Production of Sulfuric Acid

A project report Submitted to the Engineering Department

of Chemical of the University of Diyala in a partial fulfillment
for the Degree of B.Sc in Chemical Engineering

By
Ahmed Hasan M. Ali

Aya Jawad Hasan

Abdullah Burhan Raheem

Supervisor:
Assist. Prof. Dr. Ahmed D. Wehaib

April 2016

1
 Chapter One
Introduction
1.1 Introduction
Sulfuric acid : sulfuric acid is a chemical compound , colorless , odorless
, extremely , oily liquid and highly corrosive strong mineral acid with the
molecular formula H₂SO₄ and molecular weight 98.079 g/mol . which is
soluble in water at all concentrations and sometimes called oil of vitriol.

History :

Sulfuric acid has been an important item of commerce since the early to
middle 1700 s . It has been known and used since the Middle Ages . In
the eighteenth and nineteenth centuries it was produced almost entirely by
the chamber process in which oxides of nitrogen (as nitrosyl compounds)
are used as homogeneous catalysts for the oxidation of sulfur dioxide .
The product made by this process is of rather low concentration (77-78 wt
% H₂S0₄) . This is not high enough for many of the commercial uses of
the 1990 s . The chamber process is therefore considered obsolete for
primary sulfuric acid production . However, more recently modifications
to the chamber process have been used to produce sulfuric acid from
metallurgical off-gases in several European plants . During the first part
of the twentieth century , the chamber process was gradually replaced by
the contact process . The primary impetus for development of the contact
process came from a need for high strength acid and oleum to make
synthetic dyes and organic chemicals . The contact process employing
platinum catalysts began to be used on a large scale late in the nineteenth
century. The pace of its development was accelerated during World War I
in order to provide concentrated mixtures of sulfuric and nitric acid for
explosives production . In 1875 a paper by Winkler awakened interest in

2
the contact process first patented in 1831. Winkler claimed that
successful conversion of SO₂ to SO3 could only be achieved with
stoichiometric undiluted ratios of SO₂ and O₂ . Although erroneous this
belief was widely accepted for more than 20 years and was employed by
a number of firms . Meanwhile other German firms expended a
tremendous amount of time and money on research . This culminated in
1901 with Knietsch's lecture before the German Chemical Society
revealing some of the investigations carried out by the Badische Anilin-
und-Soda-Fabrik . This revealed the abandonment of Winkler's theory
and further described principles necessary for successful application of
the contact process . in 1915 an effective vanadium catalyst for the
contact process was developed and used by Badische in Germany .

This type of catalyst was employed in the United States starting in 1926
and gradually replaced platinum catalysts over the next few decades .
Vanadium catalysts have the advantages of exhibiting superior resistance
to poisoning and being relatively abundant and inexpensive compared to
platinum . After World War II , the typical size of individual contact
plants increased dramatically in the United States and around the world in
order to supply the rapidly increasing demands of the phosphate fertilizer
industry . The largest sulfur burning plants as of the mid-1990 s produce
approximately 3300 metric tons of acid per day . Plants using sulfur in
other forms especially SO₂ from smelting operations (metallurgical
plants) have also increased in size . One metallurgical plant has been built
to produce 3500 metric tons of acid per day . Another significant change
in the contact process occurred in 1963 , when Bayer AG announced the
first large-scale use of the double-contact (double-absorption) process . In
this process SO₂ gas that has been partially converted to SO 3 by catalysis
is cooled passed through sulfuric acid to remove SO 3 reheated , and then

3
passed through another one or two catalyst beds . Through these means,
overall conversions can be increased from about 98% to 99.7% , there by
reducing emissions of unconverted SO₂ to the atmosphere . Because of
world wide pressures to reduce SO₂ emissions , most plants as of the mid
— 1990 s utilize double-absorption . An early U.S . patent disclosed the
general concept of this process but apparently was not reduced to practice
at that time .

Uses of sulfuric acid :

The important uses of sulfuric acid are :-

1 - Production Fertilizer

2 - Paints and dyes .

3 – Uses in water treatment .

4 - Making Detergents

5 - A Dehydrating agent.

6 - Metal treatment and anodizing .

7- A Catalyst .

8- The acid in a car battery .

4
1.2 Physical properties of sulfuric acid:

 It is colorless, heavy and only liquid .

 It has specific gravity of 1.84 .
 It is odorless dilute solution it has sour taste.
 It is extremely corrosive to skin and all body tissues hence causes
sever burns .
 Its boiling point is 348 ℃ .
 Its M.P is 10-15 ℃ .
 It is soluble in water in all preparations with the evolution of a large
amount of heat .
 Pure H₂SO₄ is non-conductor of electricity as it is not dissociated .
 With addition of water it becomes good conductor .

There are many fingers shows the gravity with concentration , viscosity
with concentration , boiling points with concentrations and other bellow :

5
6
1.3 Chemical Properties Of H2SO4 :

The Chemical properties of Sulfuric acid are as under.

1.3.1. Reaction with water:

Sulfuric acid is a di basic acid i-e when dissolved in water one molecule
produces two hydrogen atoms.

H2SO4 2H + SO4

1.3.2. Reaction with alklies:

As it is di-basic acid as it forms two series of salts with bases i-e

hydrogen sulfates and sulfates.

H2SO4 + NaOH NaHSO4 + H2O

H2SO4 + 2 NaOH Na2SO4 + H2O

1.3.3. Reaction with Carbonates and Bicarbonates :

Sulfuric Acid reacts with Carbonates, bicarbonates to form salts and CO 2 .

H2SO4 + Na2CO3 Na2SO4 + H2O+ CO2
H2SO4 + NaHCO3 NaHSO4 + H2O + CO2

1.3.4. Reaction with Metals

Action of sulfuric acid is different under different conditions. Cold and

dilute acid gives hydrogen with metals like Zn,Mg,Al, Fe ….. etc.
Zn + H2SO4 (dil) ZnSO4 + H2
Mg + H2SO4 (dil) MgSO4 + H2
Al + H2SO4 (dil) Al2 (SO2)3 + 3 H2

7
Hot and concentrated acid gives SO2, with metals like Cu, Hg, Ag and Pb
…. Etc.

Cu + 2H2SO4 CuSO4 + H2O + SO2

Pb + H2SO4 PbSO4 + H2O + SO2

1.3.5. Dissociation of H2SO4

H2SO4 dissociates into SO 3 and H2O on boiling .

1.3.6 Oxidation Agent :

Sulfuric acid is a powerful oxidizing agent . some of its oxidizing

properties are as under :

I - It oxidizes carbon, to Co 2 and sulfur to So 2

C + H2SO4 CO2 + SO2 + H2O

S + 2 H2SO4 2 SO2 + H2O

II - it oxidizes " HI " and " HBY " to iodien " I " and bramine " Br 2"
HI + H2SO4 I2 + SO2 + H2O
HBr + H2So4 Br2 + SO2 + H2O

III - It oxidizes " H2S " to "S".

H2S + H2SO4 S + SO2 + H2O

8
 1.3.7 Drying and Dehydration Agent

Concentrated sulfuric acid has a strong attraction towards water due to

this property, the acid is not only used in drying the wet substances but
can also extract water from compounds.

Examples to Drying and Dehydration :-

- Blue copper sulfate (penta hydrated) is dehydrated and is changed into

white anbydrous copper sulfate :

CuSO4 . 5 H2O CuSO4 + 5 H2O

- Sugar is dehydrated to carbon :

C12H22O11 C + 11 H2O

- Formic acid is dehydrated to carbon monoxide :

HCOOH CO + H2O

9
1.4 Methods of Production :

There are two major processes for production of sulfuric acid ( H2SO4 ) :-
1-lead chambers
2-contact process
this processes available commercially.

