ce 152 lecture notes 3 - biological treatment

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CE 152 CE 152

SANITARY ENGINEERING IILecture 3 – Biological Treatment

Dr. Eliseo V. Ana, Jr.CE DepartmentCE Department

2nd Semester, 2010-2011

BIOLOGICAL TREATMENT

• Objectives:

– To coagulate & remove nonsettleable colloidal – To coagulate & remove nonsettleable colloidal

solids & to stabilize organic matter.

• In domestic WW treatment:

– To reduce organic content & nutrients (e.g. N, P)

– To remove toxic organic compounds

• For industrial WW treatment:• For industrial WW treatment:

– To remove or reduce concentration of organic & inorganic

compounds (toxic to microorganisms)

2


• Role of microorganisms

– Removal of carbonaceous BOD, coagulation of – Removal of carbonaceous BOD, coagulation of

nonsettleable solids & stabilization of organic

matter are accomplished biologically using a

variety of microorganisms (esp. bacteria)

– Microorganisms convert organic matter into – Microorganisms convert organic matter into

various gases & cell tissue

• Cell tissue specific gravity > water → gravity settling

3


• Microbial metabolism

– Nutrient requirements– Nutrient requirements

• Organisms must have’s:

– Source of energy

– Carbon for synthesis of new cellular material

– Inorganic elements (nutrients) – N, P, S, K, Ca, Mg

» Organic nutrients (growth factors) may also be required

» Carbon & energy sources: substrates

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• Microbial metabolism– Carbon & energy sources– Carbon & energy sources

• 2 sources: organic matter & carbon dioxide– Types of organisms based on carbon source:

» Heterotrophs: organisms that use organic carbon for cell formation (e.g. most bacteria, fungi, protozoa)

» Autotrophs: organisms that derive carbon from carbon dioxide (nitrifying bacteria)

• Conversion of CO2 to organic cell tissue is reductive process requiring net input of energy → autotrophs lower growth rate

• 2 sources of energy: light & chemical oxidation reaction– Types of organisms based on source of energy:

» Phototrophs: organisms that use light as energy source

» Chemotrophs: organisms that derive energy from chemical reactions

5



– Nutrient & growth factors– Nutrient & growth factors

• Inorganic nutrients: N, S, P, K, Mg, Ca, Fe, Na, Cl

– Could be limiting for microbial cell synthesis & growth

• Organic nutrients (growth factors): amino acids, purines• Organic nutrients (growth factors): amino acids, purines

& pyramidines, vitamins

6



– Microbial nutrition & bio treatment – Microbial nutrition & bio treatment

• Chemoheterotrophic bacteria: important in reducing

organic content (carbonaceous BOD) in WW

• Chemoautotrophic bacteria: important in converting • Chemoautotrophic bacteria: important in converting

ammonia to nitrate

7


• Microbial metabolism– Types of microbial metabolism– Types of microbial metabolism

• Chemoheterotrophic bacteria can be grouped according to metabolic type & molecular O2 req’ts:

– Obligately aerobic: organisms that are dependent on aerobic respiration to meet their energetic needs & can only exist when there is supply of molecular oxygen

– Obligately anaerobic: organisms that generate energy by fermentation & can only exist in environment devoid of oxygencan only exist in environment devoid of oxygen

– Facultative anaerobes: organisms with the ability to grow in either the presence or absence of molecular oxygen

» Fermentation is less-efficient energy yielding process →fermentative organisms are characterized by lower growth rates

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• Important microorganisms

– Bacteria– Bacteria

– Fungi

– Protozoa & rotifers

– Algae

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– Bacteria– Bacteria

• Single-celled procaryotic organisms

• Reproduction: binary fission, sexually, budding

• Forms: spherical, cylindrical, helical

• Cell composition: 80% water, 20% dry material (organic & inorganic)

• Environmental req’ts: temperature & pH• Environmental req’ts: temperature & pH

– Optimal growth occurs at narrow range

» pH: 6.5-7.5

– Increase in temp → increase in growth

10

11



– Fungi– Fungi

• Multicellular, non-photosynthetic, heterotrophic protists

• Reproduction: sexually, asexually, fission, budding, spore formation

• Strict aerobes

• Have ability to grow under low-moisture conditions & • Have ability to grow under low-moisture conditions & low pH (optimum: 5.6)

