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Section 3 multistage separation processes
- 1. MULTISTAGE SEPARATION PROCESSES CHE 452 ENG. AMAL MAGDY SECTION (3)
- 2. Flash Distillation For non ideal systems: • All calculation steps of remain the same as discussed in the previous section except calculation of ‘k’ values. • ‘k’ values for non ideal systems are predicted from much difficult equations 𝒌𝒊 = 𝜸𝒊 ∗ 𝒇𝒊 𝒐 𝝋𝒊 ∗ 𝑷 • ‘k’ values for non ideal systems especially of oil fractions are predicted from charts
- 4. Equilibrium ratio Charts Example: Determine the k-value of propane at 25 psia and 40 °F k = 3
- 5. DePriester charts 1. High Temperature DePriester Chart 2. Low Temperature DePriester Chart
- 6. If the temperature is given: • Assume Pressure • Get k-values • Calculate 𝒙𝒊 or 𝒚𝒊 • If 𝒙𝒊 or 𝒚𝒊 = 𝟏, the assumption is correct • If 𝒙𝒊 or 𝒚𝒊 ≠ 𝟏, the previous steps are repeated twice • Interpolation or extrapolation is done to get 𝑷 that corresponds to 𝒙𝒊 or 𝒚𝒊 = 1 If the pressure is given: • Assume Temperature • Get k-values • Calculate 𝒙𝒊 or 𝒚𝒊 • If 𝒙𝒊 or 𝒚𝒊 = 𝟏, the assumption is correct • If 𝒙𝒊 or 𝒚𝒊 ≠ 𝟏, the previous steps are repeated twice • Interpolation or extrapolation is done to get 𝑻 that corresponds to 𝒙𝒊 or 𝒚𝒊 = 1 DePriester charts
- 7. Flash Distillation For non ideal systems: • Overall Material Balance (OMB) F = L + V • Component Material Balance (CMB) 𝑭. 𝒙𝒇𝒊 = 𝑳. 𝒙𝒊 +𝑽. 𝒚𝒊 • Equilibrium relation 𝒚𝒊 = 𝒌𝒊 ∗ 𝒙𝒊 • k-values are predicted from charts • Operating line equation 𝑳 𝑽 = 𝒚 − 𝒙𝒇𝒊 𝒙𝒇𝒊 − 𝒙 𝒐𝒓 𝑳 𝑽 = 𝒚𝒊 − 𝒙𝒇𝒊 𝒙𝒇𝒊 − 𝒚𝒊 𝒌𝒊 𝒙𝒊 = 𝒙𝒇𝒊 𝑳 𝑽 + 𝟏 𝑳 𝑽 + 𝒌𝒊 = 𝟏 𝒂𝒏𝒅 𝒚𝒊 = 𝒙𝒇𝒊 𝑳 𝑽 + 𝟏 𝑳 𝑽 . 𝒌𝒊 + 𝟏 = 𝟏
- 8. SHEET (3)
- 9. Example (1) Calculate the product compositions and flow rates ( liquid and vapor ) from a flash tank with feed of 100 kmole of the following specifications: The flash unit operates at 40°F and 35 Psia. Use the equilibrium charts to get the k value for each component. Component Methane Ethane Propane n-butane n-pentane Hexane Heptane 𝑥𝑓 0.3396 0.0646 0.0987 0.0434 0.032 0.03 0.3917
- 10. Solution (1) • Let’s solve based on 𝒙𝒊 = 𝟏 1. Assume 𝑳 𝑽 = 1 2. 𝑥𝑖 = 𝑥𝑓𝑖 𝐿 𝑉 +1 𝐿 𝑉 + 𝑘𝑖 = 0. 3396 ∗ 1+1 1+58 + 0.0646 ∗ 1+1 1+9 + … = 0.966 ≠ 𝟏 3. Assume 𝑳 𝑽 = 0.5 4. 𝑥𝑖 = 𝑥𝑓𝑖 𝐿 𝑉 +1 𝐿 𝑉 + 𝑘𝑖 = 0. 3396 ∗ 0.5+1 0.5+58 + 0.0646 ∗ 0.5+1 0.5+9 + … = 1.344 ≠ 𝟏 5. Using extrapolation 𝑳 𝑽 = 0.955 Component Methane Ethane Propane n-butane n-pentane Hexane Heptane 𝑥𝑓 0.3396 0.0646 0.0987 0.0434 0.032 0.03 0.3917 k 58 9 2.2 5.1 0.7 0.034 0.018
- 11. Solution (1) • Using the overall material balance: L + V = 100 kmole L = 48.85 kmole V = 51.15 kmole • Apply the following law to get the liquid phase compositions: 𝒙𝒊 = 𝒙𝒇𝒊 𝑳 𝑽 + 𝟏 𝑳 𝑽 + 𝒌𝒊 • Then from the equilibrium relation get the vapor phase compositions: 𝒚𝒊 = 𝒌𝒊 ∗ 𝒙𝒊 Component Methane Ethane Propane n-butane n-pentane Hexane Heptane 𝑥𝑖 0.011 0.013 0.061 0.014 0.038 0.059 0.787 𝑦𝑖 0.653 0.114 0.135 0.07 0.026 0.002 0.014
- 12. Example (3) A vapor mixture contains 10 mole% methane, 30 mole% ethane, and rest propane at 50°C. Determine the dew composition using high temperature DePriester chart.
