Midterm%202%202007%20with%20Answers

Midterm%202%202007%20with%20Answers - ChE 211 NAME (print...

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Unformatted text preview: ChE 211 NAME (print your name) Second Midterm Exam November 20, 2007 REMARKS: (1) Your work must be presented in the space provided for each problem. Anything written outside the specified space will be disregarded. (2) Your final result(s) must be presented in the box(es) provided at the end of each problem. Any empty box(es) will be considered as "NO" answer(s). (3) Any data necessary to solve problems should be found in the problem statements or in the textbook. You are not allowed to ask the instructor for any data. Problem 1. A liquid mixture containing 35.00 mole % benzene (B), 40.00 mole % toluene (T), and 25.00 mole % xylene (X) is fed to a distillation column at a rate of 500.0 kmol/h. The bottom product contains 97.22 mole % X and no B, and 98.00 % of the X in the feed is recovered in this bottom product. The overhead product is fed to a second column. The overhead product from the second column contains 96.00 % of the B in the feed to this column. The composition of this stream is 94.91 mole % B and 5.09 mole % T. Answer the following questions, paying attention to significant figures. n1 kmol B/h n2 kmol T/h n3 kmol X/h 500.0 kmol/h 0.3500 kmol B/kmol 0.4000 kmol T/kmol 0.2500 kmol X/kmol n5 kmol/h 0.9491 kmol B/kmol 0.0509 kmol T/kmol n4 kmol/h 0.0278 kmol T/kmol 0.9722 kmol X/kmol n6 kmol B/h n7 kmol T/h n8 kmol X/h (a) Find the flow rate, n1, of benzene in the top product of the first column. (10 points) 175.0 (b) Find the total flow rate, n5, of the top product from the second column. kmol B/h (10 points) 177.0 kmol/h Problem 2. Suppose that each day the stalls of a small commercial dairy barn are washed down to remove all the manure, and this slurry is to be treated by a constructed wetlands (a natural system for wastewater remediation). A steady state process to remove solids from this slurry and prepare it for treatment in the constructed wetlands might proceed as follows. The slurry (6000. kg/day, 24.0 wt% solids) is joined by a recycle stream containing 3.20 wt% solids, and the combined stream is fed into a settling tank. A stream is withdrawn from the bottom of this tank, containing 50.0 wt% solids, and sent to composting bins. The overhead stream that leaves the settling tank contains 8.00 wt% solids. This stream is fed into a huge settling pond from the bottom of which another stream is continuously withdrawn at a rate of 2225 kg/d (50.0 wt% solids) and sent to the composting bins. The water exiting this pond (3.20 wt% solids) is ready to begin treatment in the constructed wetlands. Most of it constitutes the recycle stream, and the balance is actually sent on for further processing in the wetlands. m4 kg/d 0.0320 kg solids/kg To Wetlands 6000. kg/d 0.240 kg solids/kg m1 kg/d x1 kg solids/kg m3 kg/d 0.0800 kg solids/kg m2 kg/d 0.500 kg solids/kg To Composting Bins m5 kg/d 0.0320 kg solids/kg 2225 kg/d 0.500 kg solids/kg To Composting Bins (a) Consider the entire system as a subsystem and calculate the flow rates, m2, of the stream from (15 points) the bottom of the settling tank and, m5, of the stream to be sent to the wetlands. m2 = 442 kg/d m5 = 3333 kg/d (5 points) (b) What is the flow rate, m3, of the overhead stream leaving the settling tank. m3 = 21700 kg/d Problem 3. Stoichiometric equations are used to represent growth of microorganisms provided a “molecular formula” for the cells is available. The molecular formula for biomass is obtained by measuring the amounts of C, N, H, and O and other elements in cells. For a particular bacterial strain, the molecular formula was determined to be C4.4H7.3O1.2N0.86 ( M = 91.5). These bacteria are grown under aerobic conditions with hexadecane (C16H34, M = 226.44) as substrate together with ammonia (NH3, M =17.03). The reaction equation describing growth is: C16H34 +16.28 O2 + 1.42NH3 → 1.65C4.4H7.3O1.2N0.86 + 8.74CO2 +13.11 H2O The feed rates of hexadecane, oxygen, and ammonia are 100.0 mol C16H34/h, 2000.0 mol O2/h, and 155.0 mol NH3/h, respectively, as indicated in the flowchart. 