Module 4 Fundamentals of Wastewater Treatment Plant Operator

Module 4: Fundamentals of Wastewater Treatment Plant Operator Training

Unit 1–Preliminary Treatment 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. 2

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

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

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

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

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

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

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

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

Unit 2–Primary Treatment: Sedimentation & Flotation 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. 11

Detention Time 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 12

Detention Time: Sample 2. 1 What is the detention time when… • The flow is 3. 0 million gallons per day (MGD) or 3, 000 gal/day, and • Tank dimensions are 60 feet long by 30 feet wide by 10 feet deep? 13

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

Step #2 - Flow Given: Flow = 3. 0 million gallons per day (MGD) or 3, 000 gal/day 15

Step #3 - Calculation Detention Time, hr = Tank Volume, cu ft x 7. 5 gal/cu ft x 24 hr/day Flow, gal/day = 18, 000 cu ft X 7. 5 gal/cu ft x 24 hr/day 3, 000 gal/day = 3, 240, 000 gal/hr/day 3. 0 MGD = 1. 08 hours 16

Detention Time: Sample 2. 2 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? 17

Step #1 - Volume 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 18

Step #2 - Flow Given: Flow = 2. 5 million gallons per day (MGD) or 2, 500, 000 gal/day 19

Step #3 - Calculation 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 20

Weir Overflow Rate Formulas • Weir Overflow, gpd/ft = Flow Rate, GPD Length of Weir, ft • Length of Circular Weir = 3. 14 x Weir Diameter, ft 21

Weir Overflow Rate: Problem 2. 1 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 22

Weir Overflow Rate: Problem 2. 1 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 23

Surface Loading Rate: Problem 2. 2 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 24

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

Solids Loading Guidelines Pa DEP - Domestic Wastewater Facilities Manual Operation of Wastewater Treatment Plants, Vol I Primary Clarifiers Not considered Not 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 peak N/A 45 average, 50 peak N/A Dissolved-Air Flotation 40 #/day/sq ft (w/o polymer addition), 20 #/day sq ft (w polymer addition) 5 to 40 lbs/day/sq ft Sludge Thickening 5 to 12 #/day/sq ft 5 to 20 lbs/day/sq ft Carbonaceous Stage 26

Solids Loading 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 27

Solids Loading: Problem 2. 3 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 28

Step # 1 - Solids Applied 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 29

Step # 2 – Surface Area 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 30

Step # 3 – 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 31

Unit 3–Overview of Biological Secondary Treatment 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. 32

Trickling Filter Process AERATION TANK 33

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

Rotating Biological Contactor Process AERATION TANK 35

Complete Mix Activated Sludge 36

Contact Stabilization Schematic AERATION TANK 37

Extended Aeration Schematic AERATION TANK 38

Two Unit SBR Time Chart Time Period (hrs) Unit #1 Unit #2 0 -1 Fill Aeration 1 -2 Fill Settle 2 -3 Fill / Aeration Settle 3 -4 Fill / Aeration Draw 4 -5 Aeration Fill 5 -6 Settle Fill 6 -7 Settle Fill / Aeration 7 -8 Draw Fill / Aeration 39

Oxidation Ditch Schematic AERATION TANK 40