4-Activated Sludge Models_F12

8/18/2019 4-Activated Sludge Models_F12

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4-Activated Sludge Models_F12

1

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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2

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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3

(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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4

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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5

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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6

Mass balance on S around the aeration tank

- A mass balance for substrate entering and leaving the aeration tank


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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9

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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10

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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11

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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12

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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13

/ /

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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14

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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15

Recirculation Ratio, R

Determining the recirculation ratio, R, from the mass balance on VSS around the aeration tank.


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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16

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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17

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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22

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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23

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.

.

. .

.

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4-Activated Sludge Models_F12

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