adsorption

ADSORPTIONMain reference : 1. Seader J. D. and Henley E. J., Separation Process Principles, John Wiley, 19982. Geankoplis C. J., Transport Processes and Unit Operations, 4th Edition, Prentice Hall, 2003.

*Equilibrium relations for adsorbentsConcentration of a solute in a fluid phaseConcentration of a solute in a solid phaseData is plotted as adsorption isothermsT, P

Equilibrium relations for adsorbentsThe equilibrium isotherm places a limit on the extent to which a solute is adsorbed from a given on an adsorbent of given chemical composition and geometry for a given set of conditionsDesirable/ favorable isotherm exhibit strong adsorptionUndesirable/ unfavorable isotherm exhibit low/ weak adsorption*

*Types of Isotherms

Linear Isotherm can be used in dilute regionHenrys law is obeyed:q = Kc (12.1-1, ref 2)

c: concentration (fluid is liquid): kg adsorbate / m3 fluidp: partial pressure (fluid is a gas)q: mass, moles or volumes of adsorbate (solutes) per unit mass or per unit surface area of adsorbent: kg adsorbate (solute) / kg adsorbent (solid)K: an empirical, temperature-dependent constant (determined experimentally)

*

q = Kcn(12.1-2, ref 2)

Approximate data for many physical adsorption. Particularly useful for liquidsK = Freundlich constantn = constant (n 1)Both are determined experimentally.Freundlich isotherm

* Langmuir isotherm

q = (qo c )/ (K + c) (12.1-3, ref 2)For gasesAssumptions:Monolayer coverage on adsorbent No interactions between adsorbent moleculesAll adsorbate molecule/adsorbent interactions are the same Only a fixed number of active sites availableAdsorption is reversible and reached an equilibrium condition

*Example: Adsorption IsothermsBatch tests were performed in the laboratory using solutions of phenol in water and particles of granular activated carbon. The equilibrium data at room temperature are shown in the table below. Determine the isotherm that fits the data.Example 12.1-1 (Ref. 2)

c (kg phenol/m3 solution)q (kg phenol/kg carbon)0.3220.1500.1170.1220.0390.0940.00610.0590.00110.045

*Example: Adsorption Isotherms

Example 12.1-1Linear:q = Kcq vs cstraight line with slope KFreundlich:log q = log K + n log clog q vs log cslope: n y-axis intercept: log K Langmuir:1/q = (K/qo) (1/c) + 1/qo 1/q vs 1/cslope: K/qo y-axis intercept: 1/qo


Example 12.1-1

Chart2

0.15

0.122

0.094

0.059

0.045

q

c

q

Linear Law

Sheet1

cq1/c1/qlog clog q

0.3220.153.10559006216.6666666667-0.4921441283-0.8239087409

0.1170.1228.5470085478.1967213115-0.9318141383-0.9136401693

0.0390.09425.64102564110.6382978723-1.408935393-1.0268721464

0.00610.059163.934426229516.9491525424-2.214670165-1.2291479884

0.00110.045909.090909090922.2222222222-2.9586073148-1.3467874862

Sheet1

0

0

0

0

0

q

c

q

Linear Law

Sheet2

0

0

0

0

0

1/q

1/c

1/q

Langmuir Isotherm

Sheet3

0

0

0

0

0

log c

log q

Freundlich Isotherm


Example 12.1-1

Chart4

6.6666666667

8.1967213115

10.6382978723

16.9491525424

22.2222222222

1/q

1/c

1/q

Langmuir Isotherm

Sheet1

cq1/c1/qlog clog q

0.3220.153.10559006216.6666666667-0.4921441283-0.8239087409

0.1170.1228.5470085478.1967213115-0.9318141383-0.9136401693

0.0390.09425.64102564110.6382978723-1.408935393-1.0268721464

0.00610.059163.934426229516.9491525424-2.214670165-1.2291479884

0.00110.045909.090909090922.2222222222-2.9586073148-1.3467874862

Sheet1

0

0

0

0

0

q

c

q

Linear Law

Sheet2

0

0

0

0

0

1/q

1/c

1/q

Langmuir Isotherm

Sheet3

0

0

0

0

0

log c

log q

Freundlich Isotherm


Example 12.1-1log K = - 0.7183K = 0.199n = 0. 229A straight line produced, hence follows the Freundelich isotherm.

