1. WASTEWATER
ENGINEERING
To impart basic knowledge of waste water
characterization, collection, treatment and safe
disposal practices.
1
TARGET:
BOD and Nutrient Removal
8. Point sources
Wastes that are collected in pipes or
channels, drains and discharged to a surface
water with or without treatment
Distinguished by source
– municipal sewage or wastewater
– industrial waste waters
– combined sewers and combined sewer overflows
8
11. Nonpoint Sources
Storm/Rain water runoff discharged at
multiple points
Varies substantially with use of the land runoff
originates from
– agricultural
– urban
– suburban
– commercial
– special (e.g. golf courses)
Minimal regulations
11
16. CONSTITUENTS OF WASTEWATER
Biodegradable organics: Organic matter (BOD)- Consume
oxygen, create anaerobic (obnoxious conditions in water bodies and
environment)
Suspended solids: (SS)- Settle & degrade aerobically, thereafter
anaerobically creating (obnoxious conditions in water bodies and
environment)
Pathogens – Causes Communicable diseases
Nutrients – Nitrogen & Phosphorus. (Important if treated wastewater
is discharged into Ocean and enclosed water bodies , stagnant rivers
(lakes, reservoirs, ponds)
Dissolved solids (TDS/Salts) : Salt accumulation in agr. fields
Heavy metals – Cr, Cd, Hg, Pb : Long term effects, accumulated in
food chain
Refractory organics: CFC, Benzene, pesticides, EDCs etc.:
Carcinogenic- Long Term Health Effects , accumulated in food chain
16
17. Organics in Water
CHNO
Natural Sources
Human Activities
Biodegradable Non or slowly-Biodegradable
Can be utilized by
naturally occurring
Microorganisms
Starch Fats Protein Alcohols,
acetic acids, aldehydes, esters
Resistant to biological
degradation: Exceptionally strong
bonds
Tannic acid, lignic acid, cellulose,
phenols,Polysaccharides, benzenes,
detergents, insecticides, pesticides
17
20. Oxygen Demanding Organics
When organic substances are degraded by
bacteria in water, oxygen is consumed
organic C + O2 → CO2
Organic C : Carbohydrates,
Fats, proteins etc.,
Oxygen Consumption Equivalent
BOD: Biodegradable Organics
COD for Biodegradable and non-
bio-degradable Organics
Carbon Equivalent: TOC
20
21. Equivalent of Organics
Oxygen Equivalent/Demand of Total
Organics: COD
Oxygen Equivalent/Demand of Biodegradable
Organics: BOD
Oxygen Equivalent/Demand of non-
Biodegradable Organics: COD-BOD
Carbon Equivalent of Total Organics: TOC
21
22. Oxygen Demand
If 1 mg/L Biomass or VSS C5H7NO2 is
completely oxidized to CO2 Find
1. Theoretical Oxygen Demand (ThOD)
2. Chemical Oxygen Demand (COD)
3. BODult & Biochemical Oxygen Demand (BOD5), If rate constant is 0.23 day-1
4. Total Organic Carbon (TOC)
C5H7NO2 + 5 O2 ------------------ 5CO2 +NH3 + 2 H2O
mw = 113 mw =160
ThOD = 160/113 = 1.42 mgO2/mg C5H7NO2 or 1.42 mg/L
COD = 1.42 mgO2/mg C5H7NO2
BODult = COD
BOD5 (5 days O2 Consumption) =BODult (1-e-5x0.23) = 0.68 x 1.42 = 0.97 mg
BOD5/ mg C5H7NO2
TOC =5 x 12 / 113 = 0.53 mg TOC/ mg C5H7NO2
22
23. Repeat the Problem for Glucose
C6H12O6 + 6O2 ---> 6CO2 + 6H2O + Energy
23
24. Step 1:
Write the Eq. (Oxidation to CO2 and water) C6H6 + O2 CO2 + H2O
Example - C6H6 : Benzene : 156 mg/L
L
mgO
moleO
gO
x
e
molebenzen
moleO
x
gbenzene
e
molebenzen
x
L
mgbenzene 2
480
2
2
32
2
5
.
