Activated Sludge Process
 Most extensively used aerobic suspended growth biological
treatment process
 Sewage from sedimentation tank enters into aeration tank. 20-
30% of active sludge is mixed (returned). This activated sludge
contains large concentration of highly active aerobic
microorganisms. The mixture is aerated and mixed in the tank
for about 4-8 hrs residence time
 Microorganism oxidise organic matter in the presence of
oxygen in the aeration tank and sewage is allowed to settle in
the secondary sedimentation tank.
 This settled sludge has undergone aeration and has active
microorganisms. So some portion of this active sludge is
recirculated into the aeration tank where the mixed liquor
containing the suspended solids is under continuous agitation.
 The excess sludge is disposed off from the sedimentation tank
1
2
3
Advantages
 Efficiency is high (80-95% of BOD removal and the
bacteria removal is 90-95%)
 Land area required is less than that for trickling filter
 Designed unit can be made compact
 Initial cost of construction is lower than trickling filter (TF)
 Process is free from foul and insect nuisance
 Sludge has a high fertilizing value and the effluent can be
used for irrigation
4
Disadvantages
 Skilled personnel are required
 Operation cost is high
 Large quantity of sludge is produced (may be problem for
sewage disposal)
 Very sensitive (bacterial growth affects due to the temperature
variation)
 Requires continuous oxygen supply and continuous mixing of
activated sludge and sewage .
 Also requires constant ratio of volume of activated sludge to
that of sewage.
 Starting of new plant may requires seeding of the activated
sludge from another ASP plant. It requires about 4 weeks to
form the activated sludge if not seeding is done.
5
Volume of returnced activated sludge
Desired BOD removal
(ppm)
% of returned activated
sludge
o150
o250
o300
o400
o500
o600
25
30
35
40
48
53
6
Units required for ASP
1. Primary treatment units
 Screening
 Grit chamber
 Primary sedimentation tank
2. Aeration tanks
3. Secondary sedimentation tank
4. Sludge treatment and disposal
7
Aeration tanks of ASP
 Aeration is the most important operation in the activated
sludge process so as to provide oxygen and mixing.
 The aeration facilities are designed to meet the oxygen
demand.
 A minimum DO of about 1-2 mg/l is necessary for the
proper biological treatment.
 The aeration tanks are 3-4.5 m deep.
 The detention tank is kept between 4-8 hours for
municipal sewage.
8
Methods of aeration
1. Diffused aeration
2. Mechanical aeration
3. Combined diffused and mechanical aeration
9
Diffused aeration
 Used for large plants
 A long narrow rectangular
tank with the size of
30x5x3 to 120x10x5 m
 Compressor is used to pass
air through diffusers under
pressure of 0.55 - 0.7
kg/cm2
 Air velocity= 60-90 cm/min
10
11
12
Mechanical aeration
Compressor is not used but aeration is done in the
principle that the sewage is stirred and take them to top
surface every time from where mix liquor takes
atmospheric air.
 Generally consists of large diameter impeller plates
revolving on vertical shaft at the surface of the liquid with
or without draft tubes
 Hydraulic jump is created by the impellers at the surface
causing air entrainment in the sewage
 The speed of rotation of impellers 70-110 rpm
13
14
15
16
Capacity of aeration tank
 Depends upon the following three factors
 Aeration period
 Volume of returned sludge
 Volume of flow of sewage
V = (Q + Qs) T/24
Where, V = capacity of aeration tank, m3
Q = volume of flow of sewage, m3/day
Qs = volume of returned sludge, m3/day
T = aeration period, hours
17
Oxidation Pond
 Biological or secondary treatment process
 Simple, low cost holding tanks 1-2 m deep that take oxygen
from atmosphere
Theory
When sewage is retained in the shallow depth tank open to
atmosphere, the aerobic bacteria present in the sewage
convert the sewage into stable substances by aerobic action in
the presence of free oxygen and sunlight. Also the algae grown
in the pond exerts oxygen in the process of photosynthesis in
the presence of air and sunlight. The produced carbon dioxide
is taken by the algae whereas the oxygen produced by algae is
useful for aerobic bacteria. The combined action of aerobic
bacteria and algae together for their mutual benefit is called
bacterial algal symbiosis. The solids are settle at the bottom
and the accumulated sludge is cleaned after one third of
filling.
