DESIGN OF A PACKED DISTILLATION COLUMN
NAME
: Sandaruwan R.A.N.
INDEX NO
: 090466M
DATE OF SUB: 01 / 08 / 2013
Task identification According to the determination of packed column height through the number of ideal stages and the HETP value, first the conditions and the objective of the column have to be clarrified using the data provided by the customer. So the required data has been provided as follows.
System: carbon disulphide – carbon tetrachloride Feed rate: 45kmol/hr
Feed condition: saturated liquid
Feed composition: 40mol% carbon disulphide Distillate composition: 90mol% carbon disulphide Bottom product composition: 4mol% carbon disulphide
Selection of packing material Principal requirements of a packing are, Provide a large surface area for high interfacial area between the gas and liquid Have an open structure, so low resistance to gas flow Promote uniform liquid distribution on the packing surface Promote uniform vapour gas flow across the column cross section To satisfy these requirements many diverse types and shapes have been developed. In this distillation column design, berl saddles types rings are used as packing material. Berl saddles were developed to give improved liquid distribution compared to rasching rings. These are the original type of saddle packing. They have a smaller free gas space but their aerodynamic shape is better giving a lower pressure drop and higher capacity. They are usually made up by ceramic or metal or carbon or plastic. The choice of material will depend on the nature of the fluids and the operating temperature. Ceramic packing will be the first choice for corrosive liquids and low cost. Metal packing are usually selected for non corrosive service. They have higher capacity and efficiency. Plastic are normally polypropylene. It is inexpensive and most popular when temperature does not exceed 250˚F. In the process industries, random packings re more commonly used. So for this design plastic berl saddles are selected and random packing is used. Packing size In general the largest size of packing should be used upto 50mm. small sizes are more expensive than the larger sizes. The size of packing used influences the height and diameter of a column, the pressure drop and cost of packing. When packing size is increased, the cost per unit volume of packing and the pressure drop per unit height of packing are reduced, but in other side which will reduce the mass transfer efficiency. Reduced mass transfer efficiency results in a taller column. Also use of too large a size in a smaller column can cause poor liquid distribution.
In this design packing size is selected as 25mm to maintain the column diameter below 0.3m. Pressure drop across packing Recommended design values, mm water per m packing
According to this, 60mm water per m of packing is selected as pressure drop.
Calculating equilibrium data Equilibrium data calculation for Carbon disulphide – Carbon tetrachloride vapour – Liquid system According to antion equation 𝑙𝑛 𝑃˚ = 𝐴 − 𝐵/(𝑇 + 𝐶) Where,
A, B, C
- Constants in the antion equation