7
78
1
156
24
Step 2:
Balance the Eq. C6H6 + 7.5 O2 6 CO2 + 3 H2O
Balance the Equation
1 mg/L Benzene consumes = 3 mg/L Oxygen,
Biomass – 1.42 mg/L Oxygen
Glucose- 1 mg/L Oxygen
25. 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.
25
26. 26
Biochemical Oxygen Demand
Measurement
Take sample of waste; dilute with oxygen saturated
water; add nutrients and microorganisms (seed).
Measure dissolved oxygen (DO) levels over 3 days or 5
days.
Temperature 20° C for 5 days or 27oC for 3 Days
In dark (prevents algae from growing)
Final DO concentration must be > 2 mg/L .
27. Air
Essential
nutrients
Bacteria
(seed)
Tap/Distill
ed 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
f
B
B
DO
DO
L
mg
BOD
f
i
d
)
(
)
(
)
/
(
2
1
deg
20
,
5
B1,
B2
blank
seeded
in
water
dil.
seeded
of
volume
sample
seeded
the
in
water
dil.
seeded
of
volume
f
volume
combined
total
sample
in
wastewater
of
volume
P
DOi, DOf
B1 and B2 = Initial and final DO of the control run with seed only
27
If 3.00 mL of wastewater to the
300.0 mL BOD bottles. What is f
and P
29. Voltage is applied between the
two electrodes - Threshold
diffusion current for oxygen is
generated.
Dissolved oxygen diffuses
across the membrane at a rate
proportional to the pressure of
oxygen in the water
A reduction current in
proportion to the dissolved
oxygen is generated, and then
the dissolved oxygen is
measured.
DO METER
DO reduced by Ag electrode
Oxygen 29
30. 30
Example 1
A BOD test was conducted in the laboratory
using wastewater being dumped into River
Yamuna. The samples are prepared by adding
3.00 mL of wastewater to the 300.0 mL BOD
bottles. The bottles are filled to capacity with
seeded dilution water.
31. 31
Example 1: Raw Data
Time
(Days)
Diluted Sample DO
(mg/L)
Blank Seeded Sample
DO (mg/L)
0 7.95 8.15
1 3.75 8.1
2 3 8.05
3 2.75 8
4 2.15 7.95
5 1.8 7.9
32. Example 1: Calculations
What is the BOD5 of the sample?
Plot the BOD with respect to time.
sample
diluted
the
of
ions
concentrat
DO
final
and
initial
DO
,
DO
(blank)
water
diluted
seeded
the
of
ions
concentrat
DO
final
and
initial
B
,
B
P
)]
)(
B
(B
-
)
DO
[(DO
BOD
f
i
f
i
f
i
f
i
m
f
blank
seeded
in
water
dil.
seeded
of
volume
sample
seeded
the
in
water
dil.
seeded
of
volume
f
volume
combined
total
sample
in
r
wastewate
of
volume
P
32
33. 33
Example 1: Time – Concentration
Plot
0
100
200
300
400
500
600
700
0 1 2 3 4 5 6
Time (days)
BOD
(mg/L)
35. Oxygen Demanding Organics
Degradation
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,
dt
k
L
dL
kt
L
L
0
ln kt
e
L
L
0
L or Lt is often known as BOD/Organic Matter remaining at time t
kt
t e
L
L
0
or,
k =BOD rate constant, day-1
35
36. 0 5 10 15 20
Biodegradable
Organics
(BOD)
remaining,
mg/L
Time, days
BOD
Exerted,
mg/L
0
L
t
L
)
1
(
0
0
kt
t
t e
L
L
L
y
)
( t
t BOD
y
u
BOD
u
BOD
L
0
36
37. Example: 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 . If
K =0.23 per day
359
)
1
( 5
*
)
23
.
0
(
0
5
5
e
L
y
BOD
359
683
.
0
*
0
L
L0= BODu=525.62 mg/L
mg/L
92
.
472
)
1
(
*
62
.
