effluent control 1

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Y.P. TingCN4240R Processes for Effluent Control

Requirements and Grading:

Examination: 22 Nov 2014 (Sat) 9:00 am. Closed book, 2.5hr,

Tutorial: Tutorial assignments will be given throughout the semesterGrading: Final grades will be calculated according to the following scale:

Term Paper 10%

Mid-term / End-of-term Quizzes 30%

Final Exam 60%

http://www.lyricsmode.com/lyrics/a/albert_hammond/down_by_the_river.html

McQ and short questions

Cheat sheet
http://www.lyricsmode.com/lyrics/a/albert_hammond/down_by_the_river.htmlhttp://www.lyricsmode.com/lyrics/a/albert_hammond/down_by_the_river.html


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1 Trade, Industrial and Domestic Effluent - GENERAL CONSIDERATIONS

Characteristics of wastewater

Sources of Water Pollutants and Its Environmental Impacts

Biochemical & Chemical Oxidation Demand - calculations

2. PRETREATMENT AND PRIMARY TREATMENT

Screening and Degritting

Equalisation

Sedimentation

3. PHYSICO-CHEMICAL TREATMENT SYSTEMS

Flotation

Coagulation and Flocculation

4 BIOLOGICAL TREATMENT SYSTEMS

Microbial Growth and Metabolism

Application of Growth Kinetics to Treatment Processes

Aerobic Treatment Systems

Anaerobic Sludge Digestion

5. DISINFECTION


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Effluent: An outflow or discharge of liquid waste. (cf emission)

Industries: e.g. chemical, petrochemical, refinery, pulp & paper, textiles and dyes, soaps &

detergent, food & beverage, pharmaceutical, metal finishing & electroplating, semiconductors

& microelectronics. Difference in effluent characteristics.

Wastewater (or used water (PUB))

Greywater= w/w from sinks, showers, laundry (excluding toilet wastes).

Black water = w/w in contact with fecal matter. Fecal coliform 104-107/100 mL.

Pollutants - substances that have the potential to have negative effects on the natural

environment, to cause damage to infrastructureor harm to human health.

Agricultural wastewaterpesticides, fertilizers and salts;

Municipal wastewaterhuman sewage;

Power plantswater discharged at high temperature;

Industrial wastewaterwide range of chemical pollutants and organic wastes.

Water Pollution:.. Presence of any material in water that is harmful to plants or animals, or affects its

taste and odor, or detracts from any use that can be made of it.

Alteration in the composition or condition of water directly or indirectly as a result of the

activities of men, so that it is less suitable for any or all the purposes for which it is suitable in

its natural state.

Metals are toxic

Color

Sewage treatment plant/waste

treatment plant name changed to

water reclamation plant

Indicator of bacteria

Waste consumes oxygen giving anaerobic conditions in

pipes. Hydrogen sulfide will be produced from bacteria.

H2S can be oxidized into sulphuric acid which corrodesthe pipes, "crown sewer."

Dissolved oxygen concentration

of high temp water is lower


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Uses of water in industry: Raw material, Cleaning or reagent solvent, Heat transfer

medium, Conveyance medium, Reaction product, and Fire extinguishing

medium.

Wastewatercombination of the liquid or water-carried wastes removed from

residences, institutions, and commercial and industrial establishments, together

with such groundwater, surface water, and stormwater as may be present.

Unit Operationscontaminant removal by physical forces.

Unit Processesbiological/chemical reactions.

An effluent treatment system is composed of a combination of unit operations

and unit processes designed to reduce certain constituents of w/w to an

acceptable level. For public health and the environment. Many different

combinations are possible. (See diagram).

To protect public health and the environmentneed knowledge of constituents of concern,

impacts of constituents,

transformation and long term fates of constituents,

treatment methods, and

beneficial use, and appropriate disposal of solids/effluent after treatment.

Treatment to desired level

without over treatment

Singapore, need not remove nitrogen and

phosphorus. No specific treatment to reduce it.

Of concern if disposed into a closed water body

with no dilution


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Code of Practice on Pollution Control (3rded; amendments Feb 2004)

http://www.nea.gov.sg/cms/pcd/coppc_2002.pdf

(See Appendix 9)

Public Utilities Board, Singapores national water agency.

http://www.pub.gov.sg/Pages/default.aspx

Glossary of Terms compiled by PUB

http://www.pub.gov.sg/atoz/Pages/default.aspx

Water Reclamation Plants (WRP)

http://www.pub.gov.sg/products/usedwater/Pages/WaterReclamationPlants.aspx
http://www.nea.gov.sg/cms/pcd/coppc_2002.pdfhttp://www.pub.gov.sg/Pages/default.aspxhttp://www.pub.gov.sg/atoz/Pages/default.aspxhttp://www.pub.gov.sg/products/usedwater/Pages/WaterReclamationPlants.aspxhttp://www.pub.gov.sg/products/usedwater/Pages/WaterReclamationPlants.aspxhttp://www.pub.gov.sg/atoz/Pages/default.aspxhttp://www.pub.gov.sg/Pages/default.aspxhttp://www.nea.gov.sg/cms/pcd/coppc_2002.pdf


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Inputs: (i) Point Source, (ii) Diffuse source

Point Source

Well-defined point of discharge. (Usually continuous). Can be located and identified with a particular discharger.

Principal source: (i) municipal point source, and (ii) industrial

discharges

Non-point Source

Origin of discharge is diffused, i.e. not possible to relate the

discharge to a specific and well-defined location. (Usually

transient in time).

Principal source: Agriculture, silviculture, atmospheric, urban

and suburban runoff, and groundwater.

Others: groundwater infiltration, drainage from abandonedmines and construction activities, leaching from land disposal of

solid wastes.


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Classification of Water Pollutants:

Pathogenic organisms

Oxygen-demanding substances

Plant nutrients

Toxic organics

Inorganic chemicals

Sediments

Radioactive substances

Heat

Oil

Point source

Diffuse source

Will get oxidized as substance is

unstable. Reduces other substances

eats oxygen

Might kill the bacteria, which

reduces the BOD. However not allbacteria might be killed. Shows

overlapping factors


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Preliminary treatmentremoval of w/w constituents (e.g. rags, sticks, floatables,

grit and grease) that may cause maintenance or operational problems.

