Environmental Engineering Lecture 3 Dr. Hasan Hamouda

1. Environmental Engineering • Lecture 3 • Dr. Hasan Hamouda

2. Influence in the Environment & in Humanhealth Excretion Production Dumping Wastewater housewaste Seepage Water WWT Landfill thermal recycling Sludge Soil preflooder Groundwater Drinkwater

3. municipal wastewater treatment plants

4. Industrial wastewater treatment plants

5. wastewater treatment plants Primary Sedimentation Separation Sand trap influence O2 Primary Sludge Sedimentation Tank Bio-reactor preflooder effluence CO2 excessive Sludge

6. Sludge treatment Landfill excessive Sludge agriculturalrecycling Dewatering Anaerobic Stabilization thermalrecycling

7. Balance-boundary

8. 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.

9. Measurements of organic matter

10. 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

11. 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.

12. 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 (Temperature Different)  = 1.047 or as given

13. 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(1-e-kt) = 293 (l-e-0.29x5) = 224 mg/L

14. BOD Measurements (Lab)

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

16. BOD Measurements (Lab) • Example

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

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

20. BOD Measurements (Lab)