14.4 Removal Efficiency & Process Loading Rates

10.4 Removal Efficiency & Process Loading Rates

Evaluating the operational performance of water and wastewater treatment facilities requires continuous tracking of removal efficiencies and process loading parameters. Removal efficiency quantifies how effectively a process unit removes contaminants (e.g., total suspended solids, BOD, turbidity). Loading rates measure the hydraulic, organic, or solids mass applied to a specific unit area or volume, ensuring processes are operated within established engineering design limits.

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1. Percent Removal Efficiency Calculations

Percent removal efficiency compares raw or influent contaminant concentration to treated or effluent concentration:

$\text{Percent Removal (\%)} = \frac{\text{Influent Concentration} - \text{Effluent Concentration}}{\text{Influent Concentration}} \times 100\%$ $\text{\% Removal} = \left(\frac{\text{In} - \text{Out}}{\text{In}}\right) \times 100\%$

Multi-Stage Removal Efficiency

When evaluating overall plant efficiency across sequential process units (e.g., primary settling followed by secondary biological treatment), operators must distinguish between unit process removal and overall plant removal.

Worked Example 1: Plant BOD and TSS Removal Efficiency

Problem: Raw wastewater entering a primary clarifier contains 240 mg/L of Total Suspended Solids (TSS). The primary clarifier effluent contains 120 mg/L of TSS, and the final secondary effluent discharged to the receiving stream contains 12 mg/L of TSS. Calculate:

1. The TSS removal efficiency of the primary clarifier.
2. The overall TSS removal efficiency of the treatment facility.

Solution Step-by-Step:

1. Primary Clarifier Removal Efficiency: $\text{Primary \% Removal} = \left(\frac{240\text{ mg/L} - 120\text{ mg/L}}{240\text{ mg/L}}\right) \times 100\% = \left(\frac{120}{240}\right) \times 100\% = 50.0\%$
2. Overall Plant Removal Efficiency: $\text{Overall \% Removal} = \left(\frac{240\text{ mg/L} - 12\text{ mg/L}}{240\text{ mg/L}}\right) \times 100\% = \left(\frac{228}{240}\right) \times 100\% = 95.0\%$

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2. Clarifier Hydraulic Loading Rates

Clarifiers and sedimentation basins rely on gravity settling. If hydraulic loading rates exceed design limits, vertical fluid velocity overcomes particle settling velocity, washing solids over the effluent weirs.

A. Surface Overflow Rate (SOR)

Surface Overflow Rate (also called surface loading rate) measures the volume of liquid passing through each square foot of clarifier surface area per day:

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

Typical Clarifier SOR Design Standards:

• Primary Clarifiers: 600 to 1,200 gpd/sq ft
• Secondary Clarifiers (Activated Sludge): 400 to 800 gpd/sq ft
• High-Rate Water Settlers (Tube/Plate): 1.0 to 2.0 gpm/sq ft (1,440 to 2,880 gpd/sq ft)

Worked Example 2: Circular Clarifier Overflow Rate

Problem: A circular secondary clarifier with a diameter of 50 feet receives a daily wastewater flow of 1.5 MGD. Calculate the Surface Overflow Rate in gallons per day per square foot (gpd/sq ft).

Solution Step-by-Step:

1. Calculate clarifier surface area: $\text{Surface Area} = 0.785 \times (50\text{ ft})^2 = 0.785 \times 2,500 = 1,962.5\text{ sq ft}$
2. Calculate Surface Overflow Rate: $\text{SOR} = \frac{1,500,000\text{ gpd}}{1,962.5\text{ sq ft}} = 764.33\text{ gpd/sq ft}$ Result: The SOR of 764.3 gpd/sq ft falls within standard secondary design limits (400-800 gpd/sq ft).

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B. Weir Loading Rate (WLR)

Weir Loading Rate evaluates the volume of flow passing over each linear foot of effluent weir per day. Excessive weir loading causes high localized exit velocities (weir pulling), causing light sludge pin-floc to carry over into effluent channels.

$\text{WLR (gpd/linear ft)} = \frac{\text{Flow Rate (gpd)}}{\text{Total Weir Length (feet)}}$

• For rectangular tanks with end weirs: $\text{Weir Length} = \text{Width of Tank (ft)}$.
• For circular tanks with peripheral launders: $\text{Weir Length} = \pi \times \text{Weir Diameter (ft)} = 3.1416 \times D$.

