Biochem+Eng+Hmk+3+Soln

Biochem+Eng+Hmk+3+Soln - PROBLEM SET 3 SOLUTION MASS...

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Unformatted text preview: PROBLEM SET 3 SOLUTION MASS BALANCES WITH CHEMICAL REACTIONS *Question 8: Consider an idealized model of glucose uptake and reaction by skeletal muscle: C6H12O6 + 6O2 ‐> 6CO2 + 6H2O , The average uptake rate of glucose = 36 ug 10 6 cells ⋅ h the average uptake rate of oxygen = 10 and the excretion rate of CO2 = 13 ug , 10 cells ⋅ h 6 ug . 106 cells ⋅ h Find (A) the limiting reagent, (B) the % excess, and (C) the % conversion. Relevant information: Species Molecular Weight (g/mol) Glucose 180.2 Oxygen 32.0 44.0 CO2 H2O 18.0 Answer: It is useful to draw a flow chart ? ug/106 cells/h C6H12O6 36 ug/106 cells/h C6H12O6 ? ug/106 cells/h O2 13 ug/106 cells/h CO2 10 ug/106 cells/h O2 ? ug/106 cells/h H2O 1 mol glucose Glucose in: 36 ug glucose/106 cells/h ∙ = 0.20 umol glucose/106 cells/h 180 g glucose Oxygen in: 10 ug O2/106 cells/h ∙ 1 mol O2 = 0.31 umol O2/106 cells/h 32 g O2 CO2 out: 13 ug CO2/106 cells/h ∙ 1 mol O2 = 0.30 umol CO2/106 cells/h 44 g O2 a) Each mole of glucose requires 6 moles of O2. Therefore, to react with the glucose, would require 1.2 umol O2/106 cells/h Clearly, oxygen is the limiting reactant. b) % Excess. Glucose is in excess. To react with the oxygen requires 0.31 umol O2/106 cells/h ∙ 1 mol glucose = 0.052 umol/106 cells/h glucose 6 mol O2 Glucose excess = glucose in – glucose reacted = 0.20 ‐ 0.052 = 0.148 umol/106 cells/h glucose Percent excess: ( glucose excess / glucose reacted ) ∙ 100 = 0.148 / 0.052 ∙ 100 = 285% c) % conversion applies to both glucose and oxygen. Use CO2 information to find moles reacted 1 mol glucose reacted 0.30 umol CO2 produced/106 cells/h ∙ = 0.050 umol glucose reacted/106 cells/h 6 mol CO2 produced 0.050 umol glucose reacted/106 cells/h % conversion = = 0.25 or 25% for glucose 0.200 umol glucose fed/106 cells/h 6 mol O2 reacted 0.30 umol CO2 produced/106 cells/h ∙ = 0.30 umol O2 reacted/106 cells/h 6 mol CO2 produced 0.30 umol O2 reacted/106 cells/h % conversion = = 0.97 or 97% for oxygen 0.31 umol O2 fed/106 cells/h Question 9: The vitamin company you work for produces alanine. Alanine is produced in a reactor in a continuous process. There are two separate inlet streams that contain glutamine (100 mol/min) and pyruvic acid (50 mol/min), respectively. The ratio of the molar flow rate of pyruvic acid in the outlet stream to that in the inlet stream is 0.6. The reaction is: C5H10N2O3 (glutamine) + C3H4O3 (pyruvic acid) ‐> C5H7NO4 (alpha‐ketoglutamic acid) + C3H7NO2 (alanine) a) What is the reaction rate, R, of glutamine? Pyruvic acid? Find the limiting reagent. What are the fractional conversions of glutamine and pyruvic acid? b) Find the outlet molar flow rates of alanine, alpha‐ketoglutamic acid, and any excess reactants. Glutamine Answer: Glutamine 100 mol/min Pyruvic acid 50mol/min Reaction Pyruvic acid Alpha‐Ketoglutamic Alanine a) Pyruvic acid: we are told Pyr (out) / Pyr (in) = 0.6, therefore Pyr (out) = 50 mol/min * 0.6 = 30 mol/min, and reacting Pyr = 50 mol/min – 30 mol/min = 20 mol/min. Chemical reaction says that 1 mol glutamine reacts with 1 mol pyruvic acid. Therefore 20 mol/min glutamine react. Fractional conversion for glutamine = (20 mol/min) / (100 mol/min) = 0.2 Fractional conversion for pyruvate = 20/50 = 0.4 b) Alanine produced 20 mol/min Pyr * (1 mol Ala/1 mol Pyr) = 20 mol/min alanine. Alpha‐ketoglutamic produced is also 20 mol/min. Glutamine unreacted = 100 mol/min in – 20 mol/min reacted = 80 mol/min out Pyruvic acid unreacted = 30 mol/min out (because Pyr out/Pyr in = 0.6). Question 10: Acetobacter aceti bacteria convert ethanol to acetic acid (vinegar) under aerobic (i.e. with oxygen) conditions. A continuous fermentation process for acetic acid production is shown in the Figure. The conversion reaction is as follows: C2H5OH (ethanol) + O2 ‐> CH3COOH (acetic acid) + H2O The feed stream containing ethanol enters the reactor at a rate of 1.0 kg/hr. Also, air bubbles into the reactor at a rate of 40.0 L/min. An exit off‐gas stream as well as the liquid product stream containing acetic acid and water leave the reactor. Assume reaction goes to completion. a) What is the reaction rate for this reaction? What is the limiting reactant? What are the fractional