Biochemical Oxygen Demand BOD Explained details Animation

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#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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