cheme140-Fall_2005-mt2-Radke-soln

cheme140-Fall_2005-mt2-Radke-soln - Chemical Engineering...

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Chemical Engineering 140 October 31, 2005 Midterm #2 Solutions AVERAGE: 80/205 (39%) S.D.: 35 HIGH: 182/205 (89%) Grade Disribtution: CHE 140 Midterm 2 Grade Distribution 0 5 10 15 20 0-30 31-50 51-70 71-90 91-110 111-130 131+ Regrade Policy: DO NOT WRITE ON YOUR EXAMS Please make a photocopy. Write on the photocopy anything that we may have missed. Hand in the original and the photocopy to your TA by Friday, November 18.
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October 31, 2005 Midterm #2 Solutions ( 205 ) 1. The problem is simplified below with labeled streams, their components and compositions as given in the problem statement. Component A = salicylic acid Component B = acetic anhydride Component C = acetylsalicylic acid (Aspirin) (Product) Component D = acetic acid A + B C + D MW = 138 MW = 102 MW = 180 MW = 60 Choose 100 mol basis for Stream E (reactor feed). 35) i. What is the molar composition of the stream leaving the reactor? E = 13.93 * 1.18 = 16.44 mol +5 Reactor Centrifuge Wash Tank Dryer Stream E 100 mol basis A 13.93 mol B 16.44 mol D Stream F A B Stream G B D Stream H A B C D Stream I A B ( 1.724 mol%) C D Stream J W Stream K A B C D W Stream L B D W Stream M A (11.86 wt%) C W (0.44 wt%) ( B Reactor Centrifuge Wash Tank Dryer Stream E 100 mol basis A 13.93 mol B 16.44 mol D 69.63 mol Stream F A B Stream G B D Stream H 100 mol A 2.09 mol B 4.63 mol C 11.84 mol D 81.47 mol Stream I A B ( 1.724 mol%) C D Stream J W Stream K A B C D W Stream L B D W Stream M A (11.86 wt%) C W (0.44 wt%) ol +5 D E = 100 – 13.93 – 16.44 = 69.63 m A H = 13.93 – (0.85)*13.93 = 2.09 mol B H = 16.44 – (0.85)*13.93 = 4.59 mol +5 C = 0 + (0.85)*13.93 = 11.84 mol H +5 D H = 69.63 + (0.85)*13.93 = 81.47 mol +5 for Stream H: H = 100 mol Mole fractions A: 0.0209 +2.5 B: 0.0459 +2.5 C: 0.1184
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This note was uploaded on 02/18/2010 for the course CHEME 120A taught by Professor Gilman during the Spring '10 term at University of California, Berkeley.

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cheme140-Fall_2005-mt2-Radke-soln - Chemical Engineering...

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