3-1
CHAPTER 3
DESIGN FOR DISTILLATION COLUMN
3.1
INTRODUCTION
Distillation is most probably is the widely used separation process in the chemical industries. The design of a distillation column can be divided into several procedures:
1. Specify the degree of separation required: set product specification 2. Select the operation conditions: batch or continuous: operating pressure 3. Select the type of contacting device: plate or packing 4. Determine the stage and reflux requirements: the number of equilibrium stages 5. Size the column: diameter, number or real stages 6. Design the column internals: plates, distributors, packing supports 7. Mechanical design: vessel and internal packing
The separation of liquid mixtures by distillation is depends on the differences in the volatility between the components. This is known as continuous distillation. Vapor flows up to column and liquid counter-currently down the column. The vapor and liquid are brought into contact on plates. Part of the condensate from the condenser is returned on the top of the column to provide liquid flow above the feed point (reflux), and part of the liquid from the base of the column is vaporized in the reboiler and returned to provide the flow.
3-2 3.2
Chemical Design
The purpose of this distillation column is to separate the component mixture. Basically, components which are Propanal, DPE, water, 1-Propanol, Ethylene, Carbon Monoxide, Hydrogen and Ethane are to be separated to the bottom stream. These components will go through another distillation process. The feed is fed to the distillation column at 1.82 bar and 293K. The products at the top column leave the column at 1 bar and 357.36K. The products at the bottom column leave the column at 1.6bar and 382.35K. 1-Propanol and DPE were chosen as the key components being 1-Propanol as the light key component while DPE as the heavy key component. Distillation column with perforated tray has been chosen. Basically, this is the simplest type. The vapour passes up through perforations in the plate, and the liquid is retained on the plate by the vapour flow. There is no positive vapour liquid seal, and at low flow rate liquid will weep through the holes reducing efficiency. The perforation is usually small holes.
3.2.1
Complete Diagram
The composition of the inlet and outlet streams for distillation column is shown in table 3.1: Table 3.1 Summary of the inlet and outlet composition
Feed
Component
Molar Flow Rate (kmole/h)
Top
Mole Fraction
Molar Flow Rate (kmole/h)
Bottom
Mole Fraction
Molar Flow Rate (kmole/h)
Mole Fraction
1-Propanol
257.94
0.9768
1.2283
0.4853
256.7
0.9815
Water
2.2373
0.0085
0.45993
0.1817
1.7774
0.0068
Propanal
1.4473
0.0055
0.7767
0.3068
0.6706
0.0026
3-3
Dipropyl Ether
3.2.2
2.3986
0.0092
0.0118
0.0047
2.3868
0.0091
Bubble and Dew Point Temperature
To estimate the stages, and the condenser and reboiler temperatures, procedures are required for calculating dew and bubble points. By definition, a saturated liquid is at its bubble point (any rise in temperature will cause a drop in a liquid form). It can be calculated in terms of equilibrium constant, K. Bubble Point :
π¦π =
πΎπ π₯π = 1.0
(3.1)
Dew Point
π₯π =
π¦π / πΎπ = 1.0
(3.2)
:
Table 4.2 below shows the constants of Antoine equation for each component. (RK Sinnot, 1999) where the constant value for each component is taken from HYSYS. Table 3.2: The Antoine Constant COMPONENT
a
b
c
d
e
f
1-Propanol
79.5
-8.29Γ103
0.00
-8.9096
1.82Γ10-6
2.00
Water
65.9
-7.23Γ103
0.00
-7.18
4.03Γ10-6
2.00
Propanal
80.9
-6.51Γ103
0.00
-9.82
6.79Γ10-6
2.00
96.7
-7.45Γ103
0.00
-1.24
1.08Γ10-5
2.00
Dipropyl Ether
Antoine equation: ln ππ = π΄ β
πΎπ =
ππ ππ
π΅ + π· π₯ ln π + πΈ π₯ π^πΉ π+πΆ
(3.3)
(3.4)
3-4 Estimation of feed temperature,
π₯π =
π¦π / πΎπ = 1.0
By using the goal seek method in the excel program, with constant operating pressure at feed is 1.6 bar, the calculated temperature is 363K. The data shown in Table 3.3: Table 3.3: Calculation of Bubble Point at Feed Stream COMPONENT
ln Pi
Pi (kPa)
Xi
O.P (kPa)
Ki
Yi=KiXi
1-Propanol
5.19
179.42
0.9768
182
0.99
0.962939
Water
5.07
159.48
0.0085
182
0.88
0.007448
Propanal
6.54
692.20
0.0055
182
3.80
0.020918
Dipropyl Ether
5.23
186.62
0.0092
182
1.03
0.009434
TOTAL
1.00000
Hence, the bubble point temperature is 386.36 K By using the goal seek method in the excel program, with constant operating pressure at top is 0.5 bar, the calculated temperature is 60K. The data shown in Table 3.4: Dew Point Temperature (top column)
π₯π =
π¦π / πΎπ = 1.0
Table 3.4: Calculation of Dew Point at Top Column COMPONENT
ln Pi
Pi (kPa)
Yi
O.P (kPa)
Ki
Xi=Yi/Ki
1-Propanol
3.59
36.41
0.4853
50
0.73
0.67
Water
3.54
34.59
0.1817
50
0.69
0.26
Propanal
5.44
231.56
0.3068
50
4.63
0.07
Dipropyl Ether
4.01
55.25
0.0047
50
1.10
0.004
TOTAL
1
3-5 Hence, the dew point temperature is 345.56 K By using the goal seek method in the excel program, with constant operating pressure at bottom is 1.6 bar, the calculated temperature is 376K. The data shown in Table 3.5: Bubble Point Temperature (bottom column)
π¦π =
πΎπ π₯π = 1.0
Table 3.5: Calculation of Bubble Point at Bottom Column COMPONENT
ln Pi
Pi (kPa)
Xi
O.P (kPa)
Ki
Yi=KiXi
1-Propanol
4.69
108.95
0.9815
110
0.99
0.97
Water
4.59
98.45
0.0068
110
0.89
0.01
Propanal
6.20
490.93
0.0026
110
4.46
0.01
Dipropyl Ether
4.85
127.18
0.0091
110
1.16
0.01
TOTAL
1
Hence, the bubble point temperature is 372.33 K
3.2.3
Determination of Relative Volatility
The equilibrium vaporization constant K is defined for a compound by πΎπ =
ππ ππ
(3.5)
Where, Yi = mole fraction of component i in vapour phase Xi = mole fraction of component i in liquid phase The relative volatility, Ξ± which is needed in the calculation is defined as πΌππ =
πΎπ πΎπ
(3.6)
3-6 Where i and j represent the components to be separated From Ideal system, Raoultβs law,
Pi = PiXi
(3.7)
The relative volatility of two components can be expressed as the ratio of their K value, πΌππ =
