CE 529 Hazardous Waste Management
Physical Methods - Air Stripper Design
Dr. S.K. Ong
AIR STRIPPING
•
•
•
•
vapor stripping is the transfer of a hazardous material in liquid phase to vapor phase by
passing a high volume of the vapor phase through the liquid phase. Two common stripping
processes used for hazardous waste treatment are steam stripping and air stripping
Air stripping consists of passing air through the contaminated fluid to volatile hazardous
constituents. Air stripping is cost effective for treatment of waste streams with low
concentrations of the hazardous constituents. A common application of air stripping is
remediation of groundwater contaminated with VOCs. Note that the hazardous constituents
must be able to volatilize under room temperature for air stripping to be effective.
When semi-volatile compounds are present, steam is often used to increase the temperature
of the liquid phase to enhance volatilization of the organic compounds.
Stripping can be performed by using packed towers, tray towers, spray systems, diffused
aeration systems or mechanical aeration.
Packed tower systems - a typical packed tower is shown below. Water is distributed evenly
at the top of the tower through a packing material. The packing has high surface area.
Typical packing material are shown below.
Water in the tower is broken into small droplets and made to spread over the surface of the
packing resulting in a high surface area for the exchange of the volatile organics from the
liquid phase to air or steam.
Air is introduced from the bottom and leaves from the top of the tower. VOCs are
transferred from the liquid to the vapor phase.
•
Factors affecting the efficiency of an air stripping system include: volatility of the
compounds as measured by Henry's Law constant, air flow rate, water loading rate, mass
transfer at the air-water interface, packing material and depth of packing
•
Development of air stripping equations may proceed by considering a section of the tower.
Assume that the section has a superficial cross sectional area of A and a differential thickness
of dz.
Qw , CL,in
Z
QG, CG,in
Summary of Equations and correlations for the design of packed-bed air stripping towers
Henry's Law
C = KH ' CL
(1)
Van't Hoff's Relationship
log KH = -Ho/RT + k
KH
= exp (A - B/T)
(2)
Number of Transfer Units (NTU)
( S  1 ) C L ,in
NTU  ( S ) ln[
1]
S 1
S
C L ,out S
(3)
Stripping Factor
QG
Qw
S  K 'H
(4)
Height of Transfer Units (HTU)
HTU 
Tower packing Height
Qw
LM

K L aA  L K L a
(5)
Z = HTU x NTU
(6)
Pressure loss through packing support plate, demister, inlet and outlet duct
Q
P  k p ( G ) 2
(7)
Ax
Blower Horsepower (HP/sq. ft of tower)
AirHP 
144Q A ( p1  p 2 )
33,000( E )
(8)
Overall Mass Transfer, KL
1  1 
1
KL kL k K 1
G H
(9)
One of the more common correlations to compute the kG and kL are the Onda's correlation
2
2
L2M
aw
 C 0.75 LM 0.1 LM at  0.05
 1  exp[ 1.45(
)
(
) (
)
(
)0.2 ]
at
L
at  L
2


a
L
L
t
L g
1
kG
G
Q
 5.23[ M ] 0.7 [
] 3 [ at d p ]  2
a t DG
at  G
 G DG
kL [
L
Lg
] 1 / 3  0.0051[
LM
L
]2 / 3[
]  0.5 [ at d p ] 0.4
a w L
 L DL
(10)
(11)
(12)
Others include the Sherwood and Holloway correlation
 0.3048LM
kLa
 10.764 
DL
L




