FE Environmental · knowledge area · 12–18 items

FE Environmental water and wastewater, unit by unit

Water and wastewater is tied for the largest area of the exam at 12–18 questions. It is the treatment plant from headworks to sludge: what each unit removes, how it is loaded and how its performance is computed.

Updated Oct 29, 2026 · Not affiliated with NCEES.

ExamFE Environmental
questions on the exam12–18
problems on this page18
areas in the specification15
CheckedOct 2026

What the area covers

What the specification lists, grouped seven ways:

Characteristics: physical, chemical and biological quality of water and wastewater, nutrients included.
Mass balance and loading rates: removal efficiencies through a train of units.
Physical processes: sedimentation and clarification, filtration, adsorption and activated carbon, membranes, flocculation, headworks, flow equalization, air stripping.
Chemical processes: disinfection, ion exchange, softening, coagulation, precipitation.
Biological processes: activated sludge, fixed film, lagoons, phytoremediation; aerobic, anaerobic and anoxic conditions.
Sludge: digestion, dewatering, composting, land application.
Water conservation and reuse.

18 treatment-plant problems, worked

0 / 18 answered · 0 correct

Question 1 of 18 · Water and Wastewater · 12–18 items

A plant with Q = 10,000 m³/day has influent BOD 200 mg/L and effluent BOD 20 mg/L. Yield Y = 0.50 g biomass/g substrate. The aeration basin mixed liquor contains X = 3,000 mg/L, and the basin volume is V = 5,000 m³. Endogenous decay b = 0.05 day⁻¹. Calculate the net daily sludge (biomass) production (kg/day).

Given	Flow Q = 10,000 m³/day Influent BOD = 200 mg/L Effluent BOD = 20 mg/L Yield Y = 0.50 g biomass/g substrate MLSS X = 3000 mg/L Basin volume V = 5000 m³ Endogenous decay b = 0.05 day⁻¹
Find	Net daily sludge production (kg/day)

Solution

Net biomass production = yield on substrate removed − endogenous decay (Water and wastewater: activated sludge).

P_x = YQ(S₀ − S) − bXV

P_x = 0.50 × 10,000 × 180 g/day − 0.05 × 3,000 × 5,000 g/day = 900 − 750 = 150 kg/day

Growth = 0.50 × 10,000 m³/day × 180 g/m³ = 900 kg/day; decay = 0.05 × 3,000 g/m³ × 5,000 m³ = 750 kg/day; net = 150 kg/day.

750 kg/day is the endogenous decay term alone.

900 kg/day is gross biomass growth; endogenous decay still has to be subtracted.

1,650 kg/day adds decay to growth; endogenous respiration consumes biomass.

1,800 kg/day is the BOD removed, not the biomass produced.

Question 2 of 18 · Water and Wastewater · 12–18 items

Unit Process	BOD Removal Efficiency
Primary Clarifier	35%
Secondary Treatment	85%

A wastewater treatment plant consists of a primary clarifier followed by an aeration basin and secondary clarifier. The plant receives a flow of 5.0 MGD with an influent BOD concentration of 200 mg/L. Based on the removal efficiencies provided in the table, what is the final effluent BOD concentration?

Given	Flow = 5.0 MGD Influent BOD = 200 mg/L Primary removal = 35 % Secondary removal = 85 %
Find	Final effluent BOD (mg/L)

Solution

Removals in series multiply the fractions remaining (Water and wastewater: treatment trains).

C = 200 mg/L × (1 − 0.35) × (1 − 0.85) = 19.5 mg/L

170 mg/L confuses removal with remaining: 85% of 200 mg/L is the amount removed, not left.

130 mg/L is the primary effluent, before secondary treatment.

80.0 mg/L averages the two efficiencies (60%); removals in series multiply the fractions remaining.

30.0 mg/L applies only the 85% secondary removal to the raw 200 mg/L and skips the primary clarifier.

Primary leaves 200 × 0.65 = 130 mg/L; secondary removes 85% of that, 130 × 0.15 = 19.5 mg/L.

Question 3 of 18 · Water and Wastewater · 12–18 items

A disinfection contact basin has a volume of 600 m³ and treats a peak flow of 180 m³/h. The basin has a baffling factor of 0.4. If the chlorine residual is maintained at 2.5 mg/L, what is the actual CT value achieved?

