20.1 Surface Overflow Rate & Weir Loading Rate Calculations

ABC Water Treatment Operator Class II Certification Study Guide Exam

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1 Chapter 1: Source Water Quality & Characteristics

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2 Chapter 2: Coagulation & Rapid Mixing

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4 Chapter 4: Granular Media Filtration

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13 Chapter 13: Physical & Chemical Water Quality Testing

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15 Chapter 15: Chemical Safety & Hazardous Materials

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16 Chapter 16: Plant Safety, Confined Space & Electrical Hazards

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17 Chapter 17: Regulations, Compliance & Administrative Operations

17.1 Safe Drinking Water Act (SDWA) Framework, MCLs & MCLGs 17.2 Surface Water Treatment Rules & Revised Total Coliform Rule 17.3 Chemical Standards, Public Notification Tiers & CCR 17.4 Emergency Response Planning & Plant Physical/Cyber Security 17.5 Administrative Operations: SOPs, Records, Procurement, Complaints & Optimization

18 Chapter 18: Flow, Volume, Detention Time & Velocity Calculations

18.1 Tank Volume & Storage Capacity Calculations 18.2 Flow Rate Unit Conversions & Pipe/Channel Velocity 18.3 Theoretical Detention Time & Contact Time Calculations

19 Chapter 19: Chemical Dosage & Feed Rate Calculations

19.1 The Pounds Formula & Mass Loading Calculations 19.2 Solution Dosing, Specific Gravity & Pump Drawdown Math 19.3 Fluoride Chemical Feed Calculations & Solution Dilutions

20 Chapter 20: Clarification, Filtration & Disinfection Math

20.1 Surface Overflow Rate & Weir Loading Rate Calculations 20.2 Filter Loading Rate, Backwash Rise Rate & Washwater Volume 20.3 CT Disinfection Calculations & Log-Inactivation Compliance

21 Chapter 21: Pump, Power & Hydraulic Pressure Calculations

21.1 Pressure and Head Conversion Calculations 21.2 Water Horsepower, Brake Horsepower & Electrical Power Math 21.3 Pump Affinity Laws & Impeller Performance Calculations

20.1 Surface Overflow Rate & Weir Loading Rate Calculations

Key Takeaways Explain This Topic AI

Surface Overflow Rate (SOR) defines the upward fluid velocity against which settling particles must descend: SOR (gpd/sq ft) = Total Flow (gpd) / Surface Area (sq ft); dividing by 1,440 min/day yields the hydraulic loading rate in gpm/sq ft.

Design SOR standards depend on clarification technology: conventional rectangular and circular basins operate at 500 to 1,000 gpd/sq ft (0.35 to 0.70 gpm/sq ft), solids-contact upflow units operate at 1.0 to 1.75 gpm/sq ft, and inclined tube/plate settlers operate at 2.0 to 4.0 gpm/sq ft.

Weir Overflow Rate (WOR) controls exit velocity at effluent launders to prevent shearing and lifting settled floc: WOR (gpd/linear ft) = Total Flow (gpd) / Total Weir Crest Length (ft), with maximum regulatory limits typically set at 10,000 to 20,000 gpd/linear ft.

Basin surface area geometry determines hydraulic capacity: rectangular clarifiers require Surface Area = Length × Width, while circular clarifiers require Surface Area = 0.785 × Diameter²; peripheral circular weir length equals π × Diameter (3.1416 × D).

Operating at excessive surface overflow rates leads to pin-floc carryover, shortened filter runs, and elevated settled turbidity; operational remedies include placing standby basins in service, adding polymeric coagulant aids, or reducing plant production throughput.

Last updated: September 2026

Settling Hydraulics and Surface Overflow Rate (SOR)

In conventional water treatment, clarification separates chemically coagulated and flocculated suspended solids from water by gravitational sedimentation. The fundamental hydraulic parameter governing clarifier efficiency is the Surface Overflow Rate (SOR) , also referred to as the hydraulic surface loading rate . [ Clarifier Surface Area (A) ]

Influent Flow (Q) +--------------------------+ Effluent Supernatant
========================> | ^ ^ ^ ^ ^ ^ ^ ^ | ==========================>
| | | | | | | | | | (Low Turbidity)
| Upward Fluid Velocity |
| (SOR = Q / A) |
| |
| v v v v v v v v | Sludge Blanket
| Particle Settling (Vs) | ==========================>
+--------------------------+ (Solids to Waste)
The Physics of Surface Overflow Rate

Under Hazen's classical sedimentation theory, a clarifier operates as an ideal settling basin where settling particles possess an intrinsic downward terminal settling velocity ( V s V_s V s ), while the bulk water mass moves upward toward the effluent launders with an upward fluid velocity ( V u p V_{up} V u p ):

If V s > V u p V_s > V_{up} V s > V u p , the particle settles downward into the sludge collection zone and is successfully removed.

