Standard Methods for Examination of Water & Wastewater_8 docx

Aeration, Absorption,
and Stripping

Aeration, absorption, and stripping are unit operations that rely on flow of masses
between phases. When a difference in concentration exists between two points in a
body of mass, a flow of mass occurs between the points. When the flow occurs
between two phases of masses, a transfer of mass between the phases is said to
occur. This transfer of mass between phases is called

mass transfer

. Examples of
unit operations that embody the concept of mass transfer are distillation, absorption,
dehumidification, liquid extraction, leaching, and crystallization.

Distillation

is a unit operation that separates by vaporization liquid mixtures of
miscible and volatile substances into individual components or groups of compo-
nents. The separation of water and alcohol into the respective components of liquid
air into nitrogen, oxygen, and argon; and the separation of crude petroleum into
gasoline, oil, and kerosene are examples of the distillation unit operation.

Absorption

is a unit operation that removes a solute mass or masses from a gas
phase into a liquid phase. Aeration of water dissolves air into it; thus, aeration is
absorption. Another example of absorption is the “washing” of ammonia from an
ammonia-polluted air. In this operation, ammonia is removed from the air by its
dissolution into the water.

The reverse flow of masses from the liquid phase into the gas phase is called

stripping

. In stripping, the solute molecule is removed from its solution with the
liquid into the gas phase.

Dehumidification

is the removal of a solute liquid vapor from a gas phase by
the solute condensing into its liquid phase. The removal of water vapor in air by
condensation on a cold surface is dehumidification. The reverse of dehumidification
is

humidification

. In this unit operation, the flow of the solute is from the liquid
phase evaporating into the gas phase. The end result of this movement is saturation
of the gas. For example, during heavy rains, the atmosphere may become saturated
with water vapor, the degree of this saturation being measured by the

relative
humidity

.

Liquid extraction

is the removal of a solute component from a liquid
mixture called the

raffinate

using a liquid solvent. In this operation, the solvent
preferentially dissolves the solute molecule to be extracted.

Leaching

is a unit operation where a solute molecule is removed from a solid
using a fluid extractor. This is similar to liquid extraction, except that the solute to
be removed comes from a solid rather than from a liquid as in the case of liquid
extraction. Also, the fluid extractor may be a fluid or a gas. For example, pollutants
can be leached out from solid wastes in a landfill as rain percolates down the heap.

Crystallization

is a unit operation where solute mass flows toward a point of
concentration forming crystals. The driving force for the transfer of mass from liquid
9

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420

into the solid phase is the affinity of the solute to form into a solid. An example of
this operation is the making of ice from liquid water. The formation of snow from
water vapor in the atmosphere is also a process of crystallization.
Of all the unit operations of mass transfer, this chapter will only discuss aeration,
absorption, and stripping. As mentioned, aeration is a form of absorption. Because

this operation plays a very important and significant role in water and waste-
water treatment, however, we will give it a separate heading and call it specifically
aeration.

9.1 MASS TRANSFER UNITS

The major purpose of dissolving air is to provide oxygen to be used by microor-
ganism in the process of wastewater treatment. This is exemplified by the aeration
employed in the activated sludge process. Aeration may also be employed for the
removal of iron and manganese from groundwaters. In the removal of hardness, the
presence of high concentrations of carbon dioxide may result in high cost for lime,
as CO

2

reacts with lime. Thus, excess concentrations of this gas may be removed
from the water by stripping or spraying the water into the air. H

2

S is another
compound that may be removed by stripping as benzene, carbon tetrachloride,
p-dicholorobenzene, vinyl chloride, and trichloroethylene may also be removed by
stripping. The discussions that follow address the units or method used in aeration,
absorption, and stripping.
Figure 9.1 illustrates how a pollutant may be stripped by spraying the water into
the air. As the water is sprayed, droplets are formed. This creates the condition for
the pollutant to transfer from the droplet phase to the air phase, in addition to the
direct liberation of the pollutant as the bulk mass of water breaks up into the smaller
size droplets. Figures 9.2a through 9.2d show the various types of nozzles that may

be used in sprays. Figure 9.2e is an inclined apron which may be studded with riffle
plates. At the air–water interface at the surface of the flowing water, the air transfers
between the water and air phases. The studding creates turbulence which, in aeration,
transports the water exposed at the surface to the main body or bulk of the flowing
water. The whole mass of water is aerated this way, because the mass of water
transported to the main body carries with it any air that was dissolved when it was
exposed at the surface. In stripping, the turbulent flowing water exposes the solute
at the surface triggering the process of stripping. The rate of aeration or stripping
depends upon how fast the surface is renewed (i.e., how fast the water mass from
the main body is transported to the surface for exposure to the air).
Figure 9.2f is a stack of perforated plates. The water is introduced at the top
and allowed to trickle down the plates; the trickling water is met by a countercurrent
flow of air. The process creates a droplet phase and the air-gas phase inducing a
mass transfer between the droplets and the air. Figure 9.2g is a spray tower. The
water is sprayed using spray nozzles at the top of the tower forming droplets. These
droplets are then met by a countercurrent flow of air creating the two phase for mass
transfer as in the case of the perforated plates. Figure 9.2h is a cascade aerator or
deaerator as the case may be. The cascade operates on the same principle as the
inclined apron, only that this is more effective because of the steps.

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421

Figures 9.3 and 9.4 show various types of aeration devices used in wastewater
treatment plants. Figure 9.3a is a turbine aerator with an air sparger at the bottom.
As the air emerges from the sparger, the larger bubbles that are formed are sheared
into small pieces by the turbine blade above. Figure 9.3b is a porous ceramic diffuser.
Because of the small openings through which the air passes, this type of diffuser

creates tiny bubbles. Tiny bubbles are more effective for mass transfer, since the many
bubbles produced create a large sum total areas for transfer. Figure 9.3c is a surface
aerator. Water is drawn from the bottom of the aerator and sprayed into the air creating
droplets, thus, aerating the water. Figure 9.4 shows a dome-type bubble diffuser.
The dome is porous, can have a diameter of 18 cm, and may be constructed of

FIGURE 9.1

Water spray.
Nozzle

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422

FIGURE 9.2

Nozzles for sprays and units for aeration or stripping: (a–d) nozzle types;
(e) inclined apron that may be studded with riffle plates; (f) perforated plates; (g) spray tower;
and (h) cascade.

FIGURE 9.3

Aeration units: (a) turbine aerator with an air sparger; (b) porous ceramic
diffuser; and (c) surface aerator.
Palm Beach
nozzle
(a)
Central

cone
Sacramento nozzle
(c)
Inlet conduit
Outlet
conduit
(e)
Perforated
pipe
discharging
water
downward
Perforated
pipe
discharging
air or gas
upward
Collecting pan and outlet
(g)
Removable
mouth piece
Berlin nozzle
Vanes
New York nozzle
Perforated
inlet pipe
Outlet
Collecting pan
(d)(b) (f)
Inlet conduit

Outlet
conduit
(h)
Rotor
Liquid level
Air
inlet
Bottom
(a)
Porous ceramic diffuser units (tubes)
Compressed
air
Air
bubbles
Manifold
(b)
(c)

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423

aluminum oxide. This diffuser produces very tiny bubbles. As indicated, the diffusers
are mounted on rows of pipes. Figure 9.5 shows an actual aeration in action. This
happens to be one of the activated sludge process tanks at the Back River Wastewater
Treatment Plant, Baltimore, MD.