1.4.1 The lead chambers process :

Fig. 1.5 lead chambers process

It is the older of the two processes , is used to produce much of the

acid used to make fertilizers it produces a relatively dilute acid (62% –
78% H2SO4) . In this process sulfur enters to the furnace with amount of
air ( burin sulfur ) and produce sulfur dioxide ( SO 2 ) , the hot sulfur
dioxide gas go to the ( Dust Trap ) to purification from dust and other
impurities and mixed with nitrogen oxides in specific tower then the
mixture enters the bottom of tower called " Glover Tower " where flaking
chambers acid from the top of this tower like spray to form tower acid or
10
" Glover acid " ( about 78% H2SO4 or less ) this tower contain Ceramic or
Porcelain rings as packing . From the bottom Glover Tower a mixture of
hot gases ( including sulfur dioxide and trioxide , nitrogen oxides ,
nitrogen , oxygen , acid vapor and steam ) is transferred to a lead-lined
chambers in which sulfur dioxide ( SO 2 ) oxidize to sulfur trioxide ( SO3 )
where it is reacted with more water . The chambers may be a large boxes
like rooms .
Notes the gases is colorless in the first room due to it do not contain
higher nitrogen oxides where , reciting the gas mixture to the color red
italic to brown due to higher nitrogen oxides forming the spray water
from Ceilings of rooms make on dissolved some gases and acid vapor
which released from oxidize ( SO 2 ) to ( SO3 ) and collected the acid
formed and is condenses on the walls and collects on the floor of the
chambers . The gases that interring Champers passes through each in
succession . The acid produced in the chambers , often called chambers
acid or fertilizer acid , contains 62% to 68% H2SO4 . After the gases (
including sulfur oxides , nitrogen oxides and nitrogen gas that produced
from air used in sulfur burning ) have passed through the chambers they
are passed into a reactor called the Gay-Lussac Tower where they are
washed with cooled dilute sulfuric acid , the nitrogen oxides and un
reacted sulfur dioxide ( SO 2 ) dissolve in the acid to form the nitrous
vitriol used in the Glover tower .

11
1.4.2 Contact process :

Fig. 1.6 Contact Process .

This method discovered the first one in 1831 by Englishman his name
Phillips included his patent the basic features for modern contact
process , even in 1900, it has not built a factory in a manner of contact
in the United States , even though this method has gained importance in
Europe duties by the urgent need for Oleum and acid with high
concentration for slfonation and especially in the manufacture of dyes .
And the basic principle of the process begin by burning sulfur with an
abundance of air to produce sulfur dioxide ( SO2 ) the increase in
oxygen useful to the contact process because it make the process more
efficient. In both processes ( contact process and lead chambers ) sulfur
dioxide (SO₂) is oxidized to sulfur trioxide ( SO3 ) . The sulfur dioxide

12
is obtained by burning sulfur by burning pyrites ( iron sulfides ) by
roasting nonferrous sulfide ores preparatory to smelting , or by burning
hydrogen sulfide gas H₂S . Then in contact process convert sulfur
dioxide ( SO₂ ) to sulfur trioxide ( SO3 ) with vanadium oxide V₂O5 as a
catalyst which is in the form of tiny cylinders packaged the reaction
tower . In the final stage , sulfur trioxide is converted to sulfuric acid .
The sulfur trioxide gas is absorbed into very concentrated sulfuric acid (
a 98 per cent solution of H₂SO₄ in water ) , producing a thick fuming
liquid called oleum . The oleum is mixed carefully with water , and the
sulfuric trioxide in the oleum reacts with the water to formed sulfuric
acid with 98% concentration as follows :

SO3 (g) + H2O (L) H₂SO₄ (L)

1.5 Selection of production method :

The contact process is the selected method commercially for
sulfuric acid production because it is the newer method and in which the
sulfuric acid produced purely and in high concentrations , more safety ,
less pollute , low cost and the equipment in this process is available .
The Contact Process is important because it plays a big role in
manufacturing . Without The Contact Process , manufacturing would be
much more expensive and time consuming .
1.6 Selection of production capacity :
We will take a ten tons per day as production capacity which mean
3000 tons per year .

13
 Chapter Two
Material And Energy Balance

2.1 Material balance

Fig. 2.1 Contact process with number of streams

given :
To design a 10 TPD capacity H2SO4 acid plant
Purity :
Product which is to be manufactured is assumed to have strength of 98%
acid .
10 TPD implies that we have 10 * 10 ³ / 24 = 416.666 ( kg / hr ) of Acid
With 98% purity
Basis :
1 hour of operation .
the acid that is produced per hour = ( 0.98 * 416.666 = 408.332 kg
kmoles of Sulfuric acid to be produced = 408.332 / 98 = 4.166 kmole
It’s assumed that overall absorption of the acid is 100 %

14
SO3 (g) + H2O (L) H2SO4 (L)
Then, SO3 required = 4.166 kmole
Also its assumed that the overall conversion of SO 2 to SO3 in the reactor
is 99.8%
SO2 (g) + 1⁄2 O2 (g) SO3 (g)
Then SO2 required = 4.166 / 0.998 = 4.174 kmole
Assuming 100% combustion of Sulfur ,
S(L) + O2 (g) SO2 (g)
Then S required = 4.174 kmole = 133.568 kg ( liquid )
Amount of oxygen required to convert ( 1 ) kmole of S to SO 3
= 1.5 kmole according to the equations below :
S(L) + O2 (g) SO2 (g) 1 kmole of O2 1.5 kmole
SO2 (g) + 1⁄2 O2 (g) SO3 (g) 0.5 kmole of O2
Then, amount of Oxygen required = 4.174 x 1.5 = 6.261 kmole
As cited in the literature that some amount of excess oxygen must be used
Using 20% excess
O2 required to process = 6.261 * 1.2 = 7.5132 kmole

From this, the total dry air that is coming in can be calculated as :
Air O2
100 21
X 7.5132
X = (7.5132* 100)/ 21 = 35.777 kmole of Dry air inlet to process
= 35.777 * 29 = 1037.533 kg
Q = 1036.421 kg

15
Sulfur burner :
The combustion reaction takes place inside the burner where Sulfur is
oxidized to Sulfur Dioxide .
Dry air entering to burner = 1037.533 kg / 29 ( kg / kmole )
= 35.777 kmole
Sulfur entering ( S ) = 4.174 kmole
Oxygen entering O2 = 7.5132 kmole

As mentioned before we have assumed 100% combustion of sulfur,

SO2 leaving the burner = 4.174 kmole
Oxygen leaving the burner = 7.5132 - 4.174 = 3.339 kmole

Nitrogen leaving the burner = 35.777 * 0.79 = 28.263 kmole

SO2 percentage in the stream leaving the burner = 0.116 and this value is
nearly to note in the reference by author " NORMAN SHREVE " which
was say the percentage of SO 2 not more than 0.12 .

