidation of acetaldehyde is peroxyacetyl nitrate
(PAN). Airborne acetaldehyde in the presence of nitrogen oxides can be converted to
PAN which is a strong eye irritant and plant toxicant.
The maximum allowable concentration in air has been set at 200 ppm. There have not
been any published reports of severe injury from the use of acetaldehyde, and one can
conclude that in handling acetaldehyde if logical safety precautions are taken. Mixtures of
acetaldehyde vapours with air are flammable if the concentration of acetaldehyde and
oxygen are higher than 4 and 9%, respectively.
Flammability
Flammability is determined by the flash point of a material. Flash point is the minimum
temperature at which a liquid forms a vapor above its surface in sufficient concentration
that it can be ignited. Flammable liquids have a flash point of less than 100°F.
The vapour burns, not the liquid itself. The rate at which a liquid produces flammable
vapours depends upon its vapor pressure.
The vaporization rate increases as the temperature increases. Therefore, flammable and
combustible liquids are more hazardous at elevated temperatures than at room
temperature
Acetaldehyde is very volatile, has a low flash point, oxidizes readily, and may form
highly explosive peroxides.
According to National Fire Protection Association (NFPA) hazard classifications for
flammable and combustible liquids acetaldehyde is a Class 1 Flammable Liquid
(Flash point: below 73°F (23°C) and Boiling point: below 100°F (38°C))
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Handling:
In handling acetaldehyde, one has to remember that it is an extremely reactive compound
that can easily oxidized, reduced, or polymerized, and us highly reactive with oxygen. It
has to be treated as a volatile, flammable (flashpoint <100°F), and toxic material. The
following is a list of precautions recommended when handling acetaldehyde:
1. Nitrogen or other inert gases should be used as a blanketing material whenever
exposure to air is a possibility.
2. Safety goggles should be used.
3. Transfers should be made in open air structures or using suitable gas mask or self
contained breathing equipment if necessary.
4. Drums should be stored out of doors, avoiding direct exposure to sunlight and
5. Acetaldehyde should be chilled before transferring and a nitrogen blanket should
be used.
6. Acetaldehyde (Class 1 Flammable Liquids) must be bonded and grounded when
transferring liquids
Shipping and storage:
Acetaldehyde is shipped in 5-10, or 55-gal drums, insulated tank trucks, and insulated
tank cars. Acetaldehyde in the liquid state is noncorrosive to most metals, but it can easily
be oxidized to acetic acid, especially in the vapour phase. Suitable materials of
construction are stainless steel and aluminium. Drums coated with phenolic resins have
also been used. If a darker colour and some iron contamination are not objectionable,
carbon steel may be used. Because acetaldehyde classified has a flammable liquid, it
requires a red DOT (Department Of Transportation) shipping lable. Bulk storage held at
low temperature and pressure is recommended over storage
in pressure vessel.
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References
[1]. JAMES M.CHURCH and HANAMANT K.JOSHI-Chemical Engineering Department,
Columbia University, New York, N.Y. Acetaldehyde by Dehydrogenation of Ethyl Alcohol,
Research Article
[2]. Jiang-Jen Lin and John F. Knifton, Acetaldehyde production from synthesis gas, Publication
US4525281 A, Jun 25, 1985
[3]. Graham A. Renberg, Production of acetaldehyde, Publication US 3291839 A, Dec 13 1966
[4]. Gerald C. Tustin, Leslie S. Depew, Nick A. Collins, Hydrogenating acetic acid using iron
oxide catalyst that contains palladium, absorbing acetaldehyde from gaseous product using
solvent containing acetic acid, Publication US 6121498 A, Sep 19 2000.
[5]. The Columbia Electronic Encyclopedia, Columbia University, Wacker process.
[6]. R. H. Perry and Don W. Green, “Perry’s Chemical Engineers’ Hand Book”, 7th Ed., McGraw Hill International edition, 2002. pp. 3-96, 3-150, 3-211.
[7]. Max S. Peters and Klaus Timmerhaus, “Process Plant Design and Economics For Chemical
Engineers”, Fifth edition, Mc-Graw Hill Book Company, 2011. pg.52-57
[8]. McCabe, Warren L. Smith, “Unit Operations of Chemical Engineering”, Seventh edition,
Mc-Graw Hill Book Company, 2005. pp. 836-850.
[9]. Robert E. Treybal, “ Mass- transfer Operations”, 3rd Ed. , Mc-Graw Hill Book Company,
1981. pp. 187-200, 300-309.
[10]. J. M. Smith, H. C. Van Ness and M. M. Abbot, “Introduction to Chemical engineering
Thermodynamics”, 6th Ed., Mc-Graw Hill Book Company, 2001. pp. 116-140.
[11]. United Nations commodity trade statistics www.data.un.orgIndian Standard (IS: 45031967): Specification for Shell and Tube Type Heat Exchangers, BIS 2007, New Delhi.
[12]. Kuppan T. Heat Exchanger Design Handbook, Marcel Dekker, Inc. 2000, New
York.
[13]. Standards of the Tubular Exchanger Manufacturers Association (TEMA), Inc. 18 ed.,
1999, New York.
[14]. Brownell L.E. and E.H. Young, Process Equipment Design, John Wiley and Sons, Inc.
1959. New York.
129
[15]. Effects of chromium promoter on copper catalyst in ethanol Dehydrogenation, Yau-jen Tu,
chiuping Li and Yu-Wen Chen, Department of chemical engineering, National Central
University, Chung Li 32054, Taiwan, J. chem. Tech. Biotechnol, 1994, 59, 141-147
[16]. Selective Oxidation/Dehydrogenation Reactions, Biomass Processing over Gold Catalysts ,
[17]. O. A. Simakova , Davis R.J., Murzin D.Y., 2013.
[18]. Principles and case studies of simultaneous design, WILLIAM L. LUYBEN.
[19]. Kinetics of the dehydrogenation of ethanol to acetaldehyde on unsupported catalysts ,
A.Peloso, M.Moresi, C.Mustachi, B.Soracco, Institute of Science and Technology , Genova,
Italy, The Canadian Journal of Chemical Engineering, Vol. 57, April, 1979.
[20]. Chemical and process design handbook, James G. Speight, McGraw-Hill.
[21]. Dowtherm Q Heat transfer fluid product technical data book by Dow Chemical Company.
[22] TRC Thermodynamic Tables- Hydrocarbons, Thermodynamics Research Center, Texas A
& M Univ, System, College Station, TX; “The NBS Tables of Chemical Thermodynamic
Properties. “ J. Phys. And Chem. Reference Data, vol, 11, sup.2,1982
[23] United Nations commodity trade statistics www.data.un.org.
[24] Cost estimation data, http://www.cheresources.com.
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