Chapter 4 Adsorption Operation
Chapter 3Adsorption OperationBy: In. Nurul Hasyimah Mohd
AminObjectivesAt the end of this lessons, students should beable
to:Define adsorption process and its mechanism.Explain and
differentiate physical and chemical adsorption.Discuss the
adsorption isotherms.Perform calculation related to this topic.
Adsorption: The accumulation of molecular species at the surface
of a solid or liquid rather than in the bulk is called adsorption.
IntroductionsBulkSurfaceFree materialsAdsorbed material
Surface of solids and liquids has the tendency to attract and
retain other molecules with which it is brought in to contact.
How it is happen? This is due to unbalanced residual inward
forces of attraction at the surface of solids and liquids.How it is
happen? Due to these residual forces, surface of solid or liquid
has a higher concentration of other molecular species than the
bulk. How it is happen? Adsorbate: The molecular species that
accumulate at the surface.AdsorbateAdsorbent: The material on the
surface of which adsorption takes place.AdsorbentDesorption: The
removal of adsorbate from the surfaceExamples for adsorption
If a gas like H2, O2, Cl2 etc is taken in a vessel containing
powdered charcoal, pressure of the gas slowly decreases as the gas
is adsorbed on the surface of charcoal. Air becomes dry in presence
of silica gel because adsorption of water takes place on the
surface of the gel. Aqueous solution of raw sugar becomes
colourless when passed over a bed of charcoal. The coloring matter
is adsorbed by the charcoal.Litmus solution or a solution of a dye
like methylene blue when shaken with charcoal turns colourless due
to adsorption of coloring material. AbsorptionAbsorption: The
distribution of substance uniformly throughout the bulk of the
adsorbent.Sorption: If both adsorption and absorption takes place
simultaneously is called sorption.AdsorptionAbsorptionIt is a
surface phenomenonIt is a bulk phenomenonAdsorbed species is
accumulated in the surfaceIt is uniformly distributed throughout
the bulkIt is a fast processIt is a slow processRate of adsorption
decreases graduallyAbsorption takes place at steady rateDistinction
between Adsorption & AbsorptionTypes of
AdsorptionAdsorptionPhysical Adsorption or PhysisorptionIf
accumulation of gas molecules on the surface of solids occurs due
to weak van der Waals forces of attraction, the adsorption is
called physisorption. Characteristics of Physisorption Non-specific
nature: An adsorbent does not show any preference for a gas as the
van der Waals forces are universal. Easily liquefiable gases like
CO2, SO2, NH3 etc, are readily adsorbed.Reversible nature:
Physisortion of a gas by a solid is reversible.
Increases by increase of pressure. Surface area of adsorbant:
When the surface area of the adsorbent increases, more gas is
adsorbed, ie extent of adsorption increases. Enthalpy of
adsorption: Enthalpy of adsorption of physisorption is very low
(20-40 KJ mol-1)Since adsorption is exothermic, physisorption takes
place readily at low temperature and desorption takes place at
higher temperature. Chemical Adsorption or ChemisorptionWhen atoms
or molecules of gases are held by solids on its surface by chemical
bonds, the adsorption is called chemisorption. Characteristics of
ChemisorptionHigh specificity: It is highly specific and will occur
only if chemical bond formation takes between adsorbate and
adsorbent. Irreversibility: Chemisorption is irreversible because
the chemical bond formed is difficult to break. Chemisorption
increases with temperature. Increases by increase of pressure.
Surface area of adsorbent: When the surface area of the adsorbent
increases, more gas is adsorbed, ie extent of adsorption increases.
Enthalpy of adsorption: Enthalpy of adsorption of chemisorption is
high (80-240 KJ mol-1)PhysisorptionChemisorption1Occurs due to van
der Waals force Chemical Bond2Reversible Irreversible3Not
specificSpecific4Enthalpy of adsorption is lowEnthalpy of
adsorption is high5More liquefiable gases are adsorbed readilyGases
which form compounds with adsorbent alone undergo
chemisorption6Decreases with increase of temperatureIncreases with
increase of temperature7Low temperature is favorable.High
temperature is favorable 8High pressure favors physisorption and
decrease of pressure causes desorptionHigh pressure is favorable
but decreases of pressure does not cause desorption9Results in
multimolecular layersOnly unimolecular layer are formed 10No
activation energy is neededHigh activation energy is needed11It is
instantaneousIt is a slow process Factors Influencing Adsorption1.
