Let's break down and solve **Problems 4 and 5 step-by-step**. --- # **Problem 4 – Heterogeneous Packed-Bed Reactor (Effectiveness Factor)** ## **Case I: Zero-Order Reaction in Spherical Pellet** **Governing equation (steady-state, zero-order reaction):** \[ D_e \left( \frac{1}{r^2} \frac{d}{dr} \left( r^2 \frac{dC_A}{dr} \right) \right) = k_ \] where: - \( D_e \) = effective diffusivity, - \( k_ \) = zero-order rate constant, - \( C_A \) = reactant concentration, - \( r \) = radial coordinate. **Boundary conditions:** - At \( r = \): \( \frac{dC_A}{dr} = \) (symmetry), - At \( r = R \): \( C_A = C_{A,s} \) (surface concentration). ### **Step 1: Integrate the ODE** Integrate once: \[ \frac{d}{dr} \left( r^2 \frac{dC_A}{dr} \right) = \frac{k_}{D_e} r^2 \] Integrate: \[ r^2 \frac{dC_A}{dr} = \frac{k_}{3D_e} r^3 + C_1 \] At \( r = \), symmetry gives \( C_1 = \): \[ \frac{dC_A}{dr} = \frac{k_}{3D_e} r \] Integrate again: \[ C_A(r) = \frac{k_}{6D_e} r^2 + C_2 \] At \( r = R \), \( C_A(R) = C_{A,s} \): \[ C_{A,s} = \frac{k_}{6D_e} R^2 + C_2 \implies C_2 = C_{A,s} - \frac{k_}{6D_e} R^2 \] So, \[ C_A(r) = C_{A,s} + \frac{k_}{6D_e}(r^2 - R^2) \] ### **Step 2: Find \( r_c \) where \( C_A(r_c) = \)** Set \( C_A(r_c) = \): \[ = C_{A,s} + \frac{k_}{6D_e}(r_c^2 - R^2) \] \[ \frac{k_}{6D_e}(r_c^2 - R^2) = -C_{A,s} \] \[ r_c^2 = R^2 - \frac{6D_e C_{A,s}}{k_} \] \[ r_c = R \sqrt{1 - \frac{6D_e C_{A,s}}{k_ R^2}} \] **This matches the formula given.** --- ## **Case II: First-Order Reaction in Spherical Pellet** **Governing equation (steady-state, first-order):** \[ D_e \left( \frac{1}{r^2} \frac{d}{dr} \left( r^2 \frac{dC_A}{dr} \right) \right) = k_1 C_A \] **Dimensionless Thiele modulus:** \[ \phi_1 = R \sqrt{\frac{k_1}{D_e}} \] **Effectiveness factor (\( \eta \)), for rapid external mass transfer:** \[ \eta = \frac{\text{overall reaction rate}}{\text{rate if no diffusion limitation}} \] For first-order, the solution is: \[ \eta = \frac{3}{\phi_1^2} \left( \phi_1 \coth \phi_1 - 1 \right) \] Or, \[ \eta = \frac{3}{\phi_1} \left( \frac{1}{\tanh \phi_1} - \frac{1}{\phi_1} \right) \] **This matches the formula given.** --- # **Problem 5 – Adiabatic Batch Reactor (Competing Reactions)** Given: - **Adiabatic** (no heat loss), - **Parallel reactions**: - Desired: \( A \to B \) (rate \( k_1 \)), activation energy \( E_1 \) - Undesired: \( A \to D \) (rate \( k_2 \)), activation energy \( E_2 \) - **Maximize yield of \( B \) before onset of \( D \)** ## **a) Compute the optimal temperature for maximum instantaneous yield of \( B \):** **Yield of B per A consumed:** \[ \text{Yield} = \frac{\text{rate of }A \to B}{\text{total rate of }A \text{ consumption}} = \frac{k_1}{k_1 + k_2} \] Both \( k_1 \) and \( k_2 \) are Arrhenius: \[ k_1 = A_1 e^{-E_1/RT}, \quad k_2 = A_2 e^{-E_2/RT} \] Yield: \[ Y(T) = \frac{A_1 e^{-E_1/RT}}{A_1 e^{-E_1/RT} + A_2 e^{-E_2/RT}} \] **To maximize \( Y \) with respect to \( T \):** Set \( \frac{dY}{dT} = \): Let \( x = 1/RT \), so \( Y(x) = \frac{A_1 e^{-E_1 x}}{A_1 e^{-E_1 x} + A_2 e^{-E_2 x}} \) But a more direct approach: Usually, yield is maximized at a temperature where the selectivity is maximized: \[ \frac{d}{dT} \left( \frac{k_1}{k_1 + k_2} \right) = \] This gives the following transcendental equation for \( T_{opt} \): \[ \frac{E_2 - E_1}{RT_{opt}^2} = \ln \left( \frac{A_2 E_2}{A_1 E_1} \right) \cdot \frac{1}{T_{opt}} \] But more commonly, the optimal temperature satisfies: \[ \frac{E_2 - E_1}{RT_{opt}^2} = \ln\left( \frac{A_2}{A_1} \right) + \frac{E_2 - E_1}{RT_{opt}} \] This equation must typically be solved numerically for \( T_{opt} \). ## **b) Calculate the residence time at \( T_{opt} \) to achieve maximum yield** For a batch reactor: - Rate equations: \( -\frac{dC_A}{dt} = (k_1 + k_2)C_A \) - Integrate: \( C_A = C_{A,} e^{-(k_1 + k_2)t} \) - Product B: \( \frac{dC_B}{dt} = k_1 C_A \) Integrate for yield: \[ C_B = C_{A,} \frac{k_1}{k_1 + k_2} \left( 1 - e^{-(k_1 + k_2)t} \right) \] So, to reach a desired conversion or yield, solve for \( t \): \[ \text{If you want to maximize } C_B, \text{ set } \frac{dC_B}{dt} = \] But for maximum **instantaneous** yield, it's at the beginning. For maximum **absolute** yield, you solve: \[ \frac{dC_B}{dt} = k_1 C_A - k_2 C_B = \] But with only parallel reactions (no B to D), maximize \( C_B \) with respect to time: \[ \frac{dC_B}{dt} = k_1 C_A \] \( C_B \) increases monotonically, but as \( C_A \) depletes, the rate slows. Optimal time is typically where: \[ \text{Yield} = \frac{C_B}{C_{A,}} = \frac{k_1}{k_1 + k_2} \left( 1 - e^{-(k_1 + k_2)t} \right) \] Set \( t \to \infty \) for absolute maximum, but practically, stop before significant \( D \) forms (if D can form from B, which is not the case here). --- ## **Summary of Final Answers** ### **4. Packed-Bed Reactor:** - **Zero-order:** \( r_c = R \sqrt{1 - \frac{6D_e C_{A,s}}{k_ R^2}} \) - **First-order:** \( \eta = \frac{3}{\phi_1} \left( \frac{1}{\tanh \phi_1} - \frac{1}{\phi_1} \right) \), \( \phi_1 = R \sqrt{\frac{k_1}{D_e}} \) ### **5. Batch Reactor:** a) **Optimal temperature:** \[ T_{opt} \text{ solves: } \frac{E_2 - E_1}{RT_{opt}^2} = \ln \left( \frac{A_2}{A_1} \right) + \frac{E_2 - E_1}{RT_{opt}} \] (Solve numerically.) b) **Residence time for desired yield:** \[ \text{Yield} = \frac{k_1}{k_1 + k_2} \left( 1 - e^{-(k_1 + k_2)t} \right) \] Solve for \( t \) to achieve the desired yield at \( T_{opt} \). --- **If you need numerical examples or further derivations, let me know!**