Let's break down the problem step-by-step as per the instructions and the provided data. --- # **Given Data** **Reaction:** A → B (1st order, exothermic) **Rate:** \[-r_A = k C_A,\] \[k = k_ \exp\left(-\frac{E}{RT}\right)\] **Constants:** - \(k_ = 6 \times 10^5 \ \text{s}^{-1}\) - \(E = 108 \ \text{kJ/mol} = 108,000 \ \text{J/mol}\) - \(\Delta H = -115 \ \text{kJ/mol}\) (exothermic) - Reactor length \(L = 15 \ \text{m}\) - Reactor diameter \(D = 1.2 \ \text{m}\) - Bed void fraction \(\epsilon = .40\) - Catalyst pellet radius \(R_p = 3 \ \text{mm} = .003 \ \text{m}\) - Effective diffusivity, \(D_e = 6.5 \times 10^{-10} \ \text{m}^2/\text{s}\) - Pellet thermal conductivity \(k_s = 1.4 \ \text{W}/\text{m}\cdot\text{K}\) ### **Feed Conditions** - Inlet concentration \(C_{A} = 3.2 \ \text{kmol}/\text{m}^3\) - Inlet temperature \(T_ = 580 \ \text{K}\) - Superficial velocity \(u = .55 \ \text{m/s}\) ### **Heat Transfer** - Overall heat transfer coefficient \(U = 160 \ \text{W}/\text{m}^2\cdot\text{K}\) - Coolant temperature \(T_c = 515 \ \text{K}\) --- ## **Step 1: Derive the Interparticle Mass/Energy Balance Equation** ### **a. Mass Balance (Steady-State, 1st-order, Isothermal)** For a first-order reaction inside a spherical catalyst pellet (radius \(R_p\)): \[ \frac{1}{r^2} \frac{d}{dr} \left( r^2 D_e \frac{dC_A}{dr} \right) = k C_A \] ### **b. Energy Balance (With Heat Generation)** \[ \frac{1}{r^2} \frac{d}{dr} \left( r^2 k_s \frac{dT}{dr} \right) + (-\Delta H) k C_A = \] where \(k_s\) is the thermal conductivity of the solid, and \(-\Delta H\) is the heat generated per unit mole of \(A\) reacted. --- ## **Step 2: Define Thiele Modulus (Including Temperature Dependence)** Thiele modulus (\(\phi\)) relates diffusion and reaction rates: \[ \phi = R_p \sqrt{\frac{k}{D_e}} \] Since \(k\) is temperature-dependent: \[ k = k_ e^{-E/RT} \] So, \[ \phi = R_p \sqrt{\frac{k_ e^{-E/RT}}{D_e}} \] --- ## **Step 3: Determine Effectiveness Factor with Heat Effect** The effectiveness factor (\(\eta\)) for a first-order reaction in a sphere: \[ \eta = \frac{3}{\phi^2} \left[ \phi \coth(\phi) - 1 \right] \] **With heat effect:** The temperature inside the pellet can be higher than the surface due to exothermicity. The modified effectiveness factor can be estimated using the **Damköhler number** and the energy balance, but for now, we use the above formula with the local \(T\). --- **Step 4: Calculate Maximum Catalyst Pellet Temperature** ### **a. Surface Temperature (\(T_s\))** heat generated in the pellet must be transferred to the surroundings: \[ U a (T_s - T_c) = (-\Delta H) r_A V_p \] Where: - \(U\): overall heat transfer coefficient - \(a\): external surface area per unit volume \(= \frac{3}{R_p}\) - \(T_s\): surface temperature of pellet - \(T_c\): coolant temperature - \(r_A\): reaction rate at surface - \(V_p\): pellet volume But to find **maximum temperature (center of pellet)**, consider the pellet energy balance: \[ T_{center} - T_s = \frac{(-\Delta H) r_A R_p^2}{6 k_s} \] ### **b. Calculate \(r_A\) at Surface** First, estimate \(k\) at \(T_s\) (initially assume \(T_s = T_\)), then iterate. \[ k(T_s) = k_ e^{-E/(RT_s)} \] \[ r_A = k(T_s) C_{A,s} \] ### **c. Calculate \(\Delta T_{max}\) (center to surface)** \[ \Delta T_{max} = \frac{(-\Delta H) r_A R_p^2}{6 k_s} \] --- ## **Step 5: Determine Whether Internal or External Transport Controls** **Criteria:** - **Internal diffusion