In this study, a chemical engineering plant for the production of Isopropyl Alcohol (IPA) using the Tokuyama Soda Process has been designed. The feasibility of the process was analyzed using thermodynamic principles, and a detailed heat and mass balance were performed to determine the optimal conditions for the production of IPA. The design of key equipment such as the distillation column, heat exchanger, condenser, and reactor were also carried out. A Process and Instrumentation Diagram (PID) was developed to illustrate the process flow and control systems. Finally, a cost analysis was conducted to estimate the capital and operating costs of the plant. The results of the study show that the designed plant has the potential to produce IPA efficiently and cost-effectively, meeting the quality requirements of the target market. The study demonstrates the importance of thorough analysis and design in developing efficient and effective chemical engineering processes.Efficient and Cost-effective Production of Isopropyl Alcohol using Tokuyama Soda Process: Design and Analysis of a Chemical Engineering Plant Agilan R, Mohammed Riyaz, Mohammed Shaheer, Pratheekraj P Sheety List of Tables Table no Title Page no 3.1 Thermodynamic properties of compounds 19 4.1 Material balance across reactor 24 4.2 Material balance across heat exchanger 25 4.3 Material balance across separator 27 4.4 Material balance across distillation column 28 4.5 Material balance across the entire process 28 5.1 Energy balance across reactor 30 5.2 Total energy balance across reactor 30 5.3 Energy balance across heat exchanger 31 5.4 Total energy balance across heat exchanger 32 5.5 Energy balance across condenser 32 5.6 Total energy balance across condenser 33 5.7 Energy balance across reboiler 34 5.8 Total energy balance across reboiler 34 6.1 Inlet and outlet flowstreams of heat exchanger 37 6.2 Properties of streams at average temperature of heat exchanger 37 6.3 Pipe dimensions of heat exchanger 37 6.4 Inner side heat transfer properties of heat exchanger 38 6.5 Annular side heat transfer properties of heat exchanger 38 6.6 Overall heat transfer heat transfer coefficient of heat exchanger 39 6.7 Number of tubes and hairpin bends of heat exchanger 39 6.8 Stream compositions in terms of IPA for distillation column 40 6.9 Relative volatility for distillation column 41 6.10 Equilibrium data for distillation column 41 6.11 Properties of distillation column fluids 44 6.12 Calculation of enriching section diameter for distillation column 44 6.13 Calculation of stripping section diameter for distillation column 45 6.14 Inlet and outlet flowstreams of condenser 46 6.15 Properties of streams at average temperature of condenser 46 6.16 Pipe dimensions of condenser 47 6.17 LMTD calculations for condenser 47 6.18 General specifications for condenser 48 6.19 Tube side heat transfer properties of condenser 48 6.20 Shell side heat transfer properties of condenser 49 6.21 Overall heat transfer properties of condenser 49 6.22 Pressure drop calculation for condenser 49 7.1 List of equipments and costs 51 7.2 Factors contributing to direct cost 51 7.3 Factors contributing to indirect cost 52 7.4 Factors contributing to FCI 52 7.5 Factors contributing to fixed charges 53 7.6 Factors contributing to direct production cost 54 7.7 Factors contributing to general expenses 54 List of Figures Figure no Title Page No 1.1 Isopropyl Alcohol from propylene 01 1.2 Annual Production 03 1.3 Isopropyl Alcohol Market by Region 04 2.1 Structure of Isopropyl Alcohol 07 2.2 Physical Properties 08 2.3 Indirect Hydration Process 11 2.4 Direct Hydration Process 13 2.5 Flow Diagram for Tokuyama Process 16 4.1 Material balance for Reactor 23 4.2 Material balance for Heat Exchanger 25 4.3 Material balance for Separator 26 4.4 Material balance for Distillation Column 27 5.1 Energy Balance for Reac tor 29 5.2 Energy Balance for Heat Exchanger 31 5.3 Energy Balance for Condenser 32 5.4 Energy Balance for Re -boiler 33 6.1 Plot of y vs x for R actual 42 8.1 Reference plant layout 59 8.2 Referential plant layout 60 9.1 P&ID of the plant 63 CHAPTER -1 INTRODUCTION _____________________________________________________________________________________ 1 Isopropyl alcohol was first produced by Standard oil in 1920 by hydrating propene . Its major use at the time was not rubbing alcohol but for oxidation to acetone , whose first major use was in World War I for the preparation of cordite , a smokeless, low explosive propellant. The discovery of isopropyl alcohol happened quite early though it wasn’t very significant at that time. IPA’s distillation process was discovered in the first century and it was available for consumption and uses in the medical field as well. The first industrial process happened in 1926, which was the base for different alcohols through chemical synthesis. Scrolls have been found that were made by the Ancient Egyptia ns that gives directions to produce beer from plant foods and IPA was commonly found in the drink however it was referred to as ethanol. 1.1 About the compound Isopropyl alcohol , also called 2-propanol , one of the most common members of the alcohol family of organic compounds is manufactured from propylene as shown in Figure 1.1 . Isopropyl alcohol was the first commercial synthetic alcohol. Isopropyl alcohol (IUPAC name propan -2-ol) is a colorless, flammable chemical compound (chemical formula CH 3CHOHCH 3) with a strong odor. As an isopropyl group linked to a hydroxyl group, it is the simplest example of a secondary alcohol, where the alcohol carbon atom is attached to two other carbon atoms. It is a structural isomer of 1-propanol and ethyl methyl ether. Figure 1.1 Isopropyl alcohol from propylene [1] Isopropyl alcohol is miscible in water, ethanol, ether, and chloroform. It dissolves ethyl cellulose, polyvinyl butyral, many oils, alkaloids, gums and natural resins.Unlike ethanol or methanol, isopropyl alcohol is not miscible with salt solutions and can be separated from aqueous solutions by adding a salt such as sodium chloride. The process is CHAPTER -1 INTRODUCTION _____________________________________________________________________________________ 2 colloquially called salting out , and causes concentrated isopropyl alcohol to separate into a distinct layer . It is used in the manufacture of a wide variety of i ndustrial and household chemicals, and is a common ingredient in chemicals such as antiseptics, disinfectants, and detergents. 1.2 Growth and demand 1.2.1 Market analysis of the compound The global isopropyl alcohol market reached a value of US$ 3.8 Billion in 2018, registering a CAGR of 5.7% during 2011 -2018. The market value is further anticipated to reach US$ 4.8 Billion by 2024, exhibiting a CAGR of nearly 4% during 2019 -2024. On the basis of the industry, the market has been divided into cosmetic and personal ca re, pharmaceutical, food and beverage, paints and coatings, chemical, and others. Amongst these, isopropyl alcohol finds extensive application in the chemical industry as an intermediate and cleaning agent.  Region -wise, Asia represents the largest market f or isopropyl alcohol around the world. This can be accredited to the burgeoning personal care and cosmetics industry in the region. Isopropyl alcohol is used in the production of skin cleansers and lotions, deodorants, nail paints, hair tonics and shampoos . Other major markets include North America, Europe and South America.  On assessing the import and export scenario of the market, it has been found that the United States represents the biggest importer and exporter of isopropyl alcohol.  The competitive landscape of the market has been examined with some of the key players being The Dow Chemical Company, Exxon Mobil Corporation, Shell Chemicals Europe B.V., INEOS Enterprises Group Limited and Super Chemical Technology Co. Ltd.  Figure 1.2 shows the annual production of Isopropyl Alcohol from 1990 -2018 CHAPTER -1 INTRODUCTION _____________________________________________________________________________________ 3 Figure 1.2 Annual production of Isopropyl Alcohol (1990 -2018) [2] 1.2.2 Immense growth potential in India In India, pharmaceutical industry, the largest driver for iso -propanol, shows promising growth and expected to be the primary factor in the rise of iso -propanol growth. Chemical industry, and coating and ink industry are other major end consumers of iso -propanol in the country, which are also expected to grow in the coming years.Southern and eastern regions are also expected to drive the demand for iso -propanol in the coming years due to a rise in the pharmaceutical industry in these regions. Demand for iso -propanol has recorded impressive growth rates. The only domestic producer of IPA is, Deepak Fertilizers an d Petrochemicals Corporation Limited, which has an installed capacity of 70,000 metric tons per annum and commenced production from 2006.Globally, the demand growth of Isopropanol is majorly driven by its use in cosmetics, household products and pharmaceut ical industries. North America leads production as well as consumption of Isopropanol while Asia -Pacific is the largest consumer. India and China are among the top growing markets of Isopropanol for the pharmaceutical applications as shown in figure 1.3. CHAPTER -1 INTRODUCTION _____________________________________________________________________________________ 4 Figure 1.3 Isopropyl alcohol market by region [3] 1.3 Applications  Solvent : Isopropyl alcoho l dissolves a wide range of non polar compounds. It also evaporates quickly, leaves nearly zero oil traces, compared to ethanol, and is relatively non-toxic, compared to alternative solvents .  