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10 FE practice problems: pipe flow head loss (Darcy-Weisbach), with solutions

By Dr. Mohamed Elansary, PhD Environmental EngineeringUpdated

These ten problems drill head loss in full pipes with the Darcy-Weisbach equation, a staple of the Fluid Mechanics areas in the FE Civil, Mechanical, Environmental, and Other Disciplines specifications. Each problem gives the pipe, the flow, and the friction factor or the data to find it, and asks for a head loss or a related quantity. Work each one with the FE Reference Handbook open, then open the solution to check the equation, the substitution with units, and the named mistake behind each wrong option.

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, Pipe flow, Darcy-Weisbach

Water flows at 0.0150 m³/s through a 150 mm diameter pipe that is 90 m long. The Darcy friction factor for the pipe is 0.035. The head loss due to friction over the pipe length is most nearly:

  • A 12.3 m
  • B 0.235 m
  • C 3.09 m
  • D 0.771 m

Handbook: Fluid Mechanics, Head Loss Due to Flow (Darcy-Weisbach equation), FE Reference Handbook 10.6

Show the worked solution
Answer
D (0.771 m)
Given
Q = 0.0150 m³/s, D = 150 mm, L = 90 m, f = 0.035, g = 9.807 m/s² (handbook)
Find
head loss due to friction, hf (m)
Handbook
Fluid Mechanics, Head Loss Due to Flow (Darcy-Weisbach equation), page 187
Equation
hf = f(L/D)V²/(2g), with V = Q/A and A = πD²/4
Substitute
  1. D = 150 mm = 0.150 m
  2. A = πD²/4 = π(0.150 m)²/4 = 0.01767 m²
  3. V = Q/A = (0.0150 m³/s)/(0.01767 m²) = 0.8488 m/s
  4. hf = f(L/D)V²/(2g) = 0.035 × (90 m/0.150 m) × (0.8488 m/s)²/(2 × 9.807 m/s²) = 0.771 m
Result
0.771 m, 3 significant figures
Check
units (m/m)(m/s)²/(m/s²) = m; V = 0.849 m/s is an ordinary velocity for water in a pipe, and the loss is 8.57 m per 1,000 m of pipe.
Why the others are wrong
  • A: put the radius into A = πD²/4, so V is 4 times too large and the loss is 16 times too large
  • B: used g = 32.174 ft/sec² in an SI problem
  • C: multiplied the given Darcy f by 4 as if it were a Fanning factor; the Darcy f goes into hf = f(L/D)V²/(2g) as given

Problem 2 · FE, Pipe flow, Darcy-Weisbach

Water flows at 0.0659 m³/s through a 300 mm diameter pipe that is 270 m long. The Darcy friction factor for the pipe is 0.015. The head loss due to friction over the pipe length is most nearly:

  • A 0.642 m
  • B 1.20 m
  • C 0.0374 m
  • D 0.598 m

Handbook: Fluid Mechanics, Head Loss Due to Flow (Darcy-Weisbach equation), FE Reference Handbook 10.6

Show the worked solution
Answer
D (0.598 m)
Given
Q = 0.0659 m³/s, D = 300 mm, L = 270 m, f = 0.015, g = 9.807 m/s² (handbook)
Find
head loss due to friction, hf (m)
Handbook
Fluid Mechanics, Head Loss Due to Flow (Darcy-Weisbach equation), page 187
Equation
hf = f(L/D)V²/(2g), with V = Q/A and A = πD²/4
Substitute
  1. D = 300 mm = 0.300 m
  2. A = πD²/4 = π(0.300 m)²/4 = 0.07069 m²
  3. V = Q/A = (0.0659 m³/s)/(0.07069 m²) = 0.9323 m/s
  4. hf = f(L/D)V²/(2g) = 0.015 × (270 m/0.300 m) × (0.9323 m/s)²/(2 × 9.807 m/s²) = 0.598 m
Result
0.598 m, 3 significant figures
Check
units (m/m)(m/s)²/(m/s²) = m; V = 0.932 m/s is an ordinary velocity for water in a pipe, and the loss is 2.22 m per 1,000 m of pipe.
Why the others are wrong
  • A: did not square the velocity
  • B: used the radius instead of the diameter in L/D, which doubles the loss
  • C: used A = πD² without the 1/4, so V is 4 times too small and the loss is 16 times too small

