Adsorption
Lecture Notes
ENVE542
Air Pollution Control Technologies
By Dr. Pınar Ergenekon
GYTE 2010 Fall
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
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2
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
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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
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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
2015/4/7
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Aerosol & Particulate Research Lab
5
5
Adsorbent Material
– Silica gel
– Activated alumina
– Activated carbon
– Synthetic zeolite
Dehyrdating
purposes
• Molecular sieve
Properties of Activated Carbon
Bulk Density
22-34 lb/ft3
Heat Capacity
0.27-0.36 BTU/lboF
Pore Volume
0.56-1.20 cm3/g
Surface Area
600-1600 m2/g
Average Pore Diameter
15-25 Å
Regeneration Temperature 100-140 oC
(Steaming)
Maximum Allowable
150 oC
Temperature
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Adsorbent Material
Properties of Silica Gel
Bulk Density
44-56
lb/ft3
Heat Capacity
0.22-0.26
BTU/lboF
Pore Volume
0.37
cm3/g
Surface Area
750 m2/g
Average Pore Diameter 22 Å
Regeneration
120-250
oC
Temperature
Maximum Allowable
400 oC
Temperature
Properties of Activated Alumina
Bulk Density
Granules
38-42 lb/ft3
Pellets
54-58 lb/ft3
Specific Heat
Surface Area
0.21-0.25
BTU/lboF
0.29-0.37
cm3/g
210-360 m2/g
Average Pore Diameter
18-48 Å
Pore Volume
Regeneration Temperature 200-250 oC
(Steaming)
Maximum Allowable
500 oC
Temperature
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Adsorbent Materials
Properties of Molecular Sieves
Type
Density in bulk (lb/ft3)
Specific Heat (BTU/lboF)
Effective diameter of pores (Å)
Regeneration Temperature (oC)
Maximum Allowable Temperature
(oC)
Anhydrous
Sodium
Aluminosilicate
4A
44
0.19
4
200-300
600
Anhydrous
Calcium
Aluminosilicate
5A
44
0.19
5
200-300
600
Anhydrous
Aluminosilicat
e
13X
38
13
200-300
600
• Crystalline zeolite
• Uniform pores to selectively separate compounds by size & shape
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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
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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
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Adsorption Isotherms
• In physical adsorption Brunauer,Emmett,
and Teller (BET) is frequenlty used
V
Vm

cP
( Po  P )[ 1  ( c  1)( P / Po )
V:volume the amount of adsorbed
gas would fill at a given pressure
and temperature
Vm: volume adsorbed it a layer one
molecule thick fills the surface
Po: vapor pressure of the
adsorbate at the temperature of the
system
P: actual partial pressure of the
adsorbate
c: a parameter of the particular
adsorption process
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Adsorption Isotherms, BET
V
Vm

cP
( Po  P )[ 1  ( c  1)( P / Po )
• 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
12
evaluate Vm ENVE542 GYTE Çevre Müh.
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)
C P
f 
C P C
rd=Cdf
a
a
d
ra: rate of adsorption
Ca, Cd: constant
P: partial pressure of the adsorbate
f: is the occupied fraction of the total solid surface
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Adsorption Isotherms, Langmuir
• Since we assume a monolayer coverage,
the mass of adsorbate per unit mass of
adsorbent (a) is also proportional to f:
f 
a  Ca f
'
'
a
(C a C a / C d ) P
(C a / C d ) P  1
k1 P
a
P
a


CaP
CaP  Cd
k1 P
k2P  1
 k2P  1
k2P
k1

1
Langmuir Isotherm
k1
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Adsorption Isotherm, Freundlich
• At very low and very high adsorbate partial presssure
Langmuir isotherm equation takes the forms of:
a  k1 P
a 
k1 P
k2P
a  kP
n
P  a

•This equation is refferred to as a
Freundlich isotherm.
• 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.
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Adsorption Isotherm, Freundlich
a  kP
n
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Adsorption Isotherm Data (AID)
• 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
-0,914
0,17
-0,0158
Heptane
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Adsorption Isotherms
• Figure 12.2
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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,
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Adsorption Potential
• is defined as “the change in free energy (DGads)
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.
D G ads
 Pv 
 RT ln 

 P 
• 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
 RT
 RT
 Pv  
 Pv  
2 . 303 log 
2 . 303 log 
  


 P  i  V '
 P  j
V'
i,j denote
different
gasses
20
ENVE542 GYTE Çevre Müh.
Adsorption Potential
• is defined as “the change in free energy (DGads)
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.
D G ads
 Pv 
 RT ln 

 P 
• 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
 RT
 RT
 Pv  
 Pv  
2 . 303 log 
2 . 303 log 
  


