CE-202: Wastewater Engineering

Sudipta Sarkar

Why Study Water and Wastewater Engineering as part of Civil Engineering?

CIVIL ENGINEERING

Pollution Load > Carrying capacity of the environment

Nature has a capacity of self-purification of the natural contaminants / pollutions. This is called the carrying capacity of the environment.

When the pollution load is from man-made sources such as industry or it is higher than the carrying capacity of the environment, that is

The nature’s self-cleansing capacity fails to treat the pollutants

We need to engineer the natural treatment processes in such a way that the pollutants can be treated adequately within a smaller area in accelerated pace within a short time

Such an engineered process is called Wastewater Treatment. The engineering involved with collection, conveyance, treatment of wastewater and disposal of treated wastewater is known as Wastewater Engineering

RAW WATER

TREATED WATER

WASTEWATER

TEATED WASTEWATER

WASTEWATER TREATMENT PLANT

WATER TREATMENT PLANT

About The Course CE 202, 2014:

2. Contact Hours: L: 3 T: 1 P: 2/2

3. Examination Duration (Hrs.): Theory Practical

4. Relative Weightage: CWS 1 PRS MTE ETE PRE 15 15 30 0 40

3 0

S. No. Name of Books / Authors Year of

Publication

1. Davis, M. L., Water and Wastewater Engineering; Design

principles and Practice, McGraw Hill Education (India) Edition

2013

2. Hammer, M.J. and Hammer, M.J., “Water and Wastewater

Technology”, 6th Ed., Prentice Hall of India.

2008

3. Davis, M.L. and Cornwell, D.A., “Introduction to Environmental

Engineering”, 4th ed. McGraw Hill.

2008

4. Ronald Droste., “Theory and Practice of Water and Wastewater

Treatment”, John Wiley & Sonc

2005

5. McGhee, T.J., “Water Supply and Sewerage”, McGraw Hill. 1991

6. Peavy, H.S., Rowe, D.R. and Tchobanoglous, G., “Environmental

Engineering”, McGraw Hill.

1986

7. CPHEEO, Manual on Water Supply and Treatment- Third edition,

Ministry of Urban Development, Gov. of India

1999

http://moud.gov.in/Manual_on_Sewerage

Learning objectives

By the end of this course you will be able to: 1. Estimate the amount of sewage / wastewater to be generated from a township/ municipality/industry.

2. Analyze and characterize sewage/ wastewater for its pollutant load and suggest effective treatment technologies.

4. Use basic science of unit processes to design the unit processes for wastewater treatment.

5. Suggest effective method for the disposal of treatment residuals.

7. Understand the needs to design alternative sanitation strategies to suit the needs of people in underdeveloped areas.

6. Understand the needs to design alternative sanitation strategies to suit the needs of people in underdeveloped areas.

3. Design sanitary and storm sewers and sewer networks.

CHARACTERIZATION OF WASTEWATER

Learning Objectives

• Comprehend basic characteristics of domestic wastewater (flow rate, BOD, SS) and apply these characteristics to a preliminary engineering design.

• Calculate mass loadings within the wastewater treatment system for BOD and SS.

• Calculate and interpret mass balances for mixtures of wastewaters.

• Calculate and interpret population equivalents for industrial wastewater.

Sources of Wastewater (Major Components)

1. Domestic: food, soap and detergents, Toilets (fecal and urine), and paper.

2. Commercial: Toilets and food from restaurants and other markets and shops.

3. Industrial: highly variable types of waste , dependent on the type of the industry and the processes involved.

4. Agricultural field runoff: Organic waste, may contain fertilizers, pesticides, herbicides and insecticides.

5. Runoff from streets: sand, petroleum, tars and tire residues.

CHARACTERISTICS OF WASTEWATER

PHYSICAL CHEMICAL BIOLOGICAL

• Color • Temperature • Odor • Solids

• Organic – Biodegradable and Non-biodegradable

• pH • Alkalinity • Phosphorus • Nitrogen • Heavy metals • Dissolved gases • Priority pollutants

