14.6 Activated Sludge Process Control Math

14.6 Activated Sludge Process Control Math

Everything in Sections 8.3 and 8.4 becomes actionable only through these calculations. WPI reports that 10–16% of exam items involve calculations — the share climbs with class, from 10% at Class I to 16% on the Class IV wastewater exam — and this cluster is the densest concentration of them.

The one thing to internalize: every calculation below is built on the pounds formula from Section 8.1.

$\text{lbs} = \text{Volume (MG)} \times \text{Concentration (mg/L)} \times 8.34$

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1. Food to Microorganism Ratio (F/M)

$\text{F/M} = \frac{\text{lb BOD applied per day}}{\text{lb MLVSS under aeration}}$

Worked example. Influent to aeration is 2.5 MGD at 190 mg/L BOD. The aeration basin holds 1.1 MG at 2,400 mg/L MLSS, and MLVSS is 78% of MLSS.

• Food: 2.5 × 190 × 8.34 = 3,962 lb BOD/day
• MLVSS concentration: 2,400 × 0.78 = 1,872 mg/L
• Microorganisms: 1.1 × 1,872 × 8.34 = 17,175 lb MLVSS
• F/M = 3,962 ÷ 17,175 = 0.23

That is a conventional-range F/M. To lower F/M, waste less (build MLSS). To raise it, waste more.

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2. Mean Cell Residence Time (MCRT / Sludge Age)

$\text{MCRT (days)} = \frac{\text{lb MLSS in the system}}{\text{lb solids leaving per day (WAS + effluent)}}$

Worked example. Aeration basin 1.1 MG plus clarifier 0.25 MG, both at 2,400 mg/L MLSS. WAS is 0.030 MGD at 7,800 mg/L. Effluent is 2.5 MGD at 12 mg/L TSS.

• Solids in system: (1.1 + 0.25) × 2,400 × 8.34 = 27,022 lb
• WAS solids out: 0.030 × 7,800 × 8.34 = 1,952 lb/day
• Effluent solids out: 2.5 × 12 × 8.34 = 250 lb/day
• Total out: 1,952 + 250 = 2,202 lb/day
• MCRT = 27,022 ÷ 2,202 = 12.3 days

Cold weather rule: nitrifiers grow slowly, and more slowly still as temperature drops. Winter operation generally requires increasing MCRT — wasting less — to retain a nitrifying population.

Related: Sludge Retention Time in aeration only (Gould sludge age)

$\text{Sludge Age (days)} = \frac{\text{lb MLSS in aeration}}{\text{lb TSS entering per day}}$

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3. Sludge Volume Index (SVI)

$\text{SVI} = \frac{\text{Settled Volume after 30 min (mL/L)} \times 1{,}000}{\text{MLSS (mg/L)}}$

Interpretation is in Section 8.4: below ~70 pin floc, 80–120 ideal, above ~150 bulking.

Worked example. Settled volume 260 mL/L, MLSS 2,900 mg/L.

$\text{SVI} = \frac{260 \times 1{,}000}{2{,}900} = \textbf{89.7} \Rightarrow \text{good settling}$

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4. Return Activated Sludge (RAS)

Estimating RAS from the settleometer

$\% \text{RAS} \approx \frac{\text{Settled Volume (mL/L)}}{1{,}000 - \text{Settled Volume (mL/L)}} \times 100$

Worked example. Settled volume 300 mL/L, plant flow 2.5 MGD.

• % RAS = 300 ÷ (1,000 − 300) × 100 = 300 ÷ 700 × 100 = 42.9%
• RAS flow = 2.5 × 0.429 = 1.07 MGD

Typical RAS rates run 20–100% of influent flow, higher when SVI is poor.

RAS by solids balance

$Q_{RAS} = \frac{Q \times \text{MLSS}}{\text{RAS SS} - \text{MLSS}}$

Worked example. Q = 2.5 MGD, MLSS = 2,400 mg/L, RAS concentration = 8,000 mg/L.

$Q_{RAS} = \frac{2.5 \times 2{,}400}{8{,}000 - 2{,}400} = \frac{6{,}000}{5{,}600} = \textbf{1.07 MGD}$

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5. Waste Activated Sludge (WAS)

Wasting is how the operator sets MCRT and F/M. To hit a target MCRT:

$\text{lb to waste/day} = \frac{\text{lb MLSS in system}}{\text{Target MCRT}} - \text{lb effluent solids/day}$

Worked example. Using the numbers from Section 2 above, target MCRT of 10 days:

• Required total solids out: 27,022 ÷ 10 = 2,702 lb/day
• Less effluent solids: 2,702 − 250 = 2,452 lb/day to waste
• At a WAS concentration of 7,800 mg/L:

$Q_{WAS} = \frac{2{,}452}{7{,}800 \times 8.34} = \textbf{0.0377 MGD} \approx 26 \text{ gpm}$

Never make a large wasting change at once. Adjust wasting by roughly 10–15% per day and let the process respond. The biology takes days to equilibrate, and an aggressive change overshoots in the direction you were trying to correct.

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6. Loading Rates

Organic loading

$\text{lb BOD per 1,000 ft}^3\text{ per day} = \frac{\text{lb BOD/day}}{\text{Aeration Volume (ft}^3) \div 1{,}000}$

Typical conventional loading is 20–40 lb BOD per 1,000 ft³ per day.

Clarifier surface overflow rate

$\text{SOR} = \frac{\text{Flow (gpd)}}{\text{Surface Area (ft}^2)} \quad \left[\frac{\text{gpd}}{\text{ft}^2}\right]$

Secondary clarifiers typically run 400–800 gpd/ft² at average flow.

Solids loading rate

$\text{SLR} = \frac{\text{lb solids applied per day}}{\text{Clarifier Surface Area (ft}^2)} \quad \left[\frac{\text{lb}}{\text{day} \cdot \text{ft}^2}\right]$

Typically 20–30 lb/day/ft². Solids applied includes both influent and RAS flow carrying MLSS — forgetting the RAS contribution is the classic error.

Worked example. Clarifier 70 ft diameter. Influent 2.5 MGD, RAS 1.07 MGD, MLSS 2,400 mg/L.

• Area = π × 35² = 3,848 ft²
• Combined flow = 2.5 + 1.07 = 3.57 MGD
• Solids = 3.57 × 2,400 × 8.34 = 71,455 lb/day
• SLR = 71,455 ÷ 3,848 = 18.6 lb/day/ft²

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7. Sludge Volume and Percent Solids

$\% \text{Solids} = \frac{\text{lb dry solids}}{\text{lb total sludge}} \times 100$

$\text{lb sludge} = \text{gallons} \times 8.34 \times \text{Specific Gravity}$

Worked example. 12,000 gallons of sludge at 3.2% solids, specific gravity 1.02.

• Total weight: 12,000 × 8.34 × 1.02 = 102,082 lb
• Dry solids: 102,082 × 0.032 = 3,267 lb

Sludge volume change with thickening

$V_1 \times \%S_1 = V_2 \times \%S_2$

Worked example. Thickening 20,000 gal at 1.0% solids to 4.0%:

$V_2 = \frac{20{,}000 \times 1.0}{4.0} = \textbf{5,000 gal}$

Quadrupling the solids concentration cuts the volume to one quarter — which is the entire economic argument for thickening before digestion and dewatering (Section 10.1).