Given	Basin volume = 600 m³ Peak flow = 180 m³/h Baffling factor = 0.4 Chlorine residual = 2.5 mg/L
Find	Actual CT (mg·min/L)

Solution

CT credit uses the T₁₀ contact time (Water treatment: disinfection).

θ = V/Q = 600 m³/(180 m³/hr) = 3.33 hr = 200 min

T₁₀ = BF × θ = 0.4 × 200 = 80 min; CT = 2.5 mg/L × 80 min = 200 mg·min/L

3.33 uses T₁₀ in hours (1.33 hr); CT is reported in mg·min/L.

80 is T₁₀ in minutes; it still has to be multiplied by the 2.5 mg/L residual.

θ = 600/180 = 3.33 hr = 200 min; T₁₀ = 0.4 × 200 = 80 min; CT = 2.5 mg/L × 80 min = 200 mg·min/L.

500 uses the theoretical detention time (200 min); CT credit uses T₁₀, which accounts for short-circuiting.

1,250 divides by the baffling factor instead of multiplying.

Question 4 of 18 · Water and Wastewater · 12–18 items

Stream	Flow (MGD)	TSS (mg/L)
Influent	5.0	200
Effluent	4.95	15
Waste Sludge (WAS)	0.05	8,000
Aeration Basin (MLSS)	-	3,000

Calculate the Solids Retention Time (SRT) for the activated sludge system described by the operational data in the table. The aeration basin volume is 2.0 MG.

Given	Aeration basin volume = 2.0 MG MLSS = 3,000 mg/L WAS flow = 0.05 MGD at 8,000 mg/L Effluent flow = 4.95 MGD at 15 mg/L
Find	SRT (d)

Solution

SRT = solids in the reactor ÷ solids leaving per day (wasted + effluent) (Water and wastewater: activated sludge).

SRT = VX/(Q_wX_w + Q_eX_e) = (2.0 × 3,000)/(0.05 × 8,000 + 4.95 × 15) = 6,000/474 = 12.7 days

0.40 days is the hydraulic retention time V/Q, not the solids retention time.

6.00 days divides basin solids by the influent TSS load; SRT uses the solids leaving the system.

Solids in basin = 2.0 × 3,000 × 8.34 = 50,040 lb; out = 0.05 × 8,000 × 8.34 + 4.95 × 15 × 8.34 = 3,955 lb/d; SRT = 12.7 days.

15.0 days counts only the waste sludge and ignores the 619 lb/d of solids lost in the effluent.

40.0 days divides the basin volume by the WAS flow, ignoring the concentrations.

Question 5 of 18 · Water and Wastewater · 12–18 items

A high-rate anaerobic digester is modeled as a CSTR. At 35 °C, volatile solids (VS) destruction follows first-order kinetics with k = 0.21 d⁻¹, and the hydraulic retention time is 15 days. If the temperature drops to 20 °C, what is the expected percent VS destruction? Use a temperature coefficient θ = 1.08.

Given	Rate constant at 35 °C k = 0.21 d⁻¹ Hydraulic retention time = 15 days New temperature = 20 °C Temperature coefficient θ = 1.08
Find	Percent VS destruction at 20 °C (%)

Solution

Temperature-correct k, then apply the CSTR first-order removal (Water and wastewater: anaerobic digestion).

k₂₀ = k₃₅θ^{(20 − 35)} = 0.21 × 1.08⁻¹⁵ = 0.0662 d⁻¹

removal = kτ/(1 + kτ) = 0.993/1.993 = 49.8%

k₂₀ = 0.21 × 1.08⁻¹⁵ = 0.0662 d⁻¹; CSTR: kτ = 0.993, destruction = 0.993/1.993 = 49.8%.

63.0% applies the plug-flow form 1 − e^−kτ; the digester is a completely mixed reactor.

75.9% is the destruction at 35 °C; the rate must be corrected to 20 °C first.

90.9% applies θ with the wrong sign (1.08⁺¹⁵), raising k as the temperature falls.

95.7% uses the 35 °C rate and the plug-flow form, two errors at once.

Question 6 of 18 · Water and Wastewater · 12–18 items

An activated sludge aeration basin is designed with a maximum specific growth rate for nitrifiers of μ_max = 0.8 d⁻¹. The half-saturation constant for dissolved oxygen (DO) is K_DO = 1.5 mg L⁻¹. If the basin is operating with a DO concentration of 0.5 mg/L, what is the specific growth rate due to oxygen limitation?

