20.1 Surface Overflow Rate & Weir Loading Rate Calculations

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 ($V_s$), while the bulk water mass moves upward toward the effluent launders with an upward fluid velocity ($V_{up}$):

• If $V_s > V_{up}$, the particle settles downward into the sludge collection zone and is successfully removed.
• If $V_s < V_{up}$, 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 ($Q$) divided by the horizontal surface area ($A$) 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 $10\text{ to }16\text{ feet}$) 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 ($gpd/sq\ ft$) or gallons per minute per square foot ($gpm/sq\ ft$):

$\text{SOR }(gpd/sq\ ft) = \frac{\text{Total Plant Flow Rate }(gpd)}{\text{Clarifier Surface Area }(sq\ ft)}$

$\text{Surface Loading Rate }(gpm/sq\ ft) = \frac{\text{Flow Rate }(gpm)}{\text{Clarifier Surface Area }(sq\ ft)} = \frac{\text{SOR }(gpd/sq\ ft)}{1,440\ min/day}$

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Clarifier Surface Area Geometry Formulas

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

1. Rectangular Clarifiers

$\text{Surface Area }(sq\ ft) = \text{Length }(ft) \times \text{Width }(ft)$

2. Circular Clarifiers

$\text{Surface Area }(sq\ ft) = \frac{\pi}{4} \times [\text{Diameter }(ft)]^2 \approx 0.785 \times [\text{Diameter }(ft)]^2$

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

Table 20.1.1: Clarifier Surface Overflow Rate Design Standards

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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.

$\text{WOR }(gpd/linear\ ft) = \frac{\text{Total Flow Rate }(gpd)}{\text{Total Active Weir Crest Length }(ft)}$

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 $10,000\text{ to }15,000\ gpd/linear\ ft$.
• Heavy Lime Floc / Upflow Clarifiers: Up to $20,000\ gpd/linear\ ft$.

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

1. Circular Clarifiers with Peripheral Weirs: The weir runs around the entire internal circumference of the tank wall: $\text{Weir Length }(ft) = \pi \times \text{Diameter }(ft) \approx 3.1416 \times D$
2. Rectangular Basins with End Wall Weirs: The weir length equals the width of the basin wall.
3. 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: $\text{Active Weir Length} = 2 \times \text{Launder Length} \times \text{Number of Launders}$

Table 20.1.2: Clarifier Geometry and Loading Rate Equations

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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 $100\text{ feet}$ long, $30\text{ feet}$ wide, and has an operating water depth of $12\text{ feet}$. The plant processes a steady flow rate of $2.4\text{ MGD}$. Calculate:

1. The clarifier surface area in square feet.
2. The surface overflow rate (SOR) in gallons per day per square foot ($gpd/sq\ ft$).
3. The hydraulic surface loading rate in gallons per minute per square foot ($gpm/sq\ ft$).
4. Determine whether this operating rate complies with standard design guidelines ($500\text{ to }1,000\ gpd/sq\ ft$).

Solution Procedure:

• Step 1: Calculate Clarifier Surface Area
  $\text{Surface Area} = \text{Length} \times \text{Width}$ $\text{Surface Area} = 100\ ft \times 30\ ft = 3,000\ sq\ ft$ (Note: Depth is not used in calculating surface area or SOR).

• Step 2: Convert Flow to Gallons per Day ($gpd$) and Calculate SOR
  $\text{Flow Rate} = 2.4\ MGD = 2,400,000\ gpd$ $\text{SOR } (gpd/sq\ ft) = \frac{\text{Total Flow } (gpd)}{\text{Surface Area } (sq\ ft)}$ $\text{SOR} = \frac{2,400,000\ gpd}{3,000\ sq\ ft} = \mathbf{800\ gpd/sq\ ft}$

• Step 3: Convert SOR to $gpm/sq\ ft$
  Method A: Divide SOR by $1,440\ min/day$:
  $\text{Loading Rate } (gpm/sq\ ft) = \frac{800\ gpd/sq\ ft}{1,440\ min/day} = \mathbf{0.556\ gpm/sq\ ft}$ Method B: Convert flow to gpm first:
  $\text{Flow } (gpm) = \frac{2,400,000\ gpd}{1,440} = 1,666.67\ gpm$ $\text{Loading Rate} = \frac{1,666.67\ gpm}{3,000\ sq\ ft} = 0.556\ gpm/sq\ ft$