FIGURE 9.4

Dome-type diffusers. (Courtesy of Aerocor Co.)

FIGURE 9.5

An activated sludge aeration tank at Back River wastewater treatment plant,
Baltimore, MD.

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424

A common device used in gas absorption and stripping is the packed tower,
the elevational section of which is shown in Figure 9.6i. The device consists of a
column or tower equipped with a gas inlet and distributor at the bottom and a
liquid inlet and distributor at the top. It also consists of a liquid outlet at the bottom
and a gas outlet at the top and a supported mass of solid shapes called

tower
packing

or

filling

.
The liquid trickles down through the packing while the gas goes up the packing.
The packing causes a thin film of liquid to be created on the surfaces which are
contacted by the gases flowing by. Two phases and an interface between liquid and
gas are therefore created inducing mass transfer.

Figures 9.6a to Figure 9.6h show the various shapes of packings used in practice.
Packings are either dumped randomly into the tower or are stacked manually. Dumped
packings consists of units that are 0.6 cm to 5 cm in major dimensions; they are
mostly utilized in small towers. Stacked packings are 5 cm to 20 cm in major
dimensions and are used in large towers. The spiral partition rings single, double,
and triple are stacked. The Berl and Intalox saddles, the Raschig, Lessing, and cross-
partition rings are normally dumped packings. Large Raschig rings 5 to 7 cm in
diameter are often stacked.

9.2 INTERFACE FOR MASS TRANSFER, AND GAS
AND LIQUID BOUNDARY LAYERS

Figure 9.7a shows the formation of boundary layers in absorption operations, and
Figure 9.7b shows the formation of boundary layers in stripping operations. The right-
hand side in each of these figures represents the liquid phase as in the liquid phase of
a droplet and the left-hand side represents the gas phase as in the gas phase of the air.
Consider the absorption operation. Imagine the two phases being far apart
initially. As the phases approach each other, a point of “touching” will eventually
be reached. This point then determines a surface; being a surface, its thickness is
equal to zero. This surface is identified as the interface in the figure. This figure
shows the section cut across of the interface surface. The line representing the
interface must have a zero thickness.
From fluid mechanics, when a fluid flows parallel to a plate, a boundary layer
is formed closed to the plate surface. At the surface itself, the velocity is zero relative
to the plate, because of the no-slip condition. Considering the two phases mentioned
previously, either the liquid or the gas may be considered as analogous to the plate.
Taking the liquid phase as analogous to the plate, the gas phase would be the fluid.
The interface would then represent the surface of the plate. Because of the no-slip
condition, the relative velocity of the phases parallel to the interface at the interface
is equal to zero. In other words, the two phases are “glued” together at the interface

and they do not move relative to each other.
As in the case of the fluid flow over a plate, a boundary layer is also formed. With
the liquid phase considered as analogous to the plate, a gas phase boundary layer is
formed; with the gas phase considered as analogous to the plate, the liquid phase
boundary layer is formed (see figures). These boundary layers are also called films.

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Aeration, Absorption, and Stripping

425
FIGURE 9.6

Various types of packings (left) and an elevational cross section of a packing tower (right): (a) cross-partition ring; (b) single-spiral ring;
(c) double-spiral ring; (d) triple-spiral ring; (e) Berl saddles; (f) Intalox saddles; (g) Raschig ring; (h) Lessing ring.
(a)
(e)
(b)
(f)
(c)
(g)
(d)
(h)
(i)
Packed
section
Packed
section
Gas

outlet
Liquid
inlet
Liquid distributor
Liquid
outlet
Gas
inlet

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426

Thus, the transfer of mass between the two phases must pass through the gas and
liquid films. In environmental engineering literature, the term

film

is normally used.
The same discussions would apply to the stripping operation represented by Figure
9.7b, with the difference that the direction of flow of mass transfer is from the liquid
phase to the gas phase.
In absorption operations, the concentration in the gas phase is larger in compar-
ison to the concentration in the liquid phase. Thus, the flow of mass transfer is from
gas to liquid. The reverse is true in the case of stripping, and the direction of mass
transfer is from liquid to gas. In other words, the liquid phase is said to be “stripped”
of its solute component, decreasing the concentration of the solute in the liquid
phase and increasing the concentration of the solute in the gas phase. In absorption,
the solute is absorbed from the gas into the liquid, increasing the concentration of

the solute in the liquid phase and, of course, decreasing the concentration of the
solute in the gas phase.

9.3 MATHEMATICS OF MASS TRANSFER

Between liquid and gas phases, the transfer of mass from one phase to the other
must pass through the interfacial boundary surface. Call the concentration of the
solute at this surface as [

y

i

] referred to the gas phase. The corresponding concen-
tration referred to the liquid phase is [

x

i

].[

x

i

] and [

y

i

] are the same concentration of

FIGURE 9.7

Formation of interfacial boundary layers: (a) absorption; (b) stripping.
Buld liquid-phase concentration
Bulk gas-phase concentration
Buld liquid-phase concentration
Bulk gas-phase concentration
[y] [x]
[y
i
]
[x
i
]
[y
i
]
[x
i
]
Interface Interface
Gas-phase
boundary
layer
Liquid-
phase

boundary
layer
(a)
Gas-phase
boundary
layer
Liquid-
phase
boundary
layer
(b)

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427

the solute only that they are referred to different basis; in effect, they are equal.
Because they are equal and because the thickness of the interface is zero,

x

i

and

y

i

must be in equilibrium with respect to each other.
Consider the process of absorption. If [

y

] is the concentration in the bulk gas
phase, the driving force toward the interfacial boundary is [

y

]

−

[

y

i

] and the rate of
mass transfer is

k

y

([

y

]

−

[

y

i

]), where

k

y

is the gas film coefficient of mass transfer.
For this rate of mass transfer to exist, it must be balanced by an equal rate of mass
transfer at the liquid film. The liquid phase mass transfer rate is

k

x

([

x

i

]

−

[

x

]), where

k

x

is the liquid film coefficient of mass transfer and [

x

] is the bulk concentration of
the solute in the liquid phase. Thus,

k

y

([

y

]

−

[

y

i

])

=



k

x

([

x

i

]

−

[

x

]) (9.1)
What is known is that [

x

i

] and [

y

i

] are in equilibrium. But, determining these
values experimentally would be very difficult. Thus, instead of using them, use [ ]
and [ ], respectively. [ ] is the concentration that [

x

] would attain if it were to
reach equilibrium value. By parallel deduction, [ ] is also the concentration that
[

y

] would attain if it were to reach equilibrium value. The corresponding driving

forces are now [

y

]

