Nitric Acid Project_Final

Production of Nitric Acid for Fertilizer Applications

Prepared by: Mamdouh Al-anazi Mohammed Al-Ajlan Sultan Al-Sebeai Hussni Mandeeli

Supervised by: Dr. Farag Abd El-Salam Dr. Malik Al-Ahmad Prof. Tariq Al-Fariss

Submitted in partial Fulfillment of the Requirement for the Degree of Bachelor ofScience in chemical Engineering in the college of Engineering

Riyadh

1434 – 1435 H 2013 – 2014 G

Table of Contents Contents Acknowledgement Summery Problem Statement Objectives Scope of Work Chapter 1: Introduction 1.1 Backgrounds and History 1.2 The Nitric Acid Production Processes 1.2.1 The single Pressure Process 1.2.2 The Dual Pressure Process 1.2.3 The Process Selection 1.2.4 The Process Selection Conclusion 1.3 Main Uses of Nitric Acid 1.4 Physical Properties 1.5 Chemical Properties 1.6 Safety Properties 1.7 Market Survey 1.7.1 Economic Outlook 1.8 Preliminary Hazard Analysis 1.8.1 Summary of Previous Plant Accident 1.8.2 Inherent Safety Aspects 1.8.3 Local Safety and Environmental Regulations 1.9 Site Feasibility Study 1.9.1 Selection Criteria 1.9.2 Potential Site location Chapter 2: Material Balance 2.1 Process Description 2.1.1 Process Flow Sheet 2.1.2 Justification of Equipment Selection 2.2 Overall Material Balance 2.3 Reactor Material Balance 2.4 Oxidation Material Balance 2.5 Absorber Material Balance Chapter 3: Energy Balance 3.1 Vaporizer Energy Balance 3.2 Compressor Energy Balance 3.3 Superheater Energy Balance 3.4 Mixer Energy Balance

Page 4 5 6 6 6 7 8 9 9 11 13 13 13 14 15 15 16 16 18 18 20 22 22 22 24 25 27 28 29 30 31 32 33 34 35 36 37 38 2

3.5 Reactor Energy Balance 3.6 1st Cooler Energy Balance 3.7 Oxidation Energy Balance 3.8 2nd Cooler Energy Balance 3.9 Absorber Energy Balance Chapter 4: Design and Sizing Equipment 4.1 Sizing of Pump 4.2 Design of Vaporizer 4.3 Sizing of Compressor 4.4 Design of Superheater 4.5 Sizing of Mixer 4.6 Design of Reactor 4.7 Design of 1st Cooler 4.8 Design of Oxidation 4.9 Design of 2nd Cooler 4.10 Design of Absorber 4.11 Design of Nitric Acid Tank Chapter 5: Control Loop 5.1 Introduction 5.2 Control of Vaporizer 5.3 Control of Reactor 5.4 Control of Absorber Chapter 6: Economic Analysis 6.1 Estimation of Capital Investment Items Based on Delivered Equipment 6.2 Equipment Cost (at 2011) 6.3 Total Production Cost (TPC) Chapter 7: Process Integration Chapter 8: Safety and Loss Prevention 8.1 Plant Layout 8.2 Safety of Materials 8.3 Hazard and Operability Studies Chapter 9: Waste Treatment Conclusion Reference Appendices A: Material Safety Data Appendices B: Physical Properties Data Appendices C: Detailed Material Balance Calculation Appendices D: Detailed Energy Balance Calculation Appendices E: Equipment Design References Appendices F: Computer Simulation by CHEMCAD

39 40 41 42 43 44 45 46 56 57 67 68 71 82 83 93 101 105 106 109 110 111 112 113 114 115 117 120 122 123 126 133 140 142 143 152 162 167 181 186

Acknowledgement First, thanks are to Allah who helped us throughout this work. Then, it is our pleasure to thank those people who helped us in the completion of this project. Also, we would like to express our deep gratitude and apprec ation to our dear supervisors: Dr.Farag Abd El-Salam, Prof. Tariq Al-Fariss and Dr. Malik Al-Ahmad for their support, advise, valuable scientific knowledge and patience throughout this project. We would like to express our sincere thanks to the department of chemical engineering at the College of Engineering, King Saud University, for giving us this opportunity to carry out this project and we are grateful to our colleagues at King Saud University for their inspiration, help and encouragement without which we would not have been able to complete this project too. Last but not least we thank our dear families and relatives for their assistance and encouragement.

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Summary The results of the design project for the commercial production of nitric acid are presented. The project has been performed in two stages. The first part covers a literature review on various processes for nitric acid production, selection of the suit table processes, materials and energy balance calculations as well as the feasibility of the project. The second part presents the detailed equipment designs. From the investigation into project feasibility, it is proposed to construct a plant that will deliver 100000 tons per year of 60%(wt.) Nitric acid. This capacity is based on 8000 hours of operation per year, i.e. 330 days. It is envisaged that this nitric acid production facility will be centered within a larger chemical complex to be located in the eastern region of Saudi Arabia. Other plants on this site will include an ammonia plant and an ammonium nitrate plant. Approximately 70% of the product acid will be consumed in situ for the production of ammonium nitrate Fertilizer. The remaining acid will be available to exploit the neighboring export market. The process chosen for the nitric acid plant is the "single-pressure process" based on the technology developed by C & I Girdler. The most important use is undoubtedly in the production of ammonium nitrate for the fertilizer and explosives industries, which accounts for approximately 65% of the world production of nitric acid. So we made all our production to be used in the fertilizer.

Problem Statement It is required to design a suitable process for the production of 100,000-ton/year of nitric acid from ammonia (60% concentration).

Objectives The objectives of this design project include the following:  To integrate chemical engineering knowledge in a detailed design of chemical plants.  To design a nitric acid plant which is economically attractive, safe to workers and society and reduce harm to the environment?  To develop oneself in the applications of all the elements of knowledge and skills that have been accumulated throughout the undergraduate program for solving design related problems for typical process industrial plant.  To develop the skills for working as a team and to nurture leadership qualities.

Scope of work

This project is subjected to the designing phase of the process plant. All researches and literatures used for this project falls under the scope of the chemical compositions, current productions of nitric acid, preliminary hazards analysis, process design configurations and selections, and configuration of plant equipment (i.e. reactor, separation system, heat integration, etc.). The designing phase is then executed using manual engineering calculations .The project is aimed at achieving the objective of the plant design process.

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Chapter 1

Introduction

The initial design problem is to determine whether: ‘it is both economically and technically feasible to establish a facility to produce nitric acid in the world’. This is a diverse and complex undertaking that necessitates a full investigation into the uses, properties, market, process technology, and production economics, associated with this particular chemical. Having considered these aspects and several others, an appropriate plant to fulfill the assessed market requirements is sized and specified accordingly. [1], [10]

1.1 Backgrounds and History Nitric acid is a colorless, highly corrosive liquid and a very powerful oxidizing agent that in the highly pure state is not entirely stable and must be prepared from its azeotrope by distillation with concentrated sulfuric acid. Nitric acid gradually yellows because of decomposition to nitrogen dioxide. Solutions containing more than 80% nitric acid are called fuming nitric acids. [1], [4] Reagent-grade nitric acid is a water solution containing about 68% by weight nitric acid. This strength corresponds to the constant-boiling mixture of the acid with water, which is 68.4% by weight nitric acid and boils at 121.9°C. Nitric acid is completely miscible with water and forms a monohydrate (HNO3.H2O, melting point: - 38°C) and a dehydrate (HNO3.2H2O, melting point: -18.5°C). [7] Scholars have known nitric acid for many centuries. Probably the earliest description of its synthesis occurs in the writings of the Arabic alchemist Abu Musa Jabir Ibn Hayyan (c. 721–c. 815), better known by his Latinized name of Geber. The compound was widely used by the alchemists, although they knew nothing of its chemical composition. It was not until the middle of the seventeenth century that an improved method for making nitric acid was invented by German chemist Johann Rudolf Glauber (1604–1670). Glauber produced the acid by adding concentrated sulfuric acid (H2SO4) to saltpeter (potassium nitrate; KNO3). A similar method is still used for the laboratory preparation of nitric acid, although it has little or no commercial or industrial value. [7]

The chemical nature and composition of nitric acid were first determined in 1784 by the English chemist and physicist Henry Cavendish (1731–1810). Cavendish 8

applied an electric spark to moist air and found that a new compound - nitric acid was formed. Cavendish was later able to determine the acid’s chemical and physical properties and its chemical composition. The method of preparation most commonly used for nitric acid today was one invented in 1901 by the Russian born German chemist Friedrich Wilhelm Ostwald (1853–1932). The Ostwald process involves the oxidation of ammonia over a catalyst of platinum or a platinum-rhodium mixture. [2]

Today, nitric acid is one of the most important chemical compounds used in industry. It ranks number thirteen among all chemicals produced in the United States each year. In 2005, about 6.7 million metric tons (7.4 million short tons) of the compound were produced in the United States. [1]

1.2 The Nitric Acid Production Processes: All commercially produced nitric acid is now prepared by the oxidation of ammonia. The requirement for a nitric acid product of 60%(wt.) Immediately restricts the choice of a recommended production process. Only two processes are possible, both highly efficient, each offering distinct advantages under different market conditions. [1]

These two main Processes are:

1.2.1 The Single-Pressure Process The Single-Pressure process was developed to take full advantage of operating pressure in enabling equipment sizes to be reduced throughout the process. A single compression step is used to raise the pressure through the entire process sufficiently to favor absorption. Operating pressures range from 800 kPa as used by the Sumitomo Chemical Company Ltd. to 1100 kPa as used in the C&l Girdler single-pressure process. Increased ammonia oxidation and complete ammonia/air mixing, and uniform flow distribution can minimize increased consumption of ammonia due to the higher-pressure operation across the gauze "inside the reactor". The higher oxidation

temperature results in an increased consumption of platinum and rhodium and the need to rework the gauze every five to seven weeks .The higher temperature and the favorable pressure effect make possible a greater recovery of energy from the process. [10] The process begins with the vaporization of ammonia at 1240 kPa and 35°C using process heat "as shown in the given flow sheet, Fig. (1). Steam is then used to superheat the ammonia to 180°C, filtered air is compressed by an axial compressor to an interstage level and then, following cooling, by a centrifugal compressor to a discharge pressure of 1090 kPa. A portion of the air is diverted for acid bleaching; the remainder is circulated through a jacket surrounding the tail-gas preheater and then used for ammonia oxidation. [10] In this process; the heated air and the ammonia vapor (10.3% ammonia by volume) are then mixed and passed through the platinum/ rhodium gauze reactor where the heat of reaction (producing nitric oxide) raises the temperature to be between 927°C and 937°C. The reaction gas flows through a series of heat exchangers in which energy is recovered either as high-pressure superheated steam or as shaft horsepower from the expansion of hot tail gas. [6] Approximately 70% of the oxidation to nitrogen dioxide occurs as the gas passes through the energy recovery train and is cooled to 185°C. After further cooling to 63°C in the primary cooler/condenser, separation of approximately one third acid product as 42% strength nitric acid is achieved. The remaining gas reaches a 43% oxidation conversion to nitrogen dioxide, with approximately 20% dimerization. The gas is combined with bleached air containing additional nitrogen peroxide; it then passes through an empty oxidation vessel and the secondary cooler. [6] In cooling to 66”C, the gas provides heat to a recirculating hot water system used for vaporizing the ammonia. The gas entering the absorber is 95% oxidized to nitrogen peroxide. In the absorber deionized water is added to the top tray, and weak acid from the low-pressure condenser is added to a tray corresponding to its strength. [6], [5] Down-flowing acid and up-flowing acid alternately mix as the chemical steps of action formation and nitric oxide oxidation take place with the liberation of heat. There are three operational zones in the absorber, these are the lower zone cooled with plant cooling water, the middle zone cooled with chilled water, and the upper zone which is essentially adiabatic. High efficiency of heat removal in the middle and 10

lower zones is particularly important due to its effect on the oxidation and dimerization reactions. [6] For this process, chilled water at 7°C is used and the tail-gas exit temperature is approximately 10°C. Acid from the bottom of the absorber is bleached at 1010 kPa with partially cooled compressed air. The bleach air, containing nitrogen peroxide stripped from the acid, is then added to the main gas stream before entering the oxidation vessel. The cold gas is warmed by heat exchange with the hot compressed bleached air, and then heated to the expander inlet temperature of 620°C by two exchangers in the recovery train. The expander recovers 80% of the required compressor power. Expanded tail gas at 300°C flows through an economizer, providing heat to high-pressure boiler feed water and to low-pressure de ionized deaerator make-up water. Subsequently tail gas is exhausted to the atmosphere at 66°C and less than 1000 ppm of nitrogen oxides. [9] The chilled water (7°C) for absorption refrigeration unit, using heat, supplies the absorber recovered from the compressor and intercooler as the energy source. Heat for ammonia vaporization, as previously noted, is available at 35°C and is recovered from the secondary gas cooler. The system uses circulating condensate as the energy transfer medium. [7]

1.2.2 The Dual-Pressure Process The dual-pressure process was developed to take advantage of two factors: a) Low-pressure ammonia oxidation; b) High-pressure absorption for acid production. In addition to the higher conversion, the lower catalyst gauze temperature (associated with the low-pressure ammonia oxidation) results in a much lower rate of platinum deterioration. Both advantages are maximized at the lowest pressure. In contrast, however, absorption is best performed at the highest pressure. [7] The low-pressure ammonia oxidation is usually performed in the pressure range of 101.3 kPa to 344 kPa. High-pressure absorption is usually performed in the operating range of 800 kPa up to 1010 kPa. This process begins with the vaporization of ammonia at 550 kPa and 7”C "as shown in its flow sheet of Fig. (2)" followed by superheating to 76°C using heat from the compressed bleached air. Filtered air is

compressed in an axial compressor to 350 kPa and is mixed with the superheated ammonia vapor (1 O-l 1% ammonia by volume) prior to entering the converter/reactor. In the converter, the gases react over the platinum/rhodium gauze catalyst. [7] The gases leaving the reactor at 330 kPa and 865°C flow through a series of heat exchangers for recovery of energy, either as high-pressure superheated steam or shaft horsepower from expansion of hot tail gas. Approximately 40% of the oxidation to nitrogen dioxide occurs in the gas as it passes through the energy recovery train and is cooled to 135°C (exit from the tail-gas preheater). After further cooling to 45°C in the medium-pressure condenser, and separation of 20% of the acid product as 30% strength nitric acid, the gas reaches 50% oxidation to nitrogen peroxide with approximately 20% dimerization. [7] The gas is combined with bleach air containing additional nitrogen peroxide and is compressed in a centrifugal nitrous-gas compressor to 1025 kPa. The exit temperature of 224°C is achieved due to the combined heat effects of the compression, the further oxidation to 80% nitrogen peroxide, and the virtual disappearance of the dimer. The compressed gas flows through an empty oxidation chamber, a high-pressure boiler feed water economizer, and a low-pressure deionized water economizer, and thus is cooled to 95°C. Residence time in the system and the effect of increased pressure result in at least 95% oxidation to nitrogen peroxide, but the dimerization is low due to the temperature level. [10] The gas is then cooled to the dew point (50°C) for entry into the absorber. The dimerization increases to 48%, adding significantly to the heat removed prior to the absorber. The system uses circulating condensate as the energy transfer medium. The absorber is essentially the same as that previously described for the single-pressure process. [4] Chilled water at 15°C is used in the absorber and the outlet gas temperature is 18°C. Refrigeration for the chilled water is provided by the ammonia vaporizer which operates at 7°C. [6] Weak acid from the bottom of the absorber is let down to 330 kPa for bleaching with air from the axial compressor. This air, with nitrogen peroxide stripped from the acid, flows to the suction of the nitrous-gas compressor together with the main nitrous gas stream from the condenser. [5]

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1.2.3 Process Selection (Factors Favoring the Single-Pressure Process) The single-pressure process uses a higher ammonia conversion. This higher pressure provides advantages in terms of equipment design, e.g. smaller converter dimensions and a single heat-exchanger-train layout. The higher temperature and the favorable pressure both increase the energy recovery from the process. The singlepressure process provides an extra 10% high-level recoverable heat energy. Plant capital costs in the USA have been estimated at 8 million (USS5.1 million) for the single-pressure process and 9.1 million for the dual-pressure process. The 1.1 million higher cost of the dual-pressure process is accounted for by the larger vessels required at lower operating pressures. Estimates made for the two plants in European locations show a differential of 0.8 million, also in favour of the singlepressure process A discounted cash flow (DCF) analysis based on these US figures was performed by matching the capital cost advantage of the single pressure process against the production cost advantage of the dual pressure process. They indicate that it would take 21 years for the lower operating cost of the dual-pressure process. In this project, pressure process to finally overcome its initial capital cost disadvantage due to its smaller capital cost compared to the dual pressure process. [1] , [9] , [10]

1.2.4 Process Selection Conclusions

The single-pressure process appears to be preferred for our project. The capital cost advantage of this process surpasses the benefits of the superior operating cost structure of the dual-pressure process.

1.3. Main Uses of Nitric Acid Nitric acid is predominantly used for fertilizer manufacture. It also finds use in the manufacture of adipic acid, nitroglycerin, nitrocellulose, ammonium picrate, trinitrotoluene, nitrobenzene, silver nitrate, and various isocyanates. Nitric acid has enormously diverse applications in the chemical industry. It has commercial uses as a nitrating agent, oxidizing agent, solvent, activating agent, and

catalyst and hydrolyzing agent. The most important use is undoubtedly in the production of ammonium nitrate for the fertilizer, which accounts for approximately 65% of the world production of nitric acid. [5] Nitric acid has a number of other industrial applications. It is used for pickling stainless steels, steel refining, and in the manufacture of dyes, plastics and synthetic fibres. Most of the methods used for the recovery of uranium. Such as ion exchange and solvent extraction, use nitric acid [7]. An important point is that for most uses concerned with chemical production, the acid must be concentrated above its azeotropic point to greater than 95%(wt). Conversely, the commercial manufacture of ammonium nitrate uses nitric acid below its azeotropic point in the range 50-65%(wt). If the stronger chemical grade is to be produced, additional process equipment appropriate to super-azeotropic distillation is required. [2] , [7] , [10]

1.4. Physical Properties In its commonest form nitric acid is a pungent, colorless liquid and pure (anhydrous) that boils at 86°C and solidifies at -42°C. Those are the most common nitric acid properties. It is used in varying dilutions across many industries and chemical processes from munitions thru to agriculture, cleaning and woodworking. As a pure acid HNO3 often emits white vapor when exposed to air and as a dissolved solution can give off a vapor that is reddy-brown leading to its common name ‘red fuming acid’. When stored in a diluted form for some length of time the acid can take on a yellow tinge. Nitric acid is completely soluble in water. This mineral based acid is highly corrosive, even in dilute forms, and if splashed on skin will cause yellow blisters to be formed this should be expunged immediately with copious amounts of water. It is highly toxic. Pure anhydrous nitric acid (i.e. undiluted) is not, however, stable and even at ambient temperatures can decompose, as temperatures increase so too does the rate of the acid’s decomposition and when heated vigorously it will divide into its component form of water, oxygen and nitrogen dioxide. Care is required, therefore in its storage and handling. [3]

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1.5. Chemical Properties Nitric acid can be seen to have a number of properties that can be describes as acidic, oxidizing, reactive and as passivation. Its acidic nature means that varying degrees of corrosion can be anticipated dependent on the levels of dilution – leading to its use as a cleansing, etching and ‘ageing’ chemical in many applications. If being used as a reagent or cleanser then care needs to be taken to use vessels made from corrosion resistant alloys or metals to enable processes to take place. As a strong and powerful oxidizing agent it reacts, sometimes vigorously and violently, with numerous non-metallic substances and compounds and the resultant reaction can be an explosive one. This holds true to most metals with the exception of those classed as ‘precious’ and it is for that reason that nitric acid is used in the cleansing and assessing of precious metal purity. Depending on the level of dilution used during oxidization nitrous oxide may be formed. When used in conjunction with many metals the end result is that nitric acid will dissolve most of them and in the process creates nitrogen oxides. If combined with hydrochloric acid then Nitric acid can be used to dissolve what are known as ‘noble’ metals such as gold, platinum, iridium and others. [3]

1.6. Safety Properties Nitric acid is a strong acid and a powerful oxidizing agent. The major hazard posed by it is chemical burns as it carries out acid hydrolysis with proteins (amide) and fats (ester), which consequently decomposes living tissue (e.g. skin and flesh). Concentrated nitric acid stains human skin yellow due to its reaction with the keratin. These yellow stains turn orange when neutralized. Systemic effects are unlikely, however, and the substance is not considered a carcinogen or mutagen. The standard first aid treatment for acid spills on the skin is, as for other corrosive agents, irrigation with large quantities of water. Washing is continued for at least ten to fifteen minutes to cool the tissue surrounding the acid burn and to prevent secondary damage. Contaminated clothing is removed immediately and the underlying skin washed thoroughly.

Being a strong oxidizing agent, reactions of nitric acid with compounds such as cyanides, carbides, and metallic powders can be explosive and those with many organic compounds, such as turpentine, are violent and hypergolic (i.e. self-igniting). Hence, it should be stored away from bases and organics. [1], [3]

1.7. Market Survey

1.7.1 Economic Outlook Nitric acid is not produced in Saudi Arabia. However, worldwide annual production of nitric acid is at present approximately 34 million tons. The USA, UK, Poland and France are the largest producers. The trend in the last decade has been for growth by the larger producers, very much at the expense of the smaller ones. The global scene is a much more stable market. This can be attributed historically to consumption being more broadly based with a sizable consumption in chemical production processes. World nitric acid consumption peaked in the late 1980s before declining significantly through 1994. That decline was related primarily to economic turmoil in the Eastern bloc countries. Since then, the market has exhibited an upward trend. The largest market for nitric acid consumption is the production of ammonium nitrate (AN) and calcium ammonium nitrate (CAN). In 2010, this accounted for 80% of total world consumption of nitric acid. The major end use of AN fertilizer is in decline as a result of concerns about nitrate groundwater contamination and increased usage of solid urea, which has a higher nitrogen content (46%) than AN (34%), is less costly, and is less dangerous. Consumption of AN in explosives and blasting agent applications continues to grow, but is much more regulated since September 11, 2001. [9] The following pie chart shows (Figure 1) world Produces of nitric acid

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World Producers of Nitric Acid

%21

USA %30

Poland france USSR

%10

UK %7

%10

%4

Other Estern Europe

%18

Figure 1 World producers of nitric acid [2]

Most nitric acid is consumed captively and the merchant portion of the market accounts for less than 10% of the total. International trade is minor and has little impact on the nitric acid balance. Although there has been a relatively steady increase in capacity, the world's average operating rate increased to 76% in 2010, indicating a much stronger market balance than previously. It is estimated that AN (and CAN) production accounted for 80% of the world nitric acid market in 2010. The AN market is nearly three-quarters fertilizer and onequarter industrial applications. However, other nitric acid–based products such as nitrophosphates and potassium nitrates are also used in fertilizer applications, accounting for an additional 2.5% of total nitric acid consumption. The remaining 17.5% of nitric acid is consumed in industrial (nonfertilizer) applications. The combined production of organic compounds, such as nitrobenzene, toluene diisocyanate (TDI), adipic acid, and nitrochlorobenzenes accounted for nearly 10% of total world nitric acid consumption in 2010. Western Europe, China, the former USSR, Central Europe and the United States dominate the market statistics. Together, these five regions accounted for 81.4% of capacity, 82.0% of production, and 81.9% of consumption in 2010. Since 1994, the largest increases in capacity, production and consumption have occurred in China.

