Certainly! Let’s solve the problem step by step using the **Shell Balance approach**. --- ## **Step 1: Write the Shell Balance and Differential Equation (Part a)** ### **Shell Balance on Species A in a Thin Slab** - **Control volume:** A thin slab of thickness \( \Delta z \) at position \( z \). - **At steady state:** Rate in by diffusion = Rate out by diffusion (no accumulation, no reaction). Let \( N_{Az} \) be the flux of A (mol/(area·time)), and \( C_A \) be the concentration of A at position \( z \). #### **Shell Balance:** \[ \text{In by diffusion at } z: N_{Az}|_z \\ \text{Out by diffusion at } z + \Delta z: N_{Az}|_{z+\Delta z} \] \[ \text{Net accumulation in shell: } N_{Az}|_z - N_{Az}|_{z+\Delta z} = \] \[ \Rightarrow N_{Az}|_z = N_{Az}|_{z+\Delta z} \] #### **Fick’s Law:** \[ N_{Az} = -D_{AB} \frac{dC_A}{dz} \] #### **Plug Into the Balance:** \[ \frac{dN_{Az}}{dz} = \implies \frac{d}{dz} \left(-D_{AB} \frac{dC_A}{dz}\right) = \] If \( D_{AB} \) is constant, \[ \frac{d^2 C_A}{dz^2} = \] **Explanation:** The shell balance shows that, at steady state and with no reaction, the flux of A through any cross-section is constant. Using Fick's law, this gives a second-order differential equation for \( C_A \). --- ## **Step 2: Solve for Concentration Profile and Rate (Parts b & c)** ### **(b) Linear Profile** From step 1, \[ \frac{d^2 C_A}{dz^2} = \] Integrate once: \[ \frac{dC_A}{dz} = \text{constant} = m \] Integrate again: \[ C_A(z) = mz + b \] **Boundary Conditions:** - At \( z = \): \( C_A = C_{A} \) - At \( z = \delta \): \( C_A = C_{A\delta} \) Plug in: \[ C_A() = b = C_{A} \] \[ C_A(\delta) = m\delta + b = C_{A\delta} \] \[ m = \frac{C_{A\delta} - C_{A}}{\delta} \] So, \[ C_A(z) = C_{A} + \frac{C_{A\delta} - C_{A}}{\delta} z \] **Explanation:** The absence of reaction and steady state means the concentration profile is linear between the pill and the main stream. --- ### **(c) Rate of Diffusion** From Fick’s law: \[ N_{Az} = -D_{AB} \frac{dC_A}{dz} \] From above, \( \frac{dC_A}{dz} = \frac{C_{A\delta} - C_{A}}{\delta} \): \[ N_{Az} = -D_{AB} \left(\frac{C_{A\delta} - C_{A}}{\delta}\right) = \frac{D_{AB}(C_{A} - C_{A\delta})}{\delta} \] --- ## **Final Answer:** \[ \boxed{ N_{Az} = \frac{D_{AB}(C_{A} - C_{A\delta})}{\delta} } \] --- ### **Summary of Steps:** - **Step 1:** Shell balance and Fick's law yield \( \frac{d^2 C_A}{dz^2} = \). - **Step 2:** The solution is a linear concentration profile, and the rate is given by the above equation.

Certainly! Let's solve the problem step-by-step using the **Shell Balance approach**. --- ## **Step 1: Write the Shell Balance and Differential Equation (Part a)** ### **Shell Balance on Species A in a Thin Slab** 1. **Control Volume:** Consider a thin slab of thickness \( \Delta z \) at position \( z \). 2. **Steady State:** Rate in by diffusion = Rate out by diffusion (no accumulation, no reaction). Let \( N_{Az} \) be the flux of A (mol/(area·time)), and \( C_A \) be the concentration of A at position \( z \). #### **Shell Balance:** \[ \text{Net flux into the control volume: } N_{Az}|_z - N_{Az}|_{z+\Delta z} \] At steady state, the net accumulation is zero: \[ N_{Az}|_z - N_{Az}|_{z+\Delta z} = 0 \] Which leads to: \[ N_{Az}|_z = N_{Az}|_{z+\Delta z} \] #### **Fick’s Law:** \[ N_{Az} = -D_{AB} \frac{dC_A}{dz} \] #### **Plug Into the Balance:** Using Fick's law: \[ \frac{dN_{Az}}{dz} = -D_{AB} \frac{d^2 C_A}{dz^2} = 0 \] If \( D_{AB} \) is constant, we have: \[ \frac{d^2 C_A}{dz^2} = 0 \] **Explanation:** This shell balance indicates that, under steady state with no reactions, the flux of A through any cross-section is constant. Therefore, the concentration \( C_A \) must satisfy a second-order differential equation. --- ## **Step 2: Solve for Concentration Profile and Rate (Parts b & c)** ### **(b) Linear Profile** From step 1, we have: \[ \frac{d^2 C_A}{dz^2} = 0 \] Integrate once: \[ \frac{dC_A}{dz} = m \quad \text{(a constant)} \] Integrate again: \[ C_A(z) = mz + b \] **Boundary Conditions:** - At \( z = 0 \): \( C_A(0) = C_{A0} \) - At \( z = \delta \): \( C_A(\delta) = C_{A\delta} \) Plugging in these boundary conditions: 1. From \( C_A(0) = b \): \[ b = C_{A0} \] 2. From \( C_A(\delta) = m \delta + C_{A0} = C_{A\delta} \): \[ m = \frac{C_{A\delta} - C_{A0}}{\delta} \] Thus, the concentration profile is: \[ C_A(z) = C_{A0} + \frac{C_{A\delta} - C_{A0}}{\delta} z \] **Explanation:** The concentration profile is linear between the pill and the main stream due to the absence of reaction and steady state conditions. --- ### **(c) Rate of Diffusion** Using Fick's law: \[ N_{Az} = -D_{AB} \frac{dC_A}{dz} \] From the linear concentration profile, we have: \[ \frac{dC_A}{dz} = \frac{C_{A\delta} - C_{A0}}{\delta} \] Substituting into Fick's law: \[ N_{Az} = -D_{AB} \left(\frac{C_{A\delta} - C_{A0}}{\delta}\right) = \frac{D_{AB}(C_{A0} - C_{A\delta})}{\delta} \] --- ## **Final Answer:** The rate of diffusion is given by: \[ \boxed{N_{Az} = \frac{D_{AB}(C_{A0} - C_{A\delta})}{\delta}} \] --- ### **Summary of Steps:** - **Step 1:** Shell balance and Fick's law yield \( \frac{d^2 C_A}{dz^2} = 0 \). - **Step 2:** Solving the differential equation gives a linear concentration profile, and the rate of diffusion is as stated above.