![Solution (1)
e) Complete tower design: (For Bottom section)
1- Check weeping:
𝒖h =
[𝐾2 − 0.9(25.4 − 𝑑ℎ)]
(𝜌𝑣)
1
2
hw + how = 69.58 mm
From chart: K2=30.6 , 𝑢h =14 m/s
Actual min. vapor velocity =
𝑀𝑖𝑛.𝑣𝑎𝑝𝑜𝑟 𝑟𝑎𝑡𝑒
𝐴ℎ𝑜𝑙𝑒𝑠
=
0.7 ∗ 1.13
0.046
= 17.2 m/s < ŭh
So, no weeping occurs](https://image.slidesharecdn.com/section7-multistageseparationprocesses-210801203905/85/Section-7-multistage-separation-processes-25-320.jpg)
Section 7 multistage separation processes
- 1. MULTISTAGE SEPARATION PROCESSES CHE 452 ENG. AMAL MAGDY SECTION (7)
- 2. Design of Tray Distillation Columns • Design considerations: • Vapor bad distribution • Liquid height on plate • Weeping • Pressure drop • Downcomer flooding • Downcomer residence time • Entrainment Vapor Bad Distribution -When column diameter is large, gas is not distributed well over the plate causing bad contact that leads to bad separation. -Solution is to divide the tray into two part which is called "Split flow” Liquid height on plate -In case of small liquid height on plate, any change in that height may lead to push the liquid back and gas jets through the holes. It is also called “Coning”. Weeping -When liquid flow rate is too high, liquid weeps through plates causing bad separation efficiency & high pressure drop Downcomer flooding -It’s the accumulation of liquid (or foam) in downcomer till it reaches the above plate. This is due to the increased liquid flow rate. Downcomer residence time -It should be enough to separate liquid from gas (foam). If there is no sufficient time in down comer, this is due to high liquid flow rate. Excessive Entrainment -It takes place when gas flow rate is too high that entrains liquid droplets to the upper plate. -Due to that case, the upper plate concentration changes so, separation efficiency decreases. -Excessive entrainment may lead to flooding!!
- 3. Calculation steps It includes: • Material balance • For top and bottom section separately • Column diameter calculations • Checks in column performance
- 4. Material Balance • For the whole tower • OMB F = D + W • CMB 𝐹. 𝑥𝑓 = 𝐷. 𝑥𝐷 +𝑊. 𝑥𝑤 • For top section • V = L + D • 𝑅 = 𝐿 𝐷 • For bottom section • L’ = V’ + W • 𝐿′ 𝑉′ = 𝑠𝑙𝑜𝑝𝑒 𝑜𝑓 𝑏𝑜𝑡𝑡𝑜𝑚 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 𝑜𝑝𝑒𝑟𝑎𝑡𝑖𝑛𝑔 𝑙𝑖𝑛𝑒 • Actual number of plates = Theoretical number of plates/efficiency • 𝑁𝐴𝑆 = 𝑁𝑇𝑆 𝜂
- 5. Diameter calculations 𝑨𝒄 = 𝝅 𝟒 𝑫𝒄 𝟐 • 𝑨𝒄 : Cross sectional area of column, m2 • 𝑫𝒄 : Column diameter, m • 𝑨𝒅 : Area of downcomer, ranges from 10 – 12 % of the column cross sectional area , m2 𝑨𝒏 = 𝑨𝒄 − 𝑨𝒅 • 𝑨𝒏 : Net area available for vapor-liquid disengagement, m2 𝑨𝒂 = 𝑨𝒄 − 𝟐 𝑨𝒅 • 𝑨𝒂 : Active area (bubbling area), m2
- 6. Diameter calculations 𝑸𝒈 = 𝒖𝒈 ∗ 𝑨𝒏 • 𝑸𝒈 : Maximum gas flowrate, m3/s • 𝑨𝒏 : Net area available for vapor-liquid disengagement, m2 • 𝒖𝒈 : Maximum gas velocity ranges from 80 to 85 % of the flooding vapor velocity, m/s 𝒖𝒈 = 𝟎. 𝟖𝟓 ∗ 𝒖𝒇 • 𝒖𝒇 : Flooding vapor velocity 𝒖𝒇 = 𝑲𝟏 . 𝝆𝒍 − 𝝆𝒗 𝝆𝒗 • 𝐾1 : It is a constant determined from the following chart