If V s < V u p V_s < V_{up} V s < V u p , the particle is swept upward by hydraulic currents, escapes over the effluent weirs, and places an excessive particulate burden on downstream granular media filters.

The upward fluid velocity equals the volumetric flow rate ( Q Q Q ) divided by the horizontal surface area ( A A A ) of the basin. Therefore, Surface Overflow Rate represents the critical settling velocity : any particle whose settling velocity is equal to or greater than the SOR will be 100% captured under idealized conditions.

Exam Rule for Clarifier Depth : In ideal settling theory, the percentage of solids removed is strictly a function of clarifier surface area and flow rate— it is independent of basin depth . While adequate side water depth (typically 10 to 16 feet 10\text{ to }16\text{ feet} 10 to 16 feet ) is essential to prevent bottom scour, accommodate sludge scrapers, and provide buffer storage, doubling clarifier depth does not double its hydraulic settling capacity. Only increasing horizontal surface area decreases the SOR.

Core Formulas for Surface Overflow Rate

Depending on the units required on licensing exams, SOR is calculated in gallons per day per square foot ( g p d / s q f t gpd/sq\ ft g p d / s q f t ) or gallons per minute per square foot ( g p m / s q f t gpm/sq\ ft g p m / s q f t ): SOR ( g p d / s q f t ) = Total Plant Flow Rate ( g p d ) Clarifier Surface Area ( s q f t ) \text{SOR }(gpd/sq\ ft) = \frac{\text{Total Plant Flow Rate }(gpd)}{\text{Clarifier Surface Area }(sq\ ft)} SOR ( g p d / s q f t ) = Clarifier Surface Area ( s q f t ) Total Plant Flow Rate ( g p d ) Surface Loading Rate ( g p m / s q f t ) = Flow Rate ( g p m ) Clarifier Surface Area ( s q f t ) = SOR ( g p d / s q f t ) 1 , 440 m i n / d a y \text{Surface Loading Rate }(gpm/sq\ ft) = \frac{\text{Flow Rate }(gpm)}{\text{Clarifier Surface Area }(sq\ ft)} = \frac{\text{SOR }(gpd/sq\ ft)}{1,440\ min/day} Surface Loading Rate ( g p m / s q f t ) = Clarifier Surface Area ( s q f t ) Flow Rate ( g p m ) = 1 , 440 min / d a y SOR ( g p d / s q f t )

Clarifier Surface Area Geometry Formulas

To compute surface loading rates, operators must accurately determine the horizontal surface area ( A A A ) of the settling basin based on its geometric configuration.

1. Rectangular Clarifiers Surface Area ( s q f t ) = Length ( f t ) × Width ( f t ) \text{Surface Area }(sq\ ft) = \text{Length }(ft) \times \text{Width }(ft) Surface Area ( s q f t ) = Length ( f t ) × Width ( f t )

2. Circular Clarifiers Surface Area ( s q f t ) = π 4 × [ Diameter ( f t ) ] 2 ≈ 0.785 × [ Diameter ( f t ) ] 2 \text{Surface Area }(sq\ ft) = \frac{\pi}{4} \times [\text{Diameter }(ft)]^2 \approx 0.785 \times [\text{Diameter }(ft)]^2 Surface Area ( s q f t ) = 4 π × [ Diameter ( f t ) ] 2 ≈ 0.785 × [ Diameter ( f t ) ] 2

(Note: On WPI/ABC examinations, 0.785 × D 2 0.785 \times D^2 0.785 × D 2 is the standard formula provided on reference sheets) .

