4-Activated Sludge F11

4-Activated Sludge_F11.doc

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Primary Treatment Purpose: To remove settlable organic solids Efficiency: BOD removal - 30% (30 - 40%) SS removal - 60% (50 - 70%) Secondary Treatment Purpose: To remove soluble organics Efficiency: BOD removal - 90% (85 - 95%) SS removal - 90% (85 - 95%) Biological Treatment Basic Reaction of Aerobic System (X) more new microbes microbes Organics + O2 + nutrients --------—> CO2 + H2O Soluble air N, P BOD5 (S) Note: all naturally present in domestic sewage, except O2 Types of Secondary Treatment Systems 1. Suspended Growth Systems (Reactors) e.g. Activated Sludge processes - Conventional - Completely mixed 2. Attached Growth Systems (Reactors) e.g. Trickling Filters, Rotating Biological Contactor (RBC) Submerged Rotating Biological Contactor (SBC)


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Conventional Activated Sludge Process

MLSS

MLVSS

Diffused Aerator

Waste sludge

Log phase Endogenous phase

(Exponential growth)

BOD 5

MLVSS

O2

demand

Waste sludge

Time or distance

Completely Mixed Activated Sludge Process

MLSS

Waste sludege

Endogenous phase

BOD 5

O2 demand

MLVSS

Distance


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Completely Mixed Activated Sludge Process (3rd

DC 387; 4th DC 462)

Secondary Sedimentation Tank

Aeration Tank (Secondary Clarifier)

Q

Q + Qr Q + Qr (Q - Qw)

So V, X, S X S S Xe

X = MLSS

X = MLVSS

Qr + Qw

Qw

Qr Xr S S Xr

Terms and Definitions Hydraulic retention time (HRT), Hydraulic detention time, Aeration period, Liquid

detention time HRT = td = ө = V / Q where V = volume of the aeration tank, m3, MG Q = flow rate, m3/d, MGD MLSS, MLVSS MLSS = Mixed liquor suspended solids = SS in the aeration tank MLVSS = Mixed liquor volatile suspended solids = VSS in the aeration tank. Note: C MLSS and MLVSS are indicative of biomass C MLVSS is more indicative of true biomass than MLSS - MLSS includes inorganic constituents. MLVSS / MLSS = 0.7 - 0.8


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SVI (Sludge Volume Index) a. The SVI is the volume (in milliliters) occupied by 1 g of suspended solids (SS) after 30

min of settling.

b. It is computed by: Sludge volume after settling (mL/L) 1000 mg

SVI = ------------------------------------------------ x ------------- MLSS (mg/L) g

SVI = 50 – 150 mL/g indicates a good settling sludge 106

Xr (in mg/L) ~ ---------- SVI Unit of Xr:

g 103 mg 103 mL mg ------- ----------- ---------- = ------- mL g L L BOD (Organic) Loading BOD Loading = Q So Volumetric BOD Loading, F / V Q So Volumetric BOD Loading = ---------- V F lb BOD applied / day ----- = ------------------------------------------------- V 1000 ft3 of aeration tank capacity - This relationship has no information on microbial population density


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F/M Ratio, Food-to-Microorganism Ratio F Mass of BOD5 applied to the system / day ---- = -------------------------------------------------------------------------------------- M Mass of microbes (MLSS, MLVSS) in the system (Aeration tank) F Q So F Q So --- = ----------------- or ---- = ------------------ M (MLVSS)(V) M (MLSS)(V) Q So So = ---------- = -------- X V θ X Note: V 1 Q θ = ----- or ---- = ----- Q θ V Range of F/M value = 0.05 - 1.0 lb/lb.day 0.3 - 0.7 lb/lb.day - commonly 0.5 lb/lb day - typically Unit of F/M Ratio: lb BOD5 /d lb mg BOD5 /d mg ---------------- = --------- ------------------ = --------- lb MLVSS lb d mg MLVSS mg d Example 5-9 (3rd DC 394) Example (4th DC 468) Compute the F/M ratio for the new activated sludge plant at Gaterville. Given: Q = 0.15 m3/s, V = 970 m3, So = 84 mg/L, X = MLVSS = 2000 mg/L.


