doi: 10.

2166/9781789062304_0317

9
Aeration and mixing
Diego Rosso, Eveline I.P. Volcke, Manel Garrido-Baserba,
Coenraad Pretorius and Michael K. Stenstrom

9.1 INTRODUCTION
Chapter 9 Aeration and Mixing in the textbook Biological Wastewater Treatment: Principles, Modelling and
Design (Chen et al., 2020) introduces the fundamental quantities related to oxygen transfer and aeration, their
dynamics in relation to the biological process dynamics, the equipment required to provide aeration, and the
relation between mixing and aeration. Here we also present the energy implications of aeration and mixing.
This chapter applies all of this content through examples, questions and exercises.

9.2 LEARNING OBJECTIVES

After the successful completion of this chapter, the reader will be able to:

• Describe the equipment for aeration and mixing and its functioning.
• Size the equipment based on average and peak process conditions.
• Specify the number of aeration diffusers necessary to meet the oxygen requirement of a biological
process.
• Quantify the mixing effectiveness in a suspended-growth process.

9.3 EXAMPLES
Example 9.3.1
Aeration system design
A wastewater treatment plant requires an average mass of oxygen per day RO2 = 8,640 kgO2/d. The aeration
tanks are 5 m deep and the diffusers are to be installed 30 cm above the floor. The diffusers utilize 9 inch

© 2023 Diego Rosso. This is an Open Access book chapter distributed under a Creative Commons Attribution Non Commercial 4.0
International License (CC BY-NC-ND 4.0), (https://creativecommons.org/licenses/by-nc-nd/4.0/). The chapter is from the book Biological
Wastewater Treatment: Examples and Exercises, Carlos M. Lopez-Vazquez, Damir Brdjanovic, Eveline I.P. Volcke, Mark C.M. van Loosdrecht,
Di Wu and Guanghao Chen (Eds).

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 318

(22.86 cm) membrane disks (each with the area = 0.0373 m2) with a desired average specific airflow rate of
9.44 ·10-4 m3/s (corresponding to 2 SCFM, i.e., standard cubic feet per minute) per diffuser; for new diffusers,
this is specified by the manufacturer. The plant receives a daily peak flow rate of 1.3 times the average. The
alpha factor is assumed α = 0.35, which is a conservative (low) value typical for processes with only BOD
removal (for which α ranges between 0.25 - 0.45).

Once in operation, diffusers are characterized by an increased pressure drop (increasing DWP: dynamic
wet pressure) as well as by a reduced oxygen transfer efficiency, described by the parameters Ψ (pressure
factor, Eq. 9.1) and F (fouling factor, Eq. 9.2), respectively. For new diffusers, Ψ =1.0 and F =1.0; upon usage
Ψ > 1.0 and F < 1.0.

DWPnew _ diffuser
Ψ= (9.1)
DWPused _ diffuser

αSOTE new _ diffuser

F= (9.2)
αSOTE used _ diffuser

The blower has an efficiency of 75 % and is operated at an inlet temperature of 20 °C and an inlet pressure
of 0.9 atm.

a) Consider new diffusers. What would be the number of diffusers necessary to meet the aeration
requirements in process water? Calculate the number of diffusors based on average conditions as well
as based on peak conditions. Also, calculate the blower discharge pressure at peak airflow and the
corresponding peak blower power.
b) Consider fouled diffusers (Ψ = 1.3 and F = 0.8) but now keep the same number of diffusers as
calculated from new diffusers under peak conditions. What is the total airflow rate (AFR) and the
airflow rate per diffuser (AFRdiff) under peak conditions? Calculate again the blower discharge pressure
at peak airflow and the corresponding peak blower power.

Additional data: At standard conditions (20 oC and 1 atm) assume an air density of 1.225 kg/m3 and a ponderal
oxygen concentration in air of 23 %, i.e., 0.23 kg O2 per kg air.

Hint
Size the blower airflow rate (AFR, in m3/h) for the average load and for the peak load. Assume the diffusers
are new. The number of diffusers will be based on the peak load.

Additional information A1: estimation of SOTE from manufacturers’ clean water curves.
As an alternative to the design algorithm in Figure 9.35 in the textbook, the oxygen transfer efficiency at
standard conditions, SOTE, can be estimated from manufacturers’ clean water curves, when available. An
example is displayed in Fig. 9.1.

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 Aeration and mixing 319

11
Disc
Disc
10 Disc
Tube
Tube
9
SOTE/Z (%/m)

Tube
Panel
8 Panel

5
0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0
Flux (Nm3/m2.min)

Figure 9.1 Example of efficiency curves (expressed in % per unit depth) for various diffuser geometries over a range of air flux
(expressed as airflow per unit diffuser area).

The manufacturers’ clean water curves express the SOTE (%) per unit diffuser submergence Z (m) as a
function of the airflow rate per unit diffuser area (or air flux). The latter is expressed by Eq. 9.3:

AFR AFR diff

= (9.3)
N D ⋅ A diff A diff

in which AFR is the total airflow rate (m3/s), Adiff is the specific area of each diffuser (m2) and ND is the total
diffuser number (dimensionless). AFRdiff denotes the airflow rate per diffuser.

Using the clean water curves thus makes it possible to estimate the SOTE, given the desired airflow rate
specified by the manufacturer (AFRdiff, in m3/s), the diffuser specific area (Adiff, in m2) and the diffuser
submergence Z (m).

Additional information A2: blower curve, system curve, operating point

Each type of blower has a different blower curve, which is provided by the manufacturer and specifies the
blower discharge pressure as a function of airflow rate (see Figure 9.14 in the textbook Chen et al., 2020).

The discharge pressure of the blower (pdisch, relative to atmospheric pressure, in Pa) must always equal or
exceed the sum of the pressure losses in the system, to guarantee that the air is released (Eq. 9.4):

pdisch ≥ ρ·g·Z + hL (AFR) + DWP(AFR) (9.4)

in which g denotes the gravitational constant (g = 9.81 m/s2) and ρ the water density (kg/m3).

The pressure losses on the right-hand side of Eq. 9.4 indicate the hydrostatic head loss due to the diffuser
submergence, the head loss of the air distribution line (friction head hL, in Pa) and the dynamic wet pressure
(DWP, in Pa), i.e., the diffuser head loss, which is a function of the diffuser specific airflow rate and needs to

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be provided by the manufacturer. Figure 9.2 summarizes some DWP(AFR) curves from manufacturers, which
are typically provided for new diffusers.

For practical purposes, the line head loss hL may be considered constant, such that Eq. 9.4 simplifies to:

pdisch ≥ ρ·g·Z + hL + DWP(AFR) (9.5)

When diffusers foul, the DWP in Eq. 9.5 increases by a factor Ψ (>1, see Eq. 9.1) compared to new
diffusers.