18
19
20
Location and construction
 Located at least 300 m far from the locality
 No obstacles (tree, building) around 50-60 m
distances so that sunlight can reach to the pond.
 Water sources should be at least 15 m far away.
 Pond surrounded by earthen dykes of 1-1.5 m wide on
top and side slopes of 1:1.5 -1:3 (can be turfed for good
appearance)
 Sewage discharged at the centre of the pond form inlet
chamber through inlet pipe from bell mouth end
21
Commissining of oxidation pond
1) Culture method:
2) Natural method:
1) Culture method: (Artificial Method)
 Sewage is filled to a depth of 15 cm for the growth of
algae (seeds are introduced)
 Portion of sewage decreased is refilled
 Pond should turn completely green (at least a week)
 Application of sewage to the operation level
 Algae grow takes place from 2-3 days
22
2) Natural method:
 Algae are made to grow naturally
 Initially sewage is kept to operation level and inlet and
outlet are closed
 The reduced sewage is refilled daily
 The growth of algae takes place in 2 weeks
23
Operation and maintenance
 Does not require equipment and skilled personnel but
the fallen leaves, papers and other floating materials
should be removed
 No chemicals are sprayed to prevent form mosquitoes
and flies nuisance
 Fish farming can be done so that the larva of mosquito
and flies are fed by fish
 Settled sludge should be removed from time to time
 Efficiency of pond is high (BOD removal more than
90%, SS >90% and coliform bacteria >99%)
24
Merits and demerits of Oxi. Pond
Advantages Disadvantages
 Low initial cost if lands are
cheap
 No equipment required
 Low maintenance cost
 No skilled personnel
 Efficiency is high
 Can be used as fish farming
 Larger area of land is required
 Creates mosquito nuisance
 Creates foul smell
 Not suitable for high intensity
rainfall
25
Design Criteria
 Detention time 10-45 days (colder place require longer detention time)
Lt = La (10 -kt)
t = 1/k log10 (La/Lt) where La = influent or inlet BOD
Lt = effluent or outlet BOD
 Surface area As = 0.2-0.4 ha/1000 persons
 Volume of pond V = Q x t
 Organic loading (U) = W/As = 150-330 kg/ha/d where As = plan area
W = BOD applied = Q x La
 Rectangular in shape
 Effective depth d = 80-150 cm (no sludge depth)
 Freeboard 20-50 cm
 Length width ratio = 2-3
 Side slope = 1:1.5 - 1:3
Oxidation pond is used where at least 200 days are sunny in a year
26
Design of Activated Sludge Process (ASP)
1) Index of mass of active microorganism:
 Mix liquor suspended solids (MLSS) is generally taken
as an index of mass of microorganisms
 In the present context, mixed liquor volatile
suspended solids (MLVSS) is taken as an index of the
mass of active microorganism.
 The mixture of recycled sludge and sewage in the
aeration tank is referred to as mixed liquor
27
2) Sludge Volume Index (SVI):
 Indicates the physical state of the sludge produced in a biological
aeration system and represents the degree of concentration of sludge in
the system.
 Decides the rate of recycle of sludge required to maintain the desired
mix liquor suspended solids (MLSS) and food to microorganism ratio
(F/M) in the aeration tank or reactor to achieve the required degree of
purification
 Defined as the ratio of volume of sludge in ml to the dry weight of the
sludge in gm. i.e the volume occupied by 1 gm of MLSS in ml.
SVI = Vs/Xt (ml/g) where Vs = volume of settled sludge in ml in 1000
ml graduated cylinder
Xt = MLSS in gm/l
3) Sludge Density Index (SDI):
 Used to determine the quantity of solids concentration on returned
sludge.
 Gives suspended solids (SS) concentration of sludge
SDI = x 100 = (g/ml)
28
4) Sludeg age or sludge retention time (SRT)
 Also called as mean cell residence time (MCRT) or solid
retention time
 Defined as the time spend by mix liquor volatile suspended
solids (MLVSS) or activated sludge in aeration tank.
