Wastewater Treatment Process (from Sewage Filtration onwards)

To know and understand
about method of filtration
in sewage
To be able to solve real-life based
numericals
To know and understand about
types of filtration
To have detailed study of
Activated Sludge Method
To have detailed study of
Oxygen pond
Objective of Presentation
3

Presentation Outline
1. Sewage Filtration
– 1.1 Introduction
– 1.2 Types of Filter
– 1.3 Intermittent Sand Filter
– 1.4 Contact Beds
– 1.5 Trickling Filter
– 1.6 Two stage filter
– 1.7 Numerical on Design of Trickling Filter
7

2. Activated Sludge Process
– 2.1 Introduction and Principles
– 2.2 Process Description
– 2.3 Aeration
– 2.4 Design Criteria
– 2.5 Merits and Demerits
– 2.6 Sludge Volume Index
– 2.7 Numerical
8

3. Oxidation Pond
– 3.1 Introduction
– 3.2 Purpose
– 3.3 Mechanism
– 3.4 Types of oxidation pond
– 3.5 Operation and maintenance
– 3.6 Factors affecting oxidation pond
– 3.7 Merits and demerits
– 3.8 Construction of oxidation pond
– 3.9 Design Criteria
– 3.10 Oxidation Pond 9

1.1 Introduction
– It is the process of removing finely divided suspended matter.
– Is done in contact beds, intermittent sand filters or trickling filters.
Fig. 1.1 A trickling filter plant in the UK
Source: https://en.wikipedia.org/wiki/Trickling_filter 10

1.2 Filter Types
Common filters used for sewage filters are:
- Intermittent Sand Filters
- Contact Beds
- Trickling Filters
Fig. 1.2 Intermittent Sand Filter
Source: https://web.deu.edu.tr/atiksu/ana58/sand.html
Fig. 1.3 Contact Beds
Source: https://www.engineeringenotes.com/wp-
content/uploads/2018/04/clip_image008-7.png
Fig. 1.4 Trickling Filter
Source: https://sswm.info/taxonomy/term/3806/trickling-filter
11

1.3 Intermittent Sand Filters
1.3.1 Introduction
 They are a viable alternative to conventional methods when soil conditions are not condu
for proper treatment and disposal of wastewater through percolator beds/trenches.
 The filtering medium consists of sand.
 Sand filters can be used in sites that have shallow soil cover, inadequate permeability, hig
groundwater, and limited land area.
 Treatment is accomplished through physical and chemical means, but mainly by microorg
attached to the filter media, viz. Mechanical Straining and Bacterial Action.
 The treated wastewater is collected in under drains at the bottom of the sand filter and is
transported to a line for further treatment or disposal.
12

1.3.2 Construction
 2’-3’ of sandlayer.
 Effective size of sand- 0.2 to 0.5mm.
Finer sand  Better quality  Slow
Coarser sand  Fast  Deeper penetration of solids
(Cu = 2to 5)
 15cm to 30 cm of gravel layer is provided at the bottom of sand layer to facilitate the drai
for filtered effluent.
 The sand filter is, generally, rectangular with L/B ratio 3 to 4.
 Area: 0.2 to 0.4 ha
13

1.3.2 Construction
 3 to 4 filter beds are provided adjacent to each other.
 Suspended solids<10ppm, which is stable.
 The Topmost 25mm layer is raked for efficiency.
14

Fig. 1.5 Intermittent Sand Filter
Source: engineeringenotes.com/wp-
15

1.3.3 Working
 Sewage effluent from primary settling tank is applied intermittently on each bed throug
dosing tank provided with a siphon.
 The quantity of sewage applied each time is such that the depth of sewage flooding the
entire filter bed is about 50 to 100mm.
 During the period of sewage remains standing on the filter bed, it gradually percolates
through the bed.
 As the sewage percolates through the bed, the suspended organic matter present in th
sewage gets trapped in the voids of the top portion of the sand bed.
 Thus the organic matter is first removed by the straining action of the filter.
16

1.3.3 Working
 After the entire quantity of applied sewage has percolated down, the filter bed is kept at
for some time before the sewage is again applied.
 During the rest period the trapped organic matter is acted upon by the aerobic bacteria
present in the filter layer.
 These aerobic bacteria flourish well in the presence of free oxygen available from the
atmosphere during the rest period.
 After 24 hours, sewage is now applied over a second bed while the first bed rests  3 o
beds work in rotation.
 During the resting period, the dried sewage is scraped off.
 Hydraulic loading  80 to 110 L/d/m2 of plan of area of filter bed. 17

1.3.4 Merits
 Quality of effluent obtained from the ISF is very clean and stable and generally does no
require any further treatment before disposal except chlorination in some cases.
 Less chances of anaerobic conditions to develop, and hence there is no trouble of odor
 The operation is very simple, requiring no mechanical equipment except for dosing and
does not require constant skilled supervision.
 Small head is required for applying the sewage on the surface of the filter.
 There is no secondary sludge which is to be disposed of except for the occasional sand
scraping,
18

1.3.5 Demerits
 The land area required may be a limiting factor.
 Regular (but minimal) maintenance is required.
 Odor problems could result from open filter and configurations and may require buffer z
from inhabited areas.
 If appropriate filter media are not available locally, costs could be higher.
 Clogging of the filter media is possible.
 ISFs could be sensitive to extremely cold temperatures.
19

1.4 Contact Beds
1.4.1 Introduction
 Improved version of ISF
 The filtering medium consists of gravel, ballast or broken stones
(15mm< Size of the particles of filter media> 40mm)
 1m < Depth of filter bed >1.8m [Common Value=1.2m]
 Area of the filter bed < 2000m2
Fig. 1.6 Contact Beds
Source: https://www.engineeringenotes.com/w
20

1.4 Contact Beds
1.4.2 Operation
Filling
Contact Emptyin
g Oxidat
-ion
Step-1
Step-2 Step-3
Step-4
21

1.4 Contact Beds
1.4.2 Operation (contd.)
1. Filling : The outlet valve of the under-drain is closed and the tank is slowly filled with
sewage effluent from the primary settling tank through the dosing tank. The depth of sew
applied may be 50 to 100 mm over the top of the bed. This filling may take about 1 to 2
2. Contact : The dosing tank outlet is then closed, and the sewage admitted over the co
bed is allowed to stand for about 2 hours. During this period the fine suspended, colloid
dissolved organic matter present in the sewage gets transferred to the filter media and
in contact with the bacterial film covering the filter media.
22

1.4 Contact Beds
1.4.2 Operation (contd.)
3. Emptying : The outlet valve of the under-drain is then opened and the sewage prese
in the contact bed is withdrawn slowly without disturbing the organic film of the bed. This
operation may take about 1 to 2hours.
4. Oxidation : The contact bed is then allowed to stand empty for about 4 to 6 hours.
During this period of rest, atmospheric air enters the void spaces of the contact media, t
supplying oxygen to the aerobic bacteria, resulting in the oxidation of the organic matter
present in the film.
23