• Low N demand

» Important in industrial waste treatment

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• Important microorganisms– Protozoa– Protozoa

• Motile, microscopic, single-celled protists

• Majority are aerobic heterotrophs; few anaerobic

• Larger than bacteria & consume bacteria as energy source» Acts as polishers of effluents from biological waste-treatment by

consuming bacteria & particulate matters

– Rotifer• Aerobic, heterotrophic, multicellular• Aerobic, heterotrophic, multicellular

• Effective in consuming dispersed & flocculated bacteria & small particles of organic matter

» Presence indicative of a highly efficient biological purification process

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• Bacterial growth

– Growth patterns– Growth patterns

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– Growth patterns– Growth patterns

• Lag phase – time req’d for organisms to acclimate to new environment

• Log-growth phase – cells divide at a rate determined by their generation time & ability to process food

• Stationary phase – population remains stationary: no more • Stationary phase – population remains stationary: no more substrate or nutrients for growth; growth of new cells offset by death of old cells

• Log-death phase – death rate exceeds production of new cells

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– Growth in mixed cultures– Growth in mixed cultures

• Most biological treatment processes are comprised of

complex, interrelated, mixed biological populations

» Each with own growth curve

• Growth curve dependent on:

• Food & nutrients

• Temperature

• pH

• Aerobic or anaerobic

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• Biological treatment processes

– Trickling filters– Trickling filters

– Activated sludge

– Oxidation ponds (lagoons)

– Rotating biological contactors

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• Trickling filters

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• Trickling filters– Process description:Process description:

• Consist of a bed of highly permeable medium (stone, plastic) to which microorganisms are attached & through which WW is percolated or trickled (by a rotating arm)

• Organic material in WW is degraded by microorganisms attached to filter media

– Organic material adsorbed onto biological film/slime layer

– As slime layer increases in thickness, adsorbed matter is metabolized before it reaches near the media face → inner microbes lose ability to cling to media surface → sloughing (washing off of slime layer)surface → sloughing (washing off of slime layer)

• Flow from filter is passed through a sedimentation basin to allow solids (washed off slime) to settle out

• Recirculation – return of a portion of effluent to flow through filter– To dilute the strength of the incoming WW & to maintain the slime layer in a

moist condition20

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• Trickling filter

– Filter media:– Filter media:

• Rock or slag (25-100 mm dia), plastic packing materials

(round, square shapes)

• Depth of filter: 0.9 to 2.5 m (rock); 4 to 12 m (plastic)

• Diameter: up to 60 m (rock)

22


• Trickling filter

– Slime layer – Slime layer

• Bacteria, fungi, algae, protozoans

• Outer layer (0.1 to 0.2 mm) – aerobic microorganisms

• Inner layer near surface media: anaerobic

microorganisms

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• Trickling filters (2-stage)

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• Design

– Efficiency of filters (single-stage or 1st filter)– Efficiency of filters (single-stage or 1st filter)

• Based on National Research Council (NRC)

5.01

12.41

1

+

=

VF

QCE

in

/sm flowrate,r wastewate

ionsedimentat &ion recirculat including C,20at stage1st for removal BOD offraction

3

51

−

−

Q

E�

26

factorion recirculat

m media,filter of volume

mg/L ,BODinfluent


3

5

3

−

−

−

−

F

V

C

Q

in

2)1.01(

1

R

RF

+

+=


/sm flowrate,ion recirculat

ratioion recirculat

3

3

−

−

=−

Q

Q

/QQR

r

r


• Design

– Efficiency of filters (second stage filter)– Efficiency of filters (second stage filter)

5.0

1

2

1

12.41

1

−+

=

VF

QC

E

E

e

stage1st for removal BOD offraction

% ion,sedimentat &ion recirculat including C,20at filter stage 2ndfor removal BOD offraction

51

52

−

−

E

E�

• Effect of temperature

27

mg/L stage,1st from BODeffluent

stage1st for removal BOD offraction

5

51

−

−

eC

E

1.035 ;)20(

20 == − θθ T

T EE

EXAMPLE

Using the NRC equations, determine the BOD5

of the effluent from a single-stage, low-rate of the effluent from a single-stage, low-rate

trickling filter that has a volume of 1,443 m3, a

hydraulic loading of 1,900 m3/d, and a

recirculation factor of 2.78. The influent BOD5

is 150 mg/L.

