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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 β 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 250ΛF.
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 β Carbon tetrachloride vapour β Liquid system
According to antion equation
ππ πΛ = π΄ β π΅/(π + πΆ)
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) Carbon tetrachloride (B)
A 15.9844 15.8742
B 2690.85 2808.19
C -31.62 -45.99
Boling point of Carbon disulphide = 46ΛC
Boling point of Carbon tetrachloride =77ΛC
From Raoulβs law,
P = Pox Where, P - Partial pressure
Po β Vapour pressure
X β Molar fraction in the liquid phase
Applying Raoultβs law for A and B;
PA = P0
AxA
PB = PoBxB
From Daltons Law;
pA + pB = PT
Where, PT β Total pressure
pA + pB = PT = pO
AxA + poBxB
PT = 750 mm Hg (1bar)
xA + xB = 1
pA + pB = PT = p0
AxA + poB (1 - xA)
PT - p0
B
xA = ----------
p0
A - poB
From Daltonβs law;
PA = yAPT
yA = (pO
A/PT) xA
Sample calculation
At 46 ΛC
ππ πΛπ΄ = π΄ β π΅/(π + πΆ)
= 15.9844-2690.85/(319-31.62)
ππ πΛπ΄ = 6.621
πΛπ΄ = 750.695Hgmm
ππ πΛπ΅ = π΄ β π΅/(π + πΆ)
=15.8742-28.08.19/(319-45.99)
ππ πΛπ΅ = 5.588
πΛπ΅ =267.2Hgmm
PT - p0
B
xA = ---------- yA = (pOA/PT) xA
p0
A - poB
=0.999 = 0.998
Equilibrium data
Temperature
(ΛC) PΛA PΛB XA YA
46 750.71 267.25 1.00 1.00
47 775.48 277.47 0.95 0.98
48 800.89 288.00 0.90 0.96
49 826.95 298.86 0.85 0.94
50 853.68 310.04 0.81 0.92
51 881.07 321.55 0.77 0.90
52 909.15 333.41 0.72 0.88
53 937.92 345.61 0.68 0.85
54 967.40 358.17 0.64 0.83
55 997.59 371.09 0.60 0.80
56 1028.52 384.37 0.57 0.78
57 1060.19 398.04 0.53 0.75
58 1092.61 412.09 0.50 0.72
59 1125.80 426.54 0.46 0.69
60 1159.76 441.38 0.43 0.66
61 1194.52 456.63 0.40 0.63
62 1230.08 472.30 0.37 0.60
63 1266.45 488.39 0.34 0.57
64 1303.64 504.92 0.31 0.53
65 1341.68 521.88 0.28 0.50
66 1380.57 539.29 0.25 0.46
67 1420.32 557.16 0.22 0.42
68 1460.95 575.50 0.20 0.38
69 1502.47 594.30 0.17 0.34
70 1544.89 613.59 0.15 0.30
71 1588.22 633.37 0.12 0.26
72 1632.49 653.65 0.10 0.21
73 1677.69 674.44 0.08 0.17
74 1723.84 695.74 0.05 0.12
75 1770.96 717.57 0.03 0.07
76 1819.06 739.93 0.01 0.02
77 1868.16 762.83 -0.01 -0.03
Equilibrium curve
Reflux ratio
Determination of minimum reflux ratio (Rm)
q β line equation
In this case feed is a saturated liquid so feed is at its boiling point
If Feed at boiling point q = 1 and q β line is passing through the point of (Xf , Xf)
ππΉ= 0.4
fqq x. 1q
1 x.
