COURSE OUTCOMES CO APPLY principles of batch adsorption and
fixed-bed adsorption. CALCULATE and EXAMINE adsorption isotherms.
DEVELOP basic design of gas or liquid adsorber.
Slide 3
Introduction to adsorption. Adsorption equipments. Adsorption
Isotherms Analysis. Principles of Adsorption. Basic Equation for
Adsorption. Adsorber Design Calculation. OUTLINES
Slide 4
A DSORPTION A BSORPTION ! Absorption a fluid phase is
transferred from one medium to another. Adsorption certain
components of a fluid (liquid or gas) phase are transferred to and
held at the surface of a solid (e.g. small particles binding to a
carbon bed to improve water quality) Adsorbent the adsorbing phase
(carbon, silica gel, zeolite) Adsorbate the material adsorbed at
the surface of adsorbent. INTRODUCTION TO ADSORPTION
Slide 5
A PPLICATION OF A DSORPTION : Used in many industrial
processes: Adsorbing the desired product from fermentation broths.
Isolation of proteins. Dehumidification. odour/colour/taste
removal. gas pollutant removal (H 2 S). water softening and
deionisation. hydrocarbon fractionation. pharmaceutical
purification. INTRODUCTION TO ADSORPTION
Slide 6
* NATURE OF ADSORBENT Porous material - Large surface area per
unit mass - internal surface area greater than the external surface
area - often 500 to 1000 m 2 /g. Granular (50m - 12 mm diameter),
small pellets or beads. Suitable for packed bed use. Activated
carbon, silica gel, alumina, zeolites, clay minerals, ion exchange
resins. Separation occurs because differences in molecular weight,
shape or polarity of components. Rate of mass transfer is dependent
on the void fraction within the pores. INTRODUCTION TO
ADSORPTION
Slide 7
Zeolite structure Silica structure
Slide 8
T YPES OF A DSORPTION 1.Ion exchange Electrostatic attachment
of ionic species to site of the opposite charge at the surface of
an adsorbent 2.Physical Adsorption result of intermolecular forces
causing preferential binding of certain substances to certain
adsorbents Van der Waal forces, London dispersion force reversible
by addition of heat (via steam, hot inert gas, oven) Attachment to
the outer layer of adsorbent material 3.Chemisorption result of
chemical interaction Irreversible, mainly found in catalysis change
in the chemical form of adsorbate INTRODUCTION TO ADSORPTION
Slide 9
A DSORPTION E QUIPMENT Fixed-bed adsorbers Gas-drying equipment
Pressure-swing adsorption Fixed-bed adsorbers Gas-drying equipment
Pressure-swing adsorption
Adsorbent particles: 0.3 1.2 m deep supported on a perforated
plate Feed gas passes down through the bed. Downflow is preffered
because upflow at high rates may fluidize the particles, causing
attrition and loss of fines. The feed gas is switched to the other
bed when the conc. Of solute in exit gas reaches a certain value.
The bed is regenerate by steam / hot inert gas. FIXED-BED
ADSORBERS
Slide 12
Regeneration To remove unwanted particles from the adsorbent
surface after the adsorption process. using steam/hot inert gas.
Steam condenses in the bed, raising the temp. of the solid, provide
energy for desorption. The solvent is condensed, separated from
water. Then the bed is cooled and dried with inert gas. FIXED-BED
ADSORBERS
GAS-DRYING EQUIPMENT The equipment for drying is similar to the
shown in Fig. 25.1, but hot gas is used for regeneration. The moist
gas from the bed being generated may be vented, or much of the
water may be removed in a condenser and the gas recirculated
through a heater to the bed. For small dryers, electric heaters are
sometimes installed inside the bed to provide the energy for
regeneration. Fig. 25.1 Vapor-phase Adsorption System
PRESSURE-SWING ADSORPTION Most often, adsorption is used as a
purification process to remove small amounts of material, but,
there is a number of applications involve separations of gas
mixtures with moderate to high concentration of adsorbates. These
are called bulk separations, and they often use different operating
procedures than for gas purification. Pressure-swing adsorption
(PSA) is a bulk separation process that is used for small-scale air
separation plants and for concentration of hydrogen in process
streams.
