design of packed columns for absorption and distillation processes_prelecture slids

8/18/2019 Design of Packed Columns for Absorption and Distillation Processes_prelecture Slids

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ENSC3019/CHPR8503: Week 6 Design of Packed Columns

1

Dr Kevin [email protected]

Recommended reading:McCabeet al. , Unit Operations of Chemical Engineering,Chapter 21Treybal, R. E. Mass Transfer Operations, 3rd Edn. McGraw-Hill 1955,Chapter 9Coulson, J. M. and Richardson, J. F. Chemical Engineering, Volume 6: ParticleTechnology and Separation Processes, 5th Edn. Butterworth-Heinemann 2002


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Mass Transfer (MT) across phase interface: two-resistance model

2

Gas film Liquid filmBulk gas Bulk liquid y A,G

y A,i

x A,i

x A,L

distance

Resistances to diffusion of A:(i) in the gas phase film (ii) in the liquid phase film

At the interface: assume local equilibrium between y A and x A,no resistance to MT across the interface

( ), , A y A g A i N k y y= − ( ), , A x A i A L N k x x= −1 yk ∝ 1 xk ∝

see McCABE et al. p547; BENÍTEZ p165


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Mass-transfer coefficients:an engineering concept that allows us to simplify complex

diffusion problems.

3

( ) A y i N k y y= −

Flux(mole/m2/s) Coefficient

Driving force(concentration

difference)= ×

Since concentration could be defined in different ways,a variety of coefficients can be defined:

• k y , k x, K y , K x ……


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Summary of general forms of MT Rates for two-phase films

ky is local MTC for gas phasey i is mole fraction (of component A) in gas at the gas-liquid interface , y is bulk vapour composition

kx is local MTC for liquid phasexi is mole fraction (of component A) in liquid at thegas-liquid interface, x is bulk liquid composition

K y is overall MTC for gas phasey* is composition of vapour that would be inequilibrium with the bulk liquid of composition x

K x is overall MTC for liquid phasex* is composition of vapour that would be inequilibrium with the bulk vapour of composition y

4

( ) A y i N k y y= −

( ) A x i N k x x= −

( )* A x N K x x= −( )* A y N K y y= −

MTC=mass transfer coefficient. Subscripts A, and G, L dropped here for simplicity.

See McCabe et al. page 547-548. Or if you’re keen for more discussion look at Treybal’s Chapter 5..

m’ is local slope of equilibrium curve

i.e.

1 1 '

y y x

m

K k k

= +( ) ( )

*'i im y y x x

= − −


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Tutorial 1 Equilibrium for component A between air andwater is described by Henry’s law y*=4x . The local masstransfer coefficients are k x =2 mol m -2s-1 and ky =1 mol m -2s-1 .

(1) What is the overall mass transfer coefficient for gasphase?(2) Evaluate the flux of A between phases at a point in a

column where bulk compositions are 0.08 mole fraction inthe gas and 0.01 mole fraction in the liquid.

5


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6.1 Packed columns for absorption


Consultation hours15:00-17:00Thursdays2.49A in Civil & Mech Eng building

6


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Equipment for gas-liquid absorptionNeed intimate contact between the immisciblephases to achieve mass transfer (MT) betweenphases.Flux N A

rate of transfer per unit area of gas-liquid interface

Engineering MT equipment focuses on increasing

the interfacial area for transfer (

)


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Main equipment types

Packed columnsRandom (let to fall randomly into column during installation)Structured (engineering for lower Δ P, higher cost )

Tray columns - liquid levels on each tray

Gas sparging tanks

http://www.co2crc.com.au/imagelibrary2/vid_absorp_desorp.html


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Column internals

9GREEN, D. W. & PERRY, R. H. (eds.) ( 2008).Perry's chemical engineers' handbook, New

York: McGra w-Hill.

