64788 - PDFCOFFEE.COM

1. Home
2. 64788

64788

• Author / Uploaded
• ghatak2

• Categories
• Chemical Reactor
• Employment
• Transport
• Water
• Chemistry

Phthalic anhydride 1 CONTENTS S.No. CHAPTER NAME 1. INTRODUCTION 2. PROPERTIES 3. USES 4. SELECTION OF RAW

Views 31 Downloads 3 File size 709KB

Report DMCA / Copyright

DOWNLOAD FILE

Recommend Stories

Citation preview

Phthalic anhydride 1 CONTENTS S.No. CHAPTER NAME 1. INTRODUCTION 2. PROPERTIES 3. USES 4. SELECTION OF RAW MATERIAL 5. MATERIAL BALANCES 6. ENERGY BALANCES 7. EQUIPMENT DESIGN 8. COST ESTIMATION 9. PLANT LOCATION 10. PLANT LAYOUT 11. REFERENCES & BIBILIOGRAPHY 2 INTRODUCTION Phthalic Anhydride is an important aromatic di-carboxylic acid anhydride. It is an ortho-derivative of phthalic acid. Its abbreviation is PAN. Preparation Methods: It is normally prepared either by oxidation of Naphthalene or ortho-xylene. a) From Naphthalene: C10H8 + 4.5 O2 C8H4O3 + 2 CO2 + 2 H2O (g) b) From o-xylene: C8H10 + 3O2 C8H4O3 + 3 H2O PROPERTIES a. Physical Properties: Molecular Formula: C8H4O3 Molecular weight : 148.1 Form: Flakes Molten State: Clear liquid that resembles water Odor: Mild Characteristic Color: White Melting Point: 130.6◦ C Boiling Point: 284.5◦ C Density g/ml at 20◦C: 1.527 Vapor Pressure mm Hg at 25◦ C: 0.00052 Solubility : Slightly soluble in hot water and ether; sublimes below the Melting point Latent Heat of Vaporization: 11850 cal/mol 3 b. Chemical Properties: PAN is an aromatic di-carboxylic acid. So it exhibits the reaction characteristics of dicarboxylic acid. Some of the typical chemical reactions are: Polymerization Reactions: C8H4O3 + CH2OH – CHOH - CH2OH (PAN) (Glycerol) -H2O Glyptal (an alkyd resin) (Polyester) Alcoholysis (conversion into esters): C8H4O3 + CH3CH2CHOHCH3 (PAN) sec-Butyl hydrogen phthalate (Sec-Butyl alcohol) Friedel- Crafts acylation(Formation of Ketones): C8H4O3 + (PAN) C6H6 AlCl3 0◦C o-Benzoylbenzoic acid (Benzene) USES PAN has various applications and a few of the areas are as follows: • • • • • • • • • Insecticides Pharma Products Reinforced Plastics Paints Plasticizers for PVC, Cellulose, Resins & Cellophane Dyes, Plastics, Printing Inks, Varnishes Printing Inks Dyes SELECTION OF RAW MATERIAL 1. PAN can be manufactured from o-xylene, n-pentane and naphthalene. 4 2. Ortho-xylene is clearly preferred modern day feed stock for PAN manufacture. 3. Compared to n-pentane and naphthalene, o-xylene produces higher yields (all carbon atoms appear in the product), it is cheaper than naphthalene and provides a more efficient process. 4. Converting an existing plant to o-xylene can reduce raw material costs by 25%. 5. Advances are also being made in catalyst selectivity to improve the quality of the product and eliminate side reactions. SELECTION OF REACTOR 1. Phthalic anhydride processes can be classified according to the type of reactor used. 2. Three reactor configurations have been commercially developed. (a) Fixed-bed vapor phase reactor (b) Fluidized bed vapor phase reactor and (c) Liquid phase reactor 3. The fixed-bed vapor phase reactors have proven to be superior to fluidized-bed and liquid phase reactors. 