Table 20.1.1: Clarifier Surface Overflow Rate Design Standards

Clarifier Technology Standard Design SOR ( g p d / s q f t gpd/sq\ ft g p d / s q f t ) Equivalent Rate ( g p m / s q f t gpm/sq\ ft g p m / s q f t ) Operational Characteristics & Floc Types

Conventional Rectangular Basin 500 to 1 , 000 g p d / s q f t 500\text{ to }1,000\ gpd/sq\ ft 500 to 1 , 000 g p d / s q f t 0.35 to 0.70 g p m / s q f t 0.35\text{ to }0.70\ gpm/sq\ ft 0.35 to 0.70 g p m / s q f t Alum or ferric coagulant floc; quiescent horizontal plug flow.

Conventional Circular Center-Feed 500 to 1 , 000 g p d / s q f t 500\text{ to }1,000\ gpd/sq\ ft 500 to 1 , 000 g p d / s q f t 0.35 to 0.70 g p m / s q f t 0.35\text{ to }0.70\ gpm/sq\ ft 0.35 to 0.70 g p m / s q f t Radial flow from center feed well; peripheral effluent collection.

Solids-Contact / Upflow Sludge Blanket 1 , 440 to 2 , 500 g p d / s q f t 1,440\text{ to }2,500\ gpd/sq\ ft 1 , 440 to 2 , 500 g p d / s q f t 1.0 to 1.75 g p m / s q f t 1.0\text{ to }1.75\ gpm/sq\ ft 1.0 to 1.75 g p m / s q f t Pre-formed sludge blanket filters incoming micro-floc; lime softening.

Inclined Plate / Tube Settler Modules 2 , 880 to 5 , 760 g p d / s q f t 2,880\text{ to }5,760\ gpd/sq\ ft 2 , 880 to 5 , 760 g p d / s q f t 2.0 to 4.0 g p m / s q f t 2.0\text{ to }4.0\ gpm/sq\ ft 2.0 to 4.0 g p m / s q f t High-rate settling; 60° inclined tubes reduce settling distance to 2 inches.

Dissolved Air Flotation (DAF) 5 , 760 to 14 , 400 g p d / s q f t 5,760\text{ to }14,400\ gpd/sq\ ft 5 , 760 to 14 , 400 g p d / s q f t 4.0 to 10.0 g p m / s q f t 4.0\text{ to }10.0\ gpm/sq\ ft 4.0 to 10.0 g p m / s q f t Micro-bubbles float light algae and low-density organic floc to surface.

Weir Overflow Rate (WOR) and Effluent Launder Hydraulics

As clarified water approaches the basin exit, it flows over effluent weir plates (typically 90-degree V-notch weirs) into collection troughs termed launders . The rate of discharge over these weirs is the Weir Overflow Rate (WOR) or weir loading rate . WOR ( g p d / l i n e a r f t ) = Total Flow Rate ( g p d ) Total Active Weir Crest Length ( f t ) \text{WOR }(gpd/linear\ ft) = \frac{\text{Total Flow Rate }(gpd)}{\text{Total Active Weir Crest Length }(ft)} WOR ( g p d / l in e a r f t ) = Total Active Weir Crest Length ( f t ) Total Flow Rate ( g p d )

Why Weir Loading Matters

If the total crest length of the effluent weirs is too short for the volume of water passing through the basin, the exit velocity of the water accelerates dramatically. This localized high velocity creates upward suction currents (approach velocity) near the launders that lift settled floc off the basin floor or pull floating pin-floc over the weir plates.

State design standards (including the Recommended Standards for Water Works / 10-States Standards) enforce maximum weir loading thresholds:

Standard Clarifiers (Alum Floc) : Maximum 10 , 000 to 15 , 000 g p d / l i n e a r f t 10,000\text{ to }15,000\ gpd/linear\ ft 10 , 000 to 15 , 000 g p d / l in e a r f t .

Heavy Lime Floc / Upflow Clarifiers : Up to 20 , 000 g p d / l i n e a r f t 20,000\ gpd/linear\ ft 20 , 000 g p d / l in e a r f t . Peripheral Circular Launder: Interior Launders (Both Sides Active):

+--------------------------------+ +================================+
| ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ | | ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ |
| -------\\-------------//------ | | -----\\---------------//------ |
| \\ Effluent // | | Flow \\ Launder Crest// Flow |
| \\ Launder // | | ====> | Trough | <==== |
| | | +---------+ |
| Weir Length = π × Diameter | | Active Length = 2 × Launder |
+--------------------------------+ +================================+
Calculating Active Weir Length

Circular Clarifiers with Peripheral Weirs : The weir runs around the entire internal circumference of the tank wall: Weir Length ( f t ) = π × Diameter ( f t ) ≈ 3.1416 × D \text{Weir Length }(ft) = \pi \times \text{Diameter }(ft) \approx 3.1416 \times D Weir Length ( f t ) = π × Diameter ( f t ) ≈ 3.1416 × D

Rectangular Basins with End Wall Weirs : The weir length equals the width of the basin wall.

