Give step-by-step solution with explanation and final answer:
{give me the answer without using bullet-points}An adiabatic packed bed reactor (PBR) is used for the complex network A — B (exothermic),

Question

Give step-by-step solution with explanation and final answer:
{give me the answer without using bullet-points}An adiabatic packed bed reactor (PBR) is used for the complex network A — B (exothermic), A — C
(endothermic), and B — D (exothermic). The objective is to maximize B production and manage the
temperature profile to prevent runaway and minimize C formation.
Fao, To Governing Equations & Data
dF; _ ;

yall INSULATION Jw = vrs fori=A4,B,C,D

Cen =

an dT —X5 riAHp

pore b) dT _ —2.j=1784R;
ADIABATIC 7 A425 ¢ (Endo) V aw YP, FChi

k:

7 BD (Exo) 7 71 =k1Ca, 12 = koCy, 73 = k3Cp with k; = kjo exp(—E;/RT)

VU dessssssassss] % f=) i

 Kesesisticsies Z| Fo = 100 mol/s, Ty = 450 K

7 SESECEEOEEE I~ CATALYST BED (W)

AH AHpgy = —60, AHgy = +40, AHgg = —80 kJ/mol

Ey = 80, Ey = 120, E3 = 60 kJ/mol
Fa, Fy, Fo, Fp, T Cpa =Cpp=Cpc=Cpp = 150 J/(mol K)

1.1 Derive the specific differential mole balance equations for Fg and Fy in terms of the rate laws and flow rates.
1.2 Calculate the initial temperature gradient (dT'/dW at W = 0) given the inlet conditions and parameter values.
1.3 Sketch the qualitative concentration profiles (Fy, Fp, Fr, Fp vs. W) along

VIPSolutions · Accepted Answer

### 1.1 Derive the Specific Differential Mole Balances

The general mole balance for component $ i $ in a packed bed reactor is:
$$
\frac{dF_i}{dW} = \sum_{j=1}^3 \nu_{ij} r_j
$$
Here, $ \nu_{ij} $ is the stoichiometric coefficient of species $ i $ in reaction $ j $, and $ r_j $ is the rate of reaction $ j $.

**Reactions:**
- $ A \xrightarrow{k_1} B $ (exothermic)
- $ A \xrightarrow{k_2} C $ (endothermic)
- $ B \xrightarrow{k_3} D $ (exothermic)

**Stoichiometry Table:**

| Species | Reaction 1 ($A \to B$) | Reaction 2 ($A \to C$) | Reaction 3 ($B \to D$) |
|---------|:------------------------:|:------------------------:|:------------------------:|
| A       |     -1                   |     -1                   |                         |
| B       |     +1                   |                         |     -1                   |
| C       |                         |     +1                   |                         |
| D       |                         |                         |     +1                   |

So, the balances are:
$$
\frac{dF_A}{dW} = -r_1 - r_2
$$
$$
\frac{dF_B}{dW} = +r_1 - r_3
$$
$$
\frac{dF_C}{dW} = +r_2
$$
$$
\frac{dF_D}{dW} = +r_3
$$

Where:
- $ r_1 = k_1 C_A $
- $ r_2 = k_2 C_A $
- $ r_3 = k_3 C_B $

### 1.2 Calculate the Initial Temperature Gradient $(dT/dW)$ at the Inlet

The energy balance for an adiabatic PBR is:
$$
\frac{dT}{dW} = \frac{-\sum_{j=1}^{3} r_j \Delta H_{Rj}}{\sum_{i=A}^{D} F_i C_{P,i}}
$$

At the inlet ($ W =  $):
- $ F_{A,} = 100 $ mol/s; $ F_{B,} = F_{C,} = F_{D,} =  $
- $ T_ = 450\,\text{K} $
- $ C_{P,i} = 150\,\text{J/(mol·K)} $ for all species

**Calculate $ k_j $ values at $ T_ $:**
$$
k_j = k_{j} \exp\left(-\frac{E_j}{RT_}\right)
$$
Assume $ k_{10} = k_{20} = k_{30} = 1\,\text{mol/(kg·s)} $ (unless otherwise given).

For $ R = 8.314\,\text{J/(mol·K)} $, $ T_ = 450\,\text{K} $:

$$
k_1 = 1 \exp\left(-\frac{80,000}{8.314 \times 450}\right) \approx 1 \exp(-21.33) \approx 5.3 \times 10^{-10}
$$
$$
k_2 = 1 \exp\left(-\frac{120,000}{8.314 \times 450}\right) \approx 1 \exp(-32.04) \approx 1. \times 10^{-14}
$$
$$
k_3 = 1 \exp\left(-\frac{60,000}{8.314 \times 450}\right) \approx 1 \exp(-16.) \approx 1.1 \times 10^{-7}
$$

At $ W= $, $ C_A = \frac{F_{A,}}{v_} $, but since $ v_ $ (volumetric flow rate) is not given, leave in terms of $ C_A $:

$$
r_1 = k_1 C_{A,}
$$
$$
r_2 = k_2 C_{A,}
$$
$$
r_3 = k_3 C_{B,} = 
$$

**Plug into energy balance:**
$$
\frac{dT}{dW} \bigg|_{W=} = \frac{ - [r_1 \Delta H_{R1} + r_2 \Delta H_{R2} + r_3 \Delta H_{R3}] }{F_{A,}C_{P,A}}
$$
$$
= \frac{ - [k_1 C_{A,} (-60) + k_2 C_{A,} (40) + ] }{100 \times 150}
$$
$$
= \frac{ - C_{A,} [k_1 (-60) + k_2 (40)] }{15,000}
$$
Insert the values for $ k_1 $ and $ k_2 $ and $ C_{A,} $.

### 1.3 Qualitative Concentration Profiles

- $ F_A $ decreases monotonically as it is consumed in both reactions.
- $ F_B $ rises initially (formed from A), reaches a maximum, then falls (as it is consumed to form D).
- $ F_C $ rises as it is formed from A.
- $ F_D $ rises as it is formed from B.

### 1.4 Qualitative Temperature Profile

- The temperature initially rises due to the exothermic reaction $ A \rightarrow B $.
- As $ A \rightarrow C $ (endothermic) occurs, this can slow or reverse the temperature rise.
- The temperature may reach a maximum (hot spot), then decrease or level off as reactants are consumed and endothermic reaction dominates.

**In summary:**  
The specific mole balances for $ F_B $ and $ F_C $ are $ \frac{dF_B}{dW} = +r_1 - r_3 $ and $ \frac{dF_C}{dW} = +r_2 $, respectively, with rates as above.  
The initial temperature gradient is $ \frac{dT}{dW}\big|_{W=} = \frac{ - C_{A,} [k_1 (-60) + k_2 (40)] }{15,000} $ at the inlet.  
The qualitative profiles are: $ F_A $ decreases, $ F_B $ increases then decreases, $ F_C $ increases, $ F_D $ increases; temperature rises, peaks, then may decline or plateau.

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