Presentation1 final pasdadb bsadhroject.pptx

final year project

1. Group members

2. DESIGNING AND SIMULATION FOR THE PRODUCTION OF 30000 KG/HR OF ANILINE FROM NITROBENZENE BY HYDROGENATION

3. contents

4. INTRODUCTION Aniline basically a benzene derivative in which one hydrogen atom from benzene ring is substituted by amino group (NH 2 ). Also known as aminobenzene or phenylamine. It’s a prototypical aromatic amine. Aniline is colorless and oily liquid. It is darkening when expose to light and air. It has odor like amine and taste like burning material. Aniline is miscible in mostly organic solvent like alcohol, benzene and chloroform. When aniline reacts with acid produce salt and it is combustible in nature. Aniline in inhaled form will be fully absorbed by the lungs which causes toxicity. Its aromaticity or fish like odor tell you about the hazardous concentration of 1 ppm which came from the (OSHA PEL-TWA) is 5 ppm. Aniline vapors have greater density than air particles density which may causes the asphyxiation in the closed and low-lying areas which are not ventilated. Aniline causes irritation in eyes and skin when contact in liquid form. It causes systematic toxicity in skin when absorbed through the skin. The absorption of aniline in skin can be delayed for many hours. The molecular formula of aniline (C 6 H 7 N). Structural formula of aniline

5. ANILINE POINT The aniline point is defined as it is the minimum temperature at which the equal amounts of aniline is miscible with lubricant oil that make a single-phase mixture upon mixing. Why aniline point test The value of aniline gives approximation for the content of aromatic compound in the oil. The lower aniline point shows that mixture have a greater amount of aromatic compound in oil. How to determine aniline point The same volume of aniline and oil are in a test tube upon heating it will convert to a homogenous solution. Stop heating and cool the tube. The aniline point is recoded when tow phases are separated on a stirred smoothly temperature .  

6. Physical properties Molecular weight 93.128 g/gmol Melting point -6.3 o C Flash point 76 o C Vapor density 3.23 Boiling point 184 – 186 o C Vapor pressure 0.489 mmHg at 25 o C Specific gravity 1.022 Water solubility 36.070mg/lit at 25 o C Conversion factor 1ppm=3.8mg/m 3

7. application

8. Process selection & description 1.From nitrobenzene via Hydrogenation Reaction: C6H5NO2(g) + 3H2(g) → C2H5NH2(g) + 2H2O(g Material Requirements Nitrobenzene (10%) Hydrogen (90 %) Copper (catalyst) Nitrobenzene ( mononitrobenzene or MNB) is fed with hydrogen into a plug-flow tubular reactor containing a noble metal catalyst supported on carbon. The hydrogenation is carried out in the liquid phase and the nitrobenzene conversion to aniline is near 100% in a single pass. Yield : The yield based upon nitrobenzene is approximately 95% by weight

9. Process selection & description 2 . Manufacturing from chlorobenzene by Ammonolysis Reaction : C 6 H 5 Cl + 2 NH 3 (aqua) → C 6 H 5 NH 2 + NH 4 Cl (sol.) Material Requirements Chlorobenzene Ammonia solution (28%) Cuprous Oxide (catalyst) Chlorobenzene is charged into a series of horizontal, rotating, high pressure rolled steel autoclaves. Approximately 0.1 more of cuprous oxide and 4-5 moles of 28-30% aqueous ammonia per mole of chlorobenzene are added. The reaction is initiated at a temperature of 180 C and is later maintained at 210 to 220 C under constant agitation. If the indicated ratio of reactants is using the rate of aniline formation is about 20 times greater than the rate of phenol formation Yield: The yield of aniline is 85 to 90% based on chlorobenzene.

10. Process selection & description 3. Manufacturing from Phenol by Ammonolysis Reaction: C 6 H 5 OH + NH 3 ↔ C 6 H 5 NH 2 + H 2 O Material requirements Phenol Ammonia Alumina – based catalyst Problems The third purification step – recovery of specification aniline is considerably more difficult because aniline and phenol have almost identical vapors pressure curves. However, aniline – phenol mixtures are non-ideal as evidenced by the formation of a maximum boiling azeotrope

11. Process selection

12. Process description Conversion Reactor The feed mixture is passed into a fluidized bed reactor containing copper on silica gel as a catalyst, operation at a pressure of 20 psig (140 k N/ m 2 ). The contact time based on superficial velocity at reaction temperature and pressure and based on an unexpanded bed is 10 seconds. Excess heat of reaction is removed to maintain the temperature at 270 C by a heat transfer fluid passing through porous stainless-steel candle filter before leaving reactor.