Large land area
required
Slow rate of
loading
Less odor
problem than TF
80% BOD and 80-
90% suspended
particles are
removed
Very high
Efficiency
MERITS
DEMERITS
24

1.5 Trickling Filter
1.5.1 Introduction
 Trickling filter is a bed of crushed stone, gravel or slag of
relatively large size to which sewage coming from primary
sedimentation is applied by sprinkling on the surface of
filter bed.
 These filters are also known as sprinkling or
percolating filters.
 They have been widely used for biological treatment of sewage
since their innovation.
 It works on the principle of attached growth method.
 It oxidize and bio-flocculate the organic matter in the sewage.
Fig. 1.7 Trickling Filter Plan in The UK
Source: https://en.wikipedia.org/wiki/Trickling_filter
25

Fig. 1.8 Trickling Filter Animation
Source: https://gfycat.com/politicaldisfiguredindigobunting
26

Fig. 1.9 Scientific Diagram depicting main parts of Trickling Filter
Source: https://www.researchgate.net/figure/Typical-configuration-
and-main-parts-of-a-classic-trickling-filter-Figure-adapted-
from_fig9_266736009 27

Fig. 1.10 Trickling Filter
Source: Civildigital.com
28

Fig. 1.11 Schematic cross-section of a trickling filter
Source: https://sswm.info/taxonomy/term/3806/trickling-filter
29

Fig. 1.12 Schematic representation of cross section of biological slime layer in a trickling filter
Source: https://www.engineeringenotes.com/wp-content/uploads/2018/04/clip_image010-
6.png 30

1.5.2 Construction
a) Shape : Circular shape is most common.
b) Filter walls : Reinforced cement concrete is preferred.
c) Filter floor : It consists of concrete slab 100-150 mm thick with a slope of commonly 1%
towards main collecting channel.
d) Under drain system : It covers the entire floor of the filter and consists of drains with
semi-circular inverts.
e) Filter media : It should have high specific surface area, high percent void space, resista
abrasion.
f) Filter depth : Effectiveness of filter decreases as depth increases. Generally depth of
1.8 m to 3 m is preferred.
g) Ventilation : Minimum air flow of 0.1 m³ / min / m² is provided.
31

1.5.3 Working and Cleaning
 The wastewater in trickling filter is distributed over the top area of a vessel containing
non- submerged packing material.
 Air circulation in the void space, by either natural draft or blowers, provides oxygen for
the microorganisms growing as an attached biofilm.
 During operation, the organic material present in the wastewater is metabolised by the
biomass attached to the medium. The biological slime grows in thickness as the organic
matter abstracted from the flowing wastewater is synthesized into new cellular material.
 The thickness of the aerobic layer is limited by the depth of penetration of oxygen into
the microbial layer. 32

1.5.3 Working and Cleaning (contd.)
 The micro-organisms near the medium face enter the endogenous phase as the
substrate is metabolised before it can reach the micro-organisms near the medium face
as a result of increased thickness of the slime layer and loose their ability to cling to the
media surface. The liquid then washes the slime off the medium and a new slime layer
starts to grow. This phenomenon of losing the slime layer is called sloughing.
The sloughed off film and treated wastewater are collected by an underdrainage
which also allows circulation of air through filter. The collected liquid is passed to a
settling tank used for solid-liquid separation.
33

1.5.4 Merits
 Because of their large air-water interface, they can remove CO2, H2S, N2 and other gas
 A portion of liquid in underdrain system is recycled.
 It improves the treatment efficiency.
 To dilute the strength of incoming wastewater.
 To maintain enough wetting of slime layer.
 Prevents ponding in filter.
 Suitable for shock loads and low running cost.
34

1.5.5 Demerits
 Head loss is high.
 Cost of construction is high.
 Primary sedimentation is must.
35

1.5.6 Types of TF
Hydraulic and Organic
Loading Rates
Number of Units Used
In Series
Low rate Trickling filters
Super rate Trickling filters
High rate Trickling filters
Two Stage Trickling filters
Single Stage Trickling filters
36

1.5.6.1 Types basing on Hydraulic and Organic Loading Rates
1.5.6.1.1 Low rate Trickling Filters
 Also known as conventional trickling filters
or standard rate trickling filter.
 Low rate filters are relatively simple treatment units
those employ rock, gravel and slag as filter media.
 In low rate trickling filters, recirculation or recycling
is generally not adopted.
 A well operated low rate trickling filter in combination
with secondary settling tank may remove 75 to 90% BOD
and produce highly nitrified effluent.
 It is suitable for treatment of low to medium strength
domestic wastewaters.
Primary
Settling
Tank
Trickling
Filter
Secondar
y
Settling
Tank
37

1.5.6.1 Types basing on Hydraulic
and Organic Loading Rates
1.5.6.1.2 High rate Trickling Filters
 They are more advanced filters
than low rate trickling filters.
 The high rate trickling filter,
single stage or two stage are
recommended for medium to
relatively high strength domestic
and industrial wastewater.
 The BOD removal efficiency
is around 75 to 90% but the
effluent is only partially nitrified.
Primar
y
Clarifie
r
Filter
Second
-ary
Clarifie
r
Recirculated Flow
Alternate Return of
Final Effluent
Effluent
Influent
a. Single-Stage Trickling Filter with Recirculation Process
Tricklin
g
Filter
1
Primar
y
Settling
Tank
Second
-ary
Settling
Tank
Tricklin
g Filter
2
Tricklin
g
Filter
1
Primar
y
Settling
Tank
Second
-ary
Settling
Tank
Tricklin
g Filter
2
b. Two-Stage Trickling Filter with Effluent Recirculation
38

1.5.6.1.3 Comparison between High Rate Trickling Filters and Low Rate Trickling Filt
• The hydraulic loading rate is the total flow including recirculation applied on unit area of the filter in
a day.
39

Recirculation
An important consideration in the operation of a trickling filter is that the microorgan
(the zoogleal slime) does not move. They just hang onto the media as slime waiting
the food to trickle by. To give the microorganisms another opportunity to eat food th
missed, operators recycle clarified effluent. This is called recirculation. It is done w
the help of pump to send back a part of the filter effluent to primary settling tank or
dosing tank.
Advantages:
 Increases the contact time
 Improve sloughing
 Reduce odors
40

Basic Design Equations
a) Recirculation ratio (r): b) Recirculation Factor (F):
𝐫 =
𝐑
𝐈
𝐅 =
𝟏+𝐫
𝟏+𝟎.𝟏𝐫 𝟐
where, where,
R= re-circulated flow r= recirculation ratio
I= influent flow
r= 2 MGD or 2:1 MGD
First number>>recirculated proportion
Second number>>influent proportion
41

c) Organic Loading (U): d) BODs Removal Efficiency (E):
𝐔 =
𝐖
𝐕𝐱𝐅
𝐄 =
𝐂𝒊−𝑪𝒆
𝑪𝒊
=
𝟏
𝟏+𝟎.𝟎𝟎𝟒𝟒√U
where, NOTE:
W= BOD applied to TF (kg/day) For 2nd stage of two stage HRTF,
V= Vol. of TF (ha-m) E2=
𝐶𝑒−𝐶′𝑒
𝐶𝑖
=
𝟏
(𝟏+𝟎.𝟎𝟎𝟒𝟒√U2)(𝟏−𝐄𝟏)
42

