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10 FE practice problems: groundwater: Darcy and wells, with solutions

By Dr. Mohamed Elansary, PhD Environmental EngineeringUpdated

These ten original problems practice groundwater: Darcy and wells, a topic from the FE Environmental exam specification, using the Civil Engineering, Groundwater, Thiem Equation 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, Groundwater: Darcy and wells

A well fully penetrates a confined aquifer 27 m thick with K = 26 m/d. At steady pumping, observation wells 10 m and 200 m from the well show piezometric heads of 20.0 m and 21.2 m above the aquifer bottom. Find the pumping rate, in m³/d.

Enter your answer in m³/d, 3 significant figures.

Handbook: Civil Engineering, Groundwater, Thiem Equation, FE Reference Handbook 10.6

Show the worked solution
Answer
1,770 m³/d
Given
K = 26 m/d, b = 27 m, r1 = 10 m, r2 = 200 m, h1 = 20.0 m, h2 = 21.2 m
Find
pumping rate Q (m³/d)
Handbook
Civil Engineering, Groundwater, Thiem Equation (confined aquifer), page 299
Equation
Q = 2πT(h2 - h1)/ln(r2/r1), T = Kb
Substitute
  1. T = Kb = 26 × 27 = 702 m²/d
  2. Q = 2πT(h2 - h1)/ln(r2/r1) = 2π(702)(21.2 - 20.0)/ln(200/10) = 1,770 m³/d
Result
1,770 m³/d, 3 significant figures
Check
heads rise with distance from the pumping well; the natural log of the radius ratio sits in the denominator.
Common wrong answers
  • 883 m³/d: left out the 2 in 2πT
  • 1,350 m³/d: used the unconfined (Dupuit) formula for a confined aquifer
  • 4,070 m³/d: used log10 instead of the natural log

Problem 2 · FE Environmental, Groundwater: Darcy and wells

A well pumps an unconfined aquifer with K = 6 m/d at steady state. Observation wells 30 m and 80 m from the well show water levels of 20.1 m and 24.2 m above the impermeable base. Find the pumping rate, in m³/d.

Enter your answer in m³/d, 3 significant figures.

Handbook: Civil Engineering, Groundwater, Dupuit's Formula, FE Reference Handbook 10.6

Show the worked solution
Answer
3,490 m³/d
Given
K = 6 m/d, r1 = 30 m, r2 = 80 m, h1 = 20.1 m, h2 = 24.2 m
Find
pumping rate Q (m³/d)
Handbook
Civil Engineering, Groundwater, Dupuit's Formula (unconfined aquifer), page 299
Equation
Q = πK(h2² - h1²)/ln(r2/r1)
Substitute
  1. Q = πK(h2² - h1²)/ln(r2/r1) = π(6)(24.2² - 20.1²)/ln(80/30) = π(6)(181.6)/0.9808 = 3,490 m³/d
Result
3,490 m³/d, 3 significant figures
Check
heads rise with distance from the pumping well; the natural log of the radius ratio sits in the denominator.
Common wrong answers
  • 6,980 m³/d: used 2π (from the confined formula)
  • 323 m³/d: squared the difference instead of differencing the squares
  • 8,040 m³/d: used log10 instead of the natural log

Problem 3 · FE Environmental, Groundwater: Darcy and wells

Two observation wells 1,150 m apart along the flow direction in a sand aquifer show a head difference of 1.5 m. The aquifer is 14 m thick and 140 m wide, with K = 11.5 m/d and effective porosity 0.15. Find the groundwater discharge through the section, in m³/d.

Enter your answer in m³/d, 3 significant figures.

Handbook: Civil Engineering, Darcy's Law, FE Reference Handbook 10.6

Show the worked solution
Answer
29.4 m³/d
Given
K = 11.5 m/d, dh = 1.5 m, L = 1,150 m, n = 0.15, A = 1,960, width = 140 m, thick = 14 m
Find
discharge Q (m³/d)
Handbook
Civil Engineering, Groundwater, Darcy's Law, page 298
Equation
Q = -KA dh/dx (Darcy's law)
Substitute
  1. i = dh/dx = 1.5/1,150 = 0.001304; A = 140 m × 14 m = 1,960 m²
  2. Q = K·i·A = (11.5 m/d)(0.001304)(1,960 m²) = 29.4 m³/d
Result
29.4 m³/d, 3 significant figures
Check
the seepage velocity is faster than the Darcy velocity because water moves only through the pores (v = q/n).
Common wrong answers
  • 196 m³/d: divided by the porosity (that gives a seepage-velocity form, not the discharge)
  • 2.10 m³/d: used the aquifer width instead of the flow area

Problem 4 · FE Environmental, Groundwater: Darcy and wells

Two observation wells 250 m apart along the flow direction in a sand aquifer show a head difference of 3.3 m. The aquifer is 20 m thick and 300 m wide, with K = 37.5 m/d and effective porosity 0.40. Find the groundwater discharge through the section, in m³/d.

