Chapter 3 - Distillation Column Design PDF
Chapter 3 - Distillation Column Design PDF
Chapter 3 - Distillation Column Design PDF
CHAPTER 3
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:
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
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.
The composition of the inlet and outlet streams for distillation column is shown in table 3.1:
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.
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.
COMPONENT a b c d e f
Antoine equation:
𝐵
ln 𝑃𝑜 = 𝐴 − + 𝐷 𝑥 ln 𝑇 + 𝐸 𝑥 𝑇^𝐹 (3.3)
𝑇+𝐶
𝑃𝑜
𝐾𝑖 = (3.4)
𝑃𝑇
3-4
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:
O.P
COMPONENT ln Pi Pi (kPa) Xi Ki Yi=KiXi
(kPa)
1-Propanol 5.19 179.42 0.9768 182 0.99 0.962939
TOTAL 1.00000
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:
O.P
COMPONENT ln Pi Pi (kPa) Yi Ki Xi=Yi/Ki
(kPa)
1-Propanol 3.59 36.41 0.4853 50 0.73 0.67
TOTAL 1
3-5
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:
O.P
COMPONENT ln Pi Pi (kPa) Xi Ki Yi=KiXi
(kPa)
1-Propanol 4.69 108.95 0.9815 110 0.99 0.97
TOTAL 1
𝑌𝑖
𝐾𝑖 = (3.5)
𝑋𝑖
Pi = PiXi (3.7)
The relative volatility of two components can be expressed as the ratio of their K value,
𝐾𝐿𝐾
𝛼𝑖𝑗 = (3.8)
𝐾𝐻𝐾
Where, KLK = Light key components
KHK = Heavy key components
Table 3.6
𝑲
COMPONENT K 𝜶=
𝑲𝑯𝑲
Table 3.7
𝑲
COMPONENT K 𝜶=
𝑲𝑯𝑲
= 0.6636 (0.8534)
= 0.753
The Fenske’s 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:
X X
Log[(X LK )]d [( XHK )]b
HK LK
Nmin = (3.9)
Log αLK
0.73 0.0091
Log[( 1.1 )]d [( )]
= 0.9815 b
Log 0.753
= 17.94
= 20 stages
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:
𝛼𝑖 𝑥𝑖,𝑑
= 𝑅𝑚 + 1 (3.10)
𝛼𝑖 − 𝜃
3-8
Where,
𝛼𝑖 𝑥𝑖,𝑓
=1−𝑞 (3.11)
𝛼𝑖 − 𝜃
Where,
The value of θ must lie between the values of relative volatility of the light and heavy keys and is
found by trial and error.
𝛼𝑖 𝑥𝑖,𝑓
=0
𝛼𝑖 − 𝜃
Table 3.8
3.9 -0.07
Therefore, θ = 3.9
3-9
Table 3.9
3.9 9.35
Rm + 1 = 9.35
Rm = 8.35
𝑅𝑚
= 0.8931
𝑅𝑚 + 1
𝑅 2
= = 0.66
(𝑅 + 1) 3
Using Erbar – Maddox correlation (Erbar and Maddox, 1961) from figure 11.11 (Coulson and
Richardson, Volume 6, page 524),
𝑁𝑚
= 0.74
𝑁
18
N=
0.74
= 24.3
2 3 4 5
R
24.3 21.43 20.69 20.22
N
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.
