ammonia synthesis processing synthesis processing.pdf · ammonia synthesis reaction, from its...

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Libya Ministry of Higher Education

Bright Star University – Braga


Faculty of Technical Engineering

Department of Chemical Engineering

Ammonia Synthesis Processing

Prepared by :

Mahmoulah Youns Abd alraheem 21152895

Rahma Ahmad Hamad 21152888

Aisha Saleh Salem 21152870

Azza Abd elkader Ejeal 21152981

Saada Ali Ashtiwi 211521094

Supervised by :

Hussin Awami

DEGREE OF BACHELOR IN CHEMICAL ENGINEERING

FACULTY OF TECHNICAL ENGINEERING

CODE OF PROJECT ( CHE2017046F105 )

i

Ammonia Synthesis processing

By






Supervised By

Hussin Awami

Project Report Submitted as Partial Fulfillment of the Requirements for the

Degree of Bachelor in Chemical Engineering

Month, 2018

ii

ABSTRACT

Ammonia is produced as a result of a chemical reaction between hydrogen and nitrogen in

the mole ratio of 3:1 respectively, the source of hydrogen is natural gas and water while the

nitrogen is obtained from atmospheric air. Separation of the hydrogen needed for the

ammonia synthesis reaction, from its source is difficult. Hydrogen production method is the

main source of distinction between the various ammonia production routes.

Most of the improvements in the technology regarding the ammonia synthesis were

concerned with the hydrogen production step. Hydrogen can be produced by steam

reforming, partial oxidation and water electrolysis.

The bulk of the world ammonia production is based on steam reforming.

The major hydrogen sources are natural gas, naphtha and coal.

The mass balance and energy balance calculations for a plant capacity of 1000 Tons per

day of ammonia production is found out. Material of construction, plant layout, control,

safety and cast estimate for the above said plant is presented in this work. For a Ton of

ammonia the cost is 610 US dollars and with profit of 30 % of the payback period of the

investment money it will be 10 years.

In this project steam reforming methods to produce synthesis gas (which will form

ammonia) from natural gas is us

iii

DEDICATION

To the utmost knowledge lighthouse, to our greatest and most honored prophet Mohamed

May peace and grace from Allah be upon him. To the Spring that never stops giving, to my

mother who weaves my happiness with strings from her merciful heart. To my mother. To

whom he strives to bless comfort and welfare and never stints what he owns to push me in

the success way who taught me to promote life stairs wisely and patiently, to my dearest

father. To whose love flows in my veins, and my heart always remembers them, to my

brothers and sisters. To those who taught us letters of gold and words of jewel of the utmost

and sweetest sentences in the whole knowledge. Who reworded to us their knowledge

simply and from their thoughts made a lighthouse guides us through the knowledge and

success path, To our honored teachers and professors.

iv

ACKNOWLEDGEMENTS

I would like to express my appreciation to my supervisor, (Hussin Awami ) who has

cheerfully answered my queries, provided me with materials, checked my examples,

assisted me in a myriad ways with the writing and helpfully commented on earlier drafts of

this project. Also, I am also very grateful to my friends, family for their good humor and

support throughout the production of this project .

v

APPROVAL

This project report is submitted to the Faculty of Technical Engineering, Bright Star

University – Braga, and has been accepted as partial fulfillment of the requirement for the

degree of bachelor in Chemical Engineering. The members of the Examination Committee

are as follows:

________________________________________

Supervisor

Hussin Awami




____________________________________________

Examiner 1

Abdulftah Ibrahim




____________________________________________

Examiner 2

Raheel Gumaa



Bright Star University – Brag

vi

DECLARATION

I hereby declare that the project report is my original work except for quotations and

citations, which have been duly acknowledged. I also declare that it has not been

previously, and is not concurrently, submitted for any other degree at Bright Star University

– Braga or at any other institution.

___________________________________






vii

TABLE OF CONTENTS

ABSTRACT.................................................................................. ii

LIST OF FIGURES.......................................................................... xvi

List of Symbols.............................................................................. xix

REFERENCES.............................................................................. 104

APPENDICES.............................................................................. 106

CHAPTER 1: NTRODUCTION TO AMMONIA SYNTHESIS

1.1 History of ammonia manufacturing processes............. 1

1.2 Ammonia production....................................................... 1

1.3 Similar Processes.............................................................. 2

1.3.1 Haber-Bosch process........................................................ 2

1.3.2 KBR Advanced Ammonia Process Plus........................... 2

1.3.3 Conventional Magnetite Ammonia Synthesis Process..... 3

1.4 Ammonia Chemistry......................................................... 4

1.5 Physical and Chemical Properties..................................... 5

viii

1.6 Uses of Ammonia.............................................................. 6

1.7 Ammonia Process Theory................................................. 6

1.8 The Ammonia Process...................................................... 7

1.8.1 Desulphurization (HDS)................................................... 9

1.8.2 Primary Reforming.......................................................... 10

1.8.3 Secondary Reforming...................................................... 11

1.8.4 Shift Conversion (CO Conversion)................................... 12

1.8.5 CO2 Removal.................................................................... 13

1.8.6 Methanator........................................................................ 14

1.8.7 Synthesis Gas Compression.............................................. 14

1.8.8 Ammonia Synthesis Loop................................................. 15

CHAPTER 2: MATERIAL BALANCE AND ENERGY BALANCE

2 Material Balance............................................................... 15

2.1 Primary Reformer............................................................. 15

2.2 Secondary Reformer...................................................... 20

ix

2.3 High Temp Shift Conversion............................................ 23

2.4 Low Temp Shift Conversion............................................. 24

2.5 Water Separator (exclusion of water).............................. 26

2.6 CO2 Removal (Absorber)................................................. 27

2.7 Methanation...................................................................... 30

2.8 Cooling , Compression & Separator................................. 32

2.9 Ammonia Converter.......................................................... 33

2.10 Energy Balance................................................................. 38

2.10 Primary Reformer............................................................. 38

2.11 Secondary Reformer......................................................... 42

2.12 Product Gas Cooler........................................................... 44

2.13 High Temp Shift Converter............................................... 45

2.14 First Gas/Gas Heat Exchanger.......................................... 46

2.15 Low Temp Shift Converter............................................... 47

2.16 Water Separator................................................................. 49

x

2.17 CO2 Removal ( Absorber )......................................... 51

2.18 Second Gas/Gas Heat Exchanger...................................... 54

2.19 Methanator........................................................................ 55

2.20 Cooler................................................................................ 58

2.21 Synthesis Gas Compressor, Cooler and Separator............ 60

2.22 Gas/Gas Reactor Heat Exchanger..................................... 61

2.23 Cooling and Separator....................................................... 63

2.24 Ammonia Convertor......................................................... 65

CHAPTER 3 : HEAT EXCHANGER DESING

3.1 Heat exchanger desing...................................................... 67

3.2 Pressure drop..................................................................... 71

CHAPTER 4 : PROCESS INSTRUMENTATION AND CONTROL

4 Process Instrumentation and Control................................ 73

4.1 Instruments........................................................................ 73

4.1.1 Safe plant operation.......................................................... 74

xi

4.1.2 Production rate.................................................................. 74

4.1.3 Product quality.................................................................. 74

4.1.4 Cost................................................................................... 74

4.2 Models of Control and Measurement Systems................. 75

4.2.1 Level Control.................................................................... 75

4.2.2 Temperature Control......................................................... 75

4.2.3 Pressure Control................................................................ 76

4.2.4 Ratio Control..................................................................... 76

4.2.5 Flow Control..................................................................... 77

CHAPTER 5 : COST ESTIMATION

5 Cost Estimation................................................................. 78

5.1 Estimation of Total Capital Investment............................ 80

5.1.1 Direct cost......................................................................... 80

5.1.2 Indirect cost....................................................................... 81

5.2 Estimation of total product cost........................................ 82

xii

5.2.1 Manufacturing cost........................................................... 82

5.2.2 General Expenses.............................................................. 84

CHAPTER 6: SAFETY AND LOSS PREVENTION

6 Safety and Loss Prevention............................................... 86

6.1 Safety Concerns with Ammonia....................................... 86

6.2 Major Hazards................................................................... 87

6.3 Occupation Health & Safety............................................. 87

6.4 Ammonia Solution............................................................ 89

6.4.1 Product Identification........................................................ 89

6.4.2 Hazards Identification ...................................................... 89

6.4.3 First Aid Measures............................................................ 90

6.4.4 Fire Fighting Measures..................................................... 91

6.4.5 Accidental Release Measures............................................ 91

6.4.6 Handling and Storage........................................................ 92

6.4.7 Exposure Controls/Personal Protection............................ 92

xiii

6.4.8 Stability and Reactivity..................................................... 93

6.4.9 Toxicological Information................................................ 94

6.4.10 Ecological Information..................................................... 94

6.4.11 Disposal Considerations.................................................... 94

6.4.12 Other Information............................................................. 94

CHAPTER 7: PLANT SITE AND LAYOUT

7.1 Introduction....................................................................... 96

7.2 Plant Location and Site Selection..................................... 96

7.2.1 Marketing Area................................................................. 96

7.2.2 Raw Materials................................................................... 97

7.2.3 Transport........................................................................... 97

7.2.4 Availability of Labor......................................................... 97

7.2.5 Utilities (Services)............................................................. 98

7.2.6 Environmental Impact and Effluent Disposal................... 98

7.2.7 Local Community Considerations.................................... 98

xiv

7.2.8 Land( Site Considerations)................................................ 99

7.2.9 Climate.............................................................................. 99

7.2.10 Political and Strategic Considerations.............................. 99

7.3 Site Layout....................................................................... 99

7.4 Plant Layout..................................................................... 100

7.4.1 Costs.................................................................................. 101

7.4.2 Process Requirements....................................................... 101

7.4.3 Operation........................................................................... 101

7.4.4 Maintenance...................................................................... 101

7.4.5 Safety................................................................................ 102

7.4.6 Plant Expansion................................................................. 102

7.4.7 Modular Construction....................................................... 102

xv

TABLE LIST OF

5 Physical and chemical properties............................................ TABLE 1

16 Natural Gas Contents (W1).................................................... 2 TABLE

16 Calculation of carbon........................................................... 3 TABLE

17 Dry gas composition....................................................................... 4 TABLE

18 Carbon Balance................................................................... 5 TABLE

19 Stream outlet from Primary Reformer (W3)................................... 6 TABLE

22 Air inlet to Secondary Reformer (W4)........................................... 7 TABLE

22 Stream outlet from Secondary Reformer(W5)............................... 8 TABLE

24 Stream outlet from H.T. Shift Converter(W6)................................ 9 TABLE

26 Stream outlet from L.T. Shift Conversion(W7).............................. 10 TABLE

27 W9 Stream outlet from Separator(W9)..................................... 11 TABLE

29 KHCO3+ H2O inlet CO2 removal (W10)......................................... 12 TABLE

29 Stream outlet CO2 Removal (W12).................................................. 13 TABLE

32 Stream outlet from Methanator (W13)............................................. 14 TABLE

33 CH4+ H2O outlet from Compressor Cooler and separator (W14)... 15 TABLE

33 Steam outlet from Compressor and Cooler(W15)............................ 16 TABLE

35 Ammonia(W17).................................................................... 17 TABLE

36 Purged(W18)........................................................................ 18 TABLE

36 Recycle(W19 )..................................................................... 19 TABLE

36 Material balance at steady state (in=out).................................... 20 TABLE

37 New and Old values.............................................................. 21 TABLE

39 Constant values for heat capacity................................................... 22 TABLE

40 Calculation the Enthalpy change of the reactants (∆ H1)........... 23 TABLE

41 Calculation the Enthalpy change of the reactants (∆ H2)........... 24 TABLE

41 Calculation the Enthalpy change of the products (∆ H3).......... 25 TABLE

xvi

42 Calculation the Enthalpy change of the reactant (∆ H4)............ 26 TABLE

43 Calculation the Enthalpy change of the Products (∆ H5)........... 27 TABLE



46 Calculation the enthalpy change of the product (∆ H8)............. 30 TABLE

48 Calculation the enthalpy change of the product (∆ H9)............ 31 TABLE

49 Calculation the enthalpy change of H2O (∆ H10)........................ 32 TABLE

50 Calculation the Enthalpy change of the product (∆ H11)........... 33 TABLE

52 Calculation the Enthalpy change of the reactant (∆ H12 ).......... 34 TABLE

52 CP ( KHCO3 ) calculation...................................................... 35 TABLE

52 Calculation the enthalpy change of the product (∆ H13)............ 36 TABLE


54 Calculation the Enthalpy change of the product (∆ H15).......... 38 TABLE


57 Calculation the Enthalpy change of the reactant (∆ H17).......... 40 TABLE

58 Calculation the Enthalpy change of the water (∆ H19)............... 41 TABLE


60 Calculation the enthalpy change of the CH4+H2O (∆ H21)........ 43 TABLE


62 Calculation the enthalpy change of the product (∆ H23)........... 45 TABLE

64 Calculation the Enthalpy change of NH3 (∆ H26)........................ 46 TABLE

64 Calculation the Enthalpy change of purged (∆ H27)................... 47 TABLE

65 Calculation the Enthalpy change of Recycle (∆ H28)................ 48 TABLE

77 Some Symbols for Instrumentation................................................ 49 TABLE

xvii

LIST OF FIGURES

8 Ammonia process Block Diagram.................................. Figure1

8 Ammonia process........................................................... Figure2

9 Desulphurization (HDS)................................................. Figure3

10 Primary Reformer........................................................... Figure4

11 Operation Condition of Secondary Reformer................. Figure5

12 Shift Conversion............................................................. Figure6

13

Benfield Process............................................................. Figure7

15 Primary Reformer (M.B)................................................ Figure8

20 Secondary Reformer (M.B)............................................ Figure9

23 High Temperature Shift Converter (M.B)..................... Figure10

24 Low Temp Shift Converter (M.B)................................. Figure11

26 Water Separator (M.B).................................................. Figure12

27 CO2 REMOVAL (M.B).................................................. Figure13

xviii

30 Methanation (M.B)......................................................... Figure14

32 Cooling Compression & Separator (M.B)..................... Figure15

33 Ammonia Converter (M.B)............................................ Figure16

38 Primary Reformer (E.B)................................................. Figure17

42 Secondary Reformer (E.B)............................................. Figure18

44 Gas Cooler (E.B)............................................................. Figure19

45 High Temp Shift Converter (E.B)................................... Figure20

46 First Heat Exchanger (E.B)............................................ Figure21

47 Low Temp Shift Converter (E.B)................................... Figure22

49 Water Separator (E.B)..................................................... Figure23

51 CO2 Removal (E.B).................................................... Figure24

54 Second Heat Exchanger (E.B)........................................ Figure25

55 Methanation (E.B).......................................................... Figure26

58 cooler............................................................................. Figure27

60 Synthesis Gas Compressor, Cooler and Separator (E.B) Figure28

xix

61 Gas Reactor Heat Exchanger (E.B)............................... Figure29

63 Cooling and Separator (E.B).......................................... Figure30

65 Ammonia Convertor (E.B)............................................. Figure31

75 Level Control............................................................... Figure32

75 (a) Control of One Fluid Stream (b) By-Pass Control Figure33

76 Pressure control (a) (b) (c) (d)....................................... Figure34

77 Ratio Control................................................................. Figure35

77 Flow control (a) (b)......................................................... Figure36

79 Chemical Engineering Plant Cost Index......................... Figure37

103 Site Layout...................................................................... Figure38

xx

LIST OF SYMBOLS

NAME SYMBOI

Molecular weight M

Volume % n/n %

Number of carbon i

Number of moles n

Weight W

The Enthalpy consumption ∆ HR°

The change of Enthalpy ∆ H

The change in the temperature dt

Specific heat of the components heat capacity for gases Cp

Temperature Controller TC

Pressure Control PC

Level Control LC

xxi

Flow Control FC

Ratio Control RC

Flow Recorder FR

Flow Indicator FI

Flow Recorder and Control FRC

Level Recorder and Control LRC

Level Indicator and Control LIC

Pressure Indicator and Control PIC

1

CHAPTER 1 INTRODUCTION TO AMMONIA SYNTHESIS

1.1 History of ammonia manufacturing processes :

Before the start of World War I, most ammonia was obtained by the dry distillation of

nitrogenous vegetable and animal products; the reduction of nitrous acid and nitrites with

hydrogen; and the decomposition of ammonium salts by alkaline hydroxides or by

quicklime, the salt most generally used being the chloride (salt-ammoniac) The Haber

process, which is the production of ammonia by combining hydrogen and nitrogen, was

first patented by Fritz Haber in 1908. In 1910, Carl Bosch, while working for the German

chemical company BASF, successfully commercialized the process and secured further

patents. It was first used on an industrial scale by the Germans during World War I. Since

then, the process has often been referred to as the Haber-Bosch process.

1.2 Ammonia production :

Facilities provide the base anhydrous liquid ammonia used predominantly in fertilizers

supplying usable nitrogen for agricultural productivity. Ammonia is one of the most

abundantly-produced inorganic chemicals. There are literally dozens of large-scale

ammonia production plants throughout the industrial world, some of which produce as

much as 2,000 to 3,000 tons per day of anhydrous ammonia in liquid form. The worldwide

production in 2006 was 122,000,000 metric tons.China produced 32.0% of the worldwide

production followed by India with 8.9%, Russia with 8.2%, and the United States with

6.5%. Without such massive production, our agriculturally-dependent civilization would

face serious challenges.

2

1.3 Similar Processes :

Haber-Bosch process : 1.3.1

This process involves the direct reaction between Hydrogen and Nitrogen. Before carrying

out the process, Nitrogen and Hydrogen gas are produced. The first step in the process is to

remove sulfur compounds from the natural gas to compare and contrast similar processes to

the Haber process we first have to evaluate the Haber process. The Haber process was

invented by Fritz Haber, this was done by combing medium temperatures (500 Co), very

high pressures (50 atmospheres, 25,500kPa) and a catalyst. This process produced

Ammonia yield of approximately 10-20% . Although the Haber process was invented by

Fritz Haber it was developed into an industrial process by Carl Bosch. The reaction that is

between Nitrogen gas and Hydrogen gas to produce Ammonia gas is an exothermic

equilibrium reaction which releases 92.4 kJ/mol of energy at 298K (25Co) .