4. Fluidized bed processes have proved difficult to maintain and have suffered from erosion problems and excessive catalyst losses. 5. The construction costs of liquid-phase processes have proven prohibitive. METHOD OF PRODUCTION a) Raw Materials: Air and o-xylene b) Basis: 60,000 metric tons/year of PAN product, 98 wt% purity (or) 8400 kg/hr (approx.) of PAN product, 98 wt% purity c) Feed Requirements: o-xylene: 8480 kg/hr Air: o-xylene (wt ratio): 10:1 Air: 84800 kg/hr d) Chemical Reactions: Major reaction: C8H10 + 3O2 C8V H24OO5 3 + 355-375 °C 3 H2O Major side reaction: 5 C8H10 + 10.5O2 Minor side reaction: C8H10 + 7.5 O2 5 H2O + 8CO2 C4H2O3 + 4 H2O + 4CO2 e) Process Flow Sheet: 6 f) Process Description: The raw materials are air and o-xylene. The o-xylene feed, which may be considered pure and at 0.75 atm, is pumped to 3 atm and then vaporized in a 7 fired heater. Air, which may be assumed to contain only O2 & N2 is compressed to 3atm and heated in a heat exchanger. The hot air and vaporized o-xylene are mixed and sent to a packed bed reactor. Packed bed reactor is multitubular reactor filled with supported V2O5. 100% of oxylene is reacted in this reactor. The selectivity data is given in the following table: Fractional Conversion of o-Xylene into (Yield of) Indicated Product: T(°C) 300 320 340 360 380 400 420 Maleic anhydride 1.00 0.536 0.215 0.100 0.0463 0.0215 0.00 CO2 0.00 0.0339 0.102 0.200 0.356 0.602 1.00 Phthalic anhydride 0.00 0.425 0.683 0.700 0.598 0.377 0.00 In the selectivity data table we observe the fractional conversion of o-xylene into Phthalic anhydride and maleic anhydride are 0.7 and 0.1 at 360◦ C respectively. Also, the selectivity for complete combustion reaction is 0.2. Therefore, the reactor is maintained at 355-365◦ C. Also, the reactor is maintained at 3 atm and a contact time within the reactor is about 0.1-0.4 seconds. Since all the three reactions taking place in a reactor are highly exothermic, the temperature is controlled around 355-365◦ C using Molten salt (High Heat Transfer Salt). The reactor effluent, which is at 2 atm enters a complex series of devices known as switch condensers. The feed to switch condensers may be no higher than 180◦C, hence reactor effluent must be cooled. The net result of switch condensers is that all light gases and water leave the top stream of reactor with small amounts of both anhydrides (PAN & MAN), while large amounts of PAN and MAN leave in stream which is feed to distillation column. The stream containing large amounts of PAN is fed to distillation column which results in formation of bottom product of quality 98% PAN. g) Catalyst: It is a combination of Vanadium Pentoxide(V2O5) and Titanium dioxide(Ti2O5).It is prepared by heating a suspension containing compounds and spray coating the suspension onto a support which can be either 6mm porcelain spheres. Surface Area available for reaction is about 10-12 m2/g. h) Major Engineering Problems: 1. Explosion Hazards-minimized by adding excess air to stay below the lower explosive limit. 