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L-13
Biochemical Oxygen Demand
(BOD)
Unit- III
Industrial Waste Treatment
Growth of bacteria
w: 800 × 549 pixels.Other resolutions: 320 × 220 pixels | 640 × 439 pixels.
I. Lag phase
• During lag phase, bacteria adapt
themselves to growth conditions. It is
the period where the individual
bacteria are maturing and not yet
able to divide.able to divide.
• During the lag phase of the bacterial
growth cycle, synthesis of RNA,
enzymes and other molecules occurs.
II. Exponential phase
• Exponential phase (sometimes called the
log phase or the logarithmic phase) is a
period characterized by cell doubling.
• The number of new bacteria appearing
per unit time is proportional to theper unit time is proportional to the
present population
• For this type of exponential growth,
plotting the natural logarithm of cell
number against time produces a straight
line.
III. Stationary phase
• During stationary phase, the growth
rate slows as a result of nutrient
depletion and accumulation of toxic
products.
• This phase is reached as the bacteria• This phase is reached as the bacteria
begin to exhaust the resources that are
available to them.
• This phase is a constant value as the
rate of bacterial growth is equal to the
rate of bacterial death.
IV. Death phase
• At death phase, bacteria run out of
nutrients and die.
Binary
fission of
bacteria
Definition
• Biochemical oxygen demand or
B.O.D. is the amount of dissolved
oxygen needed by aerobic biological
organisms in a body of water ororganisms in a body of water or
wastewater sample to break down
organic material present in a given
water or wastewater sample at certain
temperature (200C) over a specific
time period (5 days).
• Used to assess the relative strength of
a waste
• The amount of oxygen required to
stabilize a waste if discharged to a
surface water.surface water.
• This is not a precise quantitative test,
although it is widely used as an
indication of the organic quality of
water.
• The BOD value is most commonly
expressed in milligrams of oxygen
consumed per litre of sample duringconsumed per litre of sample during
5 days of incubation at 20 °C and
is often used as a indicator of the
degree of organic pollution of
water.
Organic matter in waste + O2 in sample CO2 + H2O
Background
• Most natural waters and wastewaters
contain small quantities of organic
compounds.
• Aquatic microorganisms have evolved
to use some of these compounds asto use some of these compounds as
food.
• Microorganisms living in oxygenated
waters use dissolved oxygen to convert
the organic compounds into energy for
growth and reproduction.
• Populations of these
microorganisms tend to increase
in proportion to the amount of
food available.
• This microbial metabolism
creates an oxygen demand
proportional to the amount of
organic compounds useful as
food.
• Under some circumstances,
microbial metabolism can consume
dissolved oxygen faster than
atmospheric oxygen can dissolve into
the water.the water.
• Fish and aquatic animals and plants
may die when oxygen is depleted by
microbial metabolism
Types of BOD
• There are two stages of
decomposition in the BOD test:
i. a carbonaceous stage and
ii. a nitrogenous stage.ii. a nitrogenous stage.
• The carbonaceous stage, or first stage,
represents that portion of oxygen demand
involved in the conversion of organic
carbon to carbon dioxide.
• The nitrogenous stage, or second stage,• The nitrogenous stage, or second stage,
represents a combined carbonaceous plus
nitrogenous demand, when organic
nitrogen, ammonia, and nitrite are
converted to nitrate. Nitrogenous oxygen
demand generally begins after about 6
days.
TEST
• The test for biochemical oxygen
demand (BOD) is a bioassay
procedure that measures the oxygen
consumed by bacteria from the
decomposition of organic matter.decomposition of organic matter.
• The change in DO concentration is
measured over a given period of time
in water samples at a specified
temperature.
• In a nutshell, BOD gives a measure
on the impact of a waste(water) on
the oxygen content of a receiving
System(stream/river/lake).
• Wastes are broken down by• Wastes are broken down by
microbial organisms (frequently
referred to as “bugs” or
“microorganisms”), and the bugs, in
turn, require oxygen for this
monumental effort.
• Thus, in order for this test to "work",
you need
(1) a food source (Organic matter),
(2) a nice population of bugs,(2) a nice population of bugs,
(3) available oxygen (DO) to drive
the bugs(micro-organisms), and
(4) a system (Incubator) which
provides a hospitable environment
(incubation temp.) for the bugs.
Procedure
A. Dilution
• Dilution Factor. The dilution factor,
DF, is the ratio of the final volume to
the volume of sample therein. (e.g., for
the bottle method, the volume of the BOD
bottle, usually 300 mL; for the graduatedbottle, usually 300 mL; for the graduated
cylinder method, the volume of the
cylinder, usually 1,000 mL)
• DF for the bottle method = {Volume of
Diluted Mixture/Volume of Sample in
Mixture.}
B. DO fixation
• 1. Slowly siphon three portions of aerated
dilution water into three separate BOD
bottles. Avoid adding atmospheric O2 to
dilution water.