1.5.6.2 Two Stage Filter
 Recirculation is provided for each stage.
 For same volume of media, higher degree of treatment is achieved.
 Used for treatment of strong sewage
Primary
Settling
Tank
Filter 2
Filter 1
Secondary
Settling
Tank
43

WAY
TO
SUCCESS
Sort out
the given
provided by
the question
01
Understand
the given
problem
and
list out
unknown
parameter
02
Recall
Basic
Design
Equations
03
Solve for
parameters
by using
Basic
Design
Equations
04
Conclude
your
numerical
05
5 STEPS
TO SOLVE NUMERICAL
44

If question is complex, try “DIVIDE AND CONQUER” strat
45

46
LETS TRY ONE!!!
Q. The effluent from a primary settling tank is applied to a standard rate trickling filte
rate of 3MLD having BOD5 of 175 mg/l. Determine the depth and diameter of the filte
a surface loading (hydraulic loading) of 2000 l/m2/day and organic loading of 150 l/m
Also determine the efficiency of the filter using NRC formula.
Hmmm, The question seems lengthy, but lets try out our DIVIDE AND CONQUER strategy
divide it in two questions. First case can be listed as to determine the depth and diameter
of the filter and second case can be listed as to determine the efficiency of the filter using N
formula.. Lets try solving first case and then look into second one

47
Numerical
Step-1 Sorting out the given provided by the question
Given,
Sewage flow to SRTF (Q)= 3MLD (million liters per day)=3x106 l/d
Influent BOD (Ci)=175 mg/l
Surface or hydraulic loading (H)=2000 l/m2/d
Organic loading rate (U) = 150 g/m3/d = 1500 kg/ha-m
Step-2 Understanding the problem and listing out the unknown parameter
Depth of filter (d) = ?; Diameter of TF (Φ) = ?; Efficiency (E) = ?
For SRTF, there is no recirculation.
So, recirculation factor (F) = 1
Step-3 Recall BASIC DESIGN EQUATION

RECALLING……
a) Recirculation ratio (r): b) Recirculation Factor (F):
𝐫 =
𝐑
𝐈
𝐅 =
𝟏+𝐫
𝟏+𝟎.𝟏𝐫 𝟐
where, where,
R= re-circulated flow r= recirculation ratio
I= influent flow
r= 2 MGD or 2:1 MGD
First number>>recirculated proportion
Second number>>influent proportion
48

c) Organic Loading (U): d) BODs Removal Efficiency (E):
𝐔 =
𝐖
𝐕𝐱𝐅
𝐄 =
𝐂𝒊−𝑪𝒆
𝑪𝒊
=
𝟏
𝟏+𝟎.𝟎𝟎𝟒𝟒√U
where, NOTE:
W= BOD applied to TF (kg/day) For 2nd stage of two stage HRTF,
V= Vol. of TF (ha-m) E2=
𝐶𝑒−𝐶′𝑒
𝐶𝑖
=
𝟏
(𝟏+𝟎.𝟎𝟎𝟒𝟒√U2)(𝟏−𝐄𝟏)
RECALLING……
49

50
Numerical
Step-4 Solving parameters by using BASIC DESIGN EQUATION
For case-I,
BOD applied (W) = Ci*Q = 525000 g/d
If 'V' is the volume of the filter required, then U =
𝑊
𝑉𝐹
⇒ V =
𝑊
𝑈𝐹
= 3500 m3 (In this NRC formula, U must be in kg/ha-m/d)
If 'As' is the plan area of the filter required, then
As =
𝑄
𝐻
= 1500 m2;
Now, depth (d) = V/As = 2.333 m
Diameter of filter (Φ) = √(
4𝑉
𝜋𝑑
)= 43.7019 m
For case-II,
From NRC formula,
Efficiency (E) =
1
1+0.0044√U
= 85.44%

51
Numerical
Design a high rate single stage trickling filter foe treating the following
wastewater of a town having a population of 40000 persons:
a) Domestic sewage @150 lpcd having 200 mg/l of BOD
b) Industrial wastewater @0.25 million litres per day having 600 mg/l of BOD .
Assume the following:
i) BOD removal in primary clarifier =35%
ii) Permissible organic loading of filter =8000 kg/hec-m/day(excluding
recirculation sewage)
iii) Recirculation ratio =1.0
iv)Permissible surface loading =160ML/hec/day(including recirculated sewage)
Also,determine the efficiency of the filter and BOD of the effluent.

52
Numerical
Solution:
Domestic sewage= (150 X 40000) 10^-3 = 6000 m3/day
Industrial wastewater =0.25 X 10^6 litres/day =250 m3/day
Total BOD of domestic sewage = (150 X 40000 X 200)10^-6 =1200 kg/day
Total BOD of industrial wastewater= (0.25 X 10^6 X 600)10^-6 =150 kg/day
Total BOD of combined flow = 1200+150 = 1350 kg/day.
Hence BOD applied to the filter is
W= 1350 X 0.65 =877.5 kg /day.
Filter volume V=
𝑻𝒐𝒕𝒂𝒍 𝑩𝑶𝑫 𝒓𝒆𝒎𝒐𝒗𝒆𝒅(𝑾)
𝑷𝒆𝒓𝒎𝒊𝒔𝒔𝒊𝒃𝒍𝒆 𝒐𝒓𝒈𝒂𝒏𝒊𝒄 𝒍𝒐𝒂𝒅𝒊𝒏𝒈
=
𝟖𝟕𝟕.𝟓 𝑿 𝟏𝟎^𝟒
𝟖𝟎𝟎𝟎
=1096.9 m3
Total volume of wastewater influent, Q= 6000+250 = 6250 m3/day
Now,total inflow including recirculation =(1+R)Q = (1+1)Q =2 X 6250 = 12500 m3/day
Filter area =
𝑻𝒐𝒕𝒂𝒍 𝒗𝒐𝒍𝒖𝒎𝒆 𝒐𝒇 𝒇𝒍𝒐𝒘
𝐩𝐞𝐫𝐦𝐢𝐬𝐬𝐢𝐛𝐥𝐞 𝐬𝐮𝐫𝐟𝐚𝐜𝐞 𝐥𝐨𝐚𝐝𝐢𝐧𝐠
Here, permissible surface loading per day =160 ML/hectare = 16m3/m2

