Acetone Production Process from iso-propyl-alcohol (IPA)

Acetone Production Process from iso-propyl-alcohol (IPA) -...

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Unformatted text preview: T.R. EGE UNIVERSITY Chemical Engineering Department CHEMICAL ENGINEERING DESIGN PROJECT REPORT I Submitted to: Prof.Dr.Ferhan ATALAY Prepared by; Res.Assist. Nilay GİZLİ Sezai ERDEM Tuğba GÜRMEN 05078901 Ürün ARDA 05068091 M.Serkan ACARSER 05068076 Müge METİN 05078875 Sıla Ezgi GÜNGÖR 05068052 Ali KÜÇÜK March 2009 Bornova-İZMİR Date: 30.03.2009 SUMMARY The process purpose is to produce acetone from isopropyl alcohol (IPA) at the given conditions. This report is formed, some properties, manufacturing process of acetone. In manufacturing process, feed drum, vaporizer, heater, reactor, furnace, cooler, condenser, flash unit, scrubber, acetone and IPA columns are used. i INTRODUCTION Acetone (dimethyl ketone, 2-propane, CH3COCH3 ), formulation weight 58,079, is the simplest and the most important of the ketones. It is a colourless, mobile, flammable liquid with a mildly pungent and somewhat aromatic odour. It is miscible in all proportions with water and with organic solvents such as ether, methanol, ethyl alcohol, and esters. Acetone is used as a solvent for cellulose acetate and nitrocellulose, as a carrier for acetylene and as a raw material for the chemical synthesis of a wide range of products such as ketene, methyl methacrylate, bisphenol A, diacetone alcohol mesityl oxide, methyl isobutyl ketone, hexylene glycol ( 2-methyl-2, 4-pentanediol ), and isophorone. Acetone is produced in various ways; The Cumene Hydroperoxide Process for Phenol and Acetone Isopropyl Alcohol Dehydrogenation Direct Oxidation of Hydrocarbons to a Number of Oxygeanted Products Including Acetone Catalytic Oxidation of Isopropyl Alcohol Acetone as a By-Product of the Propylene Oxide Process Used by Oxirane The p-Cymene Hydroperoxide Process for p Cresol and Acetone The Diisopropylbenzene Process for Hydroquinone (or Resorcinol ) and Acetone In this report isopropyl alcohol dehydrogenation was investigated. ii TABLE OF CONTENTS: Summary i Introduction ii 1.0 Describing of Process 1 2.0 Results 2 3.0 Discussion 4 4.0 Nomenclature 7 5.0 Appendix 8 5.1 Flowchart 8 5.2 Mass Balance 9 5.2.1 Reactor 9 5.2.2 Flash Unit 10 5.2.3 Scrubber 11 5.2.4 Acetone Column 14 5.2.5 IPA Column 15 5.2.6 Feed Drum 16 5.3 Energy Balances 17 5.3.1 Feed Drum 5.3.2 Vaporizer 18 5.3.3 Pre - Heater 19 5.3.4 Reactor 20 5.3.5 Cooler 22 5.3.6 Condenser 23 5.3.7 Scrubber 26 5.3.8 Acetone Column 27 5.3.9 IPA Column References 17 30 32 1.0 DESCRIPTION OF THE PROCESS At the beginning of the process, feed including i-propyl alcohol and water, and recycle stream are mixed in feed drum. From here, this mixture is send to vaporizer to change stream’s phase as vapour. After vaporizer, mixture is heated to reaction temperature in the heater. Reactor used is a tubular flow reactor. Acetone, hydrogen gas (H2) are produced and water and i-propyl-alcohol are discharged. The mixture with acetone, hydrogen, water, ipropyl-alcohol are sent to cooler and then to condenser. After condenser the mixture is sent to flash unit. Hydrogen, acetone, i-propyl-alcohol and water are obtained as top product. This top product is sent to scrubber to remove hydrogen. The bottom product of flash unit which is formed by acetone, water, i-propyl-alcohol are mixed with the bottom product of scrubber before acetone column. In acetone column, acetone is obtained from top product with 99 wt%. İ-propyl alcohol and water and also 0,1% of acetone is sent to i-propyl-alcohol column from bottom product. The top product of this column is sent to feed drum and bottom product is thrown away as waste water. ‐ 1 ‐ 2.0 RESULTS Table1: Properties of Substances Property H2O Acetone IPA H2 Molecular Weight(kg/kmol) 18,015 