4-Activated Sludge Models_F12
Substrate Utilization Rate, r su
1 dX 1 S kSX dS m X d t Y d t Y K S K S u g s s
r su
(9)
Since k = μm/Y, kSX dS dt K S u s
(10)
Also, in a reactor So S dS dt u
(11)
Specific Substrate Utilization Rate, U
dS dt u U X
or
dS UX dt u
(3)
dS dt u So S So ( So S ) Q So ( So S ) U X
X
So X
V
X So
QSo So S F E VX So M
U
(12)
where E = (So - S)/ So = substrate removal efficiency θ = V/Q = hydraulic retention time = time = aeration time, hr F/M = Food to microorganisms ratio
Combining (10) and (11) gives So S kSX dS dt K S u s
(13)
Dividing Eq. (13) by X yields
1
4-Activated Sludge Models_F12
U
U
dS dt u X kS K s S
So S
X
kS K s S
(14)
(15)
From (7) net
m S K s S
kd
net YU k d
Y k S K s S
k d
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 th e 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
2
4-Activated Sludge Models_F12
(Xo=0)
Aeration Tank
Secondary Sedimentation X(Q+Qr) Tank
Q, Xo V, X, S
S, Xe Q-Qw
Qr, Xr
Qw, Xr
Accumulation = Inputs - Outputs ± Rxns V
dX dt
QX
o
(Q Q ) X w
e
Q X
w r
r 'V g
accidentally
intentionally
bacterial
wasted
wasted
growth
(1)
where net YkS dX r ' k X net X g dt K S d g s
r g’ = net growth rate, mg/L. d Y = yield coefficient, mg X / mg S K s = half-saturation constant, mg/L -1 k d = endogenous decay rate constant, d -1 k = maximum substrate utilization rate, d k = µm /Y µm = maximum specific growth rate µm = Y k
3
(2)
4-Activated Sludge Models_F12
Since it is assumed that there are no bacteria in inflow (i.e. Xo = 0), V
dX
dt
r ' V (Q Q ) X Q X g w e w r
Net rate of
Net rate
change in
of growth
biomass
of biomass
Net rate of biomass out of the system
Dividing Eq. (3) by V dX dt
r ' g
(Q Q ) X Q X w e w r V
At steady state, dX/dt = 0, (Q Q ) X Q X w e w r V
r ' g Since
net
dX r ' net X g dt g
(Q Q ) X Q X w e w r net X V X r ' g
By the definition of MCRT, θc c
V X (Q Q ) X Q X w e w r
Thus, net
1 c
4
(3)
4-Activated Sludge Models_F12
Since net
net
Y k S K S s Y kS K S s
k and U d
k = YU kd =
net YU k d =
d
1 c
kS Ks S 1
c
(5)
or 1 c
Y kS K S s
k
(6)
d
Solving for S yields K (1 k ) K (1 k ) s d c d c S s (Yk k ) 1 ( m k ) 1 c d c d where Y = yield coefficient, mg X / mg S K s = half-saturation constant, mg/L -1 k d = endogenous decay rate constant, d -1 k = maximum substrate utilization rate, d k = µm /Y or µm= Yk
5
4-Activated Sludge Models_F12
Mass balance on S around the aeration tank
- A mass balance for substrate entering and leaving the aeration tank
(Xo=0) Q, So
Aeration Tank
V,
Secondary Sedimentation S, (Q+Qr) Tank
S
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 V
dS dt
QSo Qr S (Q Qr ) S rsu V
(1)
where kSX dS dt u K s S
r su
(2)
r su = substrate utilization rate, mg/L-d Since R = Qr/Q, Qr = RQ V
dS dt
QSo RQS (Q RQ) S rsu V
(3)
QSo RQS Q 1 R S rsu V
QSo RQS QS RQS rsu V Q So S rsu V Divided by V dS dt
Q So S V
r su
6
4-Activated Sludge Models_F12
At steady state, dS/dt = 0, Q So S
r su
V
So S
dS dt u
Divide both sides by X
r su X
dS dt u
X
Q So S XV
So S X
U
Since net YU k d = or 1