2. To two of the three BOD bottles, add 12. To two of the three BOD bottles, add 1
ml MnS04 solution, followed by 1 ml
alkali-iodide-azide reagent. Submerge
pipette tips in sample when adding
reagents. Rinse tips well between uses.
3. Stopper carefully to exclude air
bubbles; mix by inverting bottle
several times.
4. When precipitate has settled to about
half the bottle volume, carefullyhalf the bottle volume, carefully
remove the stopper and add 1.0 ml
conc. sulfuric acid. Re-stopper and
mix by gentle inversion until the
iodine is uniformly distributed
throughout the bottle.
C. DO Measurement
5. Transfer 203 ml of sample into conical
flask and titrate with 0.0250N sodium
thiosulfate to a pale straw color.
6. Add 1-2 ml of starch solution and
continue to titrate to first disappearancecontinue to titrate to first disappearance
of the blue color. (200 ml of original
dilution water is equal to 203 ml of
dilution water plus reagents.)
•
• Determine the initial
concentration of dissolved
oxygen in one bottle of the
mixture of sample and dilution
water (DO ), and in one of thewater (DOinitial), and in one of the
bottles containing only dilution
water.
• Place the other bottles in the
incubator
• Incubate the blank dilution water and the
diluted samples for 5 days in the dark at
20 °C.
• After 5 days, remove the bottles, fix the
DO and measure the DO (DOfinal)DO and measure the DO (DOfinal)
• Calculate BOD5
• BOD5 (mg/L) = {DOinitial (mg/L) of first
bottle - DOfinal (mg/L) of second bottle}
x dilution factor
BOD
incubator
and BOD
bottle
BOD
monitors
BOD
monitors
Chemical Oxygen Demand (COD)
• Like BOD, COD provides a measure
of the amount of organic compounds
in water. The difference is that COD
is less specific since it measuresis less specific since it measures
everything that can be chemically
oxidized rather than just levels of
bio-degradable organic matter.
• COD also is different in that it reflects the
oxidation based on a specific chemical
oxidant (dichromate).
• COD value is always greater than
BOD for every wastewater sample.BOD for every wastewater sample.
Objective Questions
1. COD value is always _____________than
BOD value.
2. First stage BOD is also known as
___________________________.
3. Second stage BOD is also known as3. Second stage BOD is also known as
_______________________.
4. For BOD test sample is incubated at ____0C.
5. ______ is indicator of the degree of organic
pollution of water.
6. ______test is basis for BOD test.
Theory Questions
Q1. Define the following
i. BOD
ii. Dilution factor
iii. COD
Q2. Explain ‘BOD test’.
Q3. Explain Growth phase of bacteria.

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L 13 bod

  • 1. L-13 Biochemical Oxygen Demand (BOD) Unit- III Industrial Waste Treatment
  • 2. Growth of bacteria w: 800 × 549 pixels.Other resolutions: 320 × 220 pixels | 640 × 439 pixels.
  • 3. I. Lag phase • During lag phase, bacteria adapt themselves to growth conditions. It is the period where the individual bacteria are maturing and not yet able to divide.able to divide. • During the lag phase of the bacterial growth cycle, synthesis of RNA, enzymes and other molecules occurs.
  • 4. II. Exponential phase • Exponential phase (sometimes called the log phase or the logarithmic phase) is a period characterized by cell doubling. • The number of new bacteria appearing per unit time is proportional to theper unit time is proportional to the present population • For this type of exponential growth, plotting the natural logarithm of cell number against time produces a straight line.
  • 5. III. Stationary phase • During stationary phase, the growth rate slows as a result of nutrient depletion and accumulation of toxic products. • This phase is reached as the bacteria• This phase is reached as the bacteria begin to exhaust the resources that are available to them. • This phase is a constant value as the rate of bacterial growth is equal to the rate of bacterial death.
  • 6. IV. Death phase • At death phase, bacteria run out of nutrients and die.
  • 8. Definition • Biochemical oxygen demand or B.O.D. is the amount of dissolved oxygen needed by aerobic biological organisms in a body of water ororganisms in a body of water or wastewater sample to break down organic material present in a given water or wastewater sample at certain temperature (200C) over a specific time period (5 days).
  • 9. • Used to assess the relative strength of a waste • The amount of oxygen required to stabilize a waste if discharged to a surface water.surface water. • This is not a precise quantitative test, although it is widely used as an indication of the organic quality of water.
  • 10. • The BOD value is most commonly expressed in milligrams of oxygen consumed per litre of sample duringconsumed per litre of sample during 5 days of incubation at 20 °C and is often used as a indicator of the degree of organic pollution of water.
  • 11. Organic matter in waste + O2 in sample CO2 + H2O
  • 12. Background • Most natural waters and wastewaters contain small quantities of organic compounds. • Aquatic microorganisms have evolved to use some of these compounds asto use some of these compounds as food. • Microorganisms living in oxygenated waters use dissolved oxygen to convert the organic compounds into energy for growth and reproduction.
  • 13. • Populations of these microorganisms tend to increase in proportion to the amount of food available. • This microbial metabolism creates an oxygen demand proportional to the amount of organic compounds useful as food.