58,08 60,096 2,01 Freezing Point(°C) 0 -95 -88,5 -259,2 Boling Point(°C) 100 56,2 82,2 -252,8 Critical Temperature (°C) 647,3 508,1 508,3 33,2 Critical Pressure (bar) 220,5 47 47,6 13 Critical Volume (m3/min) 0,056 0,209 0,220 0,065 Liquid Density(kg/m3) 998 790 786 71 Heat of Vaporization(J/mol) 40683 29140 39858 904 658,25 273,84 1139,70 13,82 283,16 131,63 323,44 5,39 -242,0 20,43 -272,60 0 -228,77 62,76 -173,5 0 32,243 3,710 32,427 27,143 1,923x10-3 2,345x10-1 1,886x10-1 2,73x10-3 1,055x10-5 -1,160x10-4 6,405x10-5 -1,380x10-5 -3,596x10-8 2,204x10-8 -9,261x10-8 7,645x10-9 11 -113 0 -259 168 -33 111 -248 Constants in the liquid viscosity equation (A) Constants in the liquid viscosity equation (B) Standard Enthalpy of Formation at 298K(kJ/kmol) Standard Gibbs Energy of Formation at 298K (kJ/kmol) Constant in The Ideal Gas Heat Capacities Equation(A) Constant in The Ideal Gas Heat Capacities Equation(B) Constant in The Ideal Gas Heat Capacities Equation(C) Constant in The Ideal Gas Heat Capacities Equation(D) Minimum Temperature For Antoine Constant (°C) Maximum Temperature For Antoine Constant (°C) -2- Table 2: Calculated mol and mass values of substances Acetone 1 Basis:100kmol/h i­propyl­alcohol multiplied scale factor kmol/h ton/year kmol/h kg/h Basis:100kmol/h multiplied scale factor ton/year kmol/h ton/year kmol/h ‐ ‐ ‐ ‐ ‐ 2 ‐ ‐ ‐ ‐ ‐ 100 52644,1 3 ‐ ‐ ‐ ‐ ‐ 100 4 ‐ ‐ ‐ ‐ ‐ 5 90 6 7 Water kg/h Basis:100kmol/h Hydrogen multiplied scale factor ton/year kmol/h ton/year kmol/h kg/h Basis:100kmol/h multiplied scale factor ton/year kmol/h ton/year kmol/h 90,937 47872,96 228,797 13749,785 120448,4 44,786 7067,741 112,682 2029,966 17782,44 kg/h ton/year ‐ ‐ ‐ ‐ ‐ 251,6 15120,154 132452,6 49,25 7772,211 123,913 2232,293 19554,88 ‐ ‐ ‐ ‐ ‐ 52644,1 251,6 15120,154 132452,6 49,25 7772,211 123,913 2232,293 19554,88 ‐ ‐ ‐ ‐ ‐ 100 52644,1 251,6 15120,154 132452,6 49,25 7772,211 123,913 2232,293 19554,88 ‐ ‐ ‐ ‐ ‐ 45790,27 226,44 13151,635 115208,3 10 5264,41 25,16 1512,015 13245,26 49,25 7772,211 123,913 2232,293 19554,88 90 1584,68 226,44 455,144 3987,07 90 45790,27 226,44 13151,635 115208,3 10 5264,41 25,16 1512,015 13245,26 49,25 7772,211 123,913 2232,293 19554,88 90 1584,68 226,44 455,144 3987,07 90 45790,27 226,44 13151,635 115208,3 10 5264,41 25,16 1512,015 13245,26 49,25 7772,211 123,913 2232,293 19554,88 90 1584,68 226,44 455,144 3987,07 1,952 90 1584,68 226,44 455,144 3987,07 8 24,148 12286,04 60,756 3528,708 30911,67 0,776 408,518 9 65,789 33472,18 165,525 9613,692 84216,01 9,194 4840,098 23,132 1390,141 12177,69 45,491 7178,998 114,455 2061,907 18062,36 10 24,124 12273,83 60,696 3525,224 30880,95 0,776 408,518 1,952 117,307 1027,832 3,731 588,794 9,387 169,107 1481,407 117,307 1027,832 607,772 95913,35 1529,15 27547,709 241318 11 ‐ ‐ ‐ ‐ ‐ ‐ ‐ ‐ ‐ ‐ 12 0,024 12,211 0,06 3,485 30,722 ‐ ‐ ‐ ‐ ‐ 604,041 95324,56 1519,77 27378,603 239836,6 ‐ ‐ ‐ ‐ ‐ ‐ ‐ ‐ ‐ ‐ ‐ ‐ ‐ ‐ ‐ ‐ ‐ ‐ ‐ ‐ 90 1584,68 226,44 455,144 3987,07 13 89,913 45746,01 226,221 13138,916 115097 9,97 5248,616 25,085 1507,508 13205,52 653,263 103092,4 1643,61 29609,634 259380,4 ‐ ‐ ‐ ‐ ‐ 14 89,824 45700,73 225,997 13125,906 114983 0,907 477,482 ‐ ‐ ‐ ‐ ‐ 15 0,089 45,281 0,224 13,010 113,928 9,063 4771,134 22,803 1370,369 12004,17 653,263 103092,4 1643,61 29609,634 259380,4 ‐ ‐ ‐ ‐ ‐ 16 ‐ ‐ ‐ ‐ ‐ ‐ ‐ ‐ ‐ ‐ 17 0,089 45,281 0,224 13,010 113,928 ‐ ‐ ‐ ‐ ‐ ‐ ‐ 2,282 ‐ 137,139 1201,345 ‐ ‐ ‐ ‐ ‐ ‐ 648,799 102387,9 1632,38 29407,290 257607,9 9,063 4771,134 22,803 1370,369 12004,17 4,464 ‐ 3 ‐ ‐ 704,47 11,231 202,326 1772,447 3.0 DISCUSSION Feed drum is a kind of tank used for the mixing of the recycle stream and feed stream. Recycle stream concentration was assumed to be same with the feed stream. The temperature of the feed stream is assumed to be 250 C at 2 bar pressure, which is assumed to be constant. The temperature of recycle stream was calculated