PROJECT REPORT

ON

CALCIUM AMMONIUM NITRATE

PREPARED BY

SIDDHARTH DODIA
(U15CH030)
JIGAR PATEL

(U15CH031)
B.TECH IV
CHEMICAL ENGINEERING DEPARTMENT

GUIDED BY

Dr. CHETAN M. PATEL

ASSISTANT PROFESSOR DEPARTMENT OF
CHEMICAL ENGINEERING

Sardar Vallabhbhai National Institute of Technology Surat-395007,

Gujarat, India

1
 Sardar Vallabhbhai National Institute of Technology Surat-
395007, Gujarat, India

CERTIFICATE

This is to certify that the B.Tech IV (8th Semester) PROJECT REPORT entitled

“CALCIUM AMMONIUM NITRATE” has been presented and submitted by candidates

SIDDHARTH DODIA bearing roll no. U15CH030, and JIGAR PATEL, bearing roll no.
U15CH031, in the partial fulfillment of the requirement for the award of degree B.Tech in
Chemical Engineering.

They have successfully and satisfactorily completed their PROJECT EXAMS in all
respects. We certify that their work is comprehensive, complete and fit for evaluation.

Dr. CHETAN M. PATEL

Project Guide
Assistant Professor (SVNIT,Surat)

PROJECT EXAMINERS:
Examiner signature with date
Examiner 1

Examiner 2

Examiner 3

2
 Sardar Vallabhbhai National Institute of
Technology Surat-395 007, Gujarat, India

CHEMICAL ENGINEERING DEPARTMENT

SIDDHARTH DODIA (U15CH030) and JIGAR PATEL (U15CH031), registered in

Chemical Engineering Department of S.V.N.I.T. have successfully completed the project report

entitled “Production of CALCIUM AMMONIUM NITRATE” during year 2018-19. The

report is presented before the following members of the committee.

Sign Date

1) Examiner 1

2) Examiner 2

3) Examine 3

The Project report entitled “Production of CALCIUM AMMONIUM NITRATE” is

submitted to the Head (ChED) along with this certificate.

Place: Surat
Date:

Dr. Mausumi Mukhopadhyay

Head of Department
Chemical Engineering Department
SVNIT, Surat.

3
 ACKNOWLEDGEMENT

I express my deepest gratitude and heartiest thanks to my guide and mentor Dr Chetan M.
Patel, Assistant Professor, Department of Chemical Engineering, Sardar Vallabhbhai National
Institute of Technology, Surat, for continuous guidance, help and support during the course of
project preliminary.

I express my sincere thanks to Dr. Mausumi Mukhopadhyay, Head of Department of Chemical

Engineering, Sardar Vallabhbhai National Institute of Technology, Surat, who helped me to
accomplish my dissertation work with preciseness and excellence.

I would also like to acknowledge the contribution of all faculty and staff members of the
Department of Chemical Engineering for their continuous support towards completion of my
dissertation work.

Siddharth Dodia (U15CH030)

Jigar Patel (U15CH031)

4
 ABSTRACT

This project preliminary report is an attempt to understand the basic principles and process
involved in the manufacturing of a product at an industrial scale. Herein, we have selected
CALCIUM AMMONIUM NITRATE as our product.

This preliminary report describes information about the manufacturing process of CALCIUM
AMMONIUM NITRATE. It contains physical & chemical properties of the product along with
its safety and handling issues. Also method of industrial production of CAN practiced today are
introduced and the best & optimized method of production is chosen according to the feasibility
and context of the project. For the method chosen we have presented an attempt to the material
balance and energy balance for the entire process.

5
 Contents

ACKNOWLEDGEMENT……………………………………………………………………….4

ABSTRACT………………………………………………………………………………………5

1. INTRODUCTION…………………………………………………………………………………………………………………………………………7

2. DEMANDS AND SUPPLY…………………………………………………………………….9

3. PROCESS DESCRIPTION……………………………………………………………………12

4. MATERIAL BALANCE………………………………………………………………………21

5. ENERGY BALANCE…………………………………………………………………………26

6. KINECTICS AND THERMODYNAMICS…………………………………………………..33

7. EQUIPMENT DESIGN……………………………………………………………………….39

8. PLANT LOCATION…………………………………………………………………………..56

9. PLANT LAYOUT……………………………………………………………………………..60

10. COST ESTIMATION………………………………………………………………………...63

11. OCCUPATIONAL HEALTH & SAFETY…………………………………………………..74

REFERENCES…………………………………………………………………………………87

6
CHAPTER 1 INTRODUCTION

Calcium ammonium nitrate (CAN) is a universal nitrogenous fertilizer produced by treating

ammonium nitrate solution with powdered limestone. It is a white to grey chalky powder, with
the colour depending on the limestone used in the manufacturing process. It is made with
dolomitic limestone, the fertilizer contains 20% nitrogen, 6 % calcium and 4 % magnesium. If
the quantity of limestone is smaller than that of used ammonium nitrate, the nitrogen content can
go up to 28 %. CAN is preferred to ammonium nitrate in acid soils. The most common grade of
CAN contains about 21% nitrogen, corresponding to 60% ammonium nitrate.

Calcium nitrate contains 15.5% nitrogen and its manufacturing process involves reaction of lump
limestone with concentrated nitric acid, addition of ammonia to neutralize excess of acid,
evaporation of the resulting solution, and prilling or flaking the melt. The resulting product is a
double salt, Ca(NO3)2NH4NO3 called calcium ammonium nitrate and is more useful than the
single salt calcium nitrate.

Talking about its fertilizing effects, It can be applied on all types of soil both prior to sowing and
as top dressing. Nitrogen is absorbed easily and efficiently. Calcium and magnesium
are important to plant vital functions. Calcium ammonium nitrate enables plants grow wide
leaves of rich green colour; provide 5 - 20 % richer harvest. Long-term usage causes no soil
acidification and makes no impact on soil biological activity. Plants are enriched not only with
nitrogen but also with magnesium and calcium. Moreover, calcium ammonium nitrate by
providing plants with necessary nutrients compensates lack of light and the damage of acid soil

Its chemical formula is given by NH4NO3 + CaCO3 * MgCO3. It can be classified into different grades
which includes prilled catogary of granules 0,6-4mm, min 94% and mechanical granulated CAN
of granules 2-5mm, min 95% (e.g. Salmag) & granules 1-5mm, min 96% (e.g. Canwil).

The fertilizer should be transported in the means of transportation which protect the product from
the action of water, precipitation, direct sunlight and damage to the packaging.

7
 Calcium Ammonium Nitrate must be stored in closed, dry and clean warehouses. It may not be
stored together with the substances which may come into reaction with it e.g. with urea,
superphosphate.

Recommended dosages of CAN is tabulated below and While calculating dosage of the fertilizer
one should take into account factors such as current soil analysis, its quality class and
agronomical category, average yield per hectare in recent years, forecrop and the use of other
nutritional components.

Crops Recommended dose of fertilizer (kg/ha)

Winter rapeseed 440 – 740

Rye 220 – 370

Winter wheat 220 – 520

Spring wheat 220 – 440

Potatoes 300 – 370

Sugarbeets and
370 – 590
fodder beets

Maize 440 – 550

Grassland 370 – 660

The packaging of CAN is mainly available as

 25 kg Bags

 50 kg Bags

 500 kg Big Bags

 1.000 kg Big Bags

 Bulk

8
CHAPTER 2 DEMANDS AND SUPPLY

‘Global Calcium Nitrate Market 2017-2021’ provides an in-depth analysis of the market in terms
of revenue and emerging market trends.

The market research analysis categorizes the global calcium nitrate market into four major
application segments:

 Fertilizers

 Waste treatment

 Explosives

 Concrete

Calcium nitrate market in fertilizers

Fertilizers are the largest application segment that accounted 54.25% market share in 2016. The
market is expected to accelerate at a considerable pace under the fertilizers segment due to the
demand from the agriculture sector in APAC. The calcium nitrate market under fertilizers alone
accounted for 45% of the total calcium nitrate consumption in Europe. Every day, around
200,000 people are added to the world food demand according to the United Nations
Environment Programme (UNEP). To meet the demand, fertilizers are being used to improve the
productivity of crops.

