Program Studi Teknik Lingkungan FTSP ITS

Activated Sludge Processes

MK Unit Proses Teknik Lingkungan

Suspended Growth

•

Organisms are suspended in the treatment

basin fluid.

•

This fluid is commonly called the “mixed

liquor”.

Suspended Growth:

• Activated sludge

• Oxidation ditch/pond

Activated Sludge

• Process in which a mixture of wastewater &

microorganisms is agitated & aerated

• Leads to oxidation of dissolved organics

• After oxidation, sludge is separated from

wastewater

Activated Sludge

• Designed based on loading (the amount of organic

matter added relative to the microorganisms

available)

• Commonly called the food-to-microorganisms ratio, F/M

• _{F measured as BOD. M}

measured as volatile suspended solids

concentration

• _{F/M is the pounds of}

Activated sludge flocs

Note filmentous blcteril

Note

Vorticella

lnd other

Oxygenated systems

Settling tanks

Nitrogen removal

 _{Nitrifcltion (}Nitrosomonas lnd Nitrobacter)

NH₃ + O₂  NO₂-  NO 3

- _{Denitrifcltion}

NO₃- + orglnics _ CO

2 +

N₂

 _{Process ldlptltions}

Anoxi

c Aerobic

Phosphate removal

 _{BNR pilnts}

 _{Disclrding phosphlte lnlerobicliiy}

 _{Luxury lerobic uptlke of P in lerobic stlge}

 _{Process ldlptltions for N lnd P removli}

Anlero

bic Anoxic Aerobic

Excess biomass disposal

 _Production

 _Seplrltion

 _{Further bioiogicli treltment –}

(ln)lerobic

 Dewltering

 _{Drying –} soilr or gls helted

 _{Disposli/ benefcili use –} soii lmender/ fertiiizer

or fuei

The cost of biomlss disposli lmount to lbout hlif the cost of wlstewlter treltment. Aerltion, if used, limost up to hlif of the rest of the cost. If no lerltion, the clpitli cost , inciuding the

Type of Activated Sludge

•

Activated sludge without cell recycle

Design of Activated Sludge

•

Influent organic compounds provide the food for

the microorganisms and is called substrate (S)

•

The substrate is used by the microorganisms for

growth, to produce energy and new cell material.

•

The rate of new cell production as a result of

the use of substrate may be written

mathematically as:

•

Y is called the yield and is the mass of cells

produced per mass of substrate used (g SS/g BOD)

dt dS Y

dt dX

Monod Model for Substrate Utilization

S

K

S

s m









S

K

SX

X

dt

dX

s m











dt

dS

Y

dt

dX



)

S

K

(

Y

SX

Y

.

dt

dX

dt

dS

s m







1



:

ACTIVATED SLUDGE

Mean Cell Residence Time, θ

_c

Mean cell residence time (MCRT, θ_c) is the mass of cells in the system divided by the mass of cells wasted per day.

Consider the system:

At steady state, the amount of solids wasted per day must equal the amount produced per day:

For no recycle systems, θ_{c =} θ_h

Q

V

QX

VX

c







dt dS Y

X

V dt dS Y

VX

V dt dX

VX

c   

Mass Balance on Microorganisms:

In steady state condition  (dX/dt) V = 0, and QX₀ = 0 Accumulation = input – output – process

dibagi VX

Jika k_d diabaikan 

XV

k

V

dt

dS

Y

QX

QX

dt

dX

V



_o







_d

) S K ( Y SX dt dS s m    d s m c k S K S      1

1





c m s

K

S





XV k V S K SX QX _d s m       0 d s m _k S K S V Q       0  c  d 

Example

Solve for θ_c:

V = θ

_c

Q = 12.75 (3) = 38.25 m

3

A CSTR without cell recycle receives an influent with 600 mg/L BOD at a rate of 3 m3/day. The BOD in the effluent

must be 10 mg/L. The kinetic constants are: K_s = 500 mg/L and μ_m = 4 days-1. How large should the reactor be?

1





c m

s

K

S





days

75

.

12

4

*

10

10

500











m s c

S

S

K





Q

V

Given the conditions in the previous example, What

would the percent reduction in substrate be if the

reactor volume was 24 m

3

?

Reduction = [(600 – 16.1)/600] x 100 = 97.3%

days

8

3

24







Q

V

c



mg/L

1

.

