1. Adsorption Lecture Notes
ENVE542
Air Pollution Control Technologies
By Dr. Pınar Ergenekon
GYTE 2010 Fall

2. Useful When The pollutant gas is noncombustible or difficult to burn
The pollutant is sufficiently valuable to warrant recovery
The pollutant is in very dilute concentration
It is also used for purification of gases containing only small amounts of pollutants that are difficult to clean by other means
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3. Adsorption Process Classified as Physical and Chemical Ads.
1) Physical adsorption
The gas molecules adhere to the surface of the solid adsorbent as a result of intermolecular attractive forces (van der Waals forces) between them
The process is exothermic: the heat liberated is in the order of the the enthalpy of condensation of vapor (2-20 kJ/gmole)
The process is reversible (recovery of adsorbent material or adsorbed gas is possible) by increasing the temperature or lowering the adsorbate conc.
Physical adsorption usually directly proportional to the amount of solid surface area
Adsorbate can be adsorbed on a monolayer or a number of layers
The adsorption rate is generally quite rapid
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4. Adsorption Process 2) Chemical adsorption
Results from a chemical interaction between the adsorbate and adsorbent. Therefore formed bond is much stronger than that for physical adsorption
Heat liberated during chemisorption is in the range of kj/g mole
It is frequently irreversible. On desorption the chemical nature of the original adsorbate will have undergone a change.
Only a monomolecular layer of adsorbate appears on the adsorbing medium
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5. Aerosol & Particulate Research Lab
Adsorption Mechanism
2) Chemical adsorption
Results from a chemical interaction between the adsorbate and adsorbent. Therefore formed bond is much stronger than that for physical adsorption
Heat liberated during chemisorption is in the range of kj/g mole
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2017/4/20
Aerosol & Particulate Research Lab 5

