Acetic Acid Process Plant Design | PDF - SlideShare

05 Jun.,2025

 

Acetic Acid Process Plant Design | PDF - SlideShare

  • 1. 1 Acetic acid process plant design By Hisham Albaroudi Karen Atayi Cristian Baleca Enoch Osae Sinthujan Pushpakaran Alexander Taylor School of Chemical Engineering Faculty of Science and Engineering University of Hull June
  • 2. 2 Table of Contents 1.) Executive Summary 1 2.) Process Selection 2 2.1) Process Technology Selection 2 2.2) Process Flowsheet Development 5 2.3) Process Description 6 3.) Piping and Instrumentation Diagram (P&ID) 7 4.) Mechanical Design of Unit Operations 8 4.1) Reactor 8 4.1.1) Introduction 8 4.1.1.1) Propionic acid 9 4.1.1.2) Water – Gas shift 9 4.1.1.3) Methyl acetate 9 4.1.1.4) Methyl iodide 9 4.1.2) Reactor P&ID 10 4.1.3) Design Method 11 4.1.3.1) Reactants and product 11 4.1.3.2) Variations 12 4.1.3.3) Energy Balance and Heat of reaction 12 4.1.3.4) Choice of reactor 13 4.1.4) Reactor Specification 13 4.1.4.1) Choice of material 14 4.1.4.2) Vessel support 14 4.1.4.3) Piping sizing 14 4.1.4.4) Nozzles 15 4.1.4.5) Heat dissipation and vessel insulation 15 4.1.4.6) Shut down 16
  • 3. 3 4.1.4.7) Process safety 16 4.1.4.8) Process control 16 4.1.5) Agitator Specification 17 4.1.6) Conclusion 19 4.1.7) Engineering Drawing of reactor 21 4.2) Flash Tank 22 4.2.1) Introduction 22 4.4.2) Flash Tank P&ID 23 4.2.3) Design Method 24 4.2.4) Flash Tank Specification 25 4.2.4.1) Process controls and safety 27 4.2.5) Conclusion 28 4.1.6) Engineering Drawing of Flash Tank 29 4.3) Drying Distillation Column 30 4.3.1) Introduction 30 4.3.2) Drying Distillation Colum P&ID 31 4.3.3) Design Method 32 4.3.4) Drying Distillation Column Specification 32 4.3.5) Conclusion 33 4.1.7) Engineering Drawing of Drying Distillation Column 34 4.4) Heavy Ends Distillation column 35 4.4.1) Introduction 35 4.4.2) Heavy Ends Distillation Colum P&ID 37 4.4.3) Acetic Acid Properties within the Column 38 4.4.4) Propionic Acid Properties within the Column 38 4.4.5) Relative Volatility 39 4.4.6) Heavy – Ends Distillation Column Specification 39 4.4.7) Summary of Design Data 41
  • 4. 4 4.4.8) Conclusion 41 4.1.9) Engineering drawing of Heavy Ends Distillation Column 42 4.5) Absorption Column 43 4.5.1) Introduction 43 4.5.2) Absorption Column P&ID 44 4.5.3) Design Method 45 4.5.4) Absorption Column Specification 45 4.5.4.1) Design Considerations to Account for Drawback of Unit 45 4.5.4.2) Choice of Packing 46 4.5.4.3) Choice of Absorption Tower Equipment 46 4.5.4.4) Materials of Construction 47 4.5.4.5) External Equipment 47 4.5.4.6) Safety Control 48 4.5.4.7) Safety Considerations 48 4.5.5) Conclusion 48 4.5.6) Engineering Drawing of Absorption Column 50 4.6) Storage tank – Acetic acid 51 4.6.1) Introduction 51 4.6.2) Storage Tank – Acetic Acid P&ID 52 4.5.3) Design Method 53 4.5.4) Storage Tank Specification 53 4.5.5) Conclusion 55 4.5.6) Engineering drawing of Storage Tank – Acetic Acid 56 5.) Process Control and Instrumentation 57 5.1) Introduction to Process Control and Instrumentation 57 5.2) Objectives of Process Control 57 5.3) Implementation of Control Systems in our Design 58
  • 5. 5 5.3.1) Distillation Column Control 59 5.3.2) Reactor and Flash Tank Control 61 6.) Process Economics 62 6.1) Market Analysis 62 6.2) Costing 63 6.2.1) Feedstock price estimation 64 6.2.2) Capital Cost Estimation 65 6.2.2.1) ISBL 65 6.2.2.2) Installation Factor 66 6.2.2.3) OSBL 66 6.2.2.4) Engineering costs 66 6.2.2.5) Contingency costs 66 6.2.2.6) Fixed Capital Investment 67 6.2.3) Working Capital 67 6.2.4) Total Investment 67 6.2.5) Operating Expenditure 67 6.2.6) Revenue 67 6.2.7) Gross Profit 67 6.3) Project Financing 68 6.3.1) Financing Bank Loan 68 6.3.2) Financing Investment 70 6.3.3) Net Profit 71 6.3.4) Cumulative Cash Position 72 6.3.5) Return of Investment 73 7.) Process Safety 74 7.1) Safety legislations 75 7.2) Hazard Identification 75
  • 6. 6 7.2.1) Material Hazard 75 7.2.2) Material Toxicity 78 7.2.3) Flammability 79 7.3) Operating conditions hazard 80 7.3.1) Pressure relief strategy 80 7.3.2) High pressure response measures 81 7.3.3) Fire prevention strategy 81 7.3.4) Fire and gas detection 82 7.3.5) Noise 83 7.3.6) Loss of containment 83 7.4) Emergency Response plans 83 7.4.1) Fire 83 7.4.2) Explosion 85 7.4.3) Overpressure 85 7.4.4) Toxic release 85 7.4.5) Flooding 85 7.4.6) Earthquakes 86 7.4.7) Human error 86 7.4.8) Personal Protection Equipment 86 7.5) HAZOP 86 7.5.1) Scope of work 87 7.5.2) Term of Reference 87 7.5.3) Team Membership 88 7.5.4) Safety conclusions 89 7.5.5) Marked up P&ID 90 7.5.5) HAZOP findings 91 8.) Environmental Protection 95
  • 7. 7 8.1) Process Selection 95 8.2) Plant Location – Environmental Considerations 95 8.3) Noise 96 8.4) Odour 96 8.5) Traffic 96 8.6) Catalyst and Water Requirement 97 8.4) Methanol Feed 97 8.5) Energy Recovery 98 8.6) Storage & Handling of Raw Materials and Product 98 8.6.1) Carbon Monoxide 98 8.6.2) Methanol 98 8.6.3) Acetic Acid 99 8.7) Undesired products: By- and Co- products 99 8.7.1) Propionic Acid 100 8.7.2) Carbon Dioxide and Hydrogen 100 8.7.3) Methyl Iodide 101 8.7.4) Aqueous and Organic Discharges 101 9.) Plant Layout and Location 102 9.1) Plant location 102 9.2) Plant layout 104 9.2.1) Site Flow Plan 107 10.) Appendices APPENDIX [A] – Minutes 108 Meeting Week 13 – 9th May 109 Meeting Week 12 – 3rd May 110 Meeting Week 11 – 22nd April 111 Meeting Week 10 – 18th April 112
  • 8. 8 Meeting Week 9 – 11th April 113 Meeting Week 8 – 4th April 114 Meeting Week 7 – 7th March 115 Meeting Week 6 – 29h February 116 Meeting Week 5 – 22nd February 117 Meeting Week 4 – 18th February 118 Meeting Week 3 – 8th February 119 Meeting Week 2– 5th December 120 Meeting Week 1 – 30th November 121 APPENDIX [B] – Reactor Calculations 121 APPENDIX [C] – Flash Drum Calculations 140 APPENDIX [D] – Calculations for Drying Distillation Column 151 APPENDIX [E] – Calculations for Heavy – Ends Distillation Column 160 APPENDIX [F] – Calculations for Absorption Column 177 APPENDIX [G] – Calculations for Acetic Acid Storage Tank 189
  • 9. 9 1. Executive Summary The purpose of this document is to present a potential design to the client to build an acetic acid (CH3COOH) plant in the United Kingdom. The plant will have the capacity to produce 400,000 tonnes per annum of acetic acid base product from a feedstock of methanol and carbon monoxide. As an overview, the methanol carbonylation process is highly efficient in that it produces acetic acid with more sought after selectivity and purity. Although, the oxidation of ethylene is more environmentally friendly, it can only be operated for a capacity of up to 200,000 tonnes per year, while the oxidation of hydrocarbons route for acetic acid production is cheaper to run but it does not produce pure acetic acid and greatly affects the environment as a result of its CO2 emissions. Even though the oxidation of ethylene and methanol carbonylation processes do not pose much threat to the environment, the latter is still more environmentally friendly as it produces less waste and recycles most of its reactants. Environmental Impact Assessment has been proven successful in outlining the main environmental issues in relation to this project. The general location considerations linked to the potential pollution produced (odours, noise, traffic) has been analysed, justifying the measures that will be put in place to minimize them. The handling of raw materials and the final product both on and off site has been studied in depth in order to outline the features and add-ups that can be applied to reduce the impact on the environment: such measures are mainly related to the close monitoring and the implementation of safety measures to be applied whenever a containment vessel were to mechanically fail for any reason: it has been concluded that appropriate containment chambers and process control and instrumentation are the significant routes to apply. An eco-friendly engineering strategy has been applied to this project when dealing with significant by-products. Although the generation of highly corrosive chemicals (such as methyl-iodide) has not been possible to be prevent, other “waste” compounds produced in the system have been proved to be commercially useful (propionic acid). Certainly, a comprehensive recycling system within the plant (and especially to the reactor) is a successful strategy to ensure that non desired products are dealt with the purpose of minimizing waste. Furthermore, even when waste streams are unable to be recycled and re- used in the system, appropriate techniques to dispose of them have been developed in compliance with environmental regulations and ethical considerations i.e. flares, aqueous discharge basin and pipeline. Careful consideration of all the hazards present on the plant are outlined in the following report which highlights efficient ways of maintaining a safe environment for the production of acetic acid. In addition to environmental methodologies, principles of process control and instrumentation have been applied throughout the design stage of this project with the aim of creating a process that is ultimately safe, that complies with all the necessary safety regulations, efficient, that will not suffer unnecessary downtime to avoidable failures and maintenance being carried out on key piece of process equipment and not suffer performance impairments due to poor design, as well as being economically stable, linked to the plants efficiency, an efficient plant will bring a certain amount of economic stability in addition to ensuring unnecessary equipment or instrumentation is not put in place.
  • 10. 10 Based on market research it is possible to conclude that the acetic acid market is projected to rise the upcoming years on a global basis. Economic evaluation of this project indicates viability, the return of investment is 53% and the net profit of £1,378,000,000 is very lucrative figure for a 20-year investment. Both a bank loan and private equity investments would generate a greatly positive profit, although a bank loan would represent a significantly more profitable route. The project payback time of 2 years demonstrates that this project is highly feasible and has the potential to attract numerous investors. 2. Process Selection 2.1 Process Technology Selection The methanol carbonylation, direct oxidation of n-butane and direct oxidation of ethylene are the three most widely implemented methods to manufacture acetic acid. Methanol carbonylation also known as the “Monsanto process” was initially developed by BASF in . The process operates at 180 – 220 o C and 30 – 40 atm via the use of a rhodium catalyst, leading to energy costs set to a bare minimum. Although it operates at such low operating conditions, the process provides with high selectivity of acetic acid (Yoneda et al., ). The final product holds great purity due to the selectivities of methanol and carbon monoxide, which are 99% and 90% respectively (Yoneda et al., ). The process outlines continuous supply of methanol and carbon monoxide into the reactor. The combination of exhaust gas produced from the reactor and purification section are recovered as light-ends and recycled back into the reactor. Consequently, the acetic acid produced from the reactor is separated as a side-cut and delivered to the dehydration column (Sano et al., ). Acetic acid and water mixture are then released at the top of the column and back to the reactor while propionic acid is taken to the subsequent column. Further purification takes place and acetic acid is generated as a side-cut. Continuous recycling of overhead and bottoms found in fractional column into reactor take place (Sano et al., ). The main raw materials for this process are methanol and carbon monoxide. In the reaction process, methyl iodide is added to the rhodium complex, which consecutively migrates to a carbonyl group and reacts with CO to form the rhodium-acetyl complex (Kinnunen and Laasonen, ). The excess water readily hydrolyses the acetyl iodide (CH3COI) to produce acetic acid and hydrogen iodide in order to complete the catalytic cycle (Yoneda et al., ). However, a quantity of water (14 – 15 wt.%) is required in order to maintain stability and activity of the catalyst, thus separation of water from acetic acid requires excessive amount of energy, further limiting storage capacity (Wittcoff et al., ). Methanol carbonylation produces propionic acid as the major by-product of this process (Sunley and Watson, ), present as an impurity in methanol feed (Yoneda et al., ). In order to lower the yield of propionic acid produced, it is suggested to decrease the amount of acetaldehyde produced by the rhodium catalyst (Yoneda et al., ). The direct oxidation of hydrocarbons route occurs through pumping of an ethane and oxygen mixture at 515 K and 16 bar in a multi-tubular reactor (Smejkal et al., ). The product formed is cooled to 303 K via
  • 11. 11 two steps, initially through formation of high-density steam and subsequent separation of formed gas and liquid mixture in a flash (Smejkal et al., ). The acetic acid-water mixture produced is then separated in a rectification column and pure acetic acid is generated as the bottom product. The resulting gaseous stream, made up of ethane, ethylene and CO2, is recycled back into the system. CO2 is separated into an absorber, while ethylene and ethane are put back into the feedgas (Soliman et al., ). Nonetheless, the oxidation of n-butane requires large amounts of water and generates a dilute acetic acid solution of which concentration is highly energy intensive, as a result the yield of acetic acid produced is lower than the one obtained in other processes (Sano et al., ). A vast amount of by-products are formed, some of which are propionic and formic acids (Riegel, ). Furthermore, this particular process requires large quantities of water, hence water gas shift reaction is a major drawback as extensive CO2 is produced as a result (Wittcoff et al., ). In essence, the oxidation of hydrocarbons process is cheaper to run as a result of its feedstock, but at the cost of being less efficient as it produces more waste and a lower grade chemical. Flexibility in the process allows it to produce a purer acetic acid with high selectivity, however extensive operation expenses are necessary. The production of acetic acid through direct oxidation of ethylene was first proposed by Showa Denko K.K.. The process occurs through the mixture of ethylene and oxygen in their vapour phases at 160 – 210 o C over a solid catalyst (Xu et al., ), the acetic acid generated is of high selectivity. The reaction is initiated from the cooling of gas produced in the reactor to ambient temperature, where the products of acetic acid, water and other organic compounds are condensed and separated. The condensate transfers to the crude acetic acid tank, while the compressor pressurizes the un-condensed gas back to the reactor. Light-end products such as acetaldehyde, ethyl acetate and ethanol are removed through distillation, allowing acetic acid and light-ends compounds to migrate to the purification section where pure acetic acid is produced (Sano et al., ). This process produces large amounts of heat which is recovered as steam and used in the purification section as a source of heat (Sano et al., ).The process meets the requirements of being both competitive and environmentally friendly. Although it rivals methanol carbonylation, the process is only efficient with smaller plants of about (100-250 kt/a) (Sano et al., ) and considering the fact that the feedstock price of ethylene is more expensive than raw materials used in the other processes mentioned, thus economically it will not be as profitable as the methanol carbonylation process given that the selectivity of both is of 90%. Overall, the methanol carbonylation process is highly efficient in that it produces acetic acid with more sought after selectivity and purity. The oxidation of ethylene is more environmentally friendly, however it can only be operated for a capacity of up to 200,000 tonnes per year and the oxidation of hydrocarbons is cheaper to run but does not produce a pure acetic acid product. The oxidation of hydrocarbons highly affects the environment as a result of its emissions of CO2, whilst both the ethylene oxidation and methanol carbonylation don’t pose much threat to the environment, the latter is still more environmentally friendly as it produces less waste and recycles most of its reactants. This enables the process to be continuous and thus economically beneficial.
  • 12. 12 2.2 Process Flowsheet Development Reactor 1 2 3 Flash tank Light ends distillation column Drying distillation column Heavy ends distillation column 8 12 Scrubber 9 113 14 8 1 2 3 6 9 17 16 11 15 Pressure reduction valve 5 Figure [1]: Process Flow Diagram and Material balance of process. Steam 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 Molar Flowrate [kg/hr] Methanol 0.00 .00 0.00 56.15 56.15 1.15 56.07 0.00 56.07 Trace 0.00 0.00 0.00 0.00 0.00 Trace 57.29 Carbon monoxide .36 0.00 0.00 207.44 207.44 200.88 207.55 0.00 207.55 0.00 0.00 0.00 0.00 0.00 0.00 3.02E+02 105.97 Acetic acid 0.00 0.00 0.00 .49 .49 611.25 .05 .21 .53 8.28E+04 .25 .293 .87 184.35 .63 1.60E+02 .00 Water 0.00 0.00 720.40 234.38 234.38 3.19 210.90 23.41 210.90 Trace 0.00 0.00 0.00 0.00 0.00 Trace 214.13 Ethanol 0.00 273.14 0.00 0.00 0.00 0.00 0.00 0.00 0.00 Trace 0.00 0.00 0.00 0.00 0.00 Trace 0.00 Propionic acid 0.00 0.00 0.00 487.99 487.99 2.02 439.24 48.89 0.00 4.39E+02 0.01 439.24 0.01 439.24 0.00 Trace 2.02 Carbon dioxide 0.00 0.00 0.00 .54 .54 947.56 .36 0.00 .36 0.00 0.00 0.00 0.00 0.00 0.00 2.43E+03 .63 Hydrogen 0.00 0.00 0.00 8.69 8.69 72.79 8.69 0.00 8.69 0.00 0.00 0.00 0.00 0.00 0.00 7.99E+01 1.60 Methyl acetate 0.00 0.00 0.00 .25 .25 24.77 .89 0.00 .89 Trace 0.00 0.00 0.00 0.00 0.00 Trace .02 Hydrogen iodide 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 Trace 0.00 Methyl iodide 0.00 0.00 0.00 .05 .05 60.51 .02 459.76 .02 0.00 0.00 0.00 0.00 0.00 0.00 Trace .95 Total .36 .00 720.40 .00 .00 .24 .71 .75 .53 .17 .26 .67 .16 623.51 .71 .623 .82 Temperature [o C] 25 25.53 26.87 160 104.45 160 104.45 104.45 79.88 117.65 141.66 117.69 117.58 131.41 117.58 18.31 69.71 Pressure [bar] 31 31 31 30.40 1 30.40 1 1 1 1 2 1 1 1 1 1 59
  • 13. 13 2.3 Process Description Aspiring Consulting have decided to manufacture acetic acid using the Monsanto process. The first stage of the process involves the injection of the raw materials methanol, carbon monoxide and water into the reactor to initiate the methanol carbonylation process. The carbonylation reaction is carried out in a stirred tank reactor on a continuous basis. Liquid is then removed from the reactor through a pressure reduction valve. This then enters the flash tank, where light components of methyl acetate, methyl iodide, some water and product acid are removed as vapour through the top of the vessel. The gases are fed forward to the distillation train for further purification whilst the remaining liquid in the flash tank, containing the dissolved catalyst, is recycled to the reactor. Liquid from the reactor enters the lower half of a multiple-tray distillation column operating at conditions above atmospheric conditions. Hydrogen iodide present in the feed stream is concentrated in the acetic acid solution in the bottom of the column. This stream is recycled back to the reactor. Carbon monoxide, water, methyl iodide and some entrained hydrogen iodide comprise the overhead stream from the column, which passes through a condenser and phase separator where uncondensed gas is directed towards the scrubber. The condensate separates into two phases: a water phase consisting of some organic compounds and an organic phase (methyl iodide) containing some water. The organic phase is recycled to the reactor whilst part of the water phase is used as reflux in the distillation column and excess is recycled to the reactor. Solution of acetic acid in water containing some iodide, catalyst and by-products is withdrawn from the bottom of the column and introduced into a second multiple-tray distillation column operating at conditions above atmospheric conditions. In this column, water and remaining inerts are withdrawn overhead and directed towards to the scrubber. A portion of the condensate is returned as reflux to the column and excess is recycled to the reactor. To avoid accumulation of water in the system, a portion of the water separated in the column is discarded. Residual hydrogen iodide in the feed stream to the column concentrates at a location near the middle of the distillation column. By continually withdrawing the solution containing hydrogen iodide from the middle of the distillation column, virtually all of the hydrogen iodide is removed from the column. This solution can be recycled directly to the reactor or alternatively to the lower half of the previous distillation column, where it is concentrated and removed with the bottoms stream of that column. Acetic acid product is withdrawn from the drying column without further processing; acetic acid vapour is withdrawn from the top of the column and passed through a condenser from which it is pumped to storage. Liquid acetic acid containing residual catalyst is periodically withdrawn from the top of the column and recycled to the reactor.
  • 14. 14 3. Piping and Instrumentation Diagram (P&ID) DC-501 P-501 P-502 V-502 V-503 V-504 V-505 FI FIFI C FI C V-501 LI C LI 2" 304SS V506 TI FI C 36" 304SS V-507 TI FI C DC-401 P-401 P-402 8" 304SS V-402 V-403 V-404 V-405 FI FIFI C FI C V-401 LI C LI 4" 304SS V-406 TI FI C V-407 TI FI C 6" 304SS DC-301 P-301 P-302 V-302 V-303 V-304 V-305 FI FIFI C FI C V-301 LI C LI 10" 304SS V-306 TI FI C TI FI C V-307 Acetic acid Propionic ac id R-201 V-210 V-209 V-208 V-207 FI FI FI C FI C S-601 Fl are 6" HC V-201 LI FI C V-202 V-203 V-205 V-204 PIFI C F-201 V-206 FI C LI V-211 FI FI C 6" HC 80" CS V-308 FIFI C V-212 FIFI C CH3OHfeed H2Ofeed COfeed P-101 P-102 V-102V-101 V-104V-103 FI FIFIC FIC P-103 P-104 V-106V-105 V-108V-107 FI FIFIC FIC 1" CS V-110V-109 V-112V-111 FI FIFIC FIC 3" CS P-602 P-601 V-617 V-619 V-620 FI FIFI C FI C 12" 304SS V-602 V-604 V-411 V-410 V-409 V-408 FI FI FI C FI C 12" HC V-601 V-603 V-609 V-610 V-612 V-611 V-511 V-510 V-509 V-508 FI FI FI C FI C V-618 V-613 V-614 V-616 V-615 LIFI C V-605 V-607 V-606 V-608 P-503 P-504 LAH LAL LAH LAL LAH LAL LAH LAL Process Steam Process Steam Process Steam Process Water Process Water Process Water V-508 FI C FI V-408 FI C FI V-308 FI C FI Figure [2]: P&ID showing all unit operation. Tag Number R-201 Tag Number F-201 Tag Number DC-301 Tag Number DC-401 Tag Number DC-501 Tag Number S-601 Aspiring Consulting LTD. KEY Service Reactor Service Flask Tank Service Light Ends Distillation Column Service Drying Distillation Column Service Heavy Ends Distillation Column Service Scrubber Process P&ID Client UoH plc CS Carbon Steel Design Press 33 bara Design Press 1.1 bara Design Press N/A Design Press 1.1 bara Design Press 1.52 bara Design Press 1.5 bara HC Hastelloy C Design Temp 176 o C Design Temp 115 o C Design Temp N/A Design Temp 125 o C Design Temp 150 o C Design Temp 50 o C Location Immingham 304SS 304 Stainless Steel Height 8.95 m Height 10 m Height N/A Height 23.85 m Height 21.6 m Height 10 m
  • 15. 15 4. Mechanical Design of Unit Operations 4.1 Reactor 4.1.1 Introduction The overall reaction for the production of acetic acid is given by the kinetic equation (Cheng and Kung, ): ????????3 ???????? + ???????? → ????????3 ???????????????? ΔG = -72.79 kJ/mol ΔH = -133.82 kJ/mol The enthalpy values collected in the literature clearly indicate the exothermic nature of the process; energy is being produced within the reactor. Typical operating conditions emphasized in literature for the Monsanto process lay in the following ranges (Cheng and Kung, ): Pressure Temperature 30-60 bar 150-200 oC Table [4.1] – Ranges for operating conditions. In order to model the simulation, different values for pressure and temperature have been selected for the reactor; the values that appeared to give the highest selectivity yet maintaining a relatively lower generation of by-products are 160 o C and 30 bar respectively. The process uses a Rhodium and Iodine complex to catalyse and promote the reaction, and this catalyst is highly selective especially in relation to methanol conversion. Due to the high activation energy, the reaction would not occur without the aid of a catalyst. The catalytic reactions are listed below, and are all equilibrium limited for the purpose of the reaction simulation. This is relatable to the fact that catalysts and promoters do not affect the stoichiometry and heat of reaction: CH3OH + HI ↔ CH3I + H2O CH3I + CO ↔ CH3COI ????????3 ???????????? + ????2 ???? ↔ ????????3 ???????????????? + ???????? As far as the process simulation is concerned (Aspen PLUS), the main methanol carbonylation reaction is kinetic, and some values in the literature have been investigated in order to model the simulation. The following information has been used to model Aspen Plus simulation: ???????????????????????? ???????????????????????????????????????? ???????? ???????????????????????? ???????????????? = 400,000 ???????????????? ???????????? ???????????????????????????????????? ℎ???????????????? (???????? ???????????? ???????????????????????? ????????????????????????????????????????????????????) = 50,000 ????????/ℎ????
  • 16. 16 In accordance to the stoichiometry and molecular weight of the reactants, the required input methanol and CO have been calculated for this purpose: ???????? → 186,666 ???????????????? 0.9 = 207,407 ???????????????? ???????????????? = 25,925 ????????/ℎ???? ????????????ℎ???????????????? → 213,333 ???????????????? 0.99 = 215,488 ???????????????? ???????????????? = 26,936 ????????/ℎ???? Due to a 99% selectivity in relation to methanol, and 90% in relation to CO, not all the reactants eventually combine to produce acetic acid. Other important side reactions occurring in the system generate acetic acid, methyl acetate and propionic acid. 4.1.1.1 Propionic acid ????2 ????6 ???? + ???????? → ????3 ????6 ????2 Propionic acid is the major liquid by-product within the system. Ethanol impurity is present in the methanol streams that reacts with some of the unreacted carbon monoxide. The reaction is kinetic and its value have been researched in literature. 4.1.1.2 Water – Gas shift ????20 + ???????? → ????????2 + ????2 The reaction once again involves part of the unreacted carbon monoxide reacts with water to generate hydrogen and carbon dioxide. This reaction is equilibrium limited. 4.1.1.3 Methyl acetate ????????3 ???????? + ????????3 ???????????????? → ????????3 ????????????????????3 + ????2 ???? A fraction of the 1% unreacted methanol combines with the product (acetic acid) to generate methyl acetate and water. This reaction is equilibrium limited. 4.1.1.4 Methyl iodide ????????3 ???????? + ???????? → ????????3 ????????????????????3 + ????2 ???? A fraction of the 1% unreacted methanol combines with the hydrogen iodide to generate methyl iodide and water. This reaction is equilibrium based. An Aspen PLUS simulation has been modelled based on: • Information researched in literature • Appropriate calculations
  • 17. 17 Methanol feed 6' SS P-102 R101 PI 103 FI 103 FCV-102 FCV-101 101 101 FC FT 102 102 PCV-103 P-103 Methanol storage tank P-101 FC FT P-201 LCV-201 5.5" SS P-202 103 FI PCPT 203 FI 202 LIC FTFC PRV-201 201 PI 102 102 TI PI 103 TI PRV-201 203 PI FCV-205 205 205 FT FC H2O supply 6" SS LAH LAL 204 204 TI PITLH TLL PLH PLL PLH PLH Off to flash tank From flash tank Off-to scrubber From drying column 3/8" - SS 3" - SS 80" - SS 6" – Hastelloy 6" – Hastelloy 12" – CS CO Supply Figure [4.1]: P&ID of Reactor. R-201 4.1.2 Reactor P&ID
  • 18. 18 4.1.3 Design Method 4.1.3.1 Reactants and product Since the reactants are in different phases, liquid and gas respectively, it is advantageous to insert the gas through a sparger to facilitate even diffusion throughout the liquid. The sparger should be designed separately. Stream Density (kg/m3 ) Methanol 791 CO 1.14 Acetic Acid Table [4.2] – Composition of stream and respective densities. Table [4.3] – Material balance for reactor. Substream:MIXED CO Methanol Water CSTR-LIQ CSTR-VAP WATER-RCY SCRUB-RCY MassFlowkg/hr Methanol 0.00 .00 0.00 56.15 1.15 0.00 57.29 CO .36 0.00 0.00 207.44 200.88 0.00 105.97 AceticAcid 0.00 0.00 0.00 .49 611.25 .25 .00 Water 0.00 0.00 720.40 234.38 3.19 0.00 214.13 Ethanol 0.00 5.93 0.00 0.00 0.00 0.00 0.00 PropionicAcid 0.00 0.00 0.00 487.99 2.02 0.00 2.02 CO2 0.00 0.00 0.00 .54 947.56 0.00 .63 H2 0.00 0.00 0.00 8.69 72.79 0.00 1.60 Methylacetate 0.00 0.00 0.00 .25 24.77 0.00 .02 Hydrogenchloride 0.00 0.00 0.00 0.00 0.00 0.00 0.00 Methyliodide 0.00 0.00 0.00 16.19 0.43 0.00 13.38 TotalFlowmol/hr .00 .00 .00 .00 .76 .60 .00 TotalFlowkg/hr .36 .02 720.61 .00 .24 .50 .82 volumetricflowl/hr .00 .27 724.99 .52 .94 .02 .84 TemperatureC 25.00 25.00 25.00 160.00 160.00 220.04 69.71 Pressurebar 1.00 1.00 1.00 30.40 30.40 32.00 59.00 VaporFrac 1.00 0.00 0.00 0.00 1.00 1.00 0.00 LiquidFrac 0.00 1.00 1.00 1.00 0.00 0.00 1.00 SolidFrac 0.00 0.00 0.00 0.00 0.00 0.00 0.00 Concentration(mol/l) 25.00 25.00 1.00 14.58 0.85 1.16 17.40 Densitygm/cc 1.13 793.09 993.96 861.85 21.49 69.43 17.40 INLET OUTLET RECYCLE
  • 19. 19 4.1.3.2 Variations The material and energy balances produced refer to steady state operating conditions. There are some situations, however, in which the system does not operate under steady state conditions, including: • Start-up and shutdown • Filling and emptying • Isolation • Preventative or corrective maintenance • Failure or loss of process variation • Extreme ambient conditions (temperature, severities, pressures) 4.1.3.3 Energy Balance and Heat of reaction As acknowledged from literature, the methanol carbonylation reaction is exothermic. The heat of reaction is therefore calculated with the aid of enthalpies of each stream (reactants and products) as seen by Figure 4.2. Figure [4.2]: Energy balance of inlet, outlet and recycle stream. The negative sign of heat of reaction means that the system releases energy in the form of heat. This heat of reaction has been calculated within the simulation through the basic formula: △ ???? ???????????????????????????????? − △ ???? ???????????????????????????????????? = ???? = ???????????????????????? ???????????????????????????????????? ???????? ???????????? ???????????? ???????? ???????????????????????????? The specific enthalpies of each stream have been obtained by the simulation. As far as simulation is concerned, the water inlet stream’s purpose is to enable the catalyst and promoter activity. It is not involved in any chemical reaction and it is in fact separated later on in the process from the product stream, and recycled back into the reactor continuously. For this reason it is possible to state that its energy contribution is negligible and not relevant to take into account for energy balance purposes. The same principle is applied to the recycling streams connecting the reactor to scrubber and reactor to drying column respectively, therefore they have not been included in this energy balance. For the purpose of design optimization and practice, heat needs to be continuously removed from the system and applied to an appropriate heat integration system. CO Methanol Water CSTR-LIQ CSTR-VAP Enthalpy (cal/mol) -.54 -.93 -.29 -.00 -.50 Flowrate (mol/hr) .76 Total enthalpy (cal/hr) - -4.E+10 -2.73E+09 -2.E+11 - Total enthalpy (cal/hr) Enthalpy of reaction (cal/hr) Enthalpy of reaction (kJ/hr) - -2.E+11 -1.E+11 -5.49E+11
  • 20. 20 4.1.3.4 Choice of reactor The nature of the reaction involved in the process necessitates agitation within the system. The exothermic nature of the main reaction emphasizes that a vessel with good temperature control is desirable. Furthermore, the fact that a continuous operation of the vessel is required and a simple adaptation to two- phase reaction is possible implies that a CSTR is the most suitable choice of reactor in the Monsanto methanol carbonylation. Advantages of CSTR’s are: • Constant operation in continuous system. • High degree of temperature and process control. • Simplicity of construction. • Easy adaptation to two phase reactions (liquid-gas reaction). • Easy maintenance / clean-up operations. Assumptions that will be useful in the later stage of the reactor design can be made use on the CSTR of the process: • Steady state conditions with constant inlet (reactants) and outlet flow (products). • Uniform stream composition inside and outside the reactor. • Complete and uniform mixing. 4.1.4 Reactor Specification Reactor – Design Data Vessel volume 138.42 m3 Vessel shell diameter (internal) 4.45 m Internal pressure 30 bara = 30.59 kg/cm2 External pressure 1 bara Design pressure (10% of Operating pressure) 32.12 kg/cm2 = 33 bara = 3.3 N/mm2 Allowable stress (Hastelloy B) 351.63 N/mm2 Hydrostatic test pressure 39.77 kg/cm2 Density of material kg/cm3 Corrosion allowance (CA) 4 mm A thorough calculation of reactor design can be seen in Appendix B.