8 2. Fixed bed tubular reactor design-tube size and heat transfer must be considered to avoid too high a center temperature within each catalytic tube. 3. Catalyst development for high specificity of oxidation. 4. Choice of coolant –mercury and diphenyl are also used. If molten salt is used, then corrosion will be a problem. MATERIAL BALANCE Let us assume that 10 kmol of o-xylene is fed to a reactor. Selectivity for PAN reaction= 0.7 C8H10 + 3O2 C8H4O3 + 3 H2O o-xylene reacted = 7 kmol Oxygen reacted = 21 kmol PAN formed = 7 kmol Water (g) formed = 21 kmol Selectivity for complete combustion reaction=0.2 C8H10 + 10.5O2 5 H2O + 8CO2 o-xylene reacted = 2 kmol Oxygen reacted = 21 kmol Carbon dioxide formed = 16 kmol Water (g) formed = 10 kmol Selectivity for MAN reaction=0.1 C8H10 + 7.5 O2 C4H2O3 + 4 H2O + 4CO2 o-xylene reacted= 1 kmol Oxygen reacted =7.5 kmol 9 Carbon dioxide formed =4 kmol Water(g) formed= 4 kmol Maleic anhydride = 1 kmol Molecular weight of o-xylene =106 Therefore 10 kmol of o-xylene = 1060 kg On weight basis, Air : o-xylene = 10:1 Air fed =10,600 kg = 367.55 kmol Oxygen, Nitrogen in air are 77.18 kmol, 290.37 kmol. Excess Oxygen = 77.18 – 49.5 = 27.68 kmol Nitrogen fed = 290.37 kmol Material Balance around the reactor : Reactants entering: o-xylene = 10 kmol = 10*106 = 1060 kg Oxygen fed = 77.18 kmol = 77.18*32 = 2469.76 kg Nitrogen fed = 290.37 kmol = 290.37*28 = 8130.36 kg Total weight = 1060+10600=11660 kg Products leaving: PAN = 7 kmol = 7*148 = 1036 kg H2O(g) = 35 kmol=35*18 = 630 kg CO2 = 20 kmol =20*44 = 880 kg MAN = 1 kmol = 1*98 = 98 kg O2 = 27.68 kmol = 27.68*32 = 885.76 kg N2 = 290.37 kmol = 290.37 *28 = 8130.36 kg 10 Total weight = 11660.12 kg Law of Conservation of mass is satisfied. Material Balance around the Switch Condenser: Feed to switch condenser: PAN = 7 kmol H2O (g) = 35 kmol CO2 = 20 kmol MAN = 1 kmol O2 = 27.68 kmol N2 = 290.37 kmol Assume that 2 mol% and 0.2 mol% of MAN and PAN present in the feed leave in the top stream. Top Stream Leaving the Condenser: PAN = (0.2/100)*7 = 0.014 kmol H2O (g) = 35 kmol CO2 = 20 kmol MAN = (2/100)*1= 0.02 kmol O2 = 27.68 kmol N2 = 290.37 kmol Bottom stream Leaving the Condenser: MAN = 0.98 kmol = 0.98*98 = 96kg PA N = 6.986 kmol = 6.986*148 = 1034 kg Material Balance around the Distillation Column: Feed to Distillation Column: 11 MAN = 96 kg PA N = 1034 kg  Total (F) = 1130 kg  PAN composition (weight basis) in Distillate (D) and Residue (W) are 2% and 98% respectively Overall Material Balance: F = D + W where F= 1130 kg (wt. basis) PAN balance (wt. basis) : xF*F = xD*D + x W* W Where xF*F = 1034 kg xD = 0.98 xW = 0.02 Solving above two equations, we get D = 76.5 kg; W = 1053.5 kg Aim: To produce PAN product, 60000 metric tons/year, 98 wt% purity. Here 1 year = 300 days 60,000 tons/yr = 200 tons/day=25/3 