FE Environmental fundamental principles, growth and reactors
Fundamental principles carries 7–11 questions on the tools every other area leans on: how much water, wastewater, waste or energy a growing population will need, and how a tank, lagoon, lake or room behaves as a reactor.
Updated Oct 29, 2026 · Not affiliated with NCEES.
ExamFE Environmental
questions on the exam7–11
problems on this page15
areas in the specification15
Checked
What the area covers
Three headings, each a tool the other areas reuse:
Population projections and demand: for water, wastewater, solid waste and energy; arithmetic, geometric and logistic growth models.
Reactors: batch, completely mixed and plug-flow models of treatment units and natural water bodies.
Materials science: properties, corrosion, compatibility, stress and strain.
15 growth, demand and reactor problems
0 / 15 answered · 0 correct
Question 1 of 15 · Fundamental Principles · 7–11 items
A new gravity sewer interceptor is being designed to serve a planned residential development with a projected population of 20,000 people. The average daily per capita wastewater flow is 120 gpcd. The peak day factor is 2.0, and the peak hour factor (peak hour ÷ average day) is 3.2. What is the design flow rate for this pipe in cubic feet per second (cfs)?
Given
Projected population = 20,000
Per-capita wastewater flow = 120 gpcd
Peak day factor = 2.0
Peak hour factor (to average day) = 3.2
Find
Sewer design flow (cfs)
Solution
Size the sewer for the peak-hour flow (Water and wastewater: design flows).
Qavg = 20,000 × 120 gal/(person·day) = 2.4 MGD
Qpeak hour = 2.4 MGD × 3.2 = 7.68 MGD
Q = 7.68 MGD × 1.547 cfs/MGD = 11.9 cfs
3.71 cfs is the average daily flow (2.4 MGD) with no peaking; a gravity sewer is sized for the peak.
4.80 is the peak-day flow in MGD (2.4 × 2.0) written as cfs; it uses the treatment-plant factor and skips the unit conversion.
7.43 cfs applies the peak-day factor (4.8 MGD); a collection pipe must carry the peak hour.
23.8 cfs stacks both factors (2.4 × 2.0 × 3.2 MGD); the peak-hour factor here is already referenced to the average day.
Question 2 of 15 · Fundamental Principles · 7–11 items
Historical census data for a municipality are provided below. Based on the growth trend observed in the data, what is the estimated population for Year 30?
Year
Population
0
4,000
10
6,000
20
9,000
Given
Year 0 population = 4,000
Year 10 population = 6,000
Year 20 population = 9,000
Find
Projected population in year 30
Solution
Constant ratios between equal intervals mean geometric growth (Fundamental principles: population projection).
6,000/4,000 = 9,000/6,000 = 1.5 per decade
P30 = 9,000 × 1.5 = 13,500
11,000 adds the first decade's increase (2,000) again; the increases are not constant, so an arithmetic model does not fit.
11,333 is a straight-line fit through all three points (250 per year); the data rise by a constant ratio, not a constant amount.
11,500 adds the average increase of 2,500 per decade, an arithmetic model the data do not follow.
12,000 repeats the last decade's increase of 3,000, which is still arithmetic growth.
Each decade multiplies the population by 1.5 (6,000/4,000 = 9,000/6,000), so geometric growth gives 9,000 × 1.5 = 13,500.
Question 3 of 15 · Fundamental Principles · 7–11 items
An annual discrete growth rate is stated as 6.00% per year. For use in a continuous exponential model, convert it to the equivalent continuous rate k. What is k?
Given
Annual discrete growth rate r = 6.00 %
Find
Continuous rate k (yr−1)
Solution
Match one year of growth: ek = 1 + r.
k = ln(1 + r) = ln(1.0600) = 0.05827 yr−1
0.02531 is log10(1.06); the continuous rate uses the natural logarithm.
0.05660 is r/(1 + r), the equivalent discount rate, which is a different conversion.
k = ln(1 + r) = ln(1.06) = 0.05827 yr−1, slightly below 6% because continuous compounding grows faster at the same rate.
0.06000 uses the discrete rate unchanged; for the same annual growth, the continuous rate must be a bit lower.
0.06184 is e0.06 − 1, which converts in the opposite direction (continuous to discrete).
Question 4 of 15 · Fundamental Principles · 7–11 items
A municipality has a current population of 150,000, discrete growth of 3.0% per year, per-capita waste generation of 0.8 kg/day, and a compacted landfill density of 600 kg/m3. Year 1 waste is produced by the current population, and the population grows 3.0% each year after that. If the landfill's net capacity is 1,000,000 m3, how many complete years of waste can it accept?
