water pollution18!01!2012

7/31/2019 Water Pollution18!01!2012

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Water Pollution

Makarand GhangrekarDepartment of Civil Engineering

IIT Kharagpur


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Contributing Faculty

M.M. Ghangrekar, Ph.D.

Associate Professor

Department of Civil Engineering

(Section-2 Coordinator)

A K Gupta, Ph.D.

Associate Professor,

Civil Engineering

M K Dash, Ph.D.

Assistant Professor,

Oceans, Rivers, Atmosphere and Land Sciences

A. Mukharjee, Ph.D.

Assistant Professor, Geology and Geophysics


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Tentative time schedule for teaching of the subject Environmental Science (2-0-0)

Section-IIParticipating Faculty Load, h

1 M.M. Ghangrekar (MMG) 8

2 Mihir Das (MD) 7

3 A.K. Gupta (AKG) 6

4 A. Mukharjee (AM) 5

Sr. No. Date Faculty Topics to be covered

1 27-07-2011 MD

Water pollution (Rainfall runoff relationship, ground water contamination,

pollution prevention, water quality in lakes and reservoirs, concept of

watershade management)

2 3-08-2011 MMG

Water pollution (Remaining topics: O2 transfer, DO, BOD, Water quality

standards, sources of pollution, classification, effect, self purification)

3 10-08-2011 MMG Water and Wastewater Treatment4 17-08-2011 MMG Water and Wastewater Treatment

5 24-08-2011 MMG Water and Wastewater Treatment, Class Test 45 min

6 7-09-2011 MD Ecology

7 14-09-2011 AM Waste minimization (1h, MD); Solid and Hazardous waste management (1 h, AM)

8 21-09-2011 AM Solid and Hazardous waste management

Mid-sem from 23/09/2011 to 30/09/2011

9 12-10-2011 AKG Air Pollution

10 19-10-2011 AKG Air Pollution

11 2-11-2011 MD Noise pollution

12 9-11-2011 AKG EIA + Remaining topic if any

13 16-11-2011 AM Soil Pollution14 End semester examination from 18-11-2011


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Drinking Water Quality Standards

Bacterialogical examination For samples collected from distribution system: (IS:10500:1991)

Throughout any year, 95% of samples collected from distributionsystem should not contain any coliform organisms in 100 mL;

No sample should contain E.Coli in 100 mL.

No sample should contain more than 10 coliform organisms per 100mL; and

Coliform organisms should not be detected in 100 mL of any twoconsecutive samples.

For unpiped water supplies: The objective is to reduce thecoliform count below 10 per 100 mL and no faecal coliformshould remain present.


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Water Quality Standards

Substance or characteristic Desirable limit Permissible limit in absence

of alternative source

Colour, Hazen units, max 5 25

Odour Unobjectionable --

Taste Agreeable --

Turbidity, NTU, Max 5 10

pH 6.5 8.5 No relaxation

Total hardness, mg/L as CaCO3, max 300 600

Iron, mg/L of Fe, max 0.3 1.0

Chloride, mg/L as Cl, max 250 1000

Fluoride, mg/L as F, max 1.0 1.5


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Water Quality StandardsSubstance or characteristic Desirable limit Permissible limit in

absence of alternative

source

Residual free chlorine, mg/L, min 0.2 --

Desirable characteristics

Dissolved solids, mg/L, max 500 2000

Manganese, mg/L, max 0.1 0.3

Sulphate, mg/L, max 200 400

Nitrate, mg/L, max 45 No relaxation

Arsenic, mg/L as As, max 0.01 No relaxation

Cyanide, mg/L as CN, max 0.05 No relaxation

Lead, mg/L, max 0.05 No relaxation


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Sources of pollution

Domestic wastewater

Industrial wastewaters

Agriculture runoff

Storm water runoff

Bathing, cloth washing, recreational activities,

etc., in water bodies


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Classification and Effect of water

pollutants The various types of water pollutants can be classified in to following

major categories:

1) Organic pollutants,

2) Pathogens,

3) Nutrients and agriculture runoff,

4) Suspended solids and sediments,

5) Inorganic pollutants (salts and metals),

6) Thermal Pollution

7) Radioactive pollutants.


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ORGANIC POLLUTANTS:

a) Oxygen Demanding wastes: The wastewaters such as, domesticand municipal sewage, wastewaters from food processing industries,canning industries, slaughter houses, paper and pulp mills, tanneries,etc., have considerable concentration of biodegradable organiccompounds either in suspended, colloidal or dissolved form.