Intermediate : Isopropyl alcohol is esterified to give isopropyl acetate, another solvent. It reacts with carbon di sulfide and sodium hydroxide to give sodium isopropylxanthate, a herbicide and an ore flotation reagent.  Medical : Rubbing alcohol, hand sanitizer, and disinfecting pads typically contain a 60 – 70% solution of isopropyl alcohol or ethonal in water. Isopropyl alcohol is used as a water - drying aid for the prevention of otitis externa, it can also be used as an anesthetic.  Automotive : Isopropyl alcohol is a major ingredient in "gas dryer" fuel additives. In significant quantities water is a problem in fuel tanks, as it separates from gasoline and can freeze in the supply lines at low temperatures.  Laboratory : As a biological specimen preservative, isopropyl alcohol provides a comparatively non -toxic alternative to formaldehyde and other synthetic preservatives. Isopropyl alcohol solutions of 7 0–99% are used to preserve specimens. CHAPTER -1 INTRODUCTION _____________________________________________________________________________________ 5 Isopropyl alcohol is often used in DNA extraction. A lab worker adds it to a DNA solution to precipitate the DNA, which then forms a pellet after centrifugation. This is possible because DNA is insoluble in isopropyl alcohol. 1.4 Toxic effects of the compound  Clinical effects :  Drowsiness, ataxia and stupor.  Coma and respiratory depression.  Irritation of mucous membranes and eyes.  High risk circumstances of poisoning :  Accidental ingestion of rubbing alcohols/toiletries by children.  Occupational or accidental exposure to liquid or its vapour.  Routes of exposure:  Oral ingestion of rubbing alcohols and toilet preparations, constitute the most common route of exposure.Doses of above 20 mL may produce toxic effects.  Dermalexpos ure to the liquid and vapour. There is little absorption through intact skin, but significant delayed absorption over 4 hours postulated.  Eye exposure to liquid and vapour. Both the liquid and solvent are severely irritant.  Mode of action:  Isopropyl alcoh ol is an irritant to mucous membranes and eyes.  Isopropyl alcohol is a potent central nervous system (CNS) depressant, and in large doses causes cardiovascular depression.  Inebriation, peripheral vasodilation and hypothermia may also occur. 1.5 Objective and s cope The present project on p roduct ion of 50 tons/day of Iso propyl alcohol by hydration of propyl ene will be based on following objectives.  To execute a detailed l iterature survey of various processes used in the manufacture of Isopropyl alcohol.  To carry out thermodynamic feasibility studies for the selected process.  To study the material balance for the process.  To study the energy balance for the process. CHAPTER -1 INTRODUCTION _____________________________________________________________________________________ 6  To do the process and mechanical design of equipments used in the process.  To study the c ost estimation and economics .  To prepare the p lant location and layout report.  To prepare the PID, p ollution and safety details for the process. CHAPTER -2 LITERATURE SURVEY _____________________________________________________________________________________ 7 A detailed study about Isopropyl Alcohol and its various production methods was studied by analyzing individual sources. In this chapter we have identified the best process used for the production of Isopropyl alcohol. 2.1 About the compound Isopropyl alcohol , also called 2-propanol , one of the most common members of the alcohol family of organic compounds. Isopropyl alcohol was the first commercial synthetic alcohol. Isopropyl alcohol is miscible in water, ethanol, ether, and chloroform. It dissolves ethyl cellulose, polyvinyl butyral, many oils, alkaloids , gums and natural resins. Unlike ethanol or methanol, isopropyl alcohol is not miscible with salt solutions and can be separated from aqueous solutions by adding a salt such as sodium chloride. The process is colloquially called salting out , and causes concentrated isopropyl alcohol to separate i nto a distinct layer. It is used in the manufacture of a wide variety of industrial and household chemicals, and is a common ingredient in chemicals such as antiseptics, disinfectants, and detergents. Isopropyl alcohol (IUPAC name propan -2-ol; commonly cal led isopropanol or 2 -propanol) is a colorless, flammable chemical compound (chemical formula CH 3CHOHCH 3) with a strong odor. As an isopropyl group linked to a hydroxyl group, it is the simplest example of a secondary alcohol, where the alcohol carbon atom is attached to two other carbon atoms as shown in Figure 2.1 . It is a structural isomer of 1 -propanol and ethyl methyl ether . Figure. 2.1 Structure of Isopropyl alcohol [4] CHAPTER -2 LITERATURE SURVEY _____________________________________________________________________________________ 8 2.2 Physical properties Figure 2.2 gives us the physical properties of isopropyl alcohol Fig 2.2 Physical properties of isopropyl alcohol [1] 2.3 Chemical properties Chemical properties of isopropyl alcohol are determined by its functional hydroxyl group in the secondary position. Except for the production of acetone, most isopropyl alcohol chemistry involves the introduction of the isopropyl or isopropoxy group into o ther organic molecules by the breaking of the C –OH or the O –H bond in the isopropyl alcohol molecule. Isopropyl alcohol undergoes reactions typical of an active secondary alcohol. It can be dehydrogenated, oxidized, esterified, etherified, aminated, haloge nated, or otherwise modified at the OH moiety more readily than primary alcohols such as n -propyl or ethyl alcohol. Manufacture of the commercially important aluminum isopropoxide and isopropyl halides illustrates this reactivity. The aluminum isopropoxide reaction involves the aluminum replacement of the hydroxyl hydrogen atom and concomitant hydrogen evolution; the isopropyl halides reaction involves hydroxyl group displacement. Aluminum isopropoxide is produced in quantitative yield by refluxing isopropy l alcohol with aluminum turnings. CHAPTER -2 LITERATURE SURVEY _____________________________________________________________________________________ 9 6 (CH 3)2 CHOH + 2 Al −→ 2 ((CH 3)2 CHO) 3 Al + 3 H 2 Catalytic amounts of mercuric chloride are usually employed in this preparation. Aluminum isopropoxide is a useful Meerwein -Ponndorf -Verley reducing agent in certain est er-exchange reactions and is a precursor for aluminum glycinate, a buffering agent. Displacement of the hydroxyl group is exemplified by the production of isopropyl halides, eg, isopropyl bromide, by refluxing isopropyl alcohol with a halogen acid, eg, hyd robromic acid. (CH 3)2 CHOH + HBr −→ (CH3) 2CHBr + H 2O The order of reactivity with acid is HI >HBr>HCl. Reaction with hydrochloric acid to form isopropyl chloride is facilitated by a zinc chloride catalyst. 2.4 D ifferent processes for the manufacture of Isopropyl alcohol 2.4.1 Indirect Hydration This process is based on a two -step reaction of propylene and sulfuric acid. In the first step, mixed sulfate esters, primarily isopropyl hydrogen sulfate, but also diisopropyl sulfate, form. These are then hydrolyzed, forming the alcohol and sulfuric acid. In the process, propylene reacts with sulfuric acid (>60 wt%) in agitated reactors or absorbers at moderate (0.7 –2.8 MPa (100 –400 psig)) pressure. The isopropyl sulfate est ers form and are maintained in the liquid state at 20 –80◦C. Because the reaction is exothermic, internal cooling coils or external heat exchangers are used to control the temperature. There are two general operational CHAPTER -2 LITERATURE SURVEY _____________________________________________________________________________________ 10 modes for conducting the reaction. In the two -step strong acid process, separate reactors are used for the propylene absorption and sulfate ester hydrolysis stages. The reaction occurs at high sulfuric acid concentration (>80 wt%), at 1 –1.2 MPa (130 –160 psig) pressure, and low (eg, 20 – 30◦C) te mperature. The weak acid process is conducted in a single stage at low acid (60 –80 wt %) concentration and at higher (2.5 MPa (350 psig)) pressure and (60 –65◦C) temperature. The sulfate ester hydrolysate is stripped to give a mixture of isopropyl alcohol, isopropyl ether, and water overhead, and dilute sulfuric acid bottoms. The overhead is neutralized using sodium hydroxide and refined in a two -column distillation system. Diisopropyl ether is taken overhead in the first, ie, ether, column. This stream is g enerally recycled to the reactors to produce additional isopropyl alcohol by the following equilibrium reaction: ((CH 3)2 CH) 2 O+H 2SO 4 ←→ (CH 3)2 CHOSO 3H + (CH 3)2 CHOH Wet isopropyl alcohol (87 wt % and 91 vol %) is taken overhead in the second still. 