One environmental problem affecting the consumption of nitric acid concerns the use of ammonium nitrate fertilizers. The loss of nitrogen to groundwater because of nitrification and leaching has become a significant problem and has negatively impacted the use of nitrogen fertilizers, particularly in Western Europe. A European Community directive has set a target of reducing nitrate levels in groundwater to a maximum of 50 milligrams per liter. There is also concern about nitrate levels in groundwater in the United States. Current world production is approximately 34 million tons per annum, and over 30% is produced in the United States. Of the remaining production, about 60% is based in Europe. The USSR (6 million tons), United Kingdom (3.3 million tons), Poland (2.4 million tons) and France (1.5 million tons) are the main producers. The plant should operate on a standard 8000 hour/year basis, with approximately 330 days of production. Nitric acid market price is $400 per ton can be obtained for the product. Ammonia market price is $500 per ton; we need to 18687.08 tons for ammonia to produce 100000 tons of nitric acid. [9]

Table 1: Prices of raw materials and product

Component

Price per ton

Total Price

Ammonia

$500

$9343540

Water

$0.04

$963

Nitric acid

$400

$40000000

Total Profit = 40000000 - 9343540 - 963 = $30655497

1.8 Preliminary Hazard Analysis

1.8.1 Summary of Previous Plant Accident At about 6 a.m. on December 13, 1994, two explosions rocked the ammonium nitrate (AN) facility at the Port Neal, U.S.A Iowa nitrogen products manufacturing complex operated by Terra Industries. Four people were killed and 18 injured. 18

Approximately 5,700 tons of anhydrous ammonia was released, and releases of ammonia continued for nearly six days following the explosion. Chemicals released as a result of the blast contaminated groundwater under the facility. [10]

Figure 2 Terra's Port Neal, Iowa complex before the explosion. The ammonium nitrate plant is indicated

[2]

Figure 3 Post-explosion aerial photograph of the Port Neal plant [2]

The Port Neal plant produced an 83 percent AN solution by reacting ammonia and nitric acid in a vessel called a neutralizer. The original neutralizer was replaced in 1980 and a major modification and upgrade of the plant occurred in 1992. A scrubber and new control system were also added in September 1994. In the two days prior to the explosion, the nitric acid plant was shut down for maintenance. With the nitric acid plant not operating, the AN facility was also shut down. The accident occurred due to unsafe plant operations including poor maintenance and inadequate employee training. Specifically, during the shutdown period, the pH

of the neutralizer vessel contents dropped to an unusually low level and leaks in other equipment led to the introduction of chloride ions that catalyzed the final reaction. Unaware that the 18,000 gallon-capacity neutralizer vessel was in a highly acidic and contaminated condition, Terra employees injected superheated steam to try to keep the vessel contents from freezing due to the winter cold. The energy from injected superheated steam led to the runaway chemical reaction of the sensitized ammonium nitrate solution and resulted in the subsequent explosive detonations. [5]

1.8.2 Inherent Safety Aspects Taking into consideration the inherent safety aspects such as substitution of hazardous chemicals, safe location, plant layout, transportation and storage can reduce accidents Potential consequences: a) Hazardous Chemicals:

The following hazards may arise during nitric acid production: 

Equipment/piping failure because of corrosion



Explosion hazard due to the air ammonia mixture



Explosion of nitrite/nitrate salts

(i) Equipment/Piping Failure Corrosion protection is achieved by the well-proven use of suitable austenitic stainless steel where condensation can occur and by regular monitoring of the conditions. (ii) Explosion Hazard due to the Air Ammonia Mixture The air ammonia ratio is continuously controlled and kept below the hazardous range. Safety is ensured by the automatic closure of the ammonia control valve and separate shutdown trip valve when too high an air ammonia ratio is measured, either from each individual flow meter or indirectly from the catalyst gauze temperature. (iii) Explosion of Nitrite/Nitrate Salts Any free ammonia present in the nitrous gas will give a deposit of nitrite/nitrate in a cold spot. Local washing and well proven operating practices will prevent the hazard. [1], [6]

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b) Safe Location:

The selection of suitable location is a very important decision to make when there is new plant to build. Some of the factors that should be consider when selecting the location for a plant is for example the availability of raw materials used, also good transportation network, and availability of market, weather condition suitability and the water supply. From safety aspects, few factors should be taken in consideration before deciding whether the location is suitable. Most important is the distance from the residential area, availability of water and power supply. [4]

c) Plant Layout:

Another inherent safety is plant layout. After deciding on the plant location the overall plant layout, for example the processing areas, the absorber column, control rooms, roads and storage areas and other utilities must be planned carefully taking in consideration the future problems that might arise. Firstly and elementary layout developed first, these shows the fundamental relationship between the operating equipment and the storages area. Then the second step is the primary layout base of the flow of the material, unit operational storage, and future expansion. An efficient arrangement and coordination is very important to reduce the risk and hazards in plant by putting the element of health and safety into the design. [9]

d) Transportation:

Transportation to and from the plant is very important. Usually if the plant is big, it requires inside transportation and these kinds of vehicles should be ensured that it is safe and does not bring any hazards to the workers or the plant itself. Vehicles should be using diesel instead of petrol, as a diesel engine does not produce sparks that might ignite fire. For nitric acid transferring process, container should be constructed from insulating material. [9]

e) Storage:

Most of the large accidents in chemical or petrochemicals plants happen in the storage area. Storages room or tanks is where most plants stores whether their raw material or their products. Chemical storage areas shall be inspected at least annually and any unwanted or expired chemicals shall be removed. Adjusting the storage capacity or installing safety system will definitely reduce accident occurrence. The duration of material stored should also be taken in consideration. For example, longer store might change the material properties, which might cause to undesired accidents. Raw materials and products should not be stored for long period. [4]

1.8.3 Local Safety and Environmental Regulations (Nitrogen oxides defined as nitrogen dioxide NO2) 1) Purpose: The purpose of these standards is to prevent development of nitrogen dioxide concentrations, which could produce adverse health effects or lead to the production of significant concentrations of photochemical oxidants.

2) Standards: a) During any 30 days period, one-hour average NO2 concentration shall not exceed 660-microgram/cubic meter (0.35 ppm) more than twice at any location. b) During any 12 months period, the annual NO2 concentration shall not exceed 100-microgram/cubic meter at any location. [11]

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1.9 Site Feasibility 1.9.1 Selection Criteria Several factors influence the selection of a site for the location of a chemical plant. The following list contains a few of the important considerations, but should not be considered exhaustive. (a) Designation as a heavy industrial development area (‘light’ industry usually means assembly of electronic components, small metal fabricators, etc., and a major chemical plant would not be acceptable). (b) Prior existence of similar chemical plants and location of other industrial centres. (c) Existing roads and services, e.g. electricity, gas, water, etc. (d) Appropriate terrain, sub-surface, drainage, etc. (e) Suitable access for transportation of raw materials and chemicals, and for construction of a chemical plant. (f) Proximity to major transportation networks, e.g. roads, railways, airports, waterways and ports. This is a major consideration in the location of a plant. In some cases direct pipelines for the transportation of chemicals or utilities (e.g. water, gas, oil) may be the most economical method. The cost of transportation by tanker (road, rail or sea) is reduced if a return load can also be carried. (g) Availability of a local workforce and distance from local communities. (h) Availability of domestic water and plant cooling water. (i) Environmental discharge regulations. (j) Proximity to both the raw materials supply and the market for the product chemical. (k) Existence of services equipped to deal with a major industrial accident. (1) Climatic conditions, e.g. humidity, maximum wind velocity and its prominent direction, rainfall, etc. (m) Proposed or possible government restrictions regarding industrial development or discharge emissions. (n) Room for expansion. (0) Price of land. (p) Public opinion. (q) Possibility of earthquakes, subsidence, avalanches, etc.

(r) Availability of government regional development grants or tax incentives, subsidies, etc. [1], [9]

1.9.2 Potential Site Location The manufacture of nitric acid is categorized as a petrochemical project. The plant must therefore be sited in a special zone provided by the government. After conducting the feasibility and site survey, three existing industrial areas have been evaluated to choose the most suitable area for the acetone plant, which are Jubail Industrial City (JIC), Yanbu Industrial City (YIC) and Ras Alkhair Seaport.

Table 2: Information of site location

Factors

JIC

Site Location YIC

Type of Industrial Area Raw Materials IPA Power Water Utilities Steam Natural Gas Available Area Land Price Space for Expansion Cost of Living Seaport Transportation Railway Roadway Airport Price of Power Utilities Water Existing Infrastructure Existing Services for Industrial Accidents Training Centre Government Incentives TOTAL SCORE PERCENTAGE (%) RANKING

10 9 10 10 9 8

10 0 10 10 8 8

Ras Alkhair Seaport 10 10 10 10 6 8

9 10 0 5 10 5 10 5 6 10 10 10

8 10 0 8 10 0 10 10 6 10 10 10

10 10 0 9 10 10 10 3 6 10 10 10

10 10 166 83 2

5 10 162 81 3

5 10 167 83.5 1

In these cities, there are a lot of petrochemical plants and because the raw material (ammonia) plant is available in Ras Alkhair, we have chosen the location of our plant to be there. 24

Chapter 2

Material Balance

2.1 Process Description The process begins with the vaporization of ammonia at 1240 kPa and 35°C using process heat. Steam is then used to superheat the ammonia about 170°C, filtered air is compressed by a centrifugal compressor, discharge pressure of 1200 kPa. In our process; the air and the ammonia vapor are mixed and passed through the platinum/ rhodium gauze reactor where the heat of reaction raises the temperature to be between 650°C and 630°C. The reaction gas flows of heat exchangers to cooled down to 70 oC. Approximately 95% of the oxidation to nitrogen dioxide occurs as the gas passes in the Oxidation unit, after that cooled to 60°C, then sent to absorber to produce nitric acid (60%) purity.

26

2.1.1 The Flow Sheet of the Selected Process

Figure 4 Process flow sheet for nitric acid production by the single pressure process." our selected process"

28

2.1.2 Justification for the Equipment Selection Process Units 1. Air Compressor “Two stage compressors to achieve the high pressure required". 2. Ammonia Vaporizer: a shell and tube-type heat exchanger. This unit should contain internal baffles. This exchanger is made from mild steel. 3. Ammonia Superheater: a shell and tube-type heat exchanger of similar mechanical construction to the ammonia vaporizer. Also constructed from mild steel. 4. Mixer (usual mixing vessel). 5. Reactor: the reactor is a pressure vessel operating in the range 1050 kPa to 1100 kPa. The vessel must be designed to ensure even passage of the feed gas mixture over the platinum/rhodium catalyst gauze The catalyst gauze and accompanying platinum filter gauze are fixed in position by lateral supports across the width of the reactor. 6. Cooler: this shell and tube-type heat exchanger uses deionized water as its cooling medium. It has a design pressure of about 1200 kPa. 7. Oxidation Unit: the oxidation unit is a normal pressure vessel that takes input reaction gases and air. 8. Secondary Cooler: the secondary cooler takes the exit gases from the oxidation unit at 140°C and cools them down to 60°C, a suitable temperature for entry into the absorption column. It is a shell and tube-type heat exchanger. 9. Absorber: the absorber is usually a sieve tray-type column. It has a design pressure of 1200 kPa, and operates at a temperature range of 60oC to 30oC.

The material balances for the all units of the plant were hand calculated. A material balance for each unit presented below in tabulated form. The main equations used in the calculations are shown in their relevant sections. The detailed calculations are included in Appendix (C). The general equation for material balance is Input – Output + Generation – Consumption = Accumulation For steady state without chemical reaction Input – Output = 0

2.2 Overall Mass Balance

S1

S 12

S2 S 13

S 14

Table 3: Summery of overall mass balance

Component O2

Input S1 (Kg/h) 9142.88

Input S2 (Kg/h) -

Output S14 (Kg/h) -

Output S12 (Kg/h) 799.19

Output S13 (Kg/h) -

N2

34394.64

-

-

34490.82

-

NH3

-

2335.88

-

-

-

H2O

-

-

3007.51

646.00

5000.00

HNO3

-

-

-

-

7500.00

NO2

-

-

-

288.20

-

NO

-

-

-

156.70

-

Total

48880.91

48880.91

30

2.3 Reactor Mass Balance

S7

S8

R-201

Table 4: Summery of reactor mass balance

Component NH3

Input S7 (Kg/h) 2335.88

Output S8 (Kg/h) -

O2

9142.88

3756.58

N2

34394.64

34490.82

NO

-

3916.05

H2O

-

3709.93

Total

45873.4

45873.4

2.4 Oxidation Mass Balance

S9

S 10

R-202

Table 5: Summery of oxidation mass balance

Component NO

Input S9 (Kg/h) 3916.05

Output S10 (Kg/h) 156.70

NO2

-

5764.39

O2

3756.57

1751.55

N2

34490.82

34490.82

H2O

3709.93

3709.93

Total

45873.40

45873.40

32

2.5 Absorber Mass Balance

S 12

S 14

S 11

S 13

T-201

Table 6: Summery of absorber mass balance

Component H2O

Input S11 (Kg/h) 3709.93

Input S14 (Kg/h) 3007.51

Output S12 (Kg/h) 646.00

Output S13 (Kg/h) 5000.00

HNO3

-

-

-

7500.00

NO2

5764.40

-

288.20

-

NO

156.70

-

156.70

-

O2

1751.55

-

799.19

-

N2

34490.82

-

34490.82

-

Total

48880.91

48880.91

Chapter 3

Energy Balance

34

3.1 Vaporizer Energy Balance

S4

S5

E-101

For S4: NH3 at -15 oC and S5: NH3 at 35 oC

∑ 𝐻𝑜𝑢𝑡 = 1.78 𝐾𝐽/𝑚𝑜𝑙 ∑ 𝐻𝑖𝑛 = 0 𝑄̇ = 2902575.37 𝐾𝐽/ℎ

3.2 Compressor Energy Balance

S3

S1

C-101

For S1: Air at 101 kPa and S3: Air at 1090 kPa 𝑊 = 8879734.37 𝑘𝐽/ℎ (This is the mechanical energy required by the compressor.)

36

3.3 Superheater Energy Balance

S5

S6

E-102 For S5: NH3 at 35 oC and S6: NH3 at 177 oC 𝐻𝑖𝑛 = 0 𝐻𝑜𝑢𝑡 = 5.442 𝐾𝐽/ℎ 𝑄 = 747758.01 𝐾𝐽/ℎ

3.4 Mixer Energy Balance

S3 S7

S6

M-101

For S3: NH3 at 177 oC, S6: Air at 262 oC and S7: NH3+Air at ? 𝑄 = 0 (𝐴𝑑𝑖𝑎𝑏𝑎𝑡𝑖𝑐) 𝑇𝑜𝑢𝑡 = 250 °𝐶

38

3.5 Reactor Energy Balance

S7

S8 R-201 For S7: NH3+Air at 250 oC and S8: Air+NO+H2O at 645 oC

Table 7: Summary of enthalpy components in reactor

Component H (Air) H (NH3) H (NO) H (H2O)

𝑄̇ = −51820729.07 𝑘𝐽/ℎ

Input S7 (kJ/mol) 6.654 8.84 -

Output S8 (kJ/mol) 19 19.744 23

3.6 Heat Exchanger Energy Balance (1st Cooler) S8

S9

E-201

For S8: Air+NO+H2O at 645 oC and S9: Air+NO+H2O at 70 oC 𝐻𝑜𝑢𝑡 = 0 For air: 𝐻𝑖𝑛 = 17.693 𝑘𝐽/𝑚𝑜𝑙 For NO: 𝐻𝑖𝑛 = 18.400 𝑘𝐽/𝑚𝑜𝑙 For H2O: 𝐻𝑖𝑛 = 21.173 𝑘𝐽/𝑚𝑜𝑙 𝑄 = −30,084,503.19 𝐾𝐽/ℎ

40

3.7 Oxidation Energy Balance

S9

R-202

S 10

For S9: Air+NO+H2O at 70 oC and S10: Air+NO+H2O+NO2 at 140 oC

Table 8: Summary of enthalpy components in oxidation

Component

Input S9 (kJ/mol)

Output S10 (kJ/mol)

H (H2O) (g)

1.521

3.919

H (Air)

1.312

3.37

H (NO)

1.345

3.47

H (NO2)

-

4.5

𝑄 = −3707006.24 𝑘𝐽/ℎ

3.8 Heat Exchanger Energy Balance (2nd cooler)

S 10

S 11

E-202

For S10: Air+NO+H2O+NO2 at 140 oC and S11: Air+NO+H2O+NO2 at 60 oC 𝐻𝑜𝑢𝑡 = 0 For H2O: 𝐻𝑖𝑛 = 2.737 𝑘𝐽/𝑚𝑜𝑙 For air: 𝐻𝑖𝑛 = 2.350 𝑘𝐽/𝑚𝑜𝑙 For NO: 𝐻𝑖𝑛 = 2.4230 𝑘𝐽/𝑚𝑜𝑙 For NO2: 𝐻𝑖𝑛 = 3.1796 𝑘𝐽/𝑚𝑜𝑙 𝑄 = −3911927.94 𝐾𝐽/ℎ

42

3.9 Absorber Energy Balance

S 12

S 14 T-201

S 11 S 13

Basis: Tref = 25oC For S11: Air+NO+H2O+NO2 at 60 oC, S14: H2O at 20 oC, S12: Air+NO+H2O at 30 oC and S13: HNO3+H2O at 30 oC Table 9: Summary of enthalpy components in absorber

Component

Input S11

Input S14

Output S12

Output S13

(kJ/mol)

(kJ/mol)

(kJ/mol)

(kJ/mol)

H (H2O) (g)

1.18

-

0.166

-

H (H2O) (L)

-

-0.375

-

0.375

H (Air)

1.02

-

0.145

-

H (NO)

1.044

-

0.149

-

H (NO2)

1.32

-

-

-

H (HNO3)

-

-

-

0.55

Q= -29876241 KJ/h

Chapter 4

Design and Sizing Equipment

44

4.1 Sizing of Pump [7] Pump Specification Data Ammonia Flow rate = 2335.88 kg/h Temperature = -15°C Density at -15°C = 656.67 kg/m3 Estimating the pump diameter required Mass flowrate (G) =

2335.88 = 0.6488 Kg/s 3600

G = 0.6488 kg/s 𝐹𝑙𝑜𝑤𝑚𝑒𝑡𝑟𝑖𝑐 𝑓𝑙𝑜𝑤𝑟𝑎𝑡𝑒(𝑄) =

0.6488 656.67

Q = 0.001 m3/s Piping Specification Atypical velocity for fluid flow is 2 m/s. Determination of the pipe area: Area of pipe (A) = Area =

𝑉𝑜𝑙𝑢𝑚𝑒𝑡𝑟𝑖𝑐 𝑓𝑙𝑜𝑤𝑟𝑎𝑡𝑒 Velocity

0.001 2

Area= 0.0005 m2 4 × 𝐴𝑟𝑒𝑎 0.5 Diameter of pipe (D) = [ ] 𝜋 𝐷=[

4 × 0.0005 0.5 ] 𝜋

D = 25 mm This value is compared with the result achieved by applying the ‘Economic pipe diameter’ formula for stainless steel from Optimum diameter = 226 G0.5 ρ-0.35 Optimum diameter = 226 (0.6488)0.5 × (656.67)-0.35 Optimum diameter = 19mm Accept the larger value as a conservative estimate

4.2 Design of Heat Exchanger (Vaporizer) The objective is to design a heat exchanger (vaporizer) (shell and tube) to heat Ammonia of flow rate 2335.88 kg/h from -15oC to 35 oC using saturated steam entering at 1 atm. Table 10: Data of ammonia

Component

Mwt

N

M

NH3

Kg/Kmol 17

Kmol/h 137.404

Kg/h 2335.88

Mole fraction Y

Mass fraction X

1

1

Table 11: Physical properties of ammonia at Tave = 10oC and Pave = 1240 kPa

Component

Μ Pa.s 1.531 x 10-4

Cp KJ/kg.K 5.02

NH3

K W/m.K 0.5135

Ρ Kg/m3 623.6318

Table 12: Physical properties of standard steam at 1 atm and T ave = 100oC = 373 K

Cpl Cpv KJ/Kg.K 4.24 1.888

μl 0.27 x 10-3

μv Pa.s 1.295 x 10-5

ρl

ρv 3

Kg/m 953 0.596

kl

kv W/ m.K 0.681 0.0251

Λ KJ/Kg 2256.9

Heat duty (Q) needed to be added to gas mixture: Q = m Cpmix ΔT

[7]

Q = 163 kJ/s (kw)

For steam: Qlost = Qgain = m λ

[7]

msteam = 163/2256.9 msteam = 0.0722 kg/s

46

Finding ΔTlm:

T1 = 100 oC

T2 = 100 oC

VAPORIZER

t2 = 35 oC

t1 = -15 oC

∆Tout = T1 − t 2 = (100 − 35) = 65 oC ∆𝑇𝑖𝑛 = 𝑇2 − 𝑡1 = (100 − (−15)) = 115 𝑜𝐶

∆𝑇𝑙𝑚 =

∆𝑇𝑜𝑢𝑡 − ∆𝑇𝑖𝑛 ∆𝑇 𝑙𝑛 ∆𝑇𝑜𝑢𝑡 𝑖𝑛

[7]

∆𝑇𝑙𝑚 = 87.63 °𝐶 The heat exchanger will consist of one shell pass and one tube passes: ΔTm = 87.63 oC

According to Table A in Appendix E, Cold fluid is Ammonia and hot fluid is condensing steam. The following value of overall heat transfer coefficient U can be estimated: U= 500 W/ m2.oC

Required Area for Heat transfer: 𝑄 = 𝑈𝐴 𝛥𝑇𝑚  𝐴 = 𝑄/𝑈 𝛥𝑇𝑚 [7] 𝐴 = 3.72 𝑚2

Choosing 25 mm “OD”, 21 mm “ID” stainless steel tubes to resist corrosion problems. Placing condensing steam in the shell side. Take Tube length L=2 m [7] Calculation of Number of tubes:

𝑁𝑡 =

𝐴 3.72 = 𝜋 × 𝑑 × 𝐿 𝜋 × 25 × 10−3 × 2

𝑁𝑡 = 24 𝑡𝑢𝑏𝑒𝑠 Since the steam in the shell side is always clean, we will use triangular pitch arrangement with Pt= 1.25do:

Calculation of bundle diameter Db: 1

𝑁𝑡 𝑛1 𝐷𝑏 = 𝑑𝑜 [ ] 𝐾1

[7]

K1, n1 are constants given in Table C: K1 = 0.319 n1 = 2.142 1

159 2.142 𝐷𝑏 = 25 [ ] 0.319

[7]

