CSTR Design for Ethyl Acetate Production

CSTR Design for Ethyl Acetate Production - REPORT TO...

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Unformatted text preview: REPORT TO DEPARTMENT OF CHEMICAL ENGINEERING EGE UNIVERSITY FOR COURSE: CHE386 CONCEPTUAL DESIGN II DESIGN REPORT I CSTR DESIGN FOR ETHYL ACETATE PRODUCTION SUBMITTED TO Prof. Dr. Ferhan ATALAY SUBMISSION DATE 08/03/10 GROUP 3 05070008901 05070008103 05070008849 05060008091 05060008017 rn ARDA Berna KAYA Demet ACARGL M. Serkan ACARSER Tayfun EVCL SUMMARY This report is about the production ethyl acetate by the esterification reaction of acetic acid and ethanol. Both components are in aqueous solution; the acetic acid is 96% pure and ethanol is 96.5% pure. The reactants are fed to a CSTR at 750 C. The products are also at 750C. In the CSTR calculation which is the main part of the report, six-blade turbine is chosen, and the motor power was calculated as 4.441 kW . A CSTR, most commonly, is heated by either a jacket or an internal coil. In jacket calculations heat transfer area was found as 6.36 m2, mass flow as 60.434 kg/s, T out as 197.94 0C , h i as 574.043, h o as 921.845 and U 0 as 347.32 W/m2K. In coil calculations heat transfer area was found as 3.14 m2, mass flow as 3.537 kg/s, T out as 164.866 0C, h i as 2001.16 , h o as 903.283 and U 0 as 818.318 W/m2K. Finally, the number of coils was calculated as 6. Necessary calculations for both jacket and coil were performed and the necessary comparisons were made in discussion part. i TABLE OF CONTENTS Summary ....................................................................................................................... i 1.0 Introduction ........................................................................................................... 1 2.0 Results ..................................................................................................................... 3 3.0 Discussion ................................................................................................................ 8 4.0 Nomenclature ....................................................................................................... 11 5.0 References ............................................................................................................ 13 6.0 Appendix .............................................................................................................. 14 ii 1.0 INTRODUCTION Ethyl acetate (systematically, ethyl ethanoate, commonly abbreviated EtOAc or EA) is the organic compound with the formula CH 3 COOCH 2 CH 3 . This colorless liquid has a characteristic sweet smell (similar to pear drops) like certain glues or nail polish removers, in which it is used. Ethyl acetate is the ester of ethanol and acetic acid; it is manufactured on a large scale for use as a solvent. In 1985, about 400,000 tons were produced yearly in Japan, North America, and Europe combined.In 2004, an estimated 1.3M tons were produced worldwide. PRODUCTION Ethyl acetate is synthesized industrially mainly via the classic Fischer esterification reaction of ethanol and acetic acid. This mixture converts to the ester in about 65% yield at room temperature: CH 3 CH 2 OH + CH 3 COOH CH 3 COOCH 2 CH 3 + H 2 O The reaction can be accelerated by acid catalysis and the equilibrium can be shifted to the right by removal of water. It is also prepared