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This document provides information on calculating clarifier loading parameters such as surface area, volume, detention time, and hydraulic loading. It begins with examples of calculating surface area for rectangular and circular tanks. It then covers volume calculations for rectangular, cylindrical, and cylindrical tanks with cone bottoms. Finally, it discusses calculating detention time by dividing the tank volume by the influent flow rate. Detention time represents the average time a drop of water spends in the clarifier.

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Clarifier Calculations Prepared By Michigan Department of Environmental Quality Operator Training and Certification Unit
Clarifier Calculations Hydraulic Loading Solids Loading
Clarifier Loading Calculations = Tank Volume, MG X 24 Flow into Tank, MGD Detention Time (DT) = Flow, gallons/day Surface Area, ft2 Surface Overflow Rate (SOR) Weir Overflow Rate (WOR) = Flow, gallons/day Length of Weir, ft Solids Loading = Solids, lbs/day Surface Area, ft2
But First
Area and Volume Calculations
Surface Area Calculations Rectangles Surface Area, ft2 = Length, ft X Width, ft Example 1: If a tank is 10 ft long and 5 ft wide, what is the surface area? SA, ft2 = 10 ft X 5 ft = 50 ft2
Example 2: If a tank is 10 ft 6 inches long and 5 ft 9 inches wide, what is the surface area in sq. ft.? NOT SA, ft2 = 10.6 ft X 5.9 ft Rectangles Surface Area, ft2 = Length, ft X Width, ft Surface Area Calculations
6 inches = 6 inches 12 in/ft = 0.5 ft Converting Inches to Feet So: 10 ft 6 inches = 10.5 ft 9 inches = 9 inches 12 in/ft = 0.75 ft So: 5 ft 9 inches = 5.75 ft
Example 2: If a tank is 10 ft 6 inches long and 5 ft 9 inches wide, what is the surface area in sq. ft.? Rectangles Surface Area, ft2 = Length, ft X Width, ft Surface Area Calculations SA, ft2 = 10.5 ft X 5.75 ft = 60.4 ft2
Rectangles Surface Area, ft2 = Length, ft X Width, ft Surface Area Calculations Work Calculations on Separate Paper Answers Given on Next Slides Practice 1: If a clarifier is 25 ft long and 9 ft wide, what is the surface area in sq. ft.? Practice 2: If a clarifier is 22 ft 3 inches long and 7 ft 7 inches wide, what is the surface area in sq. ft.?
Practice 1: If a clarifier is 25 ft long and 9 ft wide, what is the surface area in sq. ft.? Rectangles Surface Area, ft2 = Length, ft X Width, ft Surface Area Calculations SA, ft2 = 25 ft X 9 ft = 225 ft2
Practice 2: If a clarifier is 22 ft 3 inches long and 7 ft 7 inches wide, what is the surface area in sq. ft.? Rectangles Surface Area, ft2 = Length, ft X Width, ft Surface Area Calculations
3 inches = 3 inches 12 in/ft = 0.25 ft Converting Inches to Feet So: 22 ft 6 inches = 22.25 ft 7 inches = 7 inches 12 in/ft = 0.58 ft So: 7 ft 7 inches = 7.58 ft
Practice 2: If a clarifier is 22 ft 3 inches long and 7 ft 7 inches wide, what is the surface area in sq. ft.? Rectangles Surface Area, ft2 = Length, ft X Width, ft Surface Area Calculations SA, ft2 = 22.25 ft X 7.58 ft = 60.4 ft2
Circles Diameter Radius = Diameter 2 Surface Area Calculations Diameter: The distance across a circle, going through the center. Radius: The distance from the center a circle to the perimeter. Radius
Circumference = distance around circle Surface Area Calculations Circles Diameter Radius Circumference
Surface Area Calculations pie Circles Diameter Radius
Surface Area Calculations pi =  Circumference Diameter = = 3.14 for any circle Circles Diameter Radius
Surface Area Calculations Surface Area =  r2 = 3.14 X radius X radius Circles Diameter Radius
Surface Area Calculations Surface Area =  r2 A = 3.14 X (D/2)2 = 3.14 X D2 4 = 0.785 X D2 = 3.14 X r2 A = 3.14 4 X D2 = 3.14 X D2 22 Circles OR OR Any on the Three Formulas Can Be Used
Surface Area Calculations Example 1: If a tank has a radius of 15 feet, what is the surface area? Surface Area =  r2 S A =  r2 = 3.14 X 15 ft. X 15 ft. = 707 ft2 Circles Diameter Radius
Surface Area Calculations Example 2: If a tank has a diameter of 25 feet, what is the surface area? Surface Area =  r2 S A =  r2 = 3.14 X 12.5 ft. X 12.5 ft. = 491 ft2 Circles Diameter Radius
Surface Area Calculations Surface Area =  r2 Circles Diameter Radius Work Calculations on Separate Paper Answers Given on Next Slides Practice 1: If a tank has a diameter of 50.5 feet, what is the surface area? Practice 2: If a tank has a diameter of 50 feet 7 inches, what is the surface area?
Surface Area Calculations Practice 1: If a tank has a diameter of 50.5 feet, what is the surface area? Surface Area =  r2 S A =  r2 = 3.14 X 25.25 ft. X 25.25 ft. = 2002 ft2 Circles Diameter Radius
Surface Area Calculations Practice 2: If a tank has a diameter of 50 feet 7 inches, what is the surface area? Surface Area =  r2 7 inches = 7 inches 12 in/ft = 0.58 ft Circles Diameter Radius 50 feet 8 inches = 50.58 ft Radius = 50.58 ft ÷ 2 = 25.29 ft
Surface Area Calculations Surface Area =  r2 S A =  r2 = 3.14 X 25.29 ft. X 25.29 ft. = 2008 ft2 Circles Diameter Radius Practice 2: If a tank has a diameter of 50 feet 7 inches, what is the surface area?
Volume Calculations Volume – Three D
Volume Calculations Volume – Three Dimensions Length Depth
Rectangular Tanks Volume = Length X Width X Height (or Depth) Example 1: If a tank is 10 feet long, 5 feet wide, and 5 feet deep, what is the volume in cubic feet? Vol. = 10 ft X 5 ft X 5 ft = 250 ft3 Volume Calculations
Rectangular Tanks Volume = Length X Width X Height (or Depth) Example 2: If a tank is 20 ft. long, 7 ft. wide, and 5.5 ft. deep, what is the volume in cubic feet? Vol. = 20 ft X 7 ft X 5.5 ft = 770 ft3 Volume Calculations
Rectangular Tanks Volume = Length X Width X Height (or Depth) Example 3: If a tank is 25 ft. long, 9 ft. 3 inches wide, and 7.5 ft. deep, what is the volume in gallons? Vol. = 25 ft X 9.25 ft X 7.5 ft = 1734 ft3 Volume Calculations
Rectangular Tanks Volume = Length X Width X Height (or Depth) There are 7.48 gallons in one cubic foot Volume Calculations OR 7.48 gal/ft3 Example 3: If a tank is 25 ft. long, 9 ft. 3 inches wide, and 7.5 ft. deep, what is the volume in gallons?
Rectangular Tanks Volume = Length X Width X Height (or Depth) Volume Calculations 1734 ft3 X 7.48 gal/ft3 = 12,970 gallons Example 3: If a tank is 25 ft. long, 9 ft. 3 inches wide, and 7.5 ft. deep, what is the volume in gallons? Vol. = 25 ft X 9.25 ft X 7.5 ft = 1734 ft3
Rectangular Tanks Volume Calculations Practice 1: If a tank is 21 feet long, 9 feet wide, and 7 feet deep, what is the volume in cubic feet? Practice 2: If a tank is 22 ft. long, 9 ft. wide, and 7.5 ft. deep, what is the volume in cubic feet? Practice 3: If a tank is 35 ft. long, 12 ft. 3 inches wide, and 9.5 ft. deep, what is the volume in gallons? Work Calculations on Separate Paper Answers Given on Next Slides
Rectangular Tanks Volume = Length X Width X Height (or Depth) Practice 1: If a tank is 21 feet long, 9 feet wide, and 7 feet deep, what is the volume in cubic feet? Vol. = 21 ft X 9 ft X 7 ft = 1323 ft3 Volume Calculations
Rectangular Tanks Volume = Length X Width X Height (or Depth) Practice 2: If a tank is 22 ft. long, 9 ft. wide, and 7.5 ft. deep, what is the volume in cubic feet? Vol. = 22 ft X 9 ft X 7.5 ft = 1485 ft3 Volume Calculations
Rectangular Tanks Volume = Length X Width X Height (or Depth) Vol. = 35 ft X 12.25 ft X 9.5 ft = 4073 ft3 Volume Calculations Practice 3: If a tank is 35 ft. long, 12 ft. 3 inches wide, and 9.5 ft. deep, what is the volume in gallons?
Rectangular Tanks Volume = Length X Width X Height (or Depth) There are 7.48 gallons in one cubic foot Volume Calculations OR 7.48 gal/ft3 Practice 3: If a tank is 35 ft. long, 12 ft. 3 inches wide, and 9.5 ft. deep, what is the volume in gallons?