Db = 187.9 mm Calculating the shell diameter Ds: 𝐷𝑠 = 𝐷𝑏 + 𝑐𝑙𝑒𝑎𝑟𝑎𝑛𝑐𝑒 [7] From Figure A in Appendix E and for split-ring floated heat type of heat exchanger we get, clearance= 48 mm 𝐷𝑠 = 187.9 + 48 = 235.64 𝑚𝑚 [7] 48

Calculating the heat transfer coefficients for tube side (hi): 𝜋 𝜋 𝐶𝑟𝑜𝑠𝑠 𝑠𝑒𝑐𝑡𝑖𝑜𝑛𝑎𝑙 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑜𝑛𝑒 𝑡𝑢𝑏𝑒 = 𝑑𝑖2 = (21)2 = 346.36 𝑚𝑚2 4 4 𝑁𝑡 𝜋 𝑇𝑜𝑡𝑎𝑙 𝑐𝑟𝑜𝑠𝑠 𝑠𝑒𝑐𝑡𝑖𝑜𝑛𝑎𝑙 𝑎𝑟𝑒𝑎 𝑓𝑜𝑟 𝑤𝑎𝑡𝑒𝑟 = ( ) × 𝑑𝑖2 = 0.008312 𝑚2 [7] 𝑁𝑜. 𝑝𝑎𝑠𝑠 4 𝑚𝑤𝑎𝑡𝑒𝑟 0.64885 𝐾𝑔⁄ = = 78.06 𝑚2 . 𝑠𝑒𝑐 [7] 𝐴 0.008312 𝐺𝑤𝑎𝑡𝑒𝑟 78.06 𝐿𝑖𝑛𝑒𝑎𝑟 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 (𝑣𝑚𝑖𝑥 ) = = = 0.125 𝑚⁄𝑠𝑒𝑐 [7] 𝜌𝑤𝑎𝑡𝑒𝑟 623.6318 𝑀𝑎𝑠𝑠 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 (𝐺𝑚𝑖𝑥 ) =

To calculated hi use the equation ℎ𝑖 𝑑𝑖 𝜇 0.14 = 𝐽ℎ × 𝑅𝑒 × 𝑃𝑟 0.33 × ( ) [7] 𝐾𝑓 𝜇𝑤 hi = inside heat transfer coefficient di = tube diameter = 21x10-3 m Kf = thermal conductivity of water fluid = 0. 5135

W/m.oC

Jh = factor for heat transfer given form Figure B in Appendix E in Appendix E by (Re)

𝑅𝑒𝑦𝑛𝑜𝑙𝑑 ′ 𝑠 𝑁𝑢𝑚𝑏𝑒𝑟 (𝑅𝑒) =

𝜌𝑣𝑑 623.6318 × 0.125 × 21 × 10−3 = = 10692.57 𝜇 1.531 × 10−4

(Turbulent Flow) 𝑃𝑟𝑎𝑛𝑑𝑡𝑙 𝑁𝑢𝑚𝑏𝑒𝑟 (𝑃𝑟) =

𝐶𝑝 𝜇 5.02 × 103 × 1.531 × 10−4 = = 1.49 𝐾𝑓 0.5125

From Figure B in Appendix E at Re= 10692.57 Jh = 4 x10-3 From (eq. hi) and neglecting (

 0.14 ) and assume it approximately 1 w

Then: ℎ𝑖 (21 × 10−3 ) = (4 × 10−3 )(10692.57)(1.49)0.33 0.5135 hi = 1192.93 W/m2.oC

[7]

[7]

Calculate heat transfer coefficients for shell side (ho): Area for cross flow of the shell side As, using Kern Method: 𝐴𝑠 = [

𝑃𝑡 − 𝑑𝑜 ] 𝐷𝑠 𝐿𝐵 𝑃𝑡

[7]

Pt = tube pitch Pt

do = tube outside diameter Ds = shell inside diameter LB = Buffle spacing (usually one fifth of shell diameter)

do

do = 25 mm Pt = 1.25 do mm = 31.25 mm 𝐷𝑠 235.9 = = 47.18 𝑚𝑚 5 5 𝑃𝑡 − 𝑑𝑜 31.25 − 25 𝐴𝑠 = [ ] 𝐷𝑠 𝐿𝐵 = [ ] (235.9)(47.18) = 2226 𝑚𝑚2 𝑃𝑡 31.25 𝐿𝐵 =

As = 0.002226 m2

Mass velocity (Gs) and liner velocity (vs): mshell = mass flow rate of steam = 0.0722 kg/sec

𝐺𝑠 = [

𝑚𝑠ℎ𝑒𝑙𝑙 0.0722 𝑘𝑔 ]=[ ] = 32.43 ⁄𝑚2 . 𝑠𝑒𝑐 𝐴𝑠 0.002226

𝐺𝑠 32.43 𝑣𝑠 = [ ] = [ ] = 54.42 𝑚⁄𝑠𝑒𝑐 𝜌𝑠 0.596

[7]

[7]

The equivalent (Hydraulic) diameter de: 𝑑𝑒 =

1.10 2 (𝑃𝑡 − 0.917𝑑𝑜2 ) 𝑑𝑜

𝑑𝑒 =

1.10 [(31.25)2 − 0.917(25)2 ] = 17.75 𝑚𝑚 = 0.01775 𝑚 25

[7]

Re & Pr for shell side: 𝛤ℎ =

𝑚𝑠 0.0722 𝐾𝑔 = = 1.5 × 10−3 ⁄𝑚. 𝑠 𝑁𝑡 𝑙 24 × 2

[7]

50

𝑅𝑒 =

𝐺𝑠 𝑑𝑒 (32.43)(0.01775) = = 45721.4 [7] 𝜇𝑣 1.259 × 10−5

𝑁𝑟 =

𝐷𝑏 187.9 = = 6.01 𝑃𝑡 31.25

[7]

The shell side coefficient ( hs or ho ) 1 −1 (953)(953 − 0.596)(9.8) 3 6 ℎ𝑠 = (0.95)(0.681) [ ] × 6.01 (0.27 × 10−3 )(1.5 × 10−3 )

hs = 13441.08 W/m2.oC

Calculating the overall heat transfer coefficient (U): 𝑑 𝑑𝑜 ln ( 𝑜⁄𝑑 ) 1 𝑑 1 1 1 1 𝑑𝑜 𝑜 𝑖 = + + + ( )+ ( ) 𝑈𝑜 ℎ𝑜 ℎ𝑜𝑑 2𝐾𝑤 ℎ𝑖 𝑑𝑖 ℎ𝑖𝑑 𝑑𝑖

[7]

Uo = overall coefficient based on outside area of the tubes ho = outside fluid film coefficient hi = inside fluid film coefficient do = tube outside diameter di = tube inside diameter Kw = thermal conductivity of tube wall material = 16 (W/m.oC) for "stainless steel" hod = outside dirt " fouling " coefficient = 3000 (W/m2.oC) (From Table B) hid = inside dirt " fouling " coefficient = 3000 (W/m2.oC) (From Table B) 25 × 10−3 ln(25⁄21) 1 1 1 1 25 1 25 = + + + ( )+ ( ) 𝑈𝑜 13441.08 3000 2 × 16 1192.93 21 3000 21 Uo = 516 W/𝒎𝟐 .oC The value of 516 W/m2.oC is well above the estimated value of 500 W/m2.oC. Hence, the present design satisfactory. (ok)

52

Calculating the pressure drop (ΔP) for side tube and shell tube: 1.The tube side ΔPt L μ −m ρvt2 ∆Pt = NP [8jf ( ) ( ) + 2.5] [ ] di μw 2

[7]

ΔPt = tube side pressure drop (N/m2) Np = Number of tube passes jf = fraction factor ' depending on Re ' Found from Figure 12.30 in the text book assuming baffle cut of 0.25 L = length of one tube = 2 m vt = flow velocity inside the tube =0.125 m/sec m = exponent value depending on type of flow For laminar (Re10000)  m = 0.14 Now, assuming viscosity ratio =1 and finding jf corresponding to Re= 5594.35, we get; ∆Pt = [8(5 × 10

−3 )

2 623.6318 × 0.1252 ( ) + 2.5] [ ] 21 × 10−3 2

ΔPt =30.74Pa = 0.0307 kpa (Acceptable)

2. The shell side ΔPs Ds L ρvs2 μ −0.14 ∆Ps = 8jf ( ) ( ) ( )( ) De LB 2 μw

[7]

ΔPs = shell side pressure drop (N/m2) jf = fraction factor ' depending on Re , Found from Figure 12.30 in the text book assuming baffle cut of 0.25 Ds = shell diameter = 235.9 mm De = equivalent diameter for shell side = 17.75 mm L = length of one tube = 2 m Re= 45721.4 vs = flow velocity in shell = 54.42 m/sec LB = Baffle spacing = 47.18 mm ρ shell = 0.596 kg/m3 Neglecting viscosity correction, we get: ∆Ps = 8(3.9 × 10

−2 )

235.9 2 × 103 0.596 × 4.422 ( )( )( ) 17.75 47.18 2

ΔPs = 155.12 kpa (acceptable)

54

Finally, we summarize our present design as follows 1. The selected heat exchanger has one shell pass and one tube passes, in which ammonia flow inside the tube and condensing steam flows inside the shell “. 2. The selected tubes are made of stainless steel with 25 mm outside diameter, 21 mm inside diameter, the total number of tube is 24, while the tube length is 2 m, and the triangular pitch applied for this tube distribution is 31.25 mm. 3. The shell diameter Ds is 235.9 mm. and the baffles used are 25% cut, while the baffle spacing is 48 mm. 4. The first estimated value for overall heat transfer coefficient Uo is 500 W/m2.oC, while the final calculated value is 516 (W/m2. oC). 5. The pressure drop (ΔPt) for the tube side is 0.0307 kPa , while the shell side (ΔPs) is 155.12 kPa .

4.3 Sizing of Compressor [7]

𝑃𝑀𝑤 101 × 1000 × 28.9 𝑘𝑔 = = 1.15 ⁄𝑚3 𝑅𝑇 298 × 8314 1 3 𝑉1 = = 0.8733 𝑚 ⁄𝑘𝑔 𝜌 𝜌=

V = 12.693/1.145 = 10 .562 m3/s = 38023.2 m3 / h From Figure C in Appendix E and for volumetric flow 38023.2 and discharge pressure 11 bars the recommended compressor is centrifugal compressor Calculation the polytropic coefficient (n) From Figure D in Appendix E and suction flow 10.562 m3/s

Ep = 0.74

Average heat capacity of mixture at T = 30 oC 𝐽 𝐶𝑝𝑎𝑣𝑒 = ∑ 𝑦𝑖 𝐶𝑝𝑖 = 29.099 ⁄𝑚𝑜𝑙. 𝐾 𝐶𝑣 = 𝐶𝑝 − 𝑅 = 29.099 − 8.314 = 20.785 𝐶𝑝⁄ 𝐶𝑣 = 1.4 (𝛾 − 1) 𝑚= = 0.386 𝛾(𝐸𝑝 ) 𝛾=

𝑛=

1 = 1.62 1−𝑚

The polytropic work: 𝑛 𝑃2 𝑊 = 𝑃1 𝑣1 [( ) 𝑛 − 1 𝑃1

(𝑛−1)⁄ 𝑛

− 1]

Wpoly =376366.4 J/kg Actual work required: Wpoly/Ep = 508731.6 J/kg Power required Pac = W × m = 12.693 × 508731.6 = 6068.531 KW Electric power: From table the approximate electrical efficiency, Ee is 0.97 Pe = Pac /Ee = 6068.531 /0.97 = 6256.2 KW

4.4 Design of Heat Exchanger (Superheater) 56

The objective is to design a heat exchanger (superheater) (shell and tube) to heat Ammonia of flow rate 2335.88 kg/hr from 350C to 177 0C using saturated steam entering at 40 bars.

Table 13: Data of ammonia

Component

Mwti kg/kmol 17

NH3

Ni kmol/h 137.404

Mi Kg/h 2335.88

Mole fraction Yi 1

Mass fraction Xi 1

Table 14: Physical properties at Tave = 106oC and P = 1240 kPa

Component

Cp kJ/kg.K 2.38

NH3

μi Pa.s 1.32 x 10-4

K W/m.K 0.0364

ρ kg/m3 7.04

Table 15: Physical properties of standard steam at 40 bars and Tave = 250.3oC = 523.45 K

Cpv kJ/kg.K 1.958

μv Pa.s 1.84 x 10-5

ρv kg/m3 20.12

Heat duty (Q) needed to be added to gas mixture: 𝑄 = 𝑚𝐶𝑝𝑚𝑖𝑥 𝛥𝑇 [7] 𝑄 = 219.28 𝑘𝐽/𝑠 (𝑘𝑤)

For steam: 𝑄𝑙𝑜𝑠𝑡 = 𝑄𝑔𝑎𝑖𝑛 = 𝑚 𝜆 [7] 𝑚𝑠𝑡𝑒𝑎𝑚 = 219.28/1712.9 𝑚𝑠𝑡𝑒𝑎𝑚 = 0.128 𝑘𝑔/𝑠

Finding ΔTlm:

kv W/ m.K 0.0388

𝝀 kJ/kg 1712.9

T1 = 250.3 oC

t2 = 177 oC

T2 = 250.3 oC

SUPERHEATER

t1 = 25oC

∆𝑇𝑜𝑢𝑡 = 𝑇1 − 𝑡2 = (250.3 − 177) = 73.3 °C ∆𝑇𝑖𝑛 = 𝑇2 − 𝑡1 = (250.3 − 35) = 215.3 °C

∆𝑇𝑙𝑚 =

∆𝑇𝑜𝑢𝑡 − ∆𝑇𝑖𝑛 [7] ∆𝑇 𝑙𝑛 ∆𝑇𝑜𝑢𝑡 𝑖𝑛

∆𝑇𝑙𝑚 = 131.79 °C

The heat exchanger will consist of one shell pass and one tube passes.

According to Table A in Appendix E, Cold fluid is Ammonia and hot fluid is condensing steam. The following value of overall heat transfer coefficient U can be estimated: U= 160 W/ m2.oC

Required Area for Heat transfer:

58

𝑄 = 𝑈𝐴 𝛥𝑇𝑚  𝐴 =

𝑄 𝑈𝛥𝑇𝑚

[7]

𝐴 = 10.39 𝑚2 Choosing 25 mm “O.D”, 21 mm “I.D” stainless steel tubes to resist corrosion problems. Placing condensing steam in the shell side. Take Tube length L = 5 m

[7]

Calculation of Number of tubes: 𝑁𝑡 =

𝐴 10.39 = 𝜋 × 𝑑 × 𝐿 𝜋 × 25 × 10−3 × 5

[7]

𝑁𝑡 = 27 𝑡𝑢𝑏𝑒𝑠 Since the gas mixture in the shell side is always clean, we will use triangular pitch arrangement with Pt= 1.25do [7]

Calculation of bundle diameter Db: 1

𝑁𝑡 𝑛1 𝐷𝑏 = 𝑑𝑜 [ ] 𝐾1

[7]

K1, n1 are constants given in Table C: K1 = 0.319 n1 = 2.142 1

27 2.142 𝐷𝑏 = 25 [ ] 0.319

[7]

Db = 198.5 mm Calculating the shell diameter Ds: 𝐷𝑠 = 𝐷𝑏 + 𝑐𝑙𝑒𝑎𝑟𝑎𝑛𝑐𝑒 [7] From Figure A in Appendix E and for split-ring floated heat type of heat exchanger we get, clearance= 49.5 mm 𝐷𝑠 = 198.5 + 49.5 = 248 𝑚𝑚

Calculating the heat transfer coefficients for tube side (hi): 𝜋 𝜋 𝐶𝑟𝑜𝑠𝑠 𝑠𝑒𝑐𝑡𝑖𝑜𝑛𝑎𝑙 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑜𝑛𝑒 𝑡𝑢𝑏𝑒 = 𝑑𝑖2 = (21)2 = 346.36 𝑚𝑚2 [7] 4 4 𝑁𝑡 𝜋 𝑇𝑜𝑡𝑎𝑙 𝑐𝑟𝑜𝑠𝑠 𝑠𝑒𝑐𝑡𝑖𝑜𝑛𝑎𝑙 𝑎𝑟𝑒𝑎 𝑓𝑜𝑟 𝑤𝑎𝑡𝑒𝑟 = ( ) × 𝑑𝑖2 = 0.009351 𝑚2 [7] 𝑁𝑜. 𝑝𝑎𝑠𝑠 4 𝑚𝑤𝑎𝑡𝑒𝑟 0.64885 𝐾𝑔⁄ = = 69.39 𝑚2 . 𝑠𝑒𝑐 𝐴 0.009351 𝐺𝑤𝑎𝑡𝑒𝑟 78.06 𝐿𝑖𝑛𝑒𝑎𝑟 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 (𝑣𝑚𝑖𝑥 ) = = = 9.85 𝑚⁄𝑠𝑒𝑐 [7] 𝜌𝑤𝑎𝑡𝑒𝑟 623.6318 𝑀𝑎𝑠𝑠 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 (𝐺𝑚𝑖𝑥 ) =

[7]

To calculated hi use the equation ℎ𝑖 𝑑𝑖 𝜇 0.14 = 𝐽ℎ × 𝑅𝑒 × 𝑃𝑟 0.33 × ( ) [7] 𝐾𝑓 𝜇𝑤 hi = inside heat transfer coefficient di = tube diameter = 21x10-3 m Kf = thermal conductivity of water fluid = 0.0364 W/m.oC Jh = factor for heat transfer given form Figure B in Appendix E by (Re) 𝑅𝑒𝑦𝑛𝑜𝑙𝑑 ′ 𝑠 𝑁𝑢𝑚𝑏𝑒𝑟 (𝑅𝑒) =

𝜌𝑣𝑑 7.04 × 9.85 × 21 × 10−3 = = 110320 𝜇 1.32 × 10−5

[7]

(Turbulent Flow) 𝐶𝑝 𝜇 2.38 × 103 × 1.32 × 10−5 𝑃𝑟𝑎𝑛𝑑𝑡𝑙 𝑁𝑢𝑚𝑏𝑒𝑟 (𝑃𝑟) = = = 0.863 𝐾𝑓 0.0364

[7]

From Figure B in Appendix E at Re = 110320 Jh = 2.8 x10-3 [7] 𝜇

From (eq. hi) and neglecting (𝜇 ) 𝑤

0.14

and assume it approximately 1

Then: ℎ𝑖 (21 × 10−3 ) = (2.8 × 10−3 )(110320)(0.863)0.33 0.0364 hi = 510 W/m2.oC

60

Calculate heat transfer coefficients for shell side (ho): Area for cross flow of the shell side As, using Kern Method: 𝐴𝑠 = [

𝑃𝑡 − 𝑑𝑜 ] 𝐷𝑠 𝐿𝐵 𝑃𝑡

[7]

Pt = tube pitch Pt

do = tube outside diameter Ds = shell inside diameter LB = Buffle spacing (usually one fifth of shell diameter) do = 25 mm Pt = 1.25 do mm = 31.25 mm 𝐷𝑠 248 = = 49.5 𝑚𝑚 5 5 𝑃𝑡 − 𝑑𝑜 31.25 − 25 𝐴𝑠 = [ ] 𝐷𝑠 𝐿𝐵 = [ ] (248)(49.5) = 2455.2 𝑚𝑚2 𝑃𝑡 31.25 𝐿𝐵 =

As = 0.0024552 m2 Mass velocity (Gs) and liner velocity (vs): mshell = mass flow rate of steam = 0.128 kg/sec 𝐺𝑠 = [

𝑚𝑠ℎ𝑒𝑙𝑙 𝐴𝑠

0.128

] = [0.0024552] = 52.13

𝑘𝑔⁄ 𝑚2 . 𝑠𝑒𝑐 [7]

𝐺𝑠 52.13 𝑣𝑠 = [ ] = = 2.59 𝑚⁄𝑠𝑒𝑐 [7] 𝜌𝑠 20.12 The equivalent (Hydraulic) diameter de: 𝑑𝑒 =

1.10 2 (𝑃𝑡 − 0.917𝑑𝑜2 ) [7] 𝑑𝑜

𝑑𝑒 =

1.10 [(31.25)2 − 0.917(25)2 ] = 17.75 𝑚𝑚 = 0.01775 𝑚 25

Re & Pr for shell side:

𝑅𝑒 =

𝜌𝑣𝑠 𝑑𝑒 20.12 × 2.59 × 0.01775 = = 50269.92 [7] 𝜇 1.84 × 10−5

do

𝐶𝑝 𝜇𝑠ℎ𝑒𝑙𝑙 1.958 × 103 × 1.84 × 10−5 𝑃𝑟 = = = 0.928 𝐾𝑓 𝑠ℎ𝑒𝑙𝑙 0.0388

[7]

The shell side coefficient ( hs or ho ) 1 ℎ𝑠 𝑑𝑜 𝜇 0.14 = 𝐽ℎ × 𝑅𝑒 × 𝑃𝑟 3 × ( ) 𝐾𝑓 𝑠ℎ𝑒𝑙𝑙 𝜇𝑤

[7]

From Figure B in Appendix E at Re= 50269.92 Jh = 3.2 x10-3

[7]

From (eq. hs) and neglecting (

 0.14 ) and assume it approximately 1 w

ℎ𝑠 (25 × 10−3 ) = (3.2 × 10−3 )(50269.92)(0.928)0.33 0.0388 hs = 343 W/m2.oC

Calculating the overall heat transfer coefficient (U):

62

𝑑 𝑑𝑜 ln ( 𝑜⁄𝑑 ) 1 𝑑 1 1 1 1 𝑑𝑜 𝑜 𝑖 = + + + ( )+ ( ) 𝑈𝑜 ℎ𝑜 ℎ𝑜𝑑 2𝐾𝑤 ℎ𝑖 𝑑𝑖 ℎ𝑖𝑑 𝑑𝑖

[7]

Uo = overall coefficient based on outside area of the tubes ho = outside fluid film coefficient hi = outside fluid film coefficient do = tube outside diameter di = tube inside diameter Kw = thermal conductivity of tube wall material = 16 (W/m.oC) for '' stainless steel'' hod = outside dirt " fouling " coefficient = 3000(W/m2.oC) (From Table B in Appendix E) hid = inside dirt " fouling " coefficient = 3000 (W/m2.oC) (From Table B in Appendix E) 25 × 10−3 ln(25⁄21) 1 1 1 1 25 1 25 = + + + ( )+ ( ) 𝑈𝑜 343 3000 2 × 16 510 21 2000 21 Uo = 163.5 W/𝑚2 .oC The value of 163.5 W/m2.oC is well above the estimated value of 160 W/m2.oC. Hence, the present design satisfactory. (Ok)

Calculating the pressure drop (ΔP) for side tube and shell tube:

1. The tube side ΔPt: L μ −m ρvt2 ∆Pt = NP [8jf ( ) ( ) + 2.5] [ ] di μw 2

[7]

ΔPt = tube side pressure drop (N/m2) Np = Number of tube passes jf = fraction factor ' depending on Re ' Found from Figure 12.30 in the text book assuming baffle cut of 0.25 L = length of one tube = 5 m vt = flow velocity inside the tube m = exponent value depending on type of flow at laminar (Re10000)  m = 0.14 Now: ∆Pt = [8(2.7 × 10−3 ) (

5 7.04 × 9.852 ) + 2.5] [ ] 21 × 10−3 2

ΔPt = 2154.82 Pa =2.154 kPa (Acceptable)

2. The shell side ΔPs:

64

Ds L ρvs2 μ −0.14 ∆Ps = 8jf ( ) ( ) ( )( ) De LB 2 μw ΔPs = shell side pressure drop (N/m2) jf = fraction factor ' depending on Re , Found from Figure 12.30 in the text book assuming baffle cut of 0.25 Ds = shell diameter = 248 mm De = equivalent diameter for shell side = 17.75 mm L = length of one tube = 5 m vs = flow velocity in shell = 2.59 m/sec LB = Baffle spacing = 49.5 mm ρ shell = 20.12 kg/m3 248 5 × 103 20.12 × 2.592 ∆Ps = 8(2.8 × 10−2 ) ( )( )( ) 17.75 49.5 2 ΔPs = 22.39 kPa (Acceptable)

Finally, we summarize our present design as follows:

[7]

1. The selected heat exchanger has one shell pass and one tube passes, in which ammonia flow inside the tube and condensing steam flows inside the shell “. 2. The selected tubes are made of stainless steel with 25 mm outside diameter, 21 mm inside diameter, the total number of tube is 27, while the tube length is 5 m, and the triangular pitch applied for this tube distribution is 31.25 mm. 3. The shell diameter Ds is 248 mm. and the baffles used are 25% cut, while the baffle spacing is (49.5 mm). 4. The first estimated value for overall heat transfer coefficient Uo is 160(W/m2 . oC) , while the final calculated value is 163.5 (W/m2. oC). 5. The pressure drop ( ΔPt ) for the tube side is 2.154 kPa , while the shell side ( ΔPs ) is 22.39 kPa .

