Ministry of High Education and Scientific Research

University of Al-Qadisiyah
College of Engineering
Chemical Engineering Department
'Isopropanol Production'
The phrase A Project Submitted to the College of Engineering of Al -
Qadisiyah University in Partial Fulfillment of the Requirements for
the Degree of Bachelor of Science in Chemical Engineering the
participants in the project, the name of the supervisor and the year of
graduation, taking into account the notes

Supervisor

Assist. Prof: Husham Mohammd Majeed

BY
Zainab Mohammed Kamal
Ghadeer Mohammd Abdul Razzaq

2021 1442
ii
‫كل الشكر والتقدير لالستاذ أ‪.‬م‪.‬هشام محمد مجيد لالشراف على هذا‬

‫البحث وتقديم كل النصح واالرشاد النا طيلة فترة االعداد‬

‫نسأل هللا ان يحفظه ويوفقه‬

‫‪iii‬‬
 Table of Contents

No. Subject page

1 ‫االهداء‬ ii

2 ‫شكر وتقدير‬ iii

3 Content iv

4 Chapter One 1
Introduction
5 Chapter Two 10
Material balance
6 Chapter Three 19
Energy balance
7 Chapter Four 27
Equipment Design
8 Chapter Five 36
Cost Estimation
9 References 46

iv
 Chapter One
Introduction

Introduction
Isopropanol (isopropyl alcohol or IPA) is an important chemical which is widely used as a
solvent, a chemical intermediate, and an ingredient in personal care products ,and
pharmaceuticals. It is ranked third in lower-alcohol production (following methanol and
ethanol) [1].
The global production of isopropanol is approximately 3*106 metric tons per annum (MTPA),
with the US, Europe, and Asia each contributing about 30% of the total production [2].
The isopropanol market has been steadily growing. Its compound annual growth rate is
approximately 7% and the growth is expected to continue especially in the Asia-Pacific region
primarily because of the increased use in the pharmaceutical sector [3].

Figure (1-1): Isopropanol Structure

1
The Structure of Isopropyl Alcohol or Isopropanol
The top image shows the structure of isopropyl alcohol or isopropanol with a model of pellets
and bars. The three gray balls represent the carbon atoms, which make up the isopropyl group,
bonded to the hydroxyl (red and white balls
Like all alcoholic beverages, it is structurally formed by alkane. In this case, propane. This
gives alcohol a lipophilic property (the ability to dissolve fats given their affinity). It is attached
to hydroxyl group (-OH), which on the other hand gives the structure a characteristic property
of water. Therefore, isopropyl alcohol can dissolve fats or stains. Note that the -OH group is
bound to the carbon in the medium (2°, that is, bound to two other carbon atoms), indicating
that this compound is a secondary alcohol.
Properties
Chemical formula C3H8O

Molar mass 60.096 g·mol−1

Appearance Colorless liquid, Odor
pungent alcoholic odor
Density 0.786 g/cm3 (20 °C)
Melting point −89 °C (−128 °F; 184 K(
Boiling point 82.6°C (180.7 °F; 355.8 K)
Solubility in water Miscible with water
Solubility Miscible with benzene, chloroform, ethanol,
ether, glycerin; soluble in acetone
log P 0.6 [3]
Acidity (pKa( 16.5 [4]
Magnetic susceptibility (x( −45.794*10 −6 cm3/mol
Refractive index (nD( 1.3776
Viscosity 1.96 cP at 25 °C [5]

Table (1-1) properties of isopropanol

Isopropyl alcohol is miscible in water, ethanol, ether, and chloroform. It dissolves ethyl
cellulose, polyvinyl butyral, many oils, alkaloids, gums and natural resins [9]. Unlike ethanol
or methanol, isopropyl alcohol is not miscible with salt solutions and can be separated from
2
aqueous solutions by adding a salt such as sodium chloride. The process is colloquially called
salting out, and causes concentrated isopropyl alcohol to separate into a distinct layer [10].
Isopropyl alcohol forms an azeotrope with water, which gives a boiling point of 80.37 °C
(176.67 °F) and a composition of 87.7 wt% (91 vol%) isopropyl alcohol. Water–isopropyl
alcohol mixtures have depressed melting points. It has a slightly bitter taste, and is not safe to
drink [10] [11].
Isopropyl alcohol becomes increasingly viscous with decreasing temperature and freezes at
−89 °C (−128 °F).
Isopropyl alcohol has a maximal absorbance at 205 nm in an ultraviolet–visible spectrum [12]
[13].

Isopropanol Naming System

In the naming system for organic compounds there are two systems: the common names system
and the internationally standardized IUPAC system.
Isopropyl alcohol
corresponds to the common name, ending in the suffix - ico , preceded by the word " alcohol "
and with the name of the alkyl group . The alkyl group consists of 3 carbon atoms, 2 methyl
parties, and one from the center attached to the -OH group; That is, the isopropyl group.

3
Isopropyl alcohol or isopropanol has other names as 2-propanol, second-propyl alcohol, among
others; But according to the IUPAC term, it is called propane-2-ol. According to this term,
it is the first "propane" because the carbon chain contains or is made up of three carbon atoms.
Second, the position of the OH group in the carbon chain is indicated with a number; In this
case it is 2.
The name ends with "ol", which is a property of the organic compounds of the alcohol family
because it contains a hydroxyl group (- OH(
The name isopropanol is considered incorrect by the IUPAC, due to the deficiency of the
isopropanol hydrocarbon .
Installation
The chemical synthesis reaction of isopropyl alcohol at the industrial level is basically the water
addition reaction; That is, the water. The starting product for synthesis or obtaining is propene,
to which water is added. Propene CH 3 -CH = CH 2 is an alkene, a %3D 1. hydrocarbon derived
from petroleum. By water, the hydrogen (H) is replaced by the hydroxyl (OH) group .