The sum of the dynamic wet pressure, static head loss and friction losses in the air distribution line (i.e. the
right-hand-side of Eq. 9.5), make up the system curve (see Figure 9.14 in Chen et al., 2020). When the
aeration system is in operation, Eq. 9.5 becomes an equality and the pressure in excess of the equality results
in bubbles detaching from the diffuser with initial velocity > 0. The system curve indicates what the
requirement of the system is in terms of aeration pressure and needs to be compared against the blower curve
representing the pressure supply.

10

9 Disc
Disc
8 Disc
Tube
Tube
7 Tube
Panel
Panel
6
DWP (kPa)

0
0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0
Flux (Nm3/m2.min)

Figure 9.2 Examples of diffuser pressure drop (DWP, i.e., dynamic wet pressure) for a range of air fluxes.

The operating point is where the system curve and blower curve meet and should correspond with the
required airflow rate in the system. Should the airflow range of a certain manufacturer be insufficient, at a
given pressure requirement, a battery of identical blowers arranged in parallel can be selected. Indeed, the
AFR of parallel blowers is cumulative, while the pressure is the same.

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In fact, it is always necessary to make sure that the actual discharge pressure of the blower is higher than
the system needs. This is because when the pressure requirements of the system approach the maximum
discharge pressure of the blowers, the blowers begin to surge; this is manifested through vibrations of the
blowers, which may cause structural failure. The surge zone is an area of operation which therefore must be
avoided. An automated system shuts down the blowers when surging begins.

Additional information A3: blower brake horsepower (BHP)

The blower brake horsepower (BHP), in short ‘blower power’, represents the force needed to brake or stop the
blower motor. This is the minimum amount of power needed to operate the blower. It equals the mechanical
work that the blower performs to impart velocity to the air. It differs from the total wire power, for it does not
include the inefficiency of the electrical motor moving the blower.

The blower brake horsepower (BHP, in kW) is a function of the required air mass flow rate and is
calculated using the following adiabatic compression formula (Metcalf & Eddy, 2014):

 
n 
Wair ⋅ R ⋅ Tin  pdisch
BHP  − 1 (9.6)
29.7 ⋅n ⋅e  pin  
 
where:
Wair = air mass flow rate (kg/s)
R = universal gas constant (8.314 J/mol.K)
Tin = absolute inlet temperature (K)
pin = absolute inlet pressure (Pa)
pdisch,abs = absolute discharge pressure (Pa)
n = 0.283 for air
e = blower efficiency

As an alternative to Eq. 9.6, it is also possible to use manufacturer curves expressing BHP in terms of the
airflow rate (examples can be found on blower manufacturers’ websites).

Solution
Note: The calculations are provided in the spreadsheet ‘Chapter 9 Design examples.xlsx’ on sheet ‘Example
9.3.1’. The results are summarized in Table 9.1.

a) New diffusers
The number of diffusers needed to meet the aeration requirements is determined by the ratio between the
required airflow rate and the diffuser manufacturer’s recommended airflow rate per diffuser, the latter being
specified by the manufacturer (Eq. 9.7).

AFR
ND = (9.7)
AFR diff

The required airflow rate (AFR) can be derived from the amount of oxygen that needs to be fed to the
aeration tank, which is characterized by the mass flow of oxygen through the blowers, WO2 (kgO2/s) (Eq. 9.8):

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1 1
=
AFR WO2 ⋅ ⋅ (9.8)
ρair yˆ O2

where ρair is the air density (= 1.225 kg/m3) and ŷO2 is the weight fraction of oxygen in air (= 23 wt%).

The required oxygen mass flow rate, WO2, in its turn follows from the required oxygen transfer rate and the
oxygen transfer efficiency (Eq. 9.9):

SOTR
WO2 = (9.9)
αFSOTE

The required average oxygen transfer rate under standard conditions, SOTR, is calculated from the average
required oxygen mass flow, RO2 = 8,640 kgO2/d, taking into account a safety factor SF (Eq. 9.10):

SOTR = SF · RO2 (9.10)

The safety factor SF (e.g., 1.2 - 1.5) conservatively addresses the need for higher OTR at peak loading
times or seasons. As an alternative to calculating the SOTR with a safety factor, it would also be possible to
use the distribution of RO2 over time, as can be calculated by a dynamic simulator.

The average SOTR being 360 kgO2/h, applying a safety factor SF of 1.3 in this example results in a peak
oxygen SOTR of 468 kgO2/h.

αFSOTE is the oxygen transfer efficiency in standard conditions in process water, taking into account the
time the diffuser has been in operation. It is proportional to the oxygen transfer efficiency in standard
conditions in process water (SOTE) through the factors α and F, and can also be written in terms of the SOTE
per unit diffuser submergence Z (m) (Eq. 9.11):

αFSOTE = α · F · SOTE = α · F · SOTE/Z · Z (9.11)

SOTE/Z is estimated from the clean water curves (see Additional information A1). The disk diffuser
proposed in this example has an area of Adiff = 0.0373 m2 and a desired average specific airflow rate AFRdiffavg
9.44·10-4 m3/s, which corresponds to an air flux of 1.52 m3/m2.min. From the given manufacturers’ clean water
curve (Figure 9.1; blue line), we find a corresponding value of SOTE/Zavg of approximately 6.20 %/m. The
aeration tanks are 5 m deep and the diffusers are to be installed 30 cm above the floor, so the diffuser
submergence Z equals 5.0 - 0.30 = 4.70 m. Taking into account the given alpha factor α = 0.35 and assuming
new diffusers (F = 1.0), the corresponding αFSOTEavg (= αSOTEavg) is calculated from Eq. 9.11 as 10.2 %.

Taking into account SOTRavg = 360 kgO2/h and αFSOTEavg 10.2 %, the required oxygen mass flow for
average load conditions is calculated (Eq. 9.9) as WO2avg = 3,530 kgO2/h, which corresponds to a required
airflow rate (Eq. 9.8) of AFRavg = 12,528 m3/h.

The required number of diffusers based on average flow conditions is calculated using Eq. 9.7 as:

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 Aeration and mixing 323

AFR avg 12,528

=
N Davg = = 3,687 (9.12)
AFR diff 9.44 ⋅ 10−4 ⋅ 3,600
avg

Under peak flow conditions, the airflow rate per diffuser will increase according to the peak factor,
resulting in a specific airflow rate AFRdiffpeak = 1.23·10-3 m3/s, which corresponds to an air flux of
1.97 m3/m2.min. This implies a decreased SOTE compared to average flow conditions; from the given
manufacturers’ clean water curve (Figure 9.1; blue line), we find a corresponding value of SOTE/Zpeak of
approximately 6.0 %/m. The associated αFSOTEpeak is calculated from Eq. 9.11 as 9.9 %.

