6.3 Return Activated Sludge (RAS) & Waste Activated Sludge (WAS) Control

6.3 Return Activated Sludge (RAS) & Waste Activated Sludge (WAS) Control

Secondary clarification and solids recycling constitute the physical foundation of the activated sludge process. The aeration basin and secondary clarifier function as a single interdependent unit: biological floc synthesized in the aeration tank must be efficiently separated and thickened in the clarifier. Regulating Return Activated Sludge (RAS) controls the distribution of solids between the clarifier and aeration basin, while regulating Waste Activated Sludge (WAS) establishes the total mass inventory, sludge age, and metabolic state of the facility.

---

1. Secondary Clarification Mechanics & Hydraulic Loading

Secondary clarifiers are gravity sedimentation basins designed to perform two distinct, simultaneous operational functions:

1. Clarification (Liquid Clarification): Producing a clarified, polished supernatant effluent by allowing biological flocs to aggregate and settle gravitationally, maintaining effluent TSS below NPDES permit limits (typically $<12\text{ to }30\text{ mg/L}$).
2. Thickening (Solids Compaction): Consolidating and concentrating settled biological solids on the basin floor into a dense underflow blanket, ensuring that recycled RAS contains a high solids concentration ($4,000\text{ to }10,000+\text{ mg/L}$).

Clarifier Design & Operational Criteria

Clarifier operational capacity is constrained by hydraulic and solids flux thresholds:

• Surface Overflow Rate (SOR): Quantifies the upward hydraulic velocity of liquid leaving the clarifier relative to its surface area:

$\text{SOR (gpd/sq ft)} = \frac{Q_{\text{in}} \text{ (gallons/day)}}{A_{\text{surface}} \text{ (sq ft)}}$

• Typical design ranges: 400 to 800 gpd/sq ft at design average flow; peak hourly limits are typically 1,000 to 1,200 gpd/sq ft.

• If SOR exceeds the settling velocity of the biological floc, particles are carried upward and wash over the effluent weirs.

• Solids Loading Rate (SLR): Represents the total mass of solids applied per square foot of clarifier surface area per day. Crucially, solids loading includes both influent wastewater flow and RAS flow:

$\text{SLR (lbs/sq ft}\cdot\text{day)} = \frac{(Q_{\text{in}} + Q_{\text{ras}}) \text{ MGD} \times \text{MLSS (mg/L)} \times 8.34}{A_{\text{surface}} \text{ (sq ft)}}$

• Typical design ranges: 20 to 30 lbs/sq ft·day for conventional systems (up to 35–40 lbs/sq ft·day at peak hourly loading).

• When SLR exceeds the thickening capacity of the sludge, solids accumulate on the clarifier floor faster than RAS pumps can withdraw them, causing the sludge blanket to rise continuously regardless of clarifier depth.

• Weir Overflow Rate (WOR): Measures the hydraulic discharge rate per linear foot of effluent perimeter weir ($Q_{\text{in}} / L_{\text{weir}}$). Maintained below 10,000 to 20,000 gpd/linear foot to prevent high localized approach velocities that pull settled floc upward over the weir plates.

---

2. Return Activated Sludge (RAS) Purpose and Pumping Control

Return Activated Sludge (RAS) is the concentrated biological slurry pumped continuously from the bottom hoppers of secondary clarifiers back to the aeration basin inlet.

Primary Operational Objectives of RAS

1. Maintain the required MLSS and MLVSS concentration in the aeration tank to sustain target F/M ratios and complete BOD/ammonia stabilization.
2. Prevent excessive accumulation of biological solids within the secondary clarifier, maintaining a stable, low sludge blanket.

