final_exam (1) - 10.37 Final Exam Spring 2007 There are 4...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
10.37 Final Exam Spring 2007 There are 4 problems . Pick any 3 of these problems to do and turn in. (If you turn in solutions to 4 problems, we will not count the problem where you earned the lowest score.) Please turn in your solution for each problem separately. Write your name on every blue book or sheet of paper you turn in. Problem 1. (100 points) A reactant stream is split to feed, in parallel, two CSTRs, one of which is twice the volume of the other. The effluents from the two CSTRs are combined. The reaction of interest follows a first-order rate law. a) (50 points). How should the feed stream be split in order to maximize the total reactant conversion in the combined effluent from the two CSTRs? Small CSTR Big CSTR Feed Effluent b) (50 points). Using the same feed, what overall conversion would be obtained by placing these same two CSTRs in series (i.e. the effluent of the first is the feed of the second)? (Consider both possible topologies – one where the large CSTR feeds the smaller, and one where the small CSTR feeds the larger.) Is this conversion superior or inferior to the best case calculated in a)? 1 Cite as: William Green, Jr., and K. Dane Wittrup, course materials for 10.37 Chemical and Biological Reaction Engineering, Spring 2007. MIT OpenCourseWare (, Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Problem 2. (100 points) An appealing partial solution to the greenhouse gas problem is to convert biomass into liquid fuels. Most of biomass is composed of linked C6 sugars (C 6 H 12 O 6 ) and C5 sugars (C 5 H 10 O 5 ), which can be broken down by an enzyme secreted by fungi. Ideally, the sugars could then be converted into ethanol or other liquids in subsequent reactions. This mechanism is proposed for the enzymatic breakdown: C 11 H 22 O 11 + Enzyme Æ Complex (reaction 1) Complex Æ Enzyme + C 6 H 12 O 6 + C 5 H 10 O 5 (reaction 2) Reaction 1 is expected to be reversible under some conditions, but reaction 2 is expected to be irreversible. Experimental rate data on this enzyme from low-conversion batch-reactor
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 11/27/2011 for the course CHEMICAL E 10.302 taught by Professor Clarkcolton during the Fall '04 term at MIT.

Page1 / 4

final_exam (1) - 10.37 Final Exam Spring 2007 There are 4...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online