16

1

8

*

4

500

1











c m

s

K

S

Now consider a CSTR with cell recycle:

Since X_e = 0:

Removal of substrate often expressed in terms of substrate removal velocity, q:

Mass balance on microorganisms:

X₀ = X_e = 0

The substrate removal velocity, q, can also be expressed as: q = μ/Y

By substitution:

But q is also equal to:

XV k XV X ) Q Q ( X Q QX dt dX

V  _o  _{w r}   _{w e}   _d

d c

d w

r _{k k}

XV Q X       1 S K S s m     since S K S Y q s m   1 

Since q = μ / Y

If we equate these two equations for q and solve for S₀ – S:

Hydraulic retention time:

Solids Separation

The success of the activated sludge process depends on the

efficiency of the secondary clarifier, which depends on the settling characteristics of the sludge (biosolids).

Some system conditions result in sludge that is very difficult to settle. In this case the return activated sludge becomes thin (low MLSS) and the concentration of organisms in the aeration tank

goes down. This produces a higher F/M ratio (same food input, but fewer organisms) and a reduced BOD removal efficiency.

One condition that commonly causes this problem is called bulking sludge. Bulking sludge occurs when a type of bacteria called

filamentous bacteria grow in large numbers in the system. This produces a very billowy floc structure with poor settling

Sludge Volume Index, SVI

SVI =

(volume of sludge after 30 min. settling, ml) x 1000

mg/L suspended solids

A mixed liquor has 4000 mg/L suspended solids. After 30

minutes of settling in a 1 L cylinder, the sludge occupied 400 ml.

SVI = (400 x 1000)/ 4000 = 100

Problem

The hydraulic retention time may be found from the following equation:

θ_h = [0.5(300 – 30)(200 + 30)] / [2 (30) (4000)] = 0.129 days = 3.1 hr An activated sludge system operates at a flow rate of 400 m3_{/day and has an}

influent BOD of 300 mg/L. The kinetic constants for the system have been determined to be: K_s = 200 mg/L, Y = 0.5 kg SS/kg BOD, μ_m = 2 day-1_{. The}

mixed liquor suspended solids concentration will be 4000 mg/L. IF the system must produce an effluent with 30 mg/L BOD, determine:

A. The volume of the aeration tank B. The sludge age (MCRT)

C. The quantity of sludge wasted per day

)

(

K

S

Y

SX

S

S

s

h m

o











SX

S

K

S

S

Y

m

s o

h



θ_c = 1/ (qY)

q = (S₀ – S) / (X θ_h ) = (300– 30) / [(4000)(0.129)]

= 0.523 (kg BOD removed/day) / (kg SS in the reactor)

θ_c = 1/ (qY) = 1 / (0.523 x 0.5) = 3.8 days

Also θ_c = (X V) / (X_r Q_w)

X_r Q_w = (X V) / θ_c = [(4000)(51.6)( 103 L/m3)( 1/106 kg/mg)] / 3.8

= 54.3 kg/day

Using the same data what MLSS is necessary to produce an effluent concentration of 15 mg BOD/L?

= [2(15)] / [0.5(200 + 15)] = 0.28 day-1

= (300 – 15) / [0.129(0.28)] = 7890 mg/L

θ_c = 1 / (q Y) = 1 / [0.28(0.5)] = 7.2 days q = (μ_m S) /[Y(K_s + S)]

Ringkasan

Persamaan-persamaan yang digunakan

dalam proses

activated sludge:

1.

TANPA RECYCLE:

d h k     1 c d o

k

)

S

S

(

Y

X









1

d s m c k S K S      1 Q V h c 







1

1









)

k

(

)

k

(

K

S

d m c c d s







h d h o

Y

Y

k

X

)

S

S

(





1

1







) S S ( X Y o h    

S

K

S

s m









S K S

k _{m m}

s h d h









1

1 _  

•

Mencari

k

_d

dan

Y

dari percobaan lab.:

•

Menclri

K

_s

dln

μ

_m

dlri percoblln

ilb.:

TANPA RECYCLE h d h o

Y

Y

k

X

)

S

S

(





1

1







S K S

k _{m m}

s h d h









1

1 _  

      h o X S S  )

( _ slope _Y1

h



1

Y k_d

intercept _

S k_{d h}

2. DENGAN RECYCLE:

d c k     1 c d o c h

k

)

S

S

(

Y

X













1

d s m c k S K S      1 Q V h 



1

1









)

k

(

)

k

(

K

S

d m c c d s







d e w w _k VX X ) Q Q ( X Q     

S

K

S

s m









) (S S X Y o h     c d h o

Y

Y

k

X

)