Given	μ_max = 0.8 d⁻¹ K_DO = 1.5 mg/L DO = 0.5 mg/L
Find	Specific growth rate (d⁻¹)

Solution

Monod (switching) function for dissolved oxygen (Water and wastewater: nitrification kinetics).

μ = μ_max·DO/(K_DO + DO) = 0.8 × 0.5/(1.5 + 0.5) = 0.20 d⁻¹

0.80 d⁻¹ is μ_max, the rate with no oxygen limitation.

0.60 d⁻¹ swaps K_DO and DO in the numerator.

0.40 d⁻¹ multiplies μ_max by DO and ignores the half-saturation term.

0.27 d⁻¹ divides by K_DO alone and drops DO from the denominator.

μ = μ_max·DO/(K_DO + DO) = 0.8 × 0.5/(1.5 + 0.5) = 0.20 d⁻¹.

Question 7 of 18 · Water and Wastewater · 12–18 items

A water treatment plant processes 15 MGD (56.8 MLD). Jar tests indicate an optimal ferric chloride dose of 25 mg/L. The plant uses a stock solution with the properties listed below. Calculate the required feed pump setting.

Property	Value
Stock Strength	40% by weight
Specific Gravity	1.45
Active Chemical	Ferric Chloride

Given	Plant flow = 15 MGD (56.8 MLD) Ferric chloride dose = 25 mg/L Stock strength = 40% by weight Stock specific gravity = 1.45
Find	Feed pump setting (L/min)

Solution

Chemical feed = dose × flow ÷ (solution strength × density) (Water treatment: chemical feed).

M = 25 mg/L × 56.8 × 10⁶ L/d = 1,420 kg/d

C_stock = 0.40 × 1.45 kg/L = 0.58 kg/L → Q = 1,420/0.58 = 2,448 L/d = 1.7 L/min

0.68 L/min accounts for the specific gravity but not the 40% strength; only 40% of the solution is ferric chloride.

0.99 L/min treats the stock as pure chemical at the density of water, ignoring both strength and specific gravity.

1,420 kg/d of FeCl₃ ÷ (0.40 × 1.45 kg/L = 0.58 kg/L) = 2,448 L/d ÷ 1,440 min/d = 1.7 L/min.

2.5 L/min uses the 40% strength but leaves out the specific gravity of 1.45.

2,400 is the feed in L/day; it was not converted to L/min.

Question 8 of 18 · Water and Wastewater · 12–18 items

A GAC contactor treats a waste stream containing 100 µg/L of PCE. The adsorption follows a Langmuir isotherm with q_max = 50 mg/g and K = 0.5 L/mg. If the contactor contains 10,000 kg of carbon and treats 2 MLD, estimate the bed life until saturation (assume the entire bed reaches equilibrium with the influent).

Given	Influent PCE = 100 µg/L q_max = 50 mg/g Langmuir K = 0.5 L/mg Carbon mass = 10,000 kg Flow = 2 MLD
Find	Bed life (d)

Solution

Bed life = equilibrium capacity at the influent concentration ÷ daily loading (Water treatment: adsorption).

q_e = q_maxKC/(1 + KC) = 50 × 0.5 × 0.1/(1 + 0.05) = 2.38 mg/g

t = (2.38 mg/g × 10⁷ g)/(0.1 mg/L × 2 × 10⁶ L/d) = 119 days

238 days uses 1 MLD instead of 2 MLD for the PCE loading.

2,500 days uses q_max = 50 mg/g; at 0.1 mg/L the carbon reaches only 2.38 mg/g.

2.45 days enters 100 µg/L as 100 mg/L, which inflates both the loading and q.

q = 50 × 0.5 × 0.1/(1 + 0.05) = 2.38 mg/g; capacity = 2.38 × 10⁷ mg; load = 0.1 mg/L × 2 × 10⁶ L/d = 2 × 10⁵ mg/d; life = 119 days.

0.119 days treats q as mg per kg of carbon instead of mg per g.

Question 9 of 18 · Water and Wastewater · 12–18 items

Operational data for a membrane filtration system is shown below. Water viscosity is 1.00 cP at 20°C and 1.31 cP at 10°C. Evaluate the condition of the membrane at Day 30.