Rectangular Tanks Volume = Length X Width X Height (or Depth) Volume Calculations 4073 ft3 X 7.48 gal/ft3 = 30,466 gallons Practice 3: If a tank is 35 ft. long, 12 ft. 3 inches wide, and 9.5 ft. deep, what is the volume in gallons? Vol. = 35 ft X 12.25 ft X 9.5 ft = 4073 ft3
Volume Calculations Round (Cylinder) Tanks  = 3.14 r = Radius of circle h = Height (or Depth) Volume – Three Dimensions V = π r2h
Volume Calculations Round (Cylinder) Tanks Example 1 Find the Volume in cubic feet of a tank having a radius of 10 feet and a depth of 8 feet. V = r2h = 3.14 X 800 ft3 = 2512 ft3 = 3.14 X 10 ft X 10 ft X 8 ft
Volume Calculations Round (Cylinder) Tanks Example 2 Find the Volume in cubic feet of a tank having a diameter of 30 feet and a depth of 8 feet. V = r2h = 3.14 X 15 ft X 15 ft X 8 ft = 3.14 X 1800 ft3 = 5652 ft3
Volume Calculations Round (Cylinder) Tanks Example 3 Find the Volume in gallons of a tank having a diameter of 50 feet and a depth of 9 feet. V = r2h = 3.14 X 25 ft X 25 ft X 9 ft = 17,662.5 ft3 = 132,116 gallons X 7.48 gal/ft3
Volume Calculations Round (Cylinder) Tanks  = 3.14 r = Radius of circle h = Height (or Depth) Volume – Three Dimensions
Volume Calculations Round (Cylinder) Tanks Practice 1 Find the Volume in cubic feet of a tank having a radius of 22 feet and a depth of 10 feet. Practice 2 Find the Volume in cubic feet of a tank having a diameter of 28 feet and a depth of 7.5 feet. Practice 3 Find the Volume in gallons of a tank having a diameter of 48 feet and a depth of 7 feet. Work Calculations on Separate Paper Answers Given on Next Slides
Volume Calculations Round (Cylinder) Tanks Practice 1 Find the Volume in cubic feet of a tank having a radius of 22 feet and a depth of 10 feet. V = r2h = 3.14 X 4840 ft3 = 15,198 ft3 = 3.14 X 22 ft X 22 ft X 10 ft
Volume Calculations Round (Cylinder) Tanks Practice 2 Find the Volume in cubic feet of a tank having a diameter of 28 feet and a depth of 7.5 feet. V = r2h = 3.14 X 14 ft X 14 ft X 7.5 ft = 3.14 X 1470 ft3 = 4616 ft3
Volume Calculations Round (Cylinder) Tanks Practice 3 Find the Volume in gallons of a tank having a diameter of 48 feet and a depth of 7 feet. V = r2h = 3.14 X 24 ft X 24 ft X 7 ft = 12,660 ft3 = 94,700 gallons X 7.48 gal/ft3
What about a Round Tank with a Cone Bottom ? Volume Calculations
Cylinder with Cone Bottom Vcylinder = r2h1 Vcone = 1/3 r2h or r2h2 3 h1 h2 Vtotal = Vcylinder + Vcone Volume Calculations
Cylinder with Cone Bottom Volume Calculations Side Water Depth For Secondary Clarifiers Volume of Cone Not Considered Insignificant Compared to Total Volume (Filled with Sludge)
Clarifier Loading Detention Time (DT) The time it takes for a drop of water to travel from inlet to outlet Hydraulic Loading
Clarifier Detention Time Must Have Detention Time Long Enough for Solids to Settle
Time, Hrs. Sedimentation Efficiency 1 2 3 4 0 20 80 100 40 60 Too Long – No Increase in Removal
Clarifier Loading Detention Time (DT) The time it takes for a drop of water to travel from inlet to outlet Hydraulic Loading Typical Design Value = 2 – 3 Hours
Detention Time = Volume Flow = 25,000 gallons 310,000 gallons/day = 0.08 Days Hydraulic Loading Detention Time = Tank Volume Influent Rate Example 1a: Calculate the Detention Time in for a clarifier with a volume of 25,000 gallons that receives a flow of 310,000 gal/day.
Hydraulic Loading Detention Time = Tank Volume Influent Rate Example 1b: Calculate the Detention Time in HOURS for a clarifier with a volume of 25,000 gallons that receives a flow of 310,000 gal/day. Detention Time = Volume Flow = 25,000 gallons 310,000 gallons/day = 0.08 Days 0.08 Days X 24 hours/days = 1.9 Hours
Clarifier Loading Detention Time (DT) The time it takes for a drop of water to travel from inlet to outlet Hydraulic Loading Detention Time = Tank Volume Flow into Tank
Clarifier Loading Detention Time (DT) The time it takes for a drop of water to travel from inlet to outlet Hydraulic Loading DT, hrs = Tank Volume, (MG or Gallons) X 24 Flow into Tank, (MG/D or Gal/D)
Detention Time Example 1. Find the detention time in hours of a circular sedimentation tank having a volume of 75,000 gallons and a flow of 900,000 gallons per day. DT, hrs = Tank Volume, gallons Flow into Tank, gallons/day X 24 hr/day DT, hrs = 75,000 gallons 900,000 gallons/day X 24 hr/day = 2 hour
Detention Time Example 2. Find the detention time in hours of a circular sedimentation tank having a volume of 55,000 gallons and a flow of 0.75 MGD. DT, hrs = Tank Volume, gallons Flow into Tank, gallons/day X 24 hr/day 0.75 MGD = 750,000 gal/day
Detention Time Example 2. Find the detention time in hours of a circular sedimentation tank having a volume of 55,000 gallons and a flow of 0.75 MGD. DT, hrs = Tank Volume, gallons Flow into Tank, gallons/day X 24 hr/day DT, hrs = 55,000 gallons 750,000 gallons/day X 24 hr/day = 1.8 hour
Detention Time Practice 1. Find the detention time in hours of a clarifier having a volume of 52,000 gallons and a flow of 520,000 gallons per day. Practice 2. Find the detention time in hours of a rectangular sedimentation tank having a volume of 16,400 gallons and a flow of 0.225 MGD. Work Calculations on Separate Paper Answers Given on Next Slides
Detention Time Practice 1. Find the detention time in hours of a clarifier having a volume of 52,000 gallons and a flow of 520,000 gallons per day. DT, hrs = Tank Volume, gallons Flow into Tank, gallons/day X 24 hr/day DT, hrs = 52,000 gallons 520,000 gallons/day X 24 hr/day = 2.4 hour
Detention Time Practice 2. Find the detention time in hours of a rectangular sedimentation tank having a volume of 16,400 gallons and a flow of 0.225 MGD. DT, hrs = Tank Volume, gallons Flow into Tank, gallons/day X 24 hr/day 0.225 MGD = 220,000 gal/day
Detention Time DT, hrs = Tank Volume, gallons Flow into Tank, gallons/day X 24 hr/day DT, hrs = 16,400 gallons 225,000 gallons/day X 24 hr/day = 1.75 hour Practice 2. Find the detention time in hours of a rectangular sedimentation tank having a volume of 16,400 gallons and a flow of 0.225 MGD.