- 7. Diameter calculations • 𝑭𝑳𝑽 = 𝑳𝑾 ′ 𝑽𝑾 ′ . 𝝆𝒗 𝝆𝒍 • 𝑭𝑳𝑽 : The liquid-vapor flow factor • 𝑳𝑾 ′ : Liquid mass flowrate, kg/s • 𝑽𝑾 ′ : Vapor mass flowrate, kg/s • This chart is used directly when surface tension of the liquid mixture (𝜎) = 0.02 N/m , else multiply K by ( 𝜎 0.02 )0.2
- 8. Liquid flow pattern • After calculating the column diameter, the arrangement of flow inside the distillation tower is determined using this chart • Notice that: liquid flow here is used as volumetric flow rate
- 9. 1. Weeping check 𝒖𝒉 > 𝒖h • 𝒖𝒉 : Actual vapor velocity, m/s • 𝒖h : minimum vapor velocity through holes, m/s 𝒖𝒉 = 𝑸𝒈𝒎𝒊𝒏 𝑨𝒉 • Ah : total hole area, m2 𝑨𝒉 𝑨𝒂 = 𝟎. 𝟗 ( 𝒅𝒉 𝒑′ )𝟐 • Ah : Total holes area, m2 • 𝒅𝒉 : Hole diameter, 2.5 – 12 mm (5 mm is the most recommended) • p' : Pitch between holes, ranges from 2.5 do to 5 do
- 10. 1. Weeping check 𝒖h = 𝑲𝟐 −𝟎.𝟗 (𝟐𝟓.𝟒 − 𝒅𝒉) 𝝆𝒗 𝟎.𝟓 • dh : hole diameter, mm • K2 : a constant depends on the height of liquid above plate • 𝝆v : Vapor density, kg/m3 Now hw and how are unknown ??!
- 11. 1. Weeping check • Weir length ( lw ) is determined using this chart • Then by knowing the minimum flowrate of liquid, the height of weir crest is calculated 𝑸𝒍𝒎𝒊𝒏 = 𝟐 𝟑 𝑪𝑫 . 𝒉𝒐𝒘 . 𝒍𝒘 𝟐 𝒈. 𝒉𝒐𝒘 • 𝑸𝒍𝒎𝒊𝒏 : minimum liquid flowrate, m3/s • CD : coefficient of discharge = 0.62 • lw : weir length, m • how : weir crest (height of liquid over the weir), m • hw : weir height • For atmospheric distillation: hw = 40 – 90 mm (40 – 50 mm is recommended) • For vacuum distillation: hw = 6 – 12 mm
- 12. 2. Entrainment • H’ should be in inch • Vg: Vapor velocity, ft/s • Qg : maximum vapor flowrate, m3/s • H: Plate spacing, mm • how it is calculated at the maximum liquid flowrate, mm • AP : Active area
- 13. 3. Calculating the pressure drop • ∆𝑃𝑃 for plate should be less than 10 cm.water (1 atm = 1033 cm.water) • hd : the dry pressure drop, m • 𝛽 : foam density = 0.4 – 0.6 ≈ 0.5 hd = 2 * 𝑽𝒈𝒉 𝟐 𝟐∗𝒈 ∗ 𝝆𝒗 𝝆𝒍 𝑽𝒈𝒉 = 𝐐𝒈𝒎𝒂𝒙 𝑨𝒉 • Vgh : maximum velocity of vapor through holes, m/s
- 14. 4. Downcomer flooding • Hdc : height of liquid in down comer, mm hdc = 3.2 * 𝑽𝑳 𝟐 𝟐∗𝒈 𝑽𝑳 = 𝐐 𝒍 𝒎𝒂𝒙 𝒍𝒘 𝒉𝒂𝒑 • hap = hw – 1 cm
- 15. 5. Downcomer residence time 𝒕 = 𝑽𝑫 𝑸𝒍 𝒎𝒂𝒙 > 𝟑 𝒔 • VD : Volume of liquid in downcomer VD = Ad * HDC
- 16. SHEET (5)
- 17. Example (1) 10,000 kg/h of acetone is to be recovered from an aqueous waste stream by continuous distillation using sieve plates. The feed contains 3.33% acetone. Acetone of at least 94% purity is required in the top, and bottom product is pure water. Reflux ratio equals 1.35 and the theoretical number of stages equal 15. The feed will be at 20°C. a. Perform material balance to get all the missing amounts in the tower. b. Estimate the number of actual plates if the overall efficiency is 60%. c. Calculate the column diameter. d. Determine the liquid flow pattern. e. Perform a complete plate design.
