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mô phỏng, tính toán cân bằng nhiệt lượng thiết bị sản xuất EG từ EO

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ETHYLENE GLYCOL PRODUCTION The production of ethylene glycol by the hydrolysis of ethylene oxide has beenselected in the following Design Thesis because it is by far the most widely used method for the production of ethylene glycol. It is relatively economical, simple reliable and can be used in plants that manufacture ethylene oxide and glycol together. For this Project design there is a modification to the common Flow sheet used for the Hydrolysis of Ethylene Oxide process that has been shown above. This modified unit arrangement will still achieve the production of Ethylene Glycol at a lower Capital cost. As stated above this reaction can be acid or based catalysed but for this work a choice of a neutral and high pressure and temperature reaction is made because it has the economic advantage of having no need for corrosion resistance and no acid separation step.

PRODUCTION OF ETHYLENE GLYCOL Manufacturing process Discrimination The different processes followed for the production of ethylene glycol includes: Ethylene carbonate process: In this method, ethylene oxide is converted to an intermediate, ethylene carbonate, by reaction with carbon dioxide, which is then hydrolyzed by water to give ethylene glycol. This process was in use in the 1970s, but this process was replaced later by combined ethylene oxide-glycol plants. Halcon Acetoxylatin Process: Two reaction steps were used in the Oxirane plant. In the first, ethylene glycol diacatate is obtained by the oxidation of ethylene in an acetic acid solution, catalyzed by tellurium and a bromine compound. The reaction complex, which is quite complicated, proceeds via a tellurium- bromoethylene complex. The oxidation, which is carried out at 90-200 °C and 20-30 atm pressure, results in a mixture of acetates due to partial hydrolysis of the diacetate. The reaction liquid effluent is withdrawn and processed to recover glycol acetates and glycol and provide the recycle streams back to oxidation. In the second step of the process, the glycol acetates are hydrolyzed to ethylene glycol and acetic acid. The process is obviously relatively more complex and will amount to huge capital cost and Literature also shows that it has operating difficulties. A plant started at Channelview in United State to produce 800 million lb/yr of ethylene glycol was shut down after difficulties in start up. Esterification: Ethylene glycol can be produced by reaction of formaldehyde with carbon monoxide. This route first produces glycolic acid which is converted by esterification and hydrogenolysis to ethylene glycol. HCHO +CO + H 2 0 HOCH 2 COOH HOCH 2 COOH +ROH HOCH 2 COOR + H 2 0 HOCH 2 COOR + 2H 2 HOCH 2 CH 2 OH + ROH Teijin Oxychlorination Process: This process produces ethylene glycol by the reaction of ethylene with thallium salts in the presence of water and chloride or bromide ions. A redox metal compound (such as copper) oxidizable with molecular oxygen is added to the reaction medium to permit the regeneration of the thallium salt. The Teijin process is still in the works and yet to be commercialized. Union Carbide Syngas Process: The following process developed by Union Carbide, Inc. Uses synthesis gas for the production of ethylene glycol. Glycerol and propylene oxide are the major byproducts. Methanol, methyl formate and water are also produced as shown below. 2CO + 3H 2 → HOCH 2 CH 2 OH An expensive rhodium based catalyst catalyzes the reaction. The process is yet to be commercialized. Union Carbide has already started work on a modified process in association with Ube Industries. It plans to set up a commercial scale plant soon. Oxidation of Ethylene: This process involves the oxidation of ethylene to ethylene glycol in an aqueous medium using an iron copper catalyst to produce Ethylene Glycol as represented in the reaction below. Fe-Cu CH 2 =CH 2 + ½ O 2 + H 2 0 HOCH 2 CH 2 OH Hydrolysis of Ethylene Oxide: The reaction chemistry is quite simple; it is either acid or thermally catalyzed. It is summarized as follows: ethylene oxide reacts with water to form ethylene glycol, and then further reacts with ethylene glycol and higher homologues in a series of consecutive reactions as shown in the following equations: CH 2 CH 2 O + H 2 O → HOCH 2 CH 2 O ethylene glycol 2CH 2 CH 2 O + H 2 O → HOCH 2 CH 2 OCH 2 CH 2 OH diethylene glycol 3CH 2 CH 2 O + H 2 O → HOCH 2 CH 2 OCH 2 CH 2 OCH 2 CH 2 O triethylene glycol Production Process Description In the process either a 0.5 to 1.0% sulphuric acid (H2SO4) catalyst is used at 50 to 70oC for 30 minutes or, in the absence of the acid, a temperature of 195oC and 185 psi for 1 hour will form the diol. The formation of higher glycols