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Water, Part 2

Water, Part 2. Wastewater Treatment Primary Chapter: 11 Supplemental Chapters: 8, 9. Activity. What’s the problem with direct discharge of untreated wastewater? Write your ideas. Share with a partner. Share with the class . Treatment Processes.

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Water, Part 2

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  1. Water, Part 2 Wastewater Treatment Primary Chapter: 11 Supplemental Chapters: 8, 9

  2. Activity What’s the problem with direct discharge of untreated wastewater? Write your ideas. Share with a partner. Share with the class.

  3. Treatment Processes Treatment process = f(amount, type/source) Discharge limits = f(type/source, discharge location, time of year)

  4. Centralized Treatment

  5. Types of Wastewater(Section 11.1.1) From where does it all come?

  6. Transport of WW

  7. Characteristics of WW(Section 11.1.2)

  8. Activity Why is it important to obtain local data on wastewater composition and flow rates rather than using typical data when designing a new or an expansion to a treatment plant? Write your ideas. Share with a partner. Share with the class.

  9. Regulations(Sections 9.3.2 and 9.3.3) What’s the primary law for WWT?

  10. Activity Who must obtain an NPDES permit? • Manufacturers • Point source dischargers • Farmers • Septic tank owners • Municipalities

  11. Typical Municipal WWTP

  12. Degrees of Treatment - Example Raw Sewage Bar Rack Preliminary Treatment Grit Chamber Equalization Basin Primary Treatment Pump Solids Handling Primary Clarifier Biological Treatment Secondary Treatment Secondary Clarifier Advancedor Tertiary Treatment Disinfection ReceivingBody

  13. Influent – How much is removed? Solids BOD

  14. After Primary Treatment

  15. After Secondary Treatment

  16. Preliminary Treatment (Section 11.2.1) Screens Grit Chamber Bar Racks Comminutors (Grinders)

  17. Primary Treatment(Section 11.2.2)

  18. Secondary Treatment(Section 11.3)

  19. Oxygen Demand Section 9.1.2

  20. Oxygen Demand Amount of oxygen required to oxidize a waste Methods Theoretical oxygen demand (ThOD) Biochemical oxygen demand (BOD) Chemical oxygen demand (COD)

  21. Theoretical Oxygen Demand Total ThOD = C-ThOD + N-ThOD C-ThOD = stoichiometric amount of O2 required to convert an organic substance to CO2, H2O, and NH3 N-ThOD = stoichiometric amount of O2 required to convert NH3 and organic N to NO3-

  22. Example 1 What is the total ThOD to oxidize completely 25 mg/L of ethanol (CH3CH2OH)? CH3CH2OH + a O2 b CO2 + c H2O

  23. Determine Volume of O2 or Air

  24. Example 2 What is the ThOD to oxidize completely 25 mg/L of serine (CH2OHCHNH2COOH)? CH2OHCHNH2COOH + a O2 b CO2 + c H2O + d NH3 NH3 + a O2 b HNO3 + c H2O

  25. Biochemical Oxygen Demand (BOD)

  26. Lab: Unseeded BOD BODt = BOD at t days (mg/L) DOi = initial dissolved oxygen (mg/L) DOf = final dissolved oxygen (mg/L) Vs= sample volume (mL) Vb = bottle volume (mL) = 300 mL DF = dilution factor = Vb/Vs

  27. Lab: Seeded BOD Bi = initial DO of blank (mg/L) Bf = final DO of blank (mg/L)

  28. In-Class Activity 10 mL of a wastewater sample are placed in a 300-mL BOD bottle with unseeded nutrient broth. The initial DO of the sample is 8.5 mg/L. The DO is 3 mg/L after 5 days. What is the 5-day BOD? A 17oC sample is initially saturated with oxygen. Saturated seeded dilution water is used to obtain a 1:25 dilution. The final DO of the seeded dilution water is 8.2 mg/L while the final DO of the diluted sample is 2.8 mg/L. What is the 5-day BOD?

  29. In-Class Activity You received the results of a BOD test of the influent to a municipal WWTP run with 300-mL bottles. The initial DOs of the samples and seeded dilution water were at saturation (9.07 mg/L). All samples were run at a dilution factor of 40:1. The 5-day DOs are shown in the table below. The client is on the phone with your boss wanting to know why he hasn’t gotten a report yet. Justify why you threw out this data and made the lab redo the test. 1B = blank (seeded dilution water), S = sample with seeded dilution water

  30. Rate of BOD Removal Relate BOD exerted (BODt or Lt) to total, or ultimate, BOD (BODu or L) Assume that the BOD reduction rate (dC/dt) is proportional to the BOD remaining (C):

  31. Rate of BOD Removal cont. Integration yields: y = BOD exerted in t days = BODt L = ultimate BOD = BODu k1 = BOD degradation rate constant = deoxygenation constant

  32. Ultimate BOD

  33. In-Class Activity continued 10 mL of a wastewater sample are placed in a 300-mL BOD bottle. The initial DO of the sample is 8.5 mg/L. The DO is 3 mg/L after 5 days. What is the 5-day BOD? 165 mg/L What is the 3-day BOD if the reaction rate constant is 0.23/d?

  34. Chemical Oxygen Demand (COD)

  35. Secondary Treatment(Section 11.3)

  36. Activated Sludge: Aeration Basin (Sections 11.3.2 - 11.3.4)

  37. Aeration Basin Design • Kinetics • Mean cell residence time & hydraulic detention time

  38. max  (d-1) max/2 Ks S (mg/L) Kinetics: Logistic Growth

  39. Aeration Basin Design Mean cell residence time & hydraulic detention time

  40. MCRT from a Reactor without Recycle V, S, X Q, So, Xo Q, S, X

  41. Wasting from Recycle Line Qe = Q-Qw, S, Xc Q, So, Xo V, S, X Vc Qr , S, Xr Qw , S, Xr

  42. Example A conventional WWTP receives 2 MGD with an average BOD of 165 mg/L to the aeration basin. The aeration basin is 100,000 ft3. The MLSS is 2,800 mg/L and the effluent SS is 25 mg/L. The WAS is 38,000 gpd from the recycle line. The SS of the recycle flow is 9,000 mg/L. What is the mean cell residence time?

  43. General Equation forMean Cell Residence Time

  44. Secondary Clarifier(Section 11.3.5)

  45. Sludge Volume Index (SVI)

  46. Other Secondary Treatment Options

  47. Sequencing Batch Reactor (SBR)

  48. Aerated Lagoons

  49. Oxidation Ditch

  50. Trickling Filters (Section 11.3.1)

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