activated sludge for class presentation

8/18/2019 Activated Sludge for Class Presentation

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Activated Sludge Design(Complete Mix Reactor)

IPRC KIGALIY3 WAS CIVIL ENGWATER AND SANITATION Wastewater treatment

technology

mailto:[email protected]



Activated Sludge System



Secondary Treatment Using Activated Sludge Process



Wastewater Treatment –

Processes• Secondary Treatment With Activated Sludge

– Effluent leaves sedimentation tank and is pumped to anaeration tank

– Effluent is mixed with air and sludge loaded with bacteria(“activated sludge” )

– Sludge contacts with raw sewage, and bacteria in the sludgethen decompose the raw sewage, lowering ODW and BOD

– Activated sludge-sewage mixture is called mixed li quor –

Mixed liquor leaves aeration tank and flows to anothersedimentation tank where suspended solids settle out for reuseas activated sludge

– Effluent is chlorinated – Activated sludge in sedimentation tank is reused



Activated sludge process • To remove suspended solids, nitrogen, and

phosphate• Food to microorganism ratio (F:M ratio): 0.25

kg BOD 5 per kg MLSS (mixed liquorsuspended solids) per day at 10 oC or 0.4 kg

BOD 5 per kg MLSS per day at 20o

C• Residence time: 2 days for high F:M ratio, 10days or more for low F:M ratio

• Optimum nutrient ratio: BOD 5:N:P =>100:5:1• 90 % removal of BOD 5 and SS• ~20 % removal of phosphate • >90 % removal of viruses and protozoa and 45

- 95 % removal of bacteria



Activated Sludge Process • The aeration tank contains a suspension of the wastewater and

microorganisms, the mixed liquor. The liquor is mixed by aeration devices(supplying also oxygen)

• A portion of the biological sludge separated from the secondary effluent bysedimentation is recycled to the aeration tank

• Types of AS Systems: Conventional, Complete-Mix, Sequencing BatchReactor, Extended Aeration, Deep Tank, Deep Shaft



Advantages/Disadvantages



p ys ca components o act vatesludge process

•

aeration tank – oxygen is introduced into the system



• aeration source – ensure that adequate oxygen is fed into the tank – provided pure oxygen or compressed air



• secondary clarifiers – activated-sludge solids separate from the surrounding

wastewater



• Activated sludgeoutflow line

– Pump activatedsludge back to theaeration tank

• Effluent outflow line – discharged effluent

into bay or tertiarytreatment plant



Schematic of activated sludge unit



Kinetics of Microbial Growth• Biochemical reaction• • Biomass concentration.•

The concentration of biomass, X (mg/L),increases as a function of time due toconversion of food to biomass:

• Where m is the specific growth rate constant (d -1).

This represents the mass of cellsproduced/mass of cells per unit of time.



Effect of substrate concentration ongrowth rate constant



Monod Kinetics• Growth rate

• Growth rate constant, m , is a function of thesubstrate concentration, S.

•

Two constants are used to describe the growthrate

m m (mg/L) is the maximum growth rate constant (the rateat which the susbtrate concentration is not limiting)

– Ks is the half-saturation constant (mg/L) (i.e.,concentration of S when m = m m/2

•



• Biomass production

•

Where k d represents the endogenousdecay rate (d -1) (i.e., microorganism deathrate).

– Substituting the growth rate constant:



• Substrate utilization

Where Y is the yield factor (mg of

biomass produced/mg of food consumed)• Y range:

– Aerobic: 0.4 - 0.8 mg/mg



• Food to microorganism ratio (F/M)• Represents the daily mass of food supplied to the

microbial biomass, X, in the mixed liquorsuspended solids, MLSS

• Units are Kg BOD 5/Kg MLSS/day



• Since the hydraulic retention time, q =V/Qo, then



Typical range of F/M ratio in activated sludge units

Treatment Proc ess F/MKg BOD 5 /K g MLSS/day

Extended aerat ion 0.03 - 0.8

Convent ional 0.8 - 2.0

High ra te > 2.0



Design parameters for activated sludge processesProcess q c d ) q h ) F/M Qr/Q X (m g/L)