1n
 L

  L DL



0.5
(13)
Shulman et al. correlation
 Lds
d L
 25.1 s M
DL
 L
G kG
QM



0.45
 L

  L DL
 d sQM 

 1.195

 G ( 1   ) 
0.36



0.5
(14)
0.667
 G 


 D 
 G G
(15)
Effective surface area using Bravo and Fair correlation
  0.5
ae

 0.498 L
at
 Z 0.4

   L
  L M
   

  L
 6 Q M


 a t  G




0.392
Symbols and Dimensional Units for Equations and Correlations
KH
=
Henry's Law Constant (atm-m3/mole)
K H'
=
Henry's Law Constant (dimensionless)
A, B
=
constants for van't hoff's equation
R
=
Universal gas constant (82.05 atm-cm3/moleoK)
T
=
Temperature (oK)
kp
=
head loss constant (0.093 ins H2O s2/ft2)
kL
=
liquid film mass transfer coefficient (m/s)
kG
=
gas phase mass transfer coefficient (m/s)
KL
=
overall mass transfer coefficient (m/s)
KLa
=
overall volumetric mass transfer coefficient (m3/s-m3)
a
=
interfacial surface area for mass transfer (m2/m3)
S
=
Stripping Factor
Q
=
Gas molar air flow rate (moles/s)
QM
=
Air flow rate per unit area (kg/m2-s)
QG
=
Air flow rate (m3/s)
QA
=
Air flow rate (ft3/min)
L
=
Liquid molar flow rate (moles/s)
LM
=
Liquid mass flux (kg/m2-s)
Qw
=
Liquid flow rate (m3/s)
NTU
=
number of transfer units
HTU
=
height of transfer units
CG
=
liquid solute concentration (mg/L)
CL
gas solute concentration (mg/L)
=
CL,in
=
influent liquid solute concentration (mg/L)
CL,out
=
effluent liquid solute concentration (mg/L)
CG,in
=
influent gas solute concentration (mg/L)
CG,out
=
effluent gas solute concentration (mg/L)
aw
=
specific wetted packing area (m2/m3)
at
=
specific total packing area (m2/m3)
L
=
liquid density (kg/m3)
µL
=
liquid viscosity (kg/m-s)
L
=
liquid surface tension (kg/s2)
c
=
critical surface tension of packing (kg/s2)
g
=
gravitational acceleration (m/s2)
DL
=
diffusivity of solute in liquid phase (m2/s)
(16)
DG
µG
Z
Ax
p1
p2
E
P
F
dp
Air HP
diffusivity of solute in gas phase (m2/s)
viscosity of air (kg/m-s)
tower of packing (m)
cross sectional area of tower (ft2)
inlet air pressure (psi)
outlet air pressure (psi)
overall air blower efficiency
pressure drop across support plate and duct works (ins H2O)
Packing factor
nominal diameter of packing (m)
Power needed for air blower (HP/ft2 of tower)
=
=
=
=
=
=
=
=
=
=
=
Equations to estimate liquid and vapor phase diffusivity
Wilke and Chang Equation to estimate diffusivity of solute in liquid
D L ,AB  B
T
Where

7.48 x10 8 (  B M B )1 / 2
Vb0.6
(A)
DL,AB
µB
T
MB
=
=
=
=
liquid diffusivity of solute A in solvent B (cm2/s)
viscosity of solvent (centipoise)
absolute temperature (K)
molecular weight of solvent B
Vb
=
=
molal volume of solute at normal boiling temperature (cm3/mole)
association parameter for solvent B
water
= 2.6
methanol
= 1.9
benzene
= 1.0
B
Hirschfelder, Bird and Spotz equation to estimate difusivity of solute in gas
(nonpolar system)
DG ,AB 
DG,AB
T
MA
MB
P
 AB
D
0.001858T 3 / 2 ( 1  1 )1 / 2
MA
P 2AB  D
MB
(B)
=
mass diffusivity of solute A in solvent B (cm2/s)
=
absolute temperature (K)
=
molecular weight of A
=
molecular weight of B
=
absolute pressure (atm)
=
collision diameter (Å)
=
collision integral for molecular diffusuion (a measure of the
intermolecular potential field of molecule A and molecule B)
 can be estimated using the following equations
  1.18Vb1 / 3  0.841Vc1 / 3  2.44(
TC 1 / 3
)
PC
(C)
where Vb = molecular volume at normal boiling point (cm3/g-mole)
Vc = critical molecular volume
Tc = critical temperature (K)
Tb = normal boiling temperature (K)
Pc = critical pressure (atm)
D is a function of
where
T
 AB
 is the Boltzmann constant (1.38 x 10-16 ergs/K)
 AB = energy of molecular interaction (ergs)
See Tables for values of D as a function of
T
 AB
or estimated from the following regression equation
 D  1.442  0.6915 ln( T )  0.2536[ln( T )] 2  3.01x10 2 [ln( T )] 3  4.966 x10 3 [ln( T )] 4
 AB
 AB
 A /  can be estimated using the following equation
for a binary system
 AB 
 A  B
2
 AB   A B
 AB
A B