Use of activated carbon to remove pollutants from aqueous
wastes. Use carbon beds up to 10 m tall, several ft in diameter,
several bed operating in parallel. Tall beds are needed to ensure
adequate treatment. ADSORPTION FROM LIQUIDS
A DSORPTION I SOTHERM S ANALYSIS Adsorption isotherm
equilibrium relationship between the concentration in the fluid
phase and the concentration in the adsorbent particles. For gas
concentration in mole % or partial pressure For liquid
concentration in mg/L (ppm) or g/L (ppb) Concentration of adsorbate
on the solid = mass adsorbed (g) per unit mass of original
adsorbent (g).
Slide 21
TYPES OF ISOTHERMS Amount of adsorbed is independent of
concentration down to very low values. Amount of adsorbed is
proportional to the concentration in the fluid. Concave upward; low
solid loadings are obtained and because it leads to quite long
mass-transfer zones in the bed. (this shape are rare)
Slide 22
TYPES OF ISOTHERMS Fig. 25.3 Adsorption isotherms for water in
air at 20 to 50 o C Nearly linear isotherm up to 50 percent
humidity, and the ultimate capacity is about twice that for the
other solids. Water is held most strongly by molecular sieves, and
the adsorption is almost irreversible, but the pore volume not as
great as for silica gel
Slide 23
A DSORPTION DATA FOR VAPORS ON ACTIVATED CARBON Sometimes
fitted to Freundlich isotherms, but data for wide range of
pressures show isotherm slopes gradually decrease as the pressure
is increased.
Slide 24
Amount of adsorbed depends on (T/V) log (f s /f), where: T:
adsorption temperature (Kelvin). V: molar volume of the liquid at
the boiling point f s : fugasity of the saturated liquid at
adsorption temperature f: fugasity of the vapor For adsorption at
atmospheric pressure; * fugasity = partial pressure = vapor
pressure Volume adsorbed is converted to mass by assuming the
adsorbed liquid has the same density as liquid at the boiling
point.
Slide 25
QUESTION 1 EXAMPLE 25.1. Adsorption on BPL carbon is used to
treat an airstream containing 0.2 percent n-hexane at 20 o C. (a)
Estimate the equilibrium capacity for a bed operating to 20 o C.
(b) How much would the capacity decrease if the heat of adsorption
raised the bed temperature to 40 o C. EXAMPLE 25.1. Adsorption on
BPL carbon is used to treat an airstream containing 0.2 percent
n-hexane at 20 o C. (a) Estimate the equilibrium capacity for a bed
operating to 20 o C. (b) How much would the capacity decrease if
the heat of adsorption raised the bed temperature to 40 o C.
Slide 26
ANSWER (a) Estimate the equilibrium capacity for a bed
operating to 20 o C. The MW n-hexane (C 6 H 14 )= 86.17, At 20 o C
(from Perrys Handbook, 7 th ed.) P =120mm Hg f s. At the normal
boiling point (68.7 o C), L =0.615 g/cm 3. The adsorption pressure
P is 760 mm Hg. (b) At 40 o C, P = 276 mm Hg ANSWER (a) Estimate
the equilibrium capacity for a bed operating to 20 o C. The MW
n-hexane (C 6 H 14 )= 86.17, At 20 o C (from Perrys Handbook, 7 th
ed.) P =120mm Hg f s. At the normal boiling point (68.7 o C), L
=0.615 g/cm 3. The adsorption pressure P is 760 mm Hg. (b) At 40 o
C, P = 276 mm Hg
Slide 27
4 TYPES OF ADSORPTION ISOTHERMS 1.Linear Isotherms - Adsorption
amount is proportional to the concentration in the fluid
2.Irreversible independent of concentration 3.Langmuir Isotherm
favorable type 4.Freundlich Isotherm strongly favorable type
Slide 28
Often been used to correlate equilibrium adsorption data for
protein. Isotherms that convex upward are called favorable. Where:
W = adsorbate loading (g absorbed/g solid) c = the concentration in
the fluid (mg/L) K = the adsorption constant K >> 1 : the
isotherm is strongly favorable. W max and K are constants
determined experimentally by plotting 1/W against 1/c LANGMUIR
ISOTHERM
Slide 29
strongly favourable Describe the adsorption of variety of
antibiotics, steroids and hormones. high adsorption at low fluid
concentration where b and m are constant -Linearize the equation:
Log W = log b + m log c -Constant determined from experimental data
by plotting log W versus log c -Slope = m, intercept = b FREUNDLICH
ISOTHERM
Slide 30
P RINCIPLES OF A DSORPTION In fixed bed adsorption, the
concentrations in the fluid phase and the solid phase change with:
a) time b) as well as the position in the bed. At first, most of
the mass transfer takes place near the inlet of the bed, where the
fluid contacts the adsorbent. After a few minutes, the solid near
the inlet is nearly saturated. Most of the mass transfer takes
place farther from the inlet. The concentration gradient become
S-shaped. The region where most of the change in concentration
occurs is called the mass-transfer zone (MTZ), and the limits are
often taken as c/c 0 values of 0.95 to 0.05.
