Chapter 3 - Distillation Column Design

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 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

Component

Feed Top Bottom

Molar Flow Rate (kmole/h) Mole Fraction Molar Flow Rate (kmole/h) 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

Dipropyl Ether 2.3986 0.0092 0.0118 0.0047 2.3868 0.0091

3.2.2 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 : 𝑦_𝑖 = 𝐾_𝑖𝑥_𝑖 = 1.0 (3.1)

Dew Point : 𝑥_𝑖 = 𝑦_𝑖/ 𝐾_𝑖 = 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 -8.29×10}3 _{0.00 -8.9096 1.82×10}-6 _2.00 Water _{65.9 -7.23×10}3 0.00 -7.18 4.03×10-6 2.00 Propanal _{80.9 -6.51×10}3 _{0.00 -9.82 6.79×10}-6 _2.00 Dipropyl Ether 96.7 -7.45×10 3 _{0.00 -1.24 1.08×10}-5 _2.00 Antoine equation: ln 𝑃𝑜 _{= 𝐴 −} 𝐵 𝑇 + 𝐶+ 𝐷 𝑥 ln 𝑇 + 𝐸 𝑥 𝑇^𝐹 (3.3) 𝐾𝑖 = 𝑃𝑜 𝑃_𝑇 (3.4)

Estimation of feed temperature, 𝑥_𝑖 = 𝑦_𝑖/ 𝐾_𝑖 = 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) 𝑥_𝑖= 𝑦_𝑖/ 𝐾_𝑖 = 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

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) 𝑦_𝑖 = 𝐾_𝑖𝑥_𝑖 = 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

𝐾𝑖 = 𝑌𝑖

𝑋𝑖 (3.5)

Where, Yi = mole fraction of component i in vapour phase

Xi = mole fraction of component i in liquid phase

The relative volatility, α which is needed in the calculation is defined as 𝛼𝑖𝑗 =

𝐾𝑖

Where i and j represent the components to be separated From Ideal system, Raoult’s law,

Pi = PiXi (3.7)

The relative volatility of two components can be expressed as the ratio of their K value, 𝛼𝑖𝑗 =

𝐾_𝐿𝐾

𝐾_𝐻𝐾 (3.8) Where, KLK = Light key components

KHK = Heavy key components

3.2.3.1 Top Column Table 3.6 COMPONENT K 𝜶 = 𝑲 𝑲_𝑯𝑲 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 COMPONENT K 𝜶 = 𝑲 𝑲𝑯𝑲 1-Propanol 0.9900 0.8534 Water 0.8900 0.7672 Propanal 4.4600 3.8448

DPE 1.1600 1.0000

Average relative volatility of the light key to heavy key;

αLK = Top α (Bottom α)

= 0.6636 (0.8534)

= 0.753

3.2.4 Minimum Number of Stages Using Fenske’s Equation

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:

N

_min

=

Log[(

X

LK

X

_HK

)]

d

[(

X

_X

HK LK

)]

b

Log α

_LK

(3.9)

=

Log[(

0.73

1.1 )]

d

[(

0.0091

_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:

𝛼_𝑖𝑥_𝑖,𝑑

Where,

αi = 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 θ is the root of the equation:

𝛼𝑖𝑥𝑖,𝑓

𝛼𝑖− 𝜃= 1 − 𝑞 (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 θ 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 𝛼𝑖𝑥𝑖,𝑓

𝛼_𝑖− 𝜃= 0 Table 3.8

Component Xi,f αi αiXi,f θ estimate (αiXi,f)/(αi - θ)

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, θ = 3.9

Table 3.9

Component Xi,d αi αiXi,d θ estimate (αiXi,d)/(αi - θ)

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 𝑅𝑚 𝑅𝑚 + 1= 0.8931 Specimen calculation, for R = 2.0

𝑅 (𝑅 + 1)=

2

3= 0.66

Using Erbar – Maddox correlation (Erbar and Maddox, 1961) from figure 11.11 (Coulson and Richardson, Volume 6, page 524),

𝑁𝑚

𝑁 = 0.74 N = 18

0.74

= 24.3 For other reflux ratios

R 2 3 4 5

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:

