ENSC3019/CHPR8503: Week 6 Design of Packed Columns 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
Dr Kevin Li
[email protected]
1
Mass Transfer (MT) across phase interface: twosee McCABE et al. p547; BENÍTEZ p165 resistance model distance
Gas film
Bulk gas
Liquid film
yA,G
Bulk liquid
xA,i xA,L
yA,i
= N A k y ( y A, g − y A,i= ) N A k x ( x A,i − x A,L ) ∝ 1 ky Resistances to diffusion of A: (i) in the gas phase film
∝ 1 kx (ii) in the liquid phase film
At the interface: assume local equilibrium between yA and xA, 2 no resistance to MT across the interface
Mass-transfer coefficients:
an engineering concept that allows us to simplify complex diffusion problems.
= N A k y ( y − yi ) Flux
(mole/m2/s)
=
Coefficient
×
Driving force (concentration difference)
Since concentration could be defined in different ways, a variety of coefficients can be defined: • ky, kx, Ky, Kx …… 3
Summary of general forms of MT Rates for two-phase films See McCabe et al. page 547-548. Or if you’re keen for more discussion look at Treybal’s Chapter 5..
= N A k y ( y − yi ) = N A k x ( xi − x ) = NA Ky ( y − y
*
= N A Kx ( x − x) *
MTC=mass transfer coefficient.
1 1 m' = + K y k y kx
ky is local MTC for gas phase yi is mole fraction (of component A) in gas at the gasliquid interface , y is bulk vapour composition kx is local MTC for liquid phase xi is mole fraction (of component A) in liquid at the gas-liquid interface, x is bulk liquid composition
)
Ky is overall MTC for gas phase y* is composition of vapour that would be in equilibrium with the bulk liquid of composition x Kx is overall MTC for liquid phase x* is composition of vapour that would be in equilibrium with the bulk vapour of composition y Subscripts A, and G, L dropped here for simplicity.
m’ is local slope of equilibrium curve i.e. m ' = ( y* − yi ) ( x − xi )
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Tutorial 1 Equilibrium for component A between air and water is described by Henry’s law y*=4x. The local mass transfer coefficients are kx=2 mol m-2s-1 and ky=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.
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6.1 Packed columns for absorption
Dr Kevin Li
[email protected]
Consultation hours 15:00-17:00Thursdays 2.49A in Civil & Mech Eng building
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Equipment for gas-liquid absorption Need intimate contact between the immiscible phases to achieve mass transfer (MT) between phases. Flux NA rate of transfer per unit area of gas-liquid interface Engineering MT equipment focuses on increasing the interfacial area for transfer (a )
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
Gas sparging tanks
Column internals Packing material, plus Liquid inlet systems Liquid & vapour distributors Liquid collecting devices Packing supports Good info at manufacturer www.sulzechemtech.com GREEN, D. W. & PERRY, R. H. (eds.) (2008). Perry's chemical engineers' handbook, New York: McGraw-Hill.
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Packed columns – random packings Raschig rings
see more images at www.tower-packing.com Metal pall rings
VSP Inner arc ring
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Structured packings www.sulzerchemtech.com
MellapakTM
www.sulzerchemtech.com
Grids
Tray columns
V-grid www.sulzerchem.com Sieve tray
High performance trays eg. Shell calming section tray
www.sulzerchem.com
Coulson and Richardson Vol 6. list some of the factors which influence choice of trays or packing in a column: 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. 14
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. 9. Packing cheaper for small columns, d < 0.6 m 15
Column internals – process design 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 Corrosion and other materials issues
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MT Rate, rA, for absorption per unit volume of See McCabe et al. page 579 packed column
rA
k y a ( y − yi )
kya is local MTC for gas phase on unit volume basis
rA
k x a ( xi − x )
kxa is local MTC for liquid phase on unit volume basis
rA
K ya ( y − y
rA
K xa ( x − x )
Coefficient
*
*
yi is mole fraction (of component A) in gas at the gasliquid interface , y is bulk vapour composition
xi is mole fraction (of component A) in liquid at the gas-liquid interface, x is bulk liquid composition
)
Kya 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
Kxa 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
a is interfacial area per unit volume of packed column 17
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 Kya here.
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Calculation of packing height (dilute gas) L, x2
y2
x+dx
y+dy y
dz x
Mass balance on component A across differential volume dz. Rate loss solute from gas = Rate gain solute by liquid
Assume: • dilute gas change in molar flow V is neglected
V, y1
x1
−Vdy = K y a ( y − y ) Adz *
a = interfacial area/unit volume of column A = cross-sectional area column (m2) ZT = total height of packed section Let’s do a dimension analysis here. How do we get this equation from NA?