53
Numerical
Solution:
Filter area =
𝟏𝟐𝟓𝟎𝟎
𝟏𝟔
=781.25 m2
Hence,dia. of filter =
𝟕𝟖𝟏.𝟐𝟓 𝑿 𝟒
𝝅
=31.54 m
However, provide a tank of 32 m dia.
Actual surface area = 𝝅
𝟑𝟐 ^𝟐
𝟒
= 804.24 m2
Hence filter depth =1096.9/804.25 =1.364 m.
Provide the depth of media equal to 1.4 m.
Efficiency, E=
𝟏𝟎𝟎
𝟏+𝟎.𝟒𝟒√U
; 𝐔 =
𝐖
𝐕𝐱𝐅
Here W =877.5 kg/day
V=Volume of filter =804.25 X 1.4 =1125.95 m3
F= Recirculation factor =
𝟏+𝑹
(𝟏+𝟎.𝟏𝑹)^𝟐
=
𝟏+𝟏
(𝟏+𝟎.𝟏 𝑿 𝟏)^𝟐
=1.653

54
Numerical
Solution:
E=
100
1+0.44√U
; U =
W
VxF
=76.8%
BOD of effluent =
100−76.8
100
X 877.5 =203.6 kg/day
Total volume of effluent =6250 m3/day
BOD concentration =
𝟐𝟎𝟑.𝟔 𝑿 𝟏𝟎𝟔(𝒎𝒈)
(𝟔𝟐𝟓𝟎 𝑿 𝟏𝟎𝟑)(𝒍)
=32.58 mg/l.

55
Numerical
Two stage filter is designed for an organic loading of 10000 kg of BOD in raw
sewage per hectare metre per day with a recirculation ratio of 1.2. This filter
treats a flow of 4 mld of raw sewage with a BOD of 220 mg/l.Here filter volume
of each stage is half of total volume.What will be BOD of the plant effluent?
Solution:
Total BOD of raw sewage =(4 X 10^6) X (220 X 10^-6) =880 kg/day
Required filter vol.(V) =
𝑻𝒐𝒕𝒂𝒍 𝑩𝑶𝑫 𝒐𝒇 𝒓𝒂𝒘 𝒔𝒆𝒘𝒂𝒈𝒆
𝒑𝒆𝒓𝒎𝒊𝒔𝒔𝒊𝒃𝒍𝒆 𝑩𝑶𝑫 𝒍𝒐𝒂𝒅𝒊𝒏𝒈
=880/10000 =880m3
Let us assume that primary clarifier removes 35% of BOD.
BOD of influent applied to filter(W)= 0.65 X 880 =572 kg.
For each filter, R=1.2 and hence F =
𝟏+𝑹
(𝟏+𝟎.𝟏𝑹)^𝟐
=
𝟏+𝟏.𝟐
(𝟏+𝟎.𝟏 𝑿 𝟏.𝟐)^𝟐
=1.754
E=
100
1+0.44√U
; U =
W
VxF
;E = 72.53%

56
Numerical
Solution:
Amount of BOD left in the effluent from the first stage filter =572(1-0.7253) =157.15
kg/day.
For the second stage filter, efficiency is given by;
E’=
100
1+
0.44
1−e
√ W′
V′xF′
Where,
W’ =157.15 kg/day
V’ =880/2 =440m3
F’ =1.754
e =E/100 =0.7253
E’=
100
1+
0.44
1−0.7253
√ 157.15
440 X 1.754
=58.05%
Total BOD of effluent from the plant=157.15(1-0.5805) =65.95 kg/day
BOD concentration of effluent=
65.95 X 10^6
=16.48 mg/L

57
2.1 Introduction
– It is the process of removing process for treating industrial wastewater and sewage using
and biological floc composed of bacteria and protozoa.
– Basic Principle: microorganisms grow within metabolizing soluted organic material, they
particle that clump together.
– Main Purpose: oxidize carbonaceous organic matter, nitrogenous matter and removing
nutrients (nitrogen and phosphorous).
Fig. 2.1 Activated Sludge Process

PHYSICAL ACTION
When air is supplied, small
particle combine and form
bigger floc. Bigger the better
😉
01
BIOCHEMICAL
ACTION
Bacteria use organic matter
and convert into inorganic
products (N2,P) i.e. BOD
reduction.
02
2.2 Actions Involved
58

Fig. 2.1 Schematic diagram of a typical activated sludge process
Source: https://www.researchgate.net/figure/Schematic-diagram-of-a-typical-activated-sludge-process_fig1_29003
59

Fig. 2.1 Completely mixed process of solids recycle.
Source: Dr. B.C.Punmia
60

61
Fig. 2.1 Graph showing Growth Phases in a Biological system
Source: https://www.google.com

62
2.3 Activated Sludge Plant involves:
• Wastewater aeration in the presence of a microbial suspension
• Solid-liquid suspension following aeration
• Discharge of clarified effluent
• Wasting of excess biomass
• Return of remaining biomass to the aeration tank

Fig. 2.4 Waste Sludge
Source: https://www.shutterstock.com 63

2.4 Biochemistry
 Involves: OXIDATION
 Oxidation: Converts organic matter to Carbon Dioxide, water and other products
 Organic Matter+ O2+ bacteria  CO2+ H2O+other products+ energy
 Nitrification: NH3 + 2O2 → NO2
- + 2H+ + H2O NO2
- + O2 → NO3
-
 Denitrification: NO3
- + C6H12O6 → N2 + CO2 + H2O + energy
 Phosphorus removal: PO4
3- + H2O + microorganisms → H3PO4 + microorganisms
 Synthesis: Synthesize the other portion of organic matter and convert it into new microbial cell
tissue using the part of energy released during oxidation.
Organic Matter+ O2+ bacteria+ energy  New microbial cell tissue+ other
 Endogenous respiration: New microbial cells begin to consume their own cell tissue to obtain
energy for cell maintenance.
New Microbial Cell+ 5O2  5CO2+2H2O+energy
64

2.5 Generalized Biological process reaction
Fig. 2.4 Generalized Biological Process Reaction
Source: https://www.google.com 65

SECONDARY
CLARIFIER
INFLUENT
PRIMARY
CLARIFIER
PRIMARY
SLUDGE
AERATION TANK
OXYGEN
SUPPLY
EFLUENT
WASTE
SLUDGE
SLUDGE
RECYCLE
SECONDARY
CLARIFIER
2.6 Process Description
66

67

2.7 Working
 The process has two distinct functions-synthesis of new cells and stabilization(oxidation
of organic matter.
 Biomass solids generated in an aeration tank is separated and thickened in a clarifier by
gravity settling.
 The activated sludge contains highly aerobic microorganisms which are in moving state
oxidizing organic matter.
 Suspended and colloidal matters tend to coagulate, settles down in secondary sedimen
tank.
 This aerated settled sludge is recycled to the aeration tank by mixing with influent from P
70

2.7 Working (contd.)
 New activated sludge is continuously being produced.
 A portion of it being utilized by returned to aeration tank and excess portion is disposed
 Treatment system consists of:
Reactors: in which microorganisms are kept in suspension.
Settling tank: in which biological flocs from reactor is separated by gravity settling.
Recycled system: to return the portion of settled sludge from clarifier bottom to the react
71