1 c
YU k d
c 1
c
YQ So S XV
k d
(4)
Solve for X 1
c
XV c XV c
YQ So S kd XV XV
YQ So S kd XV kd XV YQ So S
1 kd YQ So S c
XV
X
YQ So S
1 kd c
V
Y So S V 1
kd Q c
(5)
Y So S
1 kd c
7
4-Activated Sludge Models_F12
Multiply by Өc X
cY ( So S ) (1 k d c )
Substituting θ = V/Q, solve for V or using (5), solve for V V
YQ So S
1 k d c
X
Multiply by θc Y Q ( S o S ) V c X (1 k d c )
3
Example: Q = 10,000 m /d, So = 120 mg/L, S = 7 mg/L, -1 X (MLVSS) = 2,000 mg/L, θc = 6 days, Y = 0.6 mg VSS/mg BOD, k d = 0.06 d What is the aeration tank volume? (Solution)
Y Q ( S o S ) V c X (1 k d c )
0.6 mgVSS 10,000m3 mg 120 7 BOD 6.0 d L d mgBOD
2000mg / L1
6.0d d
0.06
3
V = 1495 m
8
4-Activated Sludge Models_F12
Determination of k and Ks
U
kS K s S 1/U
Inverse of U gives 1
U
K s S kS
K s kS
Slope = Ks/k
S kS
Intercept =1/k
K 1 1 s U k S k 1
1/S
Determination of Y and k d,
Since 1 c
YU k d
1/Өc
Slope = Y 0 U = (F/M) E Intercept = – k d
Note dX dX dX net r g’= (----)g = (----)g − (-----) d = µ X – k d X = (µ – k d) X = µ net X = µ’ X dt dt dt since dX
dS Y dt g dt u
(3)
dS kd X dt u
r g ' Y
9
(1)
4-Activated Sludge Models_F12
Divide by X r g '
net
X
Y
dS dt u X
1 c
Since
dS dt u U X
1 c
YU k d
r g '
net
X
dX dt X
net
g
1 c
Observed Growth (Cell) Yield, Y obs
dX dt g dX dS Since Y , Y dS dt g dt u dt u Yobs is defined as
Y obs
dX dt
net
g
dS dt u
(1)
where
dX dt
net
g
dS Y kd X dt u
or
YkS dX dt net X K S k d X (YU kd ) X g s net
10
(2)
4-Activated Sludge Models_F12
and
dS UX dt u
(3)
Substituting (2) and (3) into (1) yields
Y obs
dX dt
g
dS dt u 1
Since
net
c
YU kd X UX
YU k d U
(4)
1 k d c U Y
YU k d ,
(5)
Substituting (5) into (4) yields
1 c k d Y k d 1 Y Y kd k d YU kd Y c Y obs 1 1 U 1 c k d kd k d c
c
Y
Thus, Y obs
Y 1 c k d
Note: Yobs
net X X
net Yobs U
,
Thus, c
1 net
1 Yobs U
11
4-Activated Sludge Models_F12
Excess Biological (Volatile) Solids, P x
The VSS concentration (ML T ) is given by P x Yobs So S -3
Y obs
Since
Y 1 c k d
,
-1
P x
Y ( So S ) 1 k d c
Based on VSS mass flux (MT ): P x Yobs ( So S ) Q -1
Example : Determine the excess sludge (in mg/L VSS) for the given conditions. 3
Q = 10,000 m /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) 0.6 mgVSS Y obs
Y 1 c k d
mg BOD
0.06 d
0375 . mgVSS mgBOD
1 10d
0375 . mgVSS (120 7)mgBOD 42.4 mgVSS / L L mgBOD
P x
10,000 m3 1000 L kg 42.4 mgVSS P x (in kg / d ) 3 6 L d 10 mg m
420 kg VSS / d
Note
X
c Y (So S ) (1 k d c )
c
Yobs S o S
c
P x
where P x Yobs ( S o S ) Y Y obs 1 c k d
12
4-Activated Sludge Models_F12
c
X P x
XV PxQ
biomass in Aeration Tank biomass synthesized / time
biomass in Aeration Tank biomass wasted / time
XV XV θc = ---------------------------- = -------Qw X + (Q - Qw) Xe Px Q
Px Q = Qw X + (Q - Q w) 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 b y
rT r 20C T 20 where r T = reaction rate at TºC r 20ºC = reaction rate at 20ºC θ = temperature-activity coefficient T = temperature, ºC rd
Table 8.5 (3 ME 373) presents some typical values of θ for commonly used biological processes.