  • 14. • Under some circumstances, microbial metabolism can consume dissolved oxygen faster than atmospheric oxygen can dissolve into the water.the water. • Fish and aquatic animals and plants may die when oxygen is depleted by microbial metabolism
  • 15. Types of BOD • There are two stages of decomposition in the BOD test: i. a carbonaceous stage and ii. a nitrogenous stage.ii. a nitrogenous stage.
  • 16. • The carbonaceous stage, or first stage, represents that portion of oxygen demand involved in the conversion of organic carbon to carbon dioxide. • The nitrogenous stage, or second stage,• The nitrogenous stage, or second stage, represents a combined carbonaceous plus nitrogenous demand, when organic nitrogen, ammonia, and nitrite are converted to nitrate. Nitrogenous oxygen demand generally begins after about 6 days.
  • 17.
  • 18. TEST • The test for biochemical oxygen demand (BOD) is a bioassay procedure that measures the oxygen consumed by bacteria from the decomposition of organic matter.decomposition of organic matter. • The change in DO concentration is measured over a given period of time in water samples at a specified temperature.
  • 19. • In a nutshell, BOD gives a measure on the impact of a waste(water) on the oxygen content of a receiving System(stream/river/lake). • Wastes are broken down by• Wastes are broken down by microbial organisms (frequently referred to as “bugs” or “microorganisms”), and the bugs, in turn, require oxygen for this monumental effort.
  • 20. • Thus, in order for this test to "work", you need (1) a food source (Organic matter), (2) a nice population of bugs,(2) a nice population of bugs, (3) available oxygen (DO) to drive the bugs(micro-organisms), and (4) a system (Incubator) which provides a hospitable environment (incubation temp.) for the bugs.
  • 21.
  • 22. Procedure A. Dilution • Dilution Factor. The dilution factor, DF, is the ratio of the final volume to the volume of sample therein. (e.g., for the bottle method, the volume of the BOD bottle, usually 300 mL; for the graduatedbottle, usually 300 mL; for the graduated cylinder method, the volume of the cylinder, usually 1,000 mL) • DF for the bottle method = {Volume of Diluted Mixture/Volume of Sample in Mixture.}
  • 23.
  • 24.
  • 25. B. DO fixation • 1. Slowly siphon three portions of aerated dilution water into three separate BOD bottles. Avoid adding atmospheric O2 to dilution water. 2. To two of the three BOD bottles, add 12. To two of the three BOD bottles, add 1 ml MnS04 solution, followed by 1 ml alkali-iodide-azide reagent. Submerge pipette tips in sample when adding reagents. Rinse tips well between uses.
  • 26.
  • 27. 3. Stopper carefully to exclude air bubbles; mix by inverting bottle several times. 4. When precipitate has settled to about half the bottle volume, carefullyhalf the bottle volume, carefully remove the stopper and add 1.0 ml conc. sulfuric acid. Re-stopper and mix by gentle inversion until the iodine is uniformly distributed throughout the bottle.
  • 28. C. DO Measurement 5. Transfer 203 ml of sample into conical flask and titrate with 0.0250N sodium thiosulfate to a pale straw color. 6. Add 1-2 ml of starch solution and continue to titrate to first disappearancecontinue to titrate to first disappearance of the blue color. (200 ml of original dilution water is equal to 203 ml of dilution water plus reagents.) •
  • 29. • Determine the initial concentration of dissolved oxygen in one bottle of the mixture of sample and dilution water (DO ), and in one of thewater (DOinitial), and in one of the bottles containing only dilution water. • Place the other bottles in the incubator
  • 30. • Incubate the blank dilution water and the diluted samples for 5 days in the dark at 20 °C. • After 5 days, remove the bottles, fix the DO and measure the DO (DOfinal)DO and measure the DO (DOfinal) • Calculate BOD5 • BOD5 (mg/L) = {DOinitial (mg/L) of first bottle - DOfinal (mg/L) of second bottle} x dilution factor
  • 34. Chemical Oxygen Demand (COD) • Like BOD, COD provides a measure of the amount of organic compounds in water. The difference is that COD is less specific since it measuresis less specific since it measures everything that can be chemically oxidized rather than just levels of bio-degradable organic matter.
  • 35. • COD also is different in that it reflects the oxidation based on a specific chemical oxidant (dichromate). • COD value is always greater than BOD for every wastewater sample.BOD for every wastewater sample.
  • 36. Objective Questions 1. COD value is always _____________than BOD value. 2. First stage BOD is also known as ___________________________. 3. Second stage BOD is also known as3. Second stage BOD is also known as _______________________. 4. For BOD test sample is incubated at ____0C. 5. ______ is indicator of the degree of organic pollution of water. 6. ______test is basis for BOD test.
  • 37. Theory Questions Q1. Define the following i. BOD ii. Dilution factor iii. COD Q2. Explain ‘BOD test’. Q3. Explain Growth phase of bacteria.