There are many companies that engineer and construct technology for this and one of them

is the KBR (Kellogg’s brown and root)

The Kellogg’s brown and root company is a leading supplier of ammonia process

technologies. This company has a lot of processes for the synthesis of ammonia with high

technologies in placement. Some of these are

1.3.2 KBR Advanced Ammonia Process Plus :

This process is a combination of the Top – Fired Steam Methane Reformer (SMR) with a

KAAP Synthesis Converter. The auto thermal reformer that is present in this system

(KAAPplus™) uses air instead of oxygen- enriched air. This is needed because the

downstream purifier requires excess Nitrogen. An air separation unit is not need for the

system. Then all excess Nitrogen, Methane and most of the Argon are removed from the

syngas. This process also doesn’t need a separate purge gas recovery unit as the small

synloop purge is recycled to the purifier. Lastly Ruthenium is used by the KAAP

synthesis catalyst as the active ingredient on a highly-stabilized graphitic base material.

This is 10 – 20 times more active than the traditional magnetite catalyst therefore

enabling a lower synthesis loop pressure.

This Process would advantage the producer in many ways, some of them are:

3

Less space is needed as it replaces the large traditional and expensive primary reformer

with a smaller , simpler and much more reliable equipment configuration

The use of low synthesis pressure allows the use of a single- barrel syngas compressor

and reduced pipe- wall thickness; this reduces the overall plant capital cost.

The maintenance cost for this design are lower than the competing process designs as it

uses the conventional magnetite ammonia synthesis catalyst

Lastly It provides more competitive energy consumption

The two main disadvantages of this process are that even though the maintenance cost for

this design is lower than the competing process designs, the equipment used in the

ammonia plant requires he most maintenance and secondly Ruthenium metal costs have

varied significantly over the past ten years and also the cost of the initial charge of KAAP

catalyst Is important to consider as it could economically disadvantage the producer.

1.3.3 Conventional Magnetite Ammonia Synthesis Process :

This process is conducted using 2 different technologies. Firstly the process is a

conventional steam – Methane reforming, this is a top fired steam Methane primary

reformer which is operated at high pressures combined with a secondary reforming using

the stoichiometric amount of process air which helps to efficiently convert natural gas to

Hydrogen and Carbondioxide. Then for the convention Ammonia synthesis loops, a

horizontal ammonia synthesis converter is used. This has a 2 or 3 reactor stave where

each consists of a vertical downward flow in the magnetite catalyst beds. These may be

arranged as two beds in parallel but this depends on the plant capacity.

4

Maximum conversions and heat recovery intercoolers are also provided between

reactors. The advantages of this process are that it optimizes energy consumption and it’s

well established. Also because of the efficiency of the horizontal ammonia synthesis,

consists of a basket that can be rolled out of the horizontal converter vessels to trucks, it

avoids the need for scheduling erecting a heavy and expensive crane for periodic

maintenance and there is a higher loop conversion. The disadvantages to the producer of

this process would be the need for a big space for keeping the machines that require a lot

of space ( e.g. the horizontal ammonia synthesis converter and the primary reformer) and

unlike the other 2 this process is fairly older thus the waste disposal methods and quantity

of the emissions is higher.

1.4 Ammonia chemistry :

Ammonia is produced by reacting hydrogen and nitrogen together :

N2 + 3H2 ─────> 2NH3 + heat (1.1)

In the correct proportion i.e. 1 volume of nitrogen to 3 volumes of hydrogen as follows.

5

1.5 Physical and chemical properties:

Tab 1

PROPERTY DESCRIPTION

Appearance Colorless gas at ambient temperatures

Odor Pungent, suffocating

pH ( water solution conc.1%) 11.7

Freezing point -77.7 ˚C

Boiling point -33.4˚Cat 101.3 Kpa

Flammability In the region of 16 – 27 % NH₃ by vol. in air at 0 ˚C

Auto – ignition temperature 651˚C

Vapor pressure 1013 Kpa at 25 ˚C

Relative vapor density 0.6 (air =1)

Solubility in water Extremely soluble , e.g. 529g/l at 20˚C

Solubility in organic solvent Soluble in alcohol ,acetone ,chloroform

Liquid density 0.6386 g/ cm3

Gas density 0.7714g/l (at 0˚C, 101.3 Kpa).

Critical temperature 270 ˚F

6

1.6 Uses of Ammonia :

Ammonia is widely used as fertilizer, the most common of which are:

Ammonium nitrate , aqueous ammonia solutions ,ammonium sulphate,

ammonium phosphate, urea, mixed fertilizers.

In large quantities, ammonia is used as the raw material for the production of

inorganic chemicals such as nitric acid and ammonium nitrate.

In metallurgy, ammonia is used as a processing agent for the recovery of metals

from their ores i.e. copper, nickel and molybdenum.

Ammonia is used as a refrigerant, particularly in large commercial installations.

1.7 Ammonia process theory :

Ammonia is produced as a result of a chemical reaction between hydrogen and nitrogen in

the mole ratio of 3:1 . The source of hydrogen is Natural gas and water while nitrogen is

obtained from atmospheric air.

The hydrogen/nitrogen mixture reacts at a given temperature and pressure in the presence

of an iron oxide catalyst according to the following equation:

N2 + 3H2 ─────>2NH3 + heat (2.1)

7

1.8 The ammonia process ( figure 1&2 ) :

The process used for Ammonia plant can be described by the following steps :

1.8.1 Desulphurization

1.8.2 Primary Reforming

1.8.3 Secondary Reforming

1.8.4 Shift conversion (CO Conversion) :

(a) High temp Shift conversion (b) Low temp Shift conversion

1.8.5 CO2 Removal

1.8.6 Methanation

1.8.7 Synthesis Gas Compression

1.8.8 Ammonia Synthesis Loop

8

Fig. 1 Ammonia process Block Diagram

Fig. 2 Ammonia process

9

1.8.1 Desulphurization (HDS) :

The HDS consists of three different catalyst beds. as shown in figure(3).

The first bed : is 1.58 metres high and is filled with hydrogenation catalyst

(nickel/molybdenium). The function of this bed is to convert the sulphur compounds to

hydrogen sulphide and the chlorine compounds to hydrogen chloride by reaction with

H2. The bed operates at 400C and 37 kg/cm2.

RCl + H2→HCl + RH (3.1)

RSH + H2→H2S + RH (4.1)

Where R is the hydrocarbon radical

The second bed : is 3.52 metres high and is filled with dechlorination catalyst

(potassium carbonate), which absorbs the hydrogen chloride gas formed in the first bed.

K2CO3 + 2HCl → 2KCl + H2O + CO2 (5.1)

The third bed : is 3.37 meters high and is filled with desulphurization catalyst (zinc

oxide), which absorbs the hydrogen sulphide gas formed in the first bed.

At the top and the bottom of each bed, there is a layer of large alumina balls which protect

the catalyst from high velocities. They also act as a humidity absorbent.

ZnO +H2S→ZnS +H2O (6.1)

Fig. 3 Desulphurization (HDS)

10

1.8.2 Primary Reforming :

The primary reforming consists of a large number of high–nickel chromium alloy tubes

filled with nickel-containing reforming catalyst as shown in figure (4) . The overall

reaction is highly endothermicand additional heat is required to raise the temperature to

780-830 C° at the reformer outlet.

The composition of the gas leaving the primary reformer is given by the following

chemical reactions

CH4+ H2O → CO + 3H2 (7.1)

C2H6 + 2H2O →2CO + 5H2 (8.1)

Fig. 4 Primary Reformer

11

1.8.3 Secondary Reforming :

Here, the un-reformed CH4 is converted to H2 , CO as in the Primary. However, the heat

for the reactions is supplied from a different source ie. the heat of combustion of H2 and

CH4 mixed with the Process air delivered by Process Air Compressor. And the operation

condition shown in the figure (5)

The quantity of air supplied to the Secondary Reformer is based on the exact amount of N2

required for the ammonia synthesis reaction ie. 3 parts hydrogen to 1 part nitrogen.

Reaction for secondary reformer :

C H4 + 12O2 → CO + 2H2 (9.1)

Fig. 5 Operation Condition of Secondary Reformer

12

1.8.4 Shift Conversion (CO Conversion) :

The process gas from the secondary reformer contains 12-15% CO (dry gas base) and most

of the CO is converted in the shift section according to the reaction :-

CO +H2O → CO2+ H2 (10.1)

In the High Temperature Shift (HTS) conversion, the gas is passed through a bed of iron

oxide/chromium oxide catalyst at around 400°C, where the CO content is reduced to about

3% (dry gas base), limited by the shift equilibrium at the actual operating temperature.

There is a tendency to use copper containing catalyst for increased conversion. The gas

from the HTS is cooled and passed through the Low Temperature Shift (LTS) converter as

shown in the figure (6) .This LTS converter is filled with a copper oxide/zinc oxide-based

catalyst and operates at about 200-240 C°. The residual CO content in the converted gas is

about 0.2-0.4 % (dry gas base). A low residual CO content is important for the efficiency of

the process.

Fig. 6 Shift Conversion

13

1.8.5 CO2 Removal :

The CO2 is removed in a chemical or a physical absorption process. The solvents used in

chemical absorption processes are mainly aqueous amine solutions (Mono Ethanol amine

(MEA), Activated Methyl DiEthanolamine (AMDEA) or hot potassium carbonate solutions

which illustrated in figure (7). Physical solvents are glycol dimethylethers (Selexol),

propylene carbonate and others.

K2CO3 + H2O + CO2 → 2KHCO3 (11.1)

Fig. 7 Benfield Process

14

1.8.6 Methanator :

The small amounts of CO and CO2, remaining in the synthesis gas, are poisonous for the

ammonia synthesis catalyst and must be removed by conversion to CH4 in the methanator:-

CO + 3H2 CH4 + H2O (12.1)

CO2 + 4H2 CH4 + 2H2O (13.1)

The reactions take place at around 300°C in a reactor filled with a nickel containing

catalyst.

Methane is an inert gas in the synthesis reaction, but the water must be removed before

entering the converter. This is done firstly by cooling and condensation downstream of the

methanator and finally by condensation/absorption in the product ammonia in the loop or in

a make-up gas drying unit.

1.8.7 Synthesis Gas Compression :

Modern ammonia plants use centrifugal compressors for synthesis gas compression, usually

driven by steam turbines, with the steam being produced in the ammonia plant. The

refrigeration compressor, needed for condensation of product ammonia, is also usually

driven by a steam turbine.

1.8.8 Ammonia Synthesis Loop :

Ammonia synthesis takes place in the Ammonia Converter , according to the following

equation:

N2 + 3H2 2NH3 + heat (14.1)

(Three volumes of hydrogen plus one volume of nitrogen)

This is a reversible reaction and the factors affecting its equilibrium are:

operating temperature

operating pressure

H2 : N2 ratio of the makeup gas

catalyst activity

onverter inlet percentage NH3

15

CHAPTER 2

MATERIAL BALANCE

2.1 Primary Reformer ( P.R ) :

Assume the following : (operating conditions)

Temperature = 732k

pressure = 35 atm

Steam / carbon ratio 3.8 / 1 mol%

H2O

W2

W1 W3 W3

Fig. 8 Primary Reformer

Reactions in primary reformer are as follows :

CH4 + H2O CO + 3H2 (1.2)

C2H6 + 2H2O 2CO + 5H2 (2.2)

C3H8 + 3H2O 3CO + 7H2 (3.2)

CO + H2 O CO2 + H2 (4.2)

Assume : basis 100 kmol of natural gas per hour

M = molecular weight

n/n % = v/v (volume) %

PRIMARY REFORMER

16

Tab.2

Components M W1 ( Natural Gas )

v/v% kmol/hr Kg/hr

CH4 16 77.87 77.87 1245.92

C2H6 30 8.10 8.10 243

C3H8 44 4.70 4.70 206.8

N2 28 1.58 1.58 44.24

CO2 44 7.65 7.65 336.6

Ar/He 36 0.10 0.1 3.6

Total 100 100 2080.16

- Calculation of carbon required ( kmol to kg ) :

Tab.3

Natural gas Kmol of N.G Kmol of carbon

CH4 ( i = 1 )

77.87 77.87

C2H6 ( i = 2 )

8.10 16.2

C3H8 ( i = 3 )

4.70 14.1

CO2 ( i = 1 )

7.65 7.65

N2 ( i = 0 )

1.58 -

Ar/He ( i = 0 )

0.10 -

Total 100 115.82

Where i = number of carbon

n : number of moles ( kmol/hr )

M : molecular weight ( kg / kmol )

17

Weight of carbon ( kg ) in ( N.G) = n * M

= 115.82 * 12 = 1389.84 kg/hr

- calculation of steam required ( kmol to kg )

Steam / carbon = 3.8 / 1

Steam = 115.82 * 3.8 = 440.166 kmol/hr

W2 = the quantity of steam inlet

= 440.166 * 18 = 7922.088 kg/hr

Molecular weight of (H2O) = 18 kg / kmol

- overall material balance at steady state :

Input = output

W1 + W2 = W3

2080.16 + 7922.088 = W3

W3 = 10002. 248 kg/hr

- calculation of dry gas :-

Assume y = Unreacted water kg

Amount of ( Ar / He ) and N2 inlet( PR) = outlet (PR)

Dry gas basis = W3 - [ amount of N2 (44.24) and Ar / He (3.6) ]– y

= 10002.248 – ( 44.24 + 3.6 ) – y

= ( 9954.41 – y) kg/hr 1 ــــــــــــــــــــــــ

18

- Assume dry gas composition as follows :-

Tab.4

Dry Gas M V / V% kg/hr W/W%

CH4 16 30 480 30.08

CO 28 3.6 100.8 6.32

CO2 44 21 924 57.91

H2 2 45.4 90.8 5.69

Total 100 1595.6 100

W/W% = kg/hr / total (for example : 480/1595.6 = 30.08% )

CARBON BALANCE :-

Tab.5

No Reactions Xi = X1 + X2 + X3 Carbon balance

Xi = X0

DG ( kg )

1

CO2 = C + O2 X1 = [ ( 9954.41 – y) ( 0.5791 ) (12) /

44 ] = 1572.16 – 0.1579 y

1389.84 = 4088.86 –

0.4106 y

3381.05

2 CH4 = C + 2H2 X2 = [ ( 9954.41 – y ) ( 0.301 ) (12) / 16

] = 2247.2 – 0.2257 y

3 CO = C +

2

1

O2

X3 = [ ( 9954.41 – y ) ( 0.06317 ) (12) /

28 ] = 269.5 – 0.027 y

Total Xi = 4088.86 - 0.4106 y y = 6573.35 3381.05

X0 : Amount of carbon outlet

Xi : Amount of carbon inlet

y : Amount of H2O ( unreacted )

DG : Amount of Dry gas resulting of substitute ( y ) Value in equation 1

19

Calculation weight of components in Dry gas out let :-

CH4 = 0.301 * 3381.05 = 1017.69 kg/hr

N2 = (as before) = 44.24 kg/hr

CO2 = 0.5791 * 3381.05 = 1957.96 kg/hr

CO = 0.06317 * 3381.05 = 213.58 kg/hr

H2 = 0.0569 * 3381.05 = 192.38 kg/hr

Ar / He = (as before) = 3.6 kg/hr

Tab.6

Components W3 ( outlet from P.R)

kg/hr W / W % kmol/hr V / V %

CH4 1017.69 10.2 63.6 11

N2 44.24 0.44 1.6 0.3

CO2 1957.96 19.6 44.5 7.8

CO 213.58 2.2 7.6 1.3

H2 192.38 1.9 96.2 16.4

H2O 6573.35 65.72 365.2 63.1

Ar/He 3.6 0.035 0.1 0.02

Total 10002.248 100 578.8 100

V / V % = kmol/hr / total (for example : 63.6/578.8 = 11% )

20

2.2 Secondary Reformer (SR) :-

- Temp 1083 –1252 k

- Pressure 32.5 – 35 atm

W4

O2 = 21% V / V N2 = 79% V / V

W3 W5

Fig. 9 Secondary Reformer

- Assume conversion 98% W/W of CH4

- The reaction takes place in ( S.R ) is as follows :-

CH4 + 2

1 O2 CO + 2H2 (5.2)

16 2

1 (32) = 16 28 2(2) = 4

(1017.69 * 0.98) O2 CO H2

- Amount of O2 consumed ( required )

O2 = )32(2/1

16*98.0*69.1017 = 997.34 kg/hr

SECONDARY

REFORMER

21

- Air required ( W4 ) = 997.34/0.23 = 4336.26 kg/hr

- Amount of N2 in the air inlet ( SR ) = 4336.26 * 0.77 = 3339 kg/hr

- Amount of N2 inlet W3 (as before) = 10002.248 * 0.0044 = 44 kg /hr

- Total N2 inlet in ( SR ) = outlet = 44 + 3339 = N2 outlet in ( W5 )

N2 outlet in W5 = 3383 kg/hr

- Amount of ( CO ) product from the reaction

CO = 16

28*98.0*69.1017 = 1745.34 kg/hr

- Amount of ( H2 ) product from the reaction

H2 = 16

4*98.0*69.1017 = 249.34 kg/hr

- Amount of ( H2 ) inlet = 10002.248 * 0.019 = 190 kg/hr

- Total amount of H2 outlet = 249.34+ 190 = 439.34 kg/hr

- amount of ( CO ) inlet = 10002.248 * 0.022 = 220 kg/hr

- Total amount of ( CO ) outlet = ( CO ) from the reaction + CO inlet

= 1745.34 + 220 = 1965.34 kg

- Amount of ( CH4 ) unreacted outlet = 0.02 * 1017.69 = 20.35 kg/hr

- Amount of CO2 inlet = CO2 outlet = 10002.248 * 0.196 = 1960 kg/hr

- Amount of Ar / He inlet = Ar / He outlet

= 10002.248 * 0.00035 = 3.6 kg/hr

- Amount of H2O outlet = 10002.248 * 0.657 = 6571 kg/hr

22

Tab. 7

Components W4 ( Air )


CH4 - - - -

N2 3339 77 119.25 79

CO2 - - - -

CO - - - -

H2 - - - -

O2 997.34 23 31.2 21

H2O - - - -

Ar/He - - - -

Total 4336.34 100 150.45 100

Tab.8

Components W5 (outlet from S.R)


CH4 20.35 0.14 1.27 0.15

N2 3383 23.58 121 14.72

CO2 1960 13.66 44.55 5.42

CO 1965.34 13.71 70.2 8.54

H2 439.34 3.1 219.67 26.73

H2O 6571 45.81 365.06 44.42

Ar/He 3.6 0.024 0.1 0.012

Total 14342.63 100 821.85 100

23

2.3 High temp shift conversion (H.T.S.C) :-

- Temp 653 – 693 k

- Pressure 31 atm

w5 w6

Fig. 10 High Temperature Shift Converter

- The reaction takes place in ( H.T.S.C )

