Adsorption Lecture Notes

Adsorption Lecture Notes

Adsorption Lecture Notes ENVE542 Air Pollution Control Technologies By Dr. Pnar Ergenekon GYTE 2010 Fall 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 ENVE542 GYTE evre Mh. 2 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 ENVE542 GYTE evre Mh. 3 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 20-400 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 ENVE542 GYTE evre Mh.

4 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 20-400 kj/g mole 109/02/24 ENVE542 GYTE evre Mh. Aerosol & Particulate Research Lab 5 5 Adsorbent Material Silica gel Activated alumina Activated carbon Synthetic zeolite Dehyrdating purposes

Molecular sieve Properties of Activated Carbon Bulk Density 22-34 lb/ft3 Heat Capacity 0.27-0.36 BTU/lboF Pore Volume 0.56-1.20 cm3/g Surface Area 600-1600 m2/g Average Pore Diameter 15-25 Regeneration Temperature 100-140 oC (Steaming) Maximum Allowable 150 oC Temperature ENVE542 GYTE evre Mh. 6 Adsorbent Material Properties of Silica Gel Bulk Density 44-56 lb/ft3

Heat Capacity 0.22-0.26 BTU/lboF Pore Volume 0.37 cm3/ g Surface Area 750 m2/g Average Pore Diameter 22 Regeneration 120-250 o Temperature C Maximum Allowable 400 oC Temperature Properties of Activated Alumina Bulk Density Granules 38-42 lb/ft3 Pellets 54-58 lb/ft3

Specific Heat Surface Area 0.21-0.25 BTU/lboF 0.29-0.37 cm3/ g 210-360 m2/g Average Pore Diameter 18-48 Pore Volume Regeneration Temperature 200-250 oC (Steaming) Maximum Allowable 500 oC Temperature ENVE542 GYTE evre Mh. 7

Adsorbent Materials Properties of Molecular Sieves Type Density in bulk (lb/ft3) Specific Heat (BTU/lboF) Effective diameter of pores () Regeneration Temperature (oC) Maximum Allowable Temperature (oC) Anhydrous Sodium Aluminosilicate 4A 44 0.19 4 200-300 600 Anhydrous Calcium Aluminosilicate 5A 44 0.19

5 200-300 600 Anhydrous Aluminosilicat e 13X 38 13 200-300 600 Crystalline zeolite Uniform pores to selectively separate compounds by size & shape ENVE542 GYTE evre Mh. 8 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 ENVE542 GYTE evre Mh.

9 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 ENVE542 GYTE evre Mh. 10 Adsorption Isotherms In physical adsorption Brunauer,Emmett, and Teller (BET) is frequenlty used V cP Vm ( Po P)[1 (c 1)( P / Po ) 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 ENVE542 GYTE evre Mh. 11 Adsorption Isotherms, BET V cP Vm ( Po P )[1 (c 1)( P / Po ) 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 12 evaluate Vm ENVE542 GYTE evre Mh. 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) Ca P rd=Cdf f Ca P Cd ra: rate of adsorption Ca, Cd: constant P: partial pressure of the adsorbate f: is the occupied fraction of the total solid surface ENVE542 GYTE evre Mh. 13

Adsorption Isotherms, Langmuir Since we assume a monolayer coverage, the mass of adsorbate per unit mass of adsorbent (a) is also proportional to f: ' a C a f (Ca C a' / Cd ) P k1 P a (Ca / Cd ) P 1 k 2 P 1 k1 P k 2 P 1 a P k2 P 1 a k1 k1 Ca P f Ca P Cd

Langmuir Isotherm ENVE542 GYTE evre Mh. 14 Adsorption Isotherm, Freundlich At very low and very high adsorbate partial presssure Langmuir isotherm equation takes the forms of: a k1 P a kP n k1 P a k2 P P a This equation is refferred to as a Freundlich isotherm. Where the ,,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. ENVE542 GYTE evre Mh.

15 Adsorption Isotherm, Freundlich a kP n ENVE542 GYTE evre Mh. 16 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 -0,914 0,17 -0,0158 Heptane

ENVE542 GYTE evre Mh. 17 Adsorption Isotherms Figure 12.2 ENVE542 GYTE evre Mh. 18 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, ENVE542 GYTE evre Mh. 19 Adsorption Potential is defined as the change in free energy (Gads) accompanying the compression of one mole of vapor from the equilibrium partial pressure (P) to the saturated vapor pressure (P v) 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) vin cm3/gmol ads G P RT ln P RT Pv RT Pv V ' 2.303 log P V ' 2.303 log P i j i,j denote different gasses 20 ENVE542 GYTE evre Mh.