as 111,50 C. The temperature of the leaving stream was calculated as 32,890 C, by the energy balance around feed drum. In the vaporizer molten salt was used for heating. The temperature at the entrance of the unit is the temperature of the mixture leaving the feed drum, which is 32,890 C. And the leaving temperature is the bubble point temperature of the mixture, which is 109,50 C. The pressure is 2 bars, and assumed to be constant. Since the temperature leaving the vaporizer is not enough for the reaction a pre-heater was used. The unit is working at 2 bars, and assumed to be constant. The entrance and leaving temperatures are 109,50 C and 3250 C. The reactor was the starting point for the calculations. The temperature values for the entering and leaving streams were found from literature, which are 3250 C and 3500 C, respectively. The reaction taken place inside is endothermic, for this reason the reactor has to be heated. For heating, molten salt was used. The pressure is 1,8 bar, and assumed to be constant. The entrance temperature of the cooler is 3500 C and leaving is 94,70 C. For cooling, water was used. Instead of water a refrigerant may be used. Better results may get. But since it costs too much, it wasn’t chosen as the cooling material. From the temperature values it’s easily seen that the load is on the cooler not on the condenser, for this process. But in reality the unit cannot cool that much, and the load is mostly on the condenser. In this process, the mixture cooled down to its dew point. The pressure is 1,5 bar, and assumed to be constant. -4- The temperature of the entering stream is the dew point and the leaving temperature is the bubble point of the mixture. In the condenser water was used as cooling material. In the calculation of the dew and bubble points Antoine Equation was used. Trial and error was used with the help of Excel. The mixture includes acetone, i-propyl-alcohol, water and hydrogen. But hydrogen was not taken into consideration in the calculations. Since the condensation temperature of hydrogen is very low, it is not condense in the condenser. It stays in the for this reasons it has no affect on bubble and dew point calculations. Also since it does not affect the temperature calculations it’s not taken into consideration on mole and mass fraction calculations. The leaving and entering temperatures are 94,70 C and 810 C, respectively. The pressure is 1,5 bar, and assumed to be constant. Flash unit was assumed to be isothermal, for this reason temperature was not changed. It is 810 C in the entrance and exit. The pressure is 1,5 bar, and assumed to be constant. By trial and error method, (V / F) value was found to be 0,2. The entrance temperature of the unit is the bubble point of the mixture, but if it was its dew point the (V/F) value would be much higher. Scrubber was assumed to be adiabatic. The temperature of water entering the unit was assumed to be 250 C. The temperature of the off gas, including hydrogen and a very little amount of acetone, was assumed to 700 C. But this assumption is too high, a lower temperature should have been assumed, since a lot of water is used in the unit. It should have been around 400 C - 500 C. The temperature of the leaving stream was found to be 28.10 C. The pressure of the unit is 1,5 bar, and assumed to be constant. The streams leaving the scrubber and flash unit are mixed together before entering the acetone column. The temperature leaving the flash unit and scrubber are 810 C and 28.10 C, respectively. The temperature of the mixture was found to be 450 C. This result was getting by using energy balance around the mixing point. -5- The acetone column is used to separate the acetone from the mixture. The entrance temperature is 450 C. The leaving temperatures for the top and bottom product are 102,3 and 105, respectively, which are