 P  i  V '
 P  j
V'
i,j denote
different
gasses
21
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Adsorption Wave
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Adsorption Wave
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Adsorption Zone
The length of Adsorption Zone (AZ) is a function of the
rate of transfer of adsorbate from the gas to the
adsorbent
A shallow AZ indicates good adsorbent utilization and is
represented by a step breakthrough curve
The length of AZ determined the minimum depth of the
adsorbent bed
On 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 isotherm
Hence system design is based primarily on previous plant
experment and pilot scale studies.
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Analysis of an Adsorption Wave
Xsat or
Vg or Va
C or X
Vad
Db
x1
x2
• Let’s make a mass balance on the
pollutant for the overall adsorption zone
between positions 1 and 2
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Analysis of an Adsorption Wave
.
.
m g C1
 m ad X 1   ad AV ad X 1
g
 ad   true (1  f )
• 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 f 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:
.
m a C0
a
  ad AV ad X sat
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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:

C e   X sat
Assume equilibrium at point 1, Ce=C0
.
m a C0
a
  ad
 C0 
AV ad 

  

.
V ad 
ma
 a  ad A
.
( )
1/ 
(C 0 )
(  1 ) / 

V
a
 ad
( )
1/ 
(C 0 )
(  1 ) / 
• 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.
27
ENVE542 GYTE Çevre Müh.
Analysis of an Adsorption Wave
• To determine the thickness of the adsorption wave, develop a
relationship between C and x in the adsorption zone
.
.
d mp 
ma
a
dC
Mp is the rate of mass transfer
of pollutant gas onto the solid
phase
.
• Another independent equation
for mp can be developed by
general theory of mass transfer:
d m p  KA ( C  C e ) dx
.

 m a dC  KA  a ( C   X ) dx
.
X


ma C
 a  ad AV ad
.

1 
1
 C  C0

1 
 m a dC  KA  a ( C  C C 0
.

dC

dx 

1 
KA  a  C  C C 0
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ma
ENVE542




28
) dx
Analysis of an Adsorption Wave
• Integration of dx equation from x1 to x2 will yield an expression
for the adsorption zone thickness, d, 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 = h
which varies from 1 to 0. So divide each side of the dx
equation by C0 and replace dx by d.
.

dC

dx 

1 
KA  a  C  C C 0
ma
• :
d KA  a
.
ma
1

dh
 h h 
0
1





dh
 h (1  h 
0
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1
)
29
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:
d KA  a
.
 4 . 595 
ma
1
 1
ln
1  0 . 01
(  1 )
1  0 . 99
(  1 )
• Then ma/Apa can be replaced by the superfacial gas stream
velocity Va
(  1 )
d KA  a
.
ma
 4 . 595 
1
 1
ln
1  0 . 01
1  0 . 99
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(  1 )
30
Breakthrough Time
• If we assume that the time required to establish the
adsorption zone to its full thickness at the inlet is zero then,
tB 
Db  d
V ad
 d 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…
Pressure Drop Across a Fixed Bed
Figure 12.6
Ergun equation (Eq.12.13):
D Pg  d p  g
3
D (1   ) G '
2

150 (1   )  g
 1 . 75
d pG '
DP: 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)
Typical operating range:
< 20 in H2O;
20 ft/min < V < 100 ft/min
12 in<D<30 in
32
Pressure Drop Across a Fixed Bed
A simpler emprical equation by the Union
Carbide Corporation is as follows (Eq.12.14):
 V 
D P  0 . 37 L 

 100 
1 . 56
DP: pressure drop (in H2O)
L: bed depth (in)
V: superficial gas velocity ft/min
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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 online 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
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Solution
• a)Eq 12.13 D Pg  3 d 
p
g
D (1   ) G '

2
150 (1   )  g
 1 . 75
d pG '
DP: pressure drop (lb/ft2)=?
D: bed depth (ft)=2
 : void fraction=0.35
g: gas viscosity (lb/ft-hr)=0.047
dp: carbon particle diameter (ft)=0.011
Calculate G’: gas mass flux (lb/ft2-hr)
Assume that the influent gas has the properties of air , MW=29
Total mass of carbon in bed = 12x6x2x(30ft/lb)=4320 lb
Mass of benzene adsorbed/hr=4320x(10/100)= 432 lb/hr
g 
G  7500
ft
PMW
RT
3
min
x 60

1 x 29
 0 . 071 lb / ft
3
0 . 73 x 560
min
hr
x 0 . 071
lb
ft
3
x
1
12 x 6
 446 lb / hr  ft
2
35
Solution
D Pg  d p  g
150 (1   )  g
3
• a)Eq 12.13
D (1   ) G '
g 
G  7500
ft
PMW
2

RT
 1 . 75
d pG '
1 x 29

D: (ft)=2
 : 0.35
g: 0.047
dp: (ft)=0.011
 0 . 071 lb / ft
3
0 . 73 x 560
3
x 60
min
min
x 0 . 071
hr
lb
ft
3
x
1
 446 lb / hr  ft
2
12 x 6
2

(1   ) G '  150 (1   )  g


 1 . 75 
3


D
g d p  g 
d pG '