525 10
*
)
23
.
0
(
10
10
e
y
BOD
37
38. Example 2
If the BOD5 of a waste is 102 mg/L and the BOD20
(corresponds to the ultimate BOD) is 158 mg/L, what
is k (base e) ?
kt
e
L
1
0
t
BOD
kt
e
L
0
1 t
BOD
kt
L
0
1
ln t
BOD
38
40. Significance of BOD Rate
Constant -k
0
100
200
300
400
500
600
700
0 1 2 3 4 5 6
Time (days)
BOD
(mg/L)
K=0.2 day-1
K=0.6 day-1
40
41. Biological Oxygen Demand:
Temperature Dependence
Temperature dependence of biochemical oxygen
demand
As temperature increases, metabolism increases,
utilization of DO also increases
kt = k20T-20
= 1.135 if T is between 4 - 20 oC
= 1.056 if T is between 20 - 30 oC 41
42. Example 3
The BOD rate constant, k, was determined
empirically to be 0.20 days-1 at 20 oC.
– What is k if the temperature of the water increases
to 25 oC?
– What is k if the temperature of the water decreases
to 10 oC?
42
43. Example 3
20
25
-1
25 )
056
.
1
(
day
20
.
0
k
-1
day
26
.
0
25
k
20
10
10 )
135
.
1
(
20
.
0
-1
day
k
-1
day
056
.
0
10
k
43
45. Fujimoto Method
Plot BODt+1 versus BODt
Plot line of slope = 1
L0 (UBOD) – Value at intersection of the
above lines (a) and (b)
k – Use BOD equation and one of the
BOD values and corresponding t
45
46. 46
Example 1: Time – Concentration
Plot
Day 1 2 3 4 5
BOD 415 485 505 560 590
X axis – 415 mg/L, Y Axis- 485 mg/L
Plot BODt+1 versus BODt
47. 47
Find BODUlt & k
Ultimate BOD= 680 mg/L
BOD5 = 680 (1-e-5k) = 590
Gives k = 0.41 day -1
0
100
200
300
400
500
600
700
800
900
1000
1100
1200
1300
0 100 200 300 400 500 600 700 800 900 1000 1100 1200 1300
BOD (n+1) mg/L
BOD
(mg/L)
Plot of BOD & BODt+1
BOD
(t+1
(mg/L
BOD t
49. 49
Nitrogenous Oxygen Demand
So far we have dealt only with carbonaceous
demand (demand to oxidize carbon
compounds)
Many other compounds, such as proteins,
consume oxygen
Mechanism of reactions are different
50. 50
Nitrogenous Oxygen Demand
Nitrification (2 step process)
2 NH3 + 3O2 2 NO2
- + 2H+ + 2H2O
2 NO2
- + O2 2 NO3
-
– Overall reaction:
NH3 + 2O2 NO3
- + H+ + H2O
Theoretical NBOD =
N
/g
O
g
4.57
14
16
x
4
oxidized
nitrogen
of
grams
used
oxygen
of
grams
2
52. Nitrogenous Oxygen Demand
Untreated domestic wastewater
ultimate-CBOD = 250 - 350 mg/L
ultimate-NBOD = 70 - 230 mg/L
Total Kjeldahl Nitrogen (TKN) = total
concentration of organic and ammonia nitrogen
in wastewater: 15 - 50 mg/L as N
Ultimate NBOD 4.57 x TKN
What Happens if large amount of nitrifiers are present in
the treated sewage ?
TCMP: 2 chloro-6-trichloro methyl pyridine 52
53. 53
SAMPLING PRECAUTIONS
BOD
If analysis begins within 2 hours of sample collection, the
sample does not need to be cooled. Otherwise, the sample
must be cooled to <4°C immediately after it has been taken.
The time to analysis must not exceed 6 hours. If this is not
possible, the duration and temperature of storage must be
noted. Exclude all air,
The duration of bulk sampling is restricted to 24 hours.
A sample is taken using a clean dry vessel. The sampling
volume is at least one liter.