02
04
01
03
Primary
Treatment Unit
Secondary
Sedimentation
Tank
Aeration Tanks
Sludge
Treatment and
Disposal
UNITS FOR ACTIVATED SLUDGE PROCESS
72

2.8 Units Required For Activated Sludge Process
2.8.1 Primary treatment unit
• Consists of screening, grit chamber and primary sedimentation tank which is required
before aeration tank.
• Secondary sedimentation tank is required to let the settling of activated sludge
• It also returns some portion of it for mixing with effluent from primary sedimentation tank
73

2.8.2 Aeration
 process in which oxygen is supplied to waste water to clean it.
 High amount of oxygen is supplied to aeration tank because of high biological oxygen d
(BOD) in sewage.
 Aeration Tank: a chamber usually of rectangular shape where air is continuously supplie
activated sludge mix with effluent.
 Detention time is 4-8 hrs for municipal sewage
 Types:
1. Diffused aeration
2. Mechanical aeration
3. Combined aeration
74

2.8.2 Aeration
2.8.2.1. Diffused Aeration
 Used for very large plant
 Used for high strength of sewage
 Consists of rectangular tank of size 30mx5mx3m to 120mx10mx5m
 Compressed air is passed at a pressure of 0.35 to 0.7 kg/cm2
 Velocity of air should be 60 to 90 cm/minutes
 Types:
1. Ridge and furrow type
2. Spiral flow type
75

2.8.2 Aeration
2.8.2.1.1. Ridge and Furrow Type
 Rectangular of 30 to 120m length,
4.5 to 9m wide,3 to 4.5m deep.
 The diffuser plates are fixed in furrow portion.
 Air is supplied through header pipe.
 Diffuser plates are right angled to direction of flow.
 Freeboard=600mm
Fig. 2.6 Ridge and Furrow Type Aeration Tan
Source: engineeringnotes.com
76

2.8.2 Aeration
2.8.2.1.2. Spiral Flow Type
• Diffuser air is supplied from only one side of tank.
• Air passes through either plate diffuser or through
tube diffusers.
• Spiral motion is developed on only one side of tank
so that length of travel and contact period are
increased.
• Efficiency higher than ridge and furrow type.
• 25% less area of diffuser than ridge and furrow type.
• Corners are right angled.
Fig. 2.7 Spiral Flow Type Aeration Tank
77

2.8.2 Aeration
2.8.2.2. Mechanical Aeration
 Mixture of sewage and returned sludge is agitated in aeration tank by paddles
 5% of air is utilized, remaining goes to atmosphere
 Compressor is not used instead sewage is stirred and take them to top surface every tim
from where mix liquor takes atmospheric air.
 Types:
1. Simplex Aeration
2. Link-belt Aeration
3. Kessner Brush Aeration
78

2.8.2 Aeration
2.8.2.2.1. Simplex Aeration
 Also known as bio-aeration system.
 Consists of deep hopper bottom vertical flow tank of
square in plan.
 Provided with vertical uptake pipe of diameter 50-
70cm at center which is enlarged at bottom with hole.
 Top part consists nozzle on its four sides.
 Driving motor creates suction effect so the mix liquor comes up from the uptake tube an
nozzles spray from top creates aeration
 Rotation of motor=60 rev/min
 4 or more tanks are used in series to increase efficiency of aeration
Fig. 2.8 Simplex aeration
79

2.8.2 Aeration
2.8.2.2.2. Link-Belt Aeration
 The stirring and circulation of mixture is
accomplished by a rotating steel paddle wheel
which extends for full length of tank.
 Speed=45-50 rev/min with the help of driving
unit.
 Wheel rotates pushing the mix liquor down in the
tank and rises in the left channel behind baffle that
creates circulation of liquor and aeration takes place.
Fig. 2.9 Link-Belt Aeration
80

2.8.2 Aeration
2.8.2.2.3. Kessner Brush Aeration
 Adopted for small sewage treatment plant.
 Similar to spiral flow type diffused aeration
unit.
 Spiral flow is developed by rotating paddle
which displaces the mix liquor and it is
replaced by other mix liquor thus allows
aeration from the surface.
 Sides are truncated.
 Efficiency is very low, not used nowadays.
Fig. 2.10 Kessner Brush Aeration
81

2.8.2 Aeration
2.8.2.3. Combined Aeration
 Both system combined to single unit to get efficiency of
both
 Aeration done by compressed air using air diffuser,
movement of mixed liquor is done by mechanical paddles
 Movement of paddle should be clockwise or anticlockwise
as per the position of diffuser
 Detention time = 3-4 hrs.
 example: Dorr aerator Fig. 2.11 Dorr Aerator
82

Schematic diagram of Activated sludge
process
83
Source: The Biological Basis of Wastewater Treatment

Process variables governing
Treatment
1. Hydraulic retention time()
2. Volumetric BOD loading/volumetric organic loading
3. Food to micro-organisms ratio(F/M ratio)
4. Mean cell residence time(MCRT)/ Sludge retention time(SRT)/ Sludge
age
84

1.Hydraulic Retention Time()
The average time for which the sewage flowing into the reactor (or aeration
tank) remains in the reactor (or aeration tank), or in the entire system.
Hydraulic retention time can be defined as:
hydraulic retention time(θ) =
volume of reactor(V)
Influent sewage flow rate (Q)
where, v is in m3
Q is in m3
day
85

2.Volumetric BOD loading
The BOD applied for unit volume of aeration tank per day.
If S0 = BOD concentration of influent sewage mg l
Q = influent sewage flow rate(m3 day)
V = volume of aeration tank m3
Then,
volumetric BOD loading(mg l day) =
Q∗S0
V
86

3.Food to microorganisms ratio(F/M
ratio)
The BOD concentration of influent sewage represents ‘food’ and the concentration of
volatile Suspended solids(MLVSS) represents ‘microbial mass’.
Here,F = BOD(mg l) ∗ influent discharge(m3 day)
M = conc. of volatile suspended solid(mg l) ∗ volume of aeration tank(m3)
Thus,
F
M
=
S0∗Q
X∗V
F
M
=
s0
X∗θ
where =
V
Q
, is hydraulic retention time
87

3.Food to microorganisms ratio(F/M
ratio)
88
Source:thewastewaterblog.com

4.Mean cell residence
time(MCRT),𝒄
Also known as sludge retention time(SRT) or sludge age
It’s the average time for which volatile suspended solids remain in aeration tank
or the entire system.
θc =
mass of volatile suspended solid in aeration tank
mass of volatile suspended solid leaving system per day
=
V∗x
Qw∗xw+Qe∗xe
where
V = aeration tank volume m3
x = concentration of VSS in aeration tank(mg l)
Qw Qe = waste sludge and treated effluent flow rate(m3 day)
xw xe = concentration of VSS in waste sludge and treated effluent respectively
89

