Elementary Principles 94 - 4.61 (cont'd) At mixing point:...

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Unformatted text preview: 4.61 (cont'd) At mixing point: N2: (1-XI0)/4 + (1-yp)(1-fsp) n1 = n1 I: XI0 + (1-yp) n2 = n2 Total moles fed to reactor: nr = 4n1 + n2 Moles of NH3 produced: np = 2fspn1 Overall N2 conversion: (1 - X I0 ) / 4 - y p (1 - f sp )n 1 (1 - X I0 ) / 4 100% b. XI0 = 0.01 fsp = 0.20 yp = 0.10 n1 = 0.884 mol N2 n2 = 0.1 mol I nr = 3.636 mol fed np = 0.3536 mol NH3 produced N2 conversion = 71.4% c. Recycle: recover and reuse unconsumed reactants. Purge: avoid accumulation of I in the system. d. Increasing XI0 results in increasing nr, decreasing np, and has no effect on fov. Increasing fsp results in decreasing nr, increasing np, and increasing fov. Increasing yp results in decreasing nr, decreasing np, and decreasing fov. Optimal values would result in a low value of nr and fsp, and a high value of np, this would give the highest profit. XI0 0.01 0.05 0.10 0.01 0.01 0.01 0.10 0.10 0.10 fsp 0.20 0.20 0.20 0.30 0.40 0.50 0.20 0.20 0.20 yp 0.10 0.10 0.10 0.10 0.10 0.10 0.20 0.30 0.40 nr 3.636 3.893 4.214 2.776 2.252 1.900 3.000 2.379 1.981 np 0.354 0.339 0.321 0.401 0.430 0.450 0.250 0.205 0.173 fov 71.4% 71.4% 71.4% 81.1% 87.0% 90.9% 55.6% 45.5% 38.5% 4-53 ...
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This note was uploaded on 02/27/2012 for the course CHEMICAL E 312 taught by Professor Cheung during the Fall '11 term at The University of Akron.

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