Clarifier Loading Detention Time (DT) The time it takes for a drop of water to travel from inlet to outlet Hydraulic Loading DT, hrs = Tank Volume, (MG or Gallons) X 24 Flow into Tank, (MG/D or Gal/D) Typical Design Value = 2 – 3 Hours Detention Time = Tank Volume Flow into Tank
Clarifier Loading Hydraulic Loading Surface Overflow Rate (SOR) OR Surface Loading Rate (SLR) The flow in gallons per day into the clarifier per square foot of surface area
Clarifier Loading Hydraulic Loading Surface Overflow Rate (SOR) OR Surface Loading Rate (SLR) The flow in gallons per day into the clarifier per square foot of surface area SOR, gpd/ft2 = Flow, gallons/day Surface Area, ft2
Clarifier Loading Hydraulic Loading Surface Overflow Rate (SOR) SOR, gpd/ft2 = Flow, gallons/day Surface Area, ft2 Need: Flow in gallons/day Surface Area, ft2 and SA = L X W SA = r2
Clarifier Loading Hydraulic Loading Surface Overflow Rate (SOR) SOR, gpd/ft2 = Flow, gallons/day Surface Area, ft2 Typical Design Value = 400 - 800 gal/day/ft2
Clarifier Loading Hydraulic Loading - Surface Overflow Rate (SOR) SOR, gpd/ft2 = Flow, gallons/day Surface Area, ft2 Example 1. Calculate the Surface Overflow Rate for a clarifier that is 50 ft long, 15 ft wide, 12 ft deep, and receives a flow of 338,000 gallons per day. = 50 ft X 15 ft = 750 ft2 Surface Area, ft2 338,000 gallons per day 750 ft2 SOR, gpd/ft2 = = 451 gpd/ft2
Clarifier Loading Hydraulic Loading - Surface Overflow Rate (SOR) SOR, gpd/ft2 = Flow, gallons/day Surface Area, ft2 Example 2. Calculate the Surface Overflow Rate for a clarifier that has a diameter of 60 ft, and receives an influent flow of 1.65 MGD. = 1,650,000 gallons per day Flow, gallons/day = 1.65 MGD X 1,000,000 Surface Area, ft2 = r2 = 2826 ft2 = 3.14 X 30 ft X 30 ft
Clarifier Loading Hydraulic Loading - Surface Overflow Rate (SOR) SOR, gpd/ft2 = Flow, gallons/day Surface Area, ft2 Example 2. Calculate the Surface Overflow Rate for a clarifier that has a diameter of 60 ft, and receives an influent flow of 1.65 MGD. 1,650,000 gallons per day 2826 ft2 SOR, gpd/ft2 = = 584 gpd/
Clarifier Loading Hydraulic Loading - Surface Overflow Rate (SOR) Practice 1. Calculate the Surface Overflow Rate for a clarifier that is 35 ft long, 9 ft wide, 7 ft deep, and receives a flow of 235,000 gallons per day. Practice 2. Calculate the Surface Overflow Rate for a clarifier that has a diameter of 45 ft. and receives an influent flow of 0.65 MGD. Work Calculations on Separate Paper Answers Given on Next Slides
Clarifier Loading Hydraulic Loading - Surface Overflow Rate (SOR) SOR, gpd/ft2 = Flow, gallons/day Surface Area, ft2 Practice 1. Calculate the Surface Overflow Rate for a clarifier that is 35 ft long, 9 ft wide, 7 ft deep, and receives a flow of 235,000 gallons per day. = 35 ft X 9 ft = 315 ft2 Surface Area, ft2 235,000 gallons per day 315 ft2 SOR, gpd/ft2 = = 746 gpd/ft2
Clarifier Loading Hydraulic Loading - Surface Overflow Rate (SOR) SOR, gpd/ft2 = Flow, gallons/day Surface Area, ft2 Practice 2. Calculate the Surface Overflow Rate for a clarifier that has a diameter of 45 ft. and receives an influent flow of 0.65 MGD. = 650,000 gallons per day Flow, gallons/day = 0.65 MGD X 1,000,000 Surface Area, ft2 = r2 = 1590 ft2 = 3.14 X 22.5 ft X 22.5 ft
Clarifier Loading Hydraulic Loading - Surface Overflow Rate (SOR) SOR, gpd/ft2 = Flow, gallons/day Surface Area, ft2 Practice 2. Calculate the Surface Overflow Rate for a clarifier that has a diameter of 45 ft, and receives an influent flow of 0.65 MGD. 650,000 gallons per day 1590 ft2 SOR, gpd/ft2 = = 409 gpd/ft2
Clarifier Loading Hydraulic Loading Surface Overflow Rate (SOR) SOR, gpd/ft2 = Flow, gallons/day Surface Area, ft2 Typical Design Value = 300 - 1200 gal/day/ft2 (800 gal/day/ft2)
Clarifier Loading Hydraulic Loading Weir Overflow Rate (WOR) The flow in gallons per day per linear foot of weir WOR, gal/d/ft = Flow, gallons/day Length of Weir, ft Typical Design Value = ~10,000 gal/day/ft
Clarifier Loading Hydraulic Loading - Weir Overflow Rate (WOR) WOR, gal/d/ft = Flow, gallons/day Length of Weir, ft Rectangular Tanks - Length of Weir, ft
Clarifier Loading Hydraulic Loading - Weir Overflow Rate (WOR) WOR, gal/d/ft = Flow, gallons/day Length of Weir, ft Rectangular Tanks - Length of Weir, ft FLOW = Width FLOW FLOW = 2 X Width = 4 X Width
Clarifier Loading Hydraulic Loading - Weir Overflow Rate (WOR) WOR, gal/d/ft = Flow, gallons/day Length of Weir, ft Length of Weir, ft Example 1 A primary clarifier is 20 feet wide, 60 feet long, and has a SWD of 12 feet. The clarifier has two effluent troughs across the width that allow the water to flow over both sides of each trough. The average flow to the clarifier is 0.65 MGD. Calculate the weir over flow rate for this clarifier. FLOW Flow, gallons/day = 0. 65 MGD X 1,000,000 = 650,000 gal/day = 4 X Width = 4 X 20 ft = 80 ft
Clarifier Loading Hydraulic Loading - Weir Overflow Rate (WOR) WOR, gal/d/ft = Flow, gallons/day Length of Weir, ft Example 1 A primary clarifier is 12 feet wide, 40 feet long, and has a SWD of 8 feet. The clarifier has two effluent troughs across the width that allow the water to flow over both sides of each trough. The average flow to the clarifier is 0.65 MGD. Calculate the weir over flow rate for this clarifier. FLOW = 8,125 gal/day/ft = 650,000 gal/day 80 ft
Clarifier Loading Hydraulic Loading - Weir Overflow Rate (WOR) WOR, gal/d/ft = Flow, gallons/day Length of Weir, ft Practice 1 A primary clarifier is 18 feet wide, 45 feet long, and has a SWD of 9 feet. The clarifier has an effluent trough across the end. The average flow to the clarifier is 0.085 MGD. Calculate the weir over flow rate for this clarifier. Work Calculations on Separate Paper Answers Given on Next Slides Practice 2 A primary clarifier is 12 foot wide, 40 foot long, and has a SWD of 8 feet. The clarifier has two effluent troughs across the width that allow the water to flow over both sides of each trough. The average flow to the clarifier is 0.41 MGD. Calculate the weir over flow rate for this clarifier.
Clarifier Loading Hydraulic Loading - Weir Overflow Rate (WOR) WOR, gal/d/ft = Flow, gallons/day Length of Weir, ft Practice 1 A primary clarifier is 18 feet wide, 45 feet long, and has a SWD of 9 feet. The clarifier has an effluent trough across the end. The average flow to the clarifier is 0.085 MGD. Calculate the weir over flow rate for this clarifier. FLOW = 9,444 gal/day/ft 85,000 gal/day 9 ft Flow, gallons/day = 0. 085MGD X 1,000,000 = 85,000 gal/day Length of Weir, ft = Width WOR, gal/d/ft =
Clarifier Loading Hydraulic Loading - Weir Overflow Rate (WOR) WOR, gal/d/ft = Flow, gallons/day Length of Weir, ft Length of Weir, ft Practice 2 A primary clarifier is 12 foot wide, 40 foot long, and has a SWD of 8 feet. The clarifier has two effluent troughs across the width that allow the water to flow over both sides of each trough. The average flow to the clarifier is 0.41 MGD. Calculate the weir over flow rate for this clarifier. FLOW Flow, gallons/day = 0. 41 MGD X 1,000,000 = 410,000 gal/day = 4 X Width = 4 X 12 ft = 48 ft
Clarifier Loading Hydraulic Loading - Weir Overflow Rate (WOR) WOR, gal/d/ft = Flow, gallons/day Length of Weir, ft Practice 2 A primary clarifier is 12 foot wide, 40 foot long, and has a SWD of 8 feet. The clarifier has two effluent troughs across the width that allow the water to flow over both sides of each trough. The average flow to the clarifier is 0.41 MGD. Calculate the weir over flow rate for this clarifier. FLOW = 8,542 gal/day/ft = 410,000 gal/day 48 ft
WOR, gal/d/ft = Flow, gallons/day Length of Weir, ft Clarifier Loading Hydraulic Loading - Weir Overflow Rate (WOR) Weir Length for a Circular Clarifier ?
Circumference = distance around circle Surface Area Calculations Circles Diameter Radius Circumference
Circumference = distance around circle Surface Area Calculations Circles Diameter Radius Circumference = 3.14 for any circle =  = Circumference Diameter C D C =  D
Circumference = distance around circle Surface Area Calculations Circles Diameter Radius Circumference C =  D For Weir Length D = Diameter of the Weir (May Not be same as Tank)
WOR, gal/d/ft = Flow, gallons/day Length of Weir, ft Clarifier Loading Hydraulic Loading - Weir Overflow Rate (WOR) Weir Length for a Circular Clarifier
WOR, gal/d/ft = Flow, gallons/day Length of Weir, ft Clarifier Loading Hydraulic Loading - Weir Overflow Rate (WOR) Example 1. The flow to a circular clarifier is 690,000 gallons per day. The clarifier is 24 feet in diameter with the weirs set 1 foot from the outside wall, for a weir diameter of 22 feet. Calculate the weir overflow rate. 22 feet Length of Weir, ft =  X Weir Diameter = 3.14 X 22 ft = 69.08 ft
WOR, gal/d/ft = Flow, gallons/day Length of Weir, ft Clarifier Loading Hydraulic Loading - Weir Overflow Rate (WOR) Example 1. The flow to a circular clarifier is 690,000 gallons per day. The clarifier is 24 feet in diameter with the weirs set 1 foot from the outside wall, for a weir diameter of 22 feet. Calculate the weir overflow rate. 22 feet WOR, gpd/ft = 690,000 gpd 69.08 ft = 9,988 gpd/ft
WOR, gal/d/ft = Flow, gallons/day Length of Weir, ft Hydraulic Loading - Weir Overflow Rate (WOR) Example 2. The flow to a circular clarifier is 1.65 MGD. The clarifier is 60 feet in diameter with the weir set in 1 foot from the outside wall. Calculate the weir overflow rate. 58 feet Length of Weir, ft =  X Weir Diameter = 3.14 X 58 ft = 182 ft WOR, gpd/ft = 1,650,000 gpd 182 ft = 9,066 gpd/ft
