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• Suspended growth type• Unique – the name activated
sludge was originated in referringto the return sludge (biomass),since these masses ofmicroorganisms were observed tobe very “active” in removingsoluble organic matter fromsolution
•
Introduction
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Introduction
INFLUENT FROM
PRIMARYCLARIFIE
R
TREATEDEFFLUEN
T
WASTEDSLUDGE
SLUDGERETURN
Aerationtank/reactor
Sedimentationtank/secondary
settling tank/fnalclarifer
%icroorganisms(activatedsludge&biomass) are
settled in the 'nal
learsupernatant
from the'nal clari'er
is the planteuent*+cess activated sludgeis wasted from thesystem to maintain the
properfoodµorganism
Two major
components
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-rganic matter decompose in aerobic condition
.ir (o+ygen) is supplied through/Mechanica aerator or di!u"er, -r both
.ir supply also gives #i$in% in wastewater ombination of suspended solids and
microorganisms is called &#i$ed i'uor"u" ended "oid" )MLSS*+
eration tan,
-rganic %atter 0 -1 -1 0 21-0 new cells
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Mechanical Surface Aerator
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Solid separation unit inwhich the cells(biomass) from the
reactor are separated(settled) and returned tothe reactor
Secondary settling tank
/sedimentation tank
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.dvantages
• 2igh quality of euent (345 6-7removal)
• *uent quality is controlled by
sludge return
7isadvantages
• 8eed high s"ill labour
• 2igh capital, operation andmaintenance costs
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%odi'cation
•Step aeration
• 9apered aeration
• -+idation:ditch
• *+tended aeration (*.)
• Sequencing 6atch ;eactor
(S6;)
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Ste( Aeration-
<n$uent addition at intermediatepoints provides more uniform6-7 removal throughout tan"
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Ta(ered Aeration-.ir is added in proportion to 6-7e+erted
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-+idation 7itch/ plan view
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;eactor 7esign
• 9ype of reactor: plug:$ow system
: complete:mi+ system
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REACTOR DESIGN
and the conce(to. #a"" baance
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Ma"" /aance•
9he mass balance concept is based on thefundamental physical principle that mattercan neither be created nor destroyed
• . balance on a continuous processes at
steady state may be written as/
<nput 0 =eneration > -utput 0 onsumption
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.eration tan" inal lari'er
INFLUENT EFFLUENT
WASTEDSLUDGE
SLUDGERETURN
6iomass <8?U*89 0 6iomass =;-!92 > 6iomass *?U*89 0 6iomass!.S9*7 S?U7=*ood <8?U*89 : ood -8SU%*7 > ood *?U*89 0 ood !.S9*7
S?U7=*
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Schematic f C mplete-Mix
eact r
INFLUENT0 RAW
SEWAGE
TREATEDEFFLUEN
T
WASTEDSLUDGE
SLUDGERETURN
Aerationtank
FinalClarifer
@o, So, Ao
∀, S, A@o 0 @r
A, S @o – @w
Ae, S
@u, Au@r, Au
@w, Au
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%ass balance for biomass
<nput 0 =eneration > -utput 0 onsumption