Water is added to alkene propane in the presence acids, to produce isopropanol alcohol.
There are two ways to perform hydration: direct, and indirect which is carried out in polar
conditions generates isopropanol.
CH3-CH = CH 2 (propene) => CH3 CHOHCH3 (isopropanol)
In direct, gas or liquid phase hydration, propane is hydrated by acid catalysis at high pressure.
In indirect hydration, propene reacts with sulfuric acid, forming sulfate esters that produce
hydrophobic isopropyl alcohol.
Also, isopropyl alcohol is obtained by hydrogenating acetone in the liquid phase. These
processes are followed by a distillation to separate the alcohol from the water, generating
anhydrous isopropyl alcohol with a yield of about 88%.

4
Uses of isopropyl alcohol
1. Medicinal uses
In medicine, isopropyl alcohol is often used in cleansing pads, which are usually cotton or
gauze moistened with a solution of 60% to 70% alcohol diluted in water. These dressings are
used by medical professionals for tasks such as sterilizing small instruments, wiping surfaces,
and cleaning a patient's skin before an injection. When applied to the skin, the water in the
liquid quickly evaporates, which cools the skin and quickly reduces the body temperature.
2. Cleaning uses
Since isopropyl alcohol is the solvent, it has many uses as a household cleaning agent. It is
often used to clean dirt from hard-to-reach areas, such as between keys on keyboards and mouse
wheels. Since it evaporates almost instantly there is little risk of shock or damage to electrical
components, and it can even be used for laser cleaning in CD and DVD drives. The alcohol
also removes stubborn glue residue and dried ink, and stains can be removed from most natural
fibers, including cotton, silk, and wood.
3.Cars Uses
Isopropyl alcohol in cars is one of the common ingredients for fuel additives in order to prevent
the build-up of water in fuel lines, and to prevent any water in the fuel from separating and
moving into ice.
4. Laboratory uses
One of the most common uses of isopropyl alcohol in the laboratory is as a disinfectant to clean
equipment and work surfaces. When used properly, it kills a large number of bacteria and other
potential contaminants, which is why it is also used as a hand sanitizer in laboratories and
hospitals. Isopropyl alcohol can be used as a preservative agent for biological samples. It can
be used as a reaction method for a number of chemical reactions. It can also be used in place
of ethanol, another type of alcohol, to extract DNA from the cell.

5
5. Industrial
Isopropyl alcohol has a wide range of industrial uses. It is widely used in the printing industries
as solvent sensitive and cleaning equipment. The manufacture of most computer components
involves the use of isopropyl alcohol, and it is used in the manufacture of paint, as well as used
as a paint remover. Even in industries that do not use this alcohol in manufacturing, it is often
used to clean and degrease machines.

Production of isopropyl alcohol

Isopropyl alcohol is one of the most widely used solvents in the world, and it plays a role as a
chemical intermediary. The global production capacity of isopropyl alcohol was 2,153,000
metric tons in 2003, with about 74 percent of global energy concentrated in Western Europe,
Japan and the United States. Isopropyl alcohol can be produced by three methods of indirect
hydration of propylene, direct hydration of propylene and catalytic hydrogenation of acetone
from different types [4].
And from application of CD to Production of IPA
Propylene hydration to IPA is an excellent candidate for application of CD technology for the
following reasons:
)1) Direct hydration of propylene is an equilibrium-limited reaction [equations (1) and (2)]
)2) Hydration can take place in the liquid phase; catalyst pellets will remain completely wetted.
(3) The reaction will be conducted at a temperature and pressure equivalent to the boiling point
of the liquid product; thus distillation and reaction will be carried out simultaneously in the
same column.
(4) Hydration is exothermic; the heat of reaction will provide a portion of energy required for
separation of the reaction mixture by distillation.
(5) Durable heterogeneous hydration catalysts with suitable physical properties are
commercially available10–13.
(6) In a CD hydration process, water will be continuously

6
consumed by fresh propylene, and an IPA-rich stream will be continuously produced. Hence,
equilibrium limitations will be overcome, and the product stream will have a higher IPA
content than product streams from conventional processes.

A major advantage of catalytic distillation over conventional xed-bed reactors is the reduction
in capital investment 7,9,14. In addition, operating costs for production of IPA are reduced, as
there is essentially no need to cool or heat the reactor. We will show that other bene to accrue
from use of CD technique, including substantially complete
1.Hydrogenation of acetone
Isopropyl alcohol is obtained by hydrogenating acetone in the liquid phase. These processes
are followed by distillation to separate alcohol from water, and generate. Anhydrous isopropyl
alcohol with a yield of about 88%. The oxidation of isopropyl alcohol produces acetone, and
in turn, the reduction of acetone by catalytic hydrogenation would generate isopropyl alcohol.
The hydrogenation of the catalyst is achieved by reacting acetone with hydrogen gas under
high pressure and in the presence of metallic catalysts, such as Raney nickel, palladium and
ruthenium

2.Indirect humidification of propylene

Isopropanol is an organic gas by-product of natural gas processing. The indirect hydration
process of propylene was the only method used to produce isopropyl alcohol worldwide until
the first direct commercial hydration process was introduced in 1951. Indirect humidification
is also called the sulfuric acid process because it requires a reaction with sulfuric acid [2] using
the indirect process, where in the liquid phase reaction occurs in two steps. The first step is the
reaction of the feedstock with sulfuric acid at a temperature of 35 ° C to form mono- and di-
alkyl sulfates, that is, fatty acids. In a second step, the resulting mixture (fatty acids) is
separated from the residual gas in the phase separator to be subsequently diluted with water
(hydrolysis) and the formation of SBA. This raw SBA was upgraded to pure SBA to be fed to

7
the MEK plant [12]. Reportedly, an 85% alcohol yield can be achieved by weight. The main
by-product is di-sec-butyl ether, which can be recovered [13].
The reactions are as follows:

3.Direct hydration of propylene to produce isopropyl alcohol (IPA)

Isopropanol is produced by direct humidification with chemical grade propylene. The modern
path avoids the need for sulfuric acid, as in the old indirect production methods based on
refining grade propylene. Worldwide, a relatively small amount of IPA is produced by
hydrogenating acetone in the liquid phase. This process is only suitable if excess acetone is
available [14]. Direct hydration is commonly used in Europe and consists of the reaction of
propylene and water, either in the gas phase or in the liquid phase. It is used in the preparation
of ethanol from ethylene, the manufacture of isopropyl alcohol from propylene, the production
of tet-butyl alcohol from isobutene, and the production of SBA (2-butanol) from a mixture of
biotin (raffinate 2) [15].
Direct hydration of propylene is performed according to the following reversible equation:
CH;CH = CH2 + H2O → CH; CHOHCH3 Equation 1
Hydration of propylene to iso-propylene alcohol [16]. The reaction is exothermic, the reaction
temperature calculated from the available thermochemical data is approximately 12 kcal/mol
(about 50 kJ/mol) over a range of 127 to 326 ° C in the steam phase [17] .
From the operational point of view, the liquid phase process is carried out at pressures of 80-
100 bar and 150 °C whereas the vapour phase reaction is generally carried out at approximately
180 °C and 35 bar [15] or 200 °C and 25 bar over a catalyst [13] .

