ChemE - Nicole Maistrow ENGR 110 ChemE November 30, 2010 1)...

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Nicole Maistrow ENGR 110 ChemE November 30, 2010 1) To do this, I first found all of the information that wasn’t given originally in the problem. I found that since the dirty air stream contained 2% acetone, it must contain 98% air, because the other input, the pure water stream contains 100% water. Then, I knew that the acetone-rich stream contained 99% acetone, so it must contain 1% water, and since the water-rich stream contained 96% water, it must contain 4% acetone, and that the other output, the pure air stream, must contain 100% air. Next, I figured out what unknowns I needed to solve for. The flow rates of the dirty air stream and the water-rich stream were given in the problem, so I knew that my unknowns were the flow rates of the pure water stream(F P ), the pure air stream (F Pair ), and the acetone rich stream (F A ). Next, I set up four equations to solve for these unknowns. I know that F Input = F Output so I set up equations following that principle. I then solved for the flow rate of acetone, and used it in my equation to find the flow rate of water. Next, I solved for the flow rate of air, and put all of the values I
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This note was uploaded on 02/17/2011 for the course ENGR 110 taught by Professor None during the Fall '08 term at University of Michigan.

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ChemE - Nicole Maistrow ENGR 110 ChemE November 30, 2010 1)...

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