 Mathematically, θc =
Where V = volume of aerator tank or reactor
Xt = concentration of MLSS
Xr = concentration of MLVSS in returned
sludge
Q = influent sewage flow rate
Qw = wasted sludge flow rate
Xe = concentration of MLVSS in effluent
29
5) Recycle ratio or return sludge ratio
r = Qr/Q where Qr = returned sludge flow rate
where Xt = mg/l and SVI = ml/g
6) Aeration period: depends upon
 Strength of sewage and MLSS concentration
 Desired degree of purification in terms of BOD removal
 Proportion of returned activated sludge
Empirical Formula
A) American Public Health Association Formula
T (hrs) = Where T = aeration period and
La = influent BOD in mg/l
B) M/S Ames Crosta Mills and Co. Ltd (England)
T (hrs) =(La/10)3/4
C) Hydraulic Retention Time(HRT) [4-8 hrs]
HRT = T = V/Q in hrs
30
7) Volumetric BOD loading
 Defined as BOD applied per unit volume of an
aeration tank
 Also known as organic loading
VBL = QxLa/V, kg/m3/day
8) Food microorganism ratio (F/M) [0.2-0.6 per day]
F = total influent BOD = Qx La and
M = MLSS in aeration tank = Vx Xt
 F/M ratio is the main factor controlling BOD removal
 Lower the F/M value, higher will be the BOD removal
or vice versa
31
9) Tank configuration:
 Rectangular in shape Effective depth = 2.5 - 4.5 m
 FB = 0.5 m Width/depth = 1.2-2.2
 Velocity of flow ≤ 1.5 m/min
10) Air requirement
a) For diffuser type aerator
 Air required = 40-125 m3 of air/kg of BOD applied
 Standard diffuser = 0.3mx0.3mx25mm with 0.3 mm pores
 Velocity of air flow = 60-90 cm/min
b) For other types of aerator
 If F/M ≤ 0.3, air required = 75-115 m3 of air/kg of BOD removed
 If F/M ≥ 0.3, air required = 30-55 m3 of air/kg of BOD removed
32

activated sludge process. sanitary engineering

  • 1.
    Activated Sludge Process Most extensively used aerobic suspended growth biological treatment process  Sewage from sedimentation tank enters into aeration tank. 20- 30% of active sludge is mixed (returned). This activated sludge contains large concentration of highly active aerobic microorganisms. The mixture is aerated and mixed in the tank for about 4-8 hrs residence time  Microorganism oxidise organic matter in the presence of oxygen in the aeration tank and sewage is allowed to settle in the secondary sedimentation tank.  This settled sludge has undergone aeration and has active microorganisms. So some portion of this active sludge is recirculated into the aeration tank where the mixed liquor containing the suspended solids is under continuous agitation.  The excess sludge is disposed off from the sedimentation tank 1
  • 2.
  • 3.
  • 4.
    Advantages  Efficiency ishigh (80-95% of BOD removal and the bacteria removal is 90-95%)  Land area required is less than that for trickling filter  Designed unit can be made compact  Initial cost of construction is lower than trickling filter (TF)  Process is free from foul and insect nuisance  Sludge has a high fertilizing value and the effluent can be used for irrigation 4
  • 5.
    Disadvantages  Skilled personnelare required  Operation cost is high  Large quantity of sludge is produced (may be problem for sewage disposal)  Very sensitive (bacterial growth affects due to the temperature variation)  Requires continuous oxygen supply and continuous mixing of activated sludge and sewage .  Also requires constant ratio of volume of activated sludge to that of sewage.  Starting of new plant may requires seeding of the activated sludge from another ASP plant. It requires about 4 weeks to form the activated sludge if not seeding is done. 5
  • 6.
    Volume of returncedactivated sludge Desired BOD removal (ppm) % of returned activated sludge o150 o250 o300 o400 o500 o600 25 30 35 40 48 53 6
  • 7.
    Units required forASP 1. Primary treatment units  Screening  Grit chamber  Primary sedimentation tank 2. Aeration tanks 3. Secondary sedimentation tank 4. Sludge treatment and disposal 7
  • 8.
