Biochemical Oxygen Demand (BOD): Explained details (Animation) скачать в хорошем качестве

Biochemical Oxygen Demand (BOD): Explained details (Animation)

#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