CO + H2O CO2 + H2 (6.2)

- Ratio moles of steam / mole of CO initially present

= 365.06 / 70.2 = 5.2 moles of steam / moles of CO

- ( H .T .S .C ) at ( 693 k ) steam requirement in ( H .T .S .C )

- Assume the percentage conversion of CO2 = 79 % V / V

CO + H2O CO2 + H2 (6.2)

1 1 1 1

0.79 ( 70.2 ) H2O CO2 H2

- The mole conversion of CO = 55.46 k moles/hr

- The mole of CO un reacted = moles of CO inlet – mole of CO conversion

= 70.2 – 55.46 = 14.74 k moles/hr

- The moles of H2O required = 55.46 * 1 = 55.46 kmol /hr

- The moles of H2O outlet = 365.06 – 55.46 = 309.6 kmol/hr

- The moles of CO2 inlet ( H . T . S .C ) = 44.55 kmol /hr

- The total moles of CO2 outlet = 55.46 + 44.55 = 100 kmol/hr

- The moles of H2 product = 55.46 * 1 = 55.46 kmol/hr

H.T.S.C

24

- The moles of H2 inlet ( H.T.S ) = 219.67 kmol/hr

- The total moles of H2 = 55.46 + 219.67 = 275.13 kmol/hr

- The moles of CH4 inlet = moles outlet = 1.27 kmol/hr

- The moles of N2 inlet = moles outlet = 121 kmol/hr

- The moles of Ar / He inlet = moles outlet = 0.1 kmol /hr

Tab.9

Components W6 (outlet from H.T.S.C)


CH4 20.35 0.14 1.27 0.15

N2 3383 23.58 121 14.73

CO2 4400 30.67 100 12.16

CO 412.72 2.87 14.74 1.8

H2 550.26 3.83 275.13 33.47

H2O 5572.8 38.85 309.6 37.67

Ar/He 3.6 0.025 0.1 0.012

Total 14342.73 100 821.84 100

2.4. Low temp shift conversion ( L.T.S.C)

- Temp 473 k

- Pressure 31 atm

W6 W7

Fig.11 Low Temp Shift Converter

- The reaction takes place in ( L.T.S.C )

CO + H2O CO2 + H2 (7.2)

- Ratio moles of steam / mole of CO initially present ( L . T . S . C )

L.T.S.C

25

from steam requirement in shift conversio

- Assume the percentage conversion of CO = 75 % V / V

CO + H2O CO2 + H2 (7.2)

1 1 1 1

0.75 ( 14.74 ) H2O CO2 H2

- The mole conversion of CO = 11.055 kmoles/hr

- The mole of CO un reacted = moles of CO inlet – mole of CO conversion

= 14.74 – 11.055 = 3.685 k mol/hr

- The moles of H2O required = 11.055 * 1 = 11.055 kmol/hr

- The moles of H2O outlet = moles inlet – moles required

= 309.6 – 11.055 = 298.5 k mol/hr

- The moles of CO2 product = 11.055 * 1 = 11.055 k mol/hr

- The moles of CO2 inlet = 100 k mol/hr

- The total moles of CO2 outlet = 100 + 11.055 = 111.055 k mol/hr

- The moles of H2 product = 11.055 * 1 = 11.055 k mol/hr

- The moles of H2 inlet = 275.55 k mol/hr

- The total moles of H2 outlet = 275.55 + 11.055 = 286.2 k mol/hr

- The moles of CH4 inlet = moles outlet = 1.27 k mol/hr

- The moles of N2 inlet = moles outlet = 121 k mol/hr

- The moles of Ar / He inlet = moles outlet = 0.1 k mol /hr

26

Tab. 10

Components W7 (outlet from L.T.S.C)


CH4 20.35 0.14 1.27 0.15

N2 3383 23.6 121 14.72

CO2 4886.42 34.07 111.055 13.51

CO 103.18 0.72 3.685 0.45

H2 572.4 3.99 286.2 34.82

H2O 5373 37.46 298.5 36.32

Ar/He 3.6 0.025 0.1 0.01

Total 14342 100 821.81 100

2.5. Water Separator (exclusion of water)

- Temp 495 k

- Pressure 32.5 atm

W7 W9

W8 ( H2O ) condensation

Fig. 12 Water Separator

- Assume all the water inlet is separated as outlet as condensation

Water inlet = outlet as condensation

W8 = 100

46.37*14342 = 5373 kg/hr

= 5373 / 18 = 298.5 kmol/hr

SEPARATOR

H O

27

- overall material balance at steady state

W7 = W8 + W9

14342 = 5373 + W9

W9 = 14342 - 5373 = 8969 kg/hr ( weight dry basis )

Tab.11

Components W9 (outlet from Separator )


CH4 20.35 0.22 1.27 0.24

N2 3383 37.71 121 23.12

CO2 4886.42 54.48 111.055 21.22

CO 103.18 1.15 3.685 0.70

H2 572.4 6.4 286.2 54.7

Ar/He 3.6 0.04 0.1 0.02

Total 8969 100 523.31 100

2.6. CO2 Removal (Absorber) :

- Temp 480 k

- Pressure 26 atm

K2CO3 + H2O

W10

W9 W12

W11

Fig. 13 CO2 REMOVAL

CO2

REMOVAL

28

- Suppose 99 % of the CO2 is absorbed by K2CO3 solution the reaction takes place

in the absorber as following:

K2CO3 + H2O + CO2 2KHCOз (8.2)

138 18 44 200

K2CO3 H2O ( 0.99*4886.42 ) 2KHCOз

K2CO3 = Amount of K2CO3 required for reaction =

0.99*4886.42*(138/44) = 15172.3 kg/hr

H2O = Amount of H2O required for reaction =

0.99*4886.42*(18/44) = 1979 kg/hr

KHCOз = Amount of 2KHCO3 required for reaction =

0.99*4886.42*(200/44) = 21989 kg/hr

W10 = K2CO3 + H2O = 15172.3 + 1979 = 17151.3 kg/hr

W11 = KHCOз = 21989 kg/hr

= 21989 / 200 = 110 kmol/hr

- overall material balance at steady state :-

W9 + W10 = W11 + W12

8969 + 17151.3 = 21989 + W12

W12 = 4131.3 kg / hr

29

- Amount of CO2 outlet = 4886.42 * 0.01 = 48.86 kg /hr

- Amount of CO outlet = 103.18 kg/hr

- Amount of CH4 outlet = 20.35 kg/hr

- Amount of H2 outlet = 572.4 kg/hr

- Amount of Ar/He outlet = 3.6 kg/hr

- Amount of N2 outlet = 3383 kg/hr

Tab.12

Components W10 (KHCO3+ H2O inlet CO2 removal )


CH4 - - - -

N2 - - - -

CO2 - - - -

CO - - - -

H2 - - - -

Ar/He - - - -

H2O 1979 11.54 110 50

K2CO3 15172.3 88.46 110 50

Total 17151.3 100 220 100

Tab.13

Components W12 (outlet CO2 Removal )


CH4 20.35 0.49 1.27 0.30

N2 3383 81.88 121 29.27

CO2 48.86 1.18 1.11 0.27

CO 103.18 3.5 3.7 0.9

H2 572.4 13.85 286.2 69.23

Ar/He 3.6 0.087 0.1 0.023

Total 4131.39 100 413.38 100

30

2.7 Methanation :

- Temp ( 1100) k

- Pressure ( 40) atm

W12 W13

Fig. 14 Methanation

- The reaction takes place in the Methanation

CO2 + 4H2 CH4 + 2H2O (9.2)

CO + 3H2 CH4 + H2O (10.2)

- The percentage conversion for CO2 and CO 100 % to the CH4

CO2 + 4H2 CH4 + 2H2O (9.2)

44 4(2) = 8 16 2(18) = 36

48.86 H2 CH4 H2O

METHANATION

31

- Amount of H2 required for reaction (9.2) = 8.88 kg/hr

- Amount of CH4 produced for reaction (9.2) = 17.76 kg/hr

- Amount of H2O produced for reaction (9.2) = 40 kg/hr

CO + 3H2 CH4 + H2O (10.2)

28 6 16 18

103.18 H2 CH4 H2O

- Amount of H2 required for reaction (10.2) = 22. 11 kg/hr

- Amount of CH4 produced for reaction (10.2) = 59 kg/hr

- Amount of H2O produced for reaction (10.2) = 66.33 kg/hr

- Total amount of ( CH4 ) outlet = ( CH4 ) inlet + ( CH4 ) produced by reaction

= 20.35 + 17.76 + 59 = 97.11 kg /hr

- Total amount of ( H2O ) outlet ( produced by reaction ) = 40 + 66.33 =106.33 kg

/hr

- The total amount of ( H2 ) outlet = ( H2 ) inlet

- H2 required for reaction = 572.4 – ( 8.88 + 22.11 ) = 541.41 kg /hr

- The amount of ( N2 ) inlet = ( N2 ) outlet = 33883 kg/hr

- The amount of ( Ar / He ) inlet = ( Ar / He ) outlet = 3.6 kg /hr

32

Tab.14

Components W13 ( outlet from Methanator )


CO2 - - - -

CO - - - -

CH4 97.11 2.35 6.07 1.51

H2 541.41 13.105 270.70 67.04

N2 3383 81.88 121 29.96

Ar/He 3.6 0.087 0.1 0.024

H2O 106.33 2.57 5.90 1.46

Total 4131.35 100 403.77 100

2.8 Cooling , Compression & Separator :

- Temp 340 k

- Pressure 75 atm

W13 W15

W14 CH4

H2O

Fig.15 Cooling Compression & Separator

- Assume the (CH4+H2O) inlet cooling and compression are removed by condensate

- The amount of ( CH4 ) outlet = 97.11 kg/hr

- The amount of ( H2O ) outlet = 106.33 kg /hr

W14 = 97.11 + 106.33 = 203.44 kg/hr

COOLING

COMPRESSION

33

Tab.15

Components W14 (CH4+ H2O outlet from Compressor and Cooler and separator )


CH4 97.11 47.73 6.07 50.71

N2 - - - -

H2 - - - -

Ar/He - - - -

H2O 106.33 52.26 5.90 49.29

Total 203.44 100 11.97 100

Tab.16

Components W15 ( outlet from Compressor and Cooler)


CH4 - - - -

N2 3383 86.12 121 30.88

H2 541.41 13.78 270.70 69.1

Ar/He 3.6 0.091 0.1 0.02

H2O - - - -

Total 3928.01 100 391.8 100

2.9 Ammonia Converter :

- Temp 320 k

- Pressure 39 atm

W15 W16

W17 NH3 Liquid

W19 Recycle (N2 + H2 ) W18 Purged

Fig.16 Ammonia Converter

CONVERTER

25%

COOLER

&

SEPARATOR

34

- The reaction take place in ammonia converter :-

N2 + 3H2 2NH3 (11.2)

- k moles ( N2 + H2 ) entering the converter = ( 391.8 + W19 )

- k moles ( N2 + H2 ) leaving the converter = 0.75 ( 391.8 + W19 )

- k moles of Ar / He in fresh feed = 0.1

- fraction of inerts introduced to the converter should not exceed 4 k moles from the

total feed introduced ( 4 / 391.8 ) = 0.01

- kmoles of Ar / He total feed = 8.391

4 = 0.01

- kmoles of ( Ar / He )& ( N2 + H2 ) mixture leaving ( cooler and separator )

= 75.0

01.0 = 0.013

- kmoles of ( Ar / He ) purged = 0.013 W18

0.01 = 0.013 W18

W18 = 013.0

01.0 = 7.46 kmol of gases ( N2 + H2 ) &( Ar / He )

Material balance at steady state for cooler & separator to ( N2 + H2 ) + ( Ar / He )

W16 = W17 + W18 + W19

0.75 ( 391.8 + W19 ) = 0 + 7.46 + W19

293.85 + 0.75 W19 = 7.46 + W19

286.39 = 0.25 W19

W19 = 1145.5 k mol

35

W16 = 0.75 ( 391.8 + W19)

W16 = 0.75 ( 391.8 + 1145.5 ) = 1153 k mol /hr

W18 content amount from (N2 + H2) + (Ar / He)

- Amount from (N2+H2) ≈ 0

- Overall material balance at steady state for ammonia converter :-

W15 = W17 + W18

3928.01 = W17 + 268.6 :- W17 = 3659.41 kg / hr

= 215.26 k mol / hr

W19 content amount from ( N2 + H2 ) + ( Ar / He )

- Amount from ( Ar / He ) ≈ 0

( N2 : H2 ) = 1 : 3

( N2 + H2 ) = 1145.5 k mol

N2 = 3

5.1145 = 381.8 k mol/hr

H2 = 1145.5 - 381.8 = 763.7 k mol/hr

Tab.17

Components M W17 ( Ammonia)

Kg/hr K mol/hr V / V %

NH3 17 3659.41 215.26 100

N2 - - - -

H2 - - - -

Ar / He - - - -

Total 3659.41 215.26 100

36

Tab.18

Components W18 ( Purged)

Kg/hr w/w % K mol/hr V / V %

N2 - - - -

H2 - - - -

Ar / He 268.6 100 7.461 100

Total 268.6 100 7.461 100

Tab.19

Components W19 ( Recycle )

Kg/hr w/w % K mol/hr V / V %

N2 10690.4 87.5 381.8 33.34

H2 1527.4 12.50 763.7 66.66

Ar / He - - - -

Total 12217.8 100 1145.5 100

Material balance at steady state for system (Input = output)

Tab.20

Input Output

W Kg / hr W Kg / hr

W1 2080.16 W8 5373

W2 7922.088 W11 21989

W4 4336.34 W14 203.44

W10 17151.3 W17 3659.1

- - W18 268.6

Total 31490 Total 31493

- from previous calculation :

100 kmoles / hour of natural gas produces 215.26 kmoles / hours

= 87.82 ton / day

- from the statement of problem the plant of ammonia must produce

1000 ton / day of ammonia

215.26 kmol 17kg 24hr 1ton

Hr Kmol Day 1000kg

37

- scale up factor = 82.87

1000 = 11.38

- So that all streams flow must be multiplied by this scale up factor and are

shown in the table 22 :-

- new value = old value * scale up factor kg / hr

Tab.21

Streams Old value kg / hr New value kg / hr

W1 2080.16 23672.2

W2 7922.088 90153.36

W3 10002.248 113825.6

W4 4336.34 49347.5

W5 14342.63 163219.1

W6 14342.73 163220.26

W7 14342 163211.96

W8 5373 61144.74

W9 8969 102067.22

W10 17151.3 195181.8

W11 21989 250234.82

W12 4131.39 47015.2

W13 4131.35 47014.76

W14 203.44 2315.14

W15 3928.01 44700.75

W16 1153 13121.14

W17 3659.41 41644.08

W18 268.6 3056.66

W19 12217.8 139038.56

% Error = 31493 - 31490 / 31490 * 100 = 0.009

38

ENERGY BALANCE

2.10 Primary Reformer :-

∆ H2 733 K

W2

W1 W3

∆ H1 732K ∆ H3 1083 K

Fig.17 Primary Reformer

The reaction in primary reformer are

CH4 + H2O CO + H2 ∆ HR° = - 49.3 kcal (12.2)

CO + H2O CO2 + H2 ∆ HR° = + 9.8 kcal (13.2)

∆ HR° = The Enthalpy consumption

Basis temperature is 25C0 = 298 K

∆ H = Σ n T

298

CP . dt

∆ H =The change of Enthalpy

n = Number of moles for gases

dt = The change in the temperature

Cp = specific heat of the components heat capacity for gases

Cp = A + BT + CT2 + DT

3

Primary

Reformer

39

Where ( A, B , C , D ) constant values for heat capacity for gases :