Chart6

-0.8239087409

-0.9136401693

-1.0268721464

-1.2291479884

-1.3467874862

log c

log q

Freundlich Isotherm

y = 0.229x - 0.701

Sheet1

cq1/c1/qlog clog q

0.3220.153.10559006216.6666666667-0.4921441283-0.8239087409

0.1170.1228.5470085478.1967213115-0.9318141383-0.9136401693

0.0390.09425.64102564110.6382978723-1.408935393-1.0268721464

0.00610.059163.934426229516.9491525424-2.214670165-1.2291479884

0.00110.045909.090909090922.2222222222-2.9586073148-1.3467874862

Sheet1

q

c

q

Linear Law

Sheet2

1/q

1/c

1/q

Langmuir Isotherm

Sheet3

log c

log q

Freundlich Isotherm

y = 0.229x - 0.701

*Batch AdsorptionWhen quantities to be treated are of small amount.Isotherms and material balance are needed.Material balance on the adsorbate:qF M + cF S = q M + cS (12.2-1)

where:qF= initial concentration of solute adsorbed on the solidq= final concentration at equilibriumM= amount of adsorbent, kgS= volume of feed solution, m3cF= initial concentration of solute in the fluid phasec= final concentration at equilibrium in the fluid phase

Batch Adsorption*qF M + cF S = q M + cS (12.2-1)

When variable q is plotted versus c , the result is a straight line.If equilibrium isotherm is also plotted on the same graph, the intersection of both line gives the final equilibrium values of q and c.

*Example: Batch AdsorptionExample 12.2-1:A wastewater solution having a volume of 1.0 m3 contains 0.21 kg phenol/m3 of solution . A total of 1.40 kg of fresh granular activated carbon is added to the solution , which is then mixed thoroughly to reach equilibrium. Using the isotherm from Example 12.1-1, what are the final equilibrium values, and what percent of phenol extracted?

*Example: Batch AdsorptionExample 12.2-1:0(1.40) + 0.21(1.0) = q (1.40) + c (1.0)

q = 0.15- 4.17 c(a)

From the isotherm

q = 0.199 c 0.229(b)qF M + cF S = q M + cS

*Example: Batch AdsorptionExample 12.2-1:At intersection q = 0.106 kg phenol/kg carbon c = 0.062 kg phenol/m3% extracted = (cF - c)(100)/cF = (0.21-0.062)(100)/0.21 = 70.5 %

Chart1

0

0.01

0.04

0.06

0.08

0.1

0.11

0.118

0.12

0.05

0.112

q

c, kg phenol/m3 solution

q, kg phenol/kg adsorbent

Chart2

0.05

0.112

Sheet1

cq

00

00.01

0.0050.04

0.010.06

0.020.08

0.050.1

0.080.11

0.10.118

0.1150.12

0.1350.05

0.050.112

Sheet2

Sheet3

- Fixed bed adsorption design - Regeneration of adsorbents

Students should be able to :Design a fixed bed adsorption columnUnderstand the regeneration of adsorbents*

Fixed Bed Adsorption Design*Introduction and concentration profiles

Usually employ fixed bed of granular particlesThe fluid to be treated is usually passes down through the packed bed at a constant flow rate Mass transfer resistances are important in the fixed-bed process, and the process is unsteady state.The overall dynamic of the system determine the efficiency of the process, rather than just the equilibrium considerations

Fixed Bed Adsorption Design*Introduction and concentration profiles (contd)

Inlet: solid is assumed to contain no solute at the start of the process The concentration of the solute in the fluid phase and of the solid adsorbent phase change with TIME and POSITION in the fixed bed as the adsorption proceedsAs the fluid first come into contact with the inlet, most of the MASS TRANSFER and ADSORPTION takes place hereAs fluid passes thru the bed, the concentration in this fluid DROPS VERY RAPIDLY with distance in bed and REACHES ZERO well before the end of the bed reached

* After a short time, solid near entrance almost SATURATED and most of the mass transfer and adsorption now takes place at a point slightly farther from the inlet The major part of the adsorption at any time takes place in a relatively narrow adsorption or mass transfer zone As the solution continues to flow, this mass-transfer zone (S-shaped), moved down the column.This outlet concentration remains near zero until the mass transfer zone starts to reach the tower outlet at t4.Then the outlet concentration starts to rise.