If possible, the sample should not be frozen. Deep-frozen
samples result in lower measured values (up to 10% lower:
Damage to microorganisms).
54. COD (Chemical Oxygen Demand)
A measure of the oxygen equivalent of the
organic matter content of a sample that is
susceptible to oxidation by a strong chemical
oxidant.
Uses a strong chemical oxidant in an acid
solution and heat to oxidize organic carbon to
CO2 and H2O.
Oxygen demand is determined by measuring
the amount of oxidant consumed using
titrimetric or photometric methods.
54
55. 55
C H O Cr O H nCO Cr
a
H O
n a b
2 7
2
2
3
2
8 2 4
2
2
3 6 3
n a b
Where:
Stochiometry of COD
Organic
Matter
Strong
Oxidant
Potassium
Dichromate
Sulphuric
Acid
Carbon-
dioxide
HEATING 2 HOURS 150 OC
Chromic
acid
Orange
Colored
Green
Colored
56. Take reading in spectrophotometer
Take 2.5 ml sample in COD vial
Add 1.5 ml K2Cr2O7in it
Add 3.5 ml sulphuric acid reagent
Digest above solution in digester for 2 hr at 150 oC
56
57. 57
Chemical Oxygen Demand
COD test is faster than BOD analysis: used for quick
assessment of wastewater strength and treatment
performance
Using this digestion technique, it is possible to test 25
samples while using a minimum of laboratory bench
space.
Like the BOD, it does not measure oxidant demand
due to nitrogenous species
It does not distinguish between biodegradable and
non-biodegradable organic matter. As a result COD's
are always higher than BOD's.
60. 60
y = 1.5321x + 11.283
R2
= 0.7725
0
100
200
300
400
500
600
700
0 50 100 150 200 250 300 350 400
y = 1.7058x + 5.0434
R2
= 0.7466
0
20
40
60
80
100
120
0 10 20 30 40 50 60 70
BOD:COD For Inlet of
Sewage Treatment
Plant (STP) > 0.6
BOD:COD For
Outlet of STP < 0.5
BOD
COD
61. 61
What is TOC?
Total Organic Carbon
Organic contaminants (NOM’s,
insecticides/herbicides, agricultural
chemicals) – reach surface water via rainfall
runoff
Industrial organics due to spills
Domestic/Industrial wastewater effluent
62. 62
TOC vs. TC & IC
TOC = Total Carbon (TC) – Inorganic Carbon (IC)
TOC = all carbon atoms covalently bonded in organic
molecules
DOC = Carbon passed through 0.45 micrometer filter
TC is a measure of all the carbon in the sample
IC = carbonate, bicarbonate, and dissolved carbon dioxide
IC is analyzed in liquid samples by acidifying with an
inorganic acid to pH 2 or lower, then sparging for a
few minutes with a stream of gas
63. 63
TOC Measurement
TC Measurement
High temperature combustion convert TC in the sample to carbon dioxide.
This carbon dioxide flows with the carrier gas via the dehumidifier into the
NDIR sample cell.
The area of the carbon dioxide peak signal is measured and this peak area is
converted to TC concentration using a pre-prepared calibration curve.
IC Measurement
The sample is acidified with phosphoric acid and sparged to convert the IC in
the sample to carbon dioxide.
This carbon dioxide is detected by the NDIR and the sample IC concentration
is measured in the same way as TC.
TOC Measurement
Subtracting the IC concentration from the TC concentration determines the
TOC concentration.
64. 64
SHIMADZU TOC ANALYSER
Analyte = TC, IC, TOC (TC-IC)
Method – Combustion (680o C)/NDIR gas analysis
Measuring Range = 4 ppb to 4000 ppb
Avg. Analysis Time = 2 – 3 min. for both TC and IC
Shimadzu ASI-5000 – Automatic Sample Injector
78 vial or 16 vial turntables available
Rinsing between samples minimizes sample “carry-
over”
67. 67
Parameter COD BOD TOC
Oxidant Used K2Cr2O7 Oxidation by microbes O2,Heat
Suitable Use Rapid and frequent
monitoring
Effects of organic compounds
on the DO content of
receiving waters.
