Water and Wastewater Treatment (FE Civil)

NCEES FE Civil spec: Water Resources and Environmental Engineering, K. Water and wastewater treatment (e.g., biological processes, softening, drinking water treatment). FE Reference Handbook: Environmental Engineering chapter (Water Treatment Technologies, Wastewater Treatment and Technologies, Disinfection).

What the FE Civil Spec Covers

Section K covers drinking water and wastewater treatment: physical processes (screening, sedimentation, filtration), chemical processes (coagulation, softening, disinfection) and biological processes (activated sludge, trickling filters, digestion). FE problems are mostly loading rate and mass balance calculations, so a few formulas cover most of the questions.

Key Concepts

Drinking water train. A conventional surface water plant uses rapid mix with a coagulant (alum or ferric), flocculation, sedimentation, granular media filtration and disinfection. Hard groundwater may get lime-soda softening.
Wastewater train. Preliminary (screens, grit), primary clarification, secondary biological treatment (activated sludge or fixed film) with secondary clarifiers, disinfection, and sludge thickening and digestion.
Ideal settling. A basin removes every particle whose settling velocity is at least the overflow rate. Depth does not affect removal in the ideal model; surface area does.
Softening. Lime removes carbonate hardness; soda ash is also needed for noncarbonate hardness.
Disinfection. Inactivation depends on the product of residual concentration and contact time, CT. Chlorine demand is the dose minus the residual.
Activated sludge. Microorganisms (MLSS or MLVSS) consume BOD in an aerated tank. The food-to-microorganism ratio (F/M) and solids retention time (SRT) control performance. Conventional plants commonly run near an F/M of 0.2 to 0.4 per day and an SRT of roughly 5 to 15 days.

Core Formulas

Hydraulic detention time and surface overflow rate:

θ = \frac{V}{Q} v_{0} = \frac{Q}{A _{s}}

Stokes law for a discrete particle in laminar settling, and fraction removed in an ideal basin:

v_{s} = \frac{g ( ρ _{p} - ρ _{w} ) d ^{2}}{18 μ} fraction removed = \frac{v _{s}}{v _{0}} (v_{s} < v_{0})

Weir loading and filter loading rate:

WLR = \frac{Q}{L _{w e i r}} filtration rate = \frac{Q}{A _{f i l t er}}

Disinfection and chemical feed (US units):

C T = C \times t lb/day = (mg/L) (MGD) (8.34)

Example: dosing 2.5 mg/L of chlorine into a 5 MGD flow needs (2.5)(5)(8.34) = 104 lb/day. If the measured residual is 0.8 mg/L, the chlorine demand is 2.5 minus 0.8 = 1.7 mg/L, and a 30 minute contact time at that residual gives CT = 0.8 × 30 = 24 mg·min/L.

Activated sludge (S₀ influent BOD, X MLSS, Q_w and X_w waste sludge flow and concentration):

\frac{F}{M} = \frac{Q S _{0}}{V X} S R T = \frac{V X}{Q _{w} X _{w} + Q _{e} X _{e}}

Where It Is in the FE Reference Handbook

The Environmental Engineering chapter has Water Treatment Technologies (clarifiers, settling equations, filtration, lime-soda softening, coagulation, mixing), Wastewater Treatment and Technologies (activated sludge, trickling filters, digestion) and Disinfection (CT and removal requirements).

Worked Example: Sedimentation Basin Check

A water plant treats 0.25 m³/s in two identical rectangular basins, each 40 m long, 10 m wide and 3.5 m deep. Find the overflow rate, detention time, and the fraction of particles with a settling velocity of 0.20 mm/s that are removed.

Step 1: Flow per basin.

Q = 0.25 \times 86, 400 = 21, 600 m^{3} / day Q_{ba s in} = 10, 800 m^{3} / day

Step 2: Overflow rate. Surface area A = 40 × 10 = 400 m².

v_{0} = \frac{10 , 800}{400} = 27 m/day

Step 3: Detention time. V = 400 × 3.5 = 1,400 m³.

θ = \frac{1 , 400}{10 , 800} day \times 24 = 3.11 h

Step 4: Removal. Convert the settling velocity to m/day.

v_{s} = 0.20 \times 1 0^{- 3} \times 86, 400 = 17.28 m/day \frac{v _{s}}{v _{0}} = \frac{17.28}{27} = 0.64

About 64 percent of those particles settle out. To remove them all, the overflow rate would have to drop to 17.28 m/day, which means about 625 m² of surface per basin.

Common Mistakes

Using total plant flow for one basin when the flow is split among several units.
Thinking a deeper basin removes more particles. In the ideal model, only surface area matters.
Leaving settling velocity in mm/s while the overflow rate is in m/day.
Using the chlorine dose instead of the residual in CT.
Mixing MLSS and influent BOD units in F/M, or including recycle flow in the F/M numerator.

Practice Problems

Try these from the free PolyCalc FE Civil problem bank. Each one has a full worked solution.

A rectangular sedimentation basin has a surface area of 40 m² and treats a flow of 0.05 m³/s. Its surface overflow rate is most nearly: See the worked solution or practice the Environmental & Water Quality deck
A treatment basin has a working volume of 1,200 m³ and receives a steady flow of 4,800 m³/day. Its hydraulic detention time is most nearly: See the worked solution or practice the Environmental & Water Quality deck
A contact tank holds a free chlorine residual of 2.0 mg/L for a contact time of 30 minutes. The CT value achieved is most nearly: See the worked solution or practice the Environmental & Water Quality deck
An activated sludge aeration tank has a volume of 2,000 m³ and an MLSS of 3,000 mg/L. Sludge is wasted at 40 m³/day with a concentration of 10,000 mg/L. Neglecting solids lost in the effluent, the solids retention time is most nearly: See the worked solution or practice the Environmental & Water Quality deck
An aeration basin of 2,000 m³ holds an MLSS of 3,000 mg/L and receives 5,000 m³/day of wastewater with a BOD of 250 mg/L. The food-to-microorganism ratio is most nearly: See the worked solution or practice the Environmental & Water Quality deck
A wastewater plant removes 220 mg/L of suspended solids from a flow of 4,000 m³/day. The sludge is withdrawn at 4% solids with a specific gravity of 1.02. The volume of sludge produced is most nearly: See the worked solution or practice the Environmental & Water Quality deck

Related Calculators and Examples

Frequently Asked Questions

What is the surface overflow rate of a clarifier?

The overflow rate is the flow divided by the clarifier surface area, in units like m3/m2/day (m/day) or gpd/ft2. It equals the settling velocity of the smallest particle that an ideal basin removes completely; slower particles are removed in proportion to their settling velocity divided by the overflow rate.

What is the 8.34 factor in treatment calculations?

One million gallons of water weighs about 8.34 million pounds, so a concentration of 1 mg/L in a flow of 1 MGD is 8.34 lb/day. Mass loading in lb/day equals concentration in mg/L times flow in MGD times 8.34.

What is the difference between HRT and SRT in activated sludge?

Hydraulic retention time is the tank volume divided by the flow, often a few hours. Solids retention time (sludge age) is the mass of biomass in the system divided by the mass wasted per day, often several days, because clarifier recycle keeps the microorganisms in the system much longer than the water.

Back to all Water Resources and Environmental Engineering topics