Slide 31
M ASS T RANSFER Z ONE AND B REAKTHROUGH
Slide 32
Concentration Profile In Fixed Beds
Slide 33
t 1 : no part of the bed is saturated. From t 1 to t 2 : the
wave had moved down the bed. t 2 : the bed is almost saturated for
a distance L S, but is still clean at L F. Little adsorption occurs
beyond L F at time t 2, and the adsorbent is still unused. The MTZ
where adsorption takes place is the region between L S and L F. The
concentration of the adsorbate on the adsorbent is related to the
adsorbate concentration in the feed by the thermodynamic
equilibrium. Because it is difficult to determine where MTZ begins
and ends, L F can be taken where C/C F = 0.05, with L S at C/C F =
0.95. t B : the wave has moved through the bed, with the leading
point of the MTZ just reaches the end of the bed. This is known as
the breakthrough point. Rather than using C/C F = 0.05, the
breakthrough concentration can be taken as the minimum detectable
or maximum allowable solute concentration in the effluent fluid,
e.g. as dictated by downstream processing unit. t 1 : no part of
the bed is saturated. From t 1 to t 2 : the wave had moved down the
bed. t 2 : the bed is almost saturated for a distance L S, but is
still clean at L F. Little adsorption occurs beyond L F at time t
2, and the adsorbent is still unused. The MTZ where adsorption
takes place is the region between L S and L F. The concentration of
the adsorbate on the adsorbent is related to the adsorbate
concentration in the feed by the thermodynamic equilibrium. Because
it is difficult to determine where MTZ begins and ends, L F can be
taken where C/C F = 0.05, with L S at C/C F = 0.95. t B : the wave
has moved through the bed, with the leading point of the MTZ just
reaches the end of the bed. This is known as the breakthrough
point. Rather than using C/C F = 0.05, the breakthrough
concentration can be taken as the minimum detectable or maximum
allowable solute concentration in the effluent fluid, e.g. as
dictated by downstream processing unit. Concentration Profile In
Fixed Beds
Slide 34
Concentration profile in fixed beds Figure 25.6(a) Is the ratio
of solute concentration to inlet solute concentration in the
fluid.
Slide 35
B REAKTHROUGH C URVES t 1, t 2, t 3 : the exit concentration is
practically zero. Time for fluid living the bed.
Slide 36
B REAKTHROUGH C URVES
Slide 37
t b time when the concentration reaches break point The feed is
switched to a fresh adsorbent bed Break point relative
concentration c/c o of 0.05 or 0.10 Adsorption beyond the break
point would rise rapidly to about 0.50 Then, slowly approach 1.0
(concentration liq in = liq out) B REAKTHROUGH C URVES
Slide 38
t* is the ideal adsorption time for a vertical breakthrough
curve t* is also the time when c/c o reaches 0.50 Amount of
adsorbed is proportional to the rectangular area to the left of the
dashed line at t* B REAKTHROUGH C URVES
Slide 39
Solute feed rate (F A ) = superficial velocity (u o ) X
concentration (c o ) Where: W o = initial adsorbate loading W sat =
adsorbate at equilibrium with the fluid (saturation) L = length of
the bed b = bulk density of the bed B REAKTHROUGH C URVES
Slide 40
For systems with favorable isotherm, the concentration profile
in the mass-transfer zone acquires a characteristic shape and width
that do not change as the zone moves down the bed. Test with
different bed lengths have breakthrough curve of the same shape,
but with longer beds, the MTZ is a smaller fraction of the bed
length, and greater fraction of the bed is utilized. The scale-up
principles The amount of unused solid or length of unused bed does
not change with the total bed length. LENGTH OF UNUSED BED
(LUB)
Slide 41
To calculate LUB, determine the total solute adsorbed up to the
break point by integration; The break point time, t b is calculated
from the ideal time and the fraction of bed utilized: LENGTH OF
UNUSED BED (LUB)
Slide 42
QUESTION 2 EXAMPLE 25.2. The adsorption of n-butanol from air
was studied in a small (10.16 cm diameter) with 300 and 600 g
carbon, corresponding to bed lengths of 8 and 16 cm. (a) From the
following data for effluent concentration, estimate the saturation
capacity of the carbon and the fraction of the bed used at c/c 0 =
0.05. (b) Predict the break-point time for a bed length of 32 cm.
