Preliminary Design

for the Proposed

Southville Wastewater Treatment
Plant

(Secondary Treatment – Sludge management)

Submitted by:

Cesar Mariano Gapas

Winsam Amiel Cejudo

Submitted to
Prof. Rochie Amolato
Introduction

It has been a long-standing assertion that water is life, in a sense that water has been used for
many purposes in households and industries. In as much as water is important in human functions
daily, used water, otherwise known as wastewater, can be a source of contamination when released
into the water cycle without treatment. After its intended use, water may contain suspended solids,
heavy metals, pathogens, nutrients, and organic matter which induce repercussions on the
environment and the people in varying intensities.
The City of Southville has been menaced by these repercussions, inciting its citizens to file a
formal complaint against the LWD of Southville in 2020. The Supreme Court ultimately acted on this
matter, ordering the LWD to plan, design and construct a wastewater treatment facility to process
wastewater to reach an effluent acceptable for release to the environment.
Recently, two engineers were commissioned to design primary treatment processes.
Additional steps will be carried out to the effluent of the primary clarifier and the same engineers
were tasked to design components of the secondary treatment stage of the plant. Additional analysis
of the city’s wastewater was conducted by an independent laboratory. The report on wastewater
characterization contains the following measurements of wastewater constituents:

sBOD 140 mg/L VSS 60 mg/L

COD 400 mg/L TKN 35 mg/L
sCOD 220 mg/L NH4-N 25 mg/L
rbCOD 140 mg/L TP 6 mg/L
bCOD/BOD 1.6
ratio

The design MLVSS concentration is 2000 mg/L, and the secondary clarifier can produce an
effluent with 30 mg/L TSS.
.
I. Parameters and Biological Kinetic Coefficients

In the design of the secondary treatment stage, there are quantities that are important
to obtain. Among these is the hydraulic detention time which is the average time that
wastewater remains in the tank. Mean cell residence time is the average length of time that
the microorganisms will spend in the activated sludge process. From these values, the
volume of the aeration tank can be determined.
A certain portion of the flow of water going out of the aeration tank will be returned
to continuously stimulate microbial action. Therefore, the concentration of return sludge
should be evaluated. Additionally, the mass flow rate for sludge wasting should be
computed.
The oxygen demand should be determined because the presence of oxygen in the
aeration tanks is one of the primary determinants of microbial processes in aerobic condition.
This value indicates the amount of oxygen needed for the aerobic processes to push through.
Another indicator that is crucial is the food to microorganisms ratio which is the
ratio of the daily mass of BOD to microbial mass. This is important because if this value is low,
it means that the microorganisms will starve and die. High value for this would mean that the
BOD removal may not be effective due to high mass of BOD but lesser microbial population.
In the computations, several biological kinetic coefficients were assumed. The decay
rate for microorganisms (kd), maximum growth rate constant (μm) were taken from values
suggested by Metcalf & Eddy (2003).

II. Secondary Treatment Processes

Effluent from the primary treatment will head towards the secondary treatment
stage. The secondary treatment stage utilizes a completely mixed activated sludge process
where wastewater and biological sludge is agitated and aerated. CMAS is composed of
components such as the aeration tank. After the processes in the aeration tank, wastewater
will head towards the secondary clarifier to settle microbial flocs. In order for the microbial
 processes to continue in the aeration tank as more wastewater passes through the secondary
treatment stage, a portion of the activated sludge will not go through the secondary clarifier
but will return to the aeration tank to stabilize microbial population and consequently the
biological action.

III. Design Considerations and Operational Parameters

A. Aeration Tank
a. With BOD as substrate

The aeration tank will be made to condition the influent from the primary treatment. The
effluent should satisfy the requirements for the class c river. The subject to be considered for the
secondary treatment equipment will be BOD for the assumption that the city is under the PSIC Code
37000 for sewerage.

Assuming the effluent from the primary treatment is equal to its influent:
Q = 0.99 m3/s = 85,536 m3/day
SO = 140 mg/L BOD

Allowable BOD for the class C river is 50 mg/L and allowable TSS is 100 mg/L
Assuming that an estimated 30% of the TSS concentration is BOD. This percentage is estimated so
that the S value will not be below zero since allowable BOD concentration is much less than TSS
concentration.