2
𝑁𝑟 𝐵 𝑥𝑓,𝐻𝐾 x𝑏,𝐿𝐾
𝐿𝑜𝑔 = 0.2606 log (3.10)
𝑁𝑠 𝐷 𝑥𝑓,𝐿𝐾 x𝑑,𝐻𝐾
Where,
2
𝑁𝑟 2.531 0.0092 0.395
𝐿𝑜𝑔 = 0.2606 log
𝑁𝑠 261.5 0.9768 0.00382
𝑁𝑟
= 0.993
𝑁𝑠
Nr + Ns = 24
0.993Ns + Ns = 24
1.993Ns =9
3-11
Nr = 15
Where,
1 1
𝐿𝑂𝐺 µ𝑎 = 𝑉𝑖𝑠𝐴 𝑥 − (3.12)
𝑇 𝑉𝑖𝑠𝐵
Viscosity
Mole fraction Coefficient Viscosity
Component Log µ𝒂 µ𝒂 × 𝒙
feed, x (mNs/m2)
A B
Where,
= 55.44 %
Plate and overall column efficiencies will normally be between 30% to 70%. (Coulson and
Richardson’s, volume 6, page 547)
= 60.118 kg/kmol
= 50.739 kg/kmol
= 60.191 kg/kmol
3-13
3.2.8.2 Density
Top Product :
ρL = 𝑥𝐵,𝑖 𝜌𝑖 (3.14)
= 823.51 kg/m3
= 1.731 kg/m3
Bottom Product:
ρL = 𝑥𝐷,𝑖 𝜌𝑖 (4.16)
= 804.04 kg/m3
= 3.071 kg/m3
Using Sugden (1924), equation 8.23 (Coulson and Richardson’s, volume 6, page 335)
4
𝑃𝑐(𝜌𝑙 − 𝜌𝑣
𝜍= 𝑥 10−12 (3.18)
𝑀
3-14
Where,
Where,
= 177.28097
= 148.30792
65.01 969.64−4.928 4
Top Column, 𝜍 = 𝑥 10−12
21.98
= 67.965683 N/m
= 15.27159545 N/m
Where,
R = Reflux ratio
Hence,
Vn = 261.5 (2.531 + 1)
= 923.36 kmole/hr
= 923.36 – 261.5
3-16
= 661.86 kmole/hr
Where,
Hence,
Lm = 661.86 + 264.1
= 925.96kmole/hr
Where,
Hence,
Vm = 925.96 – 261.5
= 664.46kmole/hr
The equation for the operating lines below the feed plate:
𝐿𝑚 𝑊
𝑌𝑚 = 𝑋𝑚 + 1 − 𝑋𝑤 (3.24)
𝑉𝑚 𝑉𝑚
925.96 261.5
𝑌𝑚 = 𝑋𝑚 + 1 − (𝑋𝑤)
664.46 664.46
261.5
= 2.058(Xm + 1) – 664.46
(𝑋𝑤)
The equation for the operating lines above the feed plate:
𝐿𝑛 𝐷
𝑌𝑛 = 𝑋𝑛 + 1 − 𝑋𝑑 (3.25)
𝑉𝑛 𝑉𝑛
3-17
661.86 261.5
𝑌𝑛 = 𝑋𝑛 + 1 − 𝑋𝑑
923.36 923.36
𝐿𝑛 𝜌𝑉
𝐹𝐿𝑉 𝑇𝑜𝑝 = (3.26)
𝑉𝑛 𝜌𝐿
1.731
= 0.72
823.51
= 0.033
𝐿𝑚 𝜌𝑉
𝐹𝐿𝑉 𝐵𝑜𝑡𝑡𝑜𝑚 = (3.27)
𝑉𝑚 𝜌𝐿
3.071
= 1.39
804.04
= 0.09
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 Richardson’s, 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 high-
pressure drop. The equation below which is based on the Souder and Brown equation,
Lowenstein (1961), Coulson & Richarson’s 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.
𝜌𝐿 − 𝜌𝑣 0.5
𝑈𝑣 = −0.171𝑙𝑡 2 + 0.271𝑙𝑡 − 0.047 (3.28)
𝜌𝑣
0.5
969.64 − 4.928
= −0.171(0.5)2 + 0.271(0.5) − 0.047
4.928
= 2.8173 m/s
Where,
Uv = maximum allowable vapor velocity based on the gross (total) column cross
4𝑉𝑤
𝐷𝑐 = (3.29)
𝜋𝜌𝑣 𝑈𝑣
15870 𝑘𝑔 1 𝑟
𝑉𝑤 = 𝑥
𝑟 3600 𝑠
= 4.41 kg/s
3-19
4(4.41)
𝐷𝑐 =
𝜋 4.928 (0.64)
= 1.33 m
The column area can be calculated from the calculated internal column diameter
𝜋 𝐷𝑐 2
𝐴𝑐 = (3.30)
4
𝜋 (1.33)2
=
4
= 1.39 m2
Before deciding liquid flow arrangement, maximum volumetric liquid rate were determined by
the value of maximum volumetric rate
15740 𝑘𝑔 1 𝑟
𝐿= 𝑥 (3.31)
𝑟 3600 𝑠
4.372 𝑘𝑔 𝑚3
= 𝑥
𝑠 804.04 𝑘𝑔
= 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.