Primary treatmentremoval of suspended solids or organic matter from w/w.

Sedimentation, used to remove the floating & settleable solid in w/w.

Secondary treatmentremoval of biodegradable organic matter (in solution or

suspension). Biological & chemical processes. Usually biological conversion of

dissolved & colloidal organics into biomass, which can subsequently be removed by

sedimentation.

Tertiary treatmentadditional combination of unit operations & processes that are

used to further remove constituents (e.g. N, P) which are not removed by

secondary treatment.

Advanced Wastewater Treatmentoperations and processes that are used toremove more contaminants than are taken out by conventional treatment. NOT

synonymous with tertiary treatment. Tertiary treatment is a 3rd step; advanced

wastewater treatment may replace unit operations and processes in

secondary or even primary treatment.

Easy removal

Solubles

Biomass - solids


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Raw sewage

from sewers

Bar screenGritchamber Settling tank Aeration tank Settling tank

Chlorinedisinfection tank

Sludge

Sludge digester

Activated sludge

Air

pump

(kills bacteria)

To river, lake,

or ocean

Sludge drying bed

Disposed of in landfill or

ocean or applied to cropland,

pasture, or rangeland

Primary Secondary


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COD BOD SS NH3 (N) PO4(P)

600 300 500 40 -

300 150 150 50 -

100 20 20 50 3-14

15


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Processes for

speci f ic pol lutant


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Fixed-Inorganic

Volatile-Organic

Volatile has higher oxygen demand.

Bacteria eats the carbon to grow

1 question of BOD5 in finals

Nitrates are the most oxidized nitrogen.

Ammonia is the least oxidized.

Different oxygen demand.

Haemoglobin has higher affinity for nitrites

(NO2-). Competes with oxygen to transportaround the body


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Nitrogen is a component of amino acids and urea. Amino acids are the building blocks of all proteins. Proteins comprise not onlystructural components such as muscle, tissue and organs, but also enzymes and hormones essential for the functioning of all livingthings. Urea is a byproduct of protein digestion. We use the term "organic nitrogen" to describe a nitrogen compound that had itsorigin in living material. The nitrogen in protein and urea is organic nitrogen. Organic nitrogen can enter septic systems as bodily

wastes, discarded food material, or as components of cleaning agents..Amines and amides


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CHARACTERISTICS SOURCES

PHYSICAL PROPERTIES

Colour Domestics and industrial wastes, natural decay of

organic materials

Odour Decomposing wastewater, industrial wastes

Solids Domestic waster supply, domestic and industrial

wastes, soil erosion, inflow/infiltration.

Temperature Domestic and industrial wastes

CHEMICAL CONSTITUENTS

GasesHydrogen sulphide Decomposition of domestic wastes

Methane Decomposition of domestic wastes

Oxygen Domestic water supply; surface water infiltration

BIOLOGICAL CONSTITUENTS

Animals Open watercourses and treatment plants

Plants Open watercourses and treatment plants

Protists

Bacteria Domestic water supply; surface-water infiltration,

treatment plants

Viruses Domestic wastes


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CHARACTERISTICS SOURCES

CHEMICAL CONSTITUENTS

Organic

Carbohydrates Domestic, commercial and industrial wastes

Fats, oils and grease Domestic, commercial and industrial wastes

Pesticides Agricultural wastes

Phenols Industrial wastes

Proteins Domestic, commercial and industrial wastes

Priority pollutants Domestic, commercial and industrial wastes

Surfactants Domestic, commercial and industrial wastes

Volatile organic cpds Domestic, commercial and industrial wastes

Others Natural decay of organic materials

Inorganics

Alkalinity Domestic wastes, domestic water supply, groundwater infiltration

Chlorides Domestic wastes, domestic water supply, groundwater infiltration

Heavy metals Industrial wastes

Nitrogen Domestic and agricultural wastes

pH Domestic, commercial and industrial wastes

Phosphorus Domestic, commercial and industrial wastes; agricultural runoff

Priority pollutants Domestic, commercial and industrial wastes

Sulphur Domestic water supply; Domestic, commercial and industrial wastes

Priority cause it

is dangerous or

carcinogenic...


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Important contaminants of concern in effluent treatment (contd)

Contaminants Reason for importance

Priority pollutants Organic and inorganic compounds selected on the basisof their known or suspected carcinogenicity,

mutagenicity, teratogenicity, or high acute toxicity.

Refractory organics(resistant to treatment)

Compounds that resist conventional methods of

wastewater treatment. Typical examples include

surfactants, phenols and agricultural pesticides .

Heavy metals Usually added to wastewater from commercial andindustrial activities and may have to be removed if the

wastewater is to be reused.

Dissolved inorganics Inorganic constituents such as Ca2+, Na+and SO42-

added to the original domestic water supply as a result of

water use, and may have to be removed if the w/w is

to be reused.

PP includes heavy metals, pesticides, PAHs, PCBs.

Coliform bacteria:a group of bacteria commonly found in intestines of animals and

humans beings. Coliform count - hygiene indicator. Membrane filtration, and filters

placed onto selective agar. After incubation, plates are counted and suspect colonies

confirmed with confirmation tests (lactose fermentation)

Contaminated by feces


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Total solids (TS) test quantifies all solids in the watersuspended solids,

dissolved solids organic, inorganic.

Filtration does not divide solids into suspended & dissolved fraction. Some

colloids may pass through the filter, & be measured with the dissolved solids,

while some dissolved solids may absorb onto filter material. Depends on

size & nature of solids

pore size & surface characteristics of filter material.