93% of the charged propylene is converted to isopropyl alcohol in this system. The bottoms from the stripper (40 –60 wt % acid) are sent to an acid reconcentration unit for upgrading to the proper acid strength and recycling to the reactor. Because of the associa ted high energy requirements, reconcentration of the diluted sulfuric acid is a costly operation. Acid corrosion presents a problem in isopropyl alcohol factories. Steel (qv) is a satisfactory material of construction for tanks, lines, and columns where co ncentrated (>65 wt%) acid and moderate. Figure 2.3 shows the flowchart for the indirect hydration process for the manufacture of Isopropyl alcohol. CHAPTER -2 LITERATURE SURVEY _____________________________________________________________________________________ 11 Figure 2.3 Indirect hydration process for the manufacture of isopropyl alcohol [1] 2.4.2 Direct Hydration The acid -catalyzed direct hydration of propylene is exothermic and resembles the preparation of ethyl alcohol from ethylene. The equilibrium can be controlled to favor product alcohol if high pressures and low temperatures are applied. The advantage of low temperature is difficult to utilize, however, because most known catalysts require high or moderate temperatures to be effective. CHAPTER -2 LITERATURE SURVEY _____________________________________________________________________________________ 12 Propylene hydration in dilute acid solution probably proceeds according to the rate -determining formation of propyl carbonium ion. According to this mechanism, the reaction rate is proportional to the concentration of hydronium ion and is independent of the associated anion, ie,rate=k[CH3H6][H3O+]. However, the acid anion may play a marked role in hydration rat e, eg, phosphomolybdate and phosphotungstate anions exhibit hydration rates two or three times that of sulfate or phosphate. Association of the polyacid anion with the propyl carbonium ion is suggested. Protonation of propylene occurs more readily than tha t of ethylene as a result of the formation of a more stable secondary carbonium ion. Thus higher conversions are achieved in propylene hydration. There are three basic processes in commercial operation:  Vapor -phase hydration over a fixed -bed catalyst of s upported phosphoric acid (Veba - Chemie) or silica -supported tungsten oxide with zinc oxide promoter.  Mixed vapor –liquid -phase hydration at low (150◦C) temperature and high (10.13 MPa (100 atm)) pressure using a strongly acidic cation -exchange resin catalys t (Deutsche Texaco AG).  Liquid -phase hydration at high (270◦C) temperature and high (20.3 MPa (200 atm)) pressure in the presence of a soluble tungsten catalyst (Tokuyama Soda). CHAPTER -2 LITERATURE SURVEY _____________________________________________________________________________________ 13 Figure 2.4 Direct hydration process for the manufacture of isopropyl alcohol [1] Figure 2.4 shows us the flow chart for the direct hydration process. The description for the three direct hydration process is given below. CHAPTER -2 LITERATURE SURVEY _____________________________________________________________________________________ 14 In the Veba -Chemie process, a vaporized str eam of propylene and water is passed through an acidic catalyst bed (H3PO4 supported on SiO2) at 240 –260◦C and 2.5 –6.6 MPa (25 –65 atm). The gas stream from the reactor is cooled and fed to a scrubber where the remaining isopropyl alcohol is removed. Isopro pyl alcohol selectivity is ca 96% for the gas -phase process. Owing to equilibrium limitations in the gas phase at high temperature and low pressure, a low propylene conversion (5 – 6%) results and thus a large amount of unreacted propylene is recycled. Both processes involve high plant costs owing to high pressure requirements, gas recycles, and the requirement for high purity propylene (ca 99 wt %). Deutsche Texaco developed a trickle -bed process to avoid the disadvantages of the gas -phase process. In the tr ickle -bed process, a mixture of liquid water and propylene gas in a molar ratio of 12 to 15:1 is introduced at the top of a fixed -bed reactor and allowed to trickle down over a sulfonic acid ion -exchange resin. Reaction between the liquid and gas phases ta kes place at 130 –160◦C and 8–10 MPa (80 –100 atm), forming aqueous isopropyl alcohol. Propylene conversions per pass are greater than 75%, and isopropyl alcohol selectivity is 93%. Only 92 wt % propylene purity is needed for this process. Approximately 5% d iisopropyl ether and some alcohols of the higher oligomers form as by -products. The life of the cation -exchange resin is at least eight months. A liquid -phase variation of the direct hydration was developed by Tokuyama Soda . The disadvantages of the gas -phase processes are largely avoided by employing a weakly acidic aqueous catalyst solution of a silicotungstate . Preheated propylene, water, and recycled aqueous catalyst solution are pressurized and fed into a reaction chamber where they react in the liqui d state at 270◦C and 20.3 MPa (200 atm) and form aqueous isopropyl alcohol. Propylene conversions of 60–70% per pass are obtained, and selectivity to isopropyl alcohol is 98 –99 mol % of converted propylene. The catalyst is recycled and requires little repl enishment compared to other processes. The bottoms containing aqueous catalyst solution are recycled to the reactor, and the light ends are stripped of low boiling impurities, eg, diisopropyl ether and acetone. Azeotropic distillation yields dry isopropyl alcohol, and the final distillation column yields a product of more than 99.99% purity. CHAPTER -2 LITERATURE SURVEY _____________________________________________________________________________________ 15 2.5 Detailed description of the selected process Tokuyama Soda is the most economical and efficient process for direct hydration ofpropylene to isopropanol in liquid phase. The first commercial plant of this process, having a capacity of 30,000 metri c tons of isopropanol per annum. In recent years, v arious catalysts and processes for direct hydration of propylene have been proposed. It seems n ecessary for a better understanding of Tokuyama Process to review some typical hydration processes briefly. The catalysts most widely proposed for the vapour -phase direct hydration of propylene are mineral acids such as phosphoric acid and heteropoly acids on silica or diatomaceous earth, and solid acids of metal oxides . The direct hydration with a solid catalyst under a high pressure at a high mole ratio of water to propylene can be preferably employed to obtain high per pass conversions of propylene. The reaction is generally carried out by passing propylene and liquid water at a mole ratio of water to propylene between 10 and 15 over the catalyst bed at a temperature from 230 to 270 ℃ and a pressure from 200 to 300atm. For the hydration in the presence of liquid water, tungsten oxide is widely used as an effective component of catalyst. Tokuyama Soda has also proposed an active catalyst which comprises zirconium tungstate. This catalyst is suitable for both vapour phase and liquid -phase processes. Catalys t System employed in Tokuyama Process essentially comprises a dilute aqueous solution of definite kinds of polytungsten compounds within a selective pH range. This catalyst system is highly active and remarkably selective for hydrating propylene to isoprop anol. This system is also very stable because the catalyst components are strongly resistant to hydrolysis. Such high stability of the catalyst system is well maintained at all times in operation by the careful control of poisons which cause the decomposit ion of the catalyst. The catalyst life is, therefore, very long in continuous operation. Tokuyama Process gives, moreover, higher per pass conversions of propylene owing to intimate contact between dissolved propylene and the catalyst solution in the liquid phase than