Biochemical Oxygen Demand (BOD): Explained details (Animation)
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- เผยแพร่เมื่อ 16 ก.ค. 2024
- #biologicaloxygendemand #animatedChemistry #kineticschool
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
Video credit:
Joseph Redfield from Pexels,
Ivan Khmelyuk from Pexels,
Taryn Elliott from Pexels,
Kelly Lacy from Pexels
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Very well explained
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GRACIAS
thank you...thank u soooooo much for this awsome video
Most welcome 😊
BOD analysis using oxitop
th-cam.com/video/hqr6C6v7DT4/w-d-xo.html
Thanks ❤
🇮🇳
thanks
Very thank you
You are welcome
Superb
Thanks 🤗
You have to be a real NERD to even know this video exist! 😂 Hello to all my fellow nerds! 🤓
Lol I'm in class 11 and preparing for jee
Hindi me thoda samjha dete etna english kyo
So much math lol
Content is great but the way to explain is worst (with due respect)
You're just reading