 
 AB
 AB
 A /  = 0.77 Tc
= 1.15 Tb
Air Stripping Design procedure
1. Identify a target contaminant or select the contaminant whose final water quality standard is the most
difficult to achieve (usually the least volatile contaminant).
Select removal efficiency needed.
2. Select the lowest water temperature and compute Henry's law constant for the contaminant using
Equation 2.
3. Select an air to water (G/L) ratio - typical air to water ratios for groundwater are between 10 and 100.
Compute the stripping factor S with Equation 4. Most designs are for stripping factors between 3 to
5. However, S as high as 10 has been used.
4. Compute NTU from Equation 3.
5. Select a hydraulic loading rate and compute the cross-sectional area and the diameter of the tower.
Hydraulic loading rates may vary between 5 and 50 gpm/ft2 (3.4 x 10-3 to 0.034 m3/m2•s). Hydraulic
loading rates between 20 to 35 gpm/ft2 are usually used.
6. Select the packing material and the mass transfer correlations to be used - for example, Onda's
correlation.
7. Compute the wetted surface area using Equation 10. The specific total packing area is usually
obtained from the manufacturer, while the critical surface tension of the packing is assumed to be that
of water. Several other pieces of information such as the density and viscosity of air and water, etc.,
can be obtained from standard textbooks or handbooks. Note that the equations use mass flux
(kg/m2•s).
8. If liquid and gas diffusivity of contaminant are not available, estimate the diffusivity of the
contaminant in liquid and air phases with the Wilke and Chang equation and Hirschfelder, Bird, and
Spotz equation, respectively .
9. With information from items 7 and 8 above, calculate the liquid and air phase mass transfer
coefficients (kL and kG) from Equations 11 and 12.
10. The overall mass transfer KL is then computed using Equation 9. Compute HTU from Equation 5.
Assume a = aw.
11. The height of the tower required is then computed from Equation 6. A safety factor can be added to
the height of packing, if required.
12. Head losses through the packing itself can be estimated from manufacturers' literature or from
Eckert's curve (see attached).
The pressure drop across the demister packing support plate, duct work, and tower inlet and outlet is
given by Equation 7. The value of kp in Equation 15 is approximately 0.093 ins H2O-s2/ft2) or 0.004
N-s2/m4 in SI units for a full-scale tower.
The total horsepower requirement can be estimated from Equation 8 where an assumed fan and motor
efficiency of 50 and 70 percent, respectively, can be used to yield an overall efficiency of 35 percent.
13. The above procedure can be repeated, as required, for different air and liquid loading rates, packing
materials, etc., to obtain the optimum design.
Example: An aqueous waste stream is contaminated with ethylbenzene. The influent concentration of
ethylbenzene is 2 mg/L. The ethylbenzene in the effluent is to be less than 40 µg/L. Flow rate of waste
stream is 0.025 m3/s. Temperature of water is 10o C.
40
1. Removal efficiency needed = ( 2000
2000 ) *100
= 98%
2. Estimate Henry’s Law constant, KH (atm•m3/mole)
KH
=
=
exp (A - B/T)
exp(11.92 - 4994/283) = 3.26 x 10-3 atm•m3/mole)
Convert to dimensionless Henry’s Law constant KH’
KH’
=
KH/RT = 3.26 x 10-3/(8.205 x 10-5* 283)
= 0.140
3. Select QG/Qw ratio
Typical air to water ratios for groundwater are between 10 and 100
If QG/Qw ratio is too high, we get entrainment of water in the tower and may result in
flooding. Also a higher QG/Qw ratio means a higher head loss through the tower.
Assume air to water ratio of 50
Therefore stripping factor
= KH’(QG/Qw) = 50 (0.140)
(usually take values between 5 – 10, no advantage > 10)
=
7
4. Compute NTU
( S  1 ) C L ,in
NTU  ( S ) ln[
 1]
S 1
S
C L ,out S
=
7  1  2000  1 
7
ln

 
7  1  7  40  7 
=
4.39 NTUs
5. Select hydraulic loading rate.
Rates may vary from 5 to 50 gpm/ft2 (3.4 x 10-3 to 0.034 m3/m2•s)
On average between 20 to 35 gpm/ft2
Select 25 gpm/ft2 (0.017 m3/m2•s)
Cross sectional area
Diameter
Use
=
=
0.025/0.017
(1.47*4/3.142)0.5
= 4.5 ft or 1.372 m
=
=
1.47 m2
1.37 m or 4.48 ft
Cross sectional area of tower = 1.478 m2
Liquid loading rate = 0.025/[1.478] = 0.0169 m3/m2-s = 24.9 gpm/ft2
6. Select packing material. Use 1” pall rings
F= 52, a = 206 m2/m3
If mass transfer coefficient is available from experiments or literature use these values. If not, you
can estimate the mass transfer coefficient using Onda’s correlation.
To use Onda’s correlation, you need to know DL and DG. If DL and DG are given, use them if not estimate
values from Wilke Chang and Hirschfelder’s equation. Estimation of DL
D L ,AB  B
T
where

