Chapter 3 - Distillation Column Design

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 :

𝑦𝑖 =

𝐾𝑖 𝑥𝑖 = 1.0

(3.1)

Dew Point

𝑥𝑖 =

𝑦𝑖 / 𝐾𝑖 = 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

-8.29×103

0.00

-8.9096

1.82×10-6

2.00

Water

65.9

-7.23×103

0.00

-7.18

4.03×10-6

2.00

Propanal

80.9

-6.51×103

0.00

-9.82

6.79×10-6

2.00

96.7

-7.45×103

0.00

-1.24

1.08×10-5

2.00

Dipropyl Ether

Antoine equation: ln 𝑃𝑜 = 𝐴 −

𝐾𝑖 =

𝑃𝑜 𝑃𝑇

𝐵 + 𝐷 𝑥 ln 𝑇 + 𝐸 𝑥 𝑇^𝐹 𝑇+𝐶

(3.3)

(3.4)

3-4 Estimation of feed temperature,

𝑥𝑖 =

𝑦𝑖 / 𝐾𝑖 = 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)

𝑥𝑖 =

𝑦𝑖 / 𝐾𝑖 = 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)

𝑦𝑖 =

𝐾𝑖 𝑥𝑖 = 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 𝐾𝑗

(3.6)

3-6 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 =

𝐾𝐿𝐾 𝐾𝐻𝐾

(3.8)

Where, KLK = Light key components KHK = Heavy key components 3.2.3.1 Top Column

Table 3.6 𝜶=

𝑲 𝑲𝑯𝑲

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 𝜶=

𝑲 𝑲𝑯𝑲

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;

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:

Nmin

=

X X Log[(X LK )]d [( XHK )]b HK

LK

Log αLK

(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: 𝛼𝑖 𝑥𝑖,𝑑 = 𝑅𝑚 + 1 𝛼𝑖 − 𝜃

(3.10)

3-8 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:

𝛼𝑖 𝑥𝑖,𝑓 =1−𝑞 𝛼𝑖 − 𝜃

(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 𝛼𝑖 𝑥𝑖,𝑓 =0 𝛼𝑖 − 𝜃 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

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 = 0.8931 𝑅𝑚 + 1 Specimen calculation, for R = 2.0 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 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: 𝑁𝑟 𝐿𝑜𝑔 = 0.2606 log 𝑁𝑠

𝐵 𝐷

𝑥𝑓,𝐻𝐾 𝑥𝑓,𝐿𝐾

x𝑏,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 = 0.2606 log 𝑁𝑠

2.531 261.5

𝑁𝑟 = 0.993 𝑁𝑠 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: 𝐸˳ = 51 − 32.5 log (µ𝑎 𝜍𝑎 )

(3.11)

Where, µ𝑎 = the molar average liquid viscosity, mNs/m2 𝜍𝑎 = average relative volatility of the light key To find the viscosity of the flow: 𝐿𝑂𝐺 µ𝑎

= 𝑉𝑖𝑠𝐴 𝑥

1 1 − 𝑇 𝑉𝑖𝑠𝐵

(3.12)

Table 3.8 Viscosity of the mixture

Component

Mole fraction feed, x

Viscosity Coefficient A

B

Log µ𝒂

Viscosity (mNs/m2)

µ𝒂 × 𝒙

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, 𝐸˳

=

51 − 32.5 log (µ𝑎 𝜍𝑎 )

3-12 =

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

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 : 𝑥𝐵,𝑖 𝜌𝑖

ρL

=

ρL

=

0.4835(803.4) + 0.1817(100) + 0.3068(810) + 0.0047(725)

=

823.51 kg/m3

ρv

=

ρv

=

𝑅𝑀𝑀 𝐵 𝑉𝑆𝑇𝑃

(3.14)

𝑥

𝑇𝑆𝑇𝑃 𝑇𝑂𝑃

𝑥

29.167 𝑘𝑔/𝑘𝑚𝑜𝑙𝑒

𝑃𝑂𝑃

𝑥

22.4𝑚 3 /𝑘𝑚𝑜𝑙𝑒

(3.15)

𝑃𝑆𝑇𝑃 273𝐾 357.21𝐾

𝑥

1𝑏𝑎𝑟 1𝑏𝑎𝑟

1.731 kg/m3

= Bottom Product:

𝑥𝐷,𝑖 𝜌𝑖

ρL

=

ρL

=

0.9815(803.4) + 0.0068(100) +0.0026(810) + 0.0091(725)

=

804.04 kg/m3

ρv

=

ρv

=

𝑅𝑀𝑀 𝐷 𝑉𝑆𝑇𝑃

(4.16)

𝑥

𝑇𝑆𝑇𝑃 𝑇𝑂𝑃

58.988𝑘𝑔 /𝑘𝑚𝑜𝑙𝑒 22.4𝑚 3 /𝑘𝑚𝑜𝑙𝑒

𝑥 𝑥

𝑃𝑂𝑃

(4.17)

𝑃𝑆𝑇𝑃 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)

𝑃𝑐𝑕(𝜌𝑙 − 𝜌𝑣 𝜍= 𝑀

4

𝑥 10−12

(3.18)

3-14 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 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

𝑥𝐷,𝑖 𝑃𝑐𝑕𝑖 0.4853 (148.3) + 0.1817 (31.3) + 0.3068 (165.4) + 0.0047 (299.5)

3-15 =

177.28097

Pch at bottom =

𝑥𝐵,𝑖 𝑃𝑐𝑕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 4 21.98

𝑥 10−12

= 67.965683 N/m

59.04 1019.01− 0.325

Bottom Column, 𝜍 =

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

Vn

=

261.5 (2.531 + 1)