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SOLUTION

Given: single-stage, low-rate trickling filter volume: 1,443 m3

hydraulic loading: 1,900 m3/d

recirculation factor: 2.78recirculation factor: 2.78

influent BOD5 is 150 mg/L.

Req’d: using the NRC equations, determine the BOD5 of the effluent

Sol’n:

- Convert hydraulic loading rate to m3/s

3 31(1,900 / ) 0.022 /

86, 400 /Q m d m s

s d

= =

- Efficiency of single-stage filter

- BOD5 of the effluent

29

86, 400 /s d

( )( )( )( )

1 0.5

10.894

0.022 1501 4.12

1.443 2.78

E = =

+

( )( )1 0.894 150 15.8 /e

C mg L= − =


• Design

– Time of contact (Schulze)– Time of contact (Schulze)

nAQ

CDt

)/(=

mediafilter on basedconstant empirical

m applied, istewater which wasof areafilter

/dm rate, loading hydraulic

m depth,filter

1eunit volumper film activemean

d me,contact ti

2

3

−

−

−

−

≈−

−

n

A

Q

D

C

t

– Efficiency (based on Schulze & Velz):

30

( )

−=

n

t

AQ

KD

S

S

/exp

0/m(m/d) constant, empirical

mg/L ,BODinluent

mg/L ,BODeffluent

n

50

5

−

−

−

K

S

St

EXAMPLE

Determine the BOD5 of the effluent from a

low-rate trickling filter that has a diameter of low-rate trickling filter that has a diameter of

35.0 m and a depth of 1.5 m if the hydraulic

loading is 1,900 m3/d and the influent BOD5 is

150 mg/L. Assume rate constant is 2.3

(m/d)n/m and n=0.67.

31

SOLUTION

Given: low-rate trickling filter

diameter: 35.0 m, depth: 1.5 m

hydraulic loading: 1,900 m3/d hydraulic loading: 1,900 m3/d

influent BOD5: 150 mg/L

rate constant is 2.3 (m/d)n/m, n=0.67.

Req’d: determine the BOD5 of the effluent

Sol’n:

- Compute filter area

2 2(35 ) / 4 962.11A mπ= =

- Compute loading rate

- Compute effluent BOD5

32

2 2(35 ) / 4 962.11A mπ= =

( ) ( )3 2 3 2/ 1,900 / / 962.11 1.97 /Q A m d m m d m= = ⋅

( )( ) ( )

( )0.67

2.3 1.5150 exp 16.8 /

1.97tS mg L

= − =

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• Activated sludge

– Involves production of an activated mass of – Involves production of an activated mass of

microbes capable of stabilizing a waste aerobically

33

BIOLOGICAL TREATMENT• Activated sludge

– Process description• Mixture of wastewater & biological sludge (microorganisms), called

mixed liquor, is agitated & aerated in aeration tankmixed liquor, is agitated & aerated in aeration tank→ Microorganisms mixed thoroughly w/ organics under conditions that stimulate

their growth through the use of organics as food

→ As microorganisms grow & mixed by agitation of air, individual organisms clump together (flocculate) to form an active mass of microbes (biological floc) called activated sludge

2 2 3 5 7 2

bacteria

COHNS + O + nutrients CO + NH + C H NO + other products

organic matter ba

→

cterial cell

together (flocculate) to form an active mass of microbes (biological floc) called activated sludge

• Mixed liquor flows from aeration tank to secondary clarifier → settle activated sludge

• Most of settled sludge is returned to aeration tank (returned sludge) to maintain high population of microbes → rapid breakdown of organics

34


• Activated sludge– Conventional activated sludge operationConventional activated sludge operation