Packing material, plus

Liquid inlet systems

Liquid & vapour distributors

Liquid collecting devicesPacking supports

Good info at manufacturerwww.sulzechemtech.com

http://www.co2crc.com.au/imagelibrary2/vid_absorp_desorp.htmlhttp://www.co2crc.com.au/imagelibrary2/vid_absorp_desorp.htmlhttp://www.co2crc.com.au/imagelibrary2/vid_absorp_desorp.htmlhttp://www.co2crc.com.au/imagelibrary2/vid_absorp_desorp.htmlhttp://www.co2crc.com.au/imagelibrary2/vid_absorp_desorp.htmlhttp://www.sulzechemtech.com/http://www.sulzechemtech.com/http://www.co2crc.com.au/imagelibrary2/vid_absorp_desorp.html


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Packed columns – random packings

10

Metal pall rings

Raschig rings

VSP Inner arc ring

see more images at

www.tower-packing.com

http://www.tower-packing.com/http://www.tower-packing.com/http://www.tower-packing.com/http://www.tower-packing.com/


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Structured packings www.sulzerchemtech.com

Mellapak TM

www.sulzerchemtech.com

Grids

http://www.sulzerchemtech.com/http://www.sulzerchemtech.com/http://www.sulzerchemtech.com/http://www.sulzerchemtech.com/


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Tray columns

V-grid www.sulzerchem.com

Sieve tray


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High performance trayseg. Shell calming section tray

www.sulzerchem.com


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1. Tray columns can be designed to handle a wider range of liquid andgas flow rates. Packed columns are not suitable for very low liquidrates.

2. The efficiency and performance of a tray column can be moreaccurately predicted.

3. Easier to make provisions for withdrawal side streams in platecolumns.

4. Fouling & cleaning: can install manholes on trays. However, may beeasier to replace packing when fouled.

Plate columns can be designed with more assurance - some doubt thatgood liquid distribution can be maintained in a packed column.

It is easier to provided cooling or heating in a plate column – coils

directly on plates.

Coulson and Richardson Vol 6. list some of the factors which influence choice of trays or packing in a column:


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Trays/Plate columns vs. Packedcolumns5. For corrosive liquids a packed column will be cheaper

than a plate column (due to materials).

6. The liquid hold-up is lower in a packed column. Important

if amount toxic or flammable liquid needs to be keep lowfor safety.

7. Packed columns are more suitable for foaming systems

8. The pressure drop per equilibrium stage can be lower forpacked columns.

9. Packing cheaper for small columns, d < 0.6 m


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Column internals –process design

16

Process design or process tech support to operationneeds to consider:

Type of contacting device

Number equilibrium stagesHeight of packing required

Pressure drop

FoulingCorrosion and other materials issues


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MT Rate,r A, for absorption per unit volume of packed column

k ya is local MTC for gas phase onunit volume basisy i is mole fraction (of component A) in gas at the gas-liquid interface , y is bulk vapour composition

k xa is local MTC for liquid phase onunit volume basisxi is mole fraction (of component A) in liquid at thegas-liquid interface, x is bulk liquid composition

K ya is overall MTC for gas phase onunit volume basisy* is composition of vapour that would be inequilibrium with the bulk liquid of composition x

K xa is overall MTC for liquid phaseon unit volume basisx* is composition of vapour that would be inequilibrium with the bulk vapour of composition y

1

( ) A y ir k a y y= −

( ) A x ir k a x x= −

( )* A xr K a x x= −( )* A yr K a y y= −

See McCabe et al. page 579

Coefficient a is interfacial area per unit volume of packed column


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Which MTC and rate equation?

Can use any of the four basic rate equations to

design an absorption column, but the gas-film

coefficients are often used.

We’ll follow McCabe et al. and use K y a here.

18


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Calculation of packing height(dilute gas)

19

a = interfacial area/unit volume of columnA = cross-sectional area column (m 2)

ZT = total height of packed section

( )* yVdy K a y y Adz− = −

L, x 2

V, y 1 x 1

y2

dz x

x+dx y+dy

y

Mass balance on component A across differentialvolume dz.