Parameter	Day 1	Day 30
Flux (LMH)	40	30
TMP (bar)	0.4	0.5
Temperature (°C)	20	10

Given	Day 1: flux 40 LMH, TMP 0.4 bar, 20 °C Day 30: flux 30 LMH, TMP 0.5 bar, 10 °C Viscosity = 1.00 cP at 20 °C Viscosity = 1.31 cP at 10 °C
Find	Membrane condition at day 30

Solution

Compare temperature-corrected specific flux (permeability) (Water treatment: membranes).

day 1: 40/0.4 = 100 LMH/bar; day 30: (30/0.5) × (1.31/1.00) = 78.6 LMH/bar → 21% decline

Raw flux fell partly because cold water is more viscous, but even after correcting to 20 °C the permeability is 21% lower.

Temperature-corrected permeability fell from 40/0.4 = 100 to (30/0.5) × 1.31 = 78.6 LMH/bar, about 21% lower, which indicates fouling.

Both raw and temperature-corrected permeability are lower on day 30, so performance declined.

An integrity failure lets water bypass the membrane and raises permeability; the data show a decrease.

The data show a permeability loss but cannot tell whether cleaning would recover it, so irreversibility cannot be concluded.

Question 10 of 18 · Water and Wastewater · 12–18 items

A rapid sand filter operates with a hydraulic loading rate of 12 m/h. The clean bed headloss is 0.5 m, and the maximum allowable headloss is 2.5 m. Headloss accumulates at 0.04 m/h. Turbidity breakthrough consistently occurs at 40 hours. Determine the filter run time and the limiting factor.

Given	Hydraulic loading rate = 12 m/h Clean bed headloss = 0.5 m Maximum headloss = 2.5 m Headloss accumulation = 0.04 m/h Turbidity breakthrough = 40 h
Find	Filter run time and limiting factor

Solution

A filter run ends at whichever limit comes first (Water treatment: filtration).

t_HL = (2.5 − 0.5) m ÷ 0.04 m/hr = 50 hr; t_turbidity = 40 hr → run = 40 hr (turbidity limits)

Turbidity breaks through at 40 hr, not 50 hr; 50 hr is the headloss time.

50 hr is when headloss would reach 2.5 m, but the filter has already broken through at 40 hr.

62 hr divides the full 2.5 m by 0.04 m/hr and forgets the 0.5 m clean-bed headloss.

Terminal headloss would be reached after (2.5 − 0.5)/0.04 = 50 hr, but turbidity breaks through at 40 hr, so turbidity ends the run.

The run length is right, but at 40 hr the headloss is only 0.5 + 1.6 = 2.1 m, below the 2.5 m limit.

Question 11 of 18 · Water and Wastewater · 12–18 items

Adsorption data for a contaminant follows the Freundlich isotherm q_e = K_f C_e^1/n. Experimental results show that when C_e = 1 mg/L, q_e = 10 mg/g; and when C_e = 100 mg/L, q_e = 100 mg/g. Determine the isotherm equation.

Given	q_e = 10 mg/g at C_e = 1 mg/L q_e = 100 mg/g at C_e = 100 mg/L
Find	Freundlich isotherm equation

Solution

Fit both constants from two points; log q = log K_f + (1/n)log C (Water treatment: adsorption isotherms).

C_e = 1: K_f = 10; 100 = 10 × 100^1/n → 1/n = log 10/log 100 = 0.5

Dropping K_f gives q_e = 1 at C_e = 1, not the measured 10 mg/g.

At C_e = 1, q_e = K_f = 10; then 100 = 10 × 100^1/n gives 100^1/n = 10, so 1/n = 0.5.

2.0 is n, not 1/n; the exponent in the isotherm is 1/n = 0.5.

An exponent of 1.0 would give q_e = 1,000 mg/g at C_e = 100, not 100.

K_f is q_e at C_e = 1 mg/L, which is 10, not 100.

Question 12 of 18 · Water and Wastewater · 12–18 items

A membrane plant requires a total permeate production of 2.4 MLD. The design flux is 20 L/(m²·h). If each membrane module has a surface area of 25 m², how many modules are required?

Given	Permeate production = 2.4 MLD Design flux = 20 L/(m²·h) Module area = 25 m²
Find	Number of modules

Solution

Area = flow ÷ flux, then divide by module area (Water treatment: membranes).