8
 Figure (1-3) process diagram
process diagram in figure (1-3) shows a typical manufacturing of IPA at liquid phase conditions
where: (1) propylene recovery column, (2) reactor, (3) residual gas separation column, (4)
aqueous isopropanol azeotropic distillation column, (5) drying column, (6) isopropyl ether
separator, (7) isopropyl ether extraction [18]. ” An isopropanol yield of 93.5% can be realized
at 75% propylene conversion. The only important by-product is diisopropyl ether (about 5%)”
[13].

9
 Chapter Two
Material Balance

Introduction

The proposed process for production of isopropanol is based on propylene dehydration of this
material as shown in figure (1).

Figure (1) isopropanol production

10
Notation
The flowing notation will be adapted in the material balance analysis.
F: Feed to equipment
L: product from equipment
Plant production capacity
Operation Time = 300 day/year
Maintenance Time = 60 day/year
Capacity = 240205.464 ton/year
Capacity = 33361.87kg/h
Product specification = Isopropanol ( 95%wt )
Isopropanol formula = C3H8O
Molecular weight (M.wt.)= 60.096 g·mol−1
1- ED Reboiler (Condenser)

Basis=33361.87 kg/hr isopropanol (95%)

a) Overall material balance(O.M.B.)
Input=Output
F4+ F12=L10+L9
F4+ F12=33361.87+L9 …………(1)

11
b) (IPA) Component material balance
Input= Output
X*F4 =X*L9
X*F4=0.95*33361.87
X*F4 =mIPA4=31693.78 kg/h
c) (DIPE) Component material balance
Input= Output
X*F4 =X*L9
X*F4=0.02*33361.87
X*F4 =mDIPE4=667.24kg/hr
At stream 4 mIPA/mH2Ofeed =0.75
mH2Ofeed =mIPA/ 0.75
mH2Ofeed =31693.78/ 0.75=42258.37kg/h
F4=mH2O + mDIPE + mIPA
F4=42258.87++667.24+31693.78
F4=74619.39kg/h sub in equ. (1)
74619.39 + F12=33361.87+L10
F12 - L10 = -41257.52 ……………(2)
At stream 12 mDMSO=50kg/h , x DMSO=99%
𝑚𝐷𝑀𝑆𝑂
𝑥𝐷𝑀𝑆𝑂 =
𝐿12
𝑚𝐷𝑀𝑆𝑂 50𝑘𝑔/ℎ
𝐿12 = = =50.51kg/h sub. in equ. (2)
𝑥𝐷𝑀𝑆𝑂 0.99

L10 =F12 +41257.52

L10 =50.51+41257.52 =41308.03kg/h

12
 Material Stream(4) Stream(12) Stream(9) Stream(10)
name F4=74619.39Kg/hr F12=50.51 L9=33361.87Kg/hr. L10=41308.03Kg/hr.
. Kg/hr.
Weight (wt%) Weight (wt%) Weight (wt%) Weight (wt%)
(kg/hr.) (kg/hr.) (kg/hr.) (kg/hr.)
IPA 31693.78 42.5 ….. ….. 31693.78 95 ……. …….
DIPE 667.24 0.9 ….. ….. 667.24 2 ……. …….
H2O 42258.37 56.6 0.51 1 1000.86 3 41.31 0.1
DMSO ….. ….. 50 99 ….. ….. 40894.95 99
Temp.(C) 98.8 190 81.7 144.5
Press.(bar) 2 1.01 1 1.2

2-ER Reboiler (Condenser)

a) Overall material balance(O.M.B.)

Input=Output
F10=L11+L12
41308.03=L11+50.51
L11=41308.03-50.51=41257.52
b) (H2O) Component material balance

13
Input= Output
XH2O *F10 =XH2O *L11+ XH2O *L12
0.001*41308.03= XH2O11*41257.52+0.01*50.51
XH2O11= (41.30803– 0.5051)/41257.52
XH2O11=0.001
XH2O11 +XDMSO11=1
XDMSO11=1- XH2O11
XDMSO11=1- 0.001=0.999
Material Stream(12) Stream(11) Stream(10)
name F12=50.51Kg/hr. F11=41257.52Kg/hr. L10=41308.03Kg/h
r.
Weight (wt% Weight (wt%) Weight (wt%)
(kg/hr.) ) (kg/hr.) (kg/hr.)
H2O 0.5051 1 41.25752 0.1 41.31 0.1
DMSO 50 99 41216.26248 99.9 40894.95 99
Temp.(C) 190 99.6 144.5
Press.(bar) 1.01 1 1.2

3-RD Reboiler (Condenser)

a) Overall material balance(O.M.B.)