industrially using the Tishchenko reaction, by combining two equivalents of acetaldehyde in the presence of an alkoxide catalyst: 2 CH 3 CHO CH 3 COOCH 2 CH 3 By dehydrogenation of ethanol A specialized industrial route entails the catalytic dehydrogenation of ethanol. This method is less cost effective than the esterification but is applied with surplus ethanol in a chemical plant. Typically dehydrogenation is conducted with copper at an elevated temperature but below 250 C. The copper may have its surface area increased by depositing it on zinc, promoting the growth of snowflake, fractal like structures (dendrites). Surface area can be again increased by deposition onto a zeolite, typically ZSM-5. Traces of rare earth and alkali metals are beneficial to the process. Byproducts of the dehydrogenation include diethyl ether, which is thought to primarily arise due to aluminum sites in the catalyst; acetaldehyde and its aldol products; higher esters; and ketones. Separations of the byproducts is complicated by the fact that ethanol forms an azeotrope with water, as does ethyl acetate with ethanol and water, and methyl ethyl ketone (MEK, which forms from 2-butanol) with both ethanol and ethyl acetate. These azeotropes are "broken" by pressure swing distillation or membrane distillation. USES Ethyl acetate is primarily used as a solvent and diluent, being favored because of its low cost, low toxicity, and agreeable odor. For example, it is commonly used to clean circuit boards and in some nail varnish removers (acetone and acetonitrile are also used). Coffee beans and tea leaves are decaffeinated with this solvent. It is also used in paints as an activator or hardener.Ethyl acetate is present in confectionery, perfumes, and fruits. In perfumes, it evaporates quickly, leaving but the scent of the perfume on the skin. -1- LABORATORY USES In the laboratory, mixtures containing ethyl acetate are commonly used in column chromatography and extractions. Ethyl acetate is rarely selected as a reaction solvent because it is prone to hydrolysis and transesterification. In organic chemistry, especially in experiment, since ethyl acetate is very volatile and with low boiling point, it can be removed by compressed air in a hot water bath. OCCURRENCE IN WINES Ethyl acetate is the most common ester in wine, being the product of the most common volatile organic acid -- acetic acid, and the ethyl alcohol generated during the fermentation. The aroma of ethyl acetate is most vivid in younger wines and contributes towards the general perception of "fruitiness" in the wine. Sensitivity varies, with most people having a perception threshold around 120 mg/L. Excessive amounts of ethyl acetate are considered a wine fault. Exposure to oxygen can exacerbate the fault due to the oxidation of ethanol to acetaldehyde, which leaves the wine with a sharp vinegar-like taste. OTHER USES In the field of entomology, ethyl acetate is an effective asphyxiant for use in insect collecting and study. In a killing jar charged with ethyl acetate, the vapors will kill the collected (usually adult) insect quickly without destroying it. Because it is not hygroscopic, ethyl acetate also keeps the insect soft enough to allow proper mounting suitable for a collection. -2- 2.0 RESULTS Table 1. Assuming and reference data of coil, jacket and oil Coil Assuming Data coil [m/s] T in [oC] d o [in] d i [in] 3 200 1.9 