Therefore, use of the terms

FILTERABLE RESIDUES (relates more to dissolved solids)

- NON-FILTERABLE RESIDUES (relates more to suspended solids)


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Ionic

Use imhoff cone, to measure

settlement in 30mins


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Evaporation and

drying at 105 C

Evaporation

and drying at

105 C

Ignition at 500 C

TS = TVS + TFS

TS = organic + inorganic


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Laboratory procedure to determine total solids and

total volatile solids concentrations of a water or wastewater sample.

Laboratory procedure to determine the total suspended-solids and

volatile suspended-solids concentrations of a water or wastewater sample.


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.

Suspended solids (SS)

Sources:

Natural inorganic particles (clay, silt, and soil constituents)

- organic solids (plant fibers and biological solids, dead

vegetation and animals);

Domestic inorganic and organics (large amounts);

Industrial various inorganic and organic, immiscible liquids such as

oil, grease, etc.

Impact: Aesthetically displeasing;

Increase turbidity; aquatic plants unable to photosynthesize, thus

lowering dissolved oxygen (DO) level. Microbes use SS as food, thus

lowering DO level;

Adsorption sites for chemical and biological agents;

Some organic solids may be degraded to form by-products; Biologicallyactive solids may include disease-causing organisms.


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Turbidity

Measure of extent to which light is either absorbed or scattered by

suspended material in water. Determination of solids in potable water

supplies, natural waters, and secondary effluents.

Sources:

Natural surface watererosion of colloidal materials, e.g.

clay, silt, rock fragments, etc. Also, m/o contribute to turbidity;

Household & industrial wastewaters - (e.g. soaps, detergents,

emulsifying agents, etc).

ImpactColloidal matter associated with turbidity provides adsorption sites

for chemicalsthat may be harmful, or cause undesirable tastes & odours.

Also, disinfection of turbid water difficult because of adsorptive

characteristics of some colloids, & also because the solids may partially

shield the m/o from the disinfectant. In natural water, turbidity may affectlight penetration, & hence photosynthesis in streams & lakes.

Measurement(i) Formazin turbidity unitFTU; (ii) nephelometry turbidity

unitNTU


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Colour

Pure water is colourless. Colour is caused by foreign organic and inorganic

materials.

Sources:

Naturalorganic debris in water such as leaves, weeds or wood

resulting in tannins, humic acids, and humates: yellowish-brown

hues. Iron oxides (reddish), and manganese oxides (brown or

blackish)

Industrial wastewatersoperations may add substantialcolour, e.g. textiles and dyes

Impact:

Aesthetically displeasing;

Organic compounds causing colour may exert a chlorine demand,

and cause taste and odour problems. May contribute to disinfection by-product which are carcinogenic

(e.g. THMs).Tri halo methanes


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Odours

Caused by gases produced from decomposition of organic matter. Most

characteristic odour of stale w/w is H2S, produced by anaerobic m/o that

reduces sulphates to sulphides.

Major categories of offensive odors

Compound Typical formula Odor quality

Amines CH3NH2, (CH3)3N Fishy

Ammonia NH3 Ammoniacal

Diamines

NH2(CH2)4NH2,

NH2(CH2)5NH2 Decayed flesh

Hydrogen sulfide H2S Rotten eggs

Mercaptans CH3SH, CH3(CH2)3SH Skunk

Organic sulfides (CH3)2S, CH3SSCH3 Rotten cabbageSkatole C8H5NHCH3 Fecal

Threshold Odour Number: ratio by which sample has to be diluted for the

odour to become virtually unnoticeable. E.g. dilute a 50 ml sample to 200 ml

=> TON = 4.

Source of smell from n and S,

remove them to remove smell


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Temperature

Sources: Temperature of w/w is normally higher than natural andsubsurface water due to industrial operations (e.g. for cooling).

Impact: Effect on (i) aquatic life; (ii) chemical reactions & reactionrates; (iii) suitability of the water for beneficial uses; (iv)thermal stratification in lakes

Solubility of O2in waterdecreases with temperature. Increase in rates

of biochemical reaction that accompanies an increase in temperature,combined with the decrease in dissolved oxygen (D.O.) can causeserious depletion of D.O. concentration.

Warm water also leads to growth of algae, with natural secretion of oilsby algaein the mats, & decay productsof algae lead to taste & odourproblems.

Sudden changesin temperature can affect mortality of aquatic life.

(Also, species distribution, e.g. 2025 oC diatoms, 3035 oC greenalgae, > 35 oC blue green algae)

TER: T


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Water Quality parameters

Gross measuresno distinction made between individual physical,

chemical and biological species. E.g. suspended solids, odour, colour,BOD. Most easily measured and interpreted.

Specific measuresnecessary when single characteristic (e.g. toxic

compound, heavy metal ion, or species of fish) is of concern. Applies to

particular uses.


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Chemical Characteristics

Total Dissolved Solids (TDS)

Source: Solvent action of water on solids, liquids and gases results in theformation of TDS.

Impact:

Displeasing colour, odour and taste problems.

Toxicity and carcinogenicity.

Suitability of use of water for industrial operations.

pH

Range conducive to biological life is narrow & critical.

Metal specification = f (pH) e.g. carcinogenic of Cr6+vs Cr3+

Impact:

Affects aquatic animals and plants sensitive to pH.

Metal speciation (and toxicity).

Chemical reactions as well as treatment processes.

Waste Must be within pH 6-9


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Nitrogen InorganicNH3, NO3-, NO2

-

Organic -- Urea, proteins, amino acids

Phosphatesorthophosphate

PO43-, H3PO4, H2PO4

-, HPO42-

SulphurSO42-, H2S, SH

-, S

TN = ON + NH4+N + NO2N + NO3N

TKN = ON + NH4+N Kjeldahl

TKN = 40% Organic + 60% Free Ammonia

Typical concentrations:

Ammonia-N = 10-50 mg/LOrganic N = 1035 mg/L

Impacts of Nitrogen

Impairs quality of receiving stream (Eutrophication: N & P)

Taste & odour (for drinking water); DO depletion (hypoxia), algal mats; toxins & fishkills

Decreases chlorination efficiency (NH4+) Impairs suitability for water reuse

NH4+toxicity

Contaminates groundwater (nitrate);

Drinking water; Methemoglobinemia (Blue Baby Syndrome)

WHO's Guideline Value for Drinking water: nitrate= 50 mg /l ; nitrite = 3mg/l

Total kjeldahl nitrogen

Eu good

Trophos food

When algae dies


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Algae-filled Chaohu Lake,

Hefei, Anhui (2009)

Generally accepted BOD/N/P mass

ratio required for biological treatment is

100/5/1 (i.e. 100 mg/l BOD to 5 mg/l

Nitrogen to 1 mg/l phosphorus).