Problem 3 · FE, Pipe flow, Darcy-Weisbach

Oil with a kinematic viscosity of 1.6 × 10⁻⁴ m²/s flows at an average velocity of 1.80 m/s in a 25 mm diameter pipe that is 25 m long. The head loss due to friction over the pipe length is most nearly:

  • A 75.2 m
  • B 37.6 m
  • C 150 m
  • D 9.40 m

Handbook: Fluid Mechanics, Reynolds Number and Head Loss Due to Flow (Darcy-Weisbach equation), FE Reference Handbook 10.6

Show the worked solution
Answer
B (37.6 m)
Given
ν = 1.6 × 10⁻⁴ m²/s, V = 1.80 m/s, D = 25 mm, L = 25 m, g = 9.807 m/s² (handbook)
Find
head loss due to friction, hf (m)
Handbook
Fluid Mechanics, Reynolds Number; Head Loss Due to Flow (Darcy-Weisbach equation); Moody, Darcy, or Stanton Friction Factor Diagram (laminar line), pages 186, 187 and 206
Equation
Re = VD/ν; laminar flow (Re < 2,100): f = 64/Re; hf = f(L/D)V²/(2g)
Substitute
  1. D = 25 mm = 0.025 m
  2. Re = VD/ν = (1.80 m/s)(0.025 m)/(1.6 × 10⁻⁴ m²/s) = 281.3, below 2,100, so the flow is laminar
  3. f = 64/Re = 64/281.3 = 0.2276
  4. hf = f(L/D)V²/(2g) = 0.2276 × (25 m/0.025 m) × (1.80 m/s)²/(2 × 9.807 m/s²) = 37.6 m
Result
37.6 m, 3 significant figures
Check
Re = 281 is below 2,100, so the laminar f = 64/Re applies (no Moody chart reading needed); units (m/m)(m/s)²/(m/s²) = m.
Why the others are wrong
  • A: used the radius instead of the diameter in Re, which halves Re and doubles f
  • C: used the radius instead of the diameter in both Re and L/D
  • D: used f = 16/Re, the Fanning laminar factor, instead of the Darcy f = 64/Re

Problem 4 · FE, Pipe flow, Darcy-Weisbach

Oil with a kinematic viscosity of 2.1 × 10⁻⁴ m²/s flows at an average velocity of 1.30 m/s in a 100 mm diameter pipe that is 30 m long. The head loss due to friction over the pipe length is most nearly:

  • A 2.67 m
  • B 0.815 m
  • C 5.34 m
  • D 2.06 m

Handbook: Fluid Mechanics, Reynolds Number and Head Loss Due to Flow (Darcy-Weisbach equation), FE Reference Handbook 10.6

Show the worked solution
Answer
A (2.67 m)
Given
ν = 2.1 × 10⁻⁴ m²/s, V = 1.30 m/s, D = 100 mm, L = 30 m, g = 9.807 m/s² (handbook)
Find
head loss due to friction, hf (m)
Handbook
Fluid Mechanics, Reynolds Number; Head Loss Due to Flow (Darcy-Weisbach equation); Moody, Darcy, or Stanton Friction Factor Diagram (laminar line), pages 186, 187 and 206
Equation
Re = VD/ν; laminar flow (Re < 2,100): f = 64/Re; hf = f(L/D)V²/(2g)
Substitute
  1. D = 100 mm = 0.100 m
  2. Re = VD/ν = (1.30 m/s)(0.100 m)/(2.1 × 10⁻⁴ m²/s) = 619.0, below 2,100, so the flow is laminar
  3. f = 64/Re = 64/619.0 = 0.1034
  4. hf = f(L/D)V²/(2g) = 0.1034 × (30 m/0.100 m) × (1.30 m/s)²/(2 × 9.807 m/s²) = 2.67 m
Result
2.67 m, 3 significant figures
Check
Re = 619 is below 2,100, so the laminar f = 64/Re applies (no Moody chart reading needed); units (m/m)(m/s)²/(m/s²) = m.
Why the others are wrong
  • B: used g = 32.174 ft/sec² in an SI problem
  • C: used the radius instead of the diameter in Re, which halves Re and doubles f
  • D: did not square the velocity