1.61 Ref. Data k [W/mK] Cp [J/kg.K] [Pa.s] [kg/m3] 0.1105 2147.4983 0.86*10-3 910.2406 Assuming Data jacket [m/s] T in [oC] t shell [cm] t jacket [cm] 0.6 200 1.5 2.3 Jacket Ref. Data Cp oil [J/kg.K] oil [kg/m3] k oil [W/m.K] oil [Pa.s] 2204 897.6 0.109 0.73*10-3 Table 2. Agitator calculation Agitator Calculation Results D tank [m] 1.5 H tank [m] 1.95 d ag [m] 0.5 E [m] 4.5 Re 1*106 Np [from fig.] 7 P [kW] 1.537 P act [kW] P motor [kW] 3.074 4.441 -3- Table 3. Properties of the components Properties Cp [50C] [J/kgK] Cp [75C] [J/kgK] H f [J/mol] AcA 2160 2280 -486180 EtOH 2670 2960 -277630 EtAc 2020 2120 -463200 H2O 4180 4190 285840 Table 4. Density Correlations for Components (75C) Density Correlations for Components (75C) T AcA EtOH EtAc Water 348 348 348 348 MW 60.0520 46.0680 88.1050 18.0150 C1 1.449 1.629 0.900 -13.851 C2 0.25892 0.27469 0.25856 0.64038 C3 591.95 514 523.3 -0.00191 C4 0.2529 0.23178 0.278 1.8211E-06 (kmol/m3) 16.4726 16.0271 9.4389 54.4414 (kg/m3) 989.2111 738.3386 831.6123 980.7616 -4- Table 5. Viscosity Correlations for Components (75C) Viscosity Correlations for Components (75C) T AcA EtOH EtAc Water 348 348 348 348 C1 -9.0300 7.8750 14.3540 -52.8430 C2 1212.300 781.980 -154.600 3703.600 C3 -0.322 -3.0418 -3.7887 5.866 C4 -5.88E-29 C5 10 (Pa.s) 5.93*10-4 4.62*10-4 2.58*10-4 3.81*10-4 Table 6. Properties of the components after mixing, at 75C ,mix [kg/m3] 878.489 ,mix [Pa.s] 0.000429 k,mix [W/mK] 0.15 Cp,mix (Pa.s) 2898.1 Table 7. Flow rates of components F AcA [kmol/min] 0.636 F EtOH [kmol/min] 0.745 F EtAc [kmol/min] 0.453 F Water [kmol/min] 0.54 F TOTAL [kmol/min] 2.374 -5- Table 8. Concentrations of components at the exit C AcA [kmol/m3] 10.411 C EtOH [kmol/m3] 11.959 C EtAc [kmol/m3] 7.273 C Water [kmol/m3] 8.698 -rAcA = 0.152 kmol/m3min V liquid = 2.983 m3 V tank= 3.43 m3 Table 9. Mole fraction of the components X AcA 0.27 X EtOH 0.31 X EtAc 0.19 X Water 0.23 Table 10. Weight fraction of the components X AcA 0.31 X EtOH 0.28 X EtAc 0.32 X Water 0.09 -6- Table 11. Results from calculations Coil Results m [kg/s] T [0C] T out [0C] h i [W/m2.K] h ic [W/m2.K] h o [W/m2.K] U o [W/m2.K] D' [m] Perimeter of one coil [m] A req [m2] # of coil H coil [m] 3.536 35.134 164.886 2001.1634 10266.838 903.282 818.318 1.1 3.45 3.0403 6 1 Jacket Results H f [W] H R [W] H P [W] Q=H rxn [W] D ji [m] D io [m] T out [0C] G [kg/m2s] d eq [m] h i [W/m2K] h o [W/m2K] U o [W/m2K] A req [m2] A o,cal [m2] 268321 -240722 246276 273875 1.530 1.576 197.944 538.834 0.0933 574.043 921.845 347.32 6.36 9.189 -7- 3.0 DISCUSSION A reversible reaction is a chemical reaction that results in an equilibrium mixture of reactants and products. For a reaction involving two reactants and two products this can be expressed symbolically as A and B can react to form C and D or, in the reverse reaction, C and D can react to form A and B. This is distinct from reversible process in thermodynamics. The concentrations of reactants and products in an equilibrium mixture are determined by the analytical concentrations of the reagents (A and B or C and D) and the equilibrium constant K c . The magnitude of the equilibrium constant depends on the Gibbs free energy change for the reaction. So, when the free energy change is large (more than about 30 kJ mol-1), then the equilibrium constant is large (log K > 3) and the concentrations of the reactants at equilibrium are very small. Such a