Let's solve each task step by step as per the given chemical engineering process: **Process Details:** A + B → C Fresh Feed: - Stream 1: 50 mol/h A - Stream 2: 60 mol/h B Reactor: Single pass conversion of A = 50%. B in excess. Separator: 95% of C recovered as product, unreacted A & B fully recycled. --- ### 1. **Calculate reactor inlet flow rates of A & B** Let F_A and F_B be the fresh feed rates of A and B, respectively. Let R = recycle stream (of A and B, from separator output). Let F_AR and F_BR be the flow rates of A and B entering the reactor (fresh + recycle). So, F_AR = F_A + R_A F_BR = F_B + R_B But since B is in excess and we have full recycle, let's solve by drawing a material balance. Let the total reactor feed of A be F_AR. Fraction of A converted = 50%. A reacted = .5 × F_AR A unreacted = .5 × F_AR (this is recycled) Fresh A input + Recycled A = Total A to reactor So, F_AR = F_A + .5 × F_AR .5 × F_AR = F_A Therefore, F_AR = 2 × F_A = 2 × 50 = **100 mol/h** Similarly, for B: Let F_BR be B into reactor. A is limiting, so B reacted = .5 × F_AR = 50 mol/h. B in excess, so unreacted B = F_BR - 50 (this is recycled). Fresh B input + Recycled B = Total B to reactor F_BR = F_B + (F_BR - 50) F_B = 50 But F_B is given as 60 mol/h. So, F_BR = Fresh B + Recycled B Let x = B entering reactor B reacted = 50 (since A reacted = 50) Fresh B + Recycled B = x Recycle = x - 50 So, x = Fresh B + (x - 50) 50 = Fresh B But Fresh B is 60 mol/h. So, there must be an excess. Since B is in excess, all the unreacted B is recycled, but the excess B in fresh feed simply increases the recycle. So, B in reactor = Fresh B + Recycled B But since A is limiting, only 50 of B reacts. So, F_BR = 50 (to react) + Recycle (to balance) + Excess from feed But the total B entering the reactor = 60 + (Recycle) Therefore, F_BR = 60 + (F_BR - 50) So, 60 + F_BR - 50 = F_BR So, 60 - 50 = , which is satisfied. Thus, the reactor inlet flow rates are: A: **100 mol/h** B: **110 mol/h** (since for every mole of A, there must be at least 1 mole of B, but since the fresh feed is in excess, B in reactor = 60 (fresh) + 50 (recycle) = 110 mol/h) --- ### 2. **Calculate reactor outlet flow rate of A, B, C** Inlet to reactor: A = 100 mol/h B = 110 mol/h Since conversion of A is 50%, A reacted = .5 × 100 = 50 mol/h B reacted = 50 mol/h (stoichiometry 1:1) C formed = 50 mol/h Outlet: A unreacted = 100 - 50 = 50 mol/h B unreacted = 110 - 50 = 60 mol/h C = 50 mol/h So, reactor outlet: A = **50 mol/h**, B = **60 mol/h**, C = **50 mol/h** --- ### 3. **Calculate recycle flow rate** Recycle is the unreacted A and B from separator. From separator, 95% of C is removed as product, 5% of C is recycled along with all A and B. C to separator = 50 mol/h Product C = .95 × 50 = 47.5 mol/h C recycled = .05 × 50 = 2.5 mol/h Recycle stream: A = 50 mol/h B = 60 mol/h C = 2.5 mol/h Total recycle flow rate = 50 + 60 + 2.5 = **112.5 mol/h** --- ### 4. **Calculate product C flow rate** From above, product C = **47.5 mol/h** --- ### 5. **Overall conversion of fresh A** Fresh A = 50 mol/h Product C = 47.5 mol/h Since 1 mol A gives 1 mol C, Overall conversion = (A converted to product)/(A in fresh feed) = 47.5 / 50 = **95%** --- ### 6. **Effect if reactor conversion increases to 70%** Let’s repeat the calculations with single-pass conversion = 70%. Let F_AR be A to reactor. Fraction of A converted = .7 A unreacted = .3 × F_AR Fresh feed + recycle = total to reactor F_AR = F_A + .3 × F_AR .7 × F_AR = F_A F_AR = F_A / .7 = 50 / .7 ≈ **71.43 mol/h** A converted = .7 × 71.43 ≈ 50 mol/h Thus, B needed = 50 mol/h (at least); but fresh B is 60 mol/h B in reactor = 60 + (recycled B) B unreacted = 71.43 - 50 = 21.43 mol/h So, total B in reactor = 60 (fresh) + 21.43 (recycled) = **81.43 mol/h** Outlet of reactor: A unreacted = .3 × F_AR = .3 × 71.43 ≈ 21.43 mol/h B unreacted = 81.43 - 50 = 31.43 mol/h C formed = 50 mol/h Separator: C product = .95 × 50 = 47.5 mol/h C recycled = 2.5 mol/h Recycle stream: A = 21.43 B = 31.43 C = 2.5 Total recycle = **55.36 mol/h** Product C = **47.5 mol/h** Overall conversion of A = 47.5 / 50 = **95%** (no change in overall conversion when separator and recycle are 100% efficient; the recycle flow is lower, which is beneficial). --- ### **Summary Table** | Task | Value (50% conv.) | Value (70% conv.) | |------|-------------------|-------------------| | Reactor A in | 100 mol/h | 71.43 mol/h | | Reactor B in | 110 mol/h | 81.43 mol/h | | Reactor A out| 50 mol/h | 21.43 mol/h | | Reactor B out| 60 mol/h | 31.43 mol/h | | Reactor C out| 50 mol/h | 50 mol/h | | Recycle | 112.5 mol/h | 55.36 mol/h | | Product C | 47.5 mol/h | 47.5 mol/h | | Overall conv | 95% | 95% | **Conclusion:** Increasing the reactor conversion reduces recycle flow and the reactor feed, but the product rate and overall conversion of A remain the same due to full recycle and efficient separation.