• Long, rectangular basin

• Aeration time: 6-8 hrs

• Air volume: 8 m3 of air per m3 WW– To keep sludge in suspension

– Air is injected near the bottom of aeration tank

• Sludge return volume: 20-30% of WW flow– Rest of sludge (microorganisms) is wasted (waste activated sludge or WAS)

» Maintain proper amt of microbes to efficiently degrade BOD5

» Balance between growth of new microorganisms & their removal by » Balance between growth of new microorganisms & their removal by wasting

» Too much WAS, low number of microbes for effective treatment

» Too little WAS, large concentration of microbes → overflow to receiving water bodies

Mean cell residence time (θc) or solids retention time (SRT) or sludge age - the average amt of time that microorganisms are kept in the system

35



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37


• Activated sludge– ProcessesProcesses

• Complete mix

• Plug flow

• Reactors – tanks used to perform physical, chemical & biochemical reactions

– Classification:

1. Batch reactors – fill-and-draw type; mat’ls are added to the tank, mixed for sufficient time to allow reaction to occur & then drained

2. Flow reactors – continuous type operation; mat’l flows into, through & out of reactor at all times2. Flow reactors – continuous type operation; mat’l flows into, through & out of reactor at all times

Types:

Completely stirred tank reactor (CSTR)

- Complete, instantaneous mixing; composition of effluent is same as composition in tank

-

Plug-flow reactors

- Fluid particles pass through the tank in sequence; those that enter first leave first 38



– Completely mixed activated sludge process– Completely mixed activated sludge process

• Uses continuous-flow stirred-tank reactor (CSTR)

• Design

– Mass-balance application of the equations of kinetics of

microbial growth

– Mass balances req’d to design reactor

» Biomass

» Food (BOD5)

39



wastewater flowrate

return activated sludge flowrate

waste activated sludge flowrate

r

w

Q

Q

Q

−

−

−

40

0 5

5

influent soluble BOD

aeration tank & effluent soluble BOD

microorganisms conc (mixed liquor volat

S

S

X

−

−

− ile suspended solids or MLVSS) in aeration tank

microorganisms conc in effluent

microorganisms conc in return activated sludge

microorganisms conc in wasted activated sludge

volume of aerat

e

r

w

X

X

X

V

−

−

−

− ion tank

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BIOLOGICAL TREATMENT• Activated sludge

– Completely mixed activated sludge process

• Mass balance of biomass• Mass balance of biomass wastedBiomass effluent in Biomass daccumulate Biomass influent in Biomass +=+

( )

[ ]

0

Monod equation

m

d w e w r

s

SXQX V k X Q Q X Q X

K S

µ + − = − +

+

3

0

3

wastewater flowrate into the aeration tank, m /d

microbes conc (volatile suspended solids or VSS) entering aeration tank, mg/L

volume of aeration tank, m

Q

X

V

−

−

−

41

1

volume of aeration tank, m

max growth rate constant, d

so

m

V

S

µ −

−

−

− 5

5

luble BOD in aeration tank & effluent, mg/L

microbes conc (mixed liquor volatile suspended solids or MVSS) in the aeration tank, mg/L

half velocity constant (soluble BOD conc at 1/2 the max gros

X

K

−

−

1

3

wth rate), mg/L

decay rate of microbes, d

flowrate of liquid containing microbes to be wasted, m /d

microbes conc (VSS) in effluent from secondary settling tank, mg/L

microbes conc (VSS

d

w

e

r

k

Q

X

X

−−

−

−

− ) in sludge being wasted, mg/L




• Mass balance of food

Food in influent + Food consumed = Food in effluent + Food in WAS

( )( )0

mw w

s

SXQS V Q Q S Q S

Y K S

µ − = − + +

yield coefficient (decimal fraction of food mass converted to biomass)Y −

42

yield coefficient (decimal fraction of food mass converted to biomass)Y −




• Development of working design equation

Assumptions:

– The influent (X0) & effluent (Xe) biomass concentrations are

negligible compared to that in the reactor

– The influent food (S0) is immediately diluted to the reactor

concentration in accordance with the definition of CSTRconcentration in accordance with the definition of CSTR

