Biochemical Oxygen Demand Manish Kr. Semwal

Definition The Biochemical (or Biological) oxygen demand (BOD) is a measure of the amount of dissolved oxygen required to break down the organic material in a given volume of water through aerobic biological activity. ( By Micro-organism)

Definition Biochemical oxygen demand or BOD is a chemical procedure for determining the rate of uptake of dissolved oxygen by the rate biological organisms in a body of water use up oxygen. It is not an precise quantitative test, although it is widely used as an indication of the quality of water.

Method for the measuring of BOD Dilution method To ensure that all other conditions are equal, a very small amount of micro-organism seed is added to each sample being tested. Undiluted: Initial DO - Final DO = BOD Diluted: ((Initial DO - Final DO)- BOD of Seed) x Dilution Factor

Method for the measuring of BOD Manometric method This method is limited to the measurement of the oxygen consumption due only to carbonaceous oxidation. Ammonia oxidation is inhibited.

Applications Water Quality

Measures of Water Quality Some of the Most basic and Important Measures Dissolved Oxygen Biochemical Oxygen Demand Solids Nitrogen Bacteriological

Dissolved Oxygen (DO) Typically Measured by DO probe and Meter Electrochemical Half Cell Reaction

Biochemical Oxygen Demand (BOD) Amount of oxygen used by microorganisms to decompose organic matter in a water Theoretical BOD can be determined by balancing a chemical equation in which all organic matter is converted to CO2 Calculate the theoretical oxygen demand of 1.67 x 10-3 moles of glucose (C6H12O6): C6H12O6 + O2  CO2 + H2O general, unbalanced eqn C6H12O6 + 6 O2  6 CO2 + 6 H2O 1.67x 10-3moles glucose/L x 6 moles O2/ mole glucose x 32 g O2/mole O2 = 0.321 g O2/L = 321 mg O2/L

BOD Test Dark 20oC Time Standard – 5 days Ultimate

BOD = I - F I = Initial DO F = Final DO If all the DO is used up the test is invalid, as in B above To get a valid test dilute the sample, as in C above. In this case the sample was diluted by 1:10. The BOD can then be calculated by: BOD = (I – F) D D = dilution as a fraction D = volume of bottle/(volume of bottle – volume of dilution water) BOD = (8 – 4) 10 = 40 mg/L

For the BOD test to work microorganisms have to be present. Sometimes they are not naturally present in a sample so we have to add them. This is called “seeding” a sample If seed is added you may also be adding some BOD. We have to account for this in the BOD calculation: BOD = [(I – F) – (I’ – F’)(X/Y)]D Where: I’ = initial DO a bottle with only dilution water and seed F’ = final DO of bottle with only dilution water and seed X = amount of seeded dilution water in sample bottle, ml Y = amount of seeded dilution water in bottle with only seeded dilution water

Example Calculate the BOD5 of a sample under the following conditions. Seeded dilution water at 20oC was saturated with DO initially. After 5 days a BOD bottle with only seeded dilution water had a DO of 8 mg/L. The sample was diluted 1:30 with seeded dilution water. The sample was saturated with DO at 20oC initially. After five days the DO of the sample was 2 mg/L. Since a BOD bottle is 300 ml a 1:30 dilution would have 10 ml sample and 290 ml seeded dilution water. From the table, at 20oC, DOsat = 9.07 mg/L BOD5 = [(9.07 – 2) – (9.07 – 8)(290/300)] 30 = 174 mg/L

If we do a mass balance on the BOD bottle: dz/dt = -r Where: z = dissolved oxygen necessary for the microorganisms to decompose the organic matter If r is first order: dz/dt = -k1 t Separate the variables and integrate: z = z0 e-k1 t Z is defined as the amount of oxygen still to be used by the microorganisms to degrade the waste. If we define y to be the amount of oxygen which already been used to degrade the waste: L = z + y L = ultimate demand for O2

So: z = L - y By substitution: L – y = z0 e-k1 t But zo = L, so: y = L – Le-k1 t Or y = L(1-e-k1 t)

y = L(1-e-k1 t) Rearranging: (t/y)1/3 = (1/(k1 L)1/3) + (k12/3/(6 L1/3)) t y = b + m x So: k1 = 6 (slope/intercept) L = 1/(6 (slope)(intercept)2)

Example From a BOD test we have the following data: Slope = 0.545 Intercept = 0.021 k1 = 6(0.021/0.545) = 0.64 day-1 L = 1/[6(0.021)(0.545)2] = 26.7 mg/L

Nitrogenous Oxygen Demand

Solids Total Solids Residue on evaporation at 103oC TS = (Wds – Wd)/V Where: Wds = weight of dish plus solids after evaporation Wd = weight of dish alone V = volume of sample

Total Solids can be divided into two fractions: Suspended Solids Suspended solids are the solids that can not pass through a glass fiber filter with a 0.45 micron pore opening Dissolved Solids Dissolved solids are the solids that can pass through a glass fiber filter with a 0.45 micro pore size

Suspended solids SS = (Fdf – Fd)/ V Where: Fdf = weight of the Filter plus dry filtered solids Fd = weight of the clean, dry filter V = volume of sample

Volatile and Fixed Solids Volatile solids are the solids that are volatilized at 600oC Fixed solids are the solids that remain after heating to 600oC Generally the volatile solids are considered to be the organic fraction of the solids. Volatile Solids = Total Solids – Fixed Solids

Nitrogen Organic Nitrogen Ammonia Nitrates + Nitrites (NO3-, NO2-) Organic Nitrogen + Ammonia = Total Nitrogen Nitrates + Nitrites (NO3-, NO2-)

Microbiological Measurements Waterborne Pathogens Salmonella (typhoid fever) Shigella (bacillary dysentery) Hepatitus virus Entameoba Histolytica (ameobic dysentery) Giardia Lamblia (“bever fever”) Cryptosporidium Indicator Organisms coliforms MPN test

Thank You All Manish Kr. Semwal