13. Process description Heat Exchanger Nitrobenzene at room temperature of 24°C is sent to a heat exchanger to be heated to a temperature of 215°C. The heat of reaction from the reactor is used to heat the feed nitrobenzene in this heat exchanger. After this heat exchange the output stream from the reactor, containing the products and unreacted reactants, is cooled to a temperature of 163°C

14. Cooler The cooler is used to bring down the temperature of reactor output, to Bring down aniline concentration in vapour phase.Also at a temperature below 100°C, aniline, nitrobenzene and water will be in liquid state while hydrogen would be in vapor. This helps in easier separation. Using the software, a sensitivity analysis was performed on the cooler’s output stream. This stream contains aniline, water, nitrobenzene, hydrogen at a temperature of 50◦C and 1.4 bar. Change in mass flow of each of the species, in vapour phase, was studied for a temperature range of 50◦C to 150◦C. Process description

15. Process description Phase Separator The phase separator is essentially a gas-liquid separator. The reactor mixture exiting the cooler consists of 92% hydrogen in vapor phase which is separated here.

16. Process description Decanter (Component Separator) The component separator used , is assumed to be a decanter since decanter is not available in the simulation.The material stream entering the decanter consists of aniline, nitrobenzene and water. Aniline and nitrobenzene forms an organic mixture which is partially miscible with water (negligible). Therefore water is separated from the organic mixture using the decanter .

17. Process description Distillation Column Since aniline and nitrobenzene are organic solvents, distillation is the best applicable separation technique. Aniline has a boiling point of 184.1°C while nitrobenzene has a boiling point of 210.9°C. Therefore, these compounds can be easily separated using the distillation column. The no. of stages required and optimum feed entry stage is first found out using shortcut distillation column data, this data is then inputted into the rigorous distillation column. It is observed that aniline (distillate) has a purity of 99% which is as desired, however, nitrobenzene (residue) is only 84% pure .  The Second Distillation Column is used to increase the efficiency and optimize the simulation .

18. Process description We observed that at a temperature and pressure of 311.17◦C and 1.4 bar, the product stream from reactor contained aniline in large quantity Even though large amount of aniline was produced, it was a diluted stream. Therefore, its purity was improved using two distillation columns, a phase separator and a decanter. From the sensitivity analysis studies done by the DWSIM software and manual analysis performed by the us, the following observations can be concurred : Greater the heat exchanged between the hot and cold fluid in the heat exchanger, greater is the process optimization and efficiency, since the electrical load on the cooler and heater will be lower. Lower the temperature of cooler output, greater is yield of aniline. As the number of stages increases, greater separation and hence greater purity of aniline is obtained.

19. Process flow diagram

20. Material balance A  mass balance , also called a  material balance , is an application of conservation of mass to the analysis of physical systems. By accounting for materia l  entering and leaving a system, mass flows can be identified which might have been unknown, or difficult to measure without this technique. CAPACITY SELECTION Capacity selected is 30000 kg/hr. Aniline molecular weight =93 g/mol Aniline required = = 327.69 Kmol/hr. Basis: 100 Kmol/hr. of feed to the Reactor Nitrobenzene in the feed = 10 Kmol/hr. Hydrogen in the feed = 90 Kmol/hr  

21. Material balance Component Kmol/hr. Mole fraction Kg/hr. Nitrobenzene 326.6 0.1 676.4 Hydrogen 2939.6 0.9 5913.2   3266.2 1.0 6589.6 Selectivity and the conversion for our process is 99.5%. Amount of nitrobenzene converted = 10 x 0.995 = 9.95 Kmol/hr. Amount of unconverted or unreacted nitrobenzene = 10 – 9.95 = 0.25 Kmol/hr. The ratio of Hydrogen to Nitrobenzene is= 9:1 A mount of Nitrobenzene required is =326.632 kmol / hr ,  Amount of hydrogen required =9*326.632 =2939.688 kmol / hr

22. Basis : Production of Aniline (99% purity) is 30000 kg/ hr so, amount of Nitrobenzene required is=326.632 kmol / hr , =326.632*123.1092, =40211.40421 kg/hr .   The ratio of Hydrogen to Nitrobenzene is= 9:1   Amount of hydrogen required =9*326.632 =2939.688 kmol / hr   =5925.82307 kg/hr.   Hydrogen from recycle =6*326.632 =1959.792 kmol / hr =3950.548714 kg/hr . Fresh feed of Hydrogen= 3*326.632 =979.896 kmol / hr =1975.274356 kg/hr.  