Finger Launders / Double-Sided Troughs : When collection launders project out into the basin and allow water to spill over both sides, the active weir length is twice the length of the trough : Active Weir Length = 2 × Launder Length × Number of Launders \text{Active Weir Length} = 2 \times \text{Launder Length} \times \text{Number of Launders} Active Weir Length = 2 × Launder Length × Number of Launders

Table 20.1.2: Clarifier Geometry and Loading Rate Equations

Operational Parameter Geometric Configuration Mathematical Equation

Surface Area ( s q f t sq\ ft s q f t ) Rectangular Basin Area = Length ( f t ) × Width ( f t ) \text{Area} = \text{Length }(ft) \times \text{Width }(ft) Area = Length ( f t ) × Width ( f t )

Surface Area ( s q f t sq\ ft s q f t ) Circular Tank Area = 0.785 × [ Diameter ( f t ) ] 2 \text{Area} = 0.785 \times [\text{Diameter }(ft)]^2 Area = 0.785 × [ Diameter ( f t ) ] 2

Weir Length ( f t ft f t ) Circular Peripheral Length = π × Diameter = 3.1416 × D \text{Length} = \pi \times \text{Diameter} = 3.1416 \times D Length = π × Diameter = 3.1416 × D

Weir Length ( f t ft f t ) Double-Sided Finger Launder Length = 2 × Trough Length × Count \text{Length} = 2 \times \text{Trough Length} \times \text{Count} Length = 2 × Trough Length × Count

Surface Overflow Rate All Geometries ( g p d / s q f t gpd/sq\ ft g p d / s q f t ) SOR = Flow ( g p d ) ÷ Surface Area ( s q f t ) \text{SOR} = \text{Flow }(gpd) \div \text{Surface Area }(sq\ ft) SOR = Flow ( g p d ) ÷ Surface Area ( s q f t )

Weir Overflow Rate All Geometries ( g p d / f t gpd/ft g p d / f t ) WOR = Flow ( g p d ) ÷ Weir Length ( f t ) \text{WOR} = \text{Flow }(gpd) \div \text{Weir Length }(ft) WOR = Flow ( g p d ) ÷ Weir Length ( f t )

Step-by-Step Worked Multi-Step Calculations
Example 1: Rectangular Basin Surface Overflow Rate

Problem Statement : A conventional water treatment plant operates a rectangular sedimentation basin that is 100 feet 100\text{ feet} 100 feet long, 30 feet 30\text{ feet} 30 feet wide, and has an operating water depth of 12 feet 12\text{ feet} 12 feet . The plant processes a steady flow rate of 2.4 MGD 2.4\text{ MGD} 2.4 MGD . Calculate:

The clarifier surface area in square feet.

The surface overflow rate (SOR) in gallons per day per square foot ( g p d / s q f t gpd/sq\ ft g p d / s q f t ).

The hydraulic surface loading rate in gallons per minute per square foot ( g p m / s q f t gpm/sq\ ft g p m / s q f t ).

Determine whether this operating rate complies with standard design guidelines ( 500 to 1 , 000 g p d / s q f t 500\text{ to }1,000\ gpd/sq\ ft 500 to 1 , 000 g p d / s q f t ).

Solution Procedure :

Step 1: Calculate Clarifier Surface Area Surface Area = Length × Width \text{Surface Area} = \text{Length} \times \text{Width} Surface Area = Length × Width Surface Area = 100 f t × 30 f t = 3 , 000 s q f t \text{Surface Area} = 100\ ft \times 30\ ft = 3,000\ sq\ ft Surface Area = 100 f t × 30 f t = 3 , 000 s q f t

(Note: Depth is not used in calculating surface area or SOR) .