S

S

(





1

1







S K S

k _{m m}

s c d c









1

1 _  

•

Mencari k

_e

danY dari percobaan lab.:

•

Menclri K

_s

dln

μ

_m

dlri percoblln

ilb.:

RECYCLE c d h o

Y

Y

k

X

)

S

S

(





1

1







S K S

k _{m m}

s c d c









1

1 _  

      h o X S S  )

( _ slope _Y1

c



1

Y k_d

intercept _

S k_{d c}

Parameters Recycle Non-Recycle

Siudge lge & hydrluiic residence time Microorglnism in system Substrlte in system

Determining k_d &

Y

Determining K_s &

 _m

F-M rltio

Q V

h c  

 Q

V

h c  

 c d o h c k ) S S ( Y X      

1 d c

o k ) S S ( Y X     1 1 1     ) k ( ) k ( K S d m c c d s    1 1     ) k ( ) k ( K S d m c c d s    c d h o Y Y k X ) S S (   1 1    S K S

k _{m m}

s c d c     1

1 _  

      h d h o Y Y k X ) S S (   1 1    S K S

k _{m m}

s h d h     1

1 _  

Recycle Ratio

Mass Balance at Junction

•

Microorganism

•

Substrate

Xo = 0

e r o r r o

o

S

Q

S

(

Q

Q

)

S

Q







X

)

Q

Q

(

X

Q

X

Q

_{o o}



_{r r}



_o



_r

•

Sludge Production

S

_R

= (S

_o

- S

_e

).Q

_o

•

Oxygen Requirement

Y’ = oxygen coeff, mass

oxygen/mass substrate utilized = 1 – 1,42 Y

k_d‘ = endogenous respiration coeff, mass oxygen/

mass cell-day = 1,42 k_d O_n = oxygen for nitrification

= mass N x 3.84

VX

k

YS

P

_x



_R



_d

n d

R

R

Y

'

S

'k

VX

O

Soal-soal

• Suatu percobaan proses lumpur aktif dalam skala

laboratorium yang dioperasikan secara Batch, dengan waktu aerasi 24 Jam dan diketahui nilai MLVSS = 70 % dari nilai MLSS. Hitunglah nilai Y dan kd dengan acuan masa MLVSS. Data hasil

percobaan :

No. Reaktor

X_o (MLSS)

(mg/L)

X_t (MLSS)

(mg/L)

So (BOD₅)

(mg/L)

St (BOD₅)

(mg/L) 1

2 3 4

450 860 1650 3670

790 1160 1960 3875

725 725 725 725

Suatu kawasan pemukiman dengan jumlah penduduk 8000 jiwa dilayani dengan sistem IPAL terpusat. Kapasitas air buangan sebesar 225

L/orang.hari dan rata-rata BOD sebesar 425 mg/L. Konsentrasi NH₃-N sebesar 25 mg/L dan teroksidasi sebesar 95 %. Susunan sel C₅H₇NO_2. Removal BOD pada pengendap I diperkirakan sebesar 32 %. Jika

diinginkan reaktor utama adalah proses Lumpur Aktif, dengan nilai : Y = 0,81 kg VSS/ kg BOD; k_e = 0,07/hari ; Y’ = 0,73 kg O₂/kg BOD;

k_e’ = 0,16 kg O₂/kg VSS; X = 2500 mg MLSS/L ; Xr = 10.000 mg/L

MLVSS/MLSS = 0,7, umur lumpur = 10 hari , waktu tinggal hidrolik = 8 jam.

Hitunglah :

• Konsentrasi BOD effluen. • Produksi lumpur

• Kebutuhan Oksigen.

Air limbah industri rumah tangga diolah dengan Proses lumpur aktif. Karakteristik air buangan dan Parameter perencanaan ditentukan sbb:

• Q = 6.000 m3/hari.

• BOD₅ influen (So) = 320 mg/ L • BOD₅ effluen (Se) = 20 mg/ L • Umur Lumpur = 7 hari,

• Konsentrasi mikroorganisme dalam reaktor (X) = 2500 mg VSS/L

• SVI = 85

• Y = 0,6 kg VSS/kg BOD k_e = 0,04 • MLVSS = 0,75 MLSS.

1. Hitung kebutuhan Oksigen untuk proses lumpur aktif tersebut 2. Hitung nilai ratio resirkulasi (R = Qr/Qo)