Design of Packed Columns for Absorption and Distillation Processes_prelecture Slids

ENSC3019

/CHPR8503

:

Week 6 Design of Packed Columns

1 Dr Kevin Li

[email protected]

Recommended reading:

McCabe et al., Unit Operations of Chemical Engineering, Chapter 21

Treybal, R. E. Mass Transfer Operations, 3rd _{Edn. McGraw-Hill 1955, Chapter 9} Coulson, J. M. and Richardson, J. F. Chemical Engineering, Volume 6: Particle Technology and Separation Processes, 5th _{Edn. Butterworth-Heinemann 2002}

Mass Transfer (MT) across phase interface:

two-resistance model

2

Gas film Liquid film

Bulk 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

N

_A

=

k

_x

(

x

_{A i,}

−

x

_{A L,}

)

1 k_y ∝ 1 k_x ∝

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:

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

k_y is local MTC for gas phase

y_iis mole fraction (of component A) in gas at the gas-liquid interface , y is bulk vapour composition

k_x is local MTC for liquid phase

x_i is mole fraction (of component A) in liquid at the gas-liquid interface, x is bulk liquid composition

K_y is overall MTC for gas phase

y* _{is composition of vapour that would be in}

equilibrium with the bulk liquid of composition x K_x is overall MTC for liquid phase

x* _{is composition of vapour that would be in}

equilibrium 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 i} m = y − y x − x

Tutorial 1 Equilibrium for component A between air and water is described by Henry’s law y*=4x. The local mass transfer coefficients are k_x=2 mol m-2_s-1 _{and k}

y=1 mol m-2s-1. (1) What is the overall mass transfer coefficient for gas

phase?

(2) Evaluate the flux of A between phases at a point in a column where bulk compositions are 0.08 mole fraction in the gas and 0.01 mole fraction in the liquid.

6.1 Packed columns for absorption

Dr Kevin Li

[email protected]

Consultation hours 15:00-17:00Thursdays

2.49A in Civil & Mech Eng building

Equipment for gas-liquid absorption

Need intimate contact between the immiscible

phases to achieve mass transfer (MT) between

phases.

Flux

N

_A

 rate of transfer per unit area of gas-liquid interface

Engineering MT equipment focuses on increasing

Main equipment types

Packed columns

Random

(let to fall randomly into column during installation)

Structured

(engineering for lower ΔP, higher cost )

Tray columns -

liquid levels on each tray

Column internals

9

GREEN, D. W. & PERRY, R. H. (eds.) (2008).

Perry's chemical engineers' handbook, New York: McGraw-Hill.

Packing material, plus Liquid inlet systems

Liquid & vapour distributors Liquid collecting devices

Packing supports

Good info at manufacturer

Packed columns – random packings

10 Metal pall rings Raschig rings

VSP Inner arc ring

see more images at

Structured packings

www.sulzerchemtech.com

Mellapak

TM

www.sulzerchemtech.com

Tray columns

V-grid www.sulzerchem.com Sieve tray

High performance trays

eg. Shell calming section tray

14

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

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

3. Easier to make provisions for withdrawal side streams in plate columns.

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

Plate columns can be designed with more assurance - some doubt that good 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:

15

Trays/Plate columns vs. Packed

columns

5. 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 low for safety.

7. Packed columns are more suitable for foaming systems 8. The pressure drop per equilibrium stage can be lower for

packed columns.

Column internals –

process design

16

Process design or process tech support to operation needs to consider:

Type of contacting device

Number equilibrium stages Height of packing required

Pressure drop

Fouling

MT Rate, r

_A

, for absorption per unit volume of

packed column

k_ya is local MTC for gas phase on unit volume basis

y_iis 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 on unit volume basis

x_i is mole fraction (of component A) in liquid at the gas-liquid interface, x is bulk liquid composition

K_ya is overall MTC for gas phase on unit volume basis

y* _{is composition of vapour that would be in}

equilibrium with the bulk liquid of composition x

K_xa is overall MTC for liquid phase on unit volume basis

x* _{is composition of vapour that would be in}

equilibrium with the bulk vapour of composition y

17

(

)

A y i

r

=

k a y

−

y

(

)

A x i

r

=

k a x

−

x

(

*

)

A x

r

=

K a x

−

x

(

*

)

A y

r

=

K a y

−

y

See McCabe et al. page 579

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.