conversions of C2H5OH and O2? b) Determine the outlet flow rates of the elements C, H, O in the acetic acid product stream. Also determine the outlet mass flow rates of all compounds in the liquid product stream and the volumetric flow rates of all compounds in the off‐gas stream. Answer: a) Ethanol rate = 1 kg/hr * 1000 g/kg * 1 mol/46 g = 21.7 mol/hr. C2H 5OH Off‐gas Reactor Air CH 3COOH H 2O Oxygen rate = 40 L air/min * 21 L oxygen/100 L air * 1 mol/22.4 L *60 min/1 hr = 22.5 mol/hr. Note that here we used the fact that there is 21 mol% oxygen in air, and that 1 mol of ideal gas occupies ~22.4 L. Limiting reactant is ethanol since stoichiometric ratio (from reaction) is 1mol ethanol:1mol oxygen, and we have less ethanol than oxygen. If reaction goes to completion, fractional conversion of the limiting reactant is 1 (100%). For the other reactant (oxygen), it is: 21.7 mol/hr ethanol * 1 mol oxygen/1mol ethanol = 21.7 mol/hr Fractional conversion is: 21.7/22.5 = 0.96, or 96%. b) Acetic acid produced: 21.7 mol/hr ethanol reacted * 1 mol acetic acid / 1mol ethanol = 21.7 mol/hr acetic acid. Similarly, you can calculate 21.7 mol/hr H2O produced. Carbon atom flow rate in product stream: CH3COOH has 2 Carbon atoms per molecule, H2O has none. 21.7 mol/hr acetic acid * 2 Carbons/1 acetic acid = 43.2 mol of carbon atoms/hr. Hydrogen atom flow rate in product stream: CH3COOH has 4H per molecule, H2O has 2H per molecule, so 21.7 mol/hr acetic *4 Hydrogens/1 acetic acid + 21.7 mol/hr H2O* 2 Hydrogens/1 H2O = 130 mol/hr of hydrogen atoms. Oxygen atom flow rate in product stream: CH3COOH has 2 atoms O per molecule, H2O has 1 atom O per molecule, so 21.7 mol/hr*2 Oxygen/1 acetic acid + 21.7 mol/hr H2O* 1 Oxygen /1 H2O = 65.1 mol/hr of oxygen atoms. Mass flow rate of the product stream: CH3COOH (MW = 60g/mol): 21.7 mol/hr * 60 g/mol = 1302 g/hr or 1.3 kg/hr H2O (MW = 18 g/mol): 21.7 mol/hr * 18 g/mol = 391 g/hr or 0.39 kg/hr. Volumetric flow rate in off gas: 21.7 mol/hr ethanol * 1 oxygen/1 ethanol = 21.7 mol/hr oxygen reacted. Unreacted oxygen: 22.5 mol/hr in – 21.7 mol/hr oxygen reacted = 0.8 mol/hr that will end up in off‐gas. In terms of volume, this is 0.8 mol/hr * 22.4L/mol * 1 hr/60 min= 0.299 L/min. For nitrogen that came in the air, none reacts, so it all ends up in off‐gas. Therefore: 40 L air/min * 79L N2/100L air = 31.6 L/min. MASS BALANCES WITH RECYCLING *Question 11: You want to build a simple metabolic model of fat absorption and metabolism in the body. 1) Use the information to draw a compartment model that describes what happens to the fat in the body: a. The gut absorbs only 70% of the fat present in the food. b. All the fat absorbed into the gut goes straight to the liver via the portal circulation. There, 20% of the absorbed fat is removed by the liver and the rest travels to the rest of the body via the systemic circulation. c. Since part of the systemic circulation in fact goes back to the liver, some of the fat that was not removed by the liver in step 2 above will come back to the liver and not go to the other tissues. Assume that 10% of that amount comes back to the liver. Hint: build a system consisting of two subsystems, the gut and the liver. 2) What is the fraction of the initial fat in the food is delivered to the rest of the body (excluding the liver)? Answer: 1) First sketch two‐ compartment model for the system R mR F mF GUT G mG m G+mR LIVER W S L mL C mC Based on the information given, we can write the following mass balance equations for the fat . . . . . m = 0.8 (m + m ) . . m = 0.1 m . . . m = m ‐ m a) mG = 0.7 mF b) c) d) L G R C R L L R . . 2) We are looking for mC/mF. Substituting equ. (a) into equ. (b), we get . . . e) mL = 0.8 (0.7 mF + mR) Substituting equ. (c) into equ. (e), we get f) . . . mL = 0.8 (0.7 mF + 0.1 mL) . Substituting equ. (c) into equ. (d), we get . . . g) mC = mL ‐ 0.1 mL = 0.9 mL Substituting equ. (f) into equ. (g), we get . . h) mC = 0.9 (0.56/0.92) mF . and after rearranging, mL = (0.56/0.92) mF .. therefore mC/mF = 0.55 or 55% of the initial fat in the feed goes to the rest of the body. ...
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This note was uploaded on 10/11/2011 for the course BIOMEDICAL 201 taught by Professor Berth during the Fall '10 term at Rutgers.

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