These wastes undergo degradation and decomposition by bacterialactivity.

The dissolved oxygen available in the water body will be consumed foraerobic oxidation of organic matter present in the wastewater.

Hence, depletion of the DO will be a serious problem adversely affectingaquatic life, if the DO falls below 4.0 mg/L. This decrease of DO is an

index of pollution.


pollutants


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b) Synthetic Organic Compounds Synthetic organic compounds are also likely to enter the ecosystem

through various manmade activities such as production of thesecompounds, spillage during transportation, and their uses in differentapplications.

These include synthetic pesticides, synthetic detergents, foodadditives, pharmaceuticals, insecticides, paints, synthetic fibers,

plastics, solvents and volatile organic compounds (VOCs). Most of these compounds are toxic and biorefractory organics i.e.,

they are resistant to microbial degradation.

Even concentration of some of these in traces may make water unfit for

different uses. The detergents can form foams and volatile substances may cause

explosion in sewers.

Some of these compounds are exceedingly persistent and their

stability to chemical reagents is also high.


pollutants


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c) Oil Oil is a natural product which results from the plant remains fossilized over

millions of years, under marine conditions. It is a complex mixture ofhydrocarbons and degradable under bacterial action.

The biodegradation rate is different for different oils, tars being one of theslowest.

Oil enters in to water through oil spills, leak from oil pipes, and wastewaterfrom production and refineries.

Being lighter than water it spreads over the surface of water, separatingthe contact of water with air, hence resulting in reduction of DO.

This pollutant is also responsible for endangering water birds and coastalplants due to coating of oils and adversely affecting the normal activities.

It also results in reduction of light transmission through surface waters,thereby reducing the photosynthetic activity of the aquatic plants.

Oil includes polycyclic aromatic hydrocarbons (PAH), some of which areknown to be carcinogenic.


pollutants


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Mexico Oil spill, 2010

Pelican covered with oil


pollutants


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2) PATHOGENS

The pathogenic microorganisms enter in to water body throughsewage discharge as a major source or through the wastewaterfrom industries like slaughterhouses.

Viruses and bacteria can cause water borne diseases, such ascholera, typhoid, dysentery, polio and infectious hepatitis in

human.


pollutants


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3) NUTRIENTS The agriculture run-off, wastewater from fertilizer industry and sewage

contains substantial concentration of nutrients like nitrogen andphosphorous.

These wastewater supply nutrients to the plants and may stimulate thegrowth of algae and other aquatic weeds in receiving waters.

Thus, the value of the water body is degraded.

In long run, water body reduces DO, leads to eutrophication and endsup as a dead pool of water.

People swimming in eutrophic waters containing blue-green algae canhave skin and eye irritation, gastroenteritis and vomiting.

High nitrogen levels in the water supply, causes a potential risk,especially to infants under six months. This is when themethaemoglobin results in a decrease in the oxygen carrying capacityof the blood as nitrate ions in the blood readily oxidize ferrous ions in

the hemoglobin.


pollutants


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4) SUSPENDED SOLIDS AND SEDIMENTS

These comprise ofsilt, sand and minerals eroded from land.

These appear in the water through the surface runoff during rainy seasonand through municipal sewers.

This can lead to the siltation, reduces storage capacities of reservoirs.

Presence of suspended solids can block the sunlight penetration in the

water, which is required for the photosynthesis by bottom vegetation.

Deposition of the solids in the quiescent stretches of the stream orocean bottom can impair the normal aquatic life and affect the diversityof the aquatic ecosystem.

If the deposited solids are organic in nature, they will undergodecomposition leading to development of anaerobic conditions.

Finer suspended solids such as silt and coal dust may injure the gills offishes and cause asphyxiation.


pollutants


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5) INORGANIC POLLUTANTS Apart from the organic matter discharged in the water body through

sewage and industrial wastes, high concentration of heavy metals andother inorganic pollutants contaminate the water.

These compounds are non-biodegradable and persist in theenvironment.

These pollutants include mineral acids, inorganic salts, traceelements, metals, metals compounds, complexes of metals withorganic compounds, cyanides, sulphates, etc.

The accumulation of heavy metals may have adverse effect onaquatic flora and fauna and may constitute a public health problem

where contaminated organisms are used for food. Algal growth due to nitrogen and phosphorous compounds can be

observed.

