Lecture 19-23

• Primary Sedimentation Tank

• Secondary Treatment
• Activated sludge process (ASP)
Circular Basin Rectangular Basin
Weir load 125-500 m3/m/day
III) Calculate the weir load
III) Weir load = 2000 m3/day/4 m = 500 m3/m/day
 Q: In the previous problem, if the TSS
concentration is 145 mg/L and the TSS
removal efficiency is 55 %, estimate the mass
of sludge generated in the plant. The average
flow rate in the plant is 900 m3/day.
Total solids removed per day = 900 m3/day x (0.55) x 145 mg /m3
= 130 kg/day
Solid content in sludge 2 to 5%.
Calculate the sludge generation rate.
Question
Design a circular primary settling tank for a community of 45,000
people. Make suitable assumptions.
Secondary Treatment Processes
• Biological processes
Biological Processes...
q cell: derives energy from oxidation of food
sources (carbohydrate, protein & fats)

Requires…..

q microbes with the ability to degrade the waste

organics
q contact time with the organics
q favorable conditions for growth
Objectives of Biological Wastewater
Treatment
qTostabilize (oxidise) the organic matter (Soluble
and non-settleable)

qToreduce the amount of dissolved phosphorus

and nitrogen in the final effluent
 Microorganisms
qMicroorganisms = single celled organism capable of
performing all life functions independently
qBasic unit of life = cell

Enzymes Cell Wall

C, N, P Wastes
H2O H2O
O2 CO2
 Percentage by weight
q Carbon 50
q Oxygen 20
q Nitrogen 14
q Hydrogen 8
q Phosphorus 3
q Sulfur 1
q Potassium 1
q Sodium 1
q Calcium 0.5
q Magnesium 0.5
q Iron 0.2
q All other elements 0.3
q Aerobic - oxygen

q Anoxic - nitrate

q Anaerobic – no oxygen
Biological treatability
qNutritional
• Carbon source (waste to be degraded )
• N & P (100:5:1;C:N:P)
• Trace minerals
qEnvironmental
• Oxygen (terminal electron acceptor)
• Temperature
• Water
• pH
• Non-toxic
 Term Definition

Metabolic function

Aerobic process Biological treatment process that occur in the presence of oxygen

Anaerobic process The process that occur in the absence of oxygen

Anoxic process Nitrate nitrogen is converted biologically to nitrogen gas in the absence of oxygen,
Known as denitrification

Facultative process Process in which the organisms can function in the presence or absence of molecular
oxygen

Combined aerobic/ Various combinations of aerobic, anoxic, and anaerobic processes grouped together
Anoxic/Anaerobic to achieve a specific treatment objective
process
 Term Definition
Treatment processes

Suspended-growth Biological treatment processes in which the microorganisms responsible for the conversion of
processes the organic matter or other constituents in the wastewater to gases and cell tissue are maintained
suspension within the liquid

Attached-growth Biological treatment processes in which the microorganisms responsible for the conversion of the
processes organic matter or other constituents in the wastewater to gases and cell tissue are attached to
some inert medium, such as rock, slag, or specially designed ceramic or plastic materials. Also
known as fixed-film processes

Combined processes Term used to describe combined processes (combined suspended and attached growth
processes)

Lagoon processes A generic term applied to treatment processes that take place in pond or lagoons
Biological Treatment
q Inthe case of wastewater treatment, the objective of
biological treatment is:
q To stabilize(to remove – to oxidise) the organic content
q To remove nutrients such as nitrogen and phosphorus

Types:
üAttached Growth
üAerobic Processes
üSuspended Growth
üAnoxic Processes
üCombined Systems
üAnaerobic Processes
üCombined Aerobic-Anoxic-
Anaerobic Processes
üAerobic
üPond Processes üMaturation
üFacultative
üAnaerobic
Aerobic Biological treatment
Oxidation and Synthesis

28
Activated Sludge Process
Biological Process: Basic Requirements
The most commonly used biological process
 Definitions
MLSS and MLVSS (mixed liquor volatile suspended solids)
Food/Microorganism ratio (F/M)
Recycled activated sludge (RAS) and WAS (waste activated sludge)
MCRT (mean cell residence time) (sludge age) or BSRT (biological solids
retention time)
Sludge volume index (SVI)
Food to microorganism ratio (F/M)
MCRT (BSRT), Sludge Age
Average length of time that an organism remains in the AT

MCRT = Total MLVSS in AT/Total MLVSS wasted/day

Total MLVSS in aeration tank = (Volume of Aeration Tank x MLVSS)

Total MLVSS wasted/day = Sludge escaped in effluent + sludge wasted

= (QeXe ) + (QwXw )
ASP: Design and Operating Parameters
•Hydraulic Retention Time 4 - 8 hours

•MLSS 2000 – 4000 mg/L

•F/M 0.2 – 0.6 kg BOD/day/kg MLSS

•BSRT (CMRT) 3-15 day

•Qr /Qo 25-75 %

 Problem 1
The flow into an ASP is 3 MLD with an influent BOD of
175 mg/L. Design criteria is volumetric loading rate to
be 0.8 kg BOD/m3/day. Determine the volume of
aeration tank (AT) required.

Total BOD load per day = (3x106L/day) x (175 mg/L)

= 525 kg BOD/day

Volumetric loading rate = 0.8 kg BOD/m3/day

Volume of AT required = (525 kg BOD/day)/ 0.8 kg BOD/m3/day

= 656 m3
Problem 2
In problem 1, if the MLVSS concentration in
the AT is 2100 mg/L, what would be the F/M
ratio? Also determine the HRT.
Food = 525 kg BOD/day
Biomass = (2100 mg/L) x (656 m3) = 1377 kg
F/M ratio = (525 kg BOD/day)/1377 kg = 0.38

HRT = Volume of AT/flow rate

= (656 m3 ) / (3 x103 m3/day) = 5.24 hour
Problem 3
Calculate the AT volume required for a
conventional ASP treating a design flow of
20 MLD. Influent BOD is 140 mg/L and
volumetric loading rate 0.5 kg BOD/m3/day

Volume = 5600 m3
Problem 4
Determine mean cell residence time (MCRT) of a
ASP with the following details.

Flow rate = 4 MLD, Volume of aeration tank 600 m3, MLVSS =

2500 mg/L, Effluent TSS = 30 mg/L, sludge concentration =
15000 mg/L, sludge wastage rate = 12 m3/day
Total biomass in AT = 2500 mg/L x 600 m3 = 1500 kg

Sludge wasted = 15,000 mg/L x 12 m3/day = 180 kg /day

Sludge escaped in effluent = (30 mg/L) x (4x106L /day) = 120 kg/day
MCRT = (1500 kg)/( 180 + 120 kg/day) = 5 day (within range)