Taking into account SOTRpeak = 468 kgO2/h and αFSOTEpeak 9.9 %, the required oxygen mass flow (Eq.
9.9) under peak load becomes WO2peak = 4,742 kgO2/h. This corresponds to a required airflow rate (Eq. 9.8) of
AFR = 16,829 m3/h.

The required number of diffusers (Eq. 9.7) based on peak flow conditions becomes:

AFR peak 16,829

=
N D peak = = 3,810 (9.13)
peak
AFR diff 1.23 ⋅ 10−3 ⋅ 3,600

Note that the lower oxygen transfer efficiency (αFSOTE) under peak load conditions results in a higher
number of diffusers required. It is important to recognize that the number of diffusers calculated at peak
conditions, ND = 3,810, will be the total we will use from now on, so that a sufficient number of diffusers will
be installed in the tank for all operating conditions.

The procedure for calculating the number of diffusers is visualized in Figure 9.3; the resulting numerical
values for average load and peak load are summarized in Table 9.1.

Figure 9.3 Procedure for calculating the number of diffusers based on the load to be treated (oxygen demand RO2) and the
desired airflow rate per diffuser specified by the manufacturer (AFRdiff).

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The blower discharge pressure at peak airflow is calculated from Eq. 9.5, taking into account that the
blower discharge pressure should equal the pressure losses in the system (equality):

pdischpeak = ρ · g · Z + hL + DWPnew, diffuser (AFRpeak) (9.14)

From the given dynamic wet pressure (DWP) curves in Figure 9.2 (blue line), we can identify a DWP
(AFRpeak) value of approximately 3.2 kPa at the peak air flux of 1.97 m3/m2.min.

The diffuser submergence here is given as 4.70 m, which corresponds to 47.6 kPa (a 10-meter water
column corresponding to the atmospheric pressure 101.325 kPa). The line head loss (hL) can be assumed
conservatively to be 0.3 m of water column (3.04 kPa). In reality, there is a detailed procedure to calculate the
sum of the head loss from all the piping and valve elements, but because these are not known we must assume
a conservative round number.

As a result, the discharge pressure to specify a blower (i.e., the peak pressure requirements) is calculated
from Eq. 9.14 as:

pdischpeak = 47.6 + 3.04 + 3.20 = 53.84 kPa (9.15)

The blower brake horsepower (BHP) corresponding with the peak AFR is calculated using the adiabatic
compression formula (Eq. 9.6):

WO2
 peak n  ⋅ R ⋅ Tin  peak n 
Wair ⋅ R ⋅ Tin  pdisch,abs   yˆ O2

 pdisch,abs  
=BHP = peak
− 1 − 1=
29.7 ⋅ n ⋅ e  pin   29.7 ⋅ n ⋅ e  pin 
     
    (9.16)
4,742
⋅ 0.23 ⋅ 8,314 ⋅ 293.15  0.283 
3,600  53.86 + 101.325 
  − 1 kW = 359.kW
29.7 ⋅ 0.283 ⋅ 0.75  0.9 ⋅ 101.325  

b) Fouled diffusers – fixed number of diffusers

In the case of fouled diffusers, in order to keep the same desired airflow rate per diffuser (AFRdiff) as specified
by the manufacturer, it is necessary to recalculate the total number of diffusers based on the actual number
needed under peak conditions after fouling. The procedure for calculating the number of diffusers in the case
of fouling is completely analogous as for new diffusers (Eq. 9.7 and Figure 9.3); only a reduced value for the
oxygen transfer efficiency αFSOTE needs to be taken into account.

In the case of fouled diffusers, the oxygen transfer efficiency αFSOTE is decreased proportionally to the
fouling factor F = 0.8, resulting in αFSOTEavg = 8.2 % for average flow conditions and αFSOTEpeak = 7.9 %
for peak flow conditions. As a result, the airflow requirements are higher. The required oxygen mass flow for
average load conditions is now calculated (Eq. 9.9) as WO2avg = 4,412 kgO2/h, which corresponds to a required
airflow rate (Eq. 9.8) of AFRavg = 15,660 m3/h. The required oxygen mass flow for peak load conditions (Eq.
9.9) therefore becomes WO2peak = 5,927 kgO2/h, which corresponds to a required airflow rate (Eq. 9.8) of
AFRpeak = 21,037 m3/h. Note that the standard oxygen transfer rate SOTR, required for the calculation of WO2
(Eq. 9.9), remains unchanged compared to the case of new diffusers.

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However, in practice it is not possible to change the number of diffusers once installed, so the number of
diffusers is not changed with time. At best, the number of installed diffusers that discharge air could be
changed, in a so-called swing zone, could be changed. However, in this example, we will keep the number of
diffusers constant and we will consider that all diffusers are active at all times. The airflow rate and the
corresponding air flux per diffuser should be recalculated, since the number of diffusers was kept constant for
the peak conditions of new diffusers. In figures 9.1 and 9.2 we need to look for the SOTE/Z and DWP,
respectively, at this recalculated air flux per diffuser. The calculations are detailed below.

When keeping the number of diffusers constant, more air per diffuser must be discharged to compensate
for fouling. This means that, in addition to a decreased α (i.e., αF instead of α), there will be a decreased
SOTE/Z and thus a further decreased αFSOTE. The procedure for calculating the total airflow rate (AFR) and
the airflow rate per diffuser (AFRdiff) given the number of diffusers (ND) is depicted in Figure 9.4. In contrast
to the procedure for a fixed AFRdiff, the calculation needs to be performed iteratively. It will be illustrated here
for peak conditions, performing a single iteration.

In our example the number of diffusers under peak conditions was calculated as 3,810. The oxygen
transfer rate (SOTR) at peak conditions amounts to 468 kgO2/h. In order to calculate the corresponding oxygen
mass flow (Eq. 9.9), a value needs to be assumed for the oxygen transfer efficiency αFSOTE. As a first guess,
we take its value determined for fouled diffusers under peak conditions, αFSOTEpeak = 7.9 %, corresponding
with a required oxygen mass flow of WO2peak = 5,927 kgO2/h (Eq. 9.9) and a required airflow rate (Eq. 9.8) of
AFRpeak = 21,037 m3/h. The airflow rate per diffuser then follows from rearranging Eq. 9.7:

AFR 21,037 / 60
=
AFR diff = = 0.092 m3/min (9.17)
ND 3,810

which corresponds to an air flux per diffuser of AFRdiff/Adiff = 2.47 m3/m2.min for peak flow conditions. From
the given manufacturers’ clean water curve (Figure 9.1), we find a corresponding value of SOTE/Zpeak of
approximately 5.9 %/m. The associated αFSOTEpeak is calculated from Eq. 9.11 as 7.76 %, which is lower than
the value αFSOTEpeak = 7.9 % assumed initially (the ‘first guess’).