Measures total organic
carbon
Time 2.5-3 hours 5 days (standard BOD test)
Or 3 days @27oC (India)
Several min. to hour
Accuracy 5-10 % relative SD;
homogenization can
be important
15 % relative standard
deviation; not considered
highly accurate
5-10 % relative SD;
homogenization can be
important
Advantages Correlates with BOD
on waste with
constant composition.
Toxic materials do
not affect oxidant.
Short analysis time.
Most closely models the
natural environment when
used with proper “seed”
Correlates with BOD
on waste with constant
composition, but not as
closely as COD
Short analysis time
Disadvantages Interference from
chloride ions
Some organic
compounds are not
oxidized completely
Lengthy test period
Toxic materials kill
microorganisms.
Microorganisms do not
oxidize all material in waste.
Inaccuracies when used with
improper seed.
Requires expensive
equipment.
Some organic
compounds are not
oxidize completely.
Measures Total
Organic Carbon not
69. Suspended Solids
Organic and inorganic particles in water are
termed suspended solids
May be distinguished from colloids,
particles that do not settle readily
Measured by filtering a water sample,
drying and weighing the filter.
69
70. Suspended Solids
Problems
– Sedimentation
– Exert oxygen demand (Organic Solids)
– Primary transport mechanism for many metals, organics
and pathogens
– Aesthetic
– Complicates drinking water treatment
70
71. 71
Suspended Solids
Suspended Solids: SS or TSS
Drinking Waters
(Uses turbidity in place of TSS) Drinking Waters
Aesthetics, interference with other processes, treatment doses & sizing
(Turbidity)
Indian Standards Turbidity < 5 NTU
Natural Waters
Direct hazard to aquatic life: Settled & Causes Anaerobic Conditions
Siltation and hydraulic problems
Wastewaters
Measure strength & treatment efficiency
Mass balance of solids
Design of Sludge Treatment Facilities
Estimate of active biomass for process control (VSS) bacterial
populations
Regulatory control on effluent (TSS) Indian Standards < 100 mg/L
72. WASTEWATER
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=TS – SS = MB/VA - ME/VB
Suspended Solids= ME-tare wt/VB
Total Volatile Solids= (MB-MC)/VA
Volatile Suspended Solids= (ME-MF)/VB
72
75. Nitrogen-Effects
Stimulate algal growth eutrophication 0.3 mg/L
Deplete DO by nitrification O2 is consumed
Toxic to aquatic life (NH3
+) if free ammonia is
above 0.2 mg/L
NO2
- inhibit the combination of oxygen and
haemoglobin in blood. React with amines form
nitrosamines-carcinogenic
Increase chlorine dosage in disinfection
75
77. Phosphorus
0.1 % in Earth.
Phosphorus is an important component of ATP
(energy), DNA (genes), RNA (protein synthesis),
NADP (energy transfer), phospholipids (cell
membrane)
Critical for growth and survival of all living
organisms
Exists as Organic Phosphate, Polyphosphate,
Orthophosphate (Soluble inorganic Phosphates)
Under acidification and Heating – organic and
Polyphosphate converted to Orthophosphate
77
78. Phosphorus
P Sources in wastewater or surface water
– Human Waste
– Detergents
– fertilizers
The limiting nutrient in lakes, and algae growth is
linked to phosphorus inputs.
Stimulate algal growth -eutrophication 0.02 mg/L
Problems
– aesthetic
– taste and odor in drinking water
– can be toxic, especially to farm animals
– fouling
– diurnal DO cycles
78
80. Total Dissolved Solids (TDS) or
Salts
Dissolved solids, or salts, may be present as any
number of ions
– Cations: Na+, K+, Mg2+, Ca2+
– Anions: Cl-, SO4
2-, HCO3
-
Typically measures as total dissolved solids (TDS)
Water classification
– freshwater <1500 mg/L TDS
– brackish water 1500 – 5000 mg/L
– saline water >5000 mg/L
– sea water 30-34 g/L
80
81. Effects of TDS
Sources and concentration influences
– minerals
– evaporative losses
– irrigation
– industrial discharges
– sea water intrusion
Effects
– Interfere with wastewater reuse
– crop damage/soil poisoning
– Scale Formation
81
82. TDS and EC
(Electrical Conductivity)
–TDS = keEC
–where TDS is expressed in mg/L and EC
is the electrical conductivity in
microsiemens per centimeter at 25 °C.
