bod / cbod from a to z - ohio water environment association

BOD / CBOD

FROM

A TO Z

Amy Starkey

Stark County Sanitary Engineers

What

is

BOD????

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 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.

Methods

Approved Methods

• Standard Methods 18th, 19th, and 20th

editions (5210B, 5-Day BOD Test)

Two Ways in Determining DO

• Iodometric Method (Winkler DO Method)

• Membrane Electrode Method

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

Sample

Collection,

Holding Time,

and

Storage

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

Quality

Control &

Procedure

Requirements

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

Dilution Water Sources

• Distilled

• Tap

• De-mineralized

• Natural Waters

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.

Seeding

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

and

Calculations

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


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?


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


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

Interferences

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



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




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 to Make

on a Unknown

BOD Sample

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?


(%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

Questions??

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.

bod / cbod from a to z - ohio water environment association

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