πΎπΏπΎ πΎπ»πΎ
(3.8)
Where, KLK = Light key components KHK = Heavy key components 3.2.3.1 Top Column
Table 3.6 πΆ=
π² π²π―π²
COMPONENT
K
1-Propanol
0.7300
0.6636
Water
0.6900
0.6273
Propanal
4.6300
4.2091
DPE
1.1000
1.0000
3.2.3.2 Bottom Column
Table 3.7 πΆ=
π² π²π―π²
COMPONENT
K
1-Propanol
0.9900
0.8534
Water
0.8900
0.7672
Propanal
4.4600
3.8448
3-7 DPE
1.1600
1.0000
Average relative volatility of the light key to heavy key;
Ξ±LK
=
=
Top Ξ± (Bottom Ξ±)
0.6636 (0.8534)
= 0.753
3.2.4
Minimum Number of Stages Using Fenskeβs Equation
The Fenskeβs Equation (1932) can be used to estimate the minimum stages required at total reflux. The derivation of the equation for binary system and applies equally to multi-component system. The minimum number of stages will be obtained from this equation:
Nmin
=
X X Log[(X LK )]d [( XHK )]b HK
LK
Log Ξ±LK
(3.9)
0.73 0.0091 Log[( 1.1 )]d [( )] 0.9815 b = Log 0.753 = 17.94 = 20 stages
3.2.5
Minimum Reflux Ratio
Colburn (1941) and Underwood (1948) have derived equations for estimating the minimum reflux ratio for multicomponent distillations. The equation can be stated in the form: πΌπ π₯π,π = π
π + 1 πΌπ β π
(3.10)
3-8 Where, Ξ±i =
the relative volatility of component i with respect to some reference component, usually the heavy key
Rm =
the minimum reflux ratio
Xi,d =
concentration of component i in the tops at minimum reflux
and ΞΈ is the root of the equation:
πΌπ π₯π,π =1βπ πΌπ β π
(3.11)
Where, Xi,f =
the concentration of component i in the feed, and q depends on the condition of the feed
The value of ΞΈ must lie between the values of relative volatility of the light and heavy keys and is found by trial and error. As the feed at its boiling point q = 1 πΌπ π₯π,π =0 πΌπ β π Table 3.8 Component
Xi,f
Ξ±i
Ξ±iXi,f
ΞΈ estimate
(Ξ±iXi,f)/(Ξ±i - ΞΈ)
1-Propanol
0.9768
0.7600
0.7424
3.9
-0.2364
Water
0.0085
0.7000
0.0060
3.9
-0.0019
Propanal
0.0055
4.0300
0.0222
3.9
0.1705
DPE
0.0092
1.0000
0.0092
3.9
-0.0032
3.9
-0.07
Therefore, ΞΈ = 3.9
3-9 Table 3.9 Component
Xi,d
Ξ±i
Ξ±iXi,d
ΞΈ estimate
(Ξ±iXi,d)/(Ξ±i - ΞΈ)
1-Propanol
0.4853
0.76
0.3688
3.9
-0.1175
Water
0.1817
0.7
0.1272
3.9
-0.0397
Propanal
0.3068
4.03
1.2364
3.9
9.5108
DPE
0.0047
1
0.0047
3.9
-0.0016
3.9
9.35
Taking equation 3.10, Rm + 1 = 9.35 Rm = 8.35 π
π = 0.8931 π
π + 1 Specimen calculation, for R = 2.0 π
2 = = 0.66 (π
+ 1) 3 Using Erbar β Maddox correlation (Erbar and Maddox, 1961) from figure 11.11 (Coulson and Richardson, Volume 6, page 524), ππ = 0.74 π N=
18 0.74
= 24.3 For other reflux ratios R N
2
3
4
5
24.3
21.43
20.69
20.22
3-10 The optimum reflux ratio will be near to 4. Therefore, the optimum reflux ratio will be taken as 4 while the actual stage is 21.
3.2.6
Feed Point Location
Feed point location can be found using Kirkbride (1944) equation: ππ πΏππ = 0.2606 log ππ
π΅ π·
π₯π,π»πΎ π₯π,πΏπΎ
xπ,πΏπΎ xπ,π»πΎ
2
(3.10)
Where, Nr
=
no. of stages above the feed, including any partial condenser
Ns
=
no. of stages below the feed, including the reboiler
B
=
molar flow bottom product
D
=
molar flow top product
Xf,HK
=
concentration of the heavy key in the feed
Xf,LK
=
concentration of the light key in the feed
Xd,HK
=
concentration of the heavy key in the top product
Xb,HK
=
concentration of the heavy key in the bottom product
πΏππ
ππ = 0.2606 log ππ
2.531 261.5
ππ = 0.993 ππ Actual number of plates is 24 Nr + Ns
= 24
0.993Ns + Ns = 24 1.993Ns
=9
0.0092 0.9768
0.395 0.00382
2
3-11 Nr
= 15
So, feed inlet is at stage 9 from bottom.
3.2.7
Efficiency of Distillation Column
Overall column efficiency is given as: πΈΛ³ = 51 β 32.5 log (Β΅π ππ )
(3.11)
Where, Β΅π = the molar average liquid viscosity, mNs/m2 ππ = average relative volatility of the light key To find the viscosity of the flow: πΏππΊ Β΅π
= πππ π΄ π₯
1 1 β π πππ π΅
(3.12)
Table 3.8 Viscosity of the mixture
Component
Mole fraction feed, x
Viscosity Coefficient A
B
Log Β΅π
Viscosity (mNs/m2)
Β΅π Γ π
1-Propanol
0.9768
951.04
327.83
-0.32859
0.46926
0.4584
Water
0.0085
658.25
283.16
-0.54418
0.28564
0.0024
Propanal
0.0055
343.44
219.33
-0.63690
0.23073
0.0013
DPE
0.0092
410.58
219.67
-0.75852
0.17438
0.0016
1.16
0.4637
TOTAL
Where, πΈΛ³
=
51 β 32.5 log (Β΅π ππ )
3-12 =
51 β 32.5 log (0.463674405 x 0.787)
=
55.44 %
Plate and overall column efficiencies will normally be between 30% to 70%. (Coulson and Richardsonβs, volume 6, page 547) 3.2.8
Physical Properties
3.2.8.1 Relative Molar Mass (RMM) RMM = β (component mole fraction x molecular weight)
(3.13)
Table 3.9 Liquid Density Component
Mole Fraction
Molecular
Liquid Density
Weight
Feed
Distillate
Bottom
(kg/m3)
1-Propanol
60.1
0.9768
0.4853
0.9815
803.4
Water
18.015
0.0085
0.1817
0.0068
1000
Propanal
58.08
0.0055
0.3068
0.0026
810
DPE
102.18
0.0092
0.0047
0.0091
725
Feed, F
= 0.9768 (60.1) + 0.0085 (18.015) + 0.0055 (58.08) + 0.0092 (102.18) = 60.118 kg/kmol
Distillate, D
= 0.4853 (60.1) + 0.1817 (18.015) + 0.3068 (58.08) + 0.0047 (102.18) = 50.739 kg/kmol
Bottom, B
= 0.9815 (60.1) + 0.0068 (18.015) + 0.0026 (58.08) + 0.0091 (102.18) = 60.191 kg/kmol
3-13 3.2.8.2 Density Top Product : π₯π΅,π ππ
ΟL
=
ΟL
=
0.4835(803.4) + 0.1817(100) + 0.3068(810) + 0.0047(725)
=
823.51 kg/m3
Οv
=
Οv
=
π
ππ π΅ ππππ
(3.14)
π₯
ππππ πππ
π₯
29.167 ππ/πππππ
πππ
π₯
22.4π 3 /πππππ
(3.15)
ππππ 273πΎ 357.21πΎ
π₯
1πππ 1πππ
1.731 kg/m3
= Bottom Product:
π₯π·,π ππ
ΟL
=
ΟL
=
0.9815(803.4) + 0.0068(100) +0.0026(810) + 0.0091(725)
=
804.04 kg/m3
Οv
=
Οv
=
π
ππ π· ππππ
(4.16)
π₯
ππππ πππ
58.988ππ /πππππ 22.4π 3 /πππππ
π₯ π₯
πππ
(4.17)
ππππ 273πΎ 382.2πΎ
π₯
1.6πππ 1πππ
3.071 kg/m3
=
3.2.8.3 Surface Tension, Ο Using Sugden (1924), equation 8.23 (Coulson and Richardsonβs, volume 6, page 335)
πππ(ππ β ππ£ π= π
4
π₯ 10β12
(3.18)
3-14 Where, Ο
=
surface tension, MJ/m2 or (dyne/cm)
Pch
=
Sugdenβs parachor
Οv
=
Vapor density, kg/m3
ΟL
=
Liquid density, kg/m3
M
=
relative molecular weight
For mixture, Οm = Ο1x1 + Ο2x2 + β¦..