- 13. Solution (3)
- 14. Solution (3) Component 𝒚𝒊 K-value 𝒙𝒊 Methane 0.1 19 0.005 Ethane 0.3 4.5 0.067 Propane 0.6 1.65 0.364 • Assume P = 1000 kPa • 𝑥𝑖 = 0.436 • Assume P = 4000 kPa • 𝑥𝑖 = 1.25 • By interpolation P = 3078.624 kPa Component 𝒚𝒊 K-value 𝒙𝒊 Methane 0.1 5.1 0.02 Ethane 0.3 1.5 0.2 Propane 0.6 0.58 1.03
- 15. Solution (3) Component 𝒚𝒊 K-value 𝒙𝒊 Methane 0.1 6.8 0.015 Ethane 0.3 1.9 0.158 Propane 0.6 0.72 0.833 • At P = 3078.624 kPa
- 16. Steam Distillation • Steam has three benefits in this kind of distillation: 𝑚1: responsible for heating the batch to the bubble point 𝑚2: responsible for converting the hot batch into vapor (heating the batch up to the dew point) 𝑚3: the amount of steam that carries the evaporated vapors • The total amount of steam is (𝒎𝑻): 𝒎𝑻 = 𝒎𝟏 + 𝒎𝟐 + 𝒎𝟑
- 17. Steam Distillation • Calculation of steam amount: 1. Sensible heat required to heat up the batch to the boiling point (𝑻𝒃) = latent heat lost from steam (𝒎𝟏) 𝑸𝟏 = 𝒎𝒃𝒂𝒕𝒄𝒉 . 𝑪𝒑𝒃𝒂𝒕𝒄𝒉 . 𝑻𝒃 − 𝑻𝒊 = 𝒎𝟏 . 𝝀𝒔𝒕. 2. Latent heat gained by the batch = latent heat lost from steam (𝒎𝟐) 𝑸𝟐 = 𝒎𝒃𝒂𝒕𝒄𝒉 . 𝝀𝒃𝒂𝒕𝒄𝒉 = 𝒎𝟐 . 𝝀𝒔𝒕. 3. Amount of steam carrying the vapors 𝒎𝟑 = 𝑷𝒔𝒕. ∗ 𝑴𝒔𝒕. 𝑷𝑯𝑪 ∗ 𝑴𝑯𝑪 ∗ 𝒎𝑯𝑪
- 18. Example (4) 10 kg batch of ethyl-aniline (C8H11N) is to be steam distilled from small amount of non-volatile impurity. Saturated steam at 25 psia is used. Initial temperature of ethyl- aniline is 40°C and the distillation takes place at atmospheric pressure. Required data: • Heat capacity of ethyl-aniline is 0.4 kCal/kg.C • Heat capacity of steam is 0.35 kCal/kg.C • Latent heat of vaporization of ethyl-aniline is 72 kcal/kg • Latent heat of vaporization of water is 540 kcal/kg • Vapor pressures of water and ethyl aniline are given in the table below: T, °C 38.5 64.4 80.6 96 99.15 113.2 𝑃𝑤 𝑜, mmHg 51.1 199. 7 363.9 657.6 737.2 1225 𝑃𝐸𝐴 𝑜 , mmHg 1 5 10 20 22.8 40
- 19. Example (4) a. At what temperature will the distillation proceeds? b. Determine the composition of the vapor phase. c. How much steam is used?
- 20. Solution (4) • T = 98°C • yEA = 20 760 = 0.026 • yw = 740 760 = 0.974 0 200 400 600 800 1000 1200 1400 0 20 40 60 80 100 120 P, mmHg T, °C Water Ethyl-aniline Total Pressure
- 21. Solution (4) • 𝑸𝟏 = 𝒎𝒃𝒂𝒕𝒄𝒉 . 𝑪𝒑𝒃𝒂𝒕𝒄𝒉 . 𝑻𝒃 − 𝑻𝒊 = 𝒎𝟏 . 𝝀𝒔𝒕. 10 * 0.4 * (98 – 40) = m1 * 540 m1 = 0.43 kg • 𝑸𝟐 = 𝒎𝒃𝒂𝒕𝒄𝒉 . 𝝀𝒃𝒂𝒕𝒄𝒉 = 𝒎𝟐 . 𝝀𝒔𝒕. 10 * 72 = m2 * 540 m2 = 1.33 kg • 𝒎𝟑 = 𝑷𝒔𝒕. ∗ 𝑴𝒔𝒕. 𝑷𝑯𝑪 ∗ 𝑴𝑯𝑪 ∗ 𝒎𝑯𝑪 𝑚3 = 740 ∗18 20 ∗(8∗12+11∗1+14) ∗ 10 m3 = 55.04 kg mst. = m1+ m2 + m3 = 56.8 kg