Consequently, the required oxygen mass flow is recalculated as WO2peak = 6,028 kgO2/h (Eq. 9.9) and the
corresponding required airflow rate (Eq. 9.8) as AFRpeak = 21,393 m3/h, which are higher than based on the
initial assumption for αFSOTE. This is because more air needs to be released by the same number of diffusers,
and therefore the diffusers will be operating further on the right of the efficiency and head loss curves.

The higher airflow rate per diffuser also results in a higher required blower discharge pressure. The
dynamic wet pressure (DWP) can still be read from Figure 9.2, even though it needs to be compensated for
fouling. Indeed, from the given dynamic wet pressure (DWP) curves in Figure 9.2, we can identify a DWP
(AFRpeak) value of approximately 3.8 kPa at the peak air flux of 2.47 m3/m2.min. The resulting blower peak
discharge pressure is calculated from Eq. 9.14 as 55.60 kPa.

pdischpeak = ρ · g · Z + hL + Ψ · DWPnew, diffuser (AFRpeak) = 47.6 + 3.04 + 1.3 · 3.80 = 55.58 Pa (9.18)

The higher oxygen requirement WO2 and the increased peak blower discharge pressure pdischpeak further
result in a higher required blower power BHPpeak compared to the case with a higher number of diffusers,
namely (Eq. 9.6):

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WO2
 peak n  ⋅ R ⋅ Tin  peak n 
Wair ⋅ R ⋅ Tin  pdisch,abs   yˆ O2

 pdisch,abs  
=BHP =peak
− 1 − 1=
29.7 ⋅ n ⋅ e  pin   29.7 ⋅ n ⋅ e  pin 
    
     (9.19)
6,028
⋅ 0.23 ⋅ 8,314 ⋅ 293.15  0.283 
3,600  55.58 + 101.325 
  − 1 kW = 467.kW
29.7 ⋅ 0.283 ⋅ 0.75  0.9 ⋅ 101.325  

The final note on diffuser flexing: flexing, also known as purging, is the practice of inflating diffusers by
feeding airflow higher than the usual maximum range (e.g., 130 – 150 % of maximum operating airflow), as
applied in this example. Manufacturers of membrane diffusers recommend this practice as prophylaxis to
dilate the pores and delay the ensuing fouling. To carry out flexing, the blower needs to be able to discharge a
larger airflow, with the corresponding DWP(AFR). Hence, if this practice is envisioned, the higher
DWP(AFR) and its corresponding pdisch and BHP can be calculated using the procedure above with the
adjusted AFR value. When the total required airflow rate increases because of fouling, it is always necessary
to verify that the airflow per diffuser does not fall outside the range of data guaranteed by the manufacturer. In
this example, if the daily peak conditions were, for example, 1.5 times the average, it can be quickly verified
that the air flux per diffuser would fall outside figs. 9.1 and 9.2 after fouling, if the number of diffusers used
were those from the peak conditions for new diffusers. The remedy is to reset the desired airflow per diffuser
to a lower value and repeat the calculation. A spreadsheet is an adequate way to perform these calculations;
however, when there is a desire to couple them with a dynamic biokinetic model, it is necessary to employ a
simulator.

Figure 9.4 Procedure for calculating the total airflow rate (AFR) and the airflow rate per diffuser (AFRdiff) given the load to be
treated (oxygen demand RO2) and the number of diffusers (ND). The differences with the procedure to calculate ND for given
(AFRdiff) are indicated in pink.

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 Aeration and mixing 327

The results are summarized in Table 9.1.

Table 9.1 Summary of the calculations and results for Example 9.3.1. The calculations are provided in the spreadsheet ‘Chapter
9 Design examples.xlsx’ on sheet ‘Example 9.3.1’.

New New Fouled Fouled Unit

diffusers diffusers diffusers diffusers
F = 1.0 F = 1.0 F = 0.8 F = 0.8
Ψ = 1.0 Ψ = 1.0 Ψ = 1.3 Ψ = 1.3
ND constant ND constant
Parameter / Conditions Average Peak Average Peak
Required oxygen (RO2) 8,640.00 11,232.00 8,640.00 11,232.00 kgO2/d
Peak factor (SF) 1.30 1.30 -
SOTR 360.00 468.00 360.00 468.00 kgO2/d
⍺ 0.35 0.35 0.35 0.35 -
F 1.00 1.00 0.80 0.80 -
Desired AFR per diffuser (AFRdiff) 0.06 0.07 0.06 0.07 m3/min
Diffuser specific area (Adiff) 0.04 0.04 0.04 0.04 m2
3 2
Air flux per diffuser (AFRdiff / Adiff) 1.52 1.97 1.52 1.97 m /m .min
SOTE/Z 6.20 6.00 6.20 6.00 %/m
⍺SOTE/Zavg 5.00 5.00 5.00 5.00 %/ft
⍺SOTE/Zavg 0.30 0.30 0.30 0.30 %/m
Side water depth (SWD) 5.00 5.00 5.00 5.00 m
Diffuser height above floor 0.30 0.30 0.30 0.30 m
Diffuser submergence Z 4.70 4.70 4.70 4.70 m
⍺FSOTE 10.20 9.87 8.16 7.90 %
Oxygen supplied (WO2) 3,530.76 4,742.64 4,412.20 5,927.05 kgO2/h
Air density (⍴air) 1.23 1.23 1.23 1.23 kg/m3
ŷO2 0.23 0.23 0.23 0.23 kg/kg
Air flow rate (AFR) 12,528.98 16,829.25 15,660.97 21,037.56 m3/h
Number of diffusers (ND) 3,687.84 3,810.74 3,810.74 3,810.74 -
Actual AFRdiff 0.069 0.092 m3/min
Actual air flux per diffuser 1.84 2.47 m3/m2.min
Actual SOTE/Z 6.05 5.90 %/m
⍺FSOTE 7.96 7.76
Actual WO2 4,522.59 6,028.51 kgO2/h
Actual AFR 16,048.24 21,393.12 m3/h
Hydrostatic head 47.62 47.62 kPa
Air piping head loss 3.04 3.04 kPa
DWPnew,diffuserpeak 3.20 3.80 kPa
Ψ 1.00 1.30
Discharge pressure required (pdischpeak) 53.86 55.60 kPa
Blower efficiency e 0.75 0.75 -
Blower inlet temperature (Tin) 293.15 293.15 K
Blower inlet pressure (pin) 91.19 91.19 kPa
Absolute blower discharge pressure (pdisch,abs) 155.19 156.93 kPa
BHPpeak 359.50 467.33 kW

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Example 9.3.2
Diffuser specification
A wastewater treatment plant consisting of three compartments (3 CSTRs) in series needs new fine-pore
diffusers. The OUR breakdown in the three compartments is as follows: 50 %, 35 %, 15 %, corresponding
with a progressively lower oxygen requirement along the treatment train. The total airflow in the series of 3
reactors spans the range 6,800 - 13,600 m3/h.