Problem 5 · FE, Pipe flow, Darcy-Weisbach

Oil with a kinematic viscosity of 3.2 × 10⁻⁴ m²/s flows at an average velocity of 0.70 m/s in a 100 mm diameter pipe that is 130 m long. The head loss due to friction over the pipe length is most nearly:

  • A 19.0 m
  • B 2.38 m
  • C 2.90 m
  • D 9.50 m

Handbook: Fluid Mechanics, Reynolds Number and Head Loss Due to Flow (Darcy-Weisbach equation), FE Reference Handbook 10.6

Show the worked solution
Answer
D (9.50 m)
Given
ν = 3.2 × 10⁻⁴ m²/s, V = 0.70 m/s, D = 100 mm, L = 130 m, g = 9.807 m/s² (handbook)
Find
head loss due to friction, hf (m)
Handbook
Fluid Mechanics, Reynolds Number; Head Loss Due to Flow (Darcy-Weisbach equation); Moody, Darcy, or Stanton Friction Factor Diagram (laminar line), pages 186, 187 and 206
Equation
Re = VD/ν; laminar flow (Re < 2,100): f = 64/Re; hf = f(L/D)V²/(2g)
Substitute
  1. D = 100 mm = 0.100 m
  2. Re = VD/ν = (0.70 m/s)(0.100 m)/(3.2 × 10⁻⁴ m²/s) = 218.7, below 2,100, so the flow is laminar
  3. f = 64/Re = 64/218.7 = 0.2926
  4. hf = f(L/D)V²/(2g) = 0.2926 × (130 m/0.100 m) × (0.70 m/s)²/(2 × 9.807 m/s²) = 9.50 m
Result
9.50 m, 3 significant figures
Check
Re = 219 is below 2,100, so the laminar f = 64/Re applies (no Moody chart reading needed); units (m/m)(m/s)²/(m/s²) = m.
Why the others are wrong
  • A: left out the 2 in 2g, which doubles the loss
  • B: used f = 16/Re, the Fanning laminar factor, instead of the Darcy f = 64/Re
  • C: used g = 32.174 ft/sec² in an SI problem

Problem 6 · FE, Pipe flow, Darcy-Weisbach

Oil with a kinematic viscosity of 1.9 × 10⁻⁴ m²/s flows at an average velocity of 0.90 m/s in a 50 mm diameter pipe that is 75 m long. The head loss due to friction over the pipe length is most nearly:

  • A 67.0 m
  • B 16.7 m
  • C 18.6 m
  • D 4.18 m

Handbook: Fluid Mechanics, Reynolds Number and Head Loss Due to Flow (Darcy-Weisbach equation), FE Reference Handbook 10.6

Show the worked solution
Answer
B (16.7 m)
Given
ν = 1.9 × 10⁻⁴ m²/s, V = 0.90 m/s, D = 50 mm, L = 75 m, g = 9.807 m/s² (handbook)
Find
head loss due to friction, hf (m)
Handbook
Fluid Mechanics, Reynolds Number; Head Loss Due to Flow (Darcy-Weisbach equation); Moody, Darcy, or Stanton Friction Factor Diagram (laminar line), pages 186, 187 and 206
Equation
Re = VD/ν; laminar flow (Re < 2,100): f = 64/Re; hf = f(L/D)V²/(2g)
Substitute
  1. D = 50 mm = 0.050 m
  2. Re = VD/ν = (0.90 m/s)(0.050 m)/(1.9 × 10⁻⁴ m²/s) = 236.8, below 2,100, so the flow is laminar
  3. f = 64/Re = 64/236.8 = 0.2702
  4. hf = f(L/D)V²/(2g) = 0.2702 × (75 m/0.050 m) × (0.90 m/s)²/(2 × 9.807 m/s²) = 16.7 m
Result
16.7 m, 3 significant figures
Check
Re = 237 is below 2,100, so the laminar f = 64/Re applies (no Moody chart reading needed); units (m/m)(m/s)²/(m/s²) = m.
Why the others are wrong
  • A: used the radius instead of the diameter in both Re and L/D
  • C: did not square the velocity
  • D: used f = 16/Re, the Fanning laminar factor, instead of the Darcy f = 64/Re

