Let's solve **Problem 1** step by step. --- # Problem 1 Solution ## **Given Data** - Flow rate of each phase: \( Q = 60 \) m³/h (glycerin and wash acid, equal) - Glycerin (continuous phase): - Viscosity (\( \mu_G \)): 1.1 cP = 0.0011 Pa·s - Density (\( \rho_G \)): 865 kg/m³ - Acid (dispersed phase): - Viscosity (\( \mu_A \)): 1.05 cP = 0.00105 Pa·s - Density (\( \rho_A \)): 1153 kg/m³ - Diameter of droplets (\( d_d \)): 150 μm = \( 1.5 \times 10^{-4} \) m - Vessel length: \( L = 5D \) - To find: 1. The vessel diameter \( D \) 2. The height of acid overflow above the vessel floor --- ## **Step 1: Calculate Settling Velocity (Stokes' Law)** For small droplets (laminar regime, Reynolds number \( < 0.1 \)), Stokes' Law applies: \[ v_t = \frac{d_d^2 (\rho_A - \rho_G) g}{18 \mu_G} \] Where: - \( v_t \) = terminal (settling) velocity (m/s) - \( d_d \) = droplet diameter (m) - \( \rho_A, \rho_G \) = densities (kg/m³) - \( g \) = 9.81 m/s² - \( \mu_G \) = continuous phase viscosity (Pa·s) Plug in the values: \[ v_t = \frac{(1.5 \times 10^{-4})^2 \times (1153 - 865) \times 9.81}{18 \times 0.0011} \] \[ = \frac{2.25 \times 10^{-8} \times 288 \times 9.81}{0.0198} \] \[ = \frac{2.25 \times 10^{-8} \times 2825.28}{0.0198} \] \[ = \frac{6.357 \times 10^{-5}}{0.0198} \] \[ = 0.00321 \text{ m/s} \] --- ## **Step 2: Calculate Cross-sectional Area Needed** The cross-sectional area \( A \) must allow droplets to separate before leaving the vessel. The critical condition is that the upward velocity due to flow is less than the settling velocity. \[ A = \frac{Q}{v_t} \] Where: - \( Q \) = total flow rate = 60 m³/h = \( \frac{60}{3600} \) m³/s = 0.01667 m³/s (for one phase) Total flow through cross-section (since both phases are equal and mixed at inlet): \[ Q_{total} = Q_{glycerin} + Q_{acid} = 2 \times 0.01667 = 0.0333 \, \text{m}^3/\text{s} \] \[ A = \frac{Q_{total}}{v_t} = \frac{0.0333}{0.00321} \] \[ A = 10.4 \text{ m}^2 \] --- ## **Step 3: Calculate Vessel Diameter (\( D \))** \[ A = \frac{\pi}{4} D^2 \implies D = \sqrt{\frac{4A}{\pi}} \] \[ D = \sqrt{\frac{4 \times 10.4}{3.1416}} = \sqrt{13.24} \] \[ D = 3.64 \text{ m} \] --- ## **Step 4: Vessel Length (\( L = 5D \))** \[ L = 5 \times D = 5 \times 3.64 = 18.2 \text{ m} \] --- ## **Step 5: Height of Acid Overflow Above Vessel Floor** For a horizontal cylindrical vessel, the overflow height (\( h \)) determines the fraction of the cross-section occupied by acid. Since both phases are equal by volume, the overflow height should correspond to half the cross-sectional area. - For a horizontal cylinder filled halfway, - The area below the horizontal centerline is half the total area. - Height \( h = \frac{D}{2} \). \[ h = \frac{D}{2} = \frac{3.64}{2} = 1.82 \text{ m} \] --- ## **Final Answers** **(a) Size of the vessel:** - **Diameter (\( D \))** = **3.64 m** - **Length (\( L \))** = **18.2 m** **(b) Height of acid overflow above vessel floor:** - **\( h \) = 1.82 m** --- ### **Summary Table** | Parameter | Value | |----------------------------|-----------------------------| | Terminal velocity (\(v_t\))| 0.00321 m/s | | Cross-sectional area (\(A\))| 10.4 m² | | Vessel diameter (\(D\)) | 3.64 m | | Vessel length (\(L\)) | 18.2 m | | Overflow height (\(h\)) | 1.82 m | --- **If you need the worked solution for Problem 2, let me know!**