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Calculation of packing height See McCabe et al. page 580-581
Rearranging and integration of the mass balance equation:
Then, substitute Zt into above equation
Let
We now have an equation to calculate the total height of packing, ZT, based on concentration driving force (y-y*), gas flow rate and the gas phase MTC:
2
V dy Zt = * ∫ K y aA 1 y − y 20
ZT = (height transfer unit) x (number units) See McCabe et al. page 580-581
2
V dy Zt = * ∫ K y aA 1 y − y Height of transfer unit HOy Units of length.
The height of packing needed to achieve:
change in gas conc. = driving force for that section of packing.
change in gas conc. average driving force Number of transfer units NOy
Subscript Oy shows based on overall gas phase driving force.
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Operation line and equilibrium line
Graphic integration: 1/(Y – Y*) as a function of Y
Four sets of HTUs and NTUs See McCabe et al. page 583
Gas film:
Liquid film:
Overall gas:
Overall liquid:
Ny = ∫
dy y − yi
V/A Hy = k ya L/ A Hx = kxa
dx Nx = ∫ xi − x
H Oy
V/A = K ya
dy * y− y
H Ox
V/A = K xa
N Oy = ∫
N Ox
dx =∫ * x −x 23
ENSC3019
6.2 Determination of Column Height
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Values of height of transfer unit See McCabe et al. page 580-581
Values of HOy 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. This estimation is a key element of Assignment 1.
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2
V dy Evaluating the Zt = * ∫ K y aA 1 y − y integral for NOy? Simplest case - Straight operating & equilibrium lines. Can evaluate NOy by:
N Oy = y2
change in gas conc. log mean driving force
dy = N Oy ∫= * y y − ) y1 ( y − y ) −(y − y ) ( (y− y ) = y− y ( ) ln (y− y ) *
*
1
lm
lm
For details on the integration above, see Coulson & Richardson Vol2.
Log mean driving force *
y2 − y1 * y y − ( )
*
1
*
2
2
Example H2S scurbber problem & solution provided at end of these set of slides.
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Challenges and discussions 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 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 Ye=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 NOG (or Noy)required The overall coefficient for absorption KYa is 0.04 kmol/s.m3 (unit mole fraction driving force). 28
Example 13.1 solution (1) Data:
Y2
Ls, X2
1) Equilibrium expression Ye=2X 2) Top of column conditions Y2 = 0.03 x 0.01 = 0.0003 X2 = 0 Y2e = 0
absorber
Driving force = Y2-Y2e = 0.0003 3) Bottom of column conditions
Vs, Y1
X1
Y1 = 0.03 X1 = 0.013 Y1e=0.026 Driving force = Y1-Y1e = 0.004 29
Example 13.1 solution (2) Logarithmic mean driving force: 0.004 − 0.0003 Y − Y = ( e ) lm 0.004 ln 0.0003 0.0037 = = 0.00143 2.59
Mass balance on H2S in gas film:
(rate moles lost from gas) = (rate mass transfer)
Vs (Y1 − Y2 )= S KG aP (Y − Ye )lm SZ Where S is the cross section area (which is also termed as A)
KG is the pressure dependent MTC KG P = Ky
Example 13.1 solution (3) And we can rewrite in terms of lumped overall coefficient:
KG a P = KY a = 0.04 kmol/s m3 Then:
Vs (Y1 - Y2) = KY a (Y - Ye)lm Z 0.015 (0.03 – 0.0003) = 0.04 (0.00143) Z Solve for Z:
Z = 7.8 m 31
Example 13.1 solution (4) Now calculate height of transfer unit:
Which is another expression of HOy
For dilute systems, Vs ≈ V
Number of transfer units:
Z N = = 20.7 OG H OG NOG = 21 Which is another expression of32 NOy
Example 13.1 alternatives solutions If your love calculus you could solve analytically: Y2
N OG
dY =∫ Y −Y Y1 e
NOG = 21.1
Calculate HOG as before. Then calculate Z.
Z = 7.91 m 33
Example 13.1 alternatives solutions 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 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
35
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. 36
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 37
6. 3 Height Equivalent of an Ideal Stage
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
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Plate column easy to think of in # of stages, what about packed? V n+1 y n+1 n +1
n
L n+1 x n+1 Ln xn
n -1
Vn yn V n-1 y n-1
L n-1 x n-1
“Ideal stage” stage-by-stage determination
N actual =
N ideal
η 40
?
Height Equivalent to a Theoretical Plate (HETP) Column height is determined from # of theoretical plates and the height equivalent to a theoretical plate (HETP)
H packed = N ideal × HETP α=4 7 stages
1
If the HETP is 0.5 m then...
