1. Module 4: Fundamentals of Wastewater Treatment Wastewater Treatment Plant Operator Training

2. Learning Objectives Explain the general purpose of preliminary treatment. Explain the purpose of screening, grit removal, and pre-aeration. Differentiate between manually and mechanically cleaned racks and screens. Define differences between screening and comminution. Describe safe disposal of screenings and grit. Unit 1–Preliminary Treatment 2

3. Manually Cleaned Bar Screen 3 Diagram excerpted from Chapter 4: Racks, Screens, Comminutors and Grit Removal. In Operation of Wastewater Treatment Plants Volume I. AERATION TANK

4. Mechanically Cleaned Bar Screen 4 AERATION TANK Diagram excerpted from Chapter 4: Racks, Screens, Comminutors and Grit Removal. In Operation of Wastewater Treatment Plants Volume I.

5. 5 Aerated Grit Chamber AERATION TANK Diagram excerpted from Chapter 4: Racks, Screens, Comminutors and Grit Removal. In Operation of Wastewater Treatment Plants Volume I.

6. Cyclone Separator 6 AERATION TANK Diagram excerpted from Chapter 4: Racks, Screens, Comminutors and Grit Removal. In Operation of Wastewater Treatment Plants Volume I.

7. Grit Washer 7 AERATION TANK Diagram excerpted from Chapter 4: Racks, Screens, Comminutors and Grit Removal. In Operation of Wastewater Treatment Plants Volume I.

8. Comminutor 8 AERATION TANK Diagram excerpted from Chapter 4: Racks, Screens, Comminutors and Grit Removal. In Operation of Wastewater Treatment Plants Volume I.

9. Comminutor with By-Pass Screen 9 AERATION TANK Diagram excerpted from Chapter 4: Racks, Screens, Comminutors and Grit Removal. In Operation of Wastewater Treatment Plants Volume I.

10. Barminutor 10 AERATION TANK Diagram excerpted from Chapter 4: Racks, Screens, Comminutors and Grit Removal. In Operation of Wastewater Treatment Plants Volume I.

11. Learning Objectives Explain sedimentation and flotation principles. List factors that indicate a clarifier is not performing properly. Use mathematical formulas to solve for detention time, weir overflow, surface loading, and solids loading. Unit 2–Primary Treatment: Sedimentation & Flotation 11

12. Detention Time Formulas 12 Formulas Detention Time, hr. = Tank Volume, cu ft x 7.5 gal/cu ft x 24 hr/day Flow, gal/day Rectangular Tank Volume, cu ft = Length, ft x Width, ft x Height (Depth), ft Area of Circle, ft 2 = (0.785)(Diameter 2 ) or (  )(Radius 2 ) Note:  = 3.14 Circular Tank Volume, cu ft = Area x Height (Depth), ft Circular Tank Volume, cu ft = (0.785)(Diameter, ft) 2 x Height (Depth), ft or (3.14)(Radius, ft) 2 x Height (Depth), ft

13. What is the detention time when… The flow is 3.0 million gallons per day (MGD) or 3,000,000 gal/day, and Tank dimensions are 60 feet long by 30 feet wide by 10 feet deep? Detention Time: Sample 2.1 13

14. Given: Volume= 60 feet by 30 feet by 10 feet = 18,000 cu ft Step #1 - Volume 14

15. Given: Flow = 3.0 million gallons per day (MGD) or 3,000,000 gal/day Step #2 - Flow 15

16. Detention= Tank Volume, cu ft x 7.5 gal/cu ft x 24 hr/day Time, hr Flow, gal/day = 18,000 cu ft X 7.5 gal/cu ft x 24 hr/day 3,000,000 gal/day = 3,240,000 gal/hr/day 3.0 MGD = 1.08 hours Step #3 - Calculation 16

17. What is the detention time when… The flow is 2.5 million gallons per day (MGD), and Circular clarifier is 60 ft in diameter with a depth of 12 ft? Detention Time: Sample 2.2 17

18. Reminder: Circular Tank Volume, cu ft = 0.785 x (Diameter, ft) 2 x Depth, ft Volume = (0.785) x (60 ft) 2 x 12 ft deep = 33,912 cu ft Step #1 - Volume 18

19. Given: Flow = 2.5 million gallons per day (MGD) or 2,500,000 gal/day Step #2 - Flow 19

20. Detention Time, hr= 33,912 cu ft x 7.48 gal/cu ft x 24 hr/day 2.5 MGD = 6,087,882 gal/hr/day 2,500,000 gal/day = 2.44 hours Step #3 - Calculation 20

