Let's break down solve the problem step-by-step--- ## **Given Data** - **Reaction:** A → R (Second-order in A, diffusion limited) - **Rate constant (k):** \(50 \, m^6.(\,g\,cat\,s)^{-1} \) - **Concentration of A in feed (\(C_{A}\)):** \( .2\, mol\,L^{-1} = 200\, mol\,m^{-3} \) - **Superficial velocity (\(u\)):** \( 5\, m/s \) - **Particle diameter (\(d_p\)):** \( .4\, cm = .004\, m \) - **Bulk density of bed (\(\rho_b\)):** \( 1.2 \times 10^6\, g\,m^{-3} \) - **Porosity (\(\epsilon\)):** \( .4 \) - **Effective diffusivity (\(D_{eff}\)):** \( 3 \times 10^{-8}\, m^2/s \) - **Internal surface area (\(a_s\)):** \( 400\, m^2/g \) - **Required conversion (\(X\)):** \( 90\% = .9 \) - **A (cross-sectional area):** \( 4.77 \times 10^{-3}\, m^2 \) --- ## **(i) Estimate the value of the effectiveness factor** ### **Step 1: Thiele Modulus for 2nd-order reaction** For a spherical catalyst and a second-order reaction: \[ \phi = R_p \left[ k C_{A} a_s / D_{eff} \right]^{1/2} \] Where: - \( R_p = \) particle radius = \( .004 / 2 = .002\, m \) - \( k = 50\, m^6.(mol\,g\,cat\,s)^{-1} \) - \( a_s = 400\, m^2/g \) - \( C_{A} = 200\, mol\,m^{-3} \) - \( D_{eff} = 3 \times 10^{-8}\, m^2/s \) #### **Unit Consistency** - \( k \) units: \( m^6.(mol\,g\,cat\,s)^{-1} \) - \( a_s \) units: \( m^2/g \) - Multiply \( k \) and \( a_s \): \( m^6/(mol\,g\,cat\,s) \times m^2/g = m^8/(mol\,g^2\,cat\,s) \) - But we need to multiply by \( C_{A} \) (mol/m³). #### **Let’s combine:** \[ k\, a_s\, C_{A} = 50 \times 400 \times 200 = 4,000,000\, m^8/(mol\,g^2\,cat\,s) \times mol/m^3 \] But we still have \( g^2 \) in the denominator. Let's keep the calculation in terms of variables, and plug in all the numbers together. #### **Calculate Thiele Modulus:** \[ \phi = R_p \left[ k\, a_s\, C_{A} / D_{eff} \right]^{1/2} \] Plug in values: \[ \phi = .002 \left[ \frac{50 \times 400 \times 200}{3 \times 10^{-8}} \right]^{1/2} \] \[ = .002 \left[ \frac{4,000,000}{3 \times 10^{-8}} \right]^{1/2} \] \[ = .002 \left[ 1.333 \times 10^{14} \right]^{1/2} \] \[ = .002 \times 1.155 \times 10^7 \] \[ = 2.31 \times 10^4 \] ### **Step 2: Effectiveness Factor (η) for Large Thiele Modulus** For large \(\phi\): \[ \eta \approx \frac{3}{\phi^2} \] \[ \eta = \frac{3}{(2.31 \times 10^4)^2} = \frac{3}{5.34 \times 10^8} \] \[ \eta \approx 5.62 \times 10^{-9} \] --- ## **(ii) Calculate the weight of catalyst and the bed length for 90% conversion** ### **Step 1: Plug Flow Reactor Mole Balance** For second-order, internal diffusion limited: \[ -dF_A = r_A' \, dW \] \[ F_{A} - F_A = \int_^W \left( \eta k a_s C_A^2 \right) dW \] But for a PFR: \[ \frac{dF_A}{dW} = - \eta k a_s C_A^2 \] Where: - \( F_{A} = C_{A} \times u \times A \) - \( F_A = F_{A}(1-X) \) - \( X = .9 \implies F_A = .1 F_{A} \) \[ F_{A} = 200\, mol/m^3 \times 5\, m/s \times 4.77 \times 10^{-3}\, m^2 = 4.77 \, mol/s \] ### **Step 2: PFR Design Equation for Second Order** \[ \frac{dX}{dW} = \frac{-r_A'}{F_{A}} \] \[ r_A' = -\eta k a_s C_A^2 \] But \( C_A = C_{A}(1-X) \) \[ \frac{dX}{dW} = \frac{\eta k a_s \left[ C_{A}(1-X) \right]^2}{F_{A}} \] \[ dW = \frac{F_{A}}{\eta k a_s C_{A}^2} \frac{dX}{(1-X)^2} \] Integrate from \( X = \) to \( X = .9 \): \[ W = \frac{F_{A}}{\eta k a_s C_{A}^2} \int_^{.9} \frac{dX}{(1-X)^2} \] \[ \int_^{.9} \frac{dX}{(1-X)^2} = \left[ \frac{1}{1-X} \right]_^{.9} = \frac{1}{.1} - 1 = 10 - 1 = 9 \] \[ W = \frac{F_{A} \cdot 9}{\eta k a_s C_{A}^2} \] Plug in the values: - \( F_{A} = 4.77\, mol/s \) - \( \eta = 5.62 \times 10^{-9} \) - \( k = 50 \) - \( a_s = 400 \) - \( C_{A} = 200 \) \[ W = \frac{4.77 \times 9}{5.62 \times 10^{-9} \times 50 \times 400 \times (200)^2} \] Calculate denominator: - \( 5.62 \times 10^{-9} \times 50 = 2.81 \times 10^{-7} \) - \( 2.81 \times 10^{-7} \times 400 = 1.124 \times 10^{-4} \) - \( (200)^2 = 40000 \) - \( 1.124 \times 10^{-4} \times 40000 = 4.496 \) \[ W = \frac{42.93}{4.496} \approx 9.55\, g \] **So, the required weight of catalyst is approximately \( 9.55\, g \).** --- ### **Step 3: Calculate Length of the Bed** \[ \text{Bed volume} = \frac{W}{\rho_b (1 - \epsilon)} \] Given: - \( W = 9.55\, g \) - \( \rho_b = 1.2 \times 10^6\, g/m^3 \) - \( 1-\epsilon = .6 \) \[ \text{Bed volume} = \frac{9.55}{1.2 \times 10^6 \times .6} = \frac{9.55}{720,000} = 1.326 \times 10^{-5}\, m^3 \] \[ \text{Length} = \frac{\text{Bed volume}}{A} = \frac{1.326 \times 10^{-5}}{4.77 \times 10^{-3}} = 2.78 \times 10^{-3}\, m \approx 2.8\, mm \] --- ## **(iii) Which Engineer is Correct?** - **Engineer A:** Change catalyst particle size (reduce \( d_p \)) at constant velocity. - **Engineer B:** Operate at higher velocity, keep particle size the same. ### **Analysis:** - **Reducing particle size** increases surface area and decreases internal diffusion limitations, increasing the effectiveness factor \(\eta\), and thereby reducing the required reactor size. - **Increasing velocity** only increases throughput (space velocity), but does **not** affect the effectiveness factor, and may reduce conversion due to shorter residence time. **Therefore, Engineer A is correct.** Decreasing the catalyst particle size will increase the effectiveness factor and decrease the reactor size needed for the specified conversion. --- # **Summary Table** | Part | Step | Value/Conclusion | |------|-------------------------------------------------------------------|-------------------------------------------------| | (i) | Effectiveness factor (\(\eta\)) | \( 5.62 \times 10^{-9} \) | | (ii) | Required catalyst weight (\(W\)) | \( 9.55\, g \) | | | Bed length (\(L\)) | \( 2.8\, mm \) | | (iii)| Who is correct? | **Engineer A** | | | Reason | Lowering particle size increases \(\eta\) | --- **If you need more clarification on any step, let me know!**