bod / cbod from a to z - ohio water environment association
TRANSCRIPT
What is BOD? • It is a measure of the amount of oxygen consumed
by bacteria during the decomposition of organic
materials.
– Organic materials from the wastewater treatment facility
act as a food source for the bacteria.
• Directly related to Dissolved Oxygen
– The bacteria require oxygen in the form of dissolved
oxygen to decompose or eat the food source. Through a
calculation, the amount of DO depletion between the initial
day and final day of the analysis determines the BOD.
Thus, BOD directly affects the amount of Dissolved
Oxygen
– The greater the BOD = more rapid oxygen depletion = less
oxygen available to aquatic life.
What is the Difference Between
BOD and CBOD
BOD
• represents the oxidation
of carbons and
nitrogenous compounds
present in the water
CBOD
• measures the oxidation of
carbons present in water
• prevents the oxidation of reduced forms of nitrogen such
as ammonia, and organic nitrogen which exert a
nitrogenous demand.
• Should add at the beginning of the test because
nitrification will begin almost immediately if the right
organisms are present (Baird and Smith, 2002).
2-chloro-6-(trichloromethyl)pyridine
BOD verses COD
BOD
• represents the oxidation
of carbons and
nitrogenous compounds
present in the water
• Analysis completion is
done in 5-days
COD
• Is the measure of the
total amount of oxygen
required to oxidize all
organic material into
carbon dioxide and water
• analysis only takes a few
hours
BOD verses COD
• COD results are always higher than BOD
results.
– Useful in determining an unknown BOD range
for a sample but it can NOT replace the BOD
test.
Winkler Method
• Azide Modification Method – Preferred for most wastewaters
• Removes interferences caused by nitrite which is common in
wastewaters.
• Permanganate Modification
– Used when ferrous iron is present
• Azide plus Potassium Fluoride Modification
– Used when 5 mg or more of ferric iron salts/L are present
• Alum Flocculation Modification
– Used when there is interference caused by suspended solids
• Copper Sulfate-Sulfamic Acid Flocculation
Modification
– Used for biological flocs such as activated sludge mixtures which
also have a high oxygen utilization rate
Grab Samples
• Ideally samples should be analyzed within
6-hrs of collection, however if this is not
possible, then analyze samples within 48
hours of collection (40 CFR part 136).
• Store samples at < 6°C.
Composite Sampling
• Samples should be kept at or below < 6C during
compositing (limited to 24-hour period).
• start the measurement of holding time from the end of
the compositing period.
– For example if the compositing was started at 8:30 am
on Tuesday and ended at 8:30 am on Wednesday, then
the 48-hour holding time would start from the end of the
compositing period which would be 8:30am on
Wednesday.
• Store samples at < 6C
BOD Quality Controls
• Blank Control Checks
• GGA Control Checks
– Glucose Glutamic Acid
• Seed Control Checks
Blank Control Checks
• Straight dilution water
• Used to determine cleanliness of bottles
as well as the source water.
• It must have a DO uptake NO greater
than 0.2 mg/L
GGA Control Checks
• Used to check the quality of the seeding
material.
– Low results reflect poor seeding material.
• The ideal GGA range for a BOD sample is
198 + 30 mg/L.
GGA Control Checks
• GGA Needs to be pH adjusted
– Initially the pH is around 4
– Adjust between 6.5 -7.5, like any other
samples
Seed Controls
• Must have a DO uptake attributable to the
seed added to each bottle between
0.6mg/L – 1.0 mg/L.
• Most domestic wastewater, unchlorinated
or undisinfected effluents will contain a
sufficient population of microorganisms.
• Used to calculate the BOD results of
samples which are seeded
BOD / CBOD Requirements
• pH of samples should be between a pH of 6.5-7.5
• Sample temperature should be adjusted to 20° + 1°C before making dilutions
• After 5 days of incubation the final DO of samples must result in a DO depletion of at least 2.0 mg/L with a DO residual of no less than 1.0 mg/L. This is why it is recommended to make several dilutions of a sample.
– Example: The initial DO of a sample is 8.2 and after 5 days of incubation the final DO is 7.8. Then the final DO does not meet the required DO depletion of at least 2.0mg/L so a BOD result can not be calculated from this sample.