= 1.39 m2 - 0.1668 m2
= 1.2232 m2
= 1.0564 m2
= 0.10 x 1.0564 m2
= 0.10564m2
3-21
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 Richardson’s, Chemical Engineering, Volume 6,
page 572, with (Ad/Ac) x 100 is 12 percents, thus, Iw/Dc is 0.76
= 0.76 x 1.33 m
= 1.011 m
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.
𝜋(𝑑 )2
𝐴𝑙 = (3.36)
4
𝜋(0.005)2
=
4
= 1.9635 x 10-5 m2
3-22
𝑎𝑟𝑒𝑎
𝑁 = (3.37)
1 𝑜𝑙𝑒 𝑎𝑟𝑒𝑎
0.10564
=
1.9635 𝑥 10−5
= 5380.19 holes
≈ 5380 holes
Check weeping to ensure enough vapour to prevent liquid flow through hole.
15740𝑘𝑔 1 𝑟
𝑀𝑎𝑥𝑖𝑚𝑢𝑚 𝑙𝑖𝑞𝑢𝑖𝑑 𝑟𝑎𝑡𝑒 = 𝑥
𝑟 3600 𝑠
= 4.372 kg/s
= 3.06 kg/s
2
𝐿𝑤 3
𝑜𝑤 = 750 (3.38)
𝜌𝐿 𝐼𝑤
Where,
Iw = weir length, m
At maximum rate:
2
4.372 3
𝑜𝑤 = 750
804.04 𝑥 1.011
= 20.40 mm liquid
At minimum rate:
2
3.06 3
𝑜𝑤 = 750
1019.01 𝑥 0.85728
= 18.15 mm liquid
how + hw = 18.15 + 50
= 68.15 mm liquid
From Figure 11.30, in Coulson and Richardson’s, Chemical Engineering, Volume 6, page 571,
weep point correlation, K2 = 30.7
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
𝐾2 − 0.9(25.4 − 𝑑 )
𝑈 (min) = 1 (3.39)
𝜌𝑣 2
Where,
dh = hole diameter, mm
K2 = constant
30.7 − 0.9(25.4 − 5)
= 1
(3.071)2
= 8.036 m/s
4.41 𝑘𝑔 𝑚3
𝑥 0.7 𝑥
𝑠 3.071𝑘𝑔
=
0.10564
= 9.51 m/s
4.41 𝑘𝑔 𝑚3
𝑠 𝑥
3.071 𝑘𝑔
=
0.10564
= 13.59 m/s
From Figure 11.34 in Coulson and Richardson’s, Chemical Engineering, Volume 6, page 576,
for discharge coefficient for sieve plate,
𝑝𝑙𝑎𝑡𝑒 𝑡𝑖𝑐𝑘𝑛𝑒𝑠𝑠 3 𝑚𝑚
𝐴𝑡, = = 0.6
𝑜𝑙𝑒 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟 5 𝑚𝑚
𝐴
𝑎𝑛𝑑 = 0.1
𝐴𝑎
we get Co = 0.74
2
𝑈 𝜌𝑣
𝐷𝑟𝑦 𝑝𝑙𝑎𝑡𝑒, 𝑑 = 51 (3.42)
𝐶𝑜 𝜌𝐿
2
13.59 3.071
= 51
0.74 804.04
= 65.697 mm liquid
12.5 𝑥 103
𝑅𝑒𝑠𝑖𝑑𝑢𝑎𝑙 𝑒𝑎𝑑, 𝑟 = (3.43)
𝜌𝐿
12.5 𝑥 103
=
804.04
= 15.55 mm liquid
= 149.397 mm liquid
3-26
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 – 10 mm
= 50 – 10
= 40 mm
