Wastewater Engineering

1. CE-202: Wastewater Engineering
Sudipta Sarkar

2. Why Study Water and Wastewater Engineering as part of Civil Engineering?
CIVIL
ENGINEERING

5. 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

6. RAW
WATER
TREATED WATER
WASTEWATER
TEATED
WASTEWATER
WASTEWATER TREATMENT PLANT
WATER TREATMENT PLANT

9. 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 PRE15 15 30 040
3 0

13. 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

14. 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.

15. CHARACTERIZATION OF WASTEWATER

16. 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.

17. 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.

18. 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

19. 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

20. 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)

21. Approximate Sizes of Environmental Particles
1 10 1000.10.010.001
Particle Size (mm)
VIRUS
ALGAE
BACTERIA
SILTCLAY SAND
COLLOID
Dissolved
SOLIDS
Settle-able SOLIDS
SUSPENDED SOLIDS

22. 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)

23. VACUUM FILTRATION APPARATUS
Glass Fiber Filter of 1.2
micron pore size is used
to filter the wastewater
samples
Characterization of Solids in Wastewater

24. 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

26. 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

28. pH

 OHHOH2
K
14
2
10
][
]][[ 


OH
OHH
K Being pure phase, [H2O] = 1
14
10]][[ 
 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

29. 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

30. 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

31. 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.

32. 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

33. 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
gO
x
emolebenzen
moleO
x
gbenzene
emolebenzen
x
L
mgbenzene 2480
2
23225.7
78
1156


34. 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.

35. 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.

36. Biochemical Reaction
Oxygen
New Cells/CO2

37. 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.

39. 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

40. 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
DODO
LmgBOD
fi
d

)/(deg20,5
P = volumetric fraction of wastewater
used in the BOD reactor

41. 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

42. 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.

43. AirEssential
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
fBBDODO
LmgBOD
fi
d
)()(
)/(
21
deg20,5


B1, B2
blankseededinwaterdil.seededofvolume
sampleseededtheinwaterdil.seededofvolume
f
volumecombinedtotal
sampleinwastewaterofvolume
P
volumecombinedtotal
sample)inwaterdilseededofvolume-volumecombined(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

44. 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

45. 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

46. BOD reaction is a first order reaction
Rate of change in reactant concentration Amount of reactant present at
any time

L
dt
dL

kL
dt
dL
 kdt
L
dL

L= Oxygen equivalent of biodegradable
organics present at time t, mg/L
Integrating we get,   dtk
L
dL
CktL ln
At time t = 0, L= L0 L0= Oxygen equivalent of biodegradable organics
present at t=0, mg/L
kt
L
L

0
ln kt
eLL 
 0
L or Lt is often known as BOD remaining at time t
kt
t eLL 
 0or,
k =BOD rate constant, day-1

47. 0 5 10 15 20
BiodegradableOrganics(BOD)
remaining,mg/L
Time, days
BODExerted,mg/L
0L
tL
)1(00
kt
tt eLLLy 

)( tt BODy
uBOD
uBODL 0

48. 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/L92.472)1(*62.525 10*)23.0(
1010  
eyBOD

49. )101(0
Kt
t Ly 

BOD Eq. in base 10:
Capital K
k = 2.303 (K)

50. 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 
047.1Generally,

51. Effect of Temperature on BODt and BODu
L0 BODu
BODt
BODt varies with temperature but BODu,
being intrinsic property of wastewater,
does not change

52. 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.
 

53. 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.

54. 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:

56. 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

57. 0 10 20 30
BODexerted(BOD),mg/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

58. 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

59. 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.

60. 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

61. 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

62. 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

63. C H O Cr O H nCO Cr
a
H On a b      





  
   2 7
2
2
3
28 2 4
2
   
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

64. 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

65. COD and BOD - Comparatives
COD Biodegradable + Non-biodegradable
BOD Biodegradable
10 
COD
BOD
For a completely biodegradable wastewater,
1to9.0
COD
BODu
6.0
COD
BODuFor wastewater with , it is considered non-biodegradable
Theoretically, for a completely biodegradable wastewater
CODBODu 

66. 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

67. 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

68. 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.

69. 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)

70. 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

71. 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

72. 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

73. 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.

74. 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.

75. 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)

76. 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.coliPoliovirus

77. • 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

78. 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??

79. 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.

80. E. coli (indicator of faecal
contamination of waste water)
• E. coli colony SEM image of E.coli

81. 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

82. Multiple Tube Fermentation Technique and Most Probable Number
X √ X √
X X √ X √
XX√ 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

83. 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
tubestheallinsamplesofmLXtubesnegativeinsamplesofmL
tubespositiveofNumber
mL100(MPN)/NumberProbableMost
X

])()1][()()1][()()1[(
1 333322221111 qvpvqvpvqvpv
eeeeee
a
y  

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

84. 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
v
1
If there are b number of microorganism , probability that
the small sample does not contain the microorganism
b
s
V
v
P 





 1
If is very small, then
V
v







V
vb
Ps exp
V
b
is 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 

85. 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 qvpvqvpvqvpv
eeeeee
a
y  

For different dilutions or sample sizes, 1,2, 3 the probability function takes the following form