(3.19)
Where, Οm
=
surface tension mixture
Ο1 , Ο2 =
surface tension for mixture
x1 , x2
component mole fraction
=
Table 3.10 Pch Distribution
Component
Pch at top
= =
Pch
Mole Fraction
Distribution
Distillate
Bottom
1-Propanol
148.3
0.4853
0.9815
Water
31.3
0.1817
0.0068
Propanal
165.4
0.3068
0.0026
DPE
299.5
0.0047
0.0091
π₯π·,π ππππ 0.4853 (148.3) + 0.1817 (31.3) + 0.3068 (165.4) + 0.0047 (299.5)
3-15 =
177.28097
Pch at bottom =
π₯π΅,π ππππ
=
0.9815 (148.3) + 0.0068 (31.3) + 0.0026 (165.4) + 00.0091 (299.5)
=
148.30792
Calculation of surface tension: Top Column, π =
65.01 969.64β4.928 4 21.98
π₯ 10β12
= 67.965683 N/m
59.04 1019.01β 0.325
Bottom Column, π =
20.04
4
π₯ 10β12
= 15.27159545 N/m
Above feed point: Vapor flow rate: Vn
= D(R + 1)
(3.20)
Where, D
=
Distillate molar flowrate
R
=
Reflux ratio
Vn
=
261.5 (2.531 + 1)
=
923.36 kmole/hr
Hence,
Liquid down flow: Ln = Vn β D = 923.36 β 261.5
(3.21)
3-16 = 661.86 kmole/hr Below the feed point: Liquid flow rate: Lm
= Ln + F
(3.22)
Where, F
=
Feed molar flowrate
Lm
=
661.86 + 264.1
=
925.96kmole/hr
Hence,
Vapour flow rate: Vm = Lm β W
(3.23)
Where, W
=
Bottom molar flowrate
Vm
=
925.96 β 261.5
=
664.46kmole/hr
Hence,
The equation for the operating lines below the feed plate: ππ =
πΏπ ππ
ππ =
925.96 664.46
ππ + 1 β
π ππ
ππ + 1 β
= 2.058(Xm + 1) β
ππ€
(3.24)
261.5 (ππ€) 664.46
261.5 664.46
(ππ€)
The equation for the operating lines above the feed plate: ππ =
πΏπ ππ
ππ + 1 β
π· ππ
ππ
(3.25)
3-17
ππ =
661.86 923.36
ππ + 1 β
261.5 923.36
ππ
= 0.72 (Xn + 1) β 2.01 x 10-3
πΉπΏπ πππ =
πΏπ ππ
ππ ππΏ
(3.26)
1.731 823.51
= 0.72
= 0.033 where 0.72 is the slope of the top operating line.
πΉπΏπ π΅ππ‘π‘ππ =
πΏπ ππ
= 1.39
ππ ππΏ
(3.27)
3.071 804.04
= 0.09 where 1.39 is the slope of the bottom operating line.
3.2.9
Determination of Plate Spacing
The overall height of the column will depend on the plate spacing. Plate spacing from 0.15m to 1.0m are normally used. The spacing chosen will depend on the column diameter and the operating condition. Close spacing is used with small - diameter columns, and where head room is restricted, as it will be when a column is installed in a building. In this distillation column, the plate spacing is 0.5m as it is normally taken as the initial estimate recommended by Coulson and Richardsonβs, Chemical Engineering, Volume 6.
3-18 The principal factor that determines the column diameter is the vapor flowrate. The vapor velocity must be below that which would cause excessive liquid entrainment or highpressure drop. The equation below which is based on the Souder and Brown equation, Lowenstein (1961), Coulson & Richarsonβs Chemical Engineering, Volume 6, page 556, can be used to estimate the maximum allowable superficial velocity, and hence the column area and diameter of the distillation column.
ππ£ = β0.171ππ‘ 2 + 0.271ππ‘ β 0.047
ππΏ β ππ£ ππ£
= β0.171(0.5)2 + 0.271(0.5) β 0.047
0.5
969.64 β 4.928 4.928
(3.28) 0.5
= 2.8173 m/s
Where, Uv
= maximum allowable vapor velocity based on the gross (total) column cross Sectional area, m/s
lt
= plate spacing, m (range: 0.5 β 1.5)
3.2.9.1 Diameter of the column
π·π =
4ππ€ πππ£ ππ£
Where Vw is the maximum vapor rate, kg/s ππ€ =
15870 ππ 1 ππ π₯ ππ 3600 π
= 4.41 kg/s
(3.29)
3-19
π·π =
4(4.41) π 4.928 (0.64)
= 1.33 m
3.2.9.2 Column Area The column area can be calculated from the calculated internal column diameter π΄π = =
π π·π 2 4
(3.30)
π (1.33)2 4
= 1.39 m2
4.2.10 Liquid Flow Arrangement Before deciding liquid flow arrangement, maximum volumetric liquid rate were determined by the value of maximum volumetric rate πΏ=
=
15740 ππ 1 ππ π₯ ππ 3600 π
(3.31)
4.372 ππ π3 π₯ π 804.04 ππ
= 5.38 x 10-3 Dc = 1.128 m
Based in the values of maximum volumetric flow rate and the column diameter to Figure 11.28 from Coulson and Richardson, Chemical Engineering, Volume 6, page 568, the types of liquid flow rate could be considered as single pass.
3-20 Perforated plate, which is famously known as sieve tray is the simplest type of cross-flow plate. Cross flow trays are the most common used and least expensive. Sieve tray is chosen because it is consider cheaper and simpler contacting devices. The perforated trays enable designs with confident prediction of performance. According, most new designs today specify some type of perforated tray (sieve tray) instead of the traditional bubble-cap tray. Sieve tray also gives the lowest pressure drop.