Step 2: Convert Flow to Gallons per Day ( g p d gpd g p d ) and Calculate SOR Flow Rate = 2.4 M G D = 2 , 400 , 000 g p d \text{Flow Rate} = 2.4\ MGD = 2,400,000\ gpd Flow Rate = 2.4 MG D = 2 , 400 , 000 g p d SOR ( g p d / s q f t ) = Total Flow ( g p d ) Surface Area ( s q f t ) \text{SOR } (gpd/sq\ ft) = \frac{\text{Total Flow } (gpd)}{\text{Surface Area } (sq\ ft)} SOR ( g p d / s q f t ) = Surface Area ( s q f t ) Total Flow ( g p d ) SOR = 2 , 400 , 000 g p d 3 , 000 s q f t = 800 g p d / s q f t \text{SOR} = \frac{2,400,000\ gpd}{3,000\ sq\ ft} = \mathbf{800\ gpd/sq\ ft} SOR = 3 , 000 s q f t 2 , 400 , 000 g p d = 800 gpd/sq ft

Step 3: Convert SOR to g p m / s q f t gpm/sq\ ft g p m / s q f t

Method A: Divide SOR by 1 , 440 m i n / d a y 1,440\ min/day 1 , 440 min / d a y : Loading Rate ( g p m / s q f t ) = 800 g p d / s q f t 1 , 440 m i n / d a y = 0.556 g p m / s q f t \text{Loading Rate } (gpm/sq\ ft) = \frac{800\ gpd/sq\ ft}{1,440\ min/day} = \mathbf{0.556\ gpm/sq\ ft} Loading Rate ( g p m / s q f t ) = 1 , 440 min / d a y 800 g p d / s q f t = 0.556 gpm/sq ft

Method B: Convert flow to gpm first: Flow ( g p m ) = 2 , 400 , 000 g p d 1 , 440 = 1 , 666.67 g p m \text{Flow } (gpm) = \frac{2,400,000\ gpd}{1,440} = 1,666.67\ gpm Flow ( g p m ) = 1 , 440 2 , 400 , 000 g p d = 1 , 666.67 g p m Loading Rate = 1 , 666.67 g p m 3 , 000 s q f t = 0.556 g p m / s q f t \text{Loading Rate} = \frac{1,666.67\ gpm}{3,000\ sq\ ft} = 0.556\ gpm/sq\ ft Loading Rate = 3 , 000 s q f t 1 , 666.67 g p m = 0.556 g p m / s q f t

Step 4: Operational Evaluation

The calculated SOR of 800 gpd/sq ft 800\text{ gpd/sq ft} 800 gpd/sq ft ( 0.556 gpm/sq ft 0.556\text{ gpm/sq ft} 0.556 gpm/sq ft ) falls well within the recommended design range of 500 to 1 , 000 gpd/sq ft 500\text{ to }1,000\text{ gpd/sq ft} 500 to 1 , 000 gpd/sq ft for conventional alum floc clarification.

Example 2: Circular Clarifier with Peripheral Weir (SOR and WOR)

Problem Statement : A circular center-feed clarifier has a diameter of 75 feet 75\text{ feet} 75 feet , a side water depth of 14 feet 14\text{ feet} 14 feet , and a peripheral effluent weir extending around its entire rim. The facility treats a peak daily flow rate of 4.0 MGD 4.0\text{ MGD} 4.0 MGD . Determine:

The horizontal surface area of the clarifier in square feet.

The surface overflow rate in g p d / s q f t gpd/sq\ ft g p d / s q f t and g p m / s q f t gpm/sq\ ft g p m / s q f t .

The total linear length of the peripheral effluent weir.

The weir overflow rate (WOR) in gallons per day per linear foot ( g p d / l i n e a r f t gpd/linear\ ft g p d / l in e a r f t ).

Solution Procedure :

Step 1: Calculate Circular Surface Area Area ( s q f t ) = 0.785 × [ Diameter ( f t ) ] 2 \text{Area } (sq\ ft) = 0.785 \times [\text{Diameter } (ft)]^2 Area ( s q f t ) = 0.785 × [ Diameter ( f t ) ] 2 Area = 0.785 × ( 75 f t ) 2 = 0.785 × 5 , 625 = 4 , 415.63 s q f t \text{Area} = 0.785 \times (75\ ft)^2 = 0.785 \times 5,625 = \mathbf{4,415.63\ sq\ ft} Area = 0.785 × ( 75 f t ) 2 = 0.785 × 5 , 625 = 4 , 415.63 sq ft

(Using π r 2 \pi r^2 π r 2 : 3.14159 × 37.5 2 = 4 , 417.86 s q f t 3.14159 \times 37.5^2 = 4,417.86\ sq\ ft 3.14159 × 37. 5 2 = 4 , 417.86 s q f t ) .