Where, hap = height of the bottom edge of the apron above the plate
2
𝐿𝑤𝑑
𝑑𝑐 = 166 (3.45)
𝜌𝐿 𝐴𝑚
Where,
Am = either the downcomer area, Ad or the clearance area under the downcomer, Aap
whichever is smaller, m2
= 0.04 m x 1.011
= 0.04044 m2
As this less than Ad = 0.1668 m2, equation 11.92 (Coulson and Richardson’s, Volume 6, page
577) used Aap = 0.04044 m2
3-27
2
4.372
𝑑𝑐 = 166
804.04 𝑥 0.04044
= 3.00 mm
= 220.547 mm
0.2205 m <0.5 m
Sufficient residence time must be allowed in the downcomer for the entrained vapor to
disengage from the liquid stream, to prevent heavily “aerated” liquid being carried under the
downcomer. A time at least 3 seconds is recommended.
𝐴𝑑 𝑏𝑐 𝜌𝐿
𝑡𝑟 = (3.48)
𝐿𝑤𝑑
Where,
tr = residence time, s
= 6.764 s
From Figure 11.32, Coulson and Richardson’s, 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
θ = 98°
Ih/Dc = 0.18
= 180° - 98°
= 82°
= 1.832 m
= 1.832 x 0.05
= 0.0916 m2
3-29
= 1.011 + 0.05
= 1.061 m
= 2 x 0.05 x 1.061
= 0.1061 m2
= 0.8587 m2
Ah/Ap = 1.0564/0.8587
= 0.123
From Figure 11.33, Coulson and Richardson’s, Chemical Engineering, Volume 6, page 528, the
relation between hole area and pitch,
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 – 1) x (Tray spacing ) + (Tray spacing x 2) + (No. of stages – 1)
3-30
x (Plate thickness)
= 11.57 m
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
𝑃𝑖 𝐷𝑖
ℯ= 3.49
2𝐽𝑓 − 𝑃𝑖
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
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.
𝑃𝑖 𝑅𝑐 𝐶𝑠
ℯ= 3.50
2𝐽𝑓 + 𝑃𝑖 (𝐶𝑠 − 0.2)
Where,
Rc = Crown radius = Dc
Rk = Knuckle radius = 0.06 Rc
= 0.06 x 1330
= 79.8 mm
Cs = Stress concentration
1 𝑅𝑐
= 3+
4 𝑅𝑘
1 1330
= 3+
4 79.8
= 1.77
Round up to 2 mm
3-34
For welding purposes the thickness of head were taken as same thickness of the vessel, = 2
mm. It’s matching to joint factor were taken as 1.