–The correlation factor ke varies between
0.55 and 0.8.[
82
83. Sodium Adsorption Ratio (SAR)
Sodium adsorption ratio (SAR), along with pH, characterize
salt-affected soils. It is an easily measured property that gives
information on the comparative concentrations of Na+, Ca2+, and
Mg 2+ in soil solutions.
•where [Na+], [Ca2+], and [Mg2+ ] are the concentrations in meq/L of
sodium, calcium, and magnesium ions in the soil solution.
•Concentrations of sodium, calcium, and magnesium are determined by
first extracting the ions from the soil into solution.
•The solution is then analyzed to determine concentrations of the selected
ions. Na+, Ca2+, and Mg2+ concentrations are commonly determined using
atomic absorption spectrometry (AA).
84.
85. Water Related Diseases-
Pathogens
Water-related diseases (e.g. gastro-intestinal, typhoid,
shigellosis, hepatitis and cholera) – Main Concerns
Can directly affect humans by causing illness and
possible death.
Often contamination through contact with water or via
food (e.g. via irrigated agriculture, or via fish/shellfish).
85
86. Pathogenic Organisms
Many organisms that cause human or animal
diseases colonize the intestinal tract but can live
for a period of time outside the body
Carriers (who may or may not exhibit disease
symptoms) excrete these intestinal tract organisms
in very large numbers.
When water is contaminated by excreta, the
organisms can be transmitted to those who contact
the water.
86
88. Virus
• Smallest Living Organisms.
• Intracellular parasites, size ~ 20-100 nm (about
1/50th of a bacteria).
• A few out of different species of enteric viruses
present in the human gut, cause serious diseases.
• Adenovirus (Respiratory and eye infections),
Poliovirus, Hepatitis A virus, Echovirus (aseptic
meningitis), Rota virus and other virus causing
gastroenteritis, diarrhea.
88
90. Bacteria
• Single-celled entities that use soluble food and
are capable of self-reproduction. Size range:
0.5- 5 micron.
• Feces of healthy human being may contain 100
million to few billions of bacteria per gram. A
few of them are pathogens.
• Diseases: typhoid, paratyphoid, dysentery, and
cholera.
• Transmission of these pathogens can be
controlled by sanitary disposal of wastewater
and disinfection of water supplies. 90
92. Protozoa
• These parasites live attached to the human intestine
where they actively feed and reproduce.
• Common diseases: diarrhea and dysentery. Example:
Entamoeba histolytica and Giardia lambia.
• Undergo a morphologic cyst into a cyst for protection
against harsh environment outside the host. The cyst
form is infectious to other persons by the fecal-oral route
of transmission.
• The cysts have size 10-15 micron, much larger than the
bacteria.
92
94. Helminths- Parasitic Worms
• Helminths or intestinal worms do not
multiply within the human host.
• The worm burden in an infected person is
directly related with the number of infective
helminth egg ingested.
• The eggs cause infection. Size: 40-60
micron in size and heavier than water.
94
96. 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??
96
97. Indicator Organisms
Rather than testing for each 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 feces of humans or warm-blooded animals.
This microorganism may be called an indicator
organism.
Coliform bacteria are Indicator organisms
97
98. Coliforms
Total coliforms include bacteria that are found in the
soil, in water that has been influenced by surface water,
and in human or animal waste. Incubation at 35oC @24h
Fecal coliforms are the group of the total coliforms that
are considered to be present specifically in the gut and
feces of warm-blooded animals. Incubation at 45oC
@24h
Fecal coliforms are considered a more accurate
indication of animal or human waste than the total
coliforms.