Calculate the number of diffusers for the three aeration grids for the following cases:

a) Silicone tubes, 1,000 mm long, 50 mm in diameter, with an operating range of 7 - 14 m/min.

b) EPDM discs, 9 in in diameter (229 mm), with an operating range of 3.5 - 5.3 m/min.
c) Polyurethane panels, 200 mm wide, 3.6 m long, with an operating range of 1.75 - 3.50 m/min.

Solution
Since the required airflow rate is given, the number of diffusers can be calculated from Eq. 9.7. The number of
diffusers corresponding with the minimum airflow rate per diffuser at minimum airflow conditions is
calculated as:

AFR MIN
N MIN
D = MIN
(9.20)
AFR diff

while the number of diffusers corresponding with the maximum airflow rate per diffuser at maximum airflow
conditions is calculated as:

AFR MAX
N MAX
D = MAX
(9.21)
AFR diff

The actual number of diffusers, ND, should be lower than or equal to N MIN
D in order to ensure that the
airflow per diffuser remains higher than the minimum boundary of its operating range; at the same time ND,
should be higher than or equal to N MAX
D such that the airflow per diffuser remains lower than the maximum
boundary of its operating range:

N MAX
D ≤ ND ≤ ND
MIN
(9.22)

The minimum and maximum airflow are given as AFRMIN = 6,800 m3/h and AFRMAX = 13,600 m3/h. The
minimum and maximum airflow rate per diffuser can be calculated from the specified minimum and maximum
air flux for each type of diffuser, taking into account the diffuser-specific area Adiff, which in its turn is
determined by the specified diffuser geometry and dimensions. The results are summarized in Table 9.2. The
calculations are provided in the spreadsheet ‘Chapter 9 Design examples.xlsx’ on sheet ‘Example 9.3.2’.

Given that the minimum and maximum airflow rate per diffuser for the silicone tubes (case a) and the
polyurethane panels (case c) are perfectly aligned with the range of total airflow rates that needs to be
provided, the number of diffusers N MIN
D and N MAX
D is the same. The number of diffusers is found to be

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 Aeration and mixing 329

ND = 324 for the silicone tubes and ND = 90 for the polyurethane panes. Their distribution over the three
compartments is performed proportionally to the required OUR. As for the EPDM discs, it is found that
N MIN
D = 780 and N MAX
D = 1,040. As N MAX
D > N MIN
D for this case, it is not possible to satisfy both the
minimum and maximum airflow rate per diffuser requirements for the given range of airflow rates that needs
to be provided. As a result, this type of diffuser is not an option for the WWTP under study.

Table 9.2 Summary of the results for Example 9.3.2. The calculations are provided in the spreadsheet ‘Chapter 9 Design
examples.xlsx’ on sheet ‘Example 9.3.2’.

Silicone tubes EPDM discs Polyurethane panels Unit

AFRMIN 6,800 6,800 6,800 m3/h
AFRMAX 13,600 13,600 13,600 m3/h
Air fluxMIN 7.0 3.5 1.75 m/min
Air fluxMAX 14.0 5.3 3.50 m/min
Adiff 0.05 0.042 0.72 m2
AFRdiffMIN 21.00 8.7 75.6 m3/h
AFRdiffMAX 42.00 13.1 151 m3/h
NdiffMIN 324 780 90
NdiffMAX 324 1,040 90
Ndiff 1 162 45
Ndiff 2 114 32
Ndiff 3 49 14

Example 9.3.3
Aeration energy dynamics and costs
The costs for electrical energy (power) to operate the blowers varies during the day (Figure 9.5), as does the
energy demand of the WWTP itself, the latter corresponding to the load variations.

0.12
Total cost
Generation cost
Delivery cost
0.09
Costs (USD/kWh)

0.06 0.08 0.12 0.08

USD/kWh USD/kWh USD/kWh USD/kWh
0.06

0.03

0.00
23:00 08:00 12:00 18:00 23:00
Time (hh:mm)

Figure 9.5 Daily variation in power generation and delivery costs, and the resulting total energy cost.

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Reconsider the case from Example 9.3.1, for new diffusers. Calculate the power demand and associated
power costs during the day, as well as the annual power cost to operate the aeration system. Use the case of
new diffusers. To calculate costs, use the tariff structure from Figure 9.5. Assume that the period with the
highest total power cost (USD0.12/kWh, 12:00-18:00) requires peak aeration flow rate, the periods with
medium power cost (USD0.08/kWh, 8:00-12:00 and 18:00-23:00) corresponds to the average aeration flow
rate, and the period with the lowest total power cost (USD0.06/kWh, 23:00-8:00) is also the one in which only
a low aeration flow rate is required, i.e., during the night. As for the low (nightly) aeration flow rate, assume
that it equals the average airflow divided by the safety factor SF = 1.3).

Solution
The blower power demand (BHP) is calculated using the adiabatic compression formula (Eq. 9.6). The BHP
corresponding with the peak AFR has already been calculated in Example 9.3.1 as BHPpeak = 359 kW (Eq.
9.16); the BHP for low and average flow conditions is calculated analogously as BHPlow = 206 kW and
BHPavg = 268 kW, respectively. The power costs for each time interval are obtained by multiplying the
duration of each time interval (h) with the corresponding BHP value (kW) and the corresponding energy cost
(USD/kWh); the summation gives the power cost per day. The results are summarized in Table 9.3. The
calculations are provided in the spreadsheet ‘Chapter 9 Design examples.xlsx’ on sheet ‘Example 9.3.3’. It is
striking how the high-power cost during the peak period (12:00-18:00) substantially increases the overall
power costs: this period counts for a quarter of the day, but half of the aeration costs.

Table 9.3 Summary of the results for Example 9.3.3. The calculations are provided in the spreadsheet ‘Chapter 9 Design
examples.xlsx’ on sheet ‘Example 9.3.3’.

