chapter 8 - aerobic process

8/2/2019 Chapter 8 - Aerobic Process

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Lecture 8:

AEROBIC (BIOLOGICAL) PROCESS


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INTRODUCTION

Principal application of biological processes

removal of biodegradable dissolved and colloidalorganic matter

nitrification

denitrification phosphorus removal

waste stabilisation

The cheapest treatment process

Classified into two (based on growth of micoorganisms) Suspended growth activated sludge, oxidation ditch,aerated lagoon, sequencing batch reactor (SBR)

Fixed film growth - trickling filter, rotating biologicalcontactor


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MICROBIAL PROCESS DESCRIPTION

Oxidation and synthesis

(bacteria)

COHNS + O2 + nutrientsp p p

CO2 + NH3 + C5H7NO2 + otherend products

(Organic) (cells)

Endogenous respirationC5H7NO2 + 5O2 p p p 5CO2 + 2H2O + NH3 + energy

(bacteria)


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FACTORS AFFECTING AEROBIC

PROCESS

Dissolved oxygen (DO) > 0.5 mg/L (2 mg/L)

DO > 3.0 mg/L for nitrification process

Temperature affects growth rate and oxygen transfer pH between 4.0 to 9.5; optimum at 6.5 7.5

Nutrients BOD5 : N : P = 100 : 5 : 1

Free from toxic materials

Mixing (turbulence, flat propeller, blades, rpm, etc.) Time (sufficient for adsorption, permeation, oxidation,

synthesis)


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ACTIVATED SLUDGE

Suspended growth system

Wastewater stabilised biologically under aerobic conditions

Consists of two main reactors i.e. aeration tank and clarifier

Aerobic environment created in aeration tank by mechanicalaeration or diffused air

Biomass produced separated from liquid effluent in clarifier

Portion of sludge is recycled into aeration tank

Three types

Completely mixed without sludge recycle

Completely mixed with sludge recycle Plug-flow with sludge recycle


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COMMON VARIATIONS OF

ACTIVATED SLUDGE PROCESS

Step aeration

Tapered aeration

Contact stabilization

Pure-oxygen activated sludge

Oxidation ditch

High rate

Extended aeration


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Design Activated Sludge

Description Unit Design Criteria

Minimum number - 2

F/M ratio kg BOD/kg

MLSS.d

0.25 0.50

HRT h 6 16

Oxygen requirements kg O2/kg BOD rem 1.5

MLSS mg/L 1500 3000

DO in tank mg/L 1.0

Sludge yield, Y kg MLSS/kg BOD 0.8 1.0

Decay coefficient, kd d-1 0.03 0.07

Sludge age, c d 5 10

Organic loading kg/m3/d 0.3 0.7


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DESIGN AND OPERATIONAL

PARAMETERS

Types of A/S

tcday

F/M

kg BOD5/kg MLSS

VLkg BOD5 /m

3 t

hr

kg O2/kg BOD

MLSS

mg/L

Completelymixed

4-15 0.2-0.4 0.8-2.0 3-5 0.7-1.0 3000-6000

Extended

Aeration

20-30 0.5-0.15 0.16-0.40 18-24 1.2-2.0 3000-6000

R = )P(42.1f

10S)-Q(Sx

-3

o

v

f= conversion factorfor converting BOD5 to BODu (~0.68)


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ACTIVATED SLUDGE


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MASS BALANCE ANALYSIS

Aeration Tank

Q Q + QR

X

Qe

Xe

QR

XRQw

Accumulation = inflow

outflow

0 = X (Q+QR) Xe(Qe) XR(QR) XR(Qw)

XR

Aeration Tank

Q Q + QR

X

Qe

Xe

QR

XRQwXR

Accumulation = inflow

outflow

0 = XR(QR) X(Q+ QR)


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FORMULATION / EQUATIONS

dcd

t

g

obsk

Y

U

kYU

dtdS

dtdXY

U!

!!

1/

/yield,growthObserved

wwee QSSQSS

VMLSSageSludge

vv

v

!Q

Vperiod,Aeration !U

X

)S-S(

VX

)S-Q(SUn,utilizatiosubstrateSpecific eoeo

U!!

X

S

VX

QSoo

U!!