Fixed Bed Adsorption DesignBreakthrough Concentration CurveThen, the outlet conc starts to rise, and at t5 the outlet conc has risen to cb, which is called the break pointAfter the break-point time is reached, the concentration c rises very rapidly up to point cd, which is the end of the breakthrough curve, where the bed is judged ineffective.

breakthrough concentration profile in the fluid at outlet of bed

Fixed Bed Adsorption Design*Breakthrough Concentration CurveThe break-point concentration represents the maximum that can be discarded and often taken as 0.01 to 0.05 for cb/co.For a narrow MTZ, the breakthrough curve is very steep and most of the bed capacity is used at the break point (this makes efficient use of the adsorbent and lowers energy costs for regeneration)

breakthrough concentration profile in the fluid at outlet of bed

*Fixed Bed Adsorption Design

Capacity of Column and Scale-Up Design Method

Mass Transfer Zone (MTZ) width and shape depends on:the adsorption isothermflowratemass transfer rate to the particles diffusion in the pores.For systems with a favorable isotherm, similar to Freundlich and Langmuir; MTZ acquires the typical S shape. MTZ is constant in height as it moves thru d columnFor unfavorable isotherm i.e. Isotherm is linear; MTZ width increases with bed lengthA favourable isotherm for adsorption is unfavourable for effective regeneration


Capacity of Column and Scale-Up Design Method

A number of theoritical methods have been published which predict the Mass Transfer Zone (MTZ) and concentration profiles in the bed.Hence, experiments in laboratory scale are needed in order to scale up the results.

*Fixed Bed Adsorption DesignShaded area =Total or stoichiometric capacity of the packed tower(12.3-1)Time equivalent to the total or stoichiometric capacity

*Fixed Bed Adsorption DesignCrosshatched area = Usable capacity of bed up to the break-point time, tb

*Fixed Bed Adsorption Designtu : time equivalent to the usable capacity or time at which the effluent concentration reaches its maximum permissible level. (12.3-2)tu very close to tbtu/tt is the fraction of the total bed capacity or length utilized up to the break point


HB : length of bed used up to the break point( HT : Total bed length)

(12.3-3)


HUNB : Length of unused bed (mass transfer zone)

(12.3-4)

HT = HUNB + HB(12.3-5)


Design StepsDetermine the the length of bed needed to achieve the required usable capacity, HBDetermine HUNBCalculate HT

*Alternative method:Assume the breakthrough curve is symmetrical at c/co at ts.This assumes that the area below the curve between tb and ts is equal to the area above the curve between ts and td.Then ts is simply equal to tt (Eqn. 12.3-1)


Scale-up principle

1.If a system is tested with different bed length, it gives breakthrough curve of the same shape.

2. The amount of length of unused bed (HUNB) does not change with the total bed length.

3. Hence, tb is proportional to HB.

*Example: Fixed Bed Adsorption Design

Example 12.3-1A waste stream of alcohol vapour in air from a process was adsorbed by activated carbon particles in a packed bed having a diameter of 4 cm and length of 14 cm containing 79.2 g of carbon. The inlet gas stream having a concentration co of 600 ppm and a density of 0.00115 g/cm3 entered the bed at a flow rate of 754 cm3/s. Data in Table 12.3-1 give the concentrations of the breakthrough curve. The breakpoint concentration is set at c/co = 0.01. Do as follows.Determine the breakpoint time tb, the fraction of total capacity used up to the breakpoint tu/tt, and the length of the unused bed HUNB.


Example 12.3-1Table 12.3-1Time,hc/coTime, hc/co005.50.658306.00.9033.50.0026.20.93340.0306.50.9754.50.1556.80.99350.396


Example 12.3-1The plotted data from Table 12.3-1


Example 12.3-1Based on Figure 12.3-3At break point conc. 0.01: tb = 3.65 h; td = 6.95 h= A1 + A2 = 3.65 + 1.51 = 5.16 h= A1 = 3.65 htu / tt = 3.65/5.16 = 0.707= 0.707(14) = 9.9 cm= (1 - 0.707)14 = 4.1 cm


Example 12.3-1b) If the breakpoint time required for a new column is 6.0 h, what is the new total lengthtu is proportional to HBtu = 3.65 HB = 9.9 cmtb = 6 h= A1 = 6 h = (6 /3.65 )(9.9) = 16.3 cmHT= HUNB + HB = 16.3 + 4.1 = 20.4 cm(Fraction of the new bed used up to the break point)


Example 12.3-1c) Determine the saturation loading capacity of the carbon.Air flow rate= (754 cm3/s)(3600s)(0.0115g/cm3) = 3122 g air/h600 ppm = 600 g alcohol in 1 million g of air

Total alcohol adsorbed = = 9.67 g alcohol

Saturation capacity =

*Processing Variables and Adsorption Cycles

Large scale adsorption: 1) cyclic batch system -alternately saturated & then regenerated 2) continuous flow system- continuous flow of adsorbent countercurrent to a flow of feed

Bed regeneration methodTemperature-swing cyclePressure-swing cycleInert-purge gas stripping cycleDisplacement-purge cycle.