  • 21. 21 4.1.4.1 Choice of material Hastelloy B-2 – (65% Ni, 28& Mo, 5% Fe) required to hold resistance against the corrosion of hydrogen iodide and acid (Sinnott et al., , p986). Its physical and chemical properties appear to be specifically suitable as a choice for the CSTR of this system. ???????????????????????????????????? ???????????????????? ????????????????????????????ℎ = 51 ???????????? = ????????????/????2 (Alloys and Producer, ) 4.1.4.2 Vessel support A support skirt will be required for the reactor. The design of such implementation will have to be performed separately and in accordance to the vessel’s specifications. 4.1.4.3 Piping sizing Pipeline Flowrate (m3 /s) Velocity (m/s) Cross - sectional area (m2 ) Diameter (mm) Equivalent NPS (in) CO feed to reactor 6. 2.000 3. .3 80 Methanol feed to reactor 0. 2.000 0. 78.1 3 Water feed to reactor 0. 2.000 0. 11.3 0.375 Reactor to flash tank 0. 2.000 0. 139.1 6 Reactor to scrubber 0. 2.000 0. 125.9 6 Drying column to reactor 0. 2.000 0. 272.3 12 Scrubber to reactor 0. 2.000 0. 2.5 12 Table [4.4]: Representation of pipeline diameter upon data obtained from Aspen Plus simulation and conversion (mm to NPS) (Perry et al., ).
  • 22. 22 The feed of CO to the reactor pipeline represents an offset value, which is due to the high volumetric flowrate as a result of its gas-phase nature, leading to very low density. On the basis that a constant velocity of 2 m/s is assumed, the CO feed pipeline will be therefore be significantly larger. 4.1.4.4 Nozzles Divergent nozzles given a size margin of 15% and distance between nozzle entrance and flange distance assumed to be twice of the nozzle’s diameter. Dimensions of nozzle to appropriate pipeline is given in Table 4.5. Table [4.5]: Representation of nozzles in relation to pipeline upon data obtained from Aspen Plus simulation and conversion (mm to NPS) (Perry et al., ). 4.1.4.5 Heat dissipation and vessel insulation Similarly to most systems to which the laws of thermodynamics can be applied to, there is dissipation of energy in the form of heat to the surrounding areas. This rate of energy exchange is governed by Fourier’s Law, which states: ???? = ℎ???????????? ???????????????????????????????? = ???? ???? dT ???????? Where: ???? = ℎ???????????? ???????????????????????????????? k = thermal conductivity of material A = area of the vessel ???????? = temperature gradient Pipeline Nozzle Diameter (mm) Equivalent NPS (in) Nozzle entrance to flange (m) CO feed to reactor A .23 88 4.57 Methanol feed to reactor B 89.78 5 0.18 Water feed to reactor C 12.99 0.375 0.03 Reactor to flash tank D 159.95 8 0.32 Reactor to scrubber E 144.73 6 0.29 Drying column to reactor F 313.15 14 0.63 Scrubber to reactor G 64.35 2.5 0.13
  • 23. 23 ???????? = thickness of material ???? = 67,301,345 kJ/hr (???????????? ???????????????????????????????????????????????? ???????? ???????????????????????????????? [????]) Normally the issue of heat dissipation from a vessel can be overcome by applying an appropriate insulation system on the internal fitting of the vessel; in this case, however, the heat dissipated is negligible in relation to the heat of reaction produced within the reactor: ???????????????? ???????? ???????????????????????????????? ???????????????????????????????? = 5.49 × kJ/hr ???? = ℎ???????????? ???????????????????????????????????????????? = 67,301,345 = 6.7 × 107 kJ/hr Therefore by taking into account this specific system, it is fair to assume that heat dissipation is not relevant in relation to the rate of heat of reaction produced, so the conclusion is that no insulation is required for this vessel. 4.1.4.6 Shutdown Shutdown procedures are required whenever an emergency shutdown or maintenance occurs. The standard safety protocols must be followed, where the first step is to decrease production rate at constant intervals, thus allowing the progressive reduction of pressure and temperature. When conversion reaches 0, the reactor will need to get emptied. In order to do so, the vessel is continuously purged with an inert substance, nitrogen, to prevent formation of oxygen or other reactions for when the shell comes into contact with the atmosphere. 4.1.4.7 Process safety The choice of material, Hastelloy, is the first application of safety factors within the design. As a result of the highly corrosive nature of the iodides present, hydrogen and methyl iodide, Hastelloy represents the most suitable material choice. A high corrosion allowance (4mm) has thus been implemented. Parameters such as temperature and pressure, other than level of fluid within the vessel are directly related to the safety procedures, therefore appropriate process control measures have been employed to ensure a high level of process safety in the reactor. 4.1.4.8 Process control Methanol feedlines have been integrated with temperature and pressure indicators alongside a flow control valve with an appropriate transmitter due to flowrate being the most relevant parameter to control in order to allow the reaction to occur currently. A backup pump has also been employed in the methanol feed pipelines, in case of failure of the first one. Moreover, backup flow control valves have been included in both the CO and methanol feeds.
  • 24. 24 Pressure control and a transmitter have been included in the reactor to keep a constant monitoring of the pressure within the vessel, thus overpressure or under pressure is prevented which would affect the rate of reaction (30 bar). A level indicator controller, with high and low alarms, have been selected to monitor the reactor’s fluid level; they control the flow valve located in the pipeline between the liquid phase of the reactor and the flash tank in order to maintain fluid level within vessel specification, a maximum 70% of vessel’s volume is recommended). Isolation valves have been fitted around the vessel to ensure a high level of control in relation to safety and failures of the system, allowing quick shutdown of the unit operation. The pressure reduction valve located in the pipeline which connects the liquid stream of the rector to the flash tank is fitted in order to flash the content and reduce pressure from 30 bar to 1 bar. The other streams feeding from and to the reactor have been fitted with appropriate flow controllers and transmitters to monitor the overall flow within the system 4.1.5 Agitator Specification Agitators are required to increase the transfer of material within the vessel and create uniform temperature within the reactor. In order to evaluate which type of agitator is most suitable for the reaction, it is possible to observe a correlation between viscosity, volume of the tank and type of agitator. Figure [4.3]: Aspen Plus simulation data. Stream Flowrate (kg/hr) Mass fraction avg. viscosity (N-s/m^2) avg.viscosity (N-s/m^2)( fraction) CO .36 0. 1.77E-05 2.11E-06 Methanol .02 0. 0. 7.00E-05 Water 720. 3.42E-03 0. 3.12E-06 CSTR-LIQ 5.18E-01 0. 6.89E-05 CSTR-VAP .236 9.12E-03 1.83E-05 1.67E-07 WATER-RCY .5 0. 1.57E-05 2.16E-06 SCRUB-RCY .82 0. 0. 9.85E-06 Total . 1 1.56E-04 0.156 mpa.s LIQUID VISCOSITY 0.
  • 25. 25 Figure [4.4]: Aspen Plus simulation data. The values for viscosity have been collected from the simulation on Aspen PLUS. Through implementation of flowrates for all the streams connected to/from the reactor, a mass flowrate has been calculated, hence leading to calculate the total liquid viscosity of the mixture, and correlating Figure 4.3, this leads to: ???????????????????????? ???????? ???????????????????????? = 138 ????3 ???????????????????????? ???????????????????????????????????? = 7.9 × 10−3 ????????/????2 ????????????????ℎ ???????????????????????????????????????????????? = ???????????????????????????????????? ???????? ???????????????????????????? (???????????? ???????????????????????????????????????????????? ???????? ???????????????????????????????? ????) Figure [4.5]: Impeller type for different mixture viscosities (Sinnott et al., ). By analysing Figure 4.5, it is possible to evaluate that flat-blade turbine is the most suitable one in relation to the liquid viscosity of 7.9 × 10-3 Ns/m2 . Stream Flowrate (kg/hr) Mass fraction avg. density (kg/m3) avg.density (kg/m3)( fraction) CO .36 0. 1. 1.35E-01 Methanol .02 0. 793. 1.03E+02 Water 720. 3.42E-03 993.957 3.39E+00 CSTR-LIQ 5.18E-01 861. 4.47E+02 CSTR-VAP .236 9.12E-03 21. 1.96E-01 WATER-RCY .5 0. 69. 9.57E+00 SCRUB-RCY .82 0. 17. 1.43E+00 Total . 1 5.64E+02 564 kg/m3
  • 26. 26 Using ratios given in literature allows to specify the dimensions, location and characteristics of impellers and blades have been identified. In order to guarantee a good level of reaction control in relation to heat within the vessel, baffles have been integrated to the design. Due to the fact that µ < 500 mPa.s, the type of agitator should be either a propeller or a turbine. It is assumed that the agitation required is mild for such homogenous reaction. Therefore, assumed rotational speed falls within the ‘low’ category, 200 rpm is the estimated rotational velocity required (Carpenter, ). Agitator – Design Data Diameter 1.48 m Height above vessel bottom 1.48 m Blade length 0.37 m Blade width 0.45 m Baffled width 0.45 m Baffled height 8.9m Liquid depth 4.45 m Number of blades 6 Number of baffles 4 Diameter of shaft 0.053 m Power required 194 hp A thorough calculation of agitator design can be seen in Appendix B. 4.1.6 Conclusion The parameters, data and information researched about the process have been defined and simulated on appropriate software (Aspen Plus). This allowed to facilitate the collection of results in relation to mass and energy balances in order to define the main parameters of the vessel, allowing a 20% overdesign for potential expansion. A detailed mechanical design of the reactor including its minor components has been completed, defining the dimensions of turbine, blades, shaft and the power required by the agitator. In addition, a comprehensive stress analysis has outlined that the reactor’s mechanical design is suitable for industrial application. Time taken to reach steady state conditions as well as an analysis of shutdown operations, process control and safety has been implemented throughout the design. Furthermore, a cost
  • 27. 27 estimation has been evaluated in relation to the plant location, time and local currency. The final cost for the unit is approximately £800,000 – excluding delivery and installation costs.
  • 28. 28 4.1.7 Engineering Drawing of Reactor Figure [4.6]: Mechanical design of reactor
  • 29. 29 4.2 Flash Tank 4.2.1 Introduction The flash drum is a vapour liquid separator, its role is to split the mixture of the vapour-liquid mixture fed from the reactor. The vapour stream is released from the top of the drum. The liquid stream leaves through the bottom of the drum containing the Rhodium catalyst which is then recycled back to the reactor to be reused to aid reactions within the reactor. The design approach was to specify the flash tanks operating conditions and physical attributes (addition of demister and diameter and tank length) that directly affect the cost of equipment and operating costs. The orientation of the vessel will be vertical as its ideal for high flow rates, the vertical separator the process is more economical compared to the horizontal separator (Monnery and Svreck, ) A demister pad is a device with metal mesh like structure that eliminates the possibility of liquid entrainment within a pressure vessel. Entrainment is the entrapment of one phase within another, within the flash drum liquid droplets can be entrained within vapour and leave liquid droplets within the vapour stream. To prevent liquid entrainment, the velocity of vapour stream must be kept low to allow the water droplets to disengage for the vapour stream and drop back down to the liquid pool at the base of the vessel. If the operation requires a high vapour velocity the demister pad acts as an effective entrainment separator (Basic et al., ).As the vapour travels through the mesh wiring pad the stream lines are deflected, however the kinetic energy of the liquid droplet entrained within the vapour are too high to follow the streamline, they become impinged in the wires. The liquid droplets then coalesce forming a liquid layer on the surface of the wires. The droplets then detach from the pad. Due to the orientation of the vessel (vertical) the liquid droplets will be captured and be drained back and form large droplets that can drop from the upstream face of the wire mesh pad ( Al-Deffeeri et al., ). Demister pads increase the efficiency of vapour liquid separation efficiency. Flash drums that use gravity separation (without the demister) are dependent on a high residence time to separate the liquid from the vapour. The more time needed for the mixture to separate, the higher the energy cost to run the flash drum thus the plant throughput will be lower per day hence reducing revenue per day. The demister pad allows the same degree of separation to be carried out in a smaller vessel, the reduction of volume reduces the weight of the vessel which directly minimizes the cost of the vessel shell (Sinnott et al., )).The internal diameter is dependant of the vessel is dependent on the terminal velocity of the particles
  • 30. 30 4.2.2 Flash Tank P&ID F-201 V-206 FIC LI V-211 FI FIC R-201 V-210 V-209 V-208 V-207 FI FI FIC FIC R-201 DC-301 Figure [4.7]: P&ID of Flash Tank
  • 31. 31 4.2.3 Design Method This specific sizing methodology is adopted from “two phase separators within the right limits” published in the “Chemical engineering progress synopsis series” () the calculation initiates by the finding the diameter of the vessel. In order to do so, vertical terminal vapour velocity, QT, is determined by obtaining the K value using Table C1 (See Appendix C). Subsequently, QV, vapour volumetric flow rate is calculated. The internal vessel diameter, DVD, is estimated, whilst adding 6 inches to the figure obtained to accommodate the support for the mist eliminator. Referring to Table C2, hold up time and surge volume relative to a “Feed to column separator” are selected. Further referring to Table C3, low liquid level height, HLLL, is obtained, thus distance from the low liquid level, HLLL to the normal liquid level, HNLL, is estimated. This value must be minimum of 1ft. Consecutively, the height between normal liquid level, HNLL to high liquid level HHLL, must be 6 inches minimum. Henceforth, the height from high liquid level to the centre line of inlet nozzle is estimated. The disengagement height from the centre line of inlet nozzle to the bottom of demister pad is then determined and assumption of the height of the mist eliminator pad, HME, is 6 inches and 1ft is taken from the top of the mist eliminator to the tangent line of the flash drum.
  • 32. 32 Mole Flow kmol/hr Flash in Flash vapour stream Flash liquid stream METHANOL 1. 1. 0 CARBON MONOXIDE 7. 7. 0 ACETIC ACID .024 .921 164. WATER 13. 11. 1. ETHANOL 1.35E-05 1.22E-05 1.35E-06 PROPIONIC ACID 6. 5. 0. CARBON DIOXIDE 142.194 142.194 0 HYDROGEN 4. 4. 0 METHYL ACETATE 16. 16. 0 HYDROFEN CHLORIDE 7.75E-08 7.75E-08 0 METHY-IODIDE 16. 12. 3. Total Flow kmol/hr .262 .96 169. Total Flow kg/hr .71 .75 Total Flow l/min 176. Temperature C 104. 104. 104. Pressure bar 1 1 1 Vapor Frac 0. 0. 0 Liquid Frac 0. 0. 1 Solid Frac 0 0 0 Enthalpy cal/mol - - - Enthalpy cal/gm -.75 -.307 -.79 Enthalpy cal/sec - - - Entropy cal/mol-K -57. -56. -67. Entropy cal/gm-K -0. -0. -1. Density mol/cc 0. 0. 0. Density gm/cc 0. 0. 0. Average MW 59. 58. 61. Liq Vol 60F l/min .089 .876 162. Table [4.6]: Material Balance for Flash Tank. 4.2.4 Flash Tank Specification Flash Tank – Design Data Design pressure 1.1 bar Design temperature 114.89 o C Pressure 1 bar Temperature 104.445 o C Vapour volumetric flow rate 4.4 m3 /min Liquid volumetric flow rate 1.68 m3 /min Vapour density 6.24 kg/m3 Liquid density 983.31 kg/m3
  • 33. 33 The chosen material of construction is Hastelloy-B-3 (65% Ni, 28% Mo, 5% Fe) due to the corrosive nature of Hydrogen Iodide and acid. The minimum allowable diameter of the vessel has to be large enough to slow down the gas below the velocity which the particles will settle out (Sinnott et al., ). Following calculations from (Monnery and Svreck, ): ???????????????????????????????? ???????????????????????????????? = ???? ???? = 1.32 ????/???? ???????????????????????????????? ???????????????????????? ???????????????????????????????? = ???? ???? = 0.992 ????/???? ???????????????????????? ???????????????????????????????????????? ???????????????????????????????? = ???? ???? = 4.4 ????3 /???? The flash drum has a mist eliminator, therefore 6 inches are added to accommodate a support ring and rounding up to the next 6-inch increment to obtain the external diameter. Thus: ???????????????????????????????? ???????????????????????? ???????????????????????????????? = ???? ???????? = 2.53 ???? ???????????????????????????????? ???????????????????????????????? = ???? = 3.66 ???? Hold up time is the time it takes for the normal liquid level to reach the lower liquid level to empty whilst keeping a normal outlet flow rate with no feed entering the vessel. Thus: ???????????????? ???????? ???????????????? = ???? ???? = 5 ???????????????????????????? ???????????????? ???????? ???????????????????????? = ???? ???? = 8.39 ????3 The surge time is the time it takes for the normal liquid level to rise from normal liquid level to maximum when keeping normal feed flow rate and no outlet flow. Therefore: ???????????????????? ???????????????? = ???????? = 3 ???????????????????????????? The volume of liquid between the highest liquid level and the normal liquid level ???????????????????? ???????????????????????? = ???????? = 5.032 ????3 The total height of the vertical flash drum is the sum of the HLLL +HH +HS + HLIN +HD +HME + 1ft. The height of the vessel outlet at the top of the vessel must be sufficient so the liquid droplets can disengagement from the vapour. Henceforth, liquid heights within vessel: ???????????? ???????????????????????? ???????????????????? ℎ????????????ℎ???? = ???????????????? = 0. ???? ????????????????ℎ???? ???????????????????????????? ???????????????? ???????????? ???? ???????????? = ???? ???? = 1.667 ???? ????????????????ℎ???? ???????????????????????????? ???? ???????????? ???????????? ???? ???????????? = ???????? = 1 ???? ????????????????ℎ???? ???????????????? ???????????????? ???????? ????ℎ???? ???????????????????????? ???????????????? ???????? ???????????????????? ???????????????????????? = ???? ???? = 3.66 ???? ????????????????ℎ???? ???????????????? ???? ???????????? ???????? ???????????????????????? ???????????????? ???????? ???????????????????????? = ???????????????? = 4.062????
  • 34. 34 ???????????????? ???????????????????????????????????????? ℎ????????????ℎ???? = ???? ???????? = 0. ???? ????????????????ℎ???? ???????????????????????????? ???????????????? ???????????????????????????????????????? ???????????? ???????????? ???????????????????????????? ???????????????? ???????? ???????????????????????? = ???? ???????? = 0.305 ???? ???????????????????? ????????????????ℎ ???????????????? ℎ????????????ℎ???? = ???????????????? + ???? ???? + ???????? + ???????????????? + ???? ???? + ???? ???????? + ???? ???????? = 10 ???? When calculating the corrosion wall thickness of a vessel the corrosion allowance must be taken into consideration. The corrosion allowance is the amount of Hastelloy material available for corrosion without disturbing the amount of pressure the vessel can contain (Sinnott et al., ). Thus: ???????????????????????????????????? ???????????????????????????????????? = 4 ???????? Leads to: ???????????????? ????ℎ???????????????????????????? = 3.6 ???????? + 4 ???????? = 7.6 ???????? ????ℎ???????????????????????????? ???????? ???????????????????????? ℎ???????????????? = 7.6 ???????? + 4 ???????? = 11.49 ???????? The internal diameters of the Hastelloy pipes leaving and entering the flash drum were calculated using Sinnott and Towler, the method is further discussed in Appendix [C]. Pipeline Flowrate (m3 /s) Velocity (m /s) Cross sectional area (m2 ) Required diameter (m) Reactor to Flash drum 4.45 2.00 2.225 1.68 Flash drum to light ends column 4.40 2.00 2.2 1.67 Flash drum to reactor 0. 2.00 1.45 x 10-3 0.043 Table [4.7]: Pipeline sizing for Flash Tank 4.2.4.1 Process controls and safety On the left hand side of the flash drum there is a level indicator which will set off an alarm when liquid level is below a certain point, this is to avoid pump damage. When the pump is pumping air and not fluid it can lead to cavitation and produce loud noises. The pressure indicator send sends message to the alarm system if the pressure within the vessel is over the maximum pressure of 1.1 bar. The pressure within the flash tank is relatively low compared to the reactor and it’s very rare that it tends to overheat because the heat exchanger reduces the temperature before all products reach the flash drum. Going beyond this pressure can result in boiling of the liquid within the vessel and increase of temperature. The contents within the vessel are highly hazardous and flammable the build-up of pressure can cause an explosion and put all site workers at risk but this is highly unlikely. If the flash tank temperature did increase it would be due to a faulty heat exchanger feeding through high temperature streams, so to reduce the temperature the reactor must be cooled down. So a pipeline must be installed and redirected, while the faulty heat exchanger is fixed 4.2.5 Conclusion
  • 35. 35 The flash drum unit was designed with 20 % overdesign ass specified in the process specification. The P&ID provided gives detail to the indicators and control systems essential for process safety and efficient production. The Engineering drawing is cross sectional representation of the essential internal units and recommended liquid levels within the flash drum. The final cost for the unit is approximately £67,000.
  • 36. 36 4.2.6 Engineering Drawing of Flash Tank Fig [4.8]: Mechanical design of Flash Tank.
  • 37. 37 4.3 Drying column 4.3.1 Introduction This section of the design project report provides a detailed design description of the drying column in the process. The process specification requires a yield of 400,000 tonnes per year and 99% purity of acetic acid, therefore it is mandatory that the drying column is optimized to meet the client’s specification. The objective of this section is to calculate the operating condition and physical parameters required to optimise the process in order to meet the client specification; for example the column diameter, height, thickness, ends and tray sizing. The column parameters are shown in Table 4.6. Feed rate, F kg/hr .17 Feed Composition Acetic acid 99.6 kmol% Water 3.56 x10-11 kmol % Propanoic acid 0.428 kmol% Feed Temperature, oC 118 Column operating pressure, bar 1 Column reboiler Partial Reboiler Column condenser Partial Condenser Distillate composition, XD 0.992 Bottoms composition, XB 0.038 Table [4.7]: Specified column parameters.
  • 38. 38 4.3.2 Drying Distillation Column P&ID Figure [4.9]: P&ID of Drying Column DC-401
  • 39. 39 4.3.3 Design Method Kinetic and thermodynamic data were collected from literature for water and acetic acid whilst obtaining stream properties from the simulation on Aspen Plus, thus allowing to identify XD and XB in the vapour and liquid streams. The subsequent data allows the determination of the number of trays in the column by utilising the McCabe – Thiele Method for binary mixtures and thus feed position and reflux conditions are estimated. In order to propose a viable implementation of a mechanical design, dimensions of the column need to be determined, for example diameter and height, as well as selecting a suitable materials of construction, a preliminary mechanical design of the drying column, which comprises of column design, plate design and general arrangements and finally estimate a proposed cost of the column, including capital and operating cost (Sinnot et al., ). 4.3.4 Drying Distillation Column Specification The final drying distillation column specification is based on the calculations in Appendix D. The drying distillation column is represented diagrammatically in Figure 4.8. The purpose of the drying column in this acetic acid synthesis process is to increase the acetic acid purity via separating methanol, ethanol, methyl acetate and water from the product stream and recycle these undesired compounds back in to the reactor, consequently increasing the purity of the product stream. A sketch of the column and plate is shown in Figure 4.8. Drying Column – Design Data Working Pressure 1.1 bar Inside Diameter (Di) mm Material of construction 305 Stainless Steel Allowable Stress 515 N/mm2 Density of material .172 kg/m3 Design pressure 1.1 bar Height of column 23.85 m Area of column .81 m2 Thickness of column 12 mm End selection Torispherical End thickness 20 mm Number of trays 34
  • 40. 40 Feed entry 16 Plate spacing 0.7 mm Hole pitch (rectangular) diameter 5 mm Tray thickness 3.5 mm Packing size 75 mm Pipe diameter Feed 216 mm Top 178 mm Bottom 148 mm The design considerations were made based on the specification provided from the client, the following should be noted in the design. Drying Distillation Column – Design Considerations Cost of shell and trays £364,000 Cost of reboiler £22,000.00 Cost of condenser £16,500.00 Dead-weight of shell 122 kN Weight of plates 109.67 kN Weight of insulation 5.6 kN Weight of vessel 237.27 kN Wind loading 31.94 N/mm2 Bending moment .3 N/mm2 Longitudinal stresses 84.8 N/mm2 Circumferential stresses 48.4 N/mm2 4.3.5 Conclusion The following design specification on this unit complies with the necessary design intent specification. In addition, the specification fulfils the plant debottleneck allowances of a 20% overdesign.
  • 41. 41 4.3.6 Engineering Drawing of Drying Distillation Column Figure [4.10]: Mechanical Design of Drying Column
  • 42. 42 4.4 Heavy – Ends Distillation column 4.4.1 Introduction This column is designed to separate the unwanted propionic acid produced as part of the process from the desired acetic acid produced, to obtain a purity of greater than 99.9% which is required by the design brief at a capacity of 400,000 tonnes per year. Temperature = 125 o C (398K) Operating conditions Pressure = 1 atm ( Pa) Reflux ratio = 17 For the purposes of the calculations, the average internal temperature of the column was assumed to be 125 o C (398K) to ensure that the majority of the acetic acid, and only a minimal amount of propionic acid was in the vapour phase in addition to operating at 1 atm ( Pa). The reflux ratio of 17 was taken from the simulation produced as part of this project on Aspen Plus, as this gave the desired quantity and purity of acetic acid as a top product. The material chosen for the construction of the column is 304 stainless steel as both propionic and acetic acid have corrosive properties and stainless steel provides a sufficient corrosion resistance to justify its choice for the construction of the column. Grade 304 was chosen over other grades of stainless steel as the mechanical and structural advantages provided by the other, more expensive, grades is not large enough to justify the extra cost associated with them. To calculate the internal diameter of the column a tray spacing has to be assumed, it is suggested that a tray spacing of 0.5 m should be initially used to calculate the column’s diameter and if the diameter is greater than 1 m a tray spacing of between 0.3 and 0.6 m is normally appropriate (Sinnott et al., , p708-709). The calculated column diameter was greater than 1 m, so the initial assumed tray spacing on 0.5 m was carried forward throughout the calculations. The tray efficiency applied to the tray in this case is 70% as this is found to be an optimal number for the preliminary design of a distillation column (Sinnott et al., , p700). It is also suggested that 10% more tray be added in addition to tray efficiency with future expansion in mind (Branan, , p444). In addition to the spacing of the trays and tray efficiency, included in the cost analysis of the column is the costing of the trays used. This required a type of plate contactor to be chosen from a selection of sieve plate, bubble-cap plate and valve plate, each with their own advantages and disadvantages. The plate contactor chosen in this case is the valve plate, as there are weeping issues with sieve plates at low liquid flow rates and bubble-cap plates are approximately twice as expensive as valve plates. The valve plate is an ideal compromise between performance and cost when compared to the other two options. Before the tray efficiency is applied to the theoretical number of stages, there are a total of 15 rectifying stages and 8 stripping stages giving a total of 23 theoretical stages. The location of the column inlet stream
  • 43. 43 lies between the rectifying and stripping sections therefore, in this case the inlet stream is between the 15th and 16th stage, although this does not hold when the tray efficiency is applied as the number of stages changes. Although the number of stages changes, the location of the feed should maintain the same ratio of rectifying and stripping stages above and below it respectively, therefore, the initial ratio of rectifying stages to total number of stages will be applied to the actual number of stages to find the actual inlet location of the column.
  • 44. 44 4.4.2 Heavy – Ends Distillation Column P&ID DC-501 P-501 P-502 V-502 V-503 V-504 V-505 FI FIFIC FIC V-501 LIC LI V506 TI FIC V-507 TI FIC Acetic acid Propionic acid LAH LAL Process Steam Process Water DC-401 V-508 FIC FI Figure [4.11]: P&ID of Heavy-Ends Distillation Column
  • 45. 45 4.4.3 Acetic acid properties within the column The vapour density of acetic acid under the conditions of the column was calculated using the ideal gas equation (Perry, , p2-355): ρ ???????????????????????? = ???? ???? ???????????????????????????????? ????⁄ ρ ???????????????????????? = 1.84 kg/????3 The vapour pressure due to acetic acid under the conditions of the column was calculated using the Antoine equation and the parameters specific to acetic acid (Sinnott et al., , p451): ???????????????????????????? = ???? ????−( ???? ????+????⁄ ) ???????????????????????????? = 136.45 ???????????? Where: A = 7. (Dean, , p539) B = .313 (Dean, , p539) C = 222.309 (Dean, , p539) The mass flowrates and mass fractions for the inlet, bottom product and top product have been calculated using values taken from the simulation produced on Aspen Plus as displayed by the PFD: Molar flowrate [kmol/hr] Mass flowrate [kg/hr] Mass fraction Inlet 893.86 .48 0.992 Bottom product 3.07 184.17 0.296 Top product 890.79 .34 1.0 Table [4.8]: Acetic acid mass flowrates and mass flowrate for inlet, bottom product and top product. 4.4.4 Propionic acid properties within the column The liquid density of propionic acid was calculated by developing a relationship between known values of propionic acid density and temperature (CAMEO Chemicals, ) and assuming the relationship remained constant up to the operating temperature of the column. ρ ???????????????????????????????????? = 882.39 ????????/????3 The vapour pressure due to propionic acid under the conditions of the column was taken from literature and was given as (Clifford et al., ): ???????????????????????????????????????? = 59.86 ????????????
  • 46. 46 The mass flowrates and mass fractions for the inlet, bottom product and top product have been calculated using values taken from the simulation produced on Aspen Plus as displayed by the PFD: Table [4.9]: Propionic acid molar flowrates and mass flowrate for inlet, bottom product and top product. 4.4.5 Relative volatility As this is a binary mixture, the following relationship between vapour pressures can be used to calculate relative volatility (Branan, , p450): ???? = ???????????????????????????? ???????????????????????????????????????? ???? = 2. 4.4.6 Heavy Ends Distillation Column Specification The final heavy ends distillation column specification is based on the calculations in Appendix E. The heavy ends distillation column is represented diagrammatically in Figure 4.9. The mechanical properties of the column were calculated based upon the Smoker equations which are applicable to systems in which the relative volatility is close to 1 as the McCabe-Thiele method would be impractical (Sinnott et al., , p 661). Using this method and applying a 10% over design as well as a tray efficiency of 70% give an actual number of stages required to be 37. ???????????????????????? ???????? ???????????????????????? ???????????????????????? = 37 The height of the column is calculated by multiplying the tray spacing by the number of stages as well as the addition of height allowances for a condenser at the top of the column and a reboiler at the bottom of the column. The height allowance for the condenser and reboiler are suggested as 4 ft (≈1.25 m) and 6 ft (≈1.85 m) respectively (Branan, , p 444). The column height, not including either end, is calculated to be 21.6 m. ???????????????????????? ℎ????????????ℎ???? = 21.6 ???? The column diameter is calculated as a function of the maximum vapour velocity through the column, which is a function of tray spacing. The maximum vapour velocity through the column is calculated as 2.86 m/s and the column diameter is calculated as 1.83 m. Molar flowrate [kmol/hr] Mass flowrate [kg/hr] Mass fraction Inlet 5.93 438.76 0.008 Bottom product 5.93 438.76 0.704 Top product 0 0 0
  • 47. 47 ???????????????????????????? ???????????????????????? ???????????????????????????????? = 2.86 ????/???? ???????????????????????? ???????????????????????????????? = 1.83 ???? The column shell thickness chosen is a calculated minimum shell thickness required to resist the internal pressure with the addition of a corrosion allowance of 2 mm. This equates to a column shell thickness of 12 mm. ???????????????????????? ????ℎ???????????? ????ℎ???????????????????????????? = 12 ???????? The thickness of the ends also has to be calculated as this will differ from the column shell thickness as the stresses the ends of the column are put under vary from that of the column shell. The thickness of the end is equal to 18 mm including a 2 mm corrosion allowance. ???????????????????????? ???????????? ????ℎ???????????????????????????? = 18 ???????? The size of the inlet, bottom product and top product pipes are all calculated using the density of the stream, the velocity of the stream and the streams flowrate. The inlet pipe diameter was calculated to be 0.084 m which corresponds to a nominal pipe size of 4 inches including an allowance for future expansion. The bottom product pipe diameter was calculated to be 0. m which corresponds to a nominal pipe size of 3/4 inches including an allowance for future expansion. The top product pipe diameter was calculated to be 0.78 m which corresponds to a nominal pipe size of 36 inches including an allowance for future expansion. ???????????????????? ???????????????? ???????????????????????????????? = 4 ????????????ℎ???????? ???????????????????????? ???????????????????????????? ???????????????? ???????????????????????????????? = 3/4 ????????????ℎ???????? ???????????? ???????????????????????????? ???????????????? ???????????????????????????????? = 36 ????????????ℎ???????? The cost of the column shell is a function of the columns mass which is calculated to be 120,000 kg. The total cost of the column including the cost of the shell and the trays is £145,000 (, UK basis). ????ℎ???????????? ???????????????? = .6 ???????? ???????????????? ???????????? ???????? ???????????????????????? ???????????? ???????????????????? = £145,002 4.4.7 Summary of design data To follow is a summary of the full design data of the column, calculations as to how the data has been obtained is described fully in Appendix E.