1q
qy
feed theofheat latent Molar
feed of mole 1 vapourise torequiredHeat q
Top operating line equation
Dn1n .x
R1
1.x
R1
Ry
But, at minimum reflux ratio we can rewrite top operating line equation as follows,
Dn1n .xRm1
1.x
Rm1
Rmy
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 Rm1
Rm
= 0.54
Minimum reflux ratio, 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,
π¦π+1= 0.6041 π₯π + 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.π₯π = Dπ₯π + Wπ₯π€
F.π₯π = Dπ₯π + (F-D)π₯π€
D = πΉ(π₯πβπ₯π€ )
π₯πβπ₯π€
D = 45(0.4β0.04)
0.9β0.04 kmol/hr
D = 18.837 kmol/hr
Then,
W = (45 β 18.837)kmol/hr
= 26.163 kmol/hr
Material balance for rectifying section
L + D = G
R = πΏ
π·
1.5262 = πΏ
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 + π β² = πΏβ²
According to BOL,
π¦π+1 = πΏβ²
πΏβ²βπ€ π₯π β
π€
πΏβ²βπ€π₯π€
Gradient = 1.55 and W = 26.163kmol/hr
πΏβ²
πΏβ²βπ€ = 1.55
πΏβ² = 73.732kmol/hr
π β² = 73.732 β 26.163
π β² = 47.569kmol/hr
Bottom product (W) = 26.163kmol/hr
Upward vapour flow rate in stripping section (π β² ) = 47.569kmol/hr
Downward liquid flow rate in stripping section (πΏβ²) = 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)
π¦5= 0.62
π₯4 = 0.43
So, temperature of the 4th tray = 333K
Composition of the 1st tray (top plate to the feed)
π¦1 = 0.9
π₯0 = 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 Cπ2= 0.0761kg/mol
Molecular weight of CπΆπ4= 0.1538kg/mol
For rectifying section,
Density of Cπ2 at 328.5K = 1207.566kg/π3
Density of CπΆπ4 at 328.5K = 1526.596 kg/π3
For liquid
Weight fraction of Cπ2= 0.665Γ0.0761
0.665Γ0.0761+0.335Γ0.1538
= 0.4955
Basis 1000kg,
Volume of Cπ2=0.4955Γ1000
1207.566
= 0.41π3
Volume of CπΆπ4= 0.5045Γ1000
1526.596
= 0.33π3
Density of liquid mixture = 1000
0.41+0.33
= 1351.35kg/π3
For vapour mixer
According to, PV = nRT
Ο = ππ
π π
Average molecular weight of the vapour stream = 0.76Γ0.0761+0.24Γ0.1538
= 0.09475kg/mol
Ο = 1Γ105Γ94.75
8.314Γ328.5
Density of vapour mixture = 3.47kg/π3
For stripping section
Composition of 6th
plate (below plate to the feed tray)
π¦6= 0.58
π₯5 = 0.38
Temperature of the 5th tray = 334 K
Composition of 11th
plate (lowest plate of tower)
π¦12 = 0.07
π₯11 = 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 Cπ2 at 340.5K = 1187.885 kg/π3
Density of CπΆπ4 at 340.5K = 1503.208kg/π3
Similar to above calculation,
For liquid mixer
Weight fraction of the Cπ2 = 0.1225
Density of liquid mixture = 1455.8kg/π3
For vapour mixer
Average molecular weight of the mixer = 0.12855kg/mol
Density of vapour mixer = 4.541kg/π3
Column diameter for rectifying section
Gas flow rate = 47.125 kmol/hr = 47.125Γ0.09475Γ103
3600 kg/s
= 1.24 kg/s
Average molecular weight of liquid mixer = 0.665Γ0.0761+ 0.335Γ0.1538
= 0.1021kg/mol
Liquid flow rate = 28.75kmol/hr = 28.75Γ0.1021Γ103
3600kg/s
= 0.8154kg/s
Gas density at 328.5K = 3.47kg/π3
Liquid density at 328.5K = 1351.35kg/π3
πΉπΏπ= πΏπ€
ππ€
Οπ£
ΟπΏ
= 0.8154
1.24
3.47
1351.35
= 0.033
Pressure drop correlation graph
Pressure drop = 60mm water per m of packing
According to the graph,
πΎ4 = 2.2
At flooding, πΎ4= 5.6
Percentage flooding = 2.2
5.6 = 0 .627
62.7% satisfactory
Viscosity calculation
Log [viscosity] = [VISA] Γ [1
πβ
1
ππΌππ΅]
For carbon disulphide
VISA = 274.08
VISB = 200.22
T = 328.5k
Log [viscosity] = [274.08] Γ [1
328.5β
1
200.22]
Β΅1= 0.292 mNs/π2
For carbon tetrachloride
VISA = 540.15
VISB = 290.84
T = 328.5k
Log [viscosity] = [540.15] Γ [1
328.5β
1
290.84]
Β΅2= 0.6125 mNs/π2
Average viscosity of mixer
Β΅πππ₯ = ( π₯π Γ Β΅π
13)3
Β΅πππ₯ = (0.665 Γ 0.2921
3 + 0.335 Γ 0.61251
3)3
= 0.3821 mNs/π2
ππ€ = [πΎ4Οπ£(ΟπΏβΟπ£)
13.1Fπ(Β΅πΏ/ΟπΏ)0.1]