Operating Ranges and Blanket Depth Management

• Typical RAS Flow Rates: Conventional plug-flow systems operate at RAS rates equal to 25% to 75% of influent plant flow ($Q_{\text{ras}} / Q_{\text{in}} = 0.25\text{--}0.75$). Extended aeration, oxidation ditches, and complete-mix plants frequently operate at higher rates, ranging from 50% to 150%.
• Optimal Sludge Blanket Depth: Under steady-state operations, the sludge blanket depth on the clarifier floor should be maintained between 1.0 and 2.0 feet (0.3 to 0.6 meters), representing less than 20% to 25% of the total clarifier sidewall water depth (SWD).
• Core Sampling (The Sludge Judge): Operators monitor blanket depth using a Sludge Judge—a transparent, graduated plastic core sampler equipped with a weighted foot check-valve. The tube is lowered vertically through the clarifier water column to the basin floor, trapping an undisturbed cross-sectional core. The operator records the depth of the clarified liquid zone, the transition interface, and the compacted sludge blanket.

Operational Consequences of RAS Flow Imbalances

The RAS Flow Mass Balance Equation

Assuming all solids settling in the clarifier are returned via RAS (ignoring minimal effluent TSS loss), a steady-state mass balance across the secondary clarifier yields:

$\text{Mass of Solids Entering Clarifier} = \text{Mass of Solids Leaving via RAS Underflow}$ $(Q_{\text{in}} + Q_{\text{ras}}) \times \text{MLSS} = Q_{\text{ras}} \times \text{RAS}_{\text{SS}}$ $Q_{\text{in}} \times \text{MLSS} + Q_{\text{ras}} \times \text{MLSS} = Q_{\text{ras}} \times \text{RAS}_{\text{SS}}$ $Q_{\text{in}} \times \text{MLSS} = Q_{\text{ras}} \times (\text{RAS}_{\text{SS}} - \text{MLSS})$

$\mathbf{Q_{\text{ras}} = \frac{Q_{\text{in}} \times \text{MLSS}}{\text{RAS}_{\text{SS}} - \text{MLSS}}}$

$\mathbf{\frac{Q_{\text{ras}}}{Q_{\text{in}}} = \frac{\text{MLSS}}{\text{RAS}_{\text{SS}} - \text{MLSS}}}$

Worked Example:

A treatment plant receives an influent flow of $4.0\text{ MGD}$ with an aeration basin MLSS of $2,500\text{ mg/L}$. Laboratory analysis of the clarifier underflow measures a $\text{RAS}_{\text{SS}}$ concentration of $7,500\text{ mg/L}$. What RAS flow rate should the operator set?

$Q_{\text{ras}} = \frac{4.0\text{ MGD} \times 2,500\text{ mg/L}}{7,500\text{ mg/L} - 2,500\text{ mg/L}} = \frac{10,000}{5,000} = 2.0\text{ MGD}$

$\text{RAS Percentage} = \frac{2.0\text{ MGD}}{4.0\text{ MGD}} \times 100 = 50\%$

---

3. Waste Activated Sludge (WAS) Control Strategies

While RAS recycles biomass to sustain aeration inventory, Waste Activated Sludge (WAS) purges excess biomass from the system. As heterotrophic bacteria metabolize BOD, new cellular mass is synthesized continuously (roughly 0.4 to 0.7 lbs of new biomass per lb of BOD removed). Without systematic wasting, solids accumulate, clarifiers overload, and the system fails.

Primary Objectives of WAS

1. Establish and maintain the target Mean Cell Residence Time (MCRT) and F/M ratio.
2. Maintain process equilibrium between microbial growth and sludge removal.
3. Prevent solids overloading and high sludge blanket carryover in secondary clarifiers.

Comparison of WAS Control Strategies

WAS Control Approaches:
1. Constant MCRT  =======> Scientifically optimal; controls cell age; accounts for effluent loss
2. Constant MLSS  =======> Simple; ignores seasonal temperature & F/M shifts
3. Constant F/M   =======> Excellent biological control; limited by 5-day BOD lab turnaround

1. Control by Constant MCRT: The most robust and widely recommended operational methodology. The operator selects a target MCRT (e.g., 10 days for nitrification) based on season and temperature. The total required system inventory is divided by the target MCRT to determine total pounds to remove per day, subtracting incidental effluent TSS losses. Daily wasting adjusts dynamically to biomass growth.
2. Control by Constant MLSS: The operator wastes sludge to maintain a fixed MLSS concentration (e.g., $2,500\text{ mg/L}$) in the aeration basin. While straightforward for daily operations, this method fails to account for changing influent organic strengths. During low-loading periods, maintaining a constant MLSS forces the F/M ratio dangerously low, leading to over-aged sludge and pin floc.
3. Control by Constant F/M: Aims to adjust MLVSS inventory proportionally to incoming organic loading. Highly effective in theory, but limited in practice because standard 5-day BOD tests provide historical rather than real-time data, requiring plants to rely on Chemical Oxygen Demand (COD) or Total Organic Carbon (TOC) online analyzers.

