Faculty of Civil Engineering & Built Environment

ENVIRONMENTAL
ENGINEERING
BFC 32403

CHAPTER 6
WASTEWATER TREATMENT (PART 1)
 Chapter 6:
Wastewater Treatment

Topics: (6 hours lecture)

6.1 Types of Wastewater
6.2 Effluent Discharge Standard
6.3 Hydraulic and Organic Loading
6.4 Preliminary treatments
6.5 Primary treatment
6.6 Secondary Treatment (Biological Process)
6.7 Suspended Growth Biological Process
6.8 Attached Growth Biological Process
6.9 Sludge Management
 Objectives

▪ To describe the types of wastewater

▪ To understand the flow process of wastewater
treatment
▪ To eliminate the pollutants and toxicants from
wastewater
 6.1 Types of Wastewater

Wastewater is any water that has been adversely affected in quality by

anthropogenic influence

Domestic Wastewater comes from residential area, commercial outlets,

(Sewage) institutional facilities & recreational facilities. It
contains decomposable organic matter and
pathogenic agents

Industrial wastewater
Industry processes generate a wide
variety of wastewater pollutants. It
contains toxic agents ranging from metal
salts to complex synthetic organic
chemicals, in which specific toxic and
hazardous compound may exist.
 6.2 Effluent Discharge Standard
Industrial Plant

w
lo
rf
ve
Ri
Standard A Standard A

Raw water intake

Industrial Plant
Residential Area
Standard B
Water treatment plant
Standard B

Standard B
Residential Area

Treated effluent discharge standard ( )

 6.2 Effluent Discharge Standard

Extracted from
Environmental
Quality (Sewage)
Regulations 2009
(PU(A) 432)
 6.2 Effluent Discharge Standard

Extracted from
Environmental
Quality (Sewage)
Regulations 2009
(PU(A) 432)
6.2 Effluent Discharge Standard
Guideline for Sewerage
 6.2 Effluent Discharge Standard
Sewage Treatment Plant

Malaysia:
Where
sewage
goes?

IWK Sustainability Report 2012-2013

6.2 Effluent Discharge Standard

IWK STP BUNUS- sewage and sludge treatment

6.2 Effluent Discharge Standard
Guideline for Sewerage

MALAYSIAN
SEWERAGE INDUSTRY
GUIDELINES (MSIG) ,
VOLUME 4 STPS
–SECTION 5
6.2 Effluent Discharge Standard
Guideline for Sewerage

MALAYSIAN
SEWERAGE INDUSTRY
GUIDELINES (MSIG) ,
VOLUME 4 STPS
–SECTION 5
 6.2 Effluent Discharge Standard
Malaysia: Sewage Treatment Process at the STPs

Network Pump Stations (NPSs) pump sewage from low lying areas so it can flow to the Sewage Treatment Plants (STPs).

IWK Sustainability Report 2012-2013

 6.2 Effluent Discharge Standard
Malaysia :Typical Sewage Treatment Flow Process

IWK Sustainability Report 2010.

6.2 Effluent Discharge Standard

MALAYSIAN SEWERAGE INDUSTRY

GUIDELINES (MSIG) , VOLUME 4
STPS –APPENDIX A
 6.3 Hydraulic and Organic Loading
Hydraulic Load (Peak Flow)

Based on MS 1228:1991 (Malaysian Standard : Code of Practice

for Design and Installation of Sewerage Systems),

Peak flow factor = 4.7 x p -0.11

where p = equivalent population, in thousand

The peak flow is required in the design of sewerage,

pumping stations and components of the treatment plant
 6.3 Hydraulic and Organic Loading
Hydraulic Load (Peak Flow)
Equivalent Population (PE) (MS 1228:1991)
No. Type of premise/establishment Population equivalent
( recommended)

1 Residential 5 per unit *

2 Commercial ( includes entertainment/recreational 3 per 100 m gross area

centres, restaurants, cafeteria, theatres)
1 peak flow is
equivalent to
3 School/educational institutions 0.2 per school
225 L/cap.day
- Day schools/ institutions 1 per school
- Fully residential 0.2 per student for
- Partial residential non-residential student and
1 per student for residential student

4 Hospitals 4 per bed

5 Hotels ( with dining and laundry facilities) 4 per room

6 Factories (excluding process waste) 0.3 per staff

7 Market (wet type) 3 per stall

8 Petro kiosks/service stations 18 per service bay

9 Bus terminal 4 per bus bay

6.3 Hydraulic and Organic Loading
Hydraulic Load (Peak Flow)

Continued

No Type of premise/establishment Population equivalent

( recommended)
10 Taxi Terminal 4 per taxi bay
11 Mosque / Church / Temple 0.2 per person
12 Stadium 0.2 per person
13 Swimming Pool or Sports Complex 0.5 per person
14 Public Toilet 15 per toilet
15 0.2 per passenger/day
Airport
0.3 per employee
16 Laundry 10 per machine
17 Prison 1 per person
18 Golf Course 20 per hole
6.3 Hydraulic and Organic Loading
Hydraulic Load (Peak Flow)
Example 1

Calculate the peak flow for a new development area consists of

250 unit houses.