Raw sanitary w/w has a ratio 100/17/3;

after primary settling ratio is 100/23/5,

thus there is abundant N and P for

microbial growth.


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Metals

Toxic metals: aluminium,arsenic, cadmium, chromium, cobalt, copper, lead,

mercury, nickel, zinc, etc,

Essential metals: serves biological functions, e.g. copper, zinc, etc

Metal toxicity; concentration and speciation.Metals - non-biodegradable.

Zooplankton0.123ppm

Rainbow smelt1.04 ppm

Water0.000002 ppm

Phytoplankton0.0025 ppm

Lake trout4.83 ppm

Herring gull124 ppm

Herring gull eggs124 ppm

Bioaccumulation

Biomagnification

DDT(dichlorodiphenyltrichloroethane)


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Gases

Gases commonly found in untreated water

N2, O2, CO2common in the atmosphere, & found in all waters exposed toair

H2S, NH3, CH4derived from decomposition of organic matter in w/w

D.O.required for respiration of aerobic m/o. & all other aerobic life forms.

O2slightly soluble in water; actual quantity in w/w depends on:

solubility of the gas

partial pressure of the gas in the atmosphere

temperature

purity of the water (e.g. a fixed temperature, solubility of O2with chloride conc.)

H2Sformed from the decomposition of organic matter containing sulphur,or from the reduction of mineral sulphites & sulphates.

Colourless, inflammable, rotten egg smell.

Blackening of w/w & sludge usually the result of H2S + Fe FeS

CH4principal by-product. Colourless, odourless, combustible & high fuelvalue


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Standard Method of Analysis of Water and Wastewater

Standard methods for the analysis of most components in w/w.

Developed over many years & include: Sampling procedure

Sample storage

Test methods

Interference and methods of overcoming these

Analytical procedures - see STANDARD METHODS FOR THEEXAMINATION OF WATER & WASTEWATER (American Public HealthAssociation, Washington, D.C.)

Sampling programme requirements:

Sample must be truly representative of the existing conditions

Time between collection and analysis should be as short as possible. Appropriate preservation technique should be applied to slow down biological

and chemical changes.

Accurate and through sampling records

Solids or liquid


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Grab sample. A single sample

Composite sample. A mixture of grab sample at the same locationbut attimed intervals. Where quality and quantity change with time.

(Volume taken in proportion to flow rate, e.g. grab sample of 100 ml for flowof 5 L/s; and grab sample of 200 ml for flow of 10 L/s.)

Integrated sample. A mixture of grab sample at the same timebut fromdifferent locations. Gives information on average concentration ofparameters of the whole system to account for variation in depth and widthof the system.

(spatial vs tempo ral)

Preservation of samples. Analyse ASAP. Retard biological changes,

chemical changes, reduce loss of components by evaporation oradsorption. Except for refrigeration, preservation is not recommended

unless absolutely necessary. Choice of containers (glass, plastic).

Location vs time

Avoid, as concentration might be very low

Storage could cause major changes.

Especially priority pollutant

FACTORS FOR DO LEVEL SLOPE


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Under heavy loads of pollution, the

D.O. level may drop to zero. This

results in noxiousodors and very

unsightly conditions in the water. With

additional time and distancedownstream, the water will eventually

be reaerated and water quality will be

restored.

Oxygen sag curve shows the effect oforganic pollution on the D.O. levels in a

stream or river. After the organics

decompose, surface reaeration will

restore the original water quality.

St ream self-pu ri f icat ion.

Photosynthetic

Time of the day

Flow rate

Surface area/depth area


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Clean ZoneDecomposition

Zone

Septic ZoneRecovery

Zone

Clean Zone

Normal clean water organisms

(trout, perch, bass,mayfly, stonefly)

Trash fish(carp, gar,

leeches)

Fish absent,fungi,sludgeworms,bacteria

(anaerobic)

Trash fish(carp, gar,leeches)

Normal clean water organisms(trout, perch, bass,

mayfly, stonefly)

8 ppmDissolved

oxygen

(ppm)

Biological

oxygendemand

8 ppm

Types oforganisms

Pollution in Streams

DO replenishes


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Oxygen requirement of a waste

Biochemical oxygen demand (BOD)

Chemical oxygen demand (COD)

Total organic carbon (TOC)

Total oxygen demand (TOD)

BOD testmeasures the biodegradable organic carbon, & undercertain conditions, the oxidisable nitrogen present in the waste

COD testmeasures the total organic carbon, with the exception ofcertain aromatics, such as benzene, which are not completelyoxidized in the reaction. The COD test is an oxidation-reduction

reaction, so other reduced substances (e.g. sulphides, sulphites &Fe2+) will also be oxidized & reported as COD.

Ammonia also oxidized

Refractory


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Oxygen requirement of a waste

BOD

Amount of oxygen required by living organisms engaged in the utilization& stabilizationof the organic matter present in the w/w at a constant

temperature (20 oC)

Measures the biodegradable organic carbon, & under certain conditions,

the oxidisable nitrogen present in the w/w.

Theory2 Stage process

1) Stage 1Organics used by microorganisms for energy & growth. This

utilizes O2, & new m/o are produced. (UTILIZATION). Process usually

takes 18-36 hours.

2) Stage 2After the organics are removed, the cells then undergo

endogenous metabolism which eventually degrades the m/o present,

again using O2, after which only the non-biodegradable cellular residue

remains (STABILIZATION). This process usually takes about 20 days.