Adsorption Potential is defined as the change in free energy (Gads) accompanying the compression of one mole of vapor from the equilibrium partial pressure (P) to the saturated vapor pressure (P v) 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) vin cm3/gmol ads G P RT ln P RT Pv RT Pv V ' 2.303 log P V ' 2.303 log P i j i,j denote different

gasses 21 ENVE542 GYTE evre Mh. Adsorption Wave ENVE542 GYTE evre Mh. 22 Adsorption Wave ENVE542 GYTE evre Mh. 23 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. ENVE542 GYTE evre Mh. 24 Analysis of an Adsorption Wave Xsat or Vg or Va C or X Vad x1 x2 Db Lets make a mass balance on the pollutant for the overall adsorption zone between positions 1 and 2 ENVE542 GYTE evre Mh. 25 Analysis of an Adsorption Wave

. m g C1 . m ad X 1 ad AV ad X 1 g ad true (1 ) 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 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: . m a C0 ad AV ad X sat a ENVE542 GYTE evre Mh. 26 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: Ce X sat . Assume equilibrium at point 1, Ce=C0 m a C0 C0 ad AVad a . . ma Va 1/ ( 1) / Vad ( ) (C0 ) ( )1/ (C0 ) ( 1) / a ad A

ad Velocity of adsorption wave is dependent upon the shape of the equilibrium curve (characterized by and ), inlet pollutant gas concentration, the superfical velocity Va of the air and the apparent density of the adsorption bed. 27 ENVE542 GYTE evre Mh. Analysis of an Adsorption Wave To determine the thickness of the adsorption wave, develop a relationship between C and x in the adsorption zone . ma d m p dC a . Mp is the rate of mass transfer of pollutant gas onto the solid phase . Another independent equation for mp can be developed by general theory of mass transfer:

d m p KA(C Ce )dx . m a dC KA a (C X )dx . X ma C C 1C01 a ad AV ad . m a dC KA a (C C C01 )dx . ma dx KA a ENVE542 GYTE evre Mh. dC

1 C C C0 28 Analysis of an Adsorption Wave Integration of dx equation from x1 to x2 will yield an expression for the adsorption zone thickness, which equals xwhich which equals xequals which equals xx2-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 = which equals x which which equals xvaries which equals xfrom which equals x1 which equals xto which equals x0. which equals x which equals xSo which equals xdivide which equals xeach which equals xside which equals xof which equals xthe which equals xdx which equals x equation which equals xby which equals xC0 which equals xand which equals xreplace which equals xdx which equals xby which equals x. . ma dx KA a : KA a .

ma 1 dC 1 C C C0 1 d d 1

( 1 ) 0 0 ENVE542 GYTE evre Mh. 29 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 0.99. 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: KA a . ma

1 1 0.01( 1) 4.595 ln 1 1 0.99( 1) Then ma/Apa can be replaced by the superfacial gas stream velocity Va KA ( 1) 1 1 0 . 01 a 4.595 ln . ( 1) 1 1 0

. 99 ma ENVE542 GYTE evre Mh. 30 Breakthrough Time If we assume that the time required to establish the adsorption zone to its full thickness at the inlet is zero then, Db tB Vad 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 Pressure Drop Across a Fixed Bed Ergun equation (Eq.12.13): Figure 12.6 Pg 3 d p g

150(1 ) g 1.75 2 D(1 )G ' d pG' P: pressure drop (lb/ft2) D: bed depth (ft) : void fraction G: gas mass flux (lb/ft2-hr) g: 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

100 1.56 P: pressure drop (in H2O) L: bed depth (in) V: superficial gas velocity ft/min ENVE542 GYTE evre Mh. 33 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 online 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 12.13 b) Eq.12.14 and c) Figure 12.6 ENVE542 GYTE evre Mh. 34

Solution a)Eq 12.13 Pg 3 d p g 150(1 ) g 1.75 2 D(1 )G ' d pG ' P: pressure drop (lb/ft2)=? D: bed depth (ft)=2 : void fraction=0.35 g: gas viscosity (lb/ft-hr)=0.047 dp: carbon particle diameter (ft)=0.011 Calculate G: gas mass flux (lb/ft2-hr) Assume that the influent gas has the properties of air , MW=29 Total mass of carbon in bed = 12x6x2x(30ft/lb)=4320 lb Mass of benzene adsorbed/hr=4320x(10/100)= 432 lb/hr g PMW 1x 29

0.071lb / ft 3 RT 0.73x560 ft 3 min lb 1 G 7500 x60 x0.071 3 x 446lb / hr ft 2 min hr ft 12 x6 35 Solution 3 Pg d p g 150(1 ) g a)Eq 12.13 1.75

2 D(1 )G ' d pG ' g PMW 1x 29 0.071lb / ft 3 RT 0.73x560 D: (ft)=2 : 0.35 g: 0.047 dp: (ft)=0.011 ft 3 min lb 1 G 7500 x60 x0.071 3 x 446lb / hr ft 2 min

hr ft 12 x6 P (1 )G '2 150(1 ) g 3 1.75 D g d p g d pG' 150(1 0.35)(0.047) P (1 0.35)4462 1 . 75 8 3