2.9 Design Criteria
The following are design criteria that must be fulfilled while designing aeration plants:
1. No. of aeration tank, N= min. 2 (for small plants)
= 4 or more (large plants)
2. Depth of waste water in tank= 3-4.5m (usually)
= 4.5-7.5m (diffuse aeration)
= 1-6 m (surface aeration)
3. Free board (FB)= 0.3-0.6m (diffuse aeration)
= 1-1.5m (surface aeration)
4. For rectangular tank; L:B=3:1 to 5:1
B:D=3:1 to 4:1
(depends on aeration system) 90

2.9 Design Criteria
5. For air requirements;
30-55 m3 of air/kg of BOD removed for diffused aeration when
𝐅
𝐌
≥ 𝟎. 𝟑
𝐅
𝐌
≤ 𝟎. 𝟑
6. Hydraulic retention time= 4-8 hours
7. Volumetric BOD loading (VBL)= 0.3-0.7kg/m3/d
91

2.10 Merits and Demerits
2.10.1 Merits
 Effluent high quality that TF
 80 – 95% BOD removal and bacteria removal is 90 – 95%
 Land area lesser than that for TF.
 Designed unit made compact for industrial use.
 Initial cost lower than TF.
 Process is free from foul and insect nuisance.
 Sludge has a high fertilizing value/the effluent can be used for irrigation.
92

2.10 Merits and Demerits
2.10.2 Demerits
 Skilled persons required.
 Operation cost high.
 Sludge produced may create problem on sludge disposal.
 Sensitive: temperature is more or low, bacteria may die.
 Requires more oxygen supply.
 New plant requires seeding
 Requires four weeks if not seeding is done.
93

Design Criteria (Contd.)
• 5. For air requirements,
• (i) 30-55 m2 of air/kg of BOD removed for diffused aeration
• when
𝐹
𝑀
≥ 0.3
• (ii) 70-115 m2 of air/kg of BOD removed for diffused aeration
when
𝐹
𝑀
≤ 0.3
• 6. Hydraulic retention time(HTR) = 4 – 8
• 7. volumetric BOD loading (VBL) = 0.3 - 0.7 mg/𝑚3/d
94
2.11 Sludge Volume Index
 Indicates the physical state of the sludge
 Degree of concentration of sludge in the system
 Ratio of volume of sludge in ml to the dry weight of the sludge in gm.
 Mathematically, SVI = Vs/Xt (ml/gm)
Where, Vs = Volume of sludge in ml
Xt = Mix liquor suspended solids (MLSS) in mg/l.
Relation Between SVI and return ratio(r):
 Return ratio(r) =
Qr
Q
=
Xt
106
SVI
−Xt
where SVI is in ml g

2.12 Determination of SVI
Practically, SVI is determined as:
1. Take 1000 ml Imhoff cone/ fill with mix liquor
2. After 30 minutes, record the volume of settled sludge (Vs). in ml.
3. Remix and test for MLSS concentration as ‘Xt ’ in gm/ml.
4. Use above formula to find SVI.
95

2.5.2 Demerits
• Skilled persons required.
• Operation cost high.
• Sludge produced may create problem on sludge disposal.
• Sensitive: temperature is more or low, bacteria may die.
• Requires more oxygen supply.
• New plant requires seeding
• Requires four weeks if not seeding is done.
96
NUMERICAL TIME!!!!

WAY
TO
SUCCESS
Sort out
the given
provided by
the question
01
Understand
the given
problem
and
list out
unknown
parameter
02
Recall
DESIGN
CRITERIA
03
Solve
dimensions
by using
DESIGN
CRITERIA
04
Conclude
your
numerical
05
5 STEPS
TO SOLVE NUMERICAL
REMEMBER?
97

98
Numerical
Q. Design a conventional activated sludge plant to treat domestic sewage with diffus
aeration system with the following data: Population = 120000; Per capita sewage flo
BOD5 of sewage = 200 mg/l; F/M ratio = 0.2/day ; MLSS = 3000 mg/l; SVI = 100 ml/g.
LETS USE OUR ULTIMATE ALGORITHM TO SOLVE THE NUMERICAL !!!!

99
Numerical
Given,
Population (P)= 120000
Per capita sewage contribution (q)=160 lpcd
BOD of sewage (S0)=200 mg/l
Food microorganism ratio (F/M)=0.2/d
MLSS (Xt)=3000mg/l
Sludge volume index (SVI)=100ml/g
Design Activated Sludge=?
Step-3 Recall DESIGN CRITERIA

RECALLING…..
2.9 Design Criteria
The following are design criteria that must be fulfilled while designing aeration plants:
1. No. of aeration tank, N= min. 2 (for small plants)
= 4 or more (large plants)
2. Depth of waste water in tank= 3-4.5m (usually)
= 4.5-7.5m (diffuse aeration)
= 1-6 m (surface aeration)
3. Free board (FB)= 0.3-0.6m (diffuse aeration)
= 1-1.5m (surface aeration)
4. For rectangular tank; L:B=3:1 to 5:1
B:D=3:1 to 4:1
(depends on aeration system) 100

RECALLING…..
2.9 Design Criteria
5. For air requirements;
𝐅
𝐌
≥ 𝟎. 𝟑
𝐅
𝐌
≤ 𝟎. 𝟑
6. Hydraulic retention time= 4-8 hours
7. Volumetric BOD loading (VBL)= 0.3-0.7kg/m3/d
101

102
Numerical
Step-4 Solving dimensions by using DESIGN CRITERIA
1) Average sewage flow(Q) = 120000× 160 = 192 × 105
l/d = 19200 m3
/d
2) Volume of AT(V):
we have,
F
M
=
Q∗So
V∗X
or, 0.2 =
19200∗200
𝑉∗3000
or, V = 6400 m3
3) Check for hydraulic retention (HRT):
HTR =
𝑉
𝑄
=
6400
19200
day =
6400
19200
× 24hr = 8hr OK. [HTR = 4-8 hrs]
4) Checking for volumetric BOD loading (VBL):
VBL =
Q×So
V
=
19200×200
6400×103 kg/m3/d = 0.6 kg/m3/d OK. [0.3-0.7 kg/m3/d]

103
Numerical
5) Checking for return sludge ratio(r):
r =
Qr
Q
=
𝑋𝑡
(
106
𝑆𝑉𝐼
−𝑋𝑡)
where SVI in ml/g
so, r =
3000
(
106
100
−3000)
= 0.429 OK. [r = 0.25-0.5]
6) Aeration tank (AT) dimension:
Assume effective depth (d) = 3m [d = 2.5-4.5m] and FB = 0.5m (assume)
∴overall depth (D) = D+FB = 3 + 0.5 = 3.5m
assuming width (B) = 5m [B = 5-10m]
𝐵
𝐷
=
5
3.5
= 1.428 OK. [B/D =1.2-2.2]

104
Numerical
checking for horizontal velocity (Vh) in AT
Return flow (Qr) = r*Q = 0.429Q
Flow in AT (Qat) = Q+Qr = Q(1+0.429)
= 19200(1+0.429) m3/d
=
19200(1+0.429)
24∗60
m3/min = 19.05 m3/min
Now Vh =
Qat
B×d
=
19.05
5×3
= 1.27 m/min ok. [Vh ≤ 1.5m/min]
So, Length(L) = V/Av =
V
B×d
=
6400
5×3
= 426.67m but 30m ≤ L ≤ 120m
providing no of tank (Nt) = 5
length of each tank (l) = L/Nt = 426.67/5 = 86 m
Provided 5 aeration tank each of sixe 86m × 5m × 3m and thickness
of baffles walls as 0.25m minimum in parallel with flow in series.