3.2.11 Plate Design Column diameter, Dc = 1.33 m Column area, Ac
= 1.39 m2
As a first trial, take the downcomer area as 12% of the total Downcomer area, Ad = 0.12 Ac
(3.32)
= 0.12 x 1.39 m2 = 0.1668 m2 Net area, An
= Ac - Ad
(3.33)
= 1.39 m2 - 0.1668 m2 = 1.2232 m2 Active area, Aa
= Ac β 2Ad
(3.34)
= 1.39 m2 β 2(0.1668 m2) = 1.0564 m2 Assume that the hole-active area is 10% Hole area, Ah
= 0.10 Aa = 0.10 x 1.0564 m2 = 0.10564m2
(3.35)
3-21 3.2.11.1
Weir Length
With segmental downcomers the length of the weir fixes the area of the downcomer. The chord length will normally be between 0.6 to 0.85 of the column diameter. A good initial value to use is 0.77, equivalent to a downcomer area of 15%. Referring to Figure 11.31 from Coulson and Richardsonβs, Chemical Engineering, Volume 6, page 572, with (Ad/Ac) x 100 is 12 percents, thus, Iw/Dc is 0.76
Weir length, Iw = 0.76Dc = 0.76 x 1.33 m = 1.011 m
3.2.11.2
Weir Height
For column operating above atmospheric pressure, the weir-heights will normally be between 40 mm to 90 mm (1.5 to 3.5 in); 40 to 50 mm is recommended.
Take Weir height, hw = 50 mm Hole diameter, dh
= 5 mm (preferred size)
Plate thickness, t
= 3 mm (stainless steel)
For hole diameter = 5 mm, area of one hole, π΄ππ =
=
π(ππ )2 4 π(0.005)2 4
= 1.9635 x 10-5 m2
(3.36)
3-22 Number of holes per plate, ππππ 1 ππππ ππππ 0.10564 = 1.9635 π₯ 10β5
ππ =
(3.37)
= 5380.19 holes β 5380 holes
3.2.11.3
Weir Liquid Crest
Check weeping to ensure enough vapour to prevent liquid flow through hole. πππ₯πππ’π ππππ’ππ πππ‘π =
15740ππ 1 ππ π₯ ππ 3600 π
= 4.372 kg/s Minimum liquid rate, at 70% turndown = 0.7 x 4.372 kg/s = 3.06 kg/s
The weir liquid can be determine by using the equation below
πππ€
πΏπ€ = 750 ππΏ πΌπ€
2 3
Where, Iw
= weir length, m
Lw
= liquid flow rate, kg/s
ΟL
= liquid density, kg/m3
(3.38)
3-23 how
= weir crest, mm liquid
At maximum rate:
πππ€
4.372 = 750 804.04 π₯ 1.011
2 3
= 20.40 mm liquid
At minimum rate:
πππ€
3.06 = 750 1019.01 π₯ 0.85728
2 3
= 18.15 mm liquid
At minimum rate, clear liquid depth, how + hw
= 18.15 + 50 = 68.15 mm liquid
From Figure 11.30, in Coulson and Richardsonβs, Chemical Engineering, Volume 6, page 571, weep point correlation, K2 = 30.7
3.2.11.4
Weep Point
The purpose to calculate this weep point is to know the lower limit of the operating range ccurs when liquid leakage through the plate holes becomes excessive. During weeping, a minor fraction of liquid flows to the tray below through the tray perforations rather than the downcomer. This downward-flowing liquid typically has been exposed to rising vapor; so, weeping only leads to a small reduction in overall tray efficiency, to a level rarely worse than the tray point efficiency. Minimum vapor velocity through the holes based on the holes area.
3-24
ππ (min) =
πΎ2 β 0.9(25.4 β ππ )
(3.39)
1
ππ£ 2
Where, Uh
= minimum vapor velocity, m/s
dh
= hole diameter, mm
K2
= constant
=
30.7 β 0.9(25.4 β 5) 1
(3.071)2 = 8.036 m/s
π΄ππ‘π’ππ ππππππ’π π£ππππ π£ππππππ‘π¦ =
ππππππ’π π£ππππ πππ‘π π΄π
(3.40)
4.41 ππ π3 π₯ 0.7 π₯ π 3.071ππ = 0.10564 = 9.51 m/s
So, minimum operating rate will be above weep point.
3.2.12 Plate Pressure Drop Maximum vapor velocity through holes: Γπ =
πππ₯πππ’π π£ππππ ππππ€πππ‘π π΄π
(3.41)
3-25 4.41 ππ π3 π₯ π 3.071 ππ = 0.10564 = 13.59 m/s
From Figure 11.34 in Coulson and Richardsonβs, Chemical Engineering, Volume 6, page 576, for discharge coefficient for sieve plate, π΄π‘,
πππ
ππππ‘π π‘πππππππ π 3 ππ = = 0.6 ππππ ππππππ‘ππ 5 ππ π΄π = 0.1 π΄π
we get Co = 0.74 ππ πΆπ
π·ππ¦ ππππ‘π, ππ = 51
2
13.59 = 51 0.74
ππ£ ππΏ 2
(3.42) 3.071 804.04
= 65.697 mm liquid
π
ππ πππ’ππ ππππ, ππ =
12.5 π₯ 103 ππΏ
=
12.5 π₯ 103 804.04
(3.43)
= 15.55 mm liquid
Pressure drop per plate, ht
= hd + (hw + how) + hr = 65.697 + (50 + 18.15) + 15.55 = 149.397 mm liquid
(3.44)
3-26 3.2.13 Downcomer Liquid Back-Up The downcomer area and plate spacing must be such that the level of the liquid and froth in the downcomer is well below the top of the outlet weir on the plate above. If the level rises above the outlet weir the column will flood. Take hap
= hw β 10 mm = 50 β 10 = 40 mm
Where, hap
πππ
= height of the bottom edge of the apron above the plate
πΏπ€π = 166 ππΏ π΄π
2
(3.45)
Where, Lwd
= liquid flowrate in downcomer, kg/s
Am
= either the downcomer area, Ad or the clearance area under the downcomer, Aap whichever is smaller, m2
Area under apron, Aap
= hap x Iw
(3.46)
= 0.04 m x 1.011 = 0.04044 m2 Where, Aap
= the clearance area under downcomer
As this less than Ad = 0.1668 m2, equation 11.92 (Coulson and Richardsonβs, Volume 6, page 577) used Aap = 0.04044 m2
3-27
πππ = 166
4.372 804.04 π₯ 0.04044
2
= 3.00 mm
3.2.14 Backup on Downcomer hb
= (hw + how) + ht + hdc
(3.47)
= (50 + 18.15) +149.397 +3.00 = 220.547 mm hb
< Β½(plate spacing + weir height)
0.2205 m
< Β½(0.5 + 0.05) m
0.2205 m
<0.5 m
So, tray spacing = 0.5 m is acceptable (to avoid flooding).
3.2.15 Residence Time Sufficient residence time must be allowed in the downcomer for the entrained vapor to disengage from the liquid stream, to prevent heavily βaeratedβ liquid being carried under the downcomer. A time at least 3 seconds is recommended. π‘π =
π΄π πππ ππΏ πΏπ€π
(3.48)
Where, tr
= residence time, s
Lwd
= liquid flowrate in downcomer, kg/s
hbc
= clear liquid back up in the downcomer, m
3-28
π‘π =
0.1668 π₯ 0.2205π₯ 804.04 4.372
= 6.764 s tr is greater than 3.0 which is recommended so tr is satisfactory.