Enter your answer in m³/d, 3 significant figures.

Handbook: Civil Engineering, Darcy's Law, FE Reference Handbook 10.6

Show the worked solution
Answer
2,970 m³/d
Given
K = 37.5 m/d, dh = 3.3 m, L = 250 m, n = 0.40, A = 6,000, width = 300 m, thick = 20 m
Find
discharge Q (m³/d)
Handbook
Civil Engineering, Groundwater, Darcy's Law, page 298
Equation
Q = -KA dh/dx (Darcy's law)
Substitute
  1. i = dh/dx = 3.3/250 = 0.01320; A = 300 m × 20 m = 6,000 m²
  2. Q = K·i·A = (37.5 m/d)(0.01320)(6,000 m²) = 2,970 m³/d
Result
2,970 m³/d, 3 significant figures
Check
the seepage velocity is faster than the Darcy velocity because water moves only through the pores (v = q/n).
Common wrong answers
  • 149 m³/d: used the aquifer width instead of the flow area
  • 7,430 m³/d: divided by the porosity (that gives a seepage-velocity form, not the discharge)

Problem 5 · FE Environmental, Groundwater: Darcy and wells

Two observation wells 1,600 m apart along the flow direction in a sand aquifer show a head difference of 8.8 m. The aquifer is 37 m thick and 90 m wide, with K = 17.5 m/d and effective porosity 0.38. Find the average seepage velocity of the groundwater, in m/d.

Enter your answer in m/d, 3 significant figures.

Handbook: Civil Engineering, Darcy's Law, FE Reference Handbook 10.6

Show the worked solution
Answer
0.253 m/d
Given
K = 17.5 m/d, dh = 8.8 m, L = 1,600 m, n = 0.38, A = 3,330, width = 90 m, thick = 37 m
Find
average seepage velocity (m/d)
Handbook
Civil Engineering, Groundwater, Darcy's Law, page 298
Equation
q = -K dh/dx; v = q/n
Substitute
  1. i = dh/dx = 8.8/1,600 = 0.005500; A = 90 m × 37 m = 3,330 m²
  2. v = q/n = K·i/n = (17.5 m/d)(0.005500)/0.38 = 0.253 m/d
Result
0.253 m/d, 3 significant figures
Check
the seepage velocity is faster than the Darcy velocity because water moves only through the pores (v = q/n).
Common wrong answers
  • 0.0963 m/d: is the Darcy (specific) discharge q, not the seepage velocity
  • 0.0366 m/d: multiplied by the porosity instead of dividing

Problem 6 · FE Environmental, Groundwater: Darcy and wells

A well fully penetrates a confined aquifer 36 m thick with K = 13 m/d. At steady pumping, observation wells 30 m and 60 m from the well show piezometric heads of 40.1 m and 40.9 m above the aquifer bottom. Find the pumping rate, in m³/d.

Enter your answer in m³/d, 3 significant figures.

Handbook: Civil Engineering, Groundwater, Thiem Equation, FE Reference Handbook 10.6

Show the worked solution
Answer
3,390 m³/d
Given
K = 13 m/d, b = 36 m, r1 = 30 m, r2 = 60 m, h1 = 40.1 m, h2 = 40.9 m
Find
pumping rate Q (m³/d)
Handbook
Civil Engineering, Groundwater, Thiem Equation (confined aquifer), page 299
Equation
Q = 2πT(h2 - h1)/ln(r2/r1), T = Kb
Substitute
  1. T = Kb = 13 × 36 = 468 m²/d
  2. Q = 2πT(h2 - h1)/ln(r2/r1) = 2π(468)(40.9 - 40.1)/ln(60/30) = 3,390 m³/d
Result
3,390 m³/d, 3 significant figures
Check
heads rise with distance from the pumping well; the natural log of the radius ratio sits in the denominator.
Common wrong answers
  • 7,810 m³/d: used log10 instead of the natural log
  • 1,700 m³/d: left out the 2 in 2πT
  • 3,820 m³/d: used the unconfined (Dupuit) formula for a confined aquifer

Problem 7 · FE Environmental, Groundwater: Darcy and wells

Two observation wells 200 m apart along the flow direction in a sand aquifer show a head difference of 5.3 m. The aquifer is 39 m thick and 370 m wide, with K = 37.5 m/d and effective porosity 0.33. Find the average seepage velocity of the groundwater, in m/d.

Enter your answer in m/d, 3 significant figures.