Vp
- Vapour pressure (mmHg)
T
- Temperature (K)
Antion constants for Chloroform and Benzene Carbon disulphide (A) 15.9844 2690.85 -31.62
A B C Boling point of Carbon disulphide
Carbon tetrachloride (B) 15.8742 2808.19 -45.99
= 46˚C
Boling point of Carbon tetrachloride
=77˚C
From Raoul’s law, P = Pox
Where,
P - Partial pressure
Po – Vapour pressure X – Molar fraction in the liquid phase Applying Raoult’s law for A and B; PA = P0AxA PB = PoBxB From Daltons Law; p A + pB = P T Where,
PT – Total pressure
pA + pB = PT = pOAxA + poBxB PT = 750 mm Hg (1bar) xA + xB = 1 pA + pB = PT = p0AxA + poB (1 - xA)
xA =
P T - p0 B ---------p0A - poB
From Dalton’s law; PA = yAPT
yA = (pOA/PT) xA
Sample calculation At 46 ˚C 𝑙𝑛 𝑃˚𝐴 = 𝐴 − 𝐵/(𝑇 + 𝐶) = 15.9844-2690.85/(319-31.62) 𝑙𝑛 𝑃˚𝐴 = 6.621 𝑃˚𝐴 = 750.695Hgmm 𝑙𝑛 𝑃˚𝐵 = 𝐴 − 𝐵/(𝑇 + 𝐶) =15.8742-28.08.19/(319-45.99) 𝑙𝑛 𝑃˚𝐵 = 5.588 𝑃˚𝐵 =267.2Hgmm xA =
P T - p0 B ---------p0A - poB
yA = (pOA/PT) xA =0.999
=
0.998
Equilibrium data
Temperature (˚C) 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77
P˚A 750.71 775.48 800.89 826.95 853.68 881.07 909.15 937.92 967.40 997.59 1028.52 1060.19 1092.61 1125.80 1159.76 1194.52 1230.08 1266.45 1303.64 1341.68 1380.57 1420.32 1460.95 1502.47 1544.89 1588.22 1632.49 1677.69 1723.84 1770.96 1819.06 1868.16
P˚B 267.25 277.47 288.00 298.86 310.04 321.55 333.41 345.61 358.17 371.09 384.37 398.04 412.09 426.54 441.38 456.63 472.30 488.39 504.92 521.88 539.29 557.16 575.50 594.30 613.59 633.37 653.65 674.44 695.74 717.57 739.93 762.83
XA
YA 1.00 0.95 0.90 0.85 0.81 0.77 0.72 0.68 0.64 0.60 0.57 0.53 0.50 0.46 0.43 0.40 0.37 0.34 0.31 0.28 0.25 0.22 0.20 0.17 0.15 0.12 0.10 0.08 0.05 0.03 0.01 -0.01
1.00 0.98 0.96 0.94 0.92 0.90 0.88 0.85 0.83 0.80 0.78 0.75 0.72 0.69 0.66 0.63 0.60 0.57 0.53 0.50 0.46 0.42 0.38 0.34 0.30 0.26 0.21 0.17 0.12 0.07 0.02 -0.03
Equilibrium curve
Reflux ratio Determination of minimum reflux ratio (Rm) q – line equation
yq
q 1 . xq . xf q 1 q 1
q
Heat required to vapourise 1 mole of feed Molar latent heat of the feed
In this case feed is a saturated liquid so feed is at its boiling point If Feed at boiling point q = 1 and q – line is passing through the point of (Xf , Xf) 𝑋𝐹 = 0.4
Top operating line equation y n 1
R 1 .x n .x 1 R 1 R D
But, at minimum reflux ratio we can rewrite top operating line equation as follows, y n 1
Rm 1 .x n .x 1 Rm 1 Rm D
Top operating line is passing through the point of (XD , XD) XD = 0.9 In minimum reflux ratio intersection of the top operating line and Q line should be at the curve.
According to the graph, Intercept of the top operating line = 0.414 Gradient of the top operating line
= 0.54
Therefore
Rm 1 Rm
Minimum reflux ratio,
= 0.54
Rm = 1.174
Operating reflux ratio When we operate high reflux ratio our plates are less than high reflux ratio. Then construction cost is less. But we have to spend lot of money on operating cost and steams. So in industry reflux ratio is between 1.2-1.5 of minimum reflux ratio. Here distill concentration is high so I think operating reflux ratio should be 1.3Rm R = 1.3Rm
