An exothermic reaction of reactant (A) forms the economically important product (B) and sellable by- product (C),:

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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An exothermic reaction of reactant (A) forms the economically important product (B) and sellable by-
product (C),:
Which was carried out isothermally and the following data was recorded:
XA | O
mol
-TA Adm³-min 1.2
1 (dm³.min
mol
-TA
FAD (dm³)
-TA
0.83
333.33
mol
min
0.1
1.8
0.56
222.22
0.2
1.8
A-→ B+C
0.56
0.3
3
0.33
0.5
3
0.33
0.65
1.2
-
0.83
222.22 133.33 133.33 333.33
1
feed flow rate = 100
pure A
=
A) Determine the reactor volumes in dm³, of a PFR with a conversion X₁
20% followed by a CSTR that reaches conversion X₂
reactor process.
0.8
0.75
1.33
533.33
65% at the end of two
0.95
0.5464
1.83
732.06
Transcribed Image Text:An exothermic reaction of reactant (A) forms the economically important product (B) and sellable by- product (C),: Which was carried out isothermally and the following data was recorded: XA | O mol -TA Adm³-min 1.2 1 (dm³.min mol -TA FAD (dm³) -TA 0.83 333.33 mol min 0.1 1.8 0.56 222.22 0.2 1.8 A-→ B+C 0.56 0.3 3 0.33 0.5 3 0.33 0.65 1.2 - 0.83 222.22 133.33 133.33 333.33 1 feed flow rate = 100 pure A = A) Determine the reactor volumes in dm³, of a PFR with a conversion X₁ 20% followed by a CSTR that reaches conversion X₂ reactor process. 0.8 0.75 1.33 533.33 65% at the end of two 0.95 0.5464 1.83 732.06
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ah, you got it. 

Now can you solve for the reactor volumes of CSTR w/ a conversion of X1=20% followed by PFR X2=65%  i have attached the work i have done so far. looks like i was wrong but i def understand better now that you explained. 

2)
A) CSTR X₁
V₂
=
=
20% Conversion & PFR X₂
=
FAO
1 - X₁
For PFR Design Equation
·FAO
FAO
V₂
X₂
=
CSTR → 100 = 125 — 1.8 * V₁ → V₁
-
2 (dFA₂)
PFR → FAO
1
1.2
S.
285.71
100
& →
=
Vcstr
FA(X₂)
dFA =
1
1.2
=
VPFR
FAO - FA
-TA
=
=
65% Conversion
100
10.20
=
100
FAO
1 - X₂
=
B) Vol PFR @ X₁ = 20% Conversion & CSTR @ X₂
80 dFA
-1.8
80 - 35
==
100
1 - 0.65
= 125-
25
1.8
Total Volume = 13.889 + 154.76 = 168.65 dm³
=
=
mol
min
13.889 dm³
[285.71-100] = 154.76
0.83
Tot Vol = 65dm³|
285.71
= 11.11 dm³
= 65% Conversion.
mol
min
= 54.22 @ 65% conv
Transcribed Image Text:2) A) CSTR X₁ V₂ = = 20% Conversion & PFR X₂ = FAO 1 - X₁ For PFR Design Equation ·FAO FAO V₂ X₂ = CSTR → 100 = 125 — 1.8 * V₁ → V₁ - 2 (dFA₂) PFR → FAO 1 1.2 S. 285.71 100 & → = Vcstr FA(X₂) dFA = 1 1.2 = VPFR FAO - FA -TA = = 65% Conversion 100 10.20 = 100 FAO 1 - X₂ = B) Vol PFR @ X₁ = 20% Conversion & CSTR @ X₂ 80 dFA -1.8 80 - 35 == 100 1 - 0.65 = 125- 25 1.8 Total Volume = 13.889 + 154.76 = 168.65 dm³ = = mol min 13.889 dm³ [285.71-100] = 154.76 0.83 Tot Vol = 65dm³| 285.71 = 11.11 dm³ = 65% Conversion. mol min = 54.22 @ 65% conv
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