4-Activated Sludge F11

4-Activated Sludge_F11.doc

Primary Treatment Purpose: Efficiency:

To remove settlable organic solids BOD removal - 30% (30 - 40%) SS removal - 60% (50 - 70%)

Secondary Treatment Purpose: Efficiency:

To remove soluble organics BOD removal - 90% (85 - 95%) SS removal - 90% (85 - 95%)

Biological Treatment Basic Reaction of Aerobic System (X) more new microbes microbes Organics + O 2 + nutrients --------—> Soluble BOD5 (S)

air

CO 2 + H2O

N, P

Note: all naturally present in domestic sewage, except O 2 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 (Exponential growth)

Endogenous phase

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

2


Completely Mixed Activated Sludge Process (3rd DC 387; 4th DC 462) Secondary Sedimentation Tank (Secondary Clarifier)

Aeration Tank

Q Q + Qr So

Q + Qr X S

V, X, S

(Q - Qw) S Xe

X = MLSS X = MLVSS Qr + Qw Qr

Xr

Qw S Xr

S

Terms and Definitions Hydraulic retention time (HRT), Hydraulic detention time, Aeration period, Liquid detention time HRT = td =  = V / Q where V = volume of the aeration tank, m 3, MG Q = flow rate, m 3 /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

C

MLSS and MLVSS are indicative of biomass 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 --- = ----------------- or M (MLVSS)(V)

F Q So ---- = -----------------M (MLSS)(V)

Q So So = ---------- = -------θ X XV Note: V θ = ----Q

or

1 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 ---------------- = --------lb MLVSS lb d

mg BOD 5 /d mg ------------------ = --------mg MLVSS mg d

Example 5-9 (3 rd DC 394) Example (4 th DC 468) Compute the F/M ratio for the new activated sludge plant at Gaterville. Given: Q = 0.15 m3 /s, V = 970 m 3, So = 84 mg/L, X = MLVSS = 2000 mg/L.

5


(Solution) F Mass (lb) of BOD 5 applied to the system / day ---- = ---------------------------------------------------------------------------------------------M Mass (lb) of microbes (MLSS, MLVSS) in the system (Aeration tank) Q So (0.15 m 3 /s) (84 mg/L) (86,400 s/d) = ---------- = -------------------------------------------------XV (2,000 mg/L)(970 m 3) 0.56 mg BOD5 /d = --------------------------mg MLVSS 0.56 lb BOD 5 /d = --------------------------lb MLVSS

Example: The wastewater flow rate and BOD 5 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 ft 3 / 7.48 gal) = 33,500 ft 3 So Q c) Volumetric BOD Loading = ---------V (300 mg/L)(1 MGD) (8.34) 75 lb BOD 5 /d = ------------------------------------- = ---------------------33,500 ft 3 1000 ft 3

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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

Aeration Tank

Q Q + Qr So

Qr + Qw

XV θc = -----------------------------------------Qw Xr + (Q − Qw) Xe

θ c

=

Qw X V

+

X  Q − Qw 

 

V

Rate of Accidentally wasted

 X e 

=

(Q - Qw) S Xe

X = MLSS X = MLVSS

Qr

+

Q + Qr X S

V, X, S

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

Rate of Intentionally wasted

Secondary Sedimentation Tank (Secondary Clarifier)

X net

 dX   dt   g

7

Xr

S

Qw S Xr


θ c

=

1 µ net

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 -------------------------------Low F/M Low sludge wasting -------------------------------

If θc is low ------------------------------High F/M High sludge wasting -------------------------------

Process Efficiency, E (%) (So – S) 100 E = -----------------So

Example 5-27 (3 rd 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 m 3 /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 m 3 /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 m 3 V Total = 2(903 m 3) = 1806 m 3 8


(1) Aeration Period V 1806 m 3 θ = --- = ---------------------------------- = 6.3 hrs Q (0.0796 m 3 /s)(3600 s/hr)

(2) F Q So (0.0796 m 3 /s)(130 mg/L)(86400 s/d) --- = -------- = -------------------------------------------------M XV (1500 mg/L)(1806 m 3) 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 ~ ------SVI

(in mg/L)

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Bacterial Growth Kinetics

µm µ µm /2

Ks S

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

The Monod Model:

µ =

µ m S Ks

+

S

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

rg

=

 dX   dt  = µ X  g

=

 dX    dt  g

r g , 

 µ m S   Ks + S  X  

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

µ =

m

Ks

S

+

S

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 dS  Substrate Utilization Rate, r su ,    dt u  dX   dt   g