– Example #2: The initial DO of a sample is 8.2 and after 5 days of incubation the final DO is 0.20 mg/L. Here the final DO does not meet the required DO residual of at least 1.0 mg/L, so again the BOD result can not be calculated
Dilution Water Source
• Must be free of heavy metals, and toxic
substances which inhibit micro-bacterial
growth.
• Must also be able to maintain no more
than a 0.20 mg/L DO depletion during the
5-day incubation period.
Reagents Added to Dilution Water
• Phosphate Buffer Solution
• Magnesium Sulfate Solution (MgSO4)
• Calcium Chloride Solution (CaCl2)
• Ferric Chloride Solution (FeCl3)
Purpose of Adding Trace Metals,
Nutrients and Buffering Dilution
Water
• Bacteria growth requires nutrients and
trace metals.
• It is buffered to ensure the pH of the
incubated samples remain in a range
suitable for bacteria growth.
Why Dilute Samples Before
Incubation?
• Because the BOD concentration in most
wastewaters exceeds the concentration of
DO available in an air-saturated sample.
Why Seed?
• To add a population of microorganisms capable
of oxidizing the biodegradable organic matter.
– Most domestic wastewater, unchlorinated or
undisinfected effluents will contain a sufficient
population of microorganisms.
Samples That may Require Seeding
• Chlorinated samples
• High temperature wastes
• Wastes with extreme pH values
Selecting a Seed Source
• Select a material to be used for seeding which
will have a BOD of at least 180 mg/L.
• Example of seed sources according to “Standard
Methods 20th Edition”
– Raw domestic Sewage prepared as stated above
– Small quantities of digester supernatant, return
activated sludge
– Commerically available seed material (Polyseed)
Seeding
• Must have a DO uptake/depletion of 2
mg/L after the 5-day incubation period,
and also result in at least 1 mg/L residual
DO (final DO).
Over Mixing the Ployseed
•Never let the vortex
touch the stir bar
•Micro-organisms in the
seed will be too tired to
get the job done in your
samples and may see
low results in the seed
factor.
Proper way to mix the Polyseed
• Mix on a speed of about 5,
or so that the vortex is not
touching the stir bar and
splashing out.
•Mix for an hour
•Let bran settle out and
transfer to another beaker to
allow to mix for up to 5 hours
on a speed setting between
1-2
Seed Calculations
• The DO uptake attributable to the seed (the seed factor or SF) added to each bottle is between 0.6mg/L – 1.0 mg/L. The SF is calculated by using the equation below:
SF = (B1 – B2) x (f)
Seed Calculations
SF = (B1 – B2) x (f)
Where :
B1 = Initial Seed Control DO (before incubation)
B2 = Final Seed Control DO (after 5-day
incubation)
f = ratio of seed in diluted sample to seed in
seed control , or better see as
f = (volume, mls of seed in diluted sample)
(volume, mls of seed in seed control)
BOD5, mg/L = (D1 – D2) – (SF)
P
Where:
D1 = DO of diluted sample
immediately after preparation,
mg/L
D2 = DO of diluted sample after 5-day
incubation period
P = decimal volumetric fraction of
sample used
SF = seed factor
General Procedure
• 300ml BOD bottles are used.
– Filled with sample, seed, dilution water to
overfilling (airtight)
– Samples brought to room temperature or
20°C + 1°C before making dilutions
• Initial DO is taken before incubation period
(5-days at 20°C)
• Final DO is taken after incubation period
Unseeded BOD Calculation
BOD5, mg/L = D1 – D2
P
Where:
D1 = initial DO of sample
D2 = final DO of sample
P = decimal volumetric fraction of
sample used
EXAMPLE!
150 mLs of a sample was added to a 300
mL BOD bottle and the initial DO of the
sample is 8.2 and the final DO is 4.2, then
what is the BOD5 mg/L?
BOD5, mg/L = D1 – D2
P D1 = 8.2
D2 = 4.2
P = 150 mLs
300 mLs
P = 0.5
BOD5, mg/L = (8.2 – 4.2)
0.5
BOD5 mg/L = 8 mg/L
SEEDED
BOD CALCULATION
BOD5, mg/L = (D1 – D2) – (SF)
P
Where D1 = initial DO of sample
D2 = final DO of sample
P = decimal volumetric fraction of
sample used
SF = the DO uptake attributable to
the seed
EXAMPLE!