Microorganisms and virus - Pathogenic organisms

PHYSICAL CHARACTERISTICS

CHARAC. SOURCE REMARKS

COLOUR DOM. & INDST., NATURAL DECAY OF ORGANIC MATTER

ODOUR INDST. WASTE & DECAY OF ORG. MATTER

Fishy, rotten egg like, rotten cabbage like, faecal matter

TEMPERATURE DOM. & INDST. WASTEWATER, HEATING WITHIN PLUMBING SYSTEM

Waste water temperature > supplied water

SOLIDS DOM. & INDST. WASTE, SILT 1000 L wastewater contains nearly 500 gm solids (TS, TDS, TSS, TVS, VSS); Imparts turbidity

PARAMETERS OF IMPORTANCE

Total Solids Turbidity Dissolved Oxygen Biochemical Oxygen Demand, Chemical Oxygen demand, Total organic Carbon, Theoretical Oxygen Demand pH Phosphorus Nitrogen Toxic Inorganic Compounds Heavy metals Total and fecal coliform (Most Probable Number)

Approximate Sizes of Environmental Particles

1 10 100 0.1 0.01 0.001

Particle Size (mm)

VIRUS

ALGAE

BACTERIA

SILT CLAY SAND

COLLOID

Dissolved SOLIDS

Settle-able SOLIDS

SUSPENDED SOLIDS

SETTLEABLE SOLIDS 1 L sample is taken in an Imhoff

cone

The sample is allowed to stand undisturbed for 2 hrs

The volume of settleable solids can be noted from the

graduations on the cone.

The solids settled at the bottom of the cone can be dried at 105⁰C and weighed

to find conc. In mg/l

Non-settleable solids = (Total suspended solids - Settleable

solids)

VACUUM FILTRATION APPARATUS

Glass Fiber Filter of 1.2 micron pore size is used to filter the wastewater samples

Characterization of Solids in Wastewater

WATER SAMPLE

VA

VB

105 Deg C

8 h

MB

550 Deg C

>1 h

MC

Evaporation Volatile solids escape

Inorganic Ash

105 Deg C

8 h

Filter

ME

550 Deg C

>1 h

MF

105 Deg C

8 h

MG

Total Solids= MB/VA

Total Dissolved Solids= MG/VB

Suspended Solids= ME/VB

Total Volatile Solids= (MB-MC)/VA

Volatile Suspended Solids= (ME-MF)/VB

1 L sewage

~ 1,000,000 mg

Water

~ 999500 mg

Solids

~ 500 mg

Dissolved ~ ½ (~ 250 mg)

Ca++, Mg++, Na+, HCO3-, SO4

2- ,Cl-,

soluble organics

Insoluble ~ ½

~ 250 mg

Settleable

~ 125 mg

Suspended, turbid WW, ~ 125 mg

pH OHHOH2

K

14

2

10][

]][[

OH

OHHK Being pure phase, [H2O] = 1

1410]][[ OHHKw

]log[ HpH

Kw is the dissociation constant of water

]log[ OHpOH

For pure water, [H+] = [OH-] 714

14

1010][

10]][[

H

HH

710log]log[ 7 HpH

If pH < 7, the wastewater is termed as acidic. pH>7 it is called alkaline. Normally domestic wastewaters tend to be slightly alkaline, from 7.0 to 7.5

Chemical Characteristics

Biodegradable fraction:

-Carbohydrates

-Protein

-Fats, Oils, and Greases

-Surfactants (detergents)

-Urea (agricultural run-off)

Organics

Domestic wastewater predominantly contains dissolved organic matters, so does many industrial wastewaters.

Bio-degradable

Non- Biodegradable

Priority pollutants

– Industrial solvents, pesticides, etc.

HOW TO QUANTIFY THESE DIVERSE GROUP OF ORGANIC CHEMICALS?