6. Properties of Activated Carbon
Adsorbent Material
Silica gel
Activated alumina
Activated carbon
Synthetic zeolite
Molecular sieve
Dehyrdating purposes
Properties of Activated Carbon
Bulk Density
22-34 lb/ft3
Heat Capacity
BTU/lboF
Pore Volume
cm3/g
Surface Area
m2/g
Average Pore Diameter
15-25 Å
Regeneration Temperature
(Steaming) oC
Maximum Allowable
Temperature
150 oC
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7. Properties of Silica Gel Properties of Activated Alumina
Adsorbent Material
Properties of Silica Gel
Bulk Density
44-56 lb/ft3
Heat Capacity
BTU/lboF
Pore Volume
0.37 cm3/g
Surface Area
750 m2/g
Average Pore Diameter
22 Å
Regeneration Temperature oC
Maximum Allowable Temperature
400 oC
Properties of Activated Alumina
Bulk Density
Granules
38-42 lb/ft3
Pellets
54-58 lb/ft3
Specific Heat
BTU/lboF
Pore Volume
cm3/g
Surface Area
m2/g
Average Pore Diameter
18-48 Å
Regeneration Temperature (Steaming) oC
Maximum Allowable Temperature
500 oC
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8. Properties of Molecular Sieves
Adsorbent Materials
Properties of Molecular Sieves
Anhydrous Sodium Aluminosilicate
Anhydrous Calcium Aluminosilicate
Anhydrous Aluminosilicate
Type 4A 5A
13X
Density in bulk (lb/ft3) 44 38
Specific Heat (BTU/lboF)
0.19 -
Effective diameter of pores (Å) 4 5 13
Regeneration Temperature (oC)
Maximum Allowable Temperature (oC)
600
Crystalline zeolite
Uniform pores to selectively separate compounds by size & shape
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9. Adsorption Isotherm
The amount of gas adsorbed per unit of adsorbent at equilibrium is measured against the partial pressure of the adsorbate in the gas phase gives equilibrium adsorption isotherm
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10. Adsorption Isotherm
In general, an adsorption isotherm relates the volume or mass adsorbed to the partial pressure or concentration of the adsorbate in the main gas stream at a given temperature
The equilibrium concentration adsorbed is very sensitive to T
There are many equations proposed to fit analyticaly the various experimental istoherms
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11. Adsorption Isotherms
In physical adsorption Brunauer,Emmett, and Teller (BET) is frequenlty used
V:volume the amount of adsorbed gas would fill at a given pressure and temperature
Vm: volume adsorbed it a layer one molecule thick fills the surface
Po: vapor pressure of the adsorbate at the temperature of the system
P: actual partial pressure of the adsorbate
c: a parameter of the particular adsorption process
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12. Adsorption Isotherms, BET
In a plot P/(Vtotal(P-Po) versus P/Po, the slope and interception of the drawn best line can be determined and c and Vm can be estimated.
When the value of P/Po less than 0.05 and greater than 0.35, BET plot is not linear. Then other techniques must be used to evaluate Vm
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13. Adsorption Isotherms, Langmuir
In Langmuir isotherm assuming a unimolecular layer can be obtained by a kinetic approach: at equilibium knowing that the rate of adsorption is equal to the rate of desorption:
ra=CaP(1-f)
rd=Cdf
ra: rate of adsorption
Ca, Cd: constant
P: partial pressure of the adsorbate
f: is the occupied fraction of the total solid surface
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14. Adsorption Isotherms, Langmuir
Since we assume a monolayer coverage, the mass of adsorbate per unit mass of adsorbent (a) is also proportional to f:
Langmuir Isotherm
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15. Adsorption Isotherm, Freundlich
At very low and very high adsorbate partial presssure Langmuir isotherm equation takes the forms of:
This equation is refferred to as a Freundlich isotherm.
Where the a,b,k, and n are constants and can be determined from the graph (taking the log of each side). Their units are dependent on the units utilized for concentration of the adsorbate.
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16. Adsorption Isotherm, Freundlich
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17. Adsorption Isotherm Data (AID)
Yaws summarized the adsorption capacity of 283 organic compounds on activated carbon:
logx=a+blogCe+c(logCe)2
x is the adsorption capacity (g of pollutant/g of carbon), Ce is the pollutant conc. in ppm, a, b, and c are correlation constants
Pollutant a b c
Benzene
-1.189
0,288
-0,0238
Ethyl arcrylate
-1.439
0,27
-0,0089
Monochlorobenzene
-0.973
0,306
-0,0335
Phenol
-0,544
0,103
-0,0109
MIBK
-0,898
0,206
-0,0206
Toluene
-0,885
0,208
-0,0202
Heptane
-0,914
0,17
-0,0158
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18. Adsorption Isotherms
Figure 12.2 18
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19. Adsorption Isotherm Data (AID)
Other excellent sources for AID are the various vendors of the adsorbent
In the absence of experimental data on a specific carbon, an equation developed by Calgon Corporation which is a modification of Dubinin-Radushkevich equation can be used to estimate the adsorption capacity,
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20. i,j denote different gasses
Adsorption Potential
is defined as “the change in free energy (DGads) accompanying the compression of one mole of vapor from the equilibrium partial pressure (P) to the saturated vapor pressure (Pv) at the temperature of adsorption T in K.
Dubinin found that when similar gases were adsorbed on the same adsorbent the adosrption potentials were very nearly equal when the amount adsorbed was determiend based on the product of the number of moles adsorbed multiplied by the specific molal volume (V’) in cm3/gmol
i,j denote different gasses 20
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21. i,j denote different gasses
Adsorption Potential
is defined as “the change in free energy (DGads) accompanying the compression of one mole of vapor from the equilibrium partial pressure (P) to the saturated vapor pressure (Pv) at the temperature of adsorption T in K.
Dubinin found that when similar gases were adsorbed on the same adsorbent the adosrption potentials were very nearly equal when the amount adsorbed was determiend based on the product of the number of moles adsorbed multiplied by the specific molal volume (V’) in cm3/gmol
i,j denote different gasses 21
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22. Adsorption Wave
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23. Adsorption Wave
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24. Adsorption Zone
The length of Adsorption Zone (AZ) is a function of the rate of transfer of adsorbate from the gas to the adsorbent
A shallow AZ indicates good adsorbent utilization and is represented by a step breakthrough curve
The length of AZ determined the minimum depth of the adsorbent bed
On the next slides we will try to analyze this zone but note that under actual plant operating conditions bed capacity will seldom exceed 30-40% of that indicated by an equilibrium isotherm
Hence system design is based primarily on previous plant experment and pilot scale studies.
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25. Analysis of an Adsorption Wave
Xsat or
Vg or Va
C or X
Vad Db x2 x1
Let’s make a mass balance on the pollutant for the overall adsorption zone between positions 1 and 2
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26. Analysis of an Adsorption Wave
Where mg and mad are the mass flow rates of the gas phase entering and solid phase leaving respectively, pad is the apparent bulk density of the granular bed and f is the fraction of void in the adsorbent
Hence the mass flow rate of carrier gas is frequently air with mass flow rate ma and density pa, the equation becomes:
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27. Analysis of an Adsorption Wave
For many dilute solutions in the gas phase, the saturation of equilibrium data relating Ce to Xsat can take a form of Freundlich type:
Velocity of adsorption wave is dependent upon the shape of the equilibrium curve (characterized by a and b), inlet pollutant gas concentration, the superfical velocity Va of the air and the apparent density of the adsorption bed.
Assume equilibrium at point 1, Ce=C0 27
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28. Analysis of an Adsorption Wave
To determine the thickness of the adsorption wave, develop a relationship between C and x in the adsorption zone
Another independent equation
for mp can be developed by
general theory of mass transfer:
Mp is the rate of mass transfer of pollutant gas onto the solid phase
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29. Analysis of an Adsorption Wave
Integration of dx equation from x1 to x2 will yield an expression for the adsorption zone thickness, d, which equals x2-x1
At x1, C = C0 and at x2, C=0. To make it convenient in terms of a mathematical viewpoint, express the equation C/C0 = h which varies from 1 to 0. So divide each side of the dx equation by C0 and replace dx by d. :
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30. Analysis of an Adsorption Wave
This integration for the limit from0 to 1 is undefined, so instead take limits close in value to 0 and 1. We may take h to be within 1 percent of the limiting values, that is, 0.01 and That is an adsorption zone width such taht C approaches within 1% of its limitin values of 0 and C0. Based on this arbitrary selection of the integration limits, Equation becomes:
Then ma/Apa can be replaced by the superfacial gas stream velocity Va
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31. Breakthrough Time
If we assume that the time required to establish the adsorption zone to its full thickness at the inlet is zero then,
d depends on an arbitrary choice of the limits of Equation 6-9. If selected limit C/C0 = 0.01 at the leading edge of the wave, breakthrough time can be described as the situation when C reaches 1% of the inlet concentration. Other percentages might be chosen…