Data for n-butanol on Columbia JXC 4/6 carbon are as follows :
EXAMPLE 25.2. The adsorption of n-butanol from air was studied in a
small (10.16 cm diameter) with 300 and 600 g carbon, corresponding
to bed lengths of 8 and 16 cm. (a) From the following data for
effluent concentration, estimate the saturation capacity of the
carbon and the fraction of the bed used at c/c 0 = 0.05. (b)
Predict the break-point time for a bed length of 32 cm. Data for
n-butanol on Columbia JXC 4/6 carbon are as follows : EXAMPLE 25.2
300 g600 g t,hc/c 0 t,hc/c 0 10.00550.0019 1.50.015.50.003
20.02760.0079 2.40.0506.50.018 2.80.1070.039 3.30.207.50.077
40.2980.15 50.568.50.24
Slide 43
ANSWER The concentration profiles are plotted in Fig 25.8, and
extended to c/c 0 =1.0 assuming the curves are symmetric about c/c
0 =0.5. Per square centimeter of bed cross section, the solute feed
rate is The total solute adsorbed is the area above the graph
multiplied by F A. For the 8 cm bed, the area is;
Slide 44
ANSWER This area corresponds to the ideal time that would be
required to adsorb the same amount if the breakthrough curve were a
vertical line. The mass of carbon per unit cross-sectional area of
bed is; 8 x 0.461 = 3.69 g/cm 2 Thus, At the break point, where c/c
0 = 0.05, and t = 2.4 h The amount adsorbed up to the break point
is then Thus 50 percent of the bed capacity is unused, which can be
represented by a length of 4 cm. Trapezoidal rule:
Slide 45
ANSWER (-cont) For the 16-cm bed the breakthrough curve has the
same initial slope as the curve for the 8-cm bed, and although data
were not taken beyond c/c 0 = 0.25, the curves are assume to be
parallel. For the entire bed, At c/c 0 = 0.05, t = 7.1 h, and At
the break point, 74 percent of the bed capacity is used, which
corresponds to an unused section of length 0.26 x 16 = 4.2 cm.
Within experimental error, the lengths of unused bed agree, and 4.1
cm is the expected value for a still longer bed.
Slide 46
ANSWER (-cont) (b) For L = 32 cm, the expected length of the
fully used bed is; 32 - 4.1 = 27.9 cm. The fraction of the bed used
is: The break-point time is
Slide 47
QUESTION 3 A waste stream of n-butanol 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 14 cm containing 79.2 g of
carbon. The density of the activated carbon is 0.461 g/cm 3. The
inlet gas stream having a concentration, C 0 of 600 ppm and a
density of 0.00115 g/cm 3 entered the bed at the solute feed rate,
F A of 0.063 g/cm 2.s. Data in Table 3.1 give the concentrations of
the fluid in the bed, C. The break point concentration is set at
C/C o = 0.05. Time (hour) Concentration of fluid, C (ppm) 3.0 3.5
4.0 4.5 5.0 5.5 6.0 6.2 6.5 6.8 0.1 1.2 18.0 93.0 237.6 394.8 541.8
559.8 585.0 595.8 QUESTION; 1.Plot a breakthrough curve.
2.Determine the break-point time. 3.Calculate the saturation
capacity of the carbon, W sat. 4.Calculate the length of unused bed
(LUB). QUESTION; 1.Plot a breakthrough curve. 2.Determine the
break-point time. 3.Calculate the saturation capacity of the
carbon, W sat. 4.Calculate the length of unused bed (LUB).