Q = 2.4 × 10⁶ L/d ÷ 24 h/d = 1.0 × 10⁵ L/h

A = 1.0 × 10⁵/20 = 5,000 m²; N = 5,000/25 = 200 modules

8 modules divides by 24 a second time after the flow is already in L/hr.

Q = 2.4 × 10⁶ L/d ÷ 24 = 100,000 L/hr; area = 100,000/20 = 5,000 m²; modules = 5,000/25 = 200.

250 divides the 5,000 m² area by the flux (20) instead of the module area (25 m²).

4,800 leaves the flow in L/d while the flux is in L/(m²·h), a factor of 24.

5,000 is the membrane area in m², not the number of modules.

Question 13 of 18 · Water and Wastewater · 12–18 items

A rectangular grit chamber is designed with a cross-sectional area of 1.0 m² and handles a peak flow of 0.6 m³/s. The detention time is 45 seconds. Evaluate the hydraulic design with respect to grit scour.

Given	Cross-sectional area = 1.0 m² Peak flow = 0.6 m³/s Detention time = 45 s
Find	Assessment of the hydraulic design

Solution

Horizontal velocity controls grit removal: about 0.3 m/s settles grit while keeping organics in suspension (Wastewater: preliminary treatment).

v_h = Q/A = 0.6 m³/s ÷ 1.0 m² = 0.6 m/s > ≈ 0.3 m/s

No depth criterion is violated by the data; the problem shows up in the velocity.

Low velocity would let organics settle with the grit; here the velocity is too high.

The 45 s detention is typical, but the horizontal velocity of 0.6 m/s is far above the target.

Peak-flow detention of 45 to 90 s is common for grit chambers; 45 s is not the problem.

v = Q/A = 0.6/1.0 = 0.6 m/s, about twice the roughly 0.3 m/s used for grit chambers, so settled grit would be scoured back out.

Question 14 of 18 · Water and Wastewater · 12–18 items

A granular activated carbon (GAC) contactor is designed to treat 1.5 MGD of groundwater containing 20 mg/L of TOC. The theoretical equilibrium adsorption capacity (q_e) at this concentration is 25 mg TOC per gram of GAC. To account for the mass transfer zone and early breakthrough, the design usage rate is based on 40% of the theoretical equilibrium capacity. Calculate the daily GAC consumption rate.

Given	Flow = 1.5 MGD TOC = 20 mg/L Equilibrium capacity q_e = 25 mg TOC/g GAC Design capacity = 40% of equilibrium
Find	Daily GAC consumption (lb/day)

Solution

Carbon usage = TOC removed ÷ usable adsorption capacity (Water treatment: adsorption).

TOC = 1.5 MGD × 8.34 × 20 mg/L = 250 lb/day

q_design = 0.40 × 0.025 lb/lb = 0.010 lb/lb → GAC = 250/0.010 = 25,000 lb/day

10,000 lb/day uses the full equilibrium capacity; only 40% of it is credited in design.

16,700 lb/day uses 60% of the capacity instead of 40%.

100,000 lb/day converts 25 mg/g as 0.0025 lb/lb; 25 mg/g is 0.025 lb/lb.

TOC load = 1.5 × 8.34 × 20 = 250 lb/day; design capacity = 0.40 × 0.025 = 0.010 lb/lb; GAC = 250/0.010 = 25,000 lb/day.

62,500 lb/day applies the 40% factor twice.

Question 15 of 18 · Water and Wastewater · 12–18 items

For an activated sludge system with μ_max = 0.6 d⁻¹, K_s = 20 mg/L, X = 3,000 mg/L, Y = 0.5 mg VSS/mg substrate, and substrate concentration S = 5 mg/L, compare the substrate removal rates predicted by the full Monod model and by the first-order approximation (valid for S ≪ K_s). Report the two rates as Monod, then first-order.

Given	μ_max = 0.6 d⁻¹ K_s = 20 mg/L X = 3,000 mg/L Y = 0.5 mg VSS/mg S S = 5 mg/L
Find	Substrate removal rates, Monod and first-order (mg/(L·d))

Solution

Monod substrate use r = μ_maxSX/[Y(K_s + S)]; the first-order form drops S from the denominator (Water and wastewater: kinetics).