Tab.22

Component A B C D

CH4 ( g ) 19.251 52.126 * 10-3

11. 974 * 10-6

- 1.131 * 10-8

C2 H6 ( g ) 5.409 17. 811 * 10-2

- 6.937 * 10-5

87.127 * 10-10

C3 H8 ( g ) - 4.224 30.626 * 10-2

- 1.586 * 10-4

32.146 * 10-9

CO2 ( g ) 19.795 73.436 * 10-3

- 5.601 * 10-5

17.153 * 10-9

O2 ( g ) 28.106 -3.680 * 10-6

17.459 * 10-6

- 1.065 * 10-8

CO ( g ) 30.869 -1.285 * 10-2

27.89 * 10-6

- 1.271 * 10-8

N2 ( g ) 31.15 -1.357 * 10-2

26.796 * 10-6

- 1.168 * 10-8

H2O (g ) 32.243 19.238 * 10-4

10.555 * 10-6

3.596 * 10-9

H2 ( g ) 27.143 92.738 * 10-8

- 1.38 * 10-5

76.451* 10-10

Ar/He( g ) 20.804 -3.211* 10-5

51.66* 10-9

-

40

- calculation the Enthalpy change of the reactant (∆ H1) from 298k to 732k

Tab.23

Components ∆ H1 = Σ n 732

298 c p . d t

∆ H CH4 = 77.87

732

298( 19.251 + 52.126 * 10

-3T + 11.97 * 10

-6 T

2 – 1.131 * 10

-8 T

3 ) dt

= 77.87 [ ( 19.251 ( 732 – 298 ) + 52.126 * 10-3

( 7322 – 298

2 ) / 2 + 11.97 * 10

-6

( 7323 – 298

3 ) /3 – 1.131 * 10

-8 ( 732

4 – 298

4 ) /4 ] = 16.09*10

5KJ

∆ H C2H6 = 8.10

732

298( 5.409 + 17.811 * 10

-2T – 6.938 * 10

-5 T

2 + 87.127 * 10

-10 T

3 ) dt

= 8.10 [ ( 5.409 ( 732 – 298 ) + 17.811 * 10-2

(7322 – 298

2) /2 – 6.938 * 10

-5

(7323 – 298

3) /3 + 87.127 * 10

-10 (732

4 – 298

4) /4 ] = 322.2*10

3KJ

∆H C3H8 = 4.70

732

298( - 4.224 + 30.626 * 10

-2 T - 1.586 * 10

-4 T

2 + 32.146 * 10

-9 T

3 ) dt

= 4.70 [ ( - 4.224 ( 732 – 298 ) + 30.626 * 10-2

(7322 – 298

2) /2 -1.586 * 10

-4

(7323 – 298

3) /3 + 32.146 * 10

-9 (732

4 – 298

4) /4 ] = 232.7*10

3KJ

∆H N2 = 1.58

732

298( 31.150 - 1.357 * 10

-2 T + 26.796 * 10

-6 T

2 – 1.168 * 10

-8 T

3 ) dt

= 1.58 [ ( 31.150 ( 732 – 298 ) - 1.357 * 10-2

(7322 – 298

2) /2+ 26.796 * 10

-6

(7323 – 298

3) /3 – 1.168 * 10

-8 (732

4 – 298

4) /4 ] = 3.0*10

4 KJ

∆ H CO2 = 7.65

732

298( 19.795 + 73.436 * 10

-3 T – 5.601 * 10

-5 T

2 + 17.153 * 10

-9 T

3 ) dt

= 7.65 [ ( 19.795 (732 – 298) + 7.343 * 10-2

(7322 – 298

2) /2 – 5.601 * 10

-5

(7323 – 298

3) /3 + 1.715 * 10

-8 (732

4 – 298

4) /4 ] = 1.48*10

5 KJ

∆ H Ar/He = 0.1

732

298( 20.804 - 3.211 * 10

-5 T + 51.665 * 10

-9 T

2 ) dt

= 0.1 [ ( 20.804 (732–298) - 3.211*10-5

(7322 – 298

2) / 2 + 51.665 * 10

-9

(7323 – 298

3) /3 ] = 1.2*10

3 KJ

Σ∆ H1 2.32*106 KJ


41

Tab.24

Components ∆ H2 = Σ n

733

298 c p . d t

∆ H H2O = 440.166 ( 32.243 + 19.238 * 10-4

T + 10.555 * 10-6

T2 + 3.596 * 10

-9 T

3 ) dt

= 440.166 [ ( 32.243 (733 – 298 )+ 19.238 *10-4

( 7332 – 298

2 ) /2 + 10.555* 10

-6

( 7333 – 298

3 ) + 3.596 * 10

-9 ( 733

4 – 298

4 ) /4 ] = 7.04*10

6 KJ

Σ∆ H2 7.04*106 KJ

- calculation the Enthalpy change of the product (∆ H3) from 298k to 1083k

Tab.25


1083

298 c p . d t

∆ H CH4 = 63.6

1083

298 ( 19.251 + 52.126 * 10

-3 + 11.974 * 10

-6 T

2 – 1.131 * 10

-8 T

3 ) dt

= 63.6 [ ( 19.251 ( 1083 – 298 ) + 52.126 * 10-3

( 10832 – 298

2 )/2 + 11.974 *10

-6

( 10833 – 298

3 ) /3 – 1.131 * 10

-8 (1083

4 – 298

4) /4 ] = 3.46*10

6 KJ

∆ H N2 = 1.6

1083

298( 31.150 - 1.357 * 10

-2 T + 26.796 * 10

-6T

2 - 1.168 * 10

-8 T

3 ) dt

= 1.6 [ ( 31.150 ( 1083 – 298 ) -1.357 * 10-2

( 10832 – 298

2 ) / 2 + 26.796 * 10

-6

( 10833 – 298

3 ) / 3 - 1.168 * 10

-8 ( 1083

4 – 298

4 ) / 4 ] = 6.14*10

5KJ

∆ H CO2 = 44.5

1083

298( 19.795 + 73.436 * 10

-3 T – 5.601 * 10

-5 T

2 + 17.153 * 10

-9 T

3 ) dt

= 44.5 [ ( 19.795 (1083– 298) + 73.436 * 10-3

(10832 – 298

2) /2 – 5.601 * 10

-5

(10833 – 298

3) /3 + 17.153 * 10

-9 (1083

4 – 298

4) /4 ] = 2.90 *10

5KJ

∆ H CO = 7.6

1083

298( 30.869 -1.285 * 10

-2 T +27.89 * 10

-6 T

2 – 1.272 * 10

-8 T

3 ) dt

= 7.6 [ (30.869 (1083– 298) -1.285 * 10-2

(10832 – 298

2) /2 + 27.89 * 10

-6

(10833 – 298

3) /3 – 1.272 * 10

-8 (1083

4 – 298

4) /4 ] =2.99 *10

3KJ

∆ H H2 = 96.2

1083

298( 27.143 + 92.738 * 10

-4 T– 1.38 * 10

-5 T

2+ 76.45* 10

-10T

3 ) dt

= 96.2 [ (27.143 ( 1083 – 298 ) + 92.738 * 10-4

( 10832 – 298

2 ) / 2 – 1.38 * 10

-5

( 10833 – 298

3 ) / 3 + 76.45* 10

-10 ( 1083

4 – 298

4 ) / 4 =2.20*10

4KJ

∆ H H2O = 365.2

1083

298( 32.243 + 19.238 * 10

-4 T + 10.555 * 10

-6 T

2 + 3.596 * 10

-9 T

3) dt

= 365.2 [ ( 32.243 ( 1083 – 298 )+ 19.238 *10-4

( 10832 – 298

2 )/2 + 10.555 *10

-6

( 10833 – 298

3 ) / 3 + 3.596 * 10

-9 ( 1083

4 – 298

4 ) / 4 ] = 1.07*10

5KJ

∆ H Ar/He = 0.1

1083

298( 20.804 - 3.211* 10

-5T + 51.665 * 10

-9 T

2 ) dt

= 0.1 [ ( 20.804 (1083 – 298) -3.211 * 10-5

(10832 – 298

2) /2 + 51.665 * 10

-9

(10833 – 298

3) /3 ] =21.37 KJ

Σ∆ H3 4.49*106 KJ

42

2.11. Secondary Reformer :-

W4

∆ H4 733K

W5

∆ H3 ∆ H5 1253K

Fig.18 Secondary Reformer

The reaction take place in the secondary reformer is

CH4 + 2

1 O2 CO + 2H2 ∆ HR° = 8.5 kcal (14.2)


Tab.26


733

298 c p . d t

∆ H N2 = 119.25

733

298( 31.150 – 1.357 * 10

-2 T+ 26.796 * 10

-6 T

2 – 1.168* 10

-8 T

3 ) dt

= 119.25[ (31.150 (733– 298) -1.357 * 10-2

(7332 – 298

2) /2 +26.796 * 10

-6

(7333 – 298

3) /3 – 1.168* 10

-8 (733

4 – 298

4) /4 ] = 1.546*10

6 KJ

∆ H O2 = 31.2

733

298( 28.106 - 3.680 * 10

-6 T + 17.459 * 10

-6 T

2 - 1.065 * 10

-9 T

3 ) dt

= 31.2 [ (28.106 (733 – 298) -3.680 * 10-6

(7332 – 298

2) /2 +17.459 * 10

-6

(7333 – 298

3) /3 – 1.065* 10

-8 (733

4 – 298

4) /4 ] =14.288*10

3KJ

Σ∆ H4 1.56*106 KJ

Secondary

Reformer

Ni catalyst

43


Tab.27


298 c p . d t

∆ H CH4 = 1.27

1253

298( 19.251 + 52.126 * 10

-3 + 11.974 * 10

-6 T

2 – 1.132 * 10

-8 T

3 ) dt

= 1.27 [ ( 19.251 ( 1253 – 298 ) + 52.126 * 10-3

( 12532 – 298

2 ) /2 + 11.974 * 10

-6

(12533 – 298

3 ) /3 – 1.132 * 10

-8 (1253

4 – 298

4) /4 ] =7.22 *10

4 KJ

∆ H N2 = 121

1253

298( 31.150 - 1.357 * 10

-2 T + 26.796 * 10

-6 T

2 - 1.168 * 10

-8 T

3 ) dt

121 [ ( 31.150 (1253 – 298 ) -1.357 * 10-2

(12532 – 298

2 ) /2 + 26.796 * 10

-6

(12533 – 298

3 ) / 3 - 1.168 * 10

-8 (1253

4 – 298

4 ) / 4 ] =6.05*10

6 KJ

∆ H CO2 = 44.55

1253

298( 19.795 + 73.436 * 10

-3 T – 5.602 * 10

-5 T

2 + 17.153 * 10

-9 T

3 ) dt

= 44.55 [ ( 19.795 (1253– 298) + 73.436 * 10-3

(12532 – 298

2) /2 – 5.602 * 10

-5

(12533 – 298

3) /3 + 17.153 * 10

-9 (1253

4 – 298

4) /4 ] = 24.699*10

7KJ

∆ H CO = 70.2

1253

298( 30.869 - 1.285 * 10

-2 T +27.89 * 10

-6 T

2 – 1.272 * 10

-8 T

3 ) dt

= 70.2 [ (30.869 (1253– 298)- 1.285 * 10-2

(12532 – 298

2) /2 +27.89 * 10

-6

(12533 – 298

3) /3 – 1.272 * 10

-8 (1253

4 – 298

4) /4 ] =54.24 *10

6KJ

∆ H H2 = 219.67

1253

298( 27.143 + 92.738 * 10

-4 T– 1.38 * 10

-5 T

2+ 76.45* 10

-10 T

3 ) dt

= 219.67 [ (27.143 (1253 – 298 ) + 92.738 * 10-4

(12532 – 298

2 ) / 2 – 1.38 * 10

-5

(12533 – 298

3 ) / 3 + 76.45* 10

-10 (1253

4 – 298

4 ) / 4 ] = 7.28*10

6 KJ

∆ H H2O = 365.06

1253

298( 32.243 + 19.238 * 10

-4 T + 10.555 * 10

-6 T

2 – 3.596 * 10

-9 T

3) dt

= 365.06[ ( 32.243 (1253 – 298 )+ 19.238 * 10-4

(12532 – 298

2 ) / 2 +10.555 *10

-6

(12533 – 298

3 ) / 3 – 3.596 * 10

-9 (1253

4 – 298

4 ) / 4 ] =18.12 *10

6 KJ

∆ H Ar/He = 0.1

1253

298( 20.804 - 3.211 * 10

-5 T + 51.665 * 10

-9 T

2 ) dt

= 0.1 [ ( 20.804 (1253 – 298) - 3.211 * 10-5

(12532 – 298

2) /2 + 51.665 * 10

-9

(12533 – 298

3) /3 ] = 2.59 *10

3KJ

Σ∆ H5 3.32 *108KJ

44

2.12 Product Gas Cooler :-

W5 W5

∆ H5 ∆ H6 653 k

Fig.19 Gas Cooler

- Calculation the enthalpy change of the product (∆ H6) from 298k to 653k

Tab.28


653

298 c p . d t

∆ H CH4 =1.27

653

298( 19.251 + 52.126 * 10

-3 + 11.974 * 10

-6 T

2 – 1.132 * 10

-8 T

3 ) dt

= 1.27[( 19.251( 653 – 298 )+52.126*10-3

( 6532 – 298

2 ) / 2 + 11.974*10

-6 (653

3 – 298

3 ) /3 –

1.132 * 10-8

(6534 – 298

4) /4 ] =2.26*10

5 KJ

∆ H N2 = 121

653

298( 31.150 - 1.357 * 10

-2 T + 26.796 * 10

-6 T

2 - 1.168 * 10

-8 T

3 ) dt

= 121 [ ( 31.150 (653 – 298 )- 1.357* 10-2

(6532 – 298

2 ) /2 + 26.796 * 10

-6 (653

3 – 298

3 ) / 3 -

1.168 * 10-8

(6534 – 298

4 ) / 4 ] =20.43 *10

6 KJ

∆ H CO2 = 44.55

653

298( 19.795 + 73.436 * 10

-3 T – 5.602 * 10

-5 T

2 + 17.153 * 10

-9 T

3 ) dt

= 44.55 [ ( 19.795 (653– 298) + 73.436 * 10-3

(6532 – 298

2) /2 – 5.602 * 10

-5(653

3 – 298

3) /3 +

17.153 * 10-9

(6534 – 298

4) /4 ] = 7.70*10

6KJ

∆ H CO = 70.2

653

298( 30.869 - 1.285 * 10

-2 T +27.89 * 10

-6 T

2 – 1.272 * 10

-8 T

3 ) dt

= 70.2 [ (30.869 (653– 298) - 1.285 * 10-2

(6532 – 298

2) / 2 + 27.89 *10

-6 (653

3 – 298

3) /3 –

1.272 * 10-8

(6534 – 298

4) /4 ] =11.73*10

6KJ

∆ H H2 = 219.67

653

298( 27.143 + 92.738 * 10

-4 T– 1.38 * 10

-5 T

2+ 76.45* 10

-10 T

3 ) dt

= 219.67 [(27.143 (653 – 298 ) + 92.738 * 10-4

(6532 – 298

2) /2 – 1.38* 10

-5(653

3 – 298

3 ) / 3 +

76.45* 10-10

(6534 – 298

4 ) / 4 ] = 25.52*10

6KJ

∆ H H2O = 365.06

653

298( 32.243 + 19.238 * 10

-4 T + 10.555 * 10

-6 T

2 – 3.596 * 10

-9 T

3 ) dt

= 365.06[( 32.243(653 – 298) +19.238*10-4

(6532 – 298

2) / 2 +10.555*10

-6 (653

3 – 298

3 ) / 3 –

3.596 * 10-9

(6534 – 298

4 ) / 4 ] = 51.10*10

6 KJ

∆ H Ar /He = 0.1

653

298( 20.804 -3.211* 10

-5 T + 51.66 5* 10

-9 T

2 ) dt

= 0.1 [( 20.804(653 – 298) + 3.211*10-5

(6532 – 298

2) /2 + 51.665 * 10

-9 (653

3 – 298

3) /3 ] =

2*103KJ

Σ∆ H6 1.167 *108KJ

Gas cooler

45

2.13. High Temp Shift Converter ( H . T . S . C ) :-

∆ H6 ∆ H7 693K

W6 W6

Fig.20 High Temp Shift Converter

The reaction take place at high temp shift converter is :

CO + H2O CO2 + H2 ∆ HR° = 9.8 kcal (15.2)


Tab.29


693

298 c p . d t

∆ H CH4 =1.27

693

298( 19.251 + 52.126 * 10

-3 + 11.974 * 10

-6 T

2 – 1.132 * 10

-8 T

3 ) dt

= 1.27 [ ( 19.251 ( 693 – 298 ) + 52. 126* 10-3( 693

2 – 298

2 ) /2 + 11.974 * 10

-4

(6933 – 298

3 ) /3 – 1.132 * 10

-8 (693

4 – 298

4) /4 ] = 1.15*10

4KJ

∆ H N2 = 121

693

298( 31.150 - 1.357 * 10

-2 T + 26.796 * 10

-6 T

2 - 1.168 * 10

-8 T

3 ) dt

= 121 [ ( 31.150 (693 – 298 ) -1.357 * 10-2

(6932 – 298

2 ) /2 + 26.796 * 10

-6

(6933 – 298

3 ) / 3 -1.168 * 10

-8 (693

4 – 298

4 ) / 4 ] =6.50 *10

7KJ

∆ H CO2 = 100

693

298( 19.795 + 73.436 * 10

-6 T – 5.602 * 10

-5 T

2 + 17.153 * 10

-9 T

3 ) dt

=100 [ ( 19.795 (693– 298) + 73.436 * 10-6

(6932 – 298

2) /2 – 5.602 * 10

-5

(6933 – 298

3) /3 + 17.153 * 10

-9 (693

4 – 298

4) /4 ] = 4.99*10

6KJ

∆ H CO = 14.74

693

298( 30.869 _ 1.285 * 10

-2 T +27.892 * 10

-6 T

2 – 1.272 * 10

-8 T

3 ) dt

= 14.74 [ (30.869 (693– 298) -1.285 * 10-2

(6932 – 298

2) /2 +27.892 * 10

-6

(6933 – 298

3) /3 – 1.272 * 10

-8 (693

4 – 298

4) /4 ] =6.23 *10

4KJ

∆ H H2 = 275.13

693

298( 27.143 + 92.738 * 10

-4 T– 1.38 * 10

-5 T

2+ 76.45* 10

-10 T

3 ) dt

=275.13 [ (27.143 (693 – 298 ) + 92.738 * 10-4

(6932 – 298

2 ) / 2 – 1.38 * 10

-5

(6933 – 298

3 ) / 3 + 76.45* 10

-10 (693

4 – 298

4 ) / 4 ] = 35.11*10

6KJ

∆ H H2O = 309.6

693

298( 32.243 + 19.238 * 10

-4 T + 10.555 * 10

-6 T

2 – 3.596 * 10

-9 T

3) dt

= 309.6 [ ( 32.243 (693 – 298 ) + 19.238 * 10-4

(6932 – 298

2 ) / 2 + 10.555 * 10

-6

H . T . S . C

46

(6933 – 298

3 ) / 3 – 3.596 * 10

-9 (693

4 – 298

4 ) / 4 ] = 4.17*10

6KJ

∆ H Ar/He = 0.1

693

298( 20.804 - 3.211 * 10

-5 T + 51.665 * 10

-9 T

2 ) dt

= 0.1 [ ( 20.804 (693 – 298) - 3.211 * 10-5

(6932 – 298

2) /2 + 51.665 * 10

-9

(6933 – 298

3) /3 ] = 9.73*10

2KJ

Σ∆ H7 1.093*108 KJ

2.14. First Gas/Gas Heat Exchanger :-

∆ H7 ∆ H8 473 k

W6 W6

Fig.21 First Heat Exchanger


Tab.30


473

298 c p . d t

∆ H CH4 = 1.27

473

298( 19.251 + 52.126 * 10

-3 + 11.974 * 10

-6 T

2 – 1.132 * 10

-8 T

3 ) dt

= 1.27 [ ( 19.251 ( 473 – 298 ) + 52.126 * 10-3

( 4732 – 298

2 ) / 2 + 11.974 * 10

-6

(4733 – 298

3 ) /3 – 1.132 * 10

-8 (473

4 – 298

4) /4 ] = 8.85*10

4 KJ

∆ H N2 = 121

473

298( 31.150 - 1.357 * 10

-2 T + 26.796 * 10

-6 T

2 - 1.168 * 10

-8 T

3 ) dt

= 121 [ ( 31.150 (473 – 298 ) + 1.357 * 10-2

(4732 – 298

2 ) / 2 + 26.7896 * 10

-6

(4733 – 298

3 ) / 3 - 1.168 * 10

-8 (473

4 – 298

4 ) / 4 ] = 8.42*10

6 KJ

∆ H CO2 = 100

473

298( 19.795 + 73.436 * 10

-3 T – 5.602 * 10

-5 T

2 + 17.153 * 10

-9 T

3 ) dt

=100 [ ( 19.795 (473– 298 ) + 73. 436 * 10-3

(4732 – 298

2) / 2 – 5.602 * 10

-5

(4733 – 298

3) /3 + 17.153 * 10

-9 (473

4 – 298

4) /4 ] = 7.91*10

5KJ

∆ H CO = 14.74

473

298( 30.869 - 1.285 * 10

-2 T +27.89 * 10

-6 T

2 – 1.272 * 10

-8 T

3 ) dt

= 14.74 [ (30.869 (473– 298) - 1.285 * 10-2

(4732 – 298

2) / 2 + 27.89 * 10

-6

(4733 –298

3) /3 – 1.272 * 10

-8 (473

4 – 298

4) /4 ] = 2.54*10

4KJ

First

Gas / Gas

Heat exchanger

47

∆ H H2 = 275.13

473

298( 27.143 + 92.738 * 10

-4 T– 1.38 * 10

-5 T

2+ 76.451* 10

-10 T

3 ) dt

= 275.13 [ (27.143 (473 – 298 ) + 92.738 * 10-4

(4732 – 298

2 ) / 2 – 1.38 * 10

-5

(4733 – 298

3 ) / 3 + 76.451* 10

-10 (473

4 – 298

4 ) / 4 ] = 1.59*10

6KJ

∆ H H2O = 309.6

473

298( 32.243 + 19.238 * 10

-4 T + 10.555 * 10

-6 T

2 – 3.596 * 10

-9 T

3 ) dt

= 309.6[ ( 32.243 (473 – 298 ) + 19.238 * 10-4

(4732 – 298

2 ) / 2 + 10.555 * 10

-6

(4733 – 298

3 ) / 3 – 3.596 * 10

-9 (473

4 – 298

4 ) / 4 ] = 1.79*10

6KJ

∆ H Ar/He = 0.1

473

298( 20.804 - 3.211 * 10

-5T + 51.665 * 10

-9 T

2 ) dt

= 0. 1 [ ( 20.804 (473 – 298) - 3.211 * 10-5

(4732 – 298

2) / 2 + 51.665 * 10

-9

(4733 – 298

3) /3 ] = 4.13*10

2KJ

Σ∆ H8 4.196*106KJ

2.15. Low Temp Shift Converter ( L . T . S . C ) :-

∆ H8 ∆ H9 495 k

W7

Fig.22 Low Temp Shift Converter

The reaction take place at low temp shift converters :

