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10 FE practice problems: reactors and population projection, with solutions

By Dr. Mohamed Elansary, PhD Environmental EngineeringUpdated

These ten original problems practice reactors and population projection, a topic from the FE Environmental exam specification, using the Environmental Engineering, Population Projection Equations part of the FE Reference Handbook 10.6. Each problem gives the situation and the values with units. Work it with the handbook PDF open and commit to an answer before you open the solution. Every solution shows the handbook page, the equation, the substitution with units, a size check, and the mistake behind each wrong option, so a wrong pick tells you exactly what to fix.

How to use these problems

Give each problem an honest attempt before you open the solution: write the given values with units, find the equation in the FE Reference Handbook PDF, and commit to an answer. Then compare line by line. If you picked a wrong option, read the note for that option; each one names the mistake that produces it.

The problems

Problem 1 · FE Environmental, Reactors and population projection

A town's census counts were 41,200 in 2005 and 46,600 in 2015. Assuming geometric (exponential) growth, the projected population in 2040 is most nearly:

  • A 71,700 people
  • B 60,100 people
  • C 63,400 people
  • D 56,100 people

Handbook: Environmental Engineering, Population Projection Equations, FE Reference Handbook 10.6

Show the worked solution
Answer
C (63,400 people)
Given
P1 = 41,200, t1 = 2005, P2 = 46,600, t2 = 2015, t3 = 2040
Find
population in 2040
Handbook
Environmental Engineering, Population Projection Equations (geometric growth), page 334
Equation
Pt = P0 e^(kΔt), k = ln(P2/P1)/Δt
Substitute
  1. k = ln(P2015/P2005)/Δt = ln(46,600/41,200)/10 = 0.01232 per year
  2. P2040 = P2015 e^(kΔt) = 46,600 × e^(0.01232 × 25) = 63,400 people
Result
63,400 people, 3 significant figures
Check
geometric growth gives more than the linear projection (60,100) when the town is growing.
Why the others are wrong
  • A: used the time since 2005 with the 2015 count
  • B: used the linear (arithmetic) projection instead of geometric growth
  • D: projected from the 2005 count instead of 2015

Problem 2 · FE Environmental, Reactors and population projection

A contaminant enters an ideal completely mixed flow reactor (CMFR) at 395 mg/L and decays by a first-order reaction with k = 1.0/d. The hydraulic residence time is 6 h. At steady state, the effluent concentration is most nearly:

  • A 308 mg/L
  • B 316 mg/L
  • C 296 mg/L
  • D 79.0 mg/L

Handbook: Environmental Engineering, Steady-State Reactor Parameters, FE Reference Handbook 10.6

Show the worked solution
Answer
B (316 mg/L)
Given
C0 = 395 mg/L, k = 1.0/d, theta_h = 6 h
Find
effluent concentration (mg/L)
Handbook
Environmental Engineering, Steady-State Reactor Parameters (first order), page 331
Equation
Ct = Co/(1 + kθ) (first order, steady state)
Substitute
  1. θ = 6 h = 6/24 d = 0.2500 d; kθ = 1.0 × 0.2500 = 0.2500
  2. Ct = 395/(1 + 0.2500) = 316 mg/L
Result
316 mg/L, 3 significant figures
Check
for the same kθ the PFR removes more (308 mg/L left) than the CMFR (316 mg/L left).
Why the others are wrong
  • A: used the plug-flow equation
  • C: used Co(1 - kθ), a zero-order-like straight line
  • D: is the amount removed, not the effluent concentration

Problem 3 · FE Environmental, Reactors and population projection

A contaminant enters an ideal completely mixed flow reactor (CMFR) at 330 mg/L and decays by a first-order reaction with k = 2.3/d. The hydraulic residence time is 39 h. At steady state, the effluent concentration is most nearly:

  • A 14.7 mg/L
  • B 7.86 mg/L
  • C 69.7 mg/L
  • D 260 mg/L

Handbook: Environmental Engineering, Steady-State Reactor Parameters, FE Reference Handbook 10.6

Show the worked solution
Answer
C (69.7 mg/L)
Given
C0 = 330 mg/L, k = 2.3/d, theta_h = 39 h
Find
effluent concentration (mg/L)
Handbook
Environmental Engineering, Steady-State Reactor Parameters (first order), page 331
Equation
Ct = Co/(1 + kθ) (first order, steady state)
Substitute
  1. θ = 39 h = 39/24 d = 1.625 d; kθ = 2.3 × 1.625 = 3.737
  2. Ct = 330/(1 + 3.737) = 69.7 mg/L
Result
69.7 mg/L, 3 significant figures
Check
for the same kθ the PFR removes more (7.86 mg/L left) than the CMFR (69.7 mg/L left).
Why the others are wrong
  • A: squared the denominator (two tanks in series)
  • B: used the plug-flow equation
  • D: is the amount removed, not the effluent concentration

Problem 4 · FE Environmental, Reactors and population projection

A town's census counts were 10,500 in 2005 and 12,400 in 2020. Assuming geometric (exponential) growth, the projected population in 2030 is most nearly:

