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Question

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VIPSolutions · Accepted Answer

Let's break down and solve the problem step by step:

---

## Given

- **Reaction:** 
  $$
  \text{o-cresol} + 2 \text{H}_2 \rightarrow \text{2-methylcyclohexanol}
  $$
- **Catalyst:** Nickel-silica
- **Reaction conditions:** 170°C, 1.74 mol cresol/(kg cat min)
- **Feed gas composition:** 67% H₂, 33% o-cresol (molar basis)
- **Total molar flow rate:** 40 mol/min
- **Packed bed reactor total pressure:** 5 atm
- **Packing void fraction:** Not given (assume standard if needed)
- **Rate constant (for part b):** $ a = .027 \, \text{kg}^{-1} $
- **Desired conversion:** 5%

---

## (a) **Plot Rate and Concentrations as a Function of Catalyst Weight**

### 1. **Write the Rate Law**
Given: zero-order in cresol, first-order in H₂
$$
r = k C_{H_2}
$$
But the specific reaction rate is given as:  
$$
\text{Specific rate} = 1.74 \frac{\text{mol cresol}}{\text{kg cat} \cdot \text{min}}
$$
(Assume this is under standard feed conditions.)

### 2. **Mole Balances for a PFR**
For a packed bed reactor:
$$
\frac{dF_A}{dW} = -r_A
$$
where $ F_A $ is the molar flow of A (o-cresol), $ W $ is the catalyst weight.

- **Initial o-cresol feed:** $ F_{A} = .33 \times 40 = 13.2 $ mol/min
- **$ X $ = conversion of o-cresol:** $ X = \frac{F_{A} - F_A}{F_{A}} $

$$
\frac{dX}{dW} = \frac{-r_A}{F_{A}}
$$

### 3. **Express $ r_A $ in Terms of $ X $**
- **Hydrogen:** $ F_{H_2,} = .67 \times 40 = 26.8 $ mol/min
- For 5% conversion: $ X = .05 $, so $ F_A = F_{A}(1 - X) $
- For every 1 mol o-cresol, 2 mol H₂ are consumed.
- At conversion $ X $:

$$
F_{H_2} = F_{H_2,} - 2F_{A}X
$$

- **Total molar flow at $ X $:**
$$
F_{tot} = F_{A}(1 - X) + F_{H_2,} - 2F_{A}X + F_P
$$
where $ F_P $ is the product flow.

But for small conversion, $ F_{tot} \approx 40 $ mol/min.

- **Partial Pressure of H₂ at conversion $ X $:**

$$
y_{H_2} = \frac{F_{H_2}}{F_{tot}}
$$
$$
P_{H_2} = y_{H_2} P_{tot}
$$
$$
C_{H_2} = \frac{P_{H_2}}{RT}
$$

But since the rate is given as a specific rate, we can use it to directly relate to W.

### 4. **Calculate Required Catalyst for 5% Conversion**

From material balance:
$$
F_{A} X = r \cdot W
$$
$$
W = \frac{F_{A} X}{r}
$$

Given $ r = 1.74 $ mol/(kg·min), $ F_{A} = 13.2 $ mol/min, $ X = .05 $:

$$
W = \frac{13.2 \times .05}{1.74} = \frac{.66}{1.74} = .379 \text{ kg}
$$

**Ratio of catalyst weight needed to achieve 5% conversion:**
- **Required catalyst: .379 kg**

#### **Plotting (conceptually):**
- As catalyst weight increases, conversion increases linearly at low conversion (since zero-order in o-cresol).
- $ X = \frac{r}{F_{A}} W $
- H₂ will decrease slightly, but for small X, $ C_{H_2} $ is nearly constant.

---

## (b) **Including Pressure Drop: Analytical Approach**

**Given:**
- $ a = .027 \, \text{kg}^{-1} $ (Ergun parameter, not rate constant!)
- Use Ergun equation for pressure drop.

### 1. **Ergun Equation (Simplified):**
For pressure drop in packed bed:
$$
\frac{dP}{dW} = -a P
$$
where $ a = .027 \, \text{kg}^{-1} $.

### 2. **Integrate to Find Pressure Profile:**
$$
\frac{dP}{P} = -a dW
$$
$$
\ln\left(\frac{P}{P_}\right) = -a W
$$
$$
P = P_ e^{-aW}
$$

### 3. **Effect on Rate:**
- As $ P $ drops, $ C_{H_2} $ drops (since $ C_{H_2} = y_{H_2} P/(RT) $)
- Thus, rate drops along the bed.