6iomass <n 0 6iomass =rowth > 6iomass -ut/io#a"" In 1 /io#a"" Gro2th 3 /io#a"" E4uent 1 /io#a"" Wa"te "ud%e
INFLUENT0 RAW
SEWAGE
TREATEDEFFLUEN
T
WASTEDSLUDGE
SLUDGERETURN
Aerationtank
FinalClarifer
@o
, So
, Ao
∀, S, A@o 0 @r
A, S @o – @w
Ae, S
@u, Au@r, Au
@w, Au
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*q BC
/io#a"" In
/io#a""
Gro2th
/io#a""E4uent
/io#a""Wa"ted
"ud%e
%ass balance for biomass – de'ned boundary (C)
INFLUENT0 RAW
SEWAGE
TREATEDEFFLUEN
T
WASTEDSLUDGE
SLUDGERETURN
Aerationtank
FinalClarifer
@o, So, Ao
∀, S, A@o 0 @r
A, S @o – @w
Ae, S
@u, Au@r, Au
@w, Au
uwewod
s
o
oo X Q X QQ X k
S K
XS k X Q +−=−+
∀+ )()(
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*q BC
/io#a"" In
/io#a""
Gro2th
/io#a""E4uent
/io#a""Wa"ted"ud%e
!hereD@o,@w > <n$uent and waste:sludge $ow rate,mE&dAo, A, Ae, Au > biomass concentrations in in$uent,
reactor, euent and clari'erunder$ow(waste sludge), respectively,"g&mE
So, S > soluble food concentration in thein$uent and reactor, respectively, "g&mE
∀ > volume of reactor, mE
F s > half saturation constant,"g&mE
"o > ma+imum growth rate constant, d:C
> :C
%ass balance for biomass
uwewod
s
ooo X Q X )QQ( ) X k
S K XS k ( X Q +−=−+
∀+
% b l f b (f d)
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%ass balance for substrate (food)
<nput 0 =eneration > -utput 0 onsumption
ood <n : ood onsumed > ood -utFood In 5 Food Con"u#ed 3 Food E4uent 1 Food Wa"ted Sud%e
INFLUENT0 RAW
SEWAGE
TREATEDEFFLUEN
T
WASTEDSLUDGE
SLUDGERETURN
Aerationtank
FinalClarifer
@o, So, Ao
∀, S, A@o 0 @r
A, S @o – @w
Ae, S
@u, Au@r, Au
@w, Au
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INFLUENT0 RAW
SEWAGE
TREATEDEFFLUEN
T
WASTEDSLUDGE
SLUDGERETURN
Aerationtank
FinalClarifer
@o, So, Ao
∀, S,A
@o 0 @r
A,S
@o – @w
Ae, S
@u, Au@r, Au
@w, Au
E'6768
FoodIn
FoodCon"u#e
d
FoodE4uent
FoodWa"ted"ud%e!hereD
G > decimal fraction of food mass converted to
biomass
S QS QQS K Y
SX k S Q
wwo
s
o
oo +−=
+∀− )(
)(
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.ll reactions occur in the reactor 9herefore, the volumeD ∀ representsthe volume of the reactor onlyH
INFLUENT0 RAW
SEWAGE
TREATEDEFFLUEN
T
WASTEDSLUDGE
SLUDGERETURN
Aerationtank
FinalClarifer
@o, So, Ao
, S, A
@o 0 @r
A,S
@o – @w
Ae, S
@u, Au@r, Au
@w, Au
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Fro# E'6 769 )#a"" baance .orbio#a""*
/io#a"" In
/io#a""
Gro2th
/io#a""E4uent
/io#a""Wa"ted"ud%e
A""u#(tion 9- 9hein$uent and euentbiomass concentrations arenegligible compared tobiomass at other points inthe system
E'6 76:
uwewod
s
o
oo X Q X QQ X k S K
XS k
X Q +−=−+∀+ )()(
uwd
s
o X Q ) X k S K
XS k ( =−
+∀
d
uw
s
o k X
X Q
S K
S k +
∀=
+
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Fro# E'6 768 )#a"" baance .or.ood0"ub"trate"*
FoodIn
FoodCon"u#e
d
FoodE4uent
FoodWa"ted"ud%e
A""u#(tion 8- 9hein$uent food concentrationSo is immediately diluted to
the reactor concentration S
because of the complete:mi+ regimeA""u#(tion :- 9he foodconcentrations in wasted
sludge outlet is negligible
E'6 76;!here G > decimal fraction of foodmass converted to biomass > (mg&? biomass&mg&? food
utili#ed)
S QS QQ
S K Y
SX k S Q
wwo
s
o
oo +−=
+
∀− )(
)(
)( S S X
Y Q
S K
S k o
o
s
o−
∀=
+
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#binin% E'6 )76:* and )76;* %i<e"-
E'6 76=
The h>drauic detention ti#e in the aerationtan,0reactor-
E'6 76?