Slide 48
ANSWER 1.Plot a breakthrough curve. Given; Packed bed, D = 4 cm
L = 14 cm adsorbent = 79.2 g carbon = 0.461 g/cm 3 Inlet gas
stream, C 0 = 600 ppm = 0.00115 g/cm 3 F A = 0.063 g/cm 2.s C/C 0 =
0.05 t (h)C C/C 0 3.00.1 1.667E-04 3.51.2 0.002 4.018.0 0.030
4.593.0 0.155 5.0237.6 0.396 5.5394.8 0.658 6.0541.8 0.903 6.2559.8
0.933 6.5585.0 0.975 6.8595.8 0.993
Slide 49
ANSWER- cont 2.Determine the break-point time. From the
breakthrough curve, breakthrough time at C/C 0 = 0.05 is t b = 4.1
h
Slide 50
ANSWER- cont 3.Calculate the saturation capacity of the carbon,
W sat. The total solute adsorbed is the area above the graph
multiplied by F A From the graph plotted, the following data is
obtained; t (h)C/C 0 f(x) = (1-C/C 0 ) 0.0001 1.7501 3.500.0020.998
5.250.5 7.0010 Simpsons Rule of integration. (pp. 872)
Slide 51
4.Calculate the length of unused bed (LUB). t (h)C/C 0 f(x) =
(1-C/C 0 ) 0.0001 4.10.050.95
Slide 52
Introduction to adsorption. Adsorption equipments. Adsorption
Isotherms Analysis. Principles of Adsorption. Basic Equation for
Adsorption. Adsorber Design Calculation. OUTLINES
Slide 53
Rate of Mass Transfer BASIC EQUATION FOR ADSORPTION Internal
and External Mass-transfer Coefficients Solution to Mass- Transfer
Equations YIELD Irreversible Adsorption Linear isotherm
Slide 54
RATE OF MASS TRANSFER Equation for mass transfer in fixed-bed
adsorption are obtained by making a solute material balance for a
section dL of the bed, as in Fig. 25.9. The rate of accumulation in
the fluid and in the difference between input and output flows. The
change in superficial velocity is neglected: Fig.25.9 Mass balance
for a section of a fixed bed (25.5) (25.6) The transfer process is
approximated using an overall volumetric coefficient and an driving
force: The mass transfer area, a is taken as the external surface
of the particles, which is 6(1-)/D P for spheres. The concentration
c* is the value that would be in equilibrium with the average
concentration W in the solid. K c a= volumetric overall
mass-transfer coefficient. (25.7) is the external void fraction of
the bed solute dissolved in the pore fluid is included with the
particle fraction 1-. Adsorption from a gas or a dilute solution -
Accumulation in the fluid is negligible compare to accumulation on
the solid.
Slide 55
Rate of Mass Transfer BASIC EQUATION FOR ADSORPTION Internal
and External Mass-transfer Coefficients Solution to Mass- Transfer
Equations YIELD Irreversible Adsorption Linear isotherm
Slide 56
INTERNAL AND EXTERNAL MASS-TRANSFER COEFFICIENTS The overall
coefficient, K c depends on the external coefficient k c, ext and
on an effective internal coefficient, k c, int. Diffusion within
the particle is actually an unsteady-state process, and the value
of k c, int decreases with time, as solute molecules must penetrate
farther and farther into the particle to reach adsorption sites.
Average effective coefficient can be used to give an approximate
fit to uptake data for spheres. (25.8) This leads to D e =
effective diffusion coefficient, depends on particle porosity, the
pore diameter, the tortuosity, and the nature of the diffusing
species. D p = diameter of particle.
Slide 57
Rate of Mass Transfer BASIC EQUATION FOR ADSORPTION Internal
and External Mass-transfer Coefficients Solution to Mass- Transfer
Equations YIELD Irreversible Adsorption Linear isotherm
Slide 58
SOLUTION TO MASS-TRANSFER EQUATIONS There are many solutions to
Eq. (25.6) and (25.7) for different isotherm shapes and controlling
steps, and all solution involve a dimensionless time, and a
parameter N representing the overall number of transfer units:
(25.9) (25.10) The term L/u 0 in Eq. (25.9) is the time to displace
fluid from external voids in the bed, which is normally negligible.
p (1- ) is the bed density, b is the ratio of the time to the ideal
time t * from Eq. (25.3). (25.3) If there were no mass-transfer
resistance, the adsorber could be operated with complete removal of
solute up to = 1.0, and the the concentration would jump from 0 to
c/c 0 = 1.0.
