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(e) Example 2-8. The space time required to achieve 80% conversion in a CSTR is 5 h. The entering volumetric flow rate and concentration of reactant A are 1 dm³/min and 2.5 molar, respectively. If possible, deter- mine (1) the rate of reaction, -A= (2) the reactor volume, V = and (4) the PFR 1 (3) the exit concentration of A, CA = space time for 80% conversion.

(e) Example 2-8. The space time required to achieve 80% conversion in a CSTR is 5 h. The entering volumetric flow rate and concentration of reactant A are 1 dm³/min and 2.5 molar, respectively. If possible, deter- mine (1) the rate of reaction, -A= (2) the reactor volume, V = and (4) the PFR 1 (3) the exit concentration of A, CA = space time for 80% conversion.

Introduction to Chemical Engineering Thermodynamics
Introduction to Chemical Engineering Thermodynamics
8th Edition
ISBN:9781259696527
Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Chapter1: Introduction
Section: Chapter Questions
Problem 1.1P
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2.8

(e) Example 2-8. The space time required to achieve 80% conversion in a CSTR is 5 h. The entering volumetric flow rate and concentration of reactant A are 1 dm³/min and 2.5 molar, respectively. If possible, deter- mine (1) the rate of reaction, - A =__ (2) the reactor volume, V = (3) the exit concentration of A, CA = ,and (4) the PFR space time for 80% conversion. mol & FAO dm³ CA0 = 0.2- V T == Vo 1 SV == T = 0.4 mol S
Transcribed Image Text:(e) Example 2-8. The space time required to achieve 80% conversion in a CSTR is 5 h. The entering volumetric flow rate and concentration of reactant A are 1 dm³/min and 2.5 molar, respectively. If possible, deter- mine (1) the rate of reaction, - A =__ (2) the reactor volume, V = (3) the exit concentration of A, CA = ,and (4) the PFR space time for 80% conversion. mol & FAO dm³ CA0 = 0.2- V T == Vo 1 SV == T = 0.4 mol S
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