For this type packing material (plastic), πΉπ= 170
ππ€ = [2.2Γ3.47(1351.35β3.47)
13.1Γ170(0.3821/1351.35)0.1]
= 10.46kg/π2π
Column area required = 1.24
10.46 = 0.1185π2
Diameter = 4
πΓ 0.1185
Diameter of the rectifying section= 0.388m
Column diameter for stripping section
Gas flow rate = 47.569kmol/hr = 47.569Γ0.12855Γ103
3600 kg/s
= 1.7 kg/s
Average molecular weight of liquid mixer = 0.22Γ0.0761+ 0.78Γ0.1538
= 0.1367kg/mol
Liquid flow rate = 73.732kmol/hr = 73.732Γ0.1367Γ103
3600kg/s
= 2.8kg/s
Gas density at 340.5K = 4.541kg/π3
Liquid density at 340.5K = 1455.8kg/π3
πΉπΏπ= πΏπ€
ππ€
Οπ£
ΟπΏ
= 2.8
1.7
4.541
1455.8
= 0.092
From the graph, πΎ4 = 1.6
At flooding, πΎ4= 3.8
Percentage flooding = 1.6
3.8 = 0 .6488
64.88% satisfactory
Viscosity calculation
Similar to above calculation, T = 340.5K
For carbon disulphide
Β΅1= 0.273mNs/π2
For carbon tetrachloride
Β΅2= 0.536mNs/π2
Average viscosity of mixer,
Β΅πππ₯ = (0.22 Γ 0.2731
3 + 0.78 Γ 0.5361
3)3
= 0.468 mNs/π2
ππ€ = [3.8Γ4.541(1455.8β4.541)
13.1Γ170(0.468/1455.8)0.1]
= 24.635kg/π2π
Column area required = 1.7
24.635 = 0.069π2
Diameter = 4
πΓ 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 Cπ2 = 351kJ/kg
Latent heat of CπΆπ4 = 194kJ/kg
πΏπ£,π ππ ππ
Cπ2 26711.1 319.3 552
CπΆπ4 29837.2 349.8 556.35
For top products
Temperature of top plate = 322K
Carbon disulphide,
πΏπ£ = 26711.1Γ [552β322
552β319.3]0.38
= 26593kJ/kmol
Carbon tetrachloride,
πΏπ£ = 29837.2Γ [556.35β322
556.35β349.8]0.38
= 31303.8kJ/kmol
Latent heat of mixtures
For rectifying section,
πΏπ£,πππ₯ = 26593Γ0.9 + 31303.8Γ0.1
= 27064.08 kJ/kmol
Therefore condenser heat load = GΓ πΏπ£,πππ₯ = 47.125Γ27064.08
= 1275.4MJ/hr
= 354.3kW
Boiler heat load calculation
Using equation,
πππ»π£ ,π + Wπ»π€ = πΏπβ1π»πΏ ,πβ1 + ππ
π»π£ = π»πΏ + ππππ₯
π»πΏ = (πΆ1 + πΆ2π + πΆ3π2 + πΆ4π
3)πππ2
π1
πΆ1 πΆ2 πΆ3 πΆ4
Cπ2 85600 -122 0.5605 -0.001452
CπΆπ4 -752700 8966.1 -30.394 0.034455
Average temperature of stripping section, T = 340.5K
Calculating the latent heat of mixture
For stripping section
Temperature of bottom plate = 348K
For Carbon disulphide,
π»πΏ = (πΆ1 + πΆ2π + πΆ3π2 + πΆ4π
3)πππ2
π1
= (85600 β 122T + 0.5605π2 β 0.001452π3)ππ340.5
298
= 2397.325kJ/kmol
πΏπ£ = 26711.1Γ [552β340.5
552β319.3]0.38
= 25758.8kJ/kmol
Carbon tetrachloride,
π»πΏ = (β752700 + 8966.1T β 30.394π2 + 0.034455π3)ππ340 .5
298
= 5671.523kJ/kmol
πΏπ£ = 29837.2Γ [556.35β340.5
556.35β349.8]0.38
= 30340.7kJ/kmol
For stripping section,
πΏπ£,πππ₯ = 25758.8Γ0.22 +30340.7Γ0.78
Latent heat of mixture = 29332.7 kJ/kmol
Enthalpy of liquid mixture(π»πΏ ,πππ₯ ) = 2397.325Γ0.22 + 5671.523Γ0.78
= 4951.2 kJ/kmol
Enthalpy of carbon disulphide, π»π£= 2397.325 + 29332.7
= 31730kJ/kmol
Enthalpy of carbon disulphide, π»π£= 5671.523 + 29332.7
= 35004.2kJ/kmol
Enthalpy of vapour mixture (π»π£,πππ₯ ) = 31730Γ0.325 + 35004.2Γ0.675
= 33940.1 kJ/kmol
For bottom product (residue),
Temperature = 348K
For Carbon disulphide,
π»πΏ = (85600 β 122T + 0.5605π2 β 0.001452π3)ππ348
298
= 2778.2kJ/kmol
Carbon tetrachloride,
π»πΏ = (β752700 + 8966.1T β 30.394π2 + 0.034455π3)ππ348
298
= 6703.6kJ/kmol
Enthalpy of liquid mixture of residue(π»π€ ) = 2778.2 Γ0.04 + 6703.6Γ0.96
= 6546 kJ/kmol
Substituting to equation,
πππ»π£,π + Wπ»π€ = πΏπβ1π»πΏ ,πβ1 + ππ
47.569Γ33940.1 + 26.163Γ6546 = 73.732Γ4951.2 + ππ
ππ = 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 & RICHARDSONβS, CHEMICAL ENGINEERING, Volume 6, 3rd
edition, R K
Sinnott
CULSON & RICHARDSONβS, CHEMICAL ENGINEERING, Volume 2, 5th edition
PERRYβS CHEMICAL ENGINEERS HAND BOOK, 8th edition, DON W. GREEN, ROBERT
H. PERRY
DISTILLATION DESIGN, HENRY Z. KISTER