Input=Output
F3+ F5=L6+L4

14
F3+ 76589.23=L6+ 74619.39
F3-L6 =-1969.84 …………(1)
b) (H2O) Component material balance
Input= Output
XH2O *F5 =XH2O *L6+ XH2O *L4
1*76589.23 =0.05*L6+ 0.425*74619.39
L6=897519.79kg/h sub in equ.(1)
F3 =L6 -1969.84
F3 =897519.79 -1969.84
F3 =895549.94kg/h
b) (propane) Component material balance
Input= Output
Xpropane *F3 =Xpropane *L6
0.04*895549.94= Xpropane *897519.79
Xpropane=0.039
∑𝑥 = 1
XDIPE+XH2O+XIPA+Xporpylene+Xpropane=1
Xpropane =1-(XPropane+XH2O+XIPA+Xporpylene)
XDIPE=1-(0.039+0.05+0.02+0.8)=0.091
Stream(5) Stream(3) Stream(4) Stream(6)
Material F5=76589.23Kg/hr F3=895549.94Kg/hr. L4=74619.39Kg/hr. L6=897519.79Kg/hr.
name
Weight (wt%) Weight (wt%) Weight (wt%) Weight (wt
(kg/hr.) (kg/hr.) (kg/hr.) (kg/hr.) %)
API ……. ……. ……. ……. 31693.78 0.425 17950.39 2
DIPE ……. ……. ……. ……. 667.24 0.9 81674.3 9.1
H2O 76589.23 100 ……. ……. 42258.37 0.566 44875.99 5
propane ……. ……. 35821.99 4 ……. ……. 35003.27 3.9
propylene ……. ……. 859727.94 96 ……. ……. 718015.83 80
Temp.(C) 25 31.1 98.8 52.9
Press.(bar) 1.05 2 2 2

15
4-Splitter Reboiler (condenser)

a) Overall
material balance(O.M.B.)
Input=Output
F6=L8+L7
897519.79=L8+ L7 ………..(1)
b) at stream 7 propylene 2%
(propylene) Component material balance
Input= Output
Xpropane *F6 =Xpropane *L7 +Xpropane *L8
0.8*897519.79= 0.02 *L7 +0.96 *L8 ………….(2)
From equ. (1) L8=897519.79- L7 sub. in equ. (2)
0.8*897519.79= 0.02 *L7 +0.96 *(897519.79- L7)
0.02 *L7 -0.96 * L7 = (0.0.8*897519.79)-(0.96*897519.79)
(0.02-0.96)L7=71801.58-861618.99
-0.94L7=-789817.41
L7=152769.326kg/h sub in equ. (1)
L8=897519.79- L7
L8=897519.79- 152769.326
L8=744750.464kg/h

16
b) (IPA) Component material balance
Input= Output
XIPA *F6 =XIPA *L7
0.02*897519.79= XIPA *152769.326
XIPA = 0.12
c) (H2O) Component material balance
Input= Output
XH2O *F6 =XH2O *L7
0.05*897519.79= XH2O *152769.326

XH2O= 0.3
d) (DIPE) Component material balance
Input= Output
XDIPE *F6 =XDIPE *L7
0.091*897519.79= XDIPE *152769.326 , XDIPE = 0.5
∑𝑥 = 1
XDIPE+XH2O+XIPA+Xporpylene+Xpropane=1
Xpropane =1-(XDIPE+XH2O+XIPA+Xporpylene)
Xpropane=1-(0.5+0.3+0.12+0.02)
Xpropane=1-0.94=0.6
Material Stream(6) Stream(7) Stream(8)
name F6=897519.79Kg/hr. L7=152769.326Kg/hr. L8=744750.464Kg/hr.

Weight (kg/hr.) (wt%) Weight(kg/hr.) (wt%) Weight(kg/hr.) (wt%)

API 17950.39 2 18332.32 12 ……. …….
DIPE 81674.3 9.1 76384.663 50 ……. …….
H2O 44875.99 5 45830.79 30 ……. …….
propane 35003.27 3.9 91661.59 60 29790.02 4
propylene 718015.83 80 3055.39 2 714960.45 96
Press.(bar) 2 2 2
Temp.(C) 52.9 58.2 44.6

17
5-Mixer

a) Overall
material balance(O.M.B.)
Input=Output
F8+ F2=L3
744750.464+ F2=895549.94
F2=150799.476kg/h
Material Stream(3) Stream(2) Stream(8)
name F3=895549.94Kg/hr. F2=150799.476Kg /hr. L8=744750.464Kg/hr.
Weight (wt%) Weight (wt%) Weight (wt%)
(kg/hr.) (kg/hr.) (kg/hr.)
propane 35821.99 4 24184.26 4 29790.02 4
propylene 859727.94 96 1007.68 96 714960.4 96
5
Temp.(C) 31.1 24.9 44.6
Press.(bar) 2 1.05

18
 Chapter Three
Energy Balance

Introduction
The calculations will be based on the first law of thermodynamic
(The total quantity of energy is constant . When energy disappear in one form, it appears in
other forms).
∆ [(H + (1/2* U^2) + Z*g) m] = Q - Ws (for open system)
Where :
Q = heat gained by system (positive)
Ws = work done by system
Assumptions :
1-Neglect kinetic and potential energy.
∆H = Q - Ws
2-For open system shaft work (Ws) =0
3-For open system with physical operation
∆H = Q
4- For open system with chemical reaction
∆H + ∆H reaction = Q
∆H = Hout - Hin
H = m Cpmean ∆T
Cpmean = ∑ (Xi * Cpi) or Cpmean = ∑ (yi * Cpi)
∆T = T – Treference , Treference = 298 K
5-For physical mixing processs we assume ideal solution therefore ∆Hmixing = 0

19
 1- ED Reboiler (Condenser)

Overall Energy balance (O.E.B.)

