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A second-order reaction is to be carried out in a real reactor that gives the following outlet concentration for a step input:For 0 ≤ t < 10 min, then CT = 10 (1-e-.1t)For 10 min ≤ t, then CT = 5+10 (1-e-.1t)(a) What model do you propose and what are your model parameters, α and β?(b) What conversion can be expected in the real reactor?(c) How would your model change and conversion change if your outlet tracer concentration was as follows?For t ≤ 10 min, then CT = 0For t ≥ 10 min, then CT = 5+10 (1-e–0.2(t-10))v0 = 1 dm3/min, k = 0.1 dm3/mol ⋅ min, CA0 = 1.25 mol/dm.3

A second-order reaction is to be carried out in a real reactor that gives the following outlet concentration for a step input:For 0 ≤ t < 10 min, then CT = 10 (1-e-.1t)For 10 min ≤ t, then CT = 5+10 (1-e-.1t)(a) What model do you propose and what are your model parameters, α and β?(b) What conversion can be expected in the real reactor?(c) How would your model change and conversion change if your outlet tracer concentration was as follows?For t ≤ 10 min, then CT = 0For t ≥ 10 min, then CT = 5+10 (1-e–0.2(t-10))v0 = 1 dm3/min, k = 0.1 dm3/mol ⋅ min, CA0 = 1.25 mol/dm.3

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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A second-order reaction is to be carried out in a real reactor that gives the following outlet concentration for a step input:
For 0 ≤ t < 10 min, then CT = 10 (1-e-.1t)
For 10 min ≤ t, then CT = 5+10 (1-e-.1t)
(a) What model do you propose and what are your model parameters, α and β?
(b) What conversion can be expected in the real reactor?
(c) How would your model change and conversion change if your outlet tracer concentration was as follows?
For t ≤ 10 min, then CT = 0
For t ≥ 10 min, then CT = 5+10 (1-e–0.2(t-10))
v0 = 1 dm3/min, k = 0.1 dm3/mol ⋅ min, CA0 = 1.25 mol/dm.3

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