– All reactions occur in the CSTR

43





Thus from mass balance of biomass,

( )0m

d w e w r

s

SXQX V k X Q Q X Q X

K S

µ + − = − +

+

mSXV k X Q X

µ − =

44

md w r

s

SXV k X Q X

K S

µ − =

+

Rearranging in terms of Monod equation,

m w rd

s

SX Q Xk

K S VX

µ = +

+

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Rearranging the mass balance of food eq’n in terms of Monod eq’n,

( )0m

s

S Q YS S

K S V X

µ = −

+

Equating the Monod expressions for biomass & food & rearranging,

45

Equating the Monod expressions for biomass & food & rearranging,

( )0w r

d

Q X Q YS S k

VX V X= − −

V

Qθ=Hydraulic detention time of reactor:

Mean cell-residence time:c

w r

VX

Q Xθ=




• Development of working design equationModification of mean-cell residence time to account

for effluent losses of biomass,

( )c

w r w e

VX

Q X Q Q Xθ =

+ −

Soluble BOD5 concentration in effluent given θc

46

Soluble BOD5 concentration in effluent given θc

( )( )

1

1

s d c

c m d

K kS

k

θ

θ µ

+=

− −Note: conc of soluble BOD5 leaving the system (S) is only affected by θc

not by BOD5 entering the aeration tank or hydraulic detention time

Note: S is soluble BOD5 not total BOD5; some BOD5 are attached to

SS; to find allowable S:

5 5 Total BOD allowed BOD in SSS = −


47





Concentration of microorganisms in aeration tank,

( )( )0

(1 )

c

d c

Y S SX

k

θ

θ θ

−=

+

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13




• Design equations:

Mean-cell residence time:c

w r

VX

Q Xθ=

( )c

w r w e

VX

Q X Q Q Xθ =

+ −

Effluent soluble BOD5:( )

( )

1

1

s d c

c m d

K kS

k

θ

θ µ

+=

− −

49

Concentration of microorganisms in aeration tank,( )( )0

(1 )

c

d c

Y S SX

k

θ

θ θ

−=

+

Hydraulic detention time of reactor: V

Qθ=

EXAMPLE

A town has been directed to upgrade its primary WWTP to a secondary plant that can meet an effluent standard of 30.0 mg/L BOD5 and 30.0 mg/L SS. They have selected a completely mixed activated sludge system.selected a completely mixed activated sludge system.

Assuming that the BOD5 of the SS may be estimated as equal to 63% 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 = 0.150 m3/s

BOD5 = 84.0 mg/L

Assume the following values for the growth constants:

Ks = 100 mg/L BOD5

µm = 2.5 1/d

kd = 0.050 1/d

Y = 0.50 mg VSS/mg BOD5 removed

50

SOLUTION

- Calculate allowable soluble BOD5 in effluent using

63% assumption BOD5 in SS63% assumption BOD5 in SS

- Calculate mean cell-residence time

5 5 -

30.0 .63 30 11.1 /

S BOD allowed BOD in SS

S mg L

=

= − × =

( )( )

1

1

s d c

c m d

K kS

k

θ

θ µ

+=

− −

( )100 1 0.05011.1

cθ+ ×

=

51

( )( )

100 1 0.05011.1

2.5 0.050 1

c

c

θ

θ

+ ×=

− −

5.0c dθ =

SOLUTION

– Calculate hydraulic detention time assuming a

value of 2,000 mg/L MLVSSvalue of 2,000 mg/L MLVSS

( )( )0

(1 )

c

d c

Y S SX

k

θ

θ θ

−=

+

( )( )5.0 0.5 84.0 11.12000

(1 0.050 5.0)θ

−=

+ ×

0.073 or 1.8d hθ =

– Calculate volume of aeration tank

52

V

Qθ=

1.80.15 3600 /

V

s h=

×3972V m=

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– Plug-flow with recycle– Plug-flow with recycle

• Fluid particles pass through the tank in sequence

– Achieved in long, narrow aeration tanks

• Assumptions

– The concentration of microorganisms in the influent to the

aeration tank is approximately the same as that in the effluent

from the aeration tank; applicable if θc/ θ > 5.from the aeration tank; applicable if θc/ θ > 5.

