4-activated sludge models_f12
TRANSCRIPT
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Substrate Utilization Rate, rsu
1 1 m su
u g s s
S dS dX kSX r X
dt Y dt Y K S K S
(9)
Since k = μm/Y,
u s
dS kSX
dt K S
(10)
Also, in a reactor
u
dS So S
dt
(11)
Specific Substrate Utilization Rate, U
u
dS
dt U
X
or
u
dS UX
dt
(3)
( ) ( )u
dS
So S So So S Q So So S dt U
X X So X V X So
QSo So S F U E
VX So M
(12)
where E = (So - S)/ So = substrate removal efficiency
θ = V/Q = hydraulic retention time = aeration time, hr
F/M = Food to microorganisms ratio
Combining (10) and (11) gives
u s
dS So S kSX
dt K S
(13)
Dividing Eq. (13) by X yields
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u
s
dS
So S kS dt U
X X K S
(14)
s
kS U K S (15)
From (7)
mnet d d
s s
S Y k S k k
K S K S
net d YU k
Completely Mixed Activated Sludge Process
Assumptions for the formulation of the mass balance equations
- The following conditions are assumed in the formulation of the mass balance equations:
(1) Steady state- Flow, biomass concentrations, and substrate concentrations are in a steady state.
dQ/dt = 0, dX/dt = 0, dS/dt = 0(2) Soluble BOD
- All substrates are soluble (filtered BOD or COD)
(3) Completely mixed system- The aeration tank is completely mixed.
- The substrate concentration in the aeration tank equals the substrate concentration in the
effluent after treatment.(4) Biological activity occurs only in the aeration tank.
(5) No microorganisms are present in the influent wastewater, Xo = 0.
(6) The mean cell residence time (MCRT) is calculated based on the biomass in the aeration tank(7) Excess activated sludge is wasted from the aeration tank rather than from the sludge
recirculation line.
Mass balance on X around the secondary treatment
- Mass balance for biomass (X) around the entire secondary treatment system
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(Xo=0) Aeration Tank Secondary
Sedimentation
Q, Xo X(Q+Qr) Tank S, Xe
V, X, S Q-Qw
Qr, Xr
Qw, Xr
Accumulation = Inputs - Outputs ± Rxns
( ) 'dX
V QX Q Q X Q X r V o w e w r g dt
(1)
accidentally intentionally bacterial
wasted wasted growth
where
'net
net
g
dX YkS r X k X g d dt K S
s
(2)
r g’ = net growth rate, mg/L. d
Y = yield coefficient, mg X / mg SK s = half-saturation constant, mg/L
k d = endogenous decay rate constant, d-1
k = maximum substrate utilization rate, d
-1
k = µm /Y
µm = maximum specific growth rate
µm = Y k
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Since it is assumed that there are no bacteria in inflow (i.e. Xo = 0),
' ( )dX
V r V Q Q X Q X g w e w r dt
(3)
Net rate of Net rate Net rate of biomass
change in of growth out of the system
biomass of biomass
Dividing Eq. (3) by V
( )'
Q Q X Q X dX w e w r r g dt V
At steady state, dX/dt = 0,
( )'
Q Q X Q X w e w r r
g V
Since
'
net
net
g
dX r X g dt
' ( ) net
r Q Q X Q X g w e w r
X V X
By the definition of MCRT, θc
( )
V X c
Q Q X Q X w e w r
Thus,
1
cnet
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Since
net
Y k S k
d K S s
andkS
U Ks S
1= =
cnet d
Y kS k YU k
d K S s
1=
cnet d YU k
(5)
or
1
c
Y kS k
d K S s
(6)
Solving for S yields
(1 ) (1 )
( ) 1 ( ) 1m
K k K k s d c s d cS
Yk k k c d c d
where
Y = yield coefficient, mg X / mg S
K s = half-saturation constant, mg/L
k d = endogenous decay rate constant, d-1 k = maximum substrate utilization rate, d
