Let's break down and solve **Problem ** step by step. ## **Given Data** - In air: 79% nitrogen (N₂), 21% oxygen (O₂) - **Equilibrium**: vapor leaving in equilibrium with liquid - **Vapor-liquid equilibrium data**: (not shown in your image, but typically required) - **Product Purity Constraints:** - N₂ vapor stream can be sold if **≥ 90% N₂** - O₂ liquid stream can be sold if **> 60% O₂** Let: - \( F \): Feed flow rate - \( x_F \): Mole fraction of N₂ in feed (\( = .79 \)) - \( V \): Vapor flow rate - \( y \): Mole fraction of N₂ in vapor - \( L \): Liquid flow rate - \( x \): Mole fraction of N₂ in liquid The fraction of the entering air to leave in the vapor stream: \[ \theta = \frac{V}{F} \] --- ## **a. Generating a Vapor Stream of 90% or Greater Nitrogen (N₂)** ### **Step 1: Is It Possible?** You need to check the vapor-liquid equilibrium (VLE) data (usually a table or a y-x diagram) to see if it's possible to have \( y \geq .90 \) N₂ in the vapor phase, starting from \( x_F = .79 \). - **If the equilibrium curve allows \( y \geq .90 \) for any achievable \( x \), then it's possible.** - **Assume VLE data is available and shows that for some liquid composition, the vapor phase can reach \( y = .90 \).** ### **Step 2: Set Up Material Balances** #### **Overall Balance:** \[ F = V + L \] #### **Component (N₂) Balance:** \[ F x_F = V y + L x \] #### **Rearrange for θ:** Since \( \theta = \frac{V}{F} \): \[ L = F - V = F(1 - \theta) \] \[ F x_F = (F \theta) y + (F (1-\theta)) x \] \[ x_F = \theta y + (1-\theta)x \] \[ \theta = \frac{x_F - x}{y - x} \] ### **Step 3: Use Equilibrium Data** - For maximum \( \theta \), use **minimum possible \( x \)** (liquid N₂ mole fraction), which corresponds to the **maximum possible N₂ vapor purity** (\( y \)). - From VLE data, for \( y = .90 \), find corresponding \( x \). **Example** (Suppose from VLE data: for \( y = .90 \), \( x = .60 \)): \[ \theta = \frac{.79 - .60}{.90 - .60} = \frac{.19}{.30} = .633 \] ### **Step 4: Final Answers** - **Is it possible?** Yes, if the VLE allows \( y = .90 \) N₂ in vapor for some \( x \). - **Highest possible θ:** \( \theta = \frac{x_F - x}{y - x} \) for \( y = .90 \), where \( x \) is the equilibrium liquid composition for this \( y \). --- ## **b. Generating a Liquid Stream of >60% Oxygen (O₂)** - \( x_{O_2} > .60 \Rightarrow x_{N_2} < .40 \) ### **Step 1: Is It Possible?** - Check VLE data: Is it possible to achieve \( x < .40 \) N₂ in the liquid phase? ### **Step 2: Use the Same Formula** For \( x = .40 \) (maximum allowed N₂ in liquid), find corresponding \( y \) (from VLE data). \[ \theta = \frac{x_F - x}{y - x} \] Suppose VLE data gives \( y = .65 \) for \( x = .40 \): \[ \theta = \frac{.79 - .40}{.65 - .40} = \frac{.39}{.25} = 1.56 \] But \( \theta \) cannot be greater than 1 (since \( V < F \)), so this means not all feed can become vapor at that composition; check calculation based on real VLE data. ### **Step 3: Final Answer** - **Is it possible?** Yes, if VLE allows \( x < .40 \), i.e., liquid stream with >60% O₂. - **Highest possible θ:** \( \theta = \frac{x_F - x}{y - x} \), with \( x = .40 \), \( y \) from VLE for this \( x \). --- ## **Summary Table** | Case | Target Purity | Equation for \( \theta \) | VLE Needed | |-----------------------|----------------------|--------------------------------------------|------------| | a) N₂ in vapor | \( y \geq .90 \) | \( \theta = \frac{x_F - x}{y - x} \) | \( x \) at \( y = .90 \) | | b) O₂ in liquid | \( x_{O_2} > .60 \) | \( \theta = \frac{x_F - x}{y - x} \) | \( y \) at \( x = .40 \) | > **Final Answers:** > Use your VLE table to plug in the correct \( x \) and \( y \) values and calculate \( \theta \) for each case. --- **If you upload the VLE data/table, I can provide the exact numbers for \( \theta \) for each part.**