M

F

%100Q

QratesludgeReturn

in

rv!

o

wastewaste

QS

XQProductionSludge

v

!

dckU!

1

)S-YQ(SPsludge,in wastesolidsVolatile eox


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EXAMPLE 1

A conventional activated-sludge plant without primary clarification operates under the

following conditions:

Design flow = 2.14 MGD

Influent BOD = 185 mg/LSS = 212 mg/L

Aeration basins: 4 unit, 40 x 40 ft2 x 15.5 ft deep

MLSS = 2600 mg/L

Recirculation flow = 1 MGD

Waste Sludge Quantity = 39,000 GPDSS in waste sludge = 8600 mg/L

Effluent BOD = 15 mg/L, SS = 15 mg/L

Calculate the: aeration period, BOD loading, F/M ratio, SS and BOD removal efficiencies,

Sludge age and Return activated-sludge age


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SOLUTION

Aeration basin volume = 4 (40)2 x 15.5 x (7.481/106) = 0.74 mil gallon

Aeration basin, dt= 0.74 mil gallon / 2.14 MGD x 24 hr = 8.3 hr

BOD loading = [(2.14 MGD x 185 mg/L x 8.34) / (4 x 402 x 15.5 /1000 ft3)]

= 33.3 lb/day / 1000 ft3

F / M = [(2.14 MGD x 185 mg/L x 8.34) / (0.74 mil gallon x 2600 mg/L x 8.34)]

= 0.21 lb BOD/day / lb MLSS.

Suspended Solid removal = [(212 15) / (212) x 100%] = 93%

BOD removal efficiency = [(185 15) / (185) x 100%] = 92%

Suspended solids in the effluent = 2.101 MGD x 15 mg/L x 8.34 = 262 lb/day

Suspended solids in waste activated sludge = 0.039 MGD x 8600 mg/L x 8.34 = 2797 lb/day

Sludge age = [(0.74 mil gallon x 2600 mg/L x 8.34) / (2797+ 262 lb/day)]

= 5.2 days

Return sludge rate = [(1.0) / (2.14) x 100%] = 48%


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Activated Sludge Treatment

Aerobic Granulation


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CONVENTIONAL NUTRIENT REMOVAL

ACTIVATED SLUDGE SYSTEM

InfluentPre-treatment Equalization

tank

Anaerobic tankDenitrification

Aerobictank

Effluent

SludgeDigester

CH4 SludgeDewatering

Sludge


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AEROBIC GRANULAR SLUDGE

Influent

Pre-treatment

Selector

Anaerobictank

Denitrification

Aerobic tank

Effluent

SludgeDigester

CH4SludgeDewatering

Sludge

SBR

Effluent

Aerobic GranularSludge Reactor


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Extended Aeration


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Common Variations of Activated Sludge Process

EXTENDED AERATION

Long detention time and low F/M ratio in aerator tomaintain culture in endogenous phase

Secondary

clarifier

Influent

Excesssludge

Effluent

Return sludge (omitted in some

system)

Reactor


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EXTENDED AERATION

Activated sludge processes with long HRT (in theorder of 24 hrs) and long sludge age

MLSS high, so insensitive to shock loads

Designed either plug flow or completely mixed Operates at low BOD loading and in endogenous

phase, thus growth rate and sludge yield are low

Normally used to handle domestic waste from smallcommunities accept variable loading better thanCMAS

Disadvantage/Challenge formation of filamentousmicroorganisms species which difficult to settle

High oxygen requirement


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EXTENDED AERATION SLUDGE

Description Unit Design

Criteria

Minimum number of aeration - 2

F/M kg BOD/kg MLSS.d 0.05 0.1

HRT h 18 24

Oxygen requirement kg O2/kg BOD rem. 2

MLSS mg/L 3000 5000

DO level in tank mg/L 2Sludge yield, Y kg sludge prod./ kg

BOD rem

0.4 0.6

Decay coefficient, kd d-1 0.06



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EXAMPLE 1

A small extended aeration plant without sludge wasting

facilities is loaded at a rate of 170 g/m3.d with an aerationperiod of 24 hr. The measured suspended solids buildup ratein the aeration tank is 40 mg/L/d. What percentage of theraw influent BOD is converted and retained as MLSS? If theMLSS concentration is allowed to increase from 2000 to 6000

mg/L before wasting solids, how long would this builduptake?