-1
k = µm /Y or µm= Yk
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Mass balance on S around the aeration tank
- A mass balance for substrate entering and leaving the aeration tank
(Xo=0) Aeration Tank Secondary
Sedimentation
Q, So S, (Q+Qr) Tank S
V, S Q-Qw
Qr, S
Qr, S Qw, S
Net rate of Rate of Rate of Rate of
change in substrate substrate entering substrate leaving substrate utilization
in aeration tank aeration tank aeration tank in aeration
tank
Accum = Inputs - Outputs ± Rxns
( )o r r su
dS V QS Q S Q Q S r V
dt (1)
where
su
u s
dS kSX r
dt K S
(2)
r su = substrate utilization rate, mg/L-d
Since R = Qr/Q, Qr = RQ
( )
1
o su
o su
dS V QS RQS Q RQ S r V
dt
QS RQS Q R S r V
(3)
o su suQS RQS QS RQS r V Q So S r V
Divided by V
su
Q So S dS r
dt V
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At steady state, dS/dt = 0,
su
u
Q So S So S dS r
V dt
Divide both sides by X
su u
dS
Q So S r So S dt U
X X X V X
Since
1
= cnet d YU k
or
1
c d YU k
1c
d
YQ So S k
XV
(4)
Solve for X
1c
d YQ So S k XV
XV
c
d
XV YQ So S k XV
c
d
XV k XV YQ So S
1c d XV k YQ So S
(5)
1 1 1
c c cd d d
YQ So S Y So S Y So S X
V V k k k
Q
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Multiply by Өc
( )
(1 )
c o
d c
Y S S X
k
Substituting θ = V/Q, solve for V
or using (5), solve for V
1
c d
YQ So S V
X k
Multiply by θc
V c Y Q S o S X k d c
( )( )1
Example: Q = 10,000 m3/d, So = 120 mg/L, S = 7 mg/L,
X (MLVSS) = 2,000 mg/L, θc = 6 days, Y = 0.6 mg VSS/mg BOD, k d = 0.06 d-1
What is the aeration tank volume?
(Solution)
V Y Q S S
X k
d mgVSS
mgBOD
m
d
mg
L BOD
mg Ld
d
c o
d c
( )
( )
.. ,
/.
.
1
6 00 6 10 000
120 7
2000 10 06
6 0
3
V = 1495 m3
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Determination of k and Ks
s
kS U
K S
Inverse of U gives
1 s s K S K S
U kS kS kS
1 1 1 s K
U k S k
Determination of Y and k d,
Since
1
c d YU k
Note
dX dX dX
r g’= (----)g net
= (----)g − (-----)d = µ X – k d X = (µ – k d) X = µ net X = µ’ X (1)
dt dt dt
since
g u
dX dS Y
dt dt
(3)
' g d u
dS r Y k X
dt
1/Өc
Slope = Y
U = (F/M) E
Intercept = – kd
0
1/U
Slope = Ks/k
Intercept =1/k
1/S
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Divide by X
' 1 g unet
dS
r dt Y
X X c
Since
u
dS
dt U
X
1
c d YU k
'1 g
net
g net
dX
dt r
X X c
Observed Growth (Cell) Yield, Yobs
Since g u
dX dS Y
dt dt
,
g
u
dX
dt Y
dS
dt
Yobs is defined as
g
net
obs
u
dX
dt Y
dS
dt
(1)
where
g
net
d
u
dX dS Y k X
dt dt
or
( )
net
net d d
g s
dX Y k S X k X YU k X
dt K S
(2)
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and
u
dS UX
dt
(3)
Substituting (2) and (3) into (1) yields
g
net
d d obs
u
dX
dt YU k X YU k Y
dS UX U
dt
(4)
Since1
c
d YU k
,
1
c d k
U
Y
(5)
Substituting (5) into (4) yields
c
1
c1
c
1 1 1
c c
d
d
d d
d obs
d d d
k
Y k Y Y k k
YU k Y Y
U k k k
Y
Thus,
c1obs
d
Y Y
k
Note:
,net obs net obs X
Y Y U X
Thus,1 1
net obs
cY U
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Excess Biological (Volatile) Solids, Px
The VSS concentration (ML-3
T-1
) is given by x obs o P Y S S
Since 1obs c d
Y
Y k ,
( )
1o
xd c
Y S S
P k
Based on VSS mass flux (MT-1
): ( ) x obs o P Y S S Q
Example : Determine the excess sludge (in mg/L VSS) for the given conditions.Q = 10,000 m
3/d, So = 120 mg/L, S = 7 mg/L
k d = 0.06/d, θc = 10 days, Y = 0.6 mg VSS/mg BOD
(Solution)
Y Y
k
mgVSS
mg BOD
d d
mgVSS
mgBODobs
c d
1
0 6
1 100 06
0375
.
