Water Pollution

Water Pollution

Makarand GhangrekarDepartment of Civil Engineering

IIT Kharagpur

Water Quality Standards

• Bacterialogical examination – For samples collected from distribution system:

• Throughout any year, 95% of samples collected from distribution system 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 100 mL; and

• Coliform organisms should not be detected in 100 mL of any two consecutive samples.

– For unpiped water supplies: The objective is to reduce the coliform count below 10 per 100 mL and no faecal coliform should remain present.

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

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

Sources of pollution

• Domestic wastewater

• Industrial wastewaters

• Agriculture runoff

• Storm water runoff

• Bathing, cloth washing, etc., in water bodies

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.

• ORGANIC POLLUTANTS:

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

– These wastes undergo degradation by bacterial activity.

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

– Hence, depletion of the DO will be a serious problem adversely affecting aquatic life, if the DO falls below 4.0 mg/L then it is an index of pollution.


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

through various manmade activities such as production of these compounds, spillage during transportation, and their uses in different applications.

– These include synthetic pesticides, synthetic detergents, food additives, pharmaceuticals, insecticides, paints, synthetic fibers, plastics, solvents and volatile organic compounds (VOCs).

– toxic and biorefractory organics(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.


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 of hydrocarbons and degradable under bacterial action.

– The biodegradation rate is different for different oils, tars being one of the slowest.

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

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

– This pollutant is also responsible for endangering water birds and coastal plants 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 are known to be carcinogenic.


Mexico Oil spill, 2010

Pelican covered with oil


2) PATHOGENS

– The pathogenic microorganisms enter in to water body through sewage discharge as a major source or through the wastewater from industries like slaughterhouses.

– Viruses and bacteria can cause water borne diseases, such as cholera, typhoid, dysentery, polio and infectious hepatitis in human.


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

contains substantial concentration of nutrients like Nitrogen and P

– Wastewater supply nutrients to the plants and may stimulate the growth 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 ends up as a dead pool of water.

– People swimming in eutrophic waters containing blue-green algae can have 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 the methaemoglobin results in a decrease in the oxygen carrying capacity of the blood as nitrate ions in the blood readily oxidize ferrous ions in the hemoglobin.


4) SUSPENDED SOLIDS AND SEDIMENTS– These comprise of silt, sand and minerals eroded from land.

– These appear in the water through the surface runoff during rainy season and 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 or ocean bottom can impair the normal aquatic life and affect the diversity of the aquatic ecosystem.

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

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


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

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

– These compounds are non-biodegradable and persist in the environment.

– These pollutants include mineral acids, inorganic salts, trace elements, metals, metals compounds, complexes of metals with organic compounds, cyanides, sulphates, etc.

– The accumulation of heavy metals may have adverse effect on aquatic 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 toxice.g. Hg, Cu, Cd, Pb, As, Se. Copper greater than 0.1 mg/L is toxic to microbes.


6) THERMAL POLLUTION – Considerable thermal pollution results due to discharge of

hot water from thermal , nuclear power plants (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 action increases due to rise in temperature, hence, resulting in rapid decrease of DO.

– The discharge of hot water leads to the thermal stratification in the water body, where hot water will remain on the top.


7) RADIOACTIVE POLLUTANTS – Radioactive materials originate from the following:

• Mining and processing of ores,

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

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

• Uses and testing of nuclear weapons

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

• The safe concentration for lifetime consumption is 1 x 10-7 micro curies per ml.


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 microbes

– 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

• If water contains as much as specific gas that it can hold in 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 be supersaturated.

• Supersaturation occurs because of:

– High photosynthetic activity (in summer)

– When high temperature water is discharged to river, the rapid rise in water temperature causes O2 supersaturation.

Oxygen transfer

• The solubility of gases in water is related to the partial pressure of the gas in the atmosphere above the water by Henry’s law.

– Pg = Kh.Xg

– Where, Pg = partial pressure of gas, atm.