reaction is sometimes considered to be an irreversible reaction, although in reality small amounts of the reactants are still expected to be present in the reacting system. A truly irreversible chemical reaction is usually achieved when one of the products exits the reacting system, for example, as does carbon dioxide (volatile) in the reaction. In this case, the reversible reaction is between acetic acid and ethyl acetate, such that: To make this endothermic reaction irreversible, the water must be removed from the system at all times. However, in CSTR, which has a closed top, this is not possible. So the reaction's conversion becomes a percentage of the equilibrium reaction, which changes between 80-90% in ideal cases. 80% of the equilibrium conversion was assumed for this report. The reaction's conversion depends on many factors, such as temperature, the weight percentage of the catalyst in reaction (the catalyst in this reaction is H 2 SO 4 , with a weight percentage of 1.91%), presence of inert in the system, purity of the components and whether the tank is perfectly mixed. In this report, the components were taken as aqueous solutions, with mole percentage of acetic acid being 96%, and that of ethanol's percentage being 96.5%. These factors also affect the reactor tank's volume, due to the CSTR design equation. With mole balance and CSTR design equation, the inlet and outlet molar rates were calculated, and the inlet water of the solutions was added to the effluent water. -8- Concentrations were calculated from the rate equation. The k values of the rate equation depend on temperature and the weight percentage of the catalyst in the system. Upon finding these k and concentration values the rate of acetic acid's formation was calculated. After finding the rate, the reaction volume was calculated as 2.983 m3. The safety factor was added to the reaction volume and the tank volume was found as 3.43 m3. After finding the volume, the diameter and the height of the tank was calculated as 1.5 and 1.95 m respectively. The diameter of the agitator was calculated as 0.5 m. To heat the coil, many options are available. One can heat the components at a separate heat exchanger before feeding to the reactor, a jacket can be used, or a certain number of coils can be installed in the reactor. Choosing the heating fluid is very important, because choosing saturated steam (hence using steam generator) might cause additional expenses. Saturated steam is used in bigger industries in order to take advantage of its heating, and to generate electricity by means of steam turbines. Hot oil is a cheaper fluid to obtain, and for this reason hot oil was used as heating fluid. The velocity of the hot oil affects the heating of the reactor, as the velocity affects the Reynold's number, the Nusselt number, and the convection heat coefficient as a result. The pipes and pumps can be picked to obtain the desired mass flow rate. The Nusselt number of the jacket and coils were calculated using the Chilton-Drew-Jebend's correlation. The wall thickness of the jacket and reactor affects heat transfer because the mass flow rate, the heat transfer area and the temperature difference change. The jacket was calculated first, which can be seen in appendix. To calculate the heat requirement for this endothermic reaction, hypothetical steps formed such that the reactor's temperature, which was 75C, was