Surface Area of the Adsorbent
Same gas is adsorbed to different extent by different solids at
identical conditions. Greater the surface area of the adsorbent
greater the volume of gas adsorbed. 2. Temperature
Adsorption of a gas generally decreases with rise in
temperature. This is because adsorption is exothermic and increases
of temperature favors the backward process which is
desorption.Heat3. PressureAdsorption of a gas by an adsorbent at
constant temperature increases with increase of pressure.
Applications Adsorption1. In Gas Masks
2. Production of High Vacuum
3. Softening of Hard waterCa2+Ca2+Ca2+Ca2+Exchange resin4.
Heterogeneous Catalysis5. Refining of Petroleum
6. Chromatographic Separation
Mechanisms of ad
Adsorbents
Adsorbents
Silica gelActivated alumina
Molecular sieve zeolitesPAC
EAC
GAC
Characteristics of Adsorbents
AdsorbentsPore structurePorosityNatureCharacteristics of
Adsorbents
Pore structure
Silica gelActivated Carbon
zeoliteCharacteristics of Adsorbents
PorosityMesoporesMacroporesMicroporesD = 2-50 nmD > 50 nmD
< 2 nm
Characteristics of Adsorbents
NatureHydrophobic interactionHydrophilic repulsionAdsorption
equilibriumIf the adsorbent an adsorbate are contacted long enough,
an equilibrium will be establishedEq between amount of the
adsorbate in solution and being adsorbed on the
adsorbentisotherms
Adsorption equilibriumqe = mass of material adsorbed per mass of
adsorbent (at eq)
Ce = concentration in solution when amount adsorbed equal to qe
(at eq) (mg/L)
qe/ Ce depend on the types of adsorption (physical or chemical
adsorption)
Adsorption equilibrium
Adsorption equilibrium
IsothermsLinear FavorableStrongly favorableHenrys LawLangmuir
isothermFreundlich isothermHenrys LawAssume that the concentration
in one phase is proportional to the concentration in the other.Can
be expressed by equation:
K is constant determined experimentally, m3/kgApply for a very
low concentration.At very low concentration the molecules adsorbed
are widely spaced over the adsorbent surface so that one molecule
has no influence on another.
Langmuir isothermAssume monolayer coverage and constant binding
energy between surface and adsorbate. Maximum adsorption occurs
when the surface is covered by a monolayer of adsorbate.
Langmuir isothermCan be expressed by:
The linear form of equation:
Mg/LL/mgMax adsorption capacity (monolayer) (g solute/g
adsorbent)Freundlich isothermFor special case: heterogeneous
surface energy. Particularly for mixed wastes.Can be expressed
by:
The linearization will give
Freundlich isothermFor freundlich isotherm, we use log-log
version
Freundlich isothermWhere qe =the amount of solute per unit
weight of adsorbentCe =the concentration at equilibriumKF
=Freundlich constant related to adsorption capacityn = Freundlich
constant related to adsorption intensity
Freundlich isothermAdsorption capacity at equilibrium, qe
Adsorption capacity at time t, qt
Example 1Batch tests were performed in the laboratoryusing
solutions of phenol in water and particlesof granular activated
carbon. The equilibriumdata at room temperature are shown in Table
4.1.Determine the isotherm that fits the data.