S = 50 – 0.30(100) = 20 mg/L
X= MLVSS = 2000 mg/L

With biomass, VSS, and substrate COD, and both biomass and substrate are in the same volume.
Y = 60 / 140 = 0.43 mg VSS/mg COD

Assuming sBOD concentration is a one-half maximum growth rate

KS = 140 mg/L
μm = 2.5 d-1 kd = 0.050 d-1
Equations to be used:

𝐾𝑆 (1+𝑘𝑑 𝜃𝐶 ) 𝜃𝐶 𝑌(𝑆𝑂 −𝑆)

𝑆= 𝑋=
𝜃𝑐 (𝜇𝑚 −𝑘𝑑 )−1 𝜃(1+𝑘𝑑 𝜃𝐶

Solving for θC with the equation for S,

𝐾 (1+𝑘𝑑 𝜃𝐶 )
𝑆 100(1+0.05𝜃𝐶 )
𝑆 = 𝜃 (𝜇 =𝜃 = 20 𝑚𝑔/𝐿 𝜃𝐶 =2.73 days
𝑐 𝑚 −𝑘𝑑 )−1 𝑐 (2.5−0.05)−1

Then using the formula for X, we get θ

𝜃𝐶 𝑌(𝑆𝑂 −𝑆) (2.73)0.43(140−20)

𝑋= = = 2000 𝑚𝑔/𝐿 𝜃 = 0.062 𝑑
𝜃(1+𝑘𝑑 𝜃𝐶 𝜃(1+0.05(3.81))

To get volume

𝑉 𝑉
𝜃= 𝑄
= 85536
= 0.062 Volume = 5,303 ≡ 5,310 m3

Aeration tank will be rectangular with dimensions in LxWxH of 25x25x9 m3 with 0.6 meter
freeboard.

For the process in aeration tank to continue certain parameters are calculated.

1. Return sludge Concentration, X’r, and mass flow rate of sludge wasting, Qw

With the estimation that VSS is 60 to 80 percent of MLSS, we divide the given MLVSS, X, with
a factor of 0.7 to estimate MLSS concentration, X’.

X’ = 2000/0.7 = 2857 mg/L ≡ 2860 mg/L

(5310)(2860)
2.73 = QwX’r = 5,562,857 g/d
𝑄𝑤 𝑋𝑟′

We consider the effluent suspended solids.

Q = Qr = 85536 m3/day
Xe = 30 mg/L TSS
 5,562,857
85536(2860)−5,562,857 −(88536− )30
𝑋𝑟 ′
85536 = Xr’ ≈ 5625 g/m3
𝑋𝑟′ −2860

Qw = 5,562,857 / 5625 = 988 ≡ 990 m3/day

Sludge Concentration, X’r = 5625g/m3

Mass flow rate of sludge wasting, Qw = 990 m3/day

2. Estimating the mass to be wasted

𝑌 0.43
Yobs = 1+𝑘 = = 0.38
𝑑 𝜃𝑐 1+0.05(2.73)

Px = YobsQ(SO-S)(10-3 kg/g) = 0.38(85536)(140-20) ≈ 3900kg/d VSS

Mass to be Wasted = (Q-Qw)(Xe) = (85536 - 990)(30) = 2,536,380 kg/day

3. Oxygen Demand

bCOD/BOD = 1.6 SO = bCOD = BOD(1.6) = 140(1.6) = 224 mg/L

effluent BOD, S = 20 mg/L

influent BOD 𝑒𝑓𝑓𝑙𝑢𝑒𝑛𝑡 𝐵𝑂𝐷 140 20

𝐼𝑛𝑓𝑙𝑢𝑒𝑛𝑡 𝑏𝐶𝑂𝐷
= 𝑒𝑓𝑓𝑙𝑢𝑒𝑛𝑡 𝑏𝐶𝑂𝐷 224
=𝑆 SbCOD = 32 mg/L bCOD
𝑏𝐶𝑂𝐷

MO2 = Q(SO – S)(10-3 kg/g) – 1.42PX = 85536(224 – 32)(10-3)-1.42(3700) ≈ 11170 kg/d O2

4. Food to microorganism ratio

F/M = (85536 m3/day*140 BOD mg/L) / (5310 m3 * 2000 mg/L )

F/M = 1.13 Which is within the usual ratio

b. With COD as substrate

The aeration tank will be made to condition the influent from the primary treatment. The
effluent should satisfy the requirements for the class c river. The subject to be considered for the
secondary treatment equipment will be COD since we will assume the PSIC Code 39000 for waste
management services.