−

[ ] and [ ]

−

[

x

], respectively. To comprehend the physical
meaning of the driving forces, refer to Figure 9.8.
As shown, [

x

i

], [

y

i

] is on the equilibrium curve. The

equilibrium curve

is the
relationship between the concentration [

x

] in the liquid phase and the concentration
[

y

] in the gas phase when there is no net mass transfer between the phases. For any
given pair of values [

x

] and [

y

] in the liquid and gas phase, respectively, the point
([

x

i

], [

y

i

]) represents the “distance” that ([

x

], [

y

]) will have to “move” to attain their
equilibrium values concurrently at the equilibrium curve. This distance, represented
by the line segment ([

x

], [

y

]

→

[

x

i

], [

y

i

]), is the actual driving force for mass transfer;

FIGURE 9.8

Relationship among the various mole fractions.
x
*
y
*
x
*
y
*
y
*
x
*
[x
i
], [y]

[x
*
], [y]
[x], [y
*
]
[x]
[y]
[x
i
], [y
i
]
Operating line
Equilibrium curve

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428

however, as mentioned before, ([

x

i

], [

y

i

]) is impossible to determine experimentally.
Thus, locate the point ([

x

], [ ]) to view the surrogate driving force in the gas film
and locate point ([ ], [

y

]) to view the surrogate driving force in the liquid film.
As seen, [

y] − [ ] is greater than [y] − [y
i
]; however, it is not the true driving
force for transfer. Also [ ] − [x] is greater than [x
i
] − [x], but, again, it is not the
true driving force for transfer. When the transfer equation is written, however, it is
prefixed with a proportionality constant. This situation is therefore taken advantage
of by using a different proportionality constant for the case of the surrogate driving
forces. Thus, using K
y
as the proportionality constant for the gas-side mass transfer
equation in the surrogate situation,
(9.2)

On the liquid side, using K
x
as the proportionality constant,
(9.3)
K
y
and K
x
are called overall mass transfer coefficients for the gas and liquid sides,
respectively. To differentiate, k
y
and k
x
are called individual mass transfer coefficients
for the respective sides.
It is instructive to determine the equation relating the overall and the individual
mass transfer coefficients. Equation (9.2) may be rearranged to obtain
(9.4)
Replacing k
y
([y] − [y
i
]) using Equation (9.1),
(9.5)
Letting ([y
i
] − [])/([x
i
] − [x]) equal m,
(9.6)

A parallel derivation for K
x
yields
(9.7)
The coordinates [x
i
] and [y
i
] are the coordinates of the point ([x
i
], [y
i
]) on the
equilibrium curve. On the other hand, the coordinates [x] and [y] are coordinates of
y
*
x
*
y
*
x
*
k
y
yy
i
–()K
y
yy
*

–()=
k
x
x
i
[] x[]–()K
x
x
*
[] x[]–()=
1
K
y

y[] y
*
[]–
k
y
y[] y
i
[]–()

y[] y
i
[]–()( y
i
[] y
*
[])–+

k
y
y[] y
i
[]–()

1
k
y

y
i
[] y
*
[]–
k
y
y[] y
i
[]–()

+== =
1
K
y

1
k
y

y
i
[] y
*
[]–
k
x
x
i
[] x[]–()

+=
y
*
1
K
y

1
k
y

m
k
x

+=
1
K
x

1
k
x

1
mk
y

+
1
mK
y

==
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© 2003 by A. P. Sincero and G. A. Sincero
point ([x], [y]) representing the concentration [x] in the liquid phase and the concentration
[y] in the gas. Because the phases are not in equilibrium, ([x], [y]) is not on the
equilibrium curve. The various values of the pair ([x], [y]) in the liquid and gas phases
can be plotted; this plot is called an operating line (see figure). Any ([x], [y]) pair is
called an operating point. For the operating point ([x], [y]) the corresponding points
on the equilibrium curve are ([x], [ ]) and ([ ], [y]), based on the surrogate equation.
Point ([ ], [ ]) can only exist if ([x], [y]) is on the equilibrium curve.
The slope between ([x], [ ]) and ([x
i
], [y
i
]) is ([y
i

] − [])/([x
i
] − [x]); this is
equal to m. Thus, m is the slope of the equilibrium curve if it is a straight line.
Parallel derivations may be performed for the stripping operation; the results are
similar. The only difference is that there will be interchange of subscripts and
superscripts. Thus, the following analogous equations will be obtained: k
x
([x] − [x
i
]) =
k
y
([y
i
] − [y]), k
x
([x] − [x
i
]) = K
x
([x] − [ ]) and k
y
([y
i
] − [y]) = K
y
([ ] − [y]). Refer
to the figure to visualize that the mass flow is from the liquid phase to the gas phase.
9.4 DIMENSIONS OF THE OVERALL MASS

TRANSFER COEFFICIENTS
K
x
([ ] − [x]) is the mass of solute passing across the interfacial area per unit time per
unit cross-sectional area of the interface. The interfacial area of mass transfer is largely
indeterminate. To make sense of the expression, first define the term destination
medium. Destination medium is the overall entity inside the control volume from or
into which the gas and liquid phases flow. For example, in an aeration basin, the
destination media are the bulk contents of the tank; in a packed tower, the destination
media are the packings inside the tower; and in a trickling filter, the destination media
are the rocks that make up the filter.
Now, invent a parameter a that defines the area of interfacial area per unit bulk
volume of the mass transfer destination medium and form the expression K
x
a([ ] − [x]).
The dimension of K
x
a([ ] − [x]) is now mass per unit time per unit volume or
M/t ⋅ L
3
, where M is the dimension of mass, t is the dimension of time, and L is the
dimension of length. From this expression, the dimension of K
x
a is per unit time or
1/t. It will be shown later in this section on absorption towers, however, that the
dimensions of K
x
a will not be 1/t if mole fraction units are used for the concentra-
tions. Both K
x

and K
x
a are also called overall mass transfer coefficients based on
the liquid side. The corresponding overall mass transfer coefficients based on the
gas side are K
y
and K
y
a, respectively.
9.5 MECHANICS OF AERATION
Oxygen is a necessary nutrient. In suspended-growth processes, such as the activated
sludge process, air must be literally forced into the liquid. The air, thus, dissolved pro-
vides the necessary oxygen nutrient for the microorganism stabilizing the wastewater.
The basic process for oxygen mass transfer from air to water is absorption. Call
the equilibrium concentration of oxygen in water at a particular temperature and
y
*
x
*
x
*
y
*
y
*
y
*
x
*
y