3.2.16 Perforated Area From Figure 11.32, Coulson and Richardsonβs, Chemical Engineering, Volume 6, page 527, for the relaxation between angle subtended by chord, chord height and chord length: Iw/Dc
= 1.011/1.33 = 0.76
ΞΈ
= 98Β°
Ih/Dc
= 0.18
Angle subtended at plate edge by unperforated strips = 180Β° - 98Β° = 82Β° Mean length, unperforated edge strips = (Dc β weir height) x Ο x ΞΈ/180Β° = (1.33 β 0.05) x Ο x 82Β°/180Β° = 1.832 m Areas of unperforated edge strips, As = mean length unperforated edge x weir height = 1.832 x 0.05 = 0.0916 m2
3-29 Mean length of calming zone = Weir length + Width of unperforated strip = 1.011 + 0.05 = 1.061 m
Area of calming zone = 2 x (weir height x mean length calming zone) = 2 x 0.05 x 1.061 = 0.1061 m2 Total area available for perforation, Ap : = Active area β (area of unperforated edge + area of calming) = 1.0564 β (0.0916 + 0.1061) = 0.8587 m2 Ah/Ap = 1.0564/0.8587 = 0.123
From Figure 11.33, Coulson and Richardsonβs, Chemical Engineering, Volume 6, page 528, the relation between hole area and pitch, Ip/dh
= 2.7; satisfactory, which is within 2.5 to 4.0
3.2.17 Column Size The column height will be calculated based on the given below. The equation determines the height of the column without taking the skirt or any support into consideration. Its determination is based on the condition in the column. Column height = (No. of stages β 1) x (Tray spacing ) + (Tray spacing x 2) + (No. of stages β 1)
3-30 x (Plate thickness) = (24 β 1)(0.5) + (24 β 1)(0.003) = 11.57 m By adding 10% safety factor so the column height are 12.7 m β 13 m
3.2.18 Plate Design Specification
Table 3.10 Summary of Plate Design Item
Value
Column Diameter, Dc
1.33 m
No. of Plates
24 plates
Plate Spacing
0.5 m
No. of Stage Feed from bottom, F1
9
No. of Stage Solvent from bottom, F2
24
Plate Thickness
0.003 m
Total Column Height, Ht
13 m
Plate Material
Stainless Steel
Downcomer Area, Ad
0.1668 m2
Column Area, Ac
1.39 m2
Net Area, An
1.2232 m2
Active Area, Aa
0.10564 m2
Hole Area, Ah
0.010564 m2
No. of Holes
5380 units
Weir Length, Iw
1.011 m
Weir Height (standard)
0.05 m
3-31 3.3
Mechanical Design
3.3.1
Introduction
Several factors need to be considered in the mechanical design of distillation column such as 1. Design pressure 2. Design temperature 3. Material of construction 4. Design stress 5. Wall thickness 6. Welded joint efficiency 7. Analysis of stresses a. Dead weight load b. Wind load c. Pressure stress d. Bending stress 8. Vessel support 9. Insulation
3.3.2
Column Design Specification
Operating pressure
= (1.82 β 1) bar = 0.82 bar
Take as 10% above operating pressure Design pressure
= 0.82 x 1.1 =0.902 bar = 0.0902 N/mm2
Design temperature
= 113.36Β°C
Take as 10% above operating temperature Operating temperature
= 113.36 x 1.1 = 124.694Β°C
3-32 3.3.3
Material of Constructions
Selection of suitable material must be taking onto account the suitability of material for fabrication (particularly welding) as well as the compatibility of the material with the process environment. In this case, the material used in the construction of the distillation column is carbon steel as the material as it is the most used material in industry. For this material, the design stress at 150Β°C is obtained from Table 13.2 for the typical design stresses for plate.
Design stress, f
= 115 N/mm2
Tensile strength
= 360 N/mm2
Join factor
=1
Diameter vessel, D
= 1330 mm
Operating pressure
= 0.0902 N/mm2
Insulation, mineral wool
= 75 mm thick
3.3.4
Vessel Thickness
The minimum thickness of column required and other designs are calculated based on equation below:
β―=
ππ π·π 2π½π β ππ
3.49
Where, e
= minimum thickness of the plate required
Pi
= internal pressure, N/mm2
Di
= internal diameter, m
f
= design stress, N/mm2
J
= joint factor
Therefore, minimum thickness required, β―=
0.0902 π₯ 1330 2 1 (115) β 0.0902
= 0.522 mm β 0.6 mm
3-33 A much thicker wall is needed at the column base to withstand the wind and dead weight loads. As a first trial, divide the column into five sections, with the thickness increasing by 2 mm per section. Try 1, 3, 5, 7 and 9 mm. The average wall thickness is 5 mm. Take the first trial as 5 mm.
3.3.5
Heads and Closure
Hemispherical, ellipsoidal and torispherical heads are collectively referred to as domed heads. They are formed by pressing or spinning, large diameters are fabricated from formed sections. Torispherical are often referred to as dished ends. After comparing the thickness of all heads, torispherical head had been chosen because of operating pressure for this below 10 bars and suitable for liquid vapor phase process in inconsistent high pressure. The thickness of torispherical head can be calculated below:
β―=
ππ π
π πΆπ 2π½π + ππ (πΆπ β 0.2)
3.50
Where, Rc
= Crown radius
= Dc
Rk
= Knuckle radius
= 0.06 Rc = 0.06 x 1330 = 79.8 mm
Cs
= Stress concentration =
1 4
3+
π
π π
π
=
1 4
3+
1330 79.8
= 1.77
β―=
0.0902 Γ 1330 Γ 1.77 2 1 (115) + 0.09(1.77 β 0.2)
= 0.923 mm
Round up to 2 mm
3.51
3-34 For welding purposes the thickness of head were taken as same thickness of the vessel, = 2 mm. Itβs matching to joint factor were taken as 1. 3.3.6
Weight Loads
The major sources of the dead weight loads are:
1.
The vessel shell
2. The vessel fittings: manhole, nozzles 3. Internal fitting: plates, heating cooling coils 4. External fittings: ladders, platforms, piping 5. Auxiliary equipment which is not self supported, condensers, agitators 6. Insulation 7. The weight of liquid to fill the vessel.
3.3.6.1 Dead Weight of Vessel Dead weight of vessel can be calculated by using equation below: Wv = 240 x Cv x Dm x (Hv + 0.8 Dm)t x 10-3 kN
(3.52)
Where, Wv
= total weight of shell, excluding internal fitting such as plates
Cv
= a factor to account for the weight of nozzles, man ways and internal supports. (In this distillation column, take Cv as 1.15)
Dm
= mean diameter of vessel = (Dc + t) m = (1.33 + 0.007) = 1.337 m
Wv
Hv
= height or length between tangent lines, m
t
= wall thickness, m
= 240 x 1.15 x 1.337(13.0 + 0.8(1.337)) x 0.007 = 36.34 kN
3-35 3.3.6.2 Weight of Plate, Wp From Nelson Guide, page 833 Chemical Engineering Volume 6; take contacting plates, 1.2 kN/m2 (for typical liquid loading). The total of weight of plate determine by multiply the value with number of plate design.