Solution

Population = 250 house x 5 PE/house = 1250 PE

Dry weather flow = 225 L/day/cap

Hydraulic flow = 1250 PE x 225 L/day/cap = 281,250 L/day

= 0.003 m 3/s

Peak flow = Peak factor x hydraulic flow

= 4.7 x p -0.11 x 0.003 m3/s
= 4.7 x (1250/1000) -0.11 x 0.003 m3/s
= 4.59 x 0.003 m 3/s
= 0.014 m3/s
6.3 Hydraulic and Organic Loading
Hydraulic Load (Peak Flow)

TUTORIAL

A new development area consists of:

i. 1 market of 20 stalls
ii. 1100 unit houses

Use flow of 225 L/day/cap, calculate:

i. PE
ii. Hydraulic flow
iii. Peak flow factor
iv. Peak flow
 6.3 Hydraulic and Organic Loading
Organic Loading

The design influent values to be adopted in the

design of a treatment plant.

MALAYSIAN SEWERAGE INDUSTRY

GUIDELINES (MSIG) , VOLUME 4 STPS
–SECTION 3
 6.4 Preliminary Treatments
Primary Screen (Coarse screen or bar screen or bar rack)
● Upon reaching the sewage treatment
plant, sewage flows through the primary
screening facility which is the first stage
of treatment.
●
● The screens must be provided upstream of
all inlet pump stations and shall be
designed to protect downstream processes
and equipment.
●
● The purposes of primary screens are:
○ To protect equipment from rags,
wood and other debris.
○ To reduce interference with in-plant
flow and performance.

MALAYSIAN SEWERAGE INDUSTRY GUIDELINES (MSIG) ,

VOLUME 4 STPS –SECTION 5
 6.4 Preliminary Treatments
Secondary Screen (Fine Screen)
● After the inlet pump station, further
screening is required to reduce the
remaining floating matter and finer
particles in the sewage that will disrupt
the treatment process downstream.

● The purposes of secondary screens

are:
○ To remove material such as plastic,
paper, cloth and other particles that
may cause problems to the
treatment process downstream.
○ To minimise blockages in sludge
handling and treatment facilities.

MALAYSIAN SEWERAGE INDUSTRY GUIDELINES (MSIG) ,

VOLUME 4 STPS –SECTION 5
 6.4 Preliminary Treatments
Grit and Grease Chamber
❖ Purpose: To minimise problems associated with grit and grease.

❖ Grit creates problems to pumps and also sludge digestion and dewatering facilities.

❖ Grease creates problems at the clarifier and is carried over in the final effluent.

❖ In grit removal system, grit or discrete particles that have subsiding velocities or
specific gravities substantially greater than those of organic putrescible solids, e.g.
eggshells, sands, gravel are removed by gravitate settlement or centrifugal separation.

❖ Same principle applies to oil and grease removal system, where free oil and grease
globules lighter than water rise through the liquid and skimmed from the top surface.

❖ The particles must be removed at an early stage of the process because:

○ Grit particles cannot be broken down by any biological treatment.
○ Grit particles are abrasive and wear down equipment.
○ Biological treatment in sewage treatment works is not designed to degrade
grease.
 6.4 Preliminary Treatments
Provision Requirement of Grit and Grease Removal System
6.4 Preliminary Treatments
Grit Chamber
6.4 Preliminary Treatments
Grease Chamber
 6.4 Preliminary Treatments
Grit Chamber

According to According to MS 1228: 1991, there are 3 types of grit removal

system and comprise either the horizontal Constant Grit Chamber, Aerated
Grit Chamber and Detritor

1) The horizontal flow grit chamber

* flow through velocity should not exceed 0.23 m/sec
* surface loading rate should not exceed 1500 m3/d/m2

2) Aerated grit chamber

* Maximum detention time to be 3 min
* Air rates should be in the range of 4.5 to 12.5 liter/sec/m of tank
* Depth to width ratio of 1:2
* Length to depth ratio of 1: 2

Horizontal-flow grit chamber is applied commonly and comply to Type 1

Sedimentation
 6.4 Preliminary Treatments
Grit Chamber: Horizontal flow grit chamber
Example 2

Design a rectangular, horizontal grit removal facility to remove grit with a specific
gravity of 2.65 and a particle settling velocity which ranges between 0.016 to 0.022
m/sec. The horizontal velocity will be maintained at 0.3 m/sec by using a
proportional weir. Determine the channel dimensions for a maximum wastewater
flow of 0.37 m3/s.

