10 FE practice problems: bernoulli and pump power, with solutions
These ten original problems practice bernoulli and pump power, a topic from the FE exam specification, using the Fluid Mechanics, Continuity and Bernoulli 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, Bernoulli and pump power
Water (ρ = 1,000 kg/m³) flows at 0.00507 m³/s through a horizontal pipe that contracts from 150 mm to 75 mm in diameter. Neglecting losses, the pressure drop from the 150 mm section to the 75 mm section is most nearly:
Handbook: Fluid Mechanics, Continuity and Bernoulli Equations, FE Reference Handbook 10.6
Show the worked solution
- Answer
- A (0.617 kPa)
- Given
- Q = 0.00507 m³/s, D1 = 150 mm, D2 = 75 mm, rho = 1,000 kg/m³
- Find
- pressure drop P1 - P2 (kPa)
- Handbook
- Fluid Mechanics, The Continuity Equation and the Bernoulli Equation, page 185
- Equation
Q = A1v1 = A2v2; P1/γ + v1²/(2g) = P2/γ + v2²/(2g) with z1 = z2- Substitute
v1 = Q/A1 = 0.00507/[π(0.150)²/4] = 0.2869 m/s; v2 = Q/A2 = 0.00507/[π(0.075)²/4] = 1.148 m/sP1 - P2 = ρ(v2² - v1²)/2 = 1,000(1.148² - 0.2869²)/2 Pa = 0.617 kPa
- Result
- 0.617 kPa, 3 significant figures
- Check
- the pressure falls where the pipe narrows because the velocity head rises (z1 = z2, no losses).
- Why the others are wrong
- B: scaled the velocity by the diameter ratio instead of the area ratio
- C: squared the velocity difference instead of differencing the squares
- D: left out the upstream velocity head
Problem 2 · FE, Bernoulli and pump power
Water (ρ = 1,000 kg/m³) flows at 0.0134 m³/s through a horizontal pipe that contracts from 300 mm to 75 mm in diameter. Neglecting losses, the pressure drop from the 300 mm section to the 75 mm section is most nearly:
Handbook: Fluid Mechanics, Continuity and Bernoulli Equations, FE Reference Handbook 10.6
Show the worked solution
- Answer
- C (4.58 kPa)
- Given
- Q = 0.0134 m³/s, D1 = 300 mm, D2 = 75 mm, rho = 1,000 kg/m³
- Find
- pressure drop P1 - P2 (kPa)
- Handbook
- Fluid Mechanics, The Continuity Equation and the Bernoulli Equation, page 185
- Equation
Q = A1v1 = A2v2; P1/γ + v1²/(2g) = P2/γ + v2²/(2g) with z1 = z2- Substitute
v1 = Q/A1 = 0.0134/[π(0.300)²/4] = 0.1896 m/s; v2 = Q/A2 = 0.0134/[π(0.075)²/4] = 3.033 m/sP1 - P2 = ρ(v2² - v1²)/2 = 1,000(3.033² - 0.1896²)/2 Pa = 4.58 kPa
- Result
- 4.58 kPa, 3 significant figures
- Check
- the pressure falls where the pipe narrows because the velocity head rises (z1 = z2, no losses).
- Why the others are wrong
- A: squared the velocity difference instead of differencing the squares
- B: did not square the velocities
- D: scaled the velocity by the diameter ratio instead of the area ratio
Problem 3 · FE, Bernoulli and pump power
A pump moves 3,220 gpm of water against a total dynamic head of 280 ft with an efficiency of 81%. Using γ = 62.4 lbf/ft³, the brake horsepower is most nearly:
Handbook: Fluid Mechanics, Pump Power Equation, FE Reference Handbook 10.6
Show the worked solution
- Answer
- A (281 hp)
- Given
- Q = 3,220 gpm, h = 280 ft, eta = 81%, gamma = 62.4 lbf/ft³
- Find
- brake horsepower (hp)
- Handbook
- Fluid Mechanics, Pump Power Equation; Units and Conversion Factors, pages 3 and 197
- Equation
W = γQh/η, 1 hp = 550 ft-lbf/sec- Substitute
Q = 3,220 gal/min ÷ (7.481 gal/ft³ × 60 sec/min) = 7.174 ft³/secW = γQh/η = (62.4)(7.174)(280)/0.81 = 154,700 ft-lbf/sec ÷ 550 = 281 hp
- Result
- 281 hp, 3 significant figures
- Check
- the water horsepower 228 hp is less than the brake horsepower, as it must be.