∆H = Q
∆H = Hout - Hin
∆H = (H9 + H10) –( H4+ H12)
H4= m Cpmean ∆T
Cpmean = XIPA*CpIPA + XH2O*CpH2O+ XDIPE*CpDIPE
Cpmean =0.425*0.15475+0.9*0.004182+0.566*0.2167
Cpmean=0.192KJ/Kg.k
T4=98.8C+273=371.8K
H4= 74619.39Kg/h*0.192KJ/Kg.K*(371.8-298)K
H4= 1057.32KJ/h
H12= m Cpmean ∆T
Cpmean = XH2O*CpH2O+ XDMSO*CpDMSO
Cpmean =0.01*0.004182+0.99*0.154
Cpmean=0.151KJ/Kg.k
T12=190C+273=463K
H12=50.51Kg/h *0.152KJ/Kg.K*(463-298)K
20
H12= 1266.79KJ/h
H10= m Cpmean ∆T
Cpmean = XH2O*CpH2O+ XDMSO*CpDMSO
Cpmean =0.01*0.004182+0.99*0.154
Cpmean=0.151KJ/Kg.k
T10=81.7C+273=354.7K
H10=33361.87Kg/h *0.151KJ/Kg.K*(354.7-298)K
H10= 285634.32KJ/h
H9= m Cpmean ∆T
Cpmean = XIPA*CpIPA + XH2O*CpH2O+ XDIPE*CpDIPE
Cpmean =0.95*0.15475+0.03*0.004182+0.02*0.2167
Cpmean=0.152KJ/Kg.k
T9=144.5C+273=417.5K
H9=41308.03Kg/h *0.152KJ/Kg.K*(417.5-298)K
H9=750319.05KJ/h
Q=∆H =(750319.05+285634.32)-( 1057.32 + 1266.79)
Q=∆H =1033629.19KJ/h

21
2-ER Reboiler (Condenser)

Overall Energy balance (O.E.B.)

∆H = Q
∆H = Hout - Hin
∆H = (H11+ H12)- H10
H12= 1266.79KJ/h , H10= 285634.32KJ/h
H11= m Cpmean ∆T
Cpmean = XH2O*CpH2O+ XDMSO*CpDMSO
Cpmean =0.001*0.004182+0.999*0.154
Cpmean=00.153KJ/Kg.k
T11=99.6C+273=372.6K
H11=41257.52Kg/h *0.153KJ/Kg.K*(372.6-298)K
H11=470905.08KJ/h
Q=∆H =(470905.08 + 1266.79)- 285634.32
Q=∆H = 186537.55KJ/h

22
3-RD Reboiler (Condenser)

Overall
Energy balance (O.E.B.)
∆H = Q
∆H = Hout - Hin
∆H = (H6 + H4) –( H3+ H5)
H4= 1057.32KJ/h
H3= m Cpmean ∆T
Cpmean = Xpropane*Cppropane + X proplene*Cp propanlene
Cpmean =0.96*0.0885 +0.0.04*0.1208
Cpmean=0.089KJ/Kg.k
T3=24.9C+273=297.9K
H3=895549.94Kg/h *0.089KJ/Kg.K*(297.9-298)K
H3=-7970.39KJ/h
H5= m Cpmean ∆T
Cpmean = XH2O*CpH2O
Cpmean =1*0.004182 = 0.004182KJ/Kg.k
T5=25C+273=298K
H5=76589.23Kg/h *0.0041KJ/Kg.K*(298-298)K
H5= 0

23
H6= m Cpmean ∆T
Cpmean= XIPA*CpIPA + XH2O*CpH2O+ XDIPE*CpDIPE + Xpropane*Cppropane+ X proplene*Cp propanlene
Cpmean =0.02*0.15475 + 0.05*0.004182+ 0.091* 0.2167+ 0.039*0.0885 +0.8*0.1208
Cpmean=0.123KJ/Kg.k
T6=30C+273=303K
H6=897519.79Kg/h *0.123KJ/Kg.K*(303-298)K
H6= 551974.67085KJ/h
Q=∆H =(551974.67085 + 1057.32) - (-7970.39 + 0)
Q=∆H =561002.38KJ

4-Splitter Reboiler (condenser)

Overall
Energy balance (O.E.B.)
∆H = Q
∆H = Hout - Hin
∆H = (H8+ H7)- H6
H6= 551974.67085KJ/h
H8= m Cpmean ∆T
Cpmean = Xpropane*Cppropane + X proplene*Cp propanlene
Cpmean =0.96*0.0885 +0.0.04*0.1208
Cpmean=0.089KJ/Kg.k
T8=44.6C+273=317.6K
24
H8=744750.464Kg/h *0.0.089KJ/Kg.K*(317.6-298)K
H8=1299142.7KJ/h
H7= m Cpmean ∆T
Cpmean = XIPA*CpIPA + XH2O*CpH2O+ XDIPE*CpDIPE + Xpropane*Cppropane + X proplene*Cp propanlene
Cpmean =0.12*0.15475+0.3*0.004182+0.5*0.2167+0.9*0.0885 +0.0.02*0.1208
Cpmean=0.238KJ/Kg.k
T7=58.2C+273=331.2K
H7=1207122.106Kg/h *0.238KJ/Kg.K*(331.2-298)K
H7=1207122.106KJ/h
Q=∆H = (1299142.7+1207122.106) - 551974.67085
Q=∆H =1954290.135KJ/h
5-Mixer

Overall
Energy balance (O.E.B.)
∆H = Q
∆H = Hout - Hin
∆H =H3 – (H8+ H2)
H8=1299142.7KJ/h
H3=-7970.39KJ/h
H2= m Cpmean ∆T
Cpmean=0.089KJ/Kg.k
25
T2=24.9C+273=296.3K
H2=150799.476kg/h*0.089KJ/Kg.K*(296.3-298)K
H2= -22815.9607188KJ/h
Q=∆H = (-7970.39) - (1299142.7 + (-22815.9607188))
Q=∆H = -1284297.1292812KJ/h
6.Heat Exchanger

Overall
Energy balance (O.E.B.)
Q=∆H +Qsteam
∆H = Hout - Hin = H2 – H1
Qsteam = m Cpwater ∆T
Tstaem=30+273=303K
Twater=5+273=278K
∆T= ( Twater -Tsteam)=278-303
∆T=-25K
Cpwater=4.182KJ/Kg.k
Qsteam = 27000kg*4.182 KJ/Kg.k * -25k

26
Qsteam = - 2822850KJ/h
H2= -22815.9607188KJ/h
H1= m Cpmean ∆T
Cpmean=0.089KJ/Kg.k
T1= -46.7+273=226.3K
H1=150799.476kg/h *0.089KJ/Kg.K*(226.3-298)K
H1= -962296.6961988KJ/h
∆H = (-22815.9607188) -(-962296.696)
∆H = 939480736KJ/h
Q=∆H +Qsteam
Q= 939480736 + (- 2822850)
Q=1883369.264KJ/h