16
 Comp. equipment Stream m ( Kg ) no. moles ( Kmole )
S Burner 1 133.568 4.174
O2 Burner 2 240.4224 7.5132

N2 Burner 2 395.682 28.263

SO2 Burner 3 133.568 4.174

O2 Burner 3 106.848 3.339

N2 Burner 3 395.682 28.263

Reactor :
O2 entering reactor = 3.339 kmole
N2 entering reactor = 28.263 kmole
SO2 entering reactor = 4.174 kmole
SO3 entering reactor = 0.0 kmole
In the reactor the reaction below is take place :
SO2 (g) + 1⁄2 O2 (g) SO3 (g
From chemical equation above we can find O2 required for this reaction :
SO2 O2
1 0.5
4.174 X
X = 2.087 kmole of O2 required for the reaction
O2 leaving the reactor = 3.339 – 2.087 = 1.252 kmole
N2 leaving the reactor = 28.263 kmole
SO2 leaving the reactor = 4.174 * 0.002 = 0.008 kmole
SO3 leaving the reactor = 4.174 * 0.998 = 4.166 kmole

17
 Comp. equipment stream m ( Kg ) no. moles ( Kmole ) state
SO2 converter 3 267.136 4.174 In to stage 1
O2 converter 3 106.848 3.339 In to stage 1
N2 converter 3 395.682 28.263 In to stage 1
SO2 converter 4 69.45536 1.08524 In to stage 2
O2 converter 4 57.42784 1.79462 In to stage 2
N2 converter 4 395.682 28.263 In to stage 2
SO3 gen. converter 4 247.1008 3.08876 In to stage 2
SO3 total converter 4 247.1008 3.08876 In to stage 2
SO2 converter 5 20.30234 0.317224 In to stage 3
O2 converter 5 45.13958 1.410612 In to stage 3
N2 converter 5 395.682 28.263 In to stage 3
SO3 gen. converter 5 61.44128 0.768016 In to stage 3
SO3 total converter 5 308.5421 3.856776 In to stage 3
SO2 converter 6 8.815488 0.137742 In to stage 4
O2 converter 6 42.26787 1.320871 In to stage 4
N2 converter 6 395.682 28.263 In to stage 4
SO3 gen. converter 6 14.35856 0.179482 In to stage 4
SO3 total converter 6 322.9006 4.036258 In to stage 4
SO2 converter 7 0.512 0.008 Out from stage 4
O2 converter 7 42.13987 1.316871 Out from stage 4
N2 converter 7 395.682 28.263 Out from stage 4
SO3 gen. converter 7 10.35152 0.129394 Out from stage 4
SO3 total converter 7 333.2522 4.165652 Out from stage 4

18
Absorber tower :
In absorber tower the sulfur trioxide ( SO 3 ) absorbed by sulfuric
acid with purity 98% and leaving the as a product with purity 100% but
we Must be cooled to avoid disintegration the acid to SO 3 gas and water
in acid boiling point . also mentioned in the literature that “ its required to
take the strength of the solvent H2SO4 for absorption of SO 3 not to increase
by more than1 – 2 % and the best absorption will occur when the
absorbing acid has the strength between the range 97.5 to 99 % ” .
SO3 (g) + H2O (L) H2SO4 (L)
From chemical equation above we can find H2O required for this reaction
:
SO3 H2O H2SO4
1 1 1
4.166 4.166 4.166
H2O = 4.166 kmole required for the reaction to produce H2SO4

Solvent acid entering A.T H2O in solven H2SO4 in solvent

100 kmole 2 kmole 98 kmole
( X + 4.166 ) kmole 4.166 kmole X
X = 204.134 kmole H2SO4 in solvent acid entering
Solvent entering = 4.166 + 204.134 = 208.3 kmole
= 840.572 Kg

19
Comp. equipment Stream m ( Kg ) no. moles ( Kmole )
SO2 absorber 7 0.512 0.008
O2 absorber 7 40.064 1.252
N2 absorber 7 395.682 28.263
SO3 absorber 7 333.28 4.166
H2SO4 absorber 8 20413.4 208.3

SO2 absorber 9 0.512 0.008

O2 absorber 9 40.064 1.252
N2 absorber 9 395.682 28.263
H2 SO4 absorber 11 20005.13 204.134
H2 O absorber 11 74.988 4.166

20
Diluter :

Dilution water is added in the tank to bring down the concentration to

the desired 98% and this is calculated as
In = Out
840.572 Kg + G + W = D + K + P
W : Water
K : amount of H2SO4 go to Absorber Tower with 98 % purity
P : H2SO4 Product with 98 % purity
H2O balance :
In = Out
W = K * 0.02 + P * 0.02
K = 840.572 Kg
P = 416.666 Kg
W = 840.572 * 0.02 + 416.666 * 0.02
= 16.811 + 8.333 = 25.144 kg water

Comp. equipment Stream m ( Kg ) no. moles ( Kmole )

H2 SO4 diluter 8 20413.4 208.3
H2 SO4 diluter 10 80728.48 823.76
H2 O diluter 10 302.598 16.811
H2 SO4 diluter 11 20005.13 204.134
H2 O diluter 11 74.988 4.166

21
2.2 Energy balance :

Fig. 2.1 Contact process with number of streams

Burner :

∆H in = m cp ∆T

25
∆H O2 , N2 , S = m ∫115 𝑎 + 𝑏𝑇 + 𝑐𝑇 + 𝑐𝑇 dT

1100
∆H O2 , N2 , SO2 = m ∫25 𝑎 + 𝑏𝑇 + 𝑐𝑇 + 𝑐𝑇 dT

We can solve the equation above by using Excel application to make the
calculation easier , we ginned the resultant show in table below :

22
Heat Exchanger A :

1100
In : ∆H O2 , N2 , SO2 = m ∫25 𝑎 + 𝑏𝑇 + 𝑐𝑇 + 𝑐𝑇 dT

410
Out : ∆H O2 , N2 , SO2 = m ∫25 𝑎 + 𝑏𝑇 + 𝑐𝑇 + 𝑐𝑇 dT

65
∆H H2O = m ∫25 𝑎 + 𝑏𝑇 + 𝑐𝑇 + 𝑐𝑇 dT

Amount of service water = 10909.5 Kg / hr

23
Reactor Stage 1 :

∆H in + ∆Hr = ∆H out

25
In : ∆H O2 , N2 , SO2 = m ∫410 𝑎 + 𝑏𝑇 + 𝑐𝑇 + 𝑐𝑇 dT

1100
Out : ∆H O2 , N2 , SO3 = m ∫25 𝑎 + 𝑏𝑇 + 𝑐𝑇 + 𝑐𝑇 dT

Heat Exchanger 1 :

601
In : ∆H O2 , N2 , SO2 = m ∫25 𝑎 + 𝑏𝑇 + 𝑐𝑇 + 𝑐𝑇 dT

24
 438
Out : ∆H O2 , N2 , SO3 = m ∫25 𝑎 + 𝑏𝑇 + 𝑐𝑇 + 𝑐𝑇 dT

65
∆H H2O = m ∫25 𝑎 + 𝑏𝑇 + 𝑐𝑇 + 𝑐𝑇 dT

Amount of service water = 48816 Kg / hr

Reactor Stage 2

∆H in + ∆Hr = ∆H out

25
In : ∆H O2 , N2 , SO2 = m ∫438 𝑎 + 𝑏𝑇 + 𝑐𝑇 + 𝑐𝑇 dT

485
Out : ∆H O2 , N2 , SO3 = m ∫25 𝑎 + 𝑏𝑇 + 𝑐𝑇 + 𝑐𝑇 dT

25
Heat Exchanger 2 :

485
In : ∆H O2 , N2 , SO2 = m ∫25 𝑎 + 𝑏𝑇 + 𝑐𝑇 + 𝑐𝑇 dT

432
Out : ∆H O2 , N2 , SO3 = m ∫25 𝑎 + 𝑏𝑇 + 𝑐𝑇 + 𝑐𝑇 dT

65
∆H H2O = m ∫25 𝑎 + 𝑏𝑇 + 𝑐𝑇 + 𝑐𝑇 dT

Amount of service water = 394.4 Kg / hr

Reactor Stage 3 :

