Biochemical Oxygen Demand (BOD): Explained details (Animation)
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Biochemical or Biological Oxygen Demand (BOD)
Chapters:
0:00 Kinetic school's intro
0:12 Biochemical or Biological Oxygen Demand
1:31 Definition of Biochemical or Biological Oxygen Demand
2:05 Why the Standard BOD test is run in the dark at 20oC for 5 days?
2:42 What is Microorganism?
3:40 Explanation of BOD
5:15 Idealized BOD curves.
6:50 Why Nitrogenous Oxygen Demand generally begins after about 8-10 days?
8:00 Explanation of Formula D1-D5/P
8:42 What is dilution factor?
9:04 Derivation of Formula BOD₅ = (D1 – D5) - f (B1 – B5)/ P
9:41 What is Seed and Why seeded is essential?
11:32 Derivation of Formula, BODt = L0 (1- e-k1t) (ln based)
13:30 Explanation of Formula, BODt = L0 (1- 10-Rt) (log based)
15:39 BOD rate (K1) depends on
15:55 Typical value of BOD5
16:24 Sources of increasing BOD Level
16:40 Significances of BOD test
17:09 Limitations of BOD Test
Biological /Biochemical Oxygen Demand (BOD):
The amount of oxygen utilized by aerobic microorganisms in breaking down the waste, is known as the Biological /Biochemical Oxygen Demand (BOD).
Formula:
Formula 1: BOD₅ = D1-D5 / P
Formula 2: BOD₅ = ((D1–D5) - f (B1–B5)) / P
Formula 3: BODt = L0 (1- e–kt)
Formula 4: BODt = L0 (1-10 –Rt)
BOD rate (K1) depends on:
• The nature of the waste
• The ability of microorganisms to degrade the waste in water
• The temperature
Typical value of BOD₅:
• For pristine water, the BOD value is, Less than 1 milligram per liter
• For moderate polluted water, the value is, 2 to 8 milligrams per liter
• and above 8 milligrams per Liter consider as severely polluted water.
Sources of increasing BOD Level:
• Effluents from Industry
• Leaves and Woody Debris
• Domestic Sewage
• Dead Fish
• Agriculture Runoff
Significances of BOD test:
• BOD test indicates the amount of organic pollution present in an aquatic ecosystem.
• It estimates the respiration rate in living organisms.
• Data from BOD test used for the development of engineering criteria for the design of wastewater treatment plants.
Exercise 1:
The dilution factor P for an unseeded mixture of waste and water is 0.030. The DO of the mixture is initially 9.0mg/L and after 5 days, it has dropped to 3.0 mg/L. The reaction rate constant k has been found to be 0.22 day-1.
a) What is the BOD₅ of the waste?
b) What would be the ultimate CBOD?
c) What would be the remaining oxygen demand after five days?
Answer:
a) BOD₅ = (D1-D5) / P = (9.0-3.0) / 0.030= 200 mg/L
b) BOD₅ = Lo (1 – e-kt)
So, L0 = BOD5 / (1- e-kt) = 200 / (1-e-0.22x5) = 300 mg/L
c) After 5 days, 200 mg/L of oxygen demand out of the total 300 mg/L would have already been used. the remaining oxygen demand would therefore be (300-200) = 100 mg/L
In Exercise 1, the wastes had an ultimate BOD equal to 300 mg/L.
At 20oC, the five-day BOD was 200 mg/L and the reaction rate constant (k) was 0.22/day.
What would the five-day BOD of this waste be at 25oC?
Answer:
k25 = k20 (T-20)
= 0.22 x (1.047) (25-20
= 0.277/day
So,
BOD₅ = Lo (1 – e-kt)
= 300 (1- e-0.277 x 5) = 225 mg/L
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