7.48 x10 8 (  B M B )1 / 2
Vb0.6

= 283o K
= 1.307 centipoise
= molecular weight = 18
= molal volume of solute at normal boiling temp
For ethylbenzene specific gravity is 0.867. MW = 106
no. of moles per cm3
= 0.867/106 moles/cm3
= 8.179 x 10-3
or Vb = 122.27 cm3/mole
= 2.6 for water
DL
=
T
B
MB
Vb
283(7.4 x 10-8) 2.6 x 18)1/2
1.307 (122.27)0.6
= 6.13 x 10–6 cm2/s = 6.13 x 10-10 m2/s
Estimation of DG
DG ,AB 
where
For air
0.001858T 3 / 2 ( 1  1 )1 / 2
MA MB
2
P AB  D
MA
MB
P
= 106
= 29
= 1atm
 A /  = 1.15 Tb
= 470.6 K
for ethylbenzene Tb = 136.2o C or 409.2 o K
 B /  = 97 K
 air = 3.617
 solute  1.18Vb1 / 3
= 1.18(122.27)1/3
= 5.86
 AB 
= (5.86 + 3.617) /2
= 4.74
 A  B
2
 AB
A B
= ( 97 )( 470.58 )


 
 AB / T  213.6 / 283
T /  AB  283 / 213.6 =1.324
= 213.6
From Table K.1, D = 1.27
Or use the following correlation D in your handout.
 D  1.442  0.6915 ln( T )  0.2536[ln( T )] 2  3.01x10 2 [ln( T )] 3  4.966 x10 3 [ln( T )] 4
 AB
 AB
 AB
 AB
 D  1.442  0.6915 ln( 1.324 )  0.2536[ln( 1.324 )] 2  3.01x10 2 [ln( 1.324 )] 3  4.966 x10 3 [ln( 1.324 )] 4
= 1.27
Therefore
DG ,AB 
0.001858( 283 )3 / 2 ( 1  1 )1 / 2
= 0.079 cm2/s
29 106
1( 4.74 )2 ( 1.27 )
= 7.9 x 10-6 m2/s
Compute Wetted Area using Onda’s correlation
L2
L2 at
L2M
aw

 1  exp[ 1.45( C )0.75 ( M )0.1 ( M )  0.05 (
)0.2 ]
at
L
at  L
2


a
L L t
L g
assume c = L = 0.0742 N/m at 10o C
LM is in terms of kg/m2-s, we have Qw = 0.025 m3/s
Therefore, LM = 0.025*1000/1.478 = 16.9 kg/m2-s
at = 206 m3/m2, L = 1.307 x 10-3 kg/m-s, L = 999.1 kg/m3, g = 9.8 m/s2

0.75 
aW
16.9 2
 0.0742 

 1  exp 1.45

 206 * 1.307 x10 3
206
 0.0742 






0.1
 16.9 2 * 206 


 ( 999.1 ) 2 ( 9.8 ) 


0.05


16.9 2


 999.1* 0.0742* 206 


0.2 



= 1 – 0.183 = 0.82
aw = 0.82 (206) = 168.9 m2/m3
1
kG
G
Q
 5.23[ M ]0.7 [
] 3 [ at d p ]  2
at DG
at  M
G DG
Compute kG
QM = air mass flow rate (kg/m2-s)
We have LM = 0.025 m3/s, there QM = 50 x 0.025 = 1.25 m3/s
Density of air = 1.2 kg/m3 at 10o C
QM = 1.25 * 1.2/1.478 = 1.01 kg/m2-s
G = 1.8 x 10-5 (kg/m-s), dp = 0.025 m


1.01
 5.23

 ( 206 )( 1.8 x105 ) 
( 206 )( 0.079 x104 )
kG
0.7


1.8 x105


 ( 1.2 )( 0.79 x104 ) 
1/ 3
( 206 * 0.025 )2 = 5.74
kG = 0.0093 m/s
Compute kL


999.1
kL 

 1.307 x10 3 x9.8 
kL [
1/ 3
(42.727) kL = 3.83
kL = 8.97 x 10-5 m/s
L 1 / 3
L
 L  0.5
]
 0.0051[ M ] 2 / 3 [
]
[ at d p ]0.4
aw L
 L DL
L g