Milligram Oxygen per liter


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Reactions occurring in the BOD bottle. Rate of reaction of first

stage ~ 10-20 times faster than 2 stage

Cell growth


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Oxygen consumption in both processes = ULTIMATE BOD

Organics + O2+ N + P new cells + CO2+ H2O + non-biodegradable

soluble residue

Cells + O2CO2+ H2O + N + P + non-biodegradable soluble residue

TOC testmeasures all carbon as CO2, & hence inorganic carbon (CO2,

HCO3-, etc) present in the w/w must be removed prior to the analysis, or

corrected for in the calculation.

TOD testmeasures organic carbon & unoxidised nitrogen & sulphur

(BOD)5Historically, furthest point in England from the sea by river was 5

days. Consequently, the amount of O2utilised by a waste stream (over 5days) seems a sensible criteria.

Bacteria C5H7NO2

Deoxygenation

mg carbon per liter

Acidification


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Y.P. Ting

CN4240R Processes for Effluent Control

Uses of BOD Test

1) Assessment of quality of river waters

2) Determination of oxygen demand of w/w, effluents & other polluted

water3) Assessment of effect of discharge to water courses

4) Design & assessment of performance of sewage works & other w/w

Treatment plant

5) Guide to the biodegradability or treatablility of a particular effluent

BOD typical range:

6 mg/l clean river water

10-15 mg/l good sewage effluent

30-60 mg/l poor sewage effluent

60-120 mg/l settled sewage

120-240 mg/l raw sewage, industrial effluent

Royal Commission Effluent Standard - 20 mg/l BOD, 30 mg/l SS

Which is easier or difficult to breakdown

What can be biologically oxidized

can be chemically oxidized.not the

other way round

Use of BOD/COD ratio =


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Y.P. Ting


P

DODOBOD

FI

Procedure for BOD test

BOD measured by determining the O2consumed from a sample placed in an

air-tight container, & kept in a controlled environment for a preselected time.

Standard test300ml bottle, sample incubated at 20 oC for 5 days. Light must

be excluded from the incubator to prevent algae growth that may produce O2inthe bottle. Because the saturation conc. of O2in water at 20

oC is ~9 mg/l,

dilution of sample with BOD-free, O2-saturated water is necessary to measure

BOD values greater than just a few mg/l.

Range of BOD covered by various dilutionssee table.

These values assume an initial DO conc. = 9 mg/l & minimum of 2 & maximum

of 7 or 8 mg/l O2being consumed. BOD of a diluted sample is calculated from:

where DOI= initial dissolved O2conc. (mg/L)

DOF= final dissolved O2conc. (mg/L)

P = decimal fraction of sample in the 300mL bottle

As BOD > DO

level. Dilution

is necessary

As it is a biological test, it is

not very accurate.

Dilution would affect results


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Y.P. Ting


Procedure for setting

up BOD test bottles.

(a) with unseeded

dilution water and (b)

seeded dilution water.

Control


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Y.P. Ting


300

mL of waste

Make up total vol to

300


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Y.P. Ting


Schematic representation of the oxygen consumed (uptake) in the BOD

determination.

D1 = DO of diluted sample 15 minutes after preparation, g/m3

D2 = DO of diluted sample after incubation, g/m3

B1 = DO of seeded dilution water before incubation, g/m3

B2 = DO of seeded dilution water after incubation, g/m3

D1-D2 = Oxygen consumer (uptake) in sample, g/m3

B1-B2 = Oxygen consumed (uptake) in blank, g/m3

P

fBBDD )()(

2121

For Seeded dilution

Fraction


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Y.P. Ting


P

fBBDD )()( 2121

BOD is computed using the following equation:

BOD =

where

BOD = biochemical oxygen demand,

D1 = DO of diluted sample 15 minutes after preparation, g/m3

D2 = DO of diluted sample after incubation at 20 oC, g/m3

B1 = DO of seeded dilution water before incubation, g/m3

B2 = DO of seeded dilution water after incubation at 20 oC, g/m3

f = ratio of seed in sample to seed in control

= (% seed in D1) / (% seed in B1)

P = decimal fraction of sample used

= (mL of sample, Vs) / 300 mL

For an unseeded sample the terms for the blank (B1 and B2) are omitted in thecomputation of the BOD.


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Y.P. Ting


Wastewater

used (mg/L)

DOI(mg/L)

DOF(mg/L)

O2

(mg/L)

P BOD5

(mg/L)

510

20

9.29.1

8.9

6.94.4

1.5

2.34.7

7.4

0.01670.033

0.067

138142

110

Example: Determine BOD5 of a w/w, where BOD is suspected to range from

50 to 200 mg/l. Prepare 3 dilutions to cover this range.

note DOIexpected

If the third sample is disregarded (DOFis less than 2 mg/L or if change in DO< 2.0), then average, BOD5of w/w is ~140mg/L

Dilution factor =

ww used /300


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Y.P. Ting


Toxicity in a waste is evidenced by so-called sliding BOD values, i.e. an

increasing calculated BOD with increasing dilution. If this situation exists, it is

necessary to determine the dilution value below which the computed BOD5are

consistent. (Highest value obtained in valid tests, i.e. the average of the highest dilution providing aminimum uptake of 2.0 mg/l DO).

Advantages

Test is biological

Simple equipment

Cheap

Labour -- non intensive

Disadvantages

Long time (5-20 days)

Toxins, heavy metals affect test

Choice & acclimatisation of culture necessaryVariables: T, pH, seed level must be controlled

Extent of nitrification not quantified

Estimates of K, (BOD)u depends on method of analysis

Cannot be used reliably for comparative purposes.

Toxicity kills bacteria. Dilution

would allow bacteria to live due to

lower conc of toxic. Necessary to

conduct this test


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Y.P. Ting


Problem: A series of BOD tests were run at four different dilutions. The results

were as

follows:

# Dilution Do initial DO final BOD (mg/l)

1. 100 10.0 2.5 750

2. 200 10.0 6.0 800

3. 400 10.0 7.5 1000

4. 600 10.0 8.0 1200

What is the BOD?