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(Solution) F Mass (lb) of BOD5 applied to the system / day ---- = ---------------------------------------------------------------------------------------------- M Mass (lb) of microbes (MLSS, MLVSS) in the system (Aeration tank) Q So (0.15 m3/s) (84 mg/L) (86,400 s/d) = ---------- = -------------------------------------------------- X V (2,000 mg/L)(970 m3) 0.56 mg BOD5/d = --------------------------- mg MLVSS 0.56 lb BOD5/d = --------------------------- lb MLVSS Example: The wastewater flow rate and BOD5 to the secondary treatment system is 1 MGD and 300 mg/L, respectively. The hydraulic retention time of the aeration tank is 6 hrs, and MLSS is 2500 mg/L. Calculate: a) BOD loading, b) volume of the aeration tank, c) volumetric BOD loading, and d) F/M ratio. (Solution) a) BOD loading = So Q = (300 mg/L)(1 MGD) (8.34)

= 2500 lb BOD/d

Note: 1 mg/L = 8.34 lb/MG 1 x 106 gal b) V = Q θ = -------------- (6 hrs) = 250,000 gal = 0.25 MG 24 hrs = (250,000 gal) (1 ft3/ 7.48 gal) = 33,500 ft3 So Q c) Volumetric BOD Loading = ---------- V (300 mg/L)(1 MGD) (8.34) 75 lb BOD5 /d = ------------------------------------- = ---------------------- 33,500 ft3 1000 ft3


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d) F So Q (300 mg/L) (1 MGD)(8.34) ---- = ----------- = ----------------------------------------- M X V (2500 mg/L)(0.25 MG)(8.34) 2500 lb BOD/d 0.48 lb = ----------------------- = ----------- 5200 lb MLSS lb d Mean cell residence time (MCRT), Solid retention time (SRT), Sludge age (θc): a. Definition - Average length of time in which a solid particles (i.e., biomass) is retained in a

system (an aeration tank). biomass (MLVSS or MLSS) in the aeration tank θc = ---------------------------------------------------------------------------------- = day sum of the biomass in the waste sludge and effluent / day

mass of solids in aeration tank θc = ----------------------------------------- mass rate of sludge wasting

(MLSS or MLVSS) (V) θc = ------------------------------------ waste sludge/day

X V θc = ------------------------------------------ Qw Xr + (Q − Qw) Xe Rate of Rate of Intentionally + Accidentally wasted wasted

c net

w we

g

X X

Q X Q Q dXXV V dt

θ = =− +



Q


So V, X, S X S S Xe

X = MLSS

X = MLVSS

Qr + Qw

Qw

Qr Xr S S Xr


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1c

net

θµ

=

At steady state, the rate of loss = the rate of growth Typical values, θc = 3 - 60 days (commonly 10 - 15 days) If θc is high If θc is low -------------------------------- ------------------------------- Low F/M High F/M Low sludge wasting High sludge wasting ------------------------------- ------------------------------- Process Efficiency, E (%) (So – S) 100 E = ------------------ So Example 5-27 (3rd DC 452). Two activated sludge aeration tanks at Turkey Run, Indiana, are operated (in parallel). Each tank has the following dimensions: 7.0 m wide by 30.0 m long by 4.3 m effective liquid depth. The plant operating parameters are as follows: Flow = 0.0796 m3/s; Soluble BOD5 after primary settling = 130 mg/L; MLSS = (1.40) (MLVSS); MLVSS = 1,500 mg/L; Settled sludge volume after 30 min = 230 mL/L; Aeration tank liquid temperature = 15°C. Determine the following: (1) Aeration period (2) F/M ratio (3) SVI, and (4) Solids concentration in return sludge, Xr

(solution)

Q = 0.0796 m3/s So = BOD5 = 130 mg/L MLVSS = 1500 mg/L MLSS = 1.4 (MLVSS) = 1.4 (1500 mg/L) = 2100 mg/L Settled sludge volume after 30 min = 230 mL/L T = 15°C V each = (7.0 m)(30.0 m)(4.3 m) = 903 m3 V Total = 2(903 m3) = 1806 m3


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(1) Aeration Period V 1806 m3 θ = --- = ---------------------------------- = 6.3 hrs Q (0.0796 m3/s)(3600 s/hr) (2) F Q So (0.0796 m3/s)(130 mg/L)(86400 s/d) --- = -------- = -------------------------------------------------- M X V (1500 mg/L)(1806 m3) 0.33 mg 0.33 mg/d BOD = -------------- = ------------------------ mg.d mg biomass (3) (Settled sludge volume after 30 min)(1000) SVI = ------------------------------------------------------------ MLSS 230 mL/L 1000 mg = ---------------- -------------- = 109.5 mL/g 2100 mg/L g Note: SVI = 50-150 mL/g for good settling (4) 1 1 1000 mg/g Xr ~ ------- = -------------- • ----------------- = 9132.4 mg/L SVI 109.5 mL/g L/1000 mL Note: 106 Xr ~ ------- (in mg/L) SVI