3.5 m 0
xB
0 xD 1
Example: 7 Theoretical Stages
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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 less capacity • Structured packing can have much improved HETP – 0.1 to 0.2 m
• Typically no fundamental prediction for HETP
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Random and Structured Packing Random - larger HETP
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
Structured - smaller HETP (better separation with smaller column height) 43
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 Recommended design velocity: 20% less than when HETP rises rapidly superficial velocity 44
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 4m
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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
Weir Liquid distributor
Wall wiper redistributor
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Recommended examples from textbooks read, understand, try to do the problems yourself 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. 47
Not presented
6.4 DETERMINATION OF COLUMN DIAMETER
-- APPLICABLE TO BOTH DISTILLATION AND ABSORPTION COLUMNS
REFERENCE FOR ASSIGNMENT 1
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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 • 𝐷𝐷 =
4𝑉𝑉 𝜋𝜋𝑢𝑢
• 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
For packed column Design considerations: Pressure drop and flooding G’x G x
∆P
dry
log Gy Loading point Flooding point G – mass flow per unit area (Gy-gas, Gx-liquid) Centre for Energy - “energy for today and tomorrow”
For packed column Design considerations: Pressure drop and flooding Gas outlet Some flooding description
Liquid inlet
•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 Liquid outlet
Gas inlet
• Filling of the voids in the packed bed with liquid
www.see.ed.ac.uk
Centre for Energy - “energy for today and tomorrow”
Design considerations: Pressure drop and flooding V
Gx
L Gy
(McCabe, Smith, Harriott) Centre for Energy - “energy for today and tomorrow”
Design considerations: Diameter of packed towers Pressure drop analysis: Eckert graph Pressure drop in inH2O/ft of packing (brackets: mm H2O/ m of packing)
Flooding line
Normally * Moderate to high pressure distillation = 0.4 to 0.75 in water / ft packing = 32 to 63 mm water / m packing
Gy : Mass flow of gas per unit area Gy = u ρv
* Vacuum Distillation = 0.1 to 0.2 in water / ft packing = 8 to 16 mm water / m packing
McCabe, Smith, Harriott
Centre for Energy - “energy for today and tomorrow”
* Absorbers and Strippers = 0.2 to 0.6 in water / ft packing = 16 to 48 mm water / m packing
Eckert graph in IS units
𝑢𝑢2 𝜙𝜙𝜙𝜙 𝜌𝜌𝑉𝑉 𝜇𝜇𝐿𝐿 0.2 𝑔𝑔 𝜌𝜌𝐿𝐿
In a flooding line, u becomes umax
u, dry column velocity (m/s); umax, 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 = ρH2O/ρL; μL, viscosity of liquid (mPa s), wL and wV, liquid and vapor mass flowrate (kg/s).
Centre for Energy - “energy for today and tomorrow”
𝑤𝑤𝐿𝐿 𝜌𝜌𝑉𝑉 𝑤𝑤𝑉𝑉 𝜌𝜌𝐿𝐿
0.5
Design considerations: Diameter of packed towers Other different graphs
Given L, V (mass flow rates)
Select pressure drop
determine u
select packing
Double check pressure drop D Sinnott
Centre for Energy - “energy for today and tomorrow”
ENSC3019/CHPR8503 Topic 3 Solid-fluid separations 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 Sections 18 & 21 of Perry’s Chemical Engineers’ Handbook, 8th Edn. McGraw-Hill
Dr Kevin Li
[email protected]
56
We will look at: Sedimentation & Settling processes
Important solid handling processes we won’t study here: • • • •
Filtration & screening processes Size reduction Solids mixing Hopper and storage vessel designs 57
Examples of solid-fluid separations
Oil and gas industry hydrocyclones Separate sand and other solids from water or other liquids Separate oil droplets from water
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Examples of solid-fluid separations
Coal-fired power station (filter-bags) – Particulates from flue gases
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Gravity classifiers Separate particles of the same density but different particle sizes.
Feed
Liquid + fine particles overflow
Coarse particles sink, picked up by 60 screw Image from http://www.zoneding.com/Product-23.html
Examples of solid-fluid separations Food and beverage industry (filter) – Separate curd (solids) – from whey (liquid)
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Properties and handling of particulate solids
Size Shape Density
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Size and shape of particles For regular shaped particles we can easily define size and shape.