Slide 59
Rate of Mass Transfer BASIC EQUATION FOR ADSORPTION Internal
and External Mass-transfer Coefficients Solution to Mass- Transfer
Equations YIELD Irreversible Adsorption Linear isotherm
Slide 60
IRREVERSIBLE ADSORPTION Irreversible adsorption with a constant
mass transfer coefficient is the simplest case to consider, since
the rate of mass transfer is then just proportional to the fluid
concentration. Strongly favorable adsorption gives almost the same
results as irreversible film, because the equilibrium concentration
in the fluid is practically zero until the solid concentration is
over one-half the saturation value. If the accumulation term for
the fluid is neglected, Eq. (25.6) and (25.7) are combine to give;
(25.12) The initial shape of the concentration profile is obtained
by integration Eq. (25.12) (25.13) (25.6) (25.7) Since the term K c
aL/u 0 is defined as N in Eq. (25.10), the concentration at the end
of the bed is given by (25.14) (25.10)
Slide 61
IRREVERSIBLE ADSORPTION The rate of mass transfer to the first
layer of particles is assumed to be constant until the particles
reach equilibrium with the fluid, and until this happens, the
concentration profile in the bed remains constant. The time to
saturate the first portion of the bed, t 1 is the equilibrium
capacity divided by the initial transfer rate (W 0 = 0 to simply
the analysis): (25.15) After this time, the concentration profile
moves steadily down the bed, keeping the same shape. The transfer
zone moves at a velocity v z, which is equal to the amount of
solute removed per unit time divided by the amount retained on the
solid per unit length of bed: (25.16) The concentration is constant
at c 0 for the saturated portion of the bed and then falls
exponentially in the mass-transfer zone, as shown Fig. 25.10. Fig.
25.10
Slide 62
IRREVERSIBLE ADSORPTION To predict the break point, Eq. (25.17)
is applied for a bed of length L with c/c 0 set at 0.05 or another
selected value. The length of the saturated bed is the product of
transfer zone velocity and the time since the zone started to move:
(25.20) (25.17) (25.19) (25.18) Substituting the equation for L sat
in Eq. (25.17) and using the dimensionless terms and N [Eq. (25.9)
and (25.10) give: (25.21)
Slide 63
IRREVERSIBLE ADSORPTION The predicted breakthrough curve is
shown as a solid line in Fig. 25.11. The slope increases with time,
and c/c 0 becomes 1.0 at N(-1)=1.0. In practice, the breakthrough
curves are usually S-shaped, because the internal diffusion
resistance is not negligible, and it increases somewhat when the
solid becomes nearly saturated. When both internal and external
resistances are significant, the breakthrough curve is S- shaped,
as shown by the dashed line in Fig 25.11. For this plot, the value
of N is based on the overall mass-transfer coefficient given by Eq.
(25.8), or it can be expressed in Halls terminology as; (25.22) Fig
25.11
Slide 64
EXAMPLE 25.3 (a) Use the breakthrough data in Example 25.2 to
determine N and K c a for the 8-cm bed, assuming irreversible
adsorption. (b) Compare K c a with the predicted k c a for the
external film. SOLUTION: (a)From Example 25.2, at c/c 0 = 0.05, W/W
sat = 0.495, -1= -0.505. Assume equal internal and external
resistances to determine N from Fig. 25.11: (b) Prediction of k c a
from Re, Sc (k c is the external coefficient): D p = 0.37 cm At 25
o C, 1 atm, / = 0.152 cm 2 /s and D v = 0.0861 cm 2 /s. Then
SOLUTION: (a)From Example 25.2, at c/c 0 = 0.05, W/W sat = 0.495,
-1= -0.505. Assume equal internal and external resistances to
determine N from Fig. 25.11: (b) Prediction of k c a from Re, Sc (k
c is the external coefficient): D p = 0.37 cm At 25 o C, 1 atm, / =
0.152 cm 2 /s and D v = 0.0861 cm 2 /s. Then = 23.0 s -1
Slide 65
EXAMPLE 25.3 (a) Use the breakthrough data in Example 25.2 to
determine N and K c a for the 8-cm bed, assuming irreversible
adsorption. (b) Compare K c a with the predicted k c a for the
external film. SOLUTION: (a)From Example 25.2, at c/c 0 = 0.05, W/W
sat = 0.495, -1= -0.505. Assume equal internal and external
resistances to determine N from Fig. 25.11: (b) Prediction of k c a
from Re, Sc (k c is the external coefficient): D p = 0.37 cm At 25
o C, 1 atm, / = 0.152 cm 2 /s and D v = 0.0861 cm 2 /s. Then
SOLUTION: (a)From Example 25.2, at c/c 0 = 0.05, W/W sat = 0.495,
-1= -0.505. Assume equal internal and external resistances to
determine N from Fig. 25.11: (b) Prediction of k c a from Re, Sc (k
c is the external coefficient): D p = 0.37 cm At 25 o C, 1 atm, / =
0.152 cm 2 /s and D v = 0.0861 cm 2 /s. Then = 23.0 s -1
Slide 66
SOLUTION: From Eq. (17.74) Since K c a is slightly less than
one-half the predicted value of k c a, the external resistance is
close to one-half the total resistance, and the calculated value of
N need to be revised. The internal coefficient can be obtained
from; If diffusion into the particle occurred only in the gas
phase, the maximum possible value of D e would be about D v /4,
which leads to; Since the measured value of k c, int is an order of
magnitude greater than this value, surface diffusion must be the
dominant transfer mechanism.