Assume:

• dilute gas change in molar flow V is neglected

Rate loss solute from gas = Rate gain solute by liquid

Let’s do a dimension analysis here.How do we get this equation from N A ?


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Calculation of packing height

20

Rearranging and integration of the mass balance equation:

2

*1

t y

V dy Z

K aA y y=

−∫

We now have an equation to calculate the total height of packing, Z T,based on concentration driving force (y-y*), gas flow rate and the gasphase MTC:

See McCabe et al. page 580-581

LetThen, substitute Z t into above equation


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ZT = (height transfer unit) x (number units)

21


2

*1 y

t d V A Z y

ya yK −= ∫

change in gas conc.average driving force

Number of transfer unitsN Oy

Subscript O y shows based on overallgas phase driving force.

Height of transfer unitH Oy

Units of length.

The height of packingneeded to achieve:

change in gas conc.

driving force=

for that section of packing.


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Operation line and equilibrium line Graphic integration: 1/(Y – Y *) as afunction of Y


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Gas film:

Liquid film:

Overall gas:

Overall liquid:

Four sets of HTUs and NTUs

23

See McCabe et al. page 583

i y y

dy N y−= ∫

i x

x

dx N

x=

−∫

*Ox

dx N

x x=

−∫

*Oy

dy N

y y−= ∫

/ y y

V A H ak

=

/ x

x

L A H

ak =

/Oy

y

V A H

K a=

/Ox

x

V A H

K a=


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ENSC3019

6.2 Determination of Column Height

24


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Values of height of transferunit

25


Values of H Oy are system dependent.Sometimes available for a particular system directly in theliterature, or could be measured in pilot-plant studies.

But, often need to estimate height of transfer units fromempirical correlations for individual MTCs or individual heightsof a transfer unit.

This estimation is a key element of Assignment 1.


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Evaluating theintegral for NOy ?

26

2

*1

t y

d V Z

A

y

ya yK −= ∫

Simplest case - Straight operating & equilibrium lines.

Can evaluate N Oy by: change in gas conc.log mean driving forceOy

N =

( ) ( ) ( )( )

( )

* *

* 1 2*

1*

2

lnlm

y y y y y y

y y

y y

− − −− =

−

−

( ) ( )2

2 1* *

1

y

Oy

y lm

dy y y N

y y y y

−= =− −∫

Log mean driving forceFor details on the integration above, seeCoulson & Richardson Vol2.

Example H2S scurbber problem& solution provided at end of

these set of slides.


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Challenges and discussions

27

What if gas is not dilute?

Where do I get values of Mass Transfer Coefficients?

Affects of temperature and pressure?

What if there’s a chemical reaction as well as absorption?

E.g. amine absorption for acid gas removal?


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Example 13.1 H2S scrubber

28

Gas from a petroleum distillation column has its H 2S concentration

reduced from 0.03 kmol H 2S /kmol inert hydrocarbon to 1 % of thisvalue by scrubbing with triethanolamine-water solvent in acountercurrent tower, operating at atmospheric pressure and 300 K.

The equilibrium relation for the solution is described by Y e=2X .

Solvent enters the tower free of H 2S and leaves containing 0.013kmol H 2S /kmol solvent. If the flow of inert gas is 0.015 kmol/s.m 2 oftower cross-section, calculate:

(a)Height of absorber required

(b)Number of transfer units N OG (or N oy)required

The overall coefficient for absorption K Ya is 0.04 kmol/s.m 3 (unit

mole fraction driving force).


