Standard Methods for Examination of Water & Wastewater_8 docx

9.2 INTERFACE FOR MASS TRANSFER, AND GAS AND LIQUID BOUNDARY LAYERS right-hand side in each of these figures represents the liquid phase as in the liquid phase of a droplet and the left

Aeration, Absorption, and Stripping

Aeration, absorption, and stripping are unit operations that rely on flow of massesbetween phases When a difference in concentration exists between two points in abody of mass, a flow of mass occurs between the points When the flow occursbetween two phases of masses, a transfer of mass between the phases is said tooccur This transfer of mass between phases is called mass transfer Examples ofunit 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 ofmiscible and volatile substances into individual components or groups of compo-nents The separation of water and alcohol into the respective components of liquidair into nitrogen, oxygen, and argon; and the separation of crude petroleum intogasoline, oil, and kerosene are examples of the distillation unit operation

Absorption is a unit operation that removes a solute mass or masses from a gasphase into a liquid phase Aeration of water dissolves air into it; thus, aeration isabsorption Another example of absorption is the “washing” of ammonia from anammonia-polluted air In this operation, ammonia is removed from the air by itsdissolution 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 theliquid into the gas phase

the solute condensing into its liquid phase The removal of water vapor in air bycondensation on a cold surface is dehumidification The reverse of dehumidification

is humidification In this unit operation, the flow of the solute is from the liquidphase 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 saturatedwith water vapor, the degree of this saturation being measured by the relative

mixture called the raffinate using a liquid solvent In this operation, the solventpreferentially dissolves the solute molecule to be extracted

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 liquidextraction Also, the fluid extractor may be a fluid or a gas For example, pollutantscan be leached out from solid wastes in a landfill as rain percolates down the heap

concentration forming crystals The driving force for the transfer of mass from liquid

9

into the solid phase is the affinity of the solute to form into a solid An example ofthis operation is the making of ice from liquid water The formation of snow fromwater 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 Becausethis 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 specificallyaeration

9.1 MASS TRANSFER UNITS

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

microor-as CO2 reacts with lime Thus, excess concentrations of this gas may be removedfrom the water by stripping or spraying the water into the air H2S is anothercompound that may be removed by stripping as benzene, carbon tetrachloride,p-dicholorobenzene, vinyl chloride, and trichloroethylene may also be removed bystripping The discussions that follow address the units or method used in aeration,absorption, and stripping

the air As the water is sprayed, droplets are formed This creates the condition forthe pollutant to transfer from the droplet phase to the air phase, in addition to thedirect liberation of the pollutant as the bulk mass of water breaks up into the smallersize 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 riffleplates At the air–water interface at the surface of the flowing water, the air transfersbetween 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 flowingwater The whole mass of water is aerated this way, because the mass of watertransported to the main body carries with it any air that was dissolved when it wasexposed 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 strippingdepends upon how fast the surface is renewed (i.e., how fast the water mass fromthe main body is transported to the surface for exposure to the air)

and allowed to trickle down the plates; the trickling water is met by a countercurrentflow of air The process creates a droplet phase and the air-gas phase inducing amass transfer between the droplets and the air Figure 9.2g is a spray tower Thewater is sprayed using spray nozzles at the top of the tower forming droplets Thesedroplets are then met by a countercurrent flow of air creating the two phase for masstransfer as in the case of the perforated plates Figure 9.2h is a cascade aerator ordeaerator as the case may be The cascade operates on the same principle as theinclined apron, only that this is more effective because of the steps

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 shearedinto 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 diffusercreates tiny bubbles Tiny bubbles are more effective for mass transfer, since the manybubbles produced create a large sum total areas for transfer Figure 9.3c is a surfaceaerator Water is drawn from the bottom of the aerator and sprayed into the air creatingdroplets, 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

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

Outlet Collecting pan (d)

Inlet conduit

Outlet conduit

Porous ceramic diffuser units (tubes)

Compressed air

Air bubbles

Manifold (b)

(c)

aluminum oxide This diffuser produces very tiny bubbles As indicated, the diffusersare mounted on rows of pipes Figure 9.5 shows an actual aeration in action Thishappens to be one of the activated sludge process tanks at the Back River WastewaterTreatment 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.