27
 Chapter Four
Equipment Design

1.Design Heat Exchanger

T1H=30ᶱC , T2H =5ᶱC
t1C=-46.7ᶱC , t2C = 24.9ᶱC
Choose horizontal Heat Exchanger 1 shell, 2 tube passes, counter current flow.
Δ𝑇𝑚 = (T1H −t2C)−(T2H –t1C)/{Ln (T1H −t2C)/(T2H –t1C)}
Δ𝑇𝑚 = 20.08 C
R=(T1H –T2H)/(𝑇2C− 𝑇1C)
R=(30-5)/(24.9-(-46.7))
𝑅 = 0.349
S= (T2C –T1C)/(𝑇2H− 𝑇1C)
S= (24.9 - (-46.7))/(30-(-46.7))
S=0.933
To calculate corrected logarithmic mean temperature (Ft) for 1 shell, 2 passes, from Figure
12.19 page 657 Chemical Engineering Vol 6
From Figure 12.19 Ft = 0.89
Choose standard dimension for stainless steel tube of heat exchanger from Table
12.3 page 645 Chemical Engineering Vol 6

Tube outside dimeter 25 mm

Wall thickness 2 mm
Tube length 1.83m
Inner diameter 14 mm

28
 Figure 12:19 Temperature correction factor: one shell pass; two or more even tube passes
Heat transfer area Calculation
Q=𝑈𝐷A𝐹𝑡 ∆𝑡
assumed UD =600 𝑊/𝑚2 .𝐶
Q=939481.73548KJ/h
Q=260.967KJ/h
A=260.967*1000W/(600 𝑊/𝑚2 .𝐶 *20.08C*0.89)
A =24.34m2
Surface area of one tube=𝜋 𝑑 𝑙=3.14*0.025m*1.83m=0.1437𝑚2
No. of tube=total area/ Surface area of one tube
No. of tube=73.13/0.1437=168
From Table 12.4 page 649 Chemical Engineering Vol 6
For 2 passes K1=0.249, n1=2.207

Db =25(168/0.249)(1/2.207)=478mm
Viscosity for mixture 0.00028 pa.s
29
 Density for mixture 576.6 kg/m3
Thermal conductivity 0.1195 w/m.k

Mass flow-rate, based on inlet conditions

Use square pitch,
Pt=1.25*do
Pt = 1.25∗25 𝑚𝑚=31.25 𝑚𝑚
Number of tube in center row
𝑁𝑟= Db/Pt
𝑁𝑟 =168mm/31.25mm=15.31
Tube-Side Coefficient
Mean water tem. =(-46.7+24.9 )/2=-10.5ᶱC
The cross sectional area =π/4di2=π/4 *(14)2 = 153.9 mm2
Tube per pass = 15.31/2 =7.966=8
Q =150799.476kg/h
Q=m/ρ=150799.476kg/h-1/(3600s*576.6kgm-3)
Q=0.1𝑚³/s
𝑇𝑢𝑏𝑒 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦= Q/A =0.01𝑚³/s/24.34 m2 = 0.01m/s
Re = 𝜌*u*di/ µ=576.6*0.01*0.014/0.00028
Re=288.3
𝑃𝑟=𝐶𝑝∗𝜇/𝐾=1.0058*0.00028/0.1195
Pr=0.0024
baffle cuts percent 35
From Figure 12:30 vol.6 JH Factor jH =0.1

N𝑢𝑑=0.01∗288.3∗ (0.0024)⅓=372.61
30
hi = 𝑁𝑢𝑑∗𝑘/𝑑𝑖= 372.61*0.1195/0.014=3180.47w/m2.C
Fouling factors: as neither fluid is heavily fouling, use 5000 W/m2 ºC for mixture in tube side
and use 3000 W/m2 ºC for water in shell side
Kw=45 W/m°C for stainless steel
Tube-side pressure drop
Neglect viscosity correction

Δ𝑃𝑡= 2 [8∗0.01∗(1.83/0.014)+2.5 ]*( 576.6*(0.01)2/2)

Δ𝑃𝑡=0.747pa=747kpa
Shell-Side Coefficient
Use pull-through floating head, no need for close clearance. Select baffle spacing (lB) = shell
diameter, 45 per cent cut. From Figure 12.10 page 646, clearance = 94 mm.
Ds= Dp + clearance
Ds=478mm + 94mm=572mm
lB= 𝐷𝑠/5= 572/5=114.4mm
𝑀𝑒𝑎𝑛 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 =(-46.7+24.9)/2=17.5 °𝐶
Physical properties at 268.55 °𝐶 from Perry's Chemical Engineers’ Handbook
Use Kern’s method to make an approximate estimate. Cross-flow Area

𝐴𝑠= (31.25−25) ∗572*114.4/31.25=13087.36mm2

Viscosity for mixture 0.000462 pa.s
Density for mixture 778 kg/m3
Thermal conductivity 0.201 w/m.k
Table 2: mixture properties in shell side

31
Mass flow-rate, based on inlet conditions
𝐺𝑠= =m/A=27000kgh-1/3600s
𝐺𝑠=573.07𝑘𝑔/𝑠.𝑚2
𝑢𝑠=𝐺𝑠/𝜌=573.07 𝑘𝑔/𝑠.𝑚2/778 𝑘𝑔/𝑚3=0.737m/s
Equivalent diameter, de=
𝑑𝑒=1.27/25*(31.252−0.785∗252) =0.02469m
𝑅𝑒=𝐺𝑠*𝑑𝑒/𝜇=0.737*0.02469/0.000462=39.386
𝑃𝑟=𝐶𝑝∗𝜇/𝐾=3.09*0.000462/0.201
Pr=0.0071
From Figure 12.30 page 674 Chemical engineering Vol 6 baffle cuts percent 35
𝑗𝑓=0.5

N𝑢𝑑=0.05∗2147.148*(0.0071)⅓
ho= 𝑁𝑢𝑑∗𝑘/𝑑𝑒=2745.57* 0.201/0.02469

ho=2235.157W/m2 ºC
Shell-side pressure drop
Take pressure drop as 50 per cent of that calculated using the inlet flow, viscosity correction
=1