Hydraulic Loading - Weir Overflow Rate (WOR) Practice 1. The flow to a circular clarifier is 0.80 MGD. The clarifier is 30 feet in diameter with the weir set in 1 foot from the outside wall. Calculate the weir overflow rate. Practice 2. The flow to a circular clarifier is 1.4 MGD. The clarifier is 54 feet in diameter with the weir set in 1 foot from the outside wall. Calculate the weir overflow rate. Practice 3. The flow to a circular clarifier is 2.1 MGD. The clarifier is 75 feet in diameter with the weir set in 18 inches from the outside wall. Calculate the weir overflow rate. Work Calculations on Separate Paper Answers Given on Next Slides
WOR, gal/d/ft = Flow, gallons/day Length of Weir, ft Hydraulic Loading - Weir Overflow Rate (WOR) Practice 1. The flow to a circular clarifier is 0.80 MGD. The clarifier is 30 feet in diameter with the weir set in 1 foot from the outside wall. Calculate the weir overflow rate. 28 feet Length of Weir, ft =  X Weir Diameter = 3.14 X 28 ft = 88 ft WOR, gpd/ft = 800,000 gpd 88 ft = 9,091 gpd/ft
WOR, gal/d/ft = Flow, gallons/day Length of Weir, ft Hydraulic Loading - Weir Overflow Rate (WOR) Practice 2. The flow to a circular clarifier is 1.4 MGD. The clarifier is 54 feet in diameter with the weir set in 1 foot from the outside wall. Calculate the weir overflow rate. 52 feet Length of Weir, ft =  X Weir Diameter = 3.14 X 52 ft = 163 ft WOR, gpd/ft = 1,400,000 gpd 163 ft = 8,589 gpd/ft
WOR, gal/d/ft = Flow, gallons/day Length of Weir, ft Hydraulic Loading - Weir Overflow Rate (WOR) Practice 3. The flow to a circular clarifier is 2.1 MGD. The clarifier is 75 feet in diameter with the weir set in 18 inches from the outside wall. Calculate the weir overflow rate. 72 feet Length of Weir, ft =  X Weir Diameter = 3.14 X 72 ft = 226 ft WOR, gpd/ft = 2,100,000 gpd 226 ft = 9,292 gpd/ft
Clarifier Loading Hydraulic Loading Weir Overflow Rate (WOR) The flow in gallons per day per linear foot of weir WOR, gal/d/ft = Flow, gallons/day Length of Weir, ft Typical Design Value = ~10,000 gal/day/ft
Solids Loading SLR, lbs/d/ft2 = Solids, lbs/day Surface Area, ft2 Clarifier Loading The pounds per day of solids in the clarifier influent per square foot of surface area Same Calculations as for Hydraulic Loading SA = L X W SA = r2 - Solids Loading Rate (SLR)
Solids Loading SLR, lbs/d/ft2 = Solids, lbs/day Surface Area, ft2 Clarifier Loading The pounds per day of solids in the clarifier influent per square foot of surface area “Pounds Equation” - Solids Loading Rate (SLR)
Pounds = Concentration Of STUFF In the Water Conc. x Flow (or Volume) x 8.34 Lbs/gallon X Quantity Of Water The STUFF Is In Weight Of The Water X “Pounds Equation”
Flow (volume) and concentration must be expressed in specific units. Flow or volume must be expressed as millions of gallons: gallons _ 1,000,000 gal/MG = MG i.e.) A tank contains 1,125,000 gallons of water. How many million gallons are there? 1,125,000 gal = 1.125 MG 1,000,000 gal/MG “Pounds Equation”
Concentration must be expressed as parts per million parts. Concentration usually reported as milligrams per liter. This unit is equivalent to ppm. 1 mg = 1 mg = 1 mg = ppm liter 1000 grams 1,000,000 mg ppm = Parts Mil Parts = Lbs. Mil Lbs. “Pounds Equation”
Lbs. = When flow (volume) is expressed as MG and conc. is in ppm, the units cancel to leave only pounds. Concentration x Flow (or volume) x 8.34 lbs/gallon Lbs. M Lbs. X X M gal Lbs. gal “Pounds Equation”
If the flow rate is entered in M gal per day (MG/D), the answer will be in lbs/day. Example: The flow to a clarifier is 1.2 MGD and the concentration of suspended solids in the flow is 2500 mg/L. How many pounds of suspended solids come into the clarifier each day? Lbs/day = conc. (mg/L) x flow (MGD) x 8.34 lbs gal Lbs/day = 2500 mg/L x 1,200,000 gal/day x 8.34 lbs 1,000,000 gal/MG gal = 2500 x 1.2 x 8.34 = 25,020 Lbs/day “Pounds Equation”
Solids Loading SLR, lbs/d/ft2 = Solids, lbs/day Surface Area, ft2 Clarifier Loading The pounds per day of solids in the clarifier influent per square foot of surface area - Solids Loading Rate (SLR) Typical Design Value = Max 30 lbs/d/ft2
Solids Loading SLR, lbs/d/ft2 = Solids, lbs/day Surface Area, ft2 Clarifier Loading - Solids Loading Rate (SLR) Example 1. Calculate the Solids Loading Rate for a clarifier with a 50 ft diameter and a depth of 12 feet, and receives a flow of 2.4 MGD with a suspended solids concentration of 1800 mg/L. SA = 3.14 X 25 ft X 25 ft = 1962.5 ft2 1800 mg/L X 2.4 MGD X 8.34 lbs/gal = 36,029 lbs/d 36,029 lbs/day 1962.5 ft2 = 18.4 lbs/d/ft2 Solids, lbs/day = SLR, lbs/d/ft2 =
Solids Loading SLR, lbs/d/ft2 = Solids, lbs/day Surface Area, ft2 Clarifier Loading - Solids Loading Rate (SLR) Example 2. Calculate the Solids Loading Rate for a clarifier with a 31 ft diameter and a depth of 9 feet, and receives a flow of 750,000 gallons per day with a suspended solids concentration of 2600 mg/L. SA = 3.14 X 15.5 ft X 15.5 ft = 754 ft2 2600 mg/L X 0.75 MGD X 8.34 lbs/gal = 16,263 lbs/d 16,263 lbs/day 754 ft2 = 21.6 lbs/d/ft2 Solids, lbs/day = SLR, lbs/d/ft2 =
Solids Loading Clarifier Loading - Solids Loading Rate (SLR) Work Calculations on Separate Paper Answers Given on Next Slides Practice 1. Calculate the Solids Loading Rate for a clarifier with a 12 ft width and a length of 50 feet, and receives a flow of 600,000 gallons per day with a suspended solids concentration of 3400 mg/L. Practice 2. Calculate the Solids Loading Rate for a clarifier with a 22 ft radius and a depth of 8.5 feet, and receives a flow of 1,450,000 gallons per day with a suspended solids concentration of 3400 mg/L. Practice 3. Calculate the Solids Loading Rate for a clarifier with a 22 ft diameter and a depth of 7 feet, and receives a flow of 0.82 MGD with a suspended solids concentration of 1950 mg/L.
Solids Loading SLR, lbs/d/ft2 = Solids, lbs/day Surface Area, ft2 Clarifier Loading - Solids Loading Rate (SLR) Practice 1. Calculate the Solids Loading Rate for a clarifier with a 12 ft width and a length of 50 feet, and receives a flow of 600,000 gallons per day with a suspended solids concentration of 3400 mg/L. SA = 12 ft X 50 ft = 600 ft2 3400 mg/L X 0.6 MGD X 8.34 lbs/gal = 17,014 lbs/d 17,014 lbs/day 600 ft2 = 28.4 lbs/d/ft2 SLR, lbs/d/ft2 =
Solids Loading SLR, lbs/d/ft2 = Solids, lbs/day Surface Area, ft2 Clarifier Loading - Solids Loading Rate (SLR) Practice 2. Calculate the Solids Loading Rate for a clarifier with a 22 ft radius and a depth of 8.5 feet, and receives a flow of 1,450,000 gallons per day with a suspended solids concentration of 3400 mg/L. SA = 3.14 X 22 ft X 22 ft = 1520 ft2 3400 mg/L X 1.45 MGD X 8.34 lbs/gal = 41,116 lbs/d 41,116 lbs/day 1520 ft2 = 27.1 lbs/d/ft2 SLR, lbs/d/ft2 =
Solids Loading SLR, lbs/d/ft2 = Solids, lbs/day Surface Area, ft2 Clarifier Loading - Solids Loading Rate (SLR) Practice 3. Calculate the Solids Loading Rate for a clarifier with a 22 ft diameter and a depth of 7 feet, and receives a flow of 0.82 MGD with a suspended solids concentration of 1950 mg/L. SA = 3.14 X 11 ft X 11 ft = 380 ft2 1950 mg/L X 0.82 MGD X 8.34 lbs/gal = 13,336 lbs/d 13,336 lbs/day 380 ft2 = 35.1 lbs/d/ft2 SLR, lbs/d/ft2 =
Solids Loading SLR, lbs/d/ft2 = Solids, lbs/day Surface Area, ft2 Clarifier Loading The pounds per day of solids in the clarifier influent per square foot of surface area - Solids Loading Rate (SLR) Typical Design Value = 25 - 30 lbs/d/ft2
Clarifier Loading Calculations DT, hrs = Tank Volume, MG X 24 Flow into Tank, MGD Typical Design Value = 2 – 3 Hours Detention Time (DT) SOR, gpd/ft2 = Flow, gallons/day Surface Area, ft2 Surface Overflow Rate (SOR) Typical Design Value = 400 - 800 gal/d/ft2 Weir Overflow Rate (WOR) WOR, gal/d/ft =Flow, gallons/day Length of Weir, ft Typical Design Value = ~10,000 gal/d/ft Solids Loading SLR, lbs/d/ft2 = Solids, lbs/day Surface Area, ft2 Typical Design Value = 25 - 30 lbs/d/ft2
Clarifier Calculations Prepared By Michigan Department of Environmental Quality Operator Training and Certification Unit

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clarifier-calculations.ppt

  • 1. Clarifier Calculations Prepared By Michigan Department of Environmental Quality Operator Training and Certification Unit
  • 3. Clarifier Loading Calculations = Tank Volume, MG X 24 Flow into Tank, MGD Detention Time (DT) = Flow, gallons/day Surface Area, ft2 Surface Overflow Rate (SOR) Weir Overflow Rate (WOR) = Flow, gallons/day Length of Weir, ft Solids Loading = Solids, lbs/day Surface Area, ft2
  • 6. Surface Area Calculations Rectangles Surface Area, ft2 = Length, ft X Width, ft Example 1: If a tank is 10 ft long and 5 ft wide, what is the surface area? SA, ft2 = 10 ft X 5 ft = 50 ft2
  • 7. Example 2: If a tank is 10 ft 6 inches long and 5 ft 9 inches wide, what is the surface area in sq. ft.? NOT SA, ft2 = 10.6 ft X 5.9 ft Rectangles Surface Area, ft2 = Length, ft X Width, ft Surface Area Calculations
  • 8. 6 inches = 6 inches 12 in/ft = 0.5 ft Converting Inches to Feet So: 10 ft 6 inches = 10.5 ft 9 inches = 9 inches 12 in/ft = 0.75 ft So: 5 ft 9 inches = 5.75 ft
  • 9. Example 2: If a tank is 10 ft 6 inches long and 5 ft 9 inches wide, what is the surface area in sq. ft.? Rectangles Surface Area, ft2 = Length, ft X Width, ft Surface Area Calculations SA, ft2 = 10.5 ft X 5.75 ft = 60.4 ft2
  • 10. Rectangles Surface Area, ft2 = Length, ft X Width, ft Surface Area Calculations Work Calculations on Separate Paper Answers Given on Next Slides Practice 1: If a clarifier is 25 ft long and 9 ft wide, what is the surface area in sq. ft.? Practice 2: If a clarifier is 22 ft 3 inches long and 7 ft 7 inches wide, what is the surface area in sq. ft.?