– The rate of soluble BOD5 utilization as the waste passes

through the aeration tank is given by

53

; average conc of microorganisms in the aeration tankm avg

u avg

s

SXr X

K S

µ= − −

+



– Plug-flow with recycle– Plug-flow with recycle

• Design equation:

( )( ) ( ) ( )

0

0

1

1 ln /

m

d

c s i

Y S Sk

S S K S S

µ

θ α

−= −

− + +

recycle ratio, /rQ Qα −

54

0

recycle ratio, /

ln log to base e

influent conc to aeration tank after dilution w/ recycle flow, mg/L

1

r

i

i

Q Q

S

S SS

α

α

α

−

−

−

+=

+


• Activated sludge– Food to microorganisms ratio (F/M)– Food to microorganisms ratio (F/M)

• F/M ratio is controlled by wasting part of microbial mass– High rate of wasting, high F/M ratio → organisms saturated w/ food →

poor efficiency

– Low rate of wasting, low F/M ratio → organisms are starved → more complete degradation

Long θc (low F/M)

0 5mg BOD /d mg

mg MLVSS mg×d

QSF

M VX= = =

– Long θc (low F/M)

» Larger, costly tanks; high oxygen req’t; high power cost

» Less sludge

• F/M values: 0.1 – 1.0 mg/mg-d

55

EXAMPLE

Compute the F/M ratio for the new activated-sludge plant in the previous example.plant in the previous example.

Q= 0.150 m3/s

S0= 84.0 mg/L

V= 970 m3

X= 2000 mg/L

Solution:Solution:– Calculate F/M

56

( ) ( )( )

( ) ( )

3

3

0.150 / 84 / 86, 400 /0.56 /

970 2000 /

m s mg L s dFmg mg d

M m mg L= = ⋅

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– Sludge return– Sludge return

• Purpose: to maintain sufficient concentration of

activated sludge in reactor basin

• Typically 50-100% of raw waste flow

57




• Return sludge flow rate (from mass balance)

– if effluent suspended solids are negligible, Xe

XX

XQXQQ

r

rwr

′−′

′−′=

g/m ion,concentrat sludgereturn maximum

g/m (MLSS), solids suspendedliquor mixed

/dm rate, flow sludgereturn

/dm rate, flowr wastewate

3

3

3

3

−′

−′

−

−

r

r

X

X

Q

Q

– effluent suspended solids are not negligible, Xe

where,58

/dm rate, flow wastingsludge

g/m ion,concentrat sludgereturn maximum

3−

−′

w

r

Q

X

XX

XQQXQXQQ

r

ewrwr

′−′

−−′−′=

)(

]/[106

LmgSVI

X r =′




• Sludge Volume Index (SVI)

– Procedure:

» Measuring MLSS (mixed liquor suspended solids) &

sludge settleability

59


• Activated sludge– Sludge return– Sludge return

• Sludge Volume Index (SVI)

– SVI

gmgMLSS

SVSVI /000,1×=

mg/L solids, suspendedliquor mixed

mL/L settling,min 30after cylinder graduated L-1in solids settled of vol

mL/g index, volumesludge

−

−

−

MLSS

SV

SVI

– SVI

» Indicator of sludge settling characteristics → control/limits MLSS & return rates

» Typical values:

• MLSS concentration: 2,000 – 3,500 mg/L; SVI: 80-150 mg/L

60

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61


62

EXAMPLE

Calculate the return sludge flow rate in a WWTP given the following data:given the following data:

Aeration tank volume = 970 m3

Flow = 0.150 m3/s

MLVSS (X) = 2000 mg/L

MLSS (X’) = 1.43(MLVSS)