  • 48. 48 Tray selection Valve plate trays Tray spacing 0.5 m No. of stages 37 Feed location Stage 24 Column height 21.6 m Column diameter 1.83 m Shell thickness 12 mm End thickness 18 mm Inlet pipe diameter 4 inches Bottom product pipe diameter 2 inches Top product pipe diameter 36 inches Column mass .6 kg Installed cost of column shell and trays £145,000 4.4.8 Conclusion The mechanical design of this unit has been completed with consideration of the 20% overdesign required by the design brief as well as the product specification. To follow will be an engineering drawing that describes all the data put forward, to give a clear visual representation of the design of the heavy ends distillation column.
  • 49. 49 4.4.9 Engineering Drawing of Heavy – Ends Distillation Column Figure [4.12]: Mechanical Design of Heavy – Ends Distillation Column.
  • 50. 50 4.5 Absorption column 4.5.1 Introduction The production of waste gas components is inevitable in the case of carbonylation. The off was waste materials (CO2, H2, CO) have to be separated from any toxic, and carcinogenic (e.g. methyl iodide) components still present in the waste stream and then burned in the flare. It is necessary that the methyl iodide is recycled back into the system, as it is highly toxic to the environment, and because it is required as a reaction promoter for the carbonylation step. The off gases are produced in the reactor phase as unwanted products are burned in the flare. While the methyl iodide is captured in a counter-current packed absorption column using methanol and acetic acid (i.e acetic acid is used for start-up, while methanol is sued throughout the life of the plant). Absorption Column – Design Data Height of transfer units 1 m Height of packing section 8 m Total height of column 10.5 m Column Diameter 0.6 m Pressure drop 0.005 bar g/m Packing – Design Data Type Intalox ceramic saddles Size 25 mm (1 inch) Packing material Ceramic Packing arrangement Dumped
  • 51. 51 4.5.2 Absorption Column P&ID Figure [4.13]: P&ID of Absorption Column S-601
  • 52. 52 4.5.3 Design Method Absorption is mass transfer procedure in which one or more soluble components from the gas mixture are dissolved using a low volatility liquid. As a result the polluting material (i.e. methyl iodide) diffuses from a gaseous sate into a liquid state, and is then recovered at the bottom of the column. The absorption rate is driven by the driving force of the absorption, and is relatively independent of equipment used (McCabe and Smith, ). The absorption unit operates using a counter current design, where the methyl iodide present in the gaseous mixture is dissolved in a liquid with a lower volatility (Sinnott et al., ). Counter current designs have the highest theoretical removal efficiencies, and is suited for high loadings of pollutant materials while it requires a lower solvent to gas ratios than alternative designs (e.g. crosscurrent, concurrent). The most common choice for pollution control gas absorbers are packed towers. A packet tower is often preferred to plate/tray towers because it can manage higher flowrates of gas, with lower pressure drops while maintaining low liquid hold-up. It is also recommended to use a packed tower when the contacting components have corrosive/acid proprieties, as cheaper corrosive materials are for the shell, and packing are available. Also packed towers are preferred when there are no high temperature deviations, and the system removes the gaseous mixture using a pressure drop. And when the diameter of the column (i.e. based on the flowrates of material) is between 0.5-0.7 m (Sinnott et al., ). Disadvantages of packed towers • High clogging and fouling potential • Replacing damaged packing • Higher waste water/solvent disposal • Removal of very small particles 4.5.4 Absorption Column Specification The waste gas stream enters through the bottom of the column and travels vertically, counter current (i.e. through the packing) to the falling solvent liquid. As a result, gaseous methyl iodide diffuses into a liquid phase. The system works based on physical absorption, and achieves high efficiencies at low temperature and pressure (Sinnott et al., ). Physical absorption is used because it relies entirely on the proprieties of the solvent and the gas stream, and their specific characteristic (e.g. volatility, density, viscosity). In order to achieve efficient absorption is it important that the design allows large contact area for the gas stream and solvent to react, the capacity required for controlling high rates of waste gas, higher gas to liquid ratios, low pressure drop and adequate distribution of solvent to gas to allow adequate pollutant diffusion (i.e. methyl iodide) 4.5.4.1 Design considerations to account for drawback of unit • A liquid distributor is used in order to maximize area covered by solvent in packing. • A higher density component (e.g. acetic acid, methanol) is used as a scrubbing liquid.
  • 53. 53 • The solvent used is present in the system, and is recycled into the reactor with the methyl iodide instead • Packing material is corrosive-resistant, meaning that damage only occurs due to physical contact; • Packing size is chosen based on the column size, so that it maximizes particle interaction, including very small particles. • Dumped packing allows easier replacement in the case of damage as opposed to structured packing. • Packing components that are not damaged can be reused, therefore reducing costs. • A low pressure drop will results in low energy requirements. 4.5.4.2 Choice of packing For this unit, the packing is the most important component. This is because the absorption efficiency is correlated to the flow capacity, and the height of the transfer units (i.e. HTU gas, HTU liquid); these factors significantly affect the tower height of the unit, and has economic implications (e.g. installation, maintenance, cleaning). When choosing the adequate packing, it is important that the following factors are taken into consideration: • The packing material has to be inert to the liquids flowing through the packing. • The packing material needs to be corrosive resistant. • Brittleness of component needs to withstand process conditions, without presenting excessive weight. • The packing must offer enough contact area, while not restricting the gas and solvent flow. • The packing must restrict the formation of excessive liquid hold-ups. • The packing material needs to be acquired at a reasonable cost. The above considerations were taken into account and 25 mm Intalox ceramic saddles were chosen. These choice poses process advantages and low cost of packing. Intalox ceramic saddles offer the best contact area, are inexpensive, and in ideal conditions could last throughout the life of the column. It is expected that the packing will have to be changed throughout the life of the plant. However due to the small costs, it still supports the choice of dumped packing rather than structured packing, or plates (Sinnot et al., ). 4.5.4.3 Choice of absorption tower equipment The packed absorption column is comprised out of: • Column Shell • Mist eliminator • Liquid distributor • Packing restrainer
  • 54. 54 • Packing support • Packing materials The mist eliminator is in the form of a layer of mesh; its main function is to collect any droplets that gather at the top of the column as mist. It needs to be installed at the top of the column, so that any of the liquid droplets collected are returned back to the column. The droplets are moved to the top of the column via a high velocity gas stream (McCabe and Smith, ). The liquid distributor chosen is a pressure drop spray nozzle distributer. The distributor is designed to wet the packing, and facilitate consistent contact between the gas mixture and the solvent without constricting the gas flow. Its main function is to spread the solvent evenly across the area of the packed bed. Some of the disadvantages of this equipment includes plugging, formation of mist, feed rate dependent liquid distribution. Therefore adequate maintenance is required in order to maximize the efficiency of the column (Sinnott et al., ). A packing support is necessary for an even distribution of the waste gas, and requires an open space between the bottom of the absorption tower and the packing. The support plates are required to support the total weight of the packing (i.e. while still allowing the material streams to travel freely), and are therefore a necessity for this system (Sinnott et al., ). A packing restrainer is required in order to prevent the high gas velocities from raising the packing into the liquid distributors. The packing restrainer used is an unattached weighted plate placed at the top of the packing, and which settles with the packed bed. The restrainer is required since the packing material is ceramic, and keeping the integrity of the packed bed; therefore preventing any extra costs. 4.5.4.4 Materials of construction The material of construction for the absorption column is Hastelloy C. Although, Hastelloy C is more expensive when compared to a stainless steel shell lined with corrosion resistant column internals (e.g. fibre-reinforced polymers), for this process route, Hastelloy C offers a better range for temperature and pressure resistance. Additionally, the corrosion allowance for Hastelloy C accounts for extended use, with a small probability of loss of containment across the column. 4.5.4.5 External equipment The external equipment for the absorption column is comprised out of: • Off gas movers • Solvent pumps • Control equipment
  • 55. 55 The waste gas movers require to be cooled down to room temperature, as increased gas temperature leads to lower absorption in the column. Methyl iodide is gaseous above 315.9 K, and is required that the methyl iodide is cooled down to 298 K. The resulting stream after cooling down will be a mixture of gas and liquid, helping to absorb the methyl iodide as it turns into liquid phase, as well as keeping the vessel temperature at room temperature in order to maximize the efficiency of the absorption column. The solvent is moved into the column using a centrifugal pump. It is recommended that the construction material used is corrosive resistant (e.g. stainless steel) and suitable for acetic acid/methanol. The scrubbing material will be pumped from the final pure acetic acid stream. It is recommended that a storage tank for acetic acid is placed in the proximity of the scrubber, in order to have access to excess scrubbing solvent if the situation requires (Sinnott et al., ). 4.5.4.6 Safety control The absorption unit requires the following control in order to operate safely on the premises of the plant: • Gas detectors located at the outlet vent for: Acetic acid, Methanol, Carbon Monoxide, Carbon Dioxide, Hydrogen, Hydrogen Iodide, Methyl Acetate, and Methyl Iodide. • Temperature indication control for the gas stream. • Level indication control for liquid stream. • Flow indication control at the inlet and outlet streams. 4.5.4.7 Safety considerations Since the components entering and exiting the absorption system are hazardous, adequate maintenance for the equipment should be done regularly (HSE Maintenance procedures, ). A safe control methodology has to be put in place. In order to keep the safe working environment the scrubber requires: • Pressure relief system to prevent pressure accumulation in the vessel that could lead loss of containment. • Adequate insulation around the column with mineral wool to prevent any unwanted temperature deviations. • Gas detectors located around the scrubbing unit. 4.5.5 Conclusion It is necessary that an efficient scrubbing system is put in place in order to prevent the release of toxic iodide into the atmosphere. It is recommended that the above dumped packing absorption tower to be connected to a stripping column, in order to achieve an efficient pollutant removal. The methyl iodide, alongside the hydrogen iodide are toxic materials and have to be recycled back into the stream, or disposed of adequately. These two components are crucial for the process to operate, and an efficient pollutant removal system is necessary. When considering the design of the absorption column, a packed absorber operating at low temperature and low pressure will offer the required removal efficiency. The
  • 56. 56 present design offers an economic alternative to other scrubbing units. The absorption unit is the safeguard that prevents any hazardous components from polluting the environment heavily, with the adequate safeguards and system management in place could last throughout the life of the plant.
  • 57. 57 4.5.6 Engineering drawing of Absorption Column Figure [4.14]: Mechanical Design of Absorption Column
  • 58. 58 4.6 Storage tank – Acetic acid 4.6.1 Introduction The tank is specified to be a fixed cone-roof cylindrical-type design and have a capacity of m3 . This figure is the sum of one week production of acetic acid ( m3 ) plus an extra m3 provided by the volume of the conical head. The maximum working level for the tank is m3 , approximately 80% the volume of the cylindrical tank and available for additional storage due to safety purposes. The storage buffer provided will allow the plant for continuous operation for up to one week in the event of unforeseen shutdown of the plant and/or disruption in distribution. The tank internal diameter is 24.9 m and the tank height is 18.8 m. The internal and external pressure loads require a wall, base and roof plate thickness of 8.8 mm to meet the British design code for pressure vessels. This thickness gives a 100% safety factor over the maximum anticipated stresses. The product inlet line is standard nominal pipe size 4, schedule number 40s. This inlet is sized for the maximum production flowrates. The product outlet line is of standard nominal pipe size 12, schedule number 120. This line is sized such that a standard-size chemical ship tanker may be filled in 8 hours. The tank must be constructed of stainless steel type 316L (‘acetic acid grade’), the specification of this material is given in Appendix F. The design data required for this unit are specified below. Acetic Acid Tank – Design Data Design tank capacity m3 Design temperature 40 o C Design pressure 18.7 kPa Working pressure 118 kPa absolute Acetic acid density kg/m3 Material of construction SS316L Design tensile strength 485 MPa Joint efficiency 85% The design tank capacity is estimated to be m3 . The maximum tank operating level will be approximately m3 to give the extra tank capacity as a reserve volume. This also ensures that a minimum of capital cost (in the form of product acid in the tank) is unused. The design temperature represents the upper limit that acid may be fed to the tank from the process. The working pressure represents the sum of atmospheric pressure and acid vapour pressure at the design temperature. Details of the calculations associated with the tank design are presented in Appendix F. The cost of this vessel is estimated (from a correlation) to be £333,000.
  • 59. 59 4.6.2 Acetic Acid Storage Tank P&ID Figure [4.15]: P&ID for Acetic acid storage tank. T-101
  • 60. 60 4.6.3 Design Method The tank dimensions are determined according to standard tank geometries as enforced by API 650. Tank shell thickness is sized according to the limitations imposed by the British design standard for pressure vessels. The tank contents are flammable, toxic and corrosive, appropriate safety features are recommended. Details of the calculations are given in Appendix G. 4.6.4 Storage Tank Specification The final tank specification is based on the calculations in Appendix G is shown in Table 4.10. The tank is represented diagrammatically in Figure 4.16. The storage tank should be constructed of 316L stainless steel. To reduce corrosion of the tank bottom exterior, application of coating is recommended between the tank and foundation. As seen by Figure 5.2, the storage tank area is enclosed by containment facilities capable of containing the contents of the tank and maximum expected rainfall in case of a storm event; moreover, additional safety of a small, deep diked area is proven through lower evaporation rate and small area of fire. The enclosed area is drained through a trap to a safe location that is protective of human health and environment and in compliance with applicable laws and regulations. A vertical tank is implemented to provide for a more economical use of land. For outdoor storage of glacial acetic acid, a heating system and tank insulation is provided. The recommended heating system consists of low-temperature electric heating pads installed between the tank exterior and the insulation in order to maintain the temperature at a desirable level. Acetic acid is a flammable solvent, thus to inhibit the accumulation of static charges, the storage tank, pumps, transfer lines, and offloading vehicle are adequately grounded and fill line enters the tank through the roof and extended downward to within 2 or 3 inches of the bottom. The storage facility is constructed so that water cannot be introduced or generation of heat occurs. In a confined space, considerable pressure caused by this reaction can result in an explosion that may rupture the storage tank. Safety features for the tank include a pressure relief-valve system on the tank roof, to be opened when draining or adding to the tank contents. An emergency relief vent is fitted to the storage tank to allow emergency flow due to excessive venting requirement from fire burning around the tank, thus eliminating opportunity for a costly tank rupture, providing emergency venting from abnormal internal pressure beyond the capability of the pressure relief vent. The operational tank venting system handles normal tank venting due to product import/export and ambient temperature variations. In the event of fire, as vapour pressure increases to a point where normal venting equipment capacity is exceeded, the hinged cover will lift relieving the pressure and protecting the tank from rupture. The pressure build up will be quite slow, therefore the cover should not open violently and cause any damage to the tank. Emitted vapours may be ignited by the fire, but should ‘flame off’ externally until brought under control by firefighting operations.. A manhole provides access to the tank for internal maintenance.
  • 61. 61 Appropriate venting systems are issued. Vents should be angled at 45o from vertical and cut off vertically to prevent rain from entering. The vents are 1 inch larger in diameter than the tank fill line. A coarse-mesh stainless steel wire screen is placed over vent openings to prevent entry of foreign objects. A blanket of inert gas, nitrogen, is provided and equipped with a pressure/vacuum conservation vent, piped away to a safe location that is protective of human health and the environment. In order to counteract the strong odour generated by acetic acid, odour masking methods are utilised within the compound. Odour control chemicals used for masking purposes are aromatic chemicals derived from aromatic chemical manufacture. Organic odour control chemicals are numerous, some of which are vanillin, methyl ionones, benzyl acetate, phenylethyl, eugenols and heliotropin, and thus any of the organic compounds mentioned are viable to be implemented in this scheme. Furthermore, this method of treatment requires no capital investment required for equipment and readily available for application. Table [4.10]: Storage tank specification. A standard metal staircase and railing skirts the outer edge of the tank providing access to the tank roof. A manhole in the tank roof provides access for internal repairs. Discrete inlet and outlet lines are required to feed into the base of the tank. A pressure relief valve is attached to the roof. This valve is opened automatically when pumping product to, or withdrawing product from the tank. The valve is shut when pumping stops so that vapour losses from the tank are contained. A bursting disc on the roof also provides emergency pressure relief for an unforeseen pressure build-up within the tank. 4.6.5 Conclusion The acetic acid storage tank required for the plant has a capacity of approximately m3 . This represents approximately 21,000 tonnes of product acid. The tank will normally contain only about 14,000 Vertical cylindrical-type tank Fixed conical roof Total tank capacity m3 Normal operational capacity (maximum) m3 Tank inside diameter 24.9 m Tank height 18.8 m Tank wall thickness 8.8 mm Material of construction SS316L Inlet and outlet at tank base Inlet line: Nominal pipe size 4, schedule number 40s Outlet line: Nominal pipe size 10, schedule number 120 Manhole, bursting disc, pressure relief valve on roof Mineral wool insulation 0.02 m
  • 62. 62 tonnes to satisfy outside product sales. The remaining capacity is in reserve in case of plant shut-down and issues regarding distribution. This provides a one week production buffer for the plant. The final specification requires a tank of 24.9 m diameter, height 18.8 m, and a plate thickness of 8.8 mm. A size delivery of m3 by means of a standard tonne chemical tanker would be desirable, as nearly 75% of the weekly production is able to be shipped. The figure obtained for the minimum stock level will be able to accommodate a scenario where if the plant shuts down for a few days, the remaining storage will keep on supplying customers. Additionally, if there is a problem in distribution, the available space in the tank will allow for continuous running of the plant until the problem is solved. Henceforth, a delivery period of one week will provide smooth production and shipment whilst accommodating customers. The cost of the tank has been calculated (Appendix G) from correlations to be approximately £333,000.
  • 63. 63 4.6.6 Engineering drawing of Acetic Acid Storage Tank Figure [4.16]: Mechanical design of Acetic acid storage tank.
  • 64. 64 5. Process Control and Instrumentation 5.1 Introduction to Process Control and Instrumentation The premise of process control and instrumentation consists of applying the philosophies of control to all aspects of a process, whether that be the design of the process, the standard operation of the process or the operation of the process under conditions that vary from the norm, for example during a start up or shut down, as the only way to have a safe system, is to have reliable control systems in place at all times. As well as safety, putting in place reliable and effective control systems also aids in improving plant efficiency and the economic stability of the process while ensuring the plants compliance with the relevant environmental and safety regulations. As the philosophies of process control and instrumentation are incorporated into every aspect of a process, the instrumentation that’s implemented into a process should be integrated within the earliest design instead of being an afterthought that’s ‘bolted on’ once the design has been completed. Implementing process control systems in this way will lead to the safety features of the process being much more effective with regards to plant safety and the safety of those in and around the plant, in addition to being much simpler because they have not been worked into the system, around an already complete process without the necessary control features put in place. 5.2 Objectives of Process Control The primary objective of the control systems and instrumentation put in place is to ensure that the process is carried out in a safe and reliable manner while producing a product that meets the required specification desired by the consumer. This can be achieved by (Sinnott et al., , p275): 1. Maintaining process variables within known safe operating limits around a specified set point. 2. Alerting operators to deviations from the set point and safe operating limits and provide a solution to the deviation either via the manipulation of process equipment or shutdown systems. 3. Preventing operators from altering process variables such that operation outside the safe operating limits is caused. The four main process variables that have been controlled throughout the design stages of this project are: 1. Temperature. 2. Pressure. 3. Level. 4. Flow. The variables have been controlled through the manipulation of various pieces of process equipment, such as control valves, via information sent to them through a series of indicators, alarms and controllers. The process equipment works to control any deviations from the desired set point, within an upper and lower limit. The control measures also have to accommodate for planned changes in the set point, for example, during
  • 65. 65 start-up and shut down in which the desired set point for each of the process variables is going to vary from that of a steady state operation. This can be seen in the reactor, for example, during its start-up process, the set point would be at its desired operating conditions to ensure that the reaction took and a product was being produced. As the reaction continues and begins to reach a steady state the heat released by the reaction will cause an increase in the temperature of the reactor and therefore, a deviation from the desired operating conditions. In this case the set point can be decreased below the desired operating conditions while the heat released by the reaction can be used to maintain the required temperature. Variations in the set can be fine-tuned using the instrumentation in place to control situations like this. In addition to process safety, operability has to be taken into consideration when implementing control features throughout the design stages of a project, as having too many indicators and alarms, etc., can reduce the operability of the process as having “too much data being thrown at operators reduces their ability to understand what is happening and respond correctly” (Hurley, ). This has been taken into consideration when implementing control features into the design of our process as it is possible to manage a combination of process variables by manipulating just one of them. Although this could lead to undesired variations in some of the process variables when manipulating others, therefore it is important to understand the relationships each of the process variables share with each other and what impact their variation would have on the process streams as well as the process equipment. An example of this can be seen in the transition from stream 4 to stream 5 of the PFD through a pressure reduction valve. The purpose of the valve is to reduce the streams pressure before it enters the flash tank, but as a consequence of reducing the pressure, the temperature of the stream is also reduced. The relationship between the two process variables is clear in this case, which would make it easier to identify the hazards a variation of pressure would cause downstream of the valve whether that be to the flash tank or other pieces of process equipment. 5.3 Implementation of Control Systems in our Design The types of control that have been taken advantage of throughout the design process of this project has been feedback, feedforward and cascade, with feedback and feedforward being the simpler forms of control and cascade being more complex. A feedback control system measures a process variable downstream of a piece of process equipment and then send information back upstream for the process equipment to manage the process variable directly. Similarly to feedback control, feedforward control only measures and alters one process variable, although the process equipment is downstream of the measuring equipment instead of being the other way round in feedback. Whereas cascade control measures a process variable then alters another to result in a change to the original measured process variable. Feedback control can be seen throughout the process P&ID, an example of the utilisation of feedback control can be seen in figure 6.1. The figure shows a set of pumps in parallel with feedback controlled valves based on the flow through the pipe. This has been implemented into the system with pump failure in mind, for example, if P-401 was the operational pump and P-402 was being used as a backup, both V-402 and V-404 would be closed to prevent any flow through P-402 under normal operating conditions. In the event of P-401 failing, V-403 and V-405 could be closed from the control room to prevent flow through P-401 while the pump
  • 66. 66 was either replaced or having maintenance carried out on it. Valves V-402 and V-404 could then be opened from the control room and P-402 would become the operational pump in this system. The ability to switch between pumps because of the control systems in place would lead to much less downtime required in the event of the operational pump failing in comparison to a system without these control measures in place. Figure [5.1]: Pumps P-401 and P-402 is parallel with flow control. This type of control measure ensures that the operational time of the plant is not reduced due to pump failure and therefore maintains a high plant economic efficiency. Cascade control can be seen in the instrumentation implemented around each of the distillation columns, more specifically around the condenser and reboiler. The temperature of the streams that are recycled from the condenser and reboiler will impair the columns performance if they are not as they should be, therefore it is important that the temperature is maintained in those streams. The cascade of control is used to control the temperature of the recycle streams by controlling the flow rate of cooling water and steam into the condenser and reboiler respectively. This system has been chosen as it is relatively easy to control the flow of cooling water and steam via the opening and closing of a valve when compared to directly controlling temperature. 5.3.1 Distillation column control Cascade control can be seen in the instrumentation implemented around each of the distillation columns, more specifically around the condenser and reboiler, as shown in Figure 5.2. The temperature of the streams that are recycled from the condenser and reboiler will impair the columns performance if they are not as they should be, therefore it is important that the temperature is maintained in those streams. The cascade of control is used to control the temperature of the recycle streams by controlling the flow rate of cooling water and steam into the condenser and reboiler respectively. This system has been chosen as it is relatively easy to control the flow of cooling water and steam via the opening and closing of a valve when compared to directly controlling temperature. This system of cascade control also allows for effective pressure control when coupled with the ability to control the flowrate of the vapour stream (Branan, , p78). Effective pressure control can be achieved through controlling the flowrate of the vapour leaving the column, i.e. less vapour leaving the column will
  • 67. 67 cause an increase in pressure throughout it. This flowrate of the vapour stream is controlled by the valves V- 308, V-408 and V-508 on the light ends, drying and heavy ends distillation columns respectively. Figure 5.2 shows V-308 on the light ends distillation column. The temperature control element of the system, as previously described, through the control of the process water flowrate, allows the control of the vapour stream composition and importantly, gives an operator the ability to keep the stream composition constant (Branan, , p78). DC-301 P-301 P-302 V-302 V-303 V-304 V-305 FI FIFI C FI C V-301 LIC LI V-306 TI FIC TI FIC V-307 FIFIC LAH LAL Process Water V-308 FIC FI V-308 F-201 DC-401 S-601 Figure [5.2]: P&ID of DC-301 and control systems
  • 68. 68 5.3.2 Reactor and Flash tank control Cascade control can also be seen on the system controlling the flow into, and out of, the flash tank and reactor respectively. Figure 5.3 describes the system around the reactor, R-201, and flash tank, F-201, as shown by the P&ID. The flowrate of the liquid leaving the reactor and therefore, entering the flash tank is controlled by the valve, V-201 which is in turn, is being controlled by the level in the reactor through the level instrumentation put in place. Similarly for the flask tank, the liquid flow out of the flask tank, which is being recycled back into the reactor, is controlled by the valve, V-206 which in turn, is being controlled by the level in the flash tank through the level instrumentation put in place. Valve V-203 is the pressure reduction valve required by the process, as it is a key piece of process equipment a backup valve, V-204, has been included for redundancy in the case of failure or in the event of maintenance being required on the operational valve. As V-203 is the valve used in normal operation, V-204 would usually be closed with V-202 and V-205 open to allow flow through V-203. Both valves V-202 and V-205 can be closed to isolate V-203 in the event of its failure and/or requirement for maintenance. R-201 V-210 V-209 V-208 V-207 FI FI FI C FI C V-201 LI FI C V-202 V-203 V-205 V-204 PIFI C F-201 V-206 FI C LI FIFI C LAH LAL S-601 DC-301 V-211 FI FI C R-201 Feed Figure [5.3]: P&ID of R-201 and F-201 and control systems
  • 69. 69 6. Process Economics 6.1 Market Analysis In the current economic climate, demand for acetic acid is great and room for continuous growth is highly probable. (Grand View Research Inc., ) predict that “Global Acetic Acid Market is expected to reach USD 13.31 billion by ” as a result of products manufactured from acetic acid, such as Vinyl acetate monomer (VAM) extensively growing in demand. Henceforth, the development of an acetic acid plant will be very profitable in today’s current market. Acetic acid is a key precursor in the production of various commercial and industrial based products. Products range from households such as vinegar, a multifunctional household product used in cooking and as an ingredient in renowned sauces like ketchup, mayonnaise and mustard; additionally it is also employed in cleaning products for the unclogging of drains and removal of lime scale deposits. Although acetic acid is a key element in the manufacture of numerous commodities, the most significant product is Vinyl acetate monomer (VAM). The versatility of VAM allows for the production of various polymers such as polyvinyl acetate, polyvinyl alcohol, polyvinyl butyral and polyvinyl formal; these polymers are employed within industry as paint, adhesives and packaging. Literature on VAM reportedly states that the worldwide demand for the monomer back in was reported to be 4.3 million tonnes per year, contributing to 86% of overall acetic acid production worldwide. The maximum global production was approximately 5 million tonnes per annum, key contributors being Asia (43%), North America (34%) and Western Europe (18%) (Chemical Week, ). An additional source (Nexant, ), highlights data that reports global production of VAM reaching approximately 4.9 million per year in . Furthermore, recent sources indicate further growth in VAM production (Merchant Research & Consulting Ltd, ) stating “, the overall VAM output climbed to around 6.5 million tonnes, registering growth of nearly 4% (259,000 tonnes)”. As previously mentioned, there is significant evidence in support of the global demand for VAM, as a result of its thriving growth in industries such as construction, automotive, paints and coating. Continuous global growth in infrastructure at extreme rates leads to further expansion of VAM production. Figure 6.1, illustrates pricing for VAM, Terephthalic acid (PTA) and Polyethylene terephthalate (PET) in /. It can be noted that the pricing history of VAM does not fluctuate. VAM price predominantly remains constant, as further observation highlights the stable market price of just above 50 throughout /. In comparison, the market price for other acetic acid commodities tend to be more fluctuating as observed by PTA, in which case a price spike is highlighted in November and an estimated 10% price increase in .
  • 70. 70 Figure [6.1]: Graph showing the market pricing of the largest acetic acid products (Hicks, ). Correspondingly, ester production also utilises acetic acid to develop acetate esters, commonly used as solvents for inks, paints and coatings. Projected growth for cellulose acetate is encouraging as key drivers such as growing cigarette consumption in Asia makes it a market for exploitation. According to (Strategyr, ), drivers such as surface coating are fast growing and its respective market growth increasing by up to 4.1% CAGR. In addition, further reports project to exceed production of 951,000 tons by (Strategyr, ). Acetic anhydride, another acetic acid based product, is correspondingly doing well in the global market. IHS stated that “Global production of acetic anhydride has been growing at an average annual rate of 4% since ” (Ihs, ). Ihs also describes that acetic acid production in the United States, Western Europe and Japan declined at average annual rates of 3%, 2% and 4% respectively (Ihs, ). Solvent glacial acetic acid is also due to witness major growth, as of right now the Asia-Pacific region is the biggest market of glacial acetic acid due to extensive growth in the automotive industry, especially China. Each year, millions of tons of acetic acid is produced and exported globally. Numerous literatures have portrayed a market analysis of acetic acid, thus allowing to support the economic viability of acetic acid production. The key drivers for its demand are mentioned above, as well as projection of acetic acid market growth where major players such as British Petroleum PLC (BP) (UK), BP PETRONAS Acetyls Sdn Bhd (Malaysia), Celanese Corporation (USA), China Petroleum & Chemical Corporation (Sinopec Corp.) (China), Daicel Chemical Industries Limited (Japan), Kyodo Sakusan Co. Ltd. (Japan), DuPont (USA), Eastman Chemical Company (USA), Jilin Chemical Industrial Company Limited (China), Kuraray Co., Ltd. (Japan), LyondellBasell Industries (The Netherlands), Millennium Chemicals Inc. (USA), Showa Denko K K (Japan), Sterling Chemicals Inc (USA) are facing tremendous benefits, thus it is possible to deduce that development of an acetic acid plant in this current market is viable and lucrative. 6.2 Costing The basis for costing is to give an indication of process feasibility. Various costs are to be considered in order to provide a full financial evaluation of this design project. The following costs will be accounted for this study, being, feedstock costs, ISBL costs, OSBL costs, engineering costs, capital costs, working capital, plant component costs and servicing costs. In addition, this costing section will provide an indication of the fixed capital investment required for the project. Fixed capital investment considers the inside battery
  • 71. 71 limits (ISBL) investment which portrays the cost of the plant and OSBL costs affiliated with construction, installation, commissioning and ground preparation. 6.2.1 Feedstock price estimation The reaction for the Monsanto process follows: ????????4 ???? + ???????? → ????2 ????4 ????2 Where, molecular weights are: CH4O = 32 kg/mol CO = 28 kg/mol C2H4O2 = 60 kg/mol The amount of methanol and carbon monoxide can then be derived through a stream analysis in Aspen. On the assumption that conversion is ≥ 99%, a 400,000 tons of acetic acid requires: ???????? → 264,794 ???????????????? ???????????? ???????????????? ???????????????? → 215,488 ???????????????? ???????????? ???????????????? ???????????????????? → .55 ???????????????? ???????????? ???????????????? Methanol required (t) Carbon Monoxide required (t) Rhodium iodide catalyst required (g) 5 Methanol cost per ton ($)(Chemical Market Reporter, ) USD 330.00 Carbon Monoxide price per ton ($) (Soliman et al., ) USD 50.00 Water required (tons) (Salisbury and Hallinan, ) .55 Cost of water per ton (Quandl, ) 0.39 Rhodium price (per g) 22. TOTAL METHANOL COST ($) USD 87,382,020.00 TOTAL CARBON MONOXIDE COST ($) USD 12,268,550.00 TOTAL RHODIUM CATALYST COST USD 112.70 TOTAL COST OF WATER USD .89 TOTAL COST OF FEED USD 99,654,314.89 Cost per ton of feed ($) USD 195.34 Table [6.1]: Cost of Feedstock.