750 or less. Because with

increase in dilution BOD

increases


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Y.P. Ting


Influence of toxic substances on BOD:

1without toxic substance,

2,3,4presence of toxic substances.

2decrease of BOD process rate,

3delay of the start,

4total stoppage of BOD process

BOD curve, depending on temperature

Grows the fastest but can't sustain


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Y.P. Ting


Biochemical oxidation is a slow process, & theoreticallytakes an infinite

timeto go to completion.

20 days, [O] about 95-99% complete.

5 days, [O] about 60-70% complete.

O2is consumed by the m/o for energy, & new cell masssynthesized.

Organics + O2 + N + P

new cells + CO2+ H2O + non-biodegradablesoluble residue

Organisms also undergo autoxidation(endogenous respiration)

Cells + O2CO2+ H2O + N + P + non-biodegradable soluble residue


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Y.P. Ting


Nitrogenous BOD (NBOD)

Non-carbonaceous matter, e.g. NH3produced hydrolysis of proteins. Some

autotrophic bacteriais capable of using O2to oxidiseNH3NO2-NO3

-

Growth of nitrifying bacteriais very slow, normally 6-10 days to reachsignificant numbers.

Nitrification: nitrosomonas

Step 1: NH4++ 3/2 O2----------------------> NO2

-+ 2 H++ H2O

nitrobacter

Step2: NO2-+ 1/2 O2 -------------------------> NO3

-

Overall: NH4++ 2 O2--------------------------> NO3

-+ 2 H++ H2O

cf photoautotroph& chemoautotroph

Self feeding bacteria, don't need

free form nitrogen.

Energy from light Energy from chemicals


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Y.P. Ting


Nitrificationinvolves the energy-generating oxidation of ammonia (NH3).

Organisms that perform these reactions are strict aerobesbecause oxygen is

the electron acceptorduring the oxidation of ammonia.

Nitrifiers such as Nitrosomonas spp.oxidize ammonia to nitrite (NO2-) via:

(a)Generation of the intermediate hydroxylamine (NH2OH)

NH3 + O2+ 2H++ 2e- ----> NH2OH + H2O + energy

The enzyme ammonia monooxygenase catalyzes this reaction

(b)Production of nitrite from hydroxylamine:

NH2OH + H2O-----------> NO2-+ 2H++ 2e-+ energy

Nitrifiers such as Nitrobacter spp. oxidize nitrite to nitrate (NO3-)

NO2-+ H2O -------------> NO3

-+ 2H++ 2e-+ energy


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Y.P. Ting


BOD = CBOD + NBOD

Exertion of the carbonaceous and

nitrogenous biochemical oxygen demand in

a waste sample.

Nitrate inhibitor:

2-chloro-6- (trichloromethyl) pyridine (TCMP)

Where effluent from

biological treatment

units are to be

analysed, presence ofnitrifiers may be high,

and hence may need to

add inhibitor, in order to

determine the treatment

efficiency of the

treatment units.

Nitrifiers bacteria are slow

growing, more than 6 days for

NBOD to be shown


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BOD curves showing both carbonaceous and nitrogenous BOD

NBOD comes earlier, due to

nitrification during treatment


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Lt= BOD remaining at time t, mg/l

Lo= ultimate BODYt= BOD exerted after time t, mg/l

k = reaction rate constant d-1(base e)

K = reaction rate constant d-1 (base 10)

Means oxygen consumed


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Kinetics formulated according to 1storder Kinetics

tt LkdtdL '

where Ltis the amount of first-stage BOD remaining in w/w at time t.

Integrating,

o

t

L

L

where L0or BODu= BOD remaining at time t = 0

(i.e. the total or ultimate BOD initially present).

303.2

'kK

= e-kt

= 10-Kt


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At any time t,

Lt= Lo(10-Kt)

BODt= yt= Lo- Lt= Lo (1-10-Kt)

where yt= amount of BOD that has been exerted at any time t

Therefore, 5-day BOD

y5= Lo-L5= Lo(1-10-5K)

Typically, for polluted water and w/w, K (base 10, 20 oC) = 0.1 d-1.K varies, depending on the waste, from 0.05 to 0.3d-1(base e) or more.

BOD test varies with temperature.

Vant Hoff-Arrhenius relationship

KT= K20(T- 20)

20-30 oC, = 1.056, Often used value, = 1.0474-20 oC, = 1.135

Range of number


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Typically, for polluted water and w/w, K (base 10, 20 oC) = 0.1 d-1.

K varies, depending on the waste, from 0.05 to 0.3d-1(base e) or more.

BOD test varies with temperature.

Vant Hoff-Arrhenius relationship

KT= K20(T- 20)

20-30 oC, = 1.056, Often used value, = 1.0474-20 oC, = 1.135

EnergyActivationaE

)11()ln(121

2

TTREa

kk

Aek RTEa


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BOD exerted = f(K),

and K = f (T)

= f (nature of organic matter)= f (ability of m/o to utilize the waste)

For a constant ultimate BOD, the value of K determines the rate of

the BOD reaction. Value of K depends on the nature of the organic

molecules. sugar, starches (easy to digest), high value of K;

complex molecules, e.g. phenol, difficult to digest, low value of

K.

k assumes microbes able to use

the waste to grow


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Q1. As the technical manager, your environmental control engineer

reported to you that the wastewater treatment plant is apparently not

performing as expected. He reported that the BOD removal efficiency

was lower that the expected 90%, since the average influent BOD5and

the average effluent BOD5 were found to be 120 mg/l and 40 mg/l

respectively. List a few key questions you would ask him in order to

ascertain if the wastewater treatment plant is indeed not performing as

expected. What follow-up action (if any) should be taken by the

engineer in order to confirm this?

Q2. A sample of the influent to a wastewater treatment plant, and the

effluent after the treatment were obtained at the same time. BOD5

analyses of the two samples yielded values of 188 mg/l and 33 mg/l

respectively.The plant was designed to produce an effluent of 30 mg/lor less when the influent BOD5is 200 mg/l.What can you conclude (if

any) concerning the adequacy of the plant?