New diffusers New diffusers Unit

F=1.0 Ψ=1.0 F=1.0 Ψ=1.0
Parameter / Conditions Low-load Average Peak
Peak factor (SF) 1.30 -
Diffuser submergence Z 4.70 4.70 4.70 m
Oxygen supplied (WO2) 2,715 3,530 4,742 kgO2/h
Air density (⍴air) 1.225 1.225 1.225 kg/m3
ŷO2 0.23 0.23 0.23 kg O2/kg air
Air flow rate (AFR) 9,637 12,528 16,829 m3/h
Discharge pressure required (pdisch) 53.86 53.86 53.86 kPa
Absolute blower discharge pressure (pdisch,abs) 155.19 155.19 155.19 kPa
Blower efficiency e 0.75 0.75 0.75
Blower inlet temperature (Tin) 293.15 293.15 293.15 K
Blower inlet pressure (pin) 91.19 91.19 91.19 kPa
BHP 206 268 360 kW
Unit cost (23:00 - 08:00) 0.06 (duration = 9 h) USD/kWh
Unit cost (08:00 - 12:00) 0.08 (duration = 4 h) USD/kWh
Unit cost (12:00 - 18:00) 0.12 (duration = 6 h) USD/kWh
Unit cost (18:00 - 23:00) 0.08 (duration = 5 h) USD/kWh
Cost (23:00 - 08:00) 111.2 (@BHPmin) USD/d
Cost (08:00 - 12:00) 85.6 (@BHPavg) USD/d
Cost (12:00 - 18:00) 258.9 (@BHPpeak)* USD/d
Cost (18:00 - 23:00) 107.1 (@BHPavg) USD/d
Total daily cost 562.7 USD/d
Total yearly cost 205,390 USD/yr
*Note that peak cost is roughly half of total

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Example 9.3.4
Mixing and energy
Quantify the minimum power requirement to maintain mixed liquor suspended with a mechanical mixer, and
the minimum airflow requirement to maintain the same mixed liquor suspended with bubbles.

Additional information A4: mixer design

Mixer design is typically based on a target power level, which may be expressed in terms of W/m3 for
mechanical mixers or air flux, Nm3/h.m2, for air-based mixing. The power level method quantifies mixing
from the mixing power input, regardless of the distribution of velocity gradients. A disadvantage of the power
level or air flux approach is that designers often treat homogeneity as the only goal of mixing, so that power
levels are often increased to achieve ever higher levels of homogeneity. However, overmixing also causes
disadvantages, from shearing biological flocs to using too much air, because mixing results in high DO
concentrations and this can compromise BNR process performance. Another consideration is that the power
level method makes no distinction between different impellers. This means that specifying a mixer in this way
creates a strong incentive for the equipment vendor to supply a cheap and inefficient impeller. Refer to
Grenville et al. (2017) for comprehensive descriptions of the unique characteristics of each impeller.

As an alternative to the power level method, mixer design can be based on flocculation, characterized by
the velocity gradient G as a key design parameter. Wahlberg et al. (1994) developed a method for this
approach and provided a set of experimental data, which can be used to determine best, average and worst-
case design requirements. More specifically, for any given HRT there is an optimum G, where flocculation is
just completed, and the minimum possible effluent turbidity is attained. Combining this method with the
dataset, an average, minimum (for the best performing flocculating system) and maximum (for the worst
flocculating system) G value can be determined, as given in Table 9.4.

Table 9.4 Example of optimum velocity gradient G for flocculation (adapted from Wahlberg et al., 1994).

Velocity gradient G (1/s)

HRT (min) Plug flow Completely mixed
Min. Avg. Max. Min. Avg. Max.
5 4.32 10.4 16.3 4.6 12.3 22.8
10 2.89 7 10.7 3.5 9.4 16.8
20 1.82 4.5 9 2.7 7.1 12.2
30 1.35 3.4 7.7 2.2 6 10.1
45 1 2.5 6.4 1.9 5 8.4
60 0.8 2 5.5 1.6 4.4 7.3

It is clear from Table 9.4 that there is considerable variation between the best, average and worst
flocculation systems in terms of optimum G value, suggesting that there is considerable benefit in doing site-
specific testing to determine the flocculation characteristics of a specific system prior to design. The plug flow
system has a lower optimum G value compared to the completely mixed system. Further optimization of the
design is possible if a tapered velocity gradient is employed in a plug flow system, or by placing a number of
completely mixed zones in series with lower G values in the downstream zones.

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From the optimal G values (1/s) for a given system (Table 9.4), the corresponding power levels at 20 oC
for mechanical mixers can be calculated using the definition of G for the mixing power Pmix (W) per unit liquid
volume V (m3) (Metcalf and Eddy, 2014):

1/2
P 
G =  mix  (9.23)
µ
 w 
V

Pmix
= G 2 ⋅ µw (9.24)
V

in which µw is the dynamic viscosity of water (µw = 1.000510-3 N.s./m2).

The G values from Table 9.4 can also be used to calculate the air flux required for mixing with air-
powered mixers. The G values correspond to mixing power Pmix (kW), calculated through Eq. 9.24, which in
the case of air mixing is the power corresponding to the isothermal work of expansion of the rising bubbles
(Schroeder, 2021):

Pmix  Z + 10.33 
=⋅
K AFR ⋅ ln   (9.25)
V  10.33 

Pmix
AFR = V (9.26)
 Z + 10.33 
K ⋅ ln  
 10.33 

Where K is a constant (1.689 in SI units) and AFR (Nm3air/ m3reactor.min) is the required airflow through the
diffusers installed at a submergence Z (m).

Solution
The minimum power requirements Pmix/V to maintain mixed liquor suspended with a mechanical mixer
correspond to the optimum G values for flocculation from Table 9.4 and are calculated from Eq. 9.24. The
results are summarized in Table 9.5. Please note that the units are W/m3.

Table 9.5 Power levels for optimum flocculation.

Power level (W/m3reactor)

HRT (min) Plug flow Completely mixed
Min. Avg. Max. Min. Avg. Max.
5 1.87E-02 1.08E-01 2.66E-01 2.12E-02 1.51E-01 5.20E-01
10 8.36E-03 4.90E-02 1.15E-01 1.23E-02 8.84E-02 2.82E-01
20 3.31E-03 2.03E-02 8.10E-02 7.29E-03 5.04E-02 1.49E-01
30 1.82E-03 1.16E-02 5.93E-02 4.84E-03 3.60E-02 1.02E-01
45 1.00E-03 6.25E-03 4.10E-02 3.61E-03 2.50E-02 7.06E-02
60 6.40E-04 4.00E-03 3.03E-02 2.56E-03 1.94E-02 5.33E-02

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 Aeration and mixing 333

Note: A key difference between the laboratory-scale flocculation test and full-scale applications is that at
the smaller scale the G value can be assumed to be constant for the entire test volume, but in a full-scale
application, local G values would be much higher in the vicinity of the impeller and much lower in corners and
at other extremes (Wahlberg et al., 1994; Pretorius et al., 2015). The designer needs to make some allowance
for this variation in G value and power level. One approach would be to use computational fluid dynamics to
gain an understanding of how much these parameters can vary throughout the reactor zone.

The minimum airflow rate per unit reactor volume required to maintain the same liquid suspended with
bubbles follows from Eq. 9.26; Table 9.6 summarizes the results for a typical diffuser submergence Z of 5 m.

Table 9.6 Optimum airflow for air-powered mixers corresponding to the same mixing as the mechanical mixers from Table
9.5.