  • A 16,400 people
  • B 11,700 people
  • C 13,900 people
  • D 13,000 people

Handbook: Environmental Engineering, Population Projection Equations, FE Reference Handbook 10.6

Show the worked solution
Answer
C (13,900 people)
Given
P1 = 10,500, t1 = 2005, P2 = 12,400, t2 = 2020, t3 = 2030
Find
population in 2030
Handbook
Environmental Engineering, Population Projection Equations (geometric growth), page 334
Equation
Pt = P0 e^(kΔt), k = ln(P2/P1)/Δt
Substitute
  1. k = ln(P2020/P2005)/Δt = ln(12,400/10,500)/15 = 0.01109 per year
  2. P2030 = P2020 e^(kΔt) = 12,400 × e^(0.01109 × 10) = 13,900 people
Result
13,900 people, 3 significant figures
Check
geometric growth gives more than the linear projection (13,700) when the town is growing.
Why the others are wrong
  • A: used the time since 2005 with the 2020 count
  • B: projected from the 2005 count instead of 2020
  • D: found k with log10 but projected with e^(kΔt)

Problem 5 · FE Environmental, Reactors and population projection

A town's census counts were 184,300 in 1990 and 202,700 in 2000. Assuming geometric (exponential) growth, the projected population in 2025 is most nearly:

  • A 234,000 people
  • B 225,000 people
  • C 257,000 people
  • D 249,000 people

Handbook: Environmental Engineering, Population Projection Equations, FE Reference Handbook 10.6

Show the worked solution
Answer
C (257,000 people)
Given
P1 = 184,300, t1 = 1990, P2 = 202,700, t2 = 2000, t3 = 2025
Find
population in 2025
Handbook
Environmental Engineering, Population Projection Equations (geometric growth), page 334
Equation
Pt = P0 e^(kΔt), k = ln(P2/P1)/Δt
Substitute
  1. k = ln(P2000/P1990)/Δt = ln(202,700/184,300)/10 = 0.009516 per year
  2. P2025 = P2000 e^(kΔt) = 202,700 × e^(0.009516 × 25) = 257,000 people
Result
257,000 people, 3 significant figures
Check
geometric growth gives more than the linear projection (249,000) when the town is growing.
Why the others are wrong
  • A: projected from the 1990 count instead of 2000
  • B: found k with log10 but projected with e^(kΔt)
  • D: used the linear (arithmetic) projection instead of geometric growth

Problem 6 · FE Environmental, Reactors and population projection

A contaminant enters an ideal plug-flow reactor (PFR) at 300 mg/L and decays by a first-order reaction with k = 2.2/d. The hydraulic residence time is 7 h. At steady state, the effluent concentration is most nearly:

  • A 68.5 mg/L
  • B 183 mg/L
  • C 158 mg/L
  • D 142 mg/L

Handbook: Environmental Engineering, Steady-State Reactor Parameters, FE Reference Handbook 10.6

Show the worked solution
Answer
C (158 mg/L)
Given
C0 = 300 mg/L, k = 2.2/d, theta_h = 7 h
Find
effluent concentration (mg/L)
Handbook
Environmental Engineering, Steady-State Reactor Parameters (first order), page 331
Equation
Ct = Co e^(-kθ) (first order, steady state)
Substitute
  1. θ = 7 h = 7/24 d = 0.2917 d; kθ = 2.2 × 0.2917 = 0.6417
  2. Ct = 300 e^(-0.6417) = 158 mg/L
Result
158 mg/L, 3 significant figures
Check
for the same kθ the PFR removes more (158 mg/L left) than the CMFR (183 mg/L left).
Why the others are wrong
  • A: used base 10 instead of e
  • B: used the completely mixed equation
  • D: is the amount removed, not the effluent concentration

Problem 7 · FE Environmental, Reactors and population projection

A town's census counts were 36,900 in 1990 and 42,900 in 2000. Assuming geometric (exponential) growth, the projected population in 2015 is most nearly:

  • A 47,300 people
  • B 51,900 people
  • C 62,500 people
  • D 53,800 people

Handbook: Environmental Engineering, Population Projection Equations, FE Reference Handbook 10.6

Show the worked solution
Answer
D (53,800 people)
Given
P1 = 36,900, t1 = 1990, P2 = 42,900, t2 = 2000, t3 = 2015
Find
population in 2015
Handbook
Environmental Engineering, Population Projection Equations (geometric growth), page 334
Equation
Pt = P0 e^(kΔt), k = ln(P2/P1)/Δt
Substitute
  1. k = ln(P2000/P1990)/Δt = ln(42,900/36,900)/10 = 0.01507 per year
  2. P2015 = P2000 e^(kΔt) = 42,900 × e^(0.01507 × 15) = 53,800 people
Result
53,800 people, 3 significant figures
Check
geometric growth gives more than the linear projection (51,900) when the town is growing.
Why the others are wrong
  • A: found k with log10 but projected with e^(kΔt)
  • B: used the linear (arithmetic) projection instead of geometric growth
  • C: used the time since 1990 with the 2000 count