4.5 Sizing of Mixer 66

𝜌 𝑎𝑡 𝑇𝑎𝑣𝑒 = 1.15

𝑘𝑔⁄ 𝑚3

Volumetric flow rate: 3 𝑉 = 𝑚/𝜌 = 12.7 / 1.15 = 11 𝑚 ⁄𝑠

Volume of mixture: Residence time in the mixture drum: τ = 1s 𝑉 = 11 × 1 = 11 𝑚3 𝐴𝑠𝑠𝑢𝑚𝑒 𝐿/𝐷 = 4 𝜋 𝑉 = ( ) 𝐷3 4 1⁄ 3

𝑉 𝐷=( ) 𝜋

𝐷 = 1.6 𝑚 𝐿 = 6.1 𝑚

4.6 Design of Reactor Introduction

The direct oxidation of ammonia over platinum 10% odrhodium catalyst is the major step in the production of nitric acid. This oxidation at temperature about 700 oC and pressure range of 59 atm. At this temperature platinum oxides are formed on the surface and some portion of it are vaporized and carried away in the gas stream.

A suitable shape for the reactor has been found to be in the form of two frustrated square pyramids, between the large cross-sections of which are placed the catalyst gauze. This shape allows the walls of reactor to be kept as cool as possible.

A major issue in the design of this type of reactor is the arrangement of catalyst layer in the form of screen supported in a manner that will ensure good distribution of flow. Because poor distribution could cause hot spot in this exothermic reaction leading to volatilization of the catalyst metal, and a direct reduction is also reduce the plant capacity.

Kinetic of reaction

The catalyst of ammonia oxidation is so rapid that the amount of catalyst required is very small and heat transfer is not feasible. Typically configuration of this system is the woven – wire gauze. This reaction completely by mass transfer, and the design of screen packs has been based on pilot –plant studies and plant experience, and the reactor type is known as the shallow-bedadiabatic one. [5]

Assumption:

68

1. The reactions are mainly controlled by diffusion. 2. The partial pressure of ammonia on the catalyst surface is negligible. 3. The design model is developed as plug flow. 4. Typical reported data are used. 5. Ratio of mass of ammonia /hr.: mass of catalyst =85. 6. Cross sectional area = 0.256× 10-3 m2 / ton HNO3, daily. Let: [5] Nw = 80 inch mesh size Aw = area /ton HNO3 = 0.256 m2/ton HNO3 dw = wire diameter = 0.003 inch =76 x10-4 cm fw = wire area per gauza cross sectional area Awr = surface area of screen / volume of one screen The density of catalyst = 0.0214 kg / cm3 ns = number of screen per gauza X= 0.95 Vg = volume of one screen = 110 cm3 YNH3 = 0.09 𝑎𝑤𝑟 = 𝜋 𝑁𝑤

2

1 2 [( ) + 𝑑𝑤 2 ] 𝑁𝑤

0.2

= 122.6 𝑐𝑚−1 = 294.39 𝑖𝑛𝑐ℎ−1 [5]

𝑓𝑤 = 𝑎𝑤𝑟 × 2𝑑𝑤 = 294.39 × 2 × 0.003 = 1.77 aw dw 122.6 × 76 × 10−4 =1− = 0.8 [5] 4 4 𝑀𝐴 ⁄𝑀𝑤 (30⁄63) × 2000 𝐵 𝐺 = 𝑚𝑎𝑠𝑠 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 = = 𝑌𝐴 × 𝑋𝐴 × 𝑎𝑤 24 × 3600 × 0.96 × 0.09 × 0.027547 𝑔 𝐺 = 4.63 𝑙𝑏𝑚⁄𝑓𝑡 2 . 𝑠 = 2.1 ⁄𝑐𝑚2 . 𝑠 £ = Porcity = 1 −



Superficial velocity

[5]

𝑢𝑠 = 2.1 ×

1 82.06 × 1209 𝑓𝑡 = 2.1 × = 890 𝑐𝑚⁄𝑠 = 28.3 [5] 𝜌 7.8 × 30 𝑠

− 𝑙𝑛(1 − 𝑋𝐴 ) × £0.352 × 𝑑𝑤 0.648 × 𝐺 0.648 × 𝑢0.0190 𝑛𝑠 = [5] (5.81761 ∗ 10−5 ) × 𝑓𝑤 × 𝑇 0.333 × (28.85 + 11.82 × 𝑦𝐴𝑜 ) 0.667 𝑛𝑠 = 19 𝑔𝑎𝑢𝑧𝑎 Ratio 

mass amonia = 85 [5] mass catalyst

The mass of ammonia / 85 = mass of catalyst 𝑚𝑐𝑎𝑡 =

17 × 137.4 = 27.5 𝑘𝑔 85

𝐴𝑟𝑒𝑎 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 = 𝐴 = 0.256 × 300 = 76.8 𝑚2 𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑐𝑎𝑡𝑎𝑙𝑦𝑠𝑡 = 𝑡ℎ𝑒 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑐𝑎𝑡𝑎𝑙𝑦𝑠𝑡/ 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑓 𝑐𝑎𝑡𝑎𝑙𝑦𝑠𝑡 𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑐𝑎𝑡𝑎𝑙𝑦𝑠𝑡 = 27.5 / 0.0214 = 1284 𝑐𝑚3 𝑁𝑐 = 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑠𝑐𝑟𝑒𝑒𝑛 =

𝑉𝑐 = 50 𝑠𝑐𝑟𝑒𝑒𝑛 [5] 𝑉𝑔 − (1 − £)

𝐻𝑒𝑖𝑔ℎ𝑡 = 0.5 × 50 + 1 𝑇𝑜𝑝 + 1 𝑏𝑜𝑡𝑡𝑜𝑚 = 27 𝑚 𝐴𝑟𝑒𝑎 = 𝜋 × 𝐷 × 𝐿

𝐷 =

76.8 = 1𝑚 𝜋 × 27

70

4.7 Design of Heat Exchanger (1st Cooler): The objective is to design a heat exchanger (shell and tube) to cool vapor mixture of flow rate 45873.40 kg/h from 645 0C to 70 0C. Cooling water will be used as the coolant fluid entering at 200C and exiting at 600C.

Table 16: Data of mixture

Component

Mwt

N

M

Mixture

Kg/Kmol 28.27

Kmol/h 1622.88

Kg/h 45873.40

Mole fraction Y

Mass fraction X

1

1

Table 17: Physical properties of mixture at Tave = 630.5 K and Pave = 12.4 bars

Component Mixture

μ Pa.s 3.03 x 10-5

Cp KJ/kg.K 1.1538

K W/m.K 0.0484

ρ Kg/m3 6.4106

Table 18: Physical Properties of cooling water at Tave = 40oC = 313 K

Component H2O

Cp KJ/kg.K 4.181

μ Pa.s 0.656 x 10-3

K W/m.K 0.571

Heat duty (Q) needed to be remove from gas mixture: Q = mCpmix ∆T [7] = 12.7426 × 1.1538 × (645 − 70) Q = 8453.887 KJ/s

For cooling water: Qlost = Qgain = m1 Cpmix ∆T + m2 λ

[7]

m1: mass flow rate of mixture without water = 42163.34 kg/h = 11.71 kg/s m2: mass flow rate of water which is condensing Calculating the remaining water in vapor pressure after 1st cooler:

ρ Kg/m3 992.25

H2 O vapor exit at 70 oC (yH2 O ) =

P vap [7] P total

By Antoine Equation, we find Pvap of water ln P vap = A −

B T+C

A =16.3872. B =3885.70. C =230.170 P vap = 31.256 kPa yH2 O =

31.256 1064

yH2 O = 0.029 m2 =

yH2 O (mass flow rate of mixture without H2 O) 1 − y H2 O

0.029 (42163.45) 1 − 0.029 kg kg m2 = 1259.25 ⁄h = 0.35 ⁄s m2 =

The remaining water in the exit liquid stream = H2O entering 1st cooler in stream 7 – H2O exiting vapor stream in stream 8 The remaining water = 1.03 – 0.35 = 0.68 kg/s Qlost = Qgain = m1 Cpmix ∆T + m2 λ

[7]

λ = 1954.79 kJ/kg mass of water = mwater =

Q 8453.887 kg = = 50.55 ⁄s Cp∆T 4.181 × 40

Finding∆𝐓𝐥𝐦 :

T1 = 645 oC

T2 = 70 oC

COOLER t2 = 60 oC

72

t1 = 20 oC

∆Tout = T1 − t 2 = (645 − 60) = 585 oC ∆Tin = T2 − t1 = (70 − 20) = 50 oC

∆Tlm =

∆Tout − ∆Tin ∆T ln ∆Toutt in

[7]

∆Tlm = 217.52 °C The heat exchanger will consist of one shell pass and two tubes pass. So we need a correction factor for ΔTlm: ΔTm = Ft (ΔTlm) 𝑅=

𝑇1 − 𝑇2 = 14.375 𝑡2 − 𝑡1

𝑆=

𝑡2 − 𝑡1 = 0.064 𝑇1 − 𝑡1

[7] [7]

From given Figure (12.19) in textbook for Ft as function of (R, S)  Ft = 0.97 [7] ΔTm = (0.97)(217.52) = 210.99 oC

According to Table A in Appendix E, Cold fluid is water and hot fluid is gas. The following value of overall heat transfer coefficient U can be estimated: U=300 W/m2.oC Required Area for Heat transfer: Q = U A ΔTm  A = Q / U ΔTm A= 133.56 m2

Choosing 25 mm “OD”, 21 mm “ID” Cupro-Nickel tubes to resist corrosion problems. Placing cooling water in the tube side since it is more corrosive. Take Tube length L=5 m Calculation of number of tubes (Nt): Nt =

A 133.56 = π × d × L π × 25 × 10−3 × 5

[7]

Nt = 341 tubes Since the gas mixture in the shell side is always clean, we will use triangular pitch arrangement with Pt= 1.25do [7]

Calculation of bundle diameter (Db): 1

Nt n1 Db = do [ ] K1

[7]

K1, n1 are constants given in Table C: K1 = 0.249 n1 = 2.207 1

341 2.207 Db = 25 [ ] 0.249 Db = 659.36 mm Calculating the shell diameter (Ds): Ds = Db + clearance

[7]

From Figure A in Appendix E and for split-ring floated heat type of heat exchanger we get, clearance= 66 mm

[7]

Ds = 659.36 + 63 = 722.36 mm

Calculating the heat transfer coefficients for tube side (hi): [7] 𝜋 𝜋 𝐶𝑟𝑜𝑠𝑠 𝑠𝑒𝑐𝑡𝑖𝑜𝑛𝑎𝑙 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑜𝑛𝑒 𝑡𝑢𝑏𝑒 = 𝑑𝑖2 = (21)2 = 346.36 𝑚𝑚2 4 4

74

𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑡𝑢𝑏𝑒 𝑝𝑒𝑟 𝑝𝑎𝑠𝑠 =

𝑁𝑡 341 = = 170.5 𝑡𝑢𝑏𝑒⁄𝑝𝑎𝑠𝑠 𝑁𝑜. 𝑝𝑎𝑠𝑠 2

𝑁𝑡 𝜋 𝑇𝑜𝑡𝑎𝑙 𝑐𝑟𝑜𝑠𝑠 𝑠𝑒𝑐𝑡𝑖𝑜𝑛𝑎𝑙 𝑎𝑟𝑒𝑎 𝑓𝑜𝑟 𝑤𝑎𝑡𝑒𝑟 = ( ) × 𝑑𝑖2 = 0.059 𝑚2 𝑁𝑜. 𝑝𝑎𝑠𝑠 4 𝑚𝑤𝑎𝑡𝑒𝑟 50.55 𝐾𝑔⁄ = = 856.78 𝑚2 . 𝑠𝑒𝑐 𝐴 0.059 𝐺𝑤𝑎𝑡𝑒𝑟 856.78 𝐿𝑖𝑛𝑒𝑎𝑟 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 (𝑣𝑚𝑖𝑥 ) = = = 1.374 𝑚⁄𝑠𝑒𝑐 𝜌𝑤𝑎𝑡𝑒𝑟 623.6318 𝑀𝑎𝑠𝑠 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 (𝐺𝑚𝑖𝑥 ) =

To calculated hi use the equation hi di μ 0.14 0.33 = Jh × Re × Pr ×( ) Kf μw hi = inside heat transfer coefficient di = tube diameter = 21x10-3 m Kf = thermal conductivity of water fluid = 0.571 W/m.oC Jh = factor for heat transfer given form Figure B in Appendix E by (Re) 𝜌𝑣𝑑 992.25 × 1.374 × 21 × 10−3 𝑅𝑒𝑦𝑛𝑜𝑙𝑑 𝑠 𝑁𝑢𝑚𝑏𝑒𝑟 (𝑅𝑒) = = = 43643.872 𝜇 0.656 × 10−3 ′

(Turbulent Flow) 𝑃𝑟𝑎𝑛𝑑𝑡𝑙 𝑁𝑢𝑚𝑏𝑒𝑟 (𝑃𝑟) =

𝐶𝑝 𝜇 4.181 × 103 × 0.656 × 10−3 = = 4.803 𝐾𝑓 0.571

From Figure B in Appendix E at Re = 43643.872 Jh = 2.9 x10 -3 μ

From (eq. hi) and neglecting (μ ) w

0.14

and assume it approximately 1

Then: hi (21 × 10−3 ) = (2.9 × 10−3 ) × (43643.872) × (4.803)0.33 0.571 hi = 5776.13 W/m2.oC

Calculate heat transfer coefficients for shell side (ho):

[7]

Area for cross flow of the shell side As, using Kern Method:

As = [

Pt − do ] Ds LB Pt

Pt = tube pitch do = tube outside diameter

Pt

Ds = shell inside diameter LB = Buffle spacing (usually one fifth of shell diameter)

do

do = 25 mm Pt = 1.25 do mm = 31.25 mm Ds 722.36 = = 144.47 mm 5 5 𝑃𝑡 − 𝑑𝑜 31.25 − 25 𝐴𝑠 = [ ] 𝐷𝑠 𝐿𝐵 = [ ] (722.36)(144.47) = 20871.87 𝑚𝑚2 𝑃𝑡 31.25 LB =

As = 0.02087187 m2 Mass velocity (Gs) and liner velocity (vs): mshell = mass flow rate of mixture = 12.7426 kg/sec 𝐺𝑠 = [

𝑚𝑠ℎ𝑒𝑙𝑙 12.7426 𝑘𝑔 ]=[ ] = 610.52 ⁄𝑚2 . 𝑠𝑒𝑐 𝐴𝑠 0.02087187

𝐺𝑠 610.52 𝑣𝑠 = [ ] = = 95.24 𝑚⁄𝑠𝑒𝑐 𝜌𝑠 6.4106 The equivalent (Hydraulic) diameter de: de =

1.10 2 (P − 0.917d2o ) do t

𝑑𝑒 =

1.10 [(31.25)2 − 0.917(25)2 ] = 17.75 𝑚𝑚 = 0.01775 𝑚 25

For shell side: Physical properties of liquid water at Tave: ρL = 60.03

kg⁄ m3

μL = 0.083 × 10−3 Pa. s k L = 0.47 W⁄m. K kg ρv = 0.34 ⁄m3

76

Γh =

Ws 12.7426 = = 4.98 × 10−3 Nt L 511 × 5

Number of tubes in centre row Nr = Nr =

Db 659.36 = = 21 Pt 31.25

2 × 21 = 14 3 1⁄ 3

ρL (ρL − ρv )g hc = 0.95 × k L [ ] μL × Γh

× Nr

−1⁄ 6 1⁄ 3

60.03(60.03 − 0.34) × 9.8 hc = 0.95 × 0.47 [ ] 0.083 × 10−3 × 4.98 × 10−3

× 14

−1⁄ 6

hc = 1264.34 W/m2 . oC

Calculating the overall heat transfer coefficient (U):

[7]

𝑑 𝑑𝑜 ln ( 𝑜⁄𝑑 ) 1 𝑑 1 1 1 1 𝑑𝑜 𝑜 𝑖 = + + + ( )+ ( ) 𝑈𝑜 ℎ𝑜 ℎ𝑜𝑑 2𝐾𝑤 ℎ𝑖 𝑑𝑖 ℎ𝑖𝑑 𝑑𝑖 Uo = overall coefficient based on outside area of the tubes ho = outside fluid film coefficient hi = inside fluid film coefficient do = tube outside diameter di = tube inside diameter Kw = thermal conductivity of tube wall material = 50 (W/m.oC) for ''Cupro -Nickel alloys'' hod = outside dirt " fouling " coefficient = 3000(W/m2.oC) (From Table B in Appendix E) hid = inside dirt " fouling " coefficient = 3000 (W/m2.oC) (From Table B in Appendix E) 25 × 10−3 ln(25⁄21) 1 1 1 1 25 1 25 = + + + ( )+ ( ) 𝑈𝑜 5776.13 3000 2 × 50 1264.34 21 3000 21 Uo = 529.53 W/m2 .oC The value of 529.53 W/m2.oC is well above the estimated value of 300 W/m2.oC. Hence, the present design satisfactory OK

Calculating the pressure drop (ΔP) for side tube and shell tube: [7] 1.The tube side ΔPt

78

L μ −m ρvt2 ∆Pt = NP [8jf ( ) ( ) + 2.5] [ ] di μw 2 ΔPt = tube side pressure drop (N/m2) Np = Number of tube passes jf = fraction factor ' depending on Re ' Found from Figure 12.30 in the text book assuming baffle cut of 0.25 L = length of one tube = 5 m vt = flow velocity inside the tube m = exponent value depending on type of flow at laminar (Re10000)  m = 0.14 Now, 5 992.25 × 0.7582 ∆Pt = 2 [8(3.5 × 10−3 ) ( ) + 2.5] [ ] 21 × 10−3 2 ΔPt = 5226.02 Pa = 5.22602 kPa (Acceptable)

2. The shell side ΔPs

Ds L ρvs2 μ −0.14 ∆Ps = 8jf ( ) ( ) ( )( ) De LB 2 μw ΔPs = shell side pressure drop (N/m2) jf = fraction factor ' depending on Re , Found from Figure 12.30 in the text book assuming baffle cut of 0.25 = 2.8 x 10-2 Ds = shell diameter = 522.64 mm De = equivalent diameter for shell side = 17.75 mm L = length of one tube = 5 m vs = flow velocity in shell = 100.88 m/sec LB = Baffle spacing = 104.53 mm ρ shell = 11.56 kg/m3 857.99 5 × 103 6.4106 × 67.512 ∆Ps = 8(2.8 × 10−2 ) ( )( )( ) 17.75 171.59 2 ΔPs = 4609.086 kPa (Too high!) This value for ΔPs is high, so we will decrease vs by quadrupling the baffle spacing, which will reduce the velocity in the shell by a factor of (1/3) i.e. new vs new = 0.333 vs old ΔPs α vs2 ΔPs new α (0.333)2 ΔPs old ΔPs new = (0.333)2 (4609.086) = 511.09 kPa But this effect of using half vs old will affect the hs as follows hs α Re0.8 hs now = ( 0.333 )0.8 hs old hs now = ( 0.333)0.8 (5776.13) = 2396.58 W/m2.oC The new value for the overall heat transfer coefficient will be Uo = 468.91 W/m2.oC which is still above the initial estimated value of 300 W/m2.oC then this trial for reducing ΔPs is satisfactory.

Finally, we summarize our present design as follows: 80

1. The selected heat exchanger has one shell pass and two tubes pass, in which cooling water flow inside the tube and mixture flows inside the shell “. 2. The selected tubes are made of Capper - Nickel alloy with 25 mm outside diameter, 21 mm inside diameter, the total number of tube is 341, while the tube length is 5 m, and the triangular pitch applied for this tube distribution is 31.25 mm. 3. The shell diameter Ds is 722.36 mm. and the baffles used are 25% cut, while the baffle spacing is (63 mm). 4. The first estimated value for overall heat transfer coefficient Uo is 529.53 (W/m2. oC) , while the final calculated value is 468.91 (W/m2. oC). 5. The pressure drop ( ΔPt ) for the tube side is 6.20309 kPa , while the shell side ( ΔPs ) is 511.09 kPa .

4.8 Design of The Oxidation Unit

Partial pressure of O2 = 0.714 atm Partial pressure of NO = 0.70 atm 𝐿𝑜𝑔 𝑘 = 𝑟𝐴 =

2993 − 11.323 = 2.529 × 10−3 𝑘𝑃𝑎−1 [14] 𝑇

𝑘 × (𝑃2 (𝑁𝑂) × 𝑃(𝑂2 )) 𝑅𝑇

[14]

2.529 × 10−3 𝑟𝐴 = ((80.047)2 × (71.94)) 8.314(343) 𝑟𝐴 = 0.408 130 × 1000 3600 5.2 × 1000 𝐹𝐴 = 3600 𝐹𝐴𝑜 =

𝑉=

𝐹𝐴𝑜 – 𝐹𝐴 = 84.8 𝑚3 −𝑟𝐴

4.9 Design of Heat Exchanger (2nd Cooler):

82

The objective is to design a heat exchanger (shell and tube) to cool vapor mixture of flow rate 45873.40 kg/h from 140 0C to 60 0C. Cooling water will be used as the coolant fluid entering at 250C and exiting at 400C.