16.9

 0.0051
 ( 168.9 )( 1.307 x10 3 ) 
2/ 3

1.307 x10 3

 ( 999.1 )( 6.13x10 10


) 
0.5
206 x0.0250.4
1
1
1
1
1




 11916
K L K H ' k G k L ( 0.140 )( 0.0093 ) 8.97 x10 5
Compute KL
KL = 8.4 x 10-5
Therefore KLa = 8.4 x 10-5 * 168.9 = 0.0142 (1/s) (assume that interfacial area = a = aw)
Compute HTU
Qw/KLaA
= 0.025/(0.0142)(1.478) = 1.19
Z = NTU* HTU= 4.39*1.19
take 20% SF =
1.2 * 5.22
= 5.22 m
= 6.26 m or 20.5 ft of packing
Pressure Losses
Use generalized Sherwood-Eckert Plot
x axis =
y axis =


where G
F
A

w
g
L
 L   A
 
 G  W




1/ 2
G2F
 A W g
= air loading rate (lb/ft2/sec) (= kg/m2•s x (2.2)/(3.28)3 = kg/m2•s x 0.2048)
= packing factor
= air density (lbf/ft3) (= kg/m3 x 0.06243)
= water density (lbf/ft3)
= gravitational constant (32.17 ft/sec2)
= liquid loading rate (lb/ft2/sec) (= kg/m2•s x 0.2048)
L = 0.0169 m3/m2-s = 24.9 gpm/ft2 = 24.9*8.33 (lb/gallon)/60
G = 1.01 * 0.2048
A = 1.2 kg/m3* 0.06243
w = 62.4 lbf/ft3
= 3.46 lb/ft2-s
= 0.207 lb/ft2-s
= 0.075 lbf/ft3
X = (3.46/0.207)(0.075/62.4)1/2
Y = (0.207)2(52)/(0.075)(62.4)(32.2)
= 0.58
= 0.015
Head losses through packing plate, duct works
From Figure 9-5, 0.3 ins per foot of packing
Head loss through packing = 0.3 x 20.5 = 6.15 ins.
Head loss through duct works is given by
p = kp (QG/A)2 = 0.093 (1.25 *3.28/1.478)2
Total Pressure drop
= 6.15 + 0.47
= 0.47 ins.
= 6.62 ins
Total Blower horsepower
= 144 *1.25 (3.28)3 * 60 [6.62*14.7/(34* 12)]/[33,000 * 0.4]
= 6.9 hp
Stripping of Ionizable compounds
•
•
•
equations that were developed earlier assume that the contaminants are non reacting with water or
does not undergo chemical reaction in water.
air stripping of compounds that react with water or undergo chemical reaction such as oxidation in the
aqueous phase require some modifications to the equations developed earlier
In the case of acid-base reactions, the pH of the solution determines the mass fraction of the various
species present. For example, ammonia exists as both as the dissolved gas species NH3 and the
+
•
+
ionized form NH4 . The ionized form NH4 cannot be stripped while dissolved gas form NH3 may be
stripped. The fraction of each species is given by the following equations.
Generalized equations for acid-base equations are represented by:
Monoprotic acid-base pair
HA <=> H+ + A-
diprotic acid-base pair
H2A <=> HA- + H+
HA- <=> A2- + H+
CT,A = [HA] + [A-]
CT,A = [H2A] + [HA-] + [A2-]
o 
1 
o 
[H ]
[ H  ]  K1
1 
K1

[ H ]  K1
2 
Note that [HA] = o CT
examples
NH4+ <==> NH3 + H+
[H  ]2
[ H  ] 2  K1 [ H  ]  K 2 K1
K1 [ H  ]
[ H  ] 2  K1 [ H  ]  K 2 K1
K1K 2
[ H  ] 2  K1 [ H  ]  K 2 K1
[A-] = 1CT
pK1 = 9.3
H2S <==> HS- + H+
HS- <==> S2- + H+
pK1 = 7.1
pK2 = 14
CT,NH3 = [NH4+] + [NH3]
CH3COOH <==> CH3COO- + H+
pK1 = 4.7
Assumptions in air stripping of ionizable compounds
• acid-base reactions are instantaneous, i.e., acid-base equilibrium is assumed to be achieved in the
mass transfer zone while the rate limiting step is the transfer of the dissolved gas species across the
film
• pH of the solution under air stripping conditions may change with depth depending on the buffering
capacity of the solution. As an approximation, one can assume that the pH is constant throughout the
column.
• A third assumption is that the overall mass transfer coefficient KL is independent of pH
• With the above three assumptions, we can rewrite our mass transfer expression:
M
= -KLa (CL - CL,eq) as in the case of ammonia
= - KLa (1CT,NH3 - 1CT,NH3,eq) = - 1KLa (CT,NH3 - CT,NH3,eq)