FE Environmental groundwater, aquifers and cleanup
Groundwater, soils and sediments carries 8–12 questions: how water and contaminants move through the subsurface, how a pumped well draws an aquifer down, and how a site gets cleaned up.
Updated Oct 29, 2026 · Not affiliated with NCEES.
ExamFE Environmental
questions on the exam8–12
problems on this page16
areas in the specification15
Checked
What the area covers
Four headings, from the aquifer to the cleanup:
Basic hydrogeology: aquifer properties, soil characteristics, the subsurface.
Groundwater flow: Darcy's law, gradient, velocity, specific capacity, transport mechanisms.
Drawdown: the Dupuit, Thiem, Theis and Jacob solutions.
Remediation of soil, sediment and groundwater: recovery, in situ and ex situ treatment.
16 aquifer and cleanup problems
0 / 16 answered · 0 correct
Question 1 of 16 · Groundwater, Soils, and Sediments · 8–12 items
In groundwater modeling, the application of Darcy's Law assumes which of the following conditions for fluid flow?
Solution
Darcy's law, q = −K·dh/dl, is linear in the gradient and valid for laminar (viscous) flow at low pore Reynolds numbers (Groundwater: Darcy's law).
Turbulent flow in large fractures or karst breaks the linear Darcy relation; head loss grows faster than the flow.
Darcy's law in its usual groundwater form treats water as an incompressible liquid.
Darcy's law is a linear relation between flux and gradient, which holds only while viscous forces dominate (pore Reynolds number below about 1 to 10).
High-velocity flow near a well screen is where inertial (non-Darcy) losses appear and the law fails.
Question 2 of 16 · Groundwater, Soils, and Sediments · 8–12 items
An aquifer contaminant has a log Kow of 3.0. The matrix has foc = 0.02, ρb = 1.5 kg/L, and θ = 0.30. Using log Koc = log Kow − 0.21 and a groundwater velocity of 0.10 m/day, calculate the retarded contaminant velocity.
Given
log Kow = 3.0
Organic carbon fraction foc = 0.02
Bulk density ρb = 1.5 kg/L
Porosity θ = 0.30
log Koc = log Kow − 0.21
Groundwater velocity v = 0.10 m/day
Find
Retarded contaminant velocity (m/day)
Solution
Sorbing solute: Koc from the correlation, Kd = Kocfoc, then the retardation factor (Groundwater: contaminant transport).
0.0081 m/day divides by Kd = 12.3 as if it were R; the retardation factor is 1 + ρbKd/θ.
0.015 m/day multiplies by the porosity in the retardation factor (R = 1 + 1.5 × 12.3 × 0.30 = 6.5) instead of dividing by it.
0.078 m/day uses log Koc = 2.79 as Koc itself, so Kd and R come out far too small.
0.10 m/day is the groundwater velocity with no retardation at all.
Question 3 of 16 · Groundwater, Soils, and Sediments · 8–12 items
Using the Cooper–Jacob time-intercept method, estimate storativity S for a confined aquifer given T = 1.5 × 10−3 m2/s, time-intercept t0 = 200 s, and well separation r = 5 m.
S = (2.25 T t0)/r2
Given
Transmissivity T = 1.5 × 10−3 m2/s
Time intercept t0 = 200 s
Distance r = 5 m
Find
Storativity S
Solution
Cooper–Jacob time-intercept method (Groundwater: well hydraulics).
Question 4 of 16 · Groundwater, Soils, and Sediments · 8–12 items
Consider a conservative solute plume with longitudinal dispersivity αL = 8 m, pore velocity v=0.10 m/day, molecular diffusion Dm = 0.001 m2/day. Compute the plume standard deviation σ after t = 120 days using DL = αL v + Dm and σ2 = 2 DL t.
Given
Longitudinal dispersivity αL = 8 m
Pore velocity v = 0.10 m/day
Molecular diffusion Dm = 0.001 m2/day
Time t = 120 days
Find
Plume standard deviation σ (m)
Solution
Longitudinal spreading of a pulse (Groundwater: dispersion).
DL = αLv + Dm = 8 × 0.10 + 0.001 = 0.801 m2/day
σ = √(2DLt) = √(2 × 0.801 × 120) = 13.9 m
0.490 m uses only molecular diffusion (Dm = 0.001 m2/day); mechanical dispersion αLv dominates here.
43.8 m uses the dispersivity αL = 8 m as if it were DL in m2/day; it must be multiplied by the pore velocity.
192 is σ2 in m2; the square root still has to be taken.
Question 5 of 16 · Groundwater, Soils, and Sediments · 8–12 items
A proposed infiltration basin site has a hydraulic conductivity of K = 5.0 × 10−4 cm/s and a hydraulic gradient of 0.015. The soil has a total porosity of n = 0.35 and an effective porosity of ne = 0.25. Calculate the average linear (porewater) velocity of the groundwater.
