A part of Sewage treatment plant or Environmental engineering II subject of Final year civil engineering students of University of Mumbai

1. Activated sludge process
EE II
Shilpa Patil

2. • Suspended growth process (aeration tank)
– If the micro-organisms are suspended in the WW during biological operation then
process is known as suspended growth processes
• Attached growth process (trickling filter)
– Process in which the micro-organisms that are attached to a surface over which
they grow is known as attached growth processes
• Combine process (facultative ponds)
– These consists of both attached growth as well as suspended growth processes
Bio-film
Or
Slime layer
Bacteria Liquid Supporting medium (Stones)
SECONDARY
BIOLOGICAL
PROCESS

3. Introduction
In a sewage treatment plant, the activated sludge process can
be used for one or several of the following purpose:
– oxidizing carbonaceous and nitrogenous matter
– removing phosphates
– driving off entrained gases carbon dioxide, ammonia, nitrogen, etc.
– generating a biological floc that is easy to settle.
– generating a liquor (wastewater) that is low in dissolved or suspended
material

4. ASP

5. Excellent Method to treat settled
sludge
Sewage from PST is mixed with activated
sludge (20-30%) which contains highly
active micro-organisms
Air is provided continuously for 4-8 hrs
Suspended or Moving micro-organisms will
oxidize OM & form precipitate

6. ASP
• Effluent from ASP is of high quality.
• Effluent from ASP has lower BOD value as compared to effluent
from TF.
• It requires less area as compared to TF

7. Various operations units of an ASP
• Primary treatment unit
– It is necessary to provide PST
before aeration tank
– It prevents deposition of
settlable solid hence
increases efficiency of ASP
– The detention period of PST is
reduced to keep sewage
fresh

8. • Aeration tank
– Depth: 3-4.5 m
– Width: 4-6 m
– DP: 4-6 hrs
• Types of aeration tank
– Ridge & furrow type
– Spiral flow type
Quantity of air required:
4000-8000 cu. M per ML
Depending upon quantity of sewage
Quantity of air required:
100 cu. M per day per kg of BOD
Depending upon BOD

9. • Methods of aeration
– Diffused aeration/Air
diffusion
– Mechanical aeration
– Combine aerators

10. • SST
- Weir loading: (150 cu. M per m per day)
- DP: 1.5-2 hrs
- L/d ratio: 5 (Circular)
7 (Rectangular)
- Depth: 3.5 -4.5 m
Activated Sludge Process

11. Design of ASP
• Aeration period (t)
– Also called as Hydraulic retention time
– Aeration period = Volume of tank (cu. M)/ Rate of sewage flow in the tank (cu. M per day)
= [V/Q]*24 in hrs
• BOD loading per unit volume of aeration tank
– Also called as Volumetric loading or Organic loading
– Organic loading = Mass of BOD per day through influent (kg)/Volume of aeration tank (cu. M)
= Q*BOD/V in kg per cu. M per day

12. Design of ASP
• F/M ratio
– Food to micro-organisms ratio
– F: Food Organic matter Oxygen demand BOD
– M: Microbial suspended solid in mixed liquor
– F/M = BOD applied/Microbial mass
= [Q*BOD]/[Xt *V] Xt: concentration of SS in sewage (mg/l)
Xt = MLSS (Mixed liquor suspended solids)
High
amount
of BOD is
removed
Low
amount
of BOD is
removed
Food=5
micro-organism=10
F/M= 0.5
Food
micro-organism
Food=10
micro-organism=5
F/M= 2
F/M ratio is inversely
proportional to BOD
removal

13. INFLUENT (Q)
BOD5=(Y0 )
Xt: concentration of SS in sewage in
aeration tank (mg/l)
Volume of tank = V
Quantity =(Q +QR)
BOD5=(Y)
Quantity QE =(Q -QW)
BOD5=(YE )
XE: concentration of SS in sewage effluent (mg/l)
Sludge
Quantity of
sludge=(QR+QW)
XR: concentration of
SS in Sludge(mg/l)
Quantity of returned sludge= QR
XR: concentration of SS in Sludge(mg/l)
Quantity of waste sludge= QW
XR: concentration of SS in Sludge(mg/l)
SludgeProcessing
Returned Sludge or Recycled Sludge
EFFLUENT (QE)

14. Design of ASP
• Sludge Age/Solid Retention Time
– It is an average time for which particles of suspended solids remain under aeration
– As recirculation is provided
– Waste organics are repeatedly recycled
– So the Retention time increases
• SRT= Mass of SS in the System(M)/Mass of solids leaving system per day (XR)
= [MLSS*V]/[Qw* XR]

15. Sludge volume index
• Physical state of sludge produced in biological aeration system
• Degree of concentration of sludge
• SVI is the volume occupied (ml) by one gm of solids in mixed liquor after settling for 30
minutes.
• Determination of SVI
1. Volume of
sludge
•1 lit of aerated sewage sample is taken in graduated
cylinder
•Sample is allowed to settle for 30 minutes
•Volume of settled sludge is noted (VOB) in ml/L
2. MLSS
determination
•Same sample is mixed thoroughly
•SS of the sample is determined (XOB) in mg/L
SVI in ml/gm
= (VOB/ XOB)*1000

16. Any questions?
Thank you