105
Numerical
7) Air requirement (Qair):
assuming diffuser type aerator, adopt air requirement (Cair) = 65 𝑚3 of
air /kg BOD applied [40-125 but 65 in general]
BOD applied (W) = Q × S0 = 192× 105
× 200 mg/d
=
192×105×200
1000×1000
= 3840 kg/d
Air required (Qair) = Cair × W = 65 × 3840 = 249600 𝑚3 /d
=
249600
24×60
= 173.33 𝑚3
/min

Numerical
Step-5 CONCLUSION
106

Numerical
2.Design a activated sludge unit treatment for a town with
following data:
Population:65000
Average sewage flow:210lpcd
BOD of raw sewage:210mg/l
BOD removal in primary treatment unit:40%
Overall BOD removal required:90%
107

Numerical
Given,
Population (P)= 65000
Per capita sewage contribution (q)=210 lpcd
BOD of sewage =210mg/l
BOD removal in PST=40%
Overall BOD removal=90%
Design Activated Sludge=?
Step-3 Recall DESIGN CRITERIA
Influent BOD at Aeration Tank=(100-40)% of 210mg/l=126 mg/l
BOD removal required in AT(SRAT)=(90-40)% of 210mg/l=105 mg/l
108

Numerical
Average sewage flow rate(Q)=p*q=65000*210(l/d)
=568.75m3 hr
1.Hydraulic Retention Time (from American Public Health Formula),
HRT=
SRAT
20
− 1 hr=(
105
20
− 1)=4.25 hrs. [4hrs-8hrs],okay.
2.Sludge return ratio(r) and sludge return rate(Qr):
Assuming r=25%
r=
Qr
Q
Qr = 0.25Q
3. Total flow to aeration tank 𝑄𝐴𝑇 = Q + Qr = 1.25Q
 𝑄𝐴𝑇 = 1.25 ∗ 568.75 = 710.94 m3 hr
4.Volume of Aeration Tank(V)=QAT ∗ HRT = 710.94 ∗ 4.25 = 3021.5m3
109

Numerical
5.Check for volumetric BOD loading:
VBL =
Q×So
V
=
13650000∗126
3021.5∗1000000
(kg m3
day)
=0.569 kg m3
day,[0.3-0.7 kg/m3
/d],okay
6.Tank Dimensions:
Assuming effective depth(d)=2.5m [d = 2.5-4.5m] ,FB=0.5m
∴overall depth (D) = d+FB = 2.5 + 0.5 = 3m
assuming width (B) = 5m [B = 5-10m]
𝐵
𝐷
=
5
3
= 1.667 OK. [B/D =1.2-2.2]
110

Numerical
Checking for horizontal velocity (Vh) in AT
Now Vh =
𝑄𝐴𝑇
B×d
=
710.94
5×2.5∗60
(m min) = 0.948 m/min ok [Vh ≤ 1.5m/min]
Now,
Length(L)=
V
B∗d
=
3021.5
5∗2.5
= 241.72m, but 30m ≤ L ≤ 120m
Thus, providing providing no of tank (Nt) = 3
length of each tank (l) = L/Nt = 241.72/3 = 80.57m
Provided 3 aeration tank each of size 81m × 5m × 2.5m.
111

Numerical
7) Air requirement (Qair):
assuming diffuser type aerator, adopt air requirement (Cair) = 65 𝑚3
of
air /kg BOD applied [40-125 but 65 in general]
BOD applied (W) = Q × S0 = 13650000× 126 mg/d
=
13650000×126
1000×1000
= 1719.9 kg/d
Air required (Qair) = Cair × W = 65 × 1719.9 = 111793.5 𝑚3 /d
=
111793.5
24×60
= 77.63 m3/min
112

Numerical
113
Step-5
CONCLUSION

3. Oxidation Pond
3.1 Introduction
– It is the large, shallow ponds designed
to treat wastewater through the
interaction of sunlight, bacteria, and algae.
– Bacteria oxidizes organic waste releasing
inorganic chemicals.
– Algae uses inorganic chemicals and sunlight
for photosynthesis releasing O2.
– Also called:
1. Redox Pond
2. Water stabilization Pond
3. Lagoon
Fig. 3.1 Oxidation Pond
Source: researchgate.net/publication/278882875
114

3. Oxidation Pond
3.2 Purpose
 Secondary treatment of wastewater, both
industrial and municipal.
 BOD reduction rate of about 90%.
 High removal rate of pathogenic bacteria
(90-99%).
 Simple and natural method.
 Low construction and operation cost for
small and isolated units.
 Ideal in tropical regions.
 Suitable for rural areas where land is available at low cost.
Fig. 3.2 Oxidation Pond
Source: https:www.theborneopost.com
115

3. Oxidation Pond
3.3 Mechanism
 First the bacteria present in the oxidation
pond will oxidize waste of the domestic as
well as industrial sewage. By doing this,
bacteria release carbon dioxide, water and
ammonia.
 The algal growth occurs in the presence of
sunlight. It utilizes the inorganic wastes
formed by the organic matter decomposition
and releases oxygen.
 As oxidation and reduction reaction co-occurs,
an oxidation pond is sometimes called “REDOX
POND”.
Fig. 3.3 Mechanism of Oxidation Pond
Source: https://biologyreader.com/oxidation-pond.html
116

3. Oxidation Pond
3.4 Process Involved
The process involved in the conversion of industrial and domestic wastewater into simple
includes:
1. Firstly, the industrial or domestic wastewater
influents enter the oxidation pond through the
inlet system.
2. Then, the bacteria utilize the biodegradable
organics and convert them into inorganic
compounds by releasing carbon dioxide.
Achromobacter, Proteus, Alcaligenes,
Pseudomonas, Thiospirillum, Rhodothecae etc.,
are the following genera of the bacteria that predominate
in the stabilization pond.
Fig. 3.3 Design of Oxidation Pond
Source: https://biologyreader.com/oxidation-pond.html
117