.
.
P mgVSS
mgBOD
mgBOD
LmgVSS L x
0375 120 742 4
. ( ). /
P in kg d mgVSS
L
m
d
L
m
kg
mg
kg VSS d
x ( / ). ,
/
42 4 10 000 1000
10
420
3
3 6
Note
( )
(1 )c o c c
obs o x
d c
Y S S X Y S S P
k
where
P Y S S
Y Y
k
x obs o
obsc d
( )
1
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/ /
c
x x
X XV
P P Q
biomass in Aeration Tank biomass in Aeration Tank
biomass synthesized time biomass wasted time
XV XVθc = ---------------------------- = --------
Qw X + (Q - Qw) Xe Px Q
Px Q = Qw X + (Q - Qw) Xe
Effect of Temperature
Temperature affects on:1) metabolic activities of the microbial population (growth rate)2) gas transfer rate
3) settling characteristics of the biological solids.
The effect of temperature on the reaction rate of a biological process is expressed by
20
20
T
T C r r
where r T = reaction rate at TºC
r 20ºC = reaction rate at 20ºC
θ = temperature-activity coefficientT = temperature, ºC
Table 8.5 (3rd
ME 373) presents some typical values of θ for commonly used biological processes.
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Example F/M Ratio (possible take home exam problem)
Knowing the following equations: 1
c d YU k
and
F U E
M
derive the F-to-M ratio equation: F Q So
M V X
(Solution)
Since1
c d YU k
and
F U E
M
1
c d
F Y E k
M
Solving for F/M yields
1
c d k F
M YE
Since F
U E M
,
F U
M E
where
u
dS
So S dt U
X X
and
So S E
So
F U So S So So
M E X So S X
Since Ө = V/Q,
F Q So
M V X
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Recirculation Ratio, R
Determining the recirculation ratio, R, from the mass balance on VSS around the aeration tank.
(Xo=0) Aeration Tank Secondary
SedimentationQ, Xo X(Q+Qr) Tank
V, X, S
Qr, Xr
Mass balance on X around the aeration tank without bacterial growth assuming(Q+Qr ) Xe and Qw Xr
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Example : Determine the recirculation ratio, R, for the given conditions below:
Operating parameters: HRT = θ = 5.3 hr, X = 2,500 mg/L, Xr = 10,000 mg/L, S=10 mg/L
Kinetics constants: Y = 0.6 mgVSS/mg BOD, k d = 0.06 d-1
, Ks = 60 mg/L, k = 5.0 /d.
(Solutions)
'
net
m g net d d
g s s
dX S Y k S r X k X k X
dt K S K S
'
0.6 5.0 10
0.06 2500
60 10
2500 9210.36857
g d
s
mgVSS mgBOD mgBOD
mgBOD mgVSS d LY k S mgVSS r k X
mgBOD mgBOD K S d L
L L
mgVSS mgVSS
L L d
' 921 5.3 203.4 /24 /
g
mgVSS hr r mg L
L d hr d
' 2500 203 /0.31
10,000 2500 /
g r
r
X r mg LQ R
Q X X mg L
or without consideration of the bacterial growth
2500 /0.33
10,000 2500 /r
r
mg LQ X R
Q X X mg L
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Example 8-1 Activated-sludge process analysis. (3rd
ME, p. 389)
An organic waste having a soluble BOD5 of 250 mg/L is to be treated with a completely-mixactivated-sludge process. The effluent BOD5 is to be equal to or less than 20 mg/L.
Design the reactor assuming that the temperature is 20°C, the flowrate is 5.0 Mgal/d, and that the
following conditions are applicable.
1. Influent volatile suspended solids to reactor are negligible.
Xo = 0
2. Return sludge concentration, Xr = 10,000 mg/L as suspended solids (SS) or 8,000 mg/L as
volatile suspended solids (VSS).