virtually dropped down to 25C; and the reaction take place in this temperature. The products and the remaining components are then heated up to 75C. Since the temperature of the entering and exiting components were 75C, the only remaining factor was H rxn in energy balance equation. The formation enthalpies at 25C were found from references. The velocity of hot oil was assumed as 0.6 m/s, and then the temperature difference was calculated between the entrance and the exit of the jacket as 2.05 0C. If a higher velocity had been assumed, the mass flow rate would increase, required heat transfer area decrease and the pressure drop would change, a pump and pipe with a bigger diameter would be needed and this would cause more expensive operations and the temperature difference will be affected. After finding the velocity and mass flow rate, using the correlations, h o was calculated as 921.845 W/m2K. The viscosity, specific heat and thermal conductivity of hot oil were taken at 200C. With these data, h i was calculated as 574.043 W/m2K. After that U o was found as 347.32 W/m2K, and the area necessary for heating was calculated as 6.36 m2. The jacket satisfies the reactor. -9- The same steps were followed for the coil, only that the number of coils was additionally calculated as 6, with the coils being a certain distance such as 20 cm away from the tank wall. Using coil for this reactor is a better option, because coil has a smaller area which lessens the required amount of hot oil as 3.536 kg/s , therefore making the reaction costly efficient. In the agitation systems, we chose open turbine agitator with six-bladed impeller with four baffles. We chose our propeller speed as 2 rps, and the motor power needed was calculated as 4.441 kW. The safety factor and efficiency were also added for finding the actual power. - 10 - 4.0 NOMENCLATURE Fi : X: T: Ci: Molar flowrate of ith component Conversion Temperature Concentration of ith component [oC] [kmol/m3] [m] [m] [rps] [kmol/min] D T : Tank diameter H T : Height of tank N: Rotational speed N P : Power number E: D': Po: Q: Distance between reactor bottom and impeller Diameter of one coil Operating power Heat taken/given from the reactor [m] [m] [W] [W] [W/m2.K] [m] [m] [kg/m2.s] [m3] [m3] [m] [kmol/m3min] U 0 : Overall heat transfer coefficient D ji : G: Inner diameter of jacket D jo : Outer diameter of jacket Mass flux Re: Reynolds number V tank : Volume of tank V liq : Volume of reaction mixture d ag : Agitator diameter -r A : Rate of reaction with respect to component A k : Specific reaction rate mi : Mass flowrate of ith component do: di: hi: ho: Outer diameter of coil Inner diameter of coil Convective heat transfer coefficient inner fluid Convective heat transfer coefficient outer fluid [kg/h] [m] [m] [W/m2.K] [W/m2.K] - 11 - Greeks : : Density Viscosity Volumetric flowrate of ith component [kg/m3] [Pa.s] [m3/min] i : : Efficiency H: Enthalpy of out/ in/ reaction Subscripts i: i: o: ag: ith component Inner Outer Agitator [W] - 12 - 5.0 REFERENCES 1. Richard M.Felder, Ronald W.Rousseau, 2000, Elementary Principles of Chemical Processes, 3rd Ed., John Wiley & Sons, Inc., USA. 2. Smer Peker, erife .Helvaci, 2003, Akikanlar Mekanii Kavramlar, Problemler, Uygulamalar, 1st Ed., Literatr Yayincilik, stanbul. 3. Perry's R. H., Chilton, 2008, Chemical Engineers' Handbook, 8th Edition, Mc GrawHill Kokagusha, Tokyo. 