Problem 7 · FE, Pipe flow, Darcy-Weisbach

Water flows at 0.0179 m³/s through a 100 mm diameter pipe that is 720 m long. The Darcy friction factor for the pipe is 0.018. The head loss due to friction over the pipe length is most nearly:

  • A 549 m
  • B 34.3 m
  • C 68.6 m
  • D 137 m

Handbook: Fluid Mechanics, Head Loss Due to Flow (Darcy-Weisbach equation), FE Reference Handbook 10.6

Show the worked solution
Answer
B (34.3 m)
Given
Q = 0.0179 m³/s, D = 100 mm, L = 720 m, f = 0.018, g = 9.807 m/s² (handbook)
Find
head loss due to friction, hf (m)
Handbook
Fluid Mechanics, Head Loss Due to Flow (Darcy-Weisbach equation), page 187
Equation
hf = f(L/D)V²/(2g), with V = Q/A and A = πD²/4
Substitute
  1. D = 100 mm = 0.100 m
  2. A = πD²/4 = π(0.100 m)²/4 = 0.007854 m²
  3. V = Q/A = (0.0179 m³/s)/(0.007854 m²) = 2.279 m/s
  4. hf = f(L/D)V²/(2g) = 0.018 × (720 m/0.100 m) × (2.279 m/s)²/(2 × 9.807 m/s²) = 34.3 m
Result
34.3 m, 3 significant figures
Check
units (m/m)(m/s)²/(m/s²) = m; V = 2.28 m/s is an ordinary velocity for water in a pipe, and the loss is 47.7 m per 1,000 m of pipe.
Why the others are wrong
  • A: put the radius into A = πD²/4, so V is 4 times too large and the loss is 16 times too large
  • C: left out the 2 in 2g, which doubles the loss
  • D: multiplied the given Darcy f by 4 as if it were a Fanning factor; the Darcy f goes into hf = f(L/D)V²/(2g) as given

Problem 8 · FE, Pipe flow, Darcy-Weisbach

Water flows at 0.0337 m³/s through a 150 mm diameter pipe that is 470 m long. The Darcy friction factor for the pipe is 0.031. The head loss due to friction over the pipe length is most nearly:

  • A 288 m
  • B 36.0 m
  • C 5.49 m
  • D 18.0 m

Handbook: Fluid Mechanics, Head Loss Due to Flow (Darcy-Weisbach equation), FE Reference Handbook 10.6

Show the worked solution
Answer
D (18.0 m)
Given
Q = 0.0337 m³/s, D = 150 mm, L = 470 m, f = 0.031, g = 9.807 m/s² (handbook)
Find
head loss due to friction, hf (m)
Handbook
Fluid Mechanics, Head Loss Due to Flow (Darcy-Weisbach equation), page 187
Equation
hf = f(L/D)V²/(2g), with V = Q/A and A = πD²/4
Substitute
  1. D = 150 mm = 0.150 m
  2. A = πD²/4 = π(0.150 m)²/4 = 0.01767 m²
  3. V = Q/A = (0.0337 m³/s)/(0.01767 m²) = 1.907 m/s
  4. hf = f(L/D)V²/(2g) = 0.031 × (470 m/0.150 m) × (1.907 m/s)²/(2 × 9.807 m/s²) = 18.0 m
Result
18.0 m, 3 significant figures
Check
units (m/m)(m/s)²/(m/s²) = m; V = 1.91 m/s is an ordinary velocity for water in a pipe, and the loss is 38.3 m per 1,000 m of pipe.
Why the others are wrong
  • A: put the radius into A = πD²/4, so V is 4 times too large and the loss is 16 times too large
  • B: left out the 2 in 2g, which doubles the loss
  • C: used g = 32.174 ft/sec² in an SI problem

Problem 9 · FE, Pipe flow, Darcy-Weisbach

A 24 in. diameter pipe that is 2,700 ft long carries 11,980 gpm of water. The Darcy friction factor for the pipe is 0.030. The head loss due to friction in the pipe is most nearly:

  • A 182 ft
  • B 90.9 ft
  • C 45.4 ft
  • D 5.35 ft

Handbook: Fluid Mechanics, Head Loss Due to Flow (Darcy-Weisbach equation), FE Reference Handbook 10.6

