Chapter 4 Adsorption Operation

Report
Chapter 3
Adsorption Operation
By: In. Nurul Hasyimah Mohd Amin
Objectives
At the end of this lessons, students should be
able to:
• Define adsorption process and its
mechanism.
• Explain and differentiate physical and
chemical adsorption.
• Discuss the adsorption isotherms.
• Perform calculation related to this topic.
Introductions
Adsorption: The accumulation of
molecular species at the surface of
a solid or liquid rather than in the
bulk is called adsorption.
Bulk
Surface
Adsorbed material
Free materials
How it is happen?
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.
Adsorbate:
The
molecular
species that accumulate at the
surface.
Adsorbate
Adsorbent: The material on the
surface of which adsorption
takes place.
Adsorbent
Desorption: The removal
adsorbate from the surface
of
Examples for adsorption…………
1. 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.
2. Air becomes dry in presence of silica gel because
adsorption of water takes place on the surface of the
gel.
3. Aqueous solution of raw sugar becomes colourless
when passed over a bed of charcoal. The coloring
matter is adsorbed by the charcoal.
4. Litmus solution or a solution of a dye like methylene
blue when shaken with charcoal turns colourless due to
adsorption of coloring material.
Absorption:
Sorption: If both adsorption and
absorption takes place
simultaneously is called sorption.
Distinction between Adsorption & Absorption
Adsorption
Absorption
It is a surface
phenomenon
It is a bulk phenomenon
Adsorbed species is
accumulated in the
surface
It is a fast process
It is uniformly distributed
throughout the bulk
Rate of adsorption
decreases gradually
Absorption takes place at
steady rate
It is a slow process
Physisorption
Chemisorption
If 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
1. Non-specific nature: An adsorbent does not show any preference
for a gas as the van der Waals’ forces are universal.
2. Easily liquefiable gases like CO2, SO2, NH3 etc, are readily
adsorbed.
3. Reversible nature: Physisortion of a gas by a solid is reversible.
4. Increases by increase of pressure.
5. Surface area of adsorbant: When the surface area of the adsorbent
increases, more gas is adsorbed, ie extent of adsorption increases.
6. 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.
When atoms or molecules of gases
are held by solids on its surface by
chemical bonds, the adsorption is
called chemisorption.
Characteristics of Chemisorption
1. High specificity: It is highly specific and will occur only
if chemical bond formation takes between adsorbate and
adsorbent.
2. Irreversibility: Chemisorption is irreversible because the
chemical bond formed is difficult to break.
3. Chemisorption increases with temperature.
4. Increases by increase of pressure.
5. Surface area of adsorbent: When the surface area of the
adsorbent increases, more gas is adsorbed, ie extent of
adsorption increases.
6. Enthalpy of adsorption: Enthalpy of adsorption of
chemisorption is high (80-240 KJ mol-1)
Physisorption
Chemisorption
1
Occurs due to van der Waals’ force
Chemical Bond
2
Reversible
Irreversible
3
Not specific
Specific
4
Enthalpy of adsorption is low
Enthalpy of adsorption is high
5
More liquefiable gases are adsorbed Gases which form compounds with
readily
adsorbent alone undergo chemisorption
6
Decreases with increase of temperature
Increases with increase of temperature
7
Low temperature is favorable.
High temperature is favorable
8
High pressure favors physisorption and High pressure is favorable but decreases of
decrease of pressure causes desorption
pressure does not cause desorption
9
Results in multimolecular layers
Only unimolecular layer are formed
10
No activation energy is needed
High activation energy is needed
11
It is instantaneous
It is a slow process
Factors Influencing
Adsorption
1. 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.
Heat
3. Pressure
Adsorption of a gas by an adsorbent at constant
temperature increases with increase of
pressure.
Applications
Adsorption
1. In Gas Masks
2. Production of High
Vacuum
3. Softening of Hard
water
Ca
2+
Ca
2+
Ca
2+
Ca
2+
Exchange
resin
4. Heterogeneous
Catalysis
5. Refining of Petroleum
6. Chromatographic Separation
Mechanisms of ad…
1
2
• diffuse
from the
bulk of
phase to
the
exterior
surface of
the
adsorbent
3
• diffuse
inside the
pore and
to the
surface of
the pore.
• adsorbed
on the
surface
Adsorbents
Adsorbents
PAC
Molecular
sieve
Silica gel
Activated
GAC
zeolites
alumina
EAC
Characteristics of Adsorbents
Adsorbents
Pore
structure
Nature
Porosity
Characteristics of Adsorbents
Pore structure
zeolite
Silica gel
Activated Carbon
Characteristics of Adsorbents
Porosity
Micropores
Mesopores
Macropores
D = 2-50 nm
D > 50 nm
D < 2 nm
Characteristics of Adsorbents
Nature
Hydrophobic
interaction
Hydrophilic
repulsion
Adsorption equilibrium
• If the adsorbent an adsorbate are
contacted long enough, an equilibrium
will be established
• Eq  between amount of the adsorbate
in solution and being adsorbed on the
adsorbent
• “isotherms”
Adsorption equilibrium
• qe = 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
Isotherms
Linear
Favorable
Strongly
favorable
Henry’s Law
Langmuir
isotherm
Freundlich
isotherm
Henry’s Law
• Assume 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/kg
• Apply 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 isotherm
• Assume monolayer coverage and constant binding
energy between surface and adsorbate.
• Maximum adsorption occurs when the surface is
covered by a monolayer of adsorbate.
Langmuir isotherm
• Can be expressed by:
QoCe
qe 
K L  Ce
• The linear form of equation:
L/mg
1
KL
1