is inevitable because ethelene oxide reacts faster with ethylene glycols faster than with water. The most important variable is the water-to-oxide ration, and the production of diethylene glycol (DEG) and triethylene glycol (TEG) can be reduced by using a large excess of water. A 90 percent yield is realized when the ethylene oxide/water molar ratio is 1:5-8. The advantage of the acid-catalyzed reaction is no high pressure, however the thermal reaction needs no corrosion resistance and no acid separation step. The crude glycols are dehydrated and then recovered individually as highly pure overhead streams from a series of vacuum-operated purification columns. Ethylene glycol (boiling point: 197.6oC) is readily vacuum distilled and separated from the diethylene glycol (boiling point: 246oC, density: 1.118, flash point: 124oC) and triethylene glycol (boiling point: 288oC, density: 1.1274, flash point: 177oC). Since ethylene glycol is produced in relatively high purity, differences in quality are not expected. The directly synthesized product meets the high quality demands (polyester grade glycol). Other Alternatives: Other Possible catalyst for this reaction includes quaternary ammonium and phosphonium salts. A more recently developed catalyst system is based on use of Pd (II) complexes. A mixture of PdCl2, LiCl and NaNO3 in acetic acid and acetic anhydride has been shown to give a 95% selectivity During this process, Pd(II) is reduced to Pd(0). If PdCl 2 -CuCl 2 -CuOCOCH 3 is used, reaction proceeds under mild conditions without formation of a precipitate. A yield of over 90% is obtained. ETHYLENE GLYCOL PRODUCTION The production of ethylene glycol by the hydrolysis of ethylene oxide has been selected in the following Design Thesis because it is by far the most widely used method for the production of ethylene glycol. It is relatively economical, simple reliable and can be used in plants that manufacture ethylene oxide and glycol together. For this Project design there is a modification to the common Flow sheet used for the Hydrolysis of Ethylene Oxide process that has been shown above. This modified unit arrangement will still achieve the production of Ethylene Glycol at a lower Capital cost. As stated above this reaction can be acid or based catalysed but for this work a choice of a neutral and high pressure and temperature reaction is made because it has the economic advantage of having no need for corrosion resistance and no acid separation step. The chemistry of the reaction is summarized as follows: ethylene oxide reacts with water to form ethylene glycol, and then further reacts with ethylene glycol and higher homologues in a series of consecutive reactions as shown in the following equations: Step 1 CH 2 CH 2 O + H 2 O → HOCH 2 CH 2 O ethylene glycol Step 2 2CH 2 CH 2 O + H 2 O → HOCH 2 CH 2 OCH 2 CH 2 OH diethylene glycol Step 3 3CH 2 CH 2 O + H 2 O → HOCH 2 CH 2 OCH 2 CH 2 OCH 2 CH 2 O triethylene glycol The aim of this work is the yield of Ethylene Glycol not those of higher glycol does this reaction is modified in such a way as to favour reaction step one and disfavour the other two steps and this is achieved by using excess of water for the reaction to dilute the concentration of the ethylene glycol produced to discourage the formation of higher glycols. Production Process Description Ethylene Oxide and water a passed into a Continous Stirring Tank Reactor which operates a temperature of 195oC and pressure 185 psi for 1 hour will form the diol. Since the formation of higher glycols is inevitable because ethelene oxide reacts faster with ethylene glycols faster than with water then the water-to-oxide ration must be manipulated to reduce if not eliminate the production of diethylene glycol (DEG) and triethylene glycol (TEG). This is done by using a large excess of water. In this work, the ethylene oxide/water molar ratio is 1: 7 and the yield of Ethylene glycol is 90%. This Plant is designed to produced 10000kg/yr of Ethylene Glycol. The crude Ethylene produced from the reactor is dehydrated and recovered as highly pure overhead stream from a distillation column with a partial condenser. The Distillate from the partial condenser is mainly water and is recycled back to the reactor and mixed with incoming water. The flow sheet for this sequence is shown below P Fig: Shows the Flow sheet for the Production of Ethylene Glycol Simulated on Aspen Hysis MATERIAL BALANCE FOR THE REACTOR Reaction Stoichiometry: CH 2 CH 2 O + H 2 O → HOCH 2 CH 2 O ethylene glycol Molecular weight of Species: CH 2 CH 2 O = (12x2+1x4+16) = 44kg/kmol HOCH 2 CH 2 O = (12x2+1x6+16x2) =62kg/kmol Material Balance for Ethylene Oxide: The target is 10,000kg/yr 10,000kg/yr of HOCH 2 CH 2 O = 10,000/62 = 1612.9kmol/yr = 5.33x10 -5 kmol/sec HOCH 2 CH 2 O The Ethylene glycol/Water splitter column can