∆H in + ∆Hr = ∆H out

25
In : ∆H O2 , N2 , SO2 = m ∫432 𝑎 + 𝑏𝑇 + 𝑐𝑇 + 𝑐𝑇 dT
26
 444
Out : ∆H O2 , N2 , SO3 = m ∫25 𝑎 + 𝑏𝑇 + 𝑐𝑇 + 𝑐𝑇 dT

Heat Exchanger 3 :

444
In : ∆H O2 , N2 , SO2 = m ∫25 𝑎 + 𝑏𝑇 + 𝑐𝑇 + 𝑐𝑇 dT

427
Out : ∆H O2 , N2 , SO3 = m ∫25 𝑎 + 𝑏𝑇 + 𝑐𝑇 + 𝑐𝑇 dT

65
∆H H2O = m ∫25 𝑎 + 𝑏𝑇 + 𝑐𝑇 + 𝑐𝑇 dT

Amount of service water = 125 Kg / hr

27
Reactor Stage 4 :

∆H in + ∆Hr = ∆H out

25
In : ∆H O2 , N2 , SO2 = m ∫427 𝑎 + 𝑏𝑇 + 𝑐𝑇 + 𝑐𝑇 dT

430
Out : ∆H O2 , N2 , SO3 = m ∫25 𝑎 + 𝑏𝑇 + 𝑐𝑇 + 𝑐𝑇 dT

28
Heat Exchanger 4 :

430
In : ∆H O2 , N2 , SO2 = m ∫25 𝑎 + 𝑏𝑇 + 𝑐𝑇 + 𝑐𝑇 dT

427
Out : ∆H O2 , N2 , SO3 = m ∫25 𝑎 + 𝑏𝑇 + 𝑐𝑇 + 𝑐𝑇 dT

65
∆H H2O = m ∫25 𝑎 + 𝑏𝑇 + 𝑐𝑇 + 𝑐𝑇 dT

Amount of service water = 56 Kg / hr

Chapter Three
Equipment Design

3.1 Design of Heat Exchanger

29
Tube Side ( Gases : SO2 & O2 & N2 )

t 1 = 1100 ◦ C

t 2 = 410 ◦ C

shell Side ( Water )

T1 = 25 ◦ C

T2 = 65 ◦ C

Q = UA ∆Tlm Ft

Where

Q = heat transferred per unit time, W .

U = Overall heat transfer Coefficient , W/m . ◦ C

A = heat transfer area, m .

∆Tlm = Log mean temperature difference, ◦ C.

30
Ft = Temperature correction factor

From Energy balance Q lost = 32857827.27 KJ/hr

𝐾𝐽 1000𝐽 ℎ
Q = = 32857827.27 * * *
ℎ𝑟 𝐾𝐽 3600𝑆

Q = 9127174.42 W

(𝑇1 −𝑡2 )−(𝑇2 −𝑡1)

∆Tlm = [3]
𝑇1−𝑡2
𝑙𝑛
𝑇2−𝑡1

(25−410)−(65−1100)
∆Tlm = 25−410
ln⁡( )
65−1100

∆Tlm = 658

Use one shell and two tube passes

𝑇1−𝑇2
R=
𝑡2−⁡𝑡1

25−65
R= = 0.057
410−1100

𝑡2−⁡𝑡1
S=
𝑇1−𝑡1

410−1100
S= = 0.64
25−1100

For R = 0.057 and S = 0.64 F=1 From fig.12.19 [3]

Or from equation :

31
 [3]

(⁡1−0.64⁡)
√(0.0032+1) ln[⁡(⁡1−0.057∗0.64⁡)⁡]
Ft = = 0.998 ≈ 1
2−0.64[0.057+1−√(0.0032+1 )
(⁡0.057−1⁡)⁡ln⁡[⁡ ⁡]
2−0.64[0.057+1+√(0.0032+1 )

From fig.12.1 the water gases system [3]

U = 80 – 250

By trial and error assume U = 90 W/m2.C

𝑄 9127174.42
A= = = 156 m2
𝑈∆𝑇𝑙𝑚⁡ 𝐹 90∗658 ∗1

Select

Tube length L = 1.83 m

Tube out Sid diameter d o= 50mm [3]

Tube thickness = 3.2mm

di = do – 2*3.2

= 50 – 2*3.2 = 43.6 mm = 0.0436m

Surface area of one tube = ᴫ D L

= ᴫ *0.05 * 1.83

= 0.287 m2

32
 𝑇𝑜𝑡𝑎𝑙⁡𝑎𝑟𝑒𝑎 ⁡
No. Of tube =
𝑎𝑟𝑒𝑎⁡𝑜𝑓⁡𝑜𝑛𝑒⁡𝑡𝑢𝑏𝑒⁡

156
Nt = = 544
0.287⁡

Tube Side coefficient hi

𝑡1 −⁡𝑡2
Mean tube temperater tm =
2

1100⁡+⁡410
= = 755 ◦ C = 1023 K
2

𝑃𝑀
𝜌v =⁡ 𝑅𝑇

P = 10 bar = 100 kpa

T = 755 ◦ C = 1023 K

1000∗124
𝜌v = = 14.57928 Kg/m3
8.314∗1023

Cp = 3.077 KJ/Kg.C

K = 0. 54 w/m.c [4]

𝜇 = 0.1 centipoise

= 1*10-4 Ns/m2

544
No. of tube per pass = = 272
2

⁡1164.5
Mass flow rate per tube m =
272∗3600
m = 0.00118 Kg/s

33
M = 𝜌 ⁡u A

Where

M = mass flow rate kg/h

𝜌 = Fluid density kg/m2

u = Fluid velocity m/s

Ac = cross section area of tube m2

⁡ᴫ ⁡ᴫ
A= d2 = ⁡(0.00436)2 = 0.149*10-4 m2
4⁡ 4⁡

𝒎 𝟎.𝟎𝟎𝟏𝟏𝟖
u= = = 23.5 m/s
𝝆𝑨 𝟏𝟒.𝟓𝟕∗𝟎.𝟏𝟒𝟗∗𝟏𝟎 −𝟒

𝜌𝑢𝑑𝑖 14.57 ∗23.5∗0.00436

Re = =⁡ = 3481⁡ > 210
𝜇 1∗10 −4

𝜇
Nu = 0.021 Re0.8 Pr0.33( )0.14 [3]
𝜇𝑤

ℎ𝑖⁡𝑑 ⁡
𝑖
Nu = Nusselt number =
𝑘

𝑐𝑝⁡𝜇
Pr = prondtl number =
𝑘

hi = inside fluid film coefficient w/m2.◦c

di = inside tube diameter

k = fluid thermal conductivity w/m.◦c

𝜌 = Fluid density kg/m3

𝜇 = fluid viscosity kg/m.s

cp = fluid heat capacity J/kg.◦c

34
u = fluid velocity m/s

Re = 3481

𝑐𝑝⁡𝜇 3.077 ∗103⁡ ∗1∗10−4

Pr = = = 0.00064
𝑘 0.54

𝜇
( )0.14 = 1 [3]
𝜇𝑤

Nu =0.021(3481)0.8 (0.00064)0.33(1)0.14

Nu = 0.266

ℎ𝑖 𝑑𝑖 ℎ𝑖 ∗0.00436
Nu = = 0.258 =
𝑘 0.54

hi = 3295.6 w/m2.◦c

Shell side coefficient ho

𝑇1 +𝑇2
Mean shell temperature Tm =
2

25+65
Tm = = 45◦ C
2

𝜌 = 990 kg/m3

𝜇 = 0.595 mP.s

= 5.95 * 10-4 Ns/m2 [6]