Then operating reflux ratio
= 1.3 * 1.174 R = 1.5262
HETP Value For the design of packed distillation columns it is simpler to treat the separation as a stage process and use the concept of the height of an equivalent equilibrium stage to convert the number of ideal stage required to a height of packing. The height of an equivalent equilibrium stage, called height of a theoretical plate (HETP), is the height of packing that will give the same separation as an equilibrium stage. The HETP for a given type and size of packing is essentially constant and independent of the system physical properties.
HETP value for a packing size of 25 mm can be assumed as a 0.46 m.
Number of ideal stages required Then top operating line equation, 𝑦𝑛 +1 = 0.6041 𝑥𝑛 + 0.3563 q = 0.4
No of ideal stage required = 12-1 = 11 Feed tray location is above the 5th tray from the top of the tower Column height Column height = number of stages * HETP = 11* 0.46 m = 5.06 m Column height = 5.1m
Column diameter Material balance F=D+W Material balance for carbon disulphide
F.𝑥𝑓 = D𝑥𝑑 + W𝑥𝑤 F.𝑥𝑓 = D𝑥𝑑 + (F-D)𝑥𝑤 D=
D=
𝐹(𝑥 𝑓 −𝑥 𝑤 ) 𝑥 𝑑 −𝑥 𝑤 45(0.4−0.04) 0.9−0.04
kmol/hr
D = 18.837 kmol/hr Then, W = (45 – 18.837)kmol/hr = 26.163 kmol/hr Material balance for rectifying section
L+D=G
R=
𝐿 𝐷
1.5262 =
𝐿 18.837
L = 28.75 kmol/hr So, 28.75 + 18.375 = G G = 47.125kmol/hr Distillate (D) = 18.837 kmol/hr Down ward liquid flow rate (L) = 28.75 kmol/hr Upward vapour flow rate (G) = 47.125 kmol/hr Material balance for stripping section W + 𝑉 ′ = 𝐿′ According to BOL,
𝑦𝑚 +1 =
𝐿′ 𝐿′ −𝑤
𝑥𝑚 −
𝑤 𝐿′ −𝑤
𝑥𝑤
Gradient = 1.55 and W = 26.163kmol/hr 𝐿′ 𝐿′ −𝑤
= 1.55
𝐿′ = 73.732kmol/hr 𝑉 ′ = 73.732 – 26.163 𝑉 ′ = 47.569kmol/hr Bottom product (W) = 26.163kmol/hr Upwardvapour flow rate in stripping section (𝑉 ′ ) = 47.569kmol/hr section Downward liquid flow rate in stripping (𝐿′ ) = 73.732kmol/hr
Vapour liquid equilibrium mixture for carbon disulphide and carbon tetrachloride
For rectifying section Composition of the 4th tray (above plate to the feed) 𝑦5 = 0.62 𝑥4 = 0.43 So, temperature of the 4th tray = 333K Composition of the 1st tray (top plate to the feed) 𝑦1 = 0.9 𝑥0 = 0.9 Temperature of the 1st tray = 324K Average temperature of the striping section =
333+324 2
K
= 328.5K Average carbon disulphide composition of vapour (y) = 0.760 Average carbon disulphide composition of liquid (x) = 0.665
Density
Molecular weight of C𝑆2 = 0.0761kg/mol Molecular weight of C𝐶𝑙4 = 0.1538kg/mol For rectifying section, Density of C𝑆2 at 328.5K = 1207.566kg/𝑚3 Density of C𝐶𝑙4 at 328.5K = 1526.596 kg/𝑚3 For liquid Weight fraction of C𝑆2 =
0.665×0.0761 0.665×0.0761+0.335×0.1538
= 0.4955 Basis 1000kg, 0.4955×1000 Volume of C𝑆2 = 1207.566
= 0.41𝑚3 Volume of C𝐶𝑙4 =
0.5045×1000 1526.596
= 0.33𝑚3 Density of liquid mixture =
1000 0.41+0.33
= 1351.35kg/𝑚3 For vapour mixer According to, PV = nRT
ρ=
𝑃𝑀 𝑅𝑇
Average molecular weight of the vapour stream = 0.76×0.0761+0.24×0.1538 = 0.09475kg/mol
ρ=
1×10 5 ×94.75 8.314×328.5
Density of vapour mixture = 3.47kg/𝑚3 For stripping section Composition of 6th plate (below plate to the feed tray) 𝑦6 = 0.58