Solution:

Settling velocity (Vs): assume 0.016 m/sec for the design

Detention time (tD): 60 sec

1) Depth of Flow (D),

Knowing tD = D/Vs
So,
60 sec = D/0.016 m/sec
D = 0.96 m
Provide 0.3 m free board and 0.25 m grit accumulation zone depth, hence the
total depth = 0.96 + 0.30 + 0.25 = 1.51 m
 6.4 Preliminary Treatments
Grit Chamber: Horizontal flow grit chamber
2) Length of Tank (L)

Knowing, L = tDVh

Where: Vh = horizontal flow velocity in the tank = 0.3 m/sec

Therefore , L = (60 sec)(0.3 m/sec) = 18 m

Provide 25 % additional length to accommodate inlet and outlet zones, hence
the length of the grit chamber = 18 X 1.25 = 22.5 m

3) Width of Tank (w)

Knowing A = Q/Vh [cross sectional area, A= Qdesign x Vh)

Where: A = cross sectional area of flow

Q = flow rate
Vh = horizontal flow velocity in the tank

Q = 0.37 m3/sec
A = (0.37 m3/sec)/(0.3 m/sec) = 1.23 m2
Then w = A/D = (1.23m2)/(0.96 m) = 1.26 m
 6.4 Preliminary Treatments
Equalization Tank (Balancing Tank)

▪ The wastewater/sewage produced often varies in overall wastewater flow rate

(e.g.: as a result of day vs. night production processes) as well as
concentration of pollutants (resulting from different types of operations being
carried out at different times).

▪ Balancing tanks are mandatory for all treatment processes that are not
designed at peak flow. The tanks are effective means of equalising
sewage flow. For extended aeration plants that are designed with a
retention time of more than 18 hours and clarifiers designed at peak
flow, the use of balancing tanks is not required.

▪ The purposes of balancing tank are:

○ To prevent flow variations entering secondary treatment processes.
○ To reduce hydraulic loading into secondary treatment processes.
○ To reduce potential overflows that may cause health hazard and
pollution.
 6.4 Preliminary Treatments
Equalization Tank (Balancing Tank)
The design requirements for balancing tanks are:
a. All balancing tanks must be completely aerated and mixed.
b. Flow control shall be by a non-mechanical constant flow device, such
as an orifice, in order to avoid double pumping.
c. Allowance must be made for an emergency overflow.
d. Bypass and drain down facilities as well as suitable access for cleaning
shall be provided.
e. A dead water depth of 0.6 - 1.0 m is normally required.
f. For plants with PE > 10 000, where balancing tank is used, minimum
one (1) unit of balancing tank shall be provided. The design flow of the
upstream and downstream unit processes are recommended as follow:
i. Where no balancing tanks is provided, design flow of unit
process at:
1. Upstream = Peak/pumped flow
2. Downstream = Peak/pumped flow
ii. Where balancing tank is provided, design flow of unit process at:
1. Upstream = Peak/pumped flow
2. Downstream = Average flow

MALAYSIAN SEWERAGE INDUSTRY GUIDELINES (MSIG) , VOLUME 4 STPS –SECTION 5

 6.5 Primary Treatment

▪ Unit process where the wastewater is allowed to

settle for a period in a settling/sedimentation tank
or clarifier

▪ Produces clarified liquid effluent in one stream

and a liquid–solid sludge (the settled solids called
primary sludge/raw sludge) in a second stream.

▪ The tank is either rectangular or circular in shape.

Typically the circular tank is used.
6.5 Primary Treatment
Rectangular Primary Tank
6.5 Primary Treatment
Circular Primary Tank
 6.5 Primary Treatment
▪ The objectives of primary treatment are:
o To produce a liquid effluent of suitably
improved quality for the next treatment
stage and
o To achieve a solids separation
resulting in a primary sludge that can
be conveniently treated and disposed
of.

▪ The benefits of primary treatment include:

o Reduction of suspended solids
o Reduction in BOD5
o Reduction in the amount of waste
activated sludge (WAS) in the
activated sludge plant
o Removal of floating material
o Partial equalization of flow rates and
organic load.
MALAYSIAN SEWERAGE INDUSTRY GUIDELINES (MSIG) ,
VOLUME 4 STPS –SECTION 5
 6.5 Primary Treatment
Example 3

Design the rectangular primary tank to treat sewage with peak flow of 7695
m3/d. The weir of 51 m length is to be included. The raw sewage has an
average of 230 mg/L BOD5 and 260 mg/L of suspended solids. By assuming the
primary treatment removes 40% of the BOD5 and 60% of the suspended solid of
the raw sewage, calculate the BOD5 and SS concentration in the primary tank
effluent. Also determine the mass of primary sludge produced per day in term of
dry mass and wet mass assuming a sludge concentration of 6% solids and solid
having a specific gravity of 1.03 kg/l