- 18. Example (1) Additional Data: 1. The maximum feed rate is 130% of feed and the minimum feed rate is 70 % of the maximum 2. L’/V’=5 3. For Top section: ρv = 2.05 kg/m3 , ρL = 733 kg/m3 , σ = 23*10-3 N/m 4. For Bottom Section: ρv = 0.72 kg/m3 , ρL = 954 kg/m3 , σ=57 *10-3 N/m 5. Take ug= 0.85 uf , Ad = 12 %Ac , Ah = 10 %Ac 6. Take the tray spacing 50 mm. 7. Hole diameter is 5 mm. 8. Mwt of acetone = 58 kg/kmole
- 19. Solution (1) a) Perform material balance to get all the missing amounts in the tower. Mwt)feed = 0.033 * 58 + (1 - 0.033)*18 = 19.32 kg/kmole F = 10,000/19.32 = 517.6 kmole/h 𝐹. 𝑥𝑓 = 𝐷. 𝑥𝐷 +𝑊. 𝑥𝑤 D=18.2kmole/h F = D + W W = 499.4 kmole/h For top section: V=D(1+R) = 42.7 kmole/h , L = D . R = 24.5 kmole/h For bottom section: L’ = V’ + W & L’/V’=5 L’ = 624.1 kmole/h , V’ =124.85 kmole/h b) Estimate the number of actual plates if the overall efficiency is 60%. NAS = 15/0.6=25
- 20. Solution (1) c) Calculations of column diameter: 𝐹𝑙𝑣 𝑏𝑜𝑡𝑡𝑜𝑚 = 𝐿𝑤 𝑉 𝑤 𝜌𝑣 𝜌𝑙 = 5 ∗ 0.72 954 = 0.14 𝐹𝑙𝑣 𝑡𝑜𝑝 = 𝐿𝑤 𝑉 𝑤 𝜌𝑣 𝜌𝑙 = 0.57 ∗ 2.05 733 = 0.03 Given that: H = 50 mm From chart: K1 bottom=7.5*10-2 K1 top=9*10-2
- 21. Solution (1) c) Calculations of column diameter cont’d: But these values are determined based on (𝜎) = 0.02 N/m, so new K1-values 𝐾1 𝑏𝑜𝑡𝑡𝑜𝑚 = ( 0.057 0.02 )0.2𝑥7.5𝑥10−2 = 9.3∗10−2 𝐾1 𝑡𝑜𝑝 = ( 0.023 0.02 )0.2𝑥9𝑥10−2 = 9.3∗10−2 𝑢𝑓 = 𝐾1 𝜌𝐿 − 𝜌𝑉 𝜌𝑉 𝑢𝑓 𝑏𝑜𝑡𝑡𝑜𝑚 = 3.38 𝑚/𝑠 , 𝑢𝑓 𝑡𝑜𝑝 = 1.76 𝑚/𝑠 Given that: ug=0.85 uf 𝑢𝑔 𝑏𝑜𝑡𝑡𝑜𝑚 = 2.87 𝑚/𝑠 , 𝑢𝑔 𝑡𝑜𝑝 = 1.5 𝑚/𝑠
- 22. Solution (1) c) Calculations of column diameter cont’d: • To get column diameter, the maximum vapor flow rate is calculated: Qg bottom = 1.3∗124.85 ∗18 0.72∗3600 = 1.13 m3 /s Qg top = 1.3 ∗ 42.7 ∗ 55.6 2.05 ∗ 3600 = 0.42 m3/s • To get net area required: 𝐀𝐧 = 𝐐𝐠 𝐔𝐠 An bottom = 0.4 m2 , An top = 0.28m2 𝑨𝒏 = 𝑨𝒄 − 𝑨𝒅 & Ad = 12 % Ac Ac bottom = 0.46 m2 , Ac top = 0.32 m2 𝑨𝒄 = 𝝅 𝟒 𝑫𝒄 𝟐 Dbottom = 0.77 m , Dtop= 0.64 m
- 23. Solution (1) d) Determine the liquid flow pattern 𝑀𝑎𝑥. 𝑣𝑜𝑙𝑢𝑚𝑒𝑡𝑟𝑖𝑐 𝑙𝑖𝑞𝑢𝑖𝑑 𝑟𝑎𝑡𝑒 = 1.3𝑥624.1𝑥18 954𝑥3600 = 4.3 ∗ 10−3 𝑚3/𝑠𝑒𝑐 So, a single pass could be used (normal flow)