Table 19: Data of mixture

Component

Mwt

N

M

Mixture

Kg/Kmol 28.27

Kmol/h 1622.88

Kg/h 45873.40

Mole fraction Y

Mass fraction X

1

1

Table 20: Physical properties of mixture at Tave = 373 K and Pave = 12.4 bars

Component Mixture

Cp KJ/kg.K 1.1085

Μ Pa.s 2.051 x 10-5

K W/m.K 0.0603

ρ Kg/m3 11.56

Table 21: Physical properties of cooling water at Tave = 32.5oC = 305 K

Component H2O

Cp KJ/kg.K 4.183

Μ Pa.s 0.771 x 10-3

Heat duty (Q) needed to be remove from gas mixture: 𝑄 = 𝑚 𝐶𝑝𝑚𝑖𝑥 𝛥𝑇 [7] = 12.7426 𝑥 1.1085 (140 – 60) 𝑄 = 1130.0138 𝐾𝐽/𝑠 For cooling water: 𝑄𝑙𝑜𝑠𝑡 = 𝑄𝑔𝑎𝑖𝑛 = 𝑚𝑤𝑎𝑡𝑒𝑟 𝐶𝑝 𝛥𝑇 [7] 𝑚𝑤𝑎𝑡𝑒𝑟 = 1130.0138 / (4.183 × (40 − 25)) 𝑚𝑤𝑎𝑡𝑒𝑟 = 18.01 𝑘𝑔/𝑠

Finding∆𝐓𝐥𝐦 :

K W/m.K 0.616

ρ Kg/m3 995.5

T1 = 140 oC

T2 = 60 oC

COOLER

t2 = 40 oC

t1 = 25 oC

∆𝑇𝑜𝑢𝑡 = 𝑇1 − 𝑡2 = (140 − 40) = 100 𝑜𝐶 ∆𝑇𝑖𝑙 = 𝑇2 − 𝑡1 = (60 − 25) = 35 𝑜𝐶

∆𝑇𝑙𝑚 =

∆𝑇𝑜𝑢𝑡 − ∆𝑇𝑖𝑛 ∆𝑇 𝑙𝑛 ∆𝑇𝑜𝑢𝑡 𝑖𝑛

[7]

∆𝑇𝑙𝑚 = 61.92 𝑜𝐶 The heat exchanger will consist of one shell pass and two tubes pass. So we need a correction factor for ΔTlm: ΔTm = Ft (ΔTlm) [7] 𝑅=

𝑇1 − 𝑇2 = 5.333 [7] 𝑡2 − 𝑡1

𝑆=

𝑡2 − 𝑡1 = 0.130 [7] 𝑇1 − 𝑡1

From given Figure (12.19) for Ft as function of (R, S)  Ft = 0.98 [7] ΔTm = (0.98)(61.92) = 60.68 oC According to A in Appendix E, Cold fluid is water and hot fluid is gas. The following value of overall heat transfer coefficient U can be estimated: 84

U= 300 W/m2.oC Required Area for Heat transfer: Q = U A ΔTm  A = Q / U ΔTm [7] A= 62.08 m2 Choosing 25 mm “OD”, 21 mm “ID” Cupro-Nickel tubes to resist corrosion problems. Placing cooling water in the tube side since it is more corrosive. Take Tube length L=5 m Calculation of number of tubes: 𝑁𝑡 =

𝐴 62.08 = 𝜋 × 𝑑 × 𝐿 𝜋 × 25 × 10−3 × 5

𝑁𝑡 = 159 𝑡𝑢𝑏𝑒𝑠 Since the gas mixture in the shell side is always clean, we will use triangular pitch arrangement with Pt= 1.25do [7] Calculation of bundle diameter Db: [7] 1

𝑁𝑡 𝑛1 𝐷𝑏 = 𝑑𝑜 [ ] 𝐾1 K1, n1 are constants given in Table C: [7] K1 = 0.249 n1 = 2.207 1

159 2.207 𝐷𝑏 = 25 [ ] 0.249 Db = 466.64 mm Calculating the shell diameter Ds: [7] 𝐷𝑠 = 𝐷𝑏 + 𝑐𝑙𝑒𝑎𝑟𝑎𝑛𝑐𝑒 From Figure A in Appendix E and for split-ring floated heat type of heat exchanger we get, clearance= 56 mm [7] 𝐷𝑠 = 466.64 + 56 = 522.64 𝑚𝑚

Calculating the heat transfer coefficients for tube side (hi): [7] 𝜋 𝜋 𝐶𝑟𝑜𝑠𝑠 𝑠𝑒𝑐𝑡𝑖𝑜𝑛𝑎𝑙 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑜𝑛𝑒 𝑡𝑢𝑏𝑒 = 𝑑𝑖2 = (21)2 = 346.36 𝑚𝑚2 4 4

𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑡𝑢𝑏𝑒 𝑝𝑒𝑟 𝑝𝑎𝑠𝑠 =

𝑁𝑡 159 = = 79.5 𝑡𝑢𝑏𝑒⁄𝑝𝑎𝑠𝑠 𝑁𝑜. 𝑝𝑎𝑠𝑠 2

𝑁𝑡 𝜋 𝑇𝑜𝑡𝑎𝑙 𝑐𝑟𝑜𝑠𝑠 𝑠𝑒𝑐𝑡𝑖𝑜𝑛𝑎𝑙 𝑎𝑟𝑒𝑎 𝑓𝑜𝑟 𝑤𝑎𝑡𝑒𝑟 = ( ) × 𝑑𝑖2 = 0.028 𝑚2 𝑁𝑜. 𝑝𝑎𝑠𝑠 4 𝑚𝑤𝑎𝑡𝑒𝑟 18.01 𝐾𝑔⁄ = = 643.21 𝑚2 . 𝑠𝑒𝑐 𝐴 0.028 𝐺𝑤𝑎𝑡𝑒𝑟 643.21 𝐿𝑖𝑛𝑒𝑎𝑟 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 (𝑣𝑚𝑖𝑥 ) = = = 0.646 𝑚⁄𝑠𝑒𝑐 𝜌𝑤𝑎𝑡𝑒𝑟 995.5 𝑀𝑎𝑠𝑠 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 (𝐺𝑚𝑖𝑥 ) =

To calculated hi use the equation ℎ𝑖 𝑑𝑖 𝜇 0.14 0.33 = 𝐽ℎ × 𝑅𝑒 × 𝑃𝑟 ×( ) 𝐾𝑓 𝜇𝑤 hi = inside heat transfer coefficient di = tube diameter = 21x10-3 m Kf = thermal conductivity of water fluid = 0.616 W/m.oC Jh = factor for heat transfer given form Figure B in Appendix E by (Re) 𝑅𝑒𝑦𝑛𝑜𝑙𝑑 ′ 𝑠 𝑁𝑢𝑚𝑏𝑒𝑟 (𝑅𝑒) =

𝜌𝑣𝑑 995.5 × 0.646 × 21 × 10−3 = = 17516.15 𝜇 0.771 × 10−3

(Turbulent Flow) 𝐶𝑝 𝜇 4.181 × 103 × 0.771 × 10−3 𝑃𝑟𝑎𝑛𝑑𝑡𝑙 𝑁𝑢𝑚𝑏𝑒𝑟 (𝑃𝑟) = = = 5.233 𝐾𝑓 0.616 From Figure B in Appendix E at Re=17516.15 Jh = 3.8 x10-3 From (eq. hi) and neglecting (

 0.14 ) and assume it approximately 1 w

Then: hi (21 × 10−3 ) = (3.8 × 10−3 ) × (17516.15 ) × (5.233)0.33 0.616 hi = 3371.09 W/m2.oC

Calculate heat transfer coefficients for shell side (ho): [7] Area for cross flow of the shell side As, using Kern Method:

86

𝐴𝑠 = [

𝑃𝑡 − 𝑑𝑜 ] 𝐷𝑠 𝐿𝐵 𝑃𝑡

Pt = tube pitch do = tube outside diameter Ds = shell inside diameter

Pt

LB = Buffle spacing (usually one fifth of shell diameter) do = 25 mm Pt = 1.25 do mm = 31.25 mm 𝐷𝑠 522.64 = = 104.53 𝑚𝑚 5 5 𝑃𝑡 − 𝑑𝑜 31.25 − 25 𝐴𝑠 = [ ] 𝐷𝑠 𝐿𝐵 = [ ] (522.64)(104.53) = 10926.31 𝑚𝑚2 𝑃𝑡 31.25 𝐿𝐵 =

As = 0.01092631 m2 Mass velocity (Gs) and liner velocity (vs): mshell = mass flow rate of mixture = 12.7426 kg/sec 𝐺𝑠 = [

𝑚𝑠ℎ𝑒𝑙𝑙 12.7426 𝑘𝑔 ]=[ ] = 1166.23 ⁄𝑚2 . 𝑠𝑒𝑐 𝐴𝑠 0.01092631

𝐺𝑠 1166.23 𝑣𝑠 = [ ] = = 100.88 𝑚⁄𝑠𝑒𝑐 𝜌𝑠 11.56 The equivalent (Hydraulic) diameter de: 𝑑𝑒 =

1.10 2 (𝑃𝑡 − 0.917𝑑𝑜2 ) 𝑑𝑜

𝑑𝑒 =

1.10 [(31.25)2 − 0.917(25)2 ] = 17.75 𝑚𝑚 = 0.01775 𝑚 25

Re & Pr for shell side 𝑅𝑒 =

𝜌𝑣𝑠 𝑑𝑒 11.56 × 100.88 × 0.01775 = = 1009242.67 𝜇 2.051 × 10−5

𝐶𝑝 𝜇𝑠ℎ𝑒𝑙𝑙 1.1085 × 103 × 2.051 × 10−5 𝑃𝑟 = = = 0.377 𝐾𝑓 𝑠ℎ𝑒𝑙𝑙 0.0603

The shell side coefficient (hs or ho):

do

1 ℎ𝑠 𝑑𝑜 𝜇 0.14 = 𝐽ℎ × 𝑅𝑒 × 𝑃𝑟 3 × ( ) 𝐾𝑓 𝑠ℎ𝑒𝑙𝑙 𝜇𝑤

From Figure B in Appendix E at 25% assumed design value for baffle cut and Re: Jh = 6.2x10-4

(

 0.14 ) Assumed to be approximately 1. We get w

ℎ𝑠 (25 × 10−3 ) = 6.2 × 10−4 × 1009242.67 × 0.3770.33 0.0603 hs = 1093.85 W/m2.oC

Calculating the overall heat transfer coefficient (U): [7] 88

𝑑 𝑑𝑜 ln ( 𝑜⁄𝑑 ) 1 𝑑 1 1 1 1 𝑑𝑜 𝑜 𝑖 = + + + ( )+ ( ) 𝑈𝑜 ℎ𝑜 ℎ𝑜𝑑 2𝐾𝑤 ℎ𝑖 𝑑𝑖 ℎ𝑖𝑑 𝑑𝑖 Uo = overall coefficient based on outside area of the tubes ho = outside fluid film coefficient hi = inside fluid film coefficient do = tube outside diameter di = tube inside diameter Kw = thermal conductivity of tube wall material = 50 (W/m.oC) for ''Cupro -Nickel alloys'' hod = outside dirt " fouling " coefficient = 3000(W/m2.oC) (From Table B) hid = inside dirt " fouling " coefficient = 3000 (W/m2.oC) (From Table B) 25 × 10−3 ln(25⁄21) 1 1 1 1 25 1 25 = + + + ( )+ ( ) 𝑈𝑜 3371.09 3000 2(50) 1093.85 21 3000 21 Uo = 463.24 W/𝑚2 .oC The value of 463.24 W/m2.oC is well above the estimated value of 300 W/m2.oC. Hence, the present design satisfactory OK

Calculating the pressure drop (ΔP) for side tube and shell tube: [7] 1.The tube side ΔPt

L μ −m ρvt2 ∆Pt = NP [8jf ( ) ( ) + 2.5] [ ] di μw 2 ΔPt = tube side pressure drop (N/m2) Np = Number of tube passes jf = fraction factor ' depending on Re ' Found from Figure 12.30 in the text book assuming baffle cut of 0.25 L = length of one tube = 5 m vt = flow velocity inside the tube m = exponent value depending on type of flow at laminar (Re10000)  m = 0.14 Now, 5 995.5 × 0.6462 ∆Pt = 2 [8(4.2 × 10−3 ) ( ) + 2.5] [ ] 21 × 10−3 2 ΔPt = 4362.09 Pa =4.36209 kPa (Acceptable)

2. The shell side ΔPs

90

Ds L ρvs2 μ −0.14 ∆Ps = 8jf ( ) ( ) ( )( ) De LB 2 μw ΔPs = shell side pressure drop (N/m2) jf = fraction factor ' depending on Re , Found from Figure 12.30 in the text book assuming baffle cut of 0.25 = 2.5 x 10-2 Ds = shell diameter = 522.64 mm De = equivalent diameter for shell side = 17.75 mm L = length of one tube = 5 m vs = flow velocity in shell = 100.88 m/sec LB = Baffle spacing = 104.53 mm ρ shell = 11.56 kg/m3 522.64 5 × 103 11.56 × 100.882 ∆Ps = 8(2.5 × 10−2 ) ( )( )( ) 17.75 104.53 2 ΔPs = 16569.19 kPa (Too high!) This value for ΔPs is high, so we will decrease vs by quadrupling the baffle spacing, which will reduce the velocity in the shell by a factor of (1/3) i.e. new vs new = 0.333 vs old ΔPs α vs2 ΔPs new α (0.333)2 ΔPs old ΔPs new = (0.333)2 (16569.19) = 1837.34 kPa But this effect of using half vs old will affect the hs as follows hs α Re0.8 hs now = ( 0.333 )0.8 hs old hs now = ( 0.333)0.8 (3371.09) = 1398.70 W/m2.oC The new value for the overall heat transfer coefficient will be Uo = 388.04 W/m2.oC which is still above the initial estimated value of 300 W/m2.oC then this trial for reducing ΔPs is satisfactory.

Finally, we summarize our present design as follows:

1. The selected heat exchanger has one shell pass and two tubes pass, in which cooling water flow inside the tube and mixture flows inside the shell “. 2. The selected tubes are made of Capper - Nickel alloy with 25 mm outside diameter, 21 mm inside diameter, the total number of tube is 159, while the tube length is 5 m, and the triangular pitch applied for this tube distribution is 31.25 mm. 3. The shell diameter Ds is 522.64 mm. and the baffles used are 25% cut, while the baffle spacing is (56 mm). 4. The first estimated value for overall heat transfer coefficient Uo is 463.24 (W/m2 . oC) , while the final calculated value is 388.04 (W/m2. oC). 5. The pressure drop ( ΔPt ) for the tube side is 4.36209 kPa , while the shell side ( ΔPs ) is 1837.34 kPa .

4.10 Design of Absorber 92

Introduction The most important step of manufacturing nitric acid is Absorption of NOx When it is compared to other absorptions operation. Absorption of NOx is the most complex. This is for several reasons:

1. NOx is a mixture of several components of NO,NO2,N2O3 and N2O4. 2. In absorption tower many reversible and irreversible reactions. 3. Simultaneous absorption of many gases occurs followed be chemical reaction. 4. Heat Generated from the reactions, which affect the absorption. 2NO2 + H2O + ½ O2  2HNO3

Figure 1 Nitric acid column

The objective

To design an absorption tower to absorb NO2 from the mixer gas stream using water to form nitric acid with 60% weight. Absorber design calculation

Tail Gases 30 ○C Y2 = 0.0048

Water 20 ○C X2 = 0

Gases 60 ○C Y1 = 0.0884

Nitric Acid 30 ○C X1 = 0.3

Figure 2 Stream temperature and NO2 molar composition

Number of Gases moles entered the absorber= Gm= 1417.09 Kmol/hr Number of water moles entered the absorber=Lm=373.19 Kmol/hr

Table 22: Compositions of income gases

Component NO2 NO O2 N2

Mole fraction 0.0884 0.0037 0.0386 0.869

Molecular weight 46 30 32 28

94

𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑀𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑊𝑒𝑖𝑔ℎ𝑡 (𝑀𝑤𝑡) = (46 × 0.0884) + (30 × 0.0037) + (32 × 0.0386) + (0.869 × 28) = 29.74

𝐾𝑔 𝐾𝑚𝑜𝑙

𝐷𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑓 𝐺𝑎𝑠𝑒𝑠 (𝑎𝑠𝑠𝑢𝑚𝑖𝑛𝑔 𝑖𝑑𝑒𝑎𝑙 𝐺𝑎𝑠) = 𝜌𝑣 =

𝑃 × 𝑀𝑤𝑡 1200000 × 29.74 = 𝑅×𝑇 8314.34 × 333

= 12.89 𝐾𝑔/𝑚3 T=60 C○ = 333 K

P= 1200000 Pa

R= 8314.34 𝑚3 × 𝑃𝑎/𝐾𝑚𝑜𝑙 × 𝐾

Density of water at 20 C○, 1200 kPa = 𝜌𝑙 = 998.23 Kg/m3 [3] Diameter Calculation 𝑈𝑣 = (−0.171 𝐿𝑡 2 + 0.21𝐿𝑡 − 0.047) (

𝜌𝐿−𝜌𝑉 0.5 𝜌𝑉

)

[7]

Lt=plat spacing=0.9 m 𝑈𝑣 = (−0.171 ×

(0.9)2

998.23 − 12.89 0.5 + 0.21 × 0.9 − 0.047) ( ) = 0.51 𝑚/𝑠 12.89

Taking 60% Uv =0.3 m/s 4𝑉𝑚

𝐷 = √𝜋𝜌

𝑚 𝑈𝑣

𝐷=√

[7]

4 × 11.71 = 1.96 ≈ 2𝑚 3.14 × 12.98 × 0.3

Design 1 Assumptions 1. N2O4 (2NO2) is the only species, which dissolved in the Liquid. 2. Neglect the reactions occurs in the column. 3. Isothermal Operation and the temperature is 45oC. 4. The design of the column is trays absorption column. Height of Absorber Equilibrium line can be specified: 𝑌 = 𝑚𝑋 From Raoult's law 𝑃 𝑠𝑎𝑡

𝑚 = 𝑃𝑡𝑜𝑡𝑎𝑙

[18]

Psat for NO2 at 45oC: From Antoine Equation: 𝐿𝑛 𝑃 𝑠𝑎𝑡 = 𝐴 −

𝐵 𝑇+∁

[7]

From appendix C page. 1123 [7] A=20.5324 B= 4141.29 C=3.65 T= 45oC= 318 k

𝐿𝑛 𝑃 𝑠𝑎𝑡 = 20.5324 −

4141.29 = 7.6572 318 + 45

Psat = 2115.944 mmHg = 282.01 kPa So, the slope of the equilibrium line 𝑚=

282.01 = 0.234 1200

96

𝑚𝐺 𝑌 − 𝑚𝑋 𝑚𝐺 ln[(1 − 𝐿 𝑚 ) (𝑌1 − 𝑚𝑋2 ) + 𝐿 𝑚 ] 𝑚 2 2 𝑚 𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑠𝑡𝑎𝑔𝑒 = 𝑁 = 𝐿𝑚 ln(𝐿 𝐺 ) 𝑚 𝑚

[19]

m = 0.235

Gm = 1417.09 kmol/hr

Lm = 373.191 Kmol/hr

Y1 = 0.0048

Y2 = 0.0884

X2 = 0

𝑁=

0.0884 − 0 ln [(1 − 0.893) (0.0048 − 0) + 0.893] 1 ln (0.893)

= 9.3 ≈ 10 𝑠𝑡𝑎𝑔𝑒𝑠

𝐻𝑖𝑔ℎ𝑡 𝑜𝑓 𝐴𝑏𝑠𝑜𝑟𝑏𝑒𝑟 = 𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑠𝑡𝑎𝑔𝑒𝑠 × 𝑃𝑙𝑎𝑡𝑒 𝑠𝑝𝑎𝑐𝑖𝑛𝑔 = 𝑁 × 𝐿𝑡 = 10 × 0.9 = 9 𝑚

Design 2: Assumptions: 1. N2O4 (2NO2) is the only species, which dissolved in the Liquid. 2. The reaction occurs in the column. 3. 2NO2 + H2O + ½ O2  2HNO3 4. Isothermal Operation and the temperature is 45oC. 5. The design of the column is trays absorption column. Height of Absorber: From fuller equation, the diffusivity of N2O4 in water: A = N2O4

B= H2O

VA= 33.5

VB=14.7

MA= 92

MB=18

T=45 C○ =318 K○

P=1200 Kpa = 11.843 atm

𝐷𝐴𝐵

1 1 0.5 1 × 10−7 × 𝑇 1.75 × (𝑀 + 𝑀 ) 𝐴 𝐵 = [17] 0.33 0.33 )2 𝑃(𝑉𝐴 + 𝑉𝐵

𝐷𝐴𝐵 =

1 1 0.5 1 × 10−7 × 3181.75 × (92 + 18) 11.843(33.50.33 + 14.70.33 )2

𝐷𝐴𝐵 = 1.6535 × 10−6 𝑚2 /𝑠 Mass transfer Coefficient can be calculated from the following Equation: 0.33

(𝜌𝑙 − 𝜌𝑣 ) × 𝜇𝐴 × 𝑔 0.42 × 𝑁𝑆𝐶 −0.5 𝐾𝑔 = ×[ ] 𝑅𝑇 𝜌𝐴2 𝑁𝑆𝐶 = 𝑆𝑐ℎ𝑚𝑖𝑑𝑡 𝑛𝑢𝑚𝑏𝑒𝑟 =

[17]

𝜇𝐴 2.3 × 10−5 = = 13.3 𝜌𝐴 × 𝐷𝐴𝐵 1 × 1.6535 × 10−6

[3]

𝜇𝐴 = 𝑉𝑖𝑠𝑐𝑜𝑠𝑖𝑡𝑦 𝑜𝑓 𝑁2 𝑂4 = 2.2 × 10−5 𝐾𝑔/𝑚. 𝑠 𝜌𝑣 = 12.89 𝑘𝑔/𝑚3 𝜌𝑙 = 998.23 𝐾𝑔/𝑚3 𝜌𝐴 = 1.0 𝑘𝑔/𝑚3 𝑁𝑆𝐶 = 13.3 𝑔 = 𝐺𝑟𝑎𝑣𝑖𝑡𝑦 𝑎𝑐𝑐𝑒𝑙𝑟𝑎𝑡𝑖𝑜𝑛 = 9.8 𝑚/𝑠 2 𝑚3 𝑎𝑡𝑚

R= 𝐺𝑎𝑠 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 = 8.2 × 10−2 𝑘𝑚𝑜𝑙𝑒 𝑘 98

[(998.23 − 12.89) × 2.2 × 10−5 × 9.8]0.33 0.42 × 13.3−0.5 𝐾𝑔 = × 8.2 × 10−2 × 318 12 𝐾𝑔 = 2.6484 × 10−3 𝑘𝑚𝑜𝑙/𝑚2 . 𝑎𝑡𝑚. 𝑠𝑒𝑐

Determine the rate reaction 𝑟𝐴 =

𝑃𝑁2 𝑂4 × 𝑠 1 1 𝐾𝑔 + 𝐻√𝐾1 × 𝐷𝐴𝐵

[4]

Kg= mass transfer coefficient, kmol/m2.atm.sec PN2O4= partial pressure of N2O4 at interface, atm H= solubility of N2O4 in liquid = 0.352 kmol/m3.atm [19, page.275] K1= reaction rate constant= 1340 sec-1 [19, page.275] 𝐾𝑔 = 𝑚𝑎𝑠𝑠 𝑡𝑟𝑎𝑛𝑠𝑓𝑒𝑟 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡, 𝑘𝑚𝑜𝑙/𝑚2 . 𝑎𝑡𝑚. 𝑠𝑒𝑐