in the case where solid catalyst systems are employed CHAPTER -2 LITERATURE SURVEY _____________________________________________________________________________________ 16 2.5.1 Description of the process flowsheet Figure 2.5 Flow diagram of Tokuyama Process for the production of isopropy l Alcohol 50 tons/day [5] Figure 2.5 is the flow diagram for the Tokuyama process where l iquid propylene is preheated and fed to a reactor under pressure. The catalyst solution recovered from the azeotrope column is preheated by heat exchange with the reactor effluent and fed to the reactor operating at 240 ～270℃ and 150～250atm. The reaction takes place in the liquid phase between dissolved propylene and water. The aqueous solution containing isopropanol and catalyst is withdrawn from the reactor and cooled by heat exchange and then flashed un der reduced pressure in a gas separator where the unconverted propylene dissolved in the solutio n is separated and recycled. The liquid is sent to the azeotropic column, from where azeotropic mixture of isopropanol and water is drawn off and sent to furth er distillation steps for purification and dehydration. The aqueous solution containing catalyst from the bottom of the azeotrope column joins with fresh water and is recycled to the reactor. The azeotrope alcohol is freed of the light impurities in the first rectifying column and then dehydrated in two further columns with benzene as the dehydrating agent. To obtain a high grade isopropanol (99.99% purity), another additional distillation step is necessary. Under the optimum reaction conditions, the range of propylene conversion per pass is about 60 ～70% and selectivity to isopropanol based on the converted CHAPTER -2 LITERATURE SURVEY _____________________________________________________________________________________ 17 propylene is as much as 98 ～ 99%. The major by -product is di -isopropyl ether. In Tokuyama Process, both the spent gas from the reactor and all the by -products drawn off at the purification section are completely utilized as fuel for heating in the system. 2.5.2 Novelty of the process Tokuyama Process has a number of advantages based on its own catalyst system and process design. The features of the process confirmed in commercial operation are as follows:  The catalyst system comprises a homogeneous dilute aqueous solution which mainta ins high activity and excellent selectivity for hydrating propylene to isopropanol.  The catalyst solution is so highly stabilized that it is circulated repeatedly in the reaction system without any additional treatments. Catalyst life is, therefore, very long.  Since the reaction is carried out in aqueous solution under high pressure, the form ation of isopropanol is favoured by equilibrium effect.  The amount of unconverted propylene to be recycled is small owing to high conversion rate of propylene, and co nsequently this process does not require highly concentrated propylene feed. Commercial propylene of about 95% concentration is normally used.  The amount of by -products is remarkably small because the reaction takes place in an aqueous soluti on of highly selective catalyst . Organic acids and aldehydes are hardly detectable in the reaction product . A high grade isopropanol is easily obtained at the purification section.  Since the reaction is carried out in aqueous solution, the heat loss incidental to evapo ration and condensation of water can be avoided.  Because of absence of corrosion in the system, easy operation and maintenance of equipments are possible.  The process is simple. It is also a completely closed system free from waste substances, and conseque ntly the process can clear all the envir onmental protective regulation CHAPTER -2 LITERATURE SURVEY _____________________________________________________________________________________ 18 CHAPTER -3 THERMODYNAMIC FEASIBILITY _____________________________________________________________________________________ 19 Thermodynamic f easibility studies are conducted to predict whether a chemical reaction is feasible at a given temperature and pressure. Gibbs free energy is a parameter used to determi ne the feasibleness of a reaction. Gibbs free energy when negative, indicates that the reaction is feasible at a given temperature and pressure, otherwise (when positive) the reaction is not feasible and will require changes in temperature/pressure in orde r to promote the reaction. Gibbs free energy is calculated using the following relation ΔG reaction = Σ (G f) products -Σ (G f) reactants Reactions involved: 1. CH 3CH=CH 2+H 2O (CH 3)2CH0H 2. 2(CH 3)2CH0H H 2O+[(CH 3)2CH] 2 Thermodynamic properties of compounds are shown in table 3.1. Table 3.1 Thermodynamic Properties of Compounds COMPOUNDS ∆H f(kJ/mol) ∆G f (kJ/mol) CH 3CH=CH 2 20.41 62.22 H2O(liq) -285.83 -228.374 (CH 3)2CH0H -306.3 -162.44 [(CH 3)2CH] 2O -318.70 -164.88 Equations used: 1. ΔG reaction = Σ (G f) products -Σ (G f) reactants 2. ΔH reaction = Σ (H f) products -Σ (H f) reactants 3. ∆G= -RT ln(K) 4. 𝐿𝑛(𝐾1 𝐾2)=−∆𝐻 𝑅𝑇[1 𝑇2−1 𝑇1] (Vant Hoff equation) The feasibility studies of the reactions are shown as follows CHAPTER -3 THERMODYNAMIC FEASIBILITY _____________________________________________________________________________________ 20 3.1 Reacti on 1 (Reaction temperature = 523 K) CH 3CH=CH 2+H2O (CH 3)2CH0H Enthalpy of reaction at room temperature ∆H reaction = ∑∆H products - ∑∆H reactants = [-306.3 -(-285.83+20.41)] = -40.88 kJ/mol Gibbs free energy at room temperature ∆G reaction = ∑∆G products - ∑∆G reactants = [-162.44 -(62.22 -228.374)] = 3.714 kJ/mol Using Gibbs free energy equation ∆G= -RT ln(K) lnK = -∆G/RT Where R = 0.008314 kJ/mol -K and T = 298K lnK1= (-3.714 )/(0.008314x298) lnK1= 1.4982 Using Vant Hoff equation, we get 𝑙𝑛(𝐾1 𝐾2)=−∆𝐻 𝑅𝑇[1 𝑇2−1 𝑇1] 𝑙𝑛(𝐾1 𝐾2)=−−(−40.88) 0.008314[1 523−1 298] lnK2= 5.5946 ∆G= -RT ln(K) ∆G= -(0.008314x523x5.5946) ∆G= -24.3335 kJ/mol Since ∆G reaction < 0, the reaction is feasible at reaction temperature Therefore, reaction 1 is thermodynamically feasible. CHAPTER -3 THERMODYNAMIC FEASIBILITY _____________________________________________________________________________________ 21 3.2 Reaction 2 (Reaction temperature =5 23K) 2(CH 3)2CH0H H2O+ [(CH 3)2CH] 2 Enthalpy of reaction at room temperature ∆H reaction = ∑∆H products - ∑∆H reactants = [-285.33 -318.70 -(2x-306.3)] = 8.57 kJ/mol Gibbs free energy at room temperature ∆G reaction = ∑∆G products - ∑∆G reactants = [-228.374 -164.88 -(2x-162.44)] = -68.374 kJ/mol Using Gibbs free energy equation ∆G= -RT ln(K) lnK = -∆G/RT Where R = 0.008314 kJ/mol -K and T = 298K lnK1= -(-68.374)/(0.008314x298) lnK1= 27.5832 Using Vant Hoff equation, we get 𝑙𝑛(𝐾1 𝐾2)=−∆𝐻 𝑅𝑇[1 𝑇2−1 𝑇1] 𝑙𝑛(𝐾1 𝐾2)=−−8.57 0.008314[1 523−1 298] lnK2= 26.0962 ∆G= -RT ln(K) ∆G= -(0.008314x523x26.0962) ∆G= -113.5046 kJ/mol Since ∆G reaction< 0, the reaction is feasible at reaction temperature Therefore, reaction 2 is thermodynamically feasible CHAPTER -3 THERMODYNAMIC FEASIBILITY _____________________________________________________________________________________ 22 CHAPTER -4 MATERIAL BALANCE _____________________________________________________________________________________ 23 Material balances play a major role in designing a process plant. In a process, it is necessary that material balance is studied for the overall process and for individual equipments in order to determine the amount of raw materials required to run the pro cess, compositions of each streams and the flow rates of each steam. Here, material balance is studied for the production of 50 tons/day of Isopropyl alcohol. The material balance is conducted for the entire plant and for individual equipments. Henceforth, all flow rates will be depicted in kmol/h 4.1 Production Rate 50 tons of Isopropyl alcohol is produced every day 2083.333 kg of Isopropyl alcohol is produced every hour Therefore, 2083.333 kg/hr of Isopropyl alcohol is produced . 4.2 Raw Material Requirement From stoichiometry, to produce 2083.33 kg/hr Isopropyl alcohol, i.e.34.967 kmol/hr we will require ex actly the same amount of propylene and water (feed). Therefore, 34.967 kmol/hr of propylene and water is required as raw material. Water is fed in 1:1 rati o to propylene Amount of water required = 34.967x 1 = 34.967 kmol /hr Therefore, 34.967 kmol /hr of air is required. 