CO + H2O CO2 + H2 ∆ Ho R = 9.8 kcal (16.2)

L . T . S. C

48


Tab.31


298 c p . d t

∆ H CH4 = 1.27

495

298( 19.251 + 52.126 * 10

-3 + 11.974 * 10

-6 T

2 – 1.132 * 10

-8 T

3 ) dt

= 1.27 [ ( 19.251 ( 495– 298 ) + 52.126 * 10-3

( 4952 – 298

2 ) /2 + 11.974 * 10

-6

(4953 – 298

3 ) /3 – 1.132 * 10

-8 (495

4 – 298

4) /4 ] = 1.01*10

4 KJ

∆ H N2 = 121

495

298( 31.150 -1.357 * 10

-2 T + 26.796 * 10

-6 T

2 - 1.168 * 10

-8 T

3 ) dt

= 121 [ ( 31.150 (495 – 298 )+ 1.357 * 10-2

(4952 – 298

2 ) /2 + 26.796 * 10

-6

(4953 – 298

3 ) / 3 - 1.168 * 10

-8 (495

4 – 298

4 ) / 4 ] = 9.55*10

5 KJ

∆ H CO2 = 111.055

495

298( 19.795 + 73.436 * 10

-3 T –5.602 *10

-5 T

2 + 17.153 * 10

-9 T

3) dt

=111.055 [ ( 19.795 (495– 298) + 73.436 * 10-3

(4952 – 298

2) /2 – 5.602 * 10

-5

(4953 – 298

3) /3 + 17.153 * 10

-9 (495

4 – 298

4) /4 ] = 1.53*10

6 KJ

∆ H CO = 3.685

495

298( 30.869- 1.285 * 10

-2 T +27.89 * 10

-6 T

2 – 1.272 * 10

-8 T

3 ) dt

= 3.685 [ (30.869 (495– 298) + 1.285 * 10-2

(4952 – 298

2) / 2 +27.89 * 10

-6

(4953 – 298

3) /3 – 1.272 * 10

-8 (495

4 – 298

4) /4 ] = 3.25*10

4 KJ

∆ H H2 = 286.2

495

298( 27.143 + 92.738 * 10

-4 T– 1.38 * 10

-5 T

2+ 76.45* 10

-10 T

3 ) dt

= 286.2 [ (27.143 (495 – 298 ) + 92.738 * 10-4

(4952 – 298

2 ) / 2 – 1.38 * 10

-5

(4953 – 298

3 ) / 3 + 76.45* 10

-10 (495

4 – 298

4 ) / 4 ] = 1.64*10

6 KJ

∆ H H2O =298.5

495

298( 32.243 + 19.238 * 10

-4 T + 10.555 * 10

-6 T

2 – 3.596 * 10

-9 T

3 ) dt

= 298.5 [ ( 32.243 (495 – 298 ) + 19.238 * 10-4

(4952 – 298

2 ) / 2 + 10.555 * 10

-6

(4953 – 298

3 ) / 3 – 3.596 * 10

-9 (495

4 – 298

4 ) / 4 ] = 2.02*10

6 KJ

∆ H Ar/He =0.1

495

298( 20.804 -3.211 * 10

-5 T + 51.665 * 10

-9 T

2 ) dt

= 0.1 [ ( 20.804 (495 – 298) - 3.211 * 10-5

(4952 – 298

2) / 2 + 51.665 * 10

-9

(4953 – 298

3) /3 ] = 5.67*10

3 KJ

Σ∆ H9 6.19*106 KJ

49

2.16. Water Separator :-

∆ H9 ∆ H11 406 k

W9

W8 ∆ H10 363K

Fig.23 Water Separator

- Calculation the enthalpy change (∆ H10) for ( H2O ) from 298k to 363k

Tab.32


298 c p . d t

∆ H H2O = 298.5

363

298( 32.243 + 19.238 * 10

-4 T + 10.555 * 10

-6 T

2 + 3.596 * 10

-9 T

3 ) dt

= 298.5 [ ( 32.243 (363 – 298 ) + 19.238 * 10-4

(3632 – 298

2 ) / 2 + 10.555 * 10

-6

(3633 – 298

3 ) / 3 + 3.596 * 10

-9 (363

4 – 298

4 ) / 4 ] = 662.91*10

3 KJ

Σ∆ H10 662.9*103 KJ

Separator

50


Tab.33


298 c p . d t

∆ H CH4 = 1.27

406

298( 19.251 + 52.126 * 10

-3 + 11.974 * 10

-6 T

2 – 1.132 * 10

-8 T

3 ) dt

= 1.27 [ ( 19.251 ( 406– 298 ) + 52.126 * 10-3

( 4062 – 298

2 ) /2 + 11.974 * 10

-6

(4063 – 298

3 ) /3 – 1.132 * 10

-8 (406

4 – 298

4) /4 ] = 5.243 *10

3 KJ

∆ H N2 = 121

406

298( 31.150 _ 1.357 * 10

-2 T + 26.796 * 10

-6 T

2 - 1.168 * 10

-8 T

3 ) dt

= 121 [ ( 31.150 (406 – 298) - 1.357 * 10-2

(406 2 – 298

2 ) / 2 + 26.796 * 10

-6

(406 3 – 298

3 ) / 3 - 1.168 * 10

-8 (406

4 – 298

4 ) / 4 ] =

5.07*10

6 KJ

∆ HCO2 = 111.055

406

298( 19.795 + 73.436 *10

-3 T – 5.602 * 10

-5 T

2 + 17.153 *10

-9 T

3) dt

=111.055 [ ( 19.795 (406 – 298) + 73.436 * 10-3

(406 2 – 298

2) /2 – 5.602 * 10

-5

(406 3 – 298

3) /3 + 17.153 * 10

-9 (406

4 – 298

4) /4 ] = 4.73*10

6KJ

∆ H CO = 3.685

406

298( 30.869 - 1.285 * 10

-2 T +27.89 * 10

-6 T

2 – 1.272 * 10

-8 T

3 ) dt

= 3.685 [ (30.869 (406– 298) -1.285 * 10-2

(4062 – 298

2) / 2 + 27.89 * 10

-6

(4063 – 298

3) /3 – 1.272 * 10

-8 (406

4 – 298

4) /4 ] = 1.53*10

4 KJ

∆ H H2 =286.2

406

298( 27.143 + 92.738 * 10

-4 T– 1.38 * 10

-5 T

2+ 76.451* 10

-10 T

3 ) dt

= 286.2 [ (27.143 (406 – 298 ) + 92.738 * 10-4 (406

2 – 298

2 ) / 2 – 1.38 * 10

-5

(406 3 – 298

3 ) / 3 + 76.451* 10

-10 (406

4 – 298

4 ) / 4 ] = 8.9 *10

5 KJ

∆ H Ar/He =0.1

406

298( 20.804 - 3.211 * 10

-5 T + 51.665 * 10

-9 T

2 ) dt

= 0.1 [ ( 20.804 (406 – 298) - 3.211 * 10-5

(406 2 – 298

2) / 2 + 51.665 * 10

-9

(406 3 – 298

3) /3 ] = 2.5 *10

2KJ

Σ∆ H11 10.06*106 KJ

51

2.17. CO2 Removal ( Absorber ) :-

W10

∆ H12 386K

∆ H11 ∆ H14

W9 W12

W 12

W11 ∆ H13 398K

Fig.24 CO2 Removal

- The reaction taken place in the ( Co2 Removal ) is as follow

K2 CO3 + H2O + CO2 2KHCO3 ∆ HR° = - 13.3206 k cal / mol (17.2)

K2CO3 ( S ) + H2O ( L ) + CO2 ( g ) 2KHCO3 ( S ) (17.2)

The specific heat for reactants : from reference " 5 " page ( 179 ) table ( 3 ) CP (K2

CO3) (s ) = 29.9 cal / mol .C˚ = 29.9 * 4.184

= 125.1016 kJ / kmol .C˚

The specific heat for H2O ( L ) from CP H2O ( L ) = 1.01 cal / g .C˚

= 1.01 * 4.184= 76.06512 kJ / kmol. C˚

CO2

Removal

52

- Now calculation the Enthalpy change of the reactant (∆ H12 ) ( K2Co3+ H2o) from 298k

to 732k

Tab.34

Components ∆ H12= Σ n 386

298 c p . dt

∆ H K2CO3 = 110

386

298 ( 125.1016 ) dt = 110 [125.1016 (386–298) ] =1.210 *10

6KJ

∆ H H2O = 110

386

298 ( 76.06512 ) dt = 110 [76.06512 (386–298) ] =0.7363 *10

6KJ

Σ∆ H12 1.94*106 KJ

CP ( KHCO3 ) can be calculated from reference " 4 " page 245

Tab.35

Element Mol . wt Heat capacity

K 39 33.5

H 1 18.0

C 12 11.7

O 48 3*25.1

Total CP ( KHCO3 ) = 138.5 kJ/ kmol.˚C

- Calculation the enthalpy change of the product (∆ H13) ( KHCO3) from 298k to 398k