Bio-degradation

1. Wastewater contains organics such as glucose

2. Microorganism utilize the organic compounds as their food

3. Assimilation of food means consumption of oxygen dissolved in water

4. Oxygen consumed by microbes is replenished by mass-transfer from air

5. If there is a mismatch in the rates, oxygen gets depleted, causing the death of aquatic species

Representation by effect Conc. of specific substances – not found Sewage – mixture of ill-defined chemicals Representation by effect is used The strength of the mixture is defined by some common factor on which all the chemicals within the mixture depend. Ex. O2 depletion from biological/chemical decomposition of the chemical mixture. For many organic - bearing wastes, instead of identifying 100’s of individual compounds, it is convenient to report the effect, in units of the mg of O2 that can be consumed/L of water. Referred as BOD / COD

Predominant elemental make-up of organic matter: C, H, O, N, P, S

In most of the organic compounds, elements like Carbon, Nitrogen, Phosphorus and Sulfur are not present in their highest oxidation state.

Oxidation is a process of transferring of electrons from one species to the other

Highest Oxidation State: +4 +5 +6

Oxygen requirement or oxygen demand for the complete oxidation of organics can serve as an aggregate measurement of the amount of organics present in a wastewater sample. .

Organic compounds can undergo an unique chemical reaction, OXIDATION.

Inorganic dissolved species in wastewater cannot undergo oxidation as most of them are present as already oxidized to the extent possible.

The species which losses electron(s) from their outer electronic shell is considered to get oxidized and the one gaining electron(s) is called an oxidant(it itself gets reduced)

Common Oxidants are Oxygen, Chromium (VI), elemental chlorine, etc.

Oxygen Demand Specifies how much oxygen is required to completely oxidize the organic matter present in wastewater sample

OHCOOOHC 2226126 666

For 1 mM glucose

180 mg/L 192 mg/L

Stoichiometry:

A wastewater containing 1mM glucose solution can be quantified in two ways:

a) A solution with glucose concentration of 180 mg/L

b) A solution with theoretical oxygen demand of 192 mg/L

Derived from the chemical formula, difficult when the formula or type of organic is unknown

Experimental determination is possible; does not require quantification of each one of the individual organic component; value is somewhat less than the theoretical oxygen demand

Steps to calculate ThOD Example - C6H6 : Benzene : 156 mg/L Step 1: Write the Eq. (Oxidation to CO2 and water) C6H6 + O2 CO2 + H2O Step 2: Balance the Eq. C6H6 + 7.5 O2 6 CO2 + 3 H2O Sequence for balancing the no. of atoms: (i) C 6, (ii) H 3; and (iii) O 7.5 78 mg/L 240 mg/L Step 3:

L

mgO

moleO

gOx

emolebenzen

moleOx

gbenzene

emolebenzenx

L

mgbenzene 2480

2

23225.7

78

1156

Oxygen Demand

Bio-Chemical (BOD)

Chemical (COD)

Measured by quantifying O2 used by the microbes for oxidizing the organics in wastewater

Strong Oxidants like Cr(VI) species are used for quick assessment

DETERMINATION OF OXYGEN DEMAND

Oxidation-reduction reactions are slow processes. Microbial oxidation reactions are essentially slow processes as they involve enzymes acting as catalyst in the complex sub-steps in the oxidation reactions.

Biochemical Oxygen Demand

• Amount of oxygen required by bacteria while

stabilizing the decomposable organic matter

under aerobic conditions.

• It involves the measurement of oxygen

consumed by living organisms.

Biochemical Reaction

Oxygen

New Cells/CO2

Biochemical Oxygen Demand

Measurement

• Take sample of waste; dilute with oxygen saturated

water; add nutrients and microorganisms (Seed: if not

present).

• Measure dissolved oxygen (DO) levels over 5 day.

• Temperature 20° C.

• In dark (prevents algae from growing), Plastic Bottle.

• Final DO concentration must be > 2 mg/L .

• Need at least 2 mg/L change in DO over 5 days.

Simple BOD Measurement

Measure DO of the sample

Put into 20oC incubator for 5 days or 27o C for 3 days

Measure DO after 3 or Five days

BOD Determination

Wastewater Dilution Water

Seed Microorganism

Make total volume to 300 mL

Measure Initial Dissolved Oxygen (DOi) concentration (mg/L)

Maintain a constant Temperature; usually 20 deg C; kept away from light;

After a Specific Time-period, Usually 5 days

Measure Final Dissolved Oxygen (DOf) concentration (mg/L)

Microbes grow in number, Utilize the food, consumes O2, concentration of O2 falls

P

DODOLmgBOD

fi

d

)/(deg20,5

P = volumetric fraction of wastewater used in the BOD reactor

BOD Calculations

• If initial DO of a sample is 8 mg/L and final DO after 5 days is 2 mg/L. What is the BOD of the Sample.

