HW2_8 - 0.65 6.0< 0.52 “3 4.0 0.39 3.0 0.25 1.0 0.13 m...

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Unformatted text preview: 0.65 6.0 <. 0.52 “3 4.0 0.39 3.0 0.25 1.0 0.13 m 0.0” 0.0 0 20 40 \. an so 100 0 17 33 50v (.7 33 100 P2-8 (a) _ 17sz _ _ rs F50 = 1000 g/ hr At a conversion of 40% ———- -- 0.15 dmahr " r? g Therefore V = (0.1511000)(0.40) a 6Odm3 P2-8 (b) 1 dm’hr At a conversion of 80%, —-—— = 0.8 ‘ ’5 g F50 = 1000 g/hr Therefore V - (0.8X1000)(0.80) = 640dm3 P2-8 (c) X M VPFR = F so I T s From the plot of ll-rs Calculate the area under the curve such that the area is equal to V/Fso = 80 / 1000 = 0.08 X=12% F SOX )(/—rs = 0.08. From guess and check we get X = 55% For the so am3 CSTR, V = 80de = 2-14 PZ-S (d) To achieve 80% conversion with a CSTR followed by a CSTR, the optimum arrangement is to have a CSTR with a volume to achieve a conversion of about 45%, or the conversion that correSponds to the minimum value of 1 /-r5. Next is a PFR with the necessary volume to achieve the 80% conversion following the CSTR. This arrangement has the smallest reactor volume to achieve 80% conversion For two CSTR’s in series, the optimum arrangement would still include a CSTR with the volume to achieve a conversion of about 45%, or the conversion that corresponds to the minimum value of I/—rs, first. A second CSTR with a volume sufficient to reach 80% would follow the first CSTR. PZ-S (e) _ s = and CC = 0,1[CS0 —CS]+ 0.001 KM +CS r kCS (0.1[CS0 - CS ]+ 0.001) ’ KM +CS _1_ KM +CS r5 kCS 0.1CSO—CS 0.001 Let us first consider when Cs is small. C30 is a constant and if we group together the constants and simplify K C then—l-= M+ S r le§ + szs 3 since CS < KM 1 K . --— = ————M——- which is consrstent with the shape ofthe graph when X is large (if CS is r, k,C§ + szs small X is large and as C5 grows X decreases). Now consider when Cs is large (X is small) As Cs gets larger Cg approaches 0: CC = 0.1[050 — CS ]+ 0.001 and CS a C30 C. K .‘r If 4: “h the“ Aha: KM+CS r kCSCC As Cs grows larger, CS >> KM 2-15 And __'_- Cs ..__‘._ r kCch kCC 5 And since Cc is becoming very small and approaching 0 at X = 0, ll-r, should be increasing with Cs (or decreasing X). This is what is observed at small values of X. At intermediate levels of C5 and X, these driving forces are competing and why the curve of 1/-rs has a minimum. P9-2 Irreversible gas phase reaction 2A + B -) 2C See Polymath program P2-9.pol. P2-9 (a) PFR volume necessary to achieve 50% conversion Mole Balance X, dX V=FA0 f X.(""A) Volume = Geometric area under the curve of (FAD/'rA) VS X) V = (—gx 400000x0.5) + (100000x 0.5) v = 150000 m3 One of the points of this problem is for the students to recognize 150,000 in3 is ridiculously reactor volume. P2-9 (b) CSTR Volume to achieve 50% conversion Mole Balance V a Elm/Y (—rA) V = 0.5 x 100000 v = 50000013 One of the points of this problem is for the students to . *3 . . . recognize 150,000 m IS ridiculously reactor volume. 500.000- 77 490,000 fee. 4A one 000 (m3) 2‘30 000 100.00L L‘ 0 0t 02030.net: . X ,r‘.‘ 39.. _ 2C 5.x": s “aw i: u ‘ ~t—-t—-t—~ a or nzugunqg uoniuanq l0 X ...
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This note was uploaded on 04/25/2010 for the course CHEN 4330 taught by Professor Staff during the Spring '08 term at Colorado.

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HW2_8 - 0.65 6.0< 0.52 “3 4.0 0.39 3.0 0.25 1.0 0.13 m...

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