𝐻√𝐾1 × 𝐷𝐴𝐵 = 0.0166 3 𝑆 = 700 𝑚 ⁄𝑚2 𝑟𝑒𝑎𝑐𝑡𝑜𝑟 𝑣𝑜𝑙𝑢𝑚𝑒 1 × 700 −𝑟𝐴 = = 1.6 𝑃𝑎 1 1 + 2.6484 × 10−3 0.0166 𝑃𝐴1

𝐺 𝑑𝑃𝐴 𝐻𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑎𝑏𝑠𝑜𝑟𝑏𝑒𝑟 (𝑍) = ∫ 𝑃×𝑓 −𝑟𝐴

[7]

𝑃𝐴2

𝑓 =

𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑙𝑖𝑞𝑢𝑖𝑑 𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑙𝑖𝑞𝑢𝑖𝑑 + 𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑔𝑎𝑠 3

3 0.0019 𝑉 ⁄𝑠 −3 𝑚 𝑙𝑖𝑞𝑢𝑖𝑑⁄ 𝑓= = 2.05 × 10 3 3 𝑡𝑜𝑡𝑎𝑙 𝑣𝑜𝑙𝑢𝑚𝑒 0.0019 𝑉 ⁄𝑠 + 0.91 𝑉 ⁄𝑠

G = molar velocity of gas based on the area of the absorber 𝐺𝑚 = 0.3936 𝐾𝑚𝑜𝑙/𝑠. 𝑚2 𝐺=

𝐺𝑚 0.3936 =𝜋 = 0.1253 𝑘𝑚𝑜𝑙⁄𝑚2 × 𝑠 2 𝐴 × 2 4

𝑃𝑁2 𝑂4 𝑎𝑡 𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛 = 0.7 𝑎𝑡𝑚 𝑃𝑁2 𝑂4 𝑎𝑡 𝑜𝑢𝑡𝑙𝑒𝑡 = 1.3 × 10−1 𝑎𝑡𝑚

𝑍=

0.1253 0.7 ln ( ) = 27.7 𝑚 ≈ 28 𝑚 −3 11.843 × 2.05 × 10 × 1.6 1.3 × 10−4

100

4.11 Design of Nitric Acid Tank 𝐷𝑖 𝐷𝑜 (𝑍 0.5 − 1) 0.5 𝑡 = ( ) (𝑍 − 1) = ( ) [ ] 2 2 𝑍 0.5

[7]

Where: t = wall thickness (mm) Di = tank inside diameter (mm) D0 = tank outside diameter (mm) Z = (FE + P)/ (FE - P) F = design tensile strength of material (MPa) E = joint efficiency P = internal pressure (MPa) The internal pressure is equivalent to the head of liquid inside the tank. This maximum head of 10.7 m represents a pressure of 142 kPa. The recommended wall thickness is: 𝑍 = 𝑍=

(𝐹𝐸 + 𝑃) [7] (𝐹𝐸 − 𝑃) [(108 × 0.8) + 0.142] [(108 × 0.8) − 0.142]

𝑍 = 1.003 15200

𝑡 = (

2

) (1 .0030.5 − 1) = 11 𝑚𝑚

Adding a corrosion allowance of 5 mm, the final recommendation is for 16 mm plating for the tank shell based on circumferential stress. Minimum thickness based upon longitudinal stress (circumferential joints) Formula as given above for circumferential stress, except that: 𝑍 = [

𝑃 ] + 1 [7] 𝐹𝐸

With the internal pressure at 140 kPa, the calculation is: 𝑃 ] + 1 𝐹𝐸 0.140 𝑍 = [ ] + 1 108 × 0.8 𝑍 = [

𝑍 = 1.0016 𝑡 = (

15200 ) (1.00160.5 − 1) = 6 𝑚𝑚 2

This indicates that circumferential stresses are more important, and they determine the minimum shell thickness for this tank. The BHP steel plate available and closest to this specification is 16 mm thickness.

Checking Calculations for Other Stresses Dead- weight stress 𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑚𝑒𝑡𝑎𝑙 = 𝑇𝑎𝑛𝑘 𝑠ℎ𝑒𝑙𝑙 + 𝑇𝑎𝑛𝑘 𝑟𝑜𝑜𝑓 𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑚𝑒𝑡𝑎𝑙 = [𝜋{15.2 + (2 × 0.016)} 0.016 × 10.71 𝜋 + 1.3 [( ) 15.2 2 × 0.0161]] [7] 4 𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑚𝑒𝑡𝑎𝑙 = 12 𝑚3 Density of SS304L = 7.8 tones/m3 𝑀𝑎𝑠𝑠 𝑜𝑓 𝑡𝑎𝑛𝑘 = 𝑉𝑜𝑙𝑢𝑚𝑒 × 𝐷𝑒𝑛𝑠𝑖𝑡𝑦 = 12 × 7.8 = 93.6 𝑡𝑜𝑛𝑒𝑠 Considering the stresses: 𝑚𝑔 Weight stress = 𝜋 𝐷𝑜 𝑡 Weight stress =

93.6 x 9.8

π (15.2 + (2 x 0.016)) 0.016

Weight stress = 1200 kPa Weight stress = 1.2 MPa 𝐴𝑥𝑖𝑎𝑙 𝑠𝑡𝑟𝑒𝑠𝑠 = Axial stress =

𝑃𝐷𝑜 4𝐽𝑡𝑠 0.140 × {15.2 + (2 × 0.016)} 4 × 1 × 0.016

Axial stress = 33 MPa Hoop stress = 2 × Axial Stress = 66 MPa Wind stresses are disregarded because of the high ratio of tank diameter to tank height. Analysis of Stresses: Upwind Total stress = Axial stress - Radial stress = 33 - 1.2 = 31.8 MPa Downwind Total stress = Hoop stress - Radial stress = 66 - 1.2 = 64.8 MPa Radial Radial stress = 0.5 × P = 70 kPa 102

Maximum stress = 70 MPa The maximum stress is approximately 35% below the design stress of 108 MPa, and therefore the shell thickness used in this design is considered acceptable. Inlet/Outlet Line Diameters The inlet and outlet line diameters are sized for a recommended liquid flowrate of 2 m/s. The inlet line sizing is determined by allowing 20% above the normal product flowrate of 11 700 kg/h. This corresponds to a volumetric flowrate of 2.9 × 10-3 m3/s. Area of pipe = Volumetric flowrate/Velocity 2.9 × 10−3 𝐴𝑟𝑒𝑎 𝑜𝑓 𝑝𝑖𝑝𝑒 = = 1.44 × 10−3 𝑚2 2 𝐴𝑟𝑒𝑎 × 4 0.5 Diameter of pipe = [ ] 𝜋 0.5

1.44 × 10−3 × 4 [ ] 𝜋

= 40mm

The nearest commercial pipe size is a nominal pipe size of 1.5, schedule number 40s (with inside diameter of 41 mm and a wall thickness of 4 mm). The outlet flowrate is calculated based upon the need to fill a standard 30 tonne tanker in 15 minutes. This corresponds to a volumetric flowrate of 2.5 x 10-2 m3/s Area of pipe = Volumetric flowrate/Velocity = 2.5 × 10-2/ 2 = 1.23 × 10-2 Diameter of pipe = [

𝐴𝑟𝑒𝑎 × 4 0.5 ] 𝜋

0.5

1.44 × 10−3 × 4 [ ] 𝜋

= 40mm

The nearest commercial pipe size is a nominal pipe size of 6, schedule number 120 (with an inside diameter of 140 mm and a wall thickness of 14 mm).

Figure 7 Tank of nitric acid [5]

104

Chapter 5

Control Loop

5.1 Introduction Control is crucial because it provides a proper control of the process variables so that the whole plant operation is within the specification limit for safety purposes. Besides, it has direct impacts on the plant economics, environment and the product specifications. According to Seborg (2004), without process control, it is impossible to operate most modern processes safely and profitably, while satisfying plant quality standards.

The three most important considerations that make plant control necessary are:

1. Safe Plant Operation 

To keep the process variables within known safe operation limits



To make sure the temperature and pressure at the desired level for reactor and columns



To detect danger as they develop and to provide alarms and automatic shutdown systems



To provide interlocks and alarms to prevent dangerous operating conditions

2. Production Rate and Quality 

To achieve the design product output composition specification for outlet stream of reactor according to the literature specification



To maintain the desired purity especially for ethyl acetate and acetic acid

3. Cost Operation 

To operate at the lowest production cost, commensurate with the other Objectives



To minimize the usage of heating and cooling utility, and thus reducing the utility cost

The types of basic control strategies that is implemented under the scope of this project are: 

Feedback control



Feedforward control



Cascade control



Ratio control 106

The simplest control loops requires: 1- Identifying the controlled variable. 2- Setting the required value of the controlled variable. 3- Measuring the controlled variable 4- Identifying the magnitude of deviation (set value – measured value). 5- Sending the deviation to the controller. 6- Adjusting the value of manipulated variable to offset the deviation. Table 23: Symbols used in control loops

Symbol

Definition Control Valve Normally Closed Gate Valve

LC

Level Control

TC

Temperature Control

FC

Flow Control

TT

Temperature Transmitter

AC

Analyzer Control

PC

Pressure Control

PT

Pressure Transmitter

Figure 8 Process control loop for nitric acid production

108

5.2 Control of Vaporizer Table 24: Control specification of vaporizer

Control Strategy Selected



Feedback control

Control Objective



To set the output temperature to 35 ⁰C

Controlled Variables



Output temperature

Measured Variables



Output temperature

Manipulated Variables



The flowrate of steam inside the vaporizer



The flowrate of ammonia inside the vaporizer

Disturbances



The Temperature of ammonia inside the vaporizer

Set Points



35⁰C inlet temperature to Superheater.

5.3 Control of Reactor Table 25: Control specification of reactor

Control Strategy Selected



Feedback control

Control Objective



To set the output temperature to 645 ⁰C



To set the reactor pressure to 1060 kPa



Output temperature



Reactor pressure



Output temperature



Reactor pressure



The flow rate of Cooling water in the

Controlled Variables

Measured Variables

Manipulated Variables

reactor 

The flow rate of vented gas



The flow rate of air and ammonia



The Temperature of air and ammonia

Disturbances

Set Points

inside the reactor



645 ⁰C inlet temperature to cooler.

110

5.4 Control of Absorber Table 26: Control specification of absorber

Control Strategy Selected



Feedback control

Control Objective



To set the output temperature to 30 ⁰C



To set the reactor pressure to 1200 kPa



Output temperature



Absorber pressure



Output temperature



Absorber pressure



The flow rate of Cooling water in the

Controlled Variables

Measured Variables

Manipulated Variables

Absorber 

The flow rate of vented tail gas



The flow rate of multi gases



The Temperature of multi gases inside the

Disturbances Set Points

absorber 

30 ⁰C product temperature

Chapter 6

Economic Analysis

112

6.1 Estimation of capital investment items based on delivered equipment [16] ITEM *DIRECT COSTS Purchased equipment cost (PEC)

PEC

Purchased equipment Installation

0.47 PEC

Instrumentation and controlled

0.18 PEC

Piping (Installed)

0.66 PEC

Electrical (Installed)

0.11 PEC

Buildings (including services)

0.18 PEC

Yard improvements

0.10 PEC

Land

0

*TOTAL DIRECT PLANT COST

3.4 PEC

*INDIRECT COSTS Engineering and supervision

0.33 PEC

Construction expenses

0.41 PEC

TOTAL DIRECT AND DIRECT PLANT COSTS

4.14 PEC

Contractor fee

0.207 PEC

Contingency

0.414 PEC

Fixed capital investment (FCI)

4.761 PEC

Working capital investment (Wc)

0.84 PEC

Total capital investment (TCI)

5.601 PEC

6.2 Equipment Cost (at 2011) [16] Equipment

Material

Purchase Cost $

Compressor

SS

1,700,000

Vaporizer

SS

38,760

Superheater

SS

78,702

Reactor

SS

227,587

Cooler 1

Cuppro-Nickel

286,868

Oxidation

SS

115,838

Cooler 2

Cuppro-Nickel

212,937

Absorption Column

SS

187,384

Nitric Acid Tank

SS

1,396,861

Σ Cost $

I0 (Index at 2011) = 586.5

4,244,937

I (Index at 2013) = 636

Cost (in 2013) = Cost (in 2011) × (I / I0) ΣCost $ (at 2013) = 4603206 The Location Factor = 1.3 PEC = Σ cost (at 2013) × 1.3 PEC = $5,984,168

114

6.3 Total Production Cost (TPC) [16] TPC = Manufacturing Cost + General Cost 1) Manufacturing Cost A) Direct Production Cost 1) Raw Material

$9343520

2) Operating labor

0.1TPC

3) Direct Supervisory

0.11TPC

4) Utilities

0.1 TPC

5) Maintenance

0.05FCI

6) Operating Supplies

0.005FCI

7) Liberty Charges

0.1TPC

8) Patients & Royalties

0

B) Fixed Charges 1) Deprecation

0.1FCI

2) Taxes

0

3) Insurance

0.005FCI

4) Rent

0

C) Plant Overhead Cost

0.05TPC

2) General Costs a) Administration

0.02TPC

b) Distribution

0.05TPC

c) Research & Development Costs

0.05TPC

d) Financing

0

Fixed Capital Investment (FCI) FCI = 4.761 × PEC FCI = 28490624$ Working Capital Investment (WC) WC = 0.84 × PEC WC = $5026702

Total Capital Investment (TCI) TCI = 5.601 × PEC TCI = $33,517,325 Total Production Cost (TPC) TPC = Manufacturing Cost + General Cost TPC = 0.58TPC + 0.16FCI + 9,343,520 TPC = $33,100,048 Revenue (total income) Revenue = Price of Nitric Acid (per ton) × Capacity Revenue = 400 × 100,000 Revenue = $40,000,000 Net Profit Net Profit = Revenue - TPC Net Profit = 400,00,000 – 33,100,048 Net Profit = $6,899,953 Return on Investment (ROI) ROI = ((Net Profit)/TCI) ×100 ROI = 20.5%

Depreciation = 0.1 × FCI = $2,849,062.40 ACF = Annual Profit + Depreciation = 6,899,953 + 2,849,062.40 = $9,749,015.40

Pay Back Period (PBP) PBP = TCI/ACF = 3.43 years

116

Chapter 7

Process Integration

INTRODUCTION TO PROCESS INTEGRATION The process industrial is among the most important manufacturing facilities. They span a wide range of industries including chemical, petroleum, gas, petrochemical, food, pharmaceutical, microelectronics, metal, textile, and forestry product. The performance of these industries is strongly dependent on their engineering and engineers. So, what are primary responsibilities of process engineers in the process industries? Many process engineers would indicate that their role in the process industries is to design and operate industrial processes and make them work faster, better, cheaper, safer, and greener. All of these tasks lead to more competitive processes with desirable profit margins and market share. Specifically, these responsibilities may be expressed through the following specific objectives: process innovation, profitability enhancement, Yield improvement, capital productivity increase, Quality control, assurance

and

enhancement

Resource

conservation,

Pollution

prevention,

safety,

Debottlenecking.

Heat Integration –Objectives Heat integration, as a part of process integration aims at: 

Designing a process in such a way that it makes optimum use of energy that is available within the process itself



Optimizing the quality level of the utilities required



Optimizing energy exchange with neighboring plants and energy recovery with the view to optimize energy and capital cost

However, our aim was to use the steam, which occurred when the reactor cooled down by water, which leaded us to use the steam to vaporize the ammonia.

Stripping Based on our project, we avoid using the stripping due to the problems, which can be caused, by stripping such as increasing the purity of nitric acid, which can be ended up by killing the plant. Also, this project had designed to use the nitric acid (60% purity) under specific conditions on other words. If we use the nitric acid more than 60% purity we can use it as bombs, and if we use it as 60% purity we can use it as fertilizer. 118

Chapter 8

Safety and Loss Prevention

120

8.1 Plant Layout There are several plant layout considerations relevant to the Bunbury site. First, the prevailing winds direction, shown with a dark arrow in Figure 9. This would indicate that the tank farm is best placed on the seaside of the site to avoid vapors drifting back across the plant area. The second major consideration is the need for road access to the ammonium nitrate dispatch/storage area. A suggested layout for the overall complex is shown in Figure 9.It features the three chemical processes flowing approximately one to the other. The joint administration and workshop facilities are located in the southwestern corner of the plant. There is a central control room for all operations, and the roadway is extended into the dispatch area. The nitric acid plant is relatively small and should not occupy more than 1 hectare. Even this small area should leave plenty of room for expansion. The plant layout includes space for a parallel production trucks should the market expand sufficiently in the future such that a second plant is required. The layout also offers plenty of room around each unit for maintenance work, with units remaining closes enough for process streams to be transferred easily. The main objectives of the plant layout are: 

To maximize safety



To prevent spread of fire



To facilitate easy operation and maintenance



To consider future expansion



To minimize the construction cost

Figure 9 Nitric acid plant layout

122

8.2 Safety of Materials Ammonia Both gaseous and liquid ammonia pose moderate health hazards to those who come into contact with them. For example, farmers who handle liquid ammonia risk the possibility of painful blistering of the skin or damage to the mucous membranes if they come into contact with the fertilizer. Ammonia fumes can irritate the mouth, nose, and throat, causing coughing and gagging responses. Higher levels of exposure may irritate the lungs, resulting in shortness of breath and producing headaches, nausea, and vomiting. Very high exposures can cause a buildup of fluid in the lungs that can result in death. Since ammonia is a common ingredient of many household products, everyone should be aware of its health risks, although the threat posed by such products is, in fact, very small. [15]

Nitric oxide Nitric oxide is considered an environmental pollutant. It oxidizes readily to form nitrogen dioxide, which, in turn, reacts with moisture in the air to form nitric acid, a component of acid rain. Acid rain is thought to be responsible for a number of environmental problems, including damage to buildings, destruction of trees, and the death of aquatic life. The nitrogen dioxide produced from nitric oxide is also a primary component of photochemical smog, a hazardous haze created by a mixture of pollutants in the presence of sunlight. Even though it is toxic in the environment, nitric oxide plays several important roles in the human body. Nitric oxide is involved in the process by which messages are transmitted from one nerve cell to the next. It also regulates blood flow by triggering the smooth muscles surrounding blood vessels to relax. This action increases blood flow and lowers blood pressure. Nitric oxide also prevents the formation of blood clots, which can break off and travel to the heart or brain, increasing the risk of heart attack or stroke. Finally, nitric oxide plays a role in memory and learning. A deficiency of the compound appears to be related to the development of learning problems. On the other hand, an excess of nitric oxide has been implicated in the development of certain diseases, such as Huntington’s

chorea, an inherited disorder characterized by unusual body movements and memory loss, and Alzheimer’s disease, a progressive disorder that results in memory loss. When used to treat a medical condition, nitric oxide is usually administered in the form of a solid or liquid medicine that decomposes in the body, releasing the compound. For example, the drug nitroglycerin is used to treat heart problems. When it enters the bloodstream, nitroglycerin begins to break down, releasing nitric oxide. The nitric oxide causes smooth muscle cells in the heart to relax, relieving the symptoms of angina, chest pain caused by an inadequate flow of blood to the heart. Other types of drugs produce nitric oxide to inhibit the buildup of fatty deposits in blood vessels, which can lead to heart attack and stroke. Patients with pulmonary hypertension, a condition in which the vessels that supply blood to the lungs are constricted, preventing normal oxygen flow, are sometimes given an inhaler with a mixture of nitric oxide and air to open blood vessels to the lungs. In spite of its many benefits, nitric oxide may also be a health hazard. If inhaled in excessive amounts, it may replace oxygen in the lungs, leading to asphyxia, suffocation resulting from an insufficient supply of oxygen. Research suggests that exposure to low concentrations of the gas over long periods of time may result in lung disease, emphysema, and chronic bronchitis. [15]

Nitrogen dioxide Nitrogen dioxide poses both safety and health hazards. As a strong oxidizing agent, it reacts readily with combustible materials, such as paper, cloth, and other organic matter to produce fires or explosions. It is also a toxic material, producing some biological effects at relatively low concentrations in the air. These effects include irritation of the eyes, nose, and throat; coughing; congestion; chest pain; and breathing difficulties. The gas is sometimes referred to as an insidious agent because its effects may go unnoticed for several hours or days, during which time more serious damage may have occurred. This damage may include pulmonary edema, a condition in which the lungs begin to fill with fluid; cyanosis, a condition in which the lips and mucous membranes turn blue because of lack of oxygen; and a variety of heart problems. Longterm exposure to nitrogen dioxide may result in chronic health problems, such as hemorrhaging (blood loss), lung damage, emphysema, chronic bronchitis, and eventually death. [15] 124

Nitric acid Nitric acid is a highly toxic material. It attacks and destroys skin and other tissues, leaving a distinctive yellow scar caused by the destruction of proteins in the skin or tissue. If swallowed, inhaled, or spilled on the skin, it can cause a number of effects, including severe corrosive burns to the mouth, throat, and stomach; severe irritation or burning of the upper respiratory system, including nose, mouth, and throat; damage to the lungs; severe breathing problems; and burns to the eye surface, conjunctivitis, and blindness. In the most severe cases, the acid can cause death. [15] (For more information, see appendix A)

8.3 Hazard and Operability Studies

Figure 10 the three nodes chosen for HAZOP study

126

Node 1: Node No. 1: Intention:

Inlet to the Oxidation Unit (R-202) Start from the Cooler (E-201) to the Oxidation Unit (R-202) To transport the multi gases to the Oxidation Unit

Guide word Deviation None No flow

Less



Causes Flow control valve failure (closed, seat of plug broken, electrical signal fails) Pipeline rupture

 

Partial plug or large leak in pipe Valve not fully open

Less Temperature

 

Less Pressure



Cold weather Temperature controller malfunctioning Pressure controller malfunctioning Pipe leakage

Less Flow







Consequences No flow into the Oxidation unit No reaction

 

Explosion Process stops



Product not according to the desire specifications Reduced production Product not according to the desire specifications



Action requires Install low flow alarm Install flow indicator and flow control valve Plant emergency shutdown procedure Check regularly (maintain), check signal of control valve, operates bypass line Install low flow alarm

 

Install temperature indicator Install low temperature alarm

May cause change in the gas phase

 

Install pressure control Install low pressure alarm



  

   

Cont’d Node 1

Guide word More

Deviation More Flow

Causes 

Consequences Valve opened at full rate

 

More Temperature

More Pressure

 

Hold weather Temperature controller malfunctioning





Pressure controller malfunctioning Partial blockage of line due to partially closed valve Relief valve fails closed



 





Action requires

Change in product quality Some of gas exit without reaction

 

Install high level alarm Install flow indicator and flow control valve

Product not according to the desired specification Reaction fail or run away



Install high Temperature alarm

Increase pressure upstream Excess pressure (may cause explosion)



Install pressure relief valve with automatic feed from temperature control system Install high pressure alarm

 

128

Node 2: Node No. 2: Intention:

Inlet to Cooler (E-202) Start from the Oxidation Unit (R-202) to Cooler (E-202) To transport the multi gases to Cooler

Guide word Deviation None No flow

Less

Less Flow

Less Temperature



Causes Flow control valve failure (closed, seat of plug broken, electrical signal fails) Pipeline rupture

  

Partial plug or large leak in pipe Valve not fully open



Cold weather Temperature controller malfunctioning Less flow temperature from oxidation unit



Pressure controller malfunctioning Pipe leakage





  

Less Pressure

 

 

Consequences No product Process stops

  

 



Action requires Install low flow alarm Install flow indicator and flow control valve Check regularly (maintain), check signal of control valve, operates bypass line

Product not according to the desire specifications Reduced production Overheated flow



Install low flow alarm

Product not according to the desire specifications cooler outlet temperature less than the desired

 

Install temperature indicator Install low temperature alarm

May cause reverse in flow

 

Install pressure control Install low pressure alarm

Cont’d Node 2

Guide word

Deviation More Flow



Causes Process rate changes

 

More Temperature

   

Consequences Reduce Heat transfer efficiency Process fails

 

Action requires Install high level alarm Install flow indicator and flow control valve



Install high Temperature alarm

Install pressure relief valve with automatic feed from temperature control system Install high pressure alarm

Hot weather Fire near the stream Temperature controller malfunctioning High flow temperature from the oxidation unit



The output flow has high temperature than desired



Fails process

Pressure controller malfunctioning Partial blockage of line due to partially closed valve Relief valve fails closed



Increase pressure upstream





Excess pressure (may cause explosion)





cooler fails

More

More Pressure

  

130

Node 3: Node No. 3: Intention:

Inlet to Absorber (T-201) Start from the Cooler (E-202) to Absorber (T-201) To transport the multi gases to Absorber (T-201)

Guide word Deviation None No flow

Less

Less Flow

Less Temperature



Causes Flow control valve failure (closed, seat of plug broken, electrical signal fails) Pipeline rupture

 

Partial plug or large leak in pipe. Valve not fully open.