Cp = 4.18 kJ/kg.◦ C

K = 0.627 w/m.◦ C

35
 𝜇
Nu = JnRe Pr0.33( )0.14 [3]
𝜇𝑤

Where

ℎ𝑜 𝑑 𝑒
Nu = Nusselt number =
𝑘

𝜌𝑢𝑠 𝑑𝑒
Re = Reynolds number =
𝜇

𝑐𝑝𝜇
Pr = Prandtl number =
𝑘

ho =out side fluid film coefficient w/m2.c

de = equivalent diameter m

k = fluid thermal conductivity w/m.◦c

𝜌 = Fluid density kg/m3

𝜇 = fluid viscosity kg/m.s

cp = fluid heat capacity J/kg.◦c

us = fluid velocity m/s

use square tube pitch p t = 1.25 d o

1.27
de = (pt2 – 0.785 do2) [3]
𝑑𝑜

Where

do, de and pt in mm

do = 50mm

Pt = 1.25(50) = 62.5mm

36
 1.27
de = (62.52 – 0.785*502) = 49.4mm
50

= 0.0494m

𝑁𝑡
Db = do( )1/n1 [3]
𝑘1

Where

Db = bundle diameter mm

Nt = total number of tube

K1 and n1 = constant

For squar tube pitch and two tube passes

K1 =0.156

N1 =2.291 Table 12.4 [3]

544 1/2.291
Db = 50( ) = 1758 mm
0.156

For bundle diameter Do = 551 mm and split ring flooting head

Ds – Db = 60 mm Fig 12.10

Ds – 551 = 60 Ds = 1818 mm = 1.818 m

𝑝𝑡 −𝑑𝑜
As =( )(Ds*⁡𝑙𝐵 ) [3]
𝑝𝑡

Where

As = Cross flow area m2

lB = baffle Spacing m

lB= (0.20 to l) shell diameter [3]

37
lB = 0.5Ds

lB =0.5 *( 1.844 ) = 0.909 m

62.5−50
AS = (1.818 ∗ 0.909) = 0.33 m2
62.5

𝑤𝑠
Gs =
𝐴𝑠

Where

Gs = mass velocity kg/m2 .C

Ws = mass flow rate kg/s

Ws = 10909.5 kg/h

= 3.03 kg/s

3.03
Gs = ⁡= 9.16 kg/m2.s
0.33

𝐺𝑠 9.16
GS = ps us us = = = 0.0092 m/s
𝜌𝑠 990

𝐺𝑠 𝑑𝑒 9.16∗0.0494
Re = = = 1189
𝜇 5.9∗10−4

For Re = 1189 and 25% baffale cut Jn = 5*10-3 [3]

𝑐𝑝𝜇 4.18∗103 ∗5.9∗10−4

Pr = = =3.93
𝑘 0.622

Nu =5*10-3(1189)(3.39)0.33 (1)

Nu = 75.36

ℎ𝑜 𝑑𝑟
Nu =
𝜇

38
 ℎ𝑜 ∗0.0494
75.36 =
0.627

ho = 956.6 w/m2.c

Over all heat transfer Coefficient U

1 1 1 𝑑𝑜 ln 𝑑𝑜 /𝑑𝑖 𝑑𝑜 1 𝑑𝑜 1
= + + + . + . [3]
𝑈° ℎ𝑜 ℎ𝑜 𝑑 2𝑘𝑤 𝑑𝑖 ℎ𝑖𝑑 𝑑𝑖 ℎ𝑖

Where

U◦ = over all heat transfer coefficient w/m2. c

ho = out side fluid film coefficient w/ m2. c

hod = out side dirt coefficient w/m2.c.

do = tube out side diameter m.

Kw = thermarl conductivity of tube wall material w/m.c

hid = 3000

hod = 5000 Table 12.2 [3]

Thermal conductivity of stainless steel

Kw = 16 w/m.c Table 12.6 [3]

1 1 1 0.05 ln 0.05/0.0436 0.05 1 0.05 1

= + + + . + .
𝑈° 1439 4000 2∗16 0.0436 3000 0.0436 3000

U◦ = 87w/m2.c

U◦ ass = 90w/m2c

Assume U = 87 W/m2.◦ c

39
 9127174.42
A= = 161.5 m2
87∗288.9∗1

161.5
Nt = = 562
0.287

562
No.of tube per pass = = 281
2

1164.5
Mass flow rate per tube =
281∗3600

⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡𝑚 = ⁡0.00115⁡𝑘𝑔/𝑠

0.00115 ⁡
u= =22.7 m/s
14.57∗0.149∗10−4

3.39∗22.7∗0.00436
Re = = 3365
1∗10−4

Nu = 0.021 (3365)0.8(0.0064)0.33(1)

Nu = 0.258

ℎ𝑖 ∗0.00436
0.258= hi = 3207 w/m. ◦ c
0.54

Shell side coefficient ho

562
Db = 50( )1/2.291 = 1784 mm
0.156)

For Db = 511 mm and split rong floating head type

Ds – Db = 58 Ds = 1844 mm [3]

= 1.844 m

40
lb = 0.5(1.884) = 0.992 m2

62.5−50
As =( )(1.884*0.922) = 0.34 m2
62.5

7.374
Gs = = 8.9 kg/m2.s
0.0324

990∗22.7∗0.0494
Re = =2387
5.9∗10−3

For Re = 2387 and 25% baffle cut

Jn = 5*10-3 Fig 12.29 [3]

Nu = 5*10-3(1508)(3.03)0.33(1)

Nu = 132.51

ho ∗0.0494 ◦
132.51 = h0 = 992.6 w/m2. C
0.627

1 1 1 0.05 ln 0.05/0.00436 0.05 1 0.05 1

= + + + . + .
𝑈° 992.6 5000 2∗16 0.00436 4000 0.00436 3207

U◦ = 86 w/m2. ◦c

Uass = 87 w/m2.◦c

87−86
Error = *100 % = 1%
87

41
Tube side pressure drop

𝑙 𝜇 𝜌𝑢2
Δpt = NP[8 jf ( )( )−0.14 +2.5][ 𝑡
] [3]
𝑑 𝑖 𝜇𝑊 2

Where

Δpt = tube side pressure drop N/m2

NP = no. of tube passese

jf = friction factor

ut = tube side velocity m/s .

for Re = 1508

jf = 2.9*10 -3 fig 12.24

1.84 3.39∗22.72
Δpt = 2[8 *2.9*10 -3 ( ) (1)-0.14 +2.5][ ] [3]
0.00436 2

= 6068 N/m2

= 6.068 kpa

= 0.88 psi < 10psia ok

Shell side pressure drop

𝐷 𝑙 𝜇 𝜌𝑢2
Δpt = 8 jf ( 𝑆 ) ( )( )−0.14 [ 𝑡
] [3]
𝑑𝑒 𝑙𝐵 𝜇𝑊 2

Where

Δps = shell side pressure drop N/m2

us = shell side velocity m/s .