Cube
Volume = l Area = 6l Sphere
3
l
2
4 d Volume = π 3 2 d Area = 4π 63 2
2
3
Relative sizes of particulate matter
Examples of real particles
Shape of irregular particles Sphericity
φs =
6 dp S p Vp
1 for a sphere 1 for cube as dp=l
Eq. 28.1 McCabe et al. p 967 dp = nominal diameter of one particle Vp = volume of one particle Sp = surface area of one particle 65
Sphericity of some materials Material
Φ
Material
Spheres, cubes, short cylinders (L=dp)
1.0
Ottawa sand
0.95
Coal dust
0.73
Crushed glass
0.65
Mica flakes
0.28
Raschig rings (L=dp) Berl saddle (L=dp)
0.330.58 0.3
McCabe et al, p164 Table 7.1
Φ
66
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
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Separation by Settling / sedimentation
Gravity Centrifugal force Heavy media Flotation Magnetic force
Screens Screen / filter
Gravity Filters Pressure Vacuum Expression Crossflow eg. membranes 70
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 71
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
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Clarifiers and thickeners
Convert dilute slurry of fine particles into a clarified liquid and a concentrated suspension. Often performed in large open tanks. Cessnock Wastewater Treatment Works http://www.epco.com.au
73
Batch sedimentation process Time (1)
B
(2)
(3)
A
A
B
C
(4)
(5)
A
A
B
D
C D
C D
D 74
Rate of separation
Clear liquid interface height
Settling time, hours 75
Flocculation particles < few microns dp settle slowly Agglomerate particles faster separation
Videos Flocculation for waste water treatment How flocculation works? https://www.youtube.com/watc h?v=5uuQ77vAV_U 76
Equipment - thickeners
http://www.filtration-and-separation.com/
77
Motion of a particle in air Eq(1) (2) (3) (4)
The forces acting on a particle in a fluid
The drag force increases as the velocity of the particle increases, until it reaches the terminal settling velocity, the sum of force become zero, ma = 0
ρ: density of particulate or fluid, kg·m-3 Fd: drag force, kg·m·s-2 ma: sum of the forces acting on the particle a: downward acceleration of the particle, m·s-2
(5)
Stokes’ Law The relationship between velocity and drag
force:
(6)
where, μ is the fluid viscosity, Pa·s or kg·m-1·s-1 Substitute eq (6) into eq (5): (7)
Which is commonly referred to as Stocks’ Law. George Gabriel Stokes
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. About 19 months for the 1 μ particle and, 5.7 days for the 10 μ particle.
Terminal settling velocity for spherical particles with specific gravity =
2, in standard air.
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: Vavg = Q / (WH)
Physical Model Easy mathematical analysis and typical model for devices
using similar devices, i.e. cyclones and electrostatic precipitators. Vavg Vt H Chamber floor
escaped
L captured
captured
Deriving the model Assumptions for plug flow
a) The horizontal velocity of the gas in the chamber is equal to Vavg everywhere and constant b) The vertical component of the velocity of the particle is equal to the terminal settling velocity due to gravity, Vt 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 / Vavg Vertical settling distance = t·Vt = Vt·L / Vavg 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 = Vt·L / (Vavg·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 Block flow/plug flow
(9)
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) (11)
or,
In class tutorial 2 Compute the efficiency-diameter relation for a gravity settler that has
H =2m, L = 10m, and Vavg = 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:
Mixed flow
Calculation results Particle diameter, μ 1 10 30 50 57.45 80 100 120
η plug flow 0.000303 0.03 0.27 0.76 1.00 1.94 3.03 4.36
η mixed flow 0.000303 0.03 0.24 0.53 0.63 0.86 0.95 0.99
57.45 μ
Plug flow settling VS mixed flow Dust gas in
Clean gas out
Plug flow gravity settler
Dust gas in
Clean gas out
Mixed 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 (PM10)
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 Vc along the path, then it has angular velocity ω =Vc/ r (12) 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? A: (18 x 18/0.3)/9.8 = 110.2
Centrifugal Separator (Cyclone) Substituting the centrifugal acceleration of the gravitational one into Stocks’ law, eq (7), and drop the buoyancy term, we find: (13)
This is the settling velocity under centrifuge
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, Wi = 0.25
tangentially
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 Wi 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 =Wi and L = NπD0 into gravity settler
equation (9) & eq (11), finding:
plug flow (14) mixed flow (15)
Further substituting the centrifugal Stokes’ law eq (13) into
above equations, finding:
plug flow (16)
mixed flow (17)
In class tutorial 3 Compute the efficiency-diameter relation for a cyclone separator that
has Wi = 0.15 m, Vc = 18 m/s, and N =5, for both block and mixed flow assumptions, assuming Stocks’ law. Particle diameter, μ 1 10 30 50 57.45 80 100 120
η plug flow
η mixed flow
For very small particles < 5 micron
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babcock.com/products
B&W's Multiclone dust collector made of a number of parallel small cyclone
An industrial multiclone dust collector
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: plug flow (18)
Other dust collectors Electrostatic precipitators (ESP)Venturi scrubber Bag filter Venturi scrubber