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Bacterial Growth Kinetics

µm

µ

µm/2

Ks

S (3rd: DC 350; 4th DC 459 Fig. 6-18) Monod growth rate constant as a function of limiting food concentration The Monod Model:

m

s

S

K S

µµ =

+

where µ = specific growth rate, d-1

µ m = maximum specific growth rate, d-1 S = rate limiting substrate (food) concentration, mg/L (e.g., soluble BOD) Ks = half-saturation coefficient, mg/L

= Monod Constant = half-velocity constant

Bacterial growth rate ,g

g

dXr

dt

mg

g

SdXr X X

dt Ks S

µµ

= = =

+

where X = bacteria concentration (in VSS), mg/L

m

s

S

K S

µµ =

+


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Substrate Utilization Rate, ,su

u

dSr

dt

g u

dX dS

dt dt

∞ −

g u

dX dSY

dt dt

= −

where Y = bacterial cell yield = yield coefficient, mg VSS/mg BOD

Bacterial Death Rate, ,d

d

dXr

dt

d d

d

dXr k X

dt

= =

where kd = death rate constant = endogenous decay coefficient, d -1

Endogenous Phase - near starvation phase Net Growth Rate = Growth Rate - Death Rate

'

net

g

g g d

dX dX dXr

dt dt dt

= = −

where , d

g d

dX dXX k X

dt dtµ

= =

' ( )

net

g d net

g

dXr k X X

dtµ µ

= = − =

In which

mnet d d

s s

S Y k Sk k

K S K S

µµ = − = −

+ +

where k = µm / Y


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= maximum substrate utilization rate constant

'

net

g net d

g s

dX Y k Sr X k X

dt K Sµ

= = = −

+

Substrate Utilization Rate, rsu

1 1 msu

u g s

s

SdS dXr X

dt Y dt Y K S

kSX

K S

µ = = =

+

=+

Specific Substrate Utilization Rate, U

( )u

dS

So S So So SdtU

X X X So

QSo So S FE

VX So M

θ θ

− − = = =

− = =

where E = (So - S)/ So = substrate removal efficiency θ = V/Q = hydraulic retention time = aeration time, hr F/M = Food to microorganisms ratio Secondary Sedimentation Tank


Q


So V, X, S X S S Xe

X = MLSS

X = MLVSS

Qr + Qw

Qw

Qr Xr S S Xr


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Mass balance on X around the secondary treatment Accum = Inputs - Outputs ± Rxns

VdX

dtQXo Q Qw Xe Qw Xr rg V

where rgYkS

Ks Skd X

= − − − +

=+

−

( ) '

'

rg’ = net growth rate, mg/L. d Y = yield coefficient, mg X / mg S Ks = half-saturation constant, mg/L kd = endogenous decay rate constant, d-1 k = maximum substrate utilization rate, d-1 k = µm /Y µm = Y k µm = maximum specific growth rate at steady state, dX/dt = 0,

(1 ) (1 )

( ) 1 ( ) 1m

K k K ks d c s d cSYk k k

c d c d

θ θ

θ θ µ

+ += =

− − − − (6-21) 4

th DC 467

where S = soluble BOD in effluent Y = yield coefficient, mg X / mg S Ks = half-saturation constant, mg/L kd = endogenous decay rate constant, d-1 k = maximum substrate utilization rate, d-1 k = µm /Y or µm= Yk Mass balance on S around the aeration tank Accum = Inputs - Outputs ± Rxns

VdS

dtQS Q S Q Q S r V

where rkSX

K S

o r r su

sus

= + − + −

=+

( )

rsu = substrate utilization rate, mg/L-d At steady state, dS/dt = 0,


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XY S S

k

c o

d c

=−

+

θ

θ θ

( )

( )1 (6-22), 4th DC 467

Substituting θ = V/Q, solve for V

V c Y Q So S

X kd c

=−

+

θ

θ

( )

( )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, kd = 0.06 d-1 What is the aeration tank volume? (Solution)

( ) ( )

( ) ( )

VY Q S S

X k

dmgVSS

mgBOD

m

d

mg

LBOD

mg Ld

d

c o

d c

=−

+

=

−

+

θ

θ

( )

( )

.. ,

/.