Calculation of packing height

(dilute gas)

19

a = interfacial area/unit volume of column

A = cross-sectional area column (m2₎

Z_T = total height of packed section

(

*

)

y

Vdy

K a y

y

Adz

−

=

−

L, x₂ V, y₁ x1 y₂ dz x x+dx y+dy y

Mass balance on component A across differential volume 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?

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 gas phase MTC:

See McCabe et al. page 580-581

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

21

See McCabe et al. page 580-581

2 * 1 y t

d

V

A

Z

y

y

a

y

K

−

=

∫

change in gas conc.

average driving force

Number of transfer units

N_Oy

Subscript Oy shows based on overall gas phase driving force.

Height of transfer unit

H_Oy

Units of length.

The height of packing needed to achieve:

change in gas conc.

driving force

=

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

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

a

k

=

/

x x

L A

H

a

k

=

/

Oy y

V

A

H

K

a

=

/

Ox x

V

A

H

K

a

=

ENSC3019

6.2 Determination of Column Height

Values of height of transfer

unit

25

See McCabe et al. page 580-581

Values of H_Oy are system dependent.

Sometimes available for a particular system directly in the literature, or could be measured in pilot-plant studies.

But, often need to estimate height of transfer units from

empirical correlations for individual MTCs or individual heights of a transfer unit.

Evaluating the

integral for N

_Oy

?

26 2 * 1 t y

d

V

Z

A

y

y

a

y

K

−

=

∫

Simplest case - Straight operating & equilibrium lines.

Can evaluate N_Oy by: change in gas conc.

log mean driving force

Oy N =

(

) (

₍

) (

₎

)

(

)

* * * _{1 2} * 1 * 2 ln lm 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 force

For details on the integration above, see Coulson & Richardson Vol2.

Example H₂S scurbber problem & solution provided at end of these set of slides.

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?

Example 13.1 H

₂

S scrubber

28

Gas from a petroleum distillation column has its H₂S concentration reduced from 0.03 kmol H₂S /kmol inert hydrocarbon to 1 % of this value by scrubbing with triethanolamine-water solvent in a

countercurrent 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 H₂S and leaves containing 0.013

kmol H₂S /kmol solvent. If the flow of inert gas is 0.015 kmol/s.m2 _of

tower 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.m3 _(unit

Example 13.1 solution (1)

29 Data:

1) Equilibrium expression Y_e=2X

2) Top of column conditions Y₂ = 0.03 x 0.01 = 0.0003

L

_s

, X

₂

Vs, Y

₁

X

1

Y

₂ absorber

3) Bottom of column conditions Y₁ = 0.03

X₁ = 0.013  Y_1e=0.026

Driving force = Y₁-Y_1e = 0.004

X₂ = 0  Y_2e = 0

Example 13.1 solution (2)

Logarithmic mean driving force:

(

)

0.004 0.0003 0.004 ln 0.0003 0.0037 0.00143 2.59 e lm Y −Y = −       = =

Mass balance on H₂S in gas film:

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

(

1 2

)

(

)

s G _{e lm}

V Y

−

Y S

=

K aP Y

−

Y

SZ

Where S is the cross section area

Example 13.1 solution (3)

31

And we can rewrite in terms of lumped overall coefficient:

Then:

Solve for Z:

_{Z = 7.8 m}

K

G

a P = K

Y

a = 0.04 kmol/s m

3

V

_s

(Y

₁

- Y

₂

) = K

_Y

a (Y - Y

_e

)

_lm

Z

Example 13.1 solution (4)

32

Now calculate height of transfer unit:

Number of transfer units:

N

_OG

= 21

20.7

OG OG

Z

N

H

=

=

Which is another expression of N_Oy Which is another

expression of H_Oy

Example 13.1 alternatives solutions

33

If your love calculus you could solve analytically:

N

_OG

= 21.1

2 1 Y OG e Y

dY

N

Y

Y

=

−

∫

Calculate H

_OG

as before. Then calculate Z.

Example 13.1 alternatives solutions

34

Could do a graphical-numerical solution (eg.

trapezoidal rule or Simpson rule to find the N

_OG

)

Looking forward:

Plate columns vs. Packed columns

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

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

• Packed columns not suitable for very low liquid rates • The efficiency and performance of a plate column can

be more accurately predicted

Looking forward:

Plate columns vs. Packed columns

• Plate columns can be designed with more assurance - some doubt that good 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.