Types of adsorption*

Commercial Methods for Adsorption*

Agitated vessel*A batch of liquid to which is added a powdered adsorbent such as activated carbonparticle diameter typically less than 1mm, to form a slurry.With good agitation and small particles, the external resistance to mass transfer from the bulk liquid to the external surface of the adsorbent particles is smallFor small adsorbent particles, the internal resistance to mass transfer within the pores of the particles is also small. Hence, the adsorption rate is rapidMain application: removal of very small amounts dissolved, and relatively large molecules , such as coloring agents, from waterSlurry adsorption system (also called contact filtration) is also operated continuously.

Cyclic fixed-bed, batch operation*A cyclic-batch operating mode using fixed-bedWidely used with both liquid and gas feeds.Adsorbent particles size ranges from 0.05 to 1.2 cmBed pressure drop decreases with increasing particle size, but the solute transport rate increases with decreasing particle size. The optimal particle size is determined mainly from these two factors To avoid jiggling or fluidizing the bed during adsorption, the flow of the liquid or gas feed is often downwardFor removal of small amounts of dissolved hydrocarbons from water, the spent adsorbent is removed from the vessel and reactivated thermally at high temperature or it is discardedApplication of fixed-bed adsorption (also called percolation), include the removal of dissolved organic compounds from water

*Temperature Swing Adsorption (Thermal means)

Regeneration process increase in temperatureIncrease in temperature leads to a decrease in the quantity adsorbedImportant note - regeneration temperature does not cause degradation of the adsorbentsMechanism passage of a hot purge gas or steamImportant characteristic to treat feeds with low concentrations of adsorbates

Temperature Swing Adsorption (Purasiv process)-Fluidized bed for adsorption-Moving bed for desorption*Purasiv process Adsorbent particles are attrition-resistanceIn the adsorption section, sieve trays are used with the raw gas passing up through the perforations and fluidizing the adsorbent particlesThe fluidized solids flow like a liquid across the tray, into the downcomer, and onto the tray belowFrom the adsorption section, the solids pass to the desorption section, where, as moving beds, they first flow down through preheating tubes and then through desorption tubes.Steam is used for indirect heating in both sets of tubes and for stripping in the desorption tubesAt the bottom of the unit, the regenerated solids are picked up by a carrier gas, which flows up through a gas-lift line to the top, where the solids settle out onto the top tray to repeat the adsorption part of the cycle.

*Temperature Swing Adsorption

Effect of temperature on the adsorption equilibrium of a single adsorbateIf the partial pressure remains constant at p1, increasing the temperature from T1 to T2 will decrease the equilibrium loading from q1 to q2.

*Bed TSA System

Cycle Steps 1- adsorption at T1 to breakthrough 2- heating of the bed to T2 (T2 > T1) 3- desorption at T2 4- cooling of the bed to T1

*Pressure Swing Adsorption (Mechanical work)

Regeneration process reducing the partial pressure of the adsorbate2 ways:Introduction of an inert gas while maintaining the total system pressureCycle time very quick (minutes or second)Operate PSA close to ambient temperature at a given partial pressure, the loading is increased as temperature decreasedPopular for performing bulk separation of gases controlled by adsorption isotherm or adsorption kinetics Use only with gases (liquid has little or no effect with change in pressure)

*Pressure Swing Adsorption

Effect of partial pressure on equilibrium loading at temperature T1Reducing the partial pressure from p1 to p2 causes the equilibrium loading to be reduced from q1 to q2

*Bed PSA SystemEach bed operates alternately :Pressurisation followed by adsorptionDesorption by depressurisation (blowdown) followed by a purgeAdsorption pressure greater than atmosphericDesorption pressure being atmosphericPressurisation feed gasPurging effluent (non-adsorbed) product gas

Inert-purge gas stripping cycle*Adsorbate is removed by passing a non-adsorbing or inert gas through the bed.Mechanism for desorption:Partial pressure (or concentration) of original adsorbate in the gas phase surrounding the adsorbent is reduced

*Displacement-Purge cycle

Removal of adsorbates by replacing them with a more preferentially adsorbed species Mechanism for desorption:Partial pressure (or concentration) of original adsorbate in the gas phase surrounding the adsorbent is reducedThere is competitive adsorption for the displacement fluidOne advantage net heat generated or consumed will be close to zero because heat of adsorption of the displacement fluid is close to the original adsorbate adsorbent temperature constant throughout the cycle

*Displacement Purge Adsorption

*THANK YOU

adsorption

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