• If initial DO of a sample is 8 mg/L and final DO after 5 days is 0 mg/L. What is the BOD of the Sample

41

Example: 5 ml wastewater is added to a 300 ml BOD flask. DOi = 8 mg/L

DOf = 2 mg/L after 5 days. What is the value of BOD5

P = 5 = 0.0167

300

BOD5 = 8 – 2 = 359 mg/L

0.0167

When seed is added, we need to consider the oxygen demand generated by the initial seed microorganism also. To evaluate this, another parallel test is run with the seed and the dilution water but without the addition of wastewater.

Normally Domestic sewage contains microorganisms; For industrial wastewater, this may not be the case; there, addition of seed microorganism may be necessary.

Air Essential nutrients

Bacteria (seed)

Distilled Water

Seeded Dilution Water

Dilution water

Dilution water

300 mL BOD bottles

Seeded Blank Seeded Sample

Waste Sample, Vs

Organic matter and no bacteria or limited number of bacteria

300 mL

300 - Vs

P

fBBDODOLmgBOD

fi

d

)()()/(

21

deg20,5

B1, B2

blank seeded inwater dil. seeded of volume

sample seeded the inwater dil. seeded of volumef

volume combined total

sample in wastewater of volumeP

volume combined total

sample) inwater dil seeded of volume -volume combined (total f 1

DOi, DOf

Valid only when seeded blank has the same volume as the combined total volume (In this case both are 300 mL)

B1 and B2 = Initial and final DO of the control run with seed only

BOD Reaction Kinetics

First Order Reaction

DO utilization curve during BOD test

DO consumed in 5 days (Oxygen equivalent of organics destroyed)

DOf

DOi

BOD5= Oxygen equivalent of Organic destroyed in 5 days = DO consumed in 5 days

1st few days – High conc. of org. matter present / Rapid rate of O2 depletion As org. matter decreases / Later- rate of O2 consumption also decreases Last phase - O2 consumption associated with decay of bacteria those grew earlier Assumption: Rate of O2 consumption Proportional to conc. of degradable org. remaining at any time 1st order reaction

BOD reaction is a first order reaction

Rate of change in reactant concentration Amount of reactant present at any time

Ldt

dL

kLdt

dL kdt

L

dL

L= Oxygen equivalent of biodegradable organics present at time t, mg/L

Integrating we get, dtkL

dLCktL ln

At time t = 0, L= L0 L0= Oxygen equivalent of biodegradable organics present at t=0, mg/L

ktL

L

0

ln kteLL 0

L or Lt is often known as BOD remaining at time t

kt

t eLL 0or,

k =BOD rate constant, day-1

0 5 10 15 20

Bio

de

grad

able

Org

anic

s (B

OD

) re

mai

nin

g, m

g/L

Time, days

BO

D E

xert

ed

, mg/

L

0L

tL

)1(00

kt

tt eLLLy

)( tt BODy

uBOD

uBODL 0

Example: In the previous example we found out that BOD5 of the

wastewater sample was 359 mg/L. Find out the ultimate BOD of the

wastewater sample. Also find out the value of BOD10 . K =0.23 per day

359)1(5*)23.0(

055 eLyBOD

359683.0*0 L

L0= BODu=525.62 mg/L

mg/L 92.472)1(*62.52510*)23.0(

1010 eyBOD

)101(0

Kt

t Ly

BOD Eq. in base 10:

Capital K k = 2.303 (K)

BOD rate constant (k)

This can be determined from experimental data. Ideally, if we have more than one data point on the BOD curve, we can find out k from the curve.

Experimental observations vary depending on variable experimental conditions, so k is estimated from a set of experimental data, by best-fitting a linearized BOD Curve.

Usual value of k (base e) is 0.23 per day at 20 deg C.