• Easier to make provisions for withdrawal side streams in plate columns.

• Fouling by solids – can easily install manholes on plates – small columns however – may be easier to replace

packing when fouled.

Looking forward

:

Plate columns vs. Packed columns

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

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

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

• Packed columns are more suitable for foaming systems • The pressure drop per equilibrium stage can be lower

for packed columns – impt. vacuum distillation • Packing cheaper for small columns, d < 0.6 m

Review:

Plate columns vs. Packed columns

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

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

• Packed columns not suitable for very low liquid rates • The efficiency and performance of a plate column can

be more accurately predicted

Plate column easy to think of in # of stages,

what about packed?

40 V_n+1 y _n+1 V_n y_n V_n-1 y_n-1 L _n+1 x _n+1 L_n x_n L_n-1 x_n-1 n + 1 n n - 1 ideal actual N N = _η

?

“Ideal stage” stage-by-stage determination

H

eight

E

quivalent to a

T

heoretical

P

late

(

HETP

)

Column height is determined from # of theoretical plates and the height equivalent to a theoretical plate (HETP)

41 0 1 0 1 x_B x_D α = 4 7 stages

Example: 7 Theoretical Stages

If the HETP is 0.5 m then...

3.5 m

packed ideal

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 less capacity

• Structured packing can have much improved HETP

– 0.1 to 0.2 m

• Typically no fundamental prediction for HETP

Random and Structured Packing

43

Plastic Tripak (Jaeger Products Co.)

Metal Tripak (Jaeger Products Co.)

Section of expanded metal packing

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

Structured packing elements for small colums with wall

wipers at the periphery

Random - larger HETP

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

Example: HETP for iso-octane/toluene with

Intalox packing

• HETP given in terms

of a flow capacity

factor

• #25, 40 50 refer to

packing sizes of 1,

1.5, 2 inches

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

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 flow and vice versa

• Liquid will also tend toward the outside

• Also means redistribution can be important

– Recommended design practice of redistribution every 3 to 4 m

Liquid distributors/redistributors

• Liquid distributor – distribute liquid evenly over column (feed and reflux from condenser)

• Redistributors - collect liquid that has migrated to the walls and redistribute it evenly over packing or even out any other maldistribution

46

Recommended examples from textbooks

read, 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

6.4 DETERMINATION OF

COLUMN DIAMETER

-- APPLICABLE TO BOTH DISTILLATION AND ABSORPTION COLUMNS

REFERENCE FOR

ASSIGNMENT 1

48 Not presented

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

• 𝐷𝐷 = _{𝜋𝜋𝑢𝑢}4𝑉𝑉

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

• Gas velocity is often determined by the viable pressure drop in the column (which is related to operation cost). • Larger velocity  higher pressure drop  higher

operation cost

• Smaller velocity  lower pressure drop  larger column diameter and higher capital cost

Centre for Energy - “energy for today and tomorrow”

P

∆

Gy

log

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

G_x G’_x

Centre for Energy - “energy for today and tomorrow” Liquid inlet

Liquid outlet Gas inlet Gas outlet

Some flooding description

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

• A rapid increase in liquid hold-up with increasing gas rate

• Formation of a continuous liquid phase above the packing support plate

• A considerable entrainment of liquid in the outlet vapour

• Filling of the voids in the packed bed with liquid

Design considerations: Pressure drop and flooding

www.see.ed.ac.uk

Centre for Energy - “energy for today and tomorrow”

(McCabe, Smith, Harriott)

Design considerations: Pressure drop and flooding

G_y G_x

L

Centre for Energy - “energy for today and tomorrow”

McCabe, Smith, Harriott

Pressure drop analysis: Eckert graph

Design considerations: Diameter of packed towers

Flooding line G_y : Mass flow of gas per unit area G_y = u ρ_v Pressure drop in inH₂O/ft of packing (brackets: mm H₂O/ m of packing)

Normally

* Moderate to high pressure distillation = 0.4 to 0.75 in water / ft packing = 32 to 63 mm water / m packing * Vacuum Distillation = 0.1 to 0.2 in water / ft packing = 8 to 16 mm water / m packing

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

Centre for Energy - “energy for today and tomorrow” 𝑢𝑢2𝜙𝜙𝜙𝜙 𝑔𝑔 𝜌𝜌𝑉𝑉 𝜌𝜌_𝐿𝐿 𝜇𝜇𝐿𝐿0.2 𝑤𝑤𝐿𝐿 𝑤𝑤𝑉𝑉 𝜌𝜌𝑉𝑉 𝜌𝜌𝐿𝐿 0.5 In a flooding line, u becomes u_max

u, dry column velocity (m/s); u_max, flooding point velocity (m/s); g, acceleration of gravity (m/s2_{); φ, packing}

factor (1/m); ψ, liquid density correction coefficient, i.e.

density of water versus density of the liquid ψ = ρ_H2O/ρ_L; μ_L, viscosity of liquid (mPa s), w_L and w_V, liquid and vapor mass flowrate (kg/s).