the bubble and dew points. Top product of the unit includes acetone i-propyl-alcohol and 99wt% of the product is acetone. This amount is assumed to be the desired acetone production rate, which is 115000 ton/year. From the bottom i-propylalcohol, water and a very little amount of, 0,1 %, acetone is discharged. The pressure is 1,1 bar, and assumed to be constant. In the distillation column, i-propyl-alcohol and water are separated. The entrance temperature is 1050 C. The leaving temperatures of the top and bottom products are both 111,50 C. The top product is recycled to the feed drum. For this reason it’s assumed to have the same concentration with the feed stream. But in reality a very little amount of acetone exists in the stream. It’s calculated but neglected on the recycle stream calculations. The bottom product is assumed to be pure water and it’s thrown away. Since its temperature is very high it cannot be recycled to the scrubber. But if a cooler is used, a recycle can be used. The pressure is 1,1 bar, and assumed to be constant. In the calculations one year is assumed to be 360 working day and 8600 hours. If it was 300 working day and 7200 hours, the results may be higher. Since approximated values are used in the calculations, some errors may occur. The values were taken in three decimal digits. If four or more decimal digits were taken, more accurate results would get. Also during the calculations of the specific heats, approximated values used. -6- 4.0 NOMENCLATURE MW = Molecular Weight [kg/kmol] n = mole[mol/h] y = mol or mass fraction of gas stream x = mol or mass fraction of liquid stream PT = Total Pressure [bar] Pi* = Vapour Pressure of Component [bar] Pv* = Vapour Pressure [bar] F = Feed Flow Rate [kmol/h] V = Flow Rate of Vapour [kmol/h] L = Flow Rate of Liquid [kmol/h] T = Temperature [°C] ∆Hvap = Latent Heat of Vaporisation [kJ/kg] TC = Critical Temperature [°C] PC = Critical Pressure [bar] Tb = Normal Boiling Point [°C] Q = Heat [kJ] m = Mass Flow Rate [kg/h] -7- 12 STACK GAS OFF GAS H2 WATER FURNACE 11 SCRUBBER NATURAL GAS AIR 14 ACETONE 17 8 1 MOLTEN SALT 4 13 IPA COLUMN CONDERSER ACETONE COLUMN 3 7 FLASH VAPORIZER 6 REACTOR FEED DRUM 2 10 COOLER HEATER 17 9 WASTE WATER 15 16 5 RECYCLE IPA -8- 17 5.0 APPENDIX 5.1 MASS BALANCES Production Rate : 115000 ton/year 5.1.1 REACTOR 4 I-propylalcohol=100 kmol/h H2O = 49.25 kmol/h R E A C T O R 5 acetone H2 H2 O i-propyl-alcohol conversion = 90 % nacetone5 = 100*0.9 = 90 kmol / h nH 5 = 100*0.9 = 90 kmol / h 2 nH O 5 = 49.25 kmol / h 2 ni − propylalcohol 5 = 100*0.1 = 10 kmol / h nTotal 5 = nacetone5 + nH 5 + nH O 5 + ni − propylalcohol 5 = 239.25 kmol / h 2 2 90 = 0.376 239.25 90 yH 5 = = 0.376 2 239.25 49.25 yH O 5 = = 0.206 2 239.25 10 yi − propylalcohol 5 = = 0.042 239.25 yacetone5 = -9- 5.1.2 FLASH UNIT 8 acetone H2 H 2O i-propyl-alcohol acetone = 90 kmol/h H2 = 90 kmol/h F L A S H 7 H2O = 49.25 kmol/h i-propyl-alcohol = 10 kmol/h 9 acetone H 2O i-propyl-alcohol • It is assumed that there is no change at temperature and pressure. P* y K i= i = i PT xi At buble point (T = 81°C) For acetone 1161 224 + 81 = 1651.6 mmHg * log Pacetone = 7.02447 − * Pacetone K acetone = 1651.6 = 1.467 ((1.5 /1.013) *760) For i-propyl-alcohol * log PIPA = 8.37895 − 1788.02 227.438 + 81 * PIPA = 381.89 mmHg K IPA = 381.89 = 0.339 1125.092 For water 1668.21 228 + 81 = 369.89 mmHg * log PH O = 7.96681 − 2 * PH O 2 K H 2O = 369.89 = 0.328 1125.092 - 10 - From trial-error; (V/F) = 0.2 F = n acetone7 + n H O 7 + n IPA 7 = 149.25 kmol/h 2 F=V+L 0.2 = V F V = 29.85 kmol/h L = 119.4 kmol/h yv = K × xL F × zF = V × yv + L × xL For acetone yv = 1.467× xL 90 = 29.85 × yv + 119.4 × xL xL = 0.551 yv = 0.809 For i-propyl-alcohol yv = 0,339 × xL 10 = 29.85 × yv + 119.4 × xL xL = 0.077 yv = 0.026 For water yv = 0.328 × xL 49.25 = 29.85 × yv + 119.4 × xL xL = 0.381 yv = 0.125 At