Metals in high concentration can be toxic to biota e.g. Hg, Cu, Cd, Pb,

As, and Se. Copper greater than 0.1 mg/L is toxic to microbes.


pollutants


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6) THERMAL POLLUTION

Considerable thermal pollution results due to discharge of hotwater from thermal power plants, nuclear power plants, andindustries where water is used as coolant.

As a result of hot water discharge, the temperature of water body

increases, which reduces the DO content of the water. This alters the spectrum of organisms, which can adopt to live

at that temperature and DO level.

When organic matter is also present, the bacterial actionincreases due to rise in temperature, hence, resulting in rapiddecrease of DO.

The discharge of hot water leads to the thermal stratification in

the water body, where hot water will remain on the top.


pollutants


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7) RADIOACTIVE POLLUTANTS

Radioactive materials originate from the following:

Mining and processing of ores,

Use in research, agriculture, medical and industrial activities, suchas I131, P32, Co60, Ca45, S35, C14, etc.

Radioactive discharge from nuclear power plants and nuclearreactors, e.g., Sr90, Cesium Cs137, Plutonium Pu248.

Uses and testing of nuclear weapons,

These isotopes are toxic to the life forms; they accumulate in thebones, teeth and can cause serious disorders.

The safe concentration for lifetime consumption is 1 x 10-7

microcuries per ml.


pollutants


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Oxygen transfer

Solubility of gases The principal atmospheric gases that go into solution are

oxygen, nitrogen, carbon dioxide. All water exposed to atmosphere will have these gases in

solution.

Other gases in solution are ammonia (NH3

), hydrogen sulphide(H2S), and methane (CH4), which are associated with microbialactivity.

The amount of particular gas dissolved in water depends upon:

Its solubility in water Its partial pressure at the air/water interface

Temperature of water

Level of salts in the water


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If water contains as much as specific gas that it can holdin the presence of an abundant supply, then the water is

said to be saturated. E.g., saturation concentration of oxygen in water at 20 oC is 9.3

mg/L.

If oxygen content is 7.5 mg/L at 20 oC, it is 80% saturated.

When oxygen content exceeds 100%, it is said to besupersaturated.

Supersaturation occurs because of:

High photosynthetic activity (in summer)

When high temperature water is discharged to river, the rapidrise in water temperature causes O2 supersaturation.

Oxygen transfer


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The solubility of gases in water is related to the partial pressure of

the gas in the atmosphere above the water by Henrys law.

Pg = Kh.Xg

Where, Pg = partial pressure of gas, atm.

Kh = Henrys law constant, atm/mole

Xg = equilibrium mole fraction of dissolved gas

Xg = Mole of gas (ng) / {mole of gas (ng) + mole of water (nw)}

For example: As air contains nearly 21% O2, the partial pressure of

O2 0.21 atm

Oxygen transfer


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Henrys law constant for common gases soluble

in water, Kh x 10-4

atm/moleTemperature N2 O2 CO2 H2S CH4

0 5.29 2.55 0.073 0.027 2.24

10 6.68 3.27 0.104 0.037 2.97

20 8.04 4.01 0.142 0.048 3.76

30 9.24 4.75 0.186 0.061 4.49

Oxygen transfer


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Example: Determine saturation concentration of O2 in water at 10oC at 1 atm.

Solution:

O2 is 21% in air (v/v), hence, Pg = 0.21 x 1 atm = 0.21 atm

Henrys law constant Kh = 3.27 x 104 atm/mole at 10 oC.

Now Pg = Kh . Xg Therefore Xg = Pg/Kh

Xg = 0.21/(3.27 x 104

) = 6.42 x 10-6

(equilibrium mole fraction) Since, 1 mole of water is 18 g, hence mole/L for water (nw) =

1000/18 = 55.6 mole/L

Now = Xg = ng / (ng + nw)

Solving for ng, ng = 3.57 x 10-4 mole/L

Saturation concentration of O2:

Cs = ng.M

Where, M = molecular wt. of oxygen

Cs = 3.57 x 10-4 mole/L x 32 g/mole x 103 mg/g = 11.4 mg/L

Oxygen transfer


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DO and BOD

Effect of Oxygen Demanding Wastes on Rivers

Depletion of dissolved oxygen is a major problem duedischarge ofoxygen demanding organic or inorganicpollutant in the surface water.

This poses threat to higher forms of aquatic life, if the

concentration of oxygen falls below a critical point. To quantify how much oxygen will be depleted, it is necessary to

know the quantity of oxygen demanding waste and how muchoxygen will be required to degrade the waste.