Monod: r = (0.6 × 5/25) × 3,000/0.5 = 720 mg/(L·day)

first-order: r = (0.6 × 3,000)/(20 × 0.5) × 5 = 900 mg/(L·day)

Monod: μ = 0.6 × 5/25 = 0.12 d⁻¹, r = 0.12 × 3,000/0.5 = 720; first-order drops S from the denominator: 0.6 × 3,000 × 5/(20 × 0.5) = 900.

The first-order form uses K_s alone in the denominator, so it cannot equal the Monod rate when S = 5 mg/L is not negligible.

The values are reversed: the full Monod rate is the lower one, 720.

360 and 450 leave out the yield; r_s = μX/Y.

180 and 225 multiply by Y instead of dividing; substrate use is growth divided by yield.

Question 16 of 18 · Water and Wastewater · 12–18 items

A wastewater influent contains NH₄-N = 35 mg/L. During nitrification, 90% of the NH₄-N is oxidized to NO₃-N. Nitrification consumes 7.14 mg alkalinity as CaCO₃ per mg N oxidized. If the influent alkalinity is 250 mg/L as CaCO₃, what is the residual alkalinity after nitrification?

Given	Influent NH₄-N = 35 mg/L Fraction nitrified = 90 % Alkalinity use = 7.14 mg CaCO₃/mg N Influent alkalinity = 250 mg/L as CaCO₃
Find	Residual alkalinity (mg/L as CaCO₃)

Solution

Nitrification consumes 7.14 mg alkalinity (as CaCO₃) per mg NH₄-N oxidized (Wastewater: nitrification).

ΔN = 35 × 0.90 = 31.5 mg/L; Alk used = 31.5 × 7.14 = 225 mg/L

Alk_residual = 250 − 225 = 25 mg/L as CaCO₃

475 mg/L adds the consumption to the influent alkalinity; nitrification destroys alkalinity.

225 mg/L is the alkalinity consumed, not what remains.

219 mg/L subtracts the 31.5 mg/L of nitrogen oxidized instead of the alkalinity it consumes (31.5 × 7.14).

138 mg/L uses 3.57 mg/mg, the alkalinity recovered by denitrification, instead of 7.14 consumed by nitrification.

N oxidized = 35 × 0.90 = 31.5 mg/L; alkalinity used = 31.5 × 7.14 = 225 mg/L; residual = 250 − 225 = 25 mg/L as CaCO₃.

Question 17 of 18 · Water and Wastewater · 12–18 items

A wastewater stream (100 m³/d) contains 50 mg/L fluoride (F⁻). The plant requires effluent F⁻ reduction to 5 mg/L using calcium chloride (CaCl₂). Given K_sp for CaF₂ is 4.0 × 10⁻¹¹ and atomic weights Ca=40, Cl=35.5, F=19, what is the theoretical daily CaCl₂ requirement?

Given	Flow = 100 m³/d Influent F⁻ = 50 mg/L Effluent F⁻ = 5 mg/L K_sp of CaF₂ = 4.0 × 10⁻¹¹ Atomic weights Ca = 40, Cl = 35.5, F = 19
Find	Daily CaCl₂ requirement (kg/d)

Solution

Calcium is consumed by CaF₂ and an excess must stay dissolved to hold fluoride at the limit (Water and wastewater: chemical precipitation).

F⁻ removed = 45 mg/L ÷ 19 g/mol = 2.37 × 10⁻³ M → Ca for CaF₂ = 1.18 × 10⁻³ M

residual: [Ca²⁺] = K_sp/[F⁻]² = 4.0 × 10⁻¹¹/(2.63 × 10⁻⁴)² = 5.78 × 10⁻⁴ M

CaCl₂ = 1.76 × 10⁻³ M × 111 g/mol × 100 m³/d = 19.6 kg/d

6.4 kg/d covers only the excess calcium that must stay dissolved (5.78 × 10⁻⁴ M) and forgets the calcium consumed by the precipitate.

13.1 kg/d is the stoichiometric calcium for 45 mg/L of fluoride; without the residual excess the effluent cannot be held at 5 mg/L.

Stoichiometric Ca = (45/19)/2 = 1.18 × 10⁻³ M; residual Ca = K_sp/[F⁻]² = 5.78 × 10⁻⁴ M; total 1.76 × 10⁻³ M × 111 g/mol × 100 m³/d = 19.6 kg/d.

