Question

A liquid-phase reaction A + B C, p = kcbc takes place in a CSTR of volume VR in the presence of a large excess of reactant B. Assume the reactor achieves 50% conversion of A at steady state, (a) What is the steady-state conversion if the original reacter is replaced by two CSTRs of volume VR/2 in series? (b) What is the conversion if the original reactor is replaced by tbree CSTRS of volume VR/3 in series? (C) What is the conversion if the original reactor is replaced by a plug-flow reactor of volume Vr?

Note: Assume that the volumetric flow rate does not change with reaction extent; when one reactant is in large excess, it is reasonable to assume constant volumetric flow rate, especially for liquid-phase reactions. You can only find KCbVr/Q that would yield the desired 50% conversion in a single CSTR!

Answer 1

solo given liquid phase. reactor je =0 . A+B & C :-N = CCCB . Given xx = 0.5 FA is constant 2 notant y la Гот, ve ) ا couhe (@) 9f Single STR ito replaced by 2-cStr's with volume V performance eam for single KSTRI » KGBYR - RA (where KCB FAo are) BA who constants ) FAO >> Kopi Te (39= link) for q 2-OSTR's in ist georitos - (1-0.5)2 series K CONR2) - BA a 2 => om @ [1-4477 7 [1-4A2 -) X-224+I-A=0

2-CSTR's in sener

ir -3A+1=0 = 3+19-4 = 0.3819, 8-618 " RAē greater than 1 XA Dor zen't be 1 = 1 Tim, = 0:3819) and reauld (1-(A2)² - 1 = get RAL-RAI . (1-2A2) 2 => ! B92 - 0.3819 (1-8A2) => A 21x2 + 1 -Xaq +0.381959 → Tag = 0.5682 Increased conversion percentage is 6.82% . (b for in series 3 ESTR's K OBVR este 2. single CSIR BA 50-667 1st KCB NR/3 = 3 = 20 .. - ^21 = 1.499 RA - 3:58A+I = 0 XA, F 0-3.385

Sve 3-CSTR's in series

and recular → 0.6667 (B-2012+1) = 2A30.31385 0-6667 XÁ2 - 2.3332A, +0.98052= 0 TER = 0.4883 ( from bave vac) 3rd reautor = 43-0.4983 » 0 - 6667A - 1333 - TI-I549 50 XAz = 0.5967 i Increased is 9.67% percentage of conversion (C) of P.FR is replaced by CSTR with "V daa performante de pre FAO Od os = a (i from een o) sa -> A = 0.6667