tons/hr=25000/3 kg/hr; On similar basis of Material Balance using 10 kmol of o-xylene, inorder to produce 60,000 tons/yr of bottom product we have D = 605.1 kg/hr PAN and MAN in D are 12.1 kg and 593 kg; respectively. PAN = 24500/3 kg/hr; MAN = 500/3 kg/hr; Total PAN to distillation column = 24562.1/3 = 55.26 kmol/hr Total PAN to distillation column = 2279/3 = 7.75 kmol/hr We know that 2% and 0.2% of MAN and PAN present in the stream leaving the reactor is discharged to atmosphere. Total PAN leaving the reactor = 55.26 /0.998 = 55.37 kmol/hr 12 Total MAN leaving the reactor = 7.75/0.98 = 7.91 kmol/hr Selectivity of MAN reaction is 0.1. Therefore, o-xylene to be charged to reactor = 7.91/0.1 = 79.1 kmol/hr Start: o-xylene fed = 80 kmol/hr Material Balance around the reactor : Reactants entering: o-xylene = 80 kmol/hr = 80*106 =8480 kg/hr On weight basis,Air: o-xylene = 10:1 Air fed = 84,800 kg/hr = 2940.4 kmol/hr Oxygen fed = 617.5 kmol/hr = 617.5*32 = 19760 kg/hr Nitrogen fed = 2322.9 kmol/hr = 22322.9*28 = 65040 kg/hr Reactants o-xylene kmol/hr 80 mol.wt 106 kg/hr 8480 O2 617.5 32 19760 N2 2322.9 28 65041.2  93281.2  Total Selectivity for PAN reaction= 0.7 C8H10 + 3O2 C8H4O3 + 3 H2O o-xylene reacted = 80*0.7 = 56 kmol/hr Oxygen reacted = 56*3 = 168 kmol/hr PAN formed = 56*1 = 56 kmol/hr Water(g) formed = 56*3 =168 kmol /hr Selectivity for complete combustion reaction=0.2 13 C8H10 + 10.5O2 5 H2O + 8CO2 o-xylene reacted = 80*0.2 =16 kmol/hr Oxygen reacted =10.5*16 = 168 kmol/hr Carbon dioxide formed = 8*16 = 128 kmol/hr Water (g) formed = 5*16 = 80 kmol/hr Selectivity for MAN reaction=0.1 C8H10 + 7.5 O2 C4H2O3 + 4 H2O + 4CO2 o-xylene reacted = 80*0.1 = 8 kmol/hr Oxygen reacted = 7.5 *8 = 60 kmol/hr Carbon dioxide formed = 4*8 = 32 kmol/hr Water (g) formed = 4*8 = 32 kmol/hr Maleic anhydride = 8*1 = 8 kmol/hr Excess oxygen = 617.5-(168+168+60) = 221.5 kmol/hr Products leaving: Products PAN H2O (g) CO2 MAN o-xylene O2 N2 kmol/hr 56 280 160 8 0 221.5 2322.9 mol.wt 148 18 44 98 106 32 28 Total kg/hr 8288 5040 7040 784 0 7088 65041.2  93281.2  Law of Conservation of mass is satisfied. Material Balance around the Switch Condenser: Feed to switch condenser: PAN = 56 kmo/hr 14 H2O = 280 kmol/hr CO2 = 160 kmol/hr MAN = 8 kmol/hr O2 = 221.5 kmol /hr N2 = 322.9 kmol/hr Assume that 2 mol% and 0.2 mol% of MAN and PAN are present in the feed leaving the top stream. Top Stream Leaving the Condenser: PAN = (0.2/100)*56 = 0.112 kmol /hr H2O(g) = 280 kmol/hr CO2 = 160 kmol/hr MAN = (2/100)*8 = 0.16 kmol/hr O2 = 221.5 kmol /hr N2 = 2322.9 kmol/hr Bottom stream Leaving the Condenser: MAN = 7.84 kmol/hr = 7.84*98 = 768.3 kg/hr PA N = 55.888 kmol/hr = 55.888*148 = 8271.4 kg/hr  Total = 9039.7 kg/hr  Material Balance around the Distillation Column: Feed to Distillation Column: MAN = 768.3 kg/hr PA N = 8271.4 kg/hr ----------------------------------Total (F) = 9039.7 kg/hr 15 -----------------------------------PAN composition (weight basis) in Distillate(D) and Residue(W) are 2% and 