midterm2 2004 solutions - Oct. 22nd, 2004 Chemical...

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Unformatted text preview: Oct. 22nd, 2004 Chemical Engineering 140 Midterm Examination #2Solutions (150) 1. Isooctane is produced by the reaction of butylene and isobutane dispersed in pure liquid sulfuric acid as shown in the flowchart below: C4H10 + C4H8 C8H18 Isooctane and AR CV2 Fresh FF feed CV3 F1 F2 n-butane product P Acid Recycle split Reactor RE SF IR Settler DF Distillation column CV1 CV4 Internal Recycle IBR Isobutane Recycle The fresh feed to the process flows at a rate of 60,000 kg/h and contains 25 mole % isobutane (C4H10), 25 mole % butylene (C4H8), and 50 mole % n-butane (also C 4H10), which is chemically inert in this process. Additional isobutane is recycled to the reactor and all the butylene fed to the reactor is consumed. A portion of the reactor effluent is recycled to the reactor, and the remainder passes to the settler, in which the sulfuric acid and hydrocarbon phases are allowed to separate. The acid is recycled to the reactor, and the hydrocarbons pass to a distillation column. The overhead from the column contains only isooctane and n-butane, and the bottoms product, which is recycled to the reactor, contains only isobutane. The stream entering the reactor contains 2 kg of H2SO4 per kg of hydrocarbon. The stream obtained by combining the fresh feed and isobutane recycle contains 5.0 moles of isobutane per mole of butylene. (Molecular weights (kg/kg-mol): H2SO4 = 98, C4H8 = 56, C4H10 = 58, C8H18 = 114.) (25) a. Find the molar flow rate (kg-mol/h) for each component in the fresh feed. Average MW of fresh feed: 0.25(58) + 0.25(56) + 0.50(58) = 57.5 kg/kg-mol Total molar flow rate of fresh feed: (60,000 kg/h)/(57.5) = 1043.5 kg-mol/h nisobutane = nbutylene = 0.25(1043.5) = 260.9 kg-mol/h nn-butane = 0.50(1043.5) = 521.7 kg-mol/h (25) b. Find the molar flow rate for each component in the product stream. Note: I used species balances throughout, but element balances also would have worked. Overall (CV1) balance on n-butane (n-butane is an inert in this system): Page 1 of 5 Oct. 22nd, 2004 Chemical Engineering 140 Midterm Examination #2Solutions & & nn - butane ,in =nn - butane ,out & & nn - butane , FF =nn - butane , P & nn - butane , P =522kgmol / hr Conversion of butylene is 100% and stoichiometry of butylene to iso-octane is 1:1: & & niso - octane,out =100% ᄡ nbutylene,in & & niso - octane, P =100% ᄡ nbutylene , FF & niso - octane , P =261kgmol / hr (30) c. Find the molar flow rate for the isobutane recycle stream. Butylene species mole balance around CV2 & & nbutylene ,in =nbutylene,out & & nbutylene , FF =nbutylene , F 1 & nbutylene , F 1 =261kgmol / hr & & 5nbutylene , F 1 =niso - butane , F 1 (given in problem statement) & niso - butane , F 1 =5 ( 261kgmol / hr ) =1304.5kgmol / hr Iso-butane species mole balance around CV2 & & niso - butane ,in =niso - butane ,out & & & niso - butane , FF +niso - butane ,IBR =niso - butane,F1 & & & niso - butane , IBR =niso - butane,F1 - niso - butane , FF =1304.5 - 260.9 & niso - butane , IBR =1044kgmol / hr (30) d. Find the molar flow rate for the acid recycle stream. Hydrocarbon (iso-butane, butylene, and n-butane) mass balances around CV3 (in=out): � 522 � kgmol � 58kg � & & mn - butane , F 1 =mn - butane, F 2 =� = � � 30264.4kg / hr � � hr � kgmol � � � 1304.5kgmol � 58kg � � & & miso - butane , F 1 =miso - butane , F 2 =� = � � 75661kg / hr � hr � � kgmol � � � 261 � kgmol � 56kg � & & mbutylene , F 1 =mbutylene , F 2 =� = � � 14610.4kg / hr � � hr � kgmol � � & & & & mhydrocarbon , F 2 =mbutylene , F 2 +miso - butane, F 2 +mn - butane , F 2 =120535.8kg / hr 2kg Acid/kg-hydrocarbon given in problem statement & & mAcid , F 2 =2mhydrocarbon , F 2 =2 ( 120535.8kg / hr ) =241071.6kg / hr � 241071.6kg � kgmol � � & nAcid , F 2 =� = � � 2459.9kgmol / hr � � hr � 98kg � � Acid species mole balance around CV3 Page 2 of 5 Oct. 22nd, 2004 Chemical Engineering 140 Midterm