98
99. E Coli
Escherichia coli (E. coli) is the major species in the fecal
coliform group.
Out of coliforms, E. coli is generally not found growing and
reproducing in the environment.
E. coli is considered the best indicator of fecal pollution and
the possible presence of pathogens.
E. Coli. persists in the environment outside the human
intestine for a longer duration than the other pathogenic
bacteria.
Untreated domestic sewage contains upwards of 3 million
coliforms per 100 mL.
99
100. Fermentation Tube Technique
Cap
Lactose
broth
Inverte
d Vial
Fermentation Tube
Wastewater
Sample
Incubation
@ 35 deg C
Negative
Positive/
Change in
color
Growth with
gas evolution
inside the
inverted vial
No growth
and no gas
evolution
100
101. 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
101
102. MPN
Broth inoculated from the 10–3 dilution shows growth, but the
broth from the 10–4 does not, it is then possible to say that there
were greater than 1 x 103 organisms per ml of the original sample
but less than 1 x 104 per ml. 102
103. Standard MPN Test
Standard MPN procedures use a minimum of
3 dilutions
three sets of tubes which show dilution of the organisms "to
extinction" – i.e., those tubes which were inoculated from the 10–2,
10–3 and 10–4 dilutions.
103
104. MPN Table
0.93 organisms being
inoculated into each of the
tubes of the middle set (of the
three sets of tubes chosen) –
i.e., those inoculated with one
ml of a 10–3 dilution.
Therefore, the most-probable
number of organisms per one
ml of the original, undiluted
sample would be 0.93 X 103
/ml or 930 /ml.
No. of Tubes
Positive in
MPN in
the
inoculu
m
of the
middle
set
of tubes
first
set
middle
set
last
set
0 0 0 ‹0.03
0 0 1 0.03
1 3 0 0.16
1 3 1 0.20
1 3 2 0.24
1 3 3 0.29
2 0 0 0.091
2 1 0 0.15
2 3 3 0.53
3 0 0 0.23
3 0 3 0.95
3 1 0 0.43
3 1 1 0.75
3 1 2 1.2
3 1 3 1.6
3 2 0 0.93
3 3 1 4.6
3 3 2 11
104
105. Multiple Tube Fermentation Technique and
Most Probable Number
Thomas’ Formula:
tubes
the
all
in
samples
of
mL
x
tubes
negative
in
samples
of
mL
tubes
positive
of
Number
mL
(MPN)/
Number
ble
Most Proba
105
106. We assume that the dilutions are 1.0, 0.1, 0.01, 0.001, 0.0001 &
0.00001 ml.
There are 5 positive tubes, so
MPN/ml = 5 /(0.033 x 0.0013)(½) = 5/0.00657 =
759/ml
For outcome (3/3, 3/3, 3/3, 2/3, 0/3, 0/3) use only (-, -, 3/3,
2/3, 0/3,–);
ml of Sample in All tubes
3 x 0.01 + 3 x 0.001 + 3 x 0.0001 = 0.0333
ml of sample in negative tubes
= 1 x 0.001 + 3 x 0.0001 = 0.0013
106
108. Problems:
1. What is the BOD conc., SS, T-N & T-P conc. of
Saharanpur Sewage, if percapita water supply is 120
l/day.
2. What is the BOD conc., SS, T-N & T-P conc. of
Haridwar Sewage, if percapita water supply is 250
l/day and 20 % BOD degradation takes place in Sewer
pipes and pumping station.
3. What is the BOD conc., SS, T-N & T-P conc. of IIT,
R Sewage, if percapita water supply is 250 l/day. &
40 % BOD & 60 % SS Removal takes place in septic
tanks outside houses.
4. Why in Q 2 and 3, we didn’t consider N & P
Removal. 108
109. Concentrations in Sewage
The raw sewage characteristics are a function of level of
water supply and per capita pollution load.
Other significant factors are
– settlement and
– decomposition in sewers under warm weather conditions,
– partially decomposed sewage from septic tanks,
– lifestyle of the population, etc.
