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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
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 \( X_1 = 20\% \) 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** \[ V_{PFR} = \int_{100}^{80} \frac{dF_A}{-1.8} = 11.11 \, \text{dm}^3 \] \[ V_{CSTR} = \frac{F_{A0} - F_A}{-r_A} = \frac{80 - 35}{0.83} = 54.22 \, \text{(at 65\% conversion)} \] \[ \text{Total Volume} = 65 \, \text{dm}^3 \]
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** \[ V_{PFR} = \int_{100}^{80} \frac{dF_A}{-1.8} = 11.11 \, \text{dm}^3 \] \[ V_{CSTR} = \frac{F_{A0} - F_A}{-r_A} = \frac{80 - 35}{0.83} = 54.22 \, \text{(at 65\% conversion)} \] \[ \text{Total Volume} = 65 \, \text{dm}^3 \]
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