Solution

From MSIG, Volume 4 STPs , Section 5

Surface loading rate = 45 m3 /day/m2
Depth of water = 2.5 m
Width : depth ratio = 1:1 to 2.5 : 1
Length : width ratio = ≥ 3:1
Detention time = 2 hours
Weir loading rate = 150 m3/day/m
 6.5 Primary Treatment
Solution (continued)

1) The surface area of the tank = 7695 m3/d / 45m 3 /d = 171 m2

2) Assume the depth of rectangular tank = 4 m and width = 6 m.

then the width to depth to ratio is 6 : 4= 1.5:1 (OK!),

therefore, the length = 171 m2/ 6 m = 28.5 m

then the length : width ratio = 28 : 6 = 4.7:1 > 3: 1 (OK!)

3) Volume of the tank = 28 m (L) x 6 m (W) x 4 m (d) = 672 m 3

Therefore, the detention time = Volume / flow rate

= 672 m 3 /7695 m3/d
= 0.087 day
= 2.0 hours (OK!)
4) Weir loading = flow rate / weir length
= 7695 m3/ / 51 m
= 150.8 m3/d/m (OK!)
 6.5 Primary Treatment
Solution (continued)

5) Primary tank effluent

BOD5 primary effluent = (1-0.40) 230 mg/L = 138 mg/L

SS primary effluent = (1-0.60) 260 mg/L = 104 mg/L

6) The production of dry solid and wet mass in the primary tank would be

SS dry mass = Flow rate x Removed SS (in mg/L)

= 7695 m3/d x (260 -104) mg/L
= 7695 m3/d x 156 mg/L
= 1200.42 kg/d

SS wet mass = 1200.42 kg/d / 0.06 = 20007 kg/d

Then , the flow of the sludge , Q sludge

= 20007 kg/d / 1.03 kg/L
= 19424 L/d
 6.5 Primary Treatment
Example 4

Evaluate the primary tank design with respect to detention time,

overflow rate and weir loading as based on MSIG, Volume 4,
STPs Section 5 requirements. Given:

Flow = 0.140 m3/s Length = 45.0 m

Width = 6.0 m Weir length = 81.0 m
Depth = 3.0 m
Solution
Length, width and depth were given. Meaning that the primary
tank is the rectangular tank

Check the,
Length to width ratio, 45 m : 6 m = 7.1 : 1 > 3 : 1 (OK!)

Width to depth ratio, 6 m : 3 m = 2 : 1 (OK!)

 6.5 Primary Treatment

i) The detention time, θ

ii) Overflow rate, Vo

iii) The weir loading,

 Faculty of Civil Engineering & Built Environment

ENVIRONMENTAL
ENGINEERING
BFC 32403
CHAPTER 6
WASTEWATER TREATMENT (PART 2)
 Chapter 6:
Wastewater Treatment

Topics: (Part 2)
6.6 Secondary Treatment (Biological Process)
6.7 Suspended Growth Biological Process
- Activated sludge (Aerobic)
- Septic tank (Anaerobic)
6.8 Attached Growth Biological Process
6.9 Sludge: By Product of WWTP
6.6 Secondary Treatment Biological Process

▪ Provide removal beyond what is achieved in primary

treatment.
o Removal of BOD (including soluble)
o Removal of suspended solids

▪ The objective is to allow the BOD to be exerted in

the treatment plant rather than in the stream.
6.6 Secondary Treatment Biological Process

Biological processes are the most important unit

operations in wastewater treatment. Because the success
of biological processes depends on the environment
provided by treatment units, design engineers need a
basic understanding of factors affecting the growth of
mixed cultures ( microbes).

The stabilization of organic material (pollutant) is

accomplished by microbes which convert colloidal and
dissolved organic matter into gases and protoplasm.
6.6 Secondary Treatment Biological Process
Roles of Microorganisms

Organic material + Microbes Gases + New cells

(colloidal &dissolved) (protoplasma)

Major requirements for microbial growth is organic, will be

a) A terminal electron acceptor measured as BOD
b) Macronutrients: in the effluent.
i) Carbon to build cells
ii) Nitrogen to build cells
iii) Phosporus for ATP (energy carrier)
and DNA has a specific gravity slightly
c) Micronutrients greater than water, it can be
i)Trace metals removed from the treated
ii) Vitamins (required for some bacteria liquid by gravity settling.
d) Appropriate environment
iv) Moisture
v) Temperature
vi) pH
 6.6 Secondary Treatment Biological Process
Roles of Microorganisms

How is this accomplished?