Step 2: Calculate Surface Overflow Rate SOR ( g p d / s q f t ) = 4 , 000 , 000 g p d 4 , 415.63 s q f t = 905.87 g p d / s q f t ≈ 906 g p d / s q f t \text{SOR } (gpd/sq\ ft) = \frac{4,000,000\ gpd}{4,415.63\ sq\ ft} = \mathbf{905.87\ gpd/sq\ ft} \approx 906\ gpd/sq\ ft SOR ( g p d / s q f t ) = 4 , 415.63 s q f t 4 , 000 , 000 g p d = 905.87 gpd/sq ft ≈ 906 g p d / s q f t Loading Rate ( g p m / s q f t ) = 905.87 1 , 440 = 0.629 g p m / s q f t \text{Loading Rate } (gpm/sq\ ft) = \frac{905.87}{1,440} = \mathbf{0.629\ gpm/sq\ ft} Loading Rate ( g p m / s q f t ) = 1 , 440 905.87 = 0.629 gpm/sq ft

Step 3: Calculate Peripheral Weir Crest Length Weir Length ( f t ) = π × Diameter ( f t ) = 3.1416 × 75 f t = 235.62 l i n e a r f e e t \text{Weir Length } (ft) = \pi \times \text{Diameter } (ft) = 3.1416 \times 75\ ft = \mathbf{235.62\ linear\ feet} Weir Length ( f t ) = π × Diameter ( f t ) = 3.1416 × 75 f t = 235.62 linear feet

Step 4: Calculate Weir Overflow Rate (WOR) WOR ( g p d / l i n e a r f t ) = Total Flow ( g p d ) Weir Length ( f t ) \text{WOR } (gpd/linear\ ft) = \frac{\text{Total Flow } (gpd)}{\text{Weir Length } (ft)} WOR ( g p d / l in e a r f t ) = Weir Length ( f t ) Total Flow ( g p d ) WOR = 4 , 000 , 000 g p d 235.62 f t = 16 , 976.5 g p d / l i n e a r f t ≈ 16 , 977 g p d / l i n e a r f t \text{WOR} = \frac{4,000,000\ gpd}{235.62\ ft} = \mathbf{16,976.5\ gpd/linear\ ft} \approx 16,977\ gpd/linear\ ft WOR = 235.62 f t 4 , 000 , 000 g p d = 16 , 976.5 gpd/linear ft ≈ 16 , 977 g p d / l in e a r f t

Step 5: Operational Evaluation

The SOR ( 906 gpd/sq ft 906\text{ gpd/sq ft} 906 gpd/sq ft ) is below the 1 , 000 gpd/sq ft 1,000\text{ gpd/sq ft} 1 , 000 gpd/sq ft ceiling. The WOR ( 16 , 977 gpd/linear ft 16,977\text{ gpd/linear ft} 16 , 977 gpd/linear ft ) satisfies the standard 20 , 000 gpd/ft 20,000\text{ gpd/ft} 20 , 000 gpd/ft regulatory maximum, confirming the clarifier will not induce localized floc scouring.

Example 3: Sizing Clarifier Surface Area for Plant Expansion

Problem Statement : A municipal water treatment facility currently treats 6.0 MGD 6.0\text{ MGD} 6.0 MGD through two existing sedimentation basins. The utility must expand treatment capacity by an additional 3.0 MGD 3.0\text{ MGD} 3.0 MGD (new total capacity: 9.0 MGD 9.0\text{ MGD} 9.0 MGD ). State environmental regulatory criteria dictate that the maximum design surface overflow rate shall not exceed 750 gpd/sq ft 750\text{ gpd/sq ft} 750 gpd/sq ft under maximum design flow. Calculate:

The minimum additional clarifier surface area required in square feet.

The minimum internal diameter if the expansion utilizes a single new circular clarifier.