42
for Re = 15686 and 25% baffle cut

jf = 4.5*10 -2 fig 12.30

1.84 1.83 990 ∗22.7 2

∆Ps = 8*4.5*10-2( )( )( )(1)-0.14
0.0494 0.992 2

= 987 N/m2

= 0.98 kpa = 0.14 psi <⁡10 psia Ok

Pipe sizing

d = 260 G0.52p-0.37 [3]

Where

d = optimum pipe diameter mm

G = fluid mass flow rate kg/m3

𝜌 = fluid density kg/m3

Inlet and out let Gases

G = 1164.5 kg/h = 0.32 kg/s

𝜌 = 3.39 kg/m3

D = 260(0.32)0.52 (3.39)-0.37 = 91.5 mm ≈ 3.5 inch

Inlet and outlet water

G = 10909.5 kg/h = 3.03 kg/s

𝜌 = 990 kg/m3

d = 260(3.03)0.52(990)-0.37 = 36 mm ≈ 1.5 inch

43
3.2 Design Reactor

Catalyst :

The catalyst, vanadium(V) oxide on silica, is generally in the form of

small pellets

The Vanadium pentoxide catalyst

Properties:

1-partical size =0.015 in

2-Bulk density =4.339 g/cm3

3-partical density =1.1729g/cm3

44
The volume can be evaluated using method as SIMPSONS one –third
Rule :

𝑋 𝐹𝐴˚
V =∫0 dx = h/3 {f(x)0 + 4f(x)1 + f(x)2}
−𝑟𝐴

H = X2 – X0

-rA = Kca

K1 = k1˚ exp – E/RT

K1˚= 5.78954*10-6 kmol/m3.hr

K2 = K2˚exp – E/RT

K2˚= 2.5714*102 Kmol/m3.hr

E1 = 20,800 Kcal /Kgmol

E2 = 47,400 kcal/kgmol

K1 = 5.78959*10-3 exp – 20000/8.314*683

= 0.02934 Kmol/m3.h

K2 = 2.5714*102 exp – 47000/8.314*683

= 0. 547 Kmol/m3.h

4.174
𝑦𝐴 ° = = 0.12
35.746

ɛA = yA˚ϐ

−1
𝜖𝐴 = ( ) (0.12) = −⁡0.12
1

CA˚= yA*˚p/RT˚

CA˚= 1.12 atm/8.314*673

45
 = 0.0207 Kmol/m3

FA˚= 4.174 Kmol/h

Simpsons rule:

X ( 1–x)/(1+ƹx) (1+x)/(1–x)=f(x)
0 1 1
0.196 0.82336 1.487
0.329 0.638012 2.28
0.588 0.4432 3.584
0.784 0.2384 8.2592
0.98 0.02266 99

H = 0.196

F(x) = 0.196/3 {1+4*1.487+2*2.28+4*3.584+2*8.2592+99}

= 8.4817

V = 4.147 / (0. 547*0.0207) * 8.4817

V = 43.46 m3
𝜋
V = *D2*L
4

𝜋
43.46 =⁡ *D2*(4D)
4

D = 2.399 m

Hence length (L) = 4*2.399

= 9.596 m

Catalyst is divided in 4 beds in ratio

46
Volume of catalyst in 1st bed = 43.46/4

= 10.8

Volume of catalyst in 2nd bed =10.8

Bulk density of catalyst = 4339 Kg/m3

Weight of catalyst in bed =10.8*4339

= 46861 kg

NOW giving allowance for space for gas movement

Upward and downward and insulation be 0.5 m

Hence diameter of reactor becomes = 2.399-0.5

= 1.899 m

Pressure at which reactor works = 103 Kg/m2

Let factor of safety = 20%

Design pressure = 1.2*103

= 123.6 Kg/m2

We use stainless steel

Allowable stress = 66,000 psi {hesse and Ruston}

= 4 494 Kgf/cm2

NOW allowable stress = p D(1/k2-1) from ( Hesse )

Where p D = design pressure

4494 = 240 (1/k2 – 1)

47
K = 1.0337

D˚= K Di

D˚=1.0337*2.339

D˚=2.478m

Thickness of shell = (2.478-2.339)/2

= 69 mm

Operating pressure : 103 Kg/m2

Design pressure : 123.6 Kg/m2

Temperature : 683-874 k˚

Allowable stress : 4494 Kgf/cm2

Diameter inside of converter : 2.339 m

Bulk density of catalyst : 4339 Kg/m3

Volume : 43.46 m3

Thickness of shell : 69mm

Volume of 1st catalyst bed : 10. 4 m3

48
3.3 Design of Absorber

Inlet gas

Component Kg/hr M . wt Kmol /hr Mol% y

SO3 333.28 80 4.166 100

49
Column diameter

𝐿 𝜌
F L.V = ( )( 𝑉 )0.5
𝐺 𝜌𝐿

Where :

FL.V = liquid vapor flow factor .

L = liquid mass flow rate kg/s .

G = gas mass flow rate kg/s .⁡

Ρ v = vapor density kg/m3 .

ρ L = liquid density kg/m3 .

vapor density

𝑃𝑀𝑎𝑣
ρ V=
𝑅𝑇

M a v = Σ Mi y i

Where :

M a v = average molecular weight kg/k mol.

M i = molecular weight of component i kg/kmol .

yi = mol fraction of component i .

M a v = (80*1)

= 80 kg/kmol

P = 1 bar = 100 kpa

T = 30 ◦ C = 303 K

100∗80
ρ V= = 3.17 kg/m3
8.314∗303⁡

50
Liquid density :

Density of H2O at 30 ◦ C

ρ L =1000 kg/m3

L = 74.988 kg/hr

L = 0.0208 kg/s

G = 333.28 kg/hr

G = 0.0926 kg/s

0.0208 3.17 0.5

F L.V = ( )( ) = 0.012
0.0926 1000

𝐾4 𝜌𝑉( 𝜌𝐿−𝜌𝑉) 0.5

VW* = [ 𝜇 ]
13.1𝐹𝑃 ( 𝐿)0.1
𝜌𝐿

Where:

VW* = gas mass flow rate per unit cross – sectional area kg/m2.s

K4 = constant

μ L = liquid viscosity kg/m.s

Fp = packing factor .

Design pressure drop for absorber from from15 to50 mmH2O /m packing

Select Δp = 42 mmH2O /Packing

For FL.V = 0.01 and ΔP = 42 mmH2O /m packing

K4 = 3 fig 11.44

K4 at flooding = 7

51
 𝐾4
Percent flooding = ( )0.5* 100%
𝐾4 ⁡𝑎𝑡⁡𝑓𝑙𝑜𝑜𝑑𝑖𝑛𝑔⁡⁡

3
= ( )0.5 *100% = 70%
6

Select 1 1/2 inch in talox saddle

FP = 170 m-1 table 11.2

Viscosity of H2O at 30 ◦ C

μL = 0.722 CP = 0.722*10-3 kg/m.s

3∗3.17(1000−3.17)
VW* = [ ]0.5 [3]
0.722∗10−3 0.1
13.1∗170⁡( )
1000

= 4.22 kg/m2.s

𝐺
A= ∗
𝑉𝑊

Where :

A = cross-sectional area of column m2 .

0.0926
A=( 4.22
) = 3.4 m2
⁡

𝜋
A= D2
4

𝜋
3.4 = D2 D = 1.7 m
4

Height of column :

Z = HOG . NOG

Where :

Z = height of packing m.

52
HOG = over all height of gas phase transfer unit m.

NOG = number of transfer unit .

𝑚𝐺
HOG = Hg + HL
𝐿

Where :

Hg = individual height of gas film transfer unit m.

HL = individual height of of liquid film transfer unit m.

m = slop of equilibrium line .

G = gas molar flow rate kmol/hr

L = liquid molar flow rate kmol/hr

Absorption with chemical reaction , vapor pressure of SO3 over the

solution can be negligible .