  • 11. Practice 1: If a clarifier is 25 ft long and 9 ft wide, what is the surface area in sq. ft.? Rectangles Surface Area, ft2 = Length, ft X Width, ft Surface Area Calculations SA, ft2 = 25 ft X 9 ft = 225 ft2
  • 12. Practice 2: If a clarifier is 22 ft 3 inches long and 7 ft 7 inches wide, what is the surface area in sq. ft.? Rectangles Surface Area, ft2 = Length, ft X Width, ft Surface Area Calculations
  • 13. 3 inches = 3 inches 12 in/ft = 0.25 ft Converting Inches to Feet So: 22 ft 6 inches = 22.25 ft 7 inches = 7 inches 12 in/ft = 0.58 ft So: 7 ft 7 inches = 7.58 ft
  • 14. Practice 2: If a clarifier is 22 ft 3 inches long and 7 ft 7 inches wide, what is the surface area in sq. ft.? Rectangles Surface Area, ft2 = Length, ft X Width, ft Surface Area Calculations SA, ft2 = 22.25 ft X 7.58 ft = 60.4 ft2
  • 15. Circles Diameter Radius = Diameter 2 Surface Area Calculations Diameter: The distance across a circle, going through the center. Radius: The distance from the center a circle to the perimeter. Radius
  • 16. Circumference = distance around circle Surface Area Calculations Circles Diameter Radius Circumference
  • 18. Surface Area Calculations pi =  Circumference Diameter = = 3.14 for any circle Circles Diameter Radius
  • 19. Surface Area Calculations Surface Area =  r2 = 3.14 X radius X radius Circles Diameter Radius
  • 20. Surface Area Calculations Surface Area =  r2 A = 3.14 X (D/2)2 = 3.14 X D2 4 = 0.785 X D2 = 3.14 X r2 A = 3.14 4 X D2 = 3.14 X D2 22 Circles OR OR Any on the Three Formulas Can Be Used
  • 21. Surface Area Calculations Example 1: If a tank has a radius of 15 feet, what is the surface area? Surface Area =  r2 S A =  r2 = 3.14 X 15 ft. X 15 ft. = 707 ft2 Circles Diameter Radius
  • 22. Surface Area Calculations Example 2: If a tank has a diameter of 25 feet, what is the surface area? Surface Area =  r2 S A =  r2 = 3.14 X 12.5 ft. X 12.5 ft. = 491 ft2 Circles Diameter Radius
  • 23. Surface Area Calculations Surface Area =  r2 Circles Diameter Radius Work Calculations on Separate Paper Answers Given on Next Slides Practice 1: If a tank has a diameter of 50.5 feet, what is the surface area? Practice 2: If a tank has a diameter of 50 feet 7 inches, what is the surface area?
  • 24. Surface Area Calculations Practice 1: If a tank has a diameter of 50.5 feet, what is the surface area? Surface Area =  r2 S A =  r2 = 3.14 X 25.25 ft. X 25.25 ft. = 2002 ft2 Circles Diameter Radius
  • 25. Surface Area Calculations Practice 2: If a tank has a diameter of 50 feet 7 inches, what is the surface area? Surface Area =  r2 7 inches = 7 inches 12 in/ft = 0.58 ft Circles Diameter Radius 50 feet 8 inches = 50.58 ft Radius = 50.58 ft ÷ 2 = 25.29 ft
  • 26. Surface Area Calculations Surface Area =  r2 S A =  r2 = 3.14 X 25.29 ft. X 25.29 ft. = 2008 ft2 Circles Diameter Radius Practice 2: If a tank has a diameter of 50 feet 7 inches, what is the surface area?
  • 28. Volume Calculations Volume – Three Dimensions Length Depth
  • 29. Rectangular Tanks Volume = Length X Width X Height (or Depth) Example 1: If a tank is 10 feet long, 5 feet wide, and 5 feet deep, what is the volume in cubic feet? Vol. = 10 ft X 5 ft X 5 ft = 250 ft3 Volume Calculations
  • 30. Rectangular Tanks Volume = Length X Width X Height (or Depth) Example 2: If a tank is 20 ft. long, 7 ft. wide, and 5.5 ft. deep, what is the volume in cubic feet? Vol. = 20 ft X 7 ft X 5.5 ft = 770 ft3 Volume Calculations
  • 31. Rectangular Tanks Volume = Length X Width X Height (or Depth) Example 3: If a tank is 25 ft. long, 9 ft. 3 inches wide, and 7.5 ft. deep, what is the volume in gallons? Vol. = 25 ft X 9.25 ft X 7.5 ft = 1734 ft3 Volume Calculations
  • 32. Rectangular Tanks Volume = Length X Width X Height (or Depth) There are 7.48 gallons in one cubic foot Volume Calculations OR 7.48 gal/ft3 Example 3: If a tank is 25 ft. long, 9 ft. 3 inches wide, and 7.5 ft. deep, what is the volume in gallons?
  • 33. Rectangular Tanks Volume = Length X Width X Height (or Depth) Volume Calculations 1734 ft3 X 7.48 gal/ft3 = 12,970 gallons Example 3: If a tank is 25 ft. long, 9 ft. 3 inches wide, and 7.5 ft. deep, what is the volume in gallons? Vol. = 25 ft X 9.25 ft X 7.5 ft = 1734 ft3
  • 34. Rectangular Tanks Volume Calculations Practice 1: If a tank is 21 feet long, 9 feet wide, and 7 feet deep, what is the volume in cubic feet? Practice 2: If a tank is 22 ft. long, 9 ft. wide, and 7.5 ft. deep, what is the volume in cubic feet? Practice 3: If a tank is 35 ft. long, 12 ft. 3 inches wide, and 9.5 ft. deep, what is the volume in gallons? Work Calculations on Separate Paper Answers Given on Next Slides
  • 35. Rectangular Tanks Volume = Length X Width X Height (or Depth) Practice 1: If a tank is 21 feet long, 9 feet wide, and 7 feet deep, what is the volume in cubic feet? Vol. = 21 ft X 9 ft X 7 ft = 1323 ft3 Volume Calculations
  • 36. Rectangular Tanks Volume = Length X Width X Height (or Depth) Practice 2: If a tank is 22 ft. long, 9 ft. wide, and 7.5 ft. deep, what is the volume in cubic feet? Vol. = 22 ft X 9 ft X 7.5 ft = 1485 ft3 Volume Calculations
  • 37. Rectangular Tanks Volume = Length X Width X Height (or Depth) Vol. = 35 ft X 12.25 ft X 9.5 ft = 4073 ft3 Volume Calculations Practice 3: If a tank is 35 ft. long, 12 ft. 3 inches wide, and 9.5 ft. deep, what is the volume in gallons?
  • 38. Rectangular Tanks Volume = Length X Width X Height (or Depth) There are 7.48 gallons in one cubic foot Volume Calculations OR 7.48 gal/ft3 Practice 3: If a tank is 35 ft. long, 12 ft. 3 inches wide, and 9.5 ft. deep, what is the volume in gallons?