Check DO

Check nutrient level

Nitrifiers could have been encountered in the

plant, increasing BOD. Effluent might include

NBOD and CBOD while influent only has CBOD

The plant is not optimized

Possibly temp, nutrients

Can't be conclusive, need to take

several readings.

Sample at same time, plant might

not be steady state.


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Example

If the BOD3of a waste is 75 mg/l and the K is 0.150 day1, what is the

ultimate BOD?

Note that the rate constant is given in base 10 (i.e. K, not k).

Substitute the given value into previous equation, solve for Lo.

Yt= L0-Lt= L0(1-10-Kt)

75 = L0(1-10-(0.150)(3)) = 0.645 Lo

Or, using base e,

k' = 2.303(K) = 0.345 and

75 = L0(1- e-(0.345)(3)) = 0.645 L0

soL0= 116 mg/l


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Example

Determine the 1-day BOD (i.e. BOD exerted after 1-day) and the ultimate

first-stage BOD for a wastewater whose 5-day 20

o

C BOD = 200 mg/lk = 0.23d-1

Lt = L0e-kt

y5= L0L5= L0(1-e-5k)

200 = L0(1- e5 * 0.23)

L0= 293 mg/l

Lt= L0e-kt

= 293(e-0.23) = 233 mg/l

BOD exerted after 1 day = y1= L0L1= 293-233 = 60 mg/l

Or, y1= L0L1= L0(1-e-k) = 293 (1-e-0.23) = 60.2 mg/l

Example:


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Example:

BOD5of w/w = 150 mg/l at 20oC, k=0.23d-1(BOD exerted after 5 days = 150

mg/l). What is BOD8 at 15oC?

First, calculate ultimate BOD, i.e. L0

yt = BODtof w/w

= approaches Loasymptotically as t increases

(i.e. total (ultimate) BOD)

y5= L0- L5= L0(1-e-5k)

150 = Lo(1 - e-5*0.23)

L0= 220 mg/l

Correct for K at 15o

C using KT= K20(T- 20)

K15= (0.23)(1.047-5) = 0.18

y8= Lo(1-e-0.18*8) = 168 mg/l


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BOD

Time (days)

Estimate Lo, then solve for K or k

Estimating parameters for Loand K, based on experimental results

A Least Square Method


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A. Least Square Method

Fit curve through set of data points so that sum of squares of residuals

(difference between the observed & fitted values) = minimum

Tabulate: Time (day) Oxygen used (mg/l)

t1 y1

: :

: : = k(L0-yn)

: :tn yn

since rate of consumption of O2amount of BOD left (i.e. L0-yn) = Ln

Residual error R =

= k(L0-yn)-y

= kLokyny

= a + byy a = kL0, -b = k

)

dt

dy(-)y-(Lk' no

dt

dy


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R2, take differentials

da

d and

db

d

Minimise sum of square of residuals, R

Ra

2=

02a

RR

Rb

2 = 02bRR

Since R = a + byy

na + by - y = 0

and a y + by2- yy = 0 n = no. of data points


77/91


B. Log-Difference Method

y = L0(1-e-kt)

dt

dy= L0ke

(-kt)= R = Differential rate of oxygen consumption

ln R = ln (L0k) kt

y = b + mx

plot ln R vs. t,

slope = -k

intercept = ln (L0k)

C. Thomas Method


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Thomas recognised that the (1-10-Kt) term is similar to the function

2.3Kt [ 1 + (2.3Kt)/6]-3

Compare

(1-10-Kt) = 2.3Kt [ 1(2.3Kt)/2 + (2.3Kt)2/6(2.3Kt)3/24+ ]

2.3Kt [ 1 + (2.3Kt)/6]-3

= 2.3Kt [ 1(2.3Kt)/2 + (2.3Kt)2/6(2.3Kt)3/21.6+ ]

So, y = L0(1-10-kt) can be linearised as:

y = 2.3 L0Kt [ 1 + (2.3Kt)/6]-3

Linearizing

)(y

t 1/3 = (2.3 L0K)-1/3 + {K2/3/(3.43 L0

1/3)}t

)(yt 1/3 vs t, slope = b intercept = a

K (reaction rate constant base 10) = 2.61a

b

L0(ultimate BOD) = (2.3Ka3)-1

Plot

Amount of BOD exerted at any time t

y = L0Lt= L0(1-10-Kt)

Example


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Example

Compute Lo& k using the Least Square Method

t (days) 2 4 6 8 10

Y (mg/l) 11 18 22 24 26

Set up computation table

T (days) y y2 y yy

2

4

6

8

Sum

11

18

22

24

75

121

324

484

576

1505

4.50

2.75

1.50

1.00

9.75

49.5

49.5

33.0

24.0

156.0

Slope y computed from

t

yyy

dt

dy nn

2'

11


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na + by - y = 0

a y + by2- yy = 0

4a + 75b9.75 = 0

75a + 1505b156 = 0

a = 7.5, b = -0.271

but -b = k , a = kL0

k= 0.271 d-1(base e),

L0= lmgb

a/7.27


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t

Example 2

Same set of data, using Thomas method

t (days) 2 4 6 8 10

y (mg/l) 11

0.57

18

0.61

22

0.65

24

069

26

0.727)(

y

t 1/3

)(yt

K =53.0

02.0*61.2*61.2

a

b

= 0.099 d-1

k = 2.303K = 2.303 * 0.099 = 0.228 d-1

L0= 1/(2.3 Ka3) = (2.3 * 0.099 * 0.533)-1

= 29.5 mg/l

k = 0.228 d-1 L0= 29.5 mg/l

1/3

a = 0.53

Slope = 0.02 = b


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Biological Processes and Energy

Microorganisms would like to obtain as much energy from a reaction as

possible. Hence, they would prefer to use oxygen as an electron acceptor.

However, not all microorganisms can use oxygen as an electron acceptor.

(When oxygen is absent, anaerobes will dominate.)