Conven t iona l 5-15 4-8 0.2-0.4 0.25-5 1,500-3,000

Complete-mix 5-15 3-5 0.2-0.6 0.25-1 3,000-6,000

Step-aerat ion 5-15 3-5 0.2-0.4 0.25-0.75 2,000-3,500

Modi f ied- aerat ion

0.2-0.5 1.5-3 1.5-5.0 0.05-0.15 200 – 500

Contact -s tabi l iza t ion

5-15 0.5-13-6

0.2-0.6 0.25-1 1,000-3,0004,000-10,000

Extended- aerat ion

20-30 18-36 0.05-0.15 0.75-1.5 3,000-6,000

High-rateaerat ion

5-10 0.5-2 0.4-1.5 1-5 4,000-10,000

Pure-oxygen 8-20 1-3 0.25-1.0 0.25-0.5 6,000-8,000



Operational characteristics of activated sludge processes

Process F lo w m o d e l Aerat ion sys tem B OD 5 rem ov aleff icienc y (%)

Conven t iona l Plug-flow Diffused air,mechanical aerators

85-95

Complete-mix Complete-mix Diffused air,mechanical aerators

85-95

Step-aerat ion Plug-flow Diffused air 85-95

Modified-aerat ion Plug-flow Diffused air 60-75

Contact -s tabi l iza t ion

Plug-flow Diffused air,mechanical aerators

80-90

Extended-aerat ion Complete-mix Diffused air,mechanical aerators

75-95

High -rate aerat ion Complete-mix Diffused air,mechanical aerators

75-90

Pure-oxygen Complete-mix Mechanical aerators 85-95



Activated Sludge Design Equations

• Influent biomass + biomass production = effluentbiomass + sludge wasted

•

• Substitute biomass production equation•

• Assume that influent and effluent biomassconcentrations are negligible and solve

•

Mass balance of biomass production



Mass balance of food substrate • Influent substrate + substrate consumed = effluent

susbtrate + sludge wasted substrate•

• Substitute substrate removal equation•

• Assume that no biochemical action takes place inclarifier. Therefore the substrate concentration in the

aeration basin is equal to the substrate concentrations inthe effluent and the waste activated sludge. Solve :

•



Overall equations – Combine the mass balance equations for food and biomass :

• • The cell residence time is:•

• and the hydraulic retention time is, q = V/Q o » Substitute and rearrange:

• • •

Compute the F/M ratio



EXAMPLEDesign of Completely Mixed Activated Sludge System

Design a completely mixed activated sludge and secondary settling facility totreat 5.71 mgd (0.25 m 3/s) of settled wastewater with 250 mg/L BOD 5. Theeffluent is to have 20 mg/L of BOD 5 or less. Assume temperature at 20 o C.• (i) Influent volatile suspended solids to reactor are negligible• (ii) ratio of mixed liquor VSS (MLVSS) and mixed liquid suspended solids

(MLSS) = 0.8• (iii) Return sludge concentration = 10,000 mg/L SS• (iv) MLVSS = 3,500 mg/L in aeration tank• (v) Design mean cell residence time qc = 10 d• (vi) Effluent contains 22 mg/L of biological solids, 65% is biodegradable•

(vii) BOD 5 = 0.68 BOD L • (viii) Wastewater contains adequate nitrogen, phosphorus and trace

nutrients for biological growth• (ix) Peak hourly flow rate is 2.5 times the average flow rate



W t t Di i f ti



Wastewater Disinfection• To inactivate pathogens in wastewater• Several choices

– Free chlorine and combined chlorine – UV – Ozone – Chlorine dioxide

• Remove disease-causing organisms fromwastewater

• 3 different disinfection process – Chlorination – UV light radiation – Ozonation



Chlorination & UV • Most common•

Advantages: low cost & effective• Disadvantages: chlorine residue could be

harmful to environment

UV• Damage the genetic structure of bacteria,

viruses and other pathogens.•

Advantages: no chemicals are used• water taste more natural• Disadvantages: high maintenance of the UV-

lamp



Ozonation

• Oxidized most pathogenic microorganisms• Advantages: safer than chlorination

fewer disinfection by-product• Disadvantage: high cost



END

activated sludge for class presentation

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