Where,
𝜋𝐷 2
𝑃𝑙𝑎𝑡𝑒 𝑎𝑟𝑒𝑎 =
4
2
3.142 × 1.33
=
4
= 1.3895 𝑚2
Where, 1.2 is factor for contacting plates including typical liquid loading in kN/m2
Where,
Pw = wind pressure, N/m2
uw = wind speed, km/h
Pw = 0.05 (160)2
= 1280 N/m2
𝐹𝑤 𝐻𝑣 2
𝐵𝑒𝑛𝑑𝑖𝑛𝑔 𝑚𝑜𝑚𝑒𝑛𝑡, 𝑀𝑥 =
2
1920 × 132
=
2
= 162240 Nm
= 162.24 kNm
𝑝𝐼 𝐷𝑖
𝜍𝐿 = (3.55)
4𝑡
0.0902 × 1330
=
4 × 9
= 3.332 N/mm2
𝑝𝐼 𝐷𝑖
𝜍 = (3.56)
2𝑡
0.0902 × 1330
=
2 × 9
= 6.665 N/mm2
𝑊
𝜍𝑤 = (3.57)
𝜋 𝐷𝑖 + 𝑡𝑖 𝑡
3-38
86749
=
𝜋 1330 + 9 × 9
= 2.29 N/mm2 (compressive)
𝑀𝑥 𝐷𝑖
𝜍𝑏 = ± + 𝑡 (3.58)
𝐼𝑣 2
Where,
Mx = Total bending moment
Do = Outside diameter
= Di + 2t
= 1330 + 2(9)
= 1348 mm
Iv = Second moment area
𝜋
= 𝐷𝑜 4 − 𝐷𝑖 4 (3.59)
64
𝜋
= 13484 − 13304
64
= 1.358 x 1011 mm4
162240 1330
𝜍𝑏 = ± + 9
1.358 x 1011 2
= ± 8.052 x 10-4 N/mm2
𝜍𝑧 = 𝜍𝐿 + 𝜍𝑤 ± 𝜍𝑏 (3.60)
For upwind,
𝜍𝑧 = 3.332 + (−2.29) + 8.052 x 10
= 1.031 N/mm2
3-39
For downwind,
𝜍𝑧 = 3.332 + −2.29 − 8.052 𝑥 10−4
= 1.041 N/mm2
The value obtained is well below the maximum allowable design stress which is 115 N/mm2
t
Critical buckling stress, σc = 2 x 104 (3.62)
Do
9
= 2 x 104
1348
= 133.531 N/mm2
The maximum compressive stress will occur when the vessel is not under pressure
𝜍𝑤 + 𝜍𝑏 = 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.
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.
𝑊
𝜍𝑤𝑠 𝑜𝑝𝑒𝑟𝑎𝑡𝑖𝑛𝑔 = (3.66)
𝜋𝑡𝑠 (𝐷𝑠 + 𝑡𝑠 )
84749
=
𝜋 × 9(1330 + 9)
= 2.291 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.
𝜋 × 3.2 × 103
𝐵𝑜𝑙𝑡 𝑠𝑝𝑎𝑐𝑖𝑛𝑔 = = 773.32 𝑚𝑚 (𝑠𝑎𝑡𝑖𝑠𝑓𝑎𝑐𝑡𝑜𝑟𝑦)
13
1 4 𝑀𝑠
𝐵𝑜𝑙𝑡 𝑎𝑟𝑒𝑎, 𝐴𝑏 = × − 𝑊 (3.69)
𝑁𝑏 𝑓𝑏 𝐷𝑏
1 4 × 162240
= × − 86749
13 𝑥 125 3.2
= 71.416mm2
4𝐴𝑏
𝐵𝑜𝑙𝑡 𝑟𝑜𝑜𝑡 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟 = (3.70)
𝜋
71.416 × 4
=
𝜋
= 9.536 mm
Where,
Fb = the compressive load on the base ring, Newtons per linear metre
Ds = skirt diameter, m
3-44
4 × 162240 86749
𝐹𝑏 = 2
+
𝜋 × 3.0 𝜋 × 3.0
= 32.157 kN/m
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)
32157 1
𝐿𝑏 = 𝑥
5 103
= 6.4314 mm
This is the minimum width required; actual width will depend on the chair design. Actual width
required (Coulson and Richardson’s, Chemical Engineering, Volume 6, page 849, figure 13.30)
= Lr + ts + 50 mm
= 76 + 16 + 50
= 142 mm
3 𝑓′𝑐
𝑡𝑏 = 𝐿𝑟 (3.73)
𝑓𝑟
Where,
Lr = the distance from the edge of the skirt to the outer edge of the ring, mm
tb = base ring thickness, mm
3-45
3 × 0.226
𝑡𝑏 = 76
140
= 5.29 mm
The chair dimensions from figure 13.30 for bolt size M24.