Handbook: Civil Engineering, Darcy's Law, FE Reference Handbook 10.6

Show the worked solution
Answer
3.01 m/d
Given
K = 37.5 m/d, dh = 5.3 m, L = 200 m, n = 0.33, A = 14,430, width = 370 m, thick = 39 m
Find
average seepage velocity (m/d)
Handbook
Civil Engineering, Groundwater, Darcy's Law, page 298
Equation
q = -K dh/dx; v = q/n
Substitute
  1. i = dh/dx = 5.3/200 = 0.02650; A = 370 m × 39 m = 14,430 m²
  2. v = q/n = K·i/n = (37.5 m/d)(0.02650)/0.33 = 3.01 m/d
Result
3.01 m/d, 3 significant figures
Check
the seepage velocity is faster than the Darcy velocity because water moves only through the pores (v = q/n).
Common wrong answers
  • 0.328 m/d: multiplied by the porosity instead of dividing
  • 0.994 m/d: is the Darcy (specific) discharge q, not the seepage velocity
  • 114 m/d: left out the hydraulic gradient

Problem 8 · FE Environmental, Groundwater: Darcy and wells

Two observation wells 300 m apart along the flow direction in a sand aquifer show a head difference of 7.4 m. The aquifer is 9 m thick and 70 m wide, with K = 31 m/d and effective porosity 0.40. Find the groundwater discharge through the section, in m³/d.

Enter your answer in m³/d, 3 significant figures.

Handbook: Civil Engineering, Darcy's Law, FE Reference Handbook 10.6

Show the worked solution
Answer
482 m³/d
Given
K = 31 m/d, dh = 7.4 m, L = 300 m, n = 0.40, A = 630, width = 70 m, thick = 9 m
Find
discharge Q (m³/d)
Handbook
Civil Engineering, Groundwater, Darcy's Law, page 298
Equation
Q = -KA dh/dx (Darcy's law)
Substitute
  1. i = dh/dx = 7.4/300 = 0.02467; A = 70 m × 9 m = 630 m²
  2. Q = K·i·A = (31 m/d)(0.02467)(630 m²) = 482 m³/d
Result
482 m³/d, 3 significant figures
Check
the seepage velocity is faster than the Darcy velocity because water moves only through the pores (v = q/n).
Common wrong answers
  • 1,200 m³/d: divided by the porosity (that gives a seepage-velocity form, not the discharge)
  • 53.5 m³/d: used the aquifer width instead of the flow area

Problem 9 · FE Environmental, Groundwater: Darcy and wells

A well pumps an unconfined aquifer with K = 27 m/d at steady state. Observation wells 10 m and 60 m from the well show water levels of 16.1 m and 20.3 m above the impermeable base. Find the pumping rate, in m³/d.

Enter your answer in m³/d, 3 significant figures.

Handbook: Civil Engineering, Groundwater, Dupuit's Formula, FE Reference Handbook 10.6

Show the worked solution
Answer
7,240 m³/d
Given
K = 27 m/d, r1 = 10 m, r2 = 60 m, h1 = 16.1 m, h2 = 20.3 m
Find
pumping rate Q (m³/d)
Handbook
Civil Engineering, Groundwater, Dupuit's Formula (unconfined aquifer), page 299
Equation
Q = πK(h2² - h1²)/ln(r2/r1)
Substitute
  1. Q = πK(h2² - h1²)/ln(r2/r1) = π(27)(20.3² - 16.1²)/ln(60/10) = π(27)(152.9)/1.792 = 7,240 m³/d
Result
7,240 m³/d, 3 significant figures
Check
heads rise with distance from the pumping well; the natural log of the radius ratio sits in the denominator.
Common wrong answers
  • 16,700 m³/d: used log10 instead of the natural log
  • 14,500 m³/d: used 2π (from the confined formula)
  • 835 m³/d: squared the difference instead of differencing the squares

Problem 10 · FE Environmental, Groundwater: Darcy and wells

A well fully penetrates a confined aquifer 24 m thick with K = 69 m/d. At steady pumping, observation wells 25 m and 200 m from the well show piezometric heads of 35.0 m and 36.5 m above the aquifer bottom. Find the pumping rate, in m³/d.

Enter your answer in m³/d, 3 significant figures.

Handbook: Civil Engineering, Groundwater, Thiem Equation, FE Reference Handbook 10.6

Show the worked solution
Answer
7,510 m³/d
Given
K = 69 m/d, b = 24 m, r1 = 25 m, r2 = 200 m, h1 = 35.0 m, h2 = 36.5 m
Find
pumping rate Q (m³/d)
Handbook
Civil Engineering, Groundwater, Thiem Equation (confined aquifer), page 299
Equation
Q = 2πT(h2 - h1)/ln(r2/r1), T = Kb
Substitute
  1. T = Kb = 69 × 24 = 1656 m²/d
  2. Q = 2πT(h2 - h1)/ln(r2/r1) = 2π(1656)(36.5 - 35.0)/ln(200/25) = 7,510 m³/d
Result
7,510 m³/d, 3 significant figures
Check
heads rise with distance from the pumping well; the natural log of the radius ratio sits in the denominator.
Common wrong answers
  • 17,300 m³/d: used log10 instead of the natural log
  • 11,200 m³/d: used the unconfined (Dupuit) formula for a confined aquifer
  • 3,750 m³/d: left out the 2 in 2πT

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

Frequently asked questions

Where is groundwater: Darcy and wells in the FE Reference Handbook?

Look in the Civil Engineering, Groundwater, Thiem Equation 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.