chapter 9 - biological waste water treatment

8/3/2019 Chapter 9 - Biological Waste Water Treatment

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Biological Wastewater Treatment

Chapter 9

Categories of wastes, Major treatment methods, Characteristics

of wastewaters, Biological waste treatment processes, Trickling

filters, RBC, Oxidation ponds, Anaerobic treatment


2/17

Chapter 7

Categories of Waste Materials1. Industrial waste

Produced by various industries, vary from industry to another

Contain HC, alcohol and aromatic organics

High C/N ratio; thus biological treatment requires

supplemental addition of N

2. Domestic waste

Treated by municipalities and produced by human and their

daily activities Include garbage, laundry waste, etc

Varies significantly with time in terms of flow and

composition3. Agricultural waste

Produced by farm animals and include waste plants and straw

Rich in C due to cellulosic material content


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Chapter 7Chapter 7

Major Waste Treatment Methods

1. Physical treatment Screening, flocculation, sedimentation, filtration and

flotation

Used for removal of insoluble materials

2. Chemical treatment

Chemical oxidants (chlorination and ozonation)

Chemical precipitation using CaCl2, FeCl3, Ca(OH)2 orAl2(SO4)3

3. Biological treatment Aerobic and anaerobic treatment of wastewater using a mixed

culture of microorganisms


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Chapter 7Chapter 7

Characteristics of Wastewater

1. Physical Characteristics

Such as color, odor, pH, temperature and solid contents

(suspended and dissolved solids)2. Chemical characteristics

Organic:

carbohydrates, lipids, HC, proteins, phenols, surfactants,

herbicides, pesticides and aromatic compounds

Inorganics:

Nitrogenous compounds (NH+4, NO-3)

Sulfur compounds (SO2-4, S2-, S0, S2-3) Phosphorous compounds (PO3-4, HPO

2+3, H2PO

-4)

Heavy metals (Ni, Pb, Cd, Fe, Cu, Zn, Hg)

Dissolved gases (H2S, NH3, CH4)

Selection of treatment method depends on the characteristics of the

water which should be known before treatment. Among these


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Chapter 7Chapter 7

Carbon Content

1. Biological Oxygen Demand (BOD) BOD5 represents the amount of dissolved oxygen (DO)

consumed when a wastewater sample is sealed with active

bacteria and incubated at 20oC for 5 days.

Since O2 consumed is proportional to the organic content of

wastewater BOD is a measure of the strength ofwastewater

The stoichiometric coefficient for this proportionality isunknown, since the composition of the organics is unknown.

Also, some nitrogen-containing or inorganic compounds will

exert an O2 demand

The carbon content (strength) of wastewater can be expressed inseveral ways:

1) BOD 2) COD 3) TOC


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Chapter 7

If the only organic compounds is glucose, O2 consumption can

be easily related to the carbon content of watewater under

aerobic conditions:

OHCOOOHC 2226126 666 ++

According to the stoichiometry of this reaction 1.07 g of O2is required for the oxidation of 1 g of glucose

Appropriate dilution is needed to obtain an accurate BOD5

measurement BOD5 is calculated using:

( ) ( )[ ] ( ) ( )[ ])O(noBlank550555055

2====

=tttt

BODBODBODBODBOD


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Chapter 7

2. Chemical Oxygen Demand (COD)

A measure of the concentration of oxidizable organic

compounds present in wastewater Almost all organics present in WW are oxidizable by certain

strong chemical oxidant ( COD > BOD5)

Typical chemical oxidation reaction:

O2 required for oxidation of organics can be calculated

The method is faster, easier and less expensive than BOD5measurement

OHCOCrHOCrOHCheat

cba 2232

72 ++++ ++

3. Total Organic Carbon (TOC)

Done by analyzer

Acidification of sample followed by carbon analyzer, infrared


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Chapter 7

Typical Wastewater Treatment Employing Biological Treatment

Primary Treatment

Removal of coarse solids and suspended matter (screening, sedimentation,

filtration)

Conditioning of the wastewater stream by pH adjustment and nutrient

addition (PO4, NH4) Secondary Treatment

Major steps of biological treatment includes biological oxidation or anaerobic

treatment of soluble and insoluble organic compounds.

Organic compounds are oxidized to CO2 and H2O by organisms under

aerobic conditions.

Unoxidized compounds and solids from aerobic are decomposed to a mixture

of CH4, CO2 and H2S under anaerobic conditions. Tertiary treatment

Includes removal of the remaining inorganic compounds (phosphate, sulfate,

ammonia) and other refractory organic compounds by one or more physical

separation methods, such as adsorption, deep-bed filtration, RO and EOD


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Chapter 7

Biological Waste Treatment Processes

Use of mixed culture either aerobic or anaerobic processes. Major aerobic processes:

1. Activated sludge processes

2. Trickling filter3. Rotating biological contactors

4. Oxidation ponds

Activated Sludge Processes Include a well-agitated and aerated continuous-flow reactor and

a settling tank

The concentrated cells from the settling tank are recycled backto the stirred-tank reactor

See the drawing of a schematic typical activated-sludge process

in the next slide.