98% respectively. Overall Material Balance: F = D + W , where F= 9039.7 kg/hr (wt basis) PAN balance (wt basis) : xF*F = xD*D + x W* W Where xF*F=8271.44 kg xD= 0.98 xW= 0.02 Solving above two equations, we get D= 611.98 kg/hr; W= 8427.71 kg/hr Rounding Off: Phthalic anhydride Product (W) = 8400 kg/hr Distillate product (D) = 640 kg/hr ENERGY BALANCE Since the gases are at high temperature and low pressure, the gases can be assumed to be ideal . Compound o-xylene air mole fraction 0.0265 0.9735 Specific heat of reacting mixture (air and o-xylene) (Cp[M1]) = 6.386 + 4.885 *10-3 *T -2.179 *10-6 *T2 + 4.765*10-9 *T3 -3093.39 T -2 ∆H net = 3020.4*1000 ∫(6.386 + 4.885*10-3*T – 2.179*10-6*T2 + 4.765*10-9*T3 – 3093.39*T-2) dT where T is from 298 K to 593 K = 3020.4*1000*2526.38 = 0.763*1010 Cal/hr = 31940.46 MJ/hr ∆H1= o-xylene stream; ∆H2= air stream; ∆H1+∆H2 =∆H net 2.508 *10-3 *T2 + 6.56 *T -4945.12 +3177.6* T -1 = 0 Solving T = 610.5 K = 337.5◦ C 16 Now, ∆H1= 80*1000*∫(-3.786 + 0.1424*T – 8.224*10-5*T2 + 1.798*10-7*T3)dT where T is from 423 K to 610.5 K ∆H2 = 27555.47 MJ/hr o-xylene stream: Assuming 1% losses, Heat transfer rate supplied by Natural gas to o-xylene stream=4385/0.99 = 4429.3 MJ/hr Calorific Value of Natural gas = 54 kJ/g Amount of Natural gas: m(54) = 4429.3 *1000 kJ/hr m=82 kg/hr Air stream: Assuming 1% losses, Heat transfer rate supplied by steam to air stream = 27555.47/0.99 = 27833.81 MJ/hr Energy Balance around the Reactor: Heat transfer rate required to raise the reacting mixture from 320◦ C to 360◦ C = 3020.4*1000* 386.08 = 0.1166 *1010 cal/hr = 4881.1 MJ/hr PAN reaction: C8H10 + 3O2 C8H4O3 + 3 H2O ∆H◦ 298 K = -371.79 + 3(-242) -19.01 = -1116.8 KJ/mol ∆H◦ 633K = -1116.8 +11.814 = -1104.98 kJ/mol ∆H net(1)= n(∆H◦ 633K ) = 56 *1000*(-1104.98)*1000=-61878.88MJ/hr MAN reaction: C8H10 + 7.5 O2 C4H2O3 + 4 H2O + 4CO2 ∆H◦ 298 K = -469.65 + 4(-242) + 4(-393.77) -19.01 = -3031.74 kJ/mol 17 ∆H◦ 633K= -3031.74 -7.69 =-3039.43 kJ/mol ∆H net(2)= n(∆H◦ 633K ) = 8*1000*-3039.43*1000 = -24315.44 MJ/hr Complete Combustion: C8H10 + 10.5O2 5 H2O + 8CO2 ∆H◦ 298 K = 8(-393.77)+5(-242)-19.01 = -4379.17 kJ/mol ∆H◦ 633K = -4379.17 + 174.785= -4204.38kJ/mol ∆H net(3) = n(∆H◦ 633K ) =16*1000*-4204.38*1000 = -67,270.16 MJ/hr Overall Heat transfer rate liberated = -61878.88 -24315.44-67270.16 = -153464.48 MJ/hr Let us allow the Products to leave the reactor at 400◦ C. ∆HO2 = 288.68 MJ/hr ∆HN2=(23 22.9*1000)(1218.53)=2830.51 MJ/hr ∆HMAN= (8000)*5392.32= 43.14 MJ/hr ∆HCO2 = (160*1000)*1945.38 =(160*1000) * 1945.38 =311.26 MJ/hr ∆HH2O = (280*1000)*1479.97=414.39MJ/hr ∆HPAN= 600.41 MJ/hr Overall Heat utilized for raising products from 360◦ C to 400◦ C = 4488.4MJ/hr Net enthalpy heat rate change with in the reactor =4881.1-153464.5+4488.4 = -144095.01 MJ/hr 144095.01 MJ/hr is to be utilized by molten salt. Energy balance around the salt cooler: Q = m Cp ∆ T 144095.01 *106 = m *1560*(395-150) m= 377 tons/hr =104.72 kg/s Energy balance around the heat exchanger – 2(HE-2): Boiler