only achieve a 99% separation for the ethylene glycol thus 5.33x10 -5 /0.99 = 5.38x10 -5 kmol/sec of ethylene glycol must be produced from the reactor to be feed into the column. From Literature, the yield of Ethylene Glycol in the reactor is 90% Yield of 90% implies Moles of ethylene oxide used up in the reaction/total moles of ethylene Oxide = 90/100 5.38x10 -5 kmol/moles of ethylene oxide = 0.9 Therefore, moles of ethylene oxide = 5.38x10 -5 /0.9 = 5.97x10 -5 kmol of ethylene Oxide is used up during the reaction but 5.97x10 -5 /0.9 = 6.63x10 -5 kmol/sec of Ethylene is actually feed into the reactor Mass of ethylene Oxide required per second = 44kg/kmol x 6.63x10 -5 kmol = 2.92x10 -3 kg Unreacted Ethylene oxide = 6.63x10 -5 kmol/sec - 5.38x10 -5 kmol/moles = 6.6x10 -6 kmol/moles Material balance for Water: From Literature, Since the feed ratio is 1: 7 therefore the moles of water feed = 7 x 6.63x10 -5 kmols = 4.64x10 - 4 kmols Mass flow rate of water required = 18kg/kmols x 4.64x10 -4 kmol = 8.35x10 -3 kg/sec Taking density of water = 1000kg/m 3 The volume of water required per year = 8.35x10 -3 /1000 = 8.35x10 -6 m 3 /sec of water Unconverted Water The reacting moles are same for water and ethylene oxide, therefore the amount of water reacting = amount in moles of ethylene oxide=5.97x10 -5 kmol. Therefore Amount of water left unconverted = Total Water fed – Amount that took part in the reaction = 4.64x10 -4 kmol – 5.97x10 -5 kmol = 4.04x10 -4 kmol =4.04x10 -4 kmol x 18kg/kmols = 7.27x10 -3 kg/sec of Water =7.27x10 -3 /1000 = 7.27x10 -6 m 3 /sec of Water ENERGY BALANCE FOR THE REACTOR Equation of the reaction C 2 H 4(g) + H 2 0 (g) = C 2 H 6 O 2(l) Standard Heat of formation data for this condition Ethylene Oxide = -52.6kj/mol Water = -241.8kj/mol Ethylene glycol = - 460kj/mol Heat of reaction = Summation of the heat of formation for product – summation of the heat of formation of the reactants = (1 x -460) – ((1 x -52.6) + (1 x -241.8)) = -165.6kj/mol Therefore –DH R (Heat of the reaction) = +165.6kj/mol The reaction is exothermic so a cooler is incorporated take out the heat produced/mol as the reaction progresses. Heat generated by the reaction (Q) = Heat taken by cooling water (Q) Q = mCPDT of water Heat generated by the reaction (Q) = 165.6 x 5.38x10 -5 (moles of Ethylene oxide produced) = 8.91x10 -3 kj/s Where Where 25 0 c – the Temperature of water available Degree of approach is taken as 10 0 c CP of water = 4.180kj/kgk Thus 8.91x10 -3 kj/s = mCPDT = m x 4.180 x (35-25) +273k Where 25 0 c – the Temperature of water available Degree of approach is taken as 10 0 c CP of water = 4.180kj/kgk m = 8.91x10 -3 / 1162.04= 7.67 x 10 -6 kg/s Thus 7.67 x 10 -6 kg/s of water must be feed into the cooler system around the reactor to maintain at operational temperature. MATERIAL BALANCE FOR THE DISTILLATION COLUNM Ethylene Glycol has a boiling point of 197.6 0 c while water has a boiling point of 100 0 c showing that water will boil over first leaving the Ethylene glycol in the bottoms. From Literature, Relative volatility of Ethylene glycol versus water α = 15.1 Specification for the column Feed rate=F=Amount of water unconverted + Unreacted Ethylene oxide + Ethylene glycol F=4.04x10 -4 kmol/s+6.6x10 -6 kmol/s+5.38x10 -6 kmol/s=4.64x10 -4 kmol/s Which implies that Water – 0.87 moles Ethylene Glycol – 0.12 moles Ethylene Oxide – 0.01 moles Bottoms (B) Specification 99% Ethylene Glycol with flow rate of 5.38x10 -5 x0.99 = 5.33x10 -5 kmol/s 1% water with flow rate of 4.04x10 -4 x0.01 = 4.04x10 -6 kmol/s Therefore Bottoms (B) = 5.33x10 -5 kmol/s + 4.04x10 -6 kmol/s = 5.734x10 -5 kmol/s Distillate (D) Specification 100% Water Distillate (D) = F – B = 4.64x10 -4 kmol/s - 5.734x10 -5 kmol/s = 4.07x10 -4 kmol/s L+D=T Where L=Flow of Condensate returned to the column D=Flow of condensate taken out as Distillate T=Flow of vapour leaving the first stage of the column Reflux Ration = L/D At 2.5reflux, L/D=2.5 L=2.5D = 2.5x4.07x10 -4 kmol/s = 1.02x10 -3 kmol/s T=2.5D+D=3.5D=1.4x10 -3 kmol/s At total Reflux; the theoretical number of trays can be calculated with relative volatility of Ethylene glycol versus water α = 15.1 Using Fenske’s Equation . variable is the water-to-oxide ration, and the production of diethylene glycol (DEG) and triethylene glycol (TEG) can be reduced by using a large excess of water. A 90 percent yield is realized. must be manipulated to reduce if not eliminate the production of diethylene glycol (DEG) and triethylene glycol (TEG). This is done by using a large excess of water. In this work, the ethylene oxide/water. available Degree of approach is taken as 10 0 c CP of water = 4.180kj/kgk Thus 8.91x10 -3 kj/s = mCPDT = m x 4.180 x (35-25) +273k Where 25 0 c – the Temperature of water available Degree of approach

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