VIPSolutionsGive step-by-step solution with explanation and final answer:/ ¥ ) I Bry Er ARSE ES pe i EE a, ET RRR AR as [I o : EB I ar BEE a ve [Ee II o i i a ll : 38 I + “= o hie Il Zo pec LL ANAL g J. : me lf = iL Von c ETE Sa | Vs Tre be
Question:
Give step-by-step solution with explanation and final answer:
/ ¥ )
I Bry Er ARSE ES pe
i EE a, ET
RRR AR as
[I o : EB
I ar BEE
a ve [Ee
II o i i a
ll : 38
I + “= o hie
Il Zo pec LL ANAL g J.
: me
lf = iL Von c ETE Sa
| Vs Tre
be
This is AI generated Response.
Asked by: Francis
Created at: 2026-01-05 13:52:04
Asked by: Francis
Created at: 2026-01-05 13:52:04
Final Answer:
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Let's break down the solution step by step based on the information and equations given in your image.
---
## **Given Data**
- **Reaction:** \( A \rightarrow B \) (First-order, irreversible)
- **Type:** Plug Flow Reactor (PFR)
- **Feed:** \( F_{A} = 5 \text{ kmol/hr} \)
- **Initial concentration:** \( C_{A} = 1.5 \text{ kmol/m}^3 \)
- **Rate constant:** \( k = .8 \text{ hr}^{-1} \)
- **Rate Law:** \( -r_A = kC_A \)
- **\( C_A = C_{A}(1 - X) \)**
- **To Find:** Reactor volume \( V \) required for a given conversion \( X \), and space time \( \tau \).
---
## **Step 1: Reactor Volume \( V \) Required for Conversion \( X \)**
### **Plug Flow Reactor Design Equation:**
\[
V = \int_^X \frac{F_{A}}{-r_A} dX
\]
### **Substitute the Expressions:**
- \( -r_A = kC_A = k C_{A}(1 - X) \)
So,
\[
V = \int_^X \frac{F_{A}}{k C_{A}(1-X)} dX
\]
### **Take Constants Out of the Integral:**
\[
V = \frac{F_{A}}{k C_{A}} \int_^X \frac{1}{1-X} dX
\]
### **Evaluate the Integral:**
\[
\int_^X \frac{1}{1-X} dX = -\ln(1-X) \Bigg|_^X = -\ln(1-X) + \ln(1-) = -\ln(1-X)
\]
### **So:**
\[
V = \frac{F_{A}}{kC_{A}} [-\ln(1-X)]
\]
---
## **Step 2: Space Time (\( \tau \))**
- **Space time:** \( \tau = \frac{V}{V_} \)
- **Where** \( V_ = \frac{F_{A}}{C_{A}} \)
---
## **Final Answers**
### **A. Reactor Volume \( V \):**
\[
\boxed{
V = \frac{F_{A}}{kC_{A}} [-\ln(1-X)]
}
\]
Plug in the values if a specific conversion \( X \) is given.
---
### **B. Space Time (\( \tau \)):**
\[
\tau = \frac{V}{V_}
\]
\[
V_ = \frac{F_{A}}{C_{A}}
\]
So,
\[
\tau = \frac{V}{F_{A}/C_{A}} = \frac{V C_{A}}{F_{A}}
\]
Substitute \( V \) from above:
\[
\tau = \frac{C_{A}}{F_{A}} \cdot \frac{F_{A}}{kC_{A}} [-\ln(1-X)] = \frac{1}{k} [-\ln(1-X)]
\]
\[
\boxed{
\tau = \frac{-\ln(1-X)}{k}
}
\]
---
### **Summary Table**
| Quantity | Formula | Plug in Values (if \( X \) is known) |
|----------|---------|--------------------------------------|
| \( V \) | \(\frac{F_{A}}{kC_{A}} [-\ln(1-X)]\) | \( F_{A} = 5,\ k = .8,\ C_{A} = 1.5 \) |
| \( \tau \) | \(\frac{-\ln(1-X)}{k}\) | \( k = .8 \) |
If you provide a value for conversion \( X \), I can compute the exact numerical answers.
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