- 24. Solution (1) e) Complete tower design: (For Bottom section) 1- Check weeping: Assume hw = 50 mm QL min = 0.7 * QL max = 0.7 * 4.3 * 10-3= 3.01*10-3 𝑚3 /𝑠 From figure: Lw/Dc = 0.76 Lw = 0.6 m 𝑸𝒍𝒎𝒊𝒏 = 2 3 𝐶𝐷 . ℎ𝑜𝑤 . 𝑙𝑤 2 𝑔. ℎ𝑜𝑤 how = 19.58 mm
- 25. Solution (1) e) Complete tower design: (For Bottom section) 1- Check weeping: 𝒖h = [𝐾2 − 0.9(25.4 − 𝑑ℎ)] (𝜌𝑣) 1 2 hw + how = 69.58 mm From chart: K2=30.6 , 𝑢h =14 m/s Actual min. vapor velocity = 𝑀𝑖𝑛.𝑣𝑎𝑝𝑜𝑟 𝑟𝑎𝑡𝑒 𝐴ℎ𝑜𝑙𝑒𝑠 = 0.7 ∗ 1.13 0.046 = 17.2 m/s < ŭh So, no weeping occurs
- 26. Solution (1) e) Complete tower design: (For Bottom section) 2- Check entrainment: Ap = Aa = Ac – 2Ad = 0.76 Ac = 0.3496 m2 Ap = π 4 Lp 2 , Lp = 0.667 m For the maximum liquid flow rate, get how = 24.84 mm Δh = 29.76 mm vg = Qg π 4 D2 = 1.13 π 4 (0.77)2 = 2.4266 m/s = 7.96 𝑓𝑡/𝑠 H’= 500-2.5(24.84+50+(29.76/2))=275.725mm=10.85” E = 0.08 >0.1 No entrainment ow w p Δh = 0.001 (0.85 h +0.47 h ) L
- 27. Solution (1) e) Complete tower design: (For Bottom section) 3- Check pressure drop: In this section all calculations are done using the maximum flowrates: We already calculate at maximum flowrate conditions how = 24.84 mm , ∆h = 29.756 mm 𝑽𝒈𝒉 = 𝐐𝒈𝒎𝒂𝒙 𝑨𝒉 & hd = 2 ∗ 𝑽𝒈𝒉 𝟐 𝟐 ∗ 𝒈 ∗ 𝝆𝒗 𝝆𝒍 vgh = 1.13 / 0.046 = 24.565 m/s & hd = 0.0464 m ∴ hp = 0.0912 m & ∆P = 853.7 Pa ∆P = 853.7 Pa * 1033 𝑐𝑚.𝑤𝑎𝑡𝑒𝑟 1.013 ∗ 105 𝑃𝑎 = 8.7 cm.water < 10 cm.water Acceptable pressure drop
- 28. Solution (1) e) Complete tower design: (For Bottom section) 4- Check downcomer flooding: Lw = 0.6 m hap = hw – 1 cm = 40 mm 𝑽𝑳 = 𝐐 𝒍 𝒎𝒂𝒙 𝒍𝒘 𝒉𝒂𝒑 & hdc = 3.2 * 𝑽𝑳 𝟐 𝟐∗𝒈 VL = 0.18 m/s & hdc = 5.2 mm Hdc = 186.7 mm H + hw = 550 mm > 2 Hdc Therefore, no flooding will take place in the downcomer
- 29. Solution (1) e) Complete tower design: (For Bottom section) 5- Check downcomer residence time: In this section all calculations are done using the maximum flowrates: VD = Ad * Hdc = 0.12 Ac * Hdc = (0.12*0.46) * 186.17/1000 = 0.01 m3 t = 𝑽𝑫 𝑸𝒍 𝒎𝒂𝒙 = 2.39 s < 3 s So, the residence time isn’t satisfactory