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H (273 K) = -20 kcal/mol, HB (273 K) = -15 kcal/mol, H (273 K)=-41 kcal/mol • = 15 cal/mol K C₁ = 30 cal/mol K CPA = CPB 3 dm k = 0.01 Ua = 20 cal/m³/s/K Tao = 450K mol·s · Cec at 300 K E E=10,000 cal/mol mc = 50g/s C = 1cal/g/K PCool

H (273 K) = -20 kcal/mol, HB (273 K) = -15 kcal/mol, H (273 K)=-41 kcal/mol • = 15 cal/mol K C₁ = 30 cal/mol K CPA = CPB 3 dm k = 0.01 Ua = 20 cal/m³/s/K Tao = 450K mol·s · Cec at 300 K E E=10,000 cal/mol mc = 50g/s C = 1cal/g/K PCool

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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Use the data in Problem P11-4A for the following reaction.

Th elementary, irreversible, organic liquid-phase reaction

A + B → C

is carried out in a flow reactor. An equal molar feed in A and B enters at 27°C, and the volumetric flow

rate is 2 dm3/s and CA0 = 0.1 kmol/m3

Additional information:

The reversible reaction (part (d) of P11-4A) is now carried out in a PFR with a heat exchanger. Plot

and then analyze X, Xe, T, Ta, Qr, Qg, and the rate, –rA, for the following cases:

(d) Countercurrent heat exchanger Ta (Ans.: At V = 10 m3 then X = 0.364 and T = 450 K)

(e) Adiabatic operation

Please share MATLAB code. 

H (273 K) = -20 kcal/mol, HB (273 K) = -15 kcal/mol, H (273 K)=-41 kcal/mol • = 15 cal/mol K C₁ = 30 cal/mol K CPA = CPB 3 dm k = 0.01 Ua = 20 cal/m³/s/K Tao = 450K mol·s · Cec at 300 K E E=10,000 cal/mol mc = 50g/s C = 1cal/g/K PCool
Transcribed Image Text:H (273 K) = -20 kcal/mol, HB (273 K) = -15 kcal/mol, H (273 K)=-41 kcal/mol • = 15 cal/mol K C₁ = 30 cal/mol K CPA = CPB 3 dm k = 0.01 Ua = 20 cal/m³/s/K Tao = 450K mol·s · Cec at 300 K E E=10,000 cal/mol mc = 50g/s C = 1cal/g/K PCool
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