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Chapter 7

A mixed culture of organisms is utilized in the bioreactor Some of the organisms may produce polymeric materials

(polysaccharides), which helps the organisms to agglomerate, i.e.floc formation.

Cell recycle from the sedimentation unit improves the volumetricrate of biological oxidation (i.e., high-density culture), and thusreduce residence time or volume of the sludge reactor for a given

feed rate.


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Chapter 7

Recycle ratio needs to be controlled to maximize BOD removal

rate

The selection of aerator and agitator is a critical factor in the

design of activated sludge processes.

aeration requirements vary depending on the strength of thewastewater and cell concentration

for typical activated sludge process, O2 requirement is about 30-

60 m3 O2/kg of BOD removal

Various aeration devices with and without mechanical agitation

can be used in activated sludge unit

The activated sludge faces many uncontrolled disturbances ininput parameters, such as waste flow and composition lead tosystem failure, i.e. less-than-adequate treatment of the waste

stream


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Chapter 7

One type of disturbances is referred to shock loading

Shock loading: sudden input (pulse) of a high concentration of

toxic compounds.

One response to disturbance is sludge buckling

Sludge buckling: formation of flocs that do not settle wellconsequently cell mass is not recycled

Buckling sludge often results from a change in the composition

of the microbial population in the treatment unit. For example,

filamentous bacteria may dominate the normal floc-formation

cells leading to small light floc.


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Chapter 7

Volume of activated sludge tank

This can be determined using the rate expressions for microbial

growth and substrate utilization and material balance for

biomass and substrate for a certain degree of BOD removal.

Since the activated sludge tank contains a mixed culture of

organisms, the actual kinetics of BOD removal are complicatedusually, interactions among various species are not known.The analysis below assumes

1. Pure culture, and2. Growth is governed by Monod equation

Monod with death (or endogeneous respiration rate) term can be

written as:

d

S

mnet k

SK

S

+=


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Chapter 7

Steady state biomass balance in an activated sludge:

( )FXFXXV r +=+ 1

or: ( )FXFXXVkSK

Srd

S

m

+=+

+1

Steady state substrate balance around activated sludge:

( )FSXVSK

S

YFSFS

S

m

SX

r

++

+=+ 1

1

/

0

(I)

(II)Similart

ochemosta

twith

recyclestreatm

Assuming no substrate utilization and no growth in the settling

tank (due to short residence time), material balance around the

settling tank yield:

Biomass: ( ) ( ) ( ) re FXFXFX ++=+ 11

Substrate: ( ) ( ) ( ) re FSFSFS ++=+ 11

(III)

(IV)


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Chapter 7

where: ;rateflowFeed

rateflowrecycleSludge=

rateflowFeed

flowsludgeExcess=

Assuming that the substrate is not separated in the settling tank

S = Se = Sr Equation IV is not needed

Re-arranging III: ( ) ( )rer

FXFXFXFX +=+ 11 (V)

Substituting in I: ( ) renet FXFXVX += 1

Defining

c

net

1

=

Used to calculate the

cellular residence time in

the tank

where c is cells residence time

( ) renetc

XFXF

VX

+==

1

1

- c is controlled by operator choice of the recycle flow rate

(VI)


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Chapter 7

Hydraulic (liquid) residence time is:

( )[ ] ( )X

XX

X

XX

F

V

net

rerec

H

+=

+==

11

Substituting Sr= S in II:

( )

(VII)

XVY

SSFg

SX

/

0

1=

(VIII)( ) ( )

( )cdcSX

g

SX

kX

SSFY

X

SSFYV

+

=

=

1

0/0/

or:

Equation (VIII) used to calculate the required volume of the

sludge tank for a certain degree of BOD removal, i.e. (S0-S)

Vcan also be expressed in terms of by substituting (V) in (VII):

+=

X

XFV rc 1


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Chapter 7

Example: An industrial waste with an inlet BOD5 of 800 mg/l must be treated to

reduce the exit BOD5 to less than 20 mg/l.

F= 400 m3/h; m = 0.2 h-1; KS = 50 mg/l BOD5; YX/S = 0.5 mg solid/mg BOD5

kd= 0.005 h-1; V= 3200 m3; = 0.4; Xe = 0.0; c = 120 h

Find S? and determine if sufficient BOD5 removal is attained in a well-mixed

activated-sludge process to meet specification. What will be X and the sludgereduction rate from this process?

Solve for S:

d

S

m

c

net kSK

S

+==

1

< 20 mg/l

)()1(

dmc

cdS

kkKS

+=

where:

( )( ) 1005.02.0120

120005.0150

+=S S = 3.57 mg/l

( )( )cdcSX

kXSSFYV

+ = 1

0/

Amount of sludge produced:

( )120005.01)57.3800(4001205.0

3200+

=

X

X= 3733 mg/l

( )

X

XX

F

V

net

reH

+==

1

1/net = c; Xe = 0

c

r

VXXF

= = sludge production rate

h201

)l/mmg/l)(1000)(3733m(3200 33=rXF

= 9.95 107 mg/h

= 99.5 kg/h

chapter 9 - biological waste water treatment

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