2 4.17 x10 0.35 (0.011)(0.071) 0.011(446) P 24.85 2 P 49.7lbf / ft 2 49.7 406.8inH 2O x 9.5inH 2O 144 14.7 psi 36 Solution D: (ft)=2 : 0.35 g: 0.047 dp: (ft)=0.011 a)Eq 12.14 Superficial gas velocity=7500/(12x6) = 104 ft/min

V P 0.37 L 100 V = 104 ft/min 1.56 104 P 0.37 x 2 100 P 9.4inH 2O 1.56 37 Solution a)Eq 12.13 ENVE542 GYTE evre Mh. 38 Solution

From Figure 12.6 P=7.8 inH2O/ft x 2 ft=15.6inH2O Regeneration of Adsorption Bed A proper system should require no more than 1-4 lb of steam per lb of recover solvent or 0.2-0.4 lb steam per lb of carbon ENVE542 GYTE evre Mh. 40 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 mR VR ( R )1/ (C R ) ( 1) / R ad A R R R

Subscript R refers to the properties and flowrate of regenaration fluid ENVE542 GYTE evre Mh. 41 Regeneration of Adsorption Bed Since, C R R X R C0 X C C R R 0 CR can be related to C0 : R Substitute this to the speed of desorption zone eq. .

mR R C0 VR R ad A ( R 1) / R If the thickness of the desorption zone is much smaller than the bed length Db, then the time for regeneration: . Db R ad ADb mR R t VR mR R C0 ENVE542 GYTE evre Mh. ( R 1) / R 42

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 ENVE542 GYTE evre Mh. 43 Example 12.2-SOLUTION Use Figure 12.3. So Calculate abscissa for Figure 12.3

P y pen tan e x14.7 psia 1880 ppm P x14.7 psia 0.0276 psia 1000,000 From Appendix B, Pv is about 16 at T = 95F. The specific molal volume then 3 72 g / gmol 112 cm V ' 3 0.64 g / cm gmol Now solve for the abscissa of Figure 12.3 using pressures in place of fugacities: Pv T 308K

16 log log 7.6 V' P 112 cc / gmol 0.0276 ENVE542 GYTE evre Mh. 44 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: 18cc 1gmol 72 g 11 .6 gn C5 x 100 gC 112 cc gmol 100 gC 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

ENVE542 GYTE evre Mh. 45 Example 12.2-SOLUTION The mass flow rate of n-pentane to be adsorbed in one hour is: 1atmx 0.00188 x5500cfm 72lb 60 min M x x 3 0.73atm ft lbmol hr x55 R lbmol R Allow about one hour for regeneration and cooling, thus assumin 100%efficiency , M 110lb / hr 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, lbC 110lbn C5

100lbC 3 x1hrx 3143lbC / bed bed volume=3200 lbC/30 lbC/ft =106.7 ft3 bed hr 3.5lbn C5 46 ENVE542 GYTE evre Mh. Example 12.2-SOLUTION Use a minimum bed depth of 2 ft and a rectangular bed with L=2W 106.7 BedArea 53.4 ft 2 2 W 2 53.4 / 2 W 5.2 ft L 2W 10.4 ft Use bed dimensions of 5.25ftx10.5ft and check the superficial velocity:

5500acfm V 100 ft / min (5.25)(10.5) On the high end of the acceptable range, OK Required carbon amount per bed=10.5ft x 5.25ft x 2ft x 30lb/ft 3=3308 lb C Total C amount=2x3308=6616 lbC Run time before needing regeneration for 3308 lb of carbon is: 3.5lbn C5 1h x x3308lbC 1.05hr (63.2 min) 100 _ lbC 110 _ lbn C5 47 ENVE542 GYTE evre Mh. Solution, Steam Requirement Steam requirement per regeneration can be calculated as follows: steam 3 _ lb stm 110lb _ nC5

x x1.05hr 330lb _ stm lb _ n C5 hr steam 0.3 _ lb stm x3308lb _ C 992lb _ stm lb _ n C5 Based on n-C5 amount Based on carbon amount 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 Solution, Bed Pressure Drop From Figure 12.6:

P4x10=7.5 in H2O x 2 ft = 15 in H2O P6x16=15 in H2O x 2 ft = 30 in H2O Seeing these high pressure values consider adding excess carbon and making the bed area larger 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 P of 3 in H2O Allow a pressure drop of 2 inH2O for the inlet and exhust piping. Thus the fan mus provide a total P of: P=3+2+15(-4.5)=24.5 in H2O If we assume a blower efficiency of 60% then 0.0001575hp x5500acfmx24.5inH 2O cfm inH 2O blowerhp 35.4hp 0.60 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)

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