𝐿𝑜𝑔 𝑁𝑟 𝑁𝑠 = 0.2606 log 𝐵 𝐷 𝑥_{𝑓,𝐻𝐾} 𝑥𝑓,𝐿𝐾 x_{𝑏,𝐿𝐾} x𝑑,𝐻𝐾 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

𝐿𝑜𝑔 𝑁𝑟 𝑁𝑠 = 0.2606 log 2.531 261.5 0.0092 0.9768 0.395 0.00382 2 𝑁𝑟 𝑁_𝑠 = 0.993

Actual number of plates is 24 Nr + Ns = 24 0.993Ns + Ns = 24 1.993Ns = 9

Nr = 15

So, feed inlet is at stage 9 from bottom.

3.2.7 Efficiency of Distillation Column Overall column efficiency is given as:

𝐸˳ = 51 − 32.5 log (µ_𝑎𝜍_𝑎) (3.11)

Where,

µ_𝑎 = the molar average liquid viscosity, mNs/m2 𝜍_𝑎 = average relative volatility of the light key To find the viscosity of the flow:

𝐿𝑂𝐺 µ_𝑎 = 𝑉𝑖𝑠𝐴 𝑥 1_𝑇 − 1

𝑉𝑖𝑠𝐵 (3.12) Table 3.8 Viscosity of the mixture

Component Mole fraction feed, x Viscosity Coefficient _{Log µ} 𝒂 Viscosity _(mNs/m2 ) µ𝒂× 𝒙 A B 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 TOTAL 1.16 0.4637

Where,

𝐸˳ = 51 − 32.5 log (µ𝑎𝜍𝑎)

= 51 – 32.5 log (0.463674405 x 0.787)

= 55.44 %

Plate and overall column efficiencies will normally be between 30% to 70%. (Coulson and Richardson’s, volume 6, page 547)

3.2.8 Physical Properties

3.2.8.1 Relative Molar Mass (RMM)

RMM = ∑ (component mole fraction x molecular weight) (3.13)

Table 3.9 Liquid Density Component Molecular

Weight

Mole Fraction Liquid Density (kg/m3) Feed Distillate Bottom

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

3.2.8.2 Density Top Product : ρL = 𝑥𝐵,𝑖𝜌𝑖 (3.14) ρL = 0.4835(803.4) + 0.1817(100) + 0.3068(810) + 0.0047(725) = 823.51 kg/m3 ρv = 𝑅𝑀𝑀𝐵 𝑉𝑆𝑇𝑃

𝑥

𝑇𝑆𝑇𝑃 𝑇𝑂𝑃

𝑥

𝑃𝑂𝑃 𝑃𝑆𝑇𝑃 (3.15) ρv

=

29.167 𝑘𝑔/𝑘𝑚𝑜𝑙𝑒 22.4𝑚3_{/𝑘𝑚𝑜𝑙𝑒}

𝑥

273𝐾 357.21𝐾

𝑥

1𝑏𝑎𝑟 1𝑏𝑎𝑟 = 1.731 kg/m3 Bottom Product: ρL = 𝑥𝐷,𝑖𝜌𝑖 (4.16) ρL = 0.9815(803.4) + 0.0068(100) +0.0026(810) + 0.0091(725) = 804.04 kg/m3 ρv = 𝑅𝑀𝑀𝐷 𝑉𝑆𝑇𝑃

𝑥

𝑇𝑆𝑇𝑃 𝑇𝑂𝑃

𝑥

𝑃𝑂𝑃 𝑃𝑆𝑇𝑃 (4.17) ρv

=

58.988𝑘𝑔/𝑘𝑚𝑜𝑙𝑒 22.4𝑚3_{/𝑘𝑚𝑜𝑙𝑒}

𝑥

273𝐾 382.2𝐾

𝑥

1.6𝑏𝑎𝑟 1𝑏𝑎𝑟 = 3.071 kg/m3 3.2.8.3 Surface Tension, σ

Using Sugden (1924), equation 8.23 (Coulson and Richardson’s, volume 6, page 335)