DP
DP
2
DP

 150 (1  0 . 35 )( 0 . 047 )




1
.
75
8
3

4 . 17 x10 0 . 35 ( 0 . 011 )( 0 . 071 ) 
0 . 011 ( 446 )

(1  0 . 35 ) 446
2
 24 . 85
2
D P  49 . 7 lbf / ft 
2
49 . 7
144
x
406 . 8 inH 2 O
14 . 7 psi
 9 . 5 inH 2 O
36
Solution
D: (ft)=2
 : 0.35
g: 0.047
dp: (ft)=0.011
• a)Eq 12.14
Superficial gas velocity=7500/(12x6) = 104 ft/min
 V 
D P  0 . 37 L 

 100 
V = 104 ft/min
1 . 56
 104 
D P  0 . 37 x 2 

 100 
1 . 56
D P  9 . 4 inH 2 O
37
Solution
• a)Eq 12.13
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38
Solution
• From Figure 12.6
 DP=7.8 inH2O/ft x 2 ft=15.6inH2O
Regeneration of Adsorption Bed
A 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
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40
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
.
VR 
mR
 R  ad A
( R )
1/  R
(C R )
(  R 1 ) /  R
• Subscript R refers to the properties and
flowrate of regenaration fluid
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Regeneration of Adsorption Bed
• Since,
CR   R X
C0  X
R

CR
C 
 R 0 
  
R

• CR can be related to C0 :
• Substitute this to the speed
of desorption zone eq.
.
m R R  C 0 
VR 


 R  ad A   
(  R 1 ) /  R
• If the thickness of the desorption zone is much smaller than
the bed length Db, then the time for regeneration:
.
t
Db
VR

 R  ad AD b m R  R   
m R R
(  R 1 ) /  R


C 
 0
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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
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Example 12.2-SOLUTION
• Use Figure 12.3. So Calculate abscissa for Figure 12.3
P  y pen tan e x14 . 7 psia
P 
1880 ppm
x14 . 7 psia  0 . 0276 psia
1000 , 000
• From Appendix B, Pv is about 16 at T = 95F. The specific molal
volume then
3
72 g / gmol
112 cm
V '

3
0 . 64 g / cm
gmol
• Now solve for the abscissa of Figure 12.3 using pressures in place
of fugacities:
T
V'
log
Pv
P

308 K
112 cc / gmol
16
log
 7 .6
0 . 0276
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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:
18 cc
100 gC
x
1 gmol
72 g
112 cc gmol

11 . 6 gn  C 5
100 gC
• 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
ENVE542 GYTE Çevre Müh.
45
Example 12.2-SOLUTION
• The mass flow rate of n-pentane to be adsorbed in one hour is:
1atmx 0 . 00188 x 5500 cfm
72 lb
60 min

M 
x
x
3
0 . 73 atm  ft
lbmol
hr
x 55 R
lbmol  R
M  110 lb / hr
• 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:
lbC
bed

110 lbn  C 5
hr
x1hrx
100 lbC
3 . 5 lbn  C 5
 3143 lbC / 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
ENVE542 GYTE Çevre Müh.
Example 12.2-SOLUTION
Use a minimum bed depth of 2 ft and a rectangular bed with L=2W
BedArea

106 . 7
 53 . 4 ft
2
2
W
2
 53 . 4 / 2
W  5 . 2 ft    L  2W  10 . 4 ft
Use bed dimensions of 5.25ftx10.5ft and check the superficial velocity:
V 
5500 acfm
 100 ft / min
( 5 . 25 )( 10 . 5 )
On the high end of the
acceptable range, OK
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:
3 . 5 lbn  C 5
100 _ lbC
x
1h
110 _ lbn  C 5
x 3308 lbC  1 . 05 hr ( 63 . 2 min)
ENVE542 GYTE Çevre Müh.
47
Solution, Steam Requirement
• Steam requirement per regeneration can be
calculated as follows:
steam 
steam 
3 _ lb  stm
lb _ n  C 5
x
0 . 3 _ lb  stm
lb _ n  C 5
110 lb _ nC 5
x1 . 05 hr  330 lb _ stm
hr
x 3308 lb _ C  992 lb _ stm
Based on n-C5
amount
Based on
carbon amount
• 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
Solution, Bed Pressure Drop
• From Figure 12.6:
DP4x10=7.5 in H2O x 2 ft = 15 in H2O
DP6x16=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 DP of 3 in H2O
• Allow a pressure drop of 2 inH2O for the inlet and exhust
piping. Thus the fan mus provide a total DP of:
 DP=3+2+15(-4.5)=24.5 in H2O
• If we assume a blower efficiency of 60% then
0 . 0001575 hp
blowerhp

cfm  inH 2 O
x 5500 acfmx 24 . 5 inH 2 O
 35 . 4 hp
0 . 60
SOLUTION
• Preliminary Specification Summary
Adsorber 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)