9
 According to Kurva Samba Sivudu, a lead analyst at Technavio for metals and
minerals research, “Calcium is essential for growth of plants. It acts as an important
component of plant cell walls. The intake of calcium improves the quality of yield, increases
the shelf life, and protects the plants from stress and diseases. The constant supply of
calcium is essential to receive a healthy yield, because calcium moves along with the water
flow.”

Calcium nitrate market in waste-water treatment

Increasing industrialization in emerging economies and in Central and Northern European

countries is driving the demand for calcium nitrate. Often, they are used in factories and
municipality waste-water treatment. It is used to reduce odor emissions caused by the formation
of hydrogen sulfide in water. Odor caused by industrial wastewater can reduce up to 90% by
treating the infected water with calcium nitrate. It is biologically friendly inorganic chemical that
does not alter minerals in the water. It reduces corrosion and prevents water from turning septic.

Calcium nitrate market in explosives

Calcium nitrate is used in the composition of explosives along with an organic fuel component
and water. The combination of the three is used as a base for the preparation of various kinds of
explosives. It is used as an additive in explosives and is helpful in explosive applications.

“Calcium nitrate has numerous advantages over other inorganic oxidizing salts such as
ammonium nitrate. Even in the absence of sensitizers and fuels, calcium nitrate exhibits
fluidity and better sensitivity. Compared to ammonium nitrate composites, calcium nitrate
composites are detonable in smaller diameters and at higher densities,” says Kurva.

10
 The top vendors highlighted by Technavio’s market research analysts in this report are:

 ADOB
 Agrium
 GFS Chemicals
 Sterling Chemicals
 Yara International

11
CHAPTER 3 PROCESS DESCRIPTION

3.1 INTRODUCTION

Calcium Ammonium Nitrarte is commercially produced by the mixing of ammonium nitrate and
calcium carbonate. Ammonia and nitric acid are the starting raw materials, in this ammonia is
formed by Haber process by the reaction between nitrogen and hydrogen in the presence of iron
catalyst at higher temperature and Nitric acid is produced by Ostwald process by reaction
between nitrogen dioxide and water.

For the production of CAN, ammonium nitrate and calcium carbonate are the main raw
materials. Therefore in this report we will discuss in deta l about the production of ammonium
nitrate which further gives us Calcium Ammonium Nitrate as a final product.

Industrially, CAN Plant is based on UHDE’s Pug mill Granulation process. Important feature of
this process is that it makes the use of pug mill for granulation of mixture in place of
conventional drum granulator. The main advantage of pug mill are lower recycle ration and
minimum residence time.

Dry lime is fed to pug mill and Ammonium Nitrate is sprayed on the recycle material into pug
mill. Temperature in Granulator is maintained approx.110°c-. Product granules are formed in
pug mill. The product granules from the granulator are than sent to drying drum where excess
moisture is removed and finally moisture content of 0.5% W/w is achieved at outlet of drying
drum. The dried granules are screened and classified to finally separate out required product size
granules ( 1mm to 4mm). The remaining material, undersize and oversize is recycled directly or
crushed and recycled back to granulator.

Thereafter product is cooled to 45°c in fluidized bed cooler cooled granules are coated with
suitable coating agent in coating drum and then sent to bagging plant.

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RAW MATERIALS & UTILITIES : Specific Consumption per MT of CAN

AMMONIUM NITRATE : 0.758 MT

DRY LIME : 0.285 MT

AMMONIA : 0.002 MT

COATING AGENT : 0.50 Kg./MT

POWER : 43 KWH

NATURAL GAS : 4.6 Sm3

L.P. STEAM (HP) : 0.060 MT

M.P. STEAM (HP) : 0.090 MT

3.2 Process description of CAN Plant.

Main Sections of CAN plant are:

1. Neutralization (Ammonia Nitrate synthesis)

2. Ammonia Nitrate Evaporation
3. Granulation
4. Drying
5. Screeninng & Crushing
6. Coating

13
1. NEUTRALIZATION (Ammonia Nitrate Synthesis)

AN is produce from two feedstock

1) 60% W/w HNO3

2) Ammonia Gas (NH3 Gas).

Both raw materials nitric acid & Ammonia at 3.5 bar pressure are preheated in heat exchanger to
180°C and supplied into the reactor & the reaction takes place.

The reaction in the AN reactor takes place in circulation of AN solution which is already cooled
down by flush evaporation in vapour separator.

In order to avoid boiling of the AN solution in the reactor, the effective pressure of the system is
maintained higher than boiling pressure by AN reactor circulation pump.

Ammonium Nitrate (NH₄ NO₃ ) is Produced from.

- Nitric Acid (HNO₃ )

- Ammonia (NH₃ )

Reaction between Nitric acid & ammonia produces ammonium Nitrate. In which the Nitric acid
is in aqueous form and ammonia is in gaseous form. This reaction shown below equation.

NH₃ + HNO₃ NH₄ NO₃

The reaction is exothermic in nature. Ammonium Nitrate is obtained as aqueous solution with a
concentration of approx. 93% wt/wt.

The Nitric Acid and the ammonia gas are fed in Reactor, and reaction takes place and
Ammonium Nitrate (AN) is produced.

14
The flow of ammonia and Nitric acid are to be controlled by flow control valve of both the
stream respectively.

The pH must be maintained around 5. Otherwise the N- losses increases due to free NH3 in the
AN solution.

The reaction in the AN- reactor is exothermic & takes place in circulated AN solution which is
already cooled down by flash evaporating in vapour separator. The process vapours from the
vapour separator contain certain contaminations of ammonium nitrate, nitric acid or ammonia.

2. AMMONIA NITRATE EVAPORATION

The evaporator is normally required to remove the majority of the water which is present in the
ammonium nitrate solution. The acceptable water content depends on the process which is to be
used in the manufacture of the finished product, but is normally below 1% for a prilled product.
A water content up to 8% is required for the feed to some granulation processes.

The ammonia nitrate produced by neutralization is stored safely. When the storing AN solution
for longer time the pH value changes to acidic by loss of ammonia so the ammonia gas is
provided to maintain the pH value of AN solution by adjusting the flow control valve of
ammonia gas.
AN solution from the storage tank is pumped by a pump to the evaporation unit.
The AN evaporation is operating normal under an absolute pressure of 0.7 bar.

The heat needed for the AN concentration is supplied by saturated heating steam with a pressure
of approx. 8 to 9 bar absolute.

The circulation of AN stream within the evaporation system is a natural one which is effected by
thermosyphon effect.(It is a method of passive heat exchange , based on neutral convection,
which circulated the fluid without the necessity of a mechanical pump).

15
The AN melt with concentration 94%W/W is pumped by pump to the granulator. The flow of
AN solution with the concentration 94% to 94.5% w/w to granulator is control by valve.

During short interruption the granulation plant three way valve will change to full recycle to the
tank. Level controller is provided to maintain level of tank.

To maintain or control the temperature in the tank, water tapings are provided on the top of tank.

16
3. GRANULATOR

Pure Ammonium Nitrate (NH4NO3 ) contains 35% Nitrogen (N2) by weight. Lime is added as
filler to get mixture of 25% Nitrogen (N2) by weight.

AN melt from evaporator is mixed with lime in the granulator. Dry lime is coming from lime bin
through lime feeding system.

The mixing takes place at a temperature of about 145°C to 150°C.

When AN and lime are mixed and agglomerated in granulator, a small amount reacts to form
calcium nitrate according to following equation.

CaCO3(s)+2NH4NO3(l)  Ca(NO3)2(s)+2NH3(g)+CO2(g)+ H2O(l)

But this minor, so, we are neglecting

4. DRYING

Drying in dryer is done by using hot air in co current direction. This hot air is about 70’C. To
increase air temp. Air heater is used to achieve dryer outlet air temp approximately 85°C. Lifters
are provided within the drum for close contact between product and hot air.

