adsorption lecture notes enve542 air pollution control technologies by dr. pınar ergenekon gyte...

AdsorptionLecture Notes

ENVE542Air Pollution Control Technologies

By Dr. Pınar ErgenekonGYTE 2010 Fall

ENVE542 GYTE Çevre Müh. 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


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


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


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

112/04/19 Aerosol & Particulate Research Lab 5


Adsorbent Material

Properties of Activated CarbonBulk Density 22-34 lb/ft3

Heat Capacity 0.27-0.36 BTU/lboFPore Volume 0.56-1.20 cm3/gSurface Area 600-1600 m2/gAverage Pore Diameter 15-25 ÅRegeneration Temperature (Steaming)

100-140 oC

Maximum Allowable Temperature

150 oC

– Silica gel – Activated alumina– Activated carbon– Synthetic zeolite

• Molecular sieve

Dehyrdating purposes


Adsorbent Material

Properties of Silica GelBulk Density 44-56

lb/ft3

Heat Capacity 0.22-0.26 BTU/lboF

Pore Volume 0.37 cm3/g

Surface Area 750 m2/gAverage Pore Diameter 22 Å

Regeneration Temperature

120-250 oC


400 oC

Properties of Activated Alumina

Bulk Density

Granules 38-42 lb/ft3

Pellets 54-58 lb/ft3

Specific Heat 0.21-0.25 BTU/lboF

Pore Volume 0.29-0.37 cm3/g

Surface Area 210-360 m2/g

Average Pore Diameter 18-48 Å

Regeneration Temperature (Steaming)

200-250 oC


500 oC


Adsorbent MaterialsProperties of Molecular Sieves

Anhydrous Sodium

Aluminosilicate

Anhydrous Calcium

Aluminosilicate

Anhydrous Aluminosilicat

eType 4A 5A 13XDensity in bulk (lb/ft3) 44 44 38Specific Heat (BTU/lboF) 0.19 0.19 -Effective diameter of pores (Å) 4 5 13Regeneration Temperature (oC) 200-300 200-300 200-300Maximum Allowable Temperature (oC)

600 600 600

• Crystalline zeolite

• Uniform pores to selectively separate compounds by size & shape


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


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


Adsorption Isotherms

• In physical adsorption Brunauer,Emmett, and Teller (BET) is frequenlty used

)/)(1(1)[( oom PPcPP

cP

V

V

V:volume the amount of adsorbed gas would fill at a given pressure and temperatureVm: volume adsorbed it a layer one molecule thick fills the surfacePo: vapor pressure of the adsorbate at the temperature of the systemP: actual partial pressure of the adsorbate c: a parameter of the particular adsorption process

http://en.wikipedia.org/wiki/File:Adsorption_Isotherms_(Langmuir_red_%26_BET_green.JPG


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

)/)(1(1)[( oom PPcPP

cP

V

V


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=Cdfra: rate of adsorption

Ca, Cd: constant

P: partial pressure of the adsorbate

f: is the occupied fraction of the total solid surface

da

a

CPC

PCf


Adsorption Isotherms, Langmuir

11

2

21

2

1

'

'

1

1

11)/(

)/(

kk

Pk

a

P

Pka

Pk

Pk

Pk

PCC

PCCCa

fCa

da

da a

a

• Since we assume a monolayer coverage, the mass of adsorbate per unit mass of adsorbent (a) is also proportional to f:

da

a

CPC

PCf

Langmuir Isotherm

Adsorption Isotherm, Freundlich

• At very low and very high adsorbate partial presssure Langmuir isotherm equation takes the forms of:


aP

Pka n

• 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.

Pk

Pka

Pka

2

1

1

•This equation is refferred to as a Freundlich isotherm.

Adsorption Isotherm, Freundlich


nPka

• 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

Adsorption Isotherm Data (AID)

• Figure 12.2


Adsorption Isotherms

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,


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 (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

ENVE542 GYTE Çevre Müh.

20

P

PRTG v

ads ln

j

v

i

v

P

P

V

RT

P

P

V

RT

log303.2'

log303.2'

i,j denote different gasses

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 (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


21

P

PRTG v

ads ln

j

v

i

v

P

P

V

RT

P

P

V

RT

log303.2'

log303.2'

i,j denote different gasses


Adsorption Wave

Adsorption Zone


The length of Adsorption Zone (AZ) is a function of the rate of transfer of adsorbate from the gas to the adsorbentA shallow AZ indicates good adsorbent utilization and is represented by a step breakthrough curveThe length of AZ determined the minimum depth of the adsorbent bedOn 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 isothermHence system design is based primarily on previous plant experment and pilot scale studies.

Analysis of an Adsorption Wave


x1x2

Db

Xsat or

Vg or Va

Vad

• Let’s make a mass balance on the pollutant for the overall adsorption zone between positions 1 and 2

C or X


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 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:

)1(

11

.1

.

truead

adadad

g

gXAVXm

Cm

satadada

aXAV

Cm

0

.


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 and ), inlet pollutant gas concentration, the superfical velocity Va of the air and the apparent density of the adsorption bed.

/)1(0

/1

.

/)1(0

/1

.

00

.

)()()()(

CV

CA

mV

CAV

Cm

XC

ad

a

ada

a

ad

adada

a

sateAssume equilibrium at point 1, Ce=C0


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:

dCm

mda

ap

..

Mp is the rate of mass transfer of pollutant gas onto the solid phase

10

.

10

.