Value of k varies with temperature.

20

20

T

T kk

047.1Generally,

Effect of Temperature on BODt and BODu

L0 BODu

BODt

BODt varies with temperature but BODu, being intrinsic property of wastewater, does not change

Nature of the waste 1000’s of naturally occurring organic compounds Not all can be degraded with equal ease Simple sugars and starches Rapidly degrade Large k Cellulose Degrade slowly Lower k Hair Almost nondegradable k ? Sewage k depends on relative proportion Typical values for k:

Sample K (20 °C) (day -1)

k (20 °C) (day -1)

Raw sewage 0.15-0.30 0.35-0.70

Well-treated sewage 0.05-0.10 0.12-0.23

Polluted river water 0.05-0.10 0.12-0.23

Lower k: Easily degradable Organic compounds -More completely removed during treatment.

For different types of wastes having same BOD5 L0 is the same only if values of K are same. Industrial Waste has smaller k, they have greater L0 but

same BOD5. It is expected to have greater impact on DO in river. Smaller fraction of BOD exerted in 1st 5 days due to lower K.

Polluted river water: BOD5 at 20°C = 50 mg/L, K = 0.115/day L0 = 68 mg/L River water temp. = 10°C K at 10°C = 0.032/ day BOD5 at 10°C = 21 mg/L Lab. Determined value of BOD5 at 20°C (50 mg/L) overestimates O2 consump- tion in the river at 10°C (21 mg/L).

)101(0

Kt

t Ly

20

20

T

T kk

K = 0.115/day

Effect of K on L0 for 2 wastewaters having same BOD5

Effect of K on BOD5 when L0

Is same.

Graphical Determination of BOD Constants, k and L0

BOD data/ plot BOD versus time/ hyperbolic first-order curve/ asymptote- L0

Difficult to fit an accurate hyperbola to scattered data Methods that linearize data preferred Thomas Graphical Method (Thomas, 1950) Relies on similarity of the series expansion of the following two functions:

Nitrogen Oxidation Up to this point we assumed that only C in organic matter is oxidized. Actually many organic compounds, such as proteins, also contain N that can be oxidized with the consumption of O2. However, mechanisms and rates of N oxidation are distinctly different from those of C oxidation. Two processes must be considered separately. O2 consumption due to oxidation of C called carbonaceous BOD (CBOD). due to N oxidation called nitrogenous BOD (NBOD) Organisms that oxidize C to obtain energy can not oxidize N. Instead, N is released into water as ammonia (NH3) At normal pH, NH3 is present as ammonium cation (NH4

+) NH3 from organics + ind. wastes + agricultural runoff (fertilizers) oxidized NO3

-

by nitrifying bacteria (nitrification) The overall reaction for ammonia oxidation : NH4

+ + 2O2 ----microorganisms

-- NO3- + H2O + 2H+

Theoretical NBOD = g of O2 used / g of N oxidized = (4x16)/14 = 4.57 g O2 / g N

0 10 20 30

BO

D e

xert

ed

(B

OD

), m

g/L

Time, days

Ultimate BOD

Conversion of Ammonium to Nitrite (Nitrosomonas)

NH4+ + 2 O2 Bacteria (Nitrosomonous) NO2

- + 2 H+ + H2O

Conversion of Nitrite to Nitrate (Nitrobacter)

NO2- + 0.5 O2 (Nitrobactor) NO3

- Inhibitor for nitrogenous BOD reaction: Methylene blue, Thiourea.

Nitrogenous BOD (NBOD)

Carbonaceous BOD (CBOD)

Carbonaceous

Rate of nitrification depends on number of nitrifying organisms Untreated sewage: Few nitrifying organisms, NBOD exerted after much of CBOD exerted due to lag in growth Well - treated sewage: High conc. of nitrifying organisms and Less lag time Same BOD Eq. K = 0.04 – 0.10 / day (As for CBOD of well-treated effluent) Same Eq. for temp. correction

LIMITATIONS OF A BOD TEST

1. Non-biodegradable organic waste is unaccounted for.

2. Wastewater with high BOD content will use up all dissolved oxygen before the 5 days is over. Proper dilution is necessary.

