Let’s break down the **flash drum sizing** problem step by step. --- ### **Given Data** - **Feed rate (F):** 2000 lb-moles/hr - **Feed composition:** 40 mol% n-hexane, 60 mol% n-octane - **Pressure:** 101.3 kPa - **Vapor product (yH):** 60 mol% n-hexane - **Liquid product (xH):** 19 mol% n-hexane - **Drum temperature:** 378 K - **V/F (fraction vaporized):** 0.51 - **Molecular weights:** - n-hexane (MW_H): 86.17 - n-octane (MW_O): 114.22 - **Densities:** - n-hexane: 0.659 g/mL - n-octane: 0.703 g/mL - **1 ft³ = 28316.85 mL** - **Liquid residence time:** 10 min --- ## **Step 1: Calculate Amount of Liquid and Vapor Products** Fraction vaporized, \( V/F = 0.51 \Rightarrow V = 0.51F, \ L = F - V = 0.49F \) - \( F = 2000 \) lb-moles/hr - \( L = 0.49 \times 2000 = 980 \) lb-moles/hr (liquid out) - \( V = 0.51 \times 2000 = 1020 \) lb-moles/hr (vapor out) --- ## **Step 2: Calculate Moles of Each Component in Liquid** **Liquid composition:** \( x_H = 0.19 \) (hexane), \( x_O = 0.81 \) (octane) - Hexane in liquid: \( 980 \times 0.19 = 186.2 \) lb-moles/hr - Octane in liquid: \( 980 \times 0.81 = 793.8 \) lb-moles/hr --- ## **Step 3: Mass of Liquid Components per Hour** - **n-hexane:** \( 186.2 \times 86.17 = 16048 \) lb/hr - **n-octane:** \( 793.8 \times 114.22 = 90709 \) lb/hr - **Total mass:** \( 16048 + 90709 = 106757 \) lb/hr --- ## **Step 4: Convert Mass to Grams per Hour** - \( 1 \) lb \( = 453.592 \) g - **Total grams/hr:** \( 106757 \times 453.592 = 48431441 \) g/hr --- ## **Step 5: Calculate Volume of Liquid per Hour** Calculate **average density** of the liquid: \[ \rho_{avg} = \frac{w_H + w_O}{\frac{w_H}{\rho_H} + \frac{w_O}{\rho_O}} \] But since we know masses and densities, calculate **total volume** directly: - \( V_H = \frac{mass_H}{\rho_H} \) - \( V_O = \frac{mass_O}{\rho_O} \) - n-hexane: \( 16048 \) lb/hr \( \times 453.592 = 7286750 \) g/hr - n-octane: \( 90709 \) lb/hr \( \times 453.592 = 41144691 \) g/hr \[ V_H = \frac{7286750}{0.659} = 11052228 \text{ mL/hr} \] \[ V_O = \frac{41144691}{0.703} = 58527735 \text{ mL/hr} \] \[ V_{total} = V_H + V_O = 11052228 + 58527735 = 69579963 \text{ mL/hr} \] --- ## **Step 6: Calculate Liquid Volume in 10 Minutes (Residence Time)** - \( 10 \) min \( = \frac{1}{6} \) hr \[ \text{Volume in 10 min} = \frac{69579963}{6} = 11596660 \text{ mL} \] --- ## **Step 7: Convert mL to ft³** \[ \text{Volume in ft}^3 = \frac{11596660}{28316.85} = 409.6 \text{ ft}^3 \] --- ## **Step 8: Final Answer (Rounded)** ### **The required flash drum liquid volume is \(\boxed{410 \ \text{ft}^3}\) (rounded to two significant digits).** --- ### **Summary Table** | Step | Value | Units | |------|-------|-------| | Feed rate | 2000 | lb-moles/hr | | Liquid out | 980 | lb-moles/hr | | Liquid mass | 106,757 | lb/hr | | Liquid mass | 48,431,441 | g/hr | | Liquid vol/hr | 69,579,963 | mL/hr | | Liquid vol/10 min | 11,596,660 | mL | | Liquid vol/10 min | 410 | ft³ | --- ## **Assumptions** - The liquid product is perfectly mixed. - The densities are valid at process conditions. - The flash drum only needs to accommodate the liquid holdup for the residence time (not including vapor space or disengagement volume—those would make the drum larger in practice). --- **If you need to account for vapor disengagement space or want to size the drum diameter/height, let me know!**