Tab.36


298 c p . dt

∆ H KHCO3 = 110

398

298 ( 138.5 ) dt = 110 [138.5 (398–298) ] =1.52*10

6KJ

Σ∆ H13 1.5*106 KJ

53


Tab.37


298 c p . d t

∆ H CH4 = 1.27

480

298( 19.251 + 52.126 * 10

-3 + 11.974 * 10

-6 T

2 – 1.132 * 10

-8 T

3 ) dt

= 1.27 [ ( 19.251 ( 480– 298 ) + 52.126 * 10-3( 480

2 – 298

2 ) /2 + 11.974 * 10

-6

(4803 – 298

3 ) /3 – 1.132 * 10

-8 (480

4 – 298

4) /4 ] = 2.55 *10

5 KJ

∆ H N2 = 121

480

298( 31.150 -1.357 * 10

-2 T + 26.796 * 10

-6 T

2 - 1.168 * 10

-8 T

3 ) dt

= 121 [ ( 31.150 (480 – 298 ) - 1.357 *10-2

(480 2 – 298

2 ) /2 + 26.796 *10

-6

(480 3 – 298

3 ) / 3 -1.168 * 10

-8 (480

4 – 298

4 ) / 4 ] =

8.77*10

5 KJ

∆ H CO2 = 1.11

480

298( 19.795 + 73.436 * 10

-3 T – 5.602 * 10

-5 T

2 + 17.153 * 10

-9T

3 ) dt

=1.11 [ ( 19.795 (480 – 298) + 73.436 * 10-3

(480 2 – 298

2) /2 – 5.602 * 10

-5

(480 3 – 298

3) /3 + 17.153 * 10

-9(480

4 – 298

4) /4 ] = 8.24*10

3 KJ

∆ H CO = 3.7

480

298( 30.869 - 1.285 * 10

-2 T +27.89 * 10

-6 T

2 – 1.272 * 10

-8 T

3 ) dt

= 3.7 [ (30.869 (480– 298) - 1.285 * 10-2

(4802 – 298

2) /2 +27.89 * 10

-6

(4803 – 298

3) /3 – 1.272 * 10

-8 (480

4 – 298

4) /4 ] = 3.78*10

4 KJ

∆ H H2 = 286.2

480

298( 27.143 + 92.738 * 10

-4 T– 1.38 * 10

-5 T

2+ 76.451* 10

-10 T

3 ) dt

= 286.2 [ (27.143 (480 – 298 ) + 92.738 * 10-4

(480 2 – 298

2 ) / 2 – 1.38 * 10

-5

(480 3 – 298

3 ) / 3 +76 .451* 10

-10 (480

4 – 298

4 ) / 4 ] = 1.43 *10

6 KJ

∆ H Ar/He = 0.1

480

298( 20.804 - 3.211 * 10

-5T + 51.665 * 10

-9T

2 ) dt

= 0.1 [ ( 20.804 (480 – 298) - 3.211 * 10-5

(480 2 – 298

2) /2 + 51.665 * 10

-9

(4803 – 480

3) /3 ] = 3.3*10

3 KJ

Σ∆ H14 2.61*106 KJ

54

2.18. Second Gas/Gas Heat Exchanger

∆ H16 outlet methanator

∆ H14 ∆ H15 343 k

W12

W13

∆ H17 388 k

Fig.25 Second Heat Exchanger

calculation the Enthalpy change of the product (∆ H15) from 298k to 343k

Tab.38


298 c p . d t

∆ H CH4 =1.27

343

298( 19.251 + 52.126 * 10

-3 + 11.974 * 10

-6 T

2 – 1.132 * 10

-8 T

3 ) dt

= 1.27 [ ( 19.251 ( 343– 298 ) + 52.126 * 10-3( 343

2 – 298

2 ) /2 + 11.974 * 10

-6

(3433 – 298

3 ) /3 – 1.132 * 10

-8 (343

4 – 298

4) /4 ] = 1.04 *10

4 KJ

∆ H N2 = 121

343

298( 31.150- 1.357 * 10

-2 T + 26.796 * 10

-6 T

2 - 1.168 * 10

-8 T

3 ) dt

= 121 [ ( 31.150 (343 – 298 )- 1.357 * 10-2

(3432 – 298

2 ) / 2 + 26.796 * 10

-6

(343 3 – 298

3 ) / 3 - 1.168 * 10

-8 (343

4 – 298

4 ) / 4 ] =

2.06*10

6 KJ

∆ H CO2 = 1.11

343

298( 19.795 + 73.436 * 10

-3 T – 5.602 * 10

-5 T

2 + 17.153 * 10

-9T

3 ) dt

=1.11 [ ( 19.795 (343 – 298) + 73.436 * 10-3

(343 2 – 298

2) /2 – 5.602 * 10

-5

(3433 – 298

3) /3 + 17.153 * 10

-9 (343

4 – 298

4) /4 ] = 1.9*10

3KJ

Second

Gas / Gas

Heat exchanger

55

∆ H CO = 3.7

343

298( 30.869 - 1.285 * 10

-2 T +27.89 * 10

-6 T

2 – 1.272 * 10

-8 T

3 ) dt

= 3.7 [ (30.869 (343– 298)- 1.285 * 10-2

(3432 – 298

2) /2 +27.89 * 10

-6

(3433 – 298

3) /3 – 1.272 * 10

-8 (343

4 – 298

4) /4 ] =6.53 *10

3KJ

∆ H H2 = 286.2

343

298( 27.143 + 92.738 * 10

-4 T– 1.38 * 10

-5 T

2+ 76.451* 10

-10 T

3 ) dt

= 286.2 [ (27.143 (343 – 298 ) + 92.738 * 10-4

(343 2 – 298

2 ) / 2 – 1.38 * 10

-5

(3433 – 298

3 ) / 3 + 76.451* 10

-10 (343

4 – 298

4 ) / 4 ] = 3.5 *10

5KJ

∆ H Ar/He = 0.1

343

298( 20.804 -3.211 * 10

-5 T + 51.665 * 10

-9 T

2 ) dt

=0.1 [ ( 20.804 (343 – 298) - 3.211 * 10-5

(343 2 – 298

2) /2 + 51.665 * 10

-9

(3433 – 480

3) /3 ] = 80.26 KJ

Σ∆ H15 2.42*106 KJ

2.19. Methanator

∆ H15 ∆ H16 1100 K

W12 W13

Fig.26 Methanation

Methanation

56

calculation the Enthalpy change of the product (∆ H16 ) from 298k to 480k

Tab.39


298 c p . d t

∆ H CH4 = 6.07

1100

298( 19.251 + 52.126 * 10

-3+ 11.974 * 10

-6 T

2 – 1.132 * 10

-8 T

3 ) dt

= 6.07[ ( 19.251 (1100– 298 )+ 52.126 *10-3( 1100

2 – 298

2 ) /2 + 11.974 * 10

-6

(11003 – 298

3 ) /3 – 1.132 * 10

-8 (1100

4 – 298

4) /4 ] =3.216 *10

6 KJ

∆ H N2 = 121

1100

298( 31.150 - 1.357 * 10

-2 T + 26.796 * 10

-6 T

2 - 1.168 * 10

-8 T

3 ) dt

= 121 [ ( 31.150 (1100 – 298 )- 1.357 * 10-2

(11002 – 298

2 ) /2 + 26.796 * 10

-6

(1100 3 – 298

3 ) / 3 - 1.168 * 10

-8 (1100

4 – 298

4 ) / 4 ] =

12.15*10

7 KJ

∆ H H2 = 270.70

1100

298( 27.143 + 92.738 * 10

-4 T– 1.38 * 10

-5 T

2+ 76.451* 10

-10 T

3 ) dt

= 270.70 [ (27.143 (1100 – 298 ) + 92.738 * 10-4

(1100 2 – 298

2 )/ 2 – 1.38* 10

-5

(11003 – 298

3 ) / 3 + 76.451* 10

-10 (1100

4 – 298

4 ) / 4 ] = 32.13*10

3 KJ

∆ H Ar/He = 0.1

1100

298( 20.804 - 3.211 * 10

-5 T + 51,665 * 10

-9 T

2 ) dt

= 0.1 [ ( 20.804 (1100 – 298) - 3.211 * 10-5

(1100 2 – 298

2) /2 + 51.665 * 10

-9

(11003 – 480

3) /3 ] = 18.05 *10

3 KJ

∆ H H2O = 5.90

1100

298( 32.243 + 19.238 * 10

-4 T + 10.555 * 10

-6 T

2 + 3.596 * 10

-9 T

3 ) dt

= 5.90[ ( 32.243 (1100 – 298 ) + 19.238*10-4

(11002 – 298

2 ) /2 + 10.555 * 10

-6

(11003 – 298

3 ) / 3 + 3.596 * 10

-9 (1100

4 – 298

4 ) / 4 ] = 2.37*10

6 KJ

Σ∆ H16 1.27*108 KJ

57

calculation the Enthalpy change of the reactant (∆ H17) from 298k to 388k

Tab.40


298 c p . d t

∆ H CH4 = 6.07

388

298( 19.251 + 52.126 * 10

-3 + 11.974 * 10

-6 T

2 – 1.132 * 10

-8 T

3 ) dt

= 6.07 [ ( 19.251 (388– 298 ) + 52.126 * 10-3

( 3882 – 298

2 ) /2 + 11.974 * 10

-6

(3883 – 298

3 ) /3 – 1.132 * 10

-8 (388

4 – 298

4) /4 ] = 2.40 *10

5 KJ

∆ H N2 = 121

388

298( 31.150- 1.357 * 10

-2 T + 26.796 * 10

-6 T

2 - 1.168 * 10

-8 T

3 ) dt

= 121 [ ( 31.150 (388 – 298 )- 1.357 * 10-2

(3882 – 298

2 ) /2 + 2.696* 10

-6

(3883 – 298

3 ) / 3 - 1.168 * 10

-8 (388

4 – 298

4 ) / 4 ] =

10.53*10

6 KJ

∆ H H2 = 270.70

388

298( 27.143 + 92.738 * 10

-4 T– 1.38 * 10

-5 T

2+ 76.451* 10

-10 T

3 ) dt

= 270.70 [ (27.143 (388 – 298 ) + 92.738 * 10-4

(388 2 – 298

2 ) / 2 – 1.38 * 10

-5

(3883 – 298

3 ) / 3 + 76.451* 10

-10 (388

4 – 298

4 ) / 4 ] = 3.352 *10

6 KJ

∆ H Ar/He = 0.1

388

298( 20.804 - 3.211 * 10

-5 T + 51.665 * 10

-9 T

2 ) dt

= 0.1 [ ( 20.804 (388 – 298) - 3.211 * 10-5

(388 2 – 298

2) /2 + 51.665 * 10

-9

(3883 – 480

3) /3 ] = 1.81 *10

3 KJ

∆ H H2O = 5.90

338

298( 32.243 + 19.238 * 10

-4 T + 10.555 * 10

-6 T

2 + 3.596 * 10

-9 T

3 ) dt

= 5.90 [ ( 32.243 (388 – 298 ) + 19.238 * 10-4

(3882 – 298

2 ) / 2 + 10.555 * 10

-6

(3883 – 298

3 ) / 3 + 3.596 * 10

-9 (388

4 – 298

4 ) / 4 ] =20.8 *10

4 KJ

Σ∆ H17 1.4*107 KJ

58

2.20. Cooler :-

Water ∆ H18 = 0 298 K

∆ H17 ∆ H20 340 k

W13

∆ H19 363 K

Water

Fig.27 Cooler

From : Reference " 5 " page ( 137 ) CP (H2O ) = 75.61 kJ / kmol C°

at 90 C°=363k

calculation the Enthalpy change of the water (∆ H19) from 298k to 363k

Tab.41


298 c p . d t

∆ H H2O = 5.90

363

298( 75.61 ) dt = 5.90 [75.61 (363–298) ] = 423.6 *10

3 KJ

Σ∆ H19 423.6*103 KJ

COOLER

59

-calculation the Enthalpy change of the product (∆ H20) from 298k to 340k

Tab.42


298 c p . d t

∆ H CH4 = 6.07

340

298( 19.251 + 52.126 * 10

-3 + 11.974 * 10

-6 T

2 – 1.132 * 10

-8 T

3 ) dt

= 6.07 [ ( 19.251 (340– 298 ) + 52.126 * 10-3

( 3402 – 298

2 ) /2 + 11.974 * 10

-6

(3403 – 298

3 ) /3 – 1.132 * 10

-8 (340

4 – 298

4) /4 ] = 10.84*10

4 KJ

∆ H N2 = 121

340

298( 31.150 -1.357 * 10

-2 T + 26.796 * 10

-6 T

2 - 1.168 * 10

-8 T

3 ) dt

= 121 [ ( 31.150 (340 – 298 ) - 1.357 * 10-2

(3402 – 298

2 ) /2 + 26.796 * 10

-6

(3403 – 298

3 ) / 3 - 1.168 * 10

-8 (340

4 – 298

4 ) / 4 ] = 4.92*10

6 KJ

∆ H H2 = 270.70

340

298( 27.143 + 92.738 * 10

-4 T– 1.38 * 10

-5 T

2+ 76.451* 10

-10 T

3 ) dt

= 270.71[ (27.143 (340 – 298 ) + 92.738 * 10-4

(340 2 – 298

2 ) / 2 – 1.38 * 10

-5

(3403 – 298

3 ) / 3 + 76.451* 10

-10 (340

4 – 298

4 ) / 4 ] = 1.64 *10

6 KJ

∆ H Ar/He = 0.1

340

298( 20.804 -3.211 * 10

-5 T + 51.665 * 10

-9 T

2 ) dt

= 0.1 [ ( 20.804 (340 – 298) - 3.211 * 10-5

(340 2 – 298

2) / 2 + 51.665 * 10

-9

(3403 – 480

3) /3 ] = 7.8*10

2 KJ

∆ H H2O = 5.90

340

298( 32.243 + 19.238 * 10

-4 T + 10.555 * 10

-6 T

2 + 3.596 * 10

-9 T

3 ) dt

= 5.90 [ ( 32.243 (340 – 298 ) + 19.238 * 10-4

(3402 – 298

2 ) / 2 + 10.555 * 10

-6

(3403 – 298

3 ) / 3 + 3.596 * 10

-9 (340

4 – 298

4 ) / 4 ] =9.6 *10

4 KJ

Σ∆ H20 1.168*107 KJ

60

2.21. Synthesis Gas Compressor, Cooler and Separator :-

∆ H20 ∆ H22 424 k

W15

W14 ∆ H21

315 k

Fig.28 Synthesis Gas Compressor, Cooler and Separator

-Calculation the enthalpy change of the (∆ H21) ( CH4 + H2O ) from 298k to 480k

Tab.43


298 c p . d t

∆ H CH4 = 6.07

315

298( 19.251 + 52.126 * 10

-3 + 11.974 * 10

-6 T

2 – 1.132 * 10

-8 T

3 ) dt

= 6.07 [ ( 19.251 (315– 298 ) + 52.126 * 10-3

( 3152 – 298

2 ) /2 + 11.974 * 10

-6

(3153 – 298

3 ) /3 – 1.132 * 10

-8 (315

4 – 298

4) /4 ] = 3.717*10

3 KJ

∆ H H2O = 5.90

315

298( 32.243 + 19.238 * 10

-4 T + 10.555 * 10

-6 T

2 + 3.596 * 10

-9 T

3 ) dt

= 5.90 [ ( 32.243 (315 – 298 ) + 19.238 * 10-4

(3152 – 298

2 ) / 2 + 10.555 * 10

-6

(3153 – 298

3 ) / 3 + 3.596 * 10

-9 (315

4 – 298

4 ) / 4 ] = 3.40 *10

3 KJ

Σ∆ H21 7.1*103 KJ

Synthesis gas

Compressor

cooler &

separator

61


Tab.44


298 c p . d t

∆ H N2 = 121

424

298( 31.150 - 1.357 * 10

-2 T + 26.796 * 10

-6 T

2 - 1.168 * 10

-8 T

3 ) dt

= 121 [ ( 31.150 (424 – 298) - 1.357 * 10-2

(4242 – 298

2 ) /2 + 26.796 * 10

-6

(4243 – 298

3 ) / 3- 1.168 * 10

-8 (424

4 – 298

4 ) / 4 ] = 6.64*10

6 KJ

∆ H H2 = 270.70

424

298( 27.143 + 92.738 * 10

-4 T– 1.38 * 10

-5 T

2+ 76.451* 10

-10 T

3 ) dt

= 270.70 [ (27.143 (424 – 298 ) + 92.738 * 10-4

(424 2 – 298

2 ) / 2 – 1.38 * 10

-5

(4243 – 298

3 ) / 3 + 76.451* 10

-10 (424

4 – 298

4 ) / 4 ] =11.09 *10

7 KJ

∆ H Ar/He = 0.1

424

298( 20.804 - 3.211* 10

-5 T + 51.665 * 10

-9 T

2 ) dt

= 0.1 [ ( 20.804 (424 – 298) - 3.211* 10-5

(424 2 – 298

2) / 2 + 51.665 * 10

-9

(4243 – 480

3) /3 ] = 2.55 *10

3 KJ

Σ∆ H22 1.175*108 KJ

2.22. Gas/Gas Reactor Heat Exchanger :-

W16 ∆ H24 773 K

∆H22 ∆ H23 320 K

W15

∆ H25 = 0 298 k

Fig.29 Gas Reactor Heat Exchanger

Gas / Gas

Reactor

Heat

Exchanger

62


Tab.45


298 c p . d t

∆ H N2 = 121

320

298( 31.150 -1.357 * 10

-2 T + 26.796 * 10

-6 T

2 - 1.168 * 10

-8 T

3 ) dt

= 121 [ ( 31.150 (320 – 298 ) - 1.357 * 10-2

(3202 – 298

2 ) /2 + 26.796* 10

-6

(3203 – 298

3 ) / 3 - 1.168 * 10

-8 (320

4 – 298

4 ) / 4 ] = 0.4

*10

3 KJ

∆ H H2 = 270.70

320

298( 27.143 + 92.738 * 10

-4 T– 1.38 * 10

-5 T

2+ 76.451* 10

-10 T

3 ) dt

= 270.70 [ (27.143 (320 – 298 ) + 92.738 * 10-4

(320 2 – 298

2 ) / 2 – 1.38 * 10

-5

(3203 – 298

3 ) / 3 + 76.451* 10

-10 (320

4 – 298

4 ) / 4 ] =0.1 *10

6 KJ

∆ H Ar/He = 0.1

320

298( 20.804 - 3.211 * 10

-5 T + 51.665 * 10

-9 T

2 ) dt

= 0.1 [ ( 20.804 (320 – 298) - 3.211 * 10-5

(320 2 – 298

2) /2 + 51.665 * 10

-9

(3203 – 480

3) /3 ] = 0.7* 10

6 KJ

Σ∆ H23 800.4*103 KJ

63

2.23 Cooling and Separator :-

∆ H27 310 K

Purged (W18)

( N2 + H2 ) Recycle

∆ H28 325 K

∆ H25 ∆ H26 217 k

W17 NH3

Fig.30 Cooling and Separator

From : Reference " 5 " page ( 140 ) CP (NH3) = 82.429 kJ / kmol C°

Cooling and

separator

64

calculation the Enthalpy change (∆ H26) (NH3 ) from 298k to 217k

Tab.46


298 c p . d t

∆ H NH3 = 215.26

217

298( 82.429 ) dt = 215.26 [82.429 (217–298) ] = -1.43*10

6 KJ

Σ∆ H26 -1.4*106 KJ

calculation the Enthalpy change of purged (∆ H27) from 298k to 310k

Tab.47


298 c p . d t

∆ H Ar/He = 7.461

310

298( 20.804 – 3.211 * 10

-5 T + 5.166 * 10

-8 T

2 ) dt

= 7.461 [ ( 20.804 (310 – 298) + 3.211 * 10-5

(310 2 – 298

2) /2 + 5.166 * 10

-8

(3103 – 480

3) /3 ] = 1.86*10

3 KJ

Σ∆ H27 1.8*103 KJ

65

calculation the Enthalpy change of Recycle (∆ H28) from 298k to 325k

Tab.48


298 c p . d t

∆ HN2 = 381.8

325

298( 31.15 + 1.356 * 10

-2 T + 2.678 * 10

-5 T

2 - 1.168 * 10

-8 T

3 ) dt

= 381.8[ ( 31.15 (325 – 298 ) + 1.356 * 10-2

(3252 – 298

2 ) /2 + 2.678 * 10

-5

(3253 – 298

3 ) / 3 - 1.168 * 10

-8 (325

4 – 298

4 ) / 4 ] = 4.3* 10

6 KJ

∆ HH2 = 763.7

325

298( 27.143 + 9.273 * 10

-3 T– 1.38 * 10

-5 T

2+ 7.645* 10

-9 T

3 ) dt

= 763.7 [ (27.143 (325 – 298 ) + 9.273 * 10-3

(325 2 – 298

2 ) / 2 – 1.38 * 10

-5

(3253 – 298

3 ) / 3 + 7.645* 10

-9 (325

4 – 298

4 ) / 4 ] = 6.7 *10

6 KJ

Σ∆ H28 11*106 KJ

2.24. Ammonia Convertor :-

∆ H28

∆ H23

∆ H24

773 k

Fig.31 Ammonia Convertor

NH3

convertor

66

calculation the Enthalpy change (∆ H24) from 298k to 2773k

Energy balance at steady state for NH3 convertor : Input = output

∆ H23+ ∆ H28 = ∆ H24

∆ H24 = 800.4*103KJ + 11*10

6KJ =11.8*10

6KJ.

Overall heat balance inlet = outlet

∆ H1 +∆ H2 +∆ H4 +∆ H12 = ∆ H10 +∆ H13 +∆ H19 +∆ H21 +∆ H24 +∆ H26 +∆ H27

12.86*106 = 12.9*10

6

Error% = (12.9*106 – 12.86*10

6 /12.9*10

6 )

67

CHAPTER 3

Heat exchanger process desing3.1

Desing an exchanger to sub-cool condensate from a ammonia condenser

From 95oc to

oc. Flow-rate of ammonia 100,000 kg/h. Brackish water will

Be used as the coolant, with a temperature rise from 25 oc to 40

oc.

Solution

1. Only the thermal desing will be considered.

2. This example illustrates kern,s method.

6

Heat capacity of ammonia= 4.145kj/kg.k

Head load = 10003600 × 4.145 (95 − 40) = 633.26 kw

Heat capcity water = 4.2 kj/kg. C0

Cooling water flow = 633.2634.2(40−25) = 10 kg/s

∆𝑡im = (95−40)−(40−25)𝑙𝑛(95−40)(40−25) = 31 C0 Use one shell pass and two tube passes

R ═ (95−40)(40−25) = 3.67

S = (40−25)(95−25) = 0.21

Ft = 0.85 ∆𝑡𝑚 = 0.85 × 31 = 26 𝑐𝑜

U = 600 w/m2c

o

68

Provisisonal area 𝐴 = 633.263×10326×600 = 278𝑚2

Choose 20 mm i.d., 16mm id., 4.88-m longtube. Allowing for tubes

(34 in × 16ft) , l − 304 Allowing for tube-sheet thickness,take

L=4.83 m

Area of one tube = 4.83 * 20 * 10-3π = 0.303m

2

Number of one tube = 2780.303 = 918

As the shell-side fluid is relatively clean use 1.25 triangular pitch.

Bundle diameter Db = 20 ( 9180.249) 1/2.207 = 826 mm

Use a split-ring floating head type.

From figure 12.10, bundle diametrical clearance D 68 mm,

Shell diameter, Ds = 826 + 68 = 894 mm.

(Note. nearest standard pipe sizes are 863.6 or 914.4 mm).

Shell size could be read from standard tube count tables.

Tube-side coefficient

Mean water temperature = 40+252 = 33 oc

Tube cross-sectional area = π4 × 162 = 201mm2 Tubes per pass = 9182 = 459

Total flow area = 459 * 201 * 10-6

= 0.092 m2

Water mass velocity = 68.90.092 = 749𝑘𝑔/𝑠𝑚2

Hi = 4200 (1.35 + 0.02∗ 33)0.750.8160.2 = 3852w/m2oc

69

The coefficient can also be calculated using equation ; this is done to illustrate

Using of this method. ℎ 𝑑𝑖𝑘𝑓 = 𝑗ℎ𝑅𝑒𝑝𝑟0.33( 𝜇𝜇𝜔)0.14

Viscosity of water = 0.8mNs/m2

Thermal conductivity = 0.59 W/m oc

Re = 𝑝𝑢𝑑𝑖𝜇 = 995×0.75×16×10−30.8×10−3 = 14925

Re = cpμ kf = 4.2 × 103 × 0.8 × 10−30.59 = 5.7

Neglect ( 𝜇𝜇𝜔) 𝐿𝑑𝑖 = 4.83×10−316 = 302 𝑗ℎ = 3.9 × 10−3

Hi = 0.5916×10−3 × 3.9 × 14925 × 5. 70.33 = 3812 w/m2co Checks reasonably well with value calculated from equation; use

Shell-side coefficient

Choose baffle spacing = 𝑑䌊5 = 8945 = 187mm

Tube pitch = 1.25 * 20 = 25mm

Cross- flow area As = (25−20)25 894 × 178 × 10−6 = 0.032 Mass velocity, Gs = 100.0003600 × 10.032 = 868kg/sm2 Equivalent diameter de = (252 − 0.917 × 202) = 14.4𝑚𝑚

Mean shell side temperature = 95+402 = 68 𝑐𝑜

Ammonia density = 600 kg/m3

70

Viscosity = 141 * 10-6

Ns/m2

Heat capcity = 4.145 KJ/kg oc

Thermal conductivity = 0.477 w/m oc

Re = G sdeμ = 868 × 14.4 × 10−3141 × 10−6 = 88646.80

Pr = cpμkf = 4.145 × 141 × 10−60.477 = 1.225 × 10−3

Choose 25 percent baffle cut, figure 12.29

Jh = 3.3 * 10-3

Without the viscosity correction term

hs = 0.47714.4 × 10−3 × 3.3 × 10−3 × 8864.80 × 5. 11/3 = 1663.70w/m

2oc

mean temperature difference across all resistances = 68 – 33 = 35 oc

across ammonia film = Uho × ∆t = 6001663.70 × 35 = 12.62 co

mean wall temperature = 68 - 12.62 = 55.37oc

μ𝜔 = 0.37 mNs/m2 ( 𝜇𝜇𝜔)0.14

Which shows that the correctiion for a low-viscosity fluid is not significant.