  • 39. Rectangular Tanks Volume = Length X Width X Height (or Depth) Volume Calculations 4073 ft3 X 7.48 gal/ft3 = 30,466 gallons Practice 3: If a tank is 35 ft. long, 12 ft. 3 inches wide, and 9.5 ft. deep, what is the volume in gallons? Vol. = 35 ft X 12.25 ft X 9.5 ft = 4073 ft3
  • 40. Volume Calculations Round (Cylinder) Tanks  = 3.14 r = Radius of circle h = Height (or Depth) Volume – Three Dimensions V = π r2h
  • 41. Volume Calculations Round (Cylinder) Tanks Example 1 Find the Volume in cubic feet of a tank having a radius of 10 feet and a depth of 8 feet. V = r2h = 3.14 X 800 ft3 = 2512 ft3 = 3.14 X 10 ft X 10 ft X 8 ft
  • 42. Volume Calculations Round (Cylinder) Tanks Example 2 Find the Volume in cubic feet of a tank having a diameter of 30 feet and a depth of 8 feet. V = r2h = 3.14 X 15 ft X 15 ft X 8 ft = 3.14 X 1800 ft3 = 5652 ft3
  • 43. Volume Calculations Round (Cylinder) Tanks Example 3 Find the Volume in gallons of a tank having a diameter of 50 feet and a depth of 9 feet. V = r2h = 3.14 X 25 ft X 25 ft X 9 ft = 17,662.5 ft3 = 132,116 gallons X 7.48 gal/ft3
  • 44. Volume Calculations Round (Cylinder) Tanks  = 3.14 r = Radius of circle h = Height (or Depth) Volume – Three Dimensions
  • 45. Volume Calculations Round (Cylinder) Tanks Practice 1 Find the Volume in cubic feet of a tank having a radius of 22 feet and a depth of 10 feet. Practice 2 Find the Volume in cubic feet of a tank having a diameter of 28 feet and a depth of 7.5 feet. Practice 3 Find the Volume in gallons of a tank having a diameter of 48 feet and a depth of 7 feet. Work Calculations on Separate Paper Answers Given on Next Slides
  • 46. Volume Calculations Round (Cylinder) Tanks Practice 1 Find the Volume in cubic feet of a tank having a radius of 22 feet and a depth of 10 feet. V = r2h = 3.14 X 4840 ft3 = 15,198 ft3 = 3.14 X 22 ft X 22 ft X 10 ft
  • 47. Volume Calculations Round (Cylinder) Tanks Practice 2 Find the Volume in cubic feet of a tank having a diameter of 28 feet and a depth of 7.5 feet. V = r2h = 3.14 X 14 ft X 14 ft X 7.5 ft = 3.14 X 1470 ft3 = 4616 ft3
  • 48. Volume Calculations Round (Cylinder) Tanks Practice 3 Find the Volume in gallons of a tank having a diameter of 48 feet and a depth of 7 feet. V = r2h = 3.14 X 24 ft X 24 ft X 7 ft = 12,660 ft3 = 94,700 gallons X 7.48 gal/ft3
  • 49. What about a Round Tank with a Cone Bottom ? Volume Calculations
  • 50. Cylinder with Cone Bottom Vcylinder = r2h1 Vcone = 1/3 r2h or r2h2 3 h1 h2 Vtotal = Vcylinder + Vcone Volume Calculations
  • 51. Cylinder with Cone Bottom Volume Calculations Side Water Depth For Secondary Clarifiers Volume of Cone Not Considered Insignificant Compared to Total Volume (Filled with Sludge)
  • 52. Clarifier Loading Detention Time (DT) The time it takes for a drop of water to travel from inlet to outlet Hydraulic Loading
  • 53. Clarifier Detention Time Must Have Detention Time Long Enough for Solids to Settle
  • 54. Time, Hrs. Sedimentation Efficiency 1 2 3 4 0 20 80 100 40 60 Too Long – No Increase in Removal
  • 55. Clarifier Loading Detention Time (DT) The time it takes for a drop of water to travel from inlet to outlet Hydraulic Loading Typical Design Value = 2 – 3 Hours
  • 56. Detention Time = Volume Flow = 25,000 gallons 310,000 gallons/day = 0.08 Days Hydraulic Loading Detention Time = Tank Volume Influent Rate Example 1a: Calculate the Detention Time in for a clarifier with a volume of 25,000 gallons that receives a flow of 310,000 gal/day.
  • 57. Hydraulic Loading Detention Time = Tank Volume Influent Rate Example 1b: Calculate the Detention Time in HOURS for a clarifier with a volume of 25,000 gallons that receives a flow of 310,000 gal/day. Detention Time = Volume Flow = 25,000 gallons 310,000 gallons/day = 0.08 Days 0.08 Days X 24 hours/days = 1.9 Hours
  • 58. Clarifier Loading Detention Time (DT) The time it takes for a drop of water to travel from inlet to outlet Hydraulic Loading Detention Time = Tank Volume Flow into Tank
  • 59. Clarifier Loading Detention Time (DT) The time it takes for a drop of water to travel from inlet to outlet Hydraulic Loading DT, hrs = Tank Volume, (MG or Gallons) X 24 Flow into Tank, (MG/D or Gal/D)
  • 60. Detention Time Example 1. Find the detention time in hours of a circular sedimentation tank having a volume of 75,000 gallons and a flow of 900,000 gallons per day. DT, hrs = Tank Volume, gallons Flow into Tank, gallons/day X 24 hr/day DT, hrs = 75,000 gallons 900,000 gallons/day X 24 hr/day = 2 hour
  • 61. Detention Time Example 2. Find the detention time in hours of a circular sedimentation tank having a volume of 55,000 gallons and a flow of 0.75 MGD. DT, hrs = Tank Volume, gallons Flow into Tank, gallons/day X 24 hr/day 0.75 MGD = 750,000 gal/day
  • 62. Detention Time Example 2. Find the detention time in hours of a circular sedimentation tank having a volume of 55,000 gallons and a flow of 0.75 MGD. DT, hrs = Tank Volume, gallons Flow into Tank, gallons/day X 24 hr/day DT, hrs = 55,000 gallons 750,000 gallons/day X 24 hr/day = 1.8 hour
  • 63. Detention Time Practice 1. Find the detention time in hours of a clarifier having a volume of 52,000 gallons and a flow of 520,000 gallons per day. Practice 2. Find the detention time in hours of a rectangular sedimentation tank having a volume of 16,400 gallons and a flow of 0.225 MGD. Work Calculations on Separate Paper Answers Given on Next Slides
  • 64. Detention Time Practice 1. Find the detention time in hours of a clarifier having a volume of 52,000 gallons and a flow of 520,000 gallons per day. DT, hrs = Tank Volume, gallons Flow into Tank, gallons/day X 24 hr/day DT, hrs = 52,000 gallons 520,000 gallons/day X 24 hr/day = 2.4 hour
  • 65. Detention Time Practice 2. Find the detention time in hours of a rectangular sedimentation tank having a volume of 16,400 gallons and a flow of 0.225 MGD. DT, hrs = Tank Volume, gallons Flow into Tank, gallons/day X 24 hr/day 0.225 MGD = 220,000 gal/day
  • 66. Detention Time DT, hrs = Tank Volume, gallons Flow into Tank, gallons/day X 24 hr/day DT, hrs = 16,400 gallons 225,000 gallons/day X 24 hr/day = 1.75 hour Practice 2. Find the detention time in hours of a rectangular sedimentation tank having a volume of 16,400 gallons and a flow of 0.225 MGD.
  • 67. Clarifier Loading Detention Time (DT) The time it takes for a drop of water to travel from inlet to outlet Hydraulic Loading DT, hrs = Tank Volume, (MG or Gallons) X 24 Flow into Tank, (MG/D or Gal/D) Typical Design Value = 2 – 3 Hours Detention Time = Tank Volume Flow into Tank
  • 68. Clarifier Loading Hydraulic Loading Surface Overflow Rate (SOR) OR Surface Loading Rate (SLR) The flow in gallons per day into the clarifier per square foot of surface area
  • 69. Clarifier Loading Hydraulic Loading Surface Overflow Rate (SOR) OR Surface Loading Rate (SLR) The flow in gallons per day into the clarifier per square foot of surface area SOR, gpd/ft2 = Flow, gallons/day Surface Area, ft2
  • 70. Clarifier Loading Hydraulic Loading Surface Overflow Rate (SOR) SOR, gpd/ft2 = Flow, gallons/day Surface Area, ft2 Need: Flow in gallons/day Surface Area, ft2 and SA = L X W SA = r2
  • 71. Clarifier Loading Hydraulic Loading Surface Overflow Rate (SOR) SOR, gpd/ft2 = Flow, gallons/day Surface Area, ft2 Typical Design Value = 400 - 800 gal/day/ft2
  • 72. Clarifier Loading Hydraulic Loading - Surface Overflow Rate (SOR) SOR, gpd/ft2 = Flow, gallons/day Surface Area, ft2 Example 1. Calculate the Surface Overflow Rate for a clarifier that is 50 ft long, 15 ft wide, 12 ft deep, and receives a flow of 338,000 gallons per day. = 50 ft X 15 ft = 750 ft2 Surface Area, ft2 338,000 gallons per day 750 ft2 SOR, gpd/ft2 = = 451 gpd/ft2
  • 73. Clarifier Loading Hydraulic Loading - Surface Overflow Rate (SOR) SOR, gpd/ft2 = Flow, gallons/day Surface Area, ft2 Example 2. Calculate the Surface Overflow Rate for a clarifier that has a diameter of 60 ft, and receives an influent flow of 1.65 MGD. = 1,650,000 gallons per day Flow, gallons/day = 1.65 MGD X 1,000,000 Surface Area, ft2 = r2 = 2826 ft2 = 3.14 X 30 ft X 30 ft
