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 LiKevin.li@uwa.edu.au
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
Dr Kevin LiKevin.li@uwa.edu.au
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
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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.
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Calculation of packing height(dilute gas)
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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
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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
See McCabe et al. page 580-581
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
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Values of height of transferunit
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See McCabe et al. page 580-581
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
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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)
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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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Example 13.1 solution (2)
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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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 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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Example 13.1 solution (4)
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.
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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
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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
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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
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Random and Structured Packing
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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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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 ”
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
Design considerations: Pressure drop and flooding
www.see.ed.ac.uk
For packed column
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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 ”
(McCabe, Smith, Harriott)
Design considerations: Pressure drop and flooding
Gy
Gx
L
V
Design considerations: Diameter of packed towers
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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 ”
McCabe, Smith, Harriott
Pressure drop analysis: Eckert graph
Design considerations: Diameter of packed towers
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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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 ”
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
Design considerations: Diameter of packed towers
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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 ”
Design considerations: Diameter of packed towers
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
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Dr Kevin LiKevin.li@uwa.edu.au
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:thomas.rufford@uwa.edu.aumailto:thomas.rufford@uwa.edu.au
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We will look at:
Sedimentation & Settling processes
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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
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Separate sand and other solidsfrom water or other liquids
Separate oil droplets from water
Examples of solid fluid separations
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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
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Gravity classifiersSeparate particles of the same density but differentparticle sizes.
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FeedLiquid + fine particles
overflow
Coarse particles sink, picked up byscrew Image from http://www.zoneding.com/Product-23.html
Examples of solid fluid separations
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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
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Properties and handling of particulate solids
Size
Shape
Density
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Size and shape of particles
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Size and shape of particlesFor regular shaped particles we can easily define size and shape.
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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
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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
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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
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Separation by
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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
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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
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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.
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Cessnock Wastewater Treatment Works
http://www.epco.com.au
Batch sedimentation process
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Batch sedimentation process
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(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
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Clearliquidinterfaceheight
Settling time, hours
Flocculation
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Flocculationparticles < few microns d p settle slowly
Agglomerate particles faster separation
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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
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