d o
ouw
k )S S ( X
Y Q
X
X Q
−−∀=∀
o
Qt ∀=
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The #ean ce5re"idence ti#e )"ud%e a%e 0the a<era%e ti#e that ce )#icroor%ani"#"*
"(end in the reactor*
"titutin% E'6 )76?* and )767* into E'6 )76=*-
E'6 767
E'6 76@
uw
c
X Q
X t
∀=
d
o
ck tX
S S Y
t −
−=
)(1
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The <ou#etric oadin% rate L i" the #a""o. /OD in the inBuent di<ided b> the <ou#eo. the reactor
The concentration o. bio#a"" in the reactor)MLSS* i" .ound b> "o<in% E'6 )76@*-
E'6 76
E'6 769
)1(
)(
cd
oc
t k t
S S Y t X
+
−=
∀=∀ oo
L
S Q
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The .ood5to5#icroor%ani"#" ratio i" u"ed toe$(re"" /OD oadin%" 2ith re%ard to the
bio#a"" in the reactor-
he recircuation ratio i"-
E'6 7699
E'6 7698
X
S Q
M
F oo
∀=
Q
Q R r =
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*+ample – .ctivated Sludge
An acti!ated sludge system is to "e used for secondarytreatment of #$%$$$ m&/d of municipal wastewater'
After primary clarification% the ()D is #*$ mg/+% and it
is desired to ha!e not more than * mg/+ ()D in the
effluent' A completely mixed reactor is to "e used% and
pilot-plant analysis has esta"lished the following
kinetic !alues, . $'* kg/kg% kd . $'$*d-#' Assuming an
M+SS concentration of &$$$ mg/+ and an underflow
concentration of #$%$$$ mg/+ from the secondary
clarifier% determine,
a the !olume of the reactor " the mass and !olume of solids that must "e wasted
each day
c the recycle ratio
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Proce"" Bo2 chart
Aerationtank
FinalClarifer
Qo = 10000 m3/d
So = 0.15 kg
BOD/m3
X (!SS" = 3.0kg/m3
0
X# = 10.0 kg/m3
Se = $0.005 kgBOD /m3
. $'* kg/kg% kd . $'$*d-#
Q%X#0r u
∀
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Soution-C Select tc > CI day, solve *q(BJ) with t > ∀&@
∀ = 1611 m3
d
o
c
k X
S S QY
t −
∀
−=
)(1
05.00.3
)005.015.0)(5.0(000,101.0 −∀
−=
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1 .t equilibrium conditions, *q(BB) applies
%ass ⇒
Kolume ⇒
uw
c
X Q
X t ∀=
c
uW
t
X X Q ∀=
daykg X QuW
/3.48310
)0.3(1611 ==
daymmkg d kg Q
W /3.48
/10/3.483 3
3==
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E . mass balance around the secondary clari'ercan be written as follows/
.ssuming that the solids in the euent arenegligible compared to the in$uents and
under$ow ( Xe LL X )/
uwr ewor o X QQ X QQ X QQ )()()( ++−=+
uwr r o X QQ X QQ )(0)( ++=+
X X
X QQX Q
u
uw
r
−
−=
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∴ 9he recirculation ratio is
d mmkg mkg
d kg mkg d mQr
/4217/3/10
/3.483))/0.3(/000,10(
3
33
33
=
−
−=
42.0000,10
4217===
Q
Q R r
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*+ample – *+tended .eration
An &'tended Aeration S) treat* 1500 m3 o+*e%age dail, %it- a BOD onentration o+300 mg/!. Cal#late t-e ol#me re#ired +ort-e aeration omartment. 2ene -ek t-e2,dra#li 4etention ime (24" and t-e
ol#metri loading. Determine t-e #antit, o+*l#dge %a*ted er da,. A**#me a !SSonentration o+ 3500 mg/! and a Solid*4etention ime (S4" o+ 5 da,*.
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Proce"" Bo2 chart
Aerationtank
FinalClarifer
Qo = 1500 m3/d
So = 0.3 kg
BOD/m3
X (!SS" = 3.5kg/m3t (S4" = 5
d0
Q%X#0
240
0
∀
L∀
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Solution
1. Find volume
C-oo*e F/ ratio a*6
0.1 kg BOD/kg !SS.da,
X
QS
M
F o
∀=
5.3
3.015001.0
×∀
×=
31286m=∀∴
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= 20.6 hoursOK (18 < HRT < 24)
2. Find HRT
!15001"# $R% ×=
oQ )t ( $R% ∀
=
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3. Find volumetric loading
= 0.35 kgBO/m3
(OK ! 0.16 < < 0.40)
∀=∀ oo
L S Q
1286
3.01500×=
L∀
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4. Find uantity o! sludge "asted
∴ "u#$%&%' o su*g+ #s%+*, Qw X u =
= 180 kg/*#'
uw X Q
X SR%
∀
=
25
5.31286×
%ass balance for biomass – de'ned boundary (1)