– Kh = Henry’s 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

• Henry’s law constant for common gases soluble in water, Kh x 10-4 atm/mole

Temperature 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

• 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

• Henry’s 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

DO and BOD

• Effect of Oxygen Demanding Wastes on Rivers

– Depletion of dissolved oxygen is a major problem due discharge of oxygen demanding organic or inorganic pollutant 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 much oxygen will be required to degrade the waste.

– Although, oxygen is getting depleted for the degradation of organic matter, it is continuously being replenished from the atmosphere and through photosynthesis, if any.

– The concentration of oxygen is determined by the relative rates of these competing processes.

• 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

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 organic matter, 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.

• The BOD test is performed for the following:

– To determine quantity of oxygen required for biochemical stabilization of organic matter

– To determine suitability of biological treatment method, depending on COD/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 end product such as CO2, sulphate (SO4), orthophosphate (PO4) and nitrate (NO3).

• The simple representation of carbonaceous BOD can be explained as below:

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

BOD

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

• Different results would be obtained at different temperatures because 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 in aerated water, pH buffer, and nutrients.

• For wastewater like sewage, within 20 days period, the oxidation of carbonaceous 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

BOD

Idealized carbonaceous oxygen demand: (a) the BOD remaining as a function of time, and (b) the oxygen consumed.

• The biochemical oxygen demand is represented as BOD5 20oC, which indicate the total amount of oxygen consumed for biochemical oxidation of organic 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 is usually necessary to dilute the samples to keep final DO level, at the end of incubation period, above 1.5 mg/L.

• Hence, according to BOD values expected for that wastewater appropriate dilution should be carried out.

• The 5 day BOD of diluted sample =

– Where, DOi and DOf are initial and final DO of diluted wastewater sample.

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

DO i−DO f

p

BOD

• 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 adequate concentration of bacterial population to carry out the biodegradation.

• Usually 1 to 2 mL of sewage per liter is sufficient to act as a seed.

• 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 as below:

• 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

(DO i−DO f )−(B i−B f )(1− p )p

BOD

• 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

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

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

• • The kinetics of BOD reaction can be formulated in accordance with

first 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 ∫0

t

dLt=−KLt . dt

BOD

• 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)

[ log Lt ]0t=−K .t

BOD

• The amount of BOD that has been exerted (amount of oxygen consumed) 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 0.05 to 0.3 per day (base e).

BOD

• The ultimate BOD (Lo) is defined as the maximum BOD exerted by the wastewater. 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 dependes upon the characteristics of the wastewater, i.e., chemical composition of the organic matter present in the wastewater and its biodegradable properties.

• Oxygen depletion is related to both the ultimate BOD and the BOD rate constant (K).

• The ultimate BOD will increase in direct proportion to the concentration of biodegradable 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

Self purification

• SELF PURIFICATION OF NATURAL STREAMS

– The amount of dissolved 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 survive begin to be reduced.

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

– Oxygen demanding wastes remove DO; plants add DO during day but remove it at night; respiration of organisms removes oxygen.

– In summer, rising temperature reduces solubility of oxygen, while lower flows reduce the rate at which oxygen enters the water from atmosphere.

• Factors Affecting Self Purification

– Dilution: When sufficient dilution water is available in the receiving water body, 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. In small current, the solid matter from the wastewater will get deposited at bed following decomposition and reduction in DO.

– Temperature: The quantity of DO available in stream water is more in cold temperature than in hot temperature. Also, as the activity of microorganisms is more at the higher temperature, hence, the self-purification will take less time 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 adding oxygen through photosynthesis.

– Oxidation: Due to oxidation of organic matter discharged in the river DO depletion occurs. This rate is faster at higher temperature and low at lower temperature.

Self purification

• 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 and its 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

• 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 is depicted by the Oxygen Sag Curve.

Self purification

• When the DO content of the stream is gradually consumed due to BOD load, atmosphere supplies oxygen continuously to the water, through the process of reaeration or reoxygenation, i.e., along with deoxygenation reaeration is continuousaly takes place.

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

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