23. Material balance Mass balance for Vaporiser : Stream1A : Pure Nitrobenzene feed in liquid phase=326.632 kmol / hr = 40211.40421 kg/ hr Stream1B: Nitrobenzene from vaporizer in vapor phase=326.632 kmol / hr = 40211.40421 kg/ hr

24. Material balance Mass balance for Reactor: Flow IN OUT Component Stream 1B(kg/hr) Stream 2(kg/hr) Stream 4(kg/hr) Stream 6(kg/hr) Nitrobenzene 40211.40421 --- --- --- Hydrogen --- 1975.27436 3950.54871 3950.54871 Water --- --- --- 11768.68161 Aniline --- --- --- 30417.99696 TOTAL(kg/hr) 40211.40421 1975.274356 3950.548714 46137.22728 TOTAL(kg/hr) 46137.22728 46137.22728

25. Material balance Mass balance for Condensor : Flow IN OUT OUT Component Stream 6(kg/ hr ) Stream 3(kg/hr) Stream 7(kg/hr) Nitrobenzene --- --- --- Hydrogen 3950.548714 3950.54871 --- Water 11768.68161 --- 11768.68161 Aniline 30417.99696 --- 30417.99696 Total (kg/hr) 46137.22728 3950.54871 42186.67857

26. Material balance Mass balance for Decanter: Flow IN OUT OUT   Component Stream 7(kg/ hr ) Stream 12(kg/ hr ) Stream 13(kg/hr) Stream 9(kg/hr) Stream 10(kg/hr) Nitrobenzene --- --- --- --- --- Hydrogen --- --- --- --- ---   Water   11768.68161   32698.53348   5117.73724   44437.73942 5147.21291   Aniline   30417.99696   7042.02739   1392.55056   7102.20424 31750.37067 TOTAL(kg/hr) 42186.67857 39740.56087 6510.28780 51539.94366 36897.58358 TOTAL(kg/ hr ) 88437.52724 88437.52724  

27. Material balance Mass balance for Distillation Column 1: Flow IN OUT OUT Component Stream 9(kg/hr) Stream 11(kg/hr) Stream 12(kg/hr) Nitrobenzene --- --- --- Hydrogen --- --- --- Water 44437.73942 11739.20594 32698.53348 Aniline 7102.20424 60.17685 7042.02739 Total (kg/hr) 51539.94366 11799.38279 39740.56087

28. Material balance Mass balance for Distillation Column 2: Flow IN OUT OUT Component Stream 10(kg/hr) Stream 13(kg/hr) Stream 14(kg/hr) Nitrobenzene --- --- --- Hydrogen --- --- --- Water 5147.21291 5117.73724 29.47567 Aniline 31750.37067 1392.55056 30387.29578 Total (kg/hr) 36897.58358 6510.28780 30387.29578

29. ENERGY BALANCE Vaporizer (H-1) Mass of stream 1a = m 1a =3965.026 Kg/hr. = 3965026 g/hr. Temperature of oil entering = T oil,1 = 400 o C Temperature of oil leaving = T oil,2 = 290 o C Temperature of stream 1a = T 1a = 25 o C Temperature of stream 1b = T 1b = 400 o C Specific heat of 1a = c p,1a = 2.0078 J/g K Latent heat of nitrobenzene = L nitro = 357.656 J/g Total heat exchange = Q = sensible heat + latent heat = m 1a [c p,1a x (T 1b - T 1a ) + L nitro ] = 3.44 x 10 6 KJ/hr. Mass of oil required = m = Q/ [c p, oil (T oil,1 - T oil,2 )] = 1164 Kg/hr. Stream# 1 1a 5 6a Temperatu o C) 25 25 25 120 Pressure (KPa) 101.3 344.5 101.3 344.5 ΔH(KJ) - - - 1.8 x 10 4