𝜍 =

𝑃𝑐𝑕(𝜌

𝑙

− 𝜌

𝑣

𝑀

4

Where,

σ = surface tension, MJ/m2 or (dyne/cm) Pch = Sugden’s parachor

ρv = Vapor density, kg/m3

ρL = Liquid density, kg/m3

M = relative molecular weight

For mixture, σm = σ1x1 + σ2x2 + ….. (3.19)

Where,

σm = surface tension mixture

σ1 , σ2 = surface tension for mixture

x1 , x2 = component mole fraction

Table 3.10 Pch Distribution Component Pch Distribution Mole Fraction 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 Pch at top = 𝑥_𝐷,𝑖𝑃𝑐𝑕_𝑖 = 0.4853 (148.3) + 0.1817 (31.3) + 0.3068 (165.4) + 0.0047 (299.5)

= 177.28097 Pch at bottom = 𝑥_𝐵,𝑖𝑃𝑐𝑕_𝑖

= 0.9815 (148.3) + 0.0068 (31.3) + 0.0026 (165.4) + 00.0091 (299.5) = 148.30792

Calculation of surface tension:

Top Column,

𝜍 =

65.01 969.64−4.928 21.98 4

𝑥 10

−12 = 67.965683N/m Bottom Column,

𝜍 =

59.04 1019.01− 0.325 _20.04 4

𝑥 10

−12 = 15.27159545 N/m

Above feed point:

Vapor flow rate: Vn = D(R + 1) (3.20)

Where,

D = Distillate molar flowrate R = Reflux ratio

Hence,

Vn = 261.5 (2.531 + 1) = 923.36 kmole/hr

Liquid down flow: Ln = Vn – D (3.21)

= 661.86 kmole/hr Below the feed point:

Liquid flow rate: Lm = Ln + F (3.22)

Where,

F = Feed molar flowrate

Hence,

Lm = 661.86 + 264.1 = 925.96kmole/hr

Vapour flow rate: Vm = Lm – W (3.23)

Where,

W = Bottom molar flowrate Hence,

Vm = 925.96 – 261.5 = 664.46kmole/hr

The equation for the operating lines below the feed plate:

𝑌𝑚 = 𝐿𝑚 𝑉𝑚 𝑋𝑚 + 1 − 𝑊 𝑉𝑚 𝑋𝑤 (3.24) 𝑌𝑚 = 925.96 664.46 𝑋𝑚 + 1 − 261.5 664.46 (𝑋𝑤) = 2.058(Xm + 1) – 261.5 664.46 (𝑋𝑤)

The equation for the operating lines above the feed plate:

𝑌𝑛 = 𝐿𝑛

𝑉𝑛 𝑋𝑛 + 1 − 𝐷

𝑌𝑛 = 661.86 923.36 𝑋𝑛 + 1 − 261.5 923.36 𝑋𝑑 = 0.72 (Xn + 1) – 2.01 x 10-3 𝐹𝐿𝑉 𝑇𝑜𝑝 = 𝐿𝑛 𝑉𝑛 𝜌_𝑉 𝜌_𝐿 (3.26) = 0.72 1.731 823.51 = 0.033

where 0.72 is the slope of the top operating line.

𝐹𝐿𝑉 𝐵𝑜𝑡𝑡𝑜𝑚 = 𝐿𝑚 𝑉𝑚 𝜌_𝑉 𝜌_𝐿 (3.27) = 1.39 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 Richardson’s, Chemical Engineering, Volume 6.

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.

𝑈_𝑣 = −0.171𝑙𝑡2+ 0.271𝑙𝑡− 0.047 𝜌_𝐿− 𝜌_𝑣 𝜌_𝑣 0.5 (3.28) = −0.171(0.5)2_{+ 0.271(0.5) − 0.047} 969.64 − 4.928 4.928 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 – 1.5)

3.2.9.1 Diameter of the column

𝐷_𝑐= 4𝑉𝑤

𝜋𝜌_𝑣 𝑈_𝑣 (3.29)

Where Vw is the maximum vapor rate, kg/s

𝑉𝑤 = 15870 𝑘𝑔 𝑕𝑟 𝑥 1 𝑕𝑟 3600 𝑠 = 4.41 kg/s

𝐷_𝑐= 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

𝐴_𝑐= 𝜋 𝐷𝑐 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

𝐿 = 15740 𝑘𝑔 𝑕𝑟 𝑥 1 𝑕𝑟 3600 𝑠 (3.31) = 4.372 𝑘𝑔 𝑠 𝑥 𝑚3 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.