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SOLUTIONS

Because the aeration period is 24 hr,

= 170 mg/L. d

= 23.5%

3

3

m

.dg/m170

tankofliter

load/dayBODy!

%100mg/L.d170

mg/L.d40

appliedBOD

buildupMLSSv!

days100mg/L.d40

mg/L2000-6000timeBuildup !!


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EXAMPLE 2

Two activated sludge aeration tanks at Taman

University, Skudai are operated in series. Each tank

has the following dimensions: 7 m wide x 30 m long

x 4.3 m effective liquid depth. The plant operating

parameters are as follows:

Flow = 0.0796 m3/s, MLVSS = 1500 mg/L, MLSS =1.40 MLVSS, Soluble BOD5 after primary settling =

130 mg/L.

Determine the aeration period and F/M ratio


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= V/Q = (7x30x4.3)/(0.0796) = 11,344 s

= 3.15 h x 2 tanks = 6.3 hrs

F/M = QSo/VX = ??

[Answer: aeration period = 6.3 h; F/M =

0.33 mg/mg. day]

SOLUTIONS


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SEQUENCING BATCH REACTOR (SBR)


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SBR PROCESS

Batch process which takes place in ONE reactor

Consists of FIVE basic operating mode


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DESIGN OF SBR

Description Unit Design

Criteria

Minimum number of reactor - 1F/M kg BOD/kg MLSS.d 0.05 0.2

Oxygen requirement kg O2/kg BOD rem.

DO level in tank mg/L ~6.5

Sludge yield, Y kg sludge prod./ kg

BOD rem

0.75 1.10

Decay coefficient, kd d-1 0.06



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SETTLE

3 min.

1 min.

IDLE

60 min.

FILL

influent

5 min.DRAIN

effluent

121 min.

REACT

air

Cycle


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FORMULATION / EQUATIONS

! suo rQS-QSdt

dS

Batch Model Reactor with Completely Stirred Tank Reactor

S)Y(K

SX-rwhere

s

m

su

!

Q

Influent = Effluent + Production Wasted (or Consumed)

Example of Substrate

SQ)SQ-(Q)(

ww !

SKYSXQSs

mo

Q

Substrate in influent Substrate consumed = Substrate in effluent Substrate in

Wasted


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BENEFITS OF SBR

Smaller size than CMFR

Be able to absorb shock load better than

CMFR and PFR allow HRT to beprolonged until desired quality is reached

Better sedimentation tank


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Activated Sludge TreatmentMembrane Bioreactor (MBR)


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MEMBRANE BIOREACTOR (MBR)

A new version of activated sludge system

Utilises microporous membranes for solid/liquid separation in lieu

of secondary clarifiers


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MBR Design Consideration

Equalization

Sufficient tank with in-line or off-line storage

Preliminary Treatment

Removal of grits and materials (screen < 2mm)

Solid Retention Time (SRT) and HRT

SRTs as 5 20 days, not strongly related to biopolymer fouling

HRT may consider in short, 4 6 h

Mixed Liquor Suspended Solids (MLSS)

Immersed MBR of MLSS from 8,000 18,000 mg/L

Dissolved Oxygen (DO)

Anoxic = 0 0.5 mg/L;Aerobic = 1.5 3 mg/L; Membrane = 2 6 mg/L

Flux rates

Consider in between 25 46 L/m2. h


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ADVANTAGES OFMBR

Secondary clarifiers and tertiary filtration processes are eliminated,thereby reducing plant size.

Quality of solids separation is not dependent on the mixed liquorsuspended solids concentration or characteristics. Since elevatedmixed liquor concentrations are possible, the aeration basin volume

can be reduced, further reducing the plant size. Can be designed with long sludge age, hence low sludge production.