stream 8; V = 29.85 kmol/h yacetone = 0.809⇒ nacetone 8 = (0.809) ×(29.85) = 24.148 kmol/h yi-propyl-alcohol = 0.026 ⇒ ywater = 0.125 ⇒ ni-propyl-alcohol 8 = (0.026) ×(29.85) = 0.766 kmol/h nwater 8 = (0.125) ×(29.85) = 3.731kmol/h - 11 - At stream 9; L = 119.4 kmol/h ⇒ xacetone = 0.551 xi-propyl-alcohol = 0.077 ⇒ ⇒ xwater = 0.381 nacetone 9 = (0.551) ×(119.4) = 65.789 kmol/h ni-propyl-alcohol 9 = (0.077) ×(119.4) = 9.194 kmol/h nwater 9 = (0.381) ×(119.4) = 45.491 kmol/h 12 5.1.3 SCRUBBER OFF-GAS H2=90 kmol/h Acetone 11 H2 O H2 = 90 kmol/h H2O = 3.731 kmol/h 8 Acetone = 24.148 kmol/h i-propyl-alcohol = 0.776 kmol/h 10 T = 810C (354.15 K); P = 1.5 bar (1.48 atm) Assume 1/1000 of inlet acetone is in off-gas. ∴ n acetone12 = 0.024148 kmol / h n acetone10 = 24.148 − 0.024148 = 24.124 kmol / h n Total8 = n acetone8 + n H 8 + n H O8 + n IPA8 2 2 n Total8 = 24.148 + 90 + 3.731 + 0.776 = 118.655 kmol / h n Total12 = n acetone12 + n H 12 2 n Total12 = 0.024148 + 90 = 90.024 kmol / h yacetone12 = 0.024148 / 90.024 = 2.68*10−4 yacetone8 = 24.148 /118.655 = 0.203 - 12 - Acetone H2 O i-propyl-alcohol = 0.776 kmol/h y acetone12 1 − A L ; A = 11 = 6 y acetone8 1 − A mV8 m= e 3598 ⎞ ⎛ ⎜10.92 − ⎟ T⎠ ⎝ P ⇒ m= e 3598 ⎞ ⎛ ⎜10.92 − ⎟ 354.15 ⎠ ⎝ 1.48 = 1.445 y acetone12 2.68*10−4 1− A = = 1.320*10−3 = y acetone8 0.203 1 − A6 From trial-error A is found as 3.523 L11 = mAV8 = 1.445*3.523*118.655 L11 = 604.041 kmol / h n H O10 = n H O8 + n H O11 2 2 2 n H O10 = 3.731 + 604.041 = 607.772 kmol / h 2 n Total10 = n acetone10 + n H O10 + n IPA10 2 n Total10 = 24.124 + 607.772 + 0.776 = 632.672 kmol / h - 13 - 5.1.4 ACETONE COLUMN 14 acetone i-propyl-alcohol A C E T O N E acetone = 89.913 kmol/h i-propyl-alcohol = 9.97 kmol/h 13 water = 653.263 kmol/h C O L U M N acetone 15 i-propyl-alcohol water nacetone 13 = nacetone 9 + nacetone 10 = 65.789 + 24.124 = 89.913 kmol/h nı-propyl-alcohol 13 = ni-propyl-alcohol 9 + nı-propyl-alcohol 10 = 9.194 + 0.776 =9.97 kmol/h nwater 13 = nwater 9 + nwater 10 = 45.491 +607.772 = 653.263 kmol/h nT 13 = nacetone 13+ nwater 13 + nı-propyl-alcohol 13 nT 13 = 89.913 + 653.263 + 9.97 = 753.146 kmol/h Assume that 1/1000 of acetone is in bottom product ∴ nacetone 15 = 89.913 = 0.089 kmol/h 1000 nacetone 14 =89.913-0.089 = 89.824 kmol/h Since acetone purity is 99% nı-propyl-alcohol 14 = 89.824 × 0.01 = 0.907 kmol/h 0.99 nı-propyl-alcohol 15 = nı-propyl-alcohol 13 - nı-propyl-alcohol 14 = 9.97-0.907 =9.063 kmol/h nwater 15 = nwater 13 = 653.263 kmol/h - 14 - 5.1.5 IPA COLUMN 17 acetone = 0.089 kmol/h i-propyl-alcohol = 9.063 kmol/h water = 653.263 kmol/h 15 acetone i-propyl-alcohol water C IO PL AU M N 16 water since all the i-propyl-alcohol is at the top product nı-propyl-alcohol 17 = nı-propyl-alcohol 15 = 9.063 kmol/h nacetone 17 = nacetone 15 = 0.089 kmol/h Assume the composition of the recycle stream is as feed stream so that; ywater=0.33 ; yIPA=0.67 nwater 17 = 9.063 × 0.33 = 4.464 kmol/h (neglecting acetone composition) 0.67 nwater 16 = nwater 15 - nwater 17 = 653.263 – 4.464 = 648.799 kmol/h - 15 - 5.1.6 FEED DRUM i-propyl-alcohol water 1 FEED DRUM i-propyl-alcohol=9.063 kmol/h water=4.464 kmol/h 2 i-propyl-alcohol=100 kmol/h water=49.25 kmol/h 17 Input = Output nı-propyl-alcohol 2 = ni-propyl-alcohol 1 + nı-propyl-alcohol 17 nı-propyl-alcohol 1 = 100 – 9.063 = 90.937 kmol/h nwater 2 = nwater 1 + nwater 17 nwater 1 = 49.25 – 4.464 = 44.786 kmol/h since 115000 tons/year acetone is wanted to produce, all of these calculations should be correlated as this amount. These new values are shown in Table 1. amount = 89.824 kmol/h * 58.08kg 1ton 8760h × × = 45700.726ton / year 1kmol 1000kg 1year Scale Factor: 115000 ton year = 2.516 45700.726 ton year - 16 - 5.2 ENERGY BALANCES 5.2.1 FEED DRUM T=25o C mi-propyl-alcohol = 13749.785 kg/h mwater = 2029.966 kg/h 1 2 FEED DRUM T=111.5o C mi-propyl-alcohol = 1370.369 