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Example 13.1 solution (1)

29

Data:

1) Equilibrium expressionYe=2X

2) Top of column conditions

Y2 = 0.03 x 0.01 = 0.0003

Ls, X 2

Vs, Y 1 X 1

Y 2

absorber

3) Bottom of column conditionsY1 = 0.03

X1 = 0.013 Y 1e =0.026

Driving force = Y 1-Y1e = 0.004

X2 = 0 Y2e = 0

Driving force = Y 2-Y2e = 0.0003


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Logarithmic mean driving force:

( ) 0.004 0.00030.004

ln0.0003

0.00370.00143

2.59

e lmY Y

−− =

= =

Mass balance on H 2S in gas film:

(rate moles lost from gas) = (rate mass transfer)

( ) ( )1 2s G e lmV Y Y S K aP Y Y SZ − = − Where S is the cross section area(which is also termed as A)

K G is the pressure dependent MTC

K G P = K y


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31

And we can rewrite in terms of lumped overall coefficient:

Then:

Solve for Z: Z = 7.8 m

K G a P = K Ya = 0.04 kmol/s m3

V s (Y 1 - Y 2 ) = K Ya (Y - Y e )lm Z

0.015 (0.03 – 0.0003) = 0.04 (0.00143) Z


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32

Now calculate height of transfer unit:

Number of transfer units:

N OG = 21

20.7OGOG

Z

N H = =

Which is another expression of NOy

Which is anotherexpression ofHOy

For dilute systems, Vs ≈ V


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Example 13.1 alternatives solutions

33

If your love calculus you could solve analytically:

N OG = 21.1 2

1

Y

OGeY

dY N

Y Y =

−∫

Calculate H OG as before. Then calculate Z.

Z = 7.91 m


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Example 13.1 alternatives solutions

34

Could do a graphical-numerical solution (eg.trapezoidal rule or Simpson rule to find the NOG)This will be illustrated later with Tutorial Example 2


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Looking forward:Plate columns vs. Packed columns

• Coulson and Richardson Vol 6. suggest the followingadvantages/disadvantages for Plate vs Packed:

• Plate columns can be designed to handle wider range ofliquid and gas flow rates

• Packed columns not suitable for very low liquid rates

• The efficiency and performance of a plate column canbe more accurately predicted

35


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Looking forward:Plate columns vs. Packed columns• Plate columns can be designed with more assurance -

some doubt that good liquid distribution can bemaintained in a packed column.

• It is easier to provided cooling or heating in a platecolumn – coils directly on plates.

• Easier to make provisions for withdrawal side streams inplate columns.

• Fouling by solids – can easily install manholes on plates –small columns however – may be easier to replacepacking when fouled.

36


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Looking forward:Plate columns vs. Packed columns

• For corrosive liquids a packed column will be cheaperthan a plate column (due to materials).

• The liquid hold-up is appreciably lower in a packed

column – important if amount toxic or flammable liquidneeds to be keep low for safety

• Packed columns are more suitable for foaming systems

• The pressure drop per equilibrium stage can be lowerfor packed columns – impt. vacuum distillation

• Packing cheaper for small columns, d < 0.6 m

37


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6. 3 Height Equivalent of an Ideal Stage


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Review:Plate columns vs. Packed columns

• Coulson and Richardson Vol 6. suggest the followingadvantages/disadvantages for Plate vs Packed:

• Plate columns can be designed to handle wider range ofliquid and gas flow rates

• Packed columns not suitable for very low liquid rates

• The efficiency and performance of a plate column canbe more accurately predicted

39


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Plate column easy to think of in # of stages, what about packed?

40

V n+1

n+1

V n

n

V n-1

n-1

n+1

n+1

n

n

n-1

n-1

n + 1

n

n - 1

idealactual

N N η

=

?“Ideal stage”

stage-by-stage

determination

H i h E i l Th i l Pl


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Height Equivalent to aTheoretical Plate(HETP)Column height is determined from # of theoretical plates

and the height equivalent to a theoretical plate (HETP)

41

0

1

0 1 B D

α = 4

7 stages

Example: 7 Theoretical Stages

If the HETP is 0.5 m then...