feed water (bfw) available at 549 kPa,90◦ C It is to be delivered at high pressure (4300 kPa) Saturated steam enthalpy (at 4300 kPa and 254◦ C)=2799.4 kJ/kg At heat exchanger-2, 144095.01*106= m(2799.4*1000) m = 51473.5 kg/hr Flow rate of boiler feed water = 51473.5 kg/hr Heat possessed by products leaving reactor at 400◦ C : 18 ∆HO2=221.5 *1000*11633.92 = 2576.9 MJ/hr ∆HN2 = (2322.9*1000)*11120.96 = 25832.9 MJ/hr ∆HMAN = 8000*40662.54 = 325.3 MJ/hr Boiling point of water = 406.7 K at 3 atm ∆HH2O = 16403.4 MJ/hr ∆HPAN=56*1000*82634.21= 4627.52 MJ/hr ∆HCO2 =160*1000*16441.09 =2630.58 MJ/hr Overall Heat possessed by products leaving the reactor = 2576.9 + 25832.9 +325.3 +2630.58 +16403.4+4627.52 = 52396.6 MJ/hr Energy balance around the heat exchanger – 3(HE-3): Energy possessed by stream after heat exchanger-3 : ∆HO2 = 221.5 *1000* (3735.32)=827.37 MJ/hr ∆HN2 =2322.9 *1000 *3651.46=8481.97 MJ/hr ∆HMAN = 8*1000*8713.958=69711664 cal/hr=291.8 MJ/hr ∆HH2O =13929.75 MJ/hr ∆HPAN=56*1000*8713.958=487981648 cal/hr=2042.59 MJ/hr ∆HCO2 = 794.86 MJ/hr Total Heat = 26368.3 MJ/hr 52397-26368=26029 MJ/hr is being utilized by heat exchanger – 3, Boiling Point of water = 427 K at 525 kPa Saturated steam enthalpy =2749.7 kJ/hr m (2749.7*1000) =26029 *106 m= 9466 kg/hr Flow rate of cool water = 9466 kg/hr Energy balance around the Switch Condenser: Energy possessed by effluent stream: Effluents leave the heat exchanger – 3 at 140◦ C ∆HO2 = 221.5 *1000* (3431.55) = 760.09 MJ/hr ∆HN2 =2322.9 *1000 *3358.4=7801.2 MJ/hr ∆HMAN= 160*9797==1.57 MJ/hr ∆HH2O =13832.94 MJ/hr(since B.P of water = 393 K at 2 atm) ∆HPAN= 112 * 20480 = 2.29 MJ/hr ∆HCO2 =160*1000*4548.34= 727.73 MJ/hr Total energy possessed by effluent stream =23125.76 MJ/hr Since Cp (MAN) =Cp (PAN) in liquid state, we have Cp (MAN) =Cp (PAN) = Cp (Feed) Specific enthalpy = 7972.15 Cal/mol = 33369.8 kJ/kmol Total enthalpy of feed = (55.888+7.84) (33,369.8) = 63.728 *33,369.8 19 = 2126.6 MJ/hr Energy remaining =26368.3 – (2126.64+23,125.8) = 1115.9 MJ/hr Let cool water be condensing medium and its temperature rise by 10◦C i.e., from 30 ◦C to 40◦C 1115.9 *1000 =m (4.18)(40-30) m=26696.2 kg/hr=26.7 T/hr 26.7 T/hr of cooling water is being used. Energy balance around the Distillation Column: In W, 55.62 and 1.71 kmol/hr of PAN and MAN are present. In D, 0.086 and 6.4 kmol/hr of PAN and MAN are present. Solidification point of PAN = 130.8◦C Bottom product is at 150◦C Specific enthalpy of bottom product (hW ) =8713.96 cal /mol =36474.9 kJ/kmol Total Enthalpy of bottom product = (55.62+1.71) (36474.9) = 57.33 * 36474.9 = 2091.1 MJ/hr ◦ Top Product is at 60 C (333K) Specific enthalpy of top product (hD) = 2317.48 cal/mol =9700 .5 kJ/kmol Total enthalpy of Top product =(6.4+0.086) (9700.5) =6.486 *9700.5 -62.92 MJ/hr Composition of MAN and PAN in D are 0.987, 0.013 Assuming PAN and MAN obey’s Raoults law, y = ax/ (1+(a-1)x); where a=sqrt( atop* a bottom) Here MAN is more volatile component. Using Antonie’s equation, Vapor pressure of MAN