Effluent suspended solids = 30 mg/LEffluent suspended solids = 30 mg/L

Wastewater temp = 18 degrees C

θc = 5 days

63

SOLUTION

– Compute anticipated MLSSLmgMLSS /860,2)000,2(43.1 ==

– Solving for return sludge concentration

• Selecting SVI based on temp & MLSS: SVI=175

– Solving for sludge wasting flowrate, Qw

• Solving for sludge wasting concentration

LmgMLSS /860,2)000,2(43.1 ==

LmgSVIX r /700,5714,51751010 66 ≈===′

LmgXX /986,343.1/ =′=

• Solving for Qw

64

LmgXX rr /986,343.1/ =′=

dmLmgd

Lmgm

X

VXQ

rc

w /3.97)/986,3)(5(

)/2000)(970( 33

===θ

Converting to m3/s: smds

dmQw /0011.0

/400,86

/3.97 33

==

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17

SOLUTION

– Assuming effluent SS are neglected

XQXQ ′−′

XX

XQXQQ

r

rwr

′−′

′−′=

smQ

mgmg

mgsmmgsmQ

r

r

/15.0

/860,2/714,5

)/714,5)(/0011.0()/860,2)(/150.0(

3

33

3333

=

−

−=

65



– Sludge production– Sludge production

• Activated sludge process converts organic & inorganic

substances into cell material

• Cell material → sludge (needs to be removed)

• Production: 0.4-0.6 kg MLVSS/kg BOD5 removed

• Amount sludge to be removed (wasted) each day:

– The difference between the increase in sludge mass & the

suspended solids (SS) lost in the effluent:

66

Mass to be wasted = increase in MLSS SS lost in effluent−



– Sludge production– Sludge production

• Amount of activated sludge generated (increased) per day:

1obs

d c

YY

k θ=

+( )( )3

0 10 /x obs

P Y Q S S kg g−= −

5

net waste activated sludge produced each day in VSS, kg/d

observed yield, kg MLVSS/kg BOD removed

x

obs

P

Y

−

−

– Typically, MLSS increase is,

• Mass of SS lost in effluent

67

( )1.25 1.667x

MLSS P= −

( )w eSS Q Q X= −

EXAMPLE

Estimate the mass of sludge to be wasted each day from an activated sludge plant with the following data:from an activated sludge plant with the following data:

Q = 0.150 m3/s

Y = 0.50 kg VSS/kg BOD5 removed

kd = 0.050 d-1

θc = 5 days

S 0 = 84 mg/LS 0 = 84 mg/L

S = 11.1 mg/L

Q w = 0.0011 m3/s

X e = 30 mg/L

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SOLUTION

• Calculate observed yield, Yobs

( )( )5

5

0.50 kg VSS/kg BOD removed0.40 kg VSS/kg BOD removed

obs

YY

θ= = =

+ +

• Calculate net waste activated sludge produced

– Assuming MLSS = 1.43 (MLVSS)

( )( )5-1

0.40 kg VSS/kg BOD removed1 1 0.050 d 5 d

obs

d c

Yk θ

= = =+ +

( )( )

( )( )( ) ( )

3

0

3 3 3 3

10 /

0.40 0.150 m /s 84.0 g/m 11.1 g/m 86,400 s/d 10 kg/g

377.9 kg/d VSS

x obs

x

x

P Y Q S S kg g

P

P

−

−

= −

= −

=

• Calculate SS lost in effluent

69

( )Increase MLSS = 1.43 377.9 kg/d 540.4 kg/d=

( ) ( )( ) ( )( )( )

3 3 3 30.150 m /s 0.011 m /s 30 g/m 86,400 s/d 10 kg/g

385.9 kg/d

w e

w e

Q Q X

Q Q X

−− = −

− =

SOLUTION

• Mass to be wasted

Mass to be wasted = 540.4 385.9 = 154.5 kg/d−

70



– Oxygen demand– Oxygen demand

• O2 is used in the reactions to degrade substrate to

produce high-energy compounds req’d for cell synthesis

& respiration

• Minimum residual in reactor: 0.5-2 mg/L DO (dissolved

oxygen) oxygen)

• Mass of O2 demand:

71

( )( )2

3

0 10 /1.42

O x

Q S S kg gM P

f

−−= −

5conversion factor for converting BOD to ultimate BODu

f −

EXAMPLE

Estimate the volume of air to be supplied (m3/d) for the activated sludge plant with the following design values:

Q = 0.150 m3/sQ = 0.150 m3/s

S0 = 84.0 mg/L

S = 11.1 mg/L

Px = 377.9 kg/d of VSS (net waste activated sludge produced/day)

Assume BOD5 is 68% of the ultimate BOD and that the oxygen transfer efficiency is 8%.