The CAN granules must be dried in dryer. The drying air temp. must be controlled in order to
avoid excessive heating of the granules.

The drying process possible, when the partial pressure of water on surface of the CAN granules
is higher than partial pressure of water in the air.

When leaving the drum granules content approximately 0.4 -1.0 %W/W moisture and temp
about 95’C .If the temperature of drying air is to low, the heater can be used for heating up of the
drying air.

17
The dust along with moist air are supplied in the cyclone separator by means of sucking air with
help of blower. The air along with minor quantity of dust are sucked and major quantity of the
dust are collected bottom of the cyclone separator and fed in to screw conveyor to granulator.

5. SCREEN & CRUSHING

The dried granules leaving drying drum are conveyed via belt to the bucket elevator.

This elevator transports the product to distributor to ensure that the product is distributed
uniformly over coarse screen. The onsize & undersize product is sent to fine screen. Fine
screens separate product less than 1mm size particles , which is sent to recycle to granulator.

On size product separated on fine screen is desired size (between 1 mm & 4 mm) are sent to
fluidized bed cooler(FBC) via belt conveyer.

Course product of more than 4.0 mm and lumps coming from the course screen are sent to lump
crusher; which is the single roller type crusher .After crushing lumps, it is sent to single role
crusher. Finally the total crushed product is recycle to granulator.

The product divertor swing gates control recycle flow in dryloop, and diverts onsize product to
hot product belt.

The screen is vibrating with the help of motorized vibrator to provide the batter screening.

6. COATING

The CAN is a hygroscopic nature. When it comes in contact with atmospheric air, the lump or
cake can be formed.

To avoid this, coating is necessarily on surface of the granules. As anti- caking agent or coating
agent /oil amine is used. Cooled products from fluidized bed cooler (FBC) is sent to Coating
drum for coating.

18
The oil amine is pumped to the spraying nozzle in coating drum.

The product from the FBC is introduced in to the coating drum when the lifters move the product
& start forming the rolling bed.

On to this bed oil amine is sprayed and due to rolling, it is distributed the amount of the oil
amine is controlled by changing of strokes the spraying pump in order to achieve 0.15 – 0.5 %
w/w oil amine content in the final product.D

The granule leaving the coating drum are supplied by means of belt conveyors to the CAN
Bagging plant.

19
 BLOCK DIAGRAM
Unreacted NH₃

NH₃ H₂ O

HNO₃ Neutralizer Evaporator

Lime

ggg
Granulator

Drying

Screening

&

Crushing

Coating

Final Product

20
PROCESS FLOW DIAGRAM

21
22
CHAPTER 4 MATERIAL BALANCE

In any chemical process industry, chemical reactions are always involved in the manufacturing
of product from raw materials. The raw materials are fed as the input to the system and the
product comes out at the output. The balancing of inputs and outputs gives information
regarding following factors The amount accumulated inside the reactor. Accounts for the
losses in vent. The amount of raw material un-reacted during the reaction. The percentage of
back reaction taking place in case of reversible reactions, etc. Material Balance calculations are
governed by the LAW OF CONSERVATION OF MASS. The general equation of mass balance
is as follows:

Input-Output = Accumulation

Thus material balance calculation aids to estimate the Raw Material requirement and the
laboratory analysis of the input and output streams around each of the equipment can be used to
find the individual flowrates of the components present in the stream.

23
1) Neutralization (Ammonia Nitrate synthesis).

NH₃ + HNO₃ NH₄ NO₃

Assumptions : i) Conversion is 80%

ii) Feed inlet = 100 kmol/hr

Assuming 100 kmol/hr of NH₃ , so according to stoichiometry the feed rate of HNO₃ will be
also 100 kmol/hr
Since conversion is 80%, 20 kmol of each reactant will remain unreacted.

INLET
Component Flow Rate Mole Fraction Flow Rate Wt. Fraction
(kmol/hr) (kg/hr)
NH₃ 100 0.4761 1700 0.2078
HNO₃ 100 0.4761 6300 0.7701
NH₄ NO₃ 0 0 0 0
H₂ O 10 0.04761 180 0.0220
Total 210 1 8180 1

OUTLET

Component Flow Rate Mole Fraction Flow Rate Wt. Fraction

(kmol/hr) (kg/hr)
NH₃ 20 0.1538 340 0.0415
HNO₃ 20 0.1538 1260 0.1540
NH₄ NO₃ 80 0.6153 6400 0.7823
H₂ O 10 0.769 180 0.022
Total 130 1 8180 1

24
2) Evaporator
Assuming 90% moisture is removed

H₂ O(90%)

NH₄ NO₃ EVAPORATOR

H₂ O

NH₄ NO₃
H₂ O(10%)
Component Inlet Flow Top Outlet Bottom Inlet Flow Top Bottom
(kmol/hr) Flow Outlet Flow (kg/hr) Outlet Outlet
(kmol/hr) (kmol/hr) Flow Flow
(kg/hr) (kg/hr)
NH₄ NO₃ 80 0 80 6400 0 6400
H₂ O 10 9 1 180 162 18
Total 90 9 81 6580 162 6418

Total Inlet = 6580 kg/hr

Total Outlet = 162 + 6418
= 6580 kg/hr

25
3) Granulator

Assumptions : Lime inlet = 100kmol/hr

NH₄ NO₃ ∙ H₂ O
Granulator
CAN
CaCO₃

Component Inlet Flow Outlet Flow Inlet Flow Outlet Flow

(kmol/hr) (kmol/hr) (kg/hr) (kg/hr)
NH₄ NO₃ 80 0 6400 0

H₂ O 1 1 18 18

CaCO₃ 80 0 8000 0

CAN 0 80 0 14400

Total 161 81 14418 14418

26
4) Screening

Assuming 16 % removal in screening (i.e granules of size greater than 4mm are removed from
the feed )

Screening
CAN CAN

(14418 kg) (12111 kg)

27
CHAPTER 5 ENERGY BALANCE

5.1 Energy Balance around Neutralizer

Chemical Reaction: - NH₃ + HNO₃ NH₄ NO₃

Component Std. heat of formation (KJ/mole)

NH₃ -46

HNO₃ -206.63

NH₄ NO₃ -350.26

H₂ O -285.82
Table 5.1 Heat of formation data

NH₃ HNO₃ NH₄ NO₃ H₂ O

A 19.99563 19.63229 31.67876 30.092

B 49.77119 153.9599 105.76799 6.832514

C -15.37599 -155.8378 -95.899 6.793435

D 1.921168 32.87955 48.7868 -2.5344

Cp (J/mol*K) 5.6713 8.8248 9.34 5.1912

Table 5.2 Constants for heat capacity calculation at 298K

From temperature 298K to 453K : ∫Cp dT= [ A*t + B*t2 /2 + C*t3 /3 + D*t4 /4]

28
INPUT

1) Heat with NH₃ = n * ΔHf + n*∫ Cp dT

= 100*(-46) + 100*(5.6713)

= -4032.87 kJ/mol

2) Heat with HNO₃ = n * ΔHf + n * ∫ Cp dT

=100*(-206.63) + 100*(8.8248)

= - 20654.175 kJ/mol

3) Heat with H₂ O = n * ΔHf + n * ∫ Cp dT

=10*(-285.82) + 10*(5.191)

= -2806.288 kJ/mol

Total Heat of Formation in input = (-4032.87) + (-20654.175) + (-2806.288)

= -27493.333 kJ/mol

OUTPUT

1) Heat with NH₃ = n * ΔHf + n * ∫ Cp dT

= 20*(-46) + 20*(5.6713)

= -806.574 kJ/mol

2) Heat with HNO₃ = n * ΔHf + n * ∫ Cp dT

= 20*(-206.63) + 20*(8.8248)

= -3956.104 kJ/mol

29
3) Heat with H₂ O = n * ΔHf + n * ∫ Cp dT

=10*(-285.82) +10*(5.1912)

= -2806.288 kJ/mol

4) Heat with NH₄ NO₃ = n * ΔHf + n * ∫ Cp dT

= 80*(-350.26) + 80*(9.3484)

= -27272.923 kJ/mol

Total Heat of Formation in output = (-806.574) + (-3956.104) + (-2806.288) + (-27272.923)

= -34841.889 kJ/mol

Thus, heat liberated within the reactor is given by

Q = Σ(𝐻𝑖,𝑜𝑢𝑡) − Σ(𝐻𝑖,𝑖𝑛)

= (-34841.889) - (-27493.333)

= -7348.55 kJ/mol

Reaction is exothermic in nature.