4. Warren L. McCabe, Julian C.Smith, Peter Harriot, 1993, Unit Operations of Chemical Engineering, 5th Ed., McGraw-Hill, Singapore 5. Incropera, P.F., DeWitt, D.P., 2007, Fundamentals of Heat and Mass Transfer, 6th Ed., John Wiley & Sons, Inc., Canada. 6. J.M.Smith, H.C.Van Ness, M.M.Abbott, 2005, Introduction to Chemical Engineering Thermodynamics, 7th Ed., McGraw-Hill, Singapore 7. Octave Levenspiel, 1999, Chemical Reaction Engineering, 3rd Ed., John Wiley & Sons, Inc.,USA. 8. Atalay, F.S., 1994, "Kinetics of the Esterification Reaction Between Ethanol and Acetic Acid", Developments in Chemical Engineering and Mineral Processing, Vol.2, p.181-184. 9. http://www.processglobe.com/Liquid_Specific_Heat.aspx - 13 - 6.0 APPENDIX Calculation of the Volume of the CSTR Feed Stream T=750C Outlet Stream T=750C Figure 1. A typical CSTR k1 CH 3 COOH + C2 H 5 OH CH 3 CO2 C2 H 5 + H 2 O k 2 A + B k1 k 2 C + D = 75 0 C , B 1.1 Treaction = k1 = 22.63*10-3 1 min , k2 1.55*10-3 m3 kmol.min Reference 8 0.5 0.5 = k1 * C A CB - k2 CC CD -rA 0.5 0.5 = 22.63*10-3 * C A CB - 1.55*10-3 CC CD ; -rA Name Acetic Acid Ethanol Ethyl Acetate Water Composition A B C D Initial FA 0 Change - FA * x 0 FB 0 - FA * x 0 = FA ( B - x ) FB 0 Remaining = FA (1 - x ) FA 0 -- FD 0 + FA * x 0 + FA * x 0 = FA ( D - x ) FD 0 FC = FA * x 0 FC = 21000 FC 1h 1000 kg ton 1 year 1 day * * * * year 365 day 24 h 60 min 1 ton kg kmol = = 0.453 kmol min 39.954 kg min , FC 39.954 * min 88.1 kg = xe *0.8 , xe 0.52 = 0.52*0.8 0.416 x = ; x = = FA * x ; 0.453 FA *0.416 , FA 1.09 kmol min FC = = 0 0 0 FA == FA *0.96 ; FA FA *0.96 ; 1.09 = 1.135 kmol min 0 0, feed 0, feed 0, feed - 14 - = 1.1 = B FB 0 FA FB 1.09*1.1 1.199 kmol min = = 0 0 FB = ( B - x ) , FB = (1.1 - 0.416 ) ; FB = FA 1.09 0.745 kmol min 0 FB 0 FB *0.965 , 1.199 FB *0.965 , FB = = 1.242 kmol min 0, feed 0, feed 0, feed 0, feed FH O , from AcA = FA 2 2 - FA = 1.135 - 1.09= 0.0454 kmol min 0 FH O , from EtOH = FB 2 2 0, feed - FB = 1.242 - 1.199= 0.0435 kmol min 0 2 0 2 FH O , feed = from AcA + FH O , from EtOH = + 0.0435 ; FD = feed = kmol min FH O , 0.0454 FH O , 0.0889 FD =A x ; FD =+ 1.09*0.416 ; FD = FD + F 0.0889 0.542 kmol min 0 0 FA =(1 - x ) , FA = (1 - 0.416 ) ; FA = kmol min FA 1.09 0.636 0 See in Table 7 FTotal = FA + FB + FC + FD , FTotal = 0.636 + 0.745 + 0.453 + 0.542 ; FTotal = 2.376 kmo min l Design Equaition of the CSTR ; V = CA = 0 FA x 0 , ( = 0 in liquid phase ) -rA FA 0 , CA = FA 0 0 0 0 FA * M w, A *1 A , CA = 0 A M w, A Reference 7 B = 1.1 CA , D = 3 FD 0.0889 0 = = 0.0815 FA 1.09 0 0 1050 kg m ; CA = 17.485 kmol m3 0 60.05 kg kmol 0 C= C A (1 - x= 17.485 (1 - 0.416 ) ; ) A CB x) = C A ( B -= 17.485 (1.1 - 0.416 ) ; 0 C= 10.211 kmol m3 A CB = 11.959 kmol m3 CC = 7.273 kmol m 3 See in Table 8 CC = = C A x 17.485*0.416 ; 0 0 CD x) CD = C A ( D += 17.485 ( 0.0815 + 0.416 ) ; = 8.698 kmol m3 -rA 22.63*10-3 *10.2110.5 *11.9590.5 - 1.55*10-3 *7.273*8.698 -rA = 0.152 kmol m3 .min - 15 - Vliquid 1.09*0.416 ; Vliquid = 2.983 m3 0.152 D = = = 3 Assuming Vtank 1.15Vliquid , H 1.3D and d ag Vtank = Vtank 1.15* 2.983 ; Vtank 3.43 m3 = *1.3* D 3 = D 2 H ; 3.43 4 4 D = 1.497 m , D 1.5 m 1.5 ; = 0.5 m d ag 3 ; = 1.5*3 4.5 m E = See in Table 8 = 1.3*1.5 ; H 1.95 m H = = d ag See in Table 2 Reference 4 E 1 = D 3 Calculation of Motor Power = = , Pact = Assuming N 120rpm 2rps= 2= 0.9 , ( safety factor ) 1.3 P , 2 mix Nd ag 878.489* 2*0.52 = = Rec mix 4.29*10-4 