13
4-Activated Sludge Models_F12
Example F/M Ratio (possible take home exam problem)
Knowing the following equations: derive the F-to-M ratio equation:
1 c
F M
YU k d Q So V X
(Solution)
F YU k d and U E c M 1
Since
F Y E k d c M 1
Solving for F/M yields 1 k d F c M
YE
F E , M
Since U F
U
M
E
where
U
F M
dS dt u
X
So S X
and E
So S So
So S So So E X So S X
U
Since Ө = V/Q, F M
Q So V X
14
and
F E M
U
4-Activated Sludge Models_F12
Recirculation Ratio, R Determining the recirculation ratio, R, from the mass balance on VSS around the aeration tank. (Xo=0) Q, Xo
Aeration Tank
Secondary Sedimentation 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 << Qr Xr
Qr X r Q Qr X Qr X r QX Qr X Qr X r X QX R
Qr Q
X X r X
Mass balance on X around the aeration tank with bacterial growth assuming (Q+Qr ) Xe and Qw Xr << Qr Xr V
dX dt
Qr X r (Q Qr ) X rg ' V
At SS, dX/dt = 0
Qr X r rg ' V (Q Qr ) X QX Qr X Qr X r Qr X QX rg ' V
rg ' V
Q
Qr ( X r X ) Q X rg ' V Q X
r g ' V X Q X r g ' Qr R Q (Xr X ) X r X
15
4-Activated Sludge Models_F12
Example : Determine the recirculation ratio, R, for the given co nditions below:
Operating parameters: HRT = θ = 5.3 hr, X = 2,500 mg/L, Xr = 10,000 mg/L, S=10 mg/L -1 Kinetics constants: Y = 0.6 mgVSS/mg BOD, k d = 0.06 d , Ks = 60 mg/L, k = 5.0 /d. (Solutions)
m S YkS dX r X k X kd X net d K S K S dt g s s net
' g
0.6mgVSS 5.0mgBOD 10mgBOD YkS mgBOD mgVSS d L 0.06 2500mgVSS ' r g kd X 60 mgBOD 10 mgBOD K S d L s L L 2500mgVSS 921mgVSS 0.36857 L L d
921 mgVSS 5.3hr 24hr / d 203.4 mg / L L d
r g '
R
Qr Q
X r g ' X r X
2500 203 mg / L 0.31 10,000 2500 mg / L
or without consideration of the bacterial growth R
Qr Q
X X r X
2500 mg / L 0.33 10,000 2500 mg / L
16
4-Activated Sludge Models_F12
rd
Example 8-1 Activated-sludge process analysis. (3 ME, p. 389)
An organic waste having a soluble BOD5 of 250 mg/L is to be treated with a completely-mix activated-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: -1
Y = 0.65 mg cell / mg BOD5 utilized, k d = 0.06 d 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/L Xe,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 BOD demand to a BOD5 demand using the factor 0.68 [e.g., the deoxygenation coefficient, K -1 (base 10) = 0.1 d .
8.
Waste contains adequate nitrogen, phosphorus, and other trace nutrients for biological growth.
17
4-Activated Sludge Models_F12
Given: ST = 20 mg/L
Q = 5 MGD So=250 mg/L
S=?
Xo = 0
X = MLVSS = 3500 mg/L V=? S=?
Xe = 20 mg/L SS 80% volatile 65% biodegradable
Qw Xr Xr = 8,000 mg/L VSS (or 10,000 mg/L SS)
θc = 10 d Y = 0.65 mg cell/mg BOD5 -1 k d = 0.06 d
Preliminary calculation
yt Lo (1 10 Kt ) BOD5 BOD L
y5 Lo
1 10 (0.1/ d )(5d ) 0.68
where y5 = BOD5 Lo = BODL
-kt
BOD5 = Lo (1-10 )
Thus, BOD5 = 0.68 BODL
Lo -1
Note: k (base e) = 2.303 (0.1 d ) -1 = 0.2303 d (base e)
BOD5 yt
Xe = 20 mg/L SS = 0.8 (20 mg/L) = 16 mg/L VSS = 0.65 (20 mg/L) = 13 mg/L X biodeg
5
20 Time (d)
18
4-Activated Sludge Models_F12
(Solution) 1.
Estimate the soluble BOD5 in the effluent.