• Create a very rich environment
for growth of a diverse
microbial community
6.6 Secondary Treatment Biological Process
Roles of Microorganisms
Bacteria Growth in Pure Cultures

Phases of Growth
6.6 Secondary Treatment Biological Process
Roles of Microorganisms

Phases of Growth
1. During lag phase, bacteria adapt themselves to growth conditions.
Individual bacteria are maturing and not yet able to divide. During the lag
phase of the bacterial growth cycle, synthesis of RNA, enzymes and other
molecules occurs. So in this phase the microorganisms are not active.
2. Exponential phase (log phase or the logarithmic phase) is a period
characterized by cell doubling.
Number of new bacteria appearing per unit time is proportional to the
present population. Exponential growth cannot continue indefinitely,
however, because the medium is soon depleted of nutrients and enriched
with wastes.
3. During stationary phase, the growth rate slows as a result of nutrient
depletion and accumulation of toxic products.
This phase is reached as the bacteria begin to exhaust the resources that
are available to them. This phase is a constant value as the rate of bacterial
growth is equal to the rate of bacterial death.
4. At death phase, bacteria run out of nutrients and die.
6.6 Secondary Treatment Biological Process
Decomposition of Organic Waste

Decomposition is the process by which organic

substances are broken down into simpler organic
matter.

It can accomplished is three ways

1) Aerobic
2) Anoxic
3) Anaerobic
6.6 Secondary Treatment Biological Process
Decomposition of Organic Waste

Aerobic Decomposition

• Occurs in the presence of oxygen as the terminal electron acceptor

organic matter + O2 CO2 + H2O + newcell

Large production because large amount of energy released

• aerobic decomposition generates a large production of sludge (dead

and living cells).Therefore, it is suitable for low strength wastewater (ie
< 500 mg/l BOD) because decomposition is rapid, efficient and has a
low odor potential.

• For high strength w/w (>1000 mg/l BOD), aerobic decomposition is not
suitable because of difficulty in supplying of enough oxygen and also
because of the amount of sludge produced.
6.6 Secondary Treatment Biological Process
Decomposition of Organic Waste
 6.6 Secondary Treatment Biological Process
Decomposition of Organic Waste

Most Abundant Microbes in Wastewater Aerobic Treatment

Rotifer
Amoeba Ciliated Protozoa

Filamentous Flagellated Protozoa Vorticella

6.6 Secondary Treatment Biological Process
Decomposition of Organic Waste

Anaerobic Decomposition (AnD)

When the NO2-3 are finished, strict anaerobic conditions occur

Here, sulfates are used and reduced to sulfides

Sulfate reduction bacteria

and CO2 is converted into methane (CH4) methanogenesis-

hydrogenotrophic methanogen

acetotrophic methanogen
 6.6 Secondary Treatment Biological Process
Decomposition of Organic Waste
•The anaerobic decomposition of organic matter considered to be a three-step
process (figure):

1. Hydrolysis of waste.

1. Acidogenesis
(fermentation) and
acetogenesis –
Conversion of complex
organic compounds to
low-molecular fatty
acids (volatile acids).

3. Methanogenesis -
Conversion of organic
acids to methane.
Processes in anaerobic degradation
6.6 Secondary Treatment Biological Process
Decomposition of Organic Waste

A consortium of
anaerobic bacteria
for transformation
of complex organic
to methane
(anaerobic
digestion)
 6.6 Secondary Treatment Biological Process
Decomposition of Organic Waste

Anaerobic digester
6.6 Secondary Treatment Biological Process
Decomposition of Organic Waste

Differences between aerobic and anaerobic

decomposition of wastewater
6.6 Secondary Treatment Biological Process
Decomposition of Organic Waste

Differences between aerobic and anaerobic decomposition in terms of

COD balance
6.6 Secondary Treatment Biological Process
Microorganism Classes

In biological/secondary treatment system, the

microorganism are divided into the followings:

1) Suspended growth- microorganisms and bacteria

treating the wastes are suspended in the wastewater
being treated.

2) Attached growth- microorganisms and bacteria

treating the wastes are attached to the media in the
reactor
6.6 Secondary Treatment Biological Process
Microorganism Classes VS Technology : Suspended
Growth

Stabilization Pond
Activated Sludge
Aerobic, anaerobic,
Extended aeration, conventional
facultative, maturation
6.6 Secondary Treatment Biological Process

Microorganism Classes VS Technology : Suspended

Growth

Aerated lagoon Constructed wetland

6.6 Secondary Treatment Biological Process

Microorganism Classes VS Technology : Attached Growth

Rotating biological contactor (RBC)

Trickling filter
6.7 Suspended Growth Biological Process
Activated Sludge (Aerobic)
6.7 Suspended Growth Biological Process
Activated Sludge (Aerobic)