𝑥5 = 0.38 Temperature of the 5th tray = 334 K Composition of 11th plate (lowest plate of tower) 𝑦12 = 0.07 𝑥11 = 0.06 Temperature of the 11th plate = 347 K Average temperature for stripping section =
334+347 2
= 340.5 K Average carbon disulphide composition of vapour (y) = 0.325 Average carbon disulphide composition of liquid (x) = 0.22 Density of C𝑆2 at 340.5K = 1187.885 kg/𝑚3 Density of C𝐶𝑙4 at 340.5K = 1503.208kg/𝑚3 Similar to above calculation, For liquid mixer Weight fraction of the C𝑆2 = 0.1225 Density of liquid mixture = 1455.8kg/𝑚3 For vapour mixer Average molecular weight of the mixer = 0.12855kg/mol Density of vapour mixer = 4.541kg/𝑚3
Column diameter for rectifying section Gas flow rate = 47.125 kmol/hr =
47.125×0.09475×103 3600
kg/s
= 1.24 kg/s Average molecular weight of liquid mixer = 0.665×0.0761+ 0.335×0.1538 = 0.1021kg/mol Liquid flow rate = 28.75kmol/hr =
28.75×0.1021×10 3 3600
kg/s
= 0.8154kg/s Gas density at 328.5K = 3.47kg/𝑚3 Liquid density at 328.5K = 1351.35kg/𝑚3
𝐹𝐿𝑉 = =
𝐿𝑤
ρ𝑣
𝑉𝑤
ρ𝐿
0.8154
3.47
1.24
1351.35
= 0.033 Pressure drop correlation graph
Pressure drop = 60mm water per m of packing According to the graph, 𝐾4 = 2.2 At flooding, 𝐾4 = 5.6 2.2
Percentage flooding =
5.6
= 0 .627
62.7% satisfactory Viscosity calculation 1 Log [viscosity] = [VISA] × [ 𝑇
−
1 𝑉𝐼𝑆𝐵
]
For carbon disulphide VISA = 274.08 VISB = 200.22 T = 328.5k 1 Log [viscosity] = [274.08] × [ 328.5
−
1 200.22
]
µ1 = 0.292 mNs/𝑚2 For carbon tetrachloride VISA = 540.15 VISB = 290.84 T = 328.5k 1 Log [viscosity] = [540.15] × [ 328.5
−
1 290.84
]
µ2 = 0.6125 mNs/𝑚2 Average viscosity of mixer 1
µ𝑚𝑖𝑥 = ( 𝑥𝑖 × µ𝑖 3 )3 1
1
µ𝑚𝑖𝑥 = (0.665 × 0.2923 + 0.335 × 0.61253 )3 = 0.3821 mNs/𝑚2
𝑉𝑤 = [
𝐾4 ρ 𝑣 (ρ 𝐿 −ρ 𝑣 ) 13.1F 𝑃 (µ 𝐿 /ρ 𝐿 )0.1
For this type packing material (plastic), 𝐹𝑝 = 170
]
𝑉𝑤 = [
2.2×3.47(1351.35−3.47) 13.1×170(0.3821/1351.35)0.1
]
= 10.46kg/𝑚2 𝑠 Column area required =
1.24 10.46
= 0.1185𝑚2 4
Diameter =
𝜋
× 0.1185
Diameter of the rectifying section= 0.388m
Column diameter for stripping section Gas flow rate = 47.569kmol/hr =
47.569×0.12855×10 3 3600
kg/s
= 1.7 kg/s Average molecular weight of liquid mixer = 0.22×0.0761+ 0.78×0.1538 = 0.1367kg/mol Liquid flow rate = 73.732kmol/hr =
73.732×0.1367×10 3 3600
= 2.8kg/s Gas density at 340.5K = 4.541kg/𝑚3 Liquid density at 340.5K = 1455.8kg/𝑚3
𝐹𝐿𝑉 = =
𝐿𝑤
ρ𝑣
𝑉𝑤
ρ𝐿
2.8
4.541
1.7
1455.8
= 0.092 From the graph, 𝐾4 = 1.6 At flooding, 𝐾4 = 3.8 Percentage flooding =
1.6 3.8
= 0 .6488
64.88% satisfactory
kg/s
Viscosity calculation Similar to above calculation, T = 340.5K For carbon disulphide
µ1 = 0.273mNs/𝑚2 For carbon tetrachloride
µ2 = 0.536mNs/𝑚2 Average viscosity of mixer, 1
1
µ𝑚𝑖𝑥 = (0.22 × 0.2733 + 0.78 × 0.5363 )3 = 0.468 mNs/𝑚2
𝑉𝑤 = [
3.8×4.541(1455.8−4.541) 13.1×170(0.468/1455.8)0.1
]
= 24.635kg/𝑚2 𝑠 Column area required =
1.7 24.635
Diameter =
= 0.069𝑚2 4 𝜋
× 0.069
Diameter of the stripping section= 0.296m
Feed tray location Feed is given to the 7th tray from the bottom, So, it can be assumed as feed is fed into column at 7.5th stage. Height of the feed location from the bottom of column = 7.5*HETP = 3.45 m