at 333 K, 423 K are 3.536 mm Hg, 186.2 mm Hg. Vapor pressure of PAN at 333 K, 423 K are 0.153 mm Hg, 17.36 mm Hg. atop =3.536/0.153 =23.11; a bottom =186.2/17.36= 10.73 a=sqrt( atop* a bottom) =sqrt(23.11*10.73) = 15.75 y = 15.75x/(1+14.75x) is equilibrium relation. Let us assume that boiling point of feed varies linearly with composition Boiling point of PAN and MAN are 560 K and 473 K For feed composition, Boiling point of feed =(55.888*560+7.84*473)/(55.888+7.84) = 549.3 K Specific enthalpy of feed if feed is saturated liquid =18,752.08 cal/mol = 78492.5 kJ/kmol Specific enthalpy of feed = 7972.15 cal/mol =33369.8 kJ/kmol Normal Heats of vaporization PAN 11850 cal/mol = 49601.73 kJ/kmol 20 MAN 5850 cal/mol = 24486.93 kJ/kmol Let us assume that latent heat of vaporization of feed varies linearly with composition λ feed = (55.888*49601.73 +7.84 *24486.93)/(55.888+7.84) = 46511.66 kJ/kmol Specific enthalpy of feed if feed is saturated vapor = Specific enthalpy of saturated liquid + Latent heat of vaporization of feed =78492.5 +46511.66 =125004.2 kJ/kmol q= (Hv - hF )/(Hv– hL ) = (125004.2-33369.8)/(125004.2- 78492.5) = 1.97 Assume that constant Molar overflow rate is prevailing. Feed line is y = qx/(q-1) – xF /(q-1) = 2.03 x -0.127 Point of intersection of feed line and equilibrium line is 2.03 x -0.127 = 15.75x / (1+14.75x) x = 0.53 => y = 0.95 For minimum reflux (Rm) Top section operating line passes through pinch point (0.53,0.95) and (0.98,0.98) Slope = (0.98- 0.95)/(0.98- 0.53) = 1/15 = Rm / ( Rm + 1)  Rm =1/14 Ropt = 1.5 Rm =1.5/14 =0.107 Mole fraction of MAN in top product =(627.2/98)/[(627.2/98) +(12.8/98)] = 0.987 Mole fraction of PAN in top product =1- 0.987 =0.013 Assuming boiling point varies linearly, Boiling point of Top product = 0.987 * 473 + 0.013 *560 =474.13 K λ top product = 0.987*24486.93 +0.013 *49601.73 = 24813.42 kJ/kmol Specific enthalpy of top product if it is saturated vapor = 12628.32 +24 813.42 =37441.74 kJ/kmol Energy balance around Total Condenser: V*HV = D*hD + Lo* hLo + Qc Using V=(R+1)D, Lo=R*D, hLo= hD Q c = (R+!) D [HV –hD] =(1+0.107)(6.4+0.086 )[37441.74-9700.5] = 199.2 MJ/hr Cooling water is available at 30◦C and let its temperature rise by 10◦C mCp∆T = 199182.16 kJ/hr m (4.18)(40-30) = 199182.16 kJ/hr m= 4765.12 kg/hr = 4.765 T/hr 21 Flow rate of cooling water = 4765.12 kg/hr = 4.765 T/hr Overall Energy balance around Distillation Column: F*hF + QB = Q c + D*hD + W* hW 2126.6 + QB =199.2 +62.92 +2091.1 QB = 226.62 MJ/hr Reboiler load : Electrical Power to be supplied to Reboiler at a rate of 226.62 MJ/hr DESIGN OF TUBULAR REACTOR Estimation of volume of Reacting mixture: 1 kmol – 22.414 m3( at STP) P1V1/T1 =P2V2/T2 1*22.414/273= 3* V2/593 V2=16.23 m3 1 kmol of o-xylene at 3 atm and 320 C =16.23 m3 1 mol of reactant mixture =30.88 gm 93,281.2 kg/hr =3020.4 kmol/hr =49021.09 m3/hr=13.62 m3/s Estimation of volume of products mixture: P1V1/T1 =P2V2/T2 1*22.414/273 =3*V2/673 V2=18.42 m3 1 kmol of products at 2 atm and 400 C =18.42 m3 1 mol of product mixture =30.6gm 93,281.2 kg/hr =3048.4 kmol/hr = 56151.53 m3/hr=15.6 m3/s Basis: 16 m3/s of gas Use 1.5” tubes of outside diameter (BWG 10no) and length =4m Wall thickness for BWG 10 no tube =0.134” Therefore,Inside diameter of tube =1.5-2*0.134=1.232”=31.3 mm Volume of each tube = π/4 * d2 *L = π/4 (31.2928*10-3)2*4 =3.0764*10-3 m3 Residence time with in reactor =0.4 sec T =volume of reactor/volumetric flow rate of gas through reactor 0.4=16/Q Q = 40 m3/sec Let ‘N” be no.of tubes 22 N(3.0764*10-3) = 40 N = 13002 = 13000 tubes (approx.) Heat transfer area =N(π*do*h) = 13000*π*38.1*10-3* 4 = 6224 m2 Triangular pitch is used to accommodate 13000 no of tubes. Triangular tube pitch ,Pt=1.25do=1.25 *38.1=47.625 mm Bundle diameter: Db =do(Nt/K1)1/n1 =38.1(13000/0.319)1/2.142 =5409.9 =5410 mm Graph: Shell inside diameter-bundle diameter vs bundle diameter By extrapolating the bundle diameter to 5.41 m, We have Shell ID –Bundle dia = 20 +(20-10)/(1.2-0.2)*(5.41-1.2) = 20+(4.21)10=62.1 mm Therefore shell ID =Bundle dia + 62.1 mm = 5410 +62.1=5472.1 mm =5474 mm Height of a reactor = 10m Volume of a reactor =(π/4)*D2*H = (π/4) (5.474)2 *10=235.34 m3 Shell Material of Construction: PAN and MAN vapors are not particularly corrosive to most steels. The shell is susceptible to attack from cooling salt only. The preferred construction material for reactor is low alloy steel (IS:3609) From Code IS 2825, f=12.6 kgf/mm2 =1260 kgf/cm2 This is a Class-I vessel. Therefore, J = weld joint efficiency factor = 1 Here t = PDi/(200fJ-P) +CA where P=3 atm = (3.04*10-2*5474)/(200*1*12.6-3.04 *10-2) + 1 =1.066 mm Since t< 12 mm, Use t =12 mm (including corrosion allowance) in order to withstand its own weight. 2:1 ellipsoidal dome end : Calculation of Dome end Thickness: t = PDoC/200f J +CA +TA Assume t=10 mm hi=Di/4 =5474/4 =1368.5 mm 23 ho =hi + t =1368.5 +10 =1378.5 mm Ro =0.82Di + t =0.82 *5474 +10 =4498.68 mm ro= 0.15 Di+T =0.15*5474 +10=831.1 mm Do =Di +2t =5474 +2*12 =5498 mm Do2/4R0 = 54982/(4*4498.68) = 1679.82m Sqrt(Doro/2)=Sqrt (5498*831.1/2)=1511.52 ho = 1378.5 mm ho is least. Therefore hE=ho=1378.5 mm hE/Do=1378.5/5498 =0.251; t/Do =10/5498 =1.819 *10-3 From graph hE/Do vs C, we have C=1.35 t = PDoC/200f J = (3.04*10-2*5498*1.35)/(200*1*12.6) = 0.089 mm So, thickness of dome = 10mm Shell side fluid is molten salt. Total Heat removed = 144095 MJ/hr Heat removed per tube = (144095.01*10-6)/(3600*13000) = 3079 W q = UA∆T Substituting A=6224 m2, ∆T=395-150 = 245°C, q = 144095.01 MJ/hr We have U = 26.25 W/m2°C Internal diameter of vessel = 5474 mm We know that Ds/5 < baffle spacing < Ds i.e., 1094.8< baffle spacing

×

Report "64788"

Your name Email Reason -Select Reason- Pornographic Defamatory Illegal/Unlawful Spam Other Terms Of Service Violation File a copyright complaint Description Close Submit Our partners will collect data and use cookies for ad personalization and measurement. Learn how we and our ad partner Google, collect and use data. Agree & close