  • 72. 72 6.2.2 Capital Cost Estimation 6.2.2.1 ISBL To calculate the total ISBL costs, the Factorial method is applied, this will provide a class 4 estimated with an accuracy of 30%. Estimation are made based on the major items required for the process and other costs estimated as factors of the equipment cost. Applying Lang Factors (Lang, ): ???? = ???? (∑ ????????) Where: C = Total plant ISBL capital cost (including engineering costs) ΣCe = Total delivered cost of all major equipment items F = an installation factor (Lang Factor) Thus, the cost of the major equipment are listed (corresponding cost include location and cost index for UK basis): Equipment Cost of Equipment, £ Material Factor Ce (incl Material Factor, £ Storage Tank 333,000.00 1.3 Reactor 756,412.00 1.55 Flash Drum 66,225.95 1.55 Light End column 542,826.88 1.55 Drying Column 363,770.96 1.3 Heavy End Column 145,002.00 1.3 Scrubber 25,000.00 1.55 2,232,237.79 3,250,000 Table [6.2]: Collated costs of equipment.
  • 73. 73 6.2.2.2 Installation Factor Installation factors then need to be accounted of, Table 6.3 contains a list of installation factors (Hand, ): Equipment Type Installation Factor Compressors 2.5 Distillation columns 4 Fired Heaters 2 Heat exchangers 3.5 Instruments 4 Miscellaneous Equipment 2.5 Pressure vessels 4 Pumps 4 Table [6.3]: Table of installation factors. Thus: ???? ∑ ????????= £82,500,000 6.2.2.3 OSBL The off-side costs (OSBL) are accountable for infrastructure, which includes laboratories for technical staff, offices, canteens and various mandatory staffing facilities. Considerations of water treatment systems, security warehousing and loading facilities and maintenance costs are also included within the OSBL. OSBL costs are estimated to be approximately 40% of the ISBL costs. Therefore: ???????????????? ???????????????????? = £33,000,000 (???????? ) 6.2.2.4 Engineering costs Engineering costs are incurred by hiring contractors, costs acquired from the design, procurement of plant equipment, supervising construction and service installations. In addition admin charges, insurance and contractors profit. The cost of engineering is 20% of the sum of ISBL and OSBL costs for a large acetic acid plant. Therefore: ???????????????????????????????????????????? ???????????????????? = £23,100,000 (???????? ) 6.2.2.5 Contingency costs The contingency cost are provisions for unexpected events that may cause financial consequences on the project such as labour disputes, price variations, FOREX Fluctuations, unforeseen weather conditions. The contingency cost for this acetic acid plant is estimated at 15% due to the use of well-known technology. Therefore: ???????????????????????????????????????????? ???????????????????? = £17,300,000
  • 74. 74 6.2.2.6 Fixed Capital Investment The capital cost for this project is the sum of ISBL, OSBL, Engineering costs and contingency costs. Therefore: ???????????????????????????? ???????????????? = £156,000,000 (???????? ) 6.2.3 Working Capital This accounts for operating expenses from the commencing of plant start-up and operations. For this project the working capital is 20% as product inventory is fairly large. Therefore: ???????????????????????????? ???????????????????????????? = £23,100,000 6.2.4 Total Investment The investment is total expense. It corresponds to the sum of fixed capital investment and working capital, Therefore: ???????????????????? ???????????????????????????????????????? = £179,000,000 (???????? ) 6.2.5 Operating expenditure OPEX for this process is assumed to be 40% of the FCI. Therefore: ???????????????? = £62,000,000 6.2.6 Revenue The annual revenue generated from the sale of acetic, Table: 6.4 illustrates the projected annual sales: TOTAL COST OF FEED (£) GBP 68,800,000 Selling price of acetic acid per lb ($) USD 0.38 Selling price of acetic acid per ton ($) USD 760.00 Total sale (revenue) ($) USD 304,000,000.00 Total Sale (revenue) (£) GBP 227,000,000 (Adapted to a temporal factor (1.06) and a location factor (1.02 for the UK) Table [6.4]: Table showing annual sales of acetic acid. 6.2.7 Gross Profit The gross profit margin is given by: ???????????????????? ???????????????????????? = ???????????????????????????? − ???????????????? ???????? ???????????????? − ???????????????? = £97,000,000 Therefore: ???????????????????? ???????????????????????? = 69.81% The margin percentage gives a good indication of how profitable the process is going to be, correspondingly a percentage of 69.81% is very economically feasible for an acetic acid plant.
  • 75. 75 6.3 Project Financing The financing of this industrial project is based upon its financial structure, exploring debt and equity will give an indication on the cash flow generated in order to finance the project. The project has a life span of 20 years, therefore to evaluate the annual present value of the project on a yearly basis, Net Present Value is calculated. The two sources of finance that are being considered is a bank loan and private investment. The equation used for this is: Discounting Factor = 1 (1 + ????) ???? Where: i = financial interest rate n = current year of project 6.3.1 Financing Bank Loan On the assumption that the Bank of England theoretically negotiate a 7.5% interest rate– which is reasonably near to the actual current values – the following table is produced: Table [6.5]: Table showing the NPV of the money over a twenty-year time period on the assumption that the discount rate is 7.5%. Year Cash flow (£) Discount Factor (1/1.075)^n Discounted cash flow (£) 0 -£179,107,337 1 -£179,107,337.40 1 £96,606,369.54 0. £89,866,390.27 2 £96,606,369.54 0. £83,596,642.11 3 £96,606,369.54 0. £77,764,318.25 4 £96,606,369.54 0. £72,338,900.69 5 £96,606,369.54 0. £67,292,000.65 6 £96,606,369.54 0. £62,597,209.90 7 £96,606,369.54 0. £58,229,962.70 8 £96,606,369.54 0. £54,167,407.16 9 £96,606,369.54 0. £50,388,285.73 10 £96,606,369.54 0. £46,872,823.94 11 £96,606,369.54 0. £43,602,626.92 12 £96,606,369.54 0. £40,560,583.18 13 £96,606,369.54 0. £37,730,775.05 14 £96,606,369.54 0. £35,098,395.40 15 £96,606,369.54 0. £32,649,670.14 16 £96,606,369.54 0. £30,371,786.17 17 £96,606,369.54 0. £28,252,824.35 18 £96,606,369.54 0. £26,281,697.07 19 £96,606,369.54 0. £24,448,090.30 20 £96,606,369.54 0. £22,742,409.58 Total NPV £805,745,462.16
  • 76. 76 The table above illustrates the credit flow in to the company over the twenty year period taking in to consideration the 7.5% interest rate from the bank of England. The NPV of the project of £806,000,000.00 indicates that on the basis that the project financing comes from a bank loan then project will be economically feasible. This value will now be compared to the one that could potentially be given by equity investors, assuming this time a 25% discount rate (=i, realistic for a well-established technology). 6.3.2 Financing Investments On the assumption that the private investment has a 25% interest rate the following table is produced: Year Cash flow (£) Discount Factor (1/1.25)^n Discounted cash flow (£) 0 -£179,107,337 1 -£179,107,337.40 1 96,606,369.54 0.8 £77,285,095.64 2 96,606,369.54 0.64 £61,828,076.51 3 96,606,369.54 0.512 £49,462,461.21 4 96,606,369.54 0. £39,569,968.97 5 96,606,369.54 0. £31,655,975.17 6 96,606,369.54 0. £25,324,780.14 7 96,606,369.54 0. £20,259,824.11 8 96,606,369.54 0. £16,207,859.29 9 96,606,369.54 0. £12,966,287.43 10 96,606,369.54 0. £10,373,029.94 11 96,606,369.54 0. £8,298,423.96 12 96,606,369.54 0. £6,638,739.16 13 96,606,369.54 0. £5,310,991.33 14 96,606,369.54 0. £4,248,793.07 15 96,606,369.54 0. £3,399,034.45 16 96,606,369.54 0. £2,719,227.56 17 96,606,369.54 0. £2,175,382.05 18 96,606,369.54 0. £1,740,305.64 19 96,606,369.54 0. £1,392,244.51 20 96,606,369.54 0. £1,113,795.61 Total NPV £202,862,958.34 Table [6.6]: Table showing the NPV of the money over a twenty-year time period on the assumption that the discount rate is 25%. The NPV on a private investment similarly indicates that the project is economically feasible on the basis of a 25% interest rate. The total is £203,000,000.00 still signifies lucrativeness of the project. Comparing the cumulative NPV of both sources, equity is very profitable assuming that the plant would reach optimal production after 12 months. However, the net present value is exponentially lower when private equity is required, therefore it is possible to conclude that securing a bank debt type of investment is a clearly more suitable choice.
  • 77. 77 6.3.3 Net Profit Gross Profit of the project is calculated by subtracting cost of feed from. In order to finally formulate the net profit, tax must be deducted from the gross value. In order to adjust the taxable value as accurately as possible, a depreciation allowance equation is used: Depreciation Allowance = FCI − SV N Where: FCI = fixed capital investment SV = scrap value N = number of years of operation An annual scrap value of 10% of ISBL investment will be assumed; the depreciation allowance obtained is therefore £7,400,000. This figure is used to determine the final net profit, with a corporation tax value taken as 21%. Year Gross Profit £) Depreciation allowance Taxable earnings (£) Payable tax (£) Net profit 0 -£179,107,337 7,387,146 NA NA -£179,107,337 1 £96,606,369.54 7,387,146 £89,219,224 £18,736,036.96 £77,870,332.59 2 £96,606,369.54 7,387,146 £89,219,224 £18,736,036.96 £77,870,332.59 3 £96,606,369.54 7,387,146 £89,219,224 £18,736,036.96 £77,870,332.59 4 £96,606,369.54 7,387,146 £89,219,224 £18,736,036.96 £77,870,332.59 5 £96,606,369.54 7,387,146 £89,219,224 £18,736,036.96 £77,870,332.59 6 £96,606,369.54 7,387,146 £89,219,224 £18,736,036.96 £77,870,332.59 7 £96,606,369.54 7,387,146 £89,219,224 £18,736,036.96 £77,870,332.59 8 £96,606,369.54 7,387,146 £89,219,224 £18,736,036.96 £77,870,332.59 9 £96,606,369.54 7,387,146 £89,219,224 £18,736,036.96 £77,870,332.59 10 £96,606,369.54 7,387,146 £89,219,224 £18,736,036.96 £77,870,332.59 11 £96,606,369.54 7,387,146 £89,219,224 £18,736,036.96 £77,870,332.59 12 £96,606,369.54 7,387,146 £89,219,224 £18,736,036.96 £77,870,332.59 13 £96,606,369.54 7,387,146 £89,219,224 £18,736,036.96 £77,870,332.59 14 £96,606,369.54 7,387,146 £89,219,224 £18,736,036.96 £77,870,332.59 15 £96,606,369.54 7,387,146 £89,219,224 £18,736,036.96 £77,870,332.59 16 £96,606,369.54 7,387,146 £89,219,224 £18,736,036.96 £77,870,332.59 17 £96,606,369.54 7,387,146 £89,219,224 £18,736,036.96 £77,870,332.59 18 £96,606,369.54 7,387,146 £89,219,224 £18,736,036.96 £77,870,332.59 19 £96,606,369.54 7,387,146 £89,219,224 £18,736,036.96 £77,870,332.59 20 £96,606,369.54 7,387,146 £89,219,224 £18,736,036.96 £77,870,332.59 TOTAL NET PROFIT £1,378,299,314.36 Table [6.7]: Table showing the net profit over a twenty-year time period.
  • 78. 78 The total net profit is approximately £1,381,000,000.00 over the 20 year period, this provides a clear representation of the profitability of this project through the Monsanto process. 6.3.4 Cumulative Cash Position An alternative way to outline the potential profit of the plant is to demonstrate the change in cash flow throughout the years. Assumptions for the application of the cumulative cash position of the project are: • For the first 3 years of operation, the plant is not going to reach optimum production. The revenue will be considered to be 25%, 50% and 75% of the maximum for years 1,2 and 3 respectively. • Also we assume that the plant will take 2 years to build, the capital investment will be spread out 30%, 50%, 20% in the 3 years of construction and operation respectively. • The cash flow is given by the deduction of the FCI and OPEX form the revenue. • ADA is the annual depreciation allowance, which has a different value each year. Year Cash Flow Cash cumulative ADA -2 -46,799,013.96 -46,799,013.96 0.00 -1 -77,998,356.61 -124,797,370.57 0.00 0 -162,359,505.20 -287,156,875.77 0.00 1 -74,218,529.53 -361,375,405.30 147,742,918.87 2 -17,276,896.51 -378,652,301.81 73,871,459.43 3 39,664,736.52 -338,987,565.29 49,247,639.62 4 96,606,369.54 -242,381,195.75 36,935,729.72 5 96,606,369.54 -145,774,826.20 29,548,583.77 6 96,606,369.54 -49,168,456.66 24,623,819.81 7 96,606,369.54 47,437,912.89 21,106,131.27 8 96,606,369.54 144,044,282.43 18,467,864.86 9 96,606,369.54 240,650,651.98 16,415,879.87 10 96,606,369.54 337,257,021.52 22,776,653.21 11 96,606,369.54 433,863,391.06 13,431,174.44 12 96,606,369.54 530,469,760.61 12,311,909.91 13 96,606,369.54 627,076,130.15 11,364,839.91 14 96,606,369.54 723,682,499.70 10,553,065.63 15 96,606,369.54 820,288,869.24 9,849,527.92 16 96,606,369.54 916,895,238.78 9,749,794.58 17 96,606,369.54 1,013,501,608.33 8,690,759.93
  • 79. 79 18 96,606,369.54 1,110,107,977.87 8,207,939.94 19 96,606,369.54 1,206,714,347.42 7,775,943.10 20 96,606,369.54 1,303,320,716.96 7,387,145.94 Table [6.8]: Table showing the net cumulative cash flow over a twenty-year time period. Figure [6.2]: A graph showing the net cumulative cash flow over a twenty-year time period. A graphical comparison of the Cumulative Cash is plotted against the time for the total life span of the project. ???????????? ???????????????? ???????????????? = ???????????????????? ???????????????????????????????????????? ???????????????????????????? ???????????????????????? ????????????ℎ ???????????????? = 2.75 ???????????????????? Where: Total investment = £179,000,000 Average annual cash flow = £65,000,000 6.3.5 Return of Investment Another important parameter to be considered when evaluating the project is the Return of Investment (ROI).Therefore: -£600,000,000.00 -£400,000,000.00 -£200,000,000.00 £0.00 £200,000,000.00 £400,000,000.00 £600,000,000.00 £800,000,000.00 £1,000,000,000.00 £1,200,000,000.00 £1,400,000,000.00 -5 0 5 10 15 20 25 CashFlow Time in years Graph of cumulative net cash flow at the end of the project life Cash cumulative ADA
  • 80. 80 ROI = ???????????? ???????????????????????? ???????????????????????? total investment × 100 = 53.9% Where: Net annual profit = £97,000,000 Total investment = £179,000,000 7. Process Safety Safety is a moral, and legal obligation that any chemical plant design should include. It is empirical that the correct levels of safety are undertaken in order to protect the health and safety of employees, any surrounding communities that might get affected by events that occur on site, the surrounding environment, and the process itself. The risks that may arise are a combination of undesired events due to multiple factors (e.g. human error, loss of containment personal injuries), which could consequently affect the integrity of the process and release hazardous components. In this case, safety risks refer to high intensity exposure (i.e. to toxic materials) over short periods of time. While environmental risks refer to low intensity exposure over long periods of time. Finally, any health hazards associated with the process are strongly related to any type of exposure to hazardous substances with short, or long-lasting effects. The above mentioned risks are strongly related to loss of containment. For the production of acetic acid loss prevention safety layers, and risk management ensure adequate response to any damage (e.g. spillage, faulty valve) caused, while minimizing response time, and ensuring an economical solution to any technical faults. The production of acetic acid through the Monsanto process is a highly efficient in terms of component selectivity. However it is highly hazardous process route due to factors such as: • High temperatures • Toxicity of components • High pressure • Single exposure toxicity • Corrosiveness of components • Flammability • Oxidising potential • Fire and Explosive hazards • Miscibility with air/water Table [7.1]: Process route associated risks. Therefore, an adequate process design needs to be compiled in order to reduce any risk associated, ideally eliminating it overall. 7.1 Safety legislations
  • 81. 81 It is mandatory that any new chemical plant design complies with the Control of Major Accident Hazards (COMAH) Regulations . The purpose of the COMAH regulations is to prevent any major accidents, and limit the consequences to people and the surrounding environment in the event of any accidents. The COMAH regulations are implemented in Great Britain, and implements the majority of the European Union Seveso III Directive . In order to determine whether the COMAH Regulations are applicable to the new acetic acid chemical plant a basic assessment was completed (COMAH Regulations, ). This classified the chemical plant at Tier 1 with an associated HT 1 (i.e. Hazard type). This means that Upper-tier duties apply (i.e. lower tier duties including regulations 8-13, 16, and 18). Regulations that have to be considered and implemented in the United Kingdom are: • Health and Safety at Work Act ; • Management of Health and Safety at Work Regulations ; • Environmental Permitting Regulations (England and Wales); • Civil Contingencies Act ; • Environmental risk tolerability . Safety provisions; • Preparing and planning for emergencies: the National Resilience Capabilities Programme ; • Chemicals (Hazard Information and Packaging for Supply) Regulations ; Regulations that belong to the European Union (EU), but have to be considered: • Council Directive 7.2 Hazard Identification Hazard identification refers to locating potential triggers such as the materials, the process, and plant lineament that will lead to the occurrence of an undesirable event. 7.2.1. Material Hazard This section will present the chemical hazards associated with the process components individually according to Regulation (EC) No /. Which will later be quantified in terms of toxicity, corrosiveness, and flammability in terms of individual equipment.
  • 82. 82 Acetic acid (C ≥ 90%) Skin corrosion Category 1A Flammability Category 3 Upper explosion limit 19.9 % (by volume) Lower explosion limit 4% (by volume Hazard Identification by sign Figure [7.1]: Acetic acid hazard sheet. Carbon Monoxide (Gas under pressure) Flammability (gas) Category 1 Acute toxicity (inhalation) Category 3 Reproductive toxicity Category 1A Repeated exposure toxicity Category 1 Upper explosion limit 74.2 % (V) Lower explosion limit 12.5% (V) Figure [7.2]: Carbon Monoxide hazard sheet. Methanol Flammability (liquid) Category 2 Acute toxicity (oral) Category 3 Acute toxicity (inhalation) Acute toxicity (dermal) Category 3 Category 3 Specific target organ toxicity (single exposure) Category 1 Upper explosion limit 74% (V) Lower explosion limit 12.5% (V) Figure [7.3]: Methanol hazard sheet.
  • 83. 83 Hydrogen Figure [7.4]: Hydrogen hazard sheet. Carbon Dioxide Gases under pressure Liquefied gas Figure [7.5]: Carbon Dioxide hazard sheet. Propionic acid Skin corrosion Category 1B Flammability (liquid) Category 3 Upper explosion limit 12.1% (V) Lower explosion limit 2.9% (V) Hazard Identification by sign Figure [7.6]: Propionic acid hazard sheet. Flammability (gas) Category 1 Gases under pressure Compressed gas Upper explosion limit 74.2 % (V) Lower explosion limit 4% (V)
  • 84. 84 Methyl Iodide Skin irritant Category 2 Carcinogenicity Category 2 Acute toxicity (oral) Category 3 Acute toxicity (inhalation) Category 3 Acute toxicity (dermal) Category 4 Specific target organ toxicity (single exposure) Category 3 Upper explosion limit 66% (V) Lower explosion limit 8.5% (V) Figure [7.7]: Methyl Iodide hazard sheet. Rhodium (powder) Flammability (solid) Category 1 Other hazards: ingestion, or inhalation of Rhodium particles. Figure [7.8]: Rhodium hazard sheet. Methyl acetate Flammability (liquid) Category 3 Eye irritant Category 3 Hazard Identification by sign Figure [7.9]: Methyl acetate hazard sheet. 7.2.2 Material Toxicity The toxicity of a component is based on the frequency of exposure and the inherent toxicity. The three toxic components of this system are: Carbon monoxide, Methanol, Methyl iodide. They have all been identified as a safety hazard for workers. Carbon monoxide affects internal organs after prolonged or repeated exposure. Methanol is skin and eye irritant, but over-exposure can lead to death. In case of exposure it is
  • 85. 85 advised to thoroughly wash the exposed area with water for a period of minimum 15 minutes, remove any contaminated clothing, and seek medical attention immediately. In case of severe inhalation, the person has to be moved to a safe area as soon as possible and be administered oxygen for breathing difficulties. In case of extreme situations, it is advised to perform mouth-to-mouth resuscitation, while keeping aware this could cause the toxic fumes to be inhaled. For methyl iodide, the emergency answer is the same as for methanol. Materials with carcinogenic proprieties such as methyl iodide, and methanol are hazardous even is small concentrations over longer exposure times. They can cause organ damage through prolonged and repeated exposure. In order to limit the occurrence of any side effects due to being exposed to these materials, it is advised to enforce safe exposure time, and adequate PPE in case of longer exposure is required. This can be quantified with regards to the minimum concentration of a hazardous material that would cause permanent health damage as a result of prolonged exposure (Table 8.2.) Also, adequate training have to be given to all the plant staff regarding different classes of toxicity and their effects on the human body. Material PPM Mg/m^3 Carbon Monoxide 50 55 Methanol 200 260 Carbon Dioxide Acetic acid 10 25 Methyl Iodide 5 28 Table [7.2]: The Permissible Exposure Limit (PEL). 7.2.3 Flammability The process components that are flammable are: CO, Hydrogen, Methanol, Acetic acid, Propionic acid, Methyl acetate, and Rhodium. The values regarding flammability can be seen in section 8.2.1. Material Hazards. It necessary to take all safety precautions regarding the handling, and management of these components in case of loss of containment. Methanol and carbon monoxide are materials with high flammability. Considering that these components are subjected to high temperature and high pressure in the reactor phase in high quantities, it is advised that no source of ignition, open flames, or electrical devices (e.g. telephones) are allowed on site in order to remove the likelihood of such scenarios. However, in the case of a ruptured vessel it is possible for the gaseous mixture to self-ignite from the static spark created. Safety precautions must be taken across all equipment containing any flammable liquids. These include adequate process control systems, tripping systems, as well as heat and fire detection alarms. It is advised that in case of open fire an alcohol-resistant foam is used. Water sprays are also recommended as a safety
  • 86. 86 precaution as some of the components reacting (e.g. methanol, acetic acid) are miscible with water. An important component for the safety system is the heat exchanger, since the capacity of the reactor does not allow efficient cooling using a water deluge system. 7.2.4 Corrosiveness Since material in the system have corrosive properties, it is necessary that all safety design considerations are taken. This includes an adequate choice of construction material, adequate choice of process equipment, as well as adequate maintenance procedures. The methyl iodide present in the system is highly corrosive, and therefore requires special considerations for use, and handling (HSE Onshore Pipeline Integrity Management, ).It therefore requires integrity management, and regular maintenance of process equipment in order to assess corrosion. 7.3 Operating conditions hazard 7.3.1 Pressure relief strategy One of the process variables that can lead to a domino effect along the whole system is the pressure. In the scenario of a vessel, or any equipment located in its proximity failing due to overpressure. The main pressure relief strategy is by installing a venting route that prevents any pressure accumulation. It is recommended that any emergency pressure relief system in place be used last, after all other systems failed, and that it is self-actuated (i.e. independent of any other components, or systems). The pressure relief mechanism set in place will function depending on how the volume fluctuates inside the vessel. When the volume materials going in increases, the pressure will subsequently increase. Possible causes for an increase in volume are: • Excess material flow; • Gas generation from the chemical reaction; • Thermal expansion of materials; • Vaporization of materials inside the vessel due to high temperature input; If any of the above variables is not between safe operating conditions, the pressure increase can lead to cracks in the vessel, loss of containment. In extreme scenarios where the pressure increase surpasses the maximum pressure capacity of the vessel, the vessel can be damaged beyond repair, or can lead to an explosion. In case of under-pressure that can be caused by removal of material or heat, a vacuum stage can develop and cause the vessel to collapse. In order to maximise the efficiency of the pressure relief system it has to be designed according to worst case scenario reaction conditions. This includes any potential changes in pressure, temperature, material, and composition. This will ensure that the pressure relief system is working between adequate parameters, and that it is sized according to the maximum output of the vessel. Safety precautions in case of high pressure include constant monitoring of the pressure levels, a pressure detector attached to an alarm to notify the plant control room of any unusual rise in pressure, vessel seals,
  • 87. 87 and regular maintenance check. In this case, the pressure detector will have to be programmed to the reaction conditions in order to account for regular changes. A relevant response plan has to be created in order to efficiently manage dangerous situations, as well as to safely evacuate all staff from the plant. 7.3.2 High pressure response measures The alarm set in place for this scenario represents one of the crucial components of the pressure vessel. This measure will allow the control team to adequately respond to the situation, and it will also notify all employees to evacuate the plant, or go to the nearest refuge point. It is recommended that the evacuation plan put in place accounts for the maximum number of people present on site at a time, in order to provide: • Sufficient safety equipment for staff and any visitors; • Adequate number of escape routes; • Sufficient fire exits; • A quick and safe evacuation; 7.3.3 Fire prevention strategy In order to effectively prevent fires on the plant it is necessary that any potential ignition sources are accounted for, and safely secured from producing flammable medium. A standard risk assessment is required for determining all potential sources of ignition, which will then be extended to the apparatus, as well as all mechanical equipment ignition risk assessment (MEIRA) and electrical equipment (Etchells, ). A prevention method required on all acetic acid plants is having machinery designed and constructed in a manner that avoids the creation of any potential ignition sources and overheating of the equipment due to the components used by set piece of equipment. Any electrical equipment present has to be used (e.g. installed, maintained, repaired) only by qualified electrical personal. The areas of concern in this situation are: • Hot surfaces • Any free flames, including hot work or repairs • Friction sparks • Electrical equipment • Static electricity • Lightning • Shockwaves from exothermic reactions It is crucial that the fire and safety strategy are completed by a competent authority. The equipment chosen for the chemical plant needs to be in conformity with the safety policy put in place (HSE Plant Modification, ). This refers to the manufacturer of the vessels, who has to take into account that:
  • 88. 88 • Adequate insulation that prevents the combustion of a potentially explosive medium; • Isolate equipment for potential explosions to ensure the safety of its surroundings; • Any electrical equipment that is installed is adequately adjusted to the plant component; Some other precautions include the presence of supplementary equipment which will act as a safety control equipment in the case of faulty primary equipment (Etchells, ). The employer is responsible for conducting an adequate risk assessment for explosive atmospheres. Some of the variables that need to be checked are: • The possibility of explosive medium formation, and their persistence; • The likelihood if ignition sources being created on site; • The equipment, the materials present and their characteristics; • The magnitude of possible hazardous scenarios; • Explosion risk assessment; • Classification into zones of hazardous areas; • All equipment is operated and maintained according to safety regulations; • The equipment is safe to use; 7.3.4 Fire and gas detection Since all equipment will operate at constant temperature, the control room will be immediately notified in case of any fires or explosions through fire, heat, and smoke detectors as part of the emergency response plan (Etchells, ). The smoke detector is required in this care, since a loss of containment would release toxic materials that could ignite from the heat source. The heat detector will be calibrated to trigger the alarm when an unusual amount of heat is present in the environment. All the equipment used on plant for identifying leakages should be regularly checked, maintained, and calibrated in order to adequately assess the integrity of the vessels (Dangerous maintenance, ). An example would be the use of a barometer for pressure measurement inside the reactor (with a separate alarm system), as any sudden pressure drop could cause the creation of a vacuum, and therefore a leakage of very flammable materials into the surrounding environment. 7.3.5 Noise It is recommended to take account for the noise resulted from equipment, as well as the reactor for employees, any surrounding communities, other industrial sites located in the proximity of the chemical plant, and the surrounding (HSE Noise at work, ). This health and safety hazard in this situation is being exposed to excessive noise across long periods of time, which can lead to permanent damage to hearing. Since the noise levels are not constant across the reactor, and surrounding equipment it is recommended that industrial sound meters be installed across the plant. In order to quantify the damage to the ear and select appropriate protection equipment, a filter network resembling the same response as the human can be installed adjacent to the noise meter (Controlling noise at work, ).
  • 89. 89 7.3.6 Loss of containment Loss of containment can be caused by multiple factors, and the gravity of the situation is mainly determined by what materials are released into the atmosphere (HSE Emergency response/spill control, ). Some of the possible originators that lead to loss of containment are: • Poor maintenance procedures that are not adequately applied. Some of them are related to improper purging, drainage, isolation, and seal checks that result in the release of toxic, and/or flammable gasses. In this situation it is recommended that the area underneath the process vessels can contain minor, as well as major spillages so that the spillage does not spread; • Pressure relief systems; • Unusual operating conditions that can cause a leakage in the vessel; 7.4 Emergency Response plans 7.4.1 Fire Emergency response plans for any fire need to account for the different classes of fire hazard. It is dependent on the type of material that caused the fire, as well as the magnitude of the fire. An appropriate response to possible fire hazards and the appropriate response can be seen in Table 7.10. Ignition fuel/medium Flammability category Response measure Carbon Monoxide 1 • Dry chemical Hydrogen Gas 1 • Dry chemical Methanol 2 • Foam • Carbon Dioxide • Dry chemical Acetic acid 3 • Foam • Carbon Dioxide • Dry chemical Propionic acid 3 • Foam • Carbon Dioxide • Dry chemical Rhodium catalyst 1 • Dry materials that can inhibit the metal fire (e.g. dry sand) Table [7.10]: Response measure to fire hazards (HSE Emergency response/spill control, )
  • 90. 90 Type of fire inhibitor Effect on fire Foam • To remove heat and oxygen from fire Dry chemical • to remove the chemical reaction and prevent oxidation Carbon dioxide • To remove oxygen from fire Table [7.11]: Description of fire inhibitor actions. It is essential that a fire source is put out as fast as possible to prevent any further complications. This requires that employees have an adequate response this scenario. In order to improve the speed of reaction of employees it is crucial that the plant undergoes practice drills as a response to the fire alarm. Safety protocols dictate that when a fire alarm goes off, all the employees have to head towards the nearest fire escape. All safety areas, and safe evacuation routes have to be identified and adequately marked. The employer has to make sure that all the employees have undergone the fire safety drill, and know how to adequately react. After the fire assembly point is reached, a roll will be taken to check if everybody left the hazardous area. The sprinkler system put in place will limit the spread of the fire until the local fire brigade arrive; it is necessary that the local fire brigade is notified of the current situation, and that they undergo training drills prior to any accidents so that they are adequately prepared for all potential risks present on site. 7.4.2 Explosion In the highly unlikely case of an explosion, the system should be immediately shut down and evacuated. If the situation allows it, all streams should be diverted to the flare in order to minimise the release of any toxic, or flammable compounds in the atmosphere as secondary explosions are possible. Any explosions happening on site can have catastrophic consequences for the plant, people, and the surrounding environment (HSE Fire and explosion, ). In the event of an explosion for this particular chemical plant, there should be set in place and emergency plan that involves the local fire brigade, and any businesses and people living in the radius of the plant. In such cases, local authorities have to be notified of the accident in order to minimise any exposure of citizens to the toxic fumes released in the atmosphere; if necessary local households and businesses have to be evacuated due to the hazardous chemicals involved. A specialized team will later analyse the aftermath, and ensure it is safe. 7.4.3 Overpressure
  • 91. 91 It is necessary that in the unlikely case of a vessel being over pressured, all employees evacuate the plant to an assembly point located at a safe distance from the hazardous area. Local authorities will then intervene, and ensure the safety of the plant. 7.4.4 Toxic release Alarms will be located across the plant to detect any toxic releases in the atmosphere. The area where the toxic material was released has to be evacuated immediately in order to avoid any unlikely scenarios, or injuries. The immediate response put in place for this case should: • Ensure adequate management of the toxic release and limit any further release; • Ensure gas masks, or breathing apparatus is accessible to employees; • Evacuate all the staff to a predetermined assembly point safe from the toxic release; • Contact a specialised team (if available) to help in the management of toxic releases; 7.4.5 Flooding In the unlikely case of a flooding due to heavy rain, or rise in river levels a safety strategy needs to be put in place in order to prevent, or minimise any damage to the plant. In the event of a severe weather warning, a trained, competent safety officer needs to be made available on-call in order to ensure that the plant is adequately prepared. His role will be to ensure a temporary flood defence is set up around the plant in the shortest time. In the case of a flash flood, all equipment should be deactivated unless a safety trip is already set in place; all the personal has to be evacuated. Competent authorities should be called to ensure safety of the plant (HSE Emergency response/spill control, ). 7.4.6 Earthquakes In the event of an earthquake, all employees need to be trained how to react effectively to ensure personal safety. If employees are indoor, they are advised to not use the elevator, and stay inside the building; stand against an interior wall, or take refuge under a table. Employees must stay as far as possible from any heavy machinery, glass, or exterior walls (HSE Emergency response/spill control, ). There is a high possibility of major leakages of toxic/ or flammable materials occurring after an earthquake. In all cases, it is necessary to turn off major equipment to prevent any complications. It is required that a skilled team check all the equipment thoroughly, and ensure any faulty equipment is fixed/removed on spot if the circumstances allow it (HSE Maintenance procedures, ). 7.4.7 Human factor
  • 92. 92 Adequate training needs to be imposed to all employees by the employer as a safety layer, to ensure there is no human error caused by insufficient training, poor documentation, and inadequate supervision. There is always the chance of process operators, or other employees making mistakes. However, the magnitude of events caused by insufficient training is much bigger than events caused by simple human errors. This is one of the most important aspects of process safety on a chemical plant, as human error can have consequences ranging from personal injury/death (i.e. inadequate reaction to hazardous situations) to major equipment failures (HSE Human errors, ). 7.4.8 Personal Protection equipment It is mandatory that after a HAZOP study has been completed, adequate PPE is chosen for the chemical plant. The PPE and equipment need to be adapted so that safe work on the chemical site is possible, as well as emergency protection equipment. It is the duty of the employer to supply the equipment, and to adequately train staff. However it is the employees responsibility to use it, and know how to use it for their, and the plants’ safety. 7.5 HAZOP The purpose of this HAZOP study is to analyse the hazardous potential of equipment on and off-site. The HAZOP is conducted to ensure that any hazardous situations are identified, and that the right safeguards are in place to prevent any unwanted scenarios from occurring; this step ensures that the plant is operable. Developing a HAZOP is important for effective plant safety considerations. The current HAZOP had been conducted on only one node, and serves as an indication of the methodology. It is a legal requirements that a full HAZOP is conducted before a chemical plant is built in order to adequately quantify all the hazards and risk present on site. It is recommended that HAZOP studies are carried across the lifetime of the plant, for each individual zone. This helps take account of any process deviations, and account for them in the current design. Usually a HAZOP study can be conducted by itself, as it is a qualitative technique. However, it is recommended that the below assessment techniques are used in conjunction with the HAZOP: • H.A.Z.I.D. – Hazard identification system • P.H.A. – Preliminary Hazard Analysis • L.O.P.A. – Layers of protection Analysis • S.M.S. – Safety Management System • E.R.P. – Emergency Response Planning It is recommended that a Fault Tree Analysis is conducted on site with the HAZOP, to maximise the number of safety considerations taken in case of a hazardous scenario. It is important that the FTA is conducted, as it can detect system failures before they take place. It is a reliable quantitative assessment as it takes account of maintenance procedures such as rate of failure, failure probability, or repair rate (HSE Plant Modification, ).