100.0 mol C16H34/h 2000.0 mol O2/h 155.0 mol NH3/h Fermenter n1 mol C16H34/h n2 mol O2/h n3 mol NH3/h n4 mol C4.4H7.3O1.2N0.86/h n5 mol CO2/h n6 mol H2O/h (a) When 68.12 g NH3 reacts, what is the mass of bacteria produced? (10 points) 425 g bacteria (b) Find the limiting reactant and calculate the percentage by which the other reactants are in excess. (10 points) Limiting Reactant (Name of reactant) (Name of reactant) Percent of Excess (Name of reactant) Hexadecane Oxygen 22.9 % Ammonia 9.2 % Problem 4. The gas-phase reaction between methanol and acetic acid to form methyl acetate and water CH3OH + CH3COOH CH3COOCH3 + H2O ξ (A) (B) (C) (D) takes place in a batch reactor and proceeds to equilibrium. When the reaction mixture comes to equilibrium, the mole fractions of the four reactive species satisfy the relation yC y D = 1 .0 0 y A yB where yi represents the mole fraction of species i. 2.000 mol/h 0.5000 mol CH3OH/mol 0.5000 mol CH3COOH/mol Reactor n mol/h yA mol CH3OH/mol yB mol CH3COOH/mol yC mol CH3COOCH3/ mol yD mol H2O/mol (a) The feed to the reactor contains 50.00 mole % CH3OH and 50.00 mole % CH3COOH. When the feed rate is 2.000 mol/h, represent mole fractions of all the individual reactive species in terms of the extent of reaction ξ at equilibrium. (12 points) yA 1.000 − ξ 2.000 yB 1.000 − ξ 2.000 yC yD ξ 2.000 ξ 2.000 (b) Find the flow rate of product methyl acetate. (8 points) 0.5000 mol CH3COOCH3/h Problem 5. Methane and oxygen react in the presence of a catalyst to form formaldehyde. In a parallel reaction, methane is oxidized to carbon dioxide and water: CH4 + O2 → HCHO + H2O ξ1 CH4 + 2O2 → CO2 + 2H2O ξ2 The feed rates to the reactor are 120.0 mol CH4/min and 110.0 mol O2/min, respectively. The fractional conversion of methane is 0.800 and the selectivity of formaldehyde production relative to carbon dioxide production is 15.0. 120.0 mol CH4/min 110.0 mol O2/min Reactor n1 mol CH4/min n2 mol O2/min n3 mol HCHO/min n4 mol CO2/min n5 mol H2O/min (a) Calculate the flow rate, n1, of methane in the reactor effluent stream. (4 points) n1 = 24.0 CH4/min (3 points) (b) Calculate the flow rate, n3, of formaldehyde in the reactor effluent stream. n3 = (b) Calculate the fractional yield of formaldehyde. 90.0 HCHO/min (3 points) 0.818 Problem 6. Ethylene oxide is produced by the catalytic oxidation of ethylene: 2 C2H4 + O2 → 2 C2H4O An undesired competing reaction is the combustion of ethylene: C2H4 + 3 O2 → 2 CO2 + 2 H2O The feed to the reactor contains 80.0 mole% ethylene and 20.0 mole% oxygen, while the fresh feed to the process contains 62.5 mole% ethylene and 37.5 mole% oxygen. The single pass conversion of limiting reactant is 40.0 %. A multiple-unit separation process is used to separate the products: all ethylene and oxygen are recycled to the reactor, ethylene oxide is sold as a product, and carbon dioxide and water are sent to a CO2 recovery system. The flow rate of the product ethylene oxide is 120.0 mol/h. n2 mol C2H4/h n3 mol O2/h n4 mol C2H4O/h Reactor n5 mol CO2/h n6 mol H2O/h n5 mol CO2/h n6 mol H2O/h Separator 120.0 mol C2H4O/h n0 mol/h 0.625 mol C2H4/mol 0.375 mol O2/mol n1 mol/h 0.800 mol C2H4/mol 0.200 mol O2/mol n7 mol/h (a) Consider the entire system as a subsystem and determine the fresh feed rate n0. (6 points) n0 = mol/h (b) What is the ratio of the recycle flow rate to the reactor feed rate (not the fresh feed rate)? (4 points) n7 /n1 = ...
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This note was uploaded on 03/11/2009 for the course CSS 335 taught by Professor Wolf during the Spring '09 term at Oregon State.

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