  • 74. Clarifier Loading Hydraulic Loading - Surface Overflow Rate (SOR) SOR, gpd/ft2 = Flow, gallons/day Surface Area, ft2 Example 2. Calculate the Surface Overflow Rate for a clarifier that has a diameter of 60 ft, and receives an influent flow of 1.65 MGD. 1,650,000 gallons per day 2826 ft2 SOR, gpd/ft2 = = 584 gpd/
  • 75. Clarifier Loading Hydraulic Loading - Surface Overflow Rate (SOR) Practice 1. Calculate the Surface Overflow Rate for a clarifier that is 35 ft long, 9 ft wide, 7 ft deep, and receives a flow of 235,000 gallons per day. Practice 2. Calculate the Surface Overflow Rate for a clarifier that has a diameter of 45 ft. and receives an influent flow of 0.65 MGD. Work Calculations on Separate Paper Answers Given on Next Slides
  • 76. Clarifier Loading Hydraulic Loading - Surface Overflow Rate (SOR) SOR, gpd/ft2 = Flow, gallons/day Surface Area, ft2 Practice 1. Calculate the Surface Overflow Rate for a clarifier that is 35 ft long, 9 ft wide, 7 ft deep, and receives a flow of 235,000 gallons per day. = 35 ft X 9 ft = 315 ft2 Surface Area, ft2 235,000 gallons per day 315 ft2 SOR, gpd/ft2 = = 746 gpd/ft2
  • 77. Clarifier Loading Hydraulic Loading - Surface Overflow Rate (SOR) SOR, gpd/ft2 = Flow, gallons/day Surface Area, ft2 Practice 2. Calculate the Surface Overflow Rate for a clarifier that has a diameter of 45 ft. and receives an influent flow of 0.65 MGD. = 650,000 gallons per day Flow, gallons/day = 0.65 MGD X 1,000,000 Surface Area, ft2 = r2 = 1590 ft2 = 3.14 X 22.5 ft X 22.5 ft
  • 78. Clarifier Loading Hydraulic Loading - Surface Overflow Rate (SOR) SOR, gpd/ft2 = Flow, gallons/day Surface Area, ft2 Practice 2. Calculate the Surface Overflow Rate for a clarifier that has a diameter of 45 ft, and receives an influent flow of 0.65 MGD. 650,000 gallons per day 1590 ft2 SOR, gpd/ft2 = = 409 gpd/ft2
  • 79. Clarifier Loading Hydraulic Loading Surface Overflow Rate (SOR) SOR, gpd/ft2 = Flow, gallons/day Surface Area, ft2 Typical Design Value = 300 - 1200 gal/day/ft2 (800 gal/day/ft2)
  • 80. Clarifier Loading Hydraulic Loading Weir Overflow Rate (WOR) The flow in gallons per day per linear foot of weir WOR, gal/d/ft = Flow, gallons/day Length of Weir, ft Typical Design Value = ~10,000 gal/day/ft
  • 81. Clarifier Loading Hydraulic Loading - Weir Overflow Rate (WOR) WOR, gal/d/ft = Flow, gallons/day Length of Weir, ft Rectangular Tanks - Length of Weir, ft
  • 82. Clarifier Loading Hydraulic Loading - Weir Overflow Rate (WOR) WOR, gal/d/ft = Flow, gallons/day Length of Weir, ft Rectangular Tanks - Length of Weir, ft FLOW = Width FLOW FLOW = 2 X Width = 4 X Width
  • 83. Clarifier Loading Hydraulic Loading - Weir Overflow Rate (WOR) WOR, gal/d/ft = Flow, gallons/day Length of Weir, ft Length of Weir, ft Example 1 A primary clarifier is 20 feet wide, 60 feet long, and has a SWD of 12 feet. The clarifier has two effluent troughs across the width that allow the water to flow over both sides of each trough. The average flow to the clarifier is 0.65 MGD. Calculate the weir over flow rate for this clarifier. FLOW Flow, gallons/day = 0. 65 MGD X 1,000,000 = 650,000 gal/day = 4 X Width = 4 X 20 ft = 80 ft
  • 84. Clarifier Loading Hydraulic Loading - Weir Overflow Rate (WOR) WOR, gal/d/ft = Flow, gallons/day Length of Weir, ft Example 1 A primary clarifier is 12 feet wide, 40 feet long, and has a SWD of 8 feet. The clarifier has two effluent troughs across the width that allow the water to flow over both sides of each trough. The average flow to the clarifier is 0.65 MGD. Calculate the weir over flow rate for this clarifier. FLOW = 8,125 gal/day/ft = 650,000 gal/day 80 ft
  • 85. Clarifier Loading Hydraulic Loading - Weir Overflow Rate (WOR) WOR, gal/d/ft = Flow, gallons/day Length of Weir, ft Practice 1 A primary clarifier is 18 feet wide, 45 feet long, and has a SWD of 9 feet. The clarifier has an effluent trough across the end. The average flow to the clarifier is 0.085 MGD. Calculate the weir over flow rate for this clarifier. Work Calculations on Separate Paper Answers Given on Next Slides Practice 2 A primary clarifier is 12 foot wide, 40 foot long, and has a SWD of 8 feet. The clarifier has two effluent troughs across the width that allow the water to flow over both sides of each trough. The average flow to the clarifier is 0.41 MGD. Calculate the weir over flow rate for this clarifier.
  • 86. Clarifier Loading Hydraulic Loading - Weir Overflow Rate (WOR) WOR, gal/d/ft = Flow, gallons/day Length of Weir, ft Practice 1 A primary clarifier is 18 feet wide, 45 feet long, and has a SWD of 9 feet. The clarifier has an effluent trough across the end. The average flow to the clarifier is 0.085 MGD. Calculate the weir over flow rate for this clarifier. FLOW = 9,444 gal/day/ft 85,000 gal/day 9 ft Flow, gallons/day = 0. 085MGD X 1,000,000 = 85,000 gal/day Length of Weir, ft = Width WOR, gal/d/ft =
  • 87. Clarifier Loading Hydraulic Loading - Weir Overflow Rate (WOR) WOR, gal/d/ft = Flow, gallons/day Length of Weir, ft Length of Weir, ft Practice 2 A primary clarifier is 12 foot wide, 40 foot long, and has a SWD of 8 feet. The clarifier has two effluent troughs across the width that allow the water to flow over both sides of each trough. The average flow to the clarifier is 0.41 MGD. Calculate the weir over flow rate for this clarifier. FLOW Flow, gallons/day = 0. 41 MGD X 1,000,000 = 410,000 gal/day = 4 X Width = 4 X 12 ft = 48 ft
  • 88. Clarifier Loading Hydraulic Loading - Weir Overflow Rate (WOR) WOR, gal/d/ft = Flow, gallons/day Length of Weir, ft Practice 2 A primary clarifier is 12 foot wide, 40 foot long, and has a SWD of 8 feet. The clarifier has two effluent troughs across the width that allow the water to flow over both sides of each trough. The average flow to the clarifier is 0.41 MGD. Calculate the weir over flow rate for this clarifier. FLOW = 8,542 gal/day/ft = 410,000 gal/day 48 ft
  • 89. WOR, gal/d/ft = Flow, gallons/day Length of Weir, ft Clarifier Loading Hydraulic Loading - Weir Overflow Rate (WOR) Weir Length for a Circular Clarifier ?
  • 90. Circumference = distance around circle Surface Area Calculations Circles Diameter Radius Circumference
  • 91. Circumference = distance around circle Surface Area Calculations Circles Diameter Radius Circumference = 3.14 for any circle =  = Circumference Diameter C D C =  D
  • 92. Circumference = distance around circle Surface Area Calculations Circles Diameter Radius Circumference C =  D For Weir Length D = Diameter of the Weir (May Not be same as Tank)
  • 93. WOR, gal/d/ft = Flow, gallons/day Length of Weir, ft Clarifier Loading Hydraulic Loading - Weir Overflow Rate (WOR) Weir Length for a Circular Clarifier
  • 94. WOR, gal/d/ft = Flow, gallons/day Length of Weir, ft Clarifier Loading Hydraulic Loading - Weir Overflow Rate (WOR) Example 1. The flow to a circular clarifier is 690,000 gallons per day. The clarifier is 24 feet in diameter with the weirs set 1 foot from the outside wall, for a weir diameter of 22 feet. Calculate the weir overflow rate. 22 feet Length of Weir, ft =  X Weir Diameter = 3.14 X 22 ft = 69.08 ft
  • 95. WOR, gal/d/ft = Flow, gallons/day Length of Weir, ft Clarifier Loading Hydraulic Loading - Weir Overflow Rate (WOR) Example 1. The flow to a circular clarifier is 690,000 gallons per day. The clarifier is 24 feet in diameter with the weirs set 1 foot from the outside wall, for a weir diameter of 22 feet. Calculate the weir overflow rate. 22 feet WOR, gpd/ft = 690,000 gpd 69.08 ft = 9,988 gpd/ft
  • 96. WOR, gal/d/ft = Flow, gallons/day Length of Weir, ft Hydraulic Loading - Weir Overflow Rate (WOR) Example 2. The flow to a circular clarifier is 1.65 MGD. The clarifier is 60 feet in diameter with the weir set in 1 foot from the outside wall. Calculate the weir overflow rate. 58 feet Length of Weir, ft =  X Weir Diameter = 3.14 X 58 ft = 182 ft WOR, gpd/ft = 1,650,000 gpd 182 ft = 9,066 gpd/ft
