Let's break down and **solve Case 1** step by step. ### **Given Data** - **Feed:** 100 kg (25 wt% acetic acid, 75 wt% water) - Acetic acid: 25 kg - Water: 75 kg - **Solvent (isopropyl ether):** 100 kg (pure) - **Temperature:** 20°C (constant) - **Objective:** Find the mass and compositions (kg and wt%) of the raffinate and extract phases. --- ## **Case 1: Single-Stage Extraction** ### **Step 1: Construct the System (Initial Streams)** - **Feed (F):** 100 kg (25% acetic acid, 75% water) - **Solvent (S):** 100 kg (100% isopropyl ether) ### **Step 2: Draw the Ternary Diagram (Graphical Method)** We need phase equilibrium data for the acetic acid-water-isopropyl ether system at 20°C. Let's assume you have access to the equilibrium diagram or tie-line data. #### **Locate Feed and Solvent Points:** - **Feed (F):** 25% acetic acid, 75% water, 0% ether - **Solvent (S):** 0% acetic acid, 0% water, 100% ether #### **Mix Feed and Solvent:** - **Total mass mixed:** \( F + S = 100 + 100 = 200 \) kg ##### **Mixture Composition:** - Acetic acid: \( \frac{25}{200} = 12.5\% \) - Water: \( \frac{75}{200} = 37.5\% \) - Isopropyl ether: \( \frac{100}{200} = 50\% \) #### **Plot the Mixture Point (M) on Ternary Diagram:** - 12.5% acetic acid, 37.5% water, 50% ether #### **Tie Line Construction (Graphical Solution):** - Draw tie line through the mixture point (M) to the equilibrium curve. - The ends of the tie line represent the compositions of the **raffinate** (aqueous phase, R) and **extract** (organic phase, E). ##### **Read the compositions from the diagram:** Let’s **assume** from the diagram (since we don’t have the actual diagram here): - **Raffinate (R):** 5% acetic acid, 95% water, 0% ether - **Extract (E):** 5% acetic acid, 10% water, 85% ether **(Note: You should use your actual ternary diagram to read these values accurately; these are sample estimates.)** --- ### **Step 3: Mass Balance for Each Component** Let: - \( R \) = mass of raffinate phase - \( E \) = mass of extract phase #### **Overall Mass Balance:** \( F + S = R + E \) \( 200 = R + E \) #### **Component Mass Balances:** **Acetic acid:** \( 25 = R \times x_{R,AA} + E \times x_{E,AA} \) **Water:** \( 75 = R \times x_{R,W} + E \times x_{E,W} \) **Isopropyl ether:** \( 100 = R \times x_{R,E} + E \times x_{E,E} \) Use the estimated compositions: - \( x_{R,AA} = 0.05 \), \( x_{R,W} = 0.95 \), \( x_{R,E} = 0 \) - \( x_{E,AA} = 0.05 \), \( x_{E,W} = 0.10 \), \( x_{E,E} = 0.85 \) --- #### **Solve the Equations:** ##### **Isopropyl Ether:** \( 100 = R \times 0 + E \times 0.85 \implies E = \frac{100}{0.85} = 117.65 \) kg ##### **Total Mass:** \( R + E = 200 \implies R = 200 - 117.65 = 82.35 \) kg ##### **Acetic Acid Balance:** \( 25 = R \times 0.05 + E \times 0.05 \) \( 25 = (82.35 \times 0.05) + (117.65 \times 0.05) \) \( 25 = 4.12 + 5.88 = 10 \) kg **There is a discrepancy - this suggests our assumed extract composition is too low in acetic acid.** #### **Let's Try a More Realistic Split:** Suppose: - \( x_{R,AA} = 0.05 \) - \( x_{E,AA} = 0.18 \) - \( x_{R,W} = 0.95 \) - \( x_{E,W} = 0.07 \) - \( x_{R,E} = 0 \) - \( x_{E,E} = 0.75 \) **Isopropyl Ether:** \( 100 = E \times 0.75 \implies E = 133.33 \) kg **Total Mass:** \( R = 200 - 133.33 = 66.67 \) kg **Acetic Acid:** \( 25 = R \times 0.05 + E \times 0.18 \) \( 25 = (66.67 \times 0.05) + (133.33 \times 0.18) \) \( 25 = 3.33 + 24 = 27.33 \) kg **Still a discrepancy (should balance to 25 kg).** **You should use the actual tie line data from your ternary diagram** to get the real answer, but the method above is how you solve it! --- ### **Step 4: Analytical Method** This is typically the **lever arm rule** on the ternary diagram, or solving the mass balances above with the **actual equilibrium compositions**. --- ## **Summary of Steps** ### **4.1. Graphical Method** - **Plot feed and solvent compositions on ternary diagram.** - **Draw tie line through mixture point.** - **Read off raffinate and extract compositions.** - **Apply mass balances to determine quantities.** ### **4.2. Analytical Method** - **Use equilibrium partition data (distribution coefficients for acetic acid) to write component balances.** - **Solve for masses and compositions algebraically as above.** --- ## **Final Answers (Format):** | Stream | Mass (kg) | Acetic Acid (wt%) | Water (wt%) | Isopropyl Ether (wt%) | |------------|-----------|-------------------|-------------|----------------------| | Raffinate | XX | XX | XX | XX | | Extract | XX | XX | XX | XX | You need the exact equilibrium data to fill in the numbers. **If you provide the tie-line data or equilibrium distribution coefficients, I can help you solve for the exact values!** --- ### **If you want to proceed with Case 2, let me know!**