Δ𝑃𝑠= [8*0.02*(0.572/0.02469)*(1.83/0.1144)*( 576.6 *0.04222)/2)

Δ𝑃𝑠= 12528.66pa=12.52866kpa

32
1/𝑈0= 1/22351.57+ 1/5000 + (0.025*In (25/14)/2*45) + 25/14(1/3000 + 1/3180.474)
1/𝑈0=0.001563
𝑈0=639.99𝑊/𝑚2. °𝐶
Near to assumed value, firm up design

Figure 12.30: Shell-side friction factors, segmental baffles

2. Reactor Design
1.Reaction kinetics:

Direct hydration of propylene is performed according to the following reversible equation:

CH═CH = CH2 + H2O ↔ CH; ═ CHOHCH3 Equation 1
Hydration of propylene to iso-propylene alcohol . The reaction is exothermic, the reaction
temperature calculated from the available thermochemical data is approximately 12 kcal/mol
(about 50 kJ/mol).
Reaction: CH3-CHOH-CH3 Cu/C→ CH3-CO-CH3+ H2
33
-rA=KCA K=KOexp[-Ea/RT]

Reacted = rA * dv
FA *dxA = rA * dv
Acid weight Catalysts
859727.94 kg x
50 g 2g
X =34389.12 kg = catalyst weight
CA = f Aₒ /vₒ
CA = concentration of component (A)
τ = Area under the curve * CAₒ
rA = rate of reaction
v = volume of reactor
Xa =convertion of component (A)
FA = mole rate of component
2.Volumetric Flow Rate Vₒ
1- For First Component (propane)
𝑃∗𝑀𝑤𝑡 101.3∗44
ρ= = = 1.76kg /m3
𝑅𝑇 8.314∗304.1
𝑚 895549.94∗0.96
V1 = = =488481.79m3/h
𝜌 1.76
34
2- For Second Component proplyene
𝑃∗𝑀𝑤𝑡 101.3∗42
ρ= = =1.68 kg /m3
𝑅𝑇 8.314∗304.1
𝑚 895549.94∗0.04
V2 = = =21322.62m3/h
𝜌 1.68

3- For third Component H2O

𝑃∗𝑀𝑤𝑡 101.3∗18
ρ= = =0.74 kg /m3
𝑅𝑇 8.314∗298
𝑚 76589.23∗1
V2 = = =104916.75m3/h
𝜌 0.73

Vₒ = V1 +V2 +V3
Vₒ =488481.79 + 21322.62 + 104916.75
Vₒ =614721.16m3/hr *(1/3600) =170.76m3/S
3.Catalyst volume Vc
From Perry (20-29) the superficial velocity
uv = is between (0.15- 6)
uv=(0.15+6)/2 = 3.075m/s
From Perry (10-17)
Z/D = (1:4)
Vₒ = uv * A
ᴨ
A= 𝐷 2
4
ᴨ
Vₒ = uv * 𝐷2
4

D= [(Vₒ*4)/ ( uv*ᴨ)](1/2)
D= [(170.76*4)/ (3.075 *ᴨ)](1/2)=8.41m
Z/D = 4 (max)
Z = 4 *D =4*8.41=33.64m
ᴨ
𝑉𝐶 = 𝐷2 ∗ 𝑍
4

35
 ᴨ
𝑉𝐶 = (8.41)2 ∗ 33.64 = 1868.69𝑚3
4
4.Total Reactor Volume
We allow 75% of volume of fluid as the free space in the fermenter.
VC
∗ 100 = 75%
Vr
𝑉𝐶 1868.69
Vr = = =2491.59 m3
0.75 0.75
ᴨ 2 ᴨ
𝐴= 𝐷 = (8.41)2 = 55.55 𝑚2
4 4
𝑉𝑟 = 𝐴 ∗ 𝐻
𝑉𝑟 2491.59
𝐻= = = 4.49 𝑚
𝐴 55.55
5.Space-time
The time needed to treat the one reactor volume is called the space time.
τ = v/vₒ
τ =2491.59m3/170.76m3/S =14.59second

Figure (1): Packed column reactor

36
 Chapter Five
Cost Estimation

Introduction
The choice of appropriate equipment often is influenced by considerations of price. A lower
efficiency or a shorter life may be compensated for by a lower price. Funds may be low at the
time of purchase and expected to be more abundant later, or the
1. Reactor Cost
Material of Construction = Carbon Steel
Height of Reactor = 5.91 m
Diameter= 1.108 m
From table (5:1) Suitable material is carbon steel
C=15000$, Capacity=4 m3, n=0.40
Ce=CS n
=15000 (4)0.40
Ce=$ 65601.72
Cost index in 2021 = 782.8$
Ce=$ 65601.72 in 2004
𝑐𝑜𝑠𝑡 𝑖𝑛𝑑𝑥 𝑖𝑛 2021
Cost in 2021 = (cost in 2004) * ( )
𝑐𝑜𝑠𝑡 𝑖𝑛𝑒𝑥 𝑖𝑛 2004
782.8
Cost in 2021= 65601.72*
111

=462639.877$
2. Heat Exchanger Cost
Shell and Tube Heat Exchanger
Area = 24.34 m2
From Figure (5:1)

37
Exchanger Cost in 2004 = 1000 $
Pressure Factor 1-10 = 1
Type Factor U-tube = 0.85
Equipment Cost in 2004 = 1000 $ * Pressure Factor * Type Factor
Equipment Cost in 2004 = 1000 $ *1*0.85
Equipment Cost in 2004 = 850 $
2004 Cost Index = 509.2$
2021 Cost Index = 675.3$
Equipment Cost in 2021 = Ce*Cost Index2021/ Cost Index 2004
Equipment Cost in 2021 = 850 $*675.3/509.2
Equipment Cost in 2021 =1127.26$