Influent soluble BOD5 BOD5 of effluent biological solids
Effluent BOD5 = escaping treatment Biodegradable biological solids (VSS) C5H7 NO2 cell
+
5O2 oxygen
5CO2 + 2H2O + NH3 + energy
g O2 (5 mol) (32 g/mol) ------------- = -------------------------- = 1.42 mg O 2 / 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
BOD5 from Xe
20 mgXe L
1.42 mgO2 0.68 BOD5 mg Xebio deg BODL
(0.65)
12.55 mg BOD5 L
19
4-Activated Sludge Models_F12
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
1.4
= (So - S) / So = (250 mg/L - 7.4 mg/L) / 250 mg/L = 0.97 = 97 %
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:
Y Q ( So S ) V c X (1 k d c )
V
0.65 mg cell 5 Mgal 10 d 250 7.4 mg / L mgBOD5 utilized d 0.06 (3500 mg / L) 1 (10 d ) d
14 . Mgal
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 obs
Y
1 k d c
(0.65 mg cell ) / ( mg BOD5 ) 0406 . mg Cell 0.06 mg BOD5 1 (10 d ) d
3.2 The biomass production rate is: Px = Yobs (So - S) 20
4-Activated Sludge Models_F12
0406 . mg cell / L
P x
mg BOD5 / L
(250 7.4) mg BOD5 / L
98..5 mg cell L
Px = Yobs (So - S) Q
4107 lb cell 4107 lbVSS 98.5 mg cell 5 Mgal (8.34) d L d d
P x
4. If sludge wasting from a recirculation line c
VX Qw X r ( Q Qw ) Xe
Rearranging for Qw yields Qw
Qw
VX c Q Xe c ( X r Xe)
(14 . MG)(3500 mgVSS / L) (10d )(5 MGD)(16 mgVSS / L) (10d )(8000 16) mgVSS / L
0051 . MGD
If sludge wasting from an aeration rank c
VX Qw X (Q Qw ) X e
c Qw X c QXe c Qw Xe VX c Qw ( X Xe) VX c QXe Qw
Qw
VX c QXe c ( X Xe)
(14 . Mgal )(3500 mgVSS / L) (10 d )(5 MGal / d )(16mgVSS / L) (10 d )(3500 16) mgVSS / L
or c
VX Qw X (Q Qw ) X e
Assuming Qw << Q gives Q − Qw = Q 21
. MGD 0118
4-Activated Sludge Models_F12
c
VX Qw X (Q) X e
Rearranging for Qw c Qw X c QXe VX c Qw X VX c QXe Qw
Qw
VX c Q Xe c X
(14 . Mgal )(3500 mgVSS / L) (10 d )( 5 MGal / d )(16 mgVSS / L) (10 d )(3500 mg VSS / L)
. MGD 0117
4. Compute the recirculation ratio (R) using a VSS mass balance around the reactor neglecting the suspended solids in the influent (Xo = 0).
Q Aeration tank X = 3500 mg VSS/L X(Q+Qr)
Qr Xr = 8,000 mg VSS/L
Xr Qr = X (Q + Qr) Xr Qr = XQ + XQr Xr 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 g ' V
at steady state, dX/dt = 0
22
4-Activated Sludge Models_F12
Qr Xr + r g’ V = X (Q + Qr) = XQ + Qr Qr (Xr – X) = XQ – r g’ V X
Qr
Q
r g ' V Q
3500 mgVSS
X r X
X rg ' X r X
where r g ' net X
1 c
L
X
350 mgVSS L . d
(8000 3500) mgVSS / L
1 10 d
V Q
14 . Mgal 6 Mgal / d
0.28 d 6.72 hrs
7. 7-1. Check the specific utilization rate,
U
U
(dS / dt ) u X
Q( So S ) V X
(250 7.4) mgBOD5 / L (0.28 d )(3500 mgVSS / L)
So S X
0.25 mgBOD5 utilized mgMLVSS. d
7.2 Check the food-to-microorganism ratio F M
Q So V X
0.76
3500 mgVSS / L 350 mgVSS / L. d
6.
0.28 d
So X
250 mgBOD5 / L (0.28 d )(3500 mgVSS / L) 0.255 mgBOD5 mgMLVSS. d
23
4-Activated Sludge Models_F12
3
7.3 Check the volumetric BOD loading rate (VLR) in lb BOD5/1000 ft ·d
250 mgBOD5 5 Mgal d 8.34 SoQ L VLR 6 V 10 gal 1 ft 3 (1.4 Mgal ) Mgal 7.48 gal
10425 lbBOD5 / d 187166 ft 3
56 lbBOD5applied 1000 ft 3.d
Note
0255 . mgBOD5 250 7.4 0.25 mgBOD5 F E M mgVSS. d mgMLVSS. d 250
U
24