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Environmental Engineering Lecture 4c

Environmental Engineering Lecture 4c. Measurements of organic matter. 1- Biochemical oxygen demand (BOD): Biochemical oxygen demand (BOD) is the most commonly used parameter to define the strength of a municipal or organic industrial wastewater.

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Environmental Engineering Lecture 4c

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  1. Environmental Engineering Lecture 4c

  2. Measurements of organic matter • 1- Biochemical oxygen demand (BOD): • Biochemical oxygen demand (BOD) is the most commonly used parameter to define the strength of a municipal or organic industrial wastewater. • The BOD test is used to determine the relative oxygen requirements to treated effluents.

  3. Measurements of organic matter

  4. The following are the most common used methods: Measurements of organic matter 1- Biochemical oxygen demand (BOD): BOD5 is the oxygen equivalent of organic matter. It is determined by measuring the dissolved oxygen used by microorganisms during the biochemical oxidation of organic matter in 5 days 2- Chemical oxygen demand (COD): It is the oxygen equivalent of organic matter. It is determined by measuring the dissolved oxygen used during the chemical oxidation of organic matter in 3 hours. COD is assumed to be equal to uBOD

  5. Measurements of organic matter 3- Total organic carbon (TOC) This method measures the organic carbon existing in the wastewater by injecting a sample of the WW in special device in which the carbon is oxidized to carbon dioxide then carbon dioxide is measured and used to quantify the amount of organic matter in the WW. This method is only used for small concentration of organic matter. 4- Theoretical oxygen (ThOD) If the chemical formula of the organic matter existing in the WW is known the ThOD may be computed as the amount of oxygen needed to oxidize the organic carbon to carbon dioxide and other end products.

  6. Measurements of organic matter If L 0 or (BOD ultimate ) or UBOD. Lt = L0 e-kt (BOD remained). BODt = L0 - Lt = L0 – L0e-kt = L0(1-e-kt) BOD5 = L0 (1-e-k5) K = 0.23d-1 usually kT = k20T-20  = 1.047 or as given

  7. Determine the 1-day BOD and ultimate BOD for a wastewater whose 5-day 20C BOD is 200 mg/L. The reaction constant K= 0.23d-1 what would have been the 5-day BOD if it had been conducted at 25C? Solution:- • BODt = UBOD (1-e-kt) =L0(1-e-kt) 200 = L0 (1-e-0.23x5) L0 = 293 mg/L (this is UBOD) • Determine the 1-day BOD:- BODt = L0 (l-e-kt) BOD1 = 293 (l-e-0.23x1) = 60.1 mg/L’ • Determine the 5-day BOD at 25C:- KT = K20 (1.047)T-20 K25 = 0.23 (1.047)25-20=0.29 BOD5 = L0 (l-e –kt ) = 293 (l-e-0.29x5) = 224 mg/L

  8. ThOD: Calculate the Theoretical Oxygen Demand (ThOD) for sugar C12 H22 O11 dissolved in water to a concentration of 100 mg/L. Solution:- C12 H22 O11 + 12O2 12 CO2 + 11 H2O ThOD = ThOD = ThOD = 112.3 mg O2 / L

  9. BOD Measurements (Lab)

  10. BOD Measurements (Lab) • If microorganisms are already present in the sample – no outside seed

  11. BOD Measurements (Lab) • Example

  12. BOD Measurements (Lab) • If wastewater does not have sufficient population of microorganisms– need outside seed

  13. BOD Measurements (Lab) Suggested WW dilution in BOD test

  14. BOD Measurements (Lab)

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