ππππ‘π ππππ = =
ππ· 2 4 3.142 Γ 1.33 4
2
= 1.3895 π2
Weight of plate
= 1.2 x 1.3895 = 1.6674 kN
Where, 1.2 is factor for contacting plates including typical liquid loading in kN/m2
Thus, for 24 plates
= 24 x 1.6674 = 40.017 kN
3.3.6.3 Weight of Insulations The mineral wool was chosen as insulation material. By referring to Coulson & Richardson, Chemical Engineering Design, Volume 6, page 833, Density of mineral wool, Ο
= 130 kg/m3
Thickness
= 75 mm = 0.075 m
Volume of insulation, Vi
= Ο x Di x Hv x thickness of insulation = (3.142)(1.33)(13.0)(0.075) = 4.074 m3
Weight of insulation, Wi
= Volume of insulation x Ο x g = 4.074 x 130 x 9.81 = 5196.07 N = 5.196 kN
Double this to allow for fittings, 10.392 kN
3-36 3.3.6.4 Total Weight The total weight is the summation of dead weight of vessel, weight of accessories and weight of insulation:
Total weight
= Wv + W Ο + Wi
(4.53)
= 36.34 + 40.017 + 10.392 = 86.749 kN 3.3.7
Wind Load
A wind loading must be designed to withstand the highest wind speed that is likely to encounter at the site during the life of the plant. From the British Standard Code of Practice BS CP 3: 1972 βBasic Data for the Design of Buildings, Chapter V Loading: Part 2 Winds Loadβ, (Sinnot, 1999), a wind speed of 160 km/h (100 mph) can be used for preliminary design. For cylindrical column, semi-empirical equation can be used to estimate the wind pressure:
Pw
= 0.05 x uw2
(3.54)
Where,
Pw
Pw
= wind pressure, N/m2
uw
= wind speed, km/h
= 0.05 (160)2 = 1280 N/m2
Mean diameter, including insulation = Di + 2(t + tins) = 1330 + 2(7 + 75) = 1494 mm = 1.494 m
Loading (per linear meter), Fw
= PwDeff = 1280 x 1.5 = 1920 N/m
3-37
π΅ππππππ ππππππ‘, ππ₯ =
=
πΉπ€ π»π£ 2 2 1920 Γ 132 2
= 162240 Nm = 162.24 kNm
3.3.8
Analysis of Stresses of Vessel
3.3.8.1 Pressure Stress Thickness is taken as 9 mm as maximum.
1. Longitudinal stresses due to pressure is given by: ππΌ π·π 4π‘ 0.0902 Γ 1330 = 4 Γ 9
ππΏ =
(3.55)
= 3.332 N/mm2
2. Circumferential stresses due to pressure are given by: ππΌ π·π 2π‘ 0.0902 Γ 1330 = 2 Γ 9
ππ =
(3.56)
= 6.665 N/mm2
3.3.8.2 Dead Weight Stress Dead weight stresses is significant for tall columns. This stress can be tensile for points below the column support or compressive for points above the support. Dead weight stresses are given by:
ππ€ =
π π π·π + π‘π π‘
(3.57)
3-38
=
86749 π 1330 + 9 Γ 9
= 2.29 N/mm2 (compressive) 3.3.8.3 Bending Stress The bending stress will be compressive or tensile, depending on location and are given by:
ππ = Β±
ππ₯ πΌπ£
π·π + π‘ 2
(3.58)
Where, Mx
= Total bending moment
Do
= Outside diameter = Di + 2t = 1330 + 2(9) = 1348 mm
Iv
= Second moment area π = π·π 4 β π·π 4 64 π = 13484 β 13304 64
(3.59)
= 1.358 x 1011 mm4
ππ = Β±
162240 1.358 x 1011
1330 + 9 2
= Β± 8.052 x 10-4 N/mm2 3.3.8.4 The Resultant Longitudinal Stress The resultant of longitudinal stress is the summation of longitudinal stresses, dead weight stresses and bending stress. ππ§ = ππΏ + ππ€ Β± ππ For upwind, ππ§ = 3.332 + (β2.29) + 8.052 x 10 = 1.031 N/mm2
(3.60)
3-39 For downwind, ππ§ = 3.332 + β2.29 β 8.052 π₯ 10β4 = 1.041 N/mm2
Therefore, the greatest difference between the principles stresses, ππ§ = ππ β ππ§ πππ€ππ€πππ
(3.61)
= 6.665 β 1.041 = 5.624 N/mm2 The value obtained is well below the maximum allowable design stress which is 115 N/mm2 3.3.8.5 The Resultant Bulking Stress Local bulking will normally occur at stress than that required buckling the complete. A column design must be checked to ensure that the maximum value of the resultant axial stress does not exceed the critical value at which buckling will occur.
Critical buckling stress, Οc = 2 x 104 = 2 x 104
t Do
(3.62)
9 1348
= 133.531 N/mm2
The maximum compressive stress will occur when the vessel is not under pressure ππ€ + ππ
= 2.29 + 8.052 x 10-4 = 2.291 N/mm2
Since the result of maximum compressive stress is below the critical buckling stress of 157.07 N/mm2. Thus, the design is satisfactory. 3.3.9
Vessel Support Design
The method used to support a vessel will depend on the size, shape and weight of the vessel; the design temperature and pressure, the vessel location and arrangement; and the internal and external fittings and attachment. Since the distillation column is a vertical vessel, skirt support is used in this design.
3-40 A skirt support consists of a cylindrical or conical shell welded to the base of the vessel. A flange at the bottom of the skirt transmits the load to the foundations. The skirt may be welded to the bottom, level of the vessel. Skirt supports are recommended for vertical vessels as they do not imposed concentrated loads on the vessel shells; they are particularly suitable for use with tall columns subject to wind loading.