4.3 Material b alance across individual e quipments 4.3.1 Material balance across Reactor Figure 4.1 R eactor with inlet and outlet streams used in the manufacture of isopropyl alcohol 50 tons/day REACTOR C3H6 H2O C3H8O H2O C3H6 CHAPTER -4 MATERIAL BALANCE _____________________________________________________________________________________ 24 Figures 4.1 Shows the inlet and outlet material stream of a reactor Weig ht of CH 3CH=CH 2 entering= 34.967kmol /hr Weight of H2O entering= 34.967 kmol /hr Reaction occurring: CH 3CH=CH 2+H2O (CH 3)2CHOH Amount of CH 3CH=CH 2recycled = 10.4901 kmol/hr Amount ofCH 3CH=CH 2 entering the reactor= 45.95286kmol/hr Moles of H2Oentering= 34.967kmol/hr Given single pass conversion for propylene in the reactor= 70% Amount of (CH 3)2CHOH formed = 0.7*(45.9586) kmol of CH 3CH=CH 2 = 34.967 kmol of (CH 3)2CHOH Amount of H 2O formed= 0.3*34.967 kmol H 2O = 10.4901 kmol of H 2O Table 4.1 Material balance across reactor Compounds Material entering (kmol/h) Inside reactor (kmol/h) Material leaving (kmol/h) Propylene 49.9529 -34.967 14.9859 Water 49.9529 -34.967 14.9859 Isopropyl alcohol 0 34.967 34.967 Total 99.9057 -34.967 64.9387 Table 4.1 gives us the result for the material balance which was conducted across a reactor CHAPTER -4 MATERIAL BALANCE _____________________________________________________________________________________ 25 4.3.2 Material Balance across Heat Exchanger Figure 4.2 Heat Exchanger with inlet and outlet streams used in the manufacture of isopropyl alcohol 50 tons/day Figure 4.2 shows the inlet and outlet material streams of the heat exchanger. Table 4.2 Material balance across heat exchanger Compounds Material entering (kmol/h) Material leaving (kmol/h) Propylene 14.9859 14.9859 Water 14.9859 14.9859 Isopropyl Alcohol 34.967 34.967 Total 64.9387 64.9387 Table 4.2 gives us the result of the material balance which was conducted across a heat exchanger HEAT EXCHANGER C3H8O H2O C3H6 C3H8O H2O C3H6 CHAPTER -4 MATERIAL BALANCE _____________________________________________________________________________________ 26 4.3.3 Materia l Balance across Separator Fig 4.3 Separator with inlet and outlet streams used in the manufacture of isopropyl alcohol 50 tons/day Figure 4.3 shows us the inlet and outlet steam of a separator Amount of C 3H6 entering= 14.98586 kmol /hr Amount ofC 3H6 Recycled=14.98586kmol /hr Amount of C 3H6leaving = 0 kmo/hr Table 4 .3 Material balance across separator Compounds Material entering (kmol/h) Recycle (kmol/h) Material leaving (kmol/h) Propylene 14.9859 14.9859 0 Water 14.9859 0 14.9859 Isopropyl Alcohol 34.967 0 34.967 Total 64.9387 14.9859 49.9528 Table 4.3 gives us the result for the material balance which was conducted across a separator SEPARATOR H2O C3H8O C3H6 (Recycle) C3H8O H2O C3H6 CHAPTER -4 MATERIAL BALANCE _____________________________________________________________________________________ 27 4.3.4 Material balance across Distillation Column Figure 4.4 Distillation column with inlet and outlet streams used in the manufacture of isopropyl alcohol 50 tons/day Figure 4.4 shows the inlet and outlet mater ial stream of a distilla tion column We need 99% pure Isopropyl alcohol (literature) i.e 1% water comes out with Isopropyl alcohol. Amount of H2O entering= 10.4901kmol/hr Amount of C3H8O entering = 34.967kmol /hr Moles of H2O leaving (top) = 0.01*34.967 = 0.3496 kmol /hr Amount of C3H8O Leaving (top) having 99% purity = 34.967 kmol/hr Amount of H2O leaving the bottom stream= 10.14043kmol/hr H2O C3H8O H2O DISTILLATION COLUMN C3H8O H2O CHAPTER -4 MATERIAL BALANCE _____________________________________________________________________________________ 28 Table 4.4 Materia l balance across distillation column Compounds Material entering (kmol/h) Top stream (kmol/h) Bottom stream (kmol/h) Propylene 0 0 0 Water 14.9859 0.34967 14.6361 Isopropyl Alcohol 34.967 34.617 0.35 Total 49.9528 34.967 14.9861 Table 4.4 gives us the result of the material bal ance which was conducted across a distillation column 4.4 Overall Material Balance The material balance across the enti re process is shown in table 4.5. Table 4.5 Material balance across the entire p rocess Compounds Material entering (kmol/h) Material leaving (kmol/h) Propylene 34.967 0 Water 49.9528 14.9859 Isopropyl Alcohol 0 34.967 Total 84.9198 49.9528 CHAPTER -5 ENERGY BALANCE _____________________________________________________________________________________ 29 As with mass, energy can be considered to be separately conserved in all but nuclear processes. In process design, energy balances are made to determine the energy requirements of the process: the heating, cooling and power required. The steady state ener gy balance is generally calculated using the relation: Energy out=Energy in + Heat of Reaction + Q (heat supplied/heat removed) In this section, we estimate the amount of utilities (coolant and duty) required by the plant for the production of 50 tons/day of Isopropyl Alcohol . Accordingly, enthalpy balance is perfo rmed over individual equipments, assuming heat losses to be negligible and all pumps and compressors to neither supply nor dissipate heat. 5.1 Reactor Figure 5.1 Reactor with inlet and outlet streams used in the manufacture of isopropyl alcohol 50 tons/day Figure 5.1 shows us the inlet and outlet energy streams across of a reactor For an isentropic process, pressure -temperature relation is given by P1= 1atm P2= 200atm R= 0.0821L.atm/K.mol Cp= 84.5495 T2= 250 ⁰C T2/T1 = (P 2/P1) (R/Cp) T1 = (523 )x(1/200)(84.5495/0.0821) T1= 246 ⁰C From separator C3H6 REACTOR C3H6 H2O C3H6 H2O IPA CHAPTER -5 ENERGY BALANCE _____________________________________________________________________________________ 30 Table 5.1 Energy balance for reactor ENTERING LEAVING Mass flow rate (kmol/hr) Specific Heat (kJ/kmol- K) T (K) Energy (kJ/s) Mass flow rate (kmol/hr) Specific Heat (kJ/kmol- K) T (K) Energy (kJ/s) C3H6 49.95286 131.0667 519 445.5704 14.98586 131.733 523 137.0925 H2O 49.95286 90.666 519 308.2254 14.98586 90.666 523 95.47286 IPA - - - - 34.967 375.4043 523 899.6766 TOTAL 753.7958 TOTAL 1132.248 Table 5.1 shows the energy balance performed across a reactor From energy balance equation, Total Energy Leaving= Total Energy Entering Energy Out = Energy In + Heat of Reaction + Q Q=0 (since the process is adiabatic) Heat of reaction=378.4521kJ/s Table 5.2 Total energy balance for reactor TOTAL ENERGY ENTERING (kJ/s) TOTAL ENERGY LEAVING (kJ/s) Reactants e nthalpies 753.79588 - Products e nthalpies - 1132.248 Heat of r eaction 378.4521 - Heat t ransferred - - TOTAL 1132.248 1132.248 Table 5.2 shows the total energy entering and leaving the reactor CHAPTER -5 ENERGY BALANCE _____________________________________________________________________________________ 31 5.2 Heat exchanger Figure 5.2 Heat exchanger with inlet and outlet streams used in the manufacture of isopropyl alcohol 50 tons/day Table 5.3 Energy balance for heat exchanger ENTERING LEAVING Mass flow rate (kmol/hr) Specific Heat (kJ/kmol- K) T (K) Energy (kJ/s) Mass flow rate (kmol/hr) Specific Heat (kJ/kmol- K) T (K) Energy (kJ/s) C3H6 14.98586 131.733 523 137.0925 14.98586 103.733 313 158.0113 H2O 14.98586 91.746 523 95.47866 10.4901 74.43 313 12.39331 IPA 34.967 375.4043 523 911.5827 34.967 169.5153 313 65.86046 TOTAL 1144.152 TOTAL 236.2651 Table 5.3 shows the energy balance performed across a heat exchanger From energy balance equation, Total Energy Leaving= Total Energy Entering Energy Out = Energy In + Heat of Reaction + Q 236.2651=1144.152+0+Q Q =-907.8869/s Heat transferred out of the product mixture to cool the gases to a temperature of 4 0⁰C and subsequently generate liquid is 907.8869kJ/s Waterat a temperature of 283K is required to cool the gases. Heat transferred out of the gases= Sensible and latent heat gained by water 907.8869 = [mC p1 (T2-T1) + m λ + mC p2 (T3-T2)] HEAT EXCHANGER C3H6 H2O IPA C3H6 H2O IPA CHAPTER -5 ENERGY BALANCE _____________________________________________________________________________________ 32 907.8869= m*[75.766*(273 -283) + 40680+ 76.32x(383 -273)] m =0.0211kmol/s m =76.2792kmol/h Mass of chilled water (coolant) required to cool down the gases is 76.2792 kmol/h Table 5.4 Total energy balance for heat exchanger TOTAL ENERGY ENTERING (kJ/s) TOTAL ENERGY LEAVING (kJ/s) Inlet energy 1144.152 - Outlet energy - 236.2651 Heat of r eaction - - Heat t ransferred - 907.8869 (cooling) TOTAL 1144.152 1144.152 Table 5.3 shows the total energy entering and leaving the Heat exchanger 5.3 Condense r Figure 5.3 Condenser with inlet and outlet streams used in the manufacture of isopropyl alcohol 50 tons/day Table 5.5 Energy balance for condenser ENTERING LEAVING Mass flow rate (kmol/hr) Specific Heat (kJ/kmol- K) T (K) Energy (kJ/s) Mass flow rate (kmol/hr) Specific Heat (kJ/kmol- K) T (K) Energy (kJ/s) H2O 0.34967 75.6576 363 0.66137 0.34967 75.2661 323 0.21931 IPA 34.967 218.5365 363 191.0391 34.967 165.0554 323 48.0957 TOTAL 191.7005 TOTAL 48.31508 Table 5.5 shows the energy balance performed across a condenser CONDENSE R H2O IPA H2O IPA CHAPTER -5 ENERGY BALANCE _____________________________________________________________________________________ 33 From energy balance equation, Total Energy Leaving= Total Energy Entering Energy Out = Energy In + Heat of Reaction + Q 48.31508 = 