▪ Process in which a mixture of wastewater and

microorganisms (biological sludge) is agitated and
aerated (disperse growth)
▪ Leads to oxidation of dissolved organics
▪ After oxidation, separate sludge from wastewater
▪ Induce microbial growth
▪ Need food, oxygen
▪ Want Mixed Liquor Suspended Solids (MLSS) of
3,000 to 6,000 mg/L
6.7 Suspended Growth Biological Process
Activated Sludge (Aerobic)
6.7 Suspended Growth Biological Process
Activated Sludge (Aerobic)
Example 5
A wastewater treatment plant to treat wastewater to meet effluent
standard of 25 mg/L BOD and 30 mg/L suspended solids. The treatment
plant flow rate is 0.029 m3/s. The effluent from the primary tank has BOD
of 240 mg/L. Using the following assumptions, estimate the required
volume of the aeration tank. BOD of effluent suspended solids is 70% of
the allowable suspended solids concentration

Given:
a. Soluble BOD5 concentration at one half the maximum growth
rate, Ks = 100 mg/L BOD
b. Decay rate of microbes, Kd = 0.025 /d,
c. max growth rate, µm = 10/d,
d. Yield coefficient, Y = 0.8 mg VSS/mg BOD5
e. Microorganism concentration of VSS concentration entering
aeration tank, MLVSS = 3000 mg/L
6.7 Suspended Growth Biological Process
Activated Sludge (Aerobic)
70%

Effluent BOD5 Effluent SS

Initial pollutant concentration

Effluent BOD5 from primary tank

Sludge age
A measure of the length of time a particle of suspended
solids is retained in the activated sludge process, (d).
6.7 Suspended Growth Biological Process
Activated Sludge (Aerobic)

Equals to t0,
hydraulic
retention time
 6.7 Suspended Growth Biological Process
Activated Sludge (Aerobic)
Example 6
A conventional activated sludge plant without primary clarification operates under the
following conditions:

Design flow : 8100 m3/d

Influent BOD : 185 mg/L
Suspended Solids : 212 mg/L
Aeration basins : 4 units, 12 m square x 4.5 m deep
Mixed liquor Suspended solids : 2600 mg/L
Recirculation flow : 3800 m3/d
Waste sludge quantity : 150 m3/d
Suspended solids in waste sludge : 8600 mg/L
Effluent BOD : 15 mg/L
Suspended Solids : 15 mg/L
Calculate :
(a) Aeration period
(b) BOD loading
(c) Return activated sludge age
(d) F/M ratio
(e) suspended solids and BOD removal efficiencies
(f) Sludge age
6.7 Suspended Growth Biological Process
Activated Sludge (Aerobic)
Solution:

(a) Aeration period

Aeration basin volume = 4 units X ( 12m x 12m x 4.5m)

= 2592 m3
Aeration period, t = V/Q
= 2592 m3/ 8100 m3/d
= 7.7 hr

(b) BOD loading = (Q x Concentration BOD )/ Volume

= (8100 m3/d x 185 mg/L)/ 2592 m3
= 578 g/m3/day

(c) Return activated sludge = (Recirculation flow/Q) x 100

= (3800 m3/d/8100 m3/d) x 100
= 47 %

(d) F/M ratio = (Q x Concentration BOD )

Volume x MLSS
= (8100 m3/d x 185 mg/L)
(2592 m3 x 2600 mg/L)
= 0.22 g BOD/day
g MLSS
6.7 Suspended Growth Biological Process
Activated Sludge (Aerobic)
(e) Suspended solids removal = (212 – 15) x 100 = 93 %
212
BOD removal = 185 – 15 x 100 = 92 %
185

(f) Sludge age

SS in the effluent = Q x Concentration SS
= 8100 m3/d x 15 mg/l
= 121 kg/d

SS in waste activated sludge = Qsludge x Concentration SS Sludge

=150 m3/d x 8600 mg/l
= 1290 kg/d

Sludge age = V x Mixed liquor suspended solid

SS effluent + SS activated sludge
= (2592 m3 x 2600 mg/l )
121 kg/d + 1290 kg/d
= 4.8 days

Sludge age (θc) = mean cell residence time (θc)

= solid retention time (SRT)

Longer sludge age (θc) would result in bigger tank and longer aeration time (↑ power consumption)
 6.7 Suspended Growth Biological Process
Activated Sludge (Aerobic)
Example 7
A sewage treatment plant treats 2000 m3/d of wastewater. The average BOD5
of the raw wastewater after primary settling is 1500 mg/L. The aeration tank
has effective dimensions of 6.0 m wide by 10.0 m long by 6.0 m deep. The
activated sludge plant operating parameters are as follows;

(i) BOD5 after secondary settling = 95 mg/L

(ii) MLVSS = 2100 mg/L
(iii) MLVSS/MLSS = 0.75
(iv) Settled sludge volume after 30 minutes = 250 mg/L