kg/h mwater = 202.326 kg/h T=32.89o C mi-propyl-alcohol = 15120.154 kg/h mwater = 2232.293 kg/h 17 Tref = 25oC ; Cp,I-propyl-alcohol = 2,497 kJ/kg ; Cp,water = 4,178 kJ/kg. For stream 1,2 and 17 calculate Cp,mix; Cp,mix = 2,497×0,87+4,178×0,13 Cp,mix =2,715 kJ/kgK mTotal,1=13749.785 + 2029.966 = 15779.75 kg/h mTotal,2=15120.154 + 2232.293 = 17352.447 kg/h mTotal,17=1370.369 + 202.326 = 1572.695 kg/h QIN = QOUT 15779.75*2,715*(25-25) + 1572.695*2,715*(111,5-25) = 17352.447*2,715×(T-25) T = 32,830C - 17 - 5.2.2 VAPORIZER T=32.83 o C mi-propyl-alcohol = 15120.154 kg/h mwater = 2232.293 kg/h 2 VAPORIZER 3 T=109.5 o C mi-propyl-alcohol = 15120.154 kg/h mwater = 2232.293 kg/h At 32.83 oC Cp i-propyl-alcohol = 145 kJ/kmol.K = 2.413 kJ /kg.K Cp water = 4.179 kJ /kg.K For Water: TC = 508.3 K Tb = 394.399 K ∆Hf = 39838 kJ/kmol ∆H vap,H O 2 ⎡ T −T ⎤ = ∆H f ⎢ c ⎥ ⎣ Tc − Tb ⎦ 0,38 ⎡ 508.3K − 382.5 K ⎤ ∆H vap,H O = 39838 ⎢ 2 ⎣ 508.3 K − 394.399 K ⎥ ⎦ For IPA : 0.38 = 41370.970kj / kmol = 2296.473 kJ / kg TC = 647.3 K Tb = 375 K ∆Hf = 40683 kJ/kmol ∆H vap,IPA ⎡ 647,3K − 382,5 K ⎤ = 40683 ⎢ ⎥ ⎣ 647,3 K − 375 K ⎦ 0,38 - 18 - = 40253,505kj / kmol = 669,82 kj / kg Q = mi-propyl-alcohol×Cp i-propyl-alcohol×∆T+mwater×Cp,water ×∆T + mwater×∆Hvap,water+mIPA×∆Hvap,IPA Q = 15120.154*2.413* (109.5 − 32.83) + 2232.293*4.179* (109.5 − 32.83) +2232.293*2296.473 + 15120.154*669.82 Q = 9.652 ×106 kJ We assume ∆T = 20 Molten Salt : Q = m × Cp,molten salt × ∆T 9.652 × 106 kJ= 1,56 kJ /kg × m × (20) m= 309.358 tons 5.2.3 PRE-HEATER T=109.5 o C mwater = 2232.293 kg/h mi-propyl-alcohol = 15120.154 kg/h 4 3 T=325 o C mi-propyl-alcohol = 15120.154 kg/h mwater = 2232.293 kg/h HEATER Tref = 109,5 oC ; Cp i-propyl-alcohol = 2.468 kJ /kg.K ; Cp water = 2.019 kJ /kg.K Q = mwater × Cp,water × ∆T + mi-propyl-alcohol ×Cp -propyl-alcohol × ∆T Q = (2232.293 ×2,468 ×(325-109.5)) + (15120.154 ×2,019 ×215,5) Q = 1.845 ×106 kJ Molten Salt : We assume ∆T = 150 Q = m × Cp,molten salt × ∆T 1.845 × 106 kJ= 1,56 kJ /kg × m × (150) m= 7.885 ton - 19 - 5.2.4 REACTOR (CH 3 )2 CHOH → (CH 3 )2 CO + H 2 Table 3: mole and Hf values of acetone, i-propyl-alcohol and H2 nin kmol/h Hf kJ/kmol nout kmol/h (CH3)2CHOH 251.6 -272.290 25.16 (CH3)2CO 0 -216.685 226.44 H2 0 0 226.44 4 T=325 o C mi-propyl-alcohol = 15120.154 kg/h mwater = 2232.293 kg/h 325 ∆Hin , IPA = − 272,29 + R E A C T O R ∫ ( 32,427 +1,886×10 −1 5 T=350 o C mi-propyl-alcohol = 1512.015 kg/h mwater = 2232.293 kg/h macetone = 13151.635 kg/h mH2 = 455.144 kg/h T + 6,405×10−5 T2 − 9,261×10−8 T3 )dT 25 ∆Hin,IPA = -272,29 + 20,104 = -252,186 kJ/kmol 350 ∆Hout, IPA = −272,29 + ∫ ( 32,427 +1,886×10 −1 T + 6,405×10−5 T2 − 9,261×10−8 T3 )dT 25 ∆Hout, IPA = -249,691 kJ/kmol 350 ∆Hout,acetone = − 216,685 + ∫ ( 71,96 + 20,1×10−2 T +12,78×10−5 T2 + 34,76×10−8 T3 ) dT 25 ∆Hout,acetone= -182,745 kJ/kmol - 20 - 350 ∆Hout,H = 2 ∫ ( 28.84×10 −3 + 0.00765×10−5 T + 0.3288×10−8 T2 − 0.8698×10−12 T3 ) dT 25 ∆Hout,H = 9.466 kj/kmol 2 ∆Hr0=(-216,685kJ/kmol) – (-272,29 kJ/kmol) ∆Hr0= 55.605 kJ/kmol ∆H r = 226.44 kmol × 55.605 kJ kmol = 12591.196 kJ 1 Q = ∑ out n i H i − ∑ in n i H i + ∆H r Q=[25.16 (-249.691)+ 226.44(-182.745)+226.44(9.466)]-[251.6(-252.186)] + 12591.196 Q=30521.67 kJ Molten Salt : Cp (molten salt between 360°C – 410°C) = 1,56 kJ/kg Q = m × Cp,molten salt × ∆T 30521.67 kJ= 1,56 kJ /kg × m × (50) m= 391.300 kg/h - 21 - 5.2.5 COOLER T=350 o C mi-propyl-alcohol = 1512.015 kg/h mwater = 2232.293 kg/h macetone = 13151.635 kg/h m H = 455.144 kg/h 6 5 COOLER 2 T=94.7o C mi-propyl-alcohol = 1512.015 kg/h mwater = 2232.293 kg/h macetone = 13151.635 kg/h m H = 455.144 kg/h 2 Tref=94.7 oC Cp,H = 12.608 kJ / kg.K 2 Cp,water = 2.035 kJ / kg.K Cp,IPA = 2.536 kJ / kg.K Cp,acetone = 1.896 kJ / kg.K Q = [(m H *C p,H ) + (m water *C p,water ) + (m IPA *C p,IPA ) + (m acetone *Cp,acetone )]* ∆T 2 2 Q = [(455.144*12.608) + (2232.293*2.035) + (1512.015*2.536) + (13151.635*1.896)]* (94.7 − 350) Q= - 10.123 ×106 kJ Water : ∆ T for the Water = (35-15)=20 Cpwater = 4.179 kJ/kg Q = m × Cp,water × ∆T 10.123 × 106 kJ= 4.179 kJ /kg × m × (20) m= 121.117 ton/h - 22 - 5.2.6 CONDENSER T=94.7o C (Tdp) mi-propyl-alcohol = 1512.015 kg/h mwater = 2232.293 kg/h macetone = 13151.635 kg/h m H = 455.144 kg/h 6 7 CONDENSER 2 log P* = A − B C + Tdp T=81o C (Tbp) mi-propyl-alcohol = 1512.015 kg/h mwater = 2232.293 kg/h macetone = 13151.635 kg/h m H = 455.144 kg/h 2 ; P* mm Hg Assumption = PT = 1.5 bar = 1125 mmHg 0 y i − propyl − alcohol × PT y acetone × PT y × PT y × PT + water +* + H2 =1 * * * Pacetone × Tdp Pwater × Tdp Pi − propyl − alcohol × Tdp PH 2 × Tdp From literature; For acetone: A=7.02447 B=1161 C=224 For water: A=7.96681 Using; B = 1668.21 yacetone =0.6 ywater=0.33 yi-propyl-alcohol= 0.07 C=228 by trial and error Tdp= 94.7 °C found For i-propyl-alcohol: A= 8.37895 B=1788.02 C=227.438 - 23 - Using; * * P = x A PA (Tbp) + x B PB (Tbp) + ... yacetone = 0.6 ywater= 0.33 yi-propyl-alcohol= 0.07 m ×Cp ×∆T + m∆Hf = Qtot by trial and error Tbp = 81°C For Acetone: At 94.7 oC and 1.5 bar Cp,Acetone = 1.297 kJ/kg.K Qacetone = m×Cp ×∆T Qacetone = 13151.635 kg × (1.297 kJ/ kg.K) ×[(81+273.15) –( 94.7+273.15)] = -233.690*103 kJ ∆H vap ⎡ T −T ⎤ = ∆H f ⎢ c ⎥ ⎣ Tc − Tb ⎦ 0,38 ∆Hf,acetone = 29140 kJ/kmol Tc = 508.1 K Tb = 341.5 K ∆H vap ⎡ 508.1K − 354 K ⎤ = 29140* ⎢ ⎣ 508.1 K − 341.5 K ⎥ ⎦ 0,38 = 28289.029kJ / kmol = 487.07kJ / kg For IPA: At 94.7 oC and 1.5 bar Cp,i-propyl-alcohol = 1.761 kJ/kg.K Q,i-propyl-alcohol = 1512.015 kg × (1.761 kJ/ kg.K) ×(354.15-367.85) = -36.478*103 kJ ∆H vap ⎡ T −T ⎤ = ∆H f ⎢ c ⎥ ⎣ Tc − Tb ⎦ 0,38 ∆Hf,i-propyl-alcohol = 39858 kJ/kmol Tc = 508.3 K Tb = 366.6 K 0,38 ∆H vap ⎡ 508,3K − 354 K ⎤ = 39858* ⎢ ⎥ = 41169,35kJ / kmol = 685,128kJ / kg ⎣ 508,3 K − 366,6 K ⎦ - 24 - For Water : At 94.7 oC and 1.5 bar Cp,water = 1.898 kJ/kg K Q water =2232.293 kg (1,898 kJ/ kg.K) ×(354.15-367.85) = -58.045*103 kJ ⎡ T −T ⎤ ∆H vap = ∆H f ⎢ c ⎥ ⎣ Tc − Tb ⎦ 0,38 ∆Hf,water = 40683 kJ/kmol Tc = 647.3 K Tb = 385.186 K ∆H vap ⎡ 647,3K − 354 K ⎤ = 40683* ⎢ ⎣ 647.3 K − 385.186 K ⎥ ⎦ 0,38 = 42442.0561kJ / kmol = 2356.845 kJ / kg For Hydrogen : At 94.7 oC and 1.5 bar C p,H = 13.225 kJ/kg K 2 Q = 455.144 kg (13, 225 kJ / kg.K ) * ( 354.15 − 367.85 ) = -82.464 *103 kJ H2 ∑m C i p,i ∆T = −410.677 *103 kJ ; i ∑ m ∆H i vap,i i QTotal = ∑ mi C p,i ∆T + ∑ mi ∆H vap,i = 12.3*106 kJ i i Water : ∆ T for the Water = (35-15)=20 Cpwater = 4.182 kJ/kg Q = m × Cp,water × ∆T 682691.799 kJ= 4.182 kJ /kg × m × (20) m= 147.058 ton/h - 25 - = 12.702*106 kJ 5.2.7 SCRUBBER 12 mwater = 27378.603 kg/h T=70o C macetone = 3.485 kg/h m H = 455.144 kg/h 2 11 T=81o C mi-propyl-alcohol = 117.307 kg/h mwater = 169.107 kg/h macetone = 3528.708 kg/h m H = 455.144 kg/h 8 2 10 T=28.1o C mi-propyl-alcohol = 117.307 kg/h mwater = 27547.709 kg/h macetone = 3525.224 kg/h Qin = Qout TRef = 25 oC ; 455.144x14.419 x (81-25)+ 3528.708x1.259x(81-25) + 169.107x4.193x(81-25) + 117.307x1.716x (81-25) = 455.144 x14,401x(70-25) + 3.485x1,229x(70-25) +3525.224x1,249x(T-25) +27547.709x4,183x(T-25) + 117.307x1,710x(T-25) 42228,319 = 18777,661 + (T – 25) x 7551,149 T = 28.1 oC - 26 - 5.2.8 ACETONE COLUMN T=102.3o C mi-propyl-alcohol = 137.139 kg/h macetone = 13125.906 kg/h 14 T=45o C mi-propyl-alcohol = 1507.508 kg/h mwater = 29609.634 kg/h macetone = 13138.916 kg/h A C E T O N E 13 C O L U M N T=105o C mi-propyl-alcohol = 1370.369 kg/h mwater = 29609.634 kg/h macetone = 13.010 kg/h 15 ∆H vap ⎡ T −T ⎤ = ∆H f ⎢ c ⎥ ⎣ Tc − Tb ⎦ 0 ,38 Before the application the formula boiling temperatures ( Tb ) for each of the component must be find at 1,1 bar pressure. For the boiling point calculation; ln P s at = A − B T will be used. CONDENSER: For acetone: Pc = 47 bar Tc = 508.1 K P = 1.0133 bar T = 329.2 K ( normal boiling point ) ln 1.0133 = A − B 329.2 ln 47 = A − B 508.1 Then; A = 10.91 and B = 3587.3 At 1.1 bar pressure, boiling point is; - 27 - ln 1.1 = 10.91 − 3587.3 Tb Tb = 331.706 K For i-propyl-alcohol Pc = 47.6 bar Tc = 508.3K P = 1.0133 bar T = 355.35 K ( normal boiling point ) ln 1.0133 = A − B 355.35 ln 47.6 = A − B 508.3 Then; A = 12.807 and B = 4546.375 At 1.1 bar pressure, boiling point is; ln 1.1 = 12.807 − 4546.375 Tb Tb = 357.653 K Substituting the results to the first equation; ∆ H acetone ⎡ 508.1 − 375.3 ⎤ = 29140 × ⎢ ⎣ 508.1 − 331.706 ⎥ ⎦ 0 ,38 ∆Hacetone = 26160,195 kJ/kmol ( 450,417 kJ/kg ) at 102,30C ⎡ 508.3 − 375.3 ⎤ ∆H IPA = 39858 × ⎢ ⎣ 508.3 − 357.653 ⎥ ⎦ 0 ,38 ∆Hi-propyl-alcohol = 38014 kJ/kmol (632,618 kJ/kg ) at 102,3 0C For the mixture; ∆Hmixture = 450.417×0.99+632.618×0.01 ∆Hmixture =452.24 kJ/kg mT =13263.045 kg For the energy balance for the mixture; Q = mT ×∆Hmixture =6 × 106 kJ - 28 - For water: Pc = 220.5 bar Tc = 647.3 K P = 1.0133 bar T = 373.15 K ( normal boiling point ) ln 1.0133 = A − B 373.15 ln 220.5 = A − B 647.3 Then; A = 12.72 and B = 4743.39 At 1.1 bar pressure, boiling point is; ln 1.1 = 12.72 − 4743.39 Tb Tb = 375.723 K REBOILER: ⎡ 508.1 − 378 ⎤ ∆H vap,acetone = 29140 × ⎢ ⎣ 508.1 − 331.706 ⎥ ⎦ 0,38 ∆H vap,acetone = 25956.795kJ / kmol = 446.915kJ / kg For Water: ∆H vap,water ⎡ 647.3 − 378 ⎤ = 40683 × ⎢ ⎣ 647.1 − 375.723 ⎥ ⎦ 0 ,38 ∆H vap,water = 40553, 043kJ / kmol = 674,872kJ / kg ⎡ 508.3 − 378 ⎤ ∆H vap,i −propyl−alcohol = 39858 × ⎢ ⎣ 508.3 − 357.653 ⎥ ⎦ 0 ,38 ∆H vap,i − propyl−alcohol = 37719.801kJ / kmol = 627.722kJ / kmol yacetone = 4.364*10-4 ; ywater = 0.955 ; yIPA = 0.045 - 29 ∆ H vap,mixture = 446, 915 × 4, 364 × 10 −4 + 674, 872 × 0, 955 + 627, 722 × 0, 045 = 672, 945 kJ kg Balance; Q=mT∆Hvap,mixture=30993.013×672,945=20,86×106 kJ 5.2.9 IPA COLUMN 17 T=105o C 15 mwater = 29609.634 kg/h macetone = 13.010 kg/h mi-propyl-alcohol = 1370.369 kg/h C IO PL AU M N 16 Same procedure is followed as in acetone column. Tb,i-propyl-alcohol = 84.653 0 C Tb,water = 102.723 0 C ∆Hf,water = 40683 kJ/kmol ∆Hf,i-propyl-alcohol = 39858 kJ/kmol ∆Hf,acetone = 29140 kJ/kmol ∆Hvap,water = 40294.194 kJ/kmol = 2236.081 kJ/kg ∆Hvap,i-propyl-alcohol = 38014 kJ/kmol = 632.618 kJ/kg ∆Hvap,acetone = 26160.195 kJ/kmol Since acetone is neglected; ywater=0.13 ; yIPA=0.87 ∆Hvap,mixture = 2236.081×0,13+632,618×0.87 = 841.068 kJ/kg For the energy balance for the mixture; Q = mT ×∆Hmixture = 1941.326 kg × 841.068 kJ/kg - 30 - T=111.5o C mi-propyl-alcohol = 1370.369 kg/h mwater = 202.326 kg/h macetone = 13.010 kg/h mwater = 29407.290 kg/h Q = 1.633*106 kJ Reboiler: ∆H vap,WATER ⎡ 647,3 − 384,5 ⎤ = 40683 × ⎢ ⎥ ⎣ 647,1 − 375, 723 ⎦ 0 ,38 = 40179,523kJ / kmol = 2230,892 kJ / kg Q=mT∆Hvap,water=2230,892 ×29407.290=65,604×106 kJ - 31 - REFERENCES Treybol, R.E, Mass-Transfer Operations, 3rd Edition, McGraw-Hill Book Company, 1980 Coulson, J.M., Richardson,J.F, Chemical Engineering Volume6, Great Britain Pergamon Press, 1977 Yaws, C., Physical Properties, McGraw-Hill Book Company, USA, 1977 Othmer-K, Encyclopaedia Of Chemical Technology Volume-1, John Willey and Sons, 1978 Foust, A.S., Wenzel, L.A., Clump, C.W., Meus, L., Anderson, L.B., Principles of Unit Operations, John Willey and Sons Inc, USA, 1960 Perry, R.H., Green, D., Perry’s Chemical Engineers’ Handbook, 5th Edition, McGrawHill International Ed., 1984 McCabe, W.L., Smith, J.C., Horriott, P., Units Operations of Chemical Engineering, McGraw-Hill International Edition, USA, 1993 Felder, R.M., Rousseau, R.W., Elementary Principles of Chemical Process, 2nd Edition, John Willey and Sons Inc, USA, 1986 ‐ 32 ‐ ...
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This note was uploaded on 02/24/2010 for the course CHEMENG che 220 taught by Professor Ferhanatalay during the Spring '09 term at Ege Üniversitesi.

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