Air flow (Nm3air/m3reactor.min)

HRT (min) Plug flow Completely mixed
Min. Avg. Max. Min. Avg. Max.
5 6.45E-02 3.74E-01 9.18E-01 7.31E-02 5.23E-01 1.80E+00
10 2.89E-02 1.69E-01 3.96E-01 4.23E-02 3.05E-01 9.75E-01
20 1.14E-02 7.00E-02 2.80E-01 2.52E-02 1.74E-01 5.14E-01
30 6.30E-03 3.99E-02 2.05E-01 1.67E-02 1.24E-01 3.52E-01
45 3.46E-03 2.16E-02 1.42E-01 1.25E-02 8.64E-02 2.44E-01
60 2.21E-03 1.38E-02 1.05E-01 8.85E-03 6.69E-02 1.84E-01

Note: Air-powered mixing has the potential to approach uniform distribution of mixer power throughout
the reactor zone. However, at these low air flows, the challenge would be to design a diffuser layout that
would achieve full floor coverage, without dead zones, while adhering to equipment limitations. In fact, when
designing an aeration system for oxygen transfer, it is immediately clear that the minimum mixing
requirements are far exceeded using the airflow rates necessary to aerate the process. This is because oxygen
transfer is very inefficient and only a small aliquot of each bubble is transferred during the bubble rise, while
the mixing benefits from the entire bubble volume expanding.

9.4 EXERCISES
Blower sizing (exercises 9.4.1-9.4.3)
Exercise 9.4.1
What is the required pressure to guarantee air discharge during operations?

Exercise 9.4.2
What happens when the pressure requirements approach the maximum discharge pressure of the blowers?

Exercise 9.4.3
Explain what diffuser flexing is. What is the required pressure to effectively perform diffuser flexing at 130 %
of the maximum process airflow?

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Oxygen transfer rate (exercises 9.4.4-9.4.6)

Exercise 9.4.4
What is the difference between the oxygen transfer rate (OTR) and the standard oxygen transfer rate (SOTR) if
the test is conducted at 15 °C and 700m AMSL (i.e., height above mean sea level)?

Exercise 9.4.5
What happens to the kLa if we use excessively slow DO sensors?

Exercise 9.4.6
Can we use plant effluent to perform a clean water test in order to determine the kLa?

Aeration system specification (exercises 9.4.7-9.4.9)

Exercise 9.4.7
What are the main types of aerators and what are their main differences?

Exercise 9.4.8
If cleaning is not an option at a given facility, what are the challenges that the operators will face?

Exercise 9.4.9
What is the cleaning frequency for fine-pore diffusers?

Aeration efficiency (exercises 9.4.10-9.4.12)

Exercise 9.4.10
Define aeration efficiency.

Exercise 9.4.11
With regard to the effect of process design on aeration efficiency, briefly describe two favourable design
criteria or configurations useful for improving the aeration efficiency in wastewater treatment plants.

Exercise 9.4.12
With regard to the effect of process operation on aeration efficiency, describe three operating factors that
adversely affect (reduce or deteriorate) the aeration efficiency in wastewater treatment plants.

Aeration energy – oxygen transfer efficiency (exercises 9.4.13-9.4.14)

Exercise 9.4.13
What is the process performance for an aeration system operating at constant airflow and variable process
loading (hence, variable DO)?

Exercise 9.4.14
What should the diurnal curves for process load, airflow rate, and blower power demand look like?

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ANNEX 1: SOLUTIONS TO EXCERSISES

Blower sizing (solutions 9.4.1-9.4.3)
Solution 9.4.1
To guarantee that the air is released, the discharge pressure of the blower (pdisch, relative to atmospheric
pressure) must exceed the pressure requirements of the system, consisting of the sum of the pressure losses in
the system:

pdisch ≥ ρ·g·Z + hL + DWP(AFR) (Eq.9.5)

The pressure losses on the right-hand side of Eq. 9.5 are the following:
• the hydrostatic head loss ρ·g·Z due to the diffuser submergence Z;
• the head loss of the air distribution line (friction head hL), which for practical purposes may be
considered independent of the airflow rate;
• the dynamic wet pressure (DWP), i.e., the diffuser head loss, which is a function of the airflow rate
(AFR) and needs to be provided by the manufacturer.

Solution 9.4.2
When the pressure requirements approach the maximum discharge pressure of the blowers, the blowers begin
to surge. The surge zone is an area of operation and must be avoided because the vibrations that the blowers
are subject to may cause structural failure. An automated system shuts down the blowers when the search
begins.

Solution 9.4.3
Diffuser flexing, also known as purging, is the practice of inflating membrane diffusers by feeding airflow
higher than the usual maximum range (e.g., 130-150 % of maximum operating airflow), in order to dilate the
pores of the membrane diffusers and therefore delay the occurrence of fouling. The required pressure for
flexing is calculated from Eq. 9.5 using the DWP related to the envisaged airflow, in this case, 130 % of the
maximum process airflow.

Oxygen transfer rate (solutions 9.4.4-9.4.6)

Solution 9.4.4
The OTR is calculated from the dissolved oxygen time series of reiteration in field conditions. In this case
field conditions are 15 °C and 700 m above sea level. The standardized OTR, SOTR, is the value corrected for
20 °C and 1 atm. For more information on this procedure, refer to e.g., ASCE (2018).

Solution 9.4.5
Answer 9.4.2.2 Using excessively slow DO sensors causes a delay between the actual dissolved oxygen and
the reading of that value by the sensor. The delay is very visible at the beginning of the re-aeration process,
i.e., when the slope of the re-aeration curve (DO vs. time) is very high and any error on the horizontal axis
corresponds to large errors on the vertical axis. This can be visibly observed because the fitting curve, which is
an ideal exponential, will differ from the experimental data, which will be more of a sigmoid. At the end of re-
aeration, when the curve is almost horizontal because it is approaching the asymptote (DO saturation), the
speed of the DO sensor is immaterial. Hence, a slow DO sensor results in underestimation of kLa.

Solution 9.4.6
No, we cannot, because plant effluent still contains residual organics that would alter the result and produce
lower kLa than tap water.

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Aeration system specification (exercises 9.4.7-9.4.9)

Solution 9.4.7
There are two main types of aeration technologies: the first is surface aeration and the second is submerged
aeration.
• Surface aeration is mechanical and relies on impellers.
Subtypes: High-speed versus low-speed surface aerators, with horizontal or vertical shafts.
• Submerged aeration relies on bubbles released throughout the floor by coarse-bubble or fine-bubble
diffusers, or by mechanical devices (turbines or jets).
Subtypes: Coarse-bubble versus fine-bubble aeration systems.

Solution 9.4.8
When cleaning is not practised or not possible, fine-pore diffusers foul and the pressure drop of these diffusers
increases. At the same time, the diffusers experience a decline in performance, measured as αFSOTE so αF
decreases over time. The consequence is an increase in the requirements for airflow, blower energy and
pressure, with potential blower failure if pressure requirements become excessive. Cleaning is always
recommended but when it cannot be practised, technologies immune from fouling, such as coarse-bubble
diffusers or mechanical aeration systems, are recommended.