Determine;
(i) BOD5 loading
(ii) The aeration period
(iii) The food to micro-organism ratio (F/M ratio)
(iv) The sludge volume index (SVI)
(v) The percentage of BOD5 removal
6.7 Suspended Growth Biological Process
Activated Sludge (Aerobic)
Solution:

i) BOD loading = Q (m3/d) x Concentration of BOD (mg/L)

Volume
3
= 2000 m /day x 1500 mg/L
( 6 m x 10 m x 6 m)
= 8,333.33 g/m3.day = 8.33 kg/m3.day

ii) Aeration period = V/Q

= 360m3/ 2000 m3/d
= 0.18 day
= 4.32 hours

iii) F/M ratio = QS/VX = 2000 m3/day x 1500 mg/L

360 m 3 x 2100 mg/L
= 3.9
6.7 Suspended Growth Biological Process
Activated Sludge (Aerobic)
Solution:

iv) SVI = Sludge volume x 1000 mg/g

MLSS

if MLSS = MLVSS/0.75 = 2100 mg/L /0.75 = 2800 mg/L then

SVI = 250 mg/L x 1000 mg/g = 89.3 mg/g SVI is used to control the rate of
2800 mg/L sludge return to the reactor basin in
activated sludge process

v) BOD removal = BOD in – BOD out x 100%

BOD in

= 1500 – 95 x 100%
1500

= 93.7 %
6.7 Suspended Growth Biological Process

Activated Sludge (Aerobic)

ASSIGNMENT 2

Calculate the BOD concentration (mg/L) and the BOD load

(kg/day) of a wastewater from a residential unit of 4 persons
assuming the per capita BOD contribution is 55g BOD per day and
the per capita discharge contribution is 225 liter per day.

Answers: BOD concentration (mg/L) = 240 mg/L

BOD load (kg/day) = 0.22 kg/day
6.7 Suspended Growth Biological Process
Septic Tank (Anaerobic)
▪ Septic tanks are commonly used for wastewater treatment for individual households
in low-density residential areas, for institutions such as schools and hospitals, and
for small housing estates.

▪ Septic tanks have been used for anaerobic treatment of raw sewage.

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6.7 Suspended Growth Biological Process
Septic Tank (Anaerobic)
❖ Household wastewater is retained for 1-3 days.
❖ A thick crust of scum is formed (helps anaerobic condition).
❖ Sludge accumulates then desludging should be done to remove the sludge.
❖ Desludging is done for every 1-5 years.
 6.7 Suspended Growth Biological Process
Septic Tank (Anaerobic)

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6.7 Suspended Growth Biological Process
Septic Tank (Anaerobic)
6.7 Suspended Growth Biological Process
Septic Tank (Anaerobic)

Based on “A guide to the Development of on-site sanitation (WHO,1992)”

The guiding principles in designing a septic tank are:

❖ To provide sufficient retention time for the sewage in the tank to allow
separation of solids and stabilization of liquid;
❖ To provide stable quiescent hydraulic conditions for efficient settlement and
flotation of solids;
❖ To ensure that the tank is large enough to store accumulated sludge and
scum;
❖ To ensure that no blockages are likely to occur and that there is adequate
ventilation of gases.
 6.8 Attached Growth Biological Process

● Consists of a reactor with

some kind of medium to
support the growth of biomass
on it.

● The media inert material of

any type or even biological
sludge itself.

● The most popular reactors that

have been employed for the
domestic wastewater treatment
system include:
○ Rotating biological
contactors
○ Trickling filters
 6.8 Attached Growth Biological Process
Rotating Biological Contactor (RBC)
● Consists of series of closely
spaced discs mounted on a
horizontal shaft and rotated
with 40 of each disc is
submerged in wastewater

● Discs: light weight plastic.

● Slime is 1 3 mm in thickness on
the discs.
6.8 Attached Growth Biological Process

Rotating Biological Contactor (RBC)

In the RBC, the medium

moves the biofilm alternately
through water and air
6.8 Attached Growth Biological Process
Trickling Filter
▪ Not a true filtering or sieving process
▪ Material only provided surface on which bacteria to
grow
▪ Can use plastic media
o Lighter, can get deeper beds (up to 12 m)
o Reduced space requirement
o Larger surface area for growth
o Greater void ratios (better air flow)
o Less prone to plugging by accumulating slime
6.8 Attached Growth Biological Process
Trickling Filter

Rotating distribution arm sprays primary effluent over circular bed of rock or
other coarse media. Air circulates in pores between rocks and “biofilm”
develops on rocks and micro-organisms degrade waste materials as they
flow past. Organisms slough off in clumps when film gets too thick
 6.8 Attached Growth Biological Process

Trickling Filter
Advantages
a) Small land area required
b) Can be operated at a range of organic and hydraulic loading
rates.