3. Industrial wastewater with no initial microorganism load shall require inoculation of ‘seed’ bacteria. The bacteria should be acclimatized to the wastewater, otherwise may generate a lower BOD number.

4. The test is for a long duration. A faster test is much required.

5. Depends on the activity of the microbes only; presence of toxic substances such as heavy metals can inhibit the growth of the microbes.

Origin 5-day BOD (mg O2/L )

River 2

Domestic wastewater 200

Pulp and paper mill 400

Commercial laundry 2000

Sugar beet factory 10000

Tannery 15000

Brewery 25000

Cherry-canning factory 55000

BOD of Selected samples

COD is measured following digestion at high temperature with strong oxidant such as chromic acid, or sulfuric acid/potassium dichromate.

Chemical Oxygen Demand (COD)

The chromate ion reacts with the COD producing a color that is measured to determine the amount of chromate ion reacted. The oxygen equivalence of chromate ion is known as COD.

It is a fast process. BOD test is a 5-day test. COD test takes not more than 3 hours altogether.

The test is done with a strong oxidizing agent. So, all the organics, whether biodegradable or non-biodegradable, shall be oxidized.

COD Biodegradable + Non-biodegradable

BOD Biodegradable

Chemical Oxygen Demand Oxygen equivalent of the organic matter that can be oxidized by a strong oxidizing agent (potassium dichromate) in an acidic medium. COD > BOD5: (a) Because more compounds can be oxidized chemically than can be oxidized biologically and (b) Because BOD5 does not equal ultimate BOD COD: 3 h BOD: 5 d

C H O Cr O H nCO Cra

H On a b

2 7

2

2

3

28 2 42

2

3 6 3

n a bWhere:

Stochiometry of COD

Organic Matter

Strong Oxidant Potassium Dichromate

Sulphuric Acid

Carbon dioxide

HEATING 2 HOURS 150 OC

Chromic acid

Orange Colored

Green Colored

Take reading in spectrophotometer

Take 2.5 ml sample in COD vial

Add 1.5 ml K2 Cr2O7in it

Add 3.5 ml sulphuric acid reagent

Digest above solution in digester for 2 hr at 150 oC

COD and BOD - Comparatives

COD Biodegradable + Non-biodegradable

BOD Biodegradable

10 COD

BOD

For a completely biodegradable wastewater,

1 to 9.0COD

BODu

6.0COD

BODuFor wastewater with , it is considered non-biodegradable

Theoretically, for a completely biodegradable wastewater

CODBODu

Theoretical Oxygen Demand

This is an oxygen demand that is calculated using stoichiometry, from the chemical formula of a compound considering that there is a complete degradation.

It also includes the complete oxidation of ammonia that is formed in the first stage of reaction.

Reaction:

3222 dNHOcHbCOaONOHC rpnm

Stage 1

Stage 2 HOHNOONH 2323 2

ThOD = BOD = COD ? Is it possible? Rare Chemical composition of ALL the substances known : ThOD Capable of being completely oxidized chemically : COD / biologically : BOD Then, ThOD = BOD = COD

Oxygen Demand: Definition and Notation (All terms have units of mg O2 / L )

BOD Biochemical O2 demand – Amount of O2 utilized by microorganisms in oxidizing carbonaceous and nitrogenous organic matter.

CBOD Carbonaceous biochemical O2 demand – BOD where electron donor is carbonaceous organic matter.

NBOD Nitrogenous biochemical O2 demand – BOD where electron donor is nitrogenous organic matter.

ThOD Theoretical O2 demand - Amount of O2 utilized by microorganisms in oxidizing carbonaceous and/ or nitrogenous organic matter, assuming all of the organic matter is subject to microbial breakdown, i.e., it is biodegradable.

BOD5 5 – day biochemical O2 demand - Amount of O2 utilized (BOD exerted) over an incubation period of 5 days. y5.

BODu Ultimate biochemical O2 demand - Amount of O2 utilized (BOD exerted) when all of the biodegradable organic matter has been oxidized. L0.

