You were contacted by a vendor that claims it has a PFR of 200 cubic feet that can achieve 80% conversion for the following process The liquid phase reaction 3R + Q → 2P will be carried out in a PFR reactor. Reactants R and Q and dilutant S will be fed at 60 ft3/min with the following molar flow rates: Species R Species Q Species S || The rate law will be 300 lbmole / min 100 lbmole / min 50 lbmole / min - TR lbmole ft³ * min = 0.1 * C * CQ where CR and Co~ Ibmole/ft³. Determine the volume of the PFR reactor based on the given conversion and the final concentrations of all species involved in the system
You were contacted by a vendor that claims it has a PFR of 200 cubic feet that can achieve 80% conversion for the following process The liquid phase reaction 3R + Q → 2P will be carried out in a PFR reactor. Reactants R and Q and dilutant S will be fed at 60 ft3/min with the following molar flow rates: Species R Species Q Species S || The rate law will be 300 lbmole / min 100 lbmole / min 50 lbmole / min - TR lbmole ft³ * min = 0.1 * C * CQ where CR and Co~ Ibmole/ft³. Determine the volume of the PFR reactor based on the given conversion and the final concentrations of all species involved in the system
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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![You were contacted by a vendor that claims it has a PFR of 200 cubic feet that can achieve 80% conversion for the following process:
The liquid phase reaction 3R + Q → 2P will be carried out in a PFR reactor. Reactants R and Q and dilutant S will be fed at 60 ft³/min with the following molar flow rates:
- Species R = 300 lbmole/min
- Species Q = 100 lbmole/min
- Species S = 50 lbmole/min
The rate law will be
\[ -r_R \left( \frac{\text{lbmole}}{\text{ft}^3 \cdot \text{min}} \right) = 0.1 \times C_R^3 \times C_Q \]
where \( C_R \) and \( C_Q \sim \text{lbmole} / \text{ft}^3 \).
Determine the volume of the PFR reactor based on the given conversion and the final concentrations of all species involved in the system.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd47fc2c3-8f4b-4085-b234-bc909be774e3%2Fb3940955-7a84-4035-b97e-996bc8d90a27%2Fxfoinod_processed.png&w=3840&q=75)
Transcribed Image Text:You were contacted by a vendor that claims it has a PFR of 200 cubic feet that can achieve 80% conversion for the following process:
The liquid phase reaction 3R + Q → 2P will be carried out in a PFR reactor. Reactants R and Q and dilutant S will be fed at 60 ft³/min with the following molar flow rates:
- Species R = 300 lbmole/min
- Species Q = 100 lbmole/min
- Species S = 50 lbmole/min
The rate law will be
\[ -r_R \left( \frac{\text{lbmole}}{\text{ft}^3 \cdot \text{min}} \right) = 0.1 \times C_R^3 \times C_Q \]
where \( C_R \) and \( C_Q \sim \text{lbmole} / \text{ft}^3 \).
Determine the volume of the PFR reactor based on the given conversion and the final concentrations of all species involved in the system.
Expert Solution
Step 1: Explanation of PFR
The PFR is a steady state reactor. Since we have allowed fresh feed the initial conversion that is Xi will be 0.
And S is dilution reactant hence will not take part in reaction
Step by step
Solved in 3 steps with 3 images
Follow-up Questions
Read through expert solutions to related follow-up questions below.
Follow-up Question
Re-write the reactor design equation for the PFR using the rate law
can you please re-do the problem now using a different rate law, anything will help
i can repost for more help
![The liquid phase reaction \(3R + Q \rightarrow 2P\) will be carried out isothermally in a 200 Cubic Feet Plug Flow Reactor (PFR).
Reactants R and Q, and dilutant S, will be fed at 60 ft³/min with the following molar flow rates:
- Species R = 300 lbmole/min
- Species Q = 100 lbmole/min
- Species S = 50 lbmole/min
The rate law will be
\[
-r_R \left( \frac{\text{lbmole}}{\text{ft}^3 \cdot \text{min}} \right) =
\]
where \(C_R\) and \(C_Q\) are approximately in lbmole/ft³.
A) Re-write the reactor design equation for the PFR using the rate law
\[
-r_R = 0.033C_{R_0}^2(1 - X_R)^4
\]](https://content.bartleby.com/qna-images/question/d47fc2c3-8f4b-4085-b234-bc909be774e3/b094352c-e589-4f8c-926c-e545f365337a/t5yf1j7_processed.png)
Transcribed Image Text:The liquid phase reaction \(3R + Q \rightarrow 2P\) will be carried out isothermally in a 200 Cubic Feet Plug Flow Reactor (PFR).
Reactants R and Q, and dilutant S, will be fed at 60 ft³/min with the following molar flow rates:
- Species R = 300 lbmole/min
- Species Q = 100 lbmole/min
- Species S = 50 lbmole/min
The rate law will be
\[
-r_R \left( \frac{\text{lbmole}}{\text{ft}^3 \cdot \text{min}} \right) =
\]
where \(C_R\) and \(C_Q\) are approximately in lbmole/ft³.
A) Re-write the reactor design equation for the PFR using the rate law
\[
-r_R = 0.033C_{R_0}^2(1 - X_R)^4
\]
Solution
by Bartleby ExpertFollow-up Question
this is great, can you please type out solution using word (equation)
its hard for me to read
Solution
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