Disadvantages/limitations
a) High capital costs and moderate operating costs
b) Requires expert design and construction.
c) Requires constant source of electricity and constant wastewater
flow.
d) Flies and odours are often problematic.
e) Not all parts and materials may be available locally.
f) Pre-treatment is required to prevent clogging.
g) Dosing system requires more complex engineering.
6.9 Sludge: By-Product of WWTP

Objectives:

▪ To understand the characteristics of solid

sludge before treatment
▪ To understand the process of treatment sludge
 6.9 Sludge: By-Product of WWTP

▪ The higher the degree of wastewater treatment, the larger

residue of sludge that must be handled.

▪ Wastes from screens and grit chamber Not true sludge, not a
fluid. It can be drained easily and is relatively stable, it can be
disposed of directly in a municipal landfill.

▪ To treat and dispose of the sludge produced from wastewater

plants in the most effective manner, it is important to know the
characteristics of the solid and sludge that will be processed.
6.9 Sludge: By-Product of WWTP
 6.9 Sludge: By-Product of WWTP

Sources of Sludge

Primary or Raw Sludge

Sludge from bottom of the primary clarifiers
contains from 3- 8 % solid which is
approximately 70 % organic. This sludge rapidly
becomes anaerobic and is highly odiferous.
 6.9 Sludge: By-Product of WWTP
Sources of Sludge

Secondary Sludge
▪ This sludge consists of microorganisms and inert
materials that have been wasted from the secondary
treatment processes.
▪ In some cases secondary sludge contains large
quantities of chemical precipitates because of aeration
tank is used as the reaction basin for the addition of
chemicals to remove phosphorus.
6.9 Sludge: By-Product of WWTP

Sludge Characteristics
 6.9 Sludge: By-Product of WWTP
Why Sludge needs to be treated?
− Still biodegradable
− Can exert oxygen demand
Sludge Treatment − May contain hazardous microbes

The basic processes for sludge treatment are as follows:

 6.9 Sludge: By-Product of WWTP
Sludge Treatment: Thickening

Flotation Gravity thickening

• Especially effective • Best with primary sludge
on activated sludge • Increases solids content
• Increases solids from 1-3% to 10%
content from 0.5 - 1%
to 3-6%
6.9 Sludge: By-Product of WWTP
Sludge Treatment: Thickening
 6.9 Sludge: By-Product of WWTP
Sludge Treatment: Stabilization

Aerobic Digestion Anaerobic Digestion

• Extension of activated ● 2 stages: Acid fermentation followed by

methane production.
sludge
● To treat the sludge that has a high organic
• Accomplished by aeration of
content and the presence of pathogens.
sludge then followed by
● Advantages:
sedimentation
○ Produce methane
• Supernatant goes back ○ Do not add oxygen
to head of plant ● As with aerobic digestion, supernatant goes
(high in BOD, TKN, TP) to headworks
• Treated sludge is 3% solids ● Under anaerobic conditions, the sludge
should be stabilized, and high priority will
be given to prevent odor and pathogens
from spreading to the surroundings.
6.9 Sludge: By-Product of WWTP
Sludge Treatment: Stabilization
 6.9 Sludge: By-Product of WWTP
Sludge Treatment: Conditioning
Purposes:
1. To prepare the sludge for the dewatering process.
2. To improve dewatering characteristic of sludge.
3. To reduce the moisture content of the sludge.

Chemical Conditioning Heat Treatment

▪ Add lime, ferric chloride, or ▪ High temperatures (175- 230oC)

alum ▪ High pressures (10 to 20 atm)
▪ Can also add polymers ▪ Advantages
▪ Chemicals are added just o bound water is released and
prior to dewatering stage sludge is easily dewatered
▪ Disadvantages
o complex process
o highly concentrated liquid
stream
 6.9 Sludge: By-Product of WWTP
Sludge Treatment: Dewatering

Sludge Drying Beds

• Most popular method essentially
• Simple
• Low maintenance
• Effected by climate

Vacuum Filtration
• Apply vacuum to
pull out water
• Force out water by
squeezing water
between two
moving filter belts
 6.9 Sludge: By-Product of WWTP
Sludge Treatment: Volume Reduction
Incineration
• Complete evaporation of water
from sludge
• Requires fuel
• Solid material is inert
• Exhaust air must be treated
prior to discharge

Wet Oxidation
• Treated sludge is wet
• Requires energy
• Solid material is inert
• Exhaust air must be
treated prior to discharge
6.9 Sludge: By-Product of WWTP
Sludge Disposal
Methods
▪ Land Spreading:
o Lawns, gardens
o Agricultural land
o Forest land
o Golf courses and other public recreational areas

▪ Municipal solid waste landfill

▪ Utilization in other materials (E.g. composting)

6.9 Sludge: By-Product of WWTP
Malaysia :Sludge Treatment Process

IWK Sustainability Report 2012-2013