Given
Current population = 150,000
Growth rate = 3.0% per year (discrete)
Per-capita waste = 0.8 kg/day
Compacted density = 600 kg/m3
Landfill net capacity = 1,000,000 m3
Find
Complete years of waste that fit
Solution
Annual volume grows geometrically; sum the series until it reaches capacity (Solid waste: landfill capacity).
ΣV = V1[(1.03)T − 1]/0.03 → 935,000 m3 at T = 11; 1,036,000 m3 at T = 12
11 complete years of waste fit in 1,000,000 m3
17 years treats the 3.0% as a yearly decline; with a shrinking population the cell would hold 17 complete years.
14 years ignores growth and also rounds 13.7 up.
13 years ignores growth: 1,000,000/73,000 = 13.7, truncated.
12 years counts the year in which the landfill fills; the 12th year's waste does not fit (1,036,000 > 1,000,000 m3).
Year-1 volume is 150,000 × 0.8 × 365/600 = 73,000 m3; the geometric sum reaches 935,000 m3 after 11 years and 1,036,000 m3 after 12, so 11 complete years fit.
Question 5 of 15 · Fundamental Principles · 7–11 items
Population is 60,000 now, growing at 2.0%/yr for 5 years. Per-capita water use is 180 L/cap·d; return flow to wastewater is 85%. Influent BOD is 250 mg/L in the wastewater. Calculate BOD mass loading (kg BOD/day) at year 5.
Given
Current population = 60,000
Growth rate = 2.0%/yr
Projection period = 5 years
Per-capita water use = 180 L/cap·d
Return flow to wastewater = 85%
Influent BOD = 250 mg/L
Find
BOD mass loading at year 5 (kg/day)
Solution
Project the population, convert to wastewater flow, then multiply by concentration (Fundamental principles: mass loading).
2,700 kg/day uses today's population and all the water use, skipping both growth and the 85% return.
2,981 kg/day projects the population but sends all water use to the sewer instead of 85%.
3,507 kg/day divides by the 0.85 return fraction instead of multiplying.
5,711 kg/day enters the 2.0% growth as 20% per year (1.205 = 2.49).
Question 6 of 15 · Fundamental Principles · 7–11 items
A population grows at 0.8% per month (discrete). Using a continuous model with the equivalent annual rate, estimate the population after 5 years starting from 200,000 (nearest 100).
Given
Monthly growth rate = 0.8 % (discrete)
Initial population = 200,000
Time = 5 yr
Find
Population after 5 years (nearest 100)
Solution
Convert the monthly discrete rate to a continuous annual rate (Fundamental principles: growth models).
k = 12·ln(1 + 0.008) = 0.0956 yr−1
P = 200,000 × e0.0956 × 5 = 322,600
k = 12 ln(1.008) = 0.0956 yr−1, so P = 200,000 × e0.0956 × 5 = 322,600 — the same as 200,000 × 1.00860.
316,300 compounds 9.6% once a year (12 × 0.8%); monthly compounding gives an effective 10.03% per year.
296,000 is simple (linear) growth, 200,000 × (1 + 0.008 × 60), with no compounding.
208,100 applies ln(1.008) as if it were a yearly rate; the 0.8% is per month, so k = 12 ln(1.008) per year.
122,600 is the increase over 5 years, not the population.
Question 7 of 15 · Fundamental Principles · 7–11 items
A utility projects population will increase 25% over the planning horizon, but conservation reduces per-capita demand by 15%. If baseline total demand is 20,000 m3/day, what is the net percent change in total demand?
Given
Population increase = 25 %
Per-capita demand reduction = 15 %
Baseline demand = 20,000 m3/day
Find
Net percent change in total demand
Solution
Demand is the product of population and per-capita use, so the changes multiply.
1.25 × 0.85 = 1.0625 → +6.25%
A 15% decrease counts only the conservation effect and ignores the 25% larger population.
Total demand = population × per-capita use, so the factors multiply: 1.25 × 0.85 = 1.0625, a 6.25% increase.
A 10% increase adds the percentages (25 − 15); percent changes of two factors multiply, they do not add.
21.25% cuts only the added population by 15% (25 × 0.85); the reduction applies to every customer's use.
40% adds the two magnitudes as if conservation also raised demand.
Question 8 of 15 · Fundamental Principles · 7–11 items
For m identical CSTRs in series sharing a total hydraulic residence time θtotal, the effluent ratio is Cout/Cin = (1 + kθtotal/m)−m. Given k = 1.20 day−1, θtotal = 4.0 day, and a required Cout/Cin of 0.02, determine the minimum integer number of tanks m.