Standard WLR Limits:

• Standard Clarifiers: ≤ 10,000 to 15,000 gpd/ft
• High-Rate Clarifiers with Finger Launders: ≤ 20,000 gpd/ft

Worked Example 3: Circular Clarifier Weir Loading Rate

Problem: A circular clarifier has a perimeter effluent weir located 2 feet inside the outer 54-foot diameter tank wall (weir diameter = 50 feet). The clarifier treats a flow of 2.2 MGD. Calculate the Weir Loading Rate in gpd per linear foot.

Solution Step-by-Step:

1. Calculate total peripheral weir length: $\text{Weir Length} = 3.1416 \times 50\text{ ft} = 157.08\text{ linear feet}$
2. Calculate Weir Loading Rate: $\text{WLR} = \frac{2,200,000\text{ gpd}}{157.08\text{ ft}} = 14,005.6\text{ gpd/ft}$

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3. Biological Process Loading Rates

Biological wastewater treatment processes (activated sludge, trickling filters, RBCs) rely on controlled loading metrics to maintain organic conversion and settling characteristics.

A. Organic Loading Rate (OLR)

Organic Loading Rate quantifies the daily organic mass (BOD or COD) applied per unit volume of biological reactor: $\text{OLR (lbs BOD/1,000 cu ft/day)} = \frac{\text{BOD Mass Applied (lbs BOD/day)}}{\text{Reactor Volume (in 1,000 cu ft units)}}$ $\text{Lbs BOD/day} = \text{Flow (MGD)} \times \text{Influent BOD (mg/L)} \times 8.34$

B. Food-to-Microorganism Ratio (F/M Ratio)

The F/M Ratio is the key operational control parameter in the activated sludge process. It compares the daily organic food supply (lbs BOD/day) entering the aeration tank to the total active biological population (lbs MLVSS) inside the aeration basin.

$\text{F/M Ratio} = \frac{\text{Lbs BOD applied per day}}{\text{Lbs MLVSS under aeration}}$ $\text{F/M Ratio} = \frac{\text{Flow (MGD)} \times \text{Primary Effluent BOD (mg/L)} \times 8.34}{\text{Aeration Volume (MG)} \times \text{MLVSS Concentration (mg/L)} \times 8.34}$ $\text{F/M Ratio} = \frac{\text{Flow (MGD)} \times \text{BOD (mg/L)}}{\text{Aeration Volume (MG)} \times \text{MLVSS (mg/L)}}$

Standard F/M Target Operational Ranges:

• Conventional Activated Sludge: 0.20 to 0.50 lbs BOD/lb MLVSS/day
• Extended Aeration: 0.05 to 0.15 lbs BOD/lb MLVSS/day
• High-Rate Activated Sludge: 0.50 to 1.50 lbs BOD/lb MLVSS/day

Worked Example 4: F/M Ratio Calculation

Problem: An activated sludge aeration tank with a volume of 0.50 MG treats a flow rate of 1.0 MGD. The primary effluent BOD concentration is 150 mg/L, and the Mixed Liquor Volatile Suspended Solids (MLVSS) concentration inside the aeration tank is 2,000 mg/L. Calculate the operational F/M ratio.

Solution Step-by-Step:

1. Calculate daily BOD mass applied (Food): $\text{Lbs BOD/day} = 1.0\text{ MGD} \times 150\text{ mg/L} \times 8.34 = 1,251\text{ lbs BOD/day}$
2. Calculate total MLVSS mass under aeration (Microorganisms): $\text{Lbs MLVSS} = 0.50\text{ MG} \times 2,000\text{ mg/L} \times 8.34 = 8,340\text{ lbs MLVSS}$
3. Calculate F/M ratio: $\text{F/M Ratio} = \frac{1,251\text{ lbs BOD/day}}{8,340\text{ lbs MLVSS}} = 0.150\text{ lbs BOD/lb MLVSS/day}$ Process Diagnosis: An F/M ratio of 0.15 is ideal for an extended aeration process or low-rate conventional system.