Show the worked solution
Answer
C (45.4 ft)
Given
Q = 11,980 gpm, D = 24 in., L = 2,700 ft, f = 0.030, g = 32.174 ft/sec² (handbook), 7.481 gal/ft³ (handbook)
Find
head loss due to friction, hf (ft)
Handbook
Fluid Mechanics, Head Loss Due to Flow (Darcy-Weisbach equation); Units and Conversion Factors (7.481 gal per ft³), pages 3 and 187
Equation
hf = f(L/D)V²/(2g), with V = Q/A and A = πD²/4
Substitute
  1. Q = (11,980 gal/min)/(7.481 gal/ft³ × 60 sec/min) = 26.69 ft³/sec
  2. D = 24 in./(12 in./ft) = 2.000 ft
  3. A = πD²/4 = π(2.000 ft)²/4 = 3.142 ft²
  4. V = Q/A = (26.69 ft³/sec)/(3.142 ft²) = 8.496 ft/sec
  5. hf = f(L/D)V²/(2g) = 0.030 × (2,700 ft/2.000 ft) × (8.496 ft/sec)²/(2 × 32.174 ft/sec²) = 45.4 ft
Result
45.4 ft, 3 significant figures
Check
units (ft/ft)(ft/sec)²/(ft/sec²) = ft; V = 8.50 ft/sec is an ordinary velocity for water in a pipe, and the loss is 1.68 ft per 100 ft of pipe.
Why the others are wrong
  • A: multiplied the given Darcy f by 4 as if it were a Fanning factor; the Darcy f goes into hf = f(L/D)V²/(2g) as given
  • B: left out the 2 in 2g, which doubles the loss
  • D: did not square the velocity

Problem 10 · FE, Pipe flow, Darcy-Weisbach

Water flows at 0.111 m³/s through a 300 mm diameter pipe that is 820 m long. The Darcy friction factor for the pipe is 0.019. The head loss due to friction over the pipe length is most nearly:

  • A 6.53 m
  • B 13.1 m
  • C 26.1 m
  • D 0.408 m

Handbook: Fluid Mechanics, Head Loss Due to Flow (Darcy-Weisbach equation), FE Reference Handbook 10.6

Show the worked solution
Answer
A (6.53 m)
Given
Q = 0.111 m³/s, D = 300 mm, L = 820 m, f = 0.019, g = 9.807 m/s² (handbook)
Find
head loss due to friction, hf (m)
Handbook
Fluid Mechanics, Head Loss Due to Flow (Darcy-Weisbach equation), page 187
Equation
hf = f(L/D)V²/(2g), with V = Q/A and A = πD²/4
Substitute
  1. D = 300 mm = 0.300 m
  2. A = πD²/4 = π(0.300 m)²/4 = 0.07069 m²
  3. V = Q/A = (0.111 m³/s)/(0.07069 m²) = 1.570 m/s
  4. hf = f(L/D)V²/(2g) = 0.019 × (820 m/0.300 m) × (1.570 m/s)²/(2 × 9.807 m/s²) = 6.53 m
Result
6.53 m, 3 significant figures
Check
units (m/m)(m/s)²/(m/s²) = m; V = 1.57 m/s is an ordinary velocity for water in a pipe, and the loss is 7.96 m per 1,000 m of pipe.
Why the others are wrong
  • B: used the radius instead of the diameter in L/D, which doubles the loss
  • C: multiplied the given Darcy f by 4 as if it were a Fanning factor; the Darcy f goes into hf = f(L/D)V²/(2g) as given
  • D: used A = πD² without the 1/4, so V is 4 times too small and the loss is 16 times too small

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

Frequently asked questions

Is the Darcy-Weisbach equation in the FE Reference Handbook?

Yes. It is in the Fluid Mechanics chapter of the FE Reference Handbook 10.6, and each solution below gives the page so you can practice finding it in the PDF.

Do I need the Moody diagram on the FE exam?

Some head-loss problems give the friction factor and some require reading it from the Moody diagram in the handbook. Practice both, and note that a laminar case uses its own formula for the friction factor.

Why do the wrong options look so close?

Each wrong option comes from a specific mistake, such as using the radius instead of the diameter or mixing units. The solution names the mistake so you can recognize it in your own work.

Sources

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