30. ENERGY BALANCE Reactor Feed entering (stream 2) = 4222.912 Kg/hr. Moles of nitrobenzene in feed = 32.236 Kmol/hr. Moles of hydrogen in feed = 128.942 Kmol/hr. Moles of aniline formed = n aniline = 31.304 Kmol/hr. Moles of water formed = n water = 62.859 Kmol/hr. Moles of cyclohexylamine formed = n CHA = 0.126 Kmol/hr. Moles of nitrobenzene unreacted = 0.806 Kmol/hr. Mole of hydrogen unreacted = 34.402 Kmol/hr. Component C p (J/mol o C) Aniline 176.031 H 2 O 1.9241 H 2 0.732 Nitrobenzene 1.488 CHA 48.549 Products 43.785 Reactants 0.8832

31. ENERGY BALANCE Heat of reaction: At 270 o C ΔH reaction at 25 C = -552000 J/mol ΔH reaction at 270 C = -541489.059 J/mol Heat to be removed during reaction = aniline produced x ΔH reaction at 270 C = -1.7x10 7 KJ/hr. Entering temperature of cooling water = T 1 = 32 o C Exit temperature of cooling water = T 2 = 49 o C Specific heat of cooling water = c p = 2.701 KJ/Kg o C The mass of cooling water is required to remove heat = m c p Δ T = 37023.33kg/hr.   Stream# 2 3 Temperature( o ) 270 270 Pressure (KPa) 344.5 235 Δ H(KJ/hr.) 3.48 x 10 4 1.430 x 10 6

32. ENERGY BALANCE Heat Exchanger(H-3) Temperature of stream 3 = T 1 = 270 o C Temperature of stream 3a = T 2 = 40 o C Molar flow rate of stream 3 = n 3 = 129.497 Kmol/hr. Molar flow rate of stream 3a = n 3a = 129.497 Kmol/hr. C p,3 = 2.76 J/g o C = 88.774 KJ/mol o C Latent heat of mixture = 26794.14 J/mol Heat taken by the mixture = n 3 [(c p ΔT + latent heat of mixture)] = 6709988.03 KJ/hr. Stream# 3a 4 7 8 9 Temperature( o ) 40 40 40 40 40 Pressure (KPa) 235         Δ H(KJ/hr.) 5.7 x 10 3 744.682 5.0 x 10 3 3.9 x 10 3 1x 10 3

33. ENERGY BALANCE Distillation Column 1 Molar flow rate of stream 13 = m 13 = 31.486 Kmol/hr. Bubble point temperature = 184 o C ΔT = 159 o C C p, mix = 130.7875 KJ/Kmol o C Q bottom = m 13 c p, mix ΔT = 654758.1 KJ/hr. Molar flow rate of stream 12 = m 12 = 0.513 Kmol/hr. Dew point temperature = 112 o C Latent heat of mixture = 39833.73 KJ/Kmol Q top = m x latent heat of mixture = 204601.704 KJ/hr. Molar flow rate of stream 9 = 32.001 Kmol/hr. ΔT = 15 o C c p = 93.9262 KJ/Kmol o C Q feed = 45085.984 KJ/hr. Heat supplied by the reboiler = Q top + Qbottom – Q feed = 814273.82 KJ/hr..

34. Energy balance Steam at 200psig: Temperature of steam = 197.027 o C Latent heat of steam = 2790.15KJ/Kg Mass of steam required = (814273.82/2790.16) = 291.838 Kg/hr . Distillation Column 2 Molar floe rate of stream 19 = m 19 = 31.269 Kmol/hr. Bubble point temperature = 183 o C ΔT =158 o C c p,mix =131.071 KJ/Kmol o C Q bottom = m 19 x c p,mix x ΔT = 647556.538 KJ/hr . Stream# 12 13 16 19 Temperature( o C) 112 184 166 183 Pressure (KPa)         ∆H(KJ/hr.) 1.5x 10 2 7.7 x 10 5 2.02 x 10 3 7.6 x 10 5

35. designing Vaporizer Vaporizers are heat exchangers which are specially designed to supply latent heat of vaporization to the fluid. In some cases, it can also preheat or super heat the fluid then this section of vaporizers will be called upon preheating zone or superheating zone and the other section in which latent heat is supplied; is known as vaporization zone but he whole assembly will be called a vaporizer. Vaporizers are called upon to fulfill the multitude of latent-heat services which are not a part of evaporative or distillation process TYPES OF VAPORIZERS Some common types of vaporizers are Vertical or Horizontal vaporizer Kettle type vaporizer Tubular low temperature vaporizer

36. designing Kettle Type Vaporizer The major distinguishing feature of a Kettle type vaporizer is its over-sized shell which allows adequate space required for the disengagement of vapors. It has a negligible pressure drop in the shell side and it is horizontal so there is no need to give hydrostatic head. These factors make it the optimum selection for our process.

