XA 0.0 mol -TA m³.sec 1 m³.sec -TA mol FAO (m³-sec) -TA mol 0.49 0.1 0.41 0.2 0.34 0.4 0.199 Reaction AB 0.6 Feed Flow Rate = 0.8- 0.117 mol sec PureA T = 500K & P = 830 kPa (8.2 atm) 0.7 0.083 0.8 0.054
XA 0.0 mol -TA m³.sec 1 m³.sec -TA mol FAO (m³-sec) -TA mol 0.49 0.1 0.41 0.2 0.34 0.4 0.199 Reaction AB 0.6 Feed Flow Rate = 0.8- 0.117 mol sec PureA T = 500K & P = 830 kPa (8.2 atm) 0.7 0.083 0.8 0.054
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
Related questions
Question
100%
E

Transcribed Image Text:(E) Fortwo CSTRS in parallel where each CSTR is 1.6 m^3, determine the conversion after the reactors.
FAO is divided equally between the two CSTRs.

Transcribed Image Text:XA 0.0
mol
-TA ³-sec
1 (m³-sec)
-TA mol
FAO (m³-sec
-TA
mol
0.49
0.1
0.41
0.2
0.34
0.4
0.199
Reaction A → B
0.6
0.117
mol
sec PureA
T = 500K & P = 830 kPa (8.2 atm)
Feed Flow Rate = 0.8
0.7
0.083
0.8
0.054
Expert Solution

Step 1: Conversion in two CSTRs in parallel
Given, feed flow rate
Tabulating data of XA and
XA | ||
0 | 1.632653 | 0 |
0.1 | 1.95122 | 0.195122 |
0.2 | 2.352941 | 0.470588 |
0.4 | 4.020101 | 1.60804 |
0.6 | 6.837607 | 4.102564 |
0.7 | 9.638554 | 6.746988 |
0.8 | 14.81481 | 11.85185 |
Two CSTRs in parallel, each with a volume of 1.6 m3 yields same conversion as single CSTR with a volume of 3.2 m3.
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