Given
Hydraulic conductivity K = 5.0 × 10−4 cm/s
Hydraulic gradient i = 0.015
Total porosity n = 0.35
Effective porosity ne = 0.25
Find
Average linear velocity (cm/s)
Solution
Average linear (seepage) velocity = Darcy flux ÷ effective porosity (Groundwater: Darcy's law).
1.9 × 10−6 cm/s multiplies the Darcy flux by ne instead of dividing.
2.6 × 10−6 cm/s multiplies the Darcy flux by the total porosity 0.35 instead of dividing by the effective porosity.
7.5 × 10−6 cm/s is the Darcy flux Ki, the flow per total area, not the velocity in the pores.
2.1 × 10−5 cm/s divides by the total porosity 0.35; dead-end pores do not carry flow, so the effective porosity applies.
q = Ki = 5.0 × 10−4 × 0.015 = 7.5 × 10−6 cm/s, and v = q/ne = 7.5 × 10−6/0.25 = 3.0 × 10−5 cm/s.
Question 6 of 16 · Groundwater, Soils, and Sediments · 8–12 items
A contaminated site consists of three horizontal soil layers with the properties listed below. Calculate the equivalent vertical hydraulic conductivity (Kz) for flow perpendicular to the bedding planes.
Kz = Σb/Σ(b/K) = 8.0/(2,000 + 400,000 + 10,000) = 1.9 × 10−5 cm/s; the 4 m middle layer controls.
2.8 × 10−5 cm/s is an unweighted harmonic mean; each layer must be weighted by its thickness.
1.3 × 10−4 cm/s is the geometric mean, which describes random heterogeneity, not flow across layers in series.
3.0 × 10−4 cm/s is the thickness-weighted arithmetic mean, which applies to flow parallel to the layers.
1.0 × 10−3 cm/s takes the most permeable (top) layer; in series flow the tightest layer controls.
Question 7 of 16 · Groundwater, Soils, and Sediments · 8–12 items
Using the Green-Ampt model, calculate the infiltration rate (f) when the cumulative infiltration (F) is 5.0 cm. The soil has Ks = 2.0 cm/hr, wetting-front suction ψ = 12.0 cm, and moisture deficit Δθ = 0.25.
6.8 cm/hr leaves out the moisture deficit Δθ, using ψ/F = 2.4.
Question 8 of 16 · Groundwater, Soils, and Sediments · 8–12 items
Sample
Classification
d10 (mm)
A
Sandy Silt
0.02
B
Clean Fine Sand
0.15
C
Coarse Gravel
5.00
Which of the following soil samples is most appropriate for estimating hydraulic conductivity using Hazen's formula (K ≈ 100 d102), and what is the estimated K? (Note: d10 in the formula is in cm, K is in cm/s).
Given
Sample A sandy silt d10 = 0.02 mm
Sample B clean fine sand d10 = 0.15 mm
Sample C coarse gravel d10 = 5.00 mm
K ≈ 100·d102 with d10 in cm and K in cm/s
Find
Appropriate sample and K (cm/s)
Solution
Hazen: K ≈ 100·d102 (d10 in cm, K in cm/s), valid for clean sands.
K = 100 × (0.015 cm)2 = 0.0225 cm/s
The arithmetic is right for A (0.002 cm), but a sandy silt is outside the clean-sand range Hazen's formula was built for.
Hazen's relation was derived for clean, uniform sands (d10 about 0.1–3 mm); for B, d10 = 0.015 cm and K = 100 × 0.0152 = 0.0225 cm/s.
The right sample, but d10 was left in millimeters (100 × 0.152); the formula needs centimeters.
Gravel with d10 = 5.00 mm is coarser than the formula's range, and its flow may not even be laminar.
This picks the wrong soil and also leaves d10 in millimeters (100 × 0.022).
Question 9 of 16 · Groundwater, Soils, and Sediments · 8–12 items
A confined aquifer has a saturated thickness of b = 25 m and a hydraulic conductivity of K = 10 m/day. The hydraulic head at a monitoring well is h = 30 m. Calculate the groundwater flow rate (Q) through a section of width W = 100 m if the hydraulic gradient is 0.01.
Given
Saturated thickness b = 25 m
Hydraulic conductivity K = 10 m/day
Hydraulic head h = 30 m
Section width W = 100 m
Hydraulic gradient i = 0.01
Find
Groundwater flow rate Q (m3/day)
Solution
Darcy flow through a confined aquifer section (Groundwater: Darcy's law).
