Let's solve the problem step by step. --- # **Given Data** - **Stoichiometry:** \( A \rightarrow pS \) - **Rate equation:** \[ -r_A = \frac{kC_A^{1/2}}{1 + K C_A} \] - **Feed molar flow of A:** \( F_{A} = 100 \) kmol/hr - **Feed concentration of A:** \( C_{A} = .25 \) kmol/m³ - **Rate constant:** \( k = 10 \) (kmol/m³)\(^{1/2}\)/hr - **Equilibrium constant:** \( K = 6 \) (m³/kmol) --- # **(a) Reactor configuration for minimum total reactor volume (90% conversion)** **Let's analyze options:** - For a first-order or complex rate law like here, **a plug flow reactor (PFR) or a series of PFR + CSTR** often gives minimal volume. - For rate laws where the rate decreases with decreasing concentration, the CSTR is less efficient. - Let's check which to use by comparing reactor volumes. ## **Step 1: Find required conversion** \[ X = .90 \] ## **Step 2: Calculate final concentration \(C_A\)** \[ \text{Conversion:} \quad X = \frac{F_{A} - F_A}{F_{A}} \] Assuming constant volumetric flow rate (liquid phase, no change in total moles): \[ C_A = C_{A}(1 - X) = .25 \times (1 - .9) = .025 \text{ kmol/m}^3 \] --- ## **Step 3: PFR Volume Calculation** For a **PFR**: \[ F_{A} \int_{X_}^{X_f} \frac{dX}{-r_A} = V \] \[ -r_A = \frac{k C_A^{1/2}}{1 + K C_A} \] But \(C_A = C_{A} (1-X)\): \[ -r_A = \frac{k [C_{A}(1-X)]^{1/2}}{1 + K C_{A}(1-X)} \] So, \[ V = F_{A} \int_{}^{.9} \frac{dX}{\frac{k [C_{A}(1-X)]^{1/2}}{1 + K C_{A}(1-X)}} \] \[ = F_{A} \int_{}^{.9} \frac{1 + K C_{A}(1-X)}{k [C_{A}(1-X)]^{1/2}} dX \] Insert values: - \( F_{A} = 100 \) kmol/hr - \( k = 10 \) (kmol/m³)\(^{1/2}\)/hr - \( C_{A} = .25 \) kmol/m³ - \( K = 6 \) m³/kmol \[ V = 100 \int_{}^{.9} \frac{1 + 6 \times .25 \times (1-X)}{10 \times [.25 \times (1-X)]^{1/2}} dX \] \[ = 100 \int_{}^{.9} \frac{1 + 1.5(1-X)}{10 \times (.25)^{1/2} \times (1-X)^{1/2}} dX \] \[ (.25)^{1/2} = .5 \] \[ = 100 \int_{}^{.9} \frac{1 + 1.5(1-X)}{10 \times .5 \times (1-X)^{1/2}} dX \] \[ = 100 \int_{}^{.9} \frac{1 + 1.5(1-X)}{5 (1-X)^{1/2}} dX \] \[ = 20 \int_{}^{.9} \frac{1 + 1.5(1-X)}{(1-X)^{1/2}} dX \] \[ = 20 \int_{}^{.9} \left[ \frac{1}{(1-X)^{1/2}} + \frac{1.5(1-X)}{(1-X)^{1/2}} \right] dX \] \[ = 20 \int_{}^{.9} \left[ (1-X)^{-1/2} + 1.5(1-X)^{1/2} \right] dX \] Let's integrate both terms: \[ \int (1-X)^{-1/2} dX = -2(1-X)^{1/2} \] \[ \int (1-X)^{1/2} dX = -\frac{2}{3}(1-X)^{3/2} \] So, \[ V = 20 \left\{ -2(1-X)^{1/2} - 1.(1-X)^{3/2} \right\} \Big|_{}^{.9} \] Insert the limits: - At \(X=\): \((1-X)^{1/2} = 1\), \((1-X)^{3/2} = 1\) - At \(X=.9\): \((1-X)^{1/2} = (.1)^{1/2} = .3162\), \((1-X)^{3/2} = (.1)^{1.5} = .03162\) \[ V = 20 \left\{ [-2(.3162) - 1(.03162)] - [-2(1) - 1(1)] \right\} \] \[ = 20 \left\{ [-.6324 - .03162] - [-2 - 1] \right\} \] \[ = 20 \left\{ -.66402 + 3 \right\} \] \[ = 20 \times 2.33598 = 46.7196 \text{ m}^3 \] So, **PFR volume required for 90% conversion is about \(\boxed{46.7 \text{ m}^3}\)** --- ## **Step 4: CSTR Volume Check (for comparison)** For a **CSTR**: \[ V = \frac{F_{A} X}{-r_A} \] At the outlet (\(X=.9\)), \(C_A = .025\) kmol/m³: \[ -r_A = \frac{10 \times (.025)^{1/2}}{1 + 6 \times .025} = \frac{10 \times .1581}{1 + .15} = \frac{1.5811}{1.15} = 1.3749 \text{ kmol/(m}^3\cdot\text{hr)} \] \[ V = \frac{100 \times .9}{1.3749} = \frac{90}{1.3749} = 65.5 \text{ m}^3 \] So, **CSTR volume is larger than PFR**. --- ## **Conclusion for Part (a):** **A single PFR gives the minimum volume.** --- # **(b) Calculate the reactor(s) volume(s) for your suggestion in (a)** From above, **the required PFR volume is:** \[ \boxed{46.7\ \text{m}^3} \] --- # **Summary Table** | Reactor Type | Volume (m³) | |--------------|-------------| | PFR | 46.7 | | CSTR | 65.5 | --- ## **Final Answers** ### **a) Reactor configuration:** - **A single Plug Flow Reactor (PFR) yields the minimum total reactor volume.** ### **b) Reactor volume(s):** - **Required PFR volume: \(\boxed{46.7\ \text{m}^3}\)** --- **If you have more parts or need a breakdown for multiple reactors in series, let me know!**