qe QoCe Qo
Max adsorption
capacity (monolayer)
(g solute/g adsorbent)
Mg/L
Ce
C
1

 eoC
C
o1
e
qe K
 .Qa o Q
a eo
qe K .Qa Qa
Freundlich isotherm
• For special case: heterogeneous surface energy.
• Particularly for mixed wastes.
• Can be expressed by:
1/ n
e
F
e
q K C
• The linearization will give
Ce
C
1

 eoC
C
o1
e
qe K
 .Qa o Q
a eo
qe K .Qa Qa
Freundlich isotherm
• For freundlich isotherm, we use log-log version
Ce
C
1

 eoC
C
o1
e
qe K
 .Qa o Q
a eo
qe K .Qa Qa
Freundlich isotherm
• Where
qe =
Ce =
KF =
n =
the amount of solute per unit
weight of adsorbent
the concentration at equilibrium
Freundlich constant related to
adsorption capacity
Freundlich constant related to
adsorption intensity
Ce
C
1

 eoC
C
o1
e
qe K
 .Qa o Q
a eo
qe K .Qa Qa
Freundlich isotherm
• Adsorption capacity at equilibrium, qe
• Adsorption capacity at time t, qt
Ce
C
1

 eoC
C
o1
e
qe K
 .Qa o Q
a eo
qe K .Qa Qa
Example 1
Batch 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 Table 4.1.
Determine the isotherm that fits the data.
Ce
C
1

 eoC
C
o1
e
qe K
 .Qa o Q
a eo
qe K .Qa Qa
Example 1
Table 4.1: Equilibrium data for example 1
Ce
C
1

 eoC
C
o1
e
qe K
 .Qa o Q
a eo
qe K .Qa Qa
Answer
Langmuir Isotherm
Ce
C
1

 eoC
C
o1
e
qe K
 .Qa o Q
a eo
qe K .Qa Qa
Answer
Freundlich Isotherm
Ce
C
1

 eoC
C
o1
e
qe K
 .Qa o Q
a eo
qe K .Qa Qa
Batch adsoprtion
• Batch 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:
qF M  CF S  qM  CS
Where M
S
: the amount of adsorbent (kg)
: the volume of feed solution (m3)
Ce
C
1

 eoC
C
o1
e
qe K
 .Qa o Q
a eo
qe K .Qa Qa
Batch adsoprtion
• Batch 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:
qF M  CF S  qM  CS
Where M
S
: the amount of adsorbent (kg)
: the volume of feed solution (m3)
Ce
C
1

