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• 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