Re 1*106 From N P vs Rec figure N P is found as 7 ; NP = P = 7 5 mix * N 3 * d ag See in Table 2 P = ( 7 *878.489* 23 *0.55 ) P = 1537.35 W Pact = = 2*1537.35 3074.7 W Pact * ( safety factor ) 3074.7 *1.3 = 0.9 4.441 kW = Pmotor = 4441.23 W Pmotor = - 16 - Design of Jacket by Heating Process Feed Stream Tinitial=750C Outlet Stream Tfinal=750C Hot oil Tin=2000C Hot oil Tout= ? Figure 2. A typical CSTR with Jacket Total energy balance between reaction zone and jacket; Q = H out - H in + H rxn , T = (75 - 75) = 0 So; H out = in = Q = rxn H 0 H A + B C + D H rxn = H R + H P + H f At 25 0 C ^ H f , A = -486180 J mol ^ H f ,C = -463200 J mol ^ ^ ^ H f = H f , Product - H f , Reactant ^ H f , B = -277630 J mol ^ H f , D = -285840 J mol 750C Q 750C ^ H R 250C ^ H P 0 25 C ^ H f Figure 3. Hypotetical step of Hrxn 3 min mol J ^ H f = {[ -463200 - 285840] - [ -486180 - 277630]} mol * 1.09 kmol * 10 kmol * 160 s 1 min H f = 268.321*103 W = At TAvg CP , A C P ,C 75 + 25 = 50 0 C 2 = 2670 J kg .K 2160 J kg .K CP , B = 4180 J kg .K 2020 J kg .K CP , D See in Table 3 - 17 - H R J kg kmol *60.05 *1.09 2160 kg .K kmol min 1 min * 25 - 75 ) K * ( 60 s J kg kmol + 2670 * 46.07 *1.199 kg .K kmol min H R = -240.722 *103 W 2160 + 2670 H P = + 2020 + 4180 J kg kmol *60.05 *0.636 kg .K kmol min J kg kmol * 46.07 *0.745 kg .K kmol min 1 min * ( 75 - 25 ) K * 60 s J kg kmol *88.1 *0.453 kg .K kmol min J kg kmol *18 *0.542 kg .K kmol min 246.276*103 W H P = 273.875*103 W H rxn =240.722 + 246.276 + 268.321) *103 = (- Q = H rxn = 273.875*103 W See in Table 11 = moil CP ,oil T Q = A= , m moil moil = 2 A ( D jo - D 2ji ) 4 2 = 0.6 m s , tshell 1.5*10-= 2.3*10-2 m Assuming = m and t jacket D ji =2* tshell = 2*1.5*10-2 D+ 1.5 + ; D ji = m 1.53 D jo = 2* t jacket = 2* 2.3*10-2 ; D j 0 = m D ji + 1.5 + 1.576 - 18 - At Tin = 200 0 C CP ,oil 0.6 2204 J kg .K , oil 897.6 kg m3 = moil 897.6 = 60.43 kg s ; moil 2 2 (1.576 - 1.53 ) 4 = 60.43*= 273.875*103 Q 2204* T T 2.056 0 C ; T = = G = ( 200 - T ) out = Tout 197.944 0 C moil 60.43 ; G = = 538.795 kg m 2 2 2 0.785 ( D jo - D ji ) 0.785 (1.5762 - 1.532 ) D -D ) (= (1.576 2 jo 2 ji 2 d eq = - 1.532 ) See in Table 11 ; d eq = 0.093 m D ji 1.53 For calculation of the properties of hot oil average temperature must be used; 200 + 197.944 = = 198.972 0 C TAvg 2 At TAvg = 198.972 0 C 200 0 C oil hi d eq koil = = 2204 J kg .K 0.73*10-3 Pa.s , koil 0.109 W m.K and CP ,oil C * = 0.027 * P ,oil oil koil d eq * G * oil 13 13 0.8 2204*0.73*10-3 0.093*538.795 hi 0.093 = 0.027 * * -3 0.109 0.109 0.73*10 hi = 574.482 W m 2 .K 0.8 For the calculation of h o density, viscosity, specific heat capacity and thermal conductivity correlations were made for each component from References 3. Density Correlation; = C 1 T 1+ 1- C3 2 C C 4 for all components except water , For water water = 1 + C2 T + C3 T 2 + C4 T 3 C At 75 0 C A kmol 1.4486 = 16.4725 0.2529 348 m3 1+ 1- 591.95 0.25892 kmol 60.05 kg m kmol For other components the density correlation results are shown in Table 4. = 989.211 kg m3 A 16.4725 * ; A 3 - 19 - mix = A * x A + B * xB + C * xC + D * xD mix = 989.211* mix 0.636 0.745 0.453 0.542 + 738.34* + 831.61* + 960.76* 2.378 2.378 2.378 2.378 3 = 878.489 kg m Viscosity Correlation; C = exp C1 + 2 + C3 *ln T + C4 * T C5 i T At 75 0 C 1212.3 (i for each component , T [ K ]) = + ( -0.322 ) *ln 348 + 0*3480 A exp -9.03 + 348 A = 5.93*10-4 Pa.s For other components the viscosity correlation results are shown in Table 5. 