Biochemical Oxygen Demand

General Considerations

* BOD is defined as the amount of oxygen required by bacteria while stabilizing decomposable organic matter under aerobic conditions at 20o C

* Decomposable means that the organic matter can serve as food for the bacteria, and energy is derived from the oxidation.

* BOD test is used to determine the pollutional strength of domestic, and industrial wastes in terms of oxygen that they will require if discharged into natural watercourses in which aerobic conditions exist.

* BOD Test is used by regulatory for evaluating sewage treatment plants and purification capacity of receiving bodies of water.

BOD Test Design

* Conditions of test is similar to those that occur in nature at 20 C.

* Samples are protected from the air to prevent re-aeration as D.O. diminishes.

* Samples of strong wastes are diluted due to limited Oxygen solubility.

* Test conditions must be suitable for living organisms to thrive.

* No toxic substances but all necessary nutrients available, N,P,& trace elements.

* Complete biological oxidation to completion, i.e., to CO2 and H2O.

Why 5 Day BOD?

* The BOD is considered complete after 20 days

* 20 Days is too long to wait.

* 5 days is a reasonable period for most of the BOD to be exerted.

* 5 Day BOD of Domestic and many industrial wastes is about 70 to 80 % of total BOD.

Nature of BOD Reaction

* First Order Reaction: Rate is proportional to the amount of oxidizable organic matter remaining at any time

* -dC/dt = kC, where C = concentration of oxidizable organic matter at start of the time interval t, and k is the rate constant for the reaction

* In BOD, it is customary to use L in place of C, where L represents the ultimate demand

* -dL/dt = kL by integration the expression is: Lt /L = e-kt = 10-kt

* In translating 5 day BOD to total BOD, the following modification is done: y = L(1 – 10-kt)

Nature of BOD Reaction

Nature of BOD Reaction

Method of Measuring BOD

* Dilution procedure is used due to limited O2 solubility

* Major Items of importance in the BOD test are: 1. Freedom from Toxic Materials. 2. Favorable pH and osmotic conditions 3. Available nutrient elements 4. Standard temperature of 20 degrees C 5. Presence of significant population of mixed organisms of soil origin

•Dilution Water is comprised of : Phosphate buffer furnishes phosphorous and pH of 7.0 Ferric chloride, magnesium sulfate and ammonium chloride supply the requirements for iron, sulfur, and nitrogen. Potassium, sodium, calcium, and magnesium salts give buffering and proper osmotic conditions.

The Need for Blanks

Since the dilution water should always be seeded to provide a uniformpopulation of organisms for biodegradation of organic matter. Therefore,A correction must be applied.

DO value of the dilution water at zero day compared to DO of dilution waterAfter 5 days will determine the amount of organic matter in dilution water.

Dilutions of Waste

* BOD is not influenced by oxygen levels as low as 0.5 mg/L

* Oxygen depletions <2.0 mg/L are not statistically reliable

* Reliable BOD values are >2.0 mg/L D.O. depletion but with at least 0.5 mg/L D.O. remaining

* These restrictions usually means a range of 2 to 7 mg/L depletion.

Incubation Bottles

* Bottles have glass stoppers ground to a point to prevent trapped air.

* Water seal to prevent air from entering the bottle during incubation.

* Bottles be free of organic matter.

* Cleaning with chromic acid or a good grade of detergent.

* If detergent is used then bottles should be rinsed with hot water to kill Nitrifying organisms.

* Four rinses with tap water and a final rinse with distilled water.

Dilution of Waste

Calculation of BOD

For per cent mixtures:

BOD(in mg/L) = [(DOb – DOi 100/%] – (DOb – Dos)

For direct pipetting:

BOD(in mg/L) = [ (DOb – DOi) vol. of bottle/ml sample] – (DOb – Dos)

Rate of Biochemical Oxidations

L Values versus Theoretical oxygen demand Values

Oxidation of glucose to carbon dioxide and water requires 192 g ofOxygen per mole or 1.065 mg of oxygen per mg of glucose

C6H12O6 + 6CO2 + 6H2O 180 192

Example: 300 mg/L of glucose = 320 mg/L Theoretical Oxygen DemandBut actual 20 day BOD yield L values = 250 to 285 mg/L

Reason for discrepancy between L values and theoretical oxygen demandIs: there remains a certain amount of organic matter quite resistant to Further biological attack which is referred to as humus, the amount of theDiscrepancy.

Questions?