  • 93. 93 7.5.1 Scope of work In the HAZOP training, the whole team completed an assessment of the node between the reactor and the flash tank. This sections was chosen as it has the highest potential to cause hazards to the process due to the high temperature and pressure values. Even though the P&ID presents all the nodes, this design project only accounted for the hazards and its effects on the node between the reactor and the flash tank. The reason to why the HAZOP is done in this stage of the design project, is to ensure that the design is safe, and that any hazardous situations are accounted for by adequate process control. It generates recommendations for plant operability, operating instructions, and can offer a better insight into how certain design alterations can minimize the costs before the construction of the plant. 7.5.2 Term of Reference Prior to conducting the HAZOP, the design team reviewed all the completed P&IDs; the critical parameters reviewed were characteristic to the node in question. The study includes all design parameters characteristic to the node, including equipment, and pipework specifications. 7.5.3 Team Membership Everybody from the design team was present during the HAZOP study, and included: Cristian Baleca Chairman Sinthujan Pushpakaran Scribe Hisham Albaroudi Team member Karen Atayi Team member Enoch Osae Team member Alexander Taylor Team member Parameter Guide words Flow No More Less Reverse Temperature More Less Pressure More Less Viscosity More Less Corrosion Yes
  • 94. 94 Table [7.12] Parameters and guide words used in HAZOP. The study was conducted by analysing how every parameter could be affected under the conditions described by the guide words for the specified node. The whole team contributed to the development of the HAZOP in order to seek solutions for the any hazards associated, and what safety barriers should be put in place in order to achieve a safe and efficient process route. The design team fulfilled the purpose of the operability study, and thus gained a better insight on what actions need to be taken across the whole plant for each individual design. It needs to be noted that the team debated any recommendations resulted from the study in order to better differentiate self-evident information that serves as obligatory guidance, and forwarded the other outcomes to be considered for individual designs. The HAZOP study generated relevant design considerations, and is an important part of any safety report for any chemical plant. 7.5.4 Safety conclusions After a careful consideration of all the hazards present on the plant, the present report shows efficient ways of maintaining a safe environment for the production of acetic acid. The present design, even though safe presents a hazardous potential that can only be diminished by an adequate safety management system that has to be put in place for every process design individually. The above section presented all the safeguards put in place for accident prevention, as well as several indications on process and plant safety that should be implemented. Considering the nature of the hazardous components (toxic, flammable, carcinogenic, corrosive) and the process conditions, it is recommended that all safety considerations are applied in order to improve the overall safety of the people on site. As any other chemical plant, it is necessary that the people working on site comprehend the nature of the hazardous process. It is an employer’s obligation to ensure that the site is safe to work, yet in the chemical industry it is every bodies responsibility to ensure a safe working environment. Safety is an important aspect of a design. Many regulations are put in place in order to minimize any potential accidents, but it is a design engineers responsibility to ensure that all safety aspects of plant safety had been considered; a safe environment is a collective responsibility, and must always be regarded as so when dealing with hazardous large scale industrial applications. Vibration More
  • 95. 95 7.5.5 Marked up P&ID Figure [7.10]: Marked up P&ID to show the nodes that were analysed during the HAZOP.
  • 96. 96 7.5.6 HAZOP findings Hazard and Operability Study:- Node No. and Title Reactor to Flash Tank node Node Design Intent Node 1 – Line between reactor and flash tank. Transfers liquid material into the flash tank for catalysis recovery and pressure reduction. Meeting Dates. Team Members. Alexander Taylor, Cristian Baleca, Enoch Osae, Hisham Albaroudi, Karen Atayi, Sinthujan Pushpakaran P&ID Nos. P&ID 1 Major Equipment Nos. R101 and FT - 301 Plant Line Nos. Materials of Construction Hastelloy - Piping Node start/end points Reactor / Flash Tank Normal Process Contents. Acetic acid, methanol, methyl acetate, methyl iodide, water, propionic acid, ethanol and traces of other organic compounds Flowrate(s) .4 kmol/hr Normal Process Temperature 160 oC Normal Process Pressure 30 Barg Design Temperature 189 oC Design Pressure 30 Barg
  • 97. 97 Line No. Parameter Guide Word Possible causes Consequences Safeguards Action No. Recommendations 1 FLOW FLOW MORE MORE Greater fluid density Control faults Damage to valves Overflow Flow indicator and Pressure indicator Flow Indicator and isolation valve 1 2 Not applicable Not applicable 1 FLOW FLOW FLOW NO NO NO Blockage Pipe failure Incorrectly fitted equipment Damage to Flash tank, no product produced Loss of containment Loss of containment, lockages, equipment failure Isolation valve, flow indicator Emergency shutdown Flow indicator 3 4 5 Flow indicator in proximity of Flash Tank Design consideration of piping with more corrosion allowance Appropriate training of personnel 1 FLOW FLOW LESS LESS Valve failure Loss of integrity Increase in pressure in the upstream valve Isolation valve Pressure indicator and flow indicator Pressure indicator and flow indicator 6 7 Not applicable Not applicable
  • 98. 98 Line No. Parameter Guide Word Possible causes Consequences Safeguards Action No. Recommendations 1 PRESSURE MORE Control valve failure Loss of integrity Back-up valve within isolation system 8 High pressure alarm linked to pressure indicator 1 PRESSURE LOW Loss of integrity Reduced pressure upstream of leak Emergency shutdown 9 High pressure alarm linked to pressure indicator Line No. Parameter Guide Word Possible causes Consequences Safeguards Action No. Recommendations 1 TEMPERATURE TEMPERATURE HIGH HIGH Fire situation Pressure reduction valve failure Damage to equipment Explosion High temperature upstream Isolation valve, emergency shutdown Emergency services Isolation valve 10 11 12 Not applicable Emergency isolation control valve Temperature indicator after valve 1 TEMPERATURE LOW LOW Faulty valve Ambient condition Loss of containment Not applicable Isolation valve Isolation valve 13 14 Low temperature alarm Low temperature alarm
  • 99. 99 Line No. Parameter Guide Word Possible causes Consequences Safeguards Action No. Recommendations 1 VISCOSITY HIGH Pressure relief valve failure Affects flow Isolation valve 15 Low temperature alarm 1 VISCOSITY LOW Pressure relief valve failure Affects flow Isolation valve 16 Low temperature alarm Line No. Parameter Guide Word Possible causes Consequences Safeguards Action No. Recommendations 1 CORROSION YES Use overtime Loss of integrity Corrosion allowance 17 Check pipeline, plant layout around node, design consideration 1 VIBRATION HIGH High velocity Failure None 18 Check pipeline, plant layout around node, design consideration Table [7.13]: HAZOP table
  • 100. 100 8. Environmental Protection Process plants are customarily subject to the release of unwanted chemicals and this project is no different. The client demands for the design of an efficient and economically viable project whilst taking into consideration environmental protection. As a result, focus on environmental protection in this design project is to ensure that only the minimal amount of hazardous wastes are produced, and where possible, all emissions and waste is neutralised before disposal. Measures to ensure acceptable and minimal disposal of wastes and emissions whilst maintaining energy requirements have been integrated throughout the design process. 8.1 Process Selection A general environmental consideration to be made on the chosen methanol carbonylation process is the lower energy requirement in comparison to its predecessor as a result of its milder operating conditions. The cobalt based process initially developed by BASF for the production of acetic acid in the late ’s required operating conditions of 700 atm and 250 o C to obtain commercially viable products (Cheng and Kung, ). The selectivity of the raw materials, methanol and carbon monoxide, respectively being 99% and 90% also pose another positive aspect due to minimal generation of by-products and waste. An optimised version of the Monsanto process, CativaTM , has been developed and scaled-up accordingly through investigation of an alternative catalyst; this route has managed to successfully reduce water requirement in order to stabilize the catalyst in the reaction, thus effectively reducing the number of by- products formed and purification stages. Although, this route is highly efficient, the process has been patented and unavailable to be implemented in this project, henceforth the Monsanto route has been chosen as it represents an optimal compromise between environmental requirements and license availability. 8.2 Plant Location – Environmental Considerations The chosen location for the plant is on an existing chemical complex in the United Kingdom of Great Britain and Northern Ireland on a land owned by the company, thus remediation and ground stabilization is not required. The selected area is designed to host chemical industries hence no impacts will be applicable to any areas of particular environmental sensitivity (SSSI) or the skyline. In addition, other environmental issues such as odour, noise, vibration, radiation and traffic are also negligible for the purpose of this environmental assessment.
  • 101. 101 8.3 Noise Noise is a type of pollution that is present on the chemical plant. As a result of its location, the plant will have minimal impact in regards to noise compared to if it was situated nearby a residential area. Sources of noise pollution include operations of large machinery (pumps, flash tank), on-site transporting of raw materials and final product, use of vehicles on site. In order to tackle issues faced by noise pollution, appropriate noise reduction methods are implemented. Noise reduction methods are more effective when implementation is executed in the design process rather than during operational procedures. Earlier the implementation stage, higher effectiveness of the feature. Good common practice to minimize such impacts involves integration of unit operations with noise reducing devices or choosing unit operations with guaranteed lower noise emission, locate buildings as far as applicably possible from the plant, designing and construction of buildings with adequate soundproofing features, employment of solid panel fencing over traditional wire fencing in order to confine noise within the site, implementation of a constant monitoring system to keep track of noise generated and limitation in the transport of raw materials and other noise generating activities during the day (Erskine and Brunt, ). 8.4 Odour Fumes and odours are natural products of chemical plants and their emissions must be monitored and reduced as low as possible. Main sources of odours within the projected chemical plant consist of effluents (aqueous discharge), waste gases (flare) and venting of methanol/acetic acid during feed/product dispatch and settlement. Likewise noise pollution, new and efficient methods to control industrial odours have been developed. Odour masking will be the technique applied in this project. It entails the elimination of odour perception by imposing a more pleasant fragrance to “mask” the ones generated by the process. These are often synthetically originated and do not alter the composition or presence of odours produced in the system. Advantages of this implementation are smooth implementation, little or no capital investment involved and high flexibility (Bergen, ). 8.5 Traffic The plant will be situated near the Humber River, therefore it can be stipulated that majority of transportation of raw materials and final product will be achieved through shipment methods instead road transport. This method of transport allows minimization of road traffic with annexed pollution and noise. Nonetheless, traffic will occur as a result of the vehicles present on site, thus transport of raw materials will only be occurring throughout daylight shifts in order to minimize the impacts.
  • 102. 102 8.6 Catalyst and Water Requirement The reaction necessitates a Rhodium catalyst with an iodide promoter which tend to unite into insoluble salts such as RhI3. In order to prevent this occurrence, percentage content of water is kept relatively high (approximately 10% w/w of combined reactants) to successfully promote the cycle. Such high usage of water is environmentally unfavourable as an additional column is required to eliminate the water, thus increasing demand for energy utility hence reducing overall sustainability of the process. Nonetheless, the water intake is fully recycled from the drying column back into the reactor and able to be reused in support of the catalyst’s action. Rhodium is continuously recycled throughout the process, thus when considering that approximately 2 kg is required in this reaction, its environmental impact is negligible assuming total recycling and life span of 1 year. Figure [8.1]: The carbonylation rates for iridium and rhodium processes depending on water concentration taken under conditions of ~ 30 % w/w methyl acetate,8.4 % w/w methyl iodide, 28 barg total pressure and 190ºC (Jones, ). 8.7 Methanol Feed An environmental consideration regarding feedstock and main reactant, ethanol, is investigated upon. Although normally obtained from Syngas (produced from oil), methanol can also be generated through biomass sources such as wood, waste and sewage. This implication may eventually lead to independence from oil, creating a more sustainable and ‘green’ process in the long term.
  • 103. 103 8.8 Energy Recovery The nature of the Rhodium-Iodide methanol carbonylation process is exothermic, meaning that the reaction releases energy under the form of heat. In an economical and environmental perspective, it is desirable to recover this energy either internally to use within the system or externally. An appropriate heat integration system has therefore been implemented in order to manage energy dissipation, thus promoting environmental awareness. Application of such heat is yet to be finalized as options on whether to use it a source of heat for streams or to any of the industrial units found within the complex is still being debated, thus economic implication will ultimately be the relevant factor to be considered. 8.9 Storage & Handling of Raw Materials and Product Environmental considerations need to be taken into account in regards to storage and handling of raw material procedures. 8.9.1 Carbon Monoxide Carbon Monoxide (CO) is delivered to the plant from the supplier through a 1 km long pipeline. Absence of a storage system eliminates risks associated with loss of containment and contamination of the ecosystem, although a rupture in the pipe line would still pose a negative impact to the environment. Henceforth, measures that can be utilised to avoid such occurrence is the combination of relief valves application and appropriate flow control systems in conjunction with suitable preventative maintenance protocol, aimed at sampling and leak detection, that shut down the flow of CO if necessary. As a result, in order to minimize the risk and impact of such event, process control and instrumentation will be executed. In case of a detected loss of containment, installation of an isolation valve would temporarily stop flow of CO until the leakage has been fixed. 8.9.2 Methanol Methanol feedstock is stored in proximity upon sea freight delivery from supplier in large silos located 500m away from the plant. Due to the flammable nature of methanol, vessels will be stored off-plant and required to be at a minimum safe distance in order to comply HSE regulations. However, the continuous reaction requires methanol feedstock to be in proximity of the reactor so that it is delivered at a reasonably high rate. Methanol is a corrosive substance, and any spillage and consequent release into soil and ground can extensively damage the ecosystem and surrounding areas. For this reason, a design implementation is integrated to the silos in order to reduce the impact of a mechanical failure, which comprises of: • A containment chamber which collects all content of the silos • Drainage system through a tap, leading streams to a safe location where impact on people and environment is minimized • Implementation aimed at preventing emissions and odours of volatile organic compounds • Pressure relief valves
  • 104. 104 In order to achieve this implication, a smaller feed tank connected to the silos is implemented. In accordance to regulations, the volume found in this tank is far smaller and continuous feed of inert nitrogen allows the vessel to hold an optimum pressure level. Furthermore, generation of oxygen is avoided, thus inhibiting the flammability of methanol. In addition, a pressure relief valve control system is integrated on the feed tank as a safety measure. Methanol is pumped to the reactor through a pipeline system on a continuous basis. The portrayed system represents the optimum compromise between efficiency and environmental protection precautions. 8.9.3 Acetic Acid The final product, acetic acid, is stored in appropriate storage tanks, which is the final step before product is collected into chemical tankers suitable for shipment and delivered to customers. As previously mentioned, acetic acid is highly corrosive in high concentrations (99.9%) thus with the potential of damaging soil and ecosystem in case of spillages. Preventative measures are therefore put in place to minimize risk of such occurrence. The storage tank design has been developed in accordance: • Integration of a containment bund which aims to contain the tank’s content in case of mechanical failure of the vessel other than any additional expected rainfall. Other than its containment purpose, this implementation is expected to lower potential evaporation rate. • Enclosed area is drained through the use of a tap onto a safe location in order to comply with regulations aiming at minimizing impacts imposed on the environment. • These implementations are effectively aimed at preventing emissions and odours of volatile organic compounds. • Pressure relief emergency valves (designed in relation to fire management). A pumping system connected to a pipeline is in place to transfer the product from the storage tank onto the chemical tanker. This system highlights a measure taken to minimize contact with the ecosystem. Environmental measures to be applied to pipeline and leakage monitoring will be identical to those implemented for the delivery of methanol. The chosen location of the plant (Immingham) will greatly affect means of transportation as the reliability of shipments through the Humber route allows elimination of any noise, traffic, pollution from road cars. 8.10 Undesired products: By- and Co- products Specific design implementations and features have been put in place to comply with the Environment Protection Act () to achieve minimal waste and emissions generated throughout the plant. The prohibition on unauthorized or harmful depositing, treatment or disposal of waste and collection, disposal or treatment of controlled waste sections of the Environmental Protection Act () are regulations to particularly comply with. An analysis of the process has emphasized that application of Rhodium iodide catalyst will generate a range of by-products who possess high and low boiling points. As a result, elimination is not an option, yet
  • 105. 105 optimization of the system is instead possible to control and minimize these parameters. The catalytic cycle operated by the Rhodium Iodide catalyst consists of 6 elementary reactions which generate some undesired products. 8.10.1 Propionic Acid Required purity of acetic acid is stipulated to be 99.9%, according to specification, thus it is possible to state that 0.1% of the final product consists of co- and by- products. The scenario would be different within the system, henceforth appropriate recycling streams and unit operations have been put in place to ensure continuous treatment of waste generation. The major liquid by-product of the reaction is propionic acid, originated by carbonylation of ethanol. Propionic acid is produced out of the reactor at 1% scale of propionic to acetic acid and successfully removed in the heavy end column. Propionic acid is not hazardous, but rather a contamination component of the final product. Handling and contact of the acid in atmosphere is not perilous to the ecosystem, although collection from the heavy end column and appropriate storage will take place. Furthermore, it is an established intermediate product for the manufacture of a variety of chemicals such as esters, pesticides and pharmaceuticals, thus turning into an asset able to be sold externally. Its commercial use would be a bonus to the process and outweigh the CAPEX and OPEX for its removal. 8.10.2 Carbon Dioxide and Hydrogen Carbon dioxide and Hydrogen are additional by-products generated by the water-gas shift reaction. Within the system, vapour phase of the product, which comprises of CO, H2, CO2 and methyl iodide (the latter requiring vessels and piping fabricated out of Hastelloy in order to tackle its corrosion levels) migrates towards the scrubber where it undergoes an absorption reaction with a mixture of acetic acid and methanol. Methyl iodide is allowed to be recovered and recycled back into the reactor whilst acetic acid and methanol purge are stripped and recycled back into the system, thus eliminating losses. Meanwhile, CO2 and H2 are disposed by directing them towards the flare. Flares are environmentally friendly systems put in place to burn excess gases that cannot be recovered in the system that allows to prevent direct release of vapours to the atmosphere. In accordance to the plant and systems, the method portrayed indicates a sustainable route for the disposition of these gases. 8.10.3 Methyl Iodide Methyl iodide is one of the most dangerous by-products originated in the process, exerting a number of negative impacts on both human health and the environment. A potential emission of this component into the ecosystem can potentially reach the soil and contaminate water streams, hence irreversibly affecting the natural ecosystem. Due to its established properties as a neurotoxin, the side effects resulting from its exposure range from nausea, diarrhoea, loss of sight, cancer (on a long-term exposure) and other neuronal impacts, thus minimization and elimination of methyl iodide release is an essential implementation (Erickson, ). In the system, methyl iodide is generated in the vapour stream of R-101 going into the
  • 106. 106 scrubber where it is stripped with some acetic acid and methanol mixture and recycled back into the reactor on a continuous basis. An effective monitoring and control system must be implemented on the reactor and scrubber in order to continuously monitor and detected any leakage, this includes sensors, transmitters and controllers. A correct operation of the control and instrumentation system will allow constant monitoring, hence in the scenario of a loss in containment, the protocol would be isolate unit and pipeline where leakage occurred in order to prevent further flow, thus further loss of containment and effectuate an emergency shutdown of the unit operation. 8.10.4 Aqueous and Organic Discharges As of all industrial processes, unit of operations found within the plant will produce streams of aqueous discharge throughout its operation and maintenance, thus a suitable treatment strategy will be implemented to tackle these effluents waste. The discharges will converge to one station, thus a suitable canal system is to be implemented to facilitate collection. To avoid contamination with acetic acid and/or methanol in the feed, the stream will be subjected to neutralisation before it is allowed to undergo further treatment for inertness in the eco-system. The stream should ultimately abide environmental regulations as pH levels, toxicity and temperature of the streams are set to an allowance level before it is disposed of in the ecosystem. At this stage, parameters of stream are continuously monitored, subsequently allowed to pass through a pipeline into a basin, which content will be delivered to a local plant for further treatment, such as neutralization and precipitation, and charged at the rate of Mogden formula.
  • 107. 107 9. Plant Layout and Location 9.1 Plant Location The recommended location for this plant should be in North East England (Grimsby, South Humberside). Immingham has several chemical engineering companies situated around the Humber shore line including the likes of Philips 66, Total Oil Refinery, Inter terminals etc. Many companies locate here in order to take advantage of the vast supply of water used for cooling purposes and the availability of cheap land. Figure [9.1]: Site location Grimsby North East England (Google maps, ) The feedstock needed for this operation are methanol (stored offside and transported via pipe from nearby plant and connected by a line) and carbon monoxide. Air Products is a major gas supplier located in Grimsby south Humberside, approximately 1 minute drive (1 km) from the chosen site location. Air Products will provide the carbon monoxide gas needed to transport the carbon monoxide. The pipeline should be 1 km long and carbon monoxide should directly feed the reactor. The short distance reduces fuel costs and labour hours to transport the carbon monoxide. The shorter the distance travelled, the less likely an accident will occur during the delivery of methanol, travelling with hazardous chemicals over great distances is perilous as it increases the likelihood of road vehicles being involved in road accidents, hence placing the public at risk. Methanol is contained off site in a large storage tank and a fed line allows the pumping of small amounts to an onsite storage tank due to greater risks imposed by storage of high quantities. The associated British ports Immingham enables access for importing and exporting materials to and from Europe to expand market size/consumers and in turn increase company profits. The port allows transportation of raw materials via ship which may be more cost effective whilst holding a larger capacity in comparison to transportation via road. A rail road running from the port to the site location allows deliveries via ferry to the site.
  • 108. 108 Figure [9.2]: Humber Estuary schematic. The chosen location for the site is 6.3 miles away from a reliable power supply, South Humber bank power station. It will be situated close to the Humber Estuary and middle drain for easy access to water for cooling. In the north east of the site, a number of universities are located around, such as the University of Hull, Lincoln University, University of York and University of Leeds. As a result, a reliable supply of talented potential engineers have the opportunity to be offered graduate schemes, summer placements and apprenticeships in order to fulfil future employment necessities. The wind rose diagram shown in Figure 9.3 show that the wind direction in Immingham is predominantly towards south east throughout the year. Although south east winds blow towards the local community as discussed in Section 8, the smell of fumes released from the site are concealed via odour masking methods.
  • 109. 109 Figure [9.3]: Wind rose diagram for Grimsby (Meteoblue, ) 9.2 Plant Layout Upon preliminary study for plant layout, the primary infrastructures to be taken consideration of are the process units in order to allow a smooth flow of production from one location to another. Starting from the left hand side, raw materials are delivered in via trucks and stored in a tank farm. Moreover, carbon monoxide pipeline concludes at the top left hand corner of the plant, in proximity of the reactor, in order to limit the time spent in the transportation of hazardous materials across the site floor, thus reducing risks of spillages, vehicle accidents, exposure to workers, etc. There are two security booths. One is located at the entrance of the car park found at the south end of the site which purpose is to routinely check that all vehicles entering the site belong to site personnel. The implementation of such security prevents threats from any external sources, some potential hazards may be sources of ignition, bombs, etc. The second security booth is located on the east side of the plant and its purpose is to check whether delivery trucks are transporting hazardous raw materials, however these checks are more extensive depending on the size of the vehicle and the contents carried. The fire station is parallel to the process units, as a result of the high probable chance of fire propagating from this zone. Fire services are situated nearby thus reducing time wasted on fire
  • 110. 110 trucks usage, avoiding catastrophe and keep personnel safe. The faster fire services can get to the hazard points, the less damage is inflicted on the site and increase in personnel protection. The storage tanks containing hazardous materials such as acetic acid must be located at least 70 m away from the site boundary (Sinnott et al., ). The storage tanks are adjacent to the main plant area, therefore a short pipeline is needed to feed acetic acid into its relative storage tank. Prevailing wind direction determine the position of the tank farm and storage tanks. For this reason, these are best placed on Humber estuary in order to avoid vapours from drifting back the plant and local community. Emergency water storage units are located on the north side of the site for easy access of the Humber estuary water. In addition, water storage is located near the process units in order to provide quick access cooling if needed. The shaded area around the process units indicated future room for expansion in the scenario that the plant was to be modified and expanded. On the west side of the site, the delivery region consisting of finishing and packaging, vehicle deliveries and rail way link to the Immingham port for European exports is present. The finishing and packaging is opposite the acetic acid storage tank to avoid movement of product for long distances, thus increasing risks towards personnel and opportunity for spillages. The control room must be located opposite the process unit area. It is located near the evacuation zone and canteen, thus operators are not faced with travel issues regarding usage of facilities. The utilities room will be located opposite the process unit area to ensure efficient transfer energy in order to reduce heat dissipation. Office, laboratories and canteen are to be situated on the South east of the site, this proves to be a safe distance from the hazardous storage tanks and process units. It is nearby the car parks, thus further reducing walking distances for personnel. The maintenance office is located near the process units so that staff do not need to carry tools around the plant. All manual valves, sample points and instruments are located at reachable heights and positions for operators to carry out maintenance. In the process unit area, sufficient headroom for easy access to equipment will be provided (Sinnott et al., ). The base of all three distillation columns and vertical vessels such as the reactor and flash drum must be elevated for sufficient NSPH to the pumps. Methanol and carbon monoxide pipe lines are elevated on site to allow trucks to pass underneath. The trade off is large pressure drop, however having functional pipes running throughout the site floor is dangerous for large moving vehicles as chances of accidents occurring increases, as potential pipe leaks can provide source of ignition .
  • 111. 111 Modular construction will be used to assemble the process unit section. Only in recent years, more sites have increasingly started to assemble sections of the plant at the manufacturing site, these include main structural steel equipment, piping and instrumentation. These are then transported to site by ferry or road (Sinnott et al., ). This method provides many advantages such as reduced construction costs, improved quality control, less skilled labour on site required. Figure 9.4 shows a layout of the plant.