COD Chemical O2 demand - Amount of chemical oxidant, expressed in O2

equivalents, required to completely oxidize a source of organic matter; COD and ThOD should be near equal.

TOC is measured using a TOC analyzer. The sample is catalytically

combusted and organic carbon is quantified using infrared detection of

carbon dioxide.

TOTAL ORGANIC CARBON (TOC)

NITROGEN

RECEIVING BODY - High NH3-N: Toxic to fish

- Low NH3 & NO3-N : Nutrient for algae & aquatic plants

- DO exertion if longer residence

time (NH4+ to NO3)

WASTE WATER TREATMENT

- Sufficient N required for optimal growth of microorganisms

- Domestic water has good C:N:P ratio

- Ind. WW: Some may lack adequate N

Significance of Nitrogen

Nitrogen • Indicator of sanitary condition

• Initially entire ‘N’ as Protein-N & Ammonia-N

Org-N converted to Ammonia-N

Oxidation to Nitrite-N

Oxidation to Nitrate-N

•Dominant species & implication

– Org. & Ammonia-N (TKN)

=>Fresh potentially dangerous

– Nitrate-N

=> Polluted long back,

Little public health threat

Nitrogen An indicator of sanitary quality Chemical tests Chloride – No evidence of how recently the contamination had occurred Nitrogen – Most of N originally present as organic (protein) N and ammonia As the time progresses, org. N is gradually converted to ammonia N Later on, if aerobic condition present, oxidation of ammonia to nitrites and nitrates occurs For ex. (a) Waters that contained mostly org. and ammonia N – considered to be recently polluted - of great potential danger (b) Waters in which most of the N was in the form of nitrates were considered to be polluted long back – offered little threat to public health Bacteriological Tests (about 1893) provides more reliable evidence concerning hygienic safety of water – has eliminated the need for extended N analysis in most water supplies

Significance of Nitrogen (a) In receiving body: 1. In high concentrations, NH3-N is toxic to fish. 2. NH3 (in low concentrations), and NO3- serve as nutrients for excessive growth of algae. Plants require, in order of abundance in plant tissue: C, N, P, and a variety of trace elements. Ex. thick slime layers on rocks, dense growth of aquatic weeds. 3. The conversion of NH4

+ to NO3- consumes large quantities of DO.

Especially where long residence times are available. (b) In wastewater treatment

1. Biological treatment depends on reproduction of the organisms. Sufficient N required for the organisms. 2. Domestic wastewater has good C:N:P ratio. 3. In some industrial wastewaters, N is deficient. It must be added.

Phosphorus (1) Vital nutrient for the growth of algae. (2) When algae die, they become an O2 – demanding organic material as bacteria seek to degrade them. (3) This O2 – demand frequently overtakes DO supply and, as a consequence, causes fish to die.

BIOLOGICAL CHARACTERISTICS

• Microorganisms : bacteria, protozoa, worms, and virus.

• Pathogenic organisms: Common waste water related diseases- hepatitis A, typhoid, polio, cholera, and dysentery.

• Major human disease transmission route:

faecal-oral: – direct(bad personal hygiene)

– indirect ( contaminated food/water)

Pathogens in wastewater • Bacteria – Single celled entities

– Size range: 0.5-5 micron

– Consumes soluble food &

capable of self-reproduction

– Diseases: typhoid, paratyphoid,

dysentery, and cholera

• Viruses – Intracellular parasite

– Size range: ~20-100 nm (approx. 1/50 th of a bacteria)

– Diseases: These are Adenovirus (Respiratory and eye infections), Poliovirus, Hepatitis A virus, Echovirus (aseptic meningitis), Rota virus and other virus causing gastroenteritis, diarrhoea.

E.co

li Po

lio v

iru

s

• Protozoa –Live attached to the human intestine where they actively feed and reproduce. –Common diseases: diarrhea and dysentery. –Example: Entamoeba histolytica and Giardia lamblia. –At some point in their life cycle they undergo a morphological

transformation into a cyst for protection against harsh environment outside the host. The cyst form is infectious to other persons by the faecal-oral route of transmission.

–The cysts have size 10-15 micron.