*
x
*
x
*
x
*
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© 2003 by A. P. Sincero and G. A. Sincero
pressure as [C
os
]. This equilibrium concentration is also called the saturation con-
centration of the dissolved oxygen (DO) and this corresponds to [ ]. Let the
concentration in the water at any given moment be [ ]; this corresponds to x. The
driving force for mass transfer is then [C
os
] − [ ]. The rate at which the concentration
of oxygen will increase is (d[]/dt). In aeration, K
x
a is normally written as K
L
a.
Thus,
(9.8)
9.5.1 EQUIPMENT SPECIFICATION
The rating of aeration equipment is reported at standard conditions defined as 20°C,
one atmosphere pressure, and 0 mg/L of dissolved oxygen concentration in tap water
(or distilled water). Under these conditions, Equation (9.8) becomes
(9.9)
(K

L
a)
20
= the K
L
a at standard conditions and [C
os,20,sp
] = the saturation DO at 20°C
and standard pressure. This equation is the standard oxygen rate (SOR). Equipment
is specified in terms of SOR. Testing is not normally done at standard conditions,
so (K
L
a)
20
must be obtained from the K
L
a obtained at the condition of testing using
the Arrhenius temperature relation,
K
L
a = (K
L
a)
20
θ
T−20
(9.10)
where
θ
is the temperature correction factor, and T is the temperature in degrees

Celsius at testing conditions. This equation assumes that the effect of pressure on
K
L
a is negligible. In wastewater treatment, the value of
θ
is usually taken as 1.024.
The ability of aeration equipment to transfer oxygen at field conditions is, using
Equation (9.8),
(9.11)
where (K
L
a)
w
and [C
os,w
] are the K
L
a and the [C
os
] of the wastewater at field condi-
tions, respectively. This equation represents the actual oxygenation rate (AOR).
(K
L
a)
w
may also be expressed in terms of its value at 20°C, (K
L
a)
w,20
, by the Arrhenius

temperature relation
(9.12)
x
*
C
C
C
dC[]
dt

K
L
aC
os
[]C[]–()=
dC[]
dt

K
L
a()
20
C
os,20,sp
[]=
dC[]
dt

K
L

a()
w
C
os,w
[]C[]–()=
K
L
a()
w
K
L
a()
w,20
θ
T −20
=
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From Eqs. (9.9), (9.11), and (9.12),
(9.13)
where and .
As mentioned before, specification of aeration equipment requires the determi-
nation of SOR. From Equation (9.13), this involves finding the values of the aeration
parameters AOR,
β
, and
α
. The parameter
α
, in turn, requires the determination of

the K
L
a values. Each wastewater is unique in its characteristics, so these parameters
should be determined experimentally. The literature reports values of (K
L
a)
w,20
in the
neighborhood of 2.5 per hour,
α
in the range of 0.7 to 0.9, and
β
in the range of
0.9 to 1.0. AOR in the neighborhood of 1.40 kg/m
3
⋅ day has also been obtained. The
determination of these parameters will addressed in the succeeding discussions.
Example 9.1 The value of (K
L
a)
w,20
for a certain industrial waste is 2.46 per
hour. What is the value of (K
L
a)
w
at 25°C?
Solution:
(K
L

a)
w
= (K
L
a)
w,20
θ
T−20
Therefore,
Ans
Example 9.2 A chemical engineer proposes to purchase an aerator for an acti-
vated sludge reactor. In order to do so, she performs a series of experiments obtaining
the following results: AOR = 1.30 kg/m
3
⋅

day,
α
= 0.91, and
β
= 0.94. If the aeration
is to be maintained to effect a dissolved oxygen concentration of 1.0 mg/L at 25°C
in the reactor, what SOR should be specified to the manufacturer of the aerator?
Solution:
at 25°C = 8.38 mg/L
[C
os,20,sp
] = 9.17 mg/L
Therefore,
Ans

SOR
AOR
α
β
C
os
[]C[]–()
C
os,20,sp
[]

θ
T 20–

=
α
K
L
a()
w,20
K
L
a()
20

=
β
C
os,w
[]

C
os
[]

=
K
L
a()
w,25
2.46 1.024()
25−20
2.77 per hour==
SOR
AOR
α
β
C
os
[]C[]–()
C
os,20,sp
[]

θ
T −20

=
C[]
SOR
1.30

0.91
0.94 8.38()1.0–
9.17



1.024()
25−20

0.768 kg/m
3
==
TX249_frame_C09.fm Page 431 Friday, June 14, 2002 4:37 PM
© 2003 by A. P. Sincero and G. A. Sincero
Example 9.3 A civil engineer performs an experiment for the purpose of deter-
mining the value of
α
of a particular wastewater. (K
L
a)
20
and (K
L
a)
w,20
were found,
respectively, to be 2.46 per hour and 2.25 per hour. Calculate
α
.
Solution:

9.5.2 DETERMINATION OF AERATION PARAMETERS
The environmental engineer specifies an aeration equipment based upon standard
laboratory tests performed by equipment manufacturers. In other words, although
the engineer can easily determine the AOR, AOR must still be converted to SOR to
match with the standard manufacturers value. To perform this conversion, the
α
and
β
parameters must be determined.
Determination of
ββ
ββ
. The determination of
β
is very simple. Take a liter jar and
fill it half with the sample. The jar is then vigorously shaken to saturate the sample
with air or oxygen and the dissolved oxygen concentration measured. Table 9.1
shows the saturated concentrations of dissolved oxygen in clean water exposed to
one atmosphere barometric pressure at various temperatures. From this table, at the
temperature corresponding to the temperature of the experiment, the saturation DO
for clean water at one atmosphere barometric pressure can be obtained. This con-
centration is [C
os
]. From this, along with the saturation DO of the sample determined
in the experiment [C
os,w
],
β
may be calculated as
(9.14)

TABLE 9.1
Saturation DO in Distilled Water under One Atmosphere
of Atmospheric Pressure
Temperature(°C) DO(mg/L) Temperature(°C) DO(mg/L) Temperature(°C) DO(mg/L)
0 14.6 1 14.2 2 13.8
3 13.5 4 13.1 5 12.8
6 12.5 7 12.2 8 11.9
9 11.6 10 11.3 11 11.1
12 10.8 13 10.6 14 10.4
15 10.2 16 10.0 17 9.7
18 9.5 19 9.4 20 9.2
21 9.0 22 8.8 23 8.7
24 8.5 25 8.4 26 8.2
27 8.1 28 7.9 29 7.8
30 7.6
α
K
L
a()
w,20
K
L
a()
20

2.25
2.46

0.91===
β

C
os,w
[]
C
os
[]