Reaction gas temperature in oxidation unit lower





 



 Less Pressure



Gas Release (line rupture)



Consequences No Absorption in the column Entire process stops as tail-gas row stops

  

Action requires Install low flow alarm Plant emergency shutdown procedure. Check signal of control valve, operates by-pass line

The production of Nitric acid will be less Decreased absorption



Install low flow alarm

Increased dissolved NOx Concentration of the product will be higher Less efficiency of NOx absorbed

 

Install temperature indicator Install low temperature alarm

 

Install pressure control Install low pressure alarm

Cont’d Node 3

Guide word Deviation More More Flow



Causes Valve opened at full rate

 

More Temperature

More Pressure



  

Higher feed rates causing larger heat of reaction

Pressure controller malfunctioning Flooding Relief valve fails closed

Consequences Possible reduction in absorption efficiency May cause flooding



Reduce absorption efficiency



Higher NOx emission due to lower absorption



More NOx absorebed

 

Action requires Install high level alarm Install flow indicator and flow control valve



Install high Temperature alarm



Install pressure relief valve with automatic feed from temperature control system Install high pressure alarm



132

Chapter 9

Waste Treatment

What Is NOx? Most of the world’s nitrogen occurs naturally in the atmosphere as an inert gas contained in air, which consists of approximately 78% N2 by volume. NOx refers to oxides of nitrogen. These generally include nitrogen monoxide, also known as nitric oxide (NO), and nitrogen dioxide (NO2). They may also include nitrous oxide (N2O), also known as laughing gas, as well as other less common combinations of nitrogen and oxygen, such as nitrogen tetroxide (N2O4) and nitrogen pent oxide (N2O5). The EPA defines nitrogen oxides as “all oxides of nitrogen except nitrous oxide.1” In most high-temperature heating applications; the majority of the NOx exiting the exhaust stack is in the form of nitric oxide (NO). What is Wrong with NOx? Health hazards: NO is poisonous to humans and can cause irritation of the eyes and throat, tightness of the chest, nausea, headache, and gradual loss of strength. Prolonged exposure to NO can cause violent coughing, difficulty in breathing, and cyanosis; it could be fatal. NO2 is a reddish-brown, highly reactive gas that has a suffocating odor and is a strong oxidizing agent. It is highly toxic and hazardous because of its ability to cause delayed chemical pneumonitis and pulmonary edema. NO2 vapors are a strong irritant to the pulmonary tract.

Ground level ozone: Ozone (O3) is present in the upper atmosphere and is desirable as it shields the earth against high-intensity ultraviolet rays from the sun. However, ozone in the lower atmosphere is undesirable. There, NO reacts with oxygen to form ozone, in addition to NO2: NO + HC + O2 + sunlight → NO2 + O3 Ozone is also a health hazard that can cause respiratory problems in humans. It is an irritant to the eye, nose, and throat. Ozone can cause damage to plants including crops and deterioration of textiles and other materials.

Acid rain: Rain is effective at removing NO2 from the atmosphere. However, NO2 decomposes on contact with water to produce nitrous acid (HNO2) and nitric acid (HNO3), which are highly corrosive . NO is the predominant form of NOx produced in industrial combustion processes. It reacts with O2 in the atmosphere to form NO2. When NO2 forms in the atmosphere and comes into contact with water in the form of 134

rain, nitric acid is formed. Acid rain is destructive to anything it contacts, including plants, trees, and man-made structures like buildings, bridges, and the like.

Smog: Another problem with NO2 is its contribution to smog. Smog is the combination of smoke and fog and occurs where there are high concentrations of pollutants combined with fog. Smog impairs visibility through the atmosphere. When sunlight contacts a mixture of NO2 and unburned hydrocarbons in the atmosphere, photochemical smog is produced.

Figure L: Acid rain

Figure K: Smog

Nitrogen Oxide Pollution

The nitrogen oxides NO and NO2 which make up NOx have long been known as precursors of acid rain and smog. While ever more stringent emission limits apply in many countries there are still a large number of nitric acid plants, which have no NOx abatement equipment installed. Nitrous oxide (N2O) is a powerful greenhouse gas being about 300 times more potent than carbon dioxide. Even though nitric acid plants represent the largest single industrial process source of this gas, most countries have not yet imposed emission limits on N2O, although steps in this direction are being made, for example, in the European Union. As nitric acid plants are point sources for N2O and NOx emissions they make good candidates for the implementation of cost effective greenhouse gas and NOx emission reduction technologies. Uhde Technologies for NOx and N2O Abatement 1. Uhde DeNOx process: NOx reduction with ammonia: It was found that one of the iron zeolites materials tested was a very efficient catalyst for the selective reduction of NOx (NO and NO2) with ammonia. This catalyst has the advantage that it can be used over a large range of temperatures, from ~200°C to over 500°C, while classical vanadium-pent oxide-based SCR catalysts for nitric acid plants generally cannot be used at temperatures much above ~400°C. A further advantage is that it generates no N2O, again in contrast to vanadium pent oxide, which is reported to produced N2O at temperatures above 350°C (Jouannic etal.1984).

This is the basis of the Uhde DeNOx process, which has a similar ammonia consumption to classical SCR/DeNOx processes, is applicable over a wider range of temperatures and, in contrast to frequently used classical SRC/DeNOx catalyst materials, contains no environmentally hazardous substances, thus simplifying handling and disposal. The Uhde DeNOx process can be very conveniently combined with N2O abatement as described below 2. EnviNOx® process variant 1: N2O decomposition with NOX reduction: The iron zeolite catalysts investigated decompose N2O into its elements. An important phenomenon is that the rate of decomposition is greatly enhanced when NOx is present in the waste gas stream. This fact, combined with the NOx reducing

136

properties of the iron zeolite catalysts, is exploited in the EnviNOx® process variant 1, which is illustrated in Figure 13. The EnviNOx® reactor is typically located in the tail gas stream on the inlet side of the tail gas turbine where the tail gas temperature is at its highest. Because of the high NOx concentration in the tail gas entering the reactor, a very large proportion of the incoming N2O is catalytically decomposed to nitrogen and oxygen in the first bed. Ammonia is mixed with the tail gas between the beds, the ammonia distribution and mixing equipment being incorporated into the reactor. In the second bed the NOX is reduced to the level required for emission to atmosphere and some further decomposition of N2O also takes place.

Figure 13 EnviNOx® process variant 1: Combined N2O and NOX abatement for nitric acid plants using N2O decomposition and NOX reduction with ammonia

EnviNOx® process variant 1 is currently suitable for tail gas temperatures between about 425°C and 520°C. High rates of N2O removal are possible with 98% typically achieved in commercial installations and NOX emissions can be reduced to low levels depending on the amount of ammonia supplied, with 5 – 40 ppmv being usual. As with the Uhde DeNOx process ammonia consumption is similar to that of classical SCR/DeNOx processes.

3. EnviNOx® process variant 2: N2O & NOX reduction by hydrocarbons & NH3

There are, of course, a great many nitric acid plants with tail gas temperatures below 425°C. The solution for a large number of these plants is the EnviNOx® process variant 2, which relies on another significant phenomenon, which was intensively investigated during development work. While N2O in nitric acid tail gas can be reduced by reaction with a hydrocarbon over certain iron zeolite catalysts, the effectiveness of the reduction is greatly enhanced if the NOx is removed almost entirely from the tail gas. Even a few ppm of NOx is sufficient to cause strong inhibition of the reactions of hydrocarbons with N2O. This property combined with the NOX reducing properties of the iron zeolite catalysts with ammonia leads to the EnviNOx® process variant 2, depicted in Figure 14.

Figure 14 EnviNOx® process variant 2: Combined N2O and NOX abatement for nitricacid plants using N2O reduction with hydrocarbons and NOX reduction with ammonia

Ammonia is mixed with the tail gas entering the EnviNOx® reactor in such an amount that NOx is completely reduced in the first DeNOx bed. The virtually NOx free tail gas is then mixed with a hydrocarbon and passed over the second catalyst bed where the N2O is reduced to a very low level by reaction with the hydrocarbon. Tail gas with nearly no NOx and a very low concentration of N2O leaves the reactor. It is important to recognize that the hydrocarbon acts as a chemical reagent and is not used as a fuel to 138

raise the temperature of the catalyst to a level at which high rates of N 2O decomposition can occur. Because of their favourable costs and availability the hydrocarbons of choice are natural gas or propane. Consumptions of both ammonia and hydrocarbon are quite moderate. The EnviNOx® process variant 2 works best at temperatures between about 340°C and ~520°C. 4. Uhde DeN2O® process: N2O decomposition without NOX abatement

If there is no requirement to lower NOx emissions, the N2O decomposition catalyst described above can be used on its own. The catalyst achieves high N2O abatement performance up to temperatures of about 600°C, substantially increasing the range of nitric acid plants which can be equipped with N2O abatement. In such plants a NOx abatement unit can also be installed if necessary, provided an appropriate tail gas temperature level between about 200°C and 500°C is available.

Conclusion

The EnviNOx® process for the combined abatement of NOx and N2O emissions from nitric acid plants has proven itself in installations around the world which are now operating at temperatures between 340°C and 510°C. Typically, N2O emissions are reduced by ~98% – 99% while NOX emission levels of 1 to ~25 ppmv are achieved, depending on the process variant. The EnviNOx® process can be applied in nitric acid plants with tail gas temperatures between about 340°C and 600°C covering an estimated 70% – 80% of all nitric acid production worldwide. For many of the nitric acid plants with tail gas temperatures outside this range, relatively simple plant modifications are possible to enable a nitrous oxide abatement system to be installed. The EnviNOx® process can thus make a useful contribution to lowering greenhouse gas emissions. EnviNOx® is an “operator friendly” technology since the EnviNOx® catalysts are easy to handle, environmentally uncontroversial materials with a long operating lifetime. Furthermore as an end-of-pipe process EnviNOx® does not interact in any way with the plant product or its precursors.

Conclusion

In conclusion, setting up 100000 tons per year of nitric acid production plant in Saudi Arabia is very feasible and attractive in order to meet high demand of ammonia in this region. Ras Alkhair industrial city is identified as the most ideal location for the new petrochemical plant. All the desired criteria for the construction of the plant in Ras Alkhair are met and the necessary facilities for plant expansion are also available.

In order to produce the required capacity 100000 tons per year of nitric acid while the amount of ammonia required 2335.88 kg/hr while the amount of air needed 43537.52 kg/hr.

Designing and sizing for equipment used are prepared in Equipment Specification Sheet which provide the design and size estimation for chosen equipment.

A highly integrated process control system is also included to the proposed plant. Control systems are essential to ensure the plant operates in safe manner and achieves the desired production. The control system proposed are all feedback and cascade model, which can correct the error continuously to serve better control objectives. All equipment has been covered and proper control strategy and design have been proposed.

In designing the proposed plant, various safety factors and procedures such as Hazard and Operability Studies (HAZOP), plant start up and shut down procedures, maintenance and inspection of each equipment as well as control system design were taken into consideration. However, it is recommended that further studies in this area is needed unceasingly to ensure operational safety of the plant. Safety aspects are also been considered in recommending the plant layout.

In responding to the environmental obligation, the plant has been designed to achieve the target of waste minimization while achieving cost minimization at the same time.

From our economic and cost estimation analysis, the expected payback period is (3.43) years. The economic analysis of the process flowsheet indicated that the ROI is about 20.5%. The project is worth investing. 140

Finally, it can be concluded that the construction of a 100,000 tons per year of nitric acid production plant in Ras Alkhair is technically feasible and economically attractive. Encouragement from the Saudi government investment with various incentives offered serves as another factor that contribute to the feasibility of the plant.

References 1- Krick Othmer, "Encyclopedia of Chemical Technology", VoL.15, 3th Edition, Jonh Wiley,1982 2- George Charles Lowrison,"Fertilizer Technology",Eillis Horwood Limited,1989 3- R. Perry and C. Chilton,"Perry's Chemical Engineer's Hand Book", 7th Ed.,Mcgra W-Hill, 1997 4- Vincent Sauchelli, "Fertilizer Nitrogen its Chemistry and Technology", Rinhold Publishing Corporation", 1964. 5- Howard F. Rase, "Chemical Reactor Design For Process Plants",Vol. one & two, John Wiley, 1977 6- Octave Leven Spiel," Chemical Reaction Engineering", 2nd Ed.,John Wiley, 1972 7- J. M. Coulson J. Richardson, "Chemical Engineering Design", Vol.6, 3rd Ed., Pergamon Press,1983 8- P.J.C. Kaasenbrood, "Chemical Reaction Engineering", Pergamon Press, 1968 9- Richard Turton, Richard C. Bailie, Wallace B. Whiting, Joseph A. Shaeiwitz, " Analysis, Synthesis, and Design of Chemical Processes ",3rd Edition, Prentice-Hall, 2009 10- W. Dekker, E. Snoeck and H. Kramers, "Chemical Engineering Science", 1959 11- (2012, january 15). Retrieved december 1, 2013, from Saudi Presidency of Meteorology and Enviroment: Http://www.pme.gov.sa 12- (2009, may 27). Retrieved December 1, 2013, from Vermont Safty Information Resources, Inc. : Http://www.siri.org 13- Richard M. Felder, Ronald W. Rousseau, " Elementary Principles of Chemical Processes ", Third edition,2005 14- Fritz Ullmann, “ Ullmann’s Encyclopedia of Industrial Chemistry “,John Wiley and Sons, 1999 15- Neil S. Chlager, Jayne Weisblatt, and David E. Newton, “Chemical Compounds”, 2001 16- Max. S. Peters, Klans D. Timmerhans, "Plan Design and Econmics for chemical Engineering 3-rd Ed, McGraw-Hill, 1990. 17- Christie J. Geankoplis, “ Transport Processes and Separation Process Principles “, 5th Edition, Prentice Hall, 2003 18- J. M. Smith, H. C. Van Ness, M. M. Abbott, "Introduction to Chemical Engineering Thermodynamics", 7th Edition, McCraw-Hill,2005 19- W. S. Norman, "Absorption, Distillation and Cooling Tower", Longmans,1962

142

Appendix A (12) (Material Safety Data)

144

146

148

150

Appendix B

(13)

(Physical Properties Data)

152

154

156

158

160

Appendix C (Detailed Mass Balance Calculation)

162

Overall Material Balance NH3 + 2O2  HNO3 + H2O Conversion: 95%

HNO3: 𝐹𝑜𝑢𝑡 =

7500 = 119.048 𝐾𝑚𝑜𝑙/ℎ 63

NH3: 𝐹𝑜𝑢𝑡 = 137.405 −

1 (119.048) = 18.357 𝐾𝑚𝑜𝑙/ℎ 1

O2: 0 = 𝐹𝑖𝑛 − 2 (119.048) => 238.096 = 𝐹𝑖𝑛 Excess 20 % 𝐹𝑖𝑛 = 285.715 𝐾𝑚𝑜𝑙/ℎ 𝐹𝑜𝑢𝑡 = 285.715 − 238.096 = 47.619𝐾𝑚𝑜𝑙/ℎ N2: 𝐹𝑜𝑢𝑡 =

285.715 × 32 × 0.79 = 1228.38 𝐾𝑚𝑜𝑙/ℎ 28 × 0.21

H2O: 1 0.05𝐹𝑖𝑛 + 277.778 = 𝐹𝑖𝑛 + (119.048) 1 𝐹𝑖𝑛 = 167.084 𝐾𝑚𝑜𝑙/ℎ

Reactor Material Balance 1) 4NH3 + 5O2  4NO + 6H2O Conversion: 95% 𝐸1 = 0.95 × 137.405 = 130.535 2) 4NH3 + 3O2  2N2 + 6H2O Conversion: 5% 𝐸2 = 0.05 × 137.405 = 6.87 NH3: 4 4 𝐹𝑜𝑢𝑡 = 137.405 − (130.535) − (6.87) = 0 4 4 O2: 5 3 𝐹𝑜𝑢𝑡 = 285.715 − (130.535) − (6.87) = 117.393 𝐾𝑚𝑜𝑙/ℎ 4 4 N2: 2 𝐹𝑜𝑢𝑡 = 1228.38 + (6.87) = 1231.815 𝐾𝑚𝑜𝑙/ℎ 4 NO: 4 𝐹𝑜𝑢𝑡 = 0 + (130.535) = 130.535 𝐾𝑚𝑜𝑙/ℎ 4 H2O: 6 6 𝐹𝑜𝑢𝑡 = 0 + (130.535) + (6.87) = 206.107 𝐾𝑚𝑜𝑙/ℎ 4 4

164

Oxidation Material Balance 2NO + O2  2NO2 Conversion: 96% 𝐸 = 0.96 × 130.535 = 125.313 NO: 2 𝐹𝑜𝑢𝑡 = 130.535 − (125.313) = 5.222 𝐾𝑚𝑜𝑙/ℎ 2 NO2: 𝐹𝑜𝑢𝑡 = 0 + 125.313 = 125.313 𝐾𝑚𝑜𝑙/ℎ O2: 1 𝐹𝑜𝑢𝑡 = 117.393 − (125.313) = 54.736 𝐾𝑚𝑜𝑙/ℎ 2 H2O: 𝐹𝑜𝑢𝑡 = 206.107 𝐾𝑚𝑜𝑙/ℎ N2: 𝐹𝑜𝑢𝑡 = 1231.815 𝐾𝑚𝑜𝑙/ℎ

Absorber Material Balance 2NO2 + H2O + ½ O2 2HNO3 Conversion = 95% 𝐸 = 0.95 × 125.313 = 119.048 𝑘𝑚𝑜𝑙/ℎ No2: 𝐹𝑜𝑢𝑡 = 125.313 – 2⁄2 × (119.048) = 6.265 𝐾𝑚𝑜𝑙/ℎ HNO3: 𝐹𝑜𝑢𝑡 = 0 + 119.0475 = 119.0475 𝑘𝑚𝑜𝑙/ℎ H2O: 𝐹𝑜𝑢𝑡 = 373.191 − 1⁄2 × (119.0475) = 313.667 𝐾𝑚𝑜𝑙/ℎ NO: 𝐹𝑜𝑢𝑡 = 5.222 𝑘𝑚𝑜𝑙/ℎ O2: 𝐹𝑜𝑢𝑡 = 54.736 − 0.5⁄2 × (119.0475) = 24.974 𝐾𝑚𝑜𝑙/ℎ N2: 𝐹𝑜𝑢𝑡 = 1231.815 𝐾𝑚𝑜𝑙/ℎ

166

Appendix D (Detailed Energy Balance Calculation)

Energy Balance

Vaporizer Energy Balance

S4

S5

E-101

For S4: NH3 at -15 oC and S5: NH3 at 35 oC

Tref = -15 oC 𝑄̇ = ∑ 𝑛𝑜𝑢𝑡 𝐻𝑜𝑢𝑡 − ∑ 𝑛𝑖𝑛 𝐻𝑖𝑛 𝑄̇ = ∑ 𝑛𝑜𝑢𝑡 𝐻𝑜𝑢𝑡 − 0 𝑇𝑏𝑜𝑖

𝐻𝑜𝑢𝑡 = 𝑛[∫

𝑇𝑜𝑢𝑡

𝐶𝑝 𝑑𝑇 + 𝜆 + ∫

𝑇𝑟𝑒𝑓

𝐶𝑝 𝑑𝑇]

𝑇𝑏𝑜𝑖

𝑇2

𝑏 𝑐 𝑑 ∫ 𝐶𝑝 𝑑𝑇 = 𝑎(𝑇2 − 𝑇1 ) + (𝑇22 − 𝑇12 ) + (𝑇23 − 𝑇13 ) + (𝑇24 − 𝑇14 ) 2 3 4 𝑇1 ∑ 𝐻𝑜𝑢𝑡 = 1000 × 137.405[1.78 + (

1133.642 × 17) + 0.07232] 1000

∑ 𝐻𝑜𝑢𝑡 = 𝑄̇ = 2902575.37 𝐾𝐽/ℎ −15

∑ 𝐻𝑖𝑛 = ∫

𝐶𝑝 𝑑𝑇 = 0

−15

168

Compressor Energy Balance

S3

S1

C-101

For S1: Air at 101 kPa and S3: Air at 1090 kPa

𝑊 = 𝑃1 𝑄1 ln

𝑃2 𝑃1

𝑃2 𝑃1 36000 1090 𝑊 = 101 ( ) ln 1.178 101 𝑊 = 𝑃1 𝑚(𝑎𝑖𝑟) ln

𝑊 = 8879734.37 kJ/h 𝑊 = 8879734.37 kJ/h (This is the mechanical energy required by the compressor)

Superheater Energy Balance

S5

S6

E-102 For S5: NH3 at 35 oC and S6: NH3 at 177 oC 𝑄 = ∑𝑛 𝐻𝑜𝑢𝑡 − ∑𝑛 𝐻𝑖𝑛 𝑇2

𝐻 = ∫ 𝐶𝑝 𝑑𝑇 𝑇1 𝑇2 𝑏 𝑐 𝑑 ∫ 𝐶𝑝 𝑑𝑇 = 𝑎(𝑇2 − 𝑇1 ) + (𝑇22 − 𝑇12 ) + (𝑇23 − 𝑇13 ) + (𝑇24 − 𝑇14 ) 2 3 4 𝑇1

