Section 6 multistage separation processes
- 1. MULTISTAGE SEPARATION PROCESSES CHE 452 ENG. AMAL MAGDY SECTION (6)
- 2. Multistage Distillation Calculations • Binary System • Sorel plate to plate calculations • Ponchon Sevarit method • Lewis plate to plate calculations • McCabe Thiele method
- 3. McCabe Thiele method • The graphical representation of Lewis method • Like Lewis Method the tower is divided into top section and bottom section • McCabe Thiele method depends on the x-y diagram only • Each section has its own operating line y x
- 4. McCabe Thiele method Calculation steps • Calculation steps are done through three main steps: 1. Material Balance: • OMB F = D + W • CMB 𝑭. 𝒙𝒇 = 𝑫. 𝒙𝑫 +𝑾. 𝒙𝒘 Get D & W
- 5. McCabe Thiele method Calculation steps • Calculation steps are done through three main steps: 2. Operating Line for Top Section: 𝒚𝒏+𝟏 = 𝑳 𝑳+𝑫 𝒙𝒏 + 𝑫 𝑳+𝑫 𝒙𝑫 Or 𝒚𝒏+𝟏 = 𝑹 𝑹 + 𝟏 ∗ 𝒙𝒏 + 𝒙𝑫 𝑹 + 𝟏 To draw this line you should have two points: At xn = 0 yn+1 = 𝒙𝑫 𝑹+𝟏 At xn = xD yn+1 = xD xD xF xW 1 R xD
- 6. McCabe Thiele method Calculation steps • Calculation steps are done through three main steps: 3. Operating Line for Bottom Section: 𝒚𝒎 ′ = 𝑳′ 𝑽′ 𝒙𝒎+𝟏 ′ − 𝑾 𝑽′ 𝒙𝒘 To draw this line you should have two points: At 𝒙𝒎+𝟏 ′ = 0 𝒚𝒎 ′ =− 𝑾 𝑽′ 𝒙𝑾 At 𝒙𝒎+𝟏 ′ = xw 𝒚𝒎 ′ = xw xD xF xW V' Wx - W
- 7. McCabe Thiele method Calculation steps • The diagram of the whole tower will be: • The number of theoretical stages = No. of stages in top section + No. of stages in top section + Reboiler • For Top Section • NTS = 6 + 𝑎 𝑏 • For Bottom Section • NTS = 1 + Reboiler xD xF xW 1 R xD a b V' Wx - W
- 8. McCabe Thiele method Calculation steps • Q-line is calculations: 𝒒 = 𝑯𝑽 − 𝒉𝒇 𝑯𝑽 − 𝒉𝑳 = 𝑯𝑽 − 𝒉𝒇 𝝀 Slope of q-line = 𝒒 𝒒 −𝟏 • Q-line direction depends on the phase of feed: 1. Subcooled Feed q > 1 2. Saturated Liquid Feed q = 1 3. Partially Vaporized Feed 0 < q < 1 4. Saturated Vapor Feed q = 0 5. Superheated Feed q < 1 1 2 3 4 5 xD xF xW
- 9. McCabe Thiele method Calculation steps • Minimum reflux ratio: It is the point that resulted from the extension of the line plotted from (xD, xD) to the intersection of q-line with the equilibrium curve xD xF xW 1 R x min D
- 10. SHEET (4)
- 11. Example (2) A continuous fractionating column is to be designed to separate 100 kgmole/h of a mixture of 60% benzene and 40% toluene into a top product containing 2% toluene and a bottom product containing 3% benzene. Feed enters the column at 30°C. Pressure in the column is atmospheric and the reflux ratio is 1.2 the minimum. Calculate: a. Flowrates of the top and the bottom products b. Number of ideal plates c. Steam consumption if saturated steam at 3 atm is available. d. Cooling water consumption if cooling water at 30°C are available and left the system at 45°C.