Figure (5:1) heat transfer factor

38
Table (5:1):purchase cost of miscellaneous

39
3. Distillation Column Cost
Material of Construction = Carbon Steel
Height of Column = 7.64 m
Diameter of column = 0.8 m
Form Figure (5:2)
Bare cost= 18*1000 =18,000 $
Purchased cost= Bare cost from fig * Material factor * Pressure factor
Material factor = C.S = 1.0
Pressure factor = 1-5 bar = 1.0
Cost of vessel = 18,000 * 1.0 * 1.0
Cost of vessel in 2004= Ce = 18,000 $
Plate diameter = 0.8 m
Plate material = C.S
Plate type = Sieve plate
From Figure (5:2)
Installed Cost = Cost form figure * Material factor
Cost per plate = 290* 1.0
Cost per plate = 290 $
No of plates = 16
Cost of plates in 2004 = 16* 290 = 4640 $
Total Cost of equipment in 2004= Ce = Cost of Column + Cost of plates
Total Cost = Ce = 18000 + 4640 = 22640 $
2004 Cost Index = 541.2
2021 Cost Index = 676.6
Equipment Cost in 2021 = Ce× (2021 Cost Index / 2004 Cost Index)
=22640*(1990/541.2)

40
Equipment Cost in 2021 = 83247.59$
Two distillation
2*83247.59*=166495.18$
4.mixer
pressure 1bar
Cost of vertical vessel(mixer) =9500 $ From figure (6-5 b) vol.6
Purchased cost=(bare cost from figure)*material factor*pressure factor .
material stainless steel (S.S)
pressure = 1 bar.
Purchased cost= 9500*2*1
=19000 $
Cost in 2021=C0 * cost index in year 2021/cost index in year 2004
= 9500*(2300/2000)
=10925 $
Cost for agitator mixer:-
Type propeller
Ce=C*(S)n
From table 7-1 vol.6
... Ce=1900*(6)0.5
=4654.03 $
... Cost of two mixer in 2021=10925*2=21850 $

41
5.Splitter Reboiler (condenser)
Cost in 2004=300$
Purchased cost =17000*1*1
Purchased cost =17000$
Cost in 2021=C0 * cost index in year 2021/cost index in year 2004
Cost in 2021=17000*(3500/300)
Cost in 2021=198333.3$

Figure (5:2): Distillation Cost

42
 Equipment's Cost($)
Reactor 462639.877
Heat Exchanger 1127.26
Distillation two 166495.18
Mixer two 21850
Splitter (condenser) 198333.3
Total 850445.617
Table (5:3): Total Equipment Cost ($)
6.Direct Cost
From Table 4(5:4)
Direct Cost = 3.6 * Equipment’s Cost
= 3.6 * 850445.617$
= 30161604.2 $
7.Indirect Cost
Indirect Cost = 1.44 * Equipment’s Cost
Indirect Cost = 1.44 * 850445.617$
= 1190623.8$
8.Fixed Capital Investment (FCI)
FCI = Direct Cost + Indirect Cost
FCI = 30161604.2 + 1190623.8
= 4252228$
9.Working Capital Investment (WCI)
WCI = 15% FCI
= 0.15*4252228$
= 637834.2 $

43
10.Total Capital Investment
Total Capital Investment = WCI + FCI
= 4252228$ + 637834.2 $
= 4890062.2 $
11.Raw Materials
Propylene cost =38.8$/y
Flow rate = 33.36 ton/h
Flow rate=240192ton/y
Cost =9319449.6 $/y
12.Utilities
Water:
Flow rate =76.58927ton/h=551442.744ton/y
Cost = 0.01 $/ton
Total Cost = 5514.42744$/y
13.Operating Labor
Minimum Wage = 0.21 $/h
Capacity = 200 ton/day
=200,000 kg/day
Operating Labour = 40 h/day
Processing Step = 7
From Figure (5:3)
Operating Labour = 40 * 7= 280 h/day
Operating Cost of Labour = 280*330*0.21 = 19404 $

44
 Table (5:4): direct cost

Figure (5:3): Operating Labor (h/day)

45
14.Total Production Cost
Variable Cost:
From Table (5:6)
Variable Cost = Raw Materials + Utilities
Variable Cost = 9319449.6 $/y + 5514.42744$/y
Variable Cost =932496.27$/y

Table (5:5): Variable & Fixed Cost

15.Fixed Cost
Function Cost ($)
Maintenance 120897.93
Operating Labor 19404
Laboratory Cost 3880.80
Supervision 3880.80
Plant Overheads 9702.00
Capital Charges 241795.86
Insurance 24179.59
Local Taxes 48359.17
Royalties 24179.59
Total 496279.74
Table (5:6): Fixed Cost
46
 Referance

1. R. Selvaraj, R. Praveenkumar & I. Ganesh Moorthy A comprehensive review of biodiesel

production methods from various feed stocks ISSN: 1759-7269 (Print) 1759-7277 (Online)
Journal homepage.
2. https://en.wikipedia.org/wiki/Isopropyl_alcohol.
3. Sevil Yücel, Pınar Terzioğlu and Didem Özçimen, Lipase Applications in Biodiesel
Production ,http://dx.doi.org/10.5772/52662.
4. Using [EMIM][OAC] as Entrainer for Isopropyl Alcohol Dehydration via Extractive
Distillation Hung-Hsing Chen, Meng-Kai Chen, and I-Lung Chien.
5. Design and optimization of isopropanol process based on two alternatives for reactive
distillation W.J. Chua, G.P. Rangaiah, K. Hidajat.
6. Hydration of Propylene to Isopropanol ,Nehemiah Diala ,University of Central Florida.
7. Strem Chemicals, Inc. www.strem.com, Catalog # 06-3123 Novozym 435.
8. Energies 2013, 6, 2052-2064; doi:10.3390/en6042052, energies ISSN 1996-1073.
9.Ching Kuan, Chia-Chi Lee, Bing-Hong Tsai, Shiow-Ling Lee, Wei-Ting Lee and Chi-Yang
Yu , Optimizing the Production of Biodiesel Using Lipase Entrapped in Biomimetic
Silica.
10. Michael Dore Gagnon, University of New Hampshire, University of New Hampshire
Scholars' Repository, Engineering lipases and solvents for trans/-esterification of used
vegetable oils.

47