NOTE:

ρair = 1.185 kg/m3

By mass, air contains 23.3% oxygen

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19

SOLUTION

– Calculate mass of oxygen required

( )( )3

0 10 /Q S S kg g−−

– Calculate volume of air

( )( )

( )( )( )( )( )

2

2

2

0

3 3 3 3

10 /1.42

0.15 / 84.0 / 11.1 / 86, 400 / 10 /1.42 377.9 /

0.68

852.8 /

O x

O

O

Q S S kg gM P

f

m s g m g m s d kg gM kg d

M kg d

−

−= −

−= −

=

852.8 /kg d

At 8% transfer efficiency,

73

( )( )( )

3

3

852.8 /3,102 /

1.185 / 0.232air net

kg dV m d

kg m= =

( )3

33,102 /38,775 /

0.08air total

m dV m d= =


• Activated sludge– Sludge problems– Sludge problems

• Bulking sludge– Poor settling characteristics; poor compactibility

– Types wrt to cause:

1. Growth of filamentous organisms (in low pH, low N & high carbohydrates)

2. Water trapped in the bacterial flow

• Rising sludge– A sludge that floats to the surface after apparently good – A sludge that floats to the surface after apparently good

settling

– Results from denitrification (reduction of nitrates/nitrites → N gas); gas is trapped causing sludge to float

– Solution:

» Increase rate of return sludge flow (Qr)

» Decrease mean cell residence time (avoid denitrification)

74


• Oxidation ponds

– Sewage lagoon; waste stabilization pond– Sewage lagoon; waste stabilization pond

– Used as water treatment systems for small communities

– Types:

• Aerobic ponds

• Facultative ponds• Facultative ponds

• Anaerobic ponds

• Maturation or tertiary ponds

• Aerated lagoons

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• Oxidation ponds– Aerobic pondsAerobic ponds

• Shallow ponds; < 1 m deep

• DO maintained throughout depth (O2 supplied by algal photosynthesis)

– Anaerobic ponds• Deep ponds that receive high organic loadings; anaerobic conditions

throughout

• Process: – Acid fermentation – complex organic materials are broken down to short-– Acid fermentation – complex organic materials are broken down to short-

chain acids & alcohols

– Methane fermentation - acids & alcohols converted into gas (methan & CO2)

• Primarily used as pre-treatment (for high-temp, high-strength wastewater)

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• Oxidation ponds

– Facultative ponds– Facultative ponds

• 1-2.5 m deep

• Aerobic upper zone (photosynthesis, surface

reaeration), facultative middle zone, anaerobic lower

zone

• Rules of thumb in design• Rules of thumb in design

– BOD5 loading rate should not exceed 22 kg/ha-d

– Detention time: 6 months

77 78


• Oxidation ponds

– Maturation & tertiary ponds (polishing pond)– Maturation & tertiary ponds (polishing pond)

• Used for polishing effluents from biological processes

• DO is furnished through photosynthesis & surface

reaeration

– Aerated lagoons– Aerated lagoons

• Ponds oxygenated by surface or diffused air aeration

79


• Rotating biological contactors (RBCs)

– RBC process consists of a series of closely-spaced discs – RBC process consists of a series of closely-spaced discs (3-3.5 m dia, plastic) mounted on a horizontal shaft & rotated

• Half of surface area is immersed in water

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• Rotating biological contactors (RBCs)

– Process:– Process:

• Microbes adhere to rotating surface & grow there until entire surface is covered (1-3 mm thick layer slime)

• As discs rotate, film of wastewater is exposed to air (absorbing O2) – treating water

• As disc complete rotation, film of water mixes w/ water in reservoir (adding O2 to reservoir water)in reservoir (adding O2 to reservoir water)

• As attached microbes pass through reservoir, they absorb organics for breakdown

• Excess growth of microbes are sheared → clarification

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ce 152 lecture notes 3 - biological treatment

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