• Biochemical Oxygen Demand

Manish Kr. Semwal

Definition 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 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 BODMethod 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 BODMethod 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 OxygenBiochemical Oxygen DemandSolidsNitrogenBacteriological

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 DOF = Final DO

If all the DO is used up the test is invalid, as in B aboveTo 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 seedF’ = final DO of bottle with only dilution water and seedX = amount of seeded dilution water in sample bottle, mlY = 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 = -rWhere: 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:

Time, Days BOD, mg/L

2 104 166 20

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 evaporationWd = weight of dish aloneV = volume of sample

Total Solids can be divided into two fractions:

Suspended Solids

Dissolved SolidsDissolved solids are the solids that can pass through a glass fiber filter with a 0.45 micro pore size

Suspended solids are the solids that can not pass through a glass fiber filter with a 0.45 micron pore opening

Suspended solids

SS = (Fdf – Fd)/ V

Where: Fdf = weight of the Filter plus dry filtered solidsFd = weight of the clean, dry filterV = 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

Microbiological Measurements

Waterborne PathogensSalmonella (typhoid fever)Shigella (bacillary dysentery)Hepatitus virusEntameoba Histolytica (ameobic dysentery)Giardia Lamblia (“bever fever”)Cryptosporidium

Indicator Organisms

coliforms

MPN test

• Thank You All

•Manish Kr. Semwal