 eoC
C
o1
e
qe K
 .Qa o Q
a eo
qe K .Qa Qa
Batch adsoprtion
• An equilibrium relation and material balance are
needed.
CF : initial feed concentration
C : final equilibrium concentration
qF : initial feed adsorbate loading
q : final equilibrium adsorbate loading
qF M  CF S  qM  CS
Ce
C
1

 eoC
C
o1
e
qe K
 .Qa o Q
a eo
qe K .Qa Qa
Example 2
A waste water (volume 1 m3) contains 0.21
kg phenol/m3. A total of 1.40 kg of fresh granular
activated carbon is added to the solution, mixed
thoroughly to reach equilibrium. Using isotherm in
Example 1, calculate the final equilibrium
values and percentage of phenol adsorbed by the
activated carbon.
Ce
C
1

 eoC
C
o1
e
qe K
 .Qa o Q
a eo
qe K .Qa Qa
Answer
q (kg solute/kg adsorbent)
0.16
0.14
0.12
0.1
0.08
0.06
0.04
0.02
0
0
0.1
0.2
c (kg solute/m 3)
0.3
0.4
Ce
C
1

 eoC
C
o1
e
qe K
 .Qa o Q
a eo
qe K .Qa Qa
Answer
At the intersection;
q = 0.106 and c = 0.062
% extracted = [(0.210 – 0.062)/(0.21)] x 100%
= 70.5%
Ce
C
1

 eoC
C
o1
e
qe K
 .Qa o Q
a eo
qe K .Qa Qa
Adsorption through fixed bed
adsorber
• The 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 absorbent
• Solute adsorbs, filling up some of the available sites until the
sites become saturated
• Fluid 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 zero
• The 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.
Ce
C
1

 eoC
C
o1
e
qe K
 .Qa o Q
a eo
qe K .Qa Qa
Breakthrough curve
Ce
C
1

 eoC
C
o1
e
qe K
 .Qa o Q
a eo
qe K .Qa Qa
Breakthrough curve
• The 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
Ce
C
1

 eoC
C
o1
e
qe K
 .Qa o Q
a eo
qe K .Qa Qa
Breakthrough curve
After the tb is reached, the c
rises very rapidly up to cd,
(end of the breakthrough
curve where the bed is
judged ineffective)
cd
cb
t1
t2
t3
t4
t5
The
outlett concentration
(break
At time
1, t2 and t3, the exit
The
break
point
point)
starts
toremains
rise
to occurs
cbnear
at t5zero
concentration
when the concentration
until the mass transfer zone
of the fluid leaving the
starts to reach the t4
bed starts rise to cb
Breakthrough curve
Ce
C
1

 eoC
C
o1
e
qe K
 .Qa o Q
a eo
qe K .Qa Qa
Breakthrough curve
• The 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:
Ce
C
1

 eoC
C
o1
e
qe K
 .Qa o Q
a eo
qe K .Qa Qa
Breakthrough curve
• The 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.
Ce
C
1

 eoC
C
o1
e
qe K
 .Qa o Q
a eo
qe K .Qa Qa
Example 3
A waste stream of alcohol vapor in air from a process was
adsorbed by activated carbon particles in a packed bed having a
diameter of 4 cm and length of bed of 14 cm containing 79.2 g of
carbon. The inlet 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 the table give the concentrations of the
breakthrough curve. The break point concentration is set at C/Co
= 0.01. Do as follows.
Ce
C
1

 eoC
C
o1
e
qe K
 .Qa o Q
a eo
qe K .Qa Qa
Example 3
Time, h
C/Co
Time, h
C/Co
0
0
5.5
0.658
3
0
6
0.903
3.5
0.002
6.2
0.933
4
0.03
6.5
0.975
4.5
0.155
6.8
0.993
5
0.396
Example 3
a) 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 3
a) The break point as c/co = 0.01, t = 3.65hr
td = 6.95
Area A1 = 3.65
Graphically numerical methods
Area A2 = 1.51
Simpson rule
tT  

0
 c
1  dt  3.65  1.51  5.16hr
 co 
Example 3
tu  
tb
0
 c
1  dt  3.65hr
 co 
The fraction capacity used to break up to break point;
t u 3.65

 0.707
tT 5.16
The unused bed = (1.0 – 0.707) x 14 cm = 4.1 m

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