- 12. Example (2) Additional Data: Relative volatility=2.5 Bubble and dew points of feed are 90 and 100°C. Liquid specific heats of benzene and toluene are 0.43 and 0.45 cal/g.°C respectively Latent heats of benzene and toluene are 7400 and 8000 cal/gmole respectively Latent heat of steam at 3 atmosphere is 517 cal/g
- 13. Solution (2) Givens: F = 100 kgmole/h , 𝑃 = 1 𝑎𝑡𝑚 , 𝑇 = 30°𝐶 & 𝑅 = 1.2 𝑅𝑚𝑖𝑛 𝑥𝑓 = 0.6 , 𝑥𝐷 = 0.98 & 𝑥𝑤 = 0.03 Answer: a. Flowrates of the top and the bottom products F = D + W D + W = 100 (1) 𝑭. 𝒙𝒇 = 𝑫. 𝒙𝑫 +𝑾. 𝒙𝒘 100 ∗ 0.6 = 0.98 𝐷 + 0.03 𝑊 (2) From (1) & (2) D = 60 kgmole/h , W = 40 kgmole/h
- 14. Solution (2) b. Number of ideal plates • Draw the equilibrium curve Using the given relative volatility, assume xA to get yA at equilibrium 𝒚𝑨 = 𝜶. 𝒙𝑨 𝟏 + (𝜶 − 𝟏) 𝒙𝑨 xA yA 0 0 0.1 0.217 0.3 0.517 0.5 0.714 0.7 0.854 0.9 0.957 1 1 0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1 y A xA
- 15. Solution (2) b. Number of ideal plates and feed location • Determine Rmin 𝒒 = 𝑯𝑽 − 𝒉𝒇 𝑯𝑽 − 𝒉𝑳 = 𝑯𝑽 − 𝒉𝒇 𝝀 𝜆 = 0.6 ∗ 7400 + 0.4 ∗ 8000 = 7640 cal/gmole 𝐶𝑃𝑚𝑖𝑥 = 𝑥𝑖 ∗ 𝐶𝑃𝑖 = 0.6* 0.43+ 0.4*0.45 = 0.438 cal/g.°C Mavg = 0.6 * 78 + 0.4 * 92 = 83.6 g/gmole 𝐻𝑉 − ℎ𝑓 = 𝜆 + 𝐶𝑃𝑚𝑖𝑥 ∗ 𝑇𝐵𝑢𝑏𝑏𝑙𝑒 − 𝑇𝑓 = 7640 + 0.438 ∗ 90 − 30 ∗ 83.6 = 9837.008 cal/gmole 𝑞 = 𝐻𝑉 − ℎ𝑓 𝜆 = 9837.008 7640 = 1.288
- 16. Solution (2) b. Number of ideal plates and feed location Slope of q-line = 𝒒 𝒒 −𝟏 = 𝟒. 𝟒𝟕𝟐 • Determine Rmin 𝑥𝐷 𝑅𝑚𝑖𝑛 + 1 = 0.51 Rmin = 0.92 R = 1.2 * Rmin = 1.104 𝑥𝐷 𝑅 + 1 = 0.466 0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1 y A xA
- 17. Solution (2) b. Number of ideal plates and feed location NTS = 17.7 + Reboiler 0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1 y A xA
- 18. Solution (2) c. Steam consumption if saturated steam at 3 atm is available. Q r = mst.lst = V’.lr lr = yr * lA + (1 - yr ) * lB lr = 0.07*7400 + 0.93*8000 = 7958 cal/gmole Slope of op. line in bottom section = 𝐿′ 𝑉′ = 1.25 L’ = V’ + W L’ = 200 kgmole/h & V’ = 160 kgmole/h Q r = V’.lr = 160 * 7958 *1000 = 1273.28*106 cal/h lst. = 517*18*1000 = 9306*103 cal/kmole Q r = mst.lst = 1273.28*106 mst. = 136.82 kmole = 2462.82 kg 0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1 y A xA
- 19. Solution (2) d. Cooling water consumption if cooling water at 30°C is available and left the system at 45°C. Qcond=V1 . lmix =mw . Cpw . ΔT V1 = Lo + D Lo = R * D = 1.104* 60 = 66.24 kmole V1 = 126.24 kmole/h lmix = (0.98*7400 + 0.02*8000) * 1000 = 7412*103 cal/kmole Qcond. = 935690*103 cal/h Cpw = 1 cal/g.°C ΔT = 15°C Qcond = mw . Cpw . ΔT mw = 62379.3 kg/h
- 20. Special cases in Binary systems 1. Enriching Section: • Feed is saturated vapor • NTS in bottom section = 0 XD XW XF