Where,
G = flowrate, kg/s
ρ = density, kg/m3
𝑃𝑠 𝑑𝑜𝑝𝑡
𝑁𝑜𝑧𝑧𝑙𝑒 𝑡𝑖𝑐𝑘𝑛𝑒𝑠𝑠, 𝑡 = 3.75
20𝜍 + 𝑃𝑠
3-46
Where,
Ps = operating pressure
σ = design stress at working temperature
0.0902 × 54.77
𝑇𝑖𝑐𝑘𝑛𝑒𝑠𝑠 𝑜𝑓 𝑛𝑜𝑧𝑧𝑙𝑒: 𝑡 =
20 125 + 0.0902
= 0.002 mm
G = 128.5 kg/h
= 0.0357 kg/s
ρmix = 1.709 kg/m3
dopt = 293 (0.0357)0.53 (1.709)-0.37
= 41.08 mm
≈ 41 mm
𝑃𝑠 𝑑𝑜𝑝𝑡
𝑁𝑜𝑧𝑧𝑙𝑒 𝑡𝑖𝑐𝑘𝑛𝑒𝑠𝑠, 𝑡 =
20𝜍 + 𝑃𝑠
0.0902 × 41.08
𝑇𝑖𝑐𝑘𝑛𝑒𝑠𝑠 𝑜𝑓 𝑛𝑜𝑧𝑧𝑙𝑒: 𝑡 =
20 125 + 0.0902
= 0.00148 mm
𝑃𝑠 𝑑𝑜𝑝𝑡
𝑁𝑜𝑧𝑧𝑙𝑒 𝑡𝑖𝑐𝑘𝑛𝑒𝑠𝑠, 𝑡 =
20𝜍 + 𝑃𝑠
0.0902 × 56.88
𝑇𝑖𝑐𝑘𝑛𝑒𝑠𝑠 𝑜𝑓 𝑛𝑜𝑧𝑧𝑙𝑒: 𝑡 =
20 125 + 0.0902
= 0.0021 mm
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 Richardson’s, Chemical Engineering, Volume 6, page 863, figure 13.37.
Feed Stream
Flange Raised face Drilling Boss
dopt d1 Bolting
D b H d4 f No d2 k d3
65 76.1 160 14 32 110 3 M12 4 14 130 100
3-48
Top Stream
Flange Raised face Drilling Boss
dopt d1 Bolting
D b H d4 f No d2 k d3
25 33.7 100 14 24 60 2 M10 4 11 75 50
Bottom Stream
Flange Raised face Drilling Boss
dopt d1 Bolting
D b H d4 f No d2 k d3
50 60.3 140 14 28 90 3 M12 4 14 110 80
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):
where,
Volume = 𝜋𝐷 2
= (3.142)(1.33)2(13)
= 72.25m3
= 4.73
Cp° = $ 53 751.28
𝑃 +1 𝐷
+ 0.00315
2[850 − 0.6 𝑃 + 1 ]
𝐹𝑃,𝑣𝑒𝑠𝑠𝑒𝑙 = (2.0)
0.003
0.0902 + 1 (1.33)
+ 0.00315
2[850 − 0.6 0.0902 + 1 ]
𝐹𝑃,𝑣𝑒𝑠𝑠𝑒𝑙 =
0.003
The bare module factor for process vessel (Turton et al., Analysis, Synthesis, and Design of
Chemical Processess, 3rd Edition, page 927):
From Table A.3 (Appendix A), the identification number for carbon steel vertical process vessels
is 18.
= $ 121 028.03
Use correlation:
Therefore,
3-52
645.5
𝑛𝑒𝑤 𝐶𝐵𝑀 = 121 028.03 ×
297
= $ 263042.40
= RM 802 279.32
Column height = 13 m; Column diameter = 1.33 m, Area = 5.56 m2; Number of trays = 24
= 3.92
Cp° = $ 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):
Where,
N = number of trays
For N≥ 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:
Therefore,
645.5
𝑛𝑒𝑤 𝐶𝐵𝑀 = 208 533.79 ×
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. Perry’s Chemical Engineer’s Handbook. 7th Ed, McGraw-Hill
International Edition.
Ludwig, E. E. 1995. Applied Process Design. Vol.2, 3rd. Ed, Gulf Publishing Company.