Centre for Energy - “energy for today and tomorrow”

Design considerations: Diameter of packed towers

Sinnott

Other different graphs

Given L, V (mass flow rates)

Select pressure drop

determine u

select packing

Double check pressure drop

ENSC3019/

CHPR8503

Topic 3

Solid-fluid separations

56

Dr Kevin Li

[email protected]

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 McCabe et al., Unit Operations of Chemical Engineering, 7th _{Edn. McGraw Hill 2005}

We will look at:

Sedimentation & Settling processes

57

Important solid handling processes we won’t

study here:

• Filtration & screening processes

• Size reduction

• Solids mixing

Examples of solid-fluid separations

Oil and gas industry  hydrocyclones

58

Separate sand and other solids from water or other liquids

Examples of solid-fluid separations

Coal-fired power station (filter-bags)

– Particulates from flue gases

Gravity classifiers

Separate particles of the same density but different

particle sizes.

60

Feed

Liquid + fine particles overflow

Coarse particles sink, picked up by

Examples of solid-fluid separations

Food and beverage industry (filter)

– Separate curd (solids)

– from whey (liquid)

Properties and handling of particulate solids

Size

Shape

Density

Size and shape of particles

For regular shaped particles we can easily define size and shape.

63

Cube

l 3 2

6

Volume

l

Area

l

=

=

Sphere

3 2

4

3

2

4

2

d

Volume

d

Area

π

π

 

=

_{ }

 

 

=

_{ }

 

Relative sizes of particulate matter

Shape of irregular particles

Sphericity 65

6

_p s 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 particle

S

_p

= surface area of one particle

1 for a sphere

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

Description of populations of

particles

• In practice, size distribution is a histogram

• Distribution curve by mass, number and surface can differ dramatically

• Which distribution we would use is dependent on the end use of the information

Differential VS cumulative

distribution

2 basic principles of

separation

To 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

70 Settling / sedimentation Screen / filter Gravity Centrifugal force Heavy media Flotation Magnetic force Screens Filters

Crossflow eg. membranes

Separation by

Gravity Pressure Vacuum Expression

Gravity sedimentation processes

Three broad functional operations

(1) Classifier

Separate solids into two fractions

(2) Clarification

Remove a relatively small quantity of suspended

particles to produce a clear effluent

(3) Thickening

To increase concentration of solids in a feed

stream

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

Clarifiers and thickeners

Convert dilute slurry of fine particles into a clarified

liquid and a concentrated suspension.

Often performed in large open tanks.

73

Cessnock Wastewater Treatment Works

Batch sedimentation process

74 (1)

B

Time

(2)

B

A

C

D

(5)

A

D

(3)

B

A

D

C

(4)

A

D

C

Rate of separation

75 Clear liquid interface height

Flocculation

particles < few microns d_p settle slowly

Agglomerate particles  faster separation

76 Flocculation for waste water treatment

How flocculation works? Videos

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

Equipment - thickeners

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

Motion of a particle in air

The forces acting on a particle in a fluid

Eq(1) (2) (3) (4)

ρ: density of particulate or fluid, kg·m-3

F_d: drag force, kg·m·s-2

ma: sum of the forces acting on the particle a: downward acceleration of the particle, m·s-2

The drag force increases as the velocity of the particle increases, until it reaches the

terminal settling velocity, the sum of force

Stokes’ Law

 The relationship between velocity and drag

force:

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)

In class tutorial 1

 Compute the terminal settling velocity in air of a spherical particle with

diameter of 1 μ and 10 μ, respectively. Density of the particle is 2000 kg·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.

 Terminal settling velocity for spherical particles with specific gravity =

Gravity Settler

 A gravity settler is simply

a long chamber through which the contaminated gas passes slowly,

allowing time for the particles to settle by gravity to the bottom.