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 – 2Ad (3.34) = 1.39 m2 – 2(0.1668 m2) = 1.0564 m2

Assume that the hole-active area is 10%

Hole area, Ah = 0.10 Aa (3.35)

= 0.10 x 1.0564 m2 = 0.10564m2

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 Richardson’s, 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,

𝐴_𝑙𝑕 = 𝜋(𝑑𝑕)2

4 (3.36)

= 𝜋(0.005)2

4 = 1.9635 x 10-5 m2

Number of holes per plate, 𝑁_𝑕 = 𝑎𝑟𝑒𝑎 1 𝑕𝑜𝑙𝑒 𝑎𝑟𝑒𝑎 (3.37) = 0.10564 1.9635 𝑥 10−5 = 5380.19 holes ≈ 5380 holes

3.2.11.3 Weir Liquid Crest

Check weeping to ensure enough vapour to prevent liquid flow through hole.

𝑀𝑎𝑥𝑖𝑚𝑢𝑚 𝑙𝑖𝑞𝑢𝑖𝑑 𝑟𝑎𝑡𝑒 = 15740𝑘𝑔 𝑕𝑟 𝑥

1 𝑕𝑟

3600 𝑠 = 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

𝑕_𝑜𝑤 = 750 _𝜌𝐿𝑤 𝐿𝐼𝑤 2 3 (3.38) Where, Iw = weir length, m

Lw = liquid flow rate, kg/s

how = weir crest, mm liquid At maximum rate: 𝑕𝑜𝑤 = 750 4.372 804.04 𝑥 1.011 2 3 = 20.40 mm liquid At minimum rate: 𝑕_𝑜𝑤 = 750 _{1019.01 𝑥 0.85728}3.06 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 Richardson’s, 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

𝑈𝑕(min) =

𝐾₂− 0.9(25.4 − 𝑑_𝑕) 𝜌_𝑣12

(3.39)

Where,

Uh = minimum vapor velocity, m/s

dh = hole diameter, mm K2 = constant = 30.7 − 0.9(25.4 − 5) (3.071)12 = 8.036 m/s 𝐴𝑐𝑡𝑢𝑎𝑙 𝑚𝑖𝑛𝑖𝑚𝑢𝑚 𝑣𝑎𝑝𝑜𝑟 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 = 𝑀𝑖𝑛𝑖𝑚𝑢𝑚 𝑣𝑎𝑝𝑜𝑟 𝑟𝑎𝑡𝑒 𝐴𝑕 (3.40) = 4.41 𝑘𝑔 𝑠 𝑥 0.7 𝑥 𝑚 3 3.071𝑘𝑔 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:

Û_𝑕 = 𝑚𝑎𝑥𝑖𝑚𝑢𝑚 𝑣𝑎𝑝𝑜𝑟 𝑓𝑙𝑜𝑤𝑟𝑎𝑡𝑒

= 4.41 𝑘𝑔 𝑠 𝑥 𝑚 3 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,

𝐴𝑡, 𝑝𝑙𝑎𝑡𝑒 𝑡𝑕𝑖𝑐𝑘𝑛𝑒𝑠𝑠 𝑕𝑜𝑙𝑒 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟 = 3 𝑚𝑚 5 𝑚𝑚 = 0.6 𝑎𝑛𝑑 𝐴𝑕 𝐴_𝑎 = 0.1 we get Co = 0.74 𝐷𝑟𝑦 𝑝𝑙𝑎𝑡𝑒, 𝑕𝑑 = 51 𝑈_𝑕 𝐶_𝑜 2 _𝜌 𝑣 𝜌_𝐿 (3.42) = 51 13.59_0.74 2 _3.071 804.04 = 65.697 mm liquid 𝑅𝑒𝑠𝑖𝑑𝑢𝑎𝑙 𝑕𝑒𝑎𝑑, 𝑕_𝑟 = 12.5 𝑥 103 𝜌_𝐿 (3.43) = 12.5 𝑥 10 3 804.04 = 15.55 mm liquid