Overall coefficient

Thermal conductivity of stainless steel (BS 316) = 50 w/m oc

Take the fouling coefficients from table 12.2;ammonia 500wm-2 o

c-1

1𝑈𝑜 = 12740 + 15000 + 20 × 10−3ln (2016)2 × 50 + 2016 × 1300 + 2016 × 13812

Uo=738 w/m2 o

cWell above assumed value of 600 w/m2o

c

Assume Uo = 650 w/m2 o

c , and proceed with another tria

71

3.2. Pressure drop

Tube-side

For Re = 14,925

Jf = 4.3*10-3

Neglecting the viscosity correction term

∆𝑃1 = 2 (8 × 4.3 × 10−3 (4.83×10−316 ) + 2.5) 995×0.7522

= 7211 N/m2

= 7.2 Kpa(1.1 psi)

Low, could consider increasing the number of tube passes.

Shell side

Linear velocity = 𝐺𝑠𝑝 = 868750 = 1.16𝑚𝑠

At Re = 36,762 , Jf = 4 * 102

Neglect viscosity correction

∆𝑝𝑠 = 8 × 4 × 10−2 (89414.4) (4.83×10−3178 ) 750 × 1.1622

= 272,019 N/m2

= 272 Kpa (39 psi) to high.

Could be reduced by increasing the baffle pitch. Doubling the pitch halves the

Shell-side velocity, which reduces the pressure drop by afactor of approximately

(1/2)2.

72

∆𝑃𝑠 = 2724 = 68 𝐾𝑝𝑎(10 𝑝𝑠𝑖)𝑎𝑐𝑐𝑒𝑝𝑡𝑎𝑏𝑙𝑒

This will reduce the shell-side heat- transfer coefficient by a factor of

(1/2) 0.8 (ho, ∞, 𝐷𝑒0.8∞𝑈𝑠0.8)

Ho = 2740× (12)0.8 = 1573 w/m

2 oc

This gives an oveall coefficient of 615 W/m2

oc still above assumed value of Of 600

w/m2o

c.

73

CHAPTER 4

Process Instrumentation and Control

Instrumentation and control system play an important role in the industrial undertaking or

process, as they sense any faults accurately and act fastly to either correct the fault or avoid

any unacceptable cases. For the efficient operation of any process, that is equipment and

apparatus must be run at their correct rates. As a result of the rabid development, many

high accuracy devices been installed for the modern plants where a high degree of precision

is required.

4.1 Instruments :

The control of chemical industrial processes is a very important matter, such the chemical

process that comes within specific terms: pressure, degree of temperature, rate of materials

reaction and other variables. The instruments are provided to monitor the key process

variables during plant operation. They may be incorporated in automatic control loops, or

used for the manual monitoring of the process operation. They may also be part of an

automatic computer data logging system. Instruments monitoring critical process variables

will be fitted with automatic alarms to alert the operators to critical and hazardous

situations.

The most control circles used in industrial unit are:

Level Control .

Temperature Control .

Pressure Control .

Ratio Control.

Flow Control .

The primary objectives of the designer when specifying instrumentation and control

schemes are :

74

4.1.1 Safe plant Operation:

a. To keep the process variables within known safe operating limits.

b. To detect dangerous situations as they develop and to provide alarms and automatic

shut-down systems.

c. To provide interlocks and alarms to prevent dangerous operating procedures.

4.1.2 Production Rate:

To achieve the design product output.

4.1.3 Product Quality:

To maintain the product composition within the specified quality standards.

4.1.4 Cost:

To operate at the lowest production cost, commensurate with the other objectives.

Tab. 49

Symbol Function

TC Temperature Controller

PC Pressure Control

LC Level Control

FC Flow Control

RC Ratio Control

FR Flow Recorder

FI Flow Indicator

FRC Flow Recorder and Control

LRC Level Recorder and Control

LIC Level Indicator and Control

PIC Pressure Indicator and Control

75

4.2 Models of Control and Measurement Systems:

4.2.1 Level Control (LC) :

This can be done through flow control from the device by setting a control valve in the

direct current of the pump for safekeeping the specific level in the device as shown in

Figure (42).

Fig. 32 Level Control

4.2.2 Temperature Control (TC) :

Temperature control can be done by means of changing the flow in the cold and hot

medium such as control of water temperature transferring to heat exchanger.

Fig. 33 (a) Control of One Fluid Stream (b) By-Pass Control

76

4.2.3 Pressure Control (PC):

Pressure control depends on the type of the device used, as in reactors, heat exchangers, and

absorption towers, pressure control can be done by releasing atmosphere air, that should not

be toxic or expensive.

Fig. 34 (a) Pressure control by direct venting, (b) Venting of Non-Condensable after a

Condenser, (c) Condenser pressure control by Controlling Coolant Flow, (d) Pressure

control of A by varying the Heat-Exchanger area (Area dependent on liquid level)

4.2.4 Ratio Control (RC) :

To get steady operation when disturbances are absent , the controlled variables must be a

continuous function of the error . With rate control , the most widely used type the

controller output is a linear function of the error signal . To maintain a fixed rate for the

follow controller of Stream (A) , a sign will be sent to control mechanism . Also for Stream

(B) rate control is made between flows . In case of any flow changes in either Streams

between (A and B) , the controller will sent a sign to control instrument (A , B) for

77

adjustment to maintain a fixed rate between the two Streams as shown in the diagram

below:

Fig. 35 Ratio Control

4.2.5 Flow Control (FC) : Flow control usually made in pumps and storage vessels. In such case an additional vessel

should be used to draw off excess flow. Pressure control is also used in compressors and

pumps working at a fixed speed where by pass control can be used as shown in the

diagram.

Fig. 36 (a) Flow control for a reciprocating pump(b)Alternative Scheme

for a centrifugal compressor or pump

78

CHAPTER 5

COST ESTIMATION

Costing Method:

- The correlation of Timms, IChemE (1988) gives a simple equation for gas phase

processes (updated to 1998).

- C = 14000*N*Q0.615

- Where (C) is the capital cost.

Capital Cost Factors:

- N = number of functional units.

- Q = plant capacity (ton per day).

Procedures:

- Considering year = (335) days due to 30 days of shut down.

- In our project there are 7 functional units.

- Applying the cost calculations.

C = 14000 * 7 * (1000*335) 0.615

C = 244978872.8 US dollar (in 1998)

dollar US4250004080. 2003

.8)(244978872 * 390

3982003

1998Cost * 1998Index Cost

2003Index Cost 2003Cost

Cost

Cost

- To update the plant cost to the next years the rate of inflation for each year (2%)

must be applied.

– Cost 2004 = 250004080.4 * 1.02 = 255004162.1 US Dollars

– Cost 2005 = 255004162.1 * 1.02 = 260104245.3 US Dollars

– Cost 2006 = 260104245.3 * 1.02 = 265306330.2 US Dollars

– Cost 2007 = 265306330.2 * 1.02 = 270612456.8 US Dollars

– Cost 2008 = 270612456.8 * 1.02 = 276024705.9 US Dollars

79

– Cost 2009 = 276024705.9 * 1.02 = 281545200.1 US Dollars

– Cost 2010 = 281545200.1 * 1.02 = 287176104.1 US Dollars

– Cost 2011 = 287176104.1 * 1.02 = 292919626.1 US Dollars

– Cost 2012 = 292919626.1 * 1.02 = 298778018.7 US Dollars

– Cost 2013 = 298778018.7 * 1.02 = 304753579 US Dollars

– Cost 2014 = 304753579 * 1.02 = 310848650.6 US Dollars

– Cost 2015 = 310848650.6 * 1.02 = 317065623.6 US Dollars

– Cost 2016 = 317065623.6 * 1.02 = 323406936.1 US Dollars

– Cost 2017 = 323406936.1 * 1.02 = 329875074.8 US Dollars

Fig.37 Chemical Engineering Plant Cost Index

80

5.1 Estimation of Total Capital Investment :

5.1.1 Direct Cost:

A. Equipment, installation, piping etc.

1. Purchased equipment (30% of fixed capital investment)

= 0.3 x 329875074.8

= 98962522.45 US Dollars.

2.Installation, including insulation and painting (30% of purchased equipment)

= 0.3 x 98962522.45

= 29688756.73 US Dollars.

3.Instrumentation and controls, installed ( 10% of purchased equipment)

= 0.1 x 98962522.45

= 9896252.245 US Dollars.

4.Piping, installed ( 20% of purchased equipment)

= 0.2 x 98962522.45

= 19792504.49 US Dollars.

5.Electrical, installed (15% of purchased equipment)

= 0.15 x 98962522.45

= 14844378.37 US Dollars.

B.Buildings ( 20% of purchased equipment cost )

= 0.2 x 98962522.45

= 19792504.49 US Dollars.

81

C. Service facilities and yard improvements: (60% of purchased equipment)

= 0.6 x 98962522.45

= 59377513.47 US Dollars.

D. Land (5% of purchased equipment)

= 0.05 x 98962522.45

=4948126.122 US Dollars.

Direct cost = 257302558.4US Dollars.

5.1.2 Indirect Cost :

1.Engineering and supervision ( 10% of direct cost)

= 0.1 x 257302558.4

= 25730255.84 US Dollars.

2.Construction expense and contractor’s fee (11% of direct cost)

= 0.11 x 257302558.4

= 28303281.42 US Dollars.

3.Contingency (6% of fixed capital investment)

= 0.06 x 329875074.8

= 19792504.49 US Dollars.

Indirect cost = 73826041.75 US Dollars.

Total capital investment = fixed capital investment + working capital

Let working capital = 15% of total capital investment

Fixed capital investment = 329875074.8 US Dollars.

Total capital investment = 388088323.3 US Dollars.

82

5.2 Estimation of Total Product Cost :

5.2.1 Manufacturing Cost :

A. Fixed charges :

1. Depreciation (10% of fixed capital investment + 2% of building)

= 0.1 x 329875074.8 + 0.02 x 19792504.49

= 33383357.57 US Dollars.

2. Local taxes (2% of fixed capital investment)

= 0.02 x 329875074.8

= 6597501.496 US Dollars.

3. Insurance ( 0.8% of fixed capital investment )

= 0.008 x 329875074.8

= 2639000.599 US Dollars.

Fixed charges = 42619859.67 US Dollars.

Let fixed charge be 20% of total product cost

Total product cost = 42619859.67/0.2

= 213099298.3 US Dollars.

B. Direct production cost:

1.Raw materials (15% of total product cost)

= 0.15 x 213099298.3

= 31964894.75 US Dollars.

83

2.Operating labor ( 11% of total product cost)

= 0.11 x 213099298.3

= 23440922.82 US Dollars.

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

= 0.1 x 23440922.82

= 2344092.282 US Dollars.

4.Utilities (10% of total product cost)

= 0.1 x 213099298.3

= 21309929.83 US Dollars.

5.Maintenance and repairs (5% of fixed capital investment)

= 0.05 x 329875074.8

= 16493753.74 US Dollars.

6.Operating supplies (15% of maintenance and repairs)

= 0.15 x 16493753.74

= 2474063.061 US Dollars.

7. Laboratory charges (15% of operating labor)

= 0.15 x 23440922.82

= 3516138.423 US Dollars.

8. Patents and royalties (3% of total product cost)

= 0.03 x 213099298.3

= 6392978.95 US Dollars.

Direct production cost = 107936773.9

84

C. Plant overhead costs (5% of total product cost)

= 0.05 x 213099298.3

= 10654964.92 US Dollars.

D. Manufacturing cost = Fixed charges + direct production cost + plant overhead cost

= 161211598.4 US Dollars.

5.2.2 General Expenses :

A . Administrative costs (5% of total product cost)

= 0.05 x 213099298.3

= 10654964.92 US Dollars.

B. Distribution and selling costs (11 % of total product cost )

= 0.11 x 213099298.3

= 23440922.82 US Dollars.

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

= 0.05 x 213099298.3

= 10654964.92 US Dollars.

D. Financing (2% of total capital investment)

= 0.02 x 388088323.3

= 7761766.466 US Dollars.

General expenses = 52512619.12 US Dollars.

Total product cost = manufacturing cost + general expenses

= 213724217.6 US Dollars.

85

Cost of the product = (213724217.6) / (335*1000)

= 610.6406216 US Dollars.

With a profit margin of 30% = 1.3 x 610.6406216

= 793.8328081US Dollars.

Gross annual earning =(793.8328081- 610.6406216) x 335 x 1000

(GAE) = 64117265.27 US Dollars.

Net annual earnings = GAE – Income tax

Income tax = 40% of GAE

Net annual earnings = 38470359.16 US Dollars.

earnings annualnet

investment capital total periodPayback

638470359.1

3388088323. periodPayback

= 10.08798285 years.

100 * investment capital total

earnings annualnet return of Rate

100 * 3388088323.

638470359.1return of Rate

= 0.410135078*100

Rate of return = 9.9127845 = 10 %

86

CHAPTER SIX

SAFETY AND LOSS PREVENTION

6.1 Safety Concerns with Ammonia :

There are several safety-related concerns with anhydrous ammonia and with aqueous

solutions of ammonia. It is a respiratory irritant that is a highly hazardous chemical.

Release could take place through a simple industrial or transportation accident, a deliberate

release caused by terrorists, or by improper handling by those using it in the illegal

synthesis of methamphetamines. Storage tanks on farms used for dispensing ammonia as

fertilizer are referred to as ''nurse'' tanks and contain approximately 2,500 pounds (1134 kg)

of anhydrous ammonia, so any farm with four or more nurse tanks needs to assess its

safety. In fact, the U.S. Environmental Protection Agency (U.S. EPA) mandates the

performance of an "Offsite Consequence Analysis" (OCA) as part of their "Risk

Management Plan" (RMP) requirements for any facility that stores more than 10,000

pounds (4,536 kg) of anhydrous liquid ammonia or 20,000 pounds (9,072 kg) of aqueous

solutions of ammonia. The RMP requirements apply for ammonia refrigeration systems or

any other ammonia storage facilities as well as farms. The U.S. Occupational Health &

Safety Administration (OSHA) has mandated very similar requirements as part of their

"Process Safety Management" (PSM) regulations for hazardous chemicals.

In transport, ammonia containers must have proper hazardous material placards and, if the

pertinent threshold quantity is exceeded, may need additional safeguards such as reporting

the shipment to industry monitoring services such as CHEMTREC or additional local

agencies. There may be restrictions on transporting hazardous materials through tunnels, or

possibly streets in high-density areas. The U.S. Department of Homeland Security (DHS),

citing its major concern as toxic release, lists anhydrous ammonia, or mixtures containing

at least 1 percent ammonia, when stored in quantities of 10,000 pounds or more, as a

chemical of interest, which falls under the Risk for Chemical Facility Anti-Terrorism

Standards (CFATS) regulations and guidance. Organizations that store or transport more

87

than the threshold quantity of 10,000 pounds, or believe they are at a higher than normal

risk, should use the Chemical Security Assessment Tool

6.2 Major Hazards :

The following credible major hazards events are identified in an ammonia production plant :

o Fire/explosion hazard due to leaks from the hydrocarbon feed system

o Fire/explosion hazard due to leaks of synthesis gas in the CO removal /synthesis gas

compression areas (75% hydrogen)

o Toxic hazard from the release of liquid ammonia from the synthesis loop

o In ammonia storage the release of liquid ammonia (by sabotage) is a credible

major hazard event. Confined explosions in ammonia plants appear to be limited to

explosions equivalent to a few hundred kg TNT. Such explosions are normally not fatal for

humans at 50-60 m distance, and thus in most cases not severe for people outside the plant

fence The usually not a hazard or only a minor hazard to the local population although

potentially most severe for the plant operators. Appropriate precautions to protect both the

operators and the local population are taken in the design and operation of the plants .The

toxic hazard of a potential large release of liquid ammonia (ie. from a storage tank( may be

much more serious for the local population. An emergency plan for this event cover-ing the

operators and the local population must be maintained.

6.3. Occupation Health & Safety :

The occupational health and safety issues associated with ammonia production and storage

are:

- Fire/explosion-injuries

- Poisoning

- Suffocation

88

Fires or explosions from the involuntary ignition of leaks are credible, especially when

these occur in the feed-gas and synthesis gas systems (hydrocarbons, hydrogen). The

most important toxic components are CO from potential leaks in the synthesis gas

generation and shift areas and NH3 from leaks in the ammonia synthesis and ammonia

handling areas. In partial oxidation plants H2S and SO2 are present in the sulphur

removal/recovery sections. Traces of carbonyls (iron and nickel) may form during

operation. Suffocation due to lack of oxygen may occur at points where the breathing air

has been diluted with inert gases. In ammonia plants CO2 and N2 are potentially

suffocating gases. ACGIH occupational exposure limits for ammonia and other

components associated with ammonia production are given in the table below. All the

figures are ppmv:

Tab.50

Component TLV-TWA (8hr) TLV-STEL (15min)

NH3 25 35

NO2 3 5

SO2 2 5

H2 S 10 15

CO 50 400

CO2 5,000 30,000

The figures are subject to updating and may vary between European countries. Full

health and safety information is given in Safety Data Sheets which must be available and

updated. General product information on ammonia is given in Appendix D. Ammonia

plants have high technological standards and need professional management, operating

and maintenance working routines and personnel. Precautions taken to prevent accidents

and injuries during operation are incorporated in the operating and safety procedures for

the plant.