13 = mix = x , i =1 i 13 i 13 mix n 0.27 * ( 5.93*10-4 ) + 0.31* ( 4.624*10-4 ) 13 13 +0.19* ( 2.58*10-4 ) + 0.23* ( 3.8*10-4 ) 13 13 mix = 4.29*10-4 Pa.s Thermal Conductivity Calculation; kmix = x A k A + xB k B + xC kC + xD k D kmix =0.27 *0.1714 + 0.31*0.12395 + 0.19*0.14229 + 0.23*0.16555 kmix = 0.15 W m.K Heat Capacity Calculation; 0.636*60.05 = 0.31 0.636*60.05 + 0.745* 46.07 + 0.453*88.1 + 0.542*18 Other values are shown in Table 9 xw, A CP ,mix = CP , A * xw, A + CP , B * xw, B + CP ,C * xw,C + CP , D * xw, D CP ,mix = ( 2.28*0.31 + 2.96*0.28 + 2.12*0.32 + 4.19*0.09 ) *1000 CP ,mix = 2591.1 J kg .K See in Table 6 - 20 - C * ho D = 0.55* P ,mix mix kmix kmix 0.25 2 d ag * N * mix * mix 23 2 120 0.25 0.5 * 60 rps *878.489 ho *1.5 2591.1*0.000429 = 0.55* * 0.15 0.15 0.000429 ho = 921.84 W m 2 .K 1 1 1 D0 =, + U 0 ho hi Di U 0 = 342.32 W m 2 .K Q= U 0 A0 Tln T1 = = T2 23 1.576 1 1 1 * = + U 0 921.84 574.482 1.53 Reference 5 ( 200 - 75)= 125 0 C = (197.944 - 75) 122.944 0 C 123.969 0 C Tln = 273.875*103 = 342.32* A0,req *123.969 A0,req = 6.36 m 2 = DH A0,calc = *1.5*1.95 A0,calc = 9.189 m 2 See in Table 11 A 0,calc >A 0,req So; the jacket satisfies our reactor design. - 21 - Design of Coil by Heating Process Feed Stream Tinitial=750C Outlet Stream Tinitial=750C Hot Oil Tout= ? Hot Oil Tin=2000C Figure 4. A typical CSTR with coil Assume m s and Tin = 3= 200 0 C di = 0.0409 m , d o 0.0482 m = moil moil m 897.6 = = = = 3 2 A ( do ) 4 0.04822 4 moil = 3.536 kg s Q = moil CP ,oil T 2204* T Q = 3.536*= 273.875*103 T 35.134 0 C ; T = = At TAvg = koil CP ,oil Reference 2 ( 200 - T ) out Tout 164.866 0 C = 200 + 164.866 = 182.433 0 C 2 = 0.110 W m.K , oil 910.248 kg m3 = 2147.464 J kg .K , oil 0.86*10-3 Pa.s 0.8 See in Table 1 CP ,oil * oil d * * oil hi d 0 = 0.023* i * koil koil oil 0.8 0.4 0.0409*3*910.248 2147.464*0.86*10-3 hi 0.0482 = 0.023* * 0.110 0.110 0.86*10-3 hi = 2001.1634 W m 2 .K 0.4 - 22 - d hic =+ 3.5 o hi 1 di hic = 10266.838 W 0.0482 2001.1634* 1 + 3.5 = 0.0409 m 2 .K 2 d ag * N * mix * mix 13 C * ho D = 0.87 * P ,mix mix kmix kmix 0.62 2 120 13 rps *878.489 0.5 * ho *1.5 2591.1*0.000429 60 = 0.87 * * 0.15 0.15 0.000429 ho = 903.282 W m 2 .K 1 1 1 d0 =, + U 0 ho hic di U 0 = 818.318 W m 2 .K Q= U 0 A0 Tln T1 = = T2 0.0482 1 1 1 * = + U 0 903.282 10266.838 0.0409 0.62 Reference 2 ( 200 - 75)= 125 0 C = (164.866 - 75) 89.866 0 C 106.468 0 C Tln = 273.875*103 = 818.318* A0,req *106.468 A0,req = 3.143 m 2 We put the coils 20 cm apart from the reactor walls ' 1.5 - 0.4 D= D - 2* 20*10-2 = D' = 1.1 m See in Table 1 ' = = = Perimeter of one coil P D= *1.1 3.45 m = = *0.0482*3.45* n A0,calc d0 P * n A0,calc = 1.914* n if A0,calc = A0,req n = 6 coils If the space between the coils is 6 cm, the space from bottom of reactor is 45 cm, and the comparison of height of the tank with the height of the coil is shown below; H coil = H bottom + d 0 * n + H SpacesBetweenCoil * ( n - 1) H coil = 45*10-2 ) + ( 0.0482*6 ) + ( 6*10-2 * ( 6 - 1) ) ( H coil = m < H tank = 1 1.95 m So; the coil satisfies our reactor design like mentioned in the discussion part. - 23 - ...
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This note was uploaded on 03/24/2010 for the course CHEMENG che 386 taught by Professor Ferhanatalay during the Spring '10 term at Ege Üniversitesi.

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