Although, oxygen is getting depleted for the degradation oforganic matter, it is continuously being replenished from theatmosphere and through photosynthesis, if any.

The concentration of oxygen is determined by the relative rates

of these competing processes.


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Estimation of organic content of the

wastewater The organic matter present in the water body

can be analyzed in laboratory by determining:

Boichemical Oxygen Demand (BOD),

Chemical Oxygen Demand (COD), and

by determination of Total Organic Carbon (TOC).

DO and BOD


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BOD

Biochemical Oxygen Demand (BOD) The BOD can be defined as the oxygen required for biochemical

oxidation of organic matter present in the water under aerobic

conditions.

The test is performed under the conditions similar to those in actual natural

water to measure indirectly the amount of biodegradable organic

matter present.

A water sample is inoculated with bacteria that consume the biodegradable

organic matter to obtain energy for their life processes.

The organisms also utilize oxygen in the process of consuming the organicmatter, the process is called as aerobic decomposition.

This oxygen consumption is measured; more is the organic matter

concentration more is the amount of oxygen utilized.


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The BOD test is performed for the following:

To determine quantity of oxygen required for biochemical stabilization oforganic matter

To determine suitability of biological treatment method, depending onCOD/BOD ratio, and sizing the treatment units.

To measure efficiency of the process

To determine compliance with wastewater discharge permits.

During the BOD test the organic matter will be converted into stable endproduct such as CO2, sulphate (SO4), orthophosphate (PO4) and nitrate(NO3).

The simple representation of carbonaceous BOD can be explained asbelow:

microorganisms

Organic matter + O2 CO2 + H2O + New Cells + Stable product

BOD


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The 20 oC temperature used is an average temperature valuetypically for slow moving streams in temperate climates.

Different results would be obtained at different temperaturesbecause biochemical reaction rates are temperature dependent.

The BOD bottles are incubated for 5 days and the initial DO and

final DO is measured for BOD determination.

The BOD bottles contain, seed (aerobic bacteria), diluted sample inaerated water, pH buffer, and nutrients.

For wastewater like sewage, within 20 days period, the oxidation ofcarbonaceous organic matter is about 95 to 99% complete.

In the first five days, the period used for BOD determination, 60 to

70% oxidation is complete.

BOD


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BOD

Idealized carbonaceous oxygen demand: (a) the BOD remaining as a

function of time, and (b) the oxygen consumed.


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The biochemical oxygen demand is represented as BOD5 20oC, which

indicate the total amount of oxygen consumed for biochemical oxidation oforganic matter for first five days at 20oC incubation temperature.

Since, the saturation value of DO for water at 20oC is only 9.1 mg/L it isusually necessary to dilute the samples to keep final DO level, at the endof incubation period, above 1.5 mg/L.

Hence, according to BOD values expected for that wastewater appropriatedilution should be carried out.

The 5 day BOD of unseeded diluted sample =

Where, DOi and DOfare initial and final DO of diluted wastewatersample.

p is the dilution fraction = Volume of wastewater (Sample)Volume of wastewater + volume dilution water

p

DODO fi

BOD


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The total volume of the BOD bottle used for test is usually 300 mL.

Sufficient amount of seed is added to the BOD bottle to ensure adequateconcentration of bacterial population to carry out the biodegradation.

Usually 1 to 2 mL of sewage per liter is considered as sufficient to act as aseed.

In such case it is necessary to subtract the oxygen demand of the seed

from the mixed sample.

Thus, the BOD of the wastewater with seeded sample can be worked out asbelow:

BODw = Where,

DOi an DOf = DO of mixture, initial and final values, respectively.

Bi and Bf = DO of blank, initial and final values, respectively

p = Vw/Vm = Volume of wastewater in mixture / Total volume of mixture

p

pBBDODO fifi )1)(()(

BOD


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Example: A test bottle containing only seeded dilution water has its DO level

drop by 1.0 mg/L in a 5-day incubation. A 300 mL BOD bottle filled

with 15 mL of wastewater and the rest seeded dilution water

experiences a drop of 7.2 mg/L in the same time period. What

would be five day BOD of the wastewater?

Solution: Dilution factor p = 15/300 = 0.05

Therefore, BOD5 = [7.2 1.0 (1 0.05)] / 0.05 = 125 mg/L

BOD


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BOD MODEL It is generally assumed that the rate at which the oxygen is

consumed is directly proportional to the concentration of degradableorganic matter remaining at any time.