.

1

6 00 6 10 000

120 7

2000 10 06

6 0

3

V = 1495 m3 Excess Biological Solids, Px

( ) ( )x obs o x obs o

P Y S S or P Y S S Q= − = −

where

1

obs

c d

YY

kθ=

+

Example: Determine the excess sludge (in mg/L VSS) for the given conditions. Q = 10,000 m3/d, So = 120 mg/L, S = 7 mg/L kd = 0.06/d, θc = 10 days, Y = 0.6 mg VSS/mg BOD


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(Solution)

YY

k

mgVSS

mg BOD

dd

mgVSS

mgBODobs

c d

=+

=

+

=1

0 6

1 100 06

0 375

θ

.

.

.

PmgVSS

mgBOD

mgBOD

LmgVSS Lx =

−

=

0 375 120 742 4

. ( ). /

3 3

3 6

42.4 10,000 10( / )

10

420 /

x

mgVSS m L kgP in kg d

L d m mg

kg VSS d

=

=

Example 5-7 (3rd DC 391); Example 6-6 (4th DC 468) The town of Gatesville has been directed to upgrade its primary WWTP to a secondary plant that can meet an effluent standard of 30 mg/L BOD5 and 30 mg/L suspended solids (SS). They have selected a completely mixed activated sludge system. Assuming that the BOD5 of the SS may be estimated as equal to 63 percent of the SS concentration, estimate the required volume of the aeration tank. The following data are available from the existing primary plant. Existing plant effluent characteristics: Flow, Q = 0.150 m3/s, BOD5 = 85.0 mg/L Assume the following values for the growth constants: Ks = 100 mg/L BOD5, µm = 2.5 d-1, kd = 0.05 d-1, Y = 0.50 mg VSS / mg BOD5. X = 2000 mg/L (MLVSS) - additional assumption Approach Use θ = V/Q ..... (5-25) Determine θc using

SK k

Yk k

s d c

c d

=+

− −

( )

( )

1

1

θ

θ

Determine θ using

XY S S

k

c o

d c

=−

+

θ

θ θ

( )

( )1


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Q


So V, X, S X S S Xe

X = MLSS

X = MLVSS

Qr + Qw

Qw

Qr Xr S S Xr

Note that So and S are soluble BOD in influent and effluent, respectively. (Solution) Note that S and So in the above eqns are soluble BOD5. BOD5 allowed = soluble BOD5 (S) + suspended solid BOD5 30 mg/L = S + 0.63 (30 mg/L SS) 1) Calculate soluble BOD5 in effluent (S) S = BOD5 allowed - suspended solid BOD5 = 30 mg/L - (0.63)(30 mg/L) = 11.1 mg/L 2) Obtain θc

SK k

Yk k

mg L d

d dmg Ls d c

c d

c

c

=+

− −=

+

− −=

−

− −

( )

( )

( / )( . )

( . . ). /

1

1

100 1 0 05

2 5 0 05 1111

1

1 1

θ

θ

θ

θ

solve for θc: θc = 5.0 d Note: since k = µm /Y, Y k = µm = 2.5 d-1


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3) Obtain θ, with X = 2000 mg/L VSS

5

1

( )

(1 )

(5.0 )(0.5 / )(84 / 11.1 / )

(1 0.05 (5.0 )

2000 /

c o

d c

Y S SX

k

d mgVSS mg BOD mg L mg L

d d

mg L

θ

θ θ

θ −

−=

+

−=

+

=

Solve for θ: θ = 0.073 d = 1.8 hrs 4) V = θ Q = (1.8 hrs)(0.15 m3/s)(3600 sec/hr)

= 972 m3 Note

( )XY S S

kY S S

Pc o

d c

cobs o

c x=−

+= − =

θ

θ θ

θ

θ

θ

θ

( )

( )1

where

P Y S S

YY

k

x obs o

obsc d

= −

=+

( )

1 θ

θθ

cx x

X

P

XV

P Q

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 - Qw) Xe

HW-3 is due 1 week from today. http://www.pocatello.us/wpc/wpc_education.htm

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4-Activated Sludge F11