  • 112. 112 9.2.1 Site Flow Plan Figure [9.4]: Plant Layout
  • 113. 113 APPENDIX [A] – Minutes Meeting Week 13 – 9th May Time: 13:15 Venue: Chemical Engineering Room 125 Chair Person: Hisham Albaroudi Secretary: Sinthujan Pushpakaran Present: Alexander Taylor, Cristian Baleca, Enoch Osae, Hisham Albaroudi, Karen Atayi, Sinthujan Pushpakaran In Attendance: Aidan Hurley 1. No apologies. 2. Hisham – Finished environmental section and about to begin design of reactor after background research. Alex – Finished unit design and has made progress on process control. Enoch – Done majority of unit design and about to begin process economics. Cristian – Positive progress regarding unit design and about to begin process safety. Karen – Had issues with calculations of unit design but now sorted. About to begin Plant layout section. 3. PFD - Fix tanks and use appropriate material balance sheet. P&ID – Fix arrangements. Material Balance – Complete. 4. Arranged an informal deadline to have a complete report ready for draft submission by the 16th of May. Send CAPEX and OPEX by Wednesday 11th of May to Enoch. 5. Informal meeting scheduled for the 16th of May at 11:00. 6. Nothing to be added. Meeting closed at 13:53
  • 114. 114 Meeting Week 12 – 3rd May Time: 11:15 Venue: Chemical Engineering Room 125 Chair Person: Cristian Baleca Secretary: Karen Atayi Present: Alexander Taylor, Cristian Baleca, Enoch Osae, Hisham Albaroudi, Karen Atayi, Sinthujan Pushpakaran In Attendance: Aidan Hurley 1. Apologies accepted, Hisham absent. 2. Still working on individual unit designs, deadline extended. 3. Individual section to be completed by the end of this week 4. Individual section to be completed by the end of this week 5. Unit description should show what did you find and calculations to be set as appendices. 6. Informal meeting scheduled for Friday 6th of May at 10:00 in Library Group Learning room 7. Informal meeting scheduled for Friday 6th of May at 10:00 in Library Group Learning room 8. Follow layout for unit design using Dropbox document. 9. Formal meeting scheduled for 9th of May at 13:15 in the Chemical Engineering Building. Meeting closed at 11.53
  • 115. 115 Meeting Week 11 – 22nd April Time: 13:15 Venue: Chemical Engineering Room 125 Chair Person: Karen Atayi Secretary: Cristian Baleca Present: Alexander Taylor, Cristian Baleca, Enoch Osae, Hisham Albaroudi, Karen Atayi, Sinthujan Pushpakaran In Attendance: Aidan Hurley 1. No apologies. 2. HAZOP development was discussed. 3. Individual P&ID development still in progress 4. Delays occurring, but individual design is progressing. 5. General report layout has been chosen. 6. Aidan advised to avoid complicating the simulation. 7. Environmental effect discussed. Topics on how to improve energy efficiency, waste treatment and how to avoid environmental impact were mentioned. 8. Formal meeting scheduled for Tuesday 3rd May at 13:15 in the Chemical Engineering Building. 9. Informal meeting scheduled for Friday 22nd April at 14:15 in the Chemical Engineering Building. Meeting closed at 14.05
  • 116. 116 Meeting Week 10 – 18th April Time: 15:15 Venue: Chemical Engineering Room 125 Chair Person: Enoch Osae Secretary: Karen Atayi Present: Alexander Taylor, Cristian Baleca, Enoch Osae, Hisham Albaroudi, Karen Atayi, Sinthujan Pushpakaran In Attendance: Aidan Hurley 10. No apologies. 11. Need to know parameters for line between reactor and flash drum for HAZOP meeting, such as pressure, temperature, content, material of construction and volume. 12. Unit design should be approximately be about 4-6 pages for each person. It should include introduction, what techniques have been used for the design, what each person has come up with, how it affects the design, a mechanical design and update on cost. 13. Unit design should be approximately be about 4-6 pages for each person. It should include introduction, what techniques have been used for the design, what each person has come up with, how it affects the design, a mechanical design and update on cost. 14. Informal deadlines for unit design and individual sections has been set for 1st May and 9th May respectively. 15. Complete P&ID to be ready on A3 for 25th April . 16. Complete P&ID to be ready on A3 for 25th April . 17. Formal meeting scheduled for Friday 22nd April at 13:15 in the Chemical Engineering Building. 18. Informal meeting scheduled for Friday 22nd April at 14:15 in the Chemical Engineering Building. Meeting closed at 15.44
  • 117. 117 Meeting Week 9 – 11th April Time: 15:15 Venue: Chemical Engineering Room 125 Chair Person: Alexander Taylor Secretary: Enoch Osae Present: Alexander Taylor, Cristian Baleca, Enoch Osae, Hisham Albaroudi, Karen Atayi, Sinthujan Pushpakaran In Attendance: Aidan Hurley 1. No apologies. 2. Minutes were accepted 3. Hisham advises everyone to use Aidan’s lecture notes to complete unit design. 4. Alex said Contents page has been developed. 5. P&ID is required from the Reactor to the flash drum and everybody should have made a start on their report section. 6. Goals for next meeting: - Complete P&ID 7. Any other business: - Aidan mentioned HAZOP components require the hazard (MSDS) data sheet, so all potential hazards can be listed. - Aidan to be absent week commencing on the 25th April . 8. Leader for HAZOP workshop – Cristian Baleca Scribe for HAZOP workshop – Sinthujan Pushpakaran 9. Informal meeting scheduled for Thursday 14th April 10. Formal meetings schedules for Monday 18th April and Friday 22nd April respectively. Meeting closed at 15.51
  • 118. 118 Meeting Week 8 – 4th April Time: 15:15 Venue: Chemical Engineering Room 125 Chair Person: Sinthujan Pushpakaran Secretary: Alexander Taylor Present: Alexander Taylor, Cristian Baleca, Enoch Osae, Hisham Albaroudi, Karen Atayi, Sinthujan Pushpakaran In Attendance: Aidan Hurley 1. No apologies. 2. Use of Aidan’s advice to improve the simulation, don’t necessarily have to have to everything on one simulation but values need to be consistent. 3. Apply changes in the simulation to the PFD. 4. Work on the design of each of our unit operations. 5. Work on the formal report. 6. Informal meeting scheduled for Tuesday 5th April at 14:15. 7. Formal meeting scheduled for Wednesday 11th April at 13:15 in the Chemical Engineering Building. 8. Goals for next meeting: - Finalise simulation and PFD. - Research into the design of unit operations. - Look into the structure of the report. 9. Nothing to be added. Meeting closed at 15.45
  • 119. 119 Meeting Week 7 – 7th March Time: 15:15 Venue: Chemical Engineering Room 125 Chair Person: Hisham Albaroudi Secretary: Sinthujan Pushpakaran Present: Alexander Taylor, Cristian Baleca, Enoch Osae, Hisham Albaroudi, Karen Atayi, Sinthujan Pushpakaran In Attendance: Aidan Hurley 1. No apologies. 2. Simulation has produced good purity. Client specified that CO, Methyl iodide and Hydrogen iodide should be coming out. Further work in the simulation to be carried out to improve purity. 3. Client proposed an idea to fit a scrubber onto the P&ID. Carry out several P&IDs for each component, including the majority of the details. 4. Informal meeting scheduled for Tuesday 8th March with further informal meetings to be scheduled after. 5. Formal meeting scheduled for Monday 4th April in the Chemical Engineering Building. 6. P&ID agreed to be completed by Friday 11th March. 7. Design of unis to each member has been assigned: Hisham Albaroudi – Reactor Cristian Baleca – Scrubber Alexander Taylor – Column Enoch Osae – Column Karen Atayi – Flash tank Sinthujan Pushpakaran – Storage tank 8. No change in action tracker. 9. Nothing to be added. Meeting closed at 15.48
  • 120. 120 Meeting Week 6 – 29h February Time: 15:15 Venue: Chemical Engineering Room 125 Chair Person: Cristian Baleca Secretary: Hisham Albaroudi Present: Alexander Taylor, Cristian Baleca, Enoch Osae, Hisham Albaroudi, Karen Atayi, Sinthujan Pushpakaran In Attendance: Aidan Hurley 1. No apologies. 2. Everyone agreed split groups and are working towards target. 3. P&ID and simulation to be completed by 7th March and 10th March respectively. 4. Work on simulation needs to be faster. Cooperation between AT, CB and HA. Reactor’s results are not as desired. Not all side reaction’s compounds are available on Aspen Plus. Only main reactions and by-products to be used. 5. P&ID agreed to be completed by Monday 7th March . 6. Informal meeting scheduled for Thursday 3rd March at 12:15. 7. Formal meeting scheduled for Monday 7th March at 15:15 in the Chemical Engineering Building. 8. P&ID to be completed by next informal meeting. Find kinetic values for remaining reaction 9. AT, CB and HA to research on kinetics of the process. 10. Nothing to be added. Meeting closed at 15.38
  • 121. 121 Meeting Week 5 – 22nd February Time: 15:15 Venue: Chemical Engineering Room 125 Chair Person: Karen Atayi Secretary: Cristian Baleca Present: Alexander Taylor, Cristian Baleca, Enoch Osae, Hisham Albaroudi, Karen Atayi, Sinthujan Pushpakaran In Attendance: Aidan Hurley 1. No apologies. 2. Hisham, Alex and Cristian working on simulation. 3. Enoch, Karen and Sinthujan working on MSDS (Material Datasheet). 4. Given until Week 7 to finish P&ID. 5. Informal meeting scheduled for Tuesday 23rd February at 10:00. 6. Look into design options and how to better run the process. 7. Water is required as a separate stream, so consider recycling. 8. Get rid of Carbon Dioxide from system, spurge for P&ID or system will have losses. 9. Consider degrees of freedom for individual phases (Columns). 10. Formal meeting schedules for Monday 29th of February . Meeting closed at 15.38
  • 122. 122 Meeting Week 4 – 18th February Time: 12:15 Venue: Chemical Engineering Room 125 Chair Person: Enoch Osae Secretary: Karen Atayi Present: Alexander Taylor, Cristian Baleca, Enoch Osae, Hisham Albaroudi, Karen Atayi, Sinthujan Pushpakaran In Attendance: Aidan Hurley 1. No apologies. 2. Minutes from last meeting accepted. 3. Alex has made progress on flowsheet. 4. Hisham is past reactor stage on ASPEN. 5. Informal meeting scheduled for Friday 19th February at 13:00 in the Chemical Engineering Building to discuss simulation and mass balance. 6. To have informal meeting after Chemical Reaction Engineering lecture. 7. Formal meeting scheduled for Monday 22nd February at 15:15 Meeting closed at 12:31
  • 123. 123 Meeting Week 3 – 8th February Time: 15:15 Venue: Chemical Engineering Room 125 Chair Person: Alexander Taylor Secretary: Enoch Osae Present: Alexander Taylor, Cristian Baleca, Enoch Osae, Hisham Albaroudi, Karen Atayi, Sinthujan Pushpakaran In Attendance: Aidan Hurley 1. No apologies. 2. Minutes from last meeting accepted. 3. Alex discussed progress on general mass balance and everybody has approved work provided. 4. Hisham provided a general flow sheet in which everyone has approved it. 5. A standard format for documents has been set e.g. Agenda and minutes. 6. Enoch developed a document to use as an action tracker and everybody approved it. 7. Goals for the next meeting: 8. Meeting amendment due to the BCCA event on Monday. 9. Informal meeting scheduled for Thursday 18th February at 11:15 in RB 207. 10. Formal meeting scheduled for Thursday 18th February at 12:15 in the Chemical Engineering Building. 11. Cristian to get kinetic values for HYSYS. Meeting closed at 15:45
  • 124. 124 Meeting Week 2– 5th December Time: 15:15 Venue: Chemical Engineering Room 125 Chair Person: Sinthujan Pushpakaran Secretary: Alexander Taylor Present: Alexander Taylor, Cristian Baleca, Enoch Osae, Hisham Albaroudi, Karen Atayi, Sinthujan Pushpakaran In Attendance: Aidan Hurley 1. Time and Date of formal meeting confirmed – Formal meetings to be held on Mondays at 15:15 in Chemical Engineering Room 125, confirmed by Aidan. 2. Rota for the position of Chair Person and Secretary agreed on by following table: 3. A dropbox group to be set up by Hisham for file sharing. 4. Goals for next meeting: 5. Continue to work on a mass balance for the process, work on Aspen HYSYS to give an insight into the process and to make progress on a process flow sheet. Meeting closed at 15:50 Date Chair Person Secretary 05/02/ Sinthujan Alexander 08/02/ Alexander Enoch 15/02/ Enoch Karen 22/02/ Karen Cristian 29/02/ Cristian Hisham 07/02/ Hisham Sinthujan 04/02/ Sinthujan Alexander 11/02/ Alexander Enoch 18/02/ Enoch Karen 25/02/ Karen Cristian 04/02/ Cristian Hisham 09/02/ Hisham Sinthujan
  • 125. 125 Meeting Week 1 – 30th November Time: 11:15 Venue: Chemical Engineering Room 125 Chair Person: Hisham Albaroudi Secretary: Sinthujan Pushpakaran Present: Alexander Taylor, Cristian Baleca, Enoch Osae, Hisham Albaroudi, Karen Atayi, Sinthujan Pushpakaran In Attendance: Aidan Hurley 1. Plant process decided – The Monsanto process was agreed to be the most feasible process for the production of acetic acid. Further look into catalysts required as they are possibly cheaper and more efficient. 2. Area of responsibility delegated: 6. Cristian Baleca (Safety) 7. Karen Atayi (Plant layout and location) 8. Alexander Taylor (Process control and instrumentation) 9. Enoch Osae (Process Economics) 10. Hisham Albaroudi (Environmental Protection) 11. Sinthujan Pushpakaran (Report and Presentation Editor) 3. Agreed regular external meetings for Thursdays at 11:15, chair of the week will book places accordingly. 4. Future meetings with client to be decided after talking with Group 6 leader. 5. Goals for next meeting: 12. Research on mass balance by next meeting. Individual review of mass balance to be finalized. Meeting closed at 11:52
  • 126. 126 APPENDIX [B] – Calculations for Reactor Reactor volume and sizing The reactor volume is a vital parameter and it is a function of the rate of reaction, inlet flowrate and conversion as represented by the following equation: ???? = ????????0 ???? ???? −???????? The volume constraint can also be rearranged in relation to the rate coefficient (k) in the following formula (Schneider et al., ): ???? = ???? ???? ???????????? ???? ???????????????? 2 (1 − ????)2 = ???? ???? ???? ????????????????(1 − ????)2 Where: Vo = volumetric flowrate of methanol = 574.105 l/min (Methanol) and l/min (CO) = 9.56 l/s + .2 l/s = .76 l/s (Data obtained from simulation) X = conversion rate; XMethanol = 0.99; XCO = 0.90; XRXN = 0.95 k = 0.11 L/mol-1 s-1 (Iwase et al., ) CAo = Caverageinitial = ???? ???? = 2.47 mole/l (Data obtained from simulation) Therefore, substituting into equation, leads to: ???? = 110,740 ???????????????????????? = 110.74 ????3 Allowing a 20% overdesign as specified by the design brief: ???? = 110.74 × 1.2 = 138.42 ????3 Applying: ???? = ( ????????2 4 ) × ℎ Where: d = diameter of the reactor h =height of the reactor It is assumed that height of reactor is twice its diameter h = 2d (Sinnott et al., , p986)
  • 127. 127 Therefore: ???? = ( ????????2 4 ) × 2???? Thus, upon rearranging: ???? = 4.45 ???? ℎ = 8.9 ???? Ellipsoidal head’s dimensions Applying: ???????????????????????? = ( ???? × ????3 24 ) + ( ???? × ????2 × ????. ???? 4 ) Where: D = internal diameter = mm S.F. = straight face = assumed to be (3 × ????ℎ???????????????????????????? ???????? ????ℎ???? ℎ????????????) = 60 mm Substituting into equation: ???????????????????????? = ( ???? × 24 ) + ( ???? × × 60 4 ) = ????????3 = 12.96 ????3 Due to the fact that ellipsoidal heads are defined as “2:1 heads”, the inside depth of the head (excluding S.F.) is 2 times the radius of the dish, therefore: ???????????????????????? ????????????ℎ ????????????????ℎ = ???? 4 = 4 = .5 ???????? = 1.13 ????
  • 128. 128 Thickness of shell Applying (Sinnott et al., , p986): ???? = ???????? ???????? 2 ∗ ???????? − 1.2???????? ⁄ + C Where: t = wall thickness, mm Pi = internal pressure, bar = 33 bar Di = internal diameter, mm = mm S = maximum allowable stress, kg/cm2 = .59 kg/cm2 (Alloys and Producer, ) C = corrosion allowance = 4 mm Substituting into equation: ???? = 33 × [(2 × .59 × 0.85) − 1.2 × (33)]⁄ + 4 = 28.25 ???????? Therefore, external diameter: ???? ???? = ???????? + 2???? = 4.45 + (2 × 28.25 × 10−3 ) = 4.51 ???? Thickness of vessel’s head Ellipsoidal head is selected due to the operating pressure being less than 150 psi, thus applying (Sinnot et al., ): Th = ???????? ???? ???? ???? 2 ∗ ???????? − 0.2???????? ⁄ Where: P = pressure, bar = 33 bar D = diameter of outside shell, mm = mm S = allowable stress of material = .59 kg/cm2 E = joint efficiency = 0.85
  • 129. 129 Figure [B-1]: Values for Radius Factor K (Edge, ). Therefore: K = ????( ???? ???? 2ℎ ???? ) = 0.81 (Graph extrapolation) ho = h + nt h = D/4 = 4.51/4 = 1.12 m nt = Ts – CA = (28.25 – 4) mm = 24.25 mm = 24.25 × 10-3 ho = 1.12 + (24.25 × 10-3 ) = 1.14 m 2ho = (2 × 1.14) = 2.28 m Do = D – 2nt = [4.51 – (2 × 24.25 × 10-3 ) = 4.46 m Therefore, ( ???? ???? 2ℎ ???? ): = ( 4.46 2.28 ) = 1.95 Therefore, K: ???? = ???? ( ???? ???? 2ℎ ???? ) = 0.88 Substituting back into thickness equation: Th = 33 × × 0.84 [(2 × .59 × 0.88) − (0.2 × 33)]⁄ = 19.80 ???????? ????ℎ = 19.80 + 4 = 23.80 ????????
  • 130. 130 Applying corrosion allowance, therefore: ????ℎ = 19.80 + 4 = 23.80 ???????? Total height Applying: ℎ + 2????ℎ = [8.9 + (2 × 23.80 × 10−3)] = 8.95 ???? Calculation of stresses Longitudinal and circumferential stresses Applying: σ ???? = ???????????? 4???? = ( 33 × 4 × 28.25 × 10−3 ) = .56 ???????????? Applying: σℎ = ???????????? 2???? = ( 33 × 2 × 28.25 × 10−3 ) = .12 ???????????? Pressure and axial stress Applying: σ ???? = [ ???????????? 4(???????? − ????????) ] = ( 33 × 4 × 24.25 × 10−3 ) = .9 ????????????
  • 131. 131 Total weight of the shell Density of ℎ???????????????????????????????? − ???? = kg/m3 (Alloys and Producer, ) Applying, dead-weight formula (Sinnott et al., ): ???????? = ???? ???? × ???? × ???? × ???? × (???????? + 0.8????)???? × 10−3 Where: Cv = a factor to account for the weight of nozzles, manways, internal supports, etc.. which is taken as 1.08 for vessels with only a few internal fittings (Sinnott et al., ) ρ = density of hastelloy b-2 = kg/m3 (Alloys and Producer, ) D = mean diameter of vessel = 4.51 m Hv = height of the vessel = 8.94 m t = thickness of the shell = 24.25 mm Substituting into equation: ???????? = 429,306 ???? Maximum bending moment Applying (Sinnott et al., ): ???? = ???????? ???? ???? Where: Fp = W = 429,306 N Hp = 8.95 m Therefore: ???? = 3,842,288 ???????? = 38.42 ???????????? Dead-weight stress Applying (Sinnott et al., ): σ ???? = ???????? ????(???????? + ????)???? Where: W = weight of the vessel = 429,306 N Di = 4.45 m t =thickness including corrosion allowance = 28.25 × 10-3
  • 132. 132 Substituting into equation: σ ???? = 1,271,151.57 ????/????2 = 12.71 ???????????? Bending stress Applying (Sinnott et al., ): σ ???? = ???? ???? ???? ( ???????? 2 + ????) Where: M = 3,842,288 Nm lv = ???? 64 (???? ???? 4 − ???????? 4 ) = 0.42 (Sinnott et al., ) Do = 4.51 m (Sinnott et al., ) Di = 4.45 m (Sinnott et al., ) Substituting into equation: σ ???? = 8,230,729 ????/????2 = 82.30 ???????????? Principal stresses Applying (Sinnott et al., ): ????1= 1 2 [ σℎ + ????z + √σℎ − σ ????)2 + 4 ] ????1= 1 2 [ σℎ + ????z - √σℎ − σ ????)2 + 4 ] ????3 = 0.5 P Where: Total longitudinal stress ????z = ????L + ????W ± ????B = negligible (Sinnott et al., ) Therefore, ????z: (.9 + 12.71 + 82.30) = .91 ????????????
  • 133. 133 Substituting into equations: ????1 = 1 2 [.12 + .91 + √(.12 − .91)2 = .12 ???????????? ????2 = 1 2 [.12 + .91 − √(.12 − .91)2 = .91 ???????????? ????3 = (0.5 × 33) = 16.5 ???????????? (???????????????????????????????????????? ???????? ????ℎ???????? ????????????????????????????) Allowable stress intensity The allowable stress check is a technique to evaluate if the mechanical design of the vessel (especially wall thickness) can withstand the principal stresses exerted on the unit operation; the following three equations are used (Sinnot et al., ): (????1 − ????2) = 990.21 ???????????? (????1 − ????3) = .62 ???????????? (????2 − ????3) = .41 ???????????? The greatest obtained value is therefore taken and compared to the maximum allowable stress of the material (Hastelloy – B). If the value is lower than the allowable stress of the material, the allowable stress intensity is within the range (Alloys and Producer, ). ???????????????????????????????? ???????????????????????????????????? ???????????????????????? (???????????????????????????????????? ????) = 51 ???????????? = .33 ???????????? ???? > > (????1 − ????3) → ???????????????????????????????????? ???????????????????????? ???????? ???????????????? ????????????ℎ???????? ????ℎ???? ???????????????????????????????????????? ???????????????????? (Sinnott et al., )
  • 134. 134 Piping sizing Pipeline Flowrate (m3 /s) Velocity (m/s) Cross sectional area (m2 ) Required diameter (m) CO feed to reactor 0. 2.00 0. 0. Methanol feed to reactor 0. 2.00 0. 0. Water feed to reactor 0. 2.00 0. 0. Reactor to flash tank 0. 2.00 0. 0. Reactor to scrubber 0. 2.00 0. 0. Drying column to reactor 0. 2.00 0. 0. Scrubber to reactor 0. 2.00 0. 0. Table [B1]: Data obtained from Aspen Plus simulation for pipe sizing. Where: Flowrate (m3 /s) = Obtained from stream summary table (Aspen Plus simulation) and divided by (60 × ) to convert from l/min to m3 /s. Velocity (m/s) = A constant value of 2 m/s for determination of piping characteristics is suggested (Perry, ) Therefore: ???????????????????? − ???????????????????????????????????? ???????????????? = ???????????????????????????????? ???????????????????????????????? Thus, diameter can be calculated upon: ???????????????????????????????? = √ 4 × ???????????????????? − ???????????????????????????????????? ???????????????? ????
  • 135. 135 Therefore diameter can be evaluated using: ???? = ???? × ????3 16 Therefore, diameter: ???? = 5.28 ???????? = 0.053 ???? Start-up Assuming that 99% is the target conversion to reach ‘steady state’ conditions, the equation to determine the required time to reach such condition reduces to: ???????????????? = 4.6???????? 1 + ???????????? Where: Ts = ???????? ???????? ???? = 96,900 2.47 = 39,227 mol VFluid = 96.89 m3 = 96,900 dm3 CAO = 2.47 mol/dm3 K = × 10−6 (L/mol.h) = 9.45 × 10−6 (dm3 /mol.s) (Golhosseini et al., ) Therefore: ???????????????? = 4.6 × 1 + (9.45 × 10−6 × ) = ???? = 36.57 ℎ???????? Heat dissipation and insulation The heat dissipation is calculated based upon Fourier’s Law by: ???? = ℎ???????????? ???????????????????????????????? = ???????? ???????? ???????? Where: Q = heat transfer = W K = thermal conductivity of material = 10 (W/m K) (Alloys and Producer, ) A = surface area of vessel = 2????????ℎ Where: r = 2.25 m h = 8.95 m Therefore, surface area of vessel: ???????????????????????????? ???????????????? = 2 × ???? × 2.25 × 8.95
  • 136. 136 = 126.74 ????2 Change in temperature calculated by: ???????? = ???????????????????????????????????? − ???????????????????????????????? Where: TINTERNAL = temperature within the reactor = 160 o C TAMBIENT = ambient temperature = assumed to be 20 o C Therefore: ???????? = 160 − 20 = 140 °???? = 413 ???? ds is the thickness of shell, which is: ???????? = 0.028 ???? Therefore, heat transfer equals to: ???? = ℎ???????????? ???????????????????????????????? = (10 × 126.74 × 413 0.028 ) = 18,684,150 ???? = 67,301,345 ????????/ℎ???? Start-up Assuming that 99% is the target conversion to reach ‘steady state’ conditions, the equation to determine the required time to reach such condition reduces to: ???????????????? = 4.6???????? 1 + ???????????? Where: Ts = ???????? ???????? ???? = 96,900 2.47 = 39,227 mol VFluid = 96.89 m3 = 96,900 dm3 CAO = 2.47 mol/dm3 K = × 10−6 (L/mol.h) = 9.45 × 10−6 (dm3 /mol.s) (Golhosseini et al., )
  • 137. 137 Therefore: ???????????????? = 4.6 × 1 + (9.45 × 10−6 × ) = ???? = 36.57 ℎ???????? Nozzle diameter = 15% Diameter for the shell’s nozzle in relation to the diameter of pipe. Nozzle to flange distance = Value obtained as twice the nozzle diameter. Agitator It is acknowledged that the following calculations to determine the agitator’s sizes and characteristic are based on calculations obtained from (Pietranski, ). Ratios obtained from (Pietranski, ): ????1 = ???????????????????????? ???????????????????????????????? ???????????????????????????????? ???????????????????????????????? = 3 Therefore: 4.45 ???? = 3 ???? = 1.48 ???? Thus: ????2 = ℎ????????????ℎ???? ???????? ???????????????????????????????? ???????????????????? ???????????????????????? ???????????????????????????????? ???????? ???????????????????????????????? = 1 Therefore, height: ℎ????????????ℎ???? = 1.48 ???? Blade length is equal to: ????3 = ???????????????????? ????????????????????ℎ ???????????????????????????????? ???????? ???????????????????????????????? = 0.25 = 0.296 ???? Therefore: ???????????????????? ????????????????????ℎ 1.48 = 0.25 = 0.37 ????
  • 138. 138 Blade width is equal to: ????4 = ???????????????????????????????? ???????? ???????????????????????????????? ???????????????????? ????????????????ℎ = 5 Therefore: 1.48 ???????????????????? ????????????????ℎ = 5 = 0.296 ???? Blade width is equal to: ????5 = ???????????????????????????????? ???????? ???????????????????????? ???????????????????????? ????????????????ℎ = 10 Therefore: 4.45 ???????????????????????? ????????????????ℎ = 10 = 0.45 ???? It is assumed the height of the baffle is equal to the vessel’s height excluding the thickness of head. Power of impeller In order to work out the power required to operate the motor, a table of ratio constants and power number for different types of impellers is used, therefore (Pietranski, ): Figure [B1]: Ratio constant and power number for different types of impellers (Pietranski, ).
  • 139. 139 Therefore, applying: ???????? = ????2 × ???? × ???? ???? Where: µ = 0.79 mPa.a d = diameter, m = 1.48 m ρ = density, kg/m3 = 546 kg/m3 N = radial velocity = 200 rpm (as previously specified) = 3.3 rps (Carpenter, ) Therefore, Re equals to: ???????? = 5.5 × 106 ???????? > 10,000 > ???????????????????????????????????? ???????????????????????? This implies that the power of the motor required does not depend on the liquid viscosity, but on density instead, therefore (Sinnott et al., ): ???? ???????????????????????????????????? = ???????? × ???? × ????3 × ????5 Where: NP = power tables = 6 (From table 2) (Pietranski, ) ρ = density = 564 kg/m3 N = radial speed (rps) = 3.3 rps D = diameter = 1.48 m Substituting into equation: ???? ???????????????????????????????????? = 8.64 × 105 = .64 ℎ???? Further allowing losses of 25% ???? ???????????????????????? = ???? ???????????????????????????????????? 1 − ???????????????????????? = .85 ℎ???? However, this value appears to be too high, thus the assumption that mild agitation is required for the process is likely to be wrong. Another calculation can be proceeded with the estimation that 1 – 2 kW/m3 power is required to operate the agitator of a large vessel.
  • 140. 140 Assuming that the volume of the tank filled with fluid is approximately 70% (Sinnott et al., ): ???????????????????????? = ???????????????????? × 0.7 Therefore: (138.41 × 0.7) = 96.89 ????3 Referring to Figure [], highlighting ratio between power and liquid quantity, power required for the agitation can be calculated (Pietranski, ): Figure [B2]: Ratio between power and liquid quantity in order to evaluate agitation required (Pietranski, ). Therefore, power required: ???????????????????? ???????????????????????????????? = ???????????????????????? × 1.5 = 1.45 × 102 ???????? = 194 ℎ???? Therefore (Pietranski, ): ???????????????????? = ℎ???? ???????????? 96.89 ????3 = .82 ???????????? = 21.312 × 103 ???????????? Thus, ratio required is: ???????????????????? = 9.10 Hence, intensive agitation is required for this design. Agitator’s shaft It is acknowledged that the following calculations are based on example work (Anon, ): Given that: ℎ???? = 194 ???? = ???????????? = 200
  • 141. 141 Therefore, torque on shaft: ???????? = ( ℎ???? × 75 × 60) 2 × ???? × ???? ) = ( 194 × 75 × 60 2 × ???? × 200 ) = 694.711 ???????????? Polar modulus given by: ???? ???? = ???? ???? ???????? Where: Tm = maximum torque = 1.5Tc = .07 kgm Shear stress = 51 ksi = .59 kg/cm2 Therefore: ???? ???? = ( 1.5 × 694.711 × 100 .59 ) = 29.06 ????????3 Therefore diameter can be evaluated using: ???? = ???? × ????3 16 Therefore, diameter: ???? = 5.28 ???????? = 0.053 ???? Reactor and agitator cost The overall cost of the unit is worked out by: ???????????????? ???????? ???????????????? = ???????????????? ???????? ???????????????????????????? + ???????????????? ???????? ???????????????????????????????? Equation used to determine cost of unit is (Sinnott et al., ): ???????? = ???? + ???????? ???? According to this equation, the estimated cost is in USD for the year (January) based on U.S Gulf Coast location, so further currency, location and time factor are taken into account .
  • 142. 142 Reactor (jacket, agitated) Applying: ???????? = ???? + ???????? ???? Where: S = volume, m3 a = 14,000 b = 15,400 n = 0.7 Substituting into equation: ???????? = 14,000 + (15,400 × 138.410.7 ) ????. ???? $ 499,666 Agitator Applying: ???????? = ???? + ???????? ???? Where: S (power kW) = 1.45 × 102 a = 4,300 b = 1,920 n = 0.8 Substituting into equation: ???????? = 4,300 + [ × (1.45 × 102 )0.8 ] = ????. ???? $107,195 Therefore, total cost of reactor: ???????????????? ???????? ???????????????????????????? = $499,666 + $107,195 = $606,861 (????. ????. ???????????????? ???????????????????? ????????????????????, )
  • 143. 143 The cost of the column is based on a U.S Gulf Coast basis and has to be changed to a UK basis using cost indexes and location factors. Although a cost index for was not available, the cost index for will be taken as an estimate instead (Sinnott et al.,, p324): Therefore, applying: ???????????????? ???????? ???????????????? ???? = ???????????????? ???????? ???????????????? ???? × ???????????????? ???????????????????? ???????? ???????????????? ???? ???????????????? ???????????????????? ???????? ???????????????? ???? Where: Cost index in year A = 567.3 (Chemical Engineering, ) Cost index in year B = 499.6 (Chemical Engineering, ) Therefore, the total cost of the column is: ???????????????? ???????? ???????????????? ???? = 567.3 499.6 × $606,861 = $689,096 (????. ????. ???????????????? ????????????????????, ????????????????????) Applying location factors (Sinnott et al., , p327): ???????????????? ???????? ???????????????????? ???????? ???????????????????????????????? ???? = ???????????????? ???????? ???????????????????? ???????? ???????????????? × ???????????? Where: LFA (United Kingdom) = 1.02 Therefore, the total cost of the column on a United Kingdom, basis is: ???????????????? ???????? ???????????????????? ???????? ???????????????????????????????? ???? = 689,096 × 1.02 = ???????? $702,878 Currency conversion The unit of currency also has to be changed from dollars (USD) to pounds (GBP). The exchange rate, as of May , is 1. USD per 1 GBP (HM Revenue & Customs, ). Therefore, total cost of the column on a UK basis is: 702,878 1. = £488,008
  • 144. 144 Material factor The unit operation’s cost estimate is based on a carbon steel manufacturing material. The material factor for a carbon steel is 1. Therefore, this value must be related to the material factor for Hastelloy in order to adapt it. ???????????????????????????????? ???????????????????????? (????????????????????????????????????) = 1.55 Therefore, cost of unit in Hastelloy: ???????????????? = £488,008 × 1.55 = £756,412 ≈ £ 800,000
  • 145. 145 APPENDIX [C] – Calculations for Flash Tank The mechanical design method was adopted from “Two phase separator within the right limits” (Svrcek et al., ) Data from simulation WV = .97 lb/hr WL = .87 lb/hr ρV = 0. lb/ft3 ρL = 61. lb/ft3 Vapour velocity Applying: ???? ???? = ????( ???? ???? − ???? ???? ???? ???? )0.5 Where: UT = terminal velocity, m/s ρL = liquid density, kg/m3 ρV = vapour density, kg/m3 K = Table [C1]: Separator K values with a mist eliminator (Svrcek et al., )
  • 146. 146 Calculating K from Table C1, where P = 14.5 psig, K equals to: ???? = 0. + 0.(????) + 0.046ln(????) = 0. + 0.(14.5) + 0.046ln(14.5) = 0.347 Therefore, UT equals to: ???? ???? = 0.347( 61. − 0. 0. )0.5 = 4.34 ????????/???? = 1.32 ????/???? Hence, vapour velocity calculated via: ???? ???? = 0.75 × ???? ???? Therefore: ???? ???? = 0.75(4.34) = 3.256 ????????/???? = 0.992 ????/???? Vapour volumetric flowrate calculated via: ???? ???? = ???????? × ???? ???? = .97 × 0. = 155.46 ????????3 /???? = 4.4 ????3 /????
  • 147. 147 Vessel diameter Applying: ???? ???????? = ( 4???? ???? ???????? ???? )0.5 Where: DVD = Vessel diameter (internal), m D = diameter, m = ( 4(155.46) ????(3.256) )0.5 = 7.8 ???????? The drum has a mist eliminator, therefore a range between 3-6 inches is added to accommodate a support ring, therefore: 7.8 + 0.5 = 8.3 ???????? = 2.53 ???? A further increment of 6 inches is implemented, therefore final diameter: ???? = 3.66 ???? Liquid volumetric flowrate calculated via: ???? ???? = ???????? 60 × ???? ???? Where: QL = liquid volumetric flowrate, m3 /s = .97 60 × 61. = 59.23 ????????3 /???????????? = 1.68 ????3 /????????????
  • 148. 148 Hold up time By referring to Table [], feed to column hold up time is therefore 5 mins. Hold up volume Applying: ???????? = ???? ???? × ???? ???? Where: TH = hold up time VH = hold up volume QL = liquid volumetric flowrate, m3 /s Therefore: = 5 × 59.23 = 296.16 ????????3 = 8.39 ????3 Surge volume Surge volume can be calculated via: ???????? = ???????? × ???? ???? Where: Vs = surge volume, m3 Ts = surge time = 3 QL = liquid volumetric flowrate, m3 /s Therefore: = 3 × 59.23 = 177.69 ????????3 = 5.032 ????3
  • 149. 149 Table [C2]: Liquid hold up and surge times (Svrcek et al., ) Low liquid level height, HLLL Low liquid level height can be calculated by referring to Table C3[]. Knowing that Vessel diameter = 12 ft Pressure < 300 psia Therefore: ???????????????? = 6 ????????????ℎ???????? = 0.5 ???????? = 0. ???? Table [C3]: Low liquid level height (Svrcek et al., ) Height between HLLL to HNLL By applying: ???? ???? = ???????? ( ???? 4)???????????? 2
  • 150. 150 Where: VH = hold up volume, m3 Dvd = vessel diameter, m Therefore: ???? ???? = 296.16 ( ???? 4) × (8.3)2 = 5.47 ???????? = 1.667 ????
  • 151. 151 Height from HNLL to HHLL The position of high level column is calculated via: ???????? = ???????? ( ???? 4) × ???? ???????? 2 Where: HS = height between HNLL to HHLL VS = surge volume, m3 Dvd = vessel diameter, m = 177.69 ???? 4 × 8.32 = 3.28 ???????? = 1 ???? Height from high liquid level to the centre line of inlet nozzle Applying: ???????????????? = 12 + ???? ???? Where: dN = inlet nozzle sizing ???? ???? = ( 4???? ???? ???? 60 √???? ???? )0.5 Where, QM is the mixture velocity and it equals to: ???? ???? = ???? ???? + ???? ???? And QL calculated via: ???? ???? = 4 × ???????????????????????????????? ???? × (???????????????? ????????????????????????????????) = 4 × 59.23 ???? × 8.3 = 9.086 ????/???? Thus, QM equals to: = 9.08 + 3.256 = 12.342 ????????3 /???? Thus, ρM is calculated via: ???? ???? = ???? ???? ???? + ???? ????(1 − ????)