Xr (as SS) = 10,000 mg/L , Xr (as VSS) = 8,000 mg/L
Note: VSS/SS = 0.8; i.e., 0.8 (10,000 mg/L) = 8000 mg/L VSS
3. Mixed-liquor volatile suspended solids (MLVSS or X) = 3,500 mg/L
MLVSS / MLSS = 0.8
4. Mean cell-residence time θc = 10 days.
5. Hydraulic regime of reactor is complete mix.
6. Kinetic coefficients:
Y = 0.65 mg cell / mg BOD5 utilized, k d = 0.06 d-1
7. It is estimated that the effluent will contain about 20 mg/L of biological solids, of which
80 % is volatile and 65 % is biodegradable.
Xe = 20 mg/LXe,VSS = 0.8 (20 mg/L) = 16 mg/L
Xe, biodeg = 0.65 (20 mg/L) = 13 mg/L
Assume that the biodegradable biological solids can be converted from ultimate BODdemand to a BOD5 demand using the factor 0.68 [e.g., the deoxygenation coefficient, K
(base 10) = 0.1 d-1
.
8. Waste contains adequate nitrogen, phosphorus, and other trace nutrients for biological
growth.
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Given:
Q = 5 MGD ST = 20 mg/L
So=250 mg/L S = ?
Xo = 0 X = MLVSS = 3500 mg/L
V=? Xe = 20 mg/L SS
S=? 80% volatile
65% biodegradable
Qw
Xr
Xr = 8,000 mg/L VSS (or 10,000 mg/L SS)
θc = 10 dY = 0.65 mg cell/mg BOD5
k d = 0.06 d-1
Preliminary calculation
(1 10 ) Kt t o y L
BOD
BOD
y
Lo L
d d 5 5 01 51 10 0 68 ( . / )( ) .
wherey5 = BOD5
Lo = BODL
Thus, BOD5 = 0.68 BODL
Note: k (base e) = 2.303 (0.1 d-1
)
= 0.2303 d-1
(base e)
Xe = 20 mg/L SS= 0.8 (20 mg/L) = 16 mg/L VSS= 0.65 (20 mg/L) = 13 mg/L X biodeg
Lo
BOD5
yt
5 20
Time (d)
BOD5 = Lo (1-10-kt
)
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(Solution)
1. Estimate the soluble BOD5 in the effluent.
55 5Influent soluble BODEffluent BOD = BOD of effluent biological solidsescaping treatment
Biodegradable
biologicalsolids (VSS)
C5H7 NO2 + 5O2 5CO2 + 2H2O + NH3 + energy
cell oxygen
g O2 (5 mol) (32 g/mol)------------- = -------------------------- = 1.42 mg O2 / mg X biodeg
g X biodeg (1 mol) (113 g/mol)
X biodeg = 0.65 X
BOD5 = 0.68 (BODL)
1.1
BOD5 from Xe
= (20 mg Xe/L)(0.65) (1.42 mg O2 / mg Xe biodeg ) (0.68) = 12.55 mg/L BOD5
I
(20 mg Xe/L)(0.65) = X biodeg
I
(Xe biodeg)(1.42 O2 / Xe biodeg) = BODL
I
(BODL)(0.68) = BOD5
2 55
deg
5
20 1.42 0.68(0.65)
12.55
bio L
mgXe mgO BOD BOD from Xe
L mg Xe BOD
mg BOD
L
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Allowed BOD5 in effluent = 20 mg/L BOD5 = (S) + (12.55 mg/L BOD5 )
1.2
Soluble BOD5 in effluent = S = 20 - 12.6 = 7.4 mg/L soluble BOD5
1.3 The biological treatment efficiency based on soluble BOD5 would be
Es = (So - S) / So = (250 mg/L - 7.4 mg/L) / 250 mg/L
= 0.97
= 97 %
1.4 The overall plant efficiency would be
Eoverall = (250 - 20) / 250 = 0.92 = 92 %
2. Compute the reactor volume. The volume of the reactor can be determined using Eq. 8-42
by substituting V/Q for θ and rearranging the equation as follows:
V Y Q S S
X k
c o
d c
( )
( )1
V
mg
mgBOD utilized
Mgal
d d mg L
mg Ld
d
Mgal
cell
0 65 510 250 7 4
3500 10 06
10
145
.. /
( / ).
( )
.