Example 1
Table 4.1: Equilibrium data for example 1Answer
Langmuir IsothermAnswer
Freundlich IsothermBatch adsoprtionBatch adsorption is often
used to adsorb solutes from liquid solutions when the quantities
treated are small in amount.Solid will not remove the entire
contaminant unless it is infinitely good mixing.Total solute
removed is the function of amount of contaminant to mass of
solid.The material balance on the adsorbate is:
Where M: the amount of adsorbent (kg) S: the volume of feed
solution (m3)
Batch adsoprtionBatch adsorption is often used to adsorb solutes
from liquid solutions when the quantities treated are small in
amount.Solid will not remove the entire contaminant unless it is
infinitely good mixing.Total solute removed is the function of
amount of contaminant to mass of solid.The material balance on the
adsorbate is:
Where M: the amount of adsorbent (kg) S: the volume of feed
solution (m3)
Batch adsoprtionAn equilibrium relation and material balance are
needed.CF: initial feed concentrationC: final equilibrium
concentrationqF: initial feed adsorbate loadingq: final equilibrium
adsorbate loading
Example 2A waste water (volume 1 m3) contains 0.21kg phenol/m3.
A total of 1.40 kg of fresh granularactivated carbon is added to
the solution, mixedthoroughly to reach equilibrium. Using isotherm
inExample 1, calculate the final equilibriumvalues and percentage
of phenol adsorbed by theactivated carbon.
Answer
Answer
At the intersection;q = 0.106 and c = 0.062% extracted = [(0.210
0.062)/(0.21)] x 100% = 70.5%
Adsorption through fixed bed adsorberThe concentrations in the
fluid phase and the solid phase change with time and position in
the bed.As fluid enters the bed, it comes in contact with the first
few layers of absorbentSolute adsorbs, filling up some of the
available sites until the sites become saturatedFluid penetrates
further into the bed before all solute is removed. As the fluid
passes though the bed, the concentration in this fluid drops very
rapidly to almost zeroThe active region shifts down through the bed
as time goes on.As the solution continues to flow, this mass
transfer zone, which is S-shaped, moves down the column.
Breakthrough curve
Breakthrough curveThe break point occurs when the concentration
of the fluid leaving the bed as unadsorbed solute begins to emerge.
The bed has become ineffective (very small but detectable).
a breakpoint composition is set to be the maximum amount of
solute that can be acceptably lost
Breakthrough curve
t1t2t3t4t5cbcdAt time t1, t2 and t3, the exit concentration
remains near zero until the mass transfer zone starts to reach the
t4The outlet concentration (break point) starts to rise to cb at
t5The break point occurs when the concentration of the fluid
leaving the bed starts rise to cbAfter the tb is reached, the c
rises very rapidly up to cd, (end of the breakthrough curve where
the bed is judged ineffective)Breakthrough curve
Breakthrough curveThe time required for a bed to become totally
saturated is obtained by integrating as time goes to infinity:
In operation, you want to stop the process before solute breaks
through, so integration to the breakpoint time gives the "usable"
capacity:
Breakthrough curveThe length of bed used up to break point
is:
The length of unused bed is:
The total design height of a bed is determined by adding the
required usable capacity to the unused height.
Example 3A waste stream of alcohol vapor in air from a process
wasadsorbed by activated carbon particles in a packed bed having
adiameter of 4 cm and length of bed of 14 cm containing 79.2 g
ofcarbon. The inlet stream having a concentration, Co of 600 ppmand
a density of 0.00115 g/cm3 entered the bed at a flow rateof 754
cm3/s. Data in the table give the concentrations of thebreakthrough
curve. The break point concentration is set at C/Co= 0.01. Do as
follows.
Example 3
Time, hC/CoTime,
hC/Co005.50.6583060.9033.50.0026.20.93340.036.50.9754.50.1556.80.99350.396Example
3a)Determine the breakpoint time, the fraction of total capacity
used up to the break point, and the length of the unused bed. Also
determine the saturation loading capacity of the carbon.
b)If the break point time required for a new column is 6.0 h,
what is the new total length of the column required?
Example 3
Example 3a) The break point as c/co = 0.01, t = 3.65hrtd =
6.95Area A1 = 3.65Area A2 = 1.51
Graphically numerical methods Simpson rule
Example 3
The fraction capacity used to break up to break point;
The unused bed = (1.0 0.707) x 14 cm = 4.1 m
Chart10.150.1220.0940.0590.045
c (kg solute/m3)q (kg solute/kg adsorbent)
Sheet10.3220.150.1170.1220.0390.0940.00610.0590.00110.045
Sheet1
c (kg solute/m3)q (kg solute/kg adsorbent)
Sheet2
Sheet3