30
Energy Balance Across Pump P1:

P = 1 bar P=10 bar

Inlet Condition: P= 1Bar

T=180 degree C

Outlet Condition: P=10bar

T=180 degree C

Pump is at Adiabatic Reversible Condition. The Energy balance is given by

Ṁ*(H2-H1) = W + Q

Now, dU = TdS – PdV from Thermodynamics

Adding d (PV)

dH = TdS + VdP

TdS is zero as adiabatic & Reversible

Ws = W/ Ṁ

Now Volume of 60% HNO3

Density of 60% HNO3 solution at 20 degree is: [Ref: Handy math.com]

Density= 1366.7 kg/m3 Mass of 60% HNO3 = 3150 kg/h

Volume (V) = 2.299 m3

31
= 2069.1 KJ= Ws

Pump Shaft Work required is: 2069.1 KJ

32
CHAPTER 6 REACTION KINETICS AND
THERMODYNAMICS

KINETICS OF REACTION
The reaction kinetics is the study of the rates of the chemical process. Our chemical process
equation is as followed.

NH3 + HNO3 → NH4NO3

No kinetics data available as per our knowledge. So we assume our reaction is elementary and
the order of reaction is second.

We do not know about rate constant k, but from the literature we know about residence time of
the process volume in the reactor.( Liz M. Ríos Hidalgo, Luis M. Peralta Suárez, Luis E. Arteaga
Pérez, doi: 10.3303/cet1229214)

From the performance equation we can find the volume of the reactor with the known residence
time or rate constant. The reactor is CSTR, so its performance equation is as followed. One more
assumption: the reaction is constant volume process i.e.

Where,

= residence time

CA0 = Initial concentration of nitric acid

XA = conversion of nitric acid

CB0 = Initial concentration of ammonia

CA = concentration of nitric acid at time t

CB = concentration of ammonia at time t

= concentration ratio\

33
k = rate constant

FA0 = molar flow rate

V = volume of reactor or process fluid

Now , we know the residence time from the literature and it is of 15 minutes. .( Liz M. Ríos
Hidalgo, Luis M. Peralta Suárez, Luis E. Arteaga Pérez, doi: 10.3303/cet1229214).

Here wt. % of HNO3 is 60%

So the mole fraction of HNO3 in solution = (60/63)/[(60/63)+(40/18)]

= 0.3

Mole fraction of water in solution = 1 – 0.3

= 0.7

Molecular wt. of solution =[63(0.3) + 18(0.7)]

= 31.5 kg/kmol

Now, Flow rate of solution = (100 kmol/hr) * (31.5 kg/kmol)

= 3150 kg/hr

FA0 (HNO3) =3150*0.6

= 1890 kg/hr

FA0 (H₂ O) = 3150*0.4

= 1260 kg/hr

Volumetric flow rate of solution = (3150 kg/hr)/(1370 kg/m³)

= 2.299 m³/hr

HNO3 flow rate in solution = (1890 kg/hr)/(63 kg/kmol)

= 30 kmol/hr

Concentration of HNO3 =(30 kmol/hr) / (2.299 m³/hr)

= 13.0491 kmol/m³

34
So, the volume of the reactor is as follows:

CA0 = 13.0479 , FA0 = 100 , = 15 minutes

35
THERMODYNAMICS

For any reaction to take place, we must check its thermodynamic and kinetic feasibility. If the
reaction is feasible, then, one may proceed towards production.

A well-known thermodynamic property, Gibbs free energy (ΔG), gives the feasibility and the
spontaneity of the reaction.

If ΔG < 0:- reaction will occur

If ΔG > 0:- reaction will not occur

Gibbs free energy is given by equation:

Standard Gibbs free energy is given by

Heat of Formation of Individual components are found and heat of reaction is calculated by
product – reactants

Component (kJ/mole)

NH4NO3 -350.26

NH3 -46

HNO3 -206.63

36
 = -350.26 – (-46-206.63)

= -97.63 kJ/mole (Exothermic Reaction)

Now, is calculated as follows:

Component (J/kmole)

NH4NO3 151.08

NH3 192.77

HNO3 146

Now Standard Gibbs Free Energy is calculated at T= 298 K

< 0 (Reaction is feasible).

The reaction is occurring at 453 K and P=3.5 bar pressure, thus at this temperature
pressure condition Gibbs free energy is calculated, The relation between Standard
Gibbs free energy and Gibbs free energy is given by at reaction condition,

37
Where T=453K and P=3.5 bar = 3.45423 atm, Po=1 atm.

= -41933.8 + 8.314×453×ln

= -37.2651 kJ/mole (-ve value thus reaction is feasible)

Now,

is calculated using following equation

at T=453 is -7348.55 kJ/mol

Therefore,

-37.2651 = -7348.55 - ( 453 )

(-ve value thus reaction is feasible)

38
CHAPTER 7 EQUIPMENT DESIGN

(ONE MAJOR AND TWO MINOR)

MAJOR EQUIPMENT DESIGN: CSTR (Continuous Stirred Tank Reactor)

In the CSTR ammonia in gas phase reacts with aqueous nitric acid solution with continuous
mixing.

From the kinetics we know the volume of the reactor.

CA0 = 13.0479 , FA0 = 100 , = 15 minutes

Now, due to generation of vapours in the reactor we take 20% more volume.

So, now volume of reactor V = 1.916 1.2 =2.3

For diameter and height of the reactor:

Assume Sg = 10 cm/s = 0.1 m/s

Mass flow rate of ammonia at inlet: = 0.4722

Density of ammonia at given condition = 1.5942

Volumetric flow rate: Qv = 0.2962

39
Mechanical Design of reactor:

We use MOC here is SS316L due to corrosive materials.

Since, it is working under pressure it is designed as pressure vessel.

Operating Temperature = 180

Operating Pressure = 0.35

Design Pressure: P = 1.1 = 0.385

Density of MOC

Maximum Permissible Stress fp = 137.89

Corrosion allowance (assume) = 1mm

Spot radiography (assume): j = 0.85

Modulus of elasticity = 1.93053 105

Poisson’s ratio = 0.25

SHELL DESIGN

 Thickness of shell

t = 4.29 mm

So, based on market availability, 8 mm thick sheet will be required.

40
 Thickness of head
Considering Tori-spherical head,

Crown radius = 2 m (ID of shell)

Knuckle radius = 0.12 m (6% of ID of shell)

W=¼

W = 1.77

t = 6.81 mm

So, based on market availability, 10 mm thick sheet will be required.

