An exothermic reaction of reactant (A) forms the economically important product (B) and sellable by- product (C),:
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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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 \( X_1 = 20\% \) Conversion & PFR \( X_2 = 65\% \) Conversion**
\[
PFR \rightarrow F_{A0} = \frac{F_{A0}}{1-X_1} = \frac{100}{1-0.20} = 125 \, \text{mol/min}
\]
\[
CSTR \rightarrow 100 = 125 - 1.8 \times V_1 \rightarrow V_1 = \frac{25}{1.8} = 13.889 \, \text{dm}^3
\]
For PFR Design Equation:
\[
V_2 = \int_{F_A0}^{F_A0} x_2 \left( \frac{dF_A}{-r_{A2}} \right) \& \rightarrow F_A(X_2) = \frac{F_A0}{1-X_2} = \frac{100}{1-0.65} = 285.71 \, \text{mol/min}
\]
\[
V_2 = \frac{1}{1.2} \int_{100}^{285.71} dF_A = \frac{1}{1.2} [285.71 - 100] = 154.76
\]
\[
\text{Total Volume} = 13.889 + 154.76 = 168.65 \, \text{dm}^3
\]
**B) Vol PFR @ \( X_1 = 20\% \) Conversion & CSTR @ \( X_2 = 65\% \) Conversion**
\[
V_{PFR} = \int_{100}^{80} \frac{dF_A}{-1.8} = 11.11 \, \text{dm}^3
\]
\[
V_{CSTR} = \frac{F_{A0} - F_A}{-r_A} = \frac{80 - 35}{0.83} = 54.22 \, \text{(at 65\% conversion)}
\]
\[
\text{Total Volume} = 65 \, \text{dm}^3
\]](https://content.bartleby.com/qna-images/question/d47fc2c3-8f4b-4085-b234-bc909be774e3/7f679d07-5e33-44cc-ab21-f13eddddb5ac/cefkqnh_processed.png)
Transcribed Image Text:**2)**
**A) CSTR \( X_1 = 20\% \) Conversion & PFR \( X_2 = 65\% \) Conversion**
\[
PFR \rightarrow F_{A0} = \frac{F_{A0}}{1-X_1} = \frac{100}{1-0.20} = 125 \, \text{mol/min}
\]
\[
CSTR \rightarrow 100 = 125 - 1.8 \times V_1 \rightarrow V_1 = \frac{25}{1.8} = 13.889 \, \text{dm}^3
\]
For PFR Design Equation:
\[
V_2 = \int_{F_A0}^{F_A0} x_2 \left( \frac{dF_A}{-r_{A2}} \right) \& \rightarrow F_A(X_2) = \frac{F_A0}{1-X_2} = \frac{100}{1-0.65} = 285.71 \, \text{mol/min}
\]
\[
V_2 = \frac{1}{1.2} \int_{100}^{285.71} dF_A = \frac{1}{1.2} [285.71 - 100] = 154.76
\]
\[
\text{Total Volume} = 13.889 + 154.76 = 168.65 \, \text{dm}^3
\]
**B) Vol PFR @ \( X_1 = 20\% \) Conversion & CSTR @ \( X_2 = 65\% \) Conversion**
\[
V_{PFR} = \int_{100}^{80} \frac{dF_A}{-1.8} = 11.11 \, \text{dm}^3
\]
\[
V_{CSTR} = \frac{F_{A0} - F_A}{-r_A} = \frac{80 - 35}{0.83} = 54.22 \, \text{(at 65\% conversion)}
\]
\[
\text{Total Volume} = 65 \, \text{dm}^3
\]
Solution
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