    Aeration tanks ofASP  Aeration is the most important operation in the activated sludge process so as to provide oxygen and mixing.  The aeration facilities are designed to meet the oxygen demand.  A minimum DO of about 1-2 mg/l is necessary for the proper biological treatment.  The aeration tanks are 3-4.5 m deep.  The detention tank is kept between 4-8 hours for municipal sewage. 8
  • 9.
    Methods of aeration 1.Diffused aeration 2. Mechanical aeration 3. Combined diffused and mechanical aeration 9
  • 10.
    Diffused aeration  Usedfor large plants  A long narrow rectangular tank with the size of 30x5x3 to 120x10x5 m  Compressor is used to pass air through diffusers under pressure of 0.55 - 0.7 kg/cm2  Air velocity= 60-90 cm/min 10
  • 11.
  • 12.
  • 13.
    Mechanical aeration Compressor isnot used but aeration is done in the principle that the sewage is stirred and take them to top surface every time from where mix liquor takes atmospheric air.  Generally consists of large diameter impeller plates revolving on vertical shaft at the surface of the liquid with or without draft tubes  Hydraulic jump is created by the impellers at the surface causing air entrainment in the sewage  The speed of rotation of impellers 70-110 rpm 13
  • 14.
  • 15.
  • 16.
  • 17.
    Capacity of aerationtank  Depends upon the following three factors  Aeration period  Volume of returned sludge  Volume of flow of sewage V = (Q + Qs) T/24 Where, V = capacity of aeration tank, m3 Q = volume of flow of sewage, m3/day Qs = volume of returned sludge, m3/day T = aeration period, hours 17
  • 18.
    Oxidation Pond  Biologicalor secondary treatment process  Simple, low cost holding tanks 1-2 m deep that take oxygen from atmosphere Theory When sewage is retained in the shallow depth tank open to atmosphere, the aerobic bacteria present in the sewage convert the sewage into stable substances by aerobic action in the presence of free oxygen and sunlight. Also the algae grown in the pond exerts oxygen in the process of photosynthesis in the presence of air and sunlight. The produced carbon dioxide is taken by the algae whereas the oxygen produced by algae is useful for aerobic bacteria. The combined action of aerobic bacteria and algae together for their mutual benefit is called bacterial algal symbiosis. The solids are settle at the bottom and the accumulated sludge is cleaned after one third of filling. 18
  • 19.
  • 20.
  • 21.
    Location and construction Located at least 300 m far from the locality  No obstacles (tree, building) around 50-60 m distances so that sunlight can reach to the pond.  Water sources should be at least 15 m far away.  Pond surrounded by earthen dykes of 1-1.5 m wide on top and side slopes of 1:1.5 -1:3 (can be turfed for good appearance)  Sewage discharged at the centre of the pond form inlet chamber through inlet pipe from bell mouth end 21
  • 22.
    Commissining of oxidationpond 1) Culture method: 2) Natural method: 1) Culture method: (Artificial Method)  Sewage is filled to a depth of 15 cm for the growth of algae (seeds are introduced)  Portion of sewage decreased is refilled  Pond should turn completely green (at least a week)  Application of sewage to the operation level  Algae grow takes place from 2-3 days 22
  • 23.
    2) Natural method: Algae are made to grow naturally  Initially sewage is kept to operation level and inlet and outlet are closed  The reduced sewage is refilled daily  The growth of algae takes place in 2 weeks 23
  • 24.
    Operation and maintenance Does not require equipment and skilled personnel but the fallen leaves, papers and other floating materials should be removed  No chemicals are sprayed to prevent form mosquitoes and flies nuisance  Fish farming can be done so that the larva of mosquito and flies are fed by fish  Settled sludge should be removed from time to time  Efficiency of pond is high (BOD removal more than 90%, SS >90% and coliform bacteria >99%) 24
  • 25.
    Merits and demeritsof Oxi. Pond Advantages Disadvantages  Low initial cost if lands are cheap  No equipment required  Low maintenance cost  No skilled personnel  Efficiency is high  Can be used as fish farming  Larger area of land is required  Creates mosquito nuisance  Creates foul smell  Not suitable for high intensity rainfall 25
  • 26.