Pressure drop per plate, ht = hd + (hw + how) + hr (3.44)

= 65.697 + (50 + 18.15) + 15.55 = 149.397 mm liquid

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 – 10 mm

= 50 – 10 = 40 mm

Where, hap = height of the bottom edge of the apron above the plate

𝑕_𝑑𝑐 = 166 _𝜌𝐿𝑤𝑑

𝐿𝐴𝑚 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 Richardson’s, Volume 6, page

𝑕_𝑑𝑐 = 166 _{804.04 𝑥 0.04044} 4.372 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 < ½(plate spacing + weir height)

0.2205 m < ½(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 “aerated” liquid being carried under the downcomer. A time at least 3 seconds is recommended.

𝑡𝑟 =

𝐴_𝑑𝑕_𝑏𝑐𝜌_𝐿

𝐿_𝑤𝑑 (3.48) Where,

tr = residence time, s

Lwd = liquid flowrate in downcomer, kg/s

𝑡_𝑟 = 0.1668 𝑥 0.2205𝑥 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 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

Angle subtended at plate edge by unperforated strips = 180° - 98°

= 82°

Mean length, unperforated edge strips = (Dc – weir height) x π x θ/180°

= (1.33 – 0.05) x π x 82°/180° = 1.832 m

Areas of unperforated edge strips, As

= mean length unperforated edge x weir height = 1.832 x 0.05

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 – (area of unperforated edge + area of calming) = 1.0564 – (0.0916 + 0.1061)

= 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,

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.

x (Plate thickness)

= (24 – 1)(0.5) + (24 – 1)(0.003) = 11.57 m

By adding 10% safety factor so the column height are 12.7 m ≈ 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

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 – 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 = 113.36°C

Take as 10% above operating temperature Operating temperature = 113.36 x 1.1

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 150°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:

ℯ = 𝑃𝑖𝐷𝑖

2𝐽𝑓 − 𝑃_𝑖 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, ℯ = 0.0902 𝑥 1330

2 1 (115) − 0.0902 = 0.522 mm

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:

ℯ = 𝑃𝑖𝑅𝑐𝐶𝑠 2𝐽𝑓 + 𝑃_𝑖(𝐶_𝑠− 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 = 1₄ 3 + _𝑅𝑅𝑐 𝑘 = 1₄ 3 + 1330_79.8 = 1.77 ℯ = 0.0902 × 1330 × 1.77 2 1 (115) + 0.09(1.77 − 0.2) 3.51 = 0.923 mm Round up to 2 mm

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.

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

Hv = height or length between tangent lines, m

t = wall thickness, m

Wv = 240 x 1.15 x 1.337(13.0 + 0.8(1.337)) x 0.007

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.

𝑃𝑙𝑎𝑡𝑒 𝑎𝑟𝑒𝑎 = 𝜋𝐷 2 4 = 3.142 × 1.33 2 4 = 1.3895 𝑚2 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, ρ = 130 kg/m3

Thickness = 75 mm

= 0.075 m

Volume of insulation, Vi = π 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 ρ x g

= 4.074 x 130 x 9.81 = 5196.07 N

= 5.196 kN Double this to allow for fittings, 10.392 kN

3.3.6.4 Total Weight

The total weight is the summation of dead weight of vessel, weight of accessories and weight of insulation:

Total weight = Wv + W ρ + 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 “Basic Data for the Design of Buildings, Chapter V Loading: Part 2 Winds Load”, (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 = wind pressure, N/m2 uw = wind speed, km/h Pw = 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

𝐵𝑒𝑛𝑑𝑖𝑛𝑔 𝑚𝑜𝑚𝑒𝑛𝑡, 𝑀_𝑥 = 𝐹𝑤𝐻𝑣 2 2 = 1920 × 13 2 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:

𝜍𝐿 = 𝑝_𝐼𝐷_𝑖 4𝑡 (3.55) = 0.0902 × 1330 4 × 9 = 3.332 N/mm2

2. Circumferential stresses due to pressure are given by:

𝜍𝑕 = 𝑝_𝐼𝐷_𝑖 2 𝑡 (3.56) = 0.0902 × 1330 2 × 9 = 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:

𝜍_𝑤 = 𝑊

= 86749

𝜋 1330 + 9 × 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:

𝜍_𝑏= ± 𝑀𝑥 𝐼_𝑣

𝐷_𝑖

2 + 𝑡 (3.58)

Where,

Mx = Total bending moment

Do = Outside diameter

= Di + 2t

= 1330 + 2(9) = 1348 mm

Iv = Second moment area

= 𝜋 64 𝐷𝑜4− 𝐷𝑖4 (3.59) = 𝜋 64 13484− 13304 = 1.358 x 1011 mm4 𝜍𝑏= ± 162240 1.358 x 1011 1330 2 + 9 = ± 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.

𝜍_𝑧 = 𝜍_𝐿 + 𝜍_𝑤± 𝜍_𝑏 (3.60) For upwind,

𝜍_𝑧= 3.332 + (−2.29) + 8.052 x 10 = 1.031 N/mm2

For downwind,

𝜍𝑧= 3.332 + −2.29 − 8.052 𝑥 10−4

= 1.041 N/mm2

Therefore, the greatest difference between the principles stresses,

𝜍_𝑧= 𝜍_𝑕− 𝜍_𝑧 𝑑𝑜𝑤𝑛𝑤𝑖𝑛𝑑 (3.61) = 6.665 − 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, σ_c = 2 x 104 t

D_o (3.62) = 2 x 104 9

1348 = 133.531 N/mm2

The maximum compressive stress will occur when the vessel is not under pressure 𝜍_𝑤 + 𝜍_𝑏 = 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.

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

Θs = 90°

Material of construction = Carbon Steel

Design stress, f = 115 N/mm2

Young’s modulus = 200,000 N/mm2

Skirt height, hs = 4 m

Skirt thickness, ts = 9 mm

Joint factor = 0.85

3.3.9.1 Weight of the Skirt

Approximate weight, Wapprox = (π/4 x Di2 x Hv) x ρL x 9.81

= (π/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 (3.63)

= 0.5 x 1.92(13 + 4)2 = 277.44 kNm

2. Bending stress of skirt, σbs 𝐵𝑒𝑛𝑑𝑖𝑛𝑔 𝑠𝑡𝑟𝑒𝑠𝑠 𝑜𝑓 𝑠𝑘𝑖𝑟𝑡, 𝜍_𝑏𝑠 = 4 𝑀𝑠 𝜋 𝐷𝑠+ 𝑡𝑠 × 𝐷𝑠𝑡𝑠 (3.64) = 4 × 162.24 × 103 × 103 𝜋 1330 + 9 × 1330 × 9 = 12.888 N/mm2

3. Dead weight stress in skirt, σws

𝜍𝑤𝑠 𝑡𝑒𝑠𝑡 = 𝑊𝑎𝑝𝑝𝑟𝑜𝑥 𝜋 𝑡𝑠(𝐷𝑠+ 𝑡𝑠) (3.65) = 152092.43 𝜋 × 9(1128 + 9) = 4.731 N/mm2 𝜍𝑤𝑠 𝑜𝑝𝑒𝑟𝑎𝑡𝑖𝑛𝑔 = 𝑊 𝜋𝑡_𝑠(𝐷_𝑠+ 𝑡_𝑠) (3.66) = 84749 𝜋 × 9(1330 + 9) = 2.291 N/mm2

4. Resultant stress in skirt, σs

Maximum σs (tensile) = σbs + σws test (3.67)

= 12.888 + 2.416 = 15.304 N/mm2

Maximum σs (compressive) = σbs - σws operating (4.68)

= 12.888 – 2.291 = 15.179 N/mm2

5. Criteria for Design

Take the joint factor, J as 0.85 Where θs = 90°

σs (tensile) < fs J sin θs

15.304 N/mm2 < 97.75 N/mm2 σs (compressive) < 0.125 E(ts/Ds) sin θs

15.179 N/mm2 < 0.125 x 200000(9/1330) sin 90° 15.179 N/mm2 < 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 Scheiman’s short cut method. Scheiman gives the following guide rules which can be used for the selection of the anchor bolts.