=

923.36 kmole/hr

Hence,

Liquid down flow: Ln = Vn – D = 923.36 – 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 – W

(3.23)

Where, W

=

Bottom molar flowrate

Vm

=

925.96 – 261.5

=

664.46kmole/hr

Hence,

The equation for the operating lines below the feed plate: 𝑌𝑚 =

𝐿𝑚 𝑉𝑚

𝑌𝑚 =

925.96 664.46

𝑋𝑚 + 1 −

𝑊 𝑉𝑚

𝑋𝑚 + 1 −

= 2.058(Xm + 1) –

𝑋𝑤

(3.24)

261.5 (𝑋𝑤) 664.46

261.5 664.46

(𝑋𝑤)

The equation for the operating lines above the feed plate: 𝑌𝑛 =

𝐿𝑛 𝑉𝑛

𝑋𝑛 + 1 −

𝐷 𝑉𝑛

𝑋𝑑

(3.25)

3-17

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

𝜌𝑉 𝜌𝐿

(3.26)

1.731 823.51

= 0.72

= 0.033 where 0.72 is the slope of the top operating line.

𝐹𝐿𝑉 𝐵𝑜𝑡𝑡𝑜𝑚 =

𝐿𝑚 𝑉𝑚

= 1.39

𝜌𝑉 𝜌𝐿

(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 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 highpressure 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.

𝑈𝑣 = −0.171𝑙𝑡 2 + 0.271𝑙𝑡 − 0.047

𝜌𝐿 − 𝜌𝑣 𝜌𝑣

= −0.171(0.5)2 + 0.271(0.5) − 0.047

0.5

969.64 − 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 – 1.5)

3.2.9.1 Diameter of the column

𝐷𝑐 =

4𝑉𝑤 𝜋𝜌𝑣 𝑈𝑣

Where Vw is the maximum vapor rate, kg/s 𝑉𝑤 =

15870 𝑘𝑔 1 𝑕𝑟 𝑥 𝑕𝑟 3600 𝑠

= 4.41 kg/s

(3.29)

3-19

𝐷𝑐 =

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 𝐴𝑐 = =

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 1 𝑕𝑟 𝑥 𝑕𝑟 3600 𝑠

(3.31)

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.

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 = 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 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, 𝐴𝑙𝑕 =

=

𝜋(𝑑𝑕 )2 4 𝜋(0.005)2 4

= 1.9635 x 10-5 m2

(3.36)

3-22 Number of holes per plate, 𝑎𝑟𝑒𝑎 1 𝑕𝑜𝑙𝑒 𝑎𝑟𝑒𝑎 0.10564 = 1.9635 𝑥 10−5

𝑁𝑕 =

(3.37)

= 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 1 𝑕𝑟 𝑥 𝑕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

𝐿𝑤 = 750 𝜌𝐿 𝐼𝑤

2 3

Where, Iw

= weir length, m

Lw

= liquid flow rate, kg/s

ρL

= liquid density, kg/m3

(3.38)

3-23 how

= weir crest, mm liquid

At maximum rate:

𝑕𝑜𝑤

4.372 = 750 804.04 𝑥 1.011

2 3

= 20.40 mm liquid

At minimum rate:

𝑕𝑜𝑤

3.06 = 750 1019.01 𝑥 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 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 efficiency. Minimum vapor velocity through the holes based on the holes area.

3-24

𝑈𝑕 (min) =

𝐾2 − 0.9(25.4 − 𝑑𝑕 )

(3.39)

1

𝜌𝑣 2

Where, Uh

= minimum vapor velocity, m/s

dh

= hole diameter, mm

K2

= constant

=

30.7 − 0.9(25.4 − 5) 1

(3.071)2 = 8.036 m/s

𝐴𝑐𝑡𝑢𝑎𝑙 𝑚𝑖𝑛𝑖𝑚𝑢𝑚 𝑣𝑎𝑝𝑜𝑟 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 =

𝑀𝑖𝑛𝑖𝑚𝑢𝑚 𝑣𝑎𝑝𝑜𝑟 𝑟𝑎𝑡𝑒 𝐴𝑕

(3.40)

4.41 𝑘𝑔 𝑚3 𝑥 0.7 𝑥 𝑠 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 𝐴𝑕

(3.41)

3-25 4.41 𝑘𝑔 𝑚3 𝑥 𝑠 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,

𝑎𝑛𝑑

𝑝𝑙𝑎𝑡𝑒 𝑡𝑕𝑖𝑐𝑘𝑛𝑒𝑠𝑠 3 𝑚𝑚 = = 0.6 𝑕𝑜𝑙𝑒 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟 5 𝑚𝑚 𝐴𝑕 = 0.1 𝐴𝑎

we get Co = 0.74 𝑈𝑕 𝐶𝑜

𝐷𝑟𝑦 𝑝𝑙𝑎𝑡𝑒, 𝑕𝑑 = 51

2

13.59 = 51 0.74

𝜌𝑣 𝜌𝐿 2

(3.42) 3.071 804.04

= 65.697 mm liquid

𝑅𝑒𝑠𝑖𝑑𝑢𝑎𝑙 𝑕𝑒𝑎𝑑, 𝑕𝑟 =

12.5 𝑥 103 𝜌𝐿

=

12.5 𝑥 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 – 10 mm = 50 – 10 = 40 mm

Where, hap

𝑕𝑑𝑐

= height of the bottom edge of the apron above the plate

𝐿𝑤𝑑 = 166 𝜌𝐿 𝐴𝑚

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 577) used Aap = 0.04044 m2

3-27

𝑕𝑑𝑐 = 166

4.372 804.04 𝑥 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

< ½(plate spacing + weir height)

0.2205 m

< ½(0.5 + 0.05) m

0.2205 m