Type of support
= Straight cylindrical skirt
Ξs
= 90Β°
Material of construction
= Carbon Steel
Design stress, f
= 115 N/mm2
Youngβs modulus
= 200,000 N/mm2
Skirt height, hs
=4m
Skirt thickness, ts
= 9 mm
Joint factor
= 0.85
3.3.9.1 Weight of the Skirt Approximate weight, W approx = (Ο/4 x Di2 x Hv) x ΟL x 9.81 = (Ο/4 x 1.332 x 13) x 804.04 x 9.81 = 91 492 N = 91.492 kN
Weight of vessel, W
= 86.749 kN
Total weight
= 91 492 kN + 86.749 kN = 178.241 kN
3.3.9.2 Analysis of Stresses of Skirt
1. Bending moment of skirt, Ms Bending moment at base skirt, Ms
= 0.5 x Fw(Hv + Hs)2 = 0.5 x 1.92(13 + 4) = 277.44 kNm
(3.63) 2
3-41 2. Bending stress of skirt, Οbs π΅ππππππ π π‘πππ π ππ π ππππ‘, πππ = =
4 ππ π π·π + π‘π Γ π·π π‘π
(3.64)
4 Γ 162.24 Γ 103 Γ 103 π 1330 + 9 Γ 1330 Γ 9
= 12.888 N/mm2 3. Dead weight stress in skirt, Οws ππ€π π‘ππ π‘ = =
πππππππ₯ π π‘π (π·π + π‘π )
(3.65)
152092.43 π Γ 9(1128 + 9)
= 4.731 N/mm2
ππ€π ππππππ‘πππ = =
π ππ‘π (π·π + π‘π )
(3.66)
84749 π Γ 9(1330 + 9)
= 2.291 N/mm2 4. Resultant stress in skirt, Οs Maximum Οs (tensile)
= Οbs + Οws test
(3.67)
= 12.888 + 2.416 = 15.304 N/mm2 Maximum Οs (compressive)
= Οbs - Οws operating = 12.888 β 2.291 = 15.179 N/mm2
5. Criteria for Design Take the joint factor, J as 0.85 Where ΞΈs = 90Β° Οs (tensile)
<
fs J sin ΞΈs
15.304 N/mm2
<
115 x 0.85 sin 90Β°
(4.68)
3-42 15.304 N/mm2
<
Οs (compressive) <
97.75 N/mm2 0.125 E(ts/Ds) sin ΞΈs
15.179 N/mm
2
<
0.125 x 200000(9/1330) sin 90Β°
15.179 N/mm
2
<
169.17 N/mm2
Both criteria are satisfied, add 2 mm for corrosion; gives a design thickness, ts of 11 mm. The type of this equipment is assumed to be completely satisfactory thus the corrosion rate is 0.25 mm/y. Since the operation of this equipment is assumed to be operated for 20 years, thus the corrosion rate will be added:
0.25 mm/y x 20
= 5 mm
The design thickness must be added with the corrosion rate, gives actual design thickness, ts of 16 mm. 3.3.9.3 Base Ring and Anchor Bolts The loads carried by the skirt are transmitted to the foundation slab by the skirt base ring (bearing plate). The moment produced by wind and other lateral will tend to overturn vessel. A variety of base ring designs is used with skirt supports. The simplest types, suitable for small vessel, are rolled angle. The preliminary design of base ring is done by using Scheimanβs short cut method. Scheiman gives the following guide rules which can be used for the selection of the anchor bolts. Refer to Coulson and Richardsonβs, Chemical Engineering, Volume 6, page 848.
1. Bolts smaller than 25 mm diameter should not be used. 2. Minimum number of bolts = 8 3. Use multiple of 4 bolts 4. Bolts pitch should not be less than 600 mm
Pitch circle diameter, Db
= 3.2 m
Circumference of bolt circle = 2200Ο Closest multiple of 4, Nb
= 16 bolts
3-43 Bolt design stress, fb
= 125 N/mm2 (Scheiman, 1963)
Fw
= 1920 N/m
Ms
= 162.24 kNm
Number of bolts required, at minimum recommended bolt spacing = Circumference of bolt circle / 600 = 2200Ο / 600 = 11.52 β 12 bolts
π΅πππ‘ π ππππππ =
π Γ 3.2 Γ 103 = 773.32 ππ (π ππ‘ππ ππππ‘πππ¦) 13
π΅πππ‘ ππππ, π΄π =
1 4 ππ Γ β π ππ ππ π·π
=
(3.69)
1 4 Γ 162240 Γ β 86749 13 π₯ 125 3.2
= 71.416mm2 Use M24 bolts (BS 4190:1967) root area = 353 mm2
π΅πππ‘ ππππ‘ ππππππ‘ππ =
=
4π΄π π
(3.70)
71.416 Γ 4 π
= 9.536 mm
Total compressive load on the base ring per unit length: πΉπ =
4ππ ππ·π
2
+
π ππ·π
Where, Fb
= the compressive load on the base ring, Newtons per linear metre
Ds
= skirt diameter, m
(3.71)
3-44
πΉπ =
4 Γ 162240 86749 + 2 π Γ 3.0 π Γ 3.0
= 32.157 kN/m
The minimum width of the base ring: πΏπ =
πΉπ 1 π₯ ππ 103
(3.72)
Where, Lb
= base ring width, mm
fc
= the maximum allowable bearing pressure in the concrete foundation pad, which will depend on the mix sed, and will typically range from 3.5 to 7 N/mm2 (500 to 1000 psi)
πΏπ =
32157 1 π₯ 5 103
= 6.4314 mm
This is the minimum width required; actual width will depend on the chair design. Actual width required (Coulson and Richardsonβs, Chemical Engineering, Volume 6, page 849, figure 13.30) = Lr + ts + 50 mm = 76 + 16 + 50 = 142 mm
Actual bearing pressure on concrete foundation, πβ²π =
32157 142 Γ 103
= 0.226 N/mm2
The minimum thickness is given by, π‘π = πΏπ
3 πβ²π ππ
Where, Lr
= the distance from the edge of the skirt to the outer edge of the ring, mm
tb
= base ring thickness, mm
(3.73)
3-45 fβc
= actual bearing pressure on base, N/mm2
fr
= allowable design stress in the ring material, typically 140 N/mm2
π‘π = 76
3 Γ 0.226 140
= 5.29 mm
The chair dimensions from figure 13.30 for bolt size M24.
Skirt is to be welded flush with outer diameter of column shell. 3.3.10 Design of Nozzles There are three nozzles in the distillation column, which are nozzles in feed inlet, top product outlet and bottom product outlet. By assuming that the flow of the pipe is turbulent flow, therefore to determine the optimum duct diameter is: Optimum duct diameter, dopt = 293 G0.53Ο-0.37
(3.74)
Where, G
= flowrate, kg/s
Ο
= density, kg/m3
The material construction used for nozzles is carbon steel pipe.
3.3.10.1 G
Feed Stream
= 1.587 x 104 kg/h = 4.4083 kg/s
Οmix
= 778.66 kg/m3
dopt
= 293 (4.4083)0.53 (778.66)-0.37 = 54.77 mm β 55 mm
πππ§π§ππ π‘πππππππ π , π‘ =
ππ ππππ‘ 20π + ππ
3.75
3-46 Where, Ps
= operating pressure
Ο
= design stress at working temperature
ππππππππ π ππ πππ§π§ππ: π‘ =
0.0902 Γ 54.77 20 125 + 0.0902
= 0.002 mm
So, the thickness of nozzle
= corrosion allowance + 0.002 = 2 + 0.002 = 2.002 mm
3.3.10.2 G
Top Stream
= 128.5 kg/h = 0.0357 kg/s
Οmix
= 1.709 kg/m3
dopt
= 293 (0.0357)0.53 (1.709)-0.37 = 41.08 mm β 41 mm
πππ§π§ππ π‘πππππππ π , π‘ =
ππ ππππ‘ 20π + ππ
ππππππππ π ππ πππ§π§ππ: π‘ =
0.0902 Γ 41.08 20 125 + 0.0902
= 0.00148 mm
So, the thickness of nozzle
= corrosion allowance + 0.00148 = 2 + 0.00148 = 2.00148 mm
3-47 3.3.10.3 G
Bottom Stream
= 1.574 x 104 kg/h = 4.372 kg/s
Οmix
= 695.2 kg/m3
dopt
= 293 (4.372)0.53 (695.2)-0.37 = 56.88 mm β 60 mm
πππ§π§ππ π‘πππππππ π , π‘ =
ππ ππππ‘ 20π + ππ
ππππππππ π ππ πππ§π§ππ: π‘ =
0.0902 Γ 56.88 20 125 + 0.0902
= 0.0021 mm
So, the thickness of nozzle
= corrosion allowance + 0.0021 = 2 + 0.0021 = 2.0021 mm
3.3.11 Flange Design The flange class number required for a particular duty will depend on the design pressure and temperature and the flange material. The flange design is from the typical standard flange design in Coulson and Richardsonβs, Chemical Engineering, Volume 6, page 863, figure 13.37.