 Combined loading calculation:

1. Stresses due to internal pressure in circumferential direction,

= 72.971 N/mm2
2. Stresses in longitudinal direction,

a. Due to internal pressure,

= 36.38 N/mm2

b. Due to weight of vessel and contents

Straight flange Sf = 3 * Thickness of head sheet = 30 mm

B = (1.024 * OD of shell) + (0.67*Knuckle radius) + (2 Sf) = 2.188 m

Whead = π B2 t ρ / 4 = 301.92 kg

Volume of shell material = π (Do2 -Di2) h / 4 = 0.1262 m3

Wshell = 0.1262 * 8030 = 1013.38 kg

41
 Weight of mounting = 10 % of shell weight
Wmounting = 101.338 kg

Weight of fluid in shell Wfluid = 7.78 * 1370 = 10658.6 kg

Where, density of HNO3 solution (60 wt%) = 947 kg/m3

W = Wshell + (2 Whead) + Wmounting + Wfluid = 12377.15 kg

= 0.2453 N/mm2

c. Due to wind or piping

f3 = = 0 ( M = 0, bending moment due to wind load)

Total stress in the longitudinal direction,

fa = f1 + f2 + f3 = 36.134 N/mm2
3. Stress due to offset piping,

(Assume T = 500 Nm due to piping and offset)

= 0.004953 N/mm2

The equivalent stress,

fR= (ft2 - ft fa+ fa2+3 fs2)0.5 = 63.195 N/mm2

Permissible compressive stress,

fc =

= 76.13 N/mm2

fR(tensile) < ft(permissible)

fa(tensile) < ft(permissible)

fa(compressive) < fc(permissible)

All the conditions are satisfied.

42
  External pressure consideration
Design jacket pressure Pj = 0.111457 N/mm2

Working jacket pressure = 0.101325 N/mm2

Assuming thickness t = 18 mm

Corroded thickness = 18 – 1 = 17 mm

Do = 2036 mm

Maximum unsupported length = 1780 mm

Pc = 4.33183 N/m2

Factor of safety = 4

P(allowable) = 1.08295 N/mm2 > Working jacket pressure

So, thickness of sheet = 18 mm

 Jacket wall thickness

Channel width d = 200 mm

= 3.323 mm

ts ≈ 6 mm

= 2.969 mm

tj ≈ 4 mm

By considering combined loading and external pressure, 18 mm thick sheet will be required.

43
NOZZLE DESIGN

Nozzle diameter = 152.4 mm

Nozzle will be on head so,

ts = 10 mm

ts’ = 6.81 mm (Corroded thickness)

Affected area A = ts*d = 1037.84 mm2

tn’ =

tn’ = 1.25 mm

tn ≈ 3 mm

AB = 4r = 304.8 mm

AD = 3.5 ts + 2.5 tn = 42.5 mm

Assuming compensation is provided only through nozzle,

H1 = H2 = 2.5ts = 25mm

Area available for compensation

1. As = d (ts - ts’ – C.A.)
= 486.156 mm2
2. Ao = 2H1(tn – tn’ – C.A.)
= 43.75 mm2
3. Ai = 2H2(tn – 2C.A.)
= 25 mm2
Total area A’ = A – (As + Ao + Ai)

= 1037.84 – 554.906 = 482.934 mm2 > 0

Compensation pad is required.

Now,

H1 = 2.5 tn = 7.5 mm

44
H2 = 2.5ts = 25 mm

As = 486.156 mm2

Ao = 13.25 mm2

Ai = 25 mm2

A’ = A – (As + Ao + Ai) = 513.434 (this much area is required for compensation)

For collar,

Outside diameter of collar < 2d

d = 152.4 + (3*2) = 158.4

Outside diameter of collar = 300 mm < 2d

Area of collar = (300 – 158.4) * Collar thickness

513.434 = 141.6 * Collar thickness

Collar thickness = 3.626 mm

Collar with 3.626 mm thickness and 300 mm outside diameter is required for compensation.

FLANGE DESIGN

ID of flange = 2000 mm

Choosing asbestos gasket,

Gasket factor m = 2

Minimum design seating stress Y = 11.2 N/mm2

Gasket thickness N = 30 mm

Gi = 2000 mm

Go = Gi + 2N = 2060 mm

Basic gasket seating width,

b0 = = 15 mm > 6.3 mm

45
 b = 2.5 = 9.68 mm

G = ( Gi + Go ) / 2 = 2030 mm

Wm atm = π b G Y = 691065.27 N

Wm opt = 2π G m b P + (π G2 P) / 4 = 1341140.378 N

Am atm = Wm atm/ f atm, Am opt = Wm opt/f opt

F atm = Stress in bolts under atmospheric condition = 58 x 106 N/m2

F opt = Stress in bolts under operating condition = 54 x 106 N/m2

Am atm = 11914.92 mm2

Am opt = 24835.93 mm2

Number of bolts = G / 25.4 = 79.92 ≈ 80

For M-30 size bolts,

Area of one bolt = 0.51 * (30)2.09 = 623.38 mm2

Total area of bolts Ab’ = 80 * 623.38 = 49870.48 mm2

Maximum bolt area permitted for gasket (Ab) = (2πYGN)/f = 73852.8 mm2

Ab > Ab’ > max (Am opt, Am atm)

This condition is satisfied. So, 80 bolts of M – 30 is required.

Thickness of flange = G(P/(Kf))0.5 + C.A.

Where, B = G0+2 (Bolt diameter) + 12 = 2132 mm

hG = (B-G)/2 = 51 mm

H = (π G2 P) / 4 = 1246070.707 N

K = 1/(0.3+((1.5Wm opt hG)/(HG)) = 2.94

46
Thickness of flange = 63.55 mm

Flange diameter = B + 2 (Bolt diameter) = 2192 mm

Bolt pitch,

B = (bp * Number of bolts) / π

Bolt pitch bp = 83.681 mm

MINOR EQUIPMENT : 1 : HEAT EXCHANGER : HE3

Fluid Side and Passes:

For one phase heat exchanger, the hotter fluid is generally taken to shell side.
Here, ammonium nitrate solution is considerably viscous so we take it as the shell
side fluid. No de-superheating or subcooling is there. Water is being used as a
cooling medium. 1-2 pass shell and tube heat exchanger with counter current flow
is chosen.
Shell side: Ammonium Nitrate
Tube side: Cooling water

Material selection for the condenser:

NH4NO3 is corrosive in nature. Hence, material for shell and tube side should be chosen
which has corrosion resistance property. Stainless steel grade 304 is to be used for both and shell
and tube material

Hot Fluid: (NH4NO3 SOLUTION)

Mass flow rate = 6400 kg/h

NH4NO3 feed temperature = 180

NH4NO3 outlet temperature = 140

Density = 1540

47
Cp = 2.0125

µ = viscosity = 0.0026386 Pa.s

kf = thermal conductivity = 0.67

= 143.111

Cold Fluid : Cooling Water

Here, Cooling water entering temperature = 30 °C

Cooling water outlet temperature = 60 °C

°C

Cp = 4.2

Density = 994

µ = viscosity = 0.00079 Pa.s

kf = thermal conductivity = 0.58

LMTD Calculation:

= 113.79 °C

48
For boiling and condensation FT = 0.98

LMTDc = FT × LMTD = 111.5142

Overall heat transfer coefficient calculation:

For NH4NO3 as a hot fluid and water as cold fluid, Overall film coefficient

ranges from 700-1000 W/m2 oC for condenser.

We take overall film coefficient = 800 W/m2 oC for the preliminary design.

Heat transfer area:

U = overall heat transfer co-efficient (W/m2 K)

Assume U = 800 W/m2K

Standard tube dimensions are chosen as follow.

do = 20mm, di = 16 mm and Length of tube = 3 m
Tube sheet thickness of 25 mm is taken each side. Hence
effective Tube length L = 3 – 0.05 = 2.95 m

Area for single tube =

=

= 0.1884 m2

49
We take 10 tubes for safety purpose.

With triangular pitch,

For, 2 tube pass K1 = 0.249, n1 = 2.207

So, Bundle diameter

Db = 106.59 mm

Shell ID = Db + clearance

= 106.59 + 58 = 164.59 mm

TUBE SIDE COEFFICIENT:

Water: cross sectional area =

No of tubes per pass =

Tube side flow area =

For water density = 993 kg/m3

Water velocity =

At mean temp =

Cp = 4200 J/kg K
kf = 0.58 W/m K
When water is in tube side, more accurate film coefficient relation is given as,

50
di = inner diameter of tube = 16 mm
From this equation,
, ,
hi = 4354.267 W/m2K

SHELL SIDE COEFFICIENT:

Volumetric flowrate = 6400/1540

= 4.15m³/hr
Cross sectional area = 3.14*(0.16459)² /4
= 0.0213 m²
Velocity = 4.15/0.0213 = 0.0541 m/s

, ,

ho = 5214.99 W/m2 K

FOR OVERALL HEAT TRANSFER COEFFICIENT:

Here both the fluids are not non-corrosive so fouling co-efficient = 3000 W/m2 K
Thermal conductivity of tube wall material = 16 w/m K

51
Uo = 785.2632 W/m2 K

Which is near to U assumed = 800 W/m2 K

PRESSURE DROP:

Tube side,

From graph of jf vs. Re

jf = 3.5 × 10-3

∆Pt = 4.1389 kPa.