The kinetics of BOD reaction can be formulated in accordance withfirst order reaction kinetics as:

d Lt / d t = - K Lt

Where, Lt = amount of first order BOD remaining in wastewater at

time t, and K = BOD reaction rate constant, time-1

Integrating =t

tt dtKLdL0

.

BOD


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i.e.,

Lt / Lo = e-K.t = 10-K.t

Where Lo or BODL at time t = 0, i.e., ultimate first stage BOD initially

present in the sample.

The relation between K(base e) and K (base 10) is

K(base 10) = K(base e) / 2.303

The amount of BOD remaining at time t equals

Lt = Lo (e-k.t)

[ ] tKL tt .log 0 =

BOD

BOD


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The amount of BOD that has been exerted (amount of oxygenconsumed) at any time t is given by:

Yt = Lo Lt = Lo (1 e-k.t)

And the five day BOD is equal to:

Y5 = Lo L5 = Lo (1 e-5k)

For polluted water and wastewater, a typical value of K (base e,20oC) is 0.23 per day and K (base 10, 20oC) is 0.10 per day.

These values vary widely for the wastewater in the range from 0.05to 0.3 per day (base 10).

BOD

BOD


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The ultimate BOD (Lo) is defined as the maximum BOD exerted by thewastewater. It is difficult to assign exact time to achieve ultimate BOD,and theoretically it takes infinite time.

The time required to achieve the ultimate BOD depends upon thecharacteristics of the wastewater, i.e., chemical composition of the organicmatter present in the wastewater and its biodegradable properties.

Oxygen depletion is related to both the ultimate BOD and the BOD rateconstant (K).

The ultimate BOD will increase in direct proportion to the concentration ofbiodegradable organic matter.

The BOD reaction rate constant is dependent on the following: The nature of the waste

The ability of the organisms in the system to utilize the waste

The temperature

BOD


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Self purification

SELF PURIFICATION OF NATURAL STREAMS

The amount ofdissolve Oxygen (DO) in water is one of the

most commonly used indicators of a river health.

As DO drops below 4 or 5 mg/L the forms of life that can survivebegin to be reduced.

A number of factors affect the amount of DO available in a river.

Oxygen demanding wastes remove DO; plants add DO duringday but remove it at night; respiration of organisms removes

oxygen.

In summer, rising temperature reduces solubility of oxygen, whilelower flows reduce the rate at which oxygen enters the waterfrom atmosphere.


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Factors Affecting Self Purification

Dilution: When sufficient dilution water is available in the receiving waterbody, where the wastewater is discharged, the DO level in the receiving

stream may not reach to zero due to availability of DO. Current: When strong water current is available, the discharged wastewater

will be thoroughly mixed with stream water preventing deposition of solids. Insmall current, the solid matter from the wastewater will get deposited at bedfollowing decomposition and reduction in DO.

Temperature: The quantity of DO available in stream water is more in coldtemperature than in hot temperature. Also, as the activity of microorganismsis more at the higher temperature, hence, the self-purification will take lesstime at hot temperature than in winter.

Sunlight:Algae produces oxygen in presence of sunlight due to

photosynthesis. Therefore, sunlight helps in purification of stream by addingoxygen through photosynthesis.

Oxidation: Due to oxidation of organic matter discharged in the river DOdepletion occurs. This rate is faster at higher temperature and low at lowertemperature.

Self purification


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Figure *.* Oxygen sag curve

The variation of oxygen deficit (D) with the distance along the

stream, and hence the time of flow from the point of pollution isdepicted by the Oxygen Sag Curve.

Self purification


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The oxygen sag or oxygen deficit in the stream at any point of time during

self purification process is the difference between the saturation DO content

and actual DO content at that time.

Oxygen deficit, D = Saturation DO Actual DO

The saturation DO value for fresh water depends upon the temperature andits value varies from 14.62 mg/L at 0oC to 7.63 mg/L at 30oC, respectively.

Deoxygenation: When wastewater is discharged in to the stream, the DO

level in the stream goes on depleting. This depletion of DO content is

known as deoxygenation.

Self purification


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When the DO content of the stream is gradually consumed dueto BOD load, atmosphere supplies oxygen continuously to thewater, through the process of reaeration or reoxygenation, i.e.,

along with deoxygenation reaeration is continuousaly takesplace.

The rate of reoxygenation depends upon:

Depth of water in the stream (more for shallow depth)

Velocity of flow in the stream (less for stagnant water)

Oxygen deficit below saturation DO;

Temperature of water

Self purification

water pollution18!01!2012

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