32. Pressure Drop Across a Fixed Bed
Ergun equation (Eq.12.13):
Figure 12.6
P: pressure drop (lb/ft2)
D: bed depth (ft)
e : void fraction
G’: gas mass flux (lb/ft2-hr)
mg: gas viscosity (lb/ft-hr)
dp: carbon particle diameter (ft)
Typical operating range:
< 20 in H2O;
20 ft/min < V < 100 ft/min
12 in<D<30 in

33. Pressure Drop Across a Fixed Bed
A simpler emprical equation by the Union Carbide Corporation is as follows (Eq.12.14):
P: pressure drop (in H2O)
L: bed depth (in)
V: superficial gas velocity ft/min
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34. Example 12.1
An activated-carbon bed that is 12 ft x 6 ft x 2 ft deep is used in a benzene recovery system. The system is on-line for 1 hour and is then regereranted for one hour. The influent gas stream flow at 7500 acfm and contains 5000 ppmv benzene at 1 atm and 100F. The operating cpacity of the bed is 10 lbm benzene per 100 lbm carbon. The physical properties of the carbon are as follows: bulk density: 30 lbm/ft3, void fraction=0.35, particle size=4x10 mesh (0.011ft)
Determine the pressure drop across the bed from a)Eq b) Eq and c) Figure 12.6
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35. Solution a)Eq 12.13 P: pressure drop (lb/ft2)=? D: bed depth (ft)=2
Calculate G’: gas mass flux (lb/ft2-hr)
Assume that the influent gas has the properties of air , MW=29
P: pressure drop (lb/ft2)=?
D: bed depth (ft)=2
e : void fraction=0.35
mg: gas viscosity (lb/ft-hr)=0.047
dp: carbon particle diameter (ft)=0.011
Total mass of carbon in bed = 12x6x2x(30ft/lb)=4320 lb
Mass of benzene adsorbed/hr=4320x(10/100)= 432 lb/hr 35

36. Solution
D: (ft)=2
e : 0.35
mg: 0.047
dp: (ft)=0.011
a)Eq 12.13 36

37. Superficial gas velocity=7500/(12x6) = 104 ft/min
Solution
D: (ft)=2
e : 0.35
mg: 0.047
dp: (ft)=0.011
a)Eq 12.14
V = 104 ft/min
Superficial gas velocity=7500/(12x6) = 104 ft/min 37