Which is acceptable.

Shell side,
Tube bundle diameter = 106.59 mm
Ds = 164.59 mm

52
From graph jf = 2.2 ×10-2

∆Ps = 113.508 kPa

Which is acceptable.

MINOR EQUIPMENT 2. STORAGE TANK DESIGN (60 WT% HNO3)

HNO3 Solution is used per day = = 110.36 m3

HNO3(60 wt% solution) is hazardous material to store in bulk or in very large amounts. And
because of this we use 500KL or 500 m3. Since it is highly corrosive in nature, it may be stored
in s.s. or fibre plastic material storage tanks.

For design purpose we use material SS316L.

Available data:

Volume = 500 m3 ; (M.V. JOSHI)

53
MOC = SS316L

Density of MOC

Maximum Permissible Stress fp = 137.89

Corrosion allowance (assume) = 2mm

Spot radiography (assume): j = 0.85

Available plates of MOC to construct tank = 6300mm × 1800mm

Available thickness = 6, 8, 10, 12, 16, 18, 20 mm etc.

Now, we keep 0.3m height for vapor phase on the liquid stored in the tank.

Pressure at the bottom the tank =

Plate thickness t: here we use design pressure, more 10% of bottom available pressure i.e.

Pd1 = 1.1 × P

Next available thickness is 6 mm which is least plate thickness. As we move upward in the tank
the pressure will decrease, so we build the whole tank with 6 mm thickness plates.

Number of plates required:

Total height = 6120 mm

No. of plates vertically required =

Total number of plates =

54
Final height of tank = 4 × 1800 = 7200 mm = 7.2 m

Here, we use flat head as a roof of tank.

55
CHAPTER 8 PLANT LOCATION

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 while selecting a suitable site, and only a
brief review of the principal factors will be given in this section. The principal factors to be
considered are:

 Marketing availability

 Raw material supply

 Transport facilities

 Availability of labor

 Availability of utilities: water, fuel, power

 Availability of suitable land

 Environmental impact and effluent disposal

 Local community consideration

 Climate

 Political Strategic considerations

Market availability
For materials that are produced in bulk quantities: such as cement, mineral acids and fertilizers,
where the cost of the product per ton is relatively low and cost of transport a significant fraction
of the sales price, the plant should be located close to the primary market. This consideration will
be less important for low volume production, high-priced products; such as pharmaceuticals. The
market of CAN in India is mainly for fertilizer utilizers.

56
Raw material
The availability and price of raw materials will often determine the site location. Plants
producing bulk chemicals are best located close to the source of the major raw material. CAN
plant should be located near to nitro compound Industries so that NH₃ gas and HNO₃ be taken
directly through pipeline.

Transport
The transport of materials and products to and from plant will be an overriding consideration in
site selection. If practicable, a site should be selected that is close 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. Air transport is convenient and efficient for the movement of personnel
and essential equipment and supplies, and proximity of the site to a major airport should be
considered. For CAN plant too the above mentioned factors should be considered.

Availability of labor
Labor will be needed for construction of the plant and its operation. Skilled construction workers
will usually be brought in from outside the site, but there should be an adequate pool ofunskilled
labor available locally; and labor suitable for training to operate the plant. Skilled tradesmen will
be needed for plant maintenance. Local trade union customs and restrictive practices will have to
be considered when assessing the availability and suitability of the labor for recruitment and
training.

Availability of Utilities (services)

The “utilities” is now generally used for the ancillary services needed in the operation of any
production process. These services will normally be supplied from a central facility; and
willinclude:

 Electricity: Power required for processes, motors, lightings and general use.

 Steam for process heating: The steams required for the process are generated in the tube

57
  Water for general use: The water required for the general purpose will be taken from
local

 Water supplies like rivers / lakes / municipality.

 Dematerialized water: Demineralized water, from which all the minerals have been
removed by ion-exchange is used where pure water is needed for the process use, in
boiler

 Effluent disposal facilities: Facilities must be provided for the effective disposal of the
effluent without any public nuisance.

Availability of Land
Sufficient suitable land must be available for the proposed plant and future expansion. The land
should be ideally flat, well drained and have load-bearing characteristics. A full site evaluation
should be made to determine the need for piling or other foundations. Land rental/ lease rate
should be bare minimum to reduce the OPEX.

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. The disposal of
toxic and harmful effluents will be covered by local regulations and the appropriate authorities
must be consulted during the initial site survey to determine the standards that must be met.

Local community consideration

The proposed plant must 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.

58
Climate
Adverse climatic conditions at site will increase costs. Abnormally low temperatures will
required the provision of additional insulation and special heating for equipment and piping.
Stronger locations will be needed at locations subject to high wind loads or earthquakes.

Political and strategic considerations

Capital grants, tax concessions, and other inducements are often given by governments 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.

Preferable Location in India for the CAN plant:

 GNFC, BHARUCH

As all the CAN utilizers are nearby, easily products can be sold to customers.

59
CHAPTER 9 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. The principal
factors are considered are:

• Economic consideration: Construction and operating costs

• Process requirements
• Convenience of operation
• Convenience of maintenance
• Safety
• Future expansion

Economic Considerations: The cost of construction can be minimized by adopting a

layout that gives the shortest run of connecting pipe between equipment, and least amount
of structural steel work. Also, shorter pipe work will reduce the pressure drop and
eventually the operating cost.

Process requirements: An example of the need to take into account process consideration
is the requirement to elevate of the base of columns to provide the necessary net positive
suction head to a pump. Also, the steam generation should be near to the main unit in this
process. Many such process requirements are necessary to be considered while deciding the
plant layout.

Convenience of Operations: Equipment that need to have frequent attention should be

located near to the control room. Sufficient working space and headroom must be provided
to allow easy access to equipment. Valves, sample points and instruments should be
located at convenient positions and heights.

Convenience of Maintenance: The plant layout should be designed in such a manner as to

take proper care with regard to repairs and maintenance of different types of machines and
equipment being used in the industry. The machines should not be installed so closely that
may create the problems of their maintenance and repairs. Location of heat exchangers
should be such that the tube bundles can be easily withdrawn for cleaning and tube

60
replacement. Equipment that requires dismantling for maintenance, such as compressors
and large pumps, should be places under cover.

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 building. Firefighting systems must be located near to the process
involving flammable liquids.

Future expansion: Equipment should be located in such a way that any future expansion of the
process becomes convenient. Space should be left on pipe rack for future needs.

Sr.
No. DESCRIPTION

1 CANTEEN
2 ADMIN
3 PARKING
4 ETP
5 COOLING WATER
6 FIRE WATER
7 OPERATOR ROOM
8 WASTE WATER
9 CAN PLANT
10 TANK FORM
11 LOADING / UNLOADING
12 TRUCK PARKING
13 EMERGENCY GENERATOR
14 CONTROL ROOM
15 QUALITY CONTROL LABORATORY
16 STORE
17 WORKSHOP
18 WAREHOUSE
19 WEIGH BRIDGE
20 SECURITY ROOM
21 CAR PARKING (FUTURE EXPANSION)
22 TRUCK PARKING (FUTURE EXPANSION)
23 FUTURE EXPANSION AREA

61
62
CHAPTER 10: COST ESTIMATION

An estimate of the capital investment for a process may vary from a predesign estimate based on
little information except the size of the proposed project to a detailed estimate prepared from
complete drawings and specifications. Between these two extremes of capital- investment
estimates, there can be numerous other estimates which vary in accuracy depending upon the
stage of development of the project. These estimates are called by a variety of names, but the
following five categories represent the accuracy range and designation normally used for design
purposes:

1. Order-of-magnitude estimate (ratio estimate) based on similar previous cost data; probable
accuracy of estimate over + 30 percent.