PA◦ = 0

PA* = PA◦ XA

𝑃𝐴∗ 𝑃𝐴°
= XA
𝑃𝑇 𝑃𝑇

𝑃°
y*A = m xA m = ( 𝐴) = 0
𝑃𝑇

m = ρV u A
𝑚
u=
𝐴𝜌𝑉

0.0926
u=
0.022∗3.17

= 13.5 m/s

53
 For u = 13.5 m/s , 1.5 inch intalox saddle 1 bar and 30 ◦ C

Hg = 2 m

HOG = Hg = 0. 5 Table 11.3

𝑦 𝑑𝑦
NOG =∫𝑦 1
2 𝑦−𝑦 ∗

For absorption with chemical reaction y*= 0

𝑦 𝑑𝑦 𝑦1
NOG= ∫𝑦 1 = Ln
2 𝑦 𝑦2

y1 = 0.01

y2 = 0.0005

0.01
NOG = Ln = 2.99
0.0005

Z = (0. 5) (2.99)

= 4.8 m

Type Packed
Length packed 4.8 m
Diameter 1.7 m
𝜋
Volume =⁡ D2L 0.0471
4

Temp. ◦ C 30 ◦ C
Pressure bar 1bar
Material of count Carbon steel

Chapter Four
54
 Process Control
4.1 Introduction
The equipment is used in the control and measurement have a significant
role in the control of the process as well as to minimize losses and to
prevent damage and reduce risk. There are no industrial process devoid of
control and measurement, because these are very important services in
each process even if it was simple. Typically in each unit in the lab, there
is a room called the control room that contains the hardware control
associated with the equipment and machines for the process of production
and the panel representing each unit, the index for each device and the
lamps and measuring devices can be of this room control of each device
and can stop the process in case there index to a significant error such as
an explosion or fire.

4.2 The most important devices of control are :

1- devices to control the temperature.

2- control devices on the pressure.

3- control devices on the flow rate.

In general control device to reduce the error due to an abnormality of the

process works is a mistake (error) is the difference between reading
device elementary TR (reading, which is intended to occur) and read
between the reading device (measuring.TM)

Error = set point – measuring give

55
E = TR – TM

And trying to control device to reduce the amount of error to less close to
zero.

4.3 The most important control system components:

1- measuring element: it is according to the type of operating
althermukmpel used to measure the temperature and albrmitr to measure
pressure and venjor aoravs to measure the flow.

2- control element: a device that compares and gives instructions for a

(final control) it receives a reading from the reading device and compares
them with a set of reading it.

There are kinds of dominants and every kind gives

frequency:

1- proportional control.

2- proportional integral.

3- proportional derivative.

4- PIB.

First the least efficient and most qualified fourth.

3- final control: a real port where it comes to him from controlling the
main works this final control of the decreasing or increasing the amount
depending on the amount of error.

56
 4- process: the process is to be controlled, such as flow, pressure or
temperature of the process … etc.

We can represent this circuit the following scheme:

Set point
controller Final process E RSP
control

Rm Measuring controlled
element

4.4 The most important symbol of control devices:

TIC = temperature control

PIC = pressure control

LIC = level control

FIC = flow control

57
4.3 control on reactor :

Fig. (4-3) Control System of reactor

58
 Chapter Five
Location , Economic and Safety Consideration

5.1 Plant Location

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 plant operation as well as the surrounding community, are
also important. The following factors should be considered in selecting a
plant site :
1. Raw materials availability
2. Markets
3. Energy availability
4. Climate
5. Transportation facilities
6. Water supply
7. Waste disposal
8. Labor supply
9. Taxation and legal restrictions
10. Site characteristics
11. Flood and fire protection
12. Community factors.

59
5.1.1: H2SO4 Plant Location
Based on these previous factors which are required in H2SO4
manufacturing plant, we select ( Al-Ramadi /Al-Mohammadiat /
industrial city Alkilo 70 ) Sulfur can be obtained from ( Al-Mishraq
Sulphur State Company ) .

5.2 Site Considerations :

The location of the plant can have a crucial effect on the profitability of a
project, and the scope for future expansion. Many factors must be
considered when selecting a suitable site, the principle factors to consider
are:
1- Marketing area :
For materials that are produced in bulk quantities; such as cement,
mineral acids, and fertilizers, where the cost of the product per tone is
relatively low and the cost of transport a significant fraction of the sales
price, the plant should be located close to the primary market. This

60
consideration will be less important for low volume production, high-
priced products; such as pharmaceuticals.
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; where this is also close to the
marketing area.
3- Transport
The transport of materials and products to and from the plant will be an
overriding consideration in site selection.
4- Availability of labor
Labor will be needed for construction of the plant and its operation.
Skilled tradesmen will be needed for plant maintenance.
5- Utilities (services)
Chemical processes invariably require large quantities of water for
cooling and general process use, and the plant must be located near a
source of water of suitable quality. Process water may be drawn from a
river, from wells, or purchased from a local authority.
6- 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. An
environmental impact assessment should be made for each new project,
or major medication or addition to an existing process.
7- Land (site considerations)
Sufficient suitable land must be available for the proposed plant and for
future expansion. The land should ideally be flat, well drained and have
suitable load-bearing characteristics.
8- Climate

61
Adverse climatic conditions at a site will increase costs. Abnormally low
temperatures will require the provision of additional insulation and
special heating for equipment and pipe runs. Stronger structures will be
needed at locations subject to high winds (cyclone/hurricane areas) or
earthquakes.

5.3 Health & Safety Information :

Sulfuric acid is a strong acid and a powerful oxidizing agent . The major
hazard posed by it is chemical burns , however, and the substance is not
considered a carcinogen or mutagen. The standard first aid treatment for
acid spills on the skin is, as for other corrosive agents, irrigation with
large quantities of water. Washing is continued for at least ten to fifteen
minutes to cool the tissue surrounding the acid burn and to prevent
secondary damage . Contaminated clothing is removed immediately and
the underlying skin washed thoroughly. Being a strong oxidizing agent,
reactions of sulfuric acid with compounds such as cyanides, carbides,
metallic powders can be explosive and those with many organic
compounds, such as turpentine, are violent and hypergolic (i.e. self-
igniting ). Hence, it should be stored away from bases and organics .

5.4 Environmental Considerations :

Vigilance is required in both the design and operation of process plant to
ensure that no harm is done to the environment. Consideration must be
given to:
1. All emissions to land, air, water.
2. The visual impact.
3. Noise.
4. Smells.
5. Waste management.

62
 6. Any other nuisances.
7. The environmental friendliness of the products.
Waste management: Waste arises mainly as byproducts or unused
reactants from the process, or as off-specification product produced
through mis-operation. The designer must consider all possible sources
of pollution and, where practicable, select processes that will eliminate or
minimize waste generation. Unused reactants can be recycled and off-
specification product reprocessed. Integrated processes can be selected:
the waste from one process becoming the raw material for another. When
waste is produced, processes must be incorporated in the design for its
treatment and safe disposal. The following techniques can be considered :
1. Dilution and dispersion.
2. Discharge to foul water sewer (with the agreement of the appropriate
authority).
3. Physical treatments: scrubbing, settling, absorption and adsorption.
4. Chemical treatment: precipitation (for example, of heavy metals),
neutralization.
5. Biological treatment: activated sludge and other processes.
6. Incineration on land, or at sea.
7. Landfill at controlled sites.
8. Sea dumping (now subject to tight international control).
Noise: Noise can cause a serious nuisance in the neighborhood of a
process plant. Care needs to be taken when selecting and specifying
equipment such as compressors, air-cooler fans, induced and forced
draught fans for furnaces, and other noisy plant. Excessive noise can also
be generated when venting through steam and other relief valves, and
from flare stacks. Such equipment should be fitted with silencers.