### 4. **Material Balance Including Pressure Drop:**
$$
F_{A} dX = r dW
$$
But now $ r = k C_{H_2} $, and $ C_{H_2} \propto P $:

$$
C_{H_2}(W) = C_{H_2,} e^{-aW}
$$
$$
r(W) = r_ e^{-aW}
$$

$$
F_{A} dX = r_ e^{-aW} dW
$$
Integrate from $ X= $ to $ X=.05 $, $ W= $ to $ W=W_f $:

$$
F_{A} \int_^{.05} dX = r_ \int_^{W_f} e^{-aW} dW
$$
$$
F_{A} (.05) = \frac{r_}{-a} (e^{-aW_f} - 1)
$$
$$
F_{A} (.05) = \frac{r_}{a} (1 - e^{-aW_f})
$$
$$
\frac{F_{A} (.05)a}{r_} = 1 - e^{-aW_f}
$$
$$
e^{-aW_f} = 1 - \frac{F_{A} (.05)a}{r_}
$$
$$
-aW_f = \ln\left(1 - \frac{F_{A} (.05)a}{r_}\right)
$$
$$
W_f = -\frac{1}{a} \ln\left(1 - \frac{F_{A} (.05)a}{r_}\right)
$$

#### Plug in values:

- $ F_{A} = 13.2 $ mol/min
- $ a = .027 $ kg⁻¹
- $ r_ = 1.74 $ mol/(kg·min)
- $ X = .05 $

$$
\frac{F_{A} (.05) a}{r_} = \frac{13.2 \times .05 \times .027}{1.74}
= \frac{.01782}{1.74}
= .01024
$$

So,

$$
W_f = -\frac{1}{.027} \ln(1 - .01024)
= -37.04 \ln(.98976)
= -37.04 \times (-.01029)
= .381 \text{ kg}
$$

**Comparing to (a):**  
- Without pressure drop: $ W = .379 $ kg  
- With pressure drop: $ W_f = .381 $ kg

**Conclusion:** Pressure drop has a very small effect at low conversion and short beds.

---

## (c) **Increase Bed Diameter vs. Increase Catalyst**

**Idea:**  
Increasing bed diameter reduces pressure drop more effectively than simply increasing catalyst weight (since pressure drop is inversely proportional to the square of the diameter in the Ergun equation).

**Mathematical Justification:**

- **Ergun equation:**
  $$
  \Delta P \propto \frac{L}{d_p^2} \left(\frac{1}{D^2}\right)
  $$
  where $ D $ is reactor diameter.

- **If you double the bed diameter, pressure drop decreases by f

VIPSolutions · Answer

# Step-by-Step Solution

## Given Information

- **Reaction:** 
  $$
  \text{o-cresol} + 2 \text{H}_2 \rightarrow \text{2-methylcyclohexanol}
  $$
- **Catalyst:** Nickel-silica
- **Conditions:** 170°C, 1.74 mol cresol/(kg cat min)
- **Feed Composition:** 67% H₂, 33% o-cresol
- **Total Flow Rate:** 40 mol/min
- **Pressure:** 5 atm
- **Rate Constant:** $ a = 0.027 \, \text{kg}^{-1} $
- **Desired Conversion:** 5%

---

## (a) Calculate Required Catalyst for 5% Conversion

1. **Initial Molar Flow Rates:**
   - o-Cresol: 
     $$
     F_A = 0.33 \times 40 = 13.2 \, \text{mol/min}
     $$
   - H₂: 
     $$
     F_{H_2} = 0.67 \times 40 = 26.8 \, \text{mol/min}
     $$

2. **Calculate Required Catalyst (W):**
   - Conversion $ X = 0.05 $:
   $$
   W = \frac{F_A \times X}{r} = \frac{13.2 \times 0.05}{1.74} = 0.379 \text{ kg}
   $$

---

## (b) Effect of Pressure Drop on Catalyst Weight

1. **Using Ergun Equation:**
   - Pressure drop affects concentration and thus reaction rate.
   - Integrate to find:
   $$
   W_f = -\frac{1}{a} \ln\left(1 - \frac{F_A \times X \times a}{r}\right)
   $$

2. **Calculate $ W_f $:**
   $$
   W_f = -\frac{1}{0.027} \ln(1 - 0.01024) \approx 0.381 \text{ kg}
   $$

---

## (c) Diameter vs. Catalyst Weight

- **Increasing Diameter:**
  - Reduces pressure drop significantly.
  - Pressure drop decreases with the square of the diameter: 
  $$
  \Delta P \propto \frac{1}{D^2}
  $$
  
- **Conclusion:** 
  - Increasing bed diameter is preferred over just increasing catalyst weight for maintaining conversion with lower pressure drop.

---

## **Final Summary**

- **Catalyst Weight for 5% Conversion:** $ 0.379 \, \text{kg} $ (without pressure drop)
- **Catalyst Weight with Pressure Drop:** $ 0.381 \, \text{kg} $
- **Recommendation:** Increase bed diameter for better efficiency.

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