    Design Criteria  Detentiontime 10-45 days (colder place require longer detention time) Lt = La (10 -kt) t = 1/k log10 (La/Lt) where La = influent or inlet BOD Lt = effluent or outlet BOD  Surface area As = 0.2-0.4 ha/1000 persons  Volume of pond V = Q x t  Organic loading (U) = W/As = 150-330 kg/ha/d where As = plan area W = BOD applied = Q x La  Rectangular in shape  Effective depth d = 80-150 cm (no sludge depth)  Freeboard 20-50 cm  Length width ratio = 2-3  Side slope = 1:1.5 - 1:3 Oxidation pond is used where at least 200 days are sunny in a year 26
  • 27.
    Design of ActivatedSludge Process (ASP) 1) Index of mass of active microorganism:  Mix liquor suspended solids (MLSS) is generally taken as an index of mass of microorganisms  In the present context, mixed liquor volatile suspended solids (MLVSS) is taken as an index of the mass of active microorganism.  The mixture of recycled sludge and sewage in the aeration tank is referred to as mixed liquor 27
  • 28.
    2) Sludge VolumeIndex (SVI):  Indicates the physical state of the sludge produced in a biological aeration system and represents the degree of concentration of sludge in the system.  Decides the rate of recycle of sludge required to maintain the desired mix liquor suspended solids (MLSS) and food to microorganism ratio (F/M) in the aeration tank or reactor to achieve the required degree of purification  Defined as the ratio of volume of sludge in ml to the dry weight of the sludge in gm. i.e the volume occupied by 1 gm of MLSS in ml. SVI = Vs/Xt (ml/g) where Vs = volume of settled sludge in ml in 1000 ml graduated cylinder Xt = MLSS in gm/l 3) Sludge Density Index (SDI):  Used to determine the quantity of solids concentration on returned sludge.  Gives suspended solids (SS) concentration of sludge SDI = x 100 = (g/ml) 28
  • 29.
    4) Sludeg ageor sludge retention time (SRT)  Also called as mean cell residence time (MCRT) or solid retention time  Defined as the time spend by mix liquor volatile suspended solids (MLVSS) or activated sludge in aeration tank.  Mathematically, θc = Where V = volume of aerator tank or reactor Xt = concentration of MLSS Xr = concentration of MLVSS in returned sludge Q = influent sewage flow rate Qw = wasted sludge flow rate Xe = concentration of MLVSS in effluent 29
  • 30.
    5) Recycle ratioor return sludge ratio r = Qr/Q where Qr = returned sludge flow rate where Xt = mg/l and SVI = ml/g 6) Aeration period: depends upon  Strength of sewage and MLSS concentration  Desired degree of purification in terms of BOD removal  Proportion of returned activated sludge Empirical Formula A) American Public Health Association Formula T (hrs) = Where T = aeration period and La = influent BOD in mg/l B) M/S Ames Crosta Mills and Co. Ltd (England) T (hrs) =(La/10)3/4 C) Hydraulic Retention Time(HRT) [4-8 hrs] HRT = T = V/Q in hrs 30
  • 31.
    7) Volumetric BODloading  Defined as BOD applied per unit volume of an aeration tank  Also known as organic loading VBL = QxLa/V, kg/m3/day 8) Food microorganism ratio (F/M) [0.2-0.6 per day] F = total influent BOD = Qx La and M = MLSS in aeration tank = Vx Xt  F/M ratio is the main factor controlling BOD removal  Lower the F/M value, higher will be the BOD removal or vice versa 31
  • 32.
    9) Tank configuration: Rectangular in shape Effective depth = 2.5 - 4.5 m  FB = 0.5 m Width/depth = 1.2-2.2  Velocity of flow ≤ 1.5 m/min 10) Air requirement a) For diffuser type aerator  Air required = 40-125 m3 of air/kg of BOD applied  Standard diffuser = 0.3mx0.3mx25mm with 0.3 mm pores  Velocity of air flow = 60-90 cm/min b) For other types of aerator  If F/M ≤ 0.3, air required = 75-115 m3 of air/kg of BOD removed  If F/M ≥ 0.3, air required = 30-55 m3 of air/kg of BOD removed 32