37. Designing External view Internal view

38. dowtherm:hot fluid Internal view The hot fluid in this particular case will be Dowtherm A. It is a eutectic mixture of 73.5% diphenyl-oxide and 26.5% biphenyl. It can be used both in the liquid and vapors phases. Its operating temperature is from 12 C to 400 C. Another advantage of Dowtherm A is that it is non-corrosive to steel.

39. Physical properties of Nitro benzene (Cold Fluid) Mass flowrate of nitrobenzene = m = 3965kg/h Pressure = 260 kPa Temperature of Feed = t 1 = 25 C Boiling Point of Nitrobenzene = t 2 = 270 C Latent Heat of vaporization = 311 kJ/kg Mean Specific heat = Cp=2.09kJ/kg C PHYSICAL PROPERTIES OF DOWTHERM A (HOT FLUID) Mass of Dowtherm A = m = ? Temperature entering = T 1 = 400 C Temperature Leaving = T 2 =290 C Specific Heat =C p = 2.7kJ/kg C Specific Gravity = sp.gr = 0.854 API = 34.19

40. Design Steps & Calculations Heat Balance Dowtherm A,   Q T = m c p ∆t So, m = 11000 kg/h Nitrobenzene,   Q 1 = mc p ∆t =2030278 kJ/h   Q 2 = m λ =1235612 kJ/h Q T = Q 1 + Q 2 = 3265891 kJ/h

41. 2. Log Mean Temperature Difference (∆T lm ) (∆T lm ) = = 189 C   R = = 0.448   S = = 0.653   F t (1) = 0.88   ∆T = 161 C

42. 3. Caloric Temperature of Hot & Cold Fluid T c = = 2.038 K c (1) = 0.3 F c (2) = 0.55   t c = t 2 + F c (t 1 -t 2 ) = 160 C   T c = T 2 + F c (T 1 -T 2 ) = 350 C

43. Heat Transfer Co-efficient Calculation Hot Fluid: Tube side, Dowtherm A Assumptions:   Length of tube = 4.87 m OD = 0.019 m BWG = 16 Passes = 2 Max allowable Heat flux (1) = 37.85 kW/m 2  

44. 4. Number of Tubes Surface Area = Q T /q = 24 m 2 Surface area per linear ft. a t ’’ (1) = 0.018 m 2 No. of tubes N t = = 81 Now, Pitch = 0.025 m (triangular) So, N t (2) = 82   (2) D.Q Kern Table 9. Page # 842 (1) D.Q Kern Table 10. Page # 843

45. 5. Flow Area for Tubes 6. Mass Velocity Flow area per tube = a t ’ (1) = 0.0002 m 2 ID (2) = 0.620 in = 0.02 m a t = = 0.008 m 2   G t = m/a t = 1375800 kg/h.m 2 (1) D.Q Kern Table 10. Page # 843 (2) D.Q Kern Table 10. Page # 843

46. 7. Heat Transfer Co- efficients     h i = = 1318 W/m 2 . C h io = h i ×ID/OD = 1090 W/m 2 . C The correction factor is negligible   At, T c = 350 C Viscosity (1) = 2.42 cp R e = ≈ 40000   J h (2) = 120 At 350 o C & 34.19 API, (3)   k(C p µ /k) 1/2 = 0.2 W/m. C   2) D.Q Kern fig 24. Page # 834 (3) D.Q Kern fig 16. Page # 826

47. 8. Cold Fluid : Shell side, Nitrobenzene Assume, h o = 300 t w = t c + ( h io / h io + h o )(T c - t c ) = 252 C ∆ (t w ) = 74 C Hence, h o (1) = 300 1) D.Q Kern fig. 15.11 Page # 474

48. 9. Overall Clean & Design Heat Transfer Co- efficients . U d (1) = = 212 W/m 2 . C R d = = 0.013 s.m 2 . C/W   U c = = 664 W/m 2 . C Total surface = N t × L × a t ’’ A = 26 m 2      

49. Tube Side Pressure Drop For, R e = 40000 ƒ = 0.0002 (1) s = 0.854 ∆P t = = 1.51 kPa   ∆P r = = 0.645 kPa     ∆P T = ∆P t + ∆P r = 2.16 kPa Shell side The pressure drop in the Shell side of a Kettle Type Vaporiser where the cold fluid is present is negligible. (1) Hence it can be used for high pressures.