Given
Rate constant k = 1.20 day−1
Total HRT θtotal = 4.0 day
Required Cout/Cin = 0.02
Find
Minimum number of identical tanks m
Solution
Test integer m in the tanks-in-series relation with the total HRT shared among the tanks (Fundamental principles: reactors).
kθtotal = 1.20 day−1 × 4.0 day = 4.8
m = 9: (1 + 4.8/9)−9 = 0.0213 > 0.02; m = 10: (1 + 4.8/10)−10 = 0.0198 ≤ 0.02
3 tanks uses the full 4.0-day HRT in every tank; the total HRT is shared, so each tank gets θtotal/m.
8 tanks gives (1 + 0.6)−8 = 0.0233, which misses the 0.02 target.
9 tanks gives (1.533)−9 = 0.0213, still above 0.02; rounding down the non-integer solution m ≈ 9.6 falls short.
(1 + 4.8/10)−10 = 0.0198 ≤ 0.02, while 9 tanks give 0.0213, so 10 is the minimum.
12 tanks meets the target (0.0175) but is not the minimum number.
Question 9 of 15 · Fundamental Principles · 7–11 items
A batch kinetics test shows a contaminant concentration falling from 120 mg/L to 48 mg/L in 6.0 hours under first-order decay. Determine the first-order rate constant k (hr−1) and then compute the PFR hydraulic residence time (hours) needed to achieve 98% removal using the extracted k and Cout/Cin=e−kθ.
Given
Initial concentration = 120 mg/L
Concentration after 6.0 h = 48 mg/L
Required removal = 98 %
Find
PFR hydraulic residence time (h)
Solution
Extract k from the batch test, then size the PFR (Fundamental principles: reactor design).
k = ln(C0/C)/t = ln(120/48)/6.0 hr = 0.1527 hr−1
θ = ln(Cin/Cout)/k = ln(50)/0.1527 = 25.6 hr
9.80 hr fits a zero-order rate (72 mg/L in 6 h = 12 mg/(L·hr)); the data were stated to follow first-order decay.
15.1 hr is the time for 90% removal (ln 10/k), not 98%.
k = ln(120/48)/6.0 = 0.1527 hr−1, and θ = ln(1/0.02)/k = 3.912/0.1527 = 25.6 hr.
59.0 hr finds k with log10 (0.0663) but θ with ln; the same logarithm base must be used in both steps.
321 hr uses the CSTR equation θ = (Cin/Cout − 1)/k; the question asks for a plug-flow reactor.
Question 10 of 15 · Fundamental Principles · 7–11 items
A pollutant concentration is 35 mg/L in a stream with a flow of 2.5 MGD (US gallons per day). Calculate the mass loading in kg/day.
Given
Concentration = 35 mg/L
Flow Q = 2.5 MGD
1 gal = 3.785 L
Find
Mass loading (kg/day)
Solution
Load = concentration × flow, with the flow converted to liters (Fundamental principles: mass balance).
Question 11 of 15 · Fundamental Principles · 7–11 items
A chlorine contact basin must achieve 95% inactivation of a pathogen that follows first-order (Chick's law) kinetics. Comparing a completely mixed basin with a long baffled basin that behaves as plug flow, at the same flow and rate constant, what is the ratio of the required volumes, VCSTR/VPFR?
Given
Required inactivation = 95%
First-order kinetics, same Q and k for both basins
Find
Volume ratio VCSTR/VPFR
Solution
Compare residence times at equal Q and k (Fundamental principles: reactor design).
CSTR: kτ = C0/C − 1 = 20 − 1 = 19
PFR: kτ = ln(C0/C) = ln 20 = 3.0
VCSTR/VPFR = 19/3.0 = 6.3
kτCSTR = 1/0.05 − 1 = 19 and kτPFR = ln(20) = 3.0, so VCSTR/VPFR = 19/3.0 = 6.3.
3.0 is kτ for the PFR (ln 20), not the volume ratio.
19 is kτ for the CSTR alone; it still has to be divided by the PFR's kτ.
1.0 assumes both reactors need the same volume; for any positive-order reaction the CSTR needs more.
0.16 is the inverse ratio, VPFR/VCSTR; the CSTR is the larger reactor.
Question 12 of 15 · Fundamental Principles · 7–11 items
A Continuous Stirred-Tank Reactor (CSTR) achieves 80% removal efficiency of a pollutant. If the flow rate to the reactor is tripled while the volume and kinetic constants remain unchanged, what is the new removal efficiency?
27% divides the old removal by three; removal does not scale linearly with residence time.