3.5 m

packed ideal H N HETP= ×


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How to determine an HETP

• Typically determined through empirical data

• General values for random packing

– 0.3 to 0.6 m

• Smaller packing can have lower values but also lesscapacity

• Structured packing can have much improved HETP

– 0.1 to 0.2 m

• Typically no fundamental prediction for HETP

42


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Random and Structured Packing

43

Plastic Tripak(Jaeger Products Co.)

Metal Tripak(Jaeger Products Co.)

Section of expandedmetal packing

Sections of expanded metal packings placedaltenatively at right angles (Denholme Co.)

Structured packing elementsfor small colums with wall

wipers at the periphery

Random - larger HETP

Structured - smaller HETP(better separation with smaller column height)


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Example: HETP for iso-octane/toluene withIntalox packing

• HETP given in termsof a flow capacityfactor

• #25, 40 50 refer topacking sizes of 1,1.5, 2 inches

44

superficial velocity

Recommended design velocity: 20% less than when HETP rises rapidly


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Wetted area key to good separation

• The better the wetted area the lower the HETP

– Thus structured packing typically better than random

• Areas of high liquid flow tend to have low vapour flowand vice versa

• Liquid will also tend toward the outside

• Also means redistribution can be important

– Recommended design practice of redistribution every 3 to4 m

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Recommended examples from textbooksread, understand, try to do the problems yourself

47

McCABE et al:Examples: 21.1, 21.2, 21.3, 21.4, 21.5, 21.6, 21.7

BENÍTEZ, J. (2009):Examples 6.1, 6.4, 6.5, 6.6, 6.7, 6.8

SEADER, J. D. & HENLEY, E. J. (2006).Example 7.1, 7.2, 7.3, 7.4, 7.6

Treybal, R. E. (1981); illustration 9.10

Try problems from the end of these chapters as well.


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6.4 DETERMINATION OFCOLUMN DIAMETER-- APPLICABLE TO BOTH DISTILLATION AND

ABSORPTION COLUMNS

REFERENCE FOR ASSIGNMENT 1

48

Not presented

f l


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Determination of Column Diameter• Column diameter D is a function of the volumetric flow

rate V and velocity u of the gas entering the column

• =4

• For a given task, gas flow rate V is known, and thenunknown parameter is velocity u .

• Gas velocity is often determined by the viable pressuredrop in the column (which is related to operation cost).

• Larger velocity higher pressure drop higheroperation cost

• Smaller velocity lower pressure drop larger

column diameter and higher capital cost


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Centre for Energy - “en e r g y f o r t o d a y a n d t o m o r r o w ”

P∆

Gylog

dry

Loading point

Flooding point

Design considerations: Pressure drop and flooding

G – mass flow per unit area (G y-gas, G x-liquid)

For packed column

Gx G’x

For packed column


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Liquid inlet

Liquid outlet Gas inlet

Gas outlet

Some flooding description

•A visual build-up of liquid on the uppersurface of the packed bed

• A rapid increase in liquid hold-up withincreasing gas rate

• Formation of a continuous liquid phase abovethe packing support plate

• A considerable entrainment of liquid inthe outlet vapour

• Filling of the voids in the packed bed with liquid


www.see.ed.ac.uk

For packed column


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(McCabe, Smith, Harriott)


Gy

Gx

L

V

Design considerations: Diameter of packed towers


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McCabe, Smith, Harriott

Pressure drop analysis: Eckert graph


Flooding line

G y :Mass flow ofgas per

unit areaG y = u ρ v

Pressure drop ininH2O/ft of packing(brackets: mm H 2O/m of packing)

Normally* Moderate to high pressuredistillation =0.4 to 0.75 in water / ftpacking= 32 to 63 mm water / mpacking

* Vacuum Distillation =0.1 to 0.2 in water / ft packing= 8 to 16 mm water / mpacking

* Absorbers and Strippers =0.2 to 0.6 in water / ft packing= 16 to 48 mm water / mpacking