32.7 kg/d assigns one Ca²⁺ per F⁻; CaF₂ needs one calcium for every two fluorides.

19,600 is the requirement in g/d; it was not converted to kilograms.

Question 18 of 18 · Water and Wastewater · 12–18 items

An MLE wastewater process treats 4.0 MGD with influent TKN of 30 mg/L. RAS flow is 2.0 MGD and internal recycle is 14.0 MGD. Assuming complete nitrification, complete denitrification capacity, and negligible biomass synthesis, calculate the effluent nitrate (NO₃-N) concentration.

Given	Influent flow = 4.0 MGD Influent TKN = 30 mg/L RAS flow = 2.0 MGD Internal recycle = 14.0 MGD
Find	Effluent NO₃-N (mg/L)

Solution

MLE with complete nitrification and denitrification: only the nitrate not recycled to the anoxic zone leaves (Wastewater: nitrogen removal).

NO₃-N_e = TKN/(1 + R_RAS + R_IR) = 30/(1 + 2.0/4.0 + 14.0/4.0) = 6.00 mg/L

R_RAS = 2.0/4.0 = 0.5 and R_IR = 14.0/4.0 = 3.5, so NO₃-N = 30/(1 + 0.5 + 3.5) = 6.00 mg/L.

6.67 mg/L leaves out the RAS: 30/(1 + 3.5).

7.50 mg/L drops the 1 for the forward flow: 30/(0.5 + 3.5).

20.0 mg/L leaves out the internal recycle: 30/(1 + 0.5).

30.0 mg/L assumes no recycle reaches the anoxic zone, so none of the nitrate is denitrified.

Answer every question to see your score

Common traps

BOD₅ is not ultimate BOD. The five-day test sees only part of the oxygen demand. With a base-e rate constant k in 1/day:

BOD₅ = BOD_u (1 − e^−5k)

A constant given in base 10 must be converted first: k_e = 2.303 k₁₀.

Overflow rate, detention time and weir loading are three different checks. The overflow rate Q/A is the settling velocity a clarifier captures, and it does not depend on depth. Detention time is V/Q. Weir loading is flow per length of effluent weir. A problem can pass one and fail another.

Hardness is reported as CaCO₃. Convert each ion through its equivalent weight before adding: Ca²⁺ 20.0, Mg²⁺ 12.2 and CaCO₃ 50.0 g/eq. Carbonate hardness is the smaller of total hardness and alkalinity; only the noncarbonate part needs soda ash in lime-soda softening.

Loads in USCS: the mg/L × MGD × 8.34 = lb/day conversion is on exam facts. Handbook search terms for this area: BOD, overflow rate, hardness, activated sludge.

Water law in one line each

The ethics area includes compliance with environmental law, and for water two statutes cover it:

Clean Water Act (CWA): discharges to surface waters. A point source needs an NPDES permit (EPA summary).
Safe Drinking Water Act (SDWA): public drinking water. EPA sets maximum contaminant levels and treatment techniques (EPA summary).

Air law is on the air page, waste law on the waste page.

Upstream and downstream of the plant

Plant hydraulics: pumps, blowers, weirs, flumes and head loss are on fluid mechanics and hydraulics.
Basins as reactors: the completely mixed and plug-flow models behind every tank are on fundamental principles.
The chemistry of softening and coagulation: equilibria and precipitation are on environmental chemistry.
Pipes and sewers for a town: distribution networks and collection systems belong to FE Civil water resources.
Sludge leaving the plant: landfilling and composting continue on solid and hazardous waste.

The other sections

Fluid Mechanics and Hydraulics12–18 items on the exam
Surface Water Resources and Hydrology9–14 items on the exam
Groundwater, Soils, and Sediments8–12 items on the exam

Water and wastewater, quick answers

How many water and wastewater questions are on the FE Environmental exam?

12–18 of 110, tied with fluid mechanics and hydraulics as the largest area.

Are regulations tested?

Yes, under ethics: the specification lists compliance with codes, standards and regulations and names the Clean Water Act and the Safe Drinking Water Act among its examples. Expect to know what a statute governs, not its section numbers.

Treatment problems on your phone

FE Environmental questions, answers and worked solutions in a study app.

Download on theApp Store Android: practice in the browser Get the FE Environmental app

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