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 acolumn or tower equipped with a gas inlet and distributor at the bottom and aliquid inlet and distributor at the top It also consists of a liquid outlet at the bottomand 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 arecontacted by the gases flowing by Two phases and an interface between liquid andgas are therefore created inducing mass transfer

Packings are either dumped randomly into the tower or are stacked manually Dumpedpackings consists of units that are 0.6 cm to 5 cm in major dimensions; they aremostly utilized in small towers Stacked packings are 5 cm to 20 cm in majordimensions 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 indiameter are often stacked

9.2 INTERFACE FOR MASS TRANSFER, AND GAS

AND LIQUID BOUNDARY LAYERS

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 apartinitially 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 isequal to zero This surface is identified as the interface in the figure This figureshows the section cut across of the interface surface The line representing theinterface 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 mentionedpreviously, 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-slipcondition, 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 interfaceand 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 Withthe liquid phase considered as analogous to the plate, a gas phase boundary layer isformed; with the gas phase considered as analogous to the plate, the liquid phaseboundary layer is formed (see figures) These boundary layers are also called films

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.

Packed section

Liquid inlet

Liquid distributor

Liquid outlet

Gas inlet

© 2003 by A P Sincero and G A Sincero

Thus, the transfer of mass between the two phases must pass through the gas andliquid films In environmental engineering literature, the term film is normally used.The same discussions would apply to the stripping operation represented by Figure9.7b, with the difference that the direction of flow of mass transfer is from the liquidphase to the gas phase

In absorption operations, the concentration in the gas phase is larger in ison to the concentration in the liquid phase Thus, the flow of mass transfer is fromgas to liquid The reverse is true in the case of stripping, and the direction of masstransfer is from liquid to gas In other words, the liquid phase is said to be “stripped”

compar-of its solute component, decreasing the concentration compar-of the solute in the liquidphase 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 ofthe solute in the liquid phase and, of course, decreasing the concentration of thesolute in the gas phase

9.3 MATHEMATICS OF MASS TRANSFER

Between liquid and gas phases, the transfer of mass from one phase to the othermust pass through the interfacial boundary surface Call the concentration of thesolute 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.

phase boundary layer (a)

Liquid-Gas-phase boundary layer

phase boundary layer (b)

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.

Equilibrium curve

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

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

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 aresimilar 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 largelyindeterminate 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, thedestination media are the bulk contents of the tank; in a packed tower, the destinationmedia are the packings inside the tower; and in a trickling filter, the destination mediaare 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

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 activatedsludge 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 Callthe equilibrium concentration of oxygen in water at a particular temperature and

pressure as [C os ] This equilibrium concentration is also called the saturation

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.

(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)20θT−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, usingEquation (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 Arrheniustemperature relation

(9.12)

x*

C C C

From Eqs (9.9), (9.11), and (9.12),

(9.13)

As mentioned before, specification of aeration equipment requires the nation of SOR From Equation (9.13), this involves finding the values of the aerationparameters 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

determi-should be determined experimentally The literature reports values of (K L a) w,20 in theneighborhood of 2.5 per hour, α in the range of 0.7 to 0.9, and β in the range of0.9 to 1.0 AOR in the neighborhood of 1.40 kg/m3 ⋅ day has also been obtained Thedetermination of these parameters will addressed in the succeeding discussions

hour What is the value of (K L a) w at 25°C?

Solution:

(K L a) w = (K L a) w,20θT−20

Therefore,

Ans

acti-vated sludge reactor In order to do so, she performs a series of experiments obtainingthe following results: AOR = 1.30 kg/m3

⋅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/LTherefore,

 (1.024)25 −20

- 0.768 kg/m3

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 D ETERMINATION OF A ERATION P ARAMETERS

The environmental engineer specifies an aeration equipment based upon standardlaboratory tests performed by equipment manufacturers In other words, althoughthe engineer can easily determine the AOR, AOR must still be converted to SOR tomatch with the standard manufacturers value To perform this conversion, the α and

β parameters must be determined

fill it half with the sample The jar is then vigorously shaken to saturate the samplewith air or oxygen and the dissolved oxygen concentration measured Table 9.1

shows the saturated concentrations of dissolved oxygen in clean water exposed toone atmosphere barometric pressure at various temperatures From this table, at thetemperature corresponding to the temperature of the experiment, the saturation DOfor 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

β [C os,w]

C os

[ ] -

=

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 thepressure of the test Thus, during the performance of the experiment, the barometricpressure 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/m2

); 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/s2

; and R (gas constant for air) = 286.9 N ⋅ m/kg ⋅ °K As written, P b

has the unit of N/m2 The equation assumes that the temperature varies linearly withaltitude 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

determine [C os] depends upon the type of aeration device used For apron, cascadeand surface aerators and for spray and plate towers, the corresponding pressureshould be taken as barometric For aerators that are submerged below the surface ofwater 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)

pur-pose of determining the value of β of a particular wastewater The [C os,w] of thewastewater after shaking the jar thoroughly is 7.5 mg/L The temperature of thewastewater is 25°C Calculate β

T o

–

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 ofAllegheny 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 graphical map

topo-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 prevailingbarometric pressure is 761 mm Hg and the water temperature 20°C?