T = Kb = 10 m/day × 25 m = 250 m2/day
Q = TiW = 250 m2/day × 0.01 × 100 m = 250 m3/day
2.5 m3/day is the flow per meter of width (T·i); it still has to be multiplied by W = 100 m.
300 m3/day uses the 30 m head as the thickness; the head at one well says nothing about the flow area.
25,000 m3/day leaves out the hydraulic gradient of 0.01.
Question 10 of 16 · Groundwater, Soils, and Sediments · 8–12 items
Calculate the longitudinal dispersion coefficient DL given longitudinal dispersivity αL = 10 m, Darcy flux q = 0.03 m/day, and effective porosity ne = 0.25.
Given
Longitudinal dispersivity αL = 10 m
Darcy flux q = 0.03 m/day
Effective porosity ne = 0.25
Find
Longitudinal dispersion coefficient DL (m2/day)
Solution
Mechanical dispersion uses the seepage velocity (Groundwater: dispersion).
v = q/ne = 0.03/0.25 = 0.12 m/day
DL = αLv = 10 m × 0.12 m/day = 1.2 m2/day
0.012 m2/day divides the pore velocity by αL instead of multiplying.
0.075 m2/day multiplies the Darcy flux by ne instead of dividing.
0.30 m2/day multiplies αL by the Darcy flux; dispersion follows the pore velocity q/ne.
v = q/ne = 0.03/0.25 = 0.12 m/day, so DL = αLv = 10 × 0.12 = 1.2 m2/day.
40 m2/day divides αL by ne and forgets the flux.
Question 11 of 16 · Groundwater, Soils, and Sediments · 8–12 items
For a confined aquifer pumped at Q = 720 m3/day with transmissivity T = 200 m2/day, compute drawdown s at radius r = 0.1 m assuming an outer radius of influence R = 1000 m using the Thiem steady-state solution.
Given
Pumping rate Q = 720 m3/d
Transmissivity T = 200 m2/d
Radius r = 0.1 m
Radius of influence R = 1,000 m
Find
Drawdown s (m)
Solution
Thiem equation for steady drawdown in a confined aquifer (Groundwater: well hydraulics).
s = [Q/(2πT)]·ln(R/r) = [720 m3/day ÷ (2π × 200 m2/day)] × ln(1,000/0.1) = 5.28 m
33.2 m leaves out 2π entirely.
14.4 m drops the 2π and uses log10(R/r) = 4 instead of ln(R/r).
10.6 m divides by πT instead of 2πT.
9.21 is the value of ln(R/r) alone, not the drawdown.
s = [Q/(2πT)]·ln(R/r) = [720/(2π × 200)] × ln(10,000) = 0.573 × 9.21 = 5.28 m.
Question 12 of 16 · Groundwater, Soils, and Sediments · 8–12 items
In groundwater well hydraulics, which statement correctly defines the specific capacity of a pumped well?
Solution
Specific capacity = Q/s (well yield per unit drawdown), roughly proportional to T but time- and well-loss-dependent (Groundwater: well hydraulics).
Drawdown grows with pumping time and includes well losses, so specific capacity declines with time and depends on well efficiency.
T × b has units of m3/day and is not a well property; specific capacity is Q/s.
Transmissivity divided by aquifer thickness is the hydraulic conductivity, not specific capacity.
Q/s does not rise with drawdown; with turbulent well losses it falls as drawdown grows.
Specific capacity is Q/s, for example in gpm per foot of drawdown; it is often used to estimate transmissivity.
Question 13 of 16 · Groundwater, Soils, and Sediments · 8–12 items
Using the aquifer parameters below, compute the mass flux (kg/day) of a contaminant: mass flux = Q × C, where Q = K i A.
Parameter
Value
K
10 m/day
i
0.002
Cross-sectional area A
50 m2
Concentration C
120 mg/L
Given
Hydraulic conductivity K = 10 m/d
Hydraulic gradient i = 0.002
Cross-sectional area A = 50 m2
Concentration C = 120 mg/L
Find
Mass flux (kg/d)
Solution
Mass flux = Darcy discharge × concentration (Groundwater: contaminant transport).
Q = KiA = 10 m/day × 0.002 × 50 m2 = 1.0 m3/day
ṁ = QC = 1.0 m3/day × 0.12 kg/m3 = 0.12 kg/day
120 is the flux in g/day; it was not converted to kilograms.
0.0024 kg/day leaves out the 50 m2 flow area.
0.00012 kg/day divides by 1,000 one time too many; 120 mg/L is already 0.12 kg/m3.
60 kg/day leaves out the hydraulic gradient, as if the water moved at the full K.