2. Study estimate (factored estimate) based on knowledge of major items of equipment;

probable accuracy of estimate up to 30 percent.

3. Preliminary estimate (budget authorization estimate; scope estimate) based on sufficient data
to permit the estimate to be budgeted; probable accuracy of estimate within +20 percent.

4. Definitive estimate (project control estimate) based on almost complete data but before
completion of drawings and specifications; probable accuracy of estimate within f 10 percent.

5. Detailed estimate (contractor’s estimate) based on complete engineering drawings,

specifications, and site surveys; probable accuracy of estimate within +5 percent. As we are on
the initial stages of design, we can use first two techniques; however, for study estimate we will
be requiring the capacity of each equipment which is possible only after designing every
equipment so we shall be using Order of magnitude here.

63
Equipment No. Of Units Cost Per Unit ( ₹ ) Total Cost ( ₹ )
Reactor 1 11,000,000 11,000,000
Pump 3 100,000 300,000
Dryer 1 15,000,000 15,000,000
Reboiler 1 4,000,000 4,000,000
Evaporator 1 13,100,000 13,100,000
Compressor 1 6,000,000 6,000,000
Heat Exchanger 3 1,000,000 3,000,000
Storage tank 4 1,000,000 4,000,000
Granulator 1 700,000 700,000
Screener 1 300,000 300,000
Total Equipment Cost 75,400,000

(The data for range of all costs calculated below, is taken from Book ‘Plant design &
economics’ by Peters and Timmerhaus)

10.1 Estimation of Total Capital Investment

I. Direct cost:

1. Purchased equipment (15-40 % of Fixed-capital investment)

Assuming 15% of fixed capital investment

So fixed capital investment = ₹75,400,000 /0.15

= ₹ 502,666,666.67

2. Installation, including insulation and painting (6-14 % of fixed capital investment)

Assuming 8% of fixed capital investment

= 0.08 × ₹ 502,666,666.67

= ₹ 40,213,333,34

64
3. Instrumentation and controls, installed (2-8% of fixed capital investment)

Assuming 6% of fixed capital investment

= 0.06 × ₹ 502,666,666.67

= ₹ 30,160,000

4. Piping, installed (3-20% of fixed capital investment)

Assuming 10% of fixed capital investment

= 0.1 × ₹ 502,666,666.67

= ₹ 50,266,666.667

5. Electrical, installed (2-10% of fixed capital investment)

Assuming 6% of fixed capital investment

= 0.06 × ₹ 502,666,666.67

= ₹ 30,160,000

6. Yard improvements: (2-5% of fixed capital investment)

Assuming 4% of fixed capital investment

= 0.04 × ₹ 502,666,666.67

= ₹ 20,106,666.67

7. Service facilities: (8-20% of fixed capital investment)

Assuming 18% of fixed capital investment

= 0.18 × ₹ 502,666,666.67

= ₹ 90,480,000

65
8. Land: (1-2% of fixed capital investment)

Assuming 2% of fixed capital investment

= 0.02 × ₹ 502,666,666.67

= ₹ 10,053,333.33

9. Buildings (3-18% of fixed capital investment)

Assuming 10% of fixed capital investment

= 0.1 × ₹ 502,666,666.67

= ₹ 50,266,666.667

Direct cost = ₹ 382,026,667

II. Indirect cost:

1. Engineering and supervision (4-21% of fixed capital investment)

Assuming 5% of fixed capital investment

= 0.1 × ₹ 502,666,666.67

= ₹ 50,266,666.667

2. Construction expense (4-16% of fixed capital investment)

Assuming 6% of fixed capital investment

= 0.05 × ₹ 502,666,666.67

= ₹ 25,133,333.33

66
3. Contractor’s fee (2-6% of fixed capital investment)

Assuming 2% of fixed capital investment

= 0.03 × ₹ 502,666,666.67

= ₹ 15,080,000

4. Contingency (5-15% of fixed capital investment)

(Assuming 8% of fixed capital investment)

= 0.06 × ₹ 502,666,666.67

= ₹ 30,160,000

Indirect Cost = ₹ 120,639,999.7

Total fixed capital investment = Direct cost + indirect cost

= ₹ 502,666,666.67

Working capital is generally 10-20% of the total capital investment

Let working capital = 15% of total fixed capital investment

= 0.15 × ₹ 502,666,666.67

= ₹ 75,400,000

Total capital investment (TCI) = Total fixed capital investment + working capital

= ₹ 502,666,666.67 +₹ 75,400,000

= ₹ 578,066,666.67

67
10.2 Estimation of Total Product Cost

I. Manufacturing cost = Total Fixed charges + Direct production cost + Plant overhead
cost

A. Fixed charges:

1. Depreciation (depends on life period, salvage value, and method of calculation- about 10% of
Fixed-capital investment for machinery and equipment and 2-3% of building value for buildings)

The assumed value is (10% of fixed capital investment + 2% of building value found in direct
costs)

= 502,666,666.67 (0.1) + 50,266,666.667 (0.02)

= 502,666,66.667+ 1,005,333.34
= ₹ 51,272,000

2. Local taxes (1-4% of fixed capital investment)

Assuming 3% of fixed capital investment

= 0.03 × 502,666,666.67

= ₹ 15,080,000

3. Insurance (0.4-1% of fixed capital investment)

Assuming 0.8% of fixed capital investment

= 0.008 × 502,666,666.67

= ₹ 40,213,333.34

Therefore, Fixed charges = ₹ 106,565,333.3 (adding above three)

68
The fixed charge is 10-20% of total product cost

Assuming 15% of total product cost

Total product cost = ₹ 106,565,333.3 /0.15

= ₹ 710,435,555.3

B. Direct production cost:

1. Raw materials (10-50% of total product cost)

Assuming 15% of total product cost

= 0.15 × ₹ 710,435,555.3

= ₹ 106,565,333.3

2. Operating labor (10-20% of total product cost)

Assuming 10 % of total product cost

= 0.10 × ₹ 710,435,555.3

= ₹ 71,043,555.53

3. Direct supervisory and clerical labor (10-25% of operating labor)

Assuming 15% of operating labor cost

= 0.15 × ₹ 710,435,555.3

= ₹ 106,565,333.3

4. Utilities (10-20% of total product cost)

Assuming 12% of total product cost

= 0.10 × ₹ 710,435,555.3

= ₹ 71,043,555.53

69
5. Maintenance and repairs (2-10% of fixed capital investment)

Assuming 4.5% of total fixed investment

= 0.045 × ₹ 710,435,555.3

= ₹ 31,969,599.99

6. Operating supplies (10-20% cost of maintenance and repairs)

Assuming 15% of maintenance and repairs

= 0.15 × ₹ 31,969,599.99

= ₹ 4,795,439.99

7. Laboratory charges (5-20% of operating labor)

Assuming 6% of operating labor

= 0.06 × ₹ 71,043,555.53

= ₹ 4,262,613.33

8. Patents and royalties (0-6% of total product cost)

Assuming 1% of total product cost

= 0.01 × ₹ 710,435,555.3

= ₹ 71,043,555.53

Total Direct Production Cost is sum of all 8 above costs which is ₹ 403,349,786.5

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C. Plant overhead costs

= 60% of sum of (Operating Labour + Direct Supervisory and Clerical Labour + Maintenance
and Repair)
= 0.6 × (₹ 710,435,555.3 + ₹ 106,565,333.3 + ₹ 31,969,599.99)
= ₹ 509,382,293.2