Slide 67
Rate of Mass Transfer BASIC EQUATION FOR ADSORPTION Internal
and External Mass-transfer Coefficients Solution to Mass- Transfer
Equations YIELD Irreversible Adsorption Linear isotherm
Slide 68
ADSORBER DESIGN CALCULATION The design of adsorber for gas or
liquid purification involves; - choosing the adsorbent and the
particle size, selecting an appropriate velocity to get the bed
area, and either determining the bed length for a given cycle time
or calculating the break-through time for a chosen length. For gas
purification/ adsorption: - 4 x 6- or 4 x 10-mesh carbon is needed
and pressure drop is not a problem. - The gas velocity is usually
between 15 and 60 cm/s (0.5 and 2 ft/s) - Because the external area
varies with 1/D p and both k c,ext and k c, int increase as D p
decreases, k c a is expected to vary with the -1.5 to -2.0 power of
D p. For liquid adsorption: - smaller particle sizes are chosen,
and the fluid velocity is much lower than with gases. - Typical
conditions for water treatment are 20 x 50-mesh carbon (D p = 0.3
to 0.8 mm) and a superficial velocity of 0.3 cm/s (0.01 ft/s or
about 4 gal/min. ft 2 ) - Even with these conditions K c a/u 0 is
smaller than for typical gas adsorption, and LUB may be 10 to 20 cm
or even as much as 1 m if internal diffusion controls. The design
of adsorber for gas or liquid purification involves; - choosing the
adsorbent and the particle size, selecting an appropriate velocity
to get the bed area, and either determining the bed length for a
given cycle time or calculating the break-through time for a chosen
length. For gas purification/ adsorption: - 4 x 6- or 4 x 10-mesh
carbon is needed and pressure drop is not a problem. - The gas
velocity is usually between 15 and 60 cm/s (0.5 and 2 ft/s) -
Because the external area varies with 1/D p and both k c,ext and k
c, int increase as D p decreases, k c a is expected to vary with
the -1.5 to -2.0 power of D p. For liquid adsorption: - smaller
particle sizes are chosen, and the fluid velocity is much lower
than with gases. - Typical conditions for water treatment are 20 x
50-mesh carbon (D p = 0.3 to 0.8 mm) and a superficial velocity of
0.3 cm/s (0.01 ft/s or about 4 gal/min. ft 2 ) - Even with these
conditions K c a/u 0 is smaller than for typical gas adsorption,
and LUB may be 10 to 20 cm or even as much as 1 m if internal
diffusion controls.
Slide 69
QUESTION 25.4 Adsorption on activated carbon is being
considered to treat a process airstream that has 0.12 volume
percent methyl ethyl ketone (MEK), C 4 H 8 O. The gas is at 25 o C
and 1 atm, and the flow is 16,000 ft 3 /min. The pressure drop
across the bed should not exceed 12 in. H 2 O. a.If BPL 4 x 10-mesh
carbon is used, predict the saturation capacity and the working
capacity if the average bed temperature is 35 o C and the
regeneration is stopped when W = 1/3 W sat. a.What gas velocity and
bed size could be used to give reasonable cycle time if the length
of unused bed is 0.5 ft? How much carbon is needed? Adsorption on
activated carbon is being considered to treat a process airstream
that has 0.12 volume percent methyl ethyl ketone (MEK), C 4 H 8 O.