Question 14 of 16 · Groundwater, Soils, and Sediments · 8–12 items
An LNAPL plume covers an area of 200 m2 with an average thickness of 2.0 m. The soil porosity is 0.30 and the average LNAPL saturation within the plume is 15% (Sn = 0.15). If the LNAPL density is 750 kg m−3, what is the estimated total mass of the LNAPL?
18,000 kg converts the 18 m3 with the density of water (1,000 kg/m3) instead of 750 kg/m3.
45,000 kg applies the 15% saturation to the bulk volume and leaves out the porosity.
90,000 kg assumes LNAPL fills all the pore space (100% saturation).
300,000 kg treats the whole 400 m3 of soil as LNAPL, ignoring porosity and saturation.
Question 15 of 16 · Groundwater, Soils, and Sediments · 8–12 items
Monitored Natural Attenuation (MNA) is proposed for a site. The distance to the property boundary is 500 ft. Hydraulic conductivity is 20 ft/day, gradient is 0.005, and porosity is 0.25. The contaminant has a first-order decay constant of k = 0.001 day−1. What is the calculated percent reduction in concentration by the time the plume reaches the boundary?
Given
Distance to boundary L = 500 ft
Hydraulic conductivity K = 20 ft/day
Hydraulic gradient i = 0.005
Porosity n = 0.25
Decay constant k = 0.001 day−1
Find
Percent reduction in concentration at the boundary
Solution
First-order decay over the advective travel time (Groundwater: natural attenuation).
v = Ki/n = (20 × 0.005)/0.25 = 0.40 ft/day; t = 500/0.40 = 1,250 days
2% uses K = 20 ft/day as the velocity, so the travel time is only 25 days.
29% is the fraction remaining, e−1.25; the question asks for the reduction.
39% uses the 500 ft distance as if it were 500 days of travel.
v = Ki/n = 20 × 0.005/0.25 = 0.40 ft/day, t = 500/0.40 = 1,250 days, C/C0 = e−1.25 = 0.29, so the reduction is 71%.
99% uses the Darcy flux Ki = 0.10 ft/day as the velocity, which quadruples the travel time.
Question 16 of 16 · Groundwater, Soils, and Sediments · 8–12 items
Layer
Thickness b (m)
Hydraulic Conductivity K (m/s)
Effective Porosity ne
Sand
2.0
1.0 × 10−4
0.25
Silty Sand
4.0
2.0 × 10−5
0.30
Silt
2.0
5.0 × 10−6
0.40
A vertical soil column consists of three saturated layers with properties listed in the table. If the vertical hydraulic gradient across the entire system is 0.05 m/m, what is the estimated time for a conservative non-reactive tracer to travel vertically through the entire 8-meter column?
Given
Sand: b = 2.0 m, K = 1.0 × 10−4 m/s, ne = 0.25
Silty sand: b = 4.0 m, K = 2.0 × 10−5 m/s, ne = 0.30
Silt: b = 2.0 m, K = 5.0 × 10−6 m/s, ne = 0.40
Vertical hydraulic gradient i = 0.05
Find
Tracer travel time through the 8 m column (days)
Solution
Series flow: one Darcy flux through all layers; travel time sums b·ne/q (Groundwater: layered media).
52 days applies the 0.05 gradient inside every layer; in series the flux is the same in each layer, while the gradient varies.
57 days applies the silt's porosity (0.40) to the whole 8 m column instead of each layer's own porosity.
116 days uses the silt's K for the whole column instead of the harmonic mean.
144 days divides the 8 m by the Darcy flux and ignores porosity; the tracer moves at q/ne in each layer.
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Common traps
Darcy flux is not the speed water moves. q = Ki is flow per unit of total cross-section. Water and a conservative tracer travel through the pores only, at q divided by the effective porosity, and a sorbing contaminant is slower still by its retardation factor.
Hydraulic conductivity is not permeability. K depends on the fluid as well as the soil; intrinsic permeability k (an area, m2 or darcys) belongs to the medium alone:
K = kρg/μ
Confined or unconfined decides the Thiem form. A confined aquifer uses transmissivity T = Kb and heads; an unconfined one uses the squares of the saturated thicknesses:
Q = 2πT (h2 − h1) / ln(r2/r1)
Q = πK (h22 − h12) / ln(r2/r1)
Both are steady-state. A drawdown that still grows with time calls for Theis or Jacob. Handbook search terms: Darcy, Thiem, Theis, transmissivity.
Remediation, sorted
In situ: pump and treat (hydraulic containment), soil vapor extraction for volatiles above the water table, air sparging below it, permeable reactive barriers, bioremediation, monitored natural attenuation.
Ex situ: excavation followed by soil washing, thermal desorption, incineration or disposal; treatment of pumped water above ground.
The choice turns on the contaminant's volatility, its partitioning to soil and the site's conductivity, which is why chemistry and hydrogeology meet in this area.