∞

 dX   dt   g

= −

−

 dS   dt   u

 dS    dt  u

Y 

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

 dX  Bacterial Death Rate, r d ,    dt d rd

=

 dX   dt  = kd X  d

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

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

' g =

r

 dX   dt   g

=

 dX   dX   dt  −  dt   g   d

 dX   dt  = µ X  g

where

 dX   dt  = kd X  d

,

net

' g =

r

 dX   dt   g

=

( µ − kd ) X

−

kd

=

µ net X

k d

In which µ net

=

µ m S Ks

+

S

=

Y k S Ks

+

−

S

where k = µm / Y 11


= maximum substrate utilization rate constant net

' g =

r

 dX   dt   g

=

µ net X

=

YkS  − kd  X   Ks + S 

Substrate Utilization Rate, r su rsu

=

=

 dS   dt   u

=

1  dX    Y  dt  g

=

1  µ m S 



Y  Ks

+

 X

S 

kSX Ks

+

S

Specific Substrate Utilization Rate, U

 dS   dt   u U = X

=

=

So − S

θ X

=

So( So − S )

θ X So

 QSo  So − S   F   VX  So  =  M  E     

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

Aeration Tank


Q Q + Qr So

Q + Qr X S

V, X, S

(Q - Qw) S Xe


Xr

Qw S Xr

S

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

dX dt

=

where

QX o

−

(Q − Qw ) X e

−

Qw X r + r g ' V

 YkS  r g ' =  − k  X d   K s + S  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, K (1 + k θ ) K (1 + k θ ) s d c s d c S = = 1 θ (Yk − k ) − 1 θ ( µ m − k ) − c d c d

th

(6-21) 4 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 V

dS dt

=

where

QS o r su

+

=

Qr S

−

(Q + Qr )S

−

r su V

kSX Ks

+

S

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

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θ Y ( S o − S ) X = c θ (1 + k d θ c )

(6-22), 4th DC 467

Substituting θ = V/Q, solve for V θ Y Q ( S o − S ) V = c X (1 + k d θ c )

Example: Q = 10,000 m 3 /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) θ Y Q ( So − S ) V = c X (1 + k d θ c )

 0.6 mgVSS   10,000m3  mg   (120 − 7) BOD (6.0 d )      L  d  mgBOD    =



  0.06   (6.0d )  d  

(2000mg / L)1 +  

V = 1495 m 3

Excess Biological Solids, P x P x

=

Yobs (So

−

S)

or

Px

=

Yobs ( So

−

S)Q

where Y obs

=

Y

1 + θ c k d

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 kd = 0.06/d, θc = 10 days, Y = 0.6 mg VSS/mg BOD

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(Solution) 0.6 mgVSS

Y obs

P x

=

=

Y 1 + θ c k d

mg BOD

=

 0.06   d 

1 + 10d 

=

mgVSS 0375 . mgBOD

 0 .375 mgVSS   (120 − 7)mgBOD      L  mgBOD  

=

42.4 mgVSS / L

3 3  42.4 mgVSS   10,000 m  10 L  kg  P x (in kg / d ) =   3   6   L d    m   10 mg 

=

420 kg VSS / d

Example 5-7 (3 rd DC 391); Example 6-6 (4 th 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 BOD 5 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 m 3 /s, BOD5 = 85.0 mg/L Assume the following values for the growth constants: Ks = 100 mg/L BOD 5, 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

S =

K s (1 + k d θ c )

θ c (Yk

−

k d ) − 1

Determine θ using

θ Y ( S o − S ) X = c θ (1 + k d θ c ) 15



Aeration Tank

Q Q + Qr So

Q + Qr X S

V, X, S

(Q - Qw) S Xe


Xr

Qw S Xr

S

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 BOD 5. BOD5 allowed = soluble BOD 5 (S) + suspended solid BOD 5 30 mg/L = S + 0.63 (30 mg/L SS) 1) Calculate soluble BOD 5 in effluent (S) S = BOD5 allowed - suspended solid BOD 5 = 30 mg/L - (0.63)(30 mg/L) = 11.1 mg/L 2) Obtain θc S =

K s (1 + k d θ c )

θ c (Yk − k d ) − 1

=

−1 (100 mg / L )(1 + 0.05 d θ c ) θ c (2.5 d −1 − 0.05 d −1 ) − 1

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

16

=

111 . mg / L


3) Obtain θ, with X = 2000 mg/L VSS θ c Y ( So

X =

−

S )

θ (1 + k dθ c )

(5.0 d )(0.5 mg VSS / mg BOD5 )(84 mg / L −11.1 mg / L)

=

1

−

θ (1 + 0.05 d (5.0 d )

2000 mg / L

=

Solve for θ: θ = 0.073 d = 1.8 hrs

4) V = θ Q = (1.8 hrs)(0.15 m 3 /s)(3600 sec/hr) = 972 m3 Note

θ Y ( So − S ) X = c θ (1 + k d θ c )

=

θ c θ

Yobs ( So

−

θ P S ) = c x θ

where P x

=

Y obs

θ c =

=

Yobs ( S o =

−

S)

Y 1 + θ c k d

X θ P x

=

XV Px Q

biomass in Aeration Tank biomass synthesized / time

XV

=

biomass in Aeration Tank biomass wasted / time

XV

θc = ---------------------------- = ---------

Qw X + (Q - Qw) Xe

Px Q

Px Q = Qw X + (Q - Q w) 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

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