The gas is at 25 o C and 1 atm, and the flow is 16,000 ft 3 /min.
The pressure drop across the bed should not exceed 12 in. H 2 O.
a.If BPL 4 x 10-mesh carbon is used, predict the saturation
capacity and the working capacity if the average bed temperature is
35 o C and the regeneration is stopped when W = 1/3 W sat. a.What
gas velocity and bed size could be used to give reasonable cycle
time if the length of unused bed is 0.5 ft? How much carbon is
needed?
Slide 70
ANSWER (a) From the handbooks, P = f s = 151 mmHg at 35 o C and
L = 0.805 g/cm 3 at 20 o C. The normal boiling point is 79.6 o C,
and the estimated density at this temperature is L = 0.75 g/cm 3.
The molecular weight is 72.1. At 35 o C. From Fig. 25.4, the volume
adsorbed is 24 cm 3 per 100 g carbon: Working capacity = W sat W 0
= 12g/ 100 g carbon = 0.12 lb/lb carbon. (b) What gas velocity and
bed size could be used to give reasonable cycle time if the length
of unused bed is 0.5 ft? How much carbon is needed? Try u 0 = 1
ft/s:
Slide 71
Amount of adsorbed depends on (T/V) log (f s /f), where: T:
adsorption temperature (Kelvin). V: molar volume of the liquid at
the boiling point f s : fugasity of the saturated liquid at
adsorption temperature f: fugasity of the vapor For adsorption at
atmospheric pressure; * fugasity = partial pressure = vapor
pressure Volume adsorbed is converted to mass by assuming the
adsorbed liquid has the same density as liquid at the boiling
point.
Slide 72
ANSWER For a circular cross section, D = 18.4 ft. a rectangular
bed 10 ft x 27 ft might be more suitable if the bed depth is only 3
to 4 ft. Try L = 4 ft. From Eq. (25.3) At 25 o C, =18.1 h If the
length of unused bed is 0.5 ft, 3.5 ft is used, and If the bed
length is 3 ft with 2.5 ft used, (25.3)
Slide 73
ANSWER Allowing for uncertainties in the calculations, a bed
length of 3 ft would be satisfactory with regeneration once per 8-h
shift. Check P using the Ergun equation, Eq. (7.22). Note that g c
is needed when fps units are used. For granular carbon, assume s =
0.7 (see Table 7.1). Assume external void fraction = 0.35 (see
Table 7.2). From handbooks, the properties of air at 25 o C are
From Perry, 7 th ed., p. 19-20, for 4 x 10-mesh carbon,
Slide 74
ANSWER For L = 3 ft, P = 9.9 in H 2 O, which is satisfactory. A
velocity of 1.5 ft/s would give P/L = 6.06 in. H 2 O/ft and require
L 2 ft to keep P < 12 in. H 2 O. However, the breakthrough time
would be reduced to 11.3/1.5 x (1.5/2.5) = 4.5 h, and the bed would
have to be regenerated twice each shift. This design might be
satisfactory but does not give as great a margin for error. The
recommended design is for two beds 10 x 27 x 3 ft placed in
horizontal cylinders. The total inventory of carbon is; m c = 2
(270 x 3)ft 3 x 30 lb/ft 3 = 48, 600 lb
Slide 75
QUESTION 25.5 Water contaminated with 1.2 ppm TCE is to be
purified in a fixed bed of 20 x 50-mesh Ambersorb 563. (a)For a bed
length of 2 ft and a flow rate of 4.5 gal/min.ft 2, estimate the
breakthrough time if the length of the unused bed is 0.6 ft.
(b)What is the effective capacity in volume treated per unit bed
volume? The adsorbent will be regenerated by steam to remove 85
percent of the TCE. The bulk density of the adsorbent is 0.53 g/cm
3. Water contaminated with 1.2 ppm TCE is to be purified in a fixed
bed of 20 x 50-mesh Ambersorb 563. (a)For a bed length of 2 ft and
a flow rate of 4.5 gal/min.ft 2, estimate the breakthrough time if
the length of the unused bed is 0.6 ft. (b)What is the effective
capacity in volume treated per unit bed volume? The adsorbent will
be regenerated by steam to remove 85 percent of the TCE. The bulk
density of the adsorbent is 0.53 g/cm 3.
Slide 76
ANSWER (a) From Fig. 25.5 From Eq. (25.3), Breakthrough time
is; (b) Bed volumes treated,