Let's solve **Question 1** step by step: --- ## **Given Data** - Feed (L): 100 g-mol/s, 90% A, 10% I (inert) - Reaction: \( 2A \rightarrow P \) - Side reaction: \( A \rightarrow B \) - Conversion of A: 80% - Yield of P: 50% - Recycle (K): 84% A, 16% I - All streams except M contain only A and I; Product (P) and B are pure. - Molar flow rates and compositions of all streams (K, L, N, O, P, B) are required. - Calculate recycle ratio and selectivity. --- ### **Step 1: Feed (L) Calculation** Given: - \( L = 100 \) g-mol/s - \( L_A = .9 \times 100 = 90 \) g-mol/s - \( L_I = .1 \times 100 = 10 \) g-mol/s --- ### **Step 2: Conversion and Yield Definitions** - **Conversion (\( X \))**: Fraction of A entering reactor that reacts. - \( X = 80\% \) - **Yield (\( Y \))**: Fraction of A converted that forms P. - \( Y = 50\% \) - Rest forms B. Let the total A entering the reactor (M) be \( M_A \). --- ### **Step 3: Reactor Input (M) and Output (N, B, P) Calculations** **Let:** - \( M_A = \) moles of A entering reactor - \( M_I = \) moles of I entering reactor #### **A. A Consumed in Reactor:** - \( \text{A consumed} = X \cdot M_A = .8 M_A \) - \( \text{A to P} = Y \cdot (\text{A consumed}) = .5 \times .8 M_A = .4 M_A \) - \( \text{A to B} = (1-Y) \cdot (\text{A consumed}) = .5 \times .8 M_A = .4 M_A \) - \( \text{A unreacted} = (1-X) M_A = .2 M_A \) #### **B. I (Inert):** - Passes through unchanged: \( M_I \) #### **C. Products:** - \( P = \frac{\text{A to P}}{2} = \frac{.4 M_A}{2} = .2 M_A \) (from stoichiometry) - \( B = .4 M_A \) - \( N = \text{Unreacted A} + I = .2 M_A + M_I \) --- ### **Step 4: Separator Output** - **P**: Pure P (\( .2 M_A \)) - **B**: Pure B (\( .4 M_A \)) - **N**: \( .2 M_A \) A + \( M_I \) I --- ### **Step 5: Mixer (K) and Fresh Feed (L) Relationships** Recycle (K) + Fresh Feed (L) = Reactor Feed (M): **Let \( K \) be recycle (from N):** - \( K_A + L_A = M_A \) - \( K_I + L_I = M_I \) Given: - \( K \) composition: 84% A, 16% I - \( K_A = .84 K \) - \( K_I = .16 K \) - \( L_A = 90 \), \( L_I = 10 \) So: - \( M_A = K_A + 90 \) - \( M_I = K_I + 10 \) --- ### **Step 6: Recycle Stream Calculation** But, **K comes from N** (N = K): - \( N = K \) - \( N_A = .2 M_A \) - \( N_I = M_I \) - \( K_A = .84 K \) - \( K_I = .16 K \) - \( K_A = N_A \Rightarrow .84 K = .2 M_A \) - \( K_I = N_I \Rightarrow .16 K = M_I \) --- ### **Step 7: Solve the Equations** #### Equation Set: 1. \( .84 K = .2 M_A \) \( K = \frac{.2 M_A}{.84} \) 2. \( .16 K = M_I \) \( K = \frac{M_I}{.16} \) 3. \( M_A = K_A + 90 \) \( M_I = K_I + 10 \) But \( K_A = .84 K \), \( K_I = .16 K \): So, - \( M_A = .84 K + 90 \) - \( M_I = .16 K + 10 \) Substitute (from above): - \( .84 K = .2 M_A \rightarrow K = \frac{.2 M_A}{.84} \) So, - \( M_A = .84 \cdot \frac{.2 M_A}{.84} + 90 \) - \( M_A = .2 M_A + 90 \) - \( .8 M_A = 90 \) - \( M_A = \frac{90}{.8} = 112.5 \) Now, \( K = \frac{.2 \times 112.5}{.84} = \frac{22.5}{.84} = 26.7857 \) Now, \( M_I = .16 K + 10 = .16 \times 26.7857 + 10 = 4.2857 + 10 = 14.286 \) Check: \( K_I = .16 \times 26.7857 = 4.2857 \) --- ### **Step 8: Final Stream Values** #### **A. L (Feed):** - Total: 100 g-mol/s - 90 A, 10 I #### **B. K (Recycle):** - Total: 26.79 g-mol/s - 22.5 A, 4.29 I #### **C. M (to Reactor):** - Total: 126.79 g-mol/s - 112.5 A, 14.29 I #### **D. N (to Recycle):** - \( N_A = .2 M_A = .2 \times 112.5 = 22.5 \) - \( N_I = M_I = 14.29 \) - \( N = 22.5 + 14.29 = 36.79 \) g-mol/s #### **E. P (Product):** - \( P = .2 M_A = 22.5 \) g-mol/s (pure P) #### **F. B (Byproduct):** - \( B = .4 M_A = 45 \) g-mol/s (pure B) #### **G. O (Total Output):** - \( O = N + P + B = 36.79 + 22.5 + 45 = 104.29 \) g-mol/s --- ### **Step 9: Recycle Ratio** Recycle Ratio (\( R \)) = Recycle / Fresh Feed = \( K / L = 26.79 / 100 = .268 \) --- ### **Step 10: Selectivity** Selectivity = \( \frac{\text{Moles of P formed}}{\text{Moles of B formed}} = \frac{22.5}{45} = .5 \) --- ## **Summary Table** | Stream | Total (g-mol/s) | A (g-mol/s) | I (g-mol/s) | P (g-mol/s) | B (g-mol/s) | |--------|:---------------:|:-----------:|:-----------:|:-----------:|:-----------:| | L | 100 | 90 | 10 | | | | K | 26.79 | 22.5 | 4.29 | | | | M | 126.79 | 112.5 | 14.29 | | | | N | 36.79 | 22.5 | 14.29 | | | | P | 22.5 | | | 22.5 | | | B | 45 | | | | 45 | --- ## **Final Answers** ### 1. **Stream flow rates and compositions are as above.** ### 2. **Recycle Ratio:** .268 ### 3. **Selectivity:** .5 --- If you need the solution for **Question 2**, let me know!

Sure! Here's a concise summary of **Question 1**'s solution: --- ### Summary of Results for Question 1 - **Feed (L):** 100 g-mol/s (90 g-mol/s A, 10 g-mol/s I) - **Recycle (K):** 26.79 g-mol/s (22.5 g-mol/s A, 4.29 g-mol/s I) - **Reactor Input (M):** 126.79 g-mol/s (112.5 g-mol/s A, 14.29 g-mol/s I) - **Separator Output (N):** 36.79 g-mol/s (22.5 g-mol/s A, 14.29 g-mol/s I) - **Product (P):** 22.5 g-mol/s (pure P) - **Byproduct (B):** 45 g-mol/s (pure B) ### Additional Calculations - **Recycle Ratio:** 0.268 - **Selectivity:** 0.5 --- If you need assistance with **Question 2**, please let me know!

## **Calculation Steps for Question 1** ### Given Data Feed (L) is 100 g-mol/s with 90% A (90 g-mol/s) and 10% I (10 g-mol/s). The reaction is \( 2A \rightarrow P \) with an 80% conversion of A and a 50% yield of P. ### Feed Input L_A = 0.9 × 100 = 90 g-mol/s L_I = 0.1 × 100 = 10 g-mol/s ### Reactor Calculations Let M_A be moles of A entering the reactor. The A consumed in the reactor is 0.8 M_A. The A converted to P is 0.5 × (0.8 M_A) = 0.4 M_A, leading to B also being 0.4 M_A. Unreacted A is 0.2 M_A. The inert remains unchanged: M_I. ### Separator Outputs P = 0.2 M_A (pure P), B = 0.4 M_A (pure B), N = 0.2 M_A + M_I. ### Recycle Stream Calculations Using K for recycle, K_A and K_I are derived from the composition of K (84% A, 16% I). Thus, K_A = 0.84K and K_I = 0.16K. The equations yield: 1. \( M_A = K_A + 90 \) 2. \( M_I = K_I + 10 \) 3. \( N_A = 0.2 M_A \) 4. \( N_I = M_I \) Substituting K values into the equations gives: 0.84 K = 0.2 M_A leads to K = 0.2 M_A / 0.84 and subsequently M_A = 112.5 g-mol/s. ### Final Values and Composition K = 26.79 g-mol/s, M_I = 14.29 g-mol/s, N_A = 22.5 g-mol/s, N_I = 14.29 g-mol/s, leading to total N = 36.79 g-mol/s. The final outputs are P = 22.5 g-mol/s and B = 45 g-mol/s. ### Recycle Ratio and Selectivity Recycle ratio is R = K / L = 26.79 / 100 = 0.268. Selectivity is calculated as 22.5 / 45 = 0.5. --- ### Summary of Results - Feed (L): 100 g-mol/s (90 g-mol/s A, 10 g-mol/s I) - Recycle (K): 26.79 g-mol/s (22.5 g-mol/s A, 4.29 g-mol/s I) - Reactor Input (M): 126.79 g-mol/s (112.5 g-mol/s A, 14.29 g-mol/s I) - Separator Output (N): 36.79 g-mol/s (22.5 g-mol/s A, 14.29 g-mol/s I) - Product (P): 22.5 g-mol/s - Byproduct (B): 45 g-mol/s - Recycle Ratio: 0.268 - Selectivity: 0.5