Case 2 Assignment Solution

Case 2 Assignment Solution - APSC 150 Biological...

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APSC 150 Biological Engineering Tutorial Solutions APSC 150: Engineering Case Studies Assignment #1-Conceptual Process Design a) Example solution (some examples of solutions that you may have proposed and rating criteria that you may have used). Methanol Removal Method Evaluation Criteria Filtration Boiling Absorption Bacteria Eat it i) Easy 5 3 ii) Needs $$$ Equipment 2 1 iii) Enviro Friendly 3 1 iv) Able to Do job 1 5 v) Skilled People 4 1 v i ) vii) Total score 14 11 Markers: Please note that assignment 1 was just meant to get the students to brainstorm alternative methods for solving the given problem. The marking of their solutions, and associated decision matrices, is going to be subjective. Assignment #2 Solution #1. a) C 5 H 7 O 2 N + 5O 2 5CO 2 + 2H 2 O + NH 3 1 mole of bugs need 5 moles of O 2 113 g of bugs need 5 × 32 = 160 g of O 2 1 g of bugs need 1.42 g of O 2 1 mg/L of bugs requires 1.42 mg/L of O 2 (COD) b) HOOCC 6 H 4 COOK + 15/2 O 2 7CO 2 +2H 2 O + KHCO 3 12 8 x C = 96 16 4 x O = 64 5 x H = 5 K = 39 204
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APSC 150 Biological Engineering Tutorial Solutions 204 g KHP need 240 g O 2 x g KHP need 500 g O 2 x = 425 425 mg/L KHP has COD of 500 mg/L #2. CH 3 OH + aO 2 + bNH 4 + + cHCO 3 - dC 5 H 7 O 2 N + eCO 2 + fH 2 O C 12 C 5 60 4H 4 H 7 7 O 16 O 2 32 32 N 14 113 If yield is 0.42 (class example): i) Calculate d d = = × × = ethanol mole biomass mole 0.12 biomass g 113 biomass mole 1 methanol mole methanol g 2 3 methanol g biomass g 0.42 Y X/S ii) Calculate b b = 0.12 iii) Charge balance c = 0.12 iv) Calculate f LHS RHS H 4.6 (7X0.12) + 2f f = 1.88 v) Calculate e C 1 + 0.12 (0.12x5) + e e = 0.52 vi) Calculate a O 1+2a+0.36 (0.12x2)+(0.52x2)+1.88 a = 0.9
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APSC 150 Biological Engineering Tutorial Solutions The biomass yield coefficient will affect the overall stoichiometry. If the yield is 0.32 mg biomass/mg methanol, then (using the same method) the stoichiometry is found to be: CH 3 OH + 1.05O 2 + 0.09NH 4 + + 0.09HCO 3 - 0.09C 5 H 7 O 2 N + 0.6CO 2 + 1.9H 2 O If the yield is 0.53 mg biomass/mg methanol the stoichiometry will be: CH 3 OH + 0.75O 2 + 0.15NH 4 + + 0.15HCO 3 - 0.15C 5
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This note was uploaded on 02/03/2010 for the course APSC APSC 150 taught by Professor Naokoellis during the Spring '09 term at UBC.

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Case 2 Assignment Solution - APSC 150 Biological...

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