Give step-by-step solution with explanation and final answer:
{give me the answer with out using bullet points}3.25 points) As a chemical process engineer at a specialty chemical production plant, yo

Question

Give step-by-step solution with explanation and final answer:
{give me the answer with out using bullet points}3.25 points) As a chemical process engineer at a specialty chemical production plant, you are
tasked with designing 2 new isothermal. liquid-phase process to form an industrial-grade solvent,
B, from a new starting feedstock, A.
A—3B
—ry = kG
k=1210 2 mol 2 min"

The new feedstock is available as pure A at a concentration (Css) of 10 M, and can be delivered to

the process at a steady-state volumetric flow rate (vs) of 5 L/min

You are given two reactors of equal volume (cach is 100 L), where one is a CSTR and one is 2

PER, and can decide between the following three configurations for the reactor network:

() Reactors in series: CSTR followed by a PEFR.
(i) Reactors in series: PER followed by a CSTR.
(ii) Reactors in parallel with half of the feed going fo cach reactor.

(2) Calculate the exit conversion of A for each of the three configurations.

(b) Explain which reactor configuration will give the highest overall conversion, and the
differences in conversion between configurations (i) 2nd (i).

One moming, you arrive at work and find out that t

VIPSolutions · Accepted Answer

Let's solve the problem step-by-step.

---

# **Given Data**

- **Reaction:** $ A \rightarrow B $
- **Rate Law:** $ -r_A = kC_A^3 $
- **Rate Constant:** $ k = 1 \times 10^{-3} \, L^2 \, mol^{-2} \, min^{-1} $
- **Feed Concentration:** $ C_{A} = 10\, mol/L $
- **Volumetric Flow Rate:** $ v_ = 5\, L/min $
- **Reactor Volumes:** $ V_1 = V_2 = 100\, L $ (each)

---

## **(a) Calculate Exit Conversion for Each Configuration**

### **1. General Definitions**

Let $ X $ be the conversion of A. For a single reactor:

#### **CSTR:**
$$
V = \frac{F_{A} X}{-r_A}
$$
$$
F_{A} = v_ C_{A}
$$
$$
C_A = C_{A}(1-X)
$$
$$
-r_A = kC_A^3 = k(C_{A}(1-X))^3
$$

So,
$$
V = \frac{v_ C_{A} X}{k (C_{A}(1-X))^3}
$$
$$
X = \frac{k V C_{A}^2 (1-X)^3}{v_}
$$

#### **PFR:**
$$
V = \int_{}^{X} \frac{F_{A}}{-r_A} dX
$$
$$
V = v_ \int_{}^{X} \frac{dX}{k C_{A}^3 (1-X)^3}
$$
$$
V = \frac{v_}{k C_{A}^3} \int_{}^{X} (1-X)^{-3} dX
$$
$$
\int (1-X)^{-3} dX = \frac{1}{2}(1-X)^{-2}
$$
$$
V = \frac{v_}{2 k C_{A}^3} \left((1-X)^{-2} - 1\right)
$$

---

### **Configuration (i): CSTR → PFR**

#### **Step 1: CSTR First**
$$
V_{CSTR} = 100\, L
$$
$$
V_{CSTR} = \frac{v_ C_{A} X_1}{k (C_{A}(1-X_1))^3}
$$
Plug in values:
$$
100 = \frac{5 \times 10 \times X_1}{1 \times 10^{-3} \times (10(1-X_1))^3}
$$
$$
100 = \frac{50 X_1}{1 \times 10^{-3} \times 100 (1-X_1)^3}
$$
$$
100 = \frac{50 X_1}{1 (1-X_1)^3}
$$
$$
100 (1-X_1)^3 = 50 X_1
$$
$$
2 (1-X_1)^3 = X_1
$$
Let $ y = 1-X_1 $:
$$
2y^3 + y - 1 = 
$$
Solve numerically for $ y $:

Try $ y = .6 $:
$$
2(.6)^3 + .6 - 1 = 2(.216) + .6 - 1 = .432 + .6 - 1 = 1.032 - 1 = .032
$$
So $ y \approx .6 \implies X_1 \approx .4 $

#### **Step 2: PFR Second**
Feed to PFR: $ C_{A1} = C_{A}(1-X_1) = 10 \times .6 = 6\, mol/L $

For PFR with feed $ C_{A1} = 6 \, mol/L $, $ V = 100 \, L $:
$$
V_{PFR} = \frac{v_}{2k C_{A1}^3} \left((1-X_2)^{-2} - 1\right)
$$
Let $ X_2 $ be the conversion in the PFR relative to its feed, so the overall conversion is:
$$
X_{total} = 1 - (1-X_1)(1-X_2)
$$