Produces a MF/UF quality effluent suitable for reuse applications oras a high quality feed water source for Reverse Osmosis treatment.Indicative output quality of MF/UF systems include SS < 1mg/L,turbidity


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EXAMPLE OFMBR APPLICATION


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Cauley Creek, GA:MBR

Constructed in 2 x 2.5 MGD phases (total 5 MGD)

Plant is optimized for meeting discharge limits of0.13 mg/L

total P and 0.5 mg/L ammonia

Staff indicated not much the plant can do to further reduce

energy consumption without potentially violating systemwarranties or permit


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Pooler, GA: MBR

2.5 mgd MBR, operated

24 hours, staffed 8hr/day

The plant has dischargelimits for ammonia, but

not for phosphorus

Cut back on the aeration

to only night times Turned off UV to save

energy


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Membrane Fouling

TMP


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BIOLOGICAL WASTE TREATMENT

APPLICATION

Treatment

Method

Mode of

Operation

Degree of

Treatment

Land

Requirements

Equipment Remarks

Activated

Sludge

Completely

mixed or plug

flow, sludge

recycle

> 90% removal

of organics

Earth or

concrete basins

(3.66 6.10 m

deep), (0.561

2.262 m3/m3.day), (5.52

34.4 m2/ 103

m3.day)

Diffused or

mechanical

aerators,

clarifiers for

sludgeseparation and

recycle.

Excess sludge

dewatered and

disposed of

Trickling Filter Continuous

application;

may employeffluent recycle

Intermediate or

high,

depending onload

5.52 34.4

m2/103 m3. day

Plastic

packing, 6.10

12.19 m);

Pre-treatment

before publicity

ownedtreatment

works (POTW)

or activated

sludge plant


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BIOLOGICAL WASTE TREATMENT

APPLICATION

Treatment

Method

Mode of

Operation

Degree of

Treatment

Land

Requirements

Equipment Remarks

RBC Multistage

continuous

Intermediate

or high

Plastic disk Solids

separationrequired

Anaerobic Complete mix

with recycle;

upflow or

downflow filter;

fluidized bed,upflow sludge

blanket

Intermediate Gas collection

required; pre-

treatment

before POTW

or activatedsludge plant


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Choose: Aerobic or AnaerobicAerobic process Anaerobic process

COD < 3000 mg/l

BOD 3000 mg/l

BOD>2000 mg/l

Slow biodegradable matters

High-strength waste

Pre-treatment process

Biodegradation 2 types:

Mineralization

Organic compounds are converted byliving organisms to minerals (non-

organic) end products MAINLY

AEROBIC PROCESS

Biotransformation

Parent organic compounds are notcompletely mineralized, a portion is

converted into other organics

ANAEROBIC PROCESSES

SEME

STE

RIIS

ESS

ION CHAPTER 8


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CONVENTIONAL ACTIVATED

SLUDGE (CAS)


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COMPLETE ACTIVATED SLUDGE


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Common Variations of Activated Sludge Process

OXIDATION DITCH

Similar to extended aeration, except the aeration is done bybrush-type aerators, thus reducing electricity

Effluent

Influent

Return sludge

Brush-type aerators


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OXIDATION DITCH (OD)


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AERATED LAGOON SYSTEM

System Advantages Disadvantages

Aerated Lagoon Lower land

requirement

Equipment

Relatively simple

construction,operation and

maintenance

Slight increase in the

sophisticated level.Need a continuous of

sludge removal (2 5

years)

Large efficiency in

BOD removal

High energy

requirementsLow possibility of odor

problem

Fast filling of the

sedimentation lagoon

with sludge (2 5

years


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HOMEWORK 1

What is the purpose of the activated sludge process in

treating wastewater?

What is the functioning of Returned Activated Sludge

(RAS) in aeration tank? What happens to the air requirement in the aeration

tank when the strength (BOD) of the incoming

wastewater increases?

With an aid of diagram shows the following symbol inthe conventional wastewater treatment plant

(completely mixed reactor). The diagram must consist

of primary treatment, biological treatment and

secondary clarifier.


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HOMEWORK 2

A conventional activated sludge plant without primary clarification operates underthe following conditions:

Design flow = 8100 m3/day

Influent BOD = 185 mg/L

SS = 212 mg/L

Aeration basins: 4 units, 12 m square x 4.5 m deep

MLSS = 2600 mg/L

Recirculation flow = 3800 m3/d

Waste sludge quantity = 150 m3/d

SS in waste sludge = 8600 mg/L

Effluent BOD = 15 mg/L, Effluent SS = 15 mg/L

Calculate the following: aeration period

BOD loading

F/M ratio

SS removal and BOD removal

Sludge age (SRT)

Return activated sludge age

chapter 8 - aerobic process

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