Table 3.12 The Summary of Flange Design Feed Stream dopt
d1
65
76.1
Flange
Raised face
D
b
H
d4
f
160
14
32
110
3
Bolting M12
Drilling
Boss
No
d2
k
d3
4
14
130
100
3-48 Top Stream Flange
dopt
d1
25
33.7
Raised face
D
b
H
d4
f
100
14
24
60
2
Bolting M10
Drilling
Boss
No
d2
k
d3
4
11
75
50
Bottom Stream Flange
dopt
d1
50
60.3
Raised face
D
b
H
d4
f
140
14
28
90
3
Bolting M12
Drilling
Boss
No
d2
k
d3
4
14
110
80
3.3.12 Summary of Mechanical Design Summary of design distillation column are shown in table below:
Table 3.13 The Summary of Mechanical Design Pressure Vessel Operating Pressure, Po
0.11 N/mm2
Design Pressure, P1
0.19 N/mm2
Operating Temperature
90 oC
Design Temperature
99 oC
Column Material
Carbon Steel
Safety Factor
10%
Design Stress
115 N/mm2 Head and Closure
Types
Torispherical Head
Crown Radius, Rc
1.128 m
Knuckle Radius, Rk
0.0677 m
Joint Factor, J
1
Cs
1.77
Minimum thickness, e
10 mm Column Weight
Dead weight of Vessel, Wv
10.52 kN
Weight of Plates, W p
32.384 kN
3-49 Weight of Insulation, W i
5.424 kN
Total Weight, W t
53.752 kN
Wind Speed, Uw
160 km/h
Bending Moment, Mx
210.0429 kN/m
Insulation Material
Mineral Wool
Insulation Thickness
75 mm Skirt Support
Type of Support
Straight Cylindrical Skirt
Material of Construction
Carbon Steel
Youngβs Modulus
200,000 N/mm2
Approximate Weight, W approx
152.092 kN
Total Weight
205.844 kN
Bending Moment, Mx
328.2 kNm
Skirt Thickness, ts
9 mm
Skirt Height, Hs
4m Stiffness Ring
Critical Buckling Pressure for Ring, Pc
140 N/mm2
3-50 3.4
Costing for Distillation Column
The purchased cost of the equipment is calculated using equation below (Turton et al., Analysis, Synthesis, and Design of Chemical Processess, 3rd Edition, page 906): log10 CpΒ° = K1 + K2 log10 (A) + K3 [log10 (A)]2 where,
3.4.1
A
= capacity or size parameter for the equipment
K1, K2, K3
= constants in Table A.1 (Appendix A)
Process Vessels
Column height = 13 m; Column diameter = 1.33 m Material of construction Volume
= carbon steel with stainless steel cladding = ππ· 2 π = (3.142)(1.33)2(13) = 72.25m3
From Table A.1 (Appendix A), K1 = 3.4974, K2 = 0.4485, K3 = 0.1074 log10 CpΒ°
= 3.4974 + 0.4485 log10 (72.25) + 0.1074 [log10 (72.25)]2 = 4.73 CpΒ° = $ 53 751.28
Pressure factors for process vessels: For pressure vessel, when t vessel οΎ 0.003m,
(1.0)
3-51
πΉπ,π£ππ π ππ
πΉπ,π£ππ π ππ
π +1 π· + 0.00315 2[850 β 0.6 π + 1 ] = 0.003
(2.0)
0.0902 + 1 (1.33) + 0.00315 2[850 β 0.6 0.0902 + 1 ] = 0.003
πΉπ,π£ππ π ππ = 8.96 Γ 10β4
The bare module factor for process vessel (Turton et al., Analysis, Synthesis, and Design of Chemical Processess, 3rd Edition, page 927): CBM
= CpΒ°FBM = CpΒ°(B1 + B2FMFp)
(3.0)
From Table A.4 (Appendix A), B1 = 2.25, B2 = 1.82 From Table A.3 (Appendix A), the identification number for carbon steel vertical process vessels is 18.
Hence, from Figure A.18 (Appendix A), material factor, FM = 1.0 CBM
= 53751.28 [2.25 + (1.82)(1.0)(8.96Γ10-4)] = $ 121 028.03
Use correlation: CEPCI for year of 2009 is 645.5 CEPCI for year of 2001 is 397 Therefore,
3-52
πππ€ πΆπ΅π = 121 028.03 Γ
645.5 297
= $ 263042.40 = RM 802 279.32
3.4.2
Sieve Tray
Column height = 13 m; Column diameter = 1.33 m, Area = 5.56 m2; Number of trays = 24 From Table A.1 (Appendix A), K1 = 2.9949, K2 = 0.4465, K3 = 0.3961 log10 CpΒ°
= 2.9949+ 0.4465 log10 (5.56) + 0.3961 [log10 (5.56)]2 = 3.92 CpΒ° = $ 8 275.15
The bare module cost for sieve trays (Turton et al., Analysis, Synthesis, and Design of Chemical Processess, 3rd Edition, page 930, Table A.5): CBM
= CpΒ°NFBMFq
(4.0)
Where, N
= number of trays
Fq
= quantity factor for trays
For Nβ₯ 20, Fq = 1 From Table A.6 (Appendix A), the identification number for stainless steel sieve trays is 61 Hence, from Figure A.19 (Appendix A), bare module factor, FBM = 1.8
3-53 CBM
= (8275.15)(24)(1.8)(1) = $ 208 533.79
Use correlation: CEPCI for year of 2010 is 645.5 CEPCI for year of 2001 is 397 Therefore, πππ€ πΆπ΅π = 208 533.79 Γ
645.5 297
= $ 453 227.48 = RM 1 382 343.81
Thus, the total cost for distillation column = RM 802 279.32 + RM 1 382 343.81 = RM 2 184 623.13
3-54 REFERENCES
R. K. Sinnot. 2003. Chemical Engineering Design. Vol 6, 3rd Ed, Elsevier Butterworth Heinemann. Felder, R. M. & Rousseau, R. W. 2000. Elementary Principles of Chemical Processes. 3rd Ed, John Wiley & Sons, Inc. Levenspiel, O. 1999. Chemical Reaction Engineering. 3rd. Ed, John Wiley & Sons, Inc. Perry, R. H. & Green, D. W. 1998. Perryβs Chemical Engineerβs Handbook. 7th Ed, McGraw-Hill International Edition.
Walas, S. M. 1988. Chemical Process Equipment. Butterworths Publishers. Ludwig, E. E. 1995. Applied Process Design. Vol.2, 3rd. Ed, Gulf Publishing Company.