38. Solution
a)Eq 12.13
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39. Solution
From Figure 12.6
DP=7.8 inH2O/ft x 2 ft=15.6inH2O

40. Regeneration of Adsorption Bed
A proper system should require no more than 1-4 lb of steam per lb of recover solvent or lb steam per lb of carbon
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41. Regeneration of Adsorption Bed
The basic format of the equation for the speed of the adsorption zone is still valid
The speed of the desorption zone VR is given by
Subscript R refers to the properties and flowrate of regenaration fluid
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42. Regeneration of Adsorption Bed
Since,
CR can be related to C0 :
Substitute this to the speed of desorption zone eq.
If the thickness of the desorption zone is much smaller than the bed length Db, then the time for regeneration:
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43. Example 12.2
Prepare a preliminary design for a carbon adsroption system to control a stream of solvent laden air from a plastics extruder local exhaust system. The exhaust stream temperature is 95F and it contains 1880 ppm of n-pentane. The plant engineer has provided the following info:
Other gases contaminants: none
PM contaminants: plant fugitive dust only.
Flow rate: 5500 acfm
Extruder exhaus pressure:-4.5 in H2O
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44. Example 12.2-SOLUTION
Use Figure So Calculate abscissa for Figure 12.3
From Appendix B, Pv is about 16 at T = 95F. The specific molal volume then
Now solve for the abscissa of Figure 12.3 using pressures in place of fugacities:
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45. Example 12.2-SOLUTION
From Figure 12.3 the volume adsorbed is about 18 ccliquid/100g carbon
18 ccliquid/100g carbon in terms of adsorption capacity:
Choosing a capacity factor of 30% carbon capacity for design=0.3(11.6)=3.5lb n-C5/100 lb C
Note that for a final design, the capacity factor woudl be determined from experimental data
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46. Example 12.2-SOLUTION
The mass flow rate of n-pentane to be adsorbed in one hour is:
Allow about one hour for regeneration and cooling, thus assumin 100%efficiency , a tow bed system will rqurei a sufficient amount of carbon in each bed to adsorb 110 lb n-pentane. The minimum of carbon amoun in each bed:
Rounding this value to 3200 lb and using a bulk carbon density of 30lb/ft3,
bed volume=3200 lbC/30 lbC/ft3=106.7 ft3 46
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47. Example 12.2-SOLUTION
Use a minimum bed depth of 2 ft and a rectangular bed with L=2W
Use bed dimensions of 5.25ftx10.5ft and check the superficial velocity:
Required carbon amount per bed=10.5ft x 5.25ft x 2ft x 30lb/ft3=3308 lb C
Total C amount=2x3308=6616 lbC
Run time before needing regeneration for 3308 lb of carbon is:
On the high end of the acceptable range, OK 47
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48. Solution, Steam Requirement
Steam requirement per regeneration can be calculated as follows:
Assume the regeneration takes 45 minutes and teh rest of the time if for bed cooling, then steam flow rate must be 992lb/0.75 hr=1323 lb/hr
Since the n-C5 mass flow rate is low choose the higher of the two calculated steam rates
If plant steam is not available then consider a package boiler purchase
Based on n-C5 amount
Based on carbon amount

49. Solution, Bed Pressure Drop
From Figure 12.6:
DP4x10=7.5 in H2O x 2 ft = 15 in H2O
DP6x16=15 in H2O x 2 ft = 30 in H2O
Seeing these high pressure values consider adding excess carbon and making the bed area larger

50. Solution, Blower Horsepower Requirment
The local exhaust system will be picking up plant fugitive dust, so a fabric filter or guard chamber should be installed before the carbon adsorption system
Assume a filter DP of 3 in H2O
Allow a pressure drop of 2 inH2O for the inlet and exhust piping. Thus the fan mus provide a total DP of:
DP=3+2+15(-4.5)=24.5 in H2O
If we assume a blower efficiency of 60% then

51. SOLUTION Preliminary Specification Summary
Adsorber bed size=5.25 ft x 10.5 ft x 2 ft
Mass of carbon per bed=3308 lb
Steam Rqd=992lb/regenx24regen/day=23800 lb/day
Fan hp=35.4 hp (buy a 40 hp motor)