50. Specification sheet Properties Shell side Tube Side Fluid Circulated Nitrobenzene Dowtherm A Inlet Temperature 25 C 400 C Outlet Temperature 270 C 290 C Pressure Drop Negligible 2.16 kPa Mass Flow Rate 3965 kg/hr 10992 kg/hr Material of Construction Carbon Steel Stainless Steel Specifications ID = 0.416 m C = 0.0064 m B = 0.26 m OD = 0.019 m 16 BWG Pitch = 0.025 m (Triangular) Length = 4.87 m Number of tubes= 82 Unit - Kettle type Vaporiser No. of Units - 1

51. Hazop study Is a systematic way to identify possible hazards in a work process. In this approach the process is broken down into steps, and every variation in work parameters is considered for each step, to see what could go wrong. HAZOP is a qualitative technique best on use of guideword and it is carried out by multi-disciplinary team(Hazop team) during a set a point. The reasons for carrying out hazard and operability study are, Primary to identify hazards To resolve these hazard Why Hazop study carried out? Hazop study method utilizes the skill, experience, knowledge and imagination of the individual expert in the field of design and operation.

52. Hazop study Hazop team composition The basic team for a process plant HAZOP study well consist of : Hazop study team leader Project engineer Process engineer Mechanical engineer Instrument /Electrical engineer HAZARD: hazard is anything which has potential to cause harm. Example: Slippery surface, moving machinery, dust etc. RISK: risk is a chance of causing harm from hazards. Example: chance of fall a person, damage by fire, flood etc.

53. HAZOP STUDY guide word ADVANTAGE: Increase the plant safety reduce the risk of production interruption Less waste is produced Less down time DISADVATAGE: Time consuming Require significant resources commitment Guide Words Meaning No Less More Part of As well as Reverse Other than Negation of design intent Quantitative decrease Quantitative increase Qualitative decrease Qualitative increase Logical Opposite of Intent Complete substitution  

54. Instrumentation Definition Instrument is a collective term for measuring instruments that are used for indicating, measuring and recording physical quantities such as flow, temperature, level, pressure and composition. Sensors Sensor is an object whose purpose to detect events or changes in its environments and then provides a corresponding output. Transmitter These are interface between process and control systems and its function to convert the sensor signal into control signal. Set point Set point is a desired process out put that an automatic control system will aim to reach.

55. instrumentation Distillation column control The main objective to control the distillation column is that to maintain the distillate composition and bottom. Temperature, composition and flowrate of feed The pressure of steam supply. The pressure of cooling water and header temperature. Control ambient condition which create change in internal reflux.

56. Cost estimation Definition A cost estimation is the approximation of the cost of a program, project, or operation. The cost estimate is the product of the cost estimating process. Fixed tube sheet, 1-2 heat exchanger Shell material =Carbon Steel (CS-202) Tubes material =Copper Heat transfer area =15.18 m 2 Pressure factor = 1 Type factor = 1.3 Purchased Cost= Bare cost × Type factor × Pressure factor = 9000 × 1.3 × 1 =$11700

57. Cost estimation Cost index 1998 = 1077.1 Cost index 2005 = 1299.6 Cost in year 1998 =$11700 Cost in year 2005 = Cost in year 1998 × = (11700 × 1299.6) / 1077.1 Cost in year 2005 = $ 14116.91 Power required = 1377.535 kW Purchased cost in year 2002 = $800000 Cost index 2005 = 1299.6 Cost index 2002 = 1116.9 Cost in year 2005 = Cost in year 2002 × Purchased cost = (800000 × 1299.6) / 1116.9 Purchased cost = 930862.2  

58. conclusion The purpose of this project is to produce the aniline from nitrobenzene. We select the first process because the production of aniline is maximum The catalyst we used here is hydrochloric acid with water and copper. Due to cheap hydrogen we use maximum 80% and 20% nitrobenzene. We neglect the other two process because of phenol formation and azeotropic formed of aniline and phenol due to same vapor pressure curve. The flow chart we draw by using MS VISIO. The whole process is divided into three sections, feed preparation, separation and purification. The capacity of production of aniline is 25000 tons/year. We have done four designs here in the project. Vaporizer design, distillation column, fluidized bed reactor and condenser. Kettle type vaporizer we used here in the process. Because the shell area is large which produce more vapor.