191.7005 + 0 + Q Q = -143.385 kJ/s Heat transferred out of the product mixture to cool the vapours to a temperature of 3 0⁰C and subsequently generate liquid is 143.385 kJ/s Waterat a temperature of 283K is required to cool the vapours. Heat transferred out of the gases= Sensible and latent heat gained by water 143.385= mC p1 (T2-T1) 143.385 = m*[75.366*(333 -283)] m = 0.03805kmol/s m = 136.9814kmol/h Mass of chilled water (coolant)required to cool down the gases is 76.2792 kmol/h Table 5.6 Total energy balance for condenser TOTAL ENERGY ENTERING (kJ/s) TOTAL ENERGY LEAVING (kJ/s) Inlet energy 191.7005 - Outlet energy - 48.31508 Heat of r eaction - - Heat t ransferred - 143.385 (C ondensation ) TOTAL 191.7005 191.7005 Table 5.6 shows the total energy entering and leaving the condenser 5.4 Reboiler Figure 5.4 Reboiler with inlet and outlet streams used in the manufacture of isopropyl alcohol 50 tons/day REBOILER H2O H2O CHAPTER -5 ENERGY BALANCE _____________________________________________________________________________________ 34 Table 5.7 Energy balance for reboiler ENTERING LEAVING Mass flow rate (kmol/hr) Specific Heat (kJ/kmol- K) T (K) Energy (kJ/s) Mass flow rate (kmol/hr) Specific Heat (kJ/kmol- K) T (K) Energy (kJ/s) H2O 10.14043 75.312 333 12.72826 10.14043 75.7224 378 22.3958 TOTAL 12.72826 TOTAL 22.3958 Table 5.7 shows the energy balance performed across a reboiler From energy balance equation, Total Energy Leaving= Total Energy Entering Energy Out = Energy In + Heat of Reaction + Q 22.3958 = 12.72826 + 0 + Q Q = 9.6674 kJ/s Reboiler duty = 9.6674 kW Table 5.8 Total energy b alance for reboiler TOTAL ENERGY ENTERING (kJ/s) TOTAL ENERGY LEAVING (kJ/s) Inlet energy 12.72826 - Outlet energy - 22.3958 Heat of r eaction - - Heat t ransferred 9.6674 - TOTAL 22.3958 22.3958 Table 5.8 shows the total energy entering and leaving the reboiler CHAPTER -6 EQUIPMENT DESIGN _____________________________________________________________________________________ 35 The design of equipment mainly deals with studying the design parameters and specifications of each equipment involve d in the manufacturing process. This chapter discusses about process and mechanical design o f equipments used in the production of Isopropyl alcohol. 6.1 Reactor design The reac tor here is a CSTR. Propylene on reaction with water yields Isopropyl alcohol. K = 0.00623 s-1 [6] For a CSTR performance equation is given by, τ = C aoxXa/-ra Reactor pressure = 200 atm Residence time ( τ) = 0.1 hours Volumetric flow rate of the reactant mixture = 23.3891m3/h Total Volume of reactor (V) = V 0xτ = 0.1x 23.3891 = 2.3891 m3 Considering free space of 1 0% Total volume = 2.3891+ (0.1 x2.3891) = 2.62801 m3 Mechanical Design Number of baffles = 4 Number of Impeller blades = 6 Assuming H/ D = 2 Substituting in V = 3.14 (D2/4) x H, where H = 2 D We get Diameter of the reactor, D m = 1.1873 m CHAPTER -6 EQUIPMENT DESIGN _____________________________________________________________________________________ 36 Height of the reactor H = 2.3746 m Diame ter of the Impeller , Dimp= 40 % Dm= 0.4749 m Baffle thickness = D m/ 12 = 0.0989 m Height of Impeller blade, h = 20 % D m = 0.2374 m Width of Impeller blade ,w = 25% Dm = 0.296825 m Bottom clearance, E = D m / 3 = 0.3957 m Wall thickness, t = 𝑃×𝐷 200 ×𝑓×𝐽−𝑃 + C (IS 2825) Where, t = thickness in mm P = operating pressure in kgf/cm2 (1 atm = 1.0332 kgf/cm2) D = diameter of column in mm (1254 mm) f = allowable stress (13) J = joint efficiency (0.85) C = corrosion alliance (min 5 mm) t = 200 ∗1.032 ∗1187 .3 (200 ×13×0.85)−200 ∗1.0332 + 5 = 122.31 mm The t hickness of the reactor is 12.231 cm Outer diameter D o= inner diameter + thickness Do= 1187.3 + 122.31 = 1309.61 mm Assuming Torispherical head, Crown radius, r c = 0.9 x D o= 0.9 x 1309.61 = 1178.649 mm CHAPTER -6 EQUIPMENT DESIGN _____________________________________________________________________________________ 37 Knuckle radius, r k = 0.17 x D m = 0.17 x 1187.3 = 201.841 mm 6.2 Heat exchanger design Heat exchanger is used to remove excess heat from the product stream (Propene, IPA and H 2O). It is a Double pipe heat exchanger where wate r is used as the cooling medium Table 6.1 Inlet and outlet flow streams of heat exchanger used in the manufacture of Isopropyl alcohol Inlet Temp. ( ⁰C) Outlet Temp. ( ⁰C) Mass flow rate, (kg/s) Tavg(⁰C) Hot fluid 250 40 0.81 145 Cold fluid 10 110 0.3798 60 Table 6.1 shows the temperature s of inlet and outlet streams of heat exchanger Table 6.2 Properties of Streams at average temperature Property Product Mixture Water Specific Heat, Cp (kJ/kg -K) 3.814 4.184 Density, ρ (kg/m3) 628.42 983.2 Thermal Conductivity, k(W/m -K) 0.0882 0.358 Viscosity, μ (kg/m -s) 0.0002125 0.000478 From heat balance the mass flow rate of cold flui d was found to be = 0.3798 kg/s Table 6.2 gives the properties of the product mixture and water at average temperature Table 6.3 Pipe Dimensions Outer diameter of outer pipe, D o 0.06 m Inner diameter of outer pipe, D i 0.052 m Outer diameter of inner pipe, do 0.0421 m Inner diameter of inner pipe, di 0.03505 m Standard Length of Pipe, L r 4.5 m Table 6.3 shows the standard pipe dimensions CHAPTER -6 EQUIPMENT DESIGN _____________________________________________________________________________________ 38 Logarithmic Mean Temperature Difference Calculations 𝑳𝑴𝑻𝑫 = (𝑻𝒉𝒊 −𝑻𝒄𝒐 )−(𝑻𝒉𝒐 −𝑻𝒄𝒊) 𝒍𝒏(𝑻𝒉𝒊 −𝑻𝒄𝒐 ) (𝑻𝒉𝒐 −𝑻𝒄𝒊) Log mean temperature, ΔT LMTD(⁰C) = 71.407oC Inner side and annular side heat tr ansfer coefficient calculations Table 6.4 Inner side heat transfer properties Flow area A = (π/4)x di2 0.000964 m2 Mass velocity G = m / A 839.922 kg/ m2-s Reynolds number NRe = (d ixG)/μ 138549.268 Prandtl number NPr= (C pxμ)/k 9.18 Nusselt number NNu =0.023 xRe0.8xPr0.33 610.816 Inside heat transfer coefficient hi = N Nuxk/d i 1537.0605 W/m2-C0 Table 6.4 shows the calculations for inner heat transfer coefficient Table 6.5 Annular side heat transfer properties Equivalent diameter De = (D i2-do2)/do2 0.02218 m Flow area A = (π/4)x (Di2-do2) 0.000731 m2 Mass veloctiy G = m / A 519.350 kg/ m2-s Reynolds number NRe = (D exG)/μ 24098.709 Prandtl number NPr= (C pxμ)/k 5.58 Nusselt number NNu =0.023x Re0.8xPr0.4 146.546 Annular heat transfer coefficient ho = N Nuxk/D e 2365.350 W/m2-C0 Table 6.4 shows the calculations for annular heat transfer coefficients CHAPTER -6 EQUIPMENT DESIGN _____________________________________________________________________________________ 39 Table 6.6 Overall heat transfer coefficient Dirt factor RD 0.000705 m2-C0/W 1/U (1/ hi) + (1/ h o) + R D 0.00177836 m2-C0/W Overall heat transfer coefficient U 523.764 W/m2-C0 Table 6.6 shows the calculations for overall heat transfer coefficient Number of tubes and hairpin bends From energy balance excess heat to be removed from product stream is, Q = 1023.7 kW Table 6.7 Number of tubes and hairpin bends Heat transfer area A = Q/(U xΔTLMTD ) 27.3713 m2 Number of tubes Nt= A/( πdoLr) 46 Number of hairpin bend s Nb = N t-1 45 Table 6.7 shows the number of hair pin bends required in the distillation column Mechanical Design Wall thickness, t = 𝑃×𝐷 200 ×𝑓×𝐽−𝑃 + C (IS 2825) Where, t = thickness in mm P = operating pressure in kgf/cm2 (1 atm = 1.0332 kgf/cm2) D = diameter of column in mm (60 mm) f = allowable stress (13) J = joint efficiency (0.85) CHAPTER -6 EQUIPMENT DESIGN _____________________________________________________________________________________ 40 C = corrosion alliance (min 5 mm) t = 1∗1.032 ∗60 (200 ×13×0.85)−1.0332 + 5 = 5.028 mm The thickness of the heat exchanger is 5.028 mm 6.3 Distillation column design Distillation process is carried out to increase the purity of Isopropyl alcohol to 99%. Table 6.8 Stream composition of distillation column in terms of IPA Table 6.8 shows the feed composit ion in the distillation column To calculate n umber of stages Feed temperature = 40oC = 313 K Table 6.9 Relative volatility Vapour pressure of IPA (mm of Hg) ln PVIPA=16.6796 – 3640.2/( -53.5+313) 14.149 Vapour pressure of H 2O (mm of Hg) ln PVW = 16.3872 – 3885.7/( -42.9+313) 7.396 Relative volatility, α α = PVIPA/ PVw 1.913 Table 6.9 shows the relative volatility calculations Equilibrium relation is given by, y*= αx/(1+x (α-1)) Table 6.10 gives the equilibrium data calculated using Raoult’s law Feed composition, x F 0.769 Distillate composition, x D 0.99 Bottom composition, x W 0.03 CHAPTER -6 EQUIPMENT DESIGN _____________________________________________________________________________________ 41 Table 6.10 Equilibrium data x y 0 0 0.03 0.056 0.126 0.215 0.222 0.35 0.318 0.468 0.414 0.576 0.51 0.666 0.606 0.746 0.702 0.817 0.798 0.878 0.894 0.942 0.99 0.997 CHAPTER -6 EQUIPMENT DESIGN _____________________________________________________________________________________ 42 Figure 6.1 plot of y vs x for R actual Figure 6.1 shows plot of y vs x for