  • 152. 152 Where, ???? is calculated via: ???? = ???? ???? ???? ???? + ???? ???? = 9.08 9.08 + 3.256 = 0.736 Therefore, ρM equals to: ???? ???? = 61.(0.736) × 0.(1 − 0.736) = 45.28 ????????/????????3 Therefore, dN equals to: ???? ???? = 1. ???????? Thus, HLIN: ???????????????? = 12 + (1.) = 13. ???????? = 4.062 ???? Mist eliminator height: Disengagement height from the centreline of inlet nozzle to bottom of the demister pad equals the external diameter, therefore: ???? ???? = 3.66 ???? = 12 ???????? An assumption of 6 inches is to be made for HME and the height between the mist eliminator and top tangent line of vessel equals to: ???? ???????? = 1 ???????? = 0. ???? Therefore, total flash drum height: ???? ???? = ???????????????? + ???? ???? + ???????? + ???????????????? + ???? ???? + ???? ???????? + ???? ???????? = 0.5 + 5.47 + 3.28 + 13. + 12 + 0.5 + 1 = 36. ???????? ≈ 36 ???????? = 10 ????
  • 153. 153 Thickness of cylinder shell: Applying: ???? = ???????? × ???????? 2???? − ???????? + ???? Where: t = wall thickness, mm Di = internal diameter, mm S = allowable stress, N/mm2 C = corrosion allowance Pi = internal pressure, bar Therefore: = (1 × ) (2 × 351.39) − 1 + 4 = 3.60 + 4 = 7.6 ???????? Thickness of Torispherical head Applying (Sinnott et al., ): ???? = 0.885 × ???????? × ???? ???? (???? × ????) − (0.1 × ????????) + ???? Where: Pi = internal pressure, bar Rc = crown radius, mm S = allowable stress, N/mm2 E = weld joint efficiency Therefore: = 0.885 × 1 × (351.39 × 0.85) − (0.1 × 1) + 4 = 11.49 ????????
  • 154. 154 Shell mass Applying: ????ℎ???????????? ???????????????? = ???? × ???????? × ???? ???? × ???? ???? × ???? Where: Dc = vessel diameter, m Lc = vessel length, m tw = wall thickness, m ρ = metal density, kg/m3 = 9,200 kg/m3 (Alloys and producer, ) Therefore: = ???? × 3.66 × 10 × 0. × 9,200 = 8,806.718 ???????? Flash Tank cost: Purchased equipment costs can be evaluated by the equation (Sinnott et al., ): ???????? = ???? + ???????? ???? Where: Ce = purchased equipment cost, £, S = characteristic size parameter a = cost constant b = cost constant n = index for that type of equipment Given data: S = .718 kg a = - 400 (Sinnott et al., ). b = 230 (Sinnott et al., ). n = 0.6 (Sinnott et al., ). Therefore: ???????? = −400 + 230(.718)0.6 = ???????? $ 53,132.34
  • 155. 155 The cost of the flash tank is based on a U.S Gulf Coast Basis and has to be changed to a UK basis using cost indexes and location factors (Sinnott et al., ). Although a cost index for was not available, the cost index for will be taken as an estimate instead. ???????????????? ???????? ???????????????? ???? = ???????????????? ???????? ???????????????? ???? × ???????????????? ???????????????????? ???????? ???????????????? ???? ???????????????? ???????????????????? ???????? ???????????????? ???? Where: Cost index in year A () = 567.3 (Chemical Engineering, ) Cost index in year B () = 499.6 (Chemical Engineering, ) Therefore, the total cost of the storage tank is: ???????????????? ???????? ???????????????? ???? = 53,132.34 × 567.3 499.6 = ???????? $ 60,332.20 ≈ ???????? $ 60,400 Applying location factors (Sinnott et al., ): ???????????????? ???????? ???????????????????? ???????? ???????????????????????????????? ???? = ???????????????? ???????? ???????????????????? ???????? ????. ???? ???????????????? ???????????????????? ???????????????????? × ???????????? Where: LFA (United Kingdom) = 1.02 (Sinnott et al., ) Therefore, total cost of the storage tank is: ???????????????? ???????? ???????????????????? ???????? ???????????????????????????????? ???? = 60,400 × 1.02 = ???????? $ 61,608 The unit operation’s cost estimate is based on a carbon steel manufacturing material. The material factor for a carbon steel is 1. Therefore, this value must be related to the material factor for Hastelloy in order to adapt it. ???????????????????????????????? ???????????????????????? (????????????????????????????????????) = 1.55 Therefore, cost of unit in Hastelloy: ???????????????? = ???????? $ 61,608 × 1.55 = $95,492.4 Unit of currency has to be changed, from dollars (USD) to pounds (GBP). The exchange rate, as of May , is 1. USD per 1 GBP, therefore total cost of the storage tank on a UK basis is (HM Revenue & Customs, ): ???????????????? ???????? (£) = ???????????????? ???????? ($) 1. = £66,300
  • 156. 156 APPENDIX [D] – Calculations for Drying Distillation Column Data from simulation F = .17 kmol/hr XB = 0.385 mol% ZF = 0.01 mol% D = 526.73 kmol/hr XD = 0.99 mol% B = 978.214 kmol/hr Reflux ratio Applying : ???????????????????????????????????????? = ???? ???? ???? + 1 Therefore : ???????????????????????????????????????? = 0.071 Figure [D1]: Vapour mole fraction against Liquid mole fraction graph. y-x diagram for ACETI-01/WATER Liquid molefraction, WATER Vapormolefraction,WATER 0.000 0.025 0.050 0.075 0.100 0.125 0.150 0.175 0.200 0.225 0.250 0.275 0.300 0.325 0.350 0.375 0.400 0.425 0.450 0.475 0.500 0.525 0.550 0.575 0.600 0.625 0.650 0.675 0.700 0.725 0.750 0.775 0.800 0.825 0.850 0.875 0.900 0.925 0.950 0.975 1.000 0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50 0.55 0.60 0.65 0.70 0.75 0.80 0.85 0.90 0.95 1.00 1.0 bar
  • 157. 157 Using the information provided, Figure [] was able to be constructed, thus allowed to determine the following: ???? ???????????? = ???? ???? ???????????????????????????????????????? − 1 = 0.99 0.75 − 1 = 0.32 Reflux ratio can be calculated via: ???? ???????????????????????? = ???? ???????????? × 1.2 Therefore: ???? ???????????????????????? = 0.32 × 1.2 = 0.384 Figure [D2]: McCabe – Thiele Representation for stages determination.
  • 158. 158 Number of stages in column Data for determination of number of stages inside column was extracted from the Figure [] and constructed using the McCabe – Thiele Graphical Method. Thus, number of stages: ???????????????????????? ???????? ???????????????????????? = 17 + 1 (????????????????????????????????) Actual number of trays Applying: ???????????????????????????? = ????????ℎ???????????????? ???? Where: ε = 0.5 (Douglas, ) Therefore, Nactual: ???????????????????????????? = ????????ℎ???????????????? ???? = 17 0.5 = 34 Height of tower Assuming a tray spacing of 0.7 m for a drying column, the following approximation was taken (Douglas, ): ???????????????????????? = 2.3 × ???????????????????????????? Therefore: ???????????????????????? = 2.3 × 34 = 78.2 ???????? ≈ 23.85 ????
  • 159. 159 Vapour velocity Due to a 20% overdesign in the design specification and applying the equation for vapour velocity equals to: ???? ???? = ????( ???? ????− ???? ???? ???? ???? )0.5 Where: UV = vapour velocity, m/s ρL = liquid density, kg/m3 ρV = vapour density, kg/m3 K = entrainment factor Where, entrainment factor: ???? = 0.171????2 + 0.27???? − 0.047 Where, T equals to: ???? = 0.701 Therefore, K equals to: ???? = 209.67 Substituting back into vapour velocity equation: ???? ???? = 209.67( 945.3 − 4.4 4.4 )0.5 Therefore: ???? ???? = .07 ????/???? Diameter of column Applying: ???? ???????????? = ( 4???? ???? × ???? ???? × ???? ???? )0.5 Therefore: ???? ???????????? = 1.8 ????
  • 160. 160 Taking into consideration safety from flow, diameter is increased by 5%, thus: ???? = 1.05 × 1.8 = 1.85 ???? Economic pipe diameter To develop a pipe diameter, capital cost and flow rate were factors of consideration, due to the impact of economic diameter on reducing annualised cost. A rule of thumb for the economic pipe diameter is used (Genereaux, ): ????????, ???????????????????????????? = ( ???? ???? )0.5 Where: G = mass flowrate, kg/s = 24.953 ρ = density, kg/m3 = 6.4 And taking into consideration, 20% overdesign, this leads to: ????????, ???????????????? ???????????????? = 216 ???????? ????, ???????????????????????? ???????????????? = 178 ???????? ????, ???????????????????????? ???????????????????????????? ???????????????? = 148 ???????? Design specification calculations Thickness of shell Applying: ???? ???? = ???????? ???????? 2???? − ???????? Where: P = Internal Design Pressure = Operating pressure + 10% allowance = 1.1 bar D = Internal Diameter of tower = 1.85 m S = Allowable stress of stainless steel at 125 o C = 103.42 N/mm2 Therefore: ???? ???? = 9.89 ???????? Applying a corrosion allowance of 2 mm, thickness of shell is: 9.89 + 2 = 11.89 ???????? ≈ 12 ????????
  • 161. 161 Thickness of flat plate Applying (Sinnott et al., ): ???????? = ????√ ???????? ???? Where: S = maximum allowable stress = 515 N/mm2 (Shanghai STAL Precision Stainless Steel Co. Ltd ( STAL ), ) D = effective plate diameter C = constant which depends upon edge support = 0.185 Therefore, TP: ???????? = 0.035 ???? Column End Selection Torispherical head is selected due to the low pressure in the distillation chamber, therefore applying: ???? ???? = 0.885???????? ???? ???? ???? − 0.1???????? Where: RC = Crown radius = Di Therefore: ???? ???? = 17.43???????? Applying a corrosion allowance of 2mm: ???? ???? = 2 + 17.43 = 19.43 ???????? = 20 ???????? Calculation of stresses Axial stress due to pressure ???????????? = ???????? ???????? 4(???? ???? − ????. ????) = 50.875 ????/????????2
  • 162. 162 Stress due to the dead-weight of vessel Applying: ???????? = ???? ???? × ???? × ???? × ???? ???? × {????[???? ???? + (0.8???? ????)(???? ???? × 10−3)]} Where: Ws = dead-weight of vessel Cw = 1.15 Dm = 1.85 m Hv = 23.85 m Therefore: ???????? = .81 ???? ≈ 122 ???????? Weight of plates ???? ???? = ???? 4 × 1.852 = 2.69 ????2 Weight of plate equals to 1.2 N, therefore: ???????? = 34 × 1.2 × 2.69 = 109.75 ???????? Weight of insulation Knowing the following data, weight of insulation can be calculated: Mineral wool density = 130 kg/m3 Volume of insulation = 23.6 m3 Therefore: ???????? = 23.6 × 130 × 9.81 = .1 ???? ≈ 30.1 ???????? Weight of vessel ???????? = 109.75 + 122 + 30.1 = 237.3 ???????? This is an approximation based on an empty vessel, weight will vary based on other factors, such as flooding of the column.
  • 163. 163 Wind loading The following calculations are made on the assumption that maximum wind speed is 100 mph, thus: ???????????????????????????????? ???????????????????????????????? = ???????????? = ???????? × ???? ???? = 1.85 + [2 × (12 × 10−3)] = 1.874 ???? Therefore, loading (per linear diameter): ???? ???? = × 1.874 = .92 ????/???? Thus: σ???? = ???????? ????(???????? + ????)???? = 31.94 ????/????????2 Hence bending moment at tangent line: ???? ???? = .3 ????/????????2 Primary stresses The longitudinal and circumferential stresses due to pressure are: σℎ = ???????????? 2???? = 84.8 ????/????????2 σ???? = ???????????? 4???? = 48.4 ????/????????2 Drying Column cost All cost estimates formulas have been sourced from (Douglas, ). Capital cost Applying: ???????? = ( ????&???? 280 ) × 120???? ???? × ????0.8 × (218 + ????????) Where: M&S = Marshall – Shift index for UK = Cc = Cost of shell and trays
  • 164. 164 Fc = Design considerations for the column = 1.6 (approximation made based on column pressure being 1atm) (Modla and Lang, ) DT = Diameter of column = 1.85 m H = Height of column = 23.85 m Therefore: ???????? = £363,770.90 Cost of Reboiler Applying: ???????? = ( ????&???? 280 ) × 328 × ( ???????? ???? ) 0.65 × ????0.65 Where: Cr = Cost of reboiler △Hv = Heat of vaporization of bottoms = 451.37 kJ/mol V = Rate of Boil-Up = ???? ???? △???? ???? = 0.882 kJ/Kmol (Litpak, ) Therefore, Cr equals to: ???????? = £22,053.50 We have to assume that cooling water is available at 90 o F and a heat transfer coefficient of 100, the heat transfer area of the condenser heat exchanger is: ???? ???? = ( △ ???? ???? × ln ( ???????? − 90 ???????? − 125 ) × ???? Where: Tb = 125 o C at 1 atm = 257 o F △HV = 451.37 kJ/mol Therefore: ???? ???? = .81 ????2
  • 165. 165 APPENDIX [E] – Calculations for Heavy – Ends Distillation Column Conversions o F → o C o C = 5/9(o F -32) (Perry, , p1-5) lb/ft3 → kg/m3 kg/m3 = (1/16.) lb/ft3 (Perry, , p1-6) ft → m m = (1/0.) ft (Perry, , p1-6) ksi → psi ksi = psi (Sinnott et al., , p982) psi → N/mm2 psi = 0. N/mm2 (Perry, , p1-4) Propionic acid density at 125 o C (398K) Temperature (oF) Density (lb/ft3) Temperature (oC) Density (kg/m3) 35 63.56 1.67 .13 40 63.37 4.44 .09 45 63.17 7.22 .89 50 62.98 10.00 .84 55 62.79 12.78 .80 60 62.6 15.56 .76 65 62.41 18.33 999.71 70 62.22 21.11 996.67 75 62.03 23.89 993.63 80 61.84 26.67 990.58 85 61.65 29.44 987.54 90 61.46 32.22 984.49 95 61.27 35.00 981.45 100 61.08 37.78 978.41 105 60.89 40.56 975.36 110 60.7 43.33 972.32 115 60.5 46.11 969.12 120 60.31 48.89 966.07 125 60.12 51.67 963.03 130 59.93 54.44 959.99 135 59.74 57.22 956.94 140 59.55 60.00 953.90 Table [E1]: Respective propionic acid densities at different temperatures (CAMEO Chemicals, ). y = -1.x + .9 950.00 960.00 970.00 980.00 990.00 .00 .00 .00 .00 0.00 10.00 20.00 30.00 40.00 50.00 60.00 70.00 Density(kg/m3) Temperature (C) Density v Temperature (Propionic acid)
  • 166. 166 Under the assumption that the relationship described by the graph between the density of propionic acid and temperature stays constant, the density can be calculated at higher temperatures using the equation of the trend line. The liquid density of propionic acid is required in later calculations. For example at 125 o C the density of propionic acid is equal to 882.39 kg/m3 . 882.39 = −1. × 125 + .9 Acetic acid vapour density at 125 o C (398K) The ideal gas law is used to calculate an estimation for the vapour density of acetic acid at 125 o C (398K) thus (Perry, , p2-355): ρ = ???? ???? ???????????????????????????????? ????⁄ Where: ρ = density, kg/m3 P = pressure, Pa Rspecific = specific gas constant, kJ/kgK T = temperature, K And specific gas constant given by: ???? ???????????????????????????????? = ???? ????⁄ Where: R = ideal gas constant = .39 kJ/kmolK (Perry, , p1-22) M = molecular mass = 60.05 kg/kmol (Perry, , p2-28) The specific gas constant for acetic acid is equal to: .39 60.05 = 138.46 ????????/???????????? Therefore, at 125 o C the vapour density of acetic acid is equal to: 138.46 × 398 = 1.84 ????????/????3
  • 167. 167 Acetic acid vapour pressure at 125 o C (398K) Vapour pressure to be calculated via (Sinnott et al., , p451): ???? = ???? ????− ???? ????+????⁄ Where, for acetic acid: A = 7. (Dean, , p539) B = .313 (Dean, , p539) C = 222.309 (Dean, , p539) T = 398 Therefore, at 125 o C (398K) the vapour pressure of acetic acid is equal to: ????7.−(.313 222.309+398⁄ ) = 136.45 ???????????? Propionic acid vapour pressure at 125 o C (398K) The vapour pressure of propionic acid at 124.94 o C ≈ 125 o C is equal to 59.86 kPa (Clifford et al., ) Relative volatility Relative volatility can be calculated via (Branan, , p450): ???????????????????????? ???????????????? ???????????????????????? ???????????????????????????????? ???????????????????????????????????? ???????????????? ???????????????????????? ????????????????????????????????⁄ = ???? ????????????????????????/???????????????????????????????????? Therefore, the relative volatility, ???? ????????????????????????/???????????????????????????????????? is equal to: 136.45 59.86 = 2.28 Reflux ratio The reflux ratio is taken as 17, obtained via the simulation created in Aspen Plus. Number of column stages The number of stages required in the column to obtain the desired distillation was calculated using the Smoker equations (Sinnott et al., , p661-662). This was chosen as the boiling points of acetic acid and propionic acid are relatively close and the relative volatility calculated is relatively low. 1. Rectifying section Applying (Sinnott et al., , p662): ???? ???? + 1⁄ = ???? Where: R = Reflux ratio = 17
  • 168. 168 Therefore: ???? = 17 17 + 1⁄ = 0. Applying (Sinnott et al., , p662): ???? ???? ???? + 1⁄ = ???? Where: ???? ???? = fraction of acetic acid in distillate stream = 0. ≈ 1 (From simulation) Therefore: ???? = 1 17 + 1⁄ = 0. Applying (Sinnott et al., , p662): ????(???? − 1)????2 + [???? + ???? (???? − 1) − ????]???? + ???? = 0 Therefore: 0.(2. − 1)????2 + [0. + 0.(2. − 1) − 2.]???? + 0. = 0 Using the quadratic formula to solve for k: ???? = −???? ± √????2 − 4???????? 2???? (0 < ???? < 1) Where: a = 0.(2. – 1) = 1. b = [0. + 0.(2. -1) – 2.] = -1.264 c = 0. Therefore: ???? = 0.046 (0 < ???? < 1) Applying (Sinnott et al., , p662): ????0 ∗ = ???? ???? − ???? ????0 ∗ = 1 − 0.046 = 0.954 Applying (Sinnott et al., , p662): ???? ???? ∗ = ???????? − ???? Where: ???????? = fraction of acetic acid in column feed = 0. (From simulation)
  • 169. 169 Therefore: ???? ???? ∗ = 0. − 0.046 = 0. Applying (Sinnott et al., , p662): ???? = 1 + (???? − 1)???? Therefore: ???? = 1 + (2. − 1)0.046 = 1. Applying (Sinnott et al., , p662): ???? = ????????(???? − 1) ???? − ????????2⁄ Therefore: ???? = 0. ∗ 1.(2. − 1) 2. − 0. ∗ 1.⁄ = 1. Thus, number of stages is calculated via (Sinnott et al., , p662): ???? = log [ ????0 ∗ (1 − ???????? ???? ∗ ) ???? ???? ∗ (1 − ????????0 ∗ )⁄ ] ???????????? ( ???? ????????2⁄ ) Therefore: ???? = log [ 0.954(1 − 1. ∗ 0.) 0.(1 − 1. ∗ 0.954)⁄ ] ???????????? (2. 0. ∗ 1.⁄ ) = 14.329 ???? ≈ 15 ????????????????????????
  • 170. 170 2. Stripping section Applying (Sinnott et al., , p662): ???? = ???? ∗ ???????? + ???? ???? − (???? + 1)???? ???? (???? + 1)(???????? − ???? ????)⁄ Where: ???? ???? = fraction of acetic acid in bottom stream = 0. (From simulation) Therefore: ???? = 17 ∗ 0. + 1 − (17 + 1)0. (17 + 1)(0. − 0.)⁄ = 1. Applying (Sinnott et al., , p662): ???? = (???????? − ???? ????)???? ???? (???? + 1)(???????? − ???? ????)⁄ Therefore: ???? = (0. − 1)0. (17 + 1)(0. − 0.) ⁄ = −0. Applying (Sinnott et al., , p662): ????(???? − 1)????2 + [???? + ???? (???? − 1) − ????]???? + ???? = 0 Therefore: 1.(2. − 1)????2 + [1. − 0.(2. − 1) − 2.]???? − 0. = 0 Using the quadratic formula to solve for k: ???? = −???? ± √????2 − 4???????? 2???? (0 < ???? < 1) Where: a = 1.(2. – 1) = 1. b = [1. + 0.(2. – 1) – 2.] = -1.279 c = -0. Therefore: ???? = 0. (0 < ???? < 1) Applying (Sinnott et al., , p662): ????0 ∗ = ???????? − ????
  • 171. 171 Therefore: ????0 ∗ = 0. − 0. = −0.006 Applying (Sinnott et al., , p662): ???? ???? ∗ = ???? ???? − ???? Therefore: ???? ???? ∗ = 0. − 0. = −0.674 Applying (Sinnott et al., , p662): ???? = 1 + (???? − 1)???? Therefore: ???? = 1 + (2. − 1)0. = 2. Applying (Sinnott et al., , p662): ???? = ????????(???? − 1) ???? − ????????2⁄ Therefore: ???? = 1. ∗ 2.(2. − 1) 2. − 1. ∗ 2.⁄ = −1.001 Thus, number of stages is calculated via (Sinnott et al., , p662): ???? = log [ ????0 ∗ (1 − ???????? ???? ∗ ) ???? ???? ∗ (1 − ????????0 ∗ )⁄ ] ???????????? ( ???? ????????2⁄ ) Therefore: ???? = log [ −0.006(1 − (−1.001 ∗ −0.674)) −0.674(1 − (−1.001 ∗ −0.006))⁄ ] ???????????? (2. 1. ∗ 2.⁄ ) = 7. ???? ≈ 8 ????????????????????????
  • 172. 172 Total number of stages = 15 + 8 = 23 stages It is also recommended that an extra 10% be added to the total number of stages in addition to accounting for plate efficiency (Branan, ). Therefore, the ideal number of stages for this distillation column without the consideration of plate efficiency is: 23 + (0.1 × 23) = 25.3 ≈ 26 ???????????????????????? Plate efficiency For preliminary designs, a plate efficiency of 70% can be assumed (Sinnott et al., , p700). Introducing the plate efficiency to the number of ideal stages gives an actual number of: 25.3 / 0.7 = 36.1 ≈ 37 ???????????????????????? Column diameter The equations used to calculate the column diameter rely on the densities of the liquid and vapour streams as previously calculated as well as plate spacing. An initial estimate of the plate spacing will be taken as 0.5m (Sinnott et al., , p 709). Therefore: ???? ????̂ = (−0.171???? ???? 2 + 0.27???? ???? − 0.047) ∗ [ ???????? − ???? ???? ???? ???? ⁄ ] 1/2 Where: uv = maximum allowable vapour velocity, m/s lt = plate spacing, m = 0.5 m Therefore: ???? ????̂ = (−0.171 ∗ 0.52 + 0.27 ∗ 0.5 − 0.047) ∗ [882.39 − 1.84 1.84⁄ ] 1 2 = 2.86 ????/???? Applying (Sinnott et al., , p 709): ???????? = √4 ???????? ̂ ???????? ???? ???? ????̂ ⁄ Where: Dc = column diameter, m Vw = maximum vapour rate, kg/s = ⁄ = 13.89 kg/s
  • 173. 173 Therefore, column diameter is: ???????? = √4 ∗ 13.89 ???? ∗ 1.84 ∗ 2.86⁄ = 1.83 ???? Column height For columns of diameter greater than 1m, a plate spacing of between 0.3 and 0.6m is suggested (Sinnott et al., , p709). As the previous calculations show, the column diameter is greater than 1m therefore, the initial estimate of 0.5m plate spacing will be used. In addition to the height due to the plate spacing, it is also suggested that an additional 4ft should be added to the top of the tower to accommodate for the condenser, as well as 6ft being added to the bottom of the column to accommodate for the reboiler (Branan, , p444). Taking all of this into consideration, the height of the column is equal to: 4 ???????? = 4 ∗ 0. = 1.219 ???? 6 ???????? = 6 ∗ 0. = 1.829 ???? Therefore: (37 ∗ 0.5) + 1.219 + 1.829 = 21.55 ≈ 21.6 ???? Material of construction The material of construction will be a grade of stainless steel as opposed to carbon steel, as stainless steel has a corrosion resistance where as carbon steel does not. This is important as the materials flowing through the column have a corrosive nature. The grade of stainless steel chosen is 304 as the advantages other grades of steel have over 304 are not proportional to the increase in cost associated with them. Column wall thickness It is suggested that for a column of diameter of between 1 and 2m, the minimum practical wall thickness for a vessel to withstand its own weight, including a corrosion allowance of 2mm, is 7mm. Although the minimum wall thickness can be calculated (Sinnott et al., , p986): ???? = ???????? ???????? 2 ∗ ???? − ???????? ⁄ Where: t = wall thickness, mm Pi = internal pressure, bar Di = internal diameter, mm S = maximum allowable stress, N/mm2
  • 174. 174 As an estimate S for 304 stainless steel at 125 o C (398K), the maximum allowable stress at 148.89 o C (300 o F) will be taken as this is known data. S for 304 stainless steel at 148.89 o C (300 o F) is equal to 103.42 N/mm2 (Sinnott et al., , p982). ????304−???????????????????????????????????? ???????????????????? at 300 oF (148.89 oC) = 15.0 ???????????? × 0. × 15.0 = 15.0 ???????????? = 103.42 ????/????????2 Therefore, including a corrosion allowance of 2 mm, the column thickness is equal to: ???? = 1. ∗ 1.83 ∗ 103 2 ∗ 103.42 − 1.⁄ = 9. + 2 = 11. ???????? ≈ 12 ???????? Column end selection and thickness The type of end selected for this column is torispherical shaped, as this is the most commonly used end closure for a pressure vessel operating up to pressures of 15 bar (Sinnott et al., , p987). The advantages of other available shaped column ends do not give a significant enough advantage to justify the added cost. The thickness of torispherical ends differs from the thickness of the shell, therefore this also has to be calculated via (Sinnott et al., , p990): ???? = 0.885 × ???????? ???? ???? ???? − 0.1 × ???????? ⁄ Where: Rc = crown radius = Di Therefore, the torispherical end thickness, including a corrosion allowance of 2 mm, is equal to: ???? = 0.885 ∗ 1. × 1.83 × 103 103.42 − 0.1 × 1.⁄ = 15.912 + 2 = 17.912
  • 175. 175 Feed location To calculate the location of the feed, the original ratio of rectifying stages and stripping stages will be applied to the final number of stages calculated as the feed will be located between the rectifying and stripping stages. Original ratio of rectifying stages to total original stages is: 15 (15 + 8)⁄ = 0.65 Applying this ratio to the final number of stages calculated gives a feed location of: 0.65 × 37 = 24.13 Therefore, the feed to the column will be approximately located at the 24th stage. Pipe sizing To calculate the pipe diameter, the mass flowrate, density and the velocity of the fluid flowing through the pipe has to be known. The mass flowrates will be taken from the simulation produced as part of this project, while the densities and velocities will be calculated. The following equation will be used to calculate the average stream density: ???? = ∑(???????? × ????????) The stream velocity used in the following calculations will be obtained via linear interpolation of the values for the optimum velocity in terms of fluid density (Sinnott et al., , p266). Fluid Density [kg/m3 ] Velocity [m/s] 2.4 800 3.0 160 4.9 16 9.4 0.16 18.0 0.016 34.0 Inlet pipe The mass flow rate into the column will be calculated by multiplying the molar flowrate, taken from the simulation, by the molecular mass of the compound. The streams conditions are as shown by the PFD, 117.69 o C, 1 bar.
  • 176. 176 Acetic acid MOLAR flowrate = 893.86 ????????????????/ℎ???? Acetic acid MASS flowrate = 893.86 × 60 = .48 ????????/ℎ???? Propionic acid MOLAR flowrate = 5.93 ????????????????/ℎ???? Propionic acid MASS flowrate = 5.93 × 74 = 438.76 ????????/ℎ???? Acetic acid mass fraction = .48 .48+438.76 = 0.992 Propionic acid mass fraction = 438.76 .48+438.76 = 0.008 ???? ???????????????????????? = 935.646 ????????/????3 (DDBST GmbH, ) ???? ???????????????????????????????????? = 890.429 ????????/????3 (Calculated from the relationship previously derived) Therefore: ???? = (935.646 × 0.992) + (890.429 × 0.008) = 935.279 ???????? ????3⁄ Therefore, inlet flow velocity: ???????????????????? ???????????????? ???????????????????????????????? = 3.0 + (2.4 − 3.0) × 935.279 − 800 − 800 = 2.90 ????/???? Therefore, mass flowrate: ???????????????? ???????????????????????????????? = .48 + 438.76 = .24 = .24 = 15.02 ????????/???? Therefore, volumetric flowrate: ???????????????????????????????????????? ???????????????????????????????? = ???????????????? ???????????????????????????????? ???????????????????????????? = 15.02 935.279 = 0.016 ????3 /????
  • 177. 177 Therefore, pipe cross-sectional area: ???????????????? ???????????????????? ???????????????????????????????????? ???????????????? = ???????????????????????????????????????? ???????????????????????????????? ???????????????????????????????? = 0.016 2.90 = 0. ????2 Therefore, pipe diameter: ???????????????? ???????????????????????????????? = √0. × (4 ????⁄ ) = 0.084 ???? A pipe diameter of 0.084 m corresponds to a nominal pipe diameter of 4 inches with an allowance for future expansion taken into consideration. Bottom product type Similarly to the inlet pipe, the mass flow rate into the column will be calculated by multiplying the molar flowrate, taken from the simulation, by the molecular mass of the compound. The streams conditions are as shown by the PFD, 131.41 o C, 1 bar. Acetic acid MOLAR flowrate = 3.07 ????????????????/ℎ???? Acetic acid MASS flowrate = 3.07 × 60 = 184.17 ????????/ℎ???? Propionic acid MOLAR flowrate = 5.93 ????????????????/ℎ???? Propionic acid MASS flowrate = 5.93 × 74 = 438.76 ????????/ℎ???? Acetic acid mass fraction = 184.17 184.17+438.76 = 0.296 Propionic acid mass fraction = 438.76 438.76+184.17 = 0.704 ???? ???????????????????????? = 917.954 ????????/????3 (DDBST GmbH, ) ???? ???????????????????????????????????? = 875.334 ????????/????3 (Calculated from the relationship previously derived) Therefore: ???? = (917.945 × 0.296) + (875.334 × 0.704) = 887.934 ????????/????3
  • 178. 178 Therefore, outlet flow velocity: ???????????????????????? ???????????????? ???????????????????????????????? = 3.0 + (2.4 − 3.0) × 887.934 − 800 − 800 = 2.93 ????/???? Therefore, mass flowrate: ???????????????? ???????????????????????????????? = 184.17 + 438.76 = 622.92 = 622.92 = 0.17 ????????/???? Therefore, volumetric flowrate: ???????????????????????????????????????? ???????????????????????????????? = ???????????????? ???????????????????????????????? ???????????????????????????? = 0.17 887.934 = 0. ????3 /???? Therefore, pipe cross-sectional area: ???????????????? ???????????????????? ???????????????????????????????????? ???????????????? = ???????????????????????????????????????? ???????????????????????????????? ???????????????????????????????? = 0. 2.93 = 0. ????2 Therefore, pipe diameter: ???????????????? ???????????????????????????????? = √0. × (4 ????⁄ ) A pipe diameter of 0. m corresponds to a nominal pipe diameter of 3/4 inches with an allowance for future expansion taken into consideration.
  • 179. 179 Top product type As this stream is virtually 100% acetic acid, the flowrate and the density will consist of the values true to acetic acid under the stream conditions. The streams conditions are as shown by the PFD, 117.58 o C, 1 bar. As the vapour density of acetic acid has already been calculated, it will be assumed that this is the density of acetic acid in the stream, therefore: ???? ???????????????????????? = 1.84 ???????? ????3⁄ Acetic acid MOLAR flowrate = 890.789 ????????????????/ℎ???? Acetic acid MASS flowrate = 890.789 × 60 = .34 ????????/ℎ???? Top product flow velocity: ???????????? ???????????????????????????? ???????????????? ???????????????????????????????? = 18.0 + (9.4 − 18.0) × 1.84 − 0.16 16 − 0.16 = 17.09 ????/???? Therefore, mass flowrate: ???????????????? ???????????????????????????????? = .34 = .34 = 14.85 ???????? ????⁄ Therefore, volumetric flowrate: ???????????????????????????????????????? ???????????????????????????????? = ???????????????? ???????????????????????????????? ???????????????????????????? = 14.85 1.84 = 8.07 ????3 /???? Therefore, pipe cross-sectional area: ???????????????? ???????????????????? − ???????????????????????????????????? ???????????????? = ???????????????????????????????????????? ???????????????????????????????? ???????????????????????????????? = 8.07 17.09 = 0.472 ????2 Therefore, pipe diameter: ???????????????? ???????????????????????????????? = √0.472 × (4 ????⁄ ) = 0.78 ???? A pipe diameter of 0.78 m corresponds to a nominal pipe diameter of 36 inches with an allowance for future expansion taken into consideration.