β 7.58.4 - 0.89

P b 101325 1 0.0065z

T o

–

Determination of αααα Integrating Equation (9.8) from t = 0 and = 0 to t = t

(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 ofaddends, 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

similar in form to Equation (9.8) It can therefore be manipulated to obtain anequation similar to Equation (9.21) This is shown below

C

C C

may now be solved as

(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 (Na2SO3) withcobalt chloride (CoCl2) added as a catalyst The sulfite converts to sulfate whenreacted 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 isnormally 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 spraytower 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 thecases 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 Theconcentration at the top of the cascade or tower would have to be zero; that at thebottom would have to be whatever is measured

wastewater The wastewater is to be aerated using a fine-bubble diffuser in theprototype aeration tank The laboratory diffuser releases air at the bottom of thetank 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 practicalpurposes, assume mass density of water = 1000 kg/m3

Assume an ambient ature of 25°C Calculate α

P b 101325 1 0.0065z

T o

–

P b 101325 1 0.0065 304.79( )

299.98 -–

0.0298 = (K L a)20(1.0246.5−20) K( L a)20 = 0.041 per min = 2.46 per hr

8.4 117,474.31101,325 -

40.74 - 2.53 per hr

Ans 9.5.3 C ALCULATION OF A CTUAL O XYGEN R EQUIREMENT , THE AOR

Consider the contents inside an activated sludge reactor laden with dissolved oxygenand pollutants subjected to aeration Also, consider the sedimentation basin thatfollows the reactor and all the associated pipings For the purpose of performing thematerial balance, let the boundaries of these items encompass the control volume.According to the Reynolds transport theorem, the total rate of increase of theconcentration of dissolved oxygen is equal to its partial (or local) rate of increaseplus 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 dissolvedoxygen in the effluent of the secondary sedimentation basin; and [ ] is theconcentration of dissolved oxygen in the wasted sludge [ ] is equal to zero Thelocal 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 ofincrease is

(9.28)

The total rate of increase is equal to the local rate of increase plus the convectiverate 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.

At steady state, the local derivative, (∂[ ]/∂t) , of the Eulerian derivative is

zero Also, as the wastewater enters the reactor, its dissolved oxygen content ispractically zero; thus, [ ] is equal to zero [ ], as it goes out of the secondarybasin must also be equal to zero Thus,

(9.30)The respiration rate is due to the consumption of substrate BOD which iscomposed 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 tive derivative is , where [ ] is the influent CBOD concentra-

convec-tion and [S] is the outgoing CBOD concentraconvec-tion from the control volume The

outgoing concentrations are those coming out from the effluent of the secondarybasin and the wasted sludge Note that the convective derivative is preceded by anegative sign The negative sign is used to precede it, since this derivative is aconvective rate of decrease, as distinguished from the convective rate of increasewhich has a positive sign preceding it

At steady state, the local derivative is equal to zero Thus, from the Reynoldstransport theorem, (Lagrangian derivative = to the Eulerian derivative):

r

r

r = ( )r s+( )r n C

The factor f s converts to its equivalent oxygen value The value of f s isnormally 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 theoxygen equivalent

Having determined the expressions for and , the equation for AOR isfinally

(9.37)

In the derivations of the equations above, was the volume of the control volumewhich is composed of the volume of the reactor, secondary clarifier, and the associatedpipings 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 substratesceases 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 tothe material balance is the volume of the reactor Therefore, may be consideredsimply as the volume of the reactor

as C5H7NO2 (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 O2per mg C5H7NO2 Therefore, f s = 1.42

The reduction in the concentration of nitrogen is also brought about by reactionwith oxygen for energy and for the requirement for synthesis NBOD is actually inthe form of NH3 Reacting with O2,

NH3 + 2O2→ HNO3+ H2O (9.39)From this reaction, the oxygen equivalent per mg NH3− N is 4.57 mg Thus, f N isequal 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) containapproximately 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