Assuming the effluent from the primary treatment is equal to its influent:
Q = 0.99 m3/s = 85,536 m3/day
SO = 400 mg/L COD

Allowable COD for the class C river is 100 mg/L and allowable TSS is 100 mg/L
Assuming that an estimated 63% of the TSS concentration is COD
S = 100 – 0.63(100) = 37 mg/L
X= MLVSS = 2000 mg/L

With biomass, VSS, and substrate COD, and both biomass and substrate are in the same volume.
Y = 60 / 400 = 0.15 mg VSS/mg COD

Assuming sBOD concentration is a one-half maximum growth rate

KS = 140 mg/L
μm = 2.5 d-1 kd = 0.050 d-1

Equations to be used:

𝐾𝑆 (1+𝑘𝑑 𝜃𝐶 ) 𝜃𝐶 𝑌(𝑆𝑂 −𝑆)

𝑆= 𝑋=
𝜃𝑐 (𝜇𝑚 −𝑘𝑑 )−1 𝜃(1+𝑘𝑑 𝜃𝐶

Solving for θC with the equation for S,

𝐾 (1+𝑘𝑑 𝜃𝐶 )
𝑆 140(1+0.05𝜃𝐶 )
𝑆 = 𝜃 (𝜇 =𝜃 = 37𝑚𝑔/𝐿 𝜃𝐶 =2.12 days
𝑐 𝑚 −𝑘𝑑 )−1 𝑐 (2.5−0.05)−1

Then using the formula for X, we get θ

𝜃𝐶 𝑌(𝑆𝑂 −𝑆) (2.12)0.15(400−37)

𝑋= 𝜃(1+𝑘𝑑 𝜃𝐶
= 𝜃(1+0.05(2.12))
= 2000 𝑚𝑔/𝐿 𝜃 = 0.052 𝑑

To get volume

𝑉 𝑉
𝜃= 𝑄
= 85536
= 0.052 Volume = 4,456 m3

Aeration tank will be rectangular with dimensions in LxWxH of 25x20x10 m 3 with 1 meter
freeboard.

For the process in aeration tank to continue certain parameters are calculated.

1. Return sludge Concentration, X’r, and mass flow rate of sludge wasting, Qw

With the estimation that VSS is 60 to 80 percent of MLSS, we divide the given MLVSS, X, with
a factor of 0.7 to estimate MLSS concentration, X’.

X’ = 2000/0.7 = 2857 mg/L ≡ 2860 mg/L

(4456)(2860)
2.12 = QwX’r = 6,011,396 g/d
𝑄𝑤 𝑋𝑟′
 We consider the effluent suspended solids.

Q = Qr = 85536 m3/day
Xe = 30 mg/L TSS

6,011,396
85536(2860)−6,011,396 −(88536− )30
𝑋𝑟 ′
85536 = Xr’ = 5620 g/m3
𝑋𝑟′ −2860

Qw = 6,016,900 / 5620 = 1070 m3/day

Sludge Concentration, X’r = 5620 g/m3

Mass flow rate of sludge wasting, Qw = 1070 m3/day

2. Estimating the mass to be wasted

𝑌 0.15
Yobs = 1+𝑘 = = 0.136
𝑑 𝜃𝑐 1+0.05(2.12)

Px = YobsQ(SO-S)(10-3 kg/g) = 0.136(85536)(400-37) = 4223 kg/d VSS

Mass to be Wasted = (Q-Qw)(Xe) = (85536-1070)(30) = 2533980 kg/day

3. Oxygen Demand

bCOD/BOD = 1.6 SO = bCOD = BOD(1.6) = 140(1.6) = 224 mg/L

effluent COD, S = 37 mg/L

influent COD 𝑒𝑓𝑓𝑙𝑢𝑒𝑛𝑡 𝐶𝑂𝐷 400 37

𝐼𝑛𝑓𝑙𝑢𝑒𝑛𝑡 𝑏𝐶𝑂𝐷
= 𝑒𝑓𝑓𝑙𝑢𝑒𝑛𝑡 𝑏𝐶𝑂𝐷 224
=𝑆 SbCOD = 20.72 mg/L
𝑏𝐶𝑂𝐷

MO2 = Q(SO – S)(10-3 kg/g) – 1.42PX = 85536(224 – 20.72)(10-3)-1.42(4223) = 11391 kg/d O2

4. Food to microorganism ratio

F/M = (85536 m3/day*400 COD mg/L) / (4456 m3 * 2000 mg/L )

F/M = 3.8
Conclusion for Aeration tank:

We design aeration tank with the use of sBOD as substrate. It will require a larger volume
for the tank which will be conservative for the treatment.

Two aeration tanks will be constructed with the dimensions of 5310 m3 of 25x25x9
(LxWxH m3) with 0.6 meter freeboard.