=
TX249_frame_C09.fm Page 432 Friday, June 14, 2002 4:37 PM
© 2003 by A. P. Sincero and G. A. Sincero
If the test is being conducted at elevation other than zero or one atmosphere of
barometric pressure, the [C
os
] obtained from Table 9.1 must be corrected for the
pressure of the test. Thus, during the performance of the experiment, the barometric
pressure at the location should be recorded. Also, the barometric pressure at the test
location may be obtained using the barometric equation. This equation relates the
barometric pressure P
b
with altitude z in the troposphere, and from fluid mechanics,
this equation is
(9.15)
where P
bo
= barometric pressure at z = 0 (= 101325 N/m
2
); B (the temperature lapse
rate) = 0.0065° K/m for the standard atmosphere; T
o
= temperature in °K at z = 0;

g = 9.81 m/s
2
; and R (gas constant for air) = 286.9 N ⋅ m/kg ⋅ °K. As written, P
b
has the unit of N/m
2
. The equation assumes that the temperature varies linearly with
altitude. Therefore, if the elevation where the test is conducted and the temperature
at mean sea level are known, the corresponding barometric pressure can be found.
Because [C
os
] varies directly as the pressure and if the condition of test is at
barometric pressure, the [C
os
] at test conditions will be
(9.16)
where [C
os,sp
] is the [C
os
] from Table 9.1 at the standard pressure of P
s
= P
bo
=
760 mm Hg.
Pressures corresponding to [C
os
]. In sizing aerators, what pressure to use to
determine [C

os
] depends upon the type of aeration device used. For apron, cascade
and surface aerators and for spray and plate towers, the corresponding pressure
should be taken as barometric. For aerators that are submerged below the surface of
water such as the bubble-diffusion and turbine type aerators, since the point of release
of the air is submerged, the pressure must correspond to the average depth of
submergence. If the submergence depth is Z
d
, the corresponding average pressure P is
(9.17)
where
γ
is the specific weight of the water or the mixed liquor, in the case of the
activated sludge process tank. The equation for [C
os
] is then revised to
(9.18)
Example 9.4 An environmental engineer performs an experiment for the pur-
pose of determining the value of
β
of a particular wastewater. The [C
os,w
] of the
wastewater after shaking the jar thoroughly is 7.5 mg/L. The temperature of the
wastewater is 25°C. Calculate
β
.
P
b
P

bo
1
Bz
T
o
–


g/RB
101325 1
0.0065z
T
o
–


9.81/ 286.9()0.0065()[]
==
101325 1
0.0065z
T
o
–


5.26
=
C
os
[]

P
b
P
s

C
os,sp
[]=
PP
b
Z
d
2

γ
+=
C
os
[]
P
P
s

C
os,sp
[]=
TX249_frame_C09.fm Page 433 Friday, June 14, 2002 4:37 PM
© 2003 by A. P. Sincero and G. A. Sincero
Solution:
Therefore,

Ans
Example 9.5 An environmental engineer performs an experiment for the pur-
pose of determining the value of
β
of a particular wastewater in the mountains of
Allegheny County, MD. On a normal day, what is the prevailing barometric pressure,
if the temperature of the air on the Chesapeake Bay is 30°C?
Solution:
In Allegheny, z = 756 m above the Chesapeake Bay, consulting the county topo-
graphical map
Ans
Example 9.6 An activated sludge reactor is aerated using a turbine aerator
located 5.5 m below the surface of the mixed liquor. What is the pressurizing pressure
if the prevailing barometric pressure is 761 mm Hg and the water temperature 20°C?
Solution:
Ans
Example 9.7 An activated sludge reactor is aerated using a turbine aerator
located 5.5 m below the surface of the mixed liquor. What is [C
os
] if the prevailing
barometric pressure is 761 mm Hg and the water temperature 20°C?
Solution: From the previous example, P = 128,426.14 N/m
2
. From Table 9.1,
[C
os
] = 9.2 mg/L at 20°C and 1 atm pressure.
Therefore,
Ans
β

C
os,w
[]
C
os
[]

C
os
[] at 25°C 8.4 mg/L==
β
7.5
8.4

0.89==
P
b
101325 1
0.0065z
T
o
–


5.26
=
P
b
101325 1
0.0065 756()

30 273+
–


5.26
92,975 N/m
2
==
PP
b
Z
d
2

γγ
+ 997 1000()==
P 0.761 13.6()1000()9.81()
5.5
2



997()9.81()+ 128 426.14 N/m
2
,==
C
os
[] in reactor
128,426.14
101,325

9.2()11.66 mg/L==
TX249_frame_C09.fm Page 434 Friday, June 14, 2002 4:37 PM
© 2003 by A. P. Sincero and G. A. Sincero
Determination of
αα
αα
. Integrating Equation (9.8) from t = 0 and = 0 to t = t
and produces
(9.19)
In this equation, because [C
os
] is a constant, only one pair of values of t and
is needed to determine K
L
a. In practice, however, there can be several pair of these
values obtained from an experiment. One way to determine K
L
a is to plot the straight
line of the relationship between t and ln([C
os
] − [])/[C
os
]. The slope of this line
determines K
L
a. A more practical and easy method, however, is to average the t’s
and the (ln([C
os
] − [])/[C

os
])’s to obtain a single pair of value. This pair is then
used to solve for K
L
a.
The averaging for the t’s and (ln[C
os
] − [])/[C
os
])’s have the same number of
addends, their sums may be simply equated. This is shown below.
(9.20)
Solving for From K
L
a,
(9.21)
From this equation, K
L
a may be calculated, which can then be corrected to obtain
(K
L
a)
20
using the temperature relation. (K
L
a)
20
is one of the factors needed to calculate
α
.

The organisms in wastewater respire, so oxygen utilization must be incorporated.
Calling the respiration rate by , Equation (9.11) is modified to
(9.22)
This equation may be rewritten as
(9.23)
where is an apparent overall mass transfer coefficient. It encompasses both
(K
L
a)
w
, the true overall mass transfer coefficient, and .
The second part of Equation (9.23), , is
similar in form to Equation (9.8). It can therefore be manipulated to obtain an
equation similar to Equation (9.21). This is shown below.
(9.24)
C
CC=
C
os
[]C[]–
C
os
[]

ln K
L
at–=
C
C
C

C

C
os
[]C[]–
C
os
[]

ln


m
m=1
m=n
∑
K
L
at
m
m=1
m=n
∑
–=
K
L
a
∑
m=1
m=n

C
os
[]C[]–
C
s
[]

ln



m
∑
m=1
m=n
t
m

–=
r
dC[]
dt

K
L
a()
w
C
os,w

[]C[]–()r–=
dC[]
dt

K
L
a()
w
C
os,w
[]C[]–()r– K
L
′
a()
w
C
os,w
[]C[]–()==
K
L
′
a()
w
r
dC()/dt[]K
L
′
a()
w
C

os,w
[]C[]–()=

C
os,w
[]C[]–
C
os,w
[]

ln


m
m=1
m=n
∑

β
C
os
[]C[]–
β
C
os
[]

ln



m
m=1
m=n
∑
K
L
′
a()
w
t
m
m=1
m=n
∑
–==
TX249_frame_C09.fm Page 435 Friday, June 14, 2002 4:37 PM
© 2003 by A. P. Sincero and G. A. Sincero
may now be solved as
(9.25)
Having solved and for n values of [ ], Equation (9.23) may be written as
(9.26)
from which (K
L
a)
w
may be solved once is determined in a separate test. For a
constant , .
Thus,
(9.27)
(K