63
 5.5 Cost Estimation :

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 economic life of
the process is expected to be limited.

5.5.1 Cost Estimation on heat exchanger

Surface area A= 161.5 m2⁡

Pressure 𝑃 = 10⁡𝑏𝑎𝑟⁡

Material of construction : Stainless steel

Type: floating head

Cost = 20000⁡$

Pressure factor =1 Fig. 6_5b [8]

Material factor =1

𝐶𝑜𝑠𝑡 = 20000⁡ ∗ 1 ∗ 1 = ⁡20000⁡$⁡⁡⁡⁡⁡𝑖𝑛⁡(2004)

(cost⁡index⁡in2012)
𝐶𝑜𝑠𝑡⁡𝑖𝑛⁡2012⁡ = ⁡ (𝑐𝑜𝑠𝑡⁡𝑖𝑛⁡2004)
(cost⁡index⁡in⁡2004)⁡
cost⁡index⁡in⁡2004 = 444.2
cost⁡index⁡in2012 = 697.32⁡⁡⁡⁡⁡⁡ [9]
697.32
𝐶𝑜𝑠𝑡⁡𝑖𝑛⁡2012 = (20000 ) ( )
444.2
= 31397⁡$

64
 5.5.2 Cost Estimation on reactor

Height ℎ = 9.596⁡𝑚⁡

Diameter 𝐷 = 2.399⁡𝑚⁡

Pressure P= 10 bar

Material of construction : Stainless steel

Cost=2500⁡$

Pressure factor =1.1 Fig. 6_5b [8]

Material factor =2

𝐶𝑜𝑠𝑡 = 2500⁡ ∗ 1.1 ∗ 2⁡ = ⁡5500⁡$⁡⁡⁡⁡⁡𝑖𝑛⁡(2004)

(cost⁡index⁡in2012)
𝐶𝑜𝑠𝑡⁡𝑖𝑛⁡2012⁡ = ⁡ (𝑐𝑜𝑠𝑡⁡𝑖𝑛⁡2004)
(cost⁡index⁡in⁡2004)⁡
cost⁡index⁡in⁡2004 = 444.2

cost⁡index⁡in2012 = 697.32⁡⁡⁡⁡⁡⁡ [9]

697.32
𝐶𝑜𝑠𝑡⁡𝑖𝑛⁡2012 = (5500) ( )
444.2
= 8600⁡$

5.5.3 Cost estimation on absorber

Height ℎ = 4.8⁡𝑚⁡

Diameter 𝐷 = 1.7⁡𝑚⁡

65
 Pressure 𝑃 = 10⁡𝑏𝑎𝑟⁡

Material of construction high silicon iron

Cost=5500⁡$

Pressure factor =1.1 Fig. 6_5 b [8]

Material factor =1

𝐶𝑜𝑠𝑡 = 5500⁡ ∗ 1.1 ∗ 1⁡ = ⁡6050⁡$⁡⁡⁡⁡⁡𝑖𝑛⁡(2004)

(cost⁡index⁡in2012)
𝐶𝑜𝑠𝑡⁡𝑖𝑛⁡2012⁡ = ⁡ (𝑐𝑜𝑠𝑡⁡𝑖𝑛⁡2004)
(cost⁡index⁡in⁡2004)⁡
cost⁡index⁡in⁡2004 = 444.2
cost⁡index⁡in2012 = 697.32⁡⁡⁡⁡⁡⁡ [9]
697.32
𝐶𝑜𝑠𝑡⁡𝑖𝑛⁡2012 = (6050) ( )
444.2
= 9400⁡$

Chapter Six
Conclusion and Recommendation

6.1 Conclusions
The case study of the manufacture of sulfuric acid emphasizes the
benefits of a systematic design based on the analysis of the
reactor/condenser/absorber . The core of the process is the chemical
reactor, and in which the reaction is take place ( convert sulfur dioxide to
sulfur trioxide ) on a vanadium oxide V₂O5 as a catalyst, as well as the
safety and technological constraints. In this process "Accidental

66
pollution" there is always a risk of accidental pollution when chemicals
are produced and handled. The more common a chemical, the more
information is available about the different hazards and the lower the risk
of accidental pollution. The highest risk for accidental pollution is during
the transportation of the product . There is also a risk of pollution from
the storage of sulfuric acid and different plants have different systems to
collect leaks and spillages depending on guidelines for the storage of
acid. Gas leaks are not normally a problem as they are handled by various
monitoring and control systems, which measure the SO2 content in the
air.

6.2 Recommendation
Concern about the toxicity of sulfuric acid in the workplace atmosphere is
focused on its potential, as an inhaled aerosol, to exert local effects on the
respiratory tract, as a consequence of low pH. Such effects can be
manifested as sensory irritation of nerve endings, acute or longer term
inflammation at various sites along the length of the respiratory tract
epithelium, and ultimately the possibility of tumor formation in the
respiratory tract, believed to be a consequence of sustained tissue
inflammation and repair processes. Human carcinogenicity data and the
findings of a recent 28-day inhalation study in rats suggest that the larynx
is a site of particular concern, in relation to epithelial inflammation,
damage and ultimately cancer.

67
The identification of a clear NOAEL for this range of potential
respiratory tract effects is difficult, from the available data. However, the
recent 28-day inhalation study in rats (using a 50% sulfuric acid aerosol)
provides evidence of slight changes in the laryngeal epithelium at the
lowest concentration tested, 0.3 mg/m3 Other experimental studies in a
range of animal species suggest respiratory tract effects on repeated
exposure to concentrations around 0.3 mg/m3, with the possibility of
effects of some health significance even at concentrations down to about
0.1 mg/m3.
Taking into account the overall database, and with the concern for
potential human carcinogenicity in mind, SCOEL concluded that long-
term exposure should be maintained below 0.1 mg/m3 in order to provide
sufficient reassurance of avoidance of possible adverse consequences for
the respiratory tract epithelium. Hence SCOEL recommends an 8h TWA
limit of 0.05 mg/m3 in order to satisfy this requirement. SCOEL
appreciates that the reliable measurement of exposures at and around the
limit value proposed is challenging. In some circumstances there might
be interference from sulfate salts also present in the atmosphere.
However, from the most recent evidence presented to SCOEL and from
the assessment made in the Annex it appears that there are measurement
techniques available that are compatible with the proposed limit.
In terms of health protection, SCOEL considered that it would be
desirable to recommend a STEL of 0.1 mg/m3 to avoid short-term irritant
effects. However, at present there is no available measurement method
which can accommodate a short-term limit at this value (see Annex)
There is no evidence that H2SO4 can penetrate undamaged skin to cause
any signs of systemic toxicity, hence there is no requirement for a ‘Sk’
notation.

68
 References

1. Sulfuric acid Wikipedia ; The free encyclopedia .

2. shreve's chemical process industries 5 t h edition
New York (1975).
3. J.M. Coulson and J. F. Richardson , Chemical Engineering
Design vol.6 , 2 nd Edition ; Butter worth Heinemod, Oxford
(1993) .
4. Robert H. Perry and Don Green , Chemical Engineering Hand

Book , 6th Edition M (G raw-Hill New York) (1984).

5. J.M. Smith and H.C. VanNess; Introduction ; 6 th Edition M(G
raw-Hill New York) (1984).
6. Hesse ,Herman C.and Rushton ,J.Henry ''Process Equipment Design
7. Perry,R.H.,''Perry's Chemical Engineers Handbook'',6th Ed., McGraw
Hill

Company.1984.

8. Levenspiel Octave ,''Chemical Reaction Engineering 2nd Ed., Wiley

Eastern Ltd .,1991

69
70