Eckert graph in IS units


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20 . 2

0 . 5

In a flooding line,u becomesumax

u , dry column velocity (m/s);umax, flooding point velocity(m/s); g, acceleration ofgravity (m/s2); φ , packingfactor (1/m); ψ , liquid densitycorrection coefficient, i.e.density of water versus densityof the liquidψ = ρ H2O/ ρ L; μ L, viscosity of liquid (mPa s),wL and w V , liquid and vapor massflowrate (kg/s).

g p



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Sinnott

Other di fferent graphs

Given L, V (mass flow rates)

Select pressure drop

determine u

select packing

Double check pressure drop

D


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ENSC3019/CHPR8503 Topic 3Solid-fluid separations

56


Recommended reading:H. Pierson & B. Perlmutter, Settle Down (Part 1).The Chemical Engineer (TCE) , 2010, June pp48-50.H. Pierson & B. Perlmutter, The solution is clear (Part 2).TCE , 2010, July/August pp53-55.

Chapters 28 & 29 of McCabeet al. , Unit Operations of Chemical Engineering , 7th Edn. McGraw Hill 2005

Sections 18 & 21 of Perry’s Chemical Engineers’ Handbook , 8th

Edn. McGraw-Hill

We will look at:

mailto:[email protected]:[email protected]


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We will look at:

Sedimentation & Settling processes

57

Important solid handling processes we won’tstudy here:• Filtration & screening processes• Size reduction• Solids mixing• Hopper and storage vessel designs

Examples of solid fluid separations


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Examples of solid-fluid separations

Oil and gas industry hydrocyclones

58

Separate sand and other solidsfrom water or other liquids

Separate oil droplets from water



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Coal-fired power station (filter-bags)

– Particulates from flue gases

59

Gravity classifiers


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Gravity classifiersSeparate particles of the same density but differentparticle sizes.

60

FeedLiquid + fine particles

overflow

Coarse particles sink, picked up byscrew Image from http://www.zoneding.com/Product-23.html



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Food and beverage industry (filter)

– Separate curd (solids)

– from whey (liquid)

61

Properties and handling of particulate solids


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Properties and handling of particulate solids

Size

Shape

Density

62

Size and shape of particles


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Size and shape of particlesFor regular shaped particles we can easily define size and shape.

63

Cubel

3

26

Volume l

Area l

=

=

Sphere 3

2

4

3 2

42

d Volume

d Area

π

π

=

=


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Relative sizes of particulate matter

Examples of real particles

Shape of irregular particles


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Shape of irregular particlesSphericity

65

6 ps

p p

d S V

φ =

Eq. 28.1 McCabe et al. p 967

d p = nominal diameter of one particle

V p = volume of one particleS p = surface area of one particle

1 for a sphere1 for cube asd p=l

Sphericity of some materials


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Sphericity of some materials

66

Material Φ Material Φ

Spheres, cubes,short cylinders(L=d p)

1.0 Ottawa sand 0.95

Raschig rings(L=d p)

0.33-0.58 Coal dust 0.73

Berl saddle(L=d p)0.3 Crushed glass 0.65

Mica flakes 0.28

McCabe et al, p164 Table 7.1


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Diff i l VS l i


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Differential VS cumulativedistribution

2 basic principles of


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2 basic principles ofseparationTo separate liquid from solids, or solids from

liquids there are only 2 mechanisms available:

(1) Use a screen or porous medium that retains

one component and allows others to pass(2) Use differences in sedimentation rates as

particles (or drops) move through a gas or

liquid

69

Separation by


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70

Settling / sedimentation

Screen / filter

Gravity

Centrifugal forceHeavy media

Flotation

Magnetic force

Screens

Filters

Crossflow eg. membranes

Separation by

GravityPressureVacuumExpression

Gravity sedimentation processes


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Gravity sedimentation processes

Three broad functional operations

(1) ClassifierSeparate solids into two fractions

(2) ClarificationRemove a relatively small quantity of suspended particles to produce a clear effluent

(3) ThickeningTo increase concentration of solids in a feedstream

71

Selecting a separation method


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Selecting a separation method

1. Define the problem

– Is liquid or solid the valuable product?