---

4. Step-by-Step WAS Calculation & Pumping Rate Determination

Calculating the daily WAS pumping rate requires converting mass waste requirements into volumetric pumping rates based on WAS solids concentration.

Operating Scenario Data:

• Plant Influent Flow ($Q_{\text{in}}$) = 2.0 MGD
• Aeration Basin Volume ($V_{\text{aer}}$) = 0.80 MG
• Aeration Basin MLSS = 2,500 mg/L
• Target System MCRT = 10.0 days
• Secondary Clarifier Volume ($V_{\text{clar}}$) = 0.20 MG (clarifier blanket contains 2,500 lbs MLSS)
• Final Effluent TSS ($\text{Eff}_{\text{TSS}}$) = 10.0 mg/L
• WAS Solids Concentration ($\text{WAS}_{\text{SS}}$) = 6,000 mg/L (0.60% solids)

Step 1: Calculate Total System MLSS Inventory

$\text{Aeration Basin Inventory} = 0.80\text{ MG} \times 2,500\text{ mg/L} \times 8.34 = 16,680\text{ lbs MLSS}$ $\text{Total System Inventory} = 16,680\text{ lbs (aeration)} + 2,500\text{ lbs (clarifier)} = 19,180\text{ lbs MLSS}$

Step 2: Determine Total Daily Solids Removal Requirement

$\text{Total Daily Removal (lbs/day)} = \frac{\text{Total System Inventory}}{\text{Target MCRT}} = \frac{19,180\text{ lbs}}{10.0\text{ days}} = 1,918\text{ lbs/day}$

Step 3: Account for Incidental Effluent Solids Loss

Solids discharged in final effluent count toward daily system solids removal: $\text{Effluent Solids Loss} = 2.0\text{ MGD} \times 10.0\text{ mg/L} \times 8.34 = 166.8\text{ lbs/day}$

Step 4: Calculate Net WAS Mass to be Wasted

$\text{Net WAS Mass} = \text{Total Required} - \text{Effluent Loss} = 1,918\text{ lbs/day} - 166.8\text{ lbs/day} = 1,751.2\text{ lbs/day}$

Step 5: Convert Required WAS Mass into Daily Pumping Volume (Gallons/Day)

Using the standard pounds equation ($lbs = \text{Volume (MG)} \times \text{Concentration} \times 8.34$):

$Q_{\text{was}} \text{ (MGD)} = \frac{\text{Net WAS (lbs/day)}}{\text{WAS}_{\text{SS}} \text{ (mg/L)} \times 8.34} = \frac{1,751.2\text{ lbs/day}}{6,000\text{ mg/L} \times 8.34} = \frac{1,751.2}{50,040} = 0.0350\text{ MGD}$

$Q_{\text{was}} \text{ (Gallons per Day)} = 0.0350\text{ MGD} \times 1,000,000 = \mathbf{35,000\text{ gpd}}$

Step 6: Convert to Operational Pumping Rates (GPM)

• If pumping continuously over 24 hours:

$\text{WAS Pump Rate} = \frac{35,000\text{ gallons/day}}{1,440\text{ minutes/day}} = \mathbf{24.3\text{ GPM}}$

• If wasting intermittently in batches (e.g., one 6-hour cycle per day):

$\text{WAS Pump Rate (6-hr run)} = \frac{35,000\text{ gallons}}{6\text{ hours} \times 60\text{ min/hr}} = \frac{35,000}{360\text{ minutes}} = \mathbf{97.2\text{ GPM}}$