Tref = 35 ○C 𝐻𝑖𝑛 = 𝑍𝑒𝑟𝑜 𝐻𝑜𝑢𝑡 = 137.405 × 1000 × 5.442 = 747758.01 𝐾𝐽/ℎ 𝑄 = 𝐻𝑜𝑢𝑡 – 𝑧𝑒𝑟𝑜 = 747758.01 𝐾𝐽/ℎ

170

Mixer Energy Balance

S3 S7

S6

M-101

For S3: NH3 at 177 oC, S6: Air at 262 oC and S7: NH3+Air at ? 𝑄 = 0 (𝐴𝑑𝑖𝑎𝑏𝑎𝑡𝑖𝑐) Energy required to heat ammonia = Energy lost by air 𝑚𝐶𝑝(𝑇𝑜𝑢𝑡 − 𝑇𝑖𝑛 ) = 𝑚𝐶𝑝(𝑇𝑜𝑢𝑡 − 𝑇𝑖𝑛 ) 𝐶𝑝 (𝑁𝐻3 ) = 2.38 𝐾𝐽/𝐾𝑔. 𝑜𝐶 𝐶𝑝(𝐴𝑖𝑟) = 1.05 𝐾𝐽/𝐾𝑔. 𝑜𝐶 2335.88 × 2.38 × (𝑇𝑜𝑢𝑡 − 177) = 43537.52 × 1.05 × (262 − 𝑇𝑜𝑢𝑡 ) 𝑇𝑜𝑢𝑡 = 250𝐶

Reactor Energy Balance

S7

S8 R-201 For S7: NH3+Air at 250 oC and S8: Air+NO+H2O at 645 oC

𝑄 = ∑ 𝑛𝑜𝑢𝑡 – 𝐻𝑜𝑢𝑡 − ∑ 𝑛𝑖𝑛 𝐻𝑖𝑛 + 𝑛(𝑃𝑟𝑜𝑑𝑢𝑐𝑡1) ∆𝐻𝑟1 + 𝑛(𝑃𝑟𝑜𝑑𝑢𝑐𝑡2) ∆𝐻𝑟2

1) 4NH3+5O2  4NO+6H2O (1) Conversion=95% 2) 4NH3+3O2  2N2+6H2O (2) Conversion=5% 𝛥𝐻𝑟 = 𝛴(𝐹𝛥𝐻𝑓)𝑃𝑟𝑜𝑑𝑢𝑐𝑡 − 𝛴(𝐹𝛥𝐻𝑓)𝑅𝑒𝑎𝑐𝑡𝑎𝑛𝑡 𝛥𝐻𝑟1 = −904.74 𝑘𝐽/𝑚𝑜𝑙 Based on (NO): 𝛥𝐻𝑟1 = −904.74/4 = −226.185 𝑘𝐽/𝑚𝑜𝑙 𝛥𝐻𝑟2 = −1266.22 𝑘𝐽/𝑚𝑜𝑙 Based on (H2O): 𝛥𝐻𝑟2 = −1266.22/6 = −211 𝑘𝐽/𝑚𝑜𝑙 𝑄 = 𝛴(𝐹𝐻)𝑜𝑢𝑡 − 𝛴(𝐹𝐻)𝑖𝑛 + 𝛥𝐻𝑟 𝑇2

𝐻 = ∫ 𝐶𝑝 𝑑𝑇 𝑇1

𝑇2

𝑏 𝑐 𝑑 ∫ 𝐶𝑝 𝑑𝑇 = 𝑎(𝑇2 − 𝑇1 ) + (𝑇22 − 𝑇12 ) + (𝑇23 − 𝑇13 ) + (𝑇24 − 𝑇14 ) 2 3 4 𝑇1

172

Tref = 25oC

Component H (Air) H (NH3) H (NO) H (H2O)

Input S7 (kJ/mol) 6.654 8.84 -

𝑄 = 21192906.91 + 𝐹(𝑁𝑂)𝑜𝑢𝑡. 𝛥𝐻𝑟1 + 𝐹(𝐻2 𝑂)𝑜𝑢𝑡. 𝛥𝐻𝑟2 𝑄 = −51820729.07 𝑘𝐽/ℎ

Output S8 (kJ/mol) 19 19.744 23

Heat Exchanger Energy Balance (1st Cooler)

S8

S9

E-201

For S8: Air+NO+H2O at 645 oC and S9: Air+NO+H2O at 70 oC 𝑄 = 𝛴(𝐹𝐻)𝑜𝑢𝑡 − 𝛴(𝐹𝐻)𝑖𝑛 Tref = 70℃ 𝐻(𝑜𝑢𝑡) = 𝑧𝑒𝑟𝑜 Input: 645

𝐻=∫

𝑐𝑝 𝑑𝑡 = 𝑎(𝑇2 − 𝑇1 ) +

70

𝑏 2 𝑐 𝑑 (𝑇2 − 𝑇12 ) + (𝑇23 − 𝑇13 ) + (𝑇24 − 𝑇14 ) 2 3 4

For air 645

𝐻 = ∫ cp dt = 28.94 × 10−3 (645 − 70) + ( 70

+

0.4147 × 10−5 )(6452 − 702 ) 2

0.3191 × 10−8 1.965 × 10−12 (6453 − 703 ) − (6454 − 704 ) = 17.693 3 4

For No: 645

𝐻 = ∫ 𝑐𝑝 𝑑𝑡 = 18.400 70

For H2O: 645

𝐻 = ∫ 𝑐𝑝 𝑑𝑡 = 21.173 70

𝑄 = −[(1318 × 1000 × 17.963) + (130.535 × 1000 × 18.4) + (206.107 × 1000 × 21.17)] 𝑄 = −30084503.19 𝐾𝐽/ℎ

174

Oxidation Energy Balance

S9

R-202

S 10

For S9: Air+NO+H2O at 70 oC and S10: Air+NO+H2O+NO2 at 140 oC 2NO+O2  2NO2 Conversion=96% 𝛥𝐻𝑟 = 𝛴(𝐹𝛥𝐻𝑓)𝑃𝑟𝑜𝑑𝑢𝑐𝑡 − 𝛴(𝐹𝛥𝐻𝑓)𝑅𝑒𝑎𝑐𝑡𝑎𝑛𝑡

𝛥𝐻𝑟 = −113.14 𝑘𝐽/𝑚𝑜𝑙 Based on (NO2): 𝛥𝐻𝑟 = −113.14/2 = −56.57 𝑘𝐽/𝑚𝑜𝑙 𝑄 = 𝛴(𝐹𝐻)𝑜𝑢𝑡 − 𝛴(𝐹𝐻)𝑖𝑛 + 𝑛𝑝𝑟𝑜𝑑 𝛥𝐻𝑟 𝑇2

𝐻 = ∫ 𝐶𝑝 𝑑𝑇 𝑇1 𝑇2 𝑏 𝑐 𝑑 ∫ 𝐶𝑝 𝑑𝑇 = 𝑎(𝑇2 − 𝑇1 ) + (𝑇22 − 𝑇12 ) + (𝑇23 − 𝑇13 ) + (𝑇24 − 𝑇14 ) 2 3 4 𝑇1

Tref = 25oC Component

Input S9 (kJ/mol)

Output S10 (kJ/mol)

H (H2O) (g)

1.521

3.919

H (Air)

1.312

3.37

H (NO)

1.345

3.47

H (NO2)

-

4.5

𝑄 = −3707006.24 𝑘𝐽/ℎ

Heat Exchanger Energy Balance (2nd cooler)

S 10

S 11

E-202 For S10: Air+NO+H2O+NO2 at 140 oC and S11: Air+NO+H2O+NO2 at 60 oC Tref =60oC 𝑄 = 𝛴(𝐹𝐻)𝑜𝑢𝑡 − 𝛴(𝐹𝐻)𝑖𝑛 𝑇2

𝐻 = ∫ 𝐶𝑝 𝑑𝑇 𝑇1 𝑇2 𝑏 𝑐 𝑑 ∫ 𝐶𝑝 𝑑𝑇 = 𝑎(𝑇2 − 𝑇1 ) + (𝑇22 − 𝑇12 ) + (𝑇23 − 𝑇13 ) + (𝑇24 − 𝑇14 ) 2 3 4 𝑇1

𝐻(𝑜𝑢𝑡) = 𝑍𝑒𝑟𝑜 Input: For H2O 140

𝐻 = ∫ 𝐶𝑝 𝑑𝑇 = 2.737 kJ/mol 60

For air 140

𝐻 = ∫ 𝐶𝑝 𝑑𝑇 = 2.350 kJ/mol 60

176

For NO 140

𝐻= ∫

𝐶𝑝 𝑑𝑇 = 2.4230𝑘𝐽/𝑚𝑜𝑙

60

For NO2 140

𝐻 = ∫ 𝐶𝑝 𝑑𝑇 = 3.1796 kJ/mol 60

𝑄 = −3911927.94 𝐾𝐽/ℎ

Absorber Energy Balance

S 12

S 14 T-201

S 11 S 13

Basis: Tref = 25oC For S11: Air+NO+H2O+NO2 at 60 oC, S14: H2O at 20 oC, S12: Air+NO+H2O at 30 oC and S13: HNO3+H2O at 30 oC 1

3NO2+H2O + 2 O2 2HNO3 Conversion =95% 𝛥𝐻𝑟 = 𝛴(𝐹𝛥𝐻𝑓)𝑃𝑟𝑜𝑑𝑢𝑐𝑡 − 𝛴(𝐹𝛥𝐻𝑓)𝑅𝑒𝑎𝑐𝑡𝑎𝑛𝑡 𝛥𝐻𝑟 = −435.6 𝑘𝐽/𝑚𝑜𝑙 Based on (HNO3): 𝛥𝐻𝑟 = −435.6/2 = −217.8 𝑘𝐽/𝑚𝑜𝑙 𝑄 = 𝛴(𝐹𝐻)𝑜𝑢𝑡 − 𝛴(𝐹𝐻)𝑖𝑛 + 𝑛𝑃𝑟𝑜𝑑𝑢𝑐𝑡 𝛥𝐻𝑟 𝑇2

𝐻 = ∫ 𝐶𝑝 𝑑𝑇 𝑇1

𝑏 𝑐 𝑑 𝐻 = 𝑎𝑇 + 𝑇 2 + 𝑇 3 + 𝑇 4 2 3 4

178

Tref = 25oC Component

Input S11

Input S14

Output S12

Output S13

(kJ/mol)

(kJ/mol)

(kJ/mol)

(kJ/mol)

H (H2O) (g)

1.18

-

0.166

-

H (H2O) (L)

-

-0.375

-

0.375

H (Air)

1.02

-

0.145

-

H (NO)

1.044

-

0.149

-

H (NO2)

1.32

-

-

-

H (HNO3)

-

-

-

0.55

𝑄 = −29876241 𝐾𝐽/ℎ

Appendix E (Equipment Design References)

180

Table A: Typical overall coefficients

Hot fluid Heat exchanger Water Organic solvents Light oils Heavy oils Gases Coolers Organic solvents Light oils Heavy oils Gases Organic solvents Water Gases Heaters Steam Steam Steam Steam Steam Dowtherm Dowtherm Flue gases Flue Condensers Aqueous vapors Organic vapors Organic (some noncondensable) Vacuum condensers Vaporizers Steam Steam Steam

Shell and tube exchangers Cold fluid

U (W/m2.oC)

Water Organic solvents Light oils Heavy oils Gases

800-1500 100-300 100-400 50-300 10-50

Water Water Water Water Brine Brine Brine

250-750 350-900 60-300 20-300 150-500 600-1200 15-250

Water Organic solvents Light oils Heavy oils Gases Heavy oils Gases Steam Hydrocarbon vapor

1500-4000 500-1000 300-900 60-450 30-300 50-300 20-300 30-100 30-100

Water Water

1000-1500 700-1000

Water Water

500-700 200-500

Aqueous solutions Light organics Heavy organics

1000-1500 900-1200 600-900

Table B: Fouling factors (coefficients), Typical values

Fluid River water Sea water Cooling water (towers) Towns water (soft) Towns water (hard) Steam condensate Steam (oil free) Steam (oil traces) Refrigerated brine Air and industrial gases Flue gases Organic vapors Organic liquids Light hydrocarbons Heavy hydrocarbons Boiling organics Condensing organics Heat transfer fluids Aqueous salt solutions

Coefficient (W/m2.oC) 3000-12000 1000-3000 3000-6000 3000-5000 1000-2000 1500-5000 4000-10000 2000-5000 3000-5000 5000-10000 2000-5000 5000 5000 5000 2000 2500 5000 5000 3000-5000

Table C: Triangular pitch constants for use

No. Passes K1 n1

1 0.319 2.142

Triangular pitch pt = 1.25do 2 4 0.249 2.207

0.175 2.285

6

8

0.0743 2.499

0.0365 2.675

182

Figure A: Shell-bundle clearance

Figure B: Shell side heat transfer factors, segmental baffles

Figure C: Compressor operating ranges

Figure D: Approximate polytrophic efficiency of centrifugal and axial compressor

184

Appendix F (Computer Simulation by CHEMCAD)

Note: No Absorber column in CHEMECAD. So, we choose Reactor & Distillation 186

CHEMCAD 5.1.3 Job Name: nitric acid Stream No. Name - - Overall - Molar flow kmol/h Mass flow kg/h Temp C Pres kPa Vapor mole fraction Enth kcal/h Tc C Pc kPa Std. sp gr. wtr = 1 Std. sp gr. air = 1 Degree API Average mol wt Actual dens kg/m3 Actual vol m3/h Std liq m3/h Std vap 0 C m3/h - - Vapor only - Molar flow kmol/h Mass flow kg/h Average mol wt Actual dens kg/m3 Actual vol m3/h Std liq m3/h Std vap 0 C m3/h Cp kcal/kmol-C Z factor Visc Pa-sec Th cond kcal/h-m-C

Page 1 Date: 12/21/2013

Time: 17:03:43

1

2

3

4

1513.4900 43537.5214 25.0000 101.3000 1.000 -2407.9 -142.3629 3580.9491 0.860 0.993 33.0631 28.7663 1.1758 37026.8361 50.6795 33922.8318

1513.4900 43537.5214 426.6417 1240.0000 1.000 4.3619E+006 -142.3629 3580.9491 0.860 0.993 33.0631 28.7663 6.1004 7136.8696 50.6795 33922.8318

137.1546 2335.8798 177.0000 1240.0000 1.000 -1.3266E+006 132.5000 11278.4833 0.619 0.588 96.9616 17.0310 5.7967 402.9700 3.7749 3074.1346

1650.6446 45873.3981 397.9546 1060.0000 1.000 3.0352E+006 -122.8054 3657.0015 0.843 0.960 36.3168 27.7912 5.2582 8724.1045 54.4544 36996.9653

1513.4900 43537.5214 28.7663 1.1758 37026.8361 50.6795 33922.8318 7.0023 0.9999 1.826e-005 0.0219

1513.4900 43537.5214 28.7663 6.1004 7136.8696 50.6795 33922.8318 7.4548 1.0051 3.386e-005 0.0447

137.1546 2335.8798 17.0310 5.7967 402.9700 3.7749 3074.1346 9.9937 0.9735 1.580e-005 0.0398

1650.6446 45873.3981 27.7912 5.2582 8724.1045 54.4544 36996.9653 7.7200 1.0042 3.227e-005 0.0460

CHEMCAD 5.1.3 Job Name: nitric acid Stream No. Name - - Overall - Molar flow kmol/h Mass flow kg/h Temp C Pres kPa Vapor mole fraction Enth kcal/h Tc C Pc kPa Std. sp gr. wtr = 1 Std. sp gr. air = 1 Degree API Average mol wt Actual dens kg/m3 Actual vol m3/h Std liq m3/h Std vap 0 C m3/h - - Vapor only - Molar flow kmol/h Mass flow kg/h Average mol wt Actual dens kg/m3 Actual vol m3/h Std liq m3/h Std vap 0 C m3/h Cp kcal/kmol-C Z factor Visc Pa-sec Th cond kcal/h-m-C - - Liquid only - Molar flow kmol/h Mass flow kg/h Average mol wt Actual dens kg/m3 Actual vol m3/h Std liq m3/h Std vap 0 C m3/h Cp kcal/kmol-C Z factor Visc Pa-sec Th cond kcal/h-m-C Surf tens N/m

Page 2 Date: 12/21/2013

Time: 17:03:43

5

6

7

8

1684.9333 45873.2316 957.1419 1060.0000 1.000 3.0352E+006 -93.0223 2905.8708 0.847 0.940 35.5966 27.2255 2.8143 16299.9875 54.2205 37765.5022

1684.9333 45873.2316 70.0000 1060.0000 0.9050 -1.0178E+007 -93.0223 2905.8708 0.847 0.940 35.5966 27.2255 11.1596 4110.6495 54.2205 37765.5022

1622.3906 45873.1642 104.1766 1060.0000 0.9819 -1.0178E+007 -73.2175 3100.2549 0.882 0.976 28.9246 28.2750 9.7690 4695.7688 52.0554 36363.6909

1622.3907 45873.1678 60.0000 1060.0000 0.8902 -1.2225E+007 -73.2175 3100.2545 0.882 0.976 28.9247 28.2750 12.1854 3764.5903 52.0554 36363.6944

1684.9333 45873.2316 27.2255 2.8143 16299.9875 54.2205 37765.5022 8.3271 1.0026 4.840e-005 0.0745

1524.9087 42990.3889 28.1921 10.4658 4107.7002 51.3372 34178.7630 7.1471 1.0009 1.986e-005 0.0250

1593.0640 45344.8463 28.4639 9.6577 4695.2154 51.5270 35706.3755 7.5261 0.9960 2.043e-005 0.0523

1444.2333 42654.5080 29.5344 11.3403 3761.3197 48.8410 32370.5377 7.3563 0.9968 1.911e-005 0.0406

160.0246 2882.8427 18.0150 977.4477 2.9494 2.8833 3586.7355 18.0725 0.0091 0.0004115 0.5648 0.0642

29.3266 528.3180 18.0150 954.7504 0.5534 0.5284 657.3158 18.2167 0.0086 0.0002687 0.5833 0.0578

178.1574 3218.6598 18.0664 984.0946 3.2707 3.2144 3993.1563 18.0877 0.0094 0.0004782 0.5542 0.0657

188

CHEMCAD 5.1.3 Job Name: nitric acid

Page 3 Date: 12/21/2013

Stream No. 9 Name - - Overall - Molar flow kmol/h 1533.2676 Mass flow kg/h 45873.2493 Temp C 83.4790 Pres kPa 1060.0000 Vapor mole fraction 0.9504 Enth kcal/h -1.2225E+007 Tc C -61.2340 Pc kPa 6126.5270 Std. sp gr. wtr = 1 0.894 Std. sp gr. air = 1 1.033 Degree API 26.6962 Average mol wt 29.9186 Actual dens kg/m3 11.3094 Actual vol m3/h 4056.2009 Std liq m3/h 51.3324 Std vap 0 C m3/h 34366.1179 - - Vapor only - Molar flow kmol/h 1457.2828 Mass flow kg/h 44273.8775 Average mol wt 30.3810 Actual dens kg/m3 10.9193 Actual vol m3/h 4054.6525 Std liq m3/h 49.8439 Std vap 0 C m3/h 32663.1003 Cp kcal/kmol-C 7.7917 Z factor 0.9948 Visc Pa-sec 1.851e-005 Th cond kcal/h-m-C 0.0249 - - Liquid only - Molar flow kmol/h 75.9847 Mass flow kg/h 1599.3718 Average mol wt 21.0496 Actual dens kg/m3 1032.9471 Actual vol m3/h 1.5484 Std liq m3/h 1.4886 Std vap 0 C m3/h 1703.0165 Cp kcal/kmol-C 19.3733 Z factor 0.0099 Visc Pa-sec 0.0003485 Th cond kcal/h-m-C 0.5132 Surf tens N/m 0.0560

Time: 17:03:43 10

11

12

166.9448 3007.5101 25.0000 101.3000 0.0000 -1.1412E+007 374.2000 22118.2302 1.001 0.622 9.8949 18.0150 996.7087 3.0174 3.0080 3741.8412

1700.2124 48880.7581 81.9617 1060.0000 0.8481 -2.3638E+007 -30.2339 5555.8245 0.900 0.993 25.6625 28.7498 12.2125 4002.5238 54.3404 38107.9591

1274.8711 35693.1056 -168.2483 1060.0000 0.0000 -4.0263E+006 -142.3805 3356.5915 0.815 0.967 42.0561 27.9974 665.4294 53.6392 43.8188 28574.5104

1442.0068 43542.4315 30.1956 10.8929 3997.3251 49.3320 32320.7496 7.7344 0.9953 1.858e-005 0.0249 166.9448 3007.5101 18.0150 996.7087 3.0174 3.0080 3741.8412 18.0248 0.0010 0.0009227 0.5215 0.0721

258.2056 5338.3266 20.6751 1026.8606 5.1987 5.0084 5787.2082 19.2078 0.0098 0.0003542 0.5190 0.0569

1274.8711 35693.1056 27.9974 665.4294 53.6392 43.8188 28574.5104 18.6700 0.0511 0.0001005 0.0777 0.0033

CHEMCAD 5.1.3 Job Name: nitric acid

Page 4 Date: 12/21/2013

Stream No. 13 Name - - Overall - Molar flow kmol/h 425.3410 Mass flow kg/h 13187.6533 Temp C 95.7593 Pres kPa 1060.0000 Vapor mole fraction 0.0000 Enth kcal/h -2.4990E+007 Tc C 311.1441 Pc kPa 12082.3953 Std. sp gr. wtr = 1 1.255 Std. sp gr. air = 1 1.071 Degree API -18.7077 Average mol wt 31.0049 Actual dens kg/m3 1164.9129 Actual vol m3/h 11.3207 Std liq m3/h 10.5216 Std vap 0 C m3/h 9533.4425 - - Vapor only - Molar flow kmol/h Mass flow kg/h Average mol wt Actual dens kg/m3 Actual vol m3/h Std liq m3/h Std vap 0 C m3/h Cp kcal/kmol-C Z factor Visc Pa-sec Th cond kcal/h-m-C - - Liquid only - Molar flow kmol/h 425.3410 Mass flow kg/h 13187.6533 Average mol wt 31.0049 Actual dens kg/m3 1164.9129 Actual vol m3/h 11.3207 Std liq m3/h 10.5216 Std vap 0 C m3/h 9533.4425 Cp kcal/kmol-C 23.8315 Z factor 0.0131 Visc Pa-sec 0.0003193 Th cond kcal/h-m-C 0.3885 Surf tens N/m 0.0415

Time: 17:03:43 14

15

137.1546 2335.8798 -15.0000 1240.0000 0.0000 -2.3156E+006 132.5000 11278.4833 0.619 0.588 96.9616 17.0310 656.6747 3.5571 3.7749 3074.1346

137.1546 2335.8798 35.0000 1240.0000 1.000 -1.5163E+006 132.5000 11278.4833 0.619 0.588 96.9616 17.0310 9.1153 256.2590 3.7749 3074.1346 137.1546 2335.8798 17.0310 9.1153 256.2590 3.7749 3074.1346 9.6843 0.9044 1.072e-005 0.0240

137.1546 2335.8798 17.0310 656.6747 3.5571 3.7749 3074.1346 19.5609 0.0190 0.0002027 0.4913 0.0299

190