controlled:** \(\eta \ll 1\) (Thiele modulus \(\phi \gg 1\)) - **External mass transfer controlled:** External mass transfer coefficient is limiting - **Reaction controlled:** \(\eta \approx 1\) (\(\phi \ll 1\)) Compare characteristic resistances (diffusion, reaction, heat transfer). --- # **Solution Steps with Calculations** ## **Step 2: Calculate Thiele Modulus (\(\phi\))** \[ k(T_) = 6 \times 10^5 \exp\left( -\frac{108,000}{8.314 \times 580} \right) \] First, calculate the exponent: \[ \frac{108,000}{8.314 \times 580} = \frac{108,000}{4822.12} \approx 22.41 \] So, \[ k(T_) = 6 \times 10^5 \exp(-22.41) \approx 6 \times 10^5 \times 1.82 \times 10^{-10} \approx 1.09 \times 10^{-4} \ \text{s}^{-1} \] Now, \[ \phi = .003 \sqrt{\frac{1.09 \times 10^{-4}}{6.5 \times 10^{-10}}} \] \[ \frac{1.09 \times 10^{-4}}{6.5 \times 10^{-10}} \approx 1.677 \times 10^5 \] \[ \sqrt{1.677 \times 10^5} \approx 409.7 \] \[ \phi = .003 \times 409.7 \approx 1.23 \] --- ## **Step 3: Effectiveness Factor (\(\eta\))** \[ \eta = \frac{3}{\phi^2} [\phi \coth(\phi) - 1] \] First, compute \(\coth(\phi)\): For \(\phi = 1.23\): \[ \coth(1.23) = \frac{\cosh(1.23)}{\sinh(1.23)} \approx \frac{1.86}{1.56} \approx 1.19 \] \[ \phi \coth(\phi) = 1.23 \times 1.19 \approx 1.46 \] \[ \eta = \frac{3}{1.23^2} (1.46 - 1) = \frac{3}{1.51} \times .46 \approx 1.99 \times .46 \approx .92 \] --- ## **Step 4: Maximum Catalyst Pellet Temperature** ### **a. Estimate \(r_A\) at surface:** \[ r_A = k(T_) C_{A} = 1.09 \times 10^{-4} \times 3.2 \times 10^3 = .349 \ \text{mol}/(\text{m}^3 \cdot \text{s}) \] ### **b. \(\Delta T_{max}\):** \[ \Delta T_{max} = \frac{(-\Delta H) r_A R_p^2}{6 k_s} \] Use \(|\Delta H| = 115,000 \ \text{J/mol}\): \[ \Delta T_{max} = \frac{115,000 \times .349 \times (.003)^2}{6 \times 1.4} \] \[ = \frac{115,000 \times .349 \times 9 \times 10^{-6}}{8.4} \] \[ = \frac{115,000 \times 3.141 \times 10^{-6}}{8.4} \] \[ = \frac{361.215}{8.4} \times 10^{-3} \] \[ = 43. \times 10^{-3} \] \[ = .043 \ \text{K} \] So, the temperature rise inside the pellet is tiny (about .04 K). --- ## **Step 5: Identify Rate-Limiting Step** - Because \(\eta \approx .92\), the reaction is **not strongly diffusion-limited** (\(\eta\) close to 1). - The temperature rise is very small, so **no significant heat transport limitations**. - **Reaction is surface-reaction controlled** (kinetic control). --- # **Summary Table** | Step | Result | |------|--------| | 1 | Mass: \(\frac{1}{r^2} \frac{d}{dr} (r^2 D_e \frac{dC_A}{dr}) = k C_A\); Energy: \(\frac{1}{r^2} \frac{d}{dr} (r^2 k_s \frac{dT}{dr}) + (-\Delta H) k C_A = \) | | 2 | \(\phi = 1.23\) | | 3 | \(\eta = .92\) | | 4 | \(\Delta T_{max} \approx .04\, \text{K}\) | | 5 | **Reaction is surface-reaction (kinetically) controlled**. | --- ## **Final Answers** 1. **Derived balances:** - Mass: \(\frac{1}{r^2} \frac{d}{dr} (r^2 D_e \frac{dC_A}{dr}) = k C_A\) - Energy: \(\frac{1}{r^2} \frac{d}{dr} (r^2 k_s \frac{dT}{dr}) + (-\Delta H) k C_A = \) 2. **Thiele modulus (temp-dependent):** \(\phi = R_p \sqrt{\frac{k_ e^{-E/RT}}{D_e}}\). For \(T_ = 580\,K\), \(\phi = 1.23\). 3. **Effectiveness factor (with heat):** \(\eta = .92\) (no significant heat effect due to low \(\Delta T_{max}\)). 4. **Maximum catalyst pellet temperature:** \(\approx .04\,K\) above surface value. 5. **Controlling step:** **Reaction is kinetically (surface-reaction) controlled**. Neither internal diffusion nor heat transport are limiting.