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WASTE WATER TREATMENT PROCESS DESIGN CONSIDERATION ON ACTIVATED SLUDGE PROCESS Drs. Subagiyo, MSi 2007

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Page 1: Process Design Consideration

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WASTE WATER TREATMENT

PROCESS DESIGN CONSIDERATIONON ACTIVATED SLUDGE PROCESS

Drs. Subagiyo, MSi

2007

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PROCESS DESIGN CONSIDERATION

Selection of the reactor type

Loading criteria Sludge production

Oxygen requirment and transfer

Control filamentous organisms

Effluent characteristics

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LOADING CRITERIA

The 2 most commonly used parameters:

The food to microorganism ratio (F/M)

The mean cell residence time qc

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F/M

F/M = So / qX

F/M = food to microorganism ratioSo = influent BOD or COD concentration mg/L (g/m3)q = hydrolic detention time of the tank aeration = V/Q, dX = concentration of volatile suspended solid in the aeration tank,

mg/L (g/m3)

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F/M RATIO

Typical value for F/M vary from 0.05 to 1.0

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SPECIFIC UTILIZATION RATE

rsu So - S

U = - =X qX

Q So - S

=

Vr qX

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THE MEAN CELL-RESIDENCE TIME

Definition based on aeration tank volume

VrX

qc =QwXw + QeXe

Qc : mean cell-recident time based on the aeration tank volume, d

Vr : aeration tank volume, Mgal (m3)

X : concentration of VSS in the aeration tank, mg/L (g/m3)

Qw : waste sludge flowrate, Mgal/d (m3 /d)

Qe : treated effluent flowrate, Mgal/d (m3 /d)

Xe : concentration of VSS in the treated effluent, mg/L (g/m3)

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THE MEAN CELL-RESIDENCE TIME(U.S. customary units)

Definition based on total system volume

Xt 1

qct =(QwXw + QeXe) 8.34 lb/Mgal. (Mg/L)

qct : mean cell-residence time based on the total system

Xt : total mass of VSS in the system, including the solids in

the aeration tank, in the setling tank, and in the sludge-

return facilities

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THE MEAN CELL-RESIDENCE TIME(SI units)

Xt 

qct =

QwXw + QeXe

qct : mean cell-residence time based on the total system

Xt : total mass of VSS in the system, including the solids in

the aeration tank, in the setling tank, and in the sludge-

return facilities

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SLUDGE PRODUCTION

It will affect the design of the sludge-

handling and disposal facilitiesnecessary for the excess (waste)sludge

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SLUDGE PRODUCTIONU.S customary units

Px = Y obs (So – S) x 8.34 lb/Mgal. (mg/L)

Px : net waste activated sludge produced each day,measured in term of VSS, lb/d (kg/d)

Yobs : observed yield, lb/lb (g/g)

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SLUDGE PRODUCTIONSI units

Px = Y obs (So  – S) x (103 g/kg)-1

Px : net waste activated sludge produced each day,measured in term of VSS, lb/d (kg/d)

Yobs : observed yield, lb/lb (g/g)

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CONTROL OF FILAMENTOUS ORGANISMS

The most common operational problem in the

activated  – sludge process

Proliferation of filamentous organisms in mixed

liquor result in porrly settling sludge

Causes the biological flocs in reactor to be

bulky and loosly packed

Filamentous bacteria, actinomycetes and

fungi

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CONTROL OF FILAMENTOUS ORGANISMS

The addition of chlorine or hydrogen peroxyde tothe return waste-activated sludge

Alteration of the DO concentration in the aerationtank 

Increase F/M ratio

The addition of major nutrient (i.e. N & P) The addition of trace nutrient & growth factors

The use of selector

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SELECTOR

Control of the growth of filamentous organisms inthe complete-mix process have been achieved bymixing the return sludge with the incoming

wastewater in a small anoxic contact tank asselector

A separate compartment as the initial contact zoneof a biological reactor where the prymary effluent

and return activatedsludge are combined.

Contact time in the selector is relatively short, 10  –  30 min.

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EFFLUENT CHARACTERISTICS

Organic content is a major parameters

of effluent quality. Soluble biodegradable organics

Suspended organic material

Nonbiodegradable organics

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Soluble biodegradable organics

Organics that escape biological treatment Organic formed as intermediate productsin

biological degradation of the waste

Cellular componen (result of cell death orlysis)

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SUSPENDED ORGANIC MATERIAL

Biological solids produced during treatment

that escaoe separation in the final settlingtank 

Colloidal organic solids in the plant influent

that escape treatment and separation

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NONDEGRADABLE ORGANICS

Those originally present

By-products of biological degradation

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PROCESS CONTROL

Control of the activated process is important

to maintain high level of treatment

performance under a wide range ofoperating condition.

The principal factors used in process control

Maintaining DO levels in aeration tanks Regulating the amount of return activated sludge

Controlling the waste activated sludge

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PROCESS CONTROL

The most commonly used parameters forcontrolling the act.process are : F/M ratio &mean cell-residence time, also MLSS.

RAS : important in maintaining the MLVSS

WAS: important in the controlling qc

OUR (oxygen uptake rate) : monitoring &controlling the activated-sludge process

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RETURN ACTIVATED-SLUDGE CONTROL

The purpose : to maintain a sufficient

concentration of activated sludge in the

aeration tank 

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TECHNIQUES TO CALCULATE THE DESIRABLERETURN-SLUDGE FLOWRATE

Settleability

Sludge-blanket level control

Secondary clarifier mass balance

Aeration tank mass balance

Sludge quality

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Settleability

The return sludge-puming rate is set so that

the flowrate is approximately equal to the

percentage ratio of volume accupied bythe settleable solids from the aeration tank effluent to the volume of the clarified liquid

(supernatan) after settling for 30 min in a

1000 mL graduated cylinder

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example

If the settleable solids occupied a volume of

275 mL after 30 min of settling,

the percentage volume would be equal to 38% ( (275mL/725 mL) x 100 ).

If the plant flow were 46 Mgal/d (2 m

3

/sec), thereturn-sludge rate should be

0,38 x 46 Mgal/d = 17.5 Mgal/d (0.76 m3/sec)

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SLUDGE VOLUME INDEX (SVI)

Volume in mL occupied by one gram of activated-sludge mixed liquor solids, dry weight, after settlingfor 30 min in a 1000 mL graduated cylinder.

In practice, it is taken to be percentage volumeoccupied by the sludge in a mixed-liquor sampleafter 30 min of settling, Ov, divided by the SSconcentration of the mixed liquor expressed asapercentage, Pw.

The percentage of return sludge is

100/ ((100/PwSVI)-1)

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example

To maintain a mixed-liquor ss conc. of

0.3 % (3000 mg/L), the percentage of sludgethat must be returned when the SVI is 100 =100 /((100/0.3 x 100)  – 1)

= 43 %

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Secondary clarifier mass balance

Accumulation = inflow  – outflow0= X(Q+Qr)(8.34) - XrQr(8.34) + XrQ’w(8.34) 

X : mixed-liquor SS, mg/L

Q : secondary influent flow, Mgal/d

Qr : return sludge flow, Mgal/d

Xr : return activated-sludge SS. Mg/L

Q’w : waste sludge flow, Mgal/d

8.34 : conversion factor (lb/Mgal. Mg/L)

Qr = (XQ – XrQ’w) / (Xr -X)

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Aeration tank mass balance

Accumulation = inflow  – outflow

0 = XrQr (8,34) –

X(Q+Qr) (8.34)

X

Qr = Q

Xr-X