15 mLs of seed was added to a 300 mL BOD
bottle and labeled as the seed control. The
initial DO was 8.2 mg/L and the final DO is
5.0 mg/L. What is the seed factor, SF if 4
mLs of seed was added to the samples?
Using the calculated SF value, what would be
the BOD5 mg/L if 150 mLs of sample was
added to a 300 mL BOD bottle along with 4
mLs of seed and the initial DO was 8.2 mg/L
and the final DO is 4.2 mg/L?
First Calculate the SF Value
SF = (B1 – B2) x (f)
B1 = 8.2
B2 = 5.0
f = 4 mls
15 mls
SF = (8.2 – 5.0) x 4 mls
15 mls
SF = 0.853
Second calculate the BOD5 mg/L of
the sample:
BOD5, mg/L = (D1 – D2) – (SF)
P
D1 = 8.2
D2 = 4.2
P = 0.5
SF = 0.853
BOD5, mg/L = (8.2 – 4.2) – 0.853
0.5
BOD5, mg/L = 6.3 mg/L
Common Interferences
• Samples with caustic alkalinity (pH > 8.5)
– Neutralize to a pH between 6.5 – 7.5
• Samples with acidity (pH < 6.0)
– Neutralize to a pH between 6.5 – 7.5
• Samples Supersaturated with DO (> 9 mg/L )
• Residual Chlorine
Adjusting pH
• With samples of high alkalinity or acidity,
use sulfuric or sodium hydroxide solutions
of a concentration of which does not dilute
the sample by more than 0.5%
– Standard Methods Recommends a
concentration of 1N acid and alkali solutions
to neutralize caustic or acidic waste samples.
Samples Supersaturated with DO
• Samples containing more than 9 mg/L DO
at 20°C
– Cold water samples
– Where photosynthesis has occurred
Reducing Super-saturation
• Bring cold samples to room temperature or
20°C in a partially filled bottle and agitate
the bottle by shaking vigorously or
aerating with clean filtered compressed
air.
Removing Residual Chlorine
• Residual chlorine can cause erroneous
results, where no depletion may occur. In
some samples chlorine will dissipate within
1 to 2 hours of standing in the light. For
samples in which chlorine residual does
not dissipate in a reasonably short time,
destroy the residual chlorine by adding
sodium sulfite (Na2SO3).
Removing Residual Chlorine
• De-chlorinating with Sodium Sulfite (Na2SO3)
– To the 100 mL sample:
• Add 1 mL of a 1:50 sulfuric acid solution
• Add 1 mL of potassium iodide solution
• 2 mL starch indicator
• Titrate against 0.025N Na2SO3
X1 = X2
100 150 Where, X1 = known amount of titrant calculated to
neutralize a 100 mL sample
X2 = amount of titrant needed to neutralize 150 mL sample
To calculate the amount of Na2SO3 solution needed to neutralize a 75mL, 150mL and 250mL sample
X1 = X2 X1 = X3 X1 = X4
100 75 100 150 100 250
(X1)*(75) = (X2)*(100) (X1)*(150) = (X3)*(100) (X1)*(250) = (X4)*(100)
(X1)*(75) = X2 (X1)*(150) = X3 (X1)*(250) = X4
100 100 100
Where:
X1 = known amount of titrant calculated to neutralize a 100mL
sample
X2 = amount of titrant needed to neutralize a 75mL sample
X3 = amount of titrant needed to neutralize a 150mL sample
X4 = amount of titrant needed to neutralize a 250mL sample
EXAMPLE #1!
If only using 150 mLs of sample for the BOD
test, and it is known that 3 mLs of 0.025N
Na2SO3 will remove the residual chlorine from
100 mLs of the same sample, then the equation
for how much sodium sulfite the analyst would
need is:
X1 = X2
100 150
X1 = 3 mls
X2 = ?
3.0 = X2
100 150
(3.0)*(150) = (X2)*(100)
(3.0)*(150) = X2
100
X2 = 4.5 mLs 0.025N Na2SO3
EXAMPLE #2!