40% triples the fraction remaining (0.2 × 3 = 0.6), which is not how a CSTR responds.
80% removal means kτ = 4; tripling Q cuts τ to a third, so kτ = 1.33 and C/C0 = 1/2.33 = 0.43, i.e., 57% removal.
80% assumes efficiency does not depend on flow, but a shorter residence time lowers removal.
92% triples the residence time (kτ = 12); a higher flow through the same volume shortens it.
Question 13 of 15 · Fundamental Principles · 7–11 items
A 1,000 m3 lake receives a steady flow of 100 m3/day with a pollutant concentration of 10.0 mg/L. The pollutant decays with a first-order rate constant k = 0.05 day−1. Additionally, sediment releases the pollutant into the water column at a constant rate of 2,000 g/day. What is the steady-state concentration of the pollutant in the lake?
Given
Lake volume V = 1,000 m3
Inflow Q = 100 m3/day
Inflow concentration = 10.0 mg/L
Decay constant k = 0.05 day−1
Sediment release S = 2,000 g/day
Find
Steady-state lake concentration (mg/L)
Solution
Steady-state mass balance on a completely mixed lake with a source and first-order decay.
Question 14 of 15 · Fundamental Principles · 7–11 items
A completely mixed reactor receives two inflows: wastewater at Q = 2,000 m3/d with C = 100 mg/L and a sidestream return at 800 m3/d with C = 500 mg/L. The reactor volume is 3,000 m3, and the substance decays at a first-order rate constant k = 0.5 d−1. What is the steady-state effluent concentration?
Given
Wastewater inflow = 2,000 m3/d at 100 mg/L
Sidestream return = 800 m3/d at 500 mg/L
Reactor volume V = 3,000 m3
Rate constant k = 0.5 d−1
Find
Effluent concentration (mg/L)
Solution
Steady-state CSTR mass balance with two inflows and first-order decay (Fundamental principles: mass balance).
171.4 mg/L counts the return load but leaves its 800 m3/d out of the outflow.
214.3 mg/L is the blended inlet concentration with no first-order removal in the reactor.
Question 15 of 15 · Fundamental Principles · 7–11 items
A well-mixed lagoon with a volume of 500 m3 initially contains a pollutant concentration of 100 mg/L. Clean water flows in at 50 m3/h, and the effluent leaves at the same rate. The pollutant decays according to a first-order rate constant k = 0.05 hr−1. What is the pollutant concentration after 4 hours?
Given
Lagoon volume V = 500 m3
Initial concentration = 100 mg/L
Clean inflow Q = 50 m3/h
Decay rate k = 0.05 h−1
Time t = 4 h
Find
Concentration after 4 h (mg/L)
Solution
Unsteady mass balance on a mixed tank with clean inflow and first-order decay.
V·dC/dt = −QC − kVC → C = C0e−(Q/V + k)t
C = 100 mg/L × e−(0.10 + 0.05) × 4 = 54.9 mg/L
40.0 mg/L linearizes the decay as 100 × (1 − 0.6); the transient solution is exponential.
Washout and decay act together: (Q/V + k) = 0.10 + 0.05 = 0.15 hr−1, so C = 100 e−0.6 = 54.9 mg/L.
62.5 mg/L uses the steady-state CSTR form 100/(1 + 0.15 × 4); the lagoon is draining a slug, so the transient exponential applies.
67.0 mg/L counts washout (0.10 hr−1) but not decay.
81.9 mg/L counts decay (0.05 hr−1) but not washout by the clean inflow.
Answer every question to see your score
The one balance behind every reactor
Every reactor problem here starts from accumulation = in − out + generation. For a completely mixed volume with first-order decay:
V dC/dt = Q Cin − Q C − k C V
At steady state the left side is zero, which is not the same as equilibrium: the reaction keeps running, inflow just replaces what it removes. A plug-flow unit behaves like a batch reactor over its detention time V/Q.
Common traps
Average is not design. Population × per-capita use gives the average daily demand; peak and design flows come from the peaking factors a problem gives. Watch the units: L/cap·d and gal/cap·d differ by a factor of 3.785.
Corrosion and compatibility: galvanic couples and materials of construction are explained once, under FE Chemical process design.
What are the fundamental principles on the FE Environmental exam?
Population projections and demand, reactors, and materials science: 7–11 questions in all.
Are reactor problems the same as on FE Chemical?
No. Here a reactor is a treatment unit or a natural water body; FE Chemical treats it as a process vessel with conversion, selectivity and energy balances.
Reactor and demand problems on your phone
FE Environmental questions, answers and worked solutions in a study app.