# Problem 2 Solution ## **Given Data** - **Castor Oil (dispersed phase)**: - Flow rate (\( Q_O \)): 2500 kg/h = \( \frac{2500}{3600} \) kg/s = 0.6944 kg/s - Density (\( \rho_O \)): 850 kg/m³ - Viscosity (\( \mu_O \)): 2 mN·s/m² = 0.002 Pa·s - **Water (continuous phase)**: - Flow rate (\( Q_W \)): 7500 kg/h = \( \frac{7500}{3600} \) kg/s = 2.0833 kg/s - Density (\( \rho_W \)): 980 kg/m³ - Viscosity (\( \mu_W \)): 0.95 mN·s/m² = 0.00095 Pa·s - **Droplet Diameter (\( d_d \))**: 150 μm = \( 1.5 \times 10^{-4} \) m - **Vessel height-to-diameter ratio**: \( H = 3D \) ### **To Find:** 1. The vessel diameter \( D \) 2. The height \( H \) of the vessel --- ## **Step 1: Calculate Settling Velocity (Stokes' Law)** Use Stokes' Law for the settling velocity of droplets in the continuous phase. \[ v_t = \frac{d_d^2 (\rho_W - \rho_O) g}{18 \mu_W} \] ### Substituting the values: - \( g = 9.81 \, \text{m/s}^2 \) \[ v_t = \frac{(1.5 \times 10^{-4})^2 (980 - 850) \times 9.81}{18 \times 0.00095} \] Calculating step-by-step: \[ = \frac{(2.25 \times 10^{-8}) \times 130 \times 9.81}{0.0171} \] \[ = \frac{(2.25 \times 10^{-8}) \times 1275.3}{0.0171} \] \[ = \frac{2.872 \times 10^{-5}}{0.0171} \approx 0.00168 \, \text{m/s} \] --- ## **Step 2: Calculate Cross-sectional Area Needed** The cross-sectional area \( A \) must allow droplets to separate effectively. \[ A = \frac{Q_{total}}{v_t} \] Where \( Q_{total} = Q_O + Q_W \). ### Calculating total flow rate: \[ Q_{total} = 0.6944 + 2.0833 = 2.7777 \, \text{kg/s} \] ### Converting to volume flow rates: For each phase: - Castor Oil: \[ Q_O = \frac{0.6944}{850} = 0.000818 \, \text{m}^3/\text{s} \] - Water: \[ Q_W = \frac{2.0833}{980} = 0.00212 \, \text{m}^3/\text{s} \] \[ Q_{total} = 0.000818 + 0.00212 = 0.002938 \, \text{m}^3/\text{s} \] ### Now calculate \( A \): \[ A = \frac{0.002938}{0.00168} \approx 1.75 \, \text{m}^2 \] --- ## **Step 3: Calculate Vessel Diameter (\( D \))** Using the area to find the diameter: \[ A = \frac{\pi}{4} D^2 \implies D = \sqrt{\frac{4A}{\pi}} \] ### Substituting the area: \[ D = \sqrt{\frac{4 \times 1.75}{\pi}} = \sqrt{\frac{7}{3.1416}} \approx \sqrt{2.23} \approx 1.49 \, \text{m} \] --- ## **Step 4: Calculate Vessel Height (\( H \))** Given that \( H = 3D \): \[ H = 3 \times 1.49 = 4.47 \, \text{m} \] --- ## **Final Answers** **1. Size of the vessel:** - **Diameter (\( D \))** = **1.49 m** - **Height (\( H \))** = **4.47 m** --- ### **Summary Table** | Parameter | Value | |----------------------------|-----------------------------| | Terminal velocity (\(v_t\))| 0.00168 m/s | | Cross-sectional area (\(A\))| 1.75 m² | | Vessel diameter (\(D\)) | 1.49 m | | Vessel height (\(H\)) | 4.47 m | --- If you have any further questions or need additional assistance, feel free to ask!