The residual chlorine in 100 mls of a
sample was removed by titrating with
0.025N Na2SO3 solution. 3 mls of the
titrant was used. Calculate the amount of
titrant necessary to remove the residual
chlorine from 75, 150, and 250 mls of the
same sample.
X1 = 3.0 mls
X2 = amount of titrant needed to neutralize a 75mL sample
X3 = amount of titrant needed to neutralize a 150mL sample
X4 = amount of titrant needed to neutralize a 250mL sample
3.0 = X2 3.0 = X3 3.0 = X4
100 75 100 150 100 250
(3.0)*(75) = (X2)*(100) (3.0)*(150) = (X3)*(100) (3.0)*(250) = (X4)*(100)
(3.0)*(75) = X2 (3.0)*(150) = X3 (3.0)*(250) = X4
100 100 100
X2 = 2.25 mls titrant X3 = 4.5 mls titrant X4 = 7.5 mls titrant
Determining Dilutions of an
Unknown
A PT Demand Standard has a BOD
concentration range of 15.0 – 250 mg/L.
What dilutions are needed to find the
result with in this range?
BOD5, mg/L = D1 – D2
(%X)
Where:
DO1 = Initial DO
DO2 = Final DO
X = % volume of sample to be
used
Determining Dilutions of an
Unknown
Step #1: Finding the dilutions on the lower end of the range (15.0 mg/L)
BOD mg/L = 15.0 mg/L
DO1 = 8.0
DO2 = 6.0 (assume a depletion of 2.0)
15.0 mg/L = 8.0 – 6.0
(%X)
(15.0 mg/L) x (%X) = 2.0
(%X) = 2.0
15.0 mg/L
(%X) = (0.13333) x 100
= 13.33% (round off to 13.3 %) = 40 mLs
Step #2: Finding the dilutions on the lower end of the range (15.0 mg/L)
BOD mg/L = 15 mg/L
DO1 = 8.0
DO2 = 1.0
15 mg/L = 8.0 – 1.0
(%X)
(15 mg/L) x (%X) = 7.0
(%X) = 7.0
15 mg/L
(%X) = (0.4666) x 100
= 46.666% (round off to 47 %) = 141 mLs or round off to 140 mLs
Finding the dilutions on the lower end of
the range (15.0 mg/L)
• Step #1 yielded 13.3% (or 40 mLs sample)
• Step #2 yielded 47% (or 141 mLs sample)
Step #1: Finding the dilutions on the higher end of the range (250.0 mg/L)
BOD mg/L = 250.0 mg/L
DO1 = 8.0
DO2 = 6.0 (assume a depletion of 2.0)
250.0 mg/L = 8.0 – 6.0
(%X)
(250.0 mg/L) x (%X) = 2.0
(%X) = 2.0
250.0 mg/L
(%X) = (0.008) x 100
= 0.8% = 2.4 mLs (round off to 2.0 mLs)
Step #2: Finding the dilutions on the higher end of the range (250.0 mg/L)
BOD mg/L = 250 mg/L
DO1 = 8.0
DO2 = 1.0
250 mg/L = 8.0 – 1.0
(%X)
(250 mg/L) x (%X) = 7.0
(%X) = 7.0
250 mg/L
(%X) = (0.028) x 100
= 2.8% (round off to 3 %) = 9 mLs
Finding the dilutions on the higher end of
the range (250.0 mg/L)
• Step #1 yielded 0.8% (or 2.0 mLs sample)
• Step #2 yielded 3% (or 9 mLs sample)
Recommended Dilutions from
“Standard Method’s 20th Edition”
• 0.0 – 1.0% for strong industrial wastes
• 1.0 – 5.0% for raw and settled
wastewaters
• 5.0 – 25 % for biologically treated effluents
• 25 – 100% for polluted river waters
References
• American Public Health Association; American Water
Works Association; Water Environment Federation (1998).
Standard Methods for the Examination of Water and Waste
Water. 20th Edition, 4500-O Oxygen (Dissolved)
• American Public Health Association; American Water
Works Association; Water Environment Federation (1998).
Standard Methods for the Examination of Water and Waste
Water. 20th Edition, 5210B 5-Day BOD Test
• Baird, R.B., & Smith, R.K.P. (2002). Third Century of
Biochemical Oxygen Dmand. Alexandria, VA: Water
Environment Federation.