19.3 Treatment and Process Sizing Workbook

Match the process to the controlling rate

Treatment calculations are not all detention-time problems. A sedimentation basin is controlled by surface overflow rate (SOR). A filter is controlled by hydraulic loading rate, headloss, and backwash. A chlorine contact basin is controlled by CT (concentration x time) and baffling. A biological reactor is controlled by organic loading, the food-to-microorganism (F/M) ratio, solids retention time (SRT), or aeration demand. The question signals the controlling rate if you read for process purpose.

Process sizing map with target design values

Start the workbook line with purpose. Settling -> SOR and area. Contact -> time and effective (baffled) volume. Mass removal -> load in, load out, removal fraction. Dosing -> is the dose active chemical, commercial product, or solution strength?

Removal, CT, and chemical feed discipline

Removal efficiency by load

Removal can apply to concentration or load when flow is constant, but the safe method tracks load: effluent load = influent load x (1 - removal). Example: 7,339 lb/day BOD in at 85% removal leaves 7,339 x 0.15 = 1,101 lb/day out; at 4.0 MGD the effluent concentration is 1,101 / (4.0 x 8.34) = 33 mg/L. With side streams, recycle, or changing flow, concentration-only shortcuts become risky.

CT for disinfection

Under the Surface Water Treatment Rule (SWTR), CT = residual disinfectant (mg/L) x contact time T10 (min). Required CT comes from the EPA tables and depends on pathogen, temperature, and pH. Example: a basin holds 0.5 MG with a baffling factor giving T10 = 30 min and a free-chlorine residual of 1.2 mg/L. CT achieved = 1.2 x 30 = 36 mg-min/L. Compare to the required CT (e.g., approximately 6 mg-min/L for 0.5-log Giardia at 10 C, pH 7) to confirm compliance.

Chemical feed workflow

1. Convert plant flow to the required time basis.
2. Multiply by target dose: lb/day = Q (MGD) x dose (mg/L) x 8.34.
3. Adjust for purity: product mass = active mass / fractional purity.
4. Convert to feed units (lb/day, gal/day, mL/min).
5. State whether the answer is active ingredient or delivered product.

Example: 2.0 MGD dosed at 8 mg/L active chlorine needs 2.0 x 8 x 8.34 = 133.4 lb/day active. If the product is 65% available chlorine, deliver 133.4 / 0.65 = 205 lb/day of product.

Sludge mass-volume relationships

Sludge problems hinge on the percent-solids identity: wet sludge volume (gal) = dry solids mass (lb) / (8.34 x percent solids fraction x specific gravity). For dilute sludge, specific gravity is near 1.0, so a shortcut is V (gal) = dry lb / (8.34 x Ps). Example: 2,000 lb/day of dry solids at 3% solids gives V = 2,000 / (8.34 x 0.03) = 7,994 gal/day. Thickening to 6% halves the volume to about 3,997 gal/day -- the inverse relationship between percent solids and volume is the single most-tested sludge concept. A common trap multiplies by the percent solids instead of dividing, producing a volume far too small to be physical.

Operational checks

Reject answers that violate operating sense: an SOR implying scour, a CT ignoring short-circuiting, or a sludge volume that inverts percent solids. Process sizing is a calculation plus an operating story. Build a four-column drill table (process, controlling rate, unknown, unit check) so you never force every treatment item into one detention-time equation.

Filtration and backwash drill

For a rapid sand filter, area = Q / hydraulic loading rate. Example: 3.0 MGD (2,083 gpm) at 4 gpm/ft2 needs 521 ft2 of filter area; split across two filters, each is about 260 ft2 (roughly 16 x 16 ft). Backwash demand is sized separately at a higher rate (commonly 15-20 gpm/ft2) for a short duration, and that backwash water becomes a recycle flow that must rejoin the plant mass balance. Always confirm whether the problem wants total filter area, area per unit, the number of filters, or the backwash flow -- the four-column table forces that read before you compute.

Activated-sludge loading worked example

The food-to-microorganism (F/M) ratio ties influent organic load to the biomass in the aeration basin: F/M = (Q x BOD) / (V x MLVSS), in consistent units of lb BOD/day per lb mixed-liquor volatile suspended solids. Example: 2.0 MGD at 200 mg/L BOD delivers 2.0 x 200 x 8.34 = 3,336 lb BOD/day. An aeration basin of 0.5 MG at 2,500 mg/L MLVSS holds 0.5 x 2,500 x 8.34 = 10,425 lb of biomass. F/M = 3,336 / 10,425 = 0.32 day^-1, comfortably inside the 0.2-0.5 conventional range.

If a computed F/M lands at 5 or 0.02, suspect a units slip -- typically forgetting the 8.34 on one of the two terms, which is the most common activated-sludge setup error on the exam.