Order of preference for electron acceptor, based on energy

considerations:(i) oxygen, (ii) nitrate, (iii) sulphate, (iv) carbon dioxide, and finally

(v) fermentation.

Note that some microorganisms (e.g. E. coli) have the ability to use several

different electron acceptors, including oxygen, nitrate and organic matter in

fermentation. Others, such as methanogens, can only carry out methane-forming reactions.


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Energy Source:

Aerobesrequire free dissolved oxygen in decomposing organic matter to

gain energy for growth and multiplication.

Anaerobesoxidize organics in the complete absence of dissolved oxygen by

using oxygen bound in other compounds, such as nitrate and sulphate.

Facultativebacteria use free dissolve oxygen when available, but can also live

in its absence by gaining energy from anaerobic reactions.

Aerobic:

Organics + oxygen CO2+ H2O + Energy

Anaerobic:

Organics + NO3- CO2+ N2+ EnergyOrganics + SO42- CO2+ H2S + Energy

Organics Organic acids + CO2+ H2O + Energy

Organic acids CH4+ CO2+ Energy



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Free Energy

kJ/mol glucose

Aerobic oxidationC6H12O6+ 6O2 6CO2+ 6H2O -2,880

Denitrification

(5C6H12O6+ 24NO3-30CO2+ 42H2O + 12N2 -2,720

Sulphate reduction

2C6H12O6+ 6SO42-+ 9H+12CO2+ 12H2O + 3H2S + 3HS - -492

Methanogenesis

C6H12O63CO2+ 3CH4 -428

Ethanol fermentation

C6H12O62CO2+ 2CH3CH2OH -244

Note: cf nitrification

Ch i l O id i D d (COD)


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Chemical Oxidation Demand (COD)

COD test - widely used as a means of measuring the organic strength of

domestic and industrial wastes. Measures a waste in terms of the total

quantity of oxygen required for oxidation to carbon dioxide and waterin accordance with the equation:

All organic compounds, with a few exceptions, can be oxidized by the action

of strong oxidizing agents under acid conditions. The amino nitrogen (with

an oxidation number of 3) will be converted to ammonia nitrogen (see

equation). However, organic nitrogen in higher oxidation states will be

converted to nitrates. (Most organic compound containing nitrogen are derived

from ammonia.)

3222 )2

3

2()

4

3

24( cNHOH

canCOO

cbanNOHC cban


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Chemical Oxidation Demand


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A measure of concentration of waste of unknown chemical composition.

(Also includes oxidation of H2S, SO32-, S2O3

2-, NO2-, Fe2+ . .)

Method:

A wet sample refluxed (2 hr 100oC) with acidified dichromate. Loss of

dichromate through oxidation of organic matter determined by titration.

Silver sulphate as catalyst - certain organic compounds, especially low

molecular weight fatty acids are not oxidised by Cr2O72-, unless Ag+ions

as catalystis present. catalyst + heat

e.g. CaHbOc+ Cr2O72-+ H+--------> Cr3++ CO2+ H2O

organic matter

Sugars, branched chain aliphatics & substituted benzene rings are

completely oxidized with little or no difficulty.

Benzene, pyridine, touleneNOT oxidised by this method.

Chlorideions interfere with the method* & interferencecan be eliminatedby using mercuric sulphate(complex chloride)

Hg2++ 2Cl- HgCl2

* 6Cl-+ Cr2O72-+ 14H+3Cl2+ 2Cr

3++ 7H2O

COD test - organic matter converted to CO2and H2O regardless of the


88/91


g 2 2 g

biological assimilabilityof the substances. E.g., glucose and lignin

are both oxidised completely. COD > BOD values and may be much

greater when significant amounts of biologically resistant organic

matter are present. Example: Wood-pulping wastes.

Limitations:

inability to differentiate between biologically oxidizable and

biologically inert organic matter.

does not provide any evidence of the rate at which the biologicallyactive material would be stabilized under conditions that exist in

nature.

Major advantage- short time required for evaluation. (3h rather than the

5 days required for the measurement of BOD.) May be used as a

substitute for the BOD test. COD data can often be interpreted in

terms of BOD values after sufficient experience has been

accumulated to establish reliable correlation factors.

Relationship between BOD, COD and TOC


89/91


Relationship between these measures depends on the organic content of

the w/w (Recall definition of BOD, COD)

e.g. TOC is related to COD through the carbon-oxygen balance

C6H12O6+ 6 O26 CO2+ 6 H2O

CH4+ 2 O2 CO2+ 2 H2O

Depending on the organic in question, the COD/TOC ratio may vary from

ZERO (when organic material is resistant to dichromate oxidation), to 5.33

(for methane)

Since the organic content undergoes changes during biological oxidation,therefore, COD/TOC ratio will also change, likewise, BOD/TOC

67.212*6

32*6

6

6 2

C

O

M

M

TOC

COD

33.512

32*2

1

2 2

C

O

M

M

TOC

COD

Compare rate constant K in waste stream with a single waste, and one where there is

multiple waste. Does K change? If so, how?


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Theoretical Oxygen Demand

Theoretical Oxygen Demand (ThOD) - stoichiometrically determinedoxygenneeded to convert all carbon molecules in pollutants to CO2, and

all NH3 and NO2to NO3by balancing equations.

ThOD calculationis possible on certain rather pure industrial wastes, but

is impractical for most wastewaters, especially those containing domestic

sewage or containing vegetable or animal wastes.

Theoretical Oxygen Demand (THOD)


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yg ( )

THOD of w/w = amount of O2required to oxidize organics to end products

Example:

THOD of Benzene Sulfonamide (C6H5SO2NH2)

1. Carbonaceous demand

C6H5SO2NH2+ 11/2 O26 CO2+ NH3+ H2S + H2O

2. Nitrogenous demandNH3+ 3/2 O2HNO2+ H2O

HNO2+ 1/2 O2HNO3

H2S + 1/2 O2S + H2O

S + 3/2 O2+ H2OH2SO4

Therefore, THOD of Benzene Sulfonamide = 9.5 moles of O2per mole B-S

effluent control 1

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