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Feed stream CAO? gmole L-¹ YAO = 0.5 Vo = 6L*s-1 nowns: CAO H V = V = ? L XA = T = 422.2 K P = 10 atm Product stream 0.00596(1-XA) 0.94671 Fao = Caovo TA = FAOXA CAoVOXA (−ra)_exit →V = 0.00596(1 – Xa)0.94671 0.5 10atm L. atm mol.k mol L УАО Rgas FAO = CAOVO = 0.1443- 0.08206 (422.2k) = 0.144 mol *6= = 0.8659- mol L

Feed stream CAO? gmole L-¹ YAO = 0.5 Vo = 6L*s-1 nowns: CAO H V = V = ? L XA = T = 422.2 K P = 10 atm Product stream 0.00596(1-XA) 0.94671 Fao = Caovo TA = FAOXA CAoVOXA (−ra)_exit →V = 0.00596(1 – Xa)0.94671 0.5 10atm L. atm mol.k mol L УАО Rgas FAO = CAOVO = 0.1443- 0.08206 (422.2k) = 0.144 mol *6= = 0.8659- mol L

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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1. 

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

not sure if i am in the right direction is X_A 96?

also, CA0 was not given so i used ideal gas to calc that 

V = TA = 0.00596(1 − XA)0.94671 FAO = CAOVO →V= FAOXA (-ra)_exit VCSTR = YAO (P УАО CAO = (G) Rgas (T) FAO = CAOVO = 0.1443 FAOXA 0.5 10atm L. atm mol.k mol L -ra) Exit CAoVOXA 0.00596(1XA) 0.9467 0.08206 = 0.144 (422.2k) mol L *6=0.8659 S S (0.865908 mol) (XA) S 0.00596(1X)0.94671. mol L mol L*S
Transcribed Image Text:V = TA = 0.00596(1 − XA)0.94671 FAO = CAOVO →V= FAOXA (-ra)_exit VCSTR = YAO (P УАО CAO = (G) Rgas (T) FAO = CAOVO = 0.1443 FAOXA 0.5 10atm L. atm mol.k mol L -ra) Exit CAoVOXA 0.00596(1XA) 0.9467 0.08206 = 0.144 (422.2k) mol L *6=0.8659 S S (0.865908 mol) (XA) S 0.00596(1X)0.94671. mol L mol L*S
Feed stream CAO? gmole * L-1 YAO = 0.5 Vo = 6L*s-1 Knowns: H V = CAO = V = ? L XA = ΧΑ УАО Rgas T A = FAOXA (-A)_exit = = T = 422.2 K P Product stream 0.00596(1X) 0.94671 FA0 = V = = 10 atm 0.08206 FAO CAOVO = 0.1443 CAovo 0.5 10atm L. atm mol.k mol L L Cao VoXA 0.00596(1XA) 0.940 (422.2k) = 0.144. *6 = 0.8659 S mol L mol S
Transcribed Image Text:Feed stream CAO? gmole * L-1 YAO = 0.5 Vo = 6L*s-1 Knowns: H V = CAO = V = ? L XA = ΧΑ УАО Rgas T A = FAOXA (-A)_exit = = T = 422.2 K P Product stream 0.00596(1X) 0.94671 FA0 = V = = 10 atm 0.08206 FAO CAOVO = 0.1443 CAovo 0.5 10atm L. atm mol.k mol L L Cao VoXA 0.00596(1XA) 0.940 (422.2k) = 0.144. *6 = 0.8659 S mol L mol S
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