Hydraulic retention time, θ
V 1.4 Mgal
θ = --- = --------------- = 0.28 d = 6.72 hrs
Q 5 Mgal / d
3. Compute the sludge-production rate on a mass basis.
3.1 The observed yield is:
Y Y
k
mg cell mg BOD
d d
mg Cell
mg BODobs
d c
1
0 65
10 06
10
04065
5
( . ) / ( ).
( )
.
3.2 The biomass production rate is:
Px = Yobs (So - S)
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P mg cell L
mg BOD Lmg BOD L
mg cell
L x
0406250 7 4
98 5
55
. /
/( . ) /
..
Px = Yobs (So - S) Q
P mg cell
L
Mgal
d
lb cell
d
lbVSS
d x
985 58 34
4107 4107.( . )
4. If sludge wasting from a recirculation line
cw r w e
VX
Q X Q Q X
( )
Rearranging for Qw yields
Q VX Q Xe
X Xew
c
c r
( )
Q MG mgVSS L d MGD mgVSS L
d mgVSS L MGDw
( . )( / ) ( )( )( / )
( )( ) /.
14 3500 10 5 16
10 8000 160051
If sludge wasting from an aeration rank
cw w e
VX
Q X Q Q X
( )
c w c c w
c w c
Q X QXe Q Xe VX
Q X Xe VX QXe
( )
Q VX QXe
X Xew
c
c
( )
Qw Mgal mgVSS L d MGal d mgVSS L
d mgVSS L MGD
( . )( / ) ( )( / )( / )
( )( ) /.14 3500 10 5 16
10 3500 160118
or
cw w e
VX
Q X Q Q X
( )
Assuming Qw
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cw e
VX
Q X Q X
( )
Rearranging for Qw
c w c
c w c
Q X QXe VX
Q X VX QXe
Q VX Q Xe
X w
c
c
Q Mgal mgVSS L d MGal d mgVSS L
d mg VSS L MGDw
( . )( / ) ( )( / )( / )
( )( / ).
14 3500 10 5 16
10 35000117
4. Compute the recirculation ratio (R) using a VSS mass balance around the reactor neglecting the
suspended solids in the influent (Xo = 0).
Xr Qr = X (Q + Qr)
Xr Qr = XQ + XQrXr Qr – X Qr = XQ
Qr (Xr – X) = XQ
R = Qr/Q = X/(Xr – X) = (3500 mg VSS.L) / (8000 – 3500) mg VSS/L = 0.78
If the bacterial growth is included,
V dX
dt QrXr X Q Qr r V g ( ) '
at steady state, dX/dt = 0
Q
Aeration tank
X = 3500 mg VSS/L
X(Q+Qr)
Qr
Xr = 8,000 mg VSS/L
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Qr Xr + r g’ V = X (Q + Qr) = XQ + Qr
Qr (Xr – X) = XQ – r g’ V
QQ
X r V
Q X X
X rg X X
mgVSS
L
mgVSS
L d d
mgVSS L
r
g
r r
'
' ..
( ) /.
3500 3500 28
8000 35000 76
where r X X d
mgVSS L mgVSS L d g net c
' / / .
1 1
103500 350
6.
V
Q
Mgal
Mgal d d hrs
14
60 28 6 72
.
/. .
7.7-1. Check the specific utilization rate,
U dS dt
X
Q So S
V X
So S
X
u
( / ) ( )
U mgBOD Ld mgVSS L
mgBOD utilized mgMLVSS d
( . ) /( . )( / )
..
250 7 40 28 3500
0 255 5
7.2 Check the food-to-microorganism ratio
5
5
250 /(0.28 )(3500 / )
0.255
.
F Q So So
M V X X
mgBOD Ld mgVSS L
mgBOD
mgMLVSS d
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7.3 Check the volumetric BOD loading rate (VLR) in lb BOD5/1000 ft3·d
5
6 3
5 5
3 3
250 58.34
10 1(1.4 )
7.48
10425 / 56
187166 1000 .
mgBOD Mgal
SoQ L d VLR
V gal ft Mgal
Mgal gal
lbBOD d lbBOD applied
ft ft d
Note
U F
M E
mgBOD
mgMLVSS d
mgBOD
mgVSS d
0255 250 7 4
250
0 255 5.
.
. .
.