• Step 4: Operational Evaluation
  The calculated SOR of $800\text{ gpd/sq ft}$ ($0.556\text{ gpm/sq ft}$) falls well within the recommended design range of $500\text{ to }1,000\text{ gpd/sq ft}$ 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 $75\text{ feet}$, a side water depth of $14\text{ feet}$, and a peripheral effluent weir extending around its entire rim. The facility treats a peak daily flow rate of $4.0\text{ MGD}$. Determine:

1. The horizontal surface area of the clarifier in square feet.
2. The surface overflow rate in $gpd/sq\ ft$ and $gpm/sq\ ft$.
3. The total linear length of the peripheral effluent weir.
4. The weir overflow rate (WOR) in gallons per day per linear foot ($gpd/linear\ ft$).

Solution Procedure:

• Step 1: Calculate Circular Surface Area
  $\text{Area } (sq\ ft) = 0.785 \times [\text{Diameter } (ft)]^2$ $\text{Area} = 0.785 \times (75\ ft)^2 = 0.785 \times 5,625 = \mathbf{4,415.63\ sq\ ft}$ (Using $\pi r^2$: $3.14159 \times 37.5^2 = 4,417.86\ sq\ ft$).

• Step 2: Calculate Surface Overflow Rate
  $\text{SOR } (gpd/sq\ ft) = \frac{4,000,000\ gpd}{4,415.63\ sq\ ft} = \mathbf{905.87\ gpd/sq\ ft} \approx 906\ gpd/sq\ ft$ $\text{Loading Rate } (gpm/sq\ ft) = \frac{905.87}{1,440} = \mathbf{0.629\ gpm/sq\ ft}$

• Step 3: Calculate Peripheral Weir Crest Length
  $\text{Weir Length } (ft) = \pi \times \text{Diameter } (ft) = 3.1416 \times 75\ ft = \mathbf{235.62\ linear\ feet}$

• Step 4: Calculate Weir Overflow Rate (WOR)
  $\text{WOR } (gpd/linear\ ft) = \frac{\text{Total Flow } (gpd)}{\text{Weir Length } (ft)}$ $\text{WOR} = \frac{4,000,000\ gpd}{235.62\ ft} = \mathbf{16,976.5\ gpd/linear\ ft} \approx 16,977\ gpd/linear\ ft$

• Step 5: Operational Evaluation
  The SOR ($906\text{ gpd/sq ft}$) is below the $1,000\text{ gpd/sq ft}$ ceiling. The WOR ($16,977\text{ gpd/linear ft}$) satisfies the standard $20,000\text{ gpd/ft}$ 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 $6.0\text{ MGD}$ through two existing sedimentation basins. The utility must expand treatment capacity by an additional $3.0\text{ MGD}$ (new total capacity: $9.0\text{ MGD}$). State environmental regulatory criteria dictate that the maximum design surface overflow rate shall not exceed $750\text{ gpd/sq ft}$ under maximum design flow. Calculate:

1. The minimum additional clarifier surface area required in square feet.
2. 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:
  $\text{Surface Area } (sq\ ft) = \frac{\text{Expansion Flow Rate } (gpd)}{\text{Design SOR } (gpd/sq\ ft)}$ $\text{Additional Area} = \frac{3,000,000\ gpd}{750\ gpd/sq\ ft} = \mathbf{4,000\ sq\ ft}$

• Step 2: Solve for Circular Clarifier Diameter
  $\text{Area} = 0.785 \times D^2 \implies D^2 = \frac{\text{Area}}{0.785}$ $D^2 = \frac{4,000\ sq\ ft}{0.785} = 5,095.54\ sq\ ft$ $D = \sqrt{5,095.54} = \mathbf{71.38\ feet} \approx 71.4\ feet$

• Step 3: Engineering Sizing Recommendation
  Clarifiers are fabricated in standard commercial increments. Selecting a $72\text{-foot}$ diameter tank provides:
  $\text{Actual Area} = 0.785 \times (72\ ft)^2 = 4,069.4\ sq\ ft$ $\text{Actual SOR} = \frac{3,000,000\ gpd}{4,069.4\ sq\ ft} = 737.2\ gpd/sq\ ft$ This provides a safe operating buffer below the $750\text{ gpd/sq ft}$ regulatory limit.

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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