 Very effective for very

dirty gases with heavy particles (metallurgical).

The average velocity equals volumetric flow rate divided by cross sectional area:

Physical Model

 Easy mathematical analysis and typical model for devices

using similar devices, i.e. cyclones and electrostatic precipitators. H V_avg V_t L captured captured escaped Chamber floor

Deriving the model



Assumptions for plug flow

a) The horizontal velocity of the gas in the chamber is

equal to V

_avg

everywhere and constant

b) The vertical component of the velocity of the

particle is equal to the terminal settling velocity due to

gravity, V

_t

c) If a particle settles to the floor, it stays there and is

not-entrained

d) Particle size distribution is uniform, no interaction

with each other

 Traverse time of particle in the flow direction is

t = 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, is

Fractional collection efficiency =

η = V_t·L / (V_avg·H) (8)

 To compute the efficiency-particle diameter relationship, we replace

the terminal settling velocity in eq (8) with the gravity–settling relations described by Stock’s law, finding

for Mixed flow (practical)



Assumption

Gas flow is totally mixed in the z direction but not in

the x direction, as most real gas flows are turbulent.



Collection efficiency

(10) or, (11)

In class tutorial 2

 Compute the efficiency-diameter relation for a gravity settler that has

H =2m, L = 10m, and V_avg = 1 m/s for both the plug and mixed flow models, assume Stocks Density of the particle is 2000 kg·m-3_{, air}

density 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 using

ratios. For a 1 micron particle in plug flow:

Particle diameter, μ η plug flow η mixed flow 1 0.000303 0.000303 10 0.03 0.03 30 0.27 0.24 50 0.76 0.53 57.45 1.00 0.63 80 1.94 0.86 100 3.03 0.95 120 4.36 0.99 57.45 μ

Calculation results

Plug flow settling VS mixed flow

Dust gas in Clean gas out

Dust gas in Clean gas out

Plug flow gravity settler

Limitation of gravity settler



Only effective for particles with diameter >100 micron

(fine sand, mineral particle) but not for particles of air

pollution (PM

₁₀

)

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

Centrifugal force

 If a body moves in a circular

path with radius r and

velocity V_c along the path, then it has angular velocity

ω =V_c/ r

 Centrifuge force =

acceleration, substitution of g

Example



A particle is travelling in a gas stream with velocity of

18 m/s and radius of 0.3 m. What is the ratio of

centrifugal force to the gravity force acting on it?

Centrifugal Separator (Cyclone)



Substituting the centrifugal acceleration of the

gravitational one into Stocks’ law, eq (7), and drop the

buoyancy term, we find:



This is the settling velocity under centrifuge

(13)

Structure of cyclones

 Similar to gravity settlers, in

the form of two concentric helices.

 Only the outer helix

contributes to collection

 Particles get into the inner

helix escape uncollected

 Dimensions are typically

based on the diameter D₀ of outer helix. Taken as ratios to D₀. Gas inlet width, W_i = 0.25

Model details

 During the outer spiral of the gas, the particles are driven

to the wall by centrifugal force, where they collect, attach to each other, and form larger agglomerates and slide down the wall by gravity and collect in the dust hopper in the

bottom.

 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 NπD₀, where N is the

number of turns that gas traverse the outer helix (normally set as N = 5), analogues to the length of gravity settler L.

Collection efficiency of cyclones

 Substitute H =W_i and L = NπD₀ into gravity settler

equation (9) & eq (11), finding:

 Further substituting the centrifugal Stokes’ law eq (13) into

above equations, finding:

plug flow (14) mixed flow (15)

In class tutorial 3

 Compute the efficiency-diameter relation for a cyclone separator that

has W_i = 0.15 m, V_c = 18 m/s, and N =5, for both block and mixed flow assumptions, assuming Stocks’ law.

Particle diameter, μ η plug flow η mixed flow 1 10 30 50 57.45 80 100 120

For very small particles < 5 micron

An industrial multiclone dust collector

diytrade.com

B&W's Multiclone dust collector made of a number of parallel small cyclone

Cut diameter



Measure of the size of the particles caught and the size

passed for a particular particle collector.



Cut diameter is the diameter of a particel for which the

efficiency curve has the value of 0.5, i.e. 50%



Substitute η = 0.5 into Stocks’ law plug flow model,

finding:

Other dust collectors



Electrostatic precipitators (ESP)Venturi scrubber



Bag filter