Solution Procedure :

Step 1: Calculate Required Additional Surface Area

Rearranging the SOR equation to solve for area: Surface Area ( s q f t ) = Expansion Flow Rate ( g p d ) Design SOR ( g p d / s q f t ) \text{Surface Area } (sq\ ft) = \frac{\text{Expansion Flow Rate } (gpd)}{\text{Design SOR } (gpd/sq\ ft)} Surface Area ( s q f t ) = Design SOR ( g p d / s q f t ) Expansion Flow Rate ( g p d ) Additional Area = 3 , 000 , 000 g p d 750 g p d / s q f t = 4 , 000 s q f t \text{Additional Area} = \frac{3,000,000\ gpd}{750\ gpd/sq\ ft} = \mathbf{4,000\ sq\ ft} Additional Area = 750 g p d / s q f t 3 , 000 , 000 g p d = 4 , 000 sq ft

Step 2: Solve for Circular Clarifier Diameter Area = 0.785 × D 2 ⟹ D 2 = Area 0.785 \text{Area} = 0.785 \times D^2 \implies D^2 = \frac{\text{Area}}{0.785} Area = 0.785 × D 2 ⟹ D 2 = 0.785 Area D 2 = 4 , 000 s q f t 0.785 = 5 , 095.54 s q f t D^2 = \frac{4,000\ sq\ ft}{0.785} = 5,095.54\ sq\ ft D 2 = 0.785 4 , 000 s q f t = 5 , 095.54 s q f t D = 5 , 095.54 = 71.38 f e e t ≈ 71.4 f e e t D = \sqrt{5,095.54} = \mathbf{71.38\ feet} \approx 71.4\ feet D = 5 , 095.54 = 71.38 feet ≈ 71.4 f ee t

Step 3: Engineering Sizing Recommendation

Clarifiers are fabricated in standard commercial increments. Selecting a 72 -foot 72\text{-foot} 72 -foot diameter tank provides: Actual Area = 0.785 × ( 72 f t ) 2 = 4 , 069.4 s q f t \text{Actual Area} = 0.785 \times (72\ ft)^2 = 4,069.4\ sq\ ft Actual Area = 0.785 × ( 72 f t ) 2 = 4 , 069.4 s q f t Actual SOR = 3 , 000 , 000 g p d 4 , 069.4 s q f t = 737.2 g p d / s q f t \text{Actual SOR} = \frac{3,000,000\ gpd}{4,069.4\ sq\ ft} = 737.2\ gpd/sq\ ft Actual SOR = 4 , 069.4 s q f t 3 , 000 , 000 g p d = 737.2 g p d / s q f t

This provides a safe operating buffer below the 750 gpd/sq ft 750\text{ gpd/sq ft} 750 gpd/sq ft regulatory limit.

Troubleshooting Clarifier Hydraulic Loading Problems

When clariﬁers experience hydraulic overload or operational imbalances, operators must quickly identify root causes and deploy targeted corrective measures.

Table 20.1.3: Clarifier Loading Troubleshooting Guide

Observed Symptom Primary Hydraulic Root Cause Operational Remedy & Corrective Action

Pin-Floc Carryover Over Weirs Surface Overflow Rate exceeds settling velocity ( V u p > V s V_{up} > V_s V u p > V s ) due to high plant flow. Place standby clarifier online to reduce surface loading; apply polymeric coagulant aid ( 0.05 to 0.2 m g / L 0.05\text{ to }0.2\ mg/L 0.05 to 0.2 m g / L ) to increase floc mass and density.

Localized Floc Boiling Near Launder Out-of-level weir plates causing uneven flow distribution; localized WOR exceeds 20 , 000 g p d / f t 20,000\ gpd/ft 20 , 000 g p d / f t . Take basin offline to level weir plates with surveying transit; adjust V-notch weir bolt slots; clean algae or debris clogging weir notches.

Thermal Stratification / Short-Circuiting Cold influent water plunges beneath warm surface water, cutting effective detention time by 70%. Install inlet energy-dissipating baffle curtains; install intermediate perforated diffusion walls to restore uniform plug flow.

Sludge Blanket Bulking / Rising Anaerobic decomposition in sludge blanket producing nitrogen/methane gas bubbles that float settled solids. Increase sludge pumping frequency and duration; verify operation of bottom scraper flights; optimize chemical dosing.