R actual Since feed is sent at a temperature below boiling point it is subcooled liquid. For subcooled liquid q is given by, q = (C pfx(T b-Tf)+λ f)/λf q = (103.06x(82.3 -40)+45900)/45900 q = 1.1 slope of q -line = q/(q -1) q/(q-1) = 1.1/(1.1 -1) = 11 From graph of y vs x to determine R m, rectification line intersects y -axis at x D/(R m+1) xD/(R m+1) = 0.43 Minimum reflux ratio, R m = (x D/0.43) -1 = (0.99/0.43) -1 = 1.3 CHAPTER -6 EQUIPMENT DESIGN _____________________________________________________________________________________ 43 Rm = 1.3 Actual reflux ratio, R actual = 2xR m = 2x1.3 = 2.6 Ractual= 2.6 Number of ideal stages = 17.8 Assuming efficiency, ɳ = 0.6 Number of actual stages = number of ideal stages/ɳ = 17.8/0.6 = 29.66 ≈ 30 Number of actual stages = 30 Feed plate location = 12 To calculate height of column Height of column, H = (Number of stages x Tray spacing) + Topcomer height + Downcomer height Assume, Tray spacing = 18 inches = 450mm Topcomer height = Downcomer height = 500mm H = (30x450) + 500 + 500 = 14500mm H = 14.5m To calculate diameter of column Table 6.11 Properties of Distillation column fluids Boiling point of IPA 82.3oC Boiling point of water 100oC Molecular weight of IPA 60 CHAPTER -6 EQUIPMENT DESIGN _____________________________________________________________________________________ 44 Molecular weight of water 18 Operating pressure 1 atm Reflux ratio 2.6 Feed flowrate, F 45.4571 kmol/h Distillate flowrate, D 34.967 kmol/h Bottom flowrate, W 10.49043 kmol/h Table 6.11 gives us the properties of the distillation column fluids Table 6.12 Calculation of Enriching section diametre Liquid flowrate, L L = RxD 90.9142 kmol/h Gas flowrate, G G = L + D 125.8812 kmol/h Density of gas, ρ g ρg= (PxM)/(RxT) 2.0442 kg/m3 Density of liquid, ρ L ρL= ρIPAxxD + ρWx (1-xD) 787.546 kg/m3 FLV FLV = (L/G)x (ρg/ρL)0.5 0.0367 CSB From C SB vs F LV plot 0.28 Unf Unf = C SB/(ρg/(ρ L-ρg))0.5) 1.6729 m/s Operating velocity, U n Un= 0.7 x U nf 1.17103 m/s Volumetric flowrate, Q Q = (G x M)/(3600 x ρg ) 1.0191 m3/s Area, A A = Q/ U n 0.8702 m2 Area to be considered, A T AT= A/0.88 0.9889 m2 Diameter, D T DT = ((4 x A T)/π)0.5 1122.383 mm Table 6.4 shows the calculations for enriching section diameter CHAPTER -6 EQUIPMENT DESIGN _____________________________________________________________________________________ 45 Table 6.13 Calculation of Stripping section diametre Liquid flowrate, L ` L` = L + qF 140.917 kmol/h Gas flowrate, G ` G` = L` - W 130.426 kmol/h Density of gas, ρ g ρg= (PxM )/(RxT) 0.6608 kg/m3 Density of liquid, ρ L ρL= ρIPAxxW + ρWx (1-xW) 993.562 kg/m3 FLV FLV = (L/G)x (ρg/ρL)0.5 0.027 CSB From C SB vs F LV plot 0.28 Unf Unf = C SB/(ρg/(ρ L-ρg))0.5) 3.29 m/s Operating velocity, U n Un = 0.7 x U nf 2.303 m/s Volumetric flowrate, Q Q = (G x M)/(3600 x ρg ) 1.0559 m3/s Area, A A = Q/ U n 0.4584 m2 Area to be considered, A T AT = A/0.88 0.5209 m2 Diameter, D T DT = ((4 x A T)/π)0.5 814.596 mm Table 6.4 shows the calculations for stripping section diameter Since the diameter of enriching section is more than that of stripping section , we consider the d iameter of the column to be 1122.383m m. Wall thickness, t = 𝑃×𝐷 200 ×𝑓×𝐽−𝑃 + C (IS 2825) Where, t = thickness in mm P = operating pressure in kgf/cm2 (1 atm = 1.0332 kgf/cm2) D = diameter of column in mm (1122.383 mm) f = allowable stress (13) J = joint efficiency (0.85) C = corrosion alliance (min 10 mm) CHAPTER -6 EQUIPMENT DESIGN _____________________________________________________________________________________ 46 t = 1∗1.032 ∗1122 .383 (200 ×13×0.85)−1.0332 + 10 = 10.5243 mm The thickness of the di stillation column is 10. 5243 mm 6.4 Condenser Design The condenser cools the product gases to room temperature. It is a shell and tube heat exchanger where water is used as the cooling fluid. Table 6.1 4 shows the inlet and outlet temperature s of product mixture (hot fluid) an d water (cold fluid) Table 6.14 Inlet and outlet flow streams of condenser used in the manufacture of Isopropyl alcohol Inlet Temp. ( ⁰C) Outlet Temp. ( ⁰C) Mass flow rate, (kg/s) Tavg(⁰C) Hot fluid 90 50 0.582783 70 Cold fluid 10 60 0.68605 35 Table 6.15 Properties of Streams at average temperature Property Product Mixture Water Specific Heat, Cp (kJ/kg -K) 3.360 4.224 Density, ρ (kg/m3) 785 1003.567 Thermal Conductivity, k(W/m -K) 0.5245 0.618239 Viscosity, μ (kg/m -s) 0.00095 0.000797232 From heat balance the mass flow rate of cold fluid was found to be = 0.68605kg/s Table 6.15 gives the properties of the product mixture and water at average temperature Table 6.16 Pipe Dimensions Triangular Pitch, P t 0.02m BWG 16 Outer diameter of inner pipe, do 0.016m Inner diameter of inner pipe, di 0.014m Standard Length of Pipe, L r 3.5 m CHAPTER -6 EQUIPMENT DESIGN _____________________________________________________________________________________ 47 Table 6.16 shows the standard pipe dimensions Logarithmic Mean Temperature Difference Calculations 𝑳𝑴𝑻𝑫 = (𝑻𝒉𝒊 −𝑻𝒄𝒐 )−(𝑻𝒉𝒐 −𝑻𝒄𝒊) 𝒍𝒏(𝑻𝒉𝒊 −𝑻𝒄𝒐 ) (𝑻𝒉𝒐 −𝑻𝒄𝒊) Assumptions: 1. 1-2 pass STHE 2. Product gases flow on the shell side and water on the tube side Table 6.17 LMTD Calculations R = (T 1 - T2) / (t2 - t1) 0.8 S = (t 2 - t1) / (T 1 - t1) 0.6 Correction Factor, F t 0.6 Log mean temperature, ΔT LMTD (⁰C) 34.76 Corrected log mean temperature, ΔT LMTD (⁰C) 20.57 Table 6.4 shows the calculations for logarithmic mean temperature Tube side and shell side heat transfer coefficient calculations Table 6.18 General Specifications Assume heat transfer Coefficient Uo 567.83 Heat transfer Area A = Q/ (U oxΔT LMTD ) 12.2766W/m2-C0 Number of tubes n = A/π d0Lr 65.40 Number of tubes (TEMA P or S) 67 Shell diameter Ds 1.372 m Baffle spacing Ds/5 0.27444 m CHAPTER -6 EQUIPMENT DESIGN _____________________________________________________________________________________ 48 Table 6.18 gives the general specification in a condenser Table 6.19 Tube side heat transfer properties Flow area At = N(π/4) d i2 0.0004 m2 Mass veloctiy G = m / At 2228.16 kg/ m2-s Reynolds number NRe = (d ixG)/μ 43379.55 Prandtl number NPr= (C pxμ)/k 4.833 Nusselt number NNu =0.023 xRe0.8xPr0.33 198.3 Inside heat transfer coefficient ht = N Nuxk/d i 8799.133 W/m2-C0 Table 6.4 shows the calculations for i nner heat transfer coefficient Table 6.20 Shell side heat transfer properties Average temperature Tcavg 35C0 Vapour temperature Thavg 70 C0 ∆Tf Thavg - Tcavg 35 C0 Bundle diameter Db = d ox[Nt/k]1/3 0.226 m Ntf Ntf = 2/3x [Db/Pt] 28.31844 Shell side heat transfer coefficient 4508.41 W/m2-C0 Table 6.4 shows the calculations for shell side heat transfer coefficient Table 6.21 Overall heat transfer coefficient Dirt factor RD 0.0008805 m2-C0/W 1/U (1/hs) + (1/ h t) + R D 0.0012159 m2-C0/W Overall heat transfer coefficient U 822.39868 W/m2-C0 Table 6.4 shows the calculations for overall heat transfer coefficient Overall heat transfer coefficient is greater than assumed heat transfer coefficient. Design for condenser is accepted as a result. CHAPTER -6 EQUIPMENT DESIGN _____________________________________________________________________________________ 49 Pressure drop calculations Table 6.22 Tube side pressure drop calculations Friction factor jF 0.0024 Number of passes Np 2 Pressure drop ΔPt = N p [8 j F (L/d i) + 2.5]U t2 ρ /2 0.1328 kPa Table 6.22 shows the pressure drop calculation across the condenser Mechanical Design Wall thickness, t = 𝑃×𝐷 200 ×𝑓×𝐽−𝑃 + C Where, t = thickness in mm P = operating pressure in kgf/cm2 (1 atm = 1.0332 kgf/cm2) D = diameter of column in mm (1372 mm) f = allowable stress (13) J = joint efficiency (0.85) C = corrosion alliance (min 5 mm) t = 1∗1.032 ∗1372 (200 ×13×0.85)−1∗1.0332 + 5 = 5.641 mm The thickness of the shell is 5.641 mm CHAPTER -6 EQUIPMENT DESIGN _____________________________________________________________________________________ 50 CHAPTER -7 COST ESTIMATION & ECONOMICS _____________________________________________________________________________________ 51 The net profit generated by a chemical plant is often a measure of its credibility. In economics and cost estimation, we study the net profit obtained from a plant and the various component costs involved in it. The rate of return and the payback period wa s also estimated. 