• Helminthes - Intestinal worms; do not multiply inside human - Worm burden in infected person is related to no. of helminthic eggs ingested. - Size of egg: 40-60 micron; heavier than water

Pathogens in wastewater

CENTRAL QUESTION OF BIOLOGICAL CHARACTERIZATION

• Human carriers exist for all enteric diseases. The fecal-oral route is the causative pathways in almost all the cases of the disease outbreak causing public health crisis situations.

• Fecal contamination in wastewater causes the presence of the pathogens. If not adequately treated, the wastewater and subsequently, the drinking water shall contain the infectious agents.

• So, in order to be safe, it is regular practice to test water for the presence of pathogens.

• But, how to assess the pathogenic quality when there are so many varieties of microorganisms in wastewater or water??

Indicator Organism •Concept: Rather than testing for each and every pathogen, it is easier to test for only

one group of microorganism whose presence is an assured evidence/ indication that the wastewater has been polluted by faeces of humans or warm-blooded animals. This microorganism may be called an indicator organism.

•Indicator organism: Escherichia coli

•Characteristics of E. coli that makes it suitable indicator:

–Non-pathogenic faecal coliform bacteria that reside in the human intestinal tract.

–Excreted in large numbers in faeces, often amounting to about 50 million per gram.

–Untreated domestic sewage contains upwards of 3 million coliforms per 100 mL.

–E. Coli. persists in the environment outside the human intestine for a longer duration than the other pathogenic bacteria.

–Virus, protozoan cysts and helminth eggs are more persistent than E. Coli.

– But, regular wastewater treatment operations kill all the other pathogens as well as E. coli. So, E.coli ’s presence is an indicator of presence of pathogens.

–Also, its absence means the faecal contamination is absent.

– It is easier to detect E.coli.

–The severity of faecal contamination is considered to be directly related with the concentration of the E.coli. bacteria in the water or wastewater.

E. coli (indicator of faecal contamination of waste water)

• E. coli colony SEM image of E.coli

Fermentation Tube Technique

Cap

Lactose broth

Inverted Vial

Fermentation Tube

Wastewater Sample

Incubation

@ 35 deg C Negative

Positive

Growth with gas evolution inside the inverted vial

No growth and no gas evolution

Multiple Tube Fermentation Technique and Most Probable Number

X √ X √

X X √ X √

X X √ X √

1 mL WW

0.1 mL WW

0.01 mL WW

HOW TO STATISTICALLY INTERPRET THE RESULTS?? √

Statistically found concentrations are termed as Most Probable Number (MPN) of the coliform bacteria present in the wastewater

Two methods of interpretation of the results of multiple tube fermentation test:

1. Thomas’ empirical method 2. Poisson’s statistical method

Multiple Tube Fermentation Technique and Most Probable Number (cont’d)

Thomas’ Formula:

100 tubes theallin samples of mL X tubesnegativein samples of mL

tubespositive ofNumber

mL 100(MPN)/ Number ProbableMost

X

])()1][()()1][()()1[(1

333322221111 qvpvqvpvqvpveeeeee

ay

Poisson’s statistical method

y = probability λ = Coliform density/ mL vi = sample portion, mL

pi = Number of positive tubes qi = Number of negative tubes

Maximize y (or ya) by trial and error for different values of λ

a= constant

Derivation of Poisson’s Formula

Consider that a small sample v is taken out of the total wastewater sample volume V.

If there is one single microorganism in volume V.

Probability that the small sample contains the microorganism V

v

Probability that the small sample does not contain the microorganism V

v1

If there are b number of microorganism , probability that the small sample does not contain the microorganism

b

sV

vP

1

If is very small, then V

v

V

vbPs exp

V

bis the density of the microorganism = λ

vPs exp

S stands for sterile

So, probability of finding a positive or fertile sample is

]exp1[1 vPP sf

In case of multiple number of tubes, if n samples of volume v is taken, the probability of finding p fertile samples is given by binomial distribution.

q= number of negative tubes = n-p

Denote

])()1][()()1][()()1[(1

333322221111 qvpvqvpvqvpveeeeee

ay

For different dilutions or sample sizes, 1,2, 3 the probability function takes the following form