Problem 8 · FE Environmental, Reactors and population projection

A town's census counts were 10,300 in 1990 and 11,500 in 2005. Assuming geometric (exponential) growth, the projected population in 2020 is most nearly:

  • A 12,100 people
  • B 14,300 people
  • C 11,500 people
  • D 12,800 people

Handbook: Environmental Engineering, Population Projection Equations, FE Reference Handbook 10.6

Show the worked solution
Answer
D (12,800 people)
Given
P1 = 10,300, t1 = 1990, P2 = 11,500, t2 = 2005, t3 = 2020
Find
population in 2020
Handbook
Environmental Engineering, Population Projection Equations (geometric growth), page 334
Equation
Pt = P0 e^(kΔt), k = ln(P2/P1)/Δt
Substitute
  1. k = ln(P2005/P1990)/Δt = ln(11,500/10,300)/15 = 0.007347 per year
  2. P2020 = P2005 e^(kΔt) = 11,500 × e^(0.007347 × 15) = 12,800 people
Result
12,800 people, 3 significant figures
Check
geometric growth gives more than the linear projection (12,700) when the town is growing.
Why the others are wrong
  • A: found k with log10 but projected with e^(kΔt)
  • B: used the time since 1990 with the 2005 count
  • C: projected from the 1990 count instead of 2005

Problem 9 · FE Environmental, Reactors and population projection

A contaminant enters an ideal completely mixed flow reactor (CMFR) at 310 mg/L and decays by a first-order reaction with k = 6.0/d. The hydraulic residence time is 39 h. At steady state, the effluent concentration is most nearly:

  • A 2.68 mg/L
  • B 1.32 mg/L
  • C 281 mg/L
  • D 28.8 mg/L

Handbook: Environmental Engineering, Steady-State Reactor Parameters, FE Reference Handbook 10.6

Show the worked solution
Answer
D (28.8 mg/L)
Given
C0 = 310 mg/L, k = 6.0/d, theta_h = 39 h
Find
effluent concentration (mg/L)
Handbook
Environmental Engineering, Steady-State Reactor Parameters (first order), page 331
Equation
Ct = Co/(1 + kθ) (first order, steady state)
Substitute
  1. θ = 39 h = 39/24 d = 1.625 d; kθ = 6.0 × 1.625 = 9.750
  2. Ct = 310/(1 + 9.750) = 28.8 mg/L
Result
28.8 mg/L, 3 significant figures
Check
for the same kθ the PFR removes more (0.0181 mg/L left) than the CMFR (28.8 mg/L left).
Why the others are wrong
  • A: squared the denominator (two tanks in series)
  • B: used θ in hours with k per day
  • C: is the amount removed, not the effluent concentration

Problem 10 · FE Environmental, Reactors and population projection

A contaminant enters an ideal completely mixed flow reactor (CMFR) at 210 mg/L and decays by a first-order reaction with k = 5.5/d. The hydraulic residence time is 6 h. At steady state, the effluent concentration is most nearly:

  • A 53.1 mg/L
  • B 6.18 mg/L
  • C 122 mg/L
  • D 88.4 mg/L

Handbook: Environmental Engineering, Steady-State Reactor Parameters, FE Reference Handbook 10.6

Show the worked solution
Answer
D (88.4 mg/L)
Given
C0 = 210 mg/L, k = 5.5/d, theta_h = 6 h
Find
effluent concentration (mg/L)
Handbook
Environmental Engineering, Steady-State Reactor Parameters (first order), page 331
Equation
Ct = Co/(1 + kθ) (first order, steady state)
Substitute
  1. θ = 6 h = 6/24 d = 0.2500 d; kθ = 5.5 × 0.2500 = 1.375
  2. Ct = 210/(1 + 1.375) = 88.4 mg/L
Result
88.4 mg/L, 3 significant figures
Check
for the same kθ the PFR removes more (53.1 mg/L left) than the CMFR (88.4 mg/L left).
Why the others are wrong
  • A: used the plug-flow equation
  • B: used θ in hours with k per day
  • C: is the amount removed, not the effluent concentration

A new problem posts every day inside the lab, with the full worked solution the same evening.

Frequently asked questions

Where is reactors and population projection in the FE Reference Handbook?

Look in the Environmental Engineering, Population Projection Equations part of FE Reference Handbook 10.6. Each solution gives the exact page, so you can practice finding it in the PDF the way you will on exam day.

Are these real FE exam questions?

No. They are original problems generated from the handbook formulas and checked by code. Real exam questions are confidential, and sharing them breaks the NCEES agreement every examinee accepts.

How do I check my answer before opening the solution?

Check the units of your result and whether its size makes sense for the situation. Each solution ends with the same kind of check, so you can compare your habit with ours.

Sources

  1. NCEES FE Environmental CBT exam specifications (PDF). Retrieved October 3, 2026.
  2. NCEES FE Reference Handbook 10.6 (free PDF in MyNCEES). Retrieved October 3, 2026.