I. Manufacturing cost = Total Fixed charges + Direct production cost + Plant overhead
cost

= ₹ 106,565,333.3 + ₹ 403,349,786.5 + ₹ 509,382,293.2

= ₹ 1,019,297,413

II. General Expenses = Administrative costs + Distribution costs + Research and

development costs + Financing

A. Administrative costs (20-30 % of operating labor cost)

Assuming 20% of operating labor cost

= 0.20 × ₹ 71,043,555.53
= ₹ 14,208,711.11

B. Distribution and selling costs (2-20% of total product cost)

Assuming 12% of total product cost

= 0.12 × ₹ 710,435,555.3
= ₹ 85,252,266.64

C. Research and development costs (about 5% of total product cost)

= 0.04 × ₹ 710,435,555.3

= ₹ 3,410,090.66

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D. Financing (0-10% of total capital investment)

Assuming 5 % of total capital investment

= 0.05 × ₹ 578,066,666.67
= ₹ 28,903,333.33

II. General Expenses = Administrative costs + Distribution costs + Research and

development + Financing

= ₹ 102,871,068.4

Total Product Cost = Manufacturing Cost + General Expenses

= ₹ 1,019,297,413 + ₹ 102,871,068.4
= ₹ 1,122,168,481

Cost of the product = Total product cost/ our plant capacity in kg

= (₹ 1,122,168,481) / (12111 × 24 × 300) kg

= ₹ 12.86 per kg

Selling price of product with a profit margin of 30% = 1.3 × ₹ 12.86

= ₹ 16.71 per kg

Annual Selling = ₹ 16.71 Per kg × (12111 × 24 × 300) kg

= ₹ 1,457,098,632

Annual earnings = Annual Selling – Total Production Cost

= ₹ 1,457,098,632 - ₹ 1,122,168,481

= ₹ 334,930,151

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For earning more than 1 crore, Income tax = 40% of Annual Profit

= 0.4 × ₹ 334,930,151
= ₹ 133,972,060.4

Net Profit = Annual earnings - Income tax

= ₹ 334,930,151 - ₹ 133,972,060.4
= ₹ 200,958,090.6

Payback Period =

= 502,666,666.67
50,266,666.667 + 200,958,090.6
2 years

Rate of Return = (Net Profit) / (Total Capital Investment)

= ₹ 200,958,090.6/ ₹ 578,066,666.67
= 0.3476

Rate of return is about 34.76 %

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CHAPTER 11 OCCUPATIONAL HEALTH & SAFETY

The chemicals which must be considered for occupational health and safety in all plants are:–
Ammonia – Nitric acid (and nitrogen oxides) – Ammonium nitrate. Other chemicals, such as
processing aids, maintenance chemicals and anticaking agents, added to improve the storage
characteristics of the product, may be used in the plant but these cannot be discussed in such a
general document. Safety data sheets should be available for all who come into actual or
potential contact with these chemicals.

Ammonia :
Ammonia is a gas at atmospheric pressure and temperature and is normally stored as a liquid. It
has a pungent, suffocating odor which is readily recognizable. The liquid gives severe cold
burns, and the vapor is toxic and corrosive to all parts of the body. ACGIH occupational
exposure limits for ammonia are 25 ppm for 8 hour TWA and 35ppm for short term exposure (15
min). Advice on the correct medical treatment for exposed persons must be available at all points
of potential contact.

Nitric Acid :
Nitric acid is a corrosive aqueous solution of a strong acid and the liquid may give off toxic
fumes of oxides of nitrogen. These, and nitric acid fume, are toxic and corrosive to all parts of
the human body. ACGIH occupational exposure limits are 2ppm for 8 hour TWA and 4ppm for
short term exposure (15 min). First aid procedures must be specified on safety data sheets but a
particular hazard is that fluid may build up in the lungs up to 48 hours after exposure.
Appropriate protective clothing must be worn for tasks which have the potential for the spillage
of nitric acid.

Ammonium Nitrate :
Ammonium nitrate does not have any specific occupational health problems. The dust arising
from ammonium nitrate (or CAN) is of low toxicity and is generally regarded as a nuisance dust
with 10mg.Nm-3 (8 hour exposure) being accepted as the permitted level provided the particle
size is above 5µm. Ammonium nitrate may decompose in a fire situation and thus stores should
be suitably designed with consideration for factors such as access to stacks, spacing between
stacks, presence of other chemicals (such as combustible materials). Oxides of nitrogen will be
emitted during a decomposition. Full health and safety data is given in Safety Data Sheets.
Guidance on Safety Data Sheets is given in reference. General product information on
ammonium nitrate and calcium ammonium nitrate is given here.

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11.1. Ammonium Nitrate (AN)
11.1.1 Identification

Chemical name : Ammonium Nitrate

Commonly used synonyms : : AN, Ammonium Nitrate Fertilizer

C.A.S. Registry number : : 6484–52–2

EINECS Number : : 299–347–8

Molecular formula : : NH4NO3

11.1.2 Hazards to Man and the Environment

To man

Ammonium nitrate is basically harmless when handled correctly.

To the environment

Ammonium nitrate is basically harmless when handled correctly.

11.1.3 Physical and Chemical Properties

Appearance : White or off-white granules or prills

Odour : : Odourless

PH water solution : >4.5

Melting point : 160-170°C depending on moisture content

Boiling point : >210°C (decomposes by dissociation)

Explosive properties : Not explosive as per EEC test A14 (67/548/EEC). The fertilizer has a
high resistance to detonation. This resistance is decreased by the presence
of contaminants and/or high temperatures

Oxidising properties : Can support combustion and oxidise. Not classified as oxidising
according to EEC Directive 88/379/EEC and test A17

Solubility in water : 1,900g.l-1 at 20°C

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Bulk density : 830 to 1,100kg.m-3

11.2 Calcium Ammonium Nitrate (CAN)

11.2.1 Identification

Chemical name : Mixture of ammonium nitrate and calcium Carbonate Calcium ammonium
nitrate, CAN

Composition : Mixture of ammonium nitrate with calcium carbonate and/or dolomite

containing not more than 80% of ammonium nitrate

11.2.2 Hazards to Man and the Environment

To man

CAN is basically harmless when handled correctly.

To the environment

CAN is basically harmless when handled correctly.

11.2.3 Physical and Chemical Properties

Appearance : White, off-white or grey granules or prills

Odour : Odourless

pH water solution : >4.5

Explosive properties : Not explosive as per EEC test A14. The fertilizer has a very high
resistance to detonation. This resistance is decreased by the
presence of contaminants and/or high temperatures

Oxidising properties : Not classified as oxidising material according to EEC Directive

88/379/EEC. Can support combustion

Solubility in water : NH4NO3 highly soluble CaCO3/MgCO3 sparingly soluble

Bulk density : 900-1,100kg.m-3

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11.3 MSDS OF CAN

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 REFERENCE

1. https://www.businesswire.com/news/home/20180406005233/en/Global-
Calcium-Ammonium-Nitrate-Market-Analysis-Forecast-2013-2023

2. https://www.businesswire.com/news/home/20171110005375/en/High-
Demand-Fertilizers-Boost-Calcium-Nitrate-Market

3. https://www.google.co.in/search?q=heat+of+formation+of+watert&rlz=1C1D
FOC_enIN769IN769&oq=heat+of+formation+of+watert&aqs=chrome..69i57
j0l5.6519j1j7&sourceid=chrome&ie=UTF-8

4. THE HEAT CAPACITY OF AMMONIUM NITRATE BY C. C.

STEPHENSON, D. R. BENTZ' AND D. A. STEVENSON RECEIVED
OCTOBER 29, 1954

5. https://webbook.nist.gov/chemistry/

6. M.V JOSHI

7. file:///C:/Users/SURESH%20A%20DODIA/Downloads/Documents/plant_des
ign_and_economics_for_chemical_engineers.pdf

8. SC“Achema”
Safety data sheet

9. Reactor Design by Andrew Rosen

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