- Why the others are wrong
- B: is the water horsepower; the efficiency was left out
- C: converted the result to kilowatts while labeling it horsepower
- D: multiplied by the efficiency instead of dividing
Problem 4 · FE, Bernoulli and pump power
Water (ρ = 1,000 kg/m³) flows at 0.104 m³/s through a horizontal pipe that contracts from 250 mm to 200 mm in diameter. Neglecting losses, the pressure drop from the 250 mm section to the 200 mm section is most nearly:
Handbook: Fluid Mechanics, Continuity and Bernoulli Equations, FE Reference Handbook 10.6
Show the worked solution
- Answer
- A (3.24 kPa)
- Given
- Q = 0.104 m³/s, D1 = 250 mm, D2 = 200 mm, rho = 1,000 kg/m³
- Find
- pressure drop P1 - P2 (kPa)
- Handbook
- Fluid Mechanics, The Continuity Equation and the Bernoulli Equation, page 185
- Equation
Q = A1v1 = A2v2; P1/γ + v1²/(2g) = P2/γ + v2²/(2g) with z1 = z2- Substitute
v1 = Q/A1 = 0.104/[π(0.250)²/4] = 2.119 m/s; v2 = Q/A2 = 0.104/[π(0.200)²/4] = 3.310 m/sP1 - P2 = ρ(v2² - v1²)/2 = 1,000(3.310² - 2.119²)/2 Pa = 3.24 kPa
- Result
- 3.24 kPa, 3 significant figures
- Check
- the pressure falls where the pipe narrows because the velocity head rises (z1 = z2, no losses).
- Why the others are wrong
- B: left out the ½ in ρv²/2
- C: did not square the velocities
- D: scaled the velocity by the diameter ratio instead of the area ratio
Problem 5 · FE, Bernoulli and pump power
Water (ρ = 1,000 kg/m³) flows at 0.0242 m³/s through a horizontal pipe that contracts from 200 mm to 125 mm in diameter. Neglecting losses, the pressure drop from the 200 mm section to the 125 mm section is most nearly:
Handbook: Fluid Mechanics, Continuity and Bernoulli Equations, FE Reference Handbook 10.6
Show the worked solution
- Answer
- D (1.65 kPa)
- Given
- Q = 0.0242 m³/s, D1 = 200 mm, D2 = 125 mm, rho = 1,000 kg/m³
- Find
- pressure drop P1 - P2 (kPa)
- Handbook
- Fluid Mechanics, The Continuity Equation and the Bernoulli Equation, page 185
- Equation
Q = A1v1 = A2v2; P1/γ + v1²/(2g) = P2/γ + v2²/(2g) with z1 = z2- Substitute
v1 = Q/A1 = 0.0242/[π(0.200)²/4] = 0.7703 m/s; v2 = Q/A2 = 0.0242/[π(0.125)²/4] = 1.972 m/sP1 - P2 = ρ(v2² - v1²)/2 = 1,000(1.972² - 0.7703²)/2 Pa = 1.65 kPa
- Result
- 1.65 kPa, 3 significant figures
- Check
- the pressure falls where the pipe narrows because the velocity head rises (z1 = z2, no losses).
- Why the others are wrong
- A: scaled the velocity by the diameter ratio instead of the area ratio
- B: squared the velocity difference instead of differencing the squares
- C: left out the upstream velocity head
Problem 6 · FE, Bernoulli and pump power
A pump delivers 22 L/s of water against a total dynamic head of 82 m. The pump efficiency is 61%. Using γ = 9.81 kN/m³, the power input to the pump is most nearly:
Handbook: Fluid Mechanics, Pump Power Equation, FE Reference Handbook 10.6
Show the worked solution
- Answer
- B (29.0 kW)
- Given
- Q = 22 L/s, h = 82 m, eta = 61%, gamma = 9.81 kN/m³
- Find
- pump input power (kW)
- Handbook
- Fluid Mechanics, Pump Power Equation, page 197
- Equation
W = γQh/η- Substitute
Q = 22 L/s = 0.022 m³/sW = γQh/η = (9.81 kN/m³)(0.022 m³/s)(82 m)/0.61 = 29.0 kW
- Result
- 29.0 kW, 3 significant figures
- Check
- the water power γQh = 17.7 kW is less than the input, as it must be with η < 1.
- Why the others are wrong
- A: is the water power; the efficiency was left out
- C: multiplied by the efficiency instead of dividing
- D: left out the unit weight of water
Problem 7 · FE, Bernoulli and pump power
A pump delivers 95 L/s of water against a total dynamic head of 114 m. The pump efficiency is 88%. Using γ = 9.81 kN/m³, the power input to the pump is most nearly:
Handbook: Fluid Mechanics, Pump Power Equation, FE Reference Handbook 10.6
Show the worked solution
- Answer
- D (121 kW)
- Given
- Q = 95 L/s, h = 114 m, eta = 88%, gamma = 9.81 kN/m³
- Find
- pump input power (kW)
- Handbook
- Fluid Mechanics, Pump Power Equation, page 197
- Equation
W = γQh/η- Substitute
Q = 95 L/s = 0.095 m³/sW = γQh/η = (9.81 kN/m³)(0.095 m³/s)(114 m)/0.88 = 121 kW
- Result
- 121 kW, 3 significant figures
- Check
- the water power γQh = 106 kW is less than the input, as it must be with η < 1.