But first, let's convert through the PFR:
$$
100 = \frac{5}{2 \times 1 \times 10^{-3} \times 6^3} \left((1-X_2)^{-2} - 1\right)
$$
$$
100 = \frac{5}{2 \times 1 \times 10^{-3} \times 216} \left((1-X_2)^{-2} - 1\right)
$$
$$
100 = \frac{5}{.432} \left((1-X_2)^{-2} - 1\right)
$$
$$
100 = 11.574 \left((1-X_2)^{-2} - 1\right)
$$
$$
\frac{100}{11.574} = (1-X_2)^{-2} - 1
$$
$$
8.636 = (1-X_2)^{-2} - 1
$$
$$
(1-X_2)^{-2} = 9.636
$$
$$
1-X_2 = \frac{1}{\sqrt{9.636}} = .322
$$
$$
X_2 = 1 - .322 = .678
$$

**Overall conversion:**
$$
X_{overall} = 1-(1-X_1)(1-X_2) = 1-(.6 \times .322) = 1-.193 = .807
$$

---

### **Configuration (ii): PFR → CSTR**

#### **Step 1: PFR First**
Feed: $ C_{A} = 10\, mol/L $, $ V = 100\, L $
$$
100 = \frac{5}{2 \times 1 \times 10^{-3} \times 10^3} \left((1-X_1)^{-2} - 1\right)
$$
$$
100 = \frac{5}{2} \left((1-X_1)^{-2} - 1\right)
$$
$$
100 = 2.5 \left((1-X_1)^{-2} - 1\right)
$$
$$
40 = (1-X_1)^{-2} - 1
$$
$$
(1-X_1)^{-2} = 41
$$
$$
1-X_1 = \frac{1}{\sqrt{41}} = .156
$$
$$
X_1 = 1 - .156 = .844
$$

#### **Step 2: CSTR Second**
Feed to CSTR: $ C_{A1} = 10 \times .156 = 1.56\, mol/L $
$$
100 = \frac{5 \times 1.56 \times X_2}{1 \times 10^{-3} \times (1.56(1-X_2))^3}
$$
$$
100 = \frac{7.8 X_2}{1 \times 10^{-3} \times (1.56 (1-X_2))^3}
$$
$$
100 = \frac{7.8 X_2}{1 \times 10^{-3} \times 3.796 (1-X_2)^3}
$$
$$
100 = \frac{7.8 X_2}{.003796 (1-X_2)^3}
$$
$$
100 \times .003796 (1-X_2)^3 = 7.8 X_2
$$
$$
.3796 (1-X_2)^3 = 7.8 X_2
$$
$$
(1-X_2)^3 = \frac{7.8 X_2}{.3796} = 20.56 X_2
$$

Let’s try $ X_2 = .5 $:
$$
(1-.5)^3 = .125;\;\; 20.56 \times .5 = 10.28
$$

Try $ X_2 = .01 $:
$$
(1-.01)^3 = .970;\;\; 20.56 \times .01 = .2056
$$

Try $ X_2 = .02 $:
$$
(1-.02)^3 = .941;\;\; 20.56 \times .02 = .4112
$$

So, the equation is only satisfied for very small $ X_2 $, essentially close to zero. The CSTR second, after such a high conversion in the PFR, does almost nothing.

Thus, overall conversion = conversion after PFR, $ X_{overall} \approx .844 $.

---

### **Configuration (iii): Parallel**

Each reactor gets half the feed: $ v_ = 2.5\, L/min $, $ C_{A} = 10\, mol/L $, $ V = 100\, L $

For each reactor (let’s use CSTR formula for both):

$$
V = \frac{v_ C_{A} X}{k (C_{A}(1-X))^3}
$$
$$
100 = \frac{2.5 \times 10 \times X}{1 \times 10^{-3} \times (10(1-X))^3}
$$
$$
100 = \frac{25 X}{1 \times (1-X)^3}
$$
$$
100 (1-X)^3 = 25 X
$$
$$
4 (1-X)^3 = X
$$
Let $ y = 1-X $:

$$
4y^3 + y - 1 = 
$$

Try $ y = .55 $:

$$
4(.55^3) + .55 - 1 = 4(.166) + .55 - 1 = .664 + .55 - 1 = 1.214 - 1 = .214
$$

Try $ y = .5 $:

$$
4(.125) + .5 - 1 = .5 + .5 - 1 = 1 - 1 = 
$$

So, $ y = .5 \implies X = .5 $

Both reactors give $ X = .5 $

Total exit stream is the combination of both:

$$
F_{A,exit} = F_{A,1}(1-X_1) + F_{A,2}(1-X_2) = 2.5 \times 10 \times .5 + 2.5 \times 10 \times .5 = 12.5 + 12.5 = 25\, mol/min
$$
Initial: $ 5 \times 10 = 50\, mol/min $

Fraction unreacted: $ 25 / 50 = .5 \implies X_{overall} = .5 $

--

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