  • 180. 180 Column mass and cost Shell mass calculated via (Sinnott et al., , p322): ????ℎ???????????? ???????????????? = ???? × ???????? × ???? ???? × ???? × ???? Where: lc= column length, m ρ304 = kg/m3 (Sinnott et al., , p322) Therefore: ????ℎ???????????? ???????????????? = ???? × 1.83 × 21.6 ∗ 12 × 10−3 × = .6 ???????? Column cost 1.0 Shell cost Applying (Sinnott et al., , p319): ???? + ???????? ???? Where: a304 = - (Sinnott et al., , p319) b304 = 600 (Sinnott et al., , p320) n304 = 0.6 (Sinnott et al., , p320) S = shell mass, kg (90 < shell mass < ) (Sinnott et al., , p320) Therefore: ???????????????? ???????????? ???????? ????ℎ???????????? = − + 600 × (.60.6) = $157,416 2.0 Trays cost Applying (Sinnott et al., , p319): ???? + ???????? ???? Where: avalve = 130 (Sinnott et al., , p319) bvalve = 146 (Sinnott et al., , p320) nvalve = 2.0 (Sinnott et al., , p320) S = diameter, m (0.5 < diameter < 5.0) (Sinnott et al., , p320)
  • 181. 181 Therefore: ???????????????? ???????????? ???????? ???? ???????????????????????? ???????????????? = 130 + 146 × (0.52.0) = $619 ???????????????? ???????????? ???????? 37 ???????????????????? = 619 × 37 = $22,901 Therefore, the total capital cost due to the column according to an U.S Gulf Coast, basis is: $22,901 + $157,416 = $180,317 The cost of the column is based on a U.S. Gulf Coast basis and has to be changed to a UK basis using cost indexes and location factors. Although a cost index for was not available, the cost index for will be taken as an estimate instead, therefore (Sinnott et al., , p324): ???????????????? ???????? ???????????????? ???? = ???????????????? ???????? ???????????????? ???? × ???????????????? ???????????????????? ???????? ???????????????? ???? ???????????????? ???????????????????? ???????? ???????????????? ???? Where: Cost index in year A = 567.3 (Chemical Engineering, ) Cost index in year B = 499.6 (Chemical Engineering, ) Therefore, total cost of the column on an U.S Gulf Coast, basis is: ???????????????? ???????? ???????????????? ???? = × 567.3 499.6 = $204,752 Applying location factor (Sinnott et al., , p327): ???????????????? ???????? ???????????????????? ???????? ???????????????????????????????? ???? = ???????????????? ???????? ???????????????????? ???????? ???????????????? × ???????????? Where: LFA (United Kingdom) = 1.02 Therefore, total cost of the column on an U.S Gulf Coast, basis is: ???????????????? ???????? ???????????????????? ???????? ???????????????????????????????? ???? = × 1.02 = $208,847 Currency conversion The unit of currency also has to be changed, from dollars (USD) to pounds (GBP). The exchange rate, as of May , is 1. USD per 1 GBP (HM Revenue & Customs, ). Therefore, total cost of the column on a UK basis is: 1.⁄ = £145,002
  • 182. 182 APPENDIX [F] – Calculations for Absorption Column Vessel thickness Applying: ???? = ???????? × ???????? 2???????? − ???????? + ???? Where: Pi = internal pressure Di = internal diameter S = maximum allowable stress on vessel E = welding efficiency c =corrosion allowance (4 mm) Therefore substituting in equation: ???? = 0.45 + 4 = 4.45 ???????? ≈ 5 ???????? Literature suggest that an absorption column should have a minimum thickness of 5 mm with a 2 mm corrosion allowance. This means that the design vessel should have a thickness of 7 mm with a 4 mm corrosion allowance. Hemispherical head thickness Applying: ???? ???????????????? = ???????? × ???????? 4 × ???? × ???? − 0.4 × ???????? + ???? Where: Pi = internal pressure Di = internal diameter S = maximum allowable stress on vessel E = welding efficiency c =corrosion allowance (4 mm) Therefore substituting in equation: 0.25 + 4 = 4.25 ???????? ≈ 5 ????????
  • 183. 183 Hemispherical head was chosen for this unit since it has a tougher shape and capable of resisting twice the pressure of a tori-spherical head while having the same thickness. Although, they are more expensive, they resist better under higher pressure. Diameter of column Applying: ???? ???? √ ???? ???? ???? ???? Where: L = liquid mass flowrate = 0.5 kg/s G = gas mass flowrate = 0.45 kg/s qG = density of gas = 1.1 kg/m3 qL = density of solvent = kg/m3 Equalling to: = 0.035 ???? Therefore percentage of flooding by referring to Figure 11.44, the table gave a value for the abscissa of (Sinnott et al., ): ????4.1 = 1.7 And at flooding (Sinnott et al., ): ????4.2 = 5 Therefore applying: ???????????????????????????????????????? ???????? ???????????????????????????????? = √????4.1/????4.2 × 100 Substituting in equation: ???????????????????????????????????????? ???????? ???????????????????????????????? = 58% The value agrees within the range of 50 – 90% thus plausible (Sinnott et al., ) Therefore applying gas flowrate per unit cross-sectional area: ???? ???? = √ ????4 × ???? ???? × (???? ???? − ???? ????) 13.1 × ???? ???? × ( ???? ???? ???? ???? )0.1
  • 184. 184 Where: k4 = constant qG = density of gas = 1.1 kg/m3 qL = density of solvent = kg/m3 FP = packing factor = 300 m-1 µL = viscosity of solvent = 1.12 cp Therefore substituting into equation: ???? ???? = 1.6 ????????/????2 ???? Therefore column area: ???? = ????/???? ???? = 0.3 ????2 Therefore diameter is calculated via: ???????????????????????????????? = √ 4 × ???? ???? = 0.6 ???? Flat-end Applying: ???? = ???????? × √ ???? × ???????? ???? × ???? Where: C = design constant D = nominal plate diameter S = maximum stress allowed E = joint efficiency Therefore substituting into equation: ???? = 9.5 ????????
  • 185. 185 Height of tower: Applying:                                       2..0 2 21.075.0 45.1exp1 a L g aL a L a a LL w L w L w l cw   Where: a = surface area per unit packing = 253 m2 /m3 σG = critical surface tension for particular packing material = 61 × 10-3 N/m σL = liquid surface tension = 27.1 × 10-3 N/m µL = viscosity of solvent = 1.12 cp ρL = density of solvent = kg/m3 Substituting into equation: ???? ???? = 80.34 ????2 /????3 Therefore, applying liquid film mass transfer coefficient equation:   4.0 2 1 3 2 3 1 .0 p LL L Lw w L L ad Da L g K L                        Where: dp = packing size = 25 × 10-3 m DL = 1.6 × 10-9 m2 /s µL = viscosity of solvent = 1.12 cp g = gravitational acceleration = 9.81 m.s-2 ρL = density of solvent = kg/m3 a = surface area per unit packing = 253 m2 /m3 Substituting into equation: ???????? = 1.3 × 10−4 ????/????
  • 186. 186 Therefore applying gas film mass transfer coefficient equation:   2 3 1 7.0 5                   p gg g g w g gG ad Da V K aD RTK    Where: DG = 1.8 × 10-5 m2 /s k5 = 5.23 µG = viscosity of gas mixture = 0.10 cp ρG = density of gas = 1.1 kg/m3 a = surface area per unit packing = 253 m2 /m3 dp = packing size = 25 × 10-3 m Tg = temperature of inlet gas = 323 K Therefore substituting into equation: ???? ???? = 6.82 × 10−4 ????????????????/????2 ???? Hence gas film transfer unit height gas: ???? ???? = ???? ???? ???? ???? ???? ???? ???? Where: Gm = G / MG G = gas mass flowrate = 0.45 kg/s MG = molecular weight of gaseous mixture KG = gas film transfer coefficient P = pressure aw = height of transfer units = 80.34 m2 /m3 Substituting into equation: ???? ???? = 1 ???? Applying liquid film transfer unit height equation: ???????? = ???? ???? ???????? ???? ???? ????????
  • 187. 187 Where: Ct = qL/ ML ρL = density of solvent = kg/m3 ML = molecular weight of solvent = 60.05 kg/mol KL = liquid film mass transfer coefficient aw = height of transfer units = 80.34 m2 /m3 Therefore HL: ???????? = 0.045 ???? Applying, (Sinnott et al., ): L m m GoG H L mG HH  Where: y1 = 0.017 Aspen Plus data y2 = 0. Aspen Plus data m = 0.78 Aspen Plus data Gm/Lm = 0.89 Aspen Plus data Therefore HOG : ???? ???????? = 1.03 ???? Applying : ???? = ???? ???????? × ???? ???????? Where: NOG = number of transfer units = 8 m HOG = height of transfer units = 1.03 m Therefore assuming 1m allowance for liquid and gas distribution: ???????????????????? ℎ????????????ℎ???? ???????? ???????????????????? = 10 ????
  • 188. 188 Total dead weight: Dead weight of vessel Applying: ???????? = ???? ???? × ???? × ???? ???? × ???? ???? × ???? × (???? ???? + 0.8 × ???? ????) × ???? × 10−3 Where: CW = factor accounting for weight of nozzles and other internal support of the vessel qm = density of vessel material Dm = mean diameter of vessel t = thickness Substituting into equation: ???????? = Dead weight of packing ???????????????????????????? ???????????????????????? = 2.4 ????3 ???????????????? ???????????????????????????? ???????????? 1 ????????????ℎ ???????????????????????????? = 673 ????????/????3 ???????????????? ????????????????ℎ???? ???????? ???????????????????????????? = 24 × 673 × 9.81 = ???? Weight of insulation ???????????????????????????? ???????????????? ???????????????????????????? = 130 ????????/????3 ???????????????????????? ???????? ???????????????????????????????????????? = ???? × ???? × ???? × ???????????????????????????????????????????? × 10−3 = 0.98 ????3 ????????????????ℎ???? ???????? ???????????????????????????????????????? = 0.98 × 130 × 9.81 = ???? Therefore total dead-weight: ???????? = + + = ????
  • 189. 189 Primary stresses: Longitudinal stress Applying: ???????? = ???? × ???????? 4???? = 3 ????/????????2 Where: P = pressure Di = internal diameter t = thickness Hoop stress Applying: ????ℎ = ???? × ???????? 2???? = 6 ????/????????2 Direct stress due to weight of vessel Applying: ???? ???? = ???????? ???? × (???????? + ????) × ???? Where: WV = total dead-weight Di = inside diameter t = thickness Therefore substituting into equation: ???? ???? = 2.17 ????/????????2
  • 190. 190 Wetting rate Applying: ???????????????????????????? ???????????????? = ???????????????????????? ???????????????????????????????????????? ???????????????????????????????? ???????????? ???????????????? ???????????????????? − ???????????????????????????????????? ???????????????? ???? Therefore: ???????????????????????????? ???????????????? = 6.14 × 10−6 ????3 /????2 ???? Pressure drop Applying: ???????????????????????????????????????? = 0.115 ???????? 0.7 Where: ???????????????????????????????????????? = 6.23 ???????? ????2 ????/ft of packing Therefore: ???????????????????????????????????????? = 0.05 ???????????? = 0.05 ???????????? Dynamic Wind Pressure for smooth cylindrical column Applying: PW = 0.05uw 2 = 281 ????/????2 Where: PW = wind pressure per unit area, N/m2 uw 2 = wind speed, km/hr It can be assumed due to the values obtained, that the stresses are negligible. Piping sizing Optimum cross sectional area for the piping is calculated using the below formula: ???????????????????? − ???????????????????????????????????? ???????????????? = ???????????????????????????????? ???????????????????????????????? And then the diameter is calculated using: ???????????????????????????????? = √ 4 × ???????????????????? − ???????????????????????????????????? ???????????????? ????
  • 191. 191 Thus, the following flowrates were collected from Aspen Plus simulation: Pipeline Flowrate (m3 /s) Velocity (m/s) Cross sectional area (m2 ) Required diameter (m) Combined inlet Gas Stream 0.78 2.00 0.39 0.7 m Off-gas 0.61 2.00 0. 0.625 m Solvent feed 0. 2.00 0.31 0.05 m Scrubber to reactor 0. 2.00 0. 0.060
  • 192. 192 Cost of packing: 1 inch Intalox ceramic saddles have an average cost of 20$/ft3 of packing. Since packing volume is 2.4 m3 , then: ???????????????? ???????? ???????????????????????????? ???????????????????????????????? = $ ≈ $ Mass of shell ????ℎ???????????? ???????????????? = ???? × ???? ???? × ???? ???? × ???? ???? × ???? = .1 ???????? Capital cost Applying (Sinnott et al., , p319): ???? + ???????? ???? Where: a = - (Sinnott et al., , p319) b = 600 (Sinnott et al., , p320) n = 0.6 (Sinnott et al., , p320) S = shell mass, kg (90 < shell mass < ) (Sinnott et al., , p320) Therefore: ???????????????? ???????????? ???????? ????ℎ???????????? = − + 600 × (.50.6) = $31,940.4 ???????????????? ???????? ℎ????????????????????ℎ???????????????????????? ℎ???????????? = $ The cost of the column is based on a U.S. Gulf Coast basis and has to be changed to a UK basis using cost indexes and location factors. Although a cost index for was not available, the cost index for will be taken as an estimate instead, therefore (Sinnott et al., , p324): ???????????????? ???????? ???????????????? ???? = ???????????????? ???????? ???????????????? ???? × ???????????????? ???????????????????? ???????? ???????????????? ???? ???????????????? ???????????????????? ???????? ???????????????? ???? Where: Cost index in year A = 567.3 (Chemical Engineering, ) Cost index in year B = 499.6 (Chemical Engineering, ) Therefore, total cost of the column on an U.S Gulf Coast, basis is: ???????????????? ???????? ???????????????? ???? = 33,140.4 × 567.3 499.6 = $37,631
  • 193. 193 Applying location factor (Sinnott et al., , p327): ???????????????? ???????? ???????????????????? ???????? ???????????????????????????????? ???? = ???????????????? ???????? ???????????????????? ???????? ???????????????? × ???????????? Where: LFA (United Kingdom) = 1.02 Therefore, total cost of the column on an U.S Gulf Coast, basis is: ???????????????? ???????? ???????????????????? ???????? ???????????????????????????????? ???? = × 1.02 = $38,384 Currency conversion The unit of currency also has to be changed, from dollars (USD) to pounds (GBP). The exchange rate, as of May , is 1. USD per 1 GBP (HM Revenue & Customs, ). Therefore, total cost of the column on a UK basis is: 1.⁄ = £26,650
  • 194. 194 APPENDIX [G] – Calculations for Acetic Acid Storage Tank Volume of storage tank Annual acetic acid production = 400,000 tonnes = 400,000,000 kg Weekly production acetic acid = 400,000,000 52 = .692 kg = 7.7 × 106 kg Therefore, volume of acetic acid can be calculated, using the density of acetic acid of kg/m3 (Sciencelab, ): ???????????????????????? = ???????????????? ???????????????????????????? = 7.7 × 106 = .324 ????3 ≈ ????3 Therefore, minimum capacity of storage tank = m3 Closest dimensions for m3 tank ( m3 ): Diameter = 24.9 m Height = 18.8 m Aspect ratio ( ???? ???? ) = 0.76 To confirm whether dimensions are correct, liquid volume formula is used (Perry, ): ???? = ????????2 ( ???? 57.30 − ????????????????????????????????) Where: ???????????? ???? = 1 − ???? ???? = 1 − 2???? ???? ???????????? ???? = 1 − 2( 18.8 24.9 ) ???????????? ???? = − 127 249 ???? = 120.7 ???? ???????????? ???? = 0.86
  • 195. 195 Substituting into main equation: ???? = [(18.8 × 12.452)][( 120.7 57.30 ) − (0.86 × − 127 249 )] = .5 ????3 Using partially filled heads equation for volume of head (Perry, ): ???? = 0.215????2 (3???? − ????) = 0.215 × 18.82 × [(3 × 12.45) − 18.8] = ????3 Therefore, total volume of storage tank: ???????????????????? ???????????????????????????? ???????????????? ???????????????????????? = + = ????3 ≈ ????3 Wall thickness Using minimum wall thickness equation to resist hydrostatic pressure (Sinnott et al., ): ???????? = ???? ???? × ???????? × ???? × ???????? 2 × ???????? × ???? × 103 Where: es = tank thickness at depth HL, mm HL = liquid depth, m ρL = liquid density, kg/m3 J = Joint factor / efficiency g = gravitational acceleration, 9.81 ms-2 ft = design stress for tank material, N/mm2 Dt = tank diameter, m Known data: Joint factor = 0.85 (Rosenfeld, ) ρL of acetic acid = kg/m3 (Sciencelab, ) ft of stainless steel 316L = 485 N/mm2 (Azom, )
  • 196. 196 Substituting in equation: ???????? = × 18.8 × 9.81 × 24.9 2 × 485 × 0.85 × = 5.8 ???????? A corrosion allowance of 3mm is added, thus a final recommendation of 8.8mm for the tank shell is proposed based on circumferential stress. Design pressure Using Clausius – Clapeyron equation to find vapour pressure of acetic at 40 o C (Senese, ): ln ( ????1 ????2 ) = ( ∆???? ???????????? ???? )( 1 ????2 − 1 ????1 ) Where: T1 = 40 o C = 313 K P2 = Pa T2 = 118 o C = 391 K (Sciencelab, ) ∆Hvap = 23.4 kJ/mol (National Institute and Standards of Technology, ) R = 8.314 J/mol K Therefore, substituting in equation: ln ( ???? ) = ( 8.314 )( 1 391 − 1 313 ) ln ( ???? ) = −1.79 ???? = ????−1.79 ???? = .7 ???????? = ???????? ≈ 17 ???????????? Therefore: Vapour pressure acetic acid at 40o C = 17 kPa Operating pressure storage tank = 17 kPa Design pressure should be 10% more than operating pressure, therefore: ???????????????????????? ???????????????????????????????? = 1.1 × 17 = 18.7 ???????????? Design temperature = 40o C
  • 197. 197 Calculation of stresses: Dead-weight stress The material for construction is Stainless steel 316L with density = 8 tonnes/m3 (Azom, ) The insulating material is mineral wool with density = 130 kg/m3 (Engineeringtoolbox, ) Dead-weight stress for metal Volume of metal = Tank shell + Tank roof Therefore: {????[24.9 + (2 × 0.) × 0. × 18.8]} + 1.3 [( ???? 4 ) × 24.92 × 0.] = 12.951 + 5.571 = 18.5 ????3 Therefore: ???????????????? ???????? ???????????????? = ???????????????????????? × ???????????????????????????? = 18.5 × 8 = 148.2 ???????????????????????? Therefore: ????????????????ℎ???? ???????????????????????? = ???????? (???????? ???? ????) = (148.2 × 9.81) {???? × [24.9 + (2 × 0.)] × 0.} = ???????????? = 2.1 ???????????? Dead-weight stress for insulation Volume of insulation = π × h × Di × ti Therefore: ???? × 18.8 × 24.9 × 0.02 = 29.4 ????3 Therefore: ???????????????? ???????? ???????????????????????????????????????? = ???????????????????????? × ???????????????????????????? = 29.4 × 130 = ???????? = 3.8 ????????????????????????
  • 198. 198 Weight of insulation is doubled to account for extra fittings, therefore: ???????????????? ???????? ???????????????????????????????????????? = 3.8 × 2 = 7.6 ???????????????????????? Therefore: ????????????????ℎ???? ???????????????????????? = ????????/???? ???? Where: ???? ???? = ???????????????????? − ???????????????????????????????????? ???????????????? = 7.6 × 9.81 ???? × (24.9)2 4 = 0.153 ????/????2 = 0. ???????????? ≈ 1.5 × 10−4 ???????????? Therefore: ???????????????????? ???????????????? ????????????????ℎ???? = ????????????????????????????????ℎ???? ???????? ???????????????? + ????????????????????????????????ℎ???? ???????? ???????????????????????????????????????? = (148.2 × 9.81 × ) + (7.6 × 9.81 × ) = ???? ???????????????????? ???????????????????????? ???????????? ???????? ????????????????????????????????ℎ???????? = + (1.5 × 10−4 ) = . ???????????? ≈ ???????????? = 2.1 ???????????? Axial stress Axial stress calculated using (Stevenson, ): ???????? ???? 4???????? ???? Where: P = Design pressure Do = Outside diameter J = Joint factor / efficiency ts = Minimum thickness
  • 199. 199 Therefore: 0.{24.9 + (2 × 0.)} (4 × 0.85 × 0.) = 15.6 ???????????? Hoop stress Hoop stress to be found using (Stevenson, ): 2 × ???????????????????? ???????????????????????? = 2 × 15.6 = 31.2 ???????????? Wind stresses Wind stresses are disregarded because of the high ratio of tank diameter to tank height. Analysis of stresses ???????? ???????????????? = ???????????????????? ???????????????????????? − ????????????????????????????????ℎ???? ???????????????????????? = 15.6 − 2.1 = 13.5 ???????????? ???????????????? ???????????????? = ???????????????? ???????????????????????? − ????????????????????????????????ℎ???? ???????????????????????? = 31.2 − 2.1 = 29.1 ???????????? ???????????????????????? ???????????????????????? = 0.5 × ???????????????????????? ???????????????????????????????? = 9.35 ???????????? Therefore: ???????????????????????????? ???????????????????????? = 33???????????? The maximum stress is approximately 93% below the design stress of 485 MPa, and therefore the shell thickness used in this design is considered acceptable.
  • 200. 200 Storage tank emissions: Total emissions are equal to the sum of the working loss and standing loss, and can be calculated via (South Coast Air Quality Management District, ): ???? ???? = ???? ???? + ???? ???? Where: Lr = total loss, lbs/yr Lw = working loss, lbs/yr Ls = standing loss, lbs/yr Using equation to work out working loss (South Coast Air Quality Management District, ): ???? ???? = 0.024 × ???? ???? × ???????????? × ???? × ???? ???? × ???? ???? Where: Lw = working loss, lbs/yr MV = average vapour molecular weight, lb/lbmole PVA = true vapour pressure of stored liquid at average liquid surface temperature, psia Q = annual throughput, Mgallon/yr KN = turn over factor, dependent on Q and C If Q/C ≤ 36 KN = 1.0 If Q/C > 36 KN = [ (180 ×????)+???? 6???? ] KP = working loss product factor KP = 0.75 for crude oil and 1.0 for other materials Therefore: Converting tonne/yr to Mgallon/yr Q = 400,000 tonne/yr = 37 Mgallon/yr Converting m3 to Mgallons C = m3 = 1.9 Mgallons Thus: ???? ???? = 37 1.9 = 19.5 Q/C is smaller than 36, thus KN = 1.0 Substituting values in working loss equation: ???? ???? = 0.024 × 60.05 × 0.21 × 37 × 1 × 1 = 11.2 ????????????/???????? = 49.8 ????/????????
  • 201. 201 Using equation to work out standing loss (South Coast Air Quality Management District, ): ???? ???? = ???? × ???????? × ???????? × ???? ???? × ???????? Where: Ls = standing loss, lbs/yr U = number of days of the year that the tank is used to store liquid material VV = vapour space volume, calculated: ???????? = (66.84 × ????) + ???????? C = tank capacity, Mgallon VF = vapour space function, depending on tank diameter WV = vapour density, lbs/ft3 KE = vapour space expansion factor KS = vented vapour saturation factor, calculated via: ???????? = 1 1 + (???????? × ????) + (???? ???? × ????) SA, SB = vapour saturation factors D = tank diameter, ft H = tank height, ft Therefore, vapour space volume is calculated using: ???????? = (66.84 × ????) + ???????? = (66.84 × 1.9) + = And vented vapour saturation factor, KS worked out through: ???????? = 1 1 + (???????? × ????) + (???? ???? × ????) Converting m to ft: Height = 18.8 m = 61.7 ft Diameter = 24.9 m = 81.7 ft Therefore: ???????? = 1 1 + (0. × 61.7) + (0. × 81.7) = 1.38
  • 202. 202 Thus, standing storage loss: ???? ???? = 365 × × 0. × 0.039 × 1.38 = .4 ????????????/???????? = .9 ????/???????? ≈ 5.1 ????????/???????? Therefore, total emissions: ???? ???? = ???? ???? + ???? ???? = 49.8 + = .8 ????/???????? = 5.1 ????????/???????? Tank Blanketing: Inert gas, N2, is utilised to blanket the fixed-roof tank for safety. It is lost in two ways, either from breathing losses from day/night temperature differential and working losses to displace changes in active level. Using vapour volume formula, breathing losses can be calculated (Branan, ): ???????? = ???? × ????2 4 × (????????????. ????????????????????????) Where: Vo = Vapour volume, scf D = Tank diameter, ft VV = avg.outage = average vapour space, ft Therefore: ???????? = ???? × 81.72 4 × = 0.189 ???????????? = 5.4 × 10−3 ????3 Therefore, daily breathing losses are worked out via (Branan, ): ???????????? = ????????{቎ 460 + ???????? + ???????????? 2 460 + ???????? + ∆ − ???????????? 2 ቏ − 1.0}
  • 203. 203 Where: TS = storage temperature, o F Tdc = daily temperature, o F DBL = daily breathing loss, scf ∆ = Adjustment for the differential between blanketing and pressure-relief settings (normally 2-4 o F) Also using displacement equivalents of inert gas to tank liquid (Branan, ): 1 ???????????? × 5.615 = 1 ???????????? ???????????????????? ???????????? Therefore, substituting into daily breathing losses equation: ???????????? = 0.189{቎ 460 + 104 + 37.4 2 460 + 95 + 3 − 37.4 2 ቏ − 1.0} = 0.015 ????????????/???? = 0.45 ????????????/????????????????ℎ And monthly working loss: ????????????????ℎ???????? ???????????????????????????? ???????????????? = 12(5.165) = 309 ????????????/????????????????ℎ Thus, total inert gas usage: ???????????????????? ???????????????????? ???????????? ???????????????????? = 309 + 0.45 = 309.45 ????????????/????????????????ℎ = 8.8 ????3 /????????????????ℎ Inlet/Outline diameters: The inlet and outlet diameters are sized for a recommended liquid flowrate of 2 m/s (Perry, ). Inlet pipe Given data from simulation Mass flowrate = .1 kg/hr Inlet line sizing is determined by allowing 20% above the normal product flowrate of kg/hr. Therefore: × 1.2 = ????????/ℎ???? Therefore, mass flowrate in seconds: ???????????????? ???????????????? = = 17.4 ????????/????
  • 204. 204 Density of acetic acid is known, therefore volumetric flowrate: ???????????????????????????????????????? ???????????????? = 17.4 = 0.017 ????3 /???? Therefore, area of pipe: ???????????????? ???????? ???????????????? = ???????????????????????????????????????? ???????????????? ???????????????????????????????? = 0.017 2 = 8.5 × 10−3 ????2 Thus, diameter of pipe: ???????????????????????????????? ???????? ???????????????? = √(8.5 × 10−3 × 4 ???? ) = 0.104 ???? = 104 ???????? The nearest commercial pipe size is a nominal pipe size of 4, schedule number 40s (with inside diameter of 102 mm and a wall thickness of 6mm) (Perry,). Outlet pipe Outlet line sizing is determined by the outlet flowrate calculated based upon the need to fill a standard tonne chemical tanker in 8 hours. This corresponds to a volumetric flowrate of 0.099 m3 /s. Converting tonne to kg 5,000,000 kg Converting hours to seconds s Therefore, mass flowrate: ???????????????? ???????????????????????????????? = 5,000,000 = 173.6 ????????/???? Acetic acid density of kg/m3 enables the working out of volumetric flowrate: ???????????????????????????????????????? ???????????????????????????????? = 173.6 = 0.166 ????3 /???? Since outlet diameter is sized for a recommended liquid flowrate of 2 m/s, area of pipe can be calculated: ???????????????? ???????? ???????????????? = ???????????????????????????????????????? ???????????????????????????????? ???????????????????????????????? = 0.166/2 = 0.083 ????2 ≈ 0.08 ????2
  • 205. 205 Therefore, diameter of pipe: ???????????????????????????????? ???????? ???????????????? = [ ???????????????? × 4 ???? ]0.5 = [0.08 × 4 ???? ]0.5 = 0.319 ???? = 319 ???????? The nearest commercial pipe size is a nominal pipe size of 12, schedule number 120 (with an inside diameter of 273mm and a wall thickness of 25mm) (Perry, ). Size of delivery Volume of storage tank = m3 Weekly production of acetic acid = m3 Thus, empty volume space of storage tank: = − = ????3 Based on the empty volume space of the storage tank, an estimation of the optimum size delivery can be achieved, thus: ???????????????????????? ???????????????????????????????????????? ???????????????????????? ???????????????? − ???????????????????? ???????????????????????? ???????????????????? ???????? ???????????????????????????? ???????????????? = ???????????????????????? ???????? ???????? ???????????????????????????????????? = − = ????3 ≈ ????3 A further 500 m3 is deducted from the figure obtained in order to set a minimum stock level in the tank, thus optimum size delivery: − 500 = ????3 Therefore, minimum stock level in the tank: − = ????3
  • 206. 206 Therefore, diameter of pipe: ???????????????????????????????? ???????? ???????????????? = [ ???????????????? × 4 ???? ]0.5 = [0.05 × 4 ???? ]0.5 = 0.252 ???? = 252 ???????? The nearest commercial pipe size is a nominal pipe size of 10, schedule number 120 (with an inside diameter of 230mm and a wall thickness of 21mm) Size of delivery Volume of storage tank = m3 Weekly production of acetic acid = m3 Thus, empty volume space of storage tank: = − = ????3 Based on the empty volume space of the storage tank, an estimation of the optimum size delivery can be achieved, thus: ???????????????????????? ???????????????????????????????????????? ???????????????????????? ???????????????? − ???????????????????? ???????????????????????? ???????????????????? ???????? ???????????????????????????? ???????????????? = ???????????????????????? ???????? ???????? ???????????????????????????????????? = − = ????3 ≈ ????3 A further 500 m3 is deducted from the figure obtained in order to set a minimum stock level in the tank, thus optimum size delivery: − 500 = ????3 Therefore, minimum stock level in the tank: − = ????3 A size delivery of m3 would be desirable, as nearly 75% of the weekly production is able to be shipped. The figure obtained for the minimum stock level will be able to accommodate a scenario where if the plant shuts down for a few days, the remaining storage will keep on supplying customers. Additionally, if there is a problem in distribution, the available space in the tank will allow for continuous running of the plant until the problem is solved. Henceforth, a delivery period of one week will provide smooth production and shipment whilst accommodating customers. Tank cost: Purchased equipment costs can be evaluated by the equation (Sinnott et al., ): ???????? = ???? + ???????? ???? Where: Ce = purchased equipment cost, £, S = characteristic size parameter a = cost constant b = cost constant n = index for that type of equipment Given data: S = m3 a = (Sinnott et al., ). b = 700 (Sinnott et al., ). n = 0.7 (Sinnott et al., ). Therefore: ???????? = + 700()0.7 = ???????? $ 409,565 ≈ ???????? $ 410,000 The cost of the storage tank is based on a U.S Gulf Coast Basis and has to be changed to a UK basis using cost indexes and location factors (Sinnott et al., ). Although a cost index for was not available, the cost index for will be taken as an estimate instead. ???????????????? ???????? ???????????????? ???? = ???????????????? ???????? ???????????????? ???? × ???????????????? ???????????????????? ???????? ???????????????? ???? ???????????????? ???????????????????? ???????? ???????????????? ???? Where: Cost index in year A () = 567.3 (Chemical Engineering, ) Cost index in year B () = 499.6 (Chemical Engineering, ) Therefore, the total cost of the storage tank is: ???????????????? ???????? ???????????????? ???? = 410,000 × 567.3 499.6 = ???????? $ 465,558 ≈ ???????? $ 470,000
  • 207. 207 Applying location factors (Sinnott et al., ): ???????????????? ???????? ???????????????????? ???????? ???????????????????????????????? ???? = ???????????????? ???????? ???????????????????? ???????? ????. ???? ???????????????? ???????????????????? ???????????????????? × ???????????? Where: LFA (United Kingdom) = 1.02 (Sinnott et al., ) Therefore, total cost of the storage tank is: ???????????????? ???????? ???????????????????? ???????? ???????????????????????????????? ???? = 470,000 × 1.02 = ???????? $ 479,400 Unit of currency has to be changed, from dollars (USD) to pounds (GBP). The exchange rate, as of May , is 1. USD per 1 GBP, therefore total cost of the storage tank on a UK basis is (HM Revenue & Customs, ): ???????????????? ???????? (£) = ???????????????? ???????? ($) 1. = £ 332,847 ≈ £ 333,000
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