Examination #2Solutions & & nAcid ,in =nAcid ,out & & nAcid , AR =nAcid , F 2 & nAcid , AR =2460kgmol / hr (40) e. Find the molar composition (i.e. the mole fractions) of the internal recycle stream from the reactor. With the information provided, is it possible to find molar flow rate(s) of this stream? Why or why not? It is not possible to find the molar flowrates of the internal recycle stream, IR, because there is one degree of freedom when just the reactor is your CV. However, there are 0 degrees of freedom using CV4, so both streams F2 and SF can be completely solved. Since IR and SF streams result from the same split, they must have the same mole fractions. Species balances around CV4 (butylene is limiting reagent; conversion =100% given in problem statement): N-butane: & & nn - butane ,in =nn - butane ,out (inert) & & nn - butane , F 2 =nn - butane ,SF =522kgmol / hr Acid: & & nAcid ,in =nAcid ,out (inert) & & nAcid , F 2 =nAcid ,SF =2459.9kgmol / hr Butylene: & & & nn - butane ,out =nn - butane ,in - nn - butane ,in 100% & nn - butane , SF =0kgmol / hr Iso-butane: & & & niso - butane ,out =niso - butane ,in - nn - butane ,in 100% & niso - butane , SF =1304.5 - 261 =1043.6kgmol / hr Iso-octane & & & niso - octane,out =niso - octane ,in +nn - butane,in 100% & niso - octane, SF =0 +261 =261kgmol / hr Total moles in SF=522+2459.9+1043.6+261=4286.2kgmol/hr Mole fractions: n yi = i nT yn - butane =0.12 y Acid =0.57 yiso - butane =0.24 yiso - octane =0.06 Page 3 of 5 Oct. 22nd, 2004 Chemical Engineering 140 Midterm Examination #2Solutions (50) 2. The endothermic, gas-phase steam-reformation reaction used in the production of hydrogen CH4 + H2O ᄡ CO + 3H2 , takes place in a plug-flow reactor on an aluminum-oxide base catalyst. Based on principles of reaction equilibrium and kinetics, you are to discuss qualitatively the design of the reactor. (10) a. At what pressure should the reactor operate and why? The reaction above shows 2 moles (CH 4 + H2O) reacting to form 4 moles (CO + 3H2). Le Châtelier's principle states that if an equilibrium is disturbed, the system will adjust itself in order to offset the disturbance. Running at high pressure puts greater stress on the right side (products' side) which has the greater number of moles, such that the equilibrium would shift to the left and favor formation of the reactants. Therefore, we should operate the reactor at low P. (25) b. Draw a qualitative graph of the logarithm of reactor volume versus conversion for two temperatures at a fixed pressure. Explain the physical basis for the two curves you sketch. Low T High T log V Conversion, X Page 4 of 5 The curves asymptote at equilibrium conversion, marked by the dotted lines. The high T curve gives a higher equilibrium conversion because for an endothermic reaction, operating at high T favors formation of products. Also, for a given volume, we see that the high T curve always gives a higher conversion than the low T curve. This is due to the kinetics, which are always more favorable for higher T. Oct. 22nd, 2004 Chemical Engineering 140 Midterm Examination #2Solutions (15) c. At what temperature should the reactor operate and why? Please address the compromise between the ideas of LeChatelier and Arrhenius and the graph in part b in your answer. The reactor should operate at high temperature. As we discussed above, Le Châtelier tells us that for an endothermic reaction, the equilibrium will favor the products if we run at high temperature. Similarly, Arrhenius tells us that kinetics are always more favorable for high temperature (k 1=k1oe-Ea/RT). Therefore, for an endothermic reaction, there is no compromise necessary between equilibrium and kinetics; they both favor high temperature! Unlike the graph from class for the exothermic case, the high and low T curves do not intersect in the plot above. The high T case always gives higher conversion for the endothermic reaction. Page 5 of 5 ...
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