3. Oxidation Pond
3.4 Process Involved (contd.)
3. The algal biomass in the oxidation pond utilizes the inorganic compounds in the presenc
sunlight and carbon dioxide released during the oxidation of organic waste. Chlorella,
Scenedermus and Microcystis are the most common genera of algae predominating in t
stabilization pond.
4. The remaining non-biodegradable or solid organic wastes settle down in the bottom of
stabilization pond as sludge. The anaerobic bacteria convert them during the night and
absence of oxygen. The anaerobic bacteria first convert the insoluble organic waste into
organic acids like ethanol. Further decomposition of organic acids by the anaerobic ba
release H2S, NH3, CH4, CO2 etc.
5. The treated water releases out through the outlet system of a stabilization pond. By em
the dredging method, one can separate the sludge deposits from the stabilization pond.
The filtration method or the combination of chemical treatment and settling process se
the algal and bacterial biomass.
118

3. Oxidation Pond
3.5 Construction of Oxidation Pond
 Pond shape
1. No corners (usually Round Corners)
To avoid dead pockets
2. Length≤ 3xBreadth
3. Length<750m
4. Depth usually 1 to 1.5 m
 Embankment
In figure
FB 0.5 1.5 to 3m
1:1 to 3:1
2:1 to 2.5:1
119

3. Oxidation Pond
Commissioning
When construction of pond is completed, some groundwork is
required to be done before commencing the operation of pond.
1. Culture Method:-
i) Sewage is filled up to 15 cm and algae seeds are added.
ii) Depth is maintained up to 15 cm for a whole week.
iii) When pond seems green, sewage is applied up to operation
level.
120

3. Oxidation Pond
2. Natural method :-
i) Sewage is filled up to operation level while inlets and outlets
are kept closed.
ii) Depth level is maintained for two weeks where algae grows by
natural process. Failure of this method suggests use of culture
method.
121

3. Oxidation Pond
3.6 Factors Affecting Pond Performance
3.6.1. Natural Factors
 Wind Action
 Temperature
 Sunlight
122

3. Oxidation Pond
3.6.2. Physical Factors
 Surface Loading Rate
 Water Depth
 Short circuiting
 Series Operation
 Parallel Operation
123

3. Oxidation Pond
3.6.3. Chemical Factors
 Oxygen
 pH
124

3. Oxidation Pond
3.7 Operation and Maintenance
 Flow regulation to prevent growth of water weeds
 Use rakes or pump to break up scum formation
 Install aerators to prevent foul odor
 Stock pond with Gambusia to control mosquito
 Parallel feeding to pond to reduce loading
 Eliminate sources of toxic discharge
 Recirculation via pump to improve mixing
125

Advantages:
Simplicity to built, operate
and maintain.
Low operational cost.
Good quality of effluent
Disadvantages:
Need large areas.
Limited and dependent on weather
Quality of effluent varies
3. Oxidation Pond
3.8 Merits and Demerits of Oxidation Pond
126

3. Oxidation Pond
3.9 Design Criteria
 Amount of sunlight
 Temperature
 Flow
 Influent BOD
 Effluent limit
Fig. 3.4 Oxidation Pond and Process
Source: M. Ramandan and V. M. Ponce, "Design and performance of
waste stabilization pond,"
127

3. Oxidation Pond
3.9 Design Criteria (contd.)
 Detention time
Lr=La(1-10-Ptxt)
where,
La=influent BOD
Lr=BOD removed
Pt=BOD removal rate at toC
T=Detention period in days
Pt=P20(1.047T-20)
P20=0.1 per day
128

3. Oxidation Pond
3.9 Design Criteria (contd.)
 Max. pond area 0.5 ha
 Desludging time (6-12 years)
Taking sludge accumulation rate 0.05-0.1m3 per capita year
 Effective depth=0.8-1.5m
 Free board=0.5m
 Width to length ratio=1:2 to 1:3
 Side slope=1:1.5 to 1:3
129

2.5.2 Demerits
• Skilled persons required.
• Operation cost high.
• Sludge produced may create problem on sludge disposal.
• Sensitive: temperature is more or low, bacteria may die.
• Requires more oxygen supply.
• New plant requires seeding
• Requires four weeks if not seeding is done.
130
NUMERICAL TIME!!!!

131
HERE WE GO AGAIN!!!
Q. Design an oxidation pond for the following data
i. Location
ii. Elevation
iii. Mean monthly temperature
iv. Population to be served
v. Sewage flow
vi. Desired effluent BOD
vii.Pond removal constant at 20o
C
24o latitude
900 m above mean sea
level
30o max and 10 ͦ min
8000
160 lpcd
30 mg/l
0.1 /d

132
Numerical
Solution:
1. Total BOD load
BOD per capita/day= (160x300)10-6 = 0.048 kg/day
Thus, Total BOD load= 8000x0.048=348 kg/day
2. Permissible areal BOD loading
Areal BOD loading at 240 latitude=225kg/ha/d
Correction factor for elevation=1+0.003x
900
100
=1.027
Correction factor for sky clearance=
100
100+
3∗1.5
10
=0.957
Corrected areal BOD loading=225x0.957/1.027 =210kg/ha/d
3. Pond area
Pond area=
𝑇𝑜𝑡𝑎𝑙 𝑎𝑝𝑝𝑙𝑖𝑒𝑑 𝐵𝑂𝐷
𝐴𝑟𝑒𝑎𝑙 𝐵𝑂𝐷 𝑙𝑜𝑎𝑑𝑖𝑛𝑔
=
384
210
= 1.83 ha

133
Numerical
4. Detention period
Pond removal constant at 10o C is
P10 = 0.1(1.047)10−20= 0.1(1.047)−10= 0.06317
∴ Detention period, t =
1
PT
log10
La
La − Lr
Here La − Lr = BOD remaining = 30
∴ t =
1
0.06317
log10
300
30
= 15.85 days ≈ 16 days
5. Pond volume and pond depth
Total inflow = 160 x 8000 = 1280 x 103 l/d = 1280 m3/d
∴ Pond volume = 1280 x 16 = 20480 m3
∴ Depth =
20480
1.83×10000
≈ 1.12 𝑚
Provide depth of 1.2 m and free board 0.5 m
Fig. 3.5 Layout plan of stabilization pond

Numerical
6. Pond system
Total pond area = 1.8 ha .
Let us assume 6 ponds with 4 primary and 2 secondary ponds of equal area arranged
parallel-series system as shown in figure.
Area of each pond = 1.83/6 = 0.305 ha = 3050 m2
Adopt rectangular pond with width to length ratio = 1:2
∴ (B)(2B)=3050
B =39.05 m (take 39.5 m)
L=79 m
Actual area of each pond = 39.5 x 79 = 3120.5 m2
134

References:
 Punmia, B. C., Jain, A. K., & Jain, A. K. (1998). Wastewater Engineering. New Delhi:
Laxmi Publications.
 Garg, S. K. (1979). Sewage Disposal and Air Pollution Engineering. New Delhi: Khanna
 https://www.biologyreader.com
 https://www.google.com
 https://en.Wikipedia.org
 https://www.britannica.com
 https://www.sswm.info
 https://biologyreader.com/
 www.engineeringenotes.com
135

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