  • 97. Hydraulic Loading - Weir Overflow Rate (WOR) Practice 1. The flow to a circular clarifier is 0.80 MGD. The clarifier is 30 feet in diameter with the weir set in 1 foot from the outside wall. Calculate the weir overflow rate. Practice 2. The flow to a circular clarifier is 1.4 MGD. The clarifier is 54 feet in diameter with the weir set in 1 foot from the outside wall. Calculate the weir overflow rate. Practice 3. The flow to a circular clarifier is 2.1 MGD. The clarifier is 75 feet in diameter with the weir set in 18 inches from the outside wall. Calculate the weir overflow rate. Work Calculations on Separate Paper Answers Given on Next Slides
  • 98. WOR, gal/d/ft = Flow, gallons/day Length of Weir, ft Hydraulic Loading - Weir Overflow Rate (WOR) Practice 1. The flow to a circular clarifier is 0.80 MGD. The clarifier is 30 feet in diameter with the weir set in 1 foot from the outside wall. Calculate the weir overflow rate. 28 feet Length of Weir, ft =  X Weir Diameter = 3.14 X 28 ft = 88 ft WOR, gpd/ft = 800,000 gpd 88 ft = 9,091 gpd/ft
  • 99. WOR, gal/d/ft = Flow, gallons/day Length of Weir, ft Hydraulic Loading - Weir Overflow Rate (WOR) Practice 2. The flow to a circular clarifier is 1.4 MGD. The clarifier is 54 feet in diameter with the weir set in 1 foot from the outside wall. Calculate the weir overflow rate. 52 feet Length of Weir, ft =  X Weir Diameter = 3.14 X 52 ft = 163 ft WOR, gpd/ft = 1,400,000 gpd 163 ft = 8,589 gpd/ft
  • 100. WOR, gal/d/ft = Flow, gallons/day Length of Weir, ft Hydraulic Loading - Weir Overflow Rate (WOR) Practice 3. The flow to a circular clarifier is 2.1 MGD. The clarifier is 75 feet in diameter with the weir set in 18 inches from the outside wall. Calculate the weir overflow rate. 72 feet Length of Weir, ft =  X Weir Diameter = 3.14 X 72 ft = 226 ft WOR, gpd/ft = 2,100,000 gpd 226 ft = 9,292 gpd/ft
  • 101. Clarifier Loading Hydraulic Loading Weir Overflow Rate (WOR) The flow in gallons per day per linear foot of weir WOR, gal/d/ft = Flow, gallons/day Length of Weir, ft Typical Design Value = ~10,000 gal/day/ft
  • 102. Solids Loading SLR, lbs/d/ft2 = Solids, lbs/day Surface Area, ft2 Clarifier Loading The pounds per day of solids in the clarifier influent per square foot of surface area Same Calculations as for Hydraulic Loading SA = L X W SA = r2 - Solids Loading Rate (SLR)
  • 103. Solids Loading SLR, lbs/d/ft2 = Solids, lbs/day Surface Area, ft2 Clarifier Loading The pounds per day of solids in the clarifier influent per square foot of surface area “Pounds Equation” - Solids Loading Rate (SLR)
  • 104. Pounds = Concentration Of STUFF In the Water Conc. x Flow (or Volume) x 8.34 Lbs/gallon X Quantity Of Water The STUFF Is In Weight Of The Water X “Pounds Equation”
  • 105. Flow (volume) and concentration must be expressed in specific units. Flow or volume must be expressed as millions of gallons: gallons _ 1,000,000 gal/MG = MG i.e.) A tank contains 1,125,000 gallons of water. How many million gallons are there? 1,125,000 gal = 1.125 MG 1,000,000 gal/MG “Pounds Equation”
  • 106. Concentration must be expressed as parts per million parts. Concentration usually reported as milligrams per liter. This unit is equivalent to ppm. 1 mg = 1 mg = 1 mg = ppm liter 1000 grams 1,000,000 mg ppm = Parts Mil Parts = Lbs. Mil Lbs. “Pounds Equation”
  • 107. Lbs. = When flow (volume) is expressed as MG and conc. is in ppm, the units cancel to leave only pounds. Concentration x Flow (or volume) x 8.34 lbs/gallon Lbs. M Lbs. X X M gal Lbs. gal “Pounds Equation”
  • 108. If the flow rate is entered in M gal per day (MG/D), the answer will be in lbs/day. Example: The flow to a clarifier is 1.2 MGD and the concentration of suspended solids in the flow is 2500 mg/L. How many pounds of suspended solids come into the clarifier each day? Lbs/day = conc. (mg/L) x flow (MGD) x 8.34 lbs gal Lbs/day = 2500 mg/L x 1,200,000 gal/day x 8.34 lbs 1,000,000 gal/MG gal = 2500 x 1.2 x 8.34 = 25,020 Lbs/day “Pounds Equation”
  • 109. Solids Loading SLR, lbs/d/ft2 = Solids, lbs/day Surface Area, ft2 Clarifier Loading The pounds per day of solids in the clarifier influent per square foot of surface area - Solids Loading Rate (SLR) Typical Design Value = Max 30 lbs/d/ft2
  • 110. Solids Loading SLR, lbs/d/ft2 = Solids, lbs/day Surface Area, ft2 Clarifier Loading - Solids Loading Rate (SLR) Example 1. Calculate the Solids Loading Rate for a clarifier with a 50 ft diameter and a depth of 12 feet, and receives a flow of 2.4 MGD with a suspended solids concentration of 1800 mg/L. SA = 3.14 X 25 ft X 25 ft = 1962.5 ft2 1800 mg/L X 2.4 MGD X 8.34 lbs/gal = 36,029 lbs/d 36,029 lbs/day 1962.5 ft2 = 18.4 lbs/d/ft2 Solids, lbs/day = SLR, lbs/d/ft2 =
  • 111. Solids Loading SLR, lbs/d/ft2 = Solids, lbs/day Surface Area, ft2 Clarifier Loading - Solids Loading Rate (SLR) Example 2. Calculate the Solids Loading Rate for a clarifier with a 31 ft diameter and a depth of 9 feet, and receives a flow of 750,000 gallons per day with a suspended solids concentration of 2600 mg/L. SA = 3.14 X 15.5 ft X 15.5 ft = 754 ft2 2600 mg/L X 0.75 MGD X 8.34 lbs/gal = 16,263 lbs/d 16,263 lbs/day 754 ft2 = 21.6 lbs/d/ft2 Solids, lbs/day = SLR, lbs/d/ft2 =
  • 112. Solids Loading Clarifier Loading - Solids Loading Rate (SLR) Work Calculations on Separate Paper Answers Given on Next Slides Practice 1. Calculate the Solids Loading Rate for a clarifier with a 12 ft width and a length of 50 feet, and receives a flow of 600,000 gallons per day with a suspended solids concentration of 3400 mg/L. Practice 2. Calculate the Solids Loading Rate for a clarifier with a 22 ft radius and a depth of 8.5 feet, and receives a flow of 1,450,000 gallons per day with a suspended solids concentration of 3400 mg/L. Practice 3. Calculate the Solids Loading Rate for a clarifier with a 22 ft diameter and a depth of 7 feet, and receives a flow of 0.82 MGD with a suspended solids concentration of 1950 mg/L.
  • 113. Solids Loading SLR, lbs/d/ft2 = Solids, lbs/day Surface Area, ft2 Clarifier Loading - Solids Loading Rate (SLR) Practice 1. Calculate the Solids Loading Rate for a clarifier with a 12 ft width and a length of 50 feet, and receives a flow of 600,000 gallons per day with a suspended solids concentration of 3400 mg/L. SA = 12 ft X 50 ft = 600 ft2 3400 mg/L X 0.6 MGD X 8.34 lbs/gal = 17,014 lbs/d 17,014 lbs/day 600 ft2 = 28.4 lbs/d/ft2 SLR, lbs/d/ft2 =
  • 114. Solids Loading SLR, lbs/d/ft2 = Solids, lbs/day Surface Area, ft2 Clarifier Loading - Solids Loading Rate (SLR) Practice 2. Calculate the Solids Loading Rate for a clarifier with a 22 ft radius and a depth of 8.5 feet, and receives a flow of 1,450,000 gallons per day with a suspended solids concentration of 3400 mg/L. SA = 3.14 X 22 ft X 22 ft = 1520 ft2 3400 mg/L X 1.45 MGD X 8.34 lbs/gal = 41,116 lbs/d 41,116 lbs/day 1520 ft2 = 27.1 lbs/d/ft2 SLR, lbs/d/ft2 =
  • 115. Solids Loading SLR, lbs/d/ft2 = Solids, lbs/day Surface Area, ft2 Clarifier Loading - Solids Loading Rate (SLR) Practice 3. Calculate the Solids Loading Rate for a clarifier with a 22 ft diameter and a depth of 7 feet, and receives a flow of 0.82 MGD with a suspended solids concentration of 1950 mg/L. SA = 3.14 X 11 ft X 11 ft = 380 ft2 1950 mg/L X 0.82 MGD X 8.34 lbs/gal = 13,336 lbs/d 13,336 lbs/day 380 ft2 = 35.1 lbs/d/ft2 SLR, lbs/d/ft2 =
  • 116. Solids Loading SLR, lbs/d/ft2 = Solids, lbs/day Surface Area, ft2 Clarifier Loading The pounds per day of solids in the clarifier influent per square foot of surface area - Solids Loading Rate (SLR) Typical Design Value = 25 - 30 lbs/d/ft2
  • 117. Clarifier Loading Calculations DT, hrs = Tank Volume, MG X 24 Flow into Tank, MGD Typical Design Value = 2 – 3 Hours Detention Time (DT) SOR, gpd/ft2 = Flow, gallons/day Surface Area, ft2 Surface Overflow Rate (SOR) Typical Design Value = 400 - 800 gal/d/ft2 Weir Overflow Rate (WOR) WOR, gal/d/ft =Flow, gallons/day Length of Weir, ft Typical Design Value = ~10,000 gal/d/ft Solids Loading SLR, lbs/d/ft2 = Solids, lbs/day Surface Area, ft2 Typical Design Value = 25 - 30 lbs/d/ft2
  • 118. Clarifier Calculations Prepared By Michigan Department of Environmental Quality Operator Training and Certification Unit