89

6.4 Ammonia Solution :

MSDS Number: A5920

Ammonium hudroxide , Less Than i0% NH3

6.4.1 Product Identification :

Synonyms: Ammonium hydroxide solutions; ammonia aqueous ammonia solution

CAS No.: 1336-21-6

Molecular Weight : 35.05

Chemical Formula: NH4OH in H2

6.4.2 Hazards Identification :

Emergency Overview:

Poison ! dangerl corrosive . may by fatal if swallowed or inhaled , mist and vapor cause .

burns to every area of contact

Potential Health Effects:

The following hazards are for concentrated solutions Hazards of less concentrated solutions

may be reduced. Degree of hazard for reduced concentrations is not currently addressed in

the available literature

Inhalation:

Vapors and mists cause irritation to the respiratory tract. Higher concentrations can cause

burns, pulmonary cdcma and death Brief exposure to 5000 ppm can be fatal

Ingestion:

Toxic May cause corrosion to the esophagus and stomach with perforation and peritonitis.

Symptoms may include pain in the mouth. chest, and abdomen, with coughing, vomiting

and collapse. Ingestion of as little as 3-4 mL may be fatal

90

Skin Contact:

Causes irritation and burns to the skin

Eye Contact:

Vapors cause irritation. Splashes cause severe pain, eye damage, and Permanent blindness.

Chronic Exposure:

Repeated exposure may cause damage to the tissues of the mucous membranes, upper

respiratory tract, eyes and skin

Aggravation of Pre-existing Conditions:

Persons with pre-existing eye disorders or impaired respiratory Function may be more

susceptible to

6.4.3 First Aid Measures:

Inhalation:

Remove to fresh air. If not breathing, give artificial respiration. If Breathing is difficult,

give oxygen. Call a physician immediately

Ingestion:

If swallowed, do not induce vomiting . Give large quantities of water. Never give anything

by mouth to an unconscious person. Get Medical attention immediately.

Skin Contact:

Immediately flush skin with plenty of water for at least 15 minutes while removing

contaminated clothing and shoes. Call a physician Immediately. Wash clothing before reuse

91

Eye Contact:

Immediately flush eyes with gentle but large stream of water for at Least l5 minutes, lifting

lower and upper eyelids occasionally. Call a Physician immediately. Immediate action is

critical to minimize Possibility of blindness.

6.4.4 Fire Fighting Measures:

Fire:

Auto ignition temperature: 651C )1204F)Flammable limits in air by volume: lel: l6; uel:

25

Explosion:

Flammable vapors may accumulate in confined spaces

Fire Extinguishing Media:

Use any means suitable for extinguishing surrounding fire. Use water Spray to blanket fire,

cool fire exposed containers, and to flush non Ignited spills or vapors away from fire.

Special Information:

In the event of a fire, wear full protective clothing and NIOSH- approved self-contained

breathing apparatus with full face piece operated in the pressure demand or other positive

pressure mode

6.4.5 Accidental Release Measures :

Ventilate area of leak or spill. Keep unnecessary and unprotected people away from area of

spill. Wear appropriate personal protective equipment as specified in Section 8. Contain

and recover liquid when possible. Do not flush caustic residues to the sewer. Residues from

spills can be diluted with water, neutralized with dilute acid such as acetic, hydrochloric or

sulfuric. Absorb neutralized caustic residue on

clay, vermiculite or other inert substance and package in a suitable container for disposal

US Regulations (CERCLA) require reporting spills and releases to soil, water and air in

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excess of reportable quantities' The toll free number for the US Coast Guard National

Response Center is (800) 424-880

6.4.6 Handling and Storage :

Keep in a tightly closed container, stored in a cool, dry, ventilated area.Protect against

physical damage. Separate from incompatibilities. Store below 25C. Protect from direct

sunlight. Containers of this material may be hazardous when empty since they retain

product residues vapors, liquid;( observe all warnings and precautions) listed for

theproduct.

6.4.7 Exposure Controls/Personal Protection

Airborne Exposure Limits:

50 ppm (NH3) OSLIA: :(- Permissible Exposure Limit (PEL

ACGIH): 25 ppm (Ntl3) (TwA) 35 ppm (STEL Threshold Limit Value (TLV

Ventilation System:

A system of local and/or general exhaust is recommended to keep employee exposures

below the Airborne Exposure Limits. Local exhaust ventilation is generally preferred

because it can control the emissions of the contaminant at its source, preventing dispersion

of it into the general rvork area. Please refer to the ACCIH document,industrial Ventilation,

A htunual of Recon mendecl Predclice\, nost recent edition. for details.

Person al Respirators (NIOStl Approved):

If the exposure limit is exceeded an engineering controls are not feasible a full face piece

respirator with an ammonia/methylamine cartridge may be worn up to 50 times the

exposure limit or the maximum use concentration specified by the appropriate regulatory

agency or respirator supplier, whichever is lowest. For emergencies or instances where the

exposure levels are not known, use a full-face piece positive-pressure, air-supplied

respirator.

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WARNING: Air purifying respirators do not protect workers in oxygen-deficient

atmospheres

Skin Protection:

Wear impervious protective clothing, including boots, gloves, lab coat,apron or coveralls,

as appropriate, to prevent skin contact. Neoprene and nitrile rubber are recommended

materials. Polyvinyl alcohol is not recommended

Eye Protection:

Use chemical safety goggles and/or full face shield where dusting or splashing of solutions

is possible. Maintain eye wash fountain and quick-drench facilities in work area

6.4.8 Stability and Reactivity:

Stability:

Stable under ordinary conditions of use and storage Hazardous Decomposition Products

Burning may produce ammonia, nitrogen oxides.

Hazardous Polymerization:

Will not occur

Incompatibilities:

Acids, acrolein, dimethyl sulfate, halogens, silver nitrate, propylene oxide, nitromethane,

silver oxide, silver permanganate, oleum, beta propiolactone. Most common metals

Conditions to Avoid:

Heat, sunlight, incompatibles, sources of ignition

94

6.4.9 Toxicological Information

For ammonium hydroxide:Oral rat LD50: 350 mg/kg; eye, rabbit, standard Dtaize,25l ug;

severe investigated as a mutagen For ammonia inhalation rat LC50: 2000 ppm/4-hr;

investigated as a tumorigen mutagen

6.4.10 Ecological Information:

Environmental fate:

This material is not expected to significantly bioaccumulate

Environmental Toxicity:

This material is expected to be very toxic to aquatic life The LC50l96-hour values for fish

are less than I mg/I. The EC50/48-hour values for daphnia are less than I mg/l

6.4.11 Disposal Considerations :

Whatever cannot be saved for recovery or recycling should be managed in an appropriate

and approved waste facility Although not a listed RCRA hazardous waste' this material

may exhibit one or more characteristics of a hazardous waste and require appropriate

analysis to determine specific disposal requirements.

Processing' use or contamination of this product may change The waste management

options. State and local disposal regulations may differ from federal

disposal regulations. Dispose of container and unused contents in accordance with federal,

state and local requirements.

6.4.12 Other Information:

NFPA Ratings: Health: 3 Flammability: I Reactivity.

Label Hazard Warning:

Poisoni dangerl corrosive . may be fatal if swallowed or inhaled . mist and vapor cause

burns to every area of contactPOISONI

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Label Precautions:

do not get in eyes, on skin, or on clothing.do not breathe vapor or mist Keep container

closed. Use only with adequate ventilation. Wash thoroughly after handling

Label First Aid:

If swallowed , do not induce vomiting . Give large quantities of water. Never give anything

by mouth to an unconscious person. If inhaled, remove to fresh air. If not breathing, give

artificial respiration. If breathing is difficult, give oxygen. In case of contact, immediately

flush eyes or skin with plenty of water for at least 15 minutes while removing contaminated

clothing and shoes. Wash clothing before reuse.immediate action is essential for eye

exposures . In all cases call a physician immediatel

Product Use:

Laboratory Reagent.

96

CHAPTER 7

PLANT SITE AND LAYOUT

7.1 Introduction :

A suitable site must be found for a new project, and the site and equipment layout planned.

Provision must be made for the ancillary buildings and services needed for plant operation,

and for the environmentally acceptable disposal of effluent. These subjects are discussed

briefly in this chapter.

7.2 Plant Location and Site Selection:

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,

and only a brief review of the principal factors will be given in this section.

The principal factors to consider are:

1.Location, with respect to the marketing area

2.Raw material supply

3.Transport facilities

4.Availability of labor

5.Availability of utilities: water, fuel, power

6.Availability of suitable land

7.Environmental impact, including effluent disposal

8.Local community considerations

9.Climate

10.Political and strategic considerations.

7.2.1 Marketing Area :

For materials that are produced in bulk quantities, such as cement, mineral acids, and

fertilizers, where the cost of the product per metric ton is relatively low and the cost of

transport is a significant fraction of the sales price, the plant should be located close to the

primary market. This consideration is much less important for low-volume production and

high-priced products, such as pharmaceuticals.

97

7.2.2 Raw Materials :

The availability and price of suitable raw materials will often determine the site location.

Plants that produce bulk chemicals are best located close to the source of the major raw

material, as long as the costs of shipping product are not greater than the cost of shipping

feed. For example, at the time of writing much of the new ethylene capacity that is being

added worldwide is being built in the Middle East, close to supplies of cheap ethane from

natural gas. Oil refineries, on the other hand, tend to be located close to major population

centers, as an oil refinery produces many grades of fuel, which are expensive to ship

separately.

7.2.3 Transport :

The transport of materials and products to and from the plant can be an overriding

consideration in site selection. If practicable, a site should be selected that is close to at

least two major forms of transport: road, rail, waterway (canal or river), or a sea port. Road

transport is increasingly used and is suitable for local distribution from a central warehouse.

Rail transport is usually cheaper for the long-distance transport of bulk chemicals. Air

transport is convenient and efficient for the movement of personnel and essential

equipment and supplies, and the proximity of the site to a major airport should be

considered.

7.2.4 Availability of Labor :

Labor will be needed for construction of the plant and its operation. Skilled construction

workers are usually brought in from outside the site area, but there should be an adequate

pool of unskilled labor available locally, and labor suitable for training to operate the plant.

Skilled craft workers such as electricians, welders, and pipe fitters will be needed for plant

maintenance. Local labor laws, trade union customs, and restrictive practices must be

considered when assessing the availability and suitability of the local labor for recruitment

and training.

98

7.2.5 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

.At some sites, the cooling water required can be taken from a river or lake, or from the sea;

at other locations cooling towers will be needed. Electrical power is needed at all sites.

Electrochemical processes (for example, chlorine manufacture or aluminum smelting)

require large quantities of power and must be located close to a cheap source of power.A

competitively priced fuel must be available on site for steam and powergeneration.

7.2.6 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. 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. An environmental impact

assessment should be made for each new project or major modification or addition to an

existing process.

7.2.7 Local Community Considerations :

The proposed plant must fit in with and be acceptable to the local community. Full

consideration must be given to the safe location of the plant so that it does not impose a

significant additional risk to the local population. Plants should generally be sited so as not

to be upwind of residential areas under the prevailing wind. On a new site, the local

community must be able to provide adequate facilities for the plant personnel: schools,

banks, housing, and recreational and cultural facilities. The local community must also be

consulted about plant water consumption and discharge and the effect of the plant on local

traffic. Some communities welcome new plant construction as a source of new jobs and

economic prosperity. More affluent communities generally do less to encourage the

building of new

manufacturing plants and in some cases may actively discourage chemical plant

construction.

99

7.2.8 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.

A full site evaluation should be made to determine the need for piling or other special

foundations. Particular care must be taken when building plants on reclaimed land near the

ocean in earthquake zones because of the poor seismic character of such land.

7.2.9 Climate :

Adverse climatic conditions at a site will increase costs. Abnormally low temperatures

require the provision of additional insulation and special heating for equipment and pipe

runs. Stronger structures are needed at locations subject to high winds (cyclone/ hurricane

areas) or earthquakes.

7.2.10 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. In a

globalized economy, there may be an advantage to be gained by locating the plant within an

area with preferential tariff agreements, such as the European Union (EU).

7.3 Site Layout :

The process units and ancillary buildings should be laid out to give the most economical

flow of materials and personnel around the site. Hazardous processes must be located at a

safe distance from other buildings. Consideration must also be given to the future

expansion of the site. The ancillary buildings and services required on a site, in addition to

the main processing units (buildings), as shown in the figure (47), which include:

Storage for raw materials and products: tank farms and warehouses;

1. Maintenance workshops;

2. Stores, for maintenance and operating supplies;

3. Laboratories for process quality control;

4. Fire stations and other emergency services;

100

5. Utilities:steam boilers, compressed air, power generation, refrigeration, transformer

stations;

6. Effluent disposal plant: waste water treatment, solid and or liquid waste Collection;

7. Offices for general administration;

8. Canteens and other amenity buildings, such as medical centers;

9. Parking lots.

When the preliminary site layout is roughed out, the process units are normally sited first

and arranged to give a smooth flow of materials through the various processing steps, from

raw material to final product storage. Process units are normally spaced at least 30m apart;

greater spacing may be needed for hazardous processes. The location of the principal

ancillary buildings should then be decided. They should be arranged so as to minimize the

time spent by personnel in traveling between buildings. Administration offices and

laboratories, in which a relatively large number of people will be working, should be

located well away from potentially hazardous processes. Control rooms are normally

located adjacent to the processing units, but those with potentially hazardous processes may

have to be sited at a safer distance. The siting of the main process units determines the

layout

of the plant roads, pipe alleys, and drains. Access roads to each building are needed for

construction and for operation and maintenance. Utility buildings should be sited to give

the most economical run of pipes to and from the process units. Cooling towers should be

sited so that, under the prevailing wind, the plume of condensate spray drifts away from the

plant area and adjacent properties. The main storage areas should be placed between the

loading and unloading facilities and the process units they serve. Storage tanks containing

hazardous materials should be sited at least 70m (200 ft) from the site boundary.

7.4 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 to be considered are:

101

1. Economic considerations: construction and operating costs;

2. The process requirements;

3. Convenience of operation;

4. Convenience of maintenance;

5. Safety;

6. Future expansion;

7. Modular construction.

7.4.1 Costs :

The cost of construction can be minimized by adopting a layout that gives the shortest run

of connecting pipe between equipment and the least amount of structural steel work;

however, this will not necessarily be the best arrangement for operation and maintenance.

7.4.2 Process Requirements :

An example of the need to take into account process considerations is the need to elevate

the base of columns to provide the necessary net positive suction head to a pump or the

operating head for a thermo siphon reboiler.

7.4.3 Operation :

Equipment that needs to have frequent operator attention should be located convenient to

the control room. Valves, sample points, and instruments should be located at convenient

positions and heights. Sufficient working space and headroom must be provided to allow

easy access to equipment. If it is anticipated that equipment will need replacement, then

sufficient space must be allowed to permit access for lifting equipment.

7.4.4 Maintenance :

Heat exchangers need to be sited so that the tube bundles can be easily withdrawn for

cleaning and tube replacement. Vessels that require frequent replacement of catalyst or

packing should be located on the outside of buildings. Equipment that requires dismantling

for maintenance, such as compressors and large pumps, should be place dunder cover.

102

7.4.5 Safety :

Blast walls may be needed to isolate potentially hazardous equipment and confine the

effects of an explosion.At least two escape routes for operators must be provided from each

level in process buildings.

7.4.6 Plant Expansion :

Equipment should be located so that it can be conveniently tied in with any future

expansion of the process. Space should be left on pipe racks for future needs, and service

pipes should be oversized to allow for future requirements.

7.4.7 Modular Construction :

In recent years there has been a move to assemble sections of a plant at the plant

manufacturer’s site. These modules include the equipment, structural steel, piping, and

instrumentation. The modules are then transported to the plant site, by road or sea.

The advantages of modular construction are:

1. Improved quality control;

2. Reduced construction cost;

3. Less need for skilled labor on site;

4. Less need for skilled personnel on overseas sites.

Some of the disadvantages are:

1. Higher design costs.

2. More structural steel work.

3. More flanged connections.

103

Fig 36 Site Layout

104

REFERENCES

[1] Balzhiser, R.E., Samuels, M.R., and Eliassen, J. D., Chemical Engineering

Thermodynamics the Study of Energy, Entropy, and Equilibrium, Prentice- Hall, Inc,

Englewood cliffs, New Jersey, 1972.

[2] Himmelblau, D. M., Basic Principles and Calculations in Chemical Engineering, 6th

Edition, Prentice-Hall, Inc., 1996.

[3] Kirk, R. E. Othmer, D. F. Encyclopaedia of Chemical Technology, John Wiley and

Sons, Inc., New York, Chichester, Brisbane, Toronto, and Singapore, 1985.

[4] Kirk, R. E. Othmer, D. F. Encyclopedia of Chemical Technology, Third Edition Vol.

15, John Wiley and Sons, Inc., new York, Chichester, Brisbane, Toronto, 1981.

[5] Kirk, R. E. Othmer, D. F. Encyclopedia of Chemical Technology, Third Edition Vol.

18, John Wiley and Sons, Inc., new York, Chichester, Brisbane, Toronto, 1982.

[6] Levenspiel, O., Chemical Reaction Engineering, Third Edition, John Wiley and Sons,

Inc., New York, Chichester, Weinheim, Brisbane, Singapore, and Toronto, 1999.

[7] Nauman, E. B., Chemical Reactor Design, John Wiley and Sons, Inc., New York,

London, and Singapore, 1987.

[8] Peters, M. S. and Timmerhaus, K. D., Plant Design and Economics for Chemical

Engineers, Third Edition, McGraw-Hill, Inc., Auckland, 1981.

[9] Smith, J. M., Van Ness, H. C., and Abbott, M. .M., Introduction to Chemical

Engineering Thermodynamics, Sixth Edition, McGraw- Hill, New York, 2001.

105

[10] Strelzoff, S., Technology and Manufacture of Ammonia, John Wiley and Sons, Inc.,

Now York, Chichester, Brisbane, and Toronto, 1981.

[11] Weast, R. C., Handbook of Chemistry and Physics, 55th

Edition, CRC Press, Inc.,

Cleveland and Ohio, 1974-1975.

[12] Weissermel, K. and Arpe, H, Industrial Organic Chemistry, Verlag Chemie,

Weinheim, New York, 1978.

[13] Perry's Chemical Engineers' Handbook - 7th Edition.

[14] Sirte oil company, "Ammonia Plant, Plant Training Office".

[15] Lifco company, "Ammonia Plant Description".

106

APPENDICES

Appendix A

Reference 4 Appendix D Page ( 765 )

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