7.1 Equipment costs Table 7.1 List of equipments and costs S.No Equipment Quantity Cost (in INR) 1 Reactor 1 25287354.6 2 Heat exchanger 2 6453333.334 3 Separator 1 1352569.817 4 Distillation column 2 30714285.72 5 Condenser 2 1430742.858 6 Reboiler 2 5460714.286 TOTAL 70699000.62 Total equipment costs (E) = 70699000.62 INR Table 7.1 Includes the list of equipments and its costs for the production of isopropyl alcohol Table 7.2 Factors contributing to direct costs Additional expenses on equipment % of equipment cost Cost (in INR) Equipment Installation 25 17674750.15 Transportation and Local Tax 5 3534950.031 Instrumentation 15 10604850.09 Packing and Installation 10 7069900.062 CHAPTER -7 COST ESTIMATION & ECONOMICS _____________________________________________________________________________________ 52 Electrical Assembly 10 7069900.062 Building (including services) 40 28279600.25 Cost of Land 30 21209700.18 Services and Maintenance 7 4948930.043 TOTAL 100392580.9 Total direct costs = 100392580.9 INR Table 7.2 shows the factors contributing to direct costs for the production of isopropyl alcohol 7.2 Indirect costs Table 7.3 Indirect costs Services % of equipment cost Cost (in INR) Engineering expense and supervision 20 14139800.12 Construction expenses 20 14139800.12 TOTAL 28279600.25 Table 7.3 shows the indirect costs for the production of isopropyl alcohol Fixed Capital Investment = Direct costs + Indirect costs = 10039 2580.9 + 28279600.25 = 128672181.1 INR CHAPTER -7 COST ESTIMATION & ECONOMICS _____________________________________________________________________________________ 53 7.3 Factors Contributing to FCI Table 7.4 Factors contributing to FCI Names % FCI Cost (in INR) Contractor fees 5 6433609.056 Contingencies 10 12867218.11 TOTAL 19300827.17 Table 7.4 shows the factors which contribute to FCI for the production of isopropyl alcohol Total FCI = 128672181.1 + 19300827.17 = 147973008.3 INR Total Capital Investment = FCI + Working Capital Investment (15% of TCI) 0.85xTCI = 147973008.3 INR Total Capital Investment = 174085892.1 INR Working Capital Investment = 26112883.82 INR 7.4 Total Product Costs Total product cost= Manufacturing cost + General expenses Where, Manufacturing cost= Direct p roduction costs + Fixed Charges CHAPTER -7 COST ESTIMATION & ECONOMICS _____________________________________________________________________________________ 54 Table 7.5 Factors contributing to fixed charges Items % of FCI Cost (in INR) Depreciation 10 14797300.83 Local taxes 3 4439190.249 Insurance 1 1479730.083 Rent 10 14797300.83 TOTAL 35513521.99 Table 7.5 shows the factors which contribute to fixed charges for the production of isopropyl alcohol Table 7.6 Factors contributing to direct production costs Items % of TPC Cost (in INR) Raw Material 30 0.3X Operating Labour 15 0.15X Direct Suprervisory and electrical labour 3 0.03X Utilities 15 0.15X R&M 5 30086499.73 Operating Supply 10 3008649.973 Labour Charges 15 451297.4959 Patents 5 0.05X Plant Overhead Cost 10 0.1X TOTAL 0.78X + 33546477.19 CHAPTER -7 COST ESTIMATION & ECONOMICS _____________________________________________________________________________________ 55 Table 7. 6 shows the factors which contribute to direct production costs for the production of Isopropyl alcohol Where Total production cost is assumed as X 7.5 General Expenses Table 7.7 General expenses Items % of TPC Cost (in INR) Administrative Cost 4 0.04X Distribution and Selling Cost 10 0.1X R &D 5 0.05X TOTAL 0.19X Table 7.5 includes the general expenses for the production of isopropyl alcohol Manufacturing Costs = Fixed Charges + Direct Production Costs = 35513521.99 + 0.78X + 33546477.19 = 69059999.18 + 0.78X Total Production Costs = Manufacturing cost + General expenses X= 2301999973 INR Therefore, Total production costs = 1516549852 INR per year 7.6 Estimation of Total Income Selling price of IP A = Rs. 129 per kg Annual working days = 365 Production rate = 18250000 kg per year CHAPTER -7 COST ESTIMATION & ECONOMICS _____________________________________________________________________________________ 56 = 50000 kg per day Total income = (Selling price of IPA x Production rate (in days) x Annual working days) = (129 x 50000 x 365) = 2354250000 INR Gross Income = Total incom e – Total production costs = 2354250000 – 2301999973 = 52250027 INR Tax rate = 30 % of Gross Income Therefore, tax paid = 15675008.21 per year Net profit = Gross income – Tax paid = 52250027 – 15675008.21 = 36575019.16 INR Rate of Return = (Net Profit/ Total Capital Investment) x100 = 36575019.16/174085892.1 = 21.01 % Payback period = (Total Capital Investment/Net Profit) = 174085892.1/36575019.16 = 4.76 years CHAPTER -8 PLANT LOCATION & LAYOUT _____________________________________________________________________________________ 57 The location of plant is a very important and crucial step once process is selected. Many factors must be considered while selecting suitable site mainly the raw materials, market, infra -structure, water resources and impact on an adjacent ecologic systems. Also the process selected plays a major role in the location of the plant . 8.1 Plant Location The important criteria are affecting site selection:  Availability of suitable land  Raw material supply  Proximity to sea  Labor availability  Proximity to infrastructure facilities like road, rail, and ports etc  Ease of effluent disposal  Availability of marketing area  The impact on the adjacent ecological systems  Availability of utilities  Climate 8.1.1 Raw materials The process has been selecte d as Tokoyama process wh ich uses ammonia to produce nitric acid . Therefore the availability and price of these raw materials is important in the site selection. It is preferred that the plant is located in close proximity where propylene is easily available. 8.1.2 Market availability Isopropyl alcohol when produced in bulk quantities the cost of the product per ton is relatively low and the cost of transport. To get a significant fraction of sales price, the plants should be located close primary market. 8.1.3 Transport The material transport is one of the effective principal factors of site location. In practice the plant should be located in a place where these two major forms of transport are available: road and water CHAPTER -8 PLANT LOCATION & LAYOUT _____________________________________________________________________________________ 58 way. Road transport is being mainly used for local distribution from warehouse. Rail transport is a type of road transport and will be cheaper for long distance transport of bulk chemicals. Water way used for international transport of raw materials and products. 8.1.4 Availability of Utilities Plant utilities are the important parameter for any plant. The plant should be located where all the utilities are available in enough quantities. Followi ng are the important utilities for nitric acid plant are  Steam  Cooling water  Electricity  Refrigeration 8.1.5 Climate The dry and sunny condition that is present throughout th e year is extremely suitable for a process plant because loss of heat by cooling of process due to rain is avoided. This also saves energy and makes it possible to obt ain consistent product quality. 8.1.6 Effluent disposal All industrial processes produce waste product, the waste has to be treated and disposed properly without causing any harm to environment. 8.2 Plant Layout Preparing a suitable plant layout for the selected site is the immediate step once the plant location is decided. The economic construction and efficient operation of process unit will depend on how well the plant and equipment specified on the process flow sheet is laid out. The principle factors of plant layout considered are, 1. Economic consideration: installation and operating costs 2. The process requirement CHAPTER -8 PLANT LOCATION & LAYOUT _____________________________________________________________________________________ 59 3. Ease of plant operation 4. Maintenance 5. Safety 6. Future expansi on A reference plan t layout is shown in figure 8.1 Figure 8.1 Reference plant layout for the production of I sopropyl alcohol 50 tons/day GATE GATE GATE GATE GATE CHAPTER -8 PLANT LOCATION & LAYOUT _____________________________________________________________________________________ 60 8.3 Site Selection Figure 8.2 shows the chosen location, Yerawada in Pune for our process plant. Figure 8.2 Referenti al plant location selected for the production of I sopropyl alcohol 50 /tons per da y CHAPTER -9 PIPING & INSTRUMENTATION DIAGRAM _____________________________________________________________________________________ 61 A P&I diagram shows information on piping, fittings, equipment, instrumentation, and process plant in a representative and sequential arrangement on the basis of product flow paths. Installation of piping and instrumentation in process flow path plays a very important role in optimization of controlled proce ss flow. Considering the above requirement the control systems has to be introduced to provide safe operation and controlled output flow. The PID should include,  Instrumentation and designations.  Mechanical equipment with names and numbers .  All valves and their identifications  Process piping, sizes and identification .  Miscellaneous -vents, drains, special fittings, sampling lines, reducers, and increasers .  Flow direction .  Computer control system input interconnections