Operational Parameters:

Sludge Concentration, X’r = 5625g/m3

Mass flow rate of sludge wasting, Qw = 990 m3/day
Mass to be Wasted = 2,536,380 kg/day
Oxygen Demand inside tank: MO2 ≈ 11170 kg/d O2
Food to Microorganism ratio: F/M = 1.13

B. Secondary Clarifier

Secondary clarifier will have the same dimensions as the primary clarifier.

The secondary clarifier will be circular with a diameter of 44.5 meters and a height of 3
meters. Two tanks of this size will be constructed for redundancy.

IV. Recommended Nutrient Removal Option

Conventional options for post-treatment denitrification involve the addition of

methanol. However, many concerns about water effluent arise when this option is
used. Methanol achieves a high rate of denitrification but plant outflow contains excessive
organic carbon content and poses danger for its flammability.
The study by Yamashita and Ikemoto in 2014 investigated the effect of the addition
of wood and iron to wastewater to remove nitrogen and phosphorus. Sulfate-reducing
bacteria Desulfovibrio sp. CMX may use wood chips or animal manure as electron donors and
carbon sources then use various types of organic substances. Sulfate-reducing bacteria allow
nitrogen removal with wood. Ferrous ions generated by iron polarization are known to
combine with phosphate to produce vivianite and other ferrous phosphates, and thereby
remove phosphorus. It was found that both denitrification and phosphate removal occurred
in a bioreactor filled with synthetic wastewater and packed with iron and wood and allowed
to react over a long term.
For this project, an anoxic tank will be made downstream of the aeration tank. The
minimum hydraulic retention time should be 24 hours. According to the result of the study,
aspen wood chips perform better, thus, aspen wood chips, iron and the sulfate-reducing
bacteria should be introduced to the anoxic tank. The resulting water effluent should be
 filtered before secondary settling so that the wood chips would not be wasted since it was
found that aspen wood chips and iron could be used for at least 1200 days without significant
effect on their denitrification abilities.

Summary and Recommendations

The engineers recommend the City of Southville to construct a secondary treatment plant
under the Completely Mixed Activated Sludge (CMAS) Process. This process needs an aeration tank
and a secondary clarifier which is similar to the primary clarifier or the sedimentation tank, but with
more solids in the secondary clarifier.

For designing the equipment of the secondary treatment, the same discharge of fluid influent
to primary treatment is considered. The influent of partially treated wastewater to the secondary
treatment will be 85,536 m3/day.

From the given concentrations of BOD and COD, it has been computed that the final
dimensions for the Aeration tank will be 25x25x9 with a volume of 5310 m3 with a 0.6 meter
freeboard. Two of this will be produced for redundancy and to avoid hampering of the operation.

The following parameters are needed for the continued operation of the treatment plant:

Sludge Concentration back to aeration tank from clarifier, X’r = 5625g/m3

Oxygen Demand inside tank: MO2 ≈ 11170 kg/d O2
Food to Microorganism ratio: F/M = 1.13

The product of this treatment are as follows:

Mass flow rate of sludge wasting, Qw = 990 m3/day

Mass to be Wasted = 2,536,380 kg/day

Finally, the dimensions of the secondary clarifier for this treatment will be identical to the
primary treatment which is circular with a volume of 3564 m3 with a 44.5-meter diameter and a
height of 3 meters.

For the removal of nutrients, an anoxic tank will be placed downstream of the aeration tank.
In this tank, aspen wood chips, iron and the sulfate-reducing bacteria will be introduced to induce
the chemical processes needed to breakdown the nutrients present in the influent. This tank should
also have a filter to keep the wood chips from leaving the tank to the secondary clarifier.
References

Metcalf, E., & Eddy, E. (2003). Wastewater engineering: Treatment and Reuse (Vol. 4). McGraw Hill.

Obade, V.D.P., Moore, R., 2018. Synthesizing water quality indicators from standardized geospatial
information to remedy water security challenges: a review. Environ. Int. 119 (OCT),
220–231

Yamashita, T., & Yamamoto-Ikemoto, R. (2014). Nitrogen and phosphorus removal from wastewater
treatment plant effluent via bacterial sulfate reduction in an anoxic bioreactor packed with
wood and iron. International journal of environmental research and public health, 11(9),
9835–9853. https://doi.org/10.3390/ijerph110909835

UP Institute of Civil Engineering. (2022). CE 132 Modules 7, & 8 [PDF].

https://uvle.upd.edu.ph/course/view.php?id=4705