Let's solve the problem **step-by-step** as requested: --- # 1. Methane Combustion in Reactors ## **Given:** - **Reaction:** CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g) - **Rate law:** \( r = k \cdot C_{\text{CH}_4} \), \( k = 1\, \text{min}^{-1} \) - **Inlet composition:** 1 mol% CH₄ in air (20% O₂, 80% N₂) - **Total inlet volumetric flowrate:** 50 dm³/min - **Operation:** Isobaric, isothermal (600°C, 1 atm) --- ## (a) Plug Flow Reactor (PFR) ### (i) **Volumetric Flow Rate as a Function of Reactor Length** As the reaction proceeds, the number of moles of gas changes because: - 1 mol CH₄ + 2 mol O₂ → 1 mol CO₂ + 2 mol H₂O (all gases) - Net change per 1 mol CH₄ reacted = (1 + 2) - (1 + 2) = (no net change in moles). Because the number of moles doesn't change (Δn = ), there is **no change in volumetric flow rate** along the reactor length (assuming ideal gas behavior and constant T, P). Thus, the volumetric flow rate remains constant. This **simplifies the design equation** because you do not need to account for changing volumetric flow with conversion. ### (ii) **Derive the PFR Design Equation** For a first-order reaction: \[ \text{Mole balance:} \quad -r_A = \dfrac{dF_A}{dV} \] where \( r_A = -k C_{A} \), \( F_{A} = C_{A} \cdot v \), \( v \) is volumetric flowrate (constant from part i). Let \( X \) be the conversion of CH₄: \[ F_{A} = F_{A}(1 - X) \] \[ dF_{A} = -F_{A} dX \] \[ \frac{dF_{A}}{dV} = -F_{A} \frac{dX}{dV} \] So, \[ F_{A} \frac{dX}{dV} = k C_{A} \] But \( C_{A} = C_{A}(1-X) \): \[ F_{A} \frac{dX}{dV} = k C_{A} (1-X) \] \[ \frac{dX}{dV} = \frac{k C_{A}}{F_{A}} (1-X) \] But \( F_{A} = C_{A} v_ \): \[ \frac{dX}{dV} = \frac{k}{v_} (1-X) \] \[ dV = \frac{v_}{k} \frac{dX}{1-X} \] Integrate from to X: \[ V = \frac{v_}{k} \int_{}^{X} \frac{dX}{1-X} = -\frac{v_}{k} \ln(1-X) \] ### (iii) **Calculate the PFR Volume for 70% Conversion** Given: - \( X = .70 \) - \( v_ = 50\, \text{dm}^3/\text{min} \) - \( k = 1\, \text{min}^{-1} \) \[ V = -\frac{50}{1} \ln(1-.70) \] \[ V = -50 \ln(.3) \] \[ \ln(.3) \approx -1.204 \] \[ V = -50 \times (-1.204) = 60.2\, \text{dm}^3 \] ### (iv) **Sketch CH₄ Concentration at Reactor Exit as a Function of Radius** In a PFR, the concentration profile is **uniform across any cross-section** (assuming perfect radial mixing, plug flow in axial direction), so at the exit, the CH₄ concentration is **constant** across the radius. The concentration at the exit is: \[ C_{A,exit} = C_{A}(1-X) \] No variation with radius. --- ## (b) Continuous Stirred Tank Reactor (CSTR) ### (i) **Calculate CSTR Volume for 70% Conversion** For a CSTR: \[ F_{A} (X) = r_A V \] \[ F_{A} X = k C_{A} V \] But \( C_{A} = C_{A}(1-X) \), \( F_{A} = C_{A} v_ \): \[ C_{A} v_ X = k C_{A} (1-X) V \] \[ v_ X = k (1-X) V \] \[ V = \frac{v_ X}{k (1-X)} \] Plug in values (\( X = .70 \), \( v_ = 50 \), \( k = 1 \)): \[ V = \frac{50 \times .70}{1 \times .30} = \frac{35}{.3} = 116.7\, \text{dm}^3 \] ### (ii) **Why are the required volumes different?** The CSTR always has a lower average reactant concentration than a PFR for the same conversion, especially for first-order or higher-order reactions. This is because in a CSTR, the reactor contents are well mixed and the reaction occurs at the exit concentration throughout. The PFR, however, starts at a high concentration and the concentration drops along the length, making it more efficient (requiring a smaller volume) for the same conversion. ### (iii) **How Does the Answer to (ii) Change for Different Rate Laws?** (a) For \( r = k \) (zero order in CH₄): Both CSTR and PFR will require the **same volume** for a given conversion, because the rate does not depend on concentration, and thus the residence time is the same regardless of reactor type. (b) For \( r = k \cdot C_{\text{CH}_4}^{-2} \) (inverse squared dependence): The CSTR will require a **much larger volume** than the PFR, because the CSTR will always operate at the lowest concentration (exit concentration), and the reaction rate decreases sharply as concentration decreases, making the CSTR very inefficient compared to the PFR. --- ## **Final Answers:** PFR volume for 70% conversion: **60.2 dm³** CSTR volume for 70% conversion: **116.7 dm³** Volumes are different due to concentration profiles inside the reactors; the difference becomes more pronounced as the reaction becomes more sensitive to concentration. For a zero-order reaction, volumes are equal. For negative-order reactions, the CSTR becomes much less efficient than the PFR.