1. Biochemical Oxygen Demand
Manish Kr. Semwal

2. 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)

3. 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.

4. 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

5. 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.

6. Applications
Water Quality

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

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

9. 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 CO H2O
1.67x 10-3moles glucose/L x 6 moles O2/ mole glucose x 32 g O2/mole O2
= g O2/L = 321 mg O2/L

10. BOD Test
Dark
20oC
Time
Standard – 5 days
Ultimate

11. 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

12. 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

13. 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

14. 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

15. 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)

16. 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)

17. 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

18. Nitrogenous Oxygen Demand

19. 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

20. 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

21. 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

22. 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

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

24. 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

25. Thank You All
Manish Kr. Semwal