21. Formulas Weir Overflow, gpd/ft = Flow Rate, GPD Length of Weir, ft Length of Circular Weir = 3.14 x Weir Diameter, ft Weir Overflow Rate 21

22. What is the weir overflow rate when… The flow rate into the unit is 3.5 MGD, and The circular clarifier has a 75 foot diameter overflow weir Weir Overflow Rate: Problem 2.1 22

23. Weir overflow rate, gpd/ft = 3,500,000 gallons/day 3.14 x 75 feet = 3,500,000 gal/day 235.5 feet = 14,862 gpd/ft Weir Overflow Rate: Problem 2.1 23

24. What is the surface loading rate when… The flow into a rectangular clarifier is 5.0 MGD The clarifier is 40 feet wide by 110 feet long by 12 feet deep – Reminder: Surface loading rate, gpd/ft 2 = Flow Rate, gpd Surface area, ft 2 Surface Loading Rate: Problem 2.2 24

25. Surface loading rate, gpd/ft 2 = 5,000,000 gallons/day 40 ft x 110 feet = 5,000,000 gal/day 4400 ft 2 = 1,136 gpd/ft 2 Surface Loading Rate: Problem 2.2 25

26. Pa DEP - Domestic Wastewater Facilities Manual Operation of Wastewater Treatment Plants, Vol I Primary ClarifiersNot consideredNot generally considered Conventional Activated Sludge Secondary Clarifiers 40 #/day/sq ft average, 50 peak 12 to 30 lbs/day/sq ft Extended Aeration Secondary Clarifiers 30 #/day/sq ft average, 50 peak N/A Nitrification Secondary Clarifiers: Separate Nitrification Stage 30 average, 50 peakN/A Carbonaceous Stage45 average, 50 peakN/A Dissolved-Air Flotation40 #/day/sq ft (w/o polymer addition), 20 #/day sq ft (w polymer addition) 5 to 40 lbs/day/sq ft Sludge Thickening5 to 12 #/day/sq ft5 to 20 lbs/day/sq ft Solids Loading Guidelines 26

27. Formulas Solids Loading, lbs/day/ft 2 = Solids Applied, lbs/day Surface Area, ft 2 – Solids Applied, lbs/day = Flow, MGD x Conc., mg/L x 8.34 lbs/gal Solids Loading 27

28. Calculate the solids loading at which a clarifier is operating given the following… The circular clarifier has a diameter of 125 feet Forward flow is 6.0 MGD and the return sludge flow is 2.0 MGD MLSS is 4.000 mg/L Solids Loading: Problem 2.3 28

29. Reminder: Solids Applied, lbs/day = Flow, MGD x MLSS conc., mg/L x 8.34 lbs/gal Solids Applied = 8 MGD x 4,000 mg/L x 8.34 lbs/gal = 266,880 lbs/day Step # 1 - Solids Applied 29

30. Given: The circular clarifier has a diameter of 125 feet Area of Circle, ft 2 = (0.785)(Diameter) 2 = (0.785)(125 ft) 2 = (0.785)(15,625 ft 2 ) = 12,266 ft 2 Step # 2 – Surface Area 30

31. Solids Loading, lbs/day/ft 2 = Solids Applied, lbs/day Surface Area, ft 2 = 266,880 lbs/day 12,266 ft 2 = 22 lbs/day/ft 2 Step # 3 – Solids Loading 31

32. Learning Objectives List four biological secondary treatment processes. Explain the principles of the trickling filter process. Identify the different types of trickling filters. Explain the principles of the rotating biological contactor (RBC) process. Explain the principles of the activated sludge process. List the three waste treatment pond classifications and explain the principles of each. Unit 3–Overview of Biological Secondary Treatment 32

33. Trickling Filter Process 33 AERATION TANK

34. Trickling Filter 34 Diagram excerpted from Chapter 6: Trickling Filters. In Operation of Wastewater Treatment Plants Volume I. AERATION TANK

35. Rotating Biological Contactor Process 35 AERATION TANK

36. Complete Mix Activated Sludge 36

37. 37 Contact Stabilization Schematic AERATION TANK

38. 38 AERATION TANK Extended Aeration Schematic

39. Two Unit SBR Time Chart 39 Time Period (hrs)Unit #1Unit #2 0 - 1FillAeration 1 - 2FillSettle 2 - 3Fill / AerationSettle 3 - 4Fill / AerationDraw 4 - 5AerationFill 5 - 6SettleFill 6 - 7SettleFill / Aeration 7 - 8DrawFill / Aeration

40. Oxidation Ditch Schematic 40 AERATION TANK