The best way to ascertain the sewage characteristics is to
conduct the composite sampling once a week for diurnal
variation on hourly basis from the nearby existing sewage
outfall or drain.
109
110. Sampling & Analysis of Sewage
Considering a four-week month, three samples are to be taken
on weekdays, whereas the fourth sample is to be taken on an off
day i.e. Sunday.
Sampling for water quality should be conducted for at least one
month during dry weather to assess pollution load quantitatively
and qualitatively.
The samples should be analyzed for the following parameters;
pH, Temperature, Colour, Odour, Alkalinity, TSS, Volatile SS,
BOD (Total & Filtered), COD (Total and Filtered), Nitrogen
(NH3, TKN, NO3), Phosphorus (Ortho-P & T-P), Total
Coliforms and Faecal Coliforms, TDS, Chloride, Sulphates,
Heavy Metals (if there is a chance of industrial contamination)
110
114. Objectives of Wastewater
Treatment
Removal of Suspended Solids by Clarification ( In
Sedimentation Tank)& Decomposition (By providing
Suitable conditions for bacteria)
Removal of Organics by Decomposition (By
providing Suitable conditions for bacteria) & Provide
conditions for separation of the wastewater from the
Bacteria.
Providing Excess Air for Nitrification and Anoxic
Conditions for Denitrification
Removal of Residual bacteria present in separated
wastewater by adding powerful oxidants such as
Chlorine. 114
115. To Bring the River Water to Bathing Quality
( River Bathing Standards)
BOD - BIO-CHEMICAL OXYGEN DEMAND
DO - DISSOLVED OXYGEN
MPN - MOST PROBABLE NUMBER
BOD 3 mg/L (MAXIMUM)
DO 5 mg/L (MINIMUM)
COLIFORM (FAECAL) 500 (DESIRABLE)
2500 (MAX. PERMISSIBLE)
PERMISSIBLE LIMIT
PARAMETERS
MPN
100 ml
115
116. 1.0 Indian Effluent Standards- Schedule-6,
Environmental Protection Rules 1986, IS, CPCB
116
117. 2.0 NRCD Standards
(Projects sponsored under National River or National
Lake Conservation Plan- MoEF
Standards for
River Ganga
117
120. 120
Parameters Standards for New STPs
(Design after notification date)*
pH 6.5-9.0
BOD, mg/l 10
COD, mg/l 50
TSS, mg/l 10
NH4-N, mg/l 5
N-total, mg/l 10
Fecal Coliform, MPN/100ml <230
PO4-P , mg/l 2
Note:
(i) These standards will be applicable for discharge in water resources as well as for land disposal. The standards for Fecal
Coliform may not be applied for use of treated sewage in industrial purposes.
(ii) Achievements of Standards for existing STPs within 02 years from date of notification.
(iii) Above standards shall applicable for the construction projects listed below:
a. The new Housing Complex / construction project of built up area greater than 20,000 sq. meters shall meet the above
prescribed standard
b. The housing complex of built up area of less than 20,000 sq. Meters shall have sewer connection with terminal Sewage
Treatment Plants and meet the aforesaid standards.
c. The concerned stakeholders shall ensure that the old Housing Complex / construction project of built up area greater than
20,000 sq. meters shall facilitate by the sewer connection with terminal Sewage Treatment Plants and confirm the
prescribed norms.
d. Dual piping system shall enforce in new housing groups so that treated sewage may be used for flushing
121. 5.0: Punjab Pollution Control
Board Standards
RSC is expressed in meq/l units. RSC should not be higher than 1 and preferably less than
+0.5 for considering the water use for irrigation. RSC index = [HCO3 + CO3] − [Ca + Mg]
123. Flow = 9 m3/s
BOD = 0 mg/L
DO = 10 mg/L
Flow = 1 m3/s
BOD = 30 mg/L
DO = 2 mg/L
BOD= 3 mg/L
Ten times dilution is needed, otherwise we have to reduce
the standards to more strict or lower level
123