L
a)
w
may now be corrected to obtain (K
L
a)
w,20
using the Arrhenius temperature
relation. Once (K
L
a)
w,20
and (K
L
a)
20
are known,
α
can be computed.
The actual laboratory experimentation involves deaerating the sample, first. This
is done by consuming the dissolved oxygen using sodium sulfate (Na
2
SO
3
) with
cobalt chloride (CoCl
2
) added as a catalyst. The sulfite converts to sulfate when
reacted with the dissolved oxygen. From the stoichiometry of the reaction, 7.9 mg/L
of the sulfate is needed per mg/L of the dissolved oxygen. Ten to 20% excess is

normally used. Cobalt chloride has been used in concentration of 1.5 mg/L to act
as catalyst. As soon as the sample is completely deoxygenated, reaeration is allowed
to take place using the type of aeration system to be employed in the prototype such
as bubble-diffusion, turbine, surface-aeration, cascade, perforated plate, and spray
tower. The increase in dissolved oxygen concentration with respect to time is mon-
itored. The data obtained are then used to calculate the aeration parameters. In the
cases of cascades, perforated plate towers, and spray towers, the time may be taken
as the time it takes the mass of water or droplets to fall through the height. The
concentration at the top of the cascade or tower would have to be zero; that at the
bottom would have to be whatever is measured.
Example 9.8 A settling column 4 m in height is used to determine the
α
of a
wastewater. The wastewater is to be aerated using a fine-bubble diffuser in the
prototype aeration tank. The laboratory diffuser releases air at the bottom of the
tank. The result of the unsteady state aeration test is shown below. Assume
β
=
0.926, = 1.0 mg/L ⋅ h and the plant is 304.79 m above mean sea level. For practical
purposes, assume mass density of water = 1000 kg/m
3
. Assume an ambient temper-
ature of 25°C. Calculate
α
.
K
L
′
a()
w

(K
L
′
a)
w

∑
m=1
m=n

β
C
os
[]C[]–
β
C
os
[]

ln



m
∑
m=1
m=n
t
m

–=
K
L
′
a()
w
C
K
L
a()
w
β
C
os
[]C[]–()
m
r
m
m=1
m=n
∑
–
m=1
m=n
∑
K
L
′
a()
w

β
C
os
[]C[]–()
m
m=1
m=n
∑
=
r
r ∑
m=1
m=n
r
m
nr=
K
L
a()
w
K
L
′
a()
w
∑
m=1
m=n
β
C

os
[]C[]–()
m
nr+
∑
m=1
m=n
β
C
os
[]C[]–()
m

=
r
TX249_frame_C09.fm Page 436 Friday, June 14, 2002 4:37 PM
© 2003 by A. P. Sincero and G. A. Sincero
Solution:
Temperature at mean sea level = 25 + 0.0065(304.79) = 26.981°C ⇒ 299.98°K
Therefore,
Tap water at 6.5°C:
[C
os
] at 6.5°C and 1 atm ( = 101,325 N/m
2
) = 12.30 mg/L
[C
os
] at 6.5°C and 117,474.31 N/m
2

Tap Water at 6.5°C Wastewater at 25°C
Time (min) [ ] (mg/L) Time (min) [ ] (mg/L)
3 0.6 3 0.9
6 1.6 6 1.8
9 3.1 9 2.4
12 4.3 12 3.3
15 5.4 15 4.0
18 6.0 18 4.7
21 7.0 21 5.3
C
C
α
K
L
a()
w,20
K
L
a()
20

K
L
aK
L
a()
20
θ
T 20–

K
L
a()
w
K
L
a()
w,20
θ
T −20
== =
K
L
a
∑
m=1
m=n

C
os
[]C[]–
C
s
[]

ln



m

∑
m=1
m=n
t
m

K
L
a()
w
–
K
L
′
a()
w
∑
m=1
m=n
β
C
os
[]C[]–()
m
nr+
∑
m=1
m=n
β
C

os
[]C[]–()
m

==
K
L
′
a()
w

∑
m=1
m=n

β
C
os
[]C[]–
β
C
os
[]

ln



m
∑

m=1
m=n
t
m

–=
P
b
101325 1
0.0065z
T
o
–


5.26
PP
b
Z
d
2

γ
+==
P
b
101325 1
0.0065 304.79()
299.98
–



5.26
97,854.31 N/m
2
==
P 97,854.31
4
2

1000()9.81()+ 117,474.31 N/m
2
==
12.30=
117,474.31
101,325



14.26 mg/L=
TX249_frame_C09.fm Page 437 Friday, June 14, 2002 4:37 PM
© 2003 by A. P. Sincero and G. A. Sincero
Therefore,
Wastewater at 25°C:
[C
os
] at 25°C and 1 atm ( = 101325 N/m
2
) = 8.4 mg/L
[C

os
] at 25°C and 117,474.31 N/m
2

Therefore,
Therefore,
Time (min) [ ] (mg/L) ln [C
os
] −−
−−
[C]//
//
[C
s
]
3 0.5 −0.0357
6 1.7 −0.1269
9 3.1 −0.2451
12 4.4 −0.3690
15 5.5 −0.4872
18 6.1 −0.5582
21 7.1 −0.6889
∑ = 84 ∑ = −2.511
Time (min) [ ] (mg/L)
ββ
ββ
[C
os
] −−
−−

[] ln
ββ
ββ
[C
os
] −−
−−
[C]//
//
ββ
ββ
[C
os
]
3 0.9 8.12 −0.1049
6 1.8 7.22 −0.2224
9 2.4 6.62 −0.3091
12 3.3 5.72 −0.4553
15 4.0 5.02 −0.5858
18 4.7 4.32 −0.7360
21 5.3 3.72 −0.8855
∑ = 84 ∑ = 40.74 ∑ = −3.299
C
K
L
a
∑
m=1
m=n
C

os
[]C[]–
C
s
[]

ln



m
∑
m=1
m=n
t
m
–
2.511–
84
– 0.0298 per min===
0.0298 K
L
a()
20
1.024
6.5−20
() K
L
a()
20

0.041 per min 2.46 per hr===
8.4
117,474.31
101,325



= 9.74 mg/L=
β
C
os
[]0.926 9.74()6.35 mg/L==
C C
K
L
′
a()
w
∑
m=1
m=n
β
C
os
[]C[]–
β
C
os
[]

ln



m
∑
m=1
m=n
t
m

–
3.299–
84
– 0.393 per min 2.36 per hr====
K
L
a()
w
K
L
′
a()
w
∑
m=1
m=n
β
C
os

[]C[]–()
m
nr+
∑
m=1
m=n
β
C
os
[]C[]–()
m

=
2.36 40.74()7()1.0()+
40.74

2.53 per hr==
TX249_frame_C09.fm Page 438 Friday, June 14, 2002 4:37 PM
© 2003 by A. P. Sincero and G. A. Sincero
Ans
9.5.3 CALCULATION OF ACTUAL OXYGEN REQUIREMENT, THE AOR
Consider the contents inside an activated sludge reactor laden with dissolved oxygen
and pollutants subjected to aeration. Also, consider the sedimentation basin that
follows the reactor and all the associated pipings. For the purpose of performing the
material balance, let the boundaries of these items encompass the control volume.
According to the Reynolds transport theorem, the total rate of increase of the
concentration of dissolved oxygen is equal to its partial (or local) rate of increase
plus its convective rate of increase.
The convective rate of increase is equal to −Q
o