- Why the others are wrong
- A: converted the result to horsepower while labeling it kW
- B: multiplied by the efficiency instead of dividing
- C: is the water power; the efficiency was left out
Problem 8 · FE, Bernoulli and pump power
A pump delivers 12 L/s of water against a total dynamic head of 40 m. The pump efficiency is 62%. Using γ = 9.81 kN/m³, the power input to the pump is most nearly:
Handbook: Fluid Mechanics, Pump Power Equation, FE Reference Handbook 10.6
Show the worked solution
- Answer
- C (7.59 kW)
- Given
- Q = 12 L/s, h = 40 m, eta = 62%, gamma = 9.81 kN/m³
- Find
- pump input power (kW)
- Handbook
- Fluid Mechanics, Pump Power Equation, page 197
- Equation
W = γQh/η- Substitute
Q = 12 L/s = 0.012 m³/sW = γQh/η = (9.81 kN/m³)(0.012 m³/s)(40 m)/0.62 = 7.59 kW
- Result
- 7.59 kW, 3 significant figures
- Check
- the water power γQh = 4.71 kW is less than the input, as it must be with η < 1.
- Why the others are wrong
- A: is the water power; the efficiency was left out
- B: converted the result to horsepower while labeling it kW
- D: multiplied by the efficiency instead of dividing
Problem 9 · FE, Bernoulli and pump power
A pump delivers 15 L/s of water against a total dynamic head of 106 m. The pump efficiency is 69%. Using γ = 9.81 kN/m³, the power input to the pump is most nearly:
Handbook: Fluid Mechanics, Pump Power Equation, FE Reference Handbook 10.6
Show the worked solution
- Answer
- B (22.6 kW)
- Given
- Q = 15 L/s, h = 106 m, eta = 69%, gamma = 9.81 kN/m³
- Find
- pump input power (kW)
- Handbook
- Fluid Mechanics, Pump Power Equation, page 197
- Equation
W = γQh/η- Substitute
Q = 15 L/s = 0.015 m³/sW = γQh/η = (9.81 kN/m³)(0.015 m³/s)(106 m)/0.69 = 22.6 kW
- Result
- 22.6 kW, 3 significant figures
- Check
- the water power γQh = 15.6 kW is less than the input, as it must be with η < 1.
- Why the others are wrong
- A: is the water power; the efficiency was left out
- C: multiplied by the efficiency instead of dividing
- D: converted the result to horsepower while labeling it kW
Problem 10 · FE, Bernoulli and pump power
A pump delivers 65 L/s of water against a total dynamic head of 113 m. The pump efficiency is 89%. Using γ = 9.81 kN/m³, the power input to the pump is most nearly:
Handbook: Fluid Mechanics, Pump Power Equation, FE Reference Handbook 10.6
Show the worked solution
- Answer
- D (81.0 kW)
- Given
- Q = 65 L/s, h = 113 m, eta = 89%, gamma = 9.81 kN/m³
- Find
- pump input power (kW)
- Handbook
- Fluid Mechanics, Pump Power Equation, page 197
- Equation
W = γQh/η- Substitute
Q = 65 L/s = 0.065 m³/sW = γQh/η = (9.81 kN/m³)(0.065 m³/s)(113 m)/0.89 = 81.0 kW
- Result
- 81.0 kW, 3 significant figures
- Check
- the water power γQh = 72.1 kW is less than the input, as it must be with η < 1.
- Why the others are wrong
- A: multiplied by the efficiency instead of dividing
- B: left out the unit weight of water
- C: is the water power; the efficiency was left out
A new problem posts every day inside the lab, with the full worked solution the same evening.
Frequently asked questions
Where is bernoulli and pump power in the FE Reference Handbook?
Look in the Fluid Mechanics, Continuity and Bernoulli 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
- NCEES FE Civil CBT exam specifications (PDF). Retrieved October 3, 2026.
- NCEES FE Environmental CBT exam specifications (PDF). Retrieved October 3, 2026.
- NCEES FE Mechanical CBT exam specifications (PDF). Retrieved October 3, 2026.
- NCEES FE Reference Handbook 10.6 (free PDF in MyNCEES). Retrieved October 3, 2026.