– How clear does liquid need to be?

2. Establish process conditions– Particle size, concentrations, flowrates

– How long do particles take to settle?

3. Make a short list of appropriate equipment types

72

Clarifiers and thickeners


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Clarifiers and thickenersConvert dilute slurry of fine particles into a clarifiedliquid and a concentrated suspension.

Often performed in large open tanks.

73

Cessnock Wastewater Treatment Works

http://www.epco.com.au

Batch sedimentation process


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Batch sedimentation process

74

(1)

B

Time

(2)

B

A

CD

(5)

A

D

(3)

B

A

DC

(4)

A

DC

Rate of separation


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Rate of separation

75

Clearliquidinterfaceheight

Settling time, hours

Flocculation


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Flocculationparticles < few microns d p settle slowly

Agglomerate particles faster separation

76

Flocculation for waste water treatment

How flocculation works?

Videos

https://www.youtube.com/watch?v=5uuQ77vAV_U

Equipment - thickeners

https://www.youtube.com/watch?v=aMcamQJxFHshttps://www.youtube.com/watch?v=xcHVjX74o0Yhttps://www.youtube.com/watch?v=xcHVjX74o0Yhttps://www.youtube.com/watch?v=aMcamQJxFHs


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Equipment thickeners

77http://www.filtration-and-separation.com/


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Motion of a particle in air

The forces acting on a particle in a fluid

Eq(1)

(2)

(3)

(4)

ρ : density of parti culate or f luid , kg·m -3

F d : drag force, kg·m·s -2 ma : sum of the forces acting on the particlea: downward acceleration of the particle, m·s -2

The drag force increases as the velocity ofthe particle increases, until it reaches theterminal settling velocity , the sum of forcebecome zero, ma = 0

(5)


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Stokes’ LawThe relationship between velocity and dragforce:

where, μ is the fluid viscosity, Pa·s or kg·m -1·s-1 Substitute eq (6) into eq (5):

Which is commonly referred to as Stocks’ Law.George Gabriel Stokes

(6)

(7)


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In class tutorial 1Compute the terminal settling velocity in air of a spherical particle withdiameter of 1 μ and 10 μ , respectively. Density of the particle is 2000kg·m -3, air density 1.2 kg·m -3, viscosity 1.8 x 10-5 kg·m -1·s-1.

V = 9.81 · (10-6) 2 · (2000-1.2)/ (18 · 1.8 x 10-5 ) = 6.05 x 10-5 m/s V = 9.81 · (10-5) 2 · (2000-1.2)/ (18 · 1.8 x 10-5 ) = 6.05 x 10-3 m/s

Estimate how long will it take for the particle to settle down to the

ground level, if it falls from a 3000 m altitude. Assume no convection,no rainfall. About 19 months for the 1 μ particle and, 5.7 days for the 10 μ particle.


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Terminal settling velocity for spherical particles with specific gravity =2, in standard air.


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Gravity Settler A gravity settler is simplya long chamber through

which the contaminatedgas passes slowly,

allowing time for theparticles to settle bygravity to the bottom.

Very effective for verydirty gases with heavy

particles (metallurgical).

The average velocity equals volumetricflow rate divided by cross sectional area:V avg = Q / ( WH )


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Physical ModelEasy mathematical analysis and typical model for devicesusing similar devices, i.e. cyclones and electrostaticprecipitators.