[] + (Q
o
− Q
w
)[ ] + Q
w
[],
where Q
o
is the inflow to the reactor; [ ] is the concentration of dissolved oxygen
in Q
o
; Q
w
is the outflow of wasted sludge; [ ] is the concentration of dissolved
oxygen in the effluent of the secondary sedimentation basin; and [ ] is the
concentration of dissolved oxygen in the wasted sludge. [ ] is equal to zero. The
local rate of increase is simply given by (
∂
[]/
∂
t).
The total rate of increase is (d[]/dt) , where is the volume of the system
which, as mentioned, is composed of the reactor, secondary basin, and the associated
pipings. From Equation (9.22), d[]/dt is given by the rate of aeration, (K
L
a)
w
([C
os,w

] −
[ ]) = AOR, and the rate of respiration by the organisms, . Thus, the total rate of
increase is
(9.28)
The total rate of increase is equal to the local rate of increase plus the convective
rate of increase, so the following equation is obtained:
(9.29)
Note that [ ] is equal to zero; thus, it is not appearing in Equation (9.29). It will
be recalled that the left-hand side of the equation, (d[]/dt) , is called the Lagrangian
derivative and the right-hand side exp-ressions, (
∂
[]/
∂
t) − Q
o
[] + (Q
o
− Q
w
)[ ],
are collectively called the Eulerian derivative.*
* As mentioned in the chapter, “Background Chemistry and Fluid Mechanics,” the Reynolds transport
theorem distinguishes the difference between the full derivative and the partial derivative. As stated in
that chapter, the environmental engineering literature is very confusing with respect to the use of these
derivatives. Some authors use the full derivative and some use the partial derivative to express the same
meaning.
2.53 K
L
a()
w,20

1.024
25−20
()K
L
a()
w,20
2.23 per hr==
α
2.23
2.46

0.91==
C
o
C
e
C
w
C
o
C
e
C
w
C
w
C
V
C
V

V
C
C
r
dC[]
dt

V
K
L
a()
w
C
os,w
[]C[]–()r–[]
V
AOR r–[]
V
==
dC[]
dt

V
V AOR r–[]
V
∂
C[]
∂
t

V
Q
o
C
o
[]– Q
o
Q
w
–()C
e
[]+==
C
w
C
V
C
V
C
o
C
e
TX249_frame_C09.fm Page 439 Friday, June 14, 2002 4:37 PM
© 2003 by A. P. Sincero and G. A. Sincero
At steady state, the local derivative, (
∂
[]/
∂
t) , of the Eulerian derivative is
zero. Also, as the wastewater enters the reactor, its dissolved oxygen content is

practically zero; thus, [ ] is equal to zero. [ ], as it goes out of the secondary
basin must also be equal to zero. Thus,
(9.30)
The respiration rate is due to the consumption of substrate BOD which is
composed of CBOD and NBOD. Let be the respiration due to CBOD and
be the respiration due to NBOD. is then
(9.31)
Now, apply the Reynolds transport theorem to the fate of the CBOD. As a counter-
part to [ ] in the case of dissolved oxygen, let [S] represent CBOD. The Lagrangian
rate of decrease (opposite to increase, thus, will have a negative sign) of CBOD is
−(d[S]/dt) . This decrease represents the consumption of CBOD substrates to pro-
duce energy by respiration, , and the consumption of the substrates for synthesis
or to replace dead cells, (syn)
s
. Thus,
(9.32)
For the Eulerian derivative, the local derivative is and the convec-
tive derivative is , where [ ] is the influent CBOD concentra-
tion and [S] is the outgoing CBOD concentration from the control volume. The
outgoing concentrations are those coming out from the effluent of the secondary
basin and the wasted sludge. Note that the convective derivative is preceded by a
negative sign. The negative sign is used to precede it, since this derivative is a
convective rate of decrease, as distinguished from the convective rate of increase
which has a positive sign preceding it.
At steady state, the local derivative is equal to zero. Thus, from the Reynolds
transport theorem, (Lagrangian derivative = to the Eulerian derivative):
(9.33)
(9.34)
is the equivalent amount of CBOD in the body of organisms in the
waste sludge. If is the concentration of organisms in the waste sludge,

where f
s
is the factor that converts the mass of microorgan-
isms in the waste sludge to the equivalent oxygen concentration . Therefore,
(9.35)
C
V
C
o
C
e
AOR r=
r
r()
s
r()
n
r
rr()
s
r()
n
+=
C
V
r()
s
dS[]
dt



V
– r()
s
syn()
s
+{}
V
=
(
∂
S[]/
∂
t)
V
–
( Q
o
S
o
[]– Q
o
S[]+ )–
S
o
dS[]
dt




V
– r()
s
syn()
s
+{}
V
Q
o
S
o
[]– Q
o
S[]+()–==
V
r()
s
Q
o
S
o
[] S[]–()
V
syn()
s
–=
V
syn()
s

X
u
[]
V
syn()
s
f
s
Q
w
X
u
[],=
V
syn()
s
V
r()
s
Q
o
S
o
[] S[]–()f
s
Q
w
X
u
[]–=

TX249_frame_C09.fm Page 440 Friday, June 14, 2002 4:37 PM
© 2003 by A. P. Sincero and G. A. Sincero
The factor f
s
converts to its equivalent oxygen value. The value of f
s
is
normally taken as 1.42; however, to do an accurate job for a given specific waste,
it should be determined experimentally.
By analogy with Equation (9.35), the respiration rate due to NBOD, may
be obtained from
(9.36)
where f
N
is the factor for converting nitrogen concentrations to oxygen equivalent,
[N
o
] and [N] are the nitrogen concentrations in the influent and effluent, respectively,
and f
n
is the factor for converting the nitrogen in the wasted sludge to the
oxygen equivalent.
Having determined the expressions for and , the equation for AOR is
finally
(9.37)
In the derivations of the equations above, was the volume of the control volume
which is composed of the volume of the reactor, secondary clarifier, and the associated
pipings. As the effluent from the reactor is introduced into the secondary clarifier, it
is true that microorganisms continue to respire. In the absence of aeration in the basin,
however, this respiration is but for a few moments and consumption of substrates

ceases. It is also true that there will be no respiration and consumption of substrates
in the associated pipings. Thus, the only volume of the control volume applicable to
the material balance is the volume of the reactor. Therefore, may be considered
simply as the volume of the reactor.
Determination of f
s
, f
N
, and f
n
. The formula for microorganisms has been given
as C
5
H
7
NO
2
(Mandt and Bell, 1982). To find the oxygen equivalent of the mass
synthesized, f
s
, react this “molecule” with oxygen as follows:
(9.38)
From this equation, the oxygen equivalent of the mass synthesized is 1.42 mg O
2
per mg C
5
H
7
NO
2