10

1

.

.

.

)(

)(

)(

CCC

dC

KA

mdx

dxCCCKAdCm

CCAV

CmX

dxXCKAdCm

dxCCKAmd

a

a

aa

adada

a

aa

ep


Analysis of an Adsorption Wave• Integration of dx equation from x1 to x2 will yield an expression

for the adsorption zone thickness, 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 = which varies from 1 to 0. So divide each side of the dx equation by C0 and replace dx by .

• :

1

01

1

0.

10

.

)1(

dd

m

KA

CCC

dC

KA

mdx

a

a

a

a


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:

• Then ma/Apa can be replaced by the superfacial gas stream velocity Va

)1(

)1(

. 99.01

01.01ln

1

1595.4

a

a

m

KA

)1(

)1(

. 99.01

01.01ln

1

1595.4

a

a

m

KA

Breakthrough Time• If we assume that the time required to establish the

adsorption zone to its full thickness at the inlet is zero then,

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…

ad

bB V

Dt

Pressure Drop Across a Fixed Bed

32

75.1'

)1(150

')1( 2

3

GdGD

dPg

p

ggp

Typical operating range: < 20 in H2O;

20 ft/min < V < 100 ft/min12 in<D<30 in

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)

Ergun equation (Eq.12.13): Figure 12.6

Pressure Drop Across a Fixed Bed


56.1

10037.0

V

LP

P: pressure drop (in H2O)L: bed depth (in)

V: superficial gas velocity ft/min

A simpler emprical equation by the Union Carbide Corporation is as follows (Eq.12.14):

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 12.13 b) Eq.12.14 and c) Figure 12.6


Solution • a)Eq 12.13

Calculate G’: gas mass flux (lb/ft2-hr)

Assume that the influent gas has the properties of air , MW=29

35

75.1'

)1(150

')1( 2

3

GdGD

dPg

p

ggp

P: pressure drop (lb/ft2)=?D: bed depth (ft)=2: void fraction=0.35g: gas viscosity (lb/ft-hr)=0.047dp: carbon particle diameter (ft)=0.011

3/071.056073.0

291ftlb

x

x

RT

PMWg

23

3

/446612

1071.0

min60

min7500 fthrlb

xx

ft

lbx

hrx

ftG

Total mass of carbon in bed = 12x6x2x(30ft/lb)=4320 lbMass of benzene adsorbed/hr=4320x(10/100)= 432 lb/hr


36

75.1'

)1(150

')1( 2

3

GdGD

dPg

p

ggp

D: (ft)=2: 0.35g: 0.047dp: (ft)=0.011

3/071.056073.0

291ftlb

x

x

RT

PMWg

23

3

/446612

1071.0

min60

min7500 fthrlb

xx

ft

lbx

hrx

ftG

OinHpsi

OinHxftlbfP

P

x

P

Gddg

G

D

P

p

g

gp

222

38

2

3

2

5.97.14

8.406

144

7.49/7.49

85.242

75.1)446(011.0

)047.0)(35.01(150

)071.0)(011.0(35.01017.4

446)35.01(

2

75.1'

)1(150')1(


37

D: (ft)=2: 0.35g: 0.047dp: (ft)=0.011

V = 104 ft/min

OinHP

xP

VLP

2

56.1

56.1

4.9

100

104237.0

10037.0

Superficial gas velocity=7500/(12x6) = 104 ft/min

Solution

• From Figure 12.6 P=7.8 inH2O/ft x 2 ft=15.6inH2O


Regeneration of Adsorption BedA 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


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

RRRRR

adR

RR C

A

mV

/)1(/1

.

)()(


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:

XC

XC RRR

0

R

CC RR

0

RRC

A

mV

adR

RRR

/)1(

0

.

RR

Cm

mAD

V

Dt

RR

RRbadR

R

b

/)1(

0

.

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


Example 12.2-SOLUTION• Use Figure 12.3. 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:


psiapsiaxppm

P

psiaxyP epen

0276.07.14000,1000

1880

7.14tan

gmol

cm

cmg

gmolgV

3

3

112

/64.0

/72'

6.70276.0

16log

/112

308log

'

gmolcc

K

P

P

V

T v

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


gC

Cgn

gmol

g

cc

gmolx

gC

cc

100

6.1172

112

1

100

18 5

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

hrlbM

hrx

lbmol

lbx

RxRlbmolftatm

cfmxatmxM

/110

min6072

5573.0

550000188.013

bedlbCClbn

lbChrxx

hr

Clbn

bed

lbC/3143

5.3

1001

110

5

5

Example 12.2-SOLUTIONUse 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:


47

ftWLftW

W

ftBedArea

4.1022.5

2/4.53

4.532

7.106

2

2

min/100)5.10)(25.5(

5500ft

acfmV On the high end of the

acceptable range, OK

min)2.63(05.13308_110

1

_100

5.3

5

5 hrlbCxClbn

hx

lbC

Clbn

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

stmlbClbxCnlb

stmlbsteam

stmlbhrxhr

nClbx

Cnlb

stmlbsteam

_992_3308_

_3.0

_33005.1_110

_

_3

5

5

5

Based on n-C5 amount

Based on carbon amount

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

hp

OinHacfmxxOinHcfmhp

blowerhp 4.3560.0

5.2455000001575.0

22

SOLUTION• Preliminary Specification SummaryAdsorber 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)

adsorption lecture notes enve542 air pollution control technologies by dr. pınar ergenekon gyte...

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