Let's solve each part step by step as requested. --- ## 1. **Governing Differential Equation (Spherical Geometry)** The steady-state mol balance over a spherical shell of thickness \( dr \) gives: \[ \text{Net rate of diffusion in} + \textRate of generation by reaction} = \] For species A: - Diffusion flux (Fick's law): \( -D_e \frac{dC_A}{dr} \) - Surface area at radius \( r \): \( 4\pi r^2 \) - Rate of reaction per unit volume: \( -k C_A \) So, net diffusive flux at \( r \) and \( r+dr \): \[ 4\pi r^2 \left(-D_e \frac{dC_A}{dr} \right)_{r} - 4\pi (r+dr)^2 \left(-D_e \frac{dC_A}{dr} \right)_{r+dr} + (4\pi r^2 dr) (-k C_A) = \] Divide by \( 4\pi dr \) and take the limit \( dr \to \): \[ \frac{1}{r^2} \frac{d}{dr}\left( r^2 \frac{dC_A}{dr} \right) - \frac{k}{D_e} C_A = \] --- ## 2. **Non-dimensionalizing & Thiele Modulus** Let \( R \) = pellet radius, \( C_{As} \) = surface concentration, \( \xi = \frac{r}{R} \), \( \phi^2 = \frac{kR^2}{D_e} \). Set \( C_A^* = \frac{C_A}{C_{As}} \): \[ \frac{1}{\xi^2} \frac{d}{d\xi} \left( \xi^2 \frac{dC_A^*}{d\xi} \right) - \phi^2 C_A^* = \] --- ## 3. **General Solution** This is a standard second-order ODE for spherical geometry with constant reaction: \[ \frac{1}{\xi^2} \frac{d}{d\xi} \left( \xi^2 \frac{dC_A^*}{d\xi} \right) - \phi^2 C_A^* = \] The solution (finite at \( \xi = \)) is: \[ C_A^*(\xi) = \frac{\sinh(\phi \xi)}{\xi \sinh\phi} \] Boundary conditions: - At \( \xi = 1 \): \( C_A^* = 1 \) - At \( \xi = \): finite concentration (as \( \frac{\sinh(\phi \xi)}{\xi} \to \phi \) as \( \xi \to \)) --- ## 4. **a) Sketch the Concentration Profile** For \( \phi = .1, 1., 10. \): - **Small \( \phi \):** Profile is nearly flat (diffusion fast vs. reaction). - **Large \( \phi \):** Steep drop near the surface (diffusion limited). You would sketch \( C_A^* \) vs. \( \xi \), showing the above behavior. --- ## 5. **b) Internal Effectiveness Factor \( \eta \)** Defined as the ratio of actual reaction rate to rate if entire pellet were at surface concentration: \[ \eta = \frac{\text{Actual rate}}{\text{Rate at } C_{As}} \] The actual rate (integrated over the volume): \[ \text{Actual rate} = \int_^R 4\pi r^2 k C_A(r) dr \] Rate at surface concentration: \[ \text{Rate at } C_{As} = \int_^R 4\pi r^2 k C_{As} dr = k C_{As} \frac{4}{3}\pi R^3 \] So, \[ \eta = \frac{3}{R^3} \int_^R r^2 \frac{C_A(r)}{C_{As}} dr \] Substitute \( \xi = \frac{r}{R} \): \[ \eta = 3 \int_^1 \xi^2 C_A^*(\xi) d\xi \] Plugging in the solution: \[ \eta = 3 \int_^1 \xi^2 \frac{\sinh(\phi \xi)}{\xi \sinh\phi} d\xi = \frac{3}{\phi^2} \left[ \phi \coth\phi - 1 \right] \] --- ## 6. **c) Large \( \phi \) Limit** For large \( \phi \), \( \coth\phi \to 1 \), so: \[ \eta \approx \frac{3}{\phi^2} [\phi - 1] \approx \frac{3}{\phi} \] Thus, for large \( \phi \), \( \eta \to \frac{3}{\phi} \). --- ## 7. **CSTR Competitive Reactions Yield** Consider \( A \rightarrow D \) (desired, order \( n_1 \)), \( A \rightarrow U \) (undesired, order \( n_2 \)). The yield is: \[ Y = \frac{\text{rate of D formation}}{\text{rate of A consumption}} = \frac{k_1 C_A^{n_1}}{k_1 C_A^{n_1} + k_2 C_A^{n_2}} \] To maximize \( Y \), if \( n_1 > n_2 \), you want to maximize \( C_A \), i.e., operate at low conversion (CSTR) or use a PFR. If \( n_1 > n_2 \), mathematically: \[ \frac{\partial Y}{\partial C_A} > \] So higher \( C_A \) gives higher yield of D. --- ## **Summary of Final Answers** 1. **Differential equation:** \[ \frac{1}{r^2} \frac{d}{dr}\left(r^2 \frac{dC_A}{dr}\right) - \frac{k}{D_e} C_A = \] 2. **Dimensionless solution:** \[ C_A^*(\xi) = \frac{\sinh(\phi \xi)}{\xi \sinh\phi} \] 3. **Effectiveness factor:** \[ \eta = \frac{3}{\phi^2} (\phi \coth\phi - 1) \] 4. **Large \( \phi \) limit:** \[ \eta \approx \frac{3}{\phi} \] 5. **Instantaneous yield for competitive reactions:** \[ Y = \frac{k_1 C_A^{n_1}}{k_1 C_A^{n_1} + k_2 C_A^{n_2}} \] 6. **For \( n_1 > n_2 \)**, a CSTR at low conversion or a PFR is favored to maximize the desired product yield. Let me know if you need any specific plot or further explanation for any step.