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What Volume (PFR) is required to obtain 96% for the single reactor process. this is what i have done so far not sure if i am in right direction regression rate law was given as a known

What Volume (PFR) is required to obtain 96% for the single reactor process. this is what i have done so far not sure if i am in right direction regression rate law was given as a known

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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What Volume (PFR) is required to obtain 96% for the single reactor process.

this is what i have done so far not sure if i am in right direction

regression rate law was given as a known 

dFA dV CAO = 0.144 gmol * L-¹ YAO = 0.5 УАО Vo = 6L * S-1 V = ? XA =? FAO = TA (dXA) = fav 1 -dXA -rAexit FAO FAO (-A) exitdXA XA = S²² = -TA dV → FAO = =f₁dv. dV .XA 1 -TAexit 0 FA0o = √² dXA XA TA = - fº av → = T = 422.2K P = 10 atm 0.00596(1-XA) 0.94671 FAO = : CAovo 1 0.00596(1 – Xa)0.94671 dXA = V
Transcribed Image Text:dFA dV CAO = 0.144 gmol * L-¹ YAO = 0.5 УАО Vo = 6L * S-1 V = ? XA =? FAO = TA (dXA) = fav 1 -dXA -rAexit FAO FAO (-A) exitdXA XA = S²² = -TA dV → FAO = =f₁dv. dV .XA 1 -TAexit 0 FA0o = √² dXA XA TA = - fº av → = T = 422.2K P = 10 atm 0.00596(1-XA) 0.94671 FAO = : CAovo 1 0.00596(1 – Xa)0.94671 dXA = V
FAO = .XA 0 167.785(1-X)-0.94671dX dXA = V
Transcribed Image Text:FAO = .XA 0 167.785(1-X)-0.94671dX dXA = V
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