Refer to Coulson and Richardson’s, 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 = 2200π Closest multiple of 4, Nb = 16 bolts

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

= 2200π / 600 = 11.52 ≈ 12 bolts 𝐵𝑜𝑙𝑡 𝑠𝑝𝑎𝑐𝑖𝑛𝑔 = 𝜋 × 3.2 × 103 13 = 773.32 𝑚𝑚 (𝑠𝑎𝑡𝑖𝑠𝑓𝑎𝑐𝑡𝑜𝑟𝑦) 𝐵𝑜𝑙𝑡 𝑎𝑟𝑒𝑎, 𝐴𝑏 = 1 𝑁_𝑏𝑓_𝑏 × 4 𝑀_𝑠 𝐷_𝑏 − 𝑊 (3.69) = 1 13 𝑥 125 × 4 × 162240 3.2 − 86749 = 71.416mm2

Use M24 bolts (BS 4190:1967) root area = 353 mm2

𝐵𝑜𝑙𝑡 𝑟𝑜𝑜𝑡 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟 = 4𝐴𝑏

𝜋 (3.70)

= 71.416 × 4

𝜋 = 9.536 mm

Total compressive load on the base ring per unit length: 𝐹_𝑏 = 4𝑀𝑠

𝜋𝐷_𝑠2+ 𝑊

𝜋𝐷_𝑠 (3.71)

Where,

Fb = the compressive load on the base ring, Newtons per linear metre

𝐹_𝑏 = 4 × 162240 𝜋 × 3.02 +

86749

𝜋 × 3.0 = 32.157 kN/m

The minimum width of the base ring: 𝐿𝑏 = 𝐹_𝑏 𝑓_𝑐 𝑥 1 103 (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)

𝐿_𝑏 = 32157 5 𝑥

1 103

= 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

Actual bearing pressure on concrete foundation, 𝑓′𝑐=

32157 142 × 103

= 0.226 N/mm2

The minimum thickness is given by, 𝑡_𝑏 = 𝐿_𝑟 3 𝑓′𝑐

𝑓_𝑟 (3.73)

Where,

Lr = the distance from the edge of the skirt to the outer edge of the ring, mm

f’c = actual bearing pressure on base, N/mm2

fr = allowable design stress in the ring material, typically 140 N/mm2

𝑡𝑏 = 76

3 × 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 G0.53ρ-0.37 (3.74)

Where,

G = flowrate, kg/s ρ = density, kg/m3

The material construction used for nozzles is carbon steel pipe.

3.3.10.1 Feed Stream G = 1.587 x 104 kg/h = 4.4083 kg/s ρmix = 778.66 kg/m3 dopt = 293 (4.4083)0.53 (778.66)-0.37 = 54.77 mm ≈ 55 mm 𝑁𝑜𝑧𝑧𝑙𝑒 𝑡𝑕𝑖𝑐𝑘𝑛𝑒𝑠𝑠, 𝑡 = 𝑃𝑠𝑑𝑜𝑝𝑡 20𝜍 + 𝑃_𝑠 3.75

Where,

Ps = operating pressure

σ = design stress at working temperature

𝑇𝑕𝑖𝑐𝑘𝑛𝑒𝑠𝑠 𝑜𝑓 𝑛𝑜𝑧𝑧𝑙𝑒: 𝑡 = 0.0902 × 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 Top Stream 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 𝑁𝑜𝑧𝑧𝑙𝑒 𝑡𝑕𝑖𝑐𝑘𝑛𝑒𝑠𝑠, 𝑡 = 𝑃𝑠𝑑𝑜𝑝𝑡 20𝜍 + 𝑃𝑠 𝑇𝑕𝑖𝑐𝑘𝑛𝑒𝑠𝑠 𝑜𝑓 𝑛𝑜𝑧𝑧𝑙𝑒: 𝑡 = 0.0902 × 41.08 20 125 + 0.0902 = 0.00148 mm