. Therefore, f
s
= 1.42.
The reduction in the concentration of nitrogen is also brought about by reaction
with oxygen for energy and for the requirement for synthesis. NBOD is actually in
the form of NH
3
. Reacting with O
2
,
NH
3
+ 2O
2
→ HNO
3
+ H
2
O (9.39)
From this reaction, the oxygen equivalent per mg NH
3
− N is 4.57 mg. Thus, f
N
is
equal to 4.57.
The nitrogen for synthesis goes with the sludge wasted, which can be expressed
in terms of the total Kjeldahl nitrogen (TKN). Bacteria (volatile solids, VS) contain
approximately 14 nitrogen. (Protein contains approximately 16% nitrogen.) Thus,
the equivalent oxygen of the nitrogen in the sludge wasted is 4.57(0.14)Q
w

[X
u
] =
0.64 Q
w
[X
u
] and the value of f
n
is 0.64.
Q
w
X
u
[]
r()
n
,
V
r()
n
f
N
Q
o
N
o
[] N[]–()f
s
Q

w
X
u
[]–=
Q
w
X
u
[]()
r()
s
r()
n
V
AOR()
V
r
V
r()
s
r()
n
+{}==
V
AOR()Q
o
S
o
[] S[]–()f
s

Q
w
X
u
[]– f
N
Q
o
N
o
[] N[]–()f
n
Q
w
X
u
[]–+=
V
V
C
5
H
7
NO
2
5O
2
5CO
2
2H

2
ONH
3
++→+
TX249_frame_C09.fm Page 441 Friday, June 14, 2002 4:37 PM
© 2003 by A. P. Sincero and G. A. Sincero
The total actual oxygen requirement, AOR, is therefore
(9.40)
This AOR is used to find SOR in order to size the aerator needed.
Example 9.9 The influent of 10,000 m
3
/day to a secondary reactor has a BOD
5
of 150 mg/L. It is desired to have an effluent BOD
5
of 5 mg/L, an MLVSS (mixed
liquor volatile suspended solids) of 3000 mg/L, and an underflow concentration of
10,000 mg/L. The effluent suspended solids concentration is 7 mg/L at 71% volatile
suspended solids content. The volume of the reactor is 1611 m
3
and the sludge is
wasted at the rate of 43.3 m
3
/day. Calculate the SOR. Assume the aerator to be of
the fine-bubble diffuser type with an
α
= 0.55; depth of submergence equals 2.44 m.
Assume
β
of liquor is 0.90. The influent TKN is 25 mg/L and the desired effluent

NH
3
− N concentration is 5.0 mg/L ⋅ f = 1.43. The average temperature of the reactor
is 25°C and is operated at an average of 1.0 mg/L of dissolved oxygen.
Solution:
Therefore,
AOR
1
V

Q
o
S
o
[] S[]–()1.42Q
w
X
u
[]– 4.57Q
o
N
o
[] N[]–()0.64Q
w
X
u
[]–+{}=
AOR
1
V

Q
o
S
o
[] S[]–()1.42Q
w
X
u
[]4.57Q
o
N
o
[] N[]–()0.64Q
w
X
u
[]–+–{}=
Q
w
X
u
[] 43.3 10,000 0.001()[]433.0kg/d==
AOR
1
1611

{10,000 0.15 0.005–()1.43()1.42 433.3()– 4.57 10,000()+=
× 0.025 0.005–( ) − 0.64(433.3)}= 2092.8 kg/d; AOR = 1.30
kg

m
3
⋅ d

SOR
AOR
α
β C
os
[]C[]–()
C
os,20,sp
[]

θ
T −20

=
At 25°C C
os,sp
[], 8.4 mg/L; C
os,20,sp
[]9.20 mg/L==
PP
b
Z
d
2

γ

assume plant is located at sea level;+=
P 101 325
2.44
2

997 9.81()[]+, 113,257.3 N/m
2
==
C
os
[]
P
P
s

C
os,sp
[]
113,257.3
101,325

8.4()9.39 mg/L== =
SOR
1.30
0.55
0.9 9.39()1.0–[]
9.20




1.024
25−20
()

2.62 kg/d ⋅ m
3
Ans==
TX249_frame_C09.fm Page 442 Friday, June 14, 2002 4:37 PM
© 2003 by A. P. Sincero and G. A. Sincero
9.5.4 TIME OF CONTACT
Having determined the overall coefficient of mass transfer (K
L
a)
w
, the differential
equation
(9.41)
may be integrated. Take note that the only variables in this equation are t and , all
the others are constant. Integrating from t = 0 to t = t and from [ ] = 0 to [ ] = [ ],
(9.42)
t is the time of contact for aeration or mass transfer. The apparatus must provide
this time if the liquid phase is to have the concentration of [ ]. For example, for a
spray tower, this time must be the time it takes for the droplets to fall from the top
of the tower to the bottom. For a cascade aerator, this time is the time for the water
to fall through the cascade from the top to the bottom.
9.5.5 SIZING OF AERATION BASINS AND RELATIONSHIP
TO CONTACT TIME
The sizing of aeration basins (for the activated sludge reactor, for example) are deter-
mined using the parameters of hydraulic detention time and mean cell retention time.
Hydraulic detention time is the average time that the particles of water in an inflow to

a basin are retained in the basin before outflow. Mean cell retention time, on the other
hand, is the average time that cells of organisms (not water) are retained in the basin.
A third parameter that is used not to size aeration basins but to design the aerators
used in the basin is the contact time (as derived previously). The size of the basin
must be such that it provides this time to effect the required length of time of contact
between the gas phase and liquid phase. There are situations where contact time for
aeration is equal to the hydraulic detention time. For example, in the case of the
trickling filter, as the water flows down the bed, a volume of this water occupies the
interstices of the bed. The total volume of these interstices multiplied by a factor to
account for the partial filling of the interstices divided by the inflow Q
o
is the
hydraulic detention time. Because contact between the water phase and the air also
occurs during the time that the water occupies the interstices, however, detention
time is equal to contact time.
In other situations, contact time is not equal to detention time. For example, in
the case of the activated sludge reactor, contact time is the rise time of the bubbles
to the surface, whereas detention time is the average time of travel of the water
between the inlet to the outlet of the reactor. Contact time refers solely to the bubbles
contacting the water during the time they were rising from the diffusers. Hydraulic
detention time, in this case, has no relation to the contact time.
dC[]
dt

K
L
a()
w
C
os,w

[]C[]–()r–=
C
C C C
t
1
K
L
a()
w

– ln
K
L
a()
w
C
os,w
[]C[]–()r–
K
L
a()
w
C
os,w
[]r–

1
K
L
a()

w

ln
K
L
a()
w
β
C
os
[]C[]–()r–
K
L
a()
w
β
C
os
[]r–

–==
C
TX249_frame_C09.fm Page 443 Friday, June 14, 2002 4:37 PM
© 2003 by A. P. Sincero and G. A. Sincero