Solution 9.4.9
Experience and documented measurements in the field suggest that the diffusers should be cleaned at least
once every 24 months, and preferably once a year.

Aeration efficiency (exercises 9.4.10-9.4.12)

Solution 9.4.10
The aeration efficiency (AE, expressed in kgO2/kWh) is defined by the oxygen transfer rate OTR (in kg/h)
relative to the power P (in kW) drawn by the aeration system (Eq. 9.4 in Chen et al., 2020): AE = OTP/R.

The AE is a measure for the efficiency of the aeration system, in contrast to the OTR, which only defines
the capacity of the aeration system

Solution 9.4.11
The aeration efficiency in wastewater treatment plants can be improved by the following design measures:
• Use of anoxic or anaerobic selectors: they result in improved (higher) alpha factors, probably by uptake of
soluble contaminants (surfactants) into the biomass.
• Relatively high sludge retention time (SRT): higher SRT systems, which operate with higher biomass
concentrations, remove or sorb the surfactants early in the process, improving the average oxygen transfer
efficiency and outweighing the increasing oxygen requirement for increasing SRT.

Solution 9.4.12
• Type of aerators used: fine-pore diffusers are the most efficient, coarse bubble diffusers are much less
efficient; surface aerators are also less efficient.
• Fouling: the oxygen transfer efficiency of fine-pore diffusers decreases over time. Cleaning fine-pore
diffusers is almost always required and restores process efficiency and reduces power costs. Other aerator
types are less prone to fouling.
• Surfactants: surface active agents accumulate at the air-water interface of rising bubbles, increasing the
rigidity of the interface and reducing internal gas circulation and overall transfer rate.

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 Aeration and mixing 337

Aeration energy – oxygen transfer efficiency (exercises 9.4.13-9.4.14)

Solution 9.4.13
An aeration system operating a constant airflow will over-deliver air when process loading is below average
and under-deliver air when the process loading is above average. Although the power demand and energy
requirement for this aeration system is constant because the airflow is constant, the process performs poorly at
peak loading (i.e., when it is most needed) and excessively well at minimum loading. This is not the correct
way to operate a process and should be avoided

Solution 9.4.14
Usually, a plant experiences a peak in process loading during the day. Responding to the peak loading, the
aeration control system will discharge more airflow, consequently demanding more blower power. However,
at peak loading, the plant receives a higher concentration of contaminants that depress the alpha factor,
requiring more airflow in proportion to reach the same oxygen transfer. Moreover, at higher airflows, the
oxygen transfer efficiency (i.e., OTE or SOTE, %), which is the ratio between the oxygen transfer rate and the
mass flow delivered to the system (OTE = OTR/WO2, Eq. 9.6 in Chen et al., 2020) is lower. The
compounding effect is such that at peak loading the process requires proportionally much more airflow and
much more power demand. This phenomenon is called peak loading amplification and is illustrated below.

Figure 9.6 Example of daily patterns of influent

flow, BOD, TKN, oxygen transfer efficiency
(αSOTE) and airflow vs. time, all normalized
against their daily averages. Note the
amplification of air requirements, due to the
compounding of flow and load with αSOTE
variations. Image from Emami et al. (2018).

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REFERENCES
ASCE - American Society of Civil Engineers (2018). ASCE/EWRI 18-18 Standard Guidelines for In-Process Oxygen Transfer
Testing, 2nd ed. Reston, VA: American Society of Civil Engineers.
Chen G.H., van Loosdrecht M.C.M., Ekama G.A. and Brdjanovic D. (eds.). (2020). Biological Wastewater Treatment: Principles,
Modelling and Design. ISBN: 9781789060355. IWA Publishing, London, UK.
Emami N., Sobhani R. and Rosso D. (2018). Diurnal variations of the energy intensity and associated greenhouse gas emissions for
activated sludge processes. Water Science & Technology, 77(7) 1838-1850. https://doi.org/10.2166/wst.2018.054
Grenville R.K., Giacomelli J.J., Padron G. and Brown D.A.R. (2017). Impeller performance in stirred tanks, Chemical Engineering,
April 2017, pp.42-51.
Metcalf & Eddy (2014). Wastewater Engineering: Treatment and Resource Recovery. 5th Edition, McGraw-Hill, New York.
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Water Environment Research, pp. 779-786

NOMENCLATURE
Symbol or
Description Unit
abbreviation
Adiff Specific area of each diffuser m2
AE Aeration efficiency in clean water kgO2/kWh
AFR Airflow rate m3/s
AFRdiff Desired airflow rate per diffuser – specified by manufacturer m3/s
BHP Blower brake horsepower kW
DO Dissolved oxygen in water kgO2/m3
DWP Dynamic wet pressure Pa
e Blower efficiency -
F Fouling factor -
g Gravitational constant (= 9.81) m/s2
G Velocity gradient 1/s
hL Hydrostatic head loss Pa
kLa Liquid side mass transfer coefficient 1/h
n Empirical constant (= 0.283 for air) -
ND Number of diffusers -
OTE Oxygen transfer efficiency in clean water %
OTR Oxygen transfer rate in clean water kgO2/h
K Empirical constant S.I. units
pin Absolute inlet pressure Pa
pdisch Discharge pressure of the blower (relative to atmospheric pressure) Pa
pdisch,abs Absolute discharge pressure of the blower Pa
Pmix Mixing power W
P Power drawn by the aeration system kW
R Universal gas constant (= 8.314) J/mol.K
RO2 Average required oxygen mass flow kgO2/s

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 Aeration and mixing 339

SF Safety factor -
SOTE Oxygen transfer efficiency in standard conditions in clean water %
SOTR Oxygen transfer rate in standard conditions in clean water kgO2/h
SRT Sludge retention time days
Tin Absolute inlet temperature K
V Water volume m3
Wair Air mass flow rate kg/s
WO2 Oxygen mass flow fed to aeration tank kgO2/s
ŷO2 Weight fraction of oxygen in air wt%
Z Diffuser submergence m

Superscripts Description
avg Average
MAX Maximum
MIN Minimum

Greek symbol Description Unit

α Ratio of process- to clean-water mass transfer -
αFSOTE Oxygen transfer efficiency in standard conditions in process water for %
used diffusers
αSOTE Oxygen transfer efficiency in process water at standard conditions in %
process water
µw Dynamic viscosity of water N.s/m2.
ρ Water density kg/m3
ρair Air density kg/m3
Ψ Pressure factor -

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 340

Aeration has been used for centuries in wastewater treatment. Still, scientist and engineers are continuously searching for
new ways to increase its efficiency, save energy, reduce costs, minimise emissions, and protect the environment (photo: D.
Brdjanovic).

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