Condenser heat load Latent heat calculation
Latent heat of C𝑆2 = 351kJ/kg Latent heat of C𝐶𝑙4 = 194kJ/kg 𝐿𝑣,𝑏 26711.1 29837.2
C𝑆2 C𝐶𝑙4
𝑇𝑏 319.3 349.8
For top products Temperature of top plate = 322K Carbon disulphide, 552−322 0.38 𝐿𝑣 = 26711.1× [ ] 552−319.3
= 26593kJ/kmol Carbon tetrachloride, 556.35−322 0.38 𝐿𝑣 = 29837.2× [ ] 556.35−349.8
= 31303.8kJ/kmol Latent heat of mixtures
For rectifying section, 𝐿𝑣,𝑚𝑖𝑥 = 26593×0.9 + 31303.8×0.1 = 27064.08 kJ/kmol Therefore condenser heat load = G× 𝐿𝑣,𝑚𝑖𝑥 = 47.125×27064.08 = 1275.4MJ/hr
𝑇𝑐 552 556.35
= 354.3kW Boiler heat load calculation Using equation, 𝑉𝑚 𝐻𝑣,𝑚 + W𝐻𝑤 = 𝐿𝑚 −1 𝐻𝐿,𝑚 −1 + 𝑄𝑅 𝐻𝑣 = 𝐻𝐿 + 𝜆𝑚𝑖𝑥 𝑇2 (𝐶1 𝑇1
𝐻𝐿 =
+ 𝐶2 𝑇 + 𝐶3 𝑇 2 + 𝐶4 𝑇 3 )𝑑𝑇
𝐶1 𝐶2 𝐶3 85600 -122 0.5605 C𝑆2 -752700 8966.1 -30.394 C𝐶𝑙4 Average temperature of stripping section, T = 340.5K
𝐶4 -0.001452 0.034455
Calculating the latent heat of mixture For stripping section Temperature of bottom plate = 348K
For Carbon disulphide, 𝐻𝐿 = =
𝑇2 (𝐶1 𝑇1
+ 𝐶2 𝑇 + 𝐶3 𝑇 2 + 𝐶4 𝑇 3 )𝑑𝑇
340 .5 (85600 − 298
122T + 0.5605𝑇 2 − 0.001452𝑇 3 )𝑑𝑇
= 2397.325kJ/kmol
552−340.5 0.38 𝐿𝑣 = 26711.1× [ ] 552−319.3
= 25758.8kJ/kmol
Carbon tetrachloride, 𝐻𝐿 =
340 .5 (−752700 + 298
8966.1T − 30.394𝑇 2 + 0.034455𝑇 3 )𝑑𝑇
= 5671.523kJ/kmol
556.35−340.5 0.38 𝐿𝑣 = 29837.2× [ ] 556.35−349.8
= 30340.7kJ/kmol
For stripping section, 𝐿𝑣,𝑚𝑖𝑥 = 25758.8×0.22 +30340.7×0.78 Latent heat of mixture = 29332.7 kJ/kmol Enthalpy of liquid mixture(𝐻𝐿,𝑚𝑖𝑥 ) = 2397.325×0.22 + 5671.523×0.78 = 4951.2 kJ/kmol Enthalpy of carbon disulphide, 𝐻𝑣 = 2397.325 + 29332.7 = 31730kJ/kmol Enthalpy of carbon disulphide, 𝐻𝑣 = 5671.523 + 29332.7 = 35004.2kJ/kmol Enthalpy of vapour mixture (𝐻𝑣,𝑚𝑖𝑥 ) = 31730×0.325 + 35004.2×0.675 = 33940.1 kJ/kmol
For bottom product (residue), Temperature = 348K For Carbon disulphide, 𝐻𝐿 =
348 (85600 − 298
122T + 0.5605𝑇 2 − 0.001452𝑇 3 )𝑑𝑇
= 2778.2kJ/kmol Carbon tetrachloride, 𝐻𝐿 =
348 (−752700 + 298
8966.1T − 30.394𝑇 2 + 0.034455𝑇 3 )𝑑𝑇
= 6703.6kJ/kmol
Enthalpy of liquid mixture of residue(𝐻𝑤 ) = 2778.2 ×0.04 + 6703.6×0.96 = 6546 kJ/kmol
Substituting to equation, 𝑉𝑚 𝐻𝑣,𝑚 + W𝐻𝑤 = 𝐿𝑚 −1 𝐻𝐿,𝑚 −1 + 𝑄𝑅 47.569×33940.1 + 26.163×6546 = 73.732×4951.2 + 𝑄𝑅 𝑄𝑅 = 1420.7MJ/hr = 394.6kW Heat load of the reboiler = 394.6kW
Final data Packing material
plastic
Packing size
25mm
Pressure drop
62water mm/ m of packing
Reflux ratio
1.5262
Number of ideal stages
11
Column height
5.1m
Diameter of rectifying
0.388m
Diameter of stripping
0.296m
Feed tray location
3.45m from bottom
Condenser heat load
354.3kW
Reboiler heat load
394.6kW
Referencess
http://www.engineeringtoolbox.com/fluids-evaporation-latent-heat-d_147.html http://www.britannica.com/EBchecked/topic/94993/carbon-disulfide-CS2 http://encyclopedia2.thefreedictionary.com/Carbon+Tetrachloride CULSON & RICHARDSON’S, CHEMICAL ENGINEERING, Volume 6, 3 rd edition, R K Sinnott CULSON & RICHARDSON’S, CHEMICAL ENGINEERING, Volume 2, 5 th edition PERRY’S CHEMICAL ENGINEERS HAND BOOK, 8th edition, DON W. GREEN, ROBERT H. PERRY DISTILLATION DESIGN, HENRY Z. KISTER