So, the thickness of nozzle = corrosion allowance + 0.00148 = 2 + 0.00148

3.3.10.3 Bottom Stream G = 1.574 x 104 kg/h = 4.372 kg/s ρmix = 695.2 kg/m3 dopt = 293 (4.372)0.53 (695.2)-0.37 = 56.88 mm ≈ 60 mm 𝑁𝑜𝑧𝑧𝑙𝑒 𝑡𝑕𝑖𝑐𝑘𝑛𝑒𝑠𝑠, 𝑡 = 𝑃𝑠𝑑𝑜𝑝𝑡 20𝜍 + 𝑃𝑠 𝑇𝑕𝑖𝑐𝑘𝑛𝑒𝑠𝑠 𝑜𝑓 𝑛𝑜𝑧𝑧𝑙𝑒: 𝑡 = 0.0902 × 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 Richardson’s, Chemical Engineering, Volume 6, page 863, figure 13.37.

Table 3.12 The Summary of Flange Design Feed Stream

dopt d1

Flange Raised face

Bolting Drilling Boss

D b H d4 f No d2 k d3

Top Stream dopt d1

Flange Raised face

Bolting Drilling Boss

D b H d4 f No d2 k d3

25 33.7 100 14 24 60 2 M10 4 11 75 50

Bottom Stream dopt d1

Flange Raised face

Bolting Drilling Boss

D b H d4 f No d2 k d3

50 60.3 140 14 28 90 3 M12 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 Insulation, Wi 5.424 kN

Total Weight, Wt 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

Young’s Modulus 200,000 N/mm2

Approximate Weight, Wapprox 152.092 kN

Total Weight 205.844 kN

Bending Moment, Mx 328.2 kNm

Skirt Thickness, ts 9 mm

Skirt Height, Hs 4 m

Stiffness Ring

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 Cp° = K1 + K2 log10 (A) + K3 [log10 (A)]2 (1.0)

where,

A = capacity or size parameter for the equipment K1, K2, K3 = constants in Table A.1 (Appendix A)

3.4.1 Process Vessels

Column height = 13 m; Column diameter = 1.33 m

Material of construction = carbon steel with stainless steel cladding Volume = 𝜋𝐷2𝑕

= (3.142)(1.33)2(13) = 72.25m3

From Table A.1 (Appendix A), K1 = 3.4974, K2 = 0.4485, K3 = 0.1074

log10 Cp° = 3.4974 + 0.4485 log10 (72.25) + 0.1074 [log10 (72.25)]2

= 4.73

Cp° = $ 53 751.28

Pressure factors for process vessels: For pressure vessel, when

t

_vessel



0

.

003

m

,

𝐹_{𝑃,𝑣𝑒𝑠𝑠𝑒𝑙} = 𝑃 + 1 𝐷 2[850 − 0.6 𝑃 + 1 ] + 0.00315 0.003 (2.0) 𝐹_{𝑃,𝑣𝑒𝑠𝑠𝑒𝑙} = 0.0902 + 1 (1.33) 2[850 − 0.6 0.0902 + 1 ] + 0.00315 0.003 𝐹𝑝,𝑣𝑒𝑠𝑠𝑒𝑙 = 8.96 × 10−4

The bare module factor for process vessel (Turton et al., Analysis, Synthesis, and Design of Chemical Processess, 3rd Edition, page 927):

CBM = Cp°FBM = Cp°(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)(8.96×10-4)]

= $ 121 028.03

Use correlation:

CEPCI for year of 2009 is 645.5 CEPCI for year of 2001 is 397 Therefore,

𝑛𝑒𝑤 𝐶_𝐵𝑀 = 121 028.03 ×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 Cp° = 2.9949+ 0.4465 log10 (5.56) + 0.3961 [log10 (5.56)]2

= 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):

CBM = Cp°NFBMFq (4.0)

Where,

N = number of trays

Fq = quantity factor for 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

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, 𝑛𝑒𝑤 𝐶𝐵𝑀 = 208 533.79 × 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

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.

Walas, S. M. 1988. Chemical Process Equipment. Butterworths Publishers.