- 21. Special cases in Binary systems 2. Stripping Section: • Feed is saturated liquid • NTS in top section = 0 XD XW XF
- 22. Special cases in Binary systems 3. Complex Feed or Multiple feeds: • Feed (1) is saturated liquid • Feed (2) is saturated vapor • The slope of the middle section operating line = 𝐿′ 𝑉′ • Where, • L’ = F1 + L • V’ = V XD XF1 XW 1 R xD (XD,XD) (XW,XW ) XF2 L’/V’
- 23. Special cases in Binary systems 4. Top Side Product: • The top product is saturated liquid • The middle section operating line is determined using: 𝑥 = 𝑦 = 𝑆 . 𝑥𝑠 + 𝐷 . 𝑥𝐷 𝑆 + 𝐷 XD XS XW 1 R xD (XD,XD) (XW,XW) XF x=y
- 24. Special cases in Binary systems 5. Bottom Side Product: • The bottom product is saturated liquid • The middle section operating line is determined using: 𝑥 = 𝑦 = 𝑆 . 𝑥𝑠 + 𝑊 . 𝑥𝑊 𝑆 + 𝑊 XW 1 R xD (XD,XD) (XW,XW ) XS x=y XD XF
- 25. Special cases in Binary systems 6. Open Steam: • The bottom section operating line is determined by matching (xw, xw) with (x,y) that calculated using the following equation: 𝑥 = 𝑦 = 𝑊 . 𝑥𝑊 𝑊 − 𝑆 XD XF XW 1 R xD (XD,XD) (XW,ys) S W x W W .
- 26. Example (3) A continuous distillation is supplied with two equimolar feed streams, one is saturated liquid containing 40% benzene and the other is saturated vapor with 20% benzene. The required top product is 98% benzene while the bottom product is 2% benzene. Reflux ratio used is 4. a. Calculate the required number of theoretical stages and the correct feed location of each feed. b. If the two streams are to be mixed before entering the tower, find the required number of stages. Equilibrium data are given below: Mole % of benzene in liquid phase 0 10 20 30 40 50 60 70 80 90 95 100 Mole % of benzene in vapor phase 0 20.8 37.2 50.7 61.9 71.3 79.1 85.7 91.2 95.9 98 100
- 27. Solution (3) Assume F1 = F2 = 100 mole F1 + F2 = D + W = 200 (1) 𝐹1 . 𝑥𝑓1 + 𝐹2 . 𝑥𝑓2 = 𝐷 . 𝑥𝐷 + 𝑊 . 𝑥𝑤 (2) From (1) & (2) D = 58.3 mole W = 141.7 mole XD XF1 XW 1 R xD (XD,XD) (XW,XW ) XF2 L’/V’
- 28. Solution (3) 𝑥𝐷 𝑅+1 = 0.196 get the top section op. line L = D * R = 233.2 mole L’ = F1 + L = 333.2 mole V’ = V = L + D = 291.5 mole Slope of the middle section op. line = 𝐿′ 𝑉′ = 1.143 Get the middle section op. line Then calculate the number of stages XD XF1 XW 1 R xD (XD,XD) (XW,XW ) XF2 L’/V’
- 29. Solution (3) The final shape will be: 0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1 y A xA
- 30. Solution (3) b. If the two feeds are mixed together: Assume F1 = F2 = 100 mole F1 + F2 = F = 200 𝐹1 . 𝑥𝑓1 + 𝐹2 . 𝑥𝑓2 = 𝐹 . 𝑥𝑓 𝑥𝑓 = 0.3 Then begin to solve as mentioned in example 2 (b)