H

V avg

V t

Lcaptured captured

escapedChamber floor


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Traverse time of particle in the flow direction ist = L / V avg

Vertical settling distance =t ·V t = V t·L / V avg

So all the particles with vertically settling distance smaller than H will

settle on the floor.The fraction of particles that will be captured, isFractional collection efficiency =

η = V t·L / ( V avg·H ) (8)To compute the efficiency-particle diameter relationship, we replacethe terminal settling velocity in eq (8) with the gravity–settlingrelations described by Stock’s law, finding

Block flow/plug flow (9)


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for Mixed flow (practical) AssumptionGas flow is totally mixed in the z direction but not inthe x direction, as most real gas flows are turbulent.Collection efficiency

(10) or,

(11)


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In class tutorial 2Compute the efficiency-diameter relation for a gravity settler that hasH =2m, L = 10m, and V avg = 1 m/s for both the plug and mixed flowmodels, assume Stocks Density of the particle is 2000 kg·m -3, airdensity 1.2 kg·m -3, viscosity 1.8 x 10-5 kg·m -1·s-1.’ law.

A: We can get the result using only one computation and then usingratios. For a 1 micron particle in plug flow:

Mixed flow


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Particle diameter plug flow mixed flow

1 0.000303 0.00030310 0.03 0.0330 0.27 0.2450 0.76 0.53

57.45 1.00 0.6380 1.94 0.86

100 3.03 0.95120 4.36 0.99

57.45

Calculation results

l fl l d fl


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Plug flow settling VS mixed flowDust gas in Clean gas out

Dust gas in Clean gas out

Plug f low gravity settler

Mixed flow gravity settler


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Limitation of gravity settlerOnly effective for particles with diameter >100 micron(fine sand, mineral particle) but not for particles of airpollution (PM 10)

To increase the collection efficiency substantially and practically,by substituting some other force for the gravity in driving theparticles from the gas stream to the collecting surface


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Example A particle is travelling in a gas stream with velocity of18 m/s and radius of 0.3 m. What is the ratio ofcentrifugal force to the gravity force acting on it?

A: (18 x 18/0.3)/9.8 = 110.2


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Centrifugal Separator (Cyclone)Substituting the centrifugal acceleration of thegravitational one into Stocks’ law, eq (7), and drop thebuoyancy term, we find:

This is the settling velocity under centrifuge

(13)


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Structure of cyclones

Similar to gravity settlers, inthe form of two concentrichelices.Only the outer helix

contributes to collectionParticles get into the innerhelix escape uncollectedDimensions are typicallybased on the diameter D0 ofouter helix. Taken as ratiosto D0. Gas inlet width, W i =0.25

tangentially


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Model detailsDuring the outer spiral of the gas, the particles are drivento the wall by centrifugal force, where they collect, attachto each other, and form larger agglomerates and slide down

the wall by gravity and collect in the dust hopper in thebottom.The inlet stream has a height W i in the radial direction,equivalent to the height H of pure gravity settler

The length of the flow path is N π D0, where N is thenumber of turns that gas traverse the outer helix (normallyset as N = 5), analogues to the length of gravity settler L.


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Collection efficiency of cyclonesSubstitute H =W i and L = N π D0 into gravity settlerequation (9) & eq (11), finding:

Further substituting the centrifugal Stokes’ law eq (13) intoabove equations, finding:

plug flow (14)

mixed flow (15)

plug flow (16)mixed flow (17)


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In class tutorial 3Compute the efficiency-diameter relation for a cyclone separator thathas W i = 0.15 m, V c = 18 m/s, and N =5, for both block and mixed flowassumptions, assuming Stocks’ law.

Particle diameter plug flow mixed flow

1103050

57.4580

100120


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For very small particles < 5 micron

An industrial multiclone dust collector

diytrade.com

B&W's Multiclone dust collectormade of a number of parallel smallcyclone

babcock.com/products


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Cut diameterMeasure of the size of the particles caught and the sizepassed for a particular particle collector.Cut diameter is the diameter of a particel for which theefficiency curve has the value of 0.5, i.e. 50%Substitute η = 0.5 into Stocks’ law plug flow model,finding:

plug flow (18)


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Other dust collectorsElectrostatic precipitators (ESP)Venturi scrubberBag filter

Venturi scrubber

design of packed columns for absorption and distillation processes_prelecture slids

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