10 FE practice problems: open channel flow, Manning, with solutions
These ten original problems practice open channel flow, Manning, a topic from the FE exam specification, using the Civil Engineering, Froude Number 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, Open channel flow, Manning
Water flows at 221 ft³/sec in a rectangular channel 5 ft wide at a depth of 3.8 ft. The Froude number is most nearly:
Handbook: Civil Engineering, Froude Number, FE Reference Handbook 10.6
Show the worked solution
- Answer
- B (1.05)
- Given
- Q = 221 ft³/sec, b = 5 ft, y = 3.8 ft, g = 32.174 ft/sec² (handbook), si =
- Find
- Froude number
- Handbook
- Civil Engineering, Hydraulics, Froude number, page 302
- Equation
Fr = V/√(g·yh), yh = A/T (= y for a rectangular channel)- Substitute
V = Q/(by) = 221/(5 × 3.8) = 11.63 ft/secFr = V/√(g·yh) with yh = y for a rectangle: 11.63/√(32.174 × 3.8) = 1.05
- Result
- 1.05, 3 significant figures
- Check
- Fr > 1, so the flow is supercritical.
- Why the others are wrong
- A: used g = 9.807 in a USCS problem
- C: is Fr², not Fr
- D: used √(2gy) instead of √(gy)
Problem 2 · FE, Open channel flow, Manning
A rectangular channel is 7 ft wide and flows 6.9 ft deep at uniform flow. Manning's n = 0.025 and the bed slope is 0.002. The discharge is most nearly:
Handbook: Civil Engineering, Manning's Equation, FE Reference Handbook 10.6
Show the worked solution
- Answer
- A (225 ft³/sec)
- Given
- b = 7 ft, y = 6.9 ft, n = 0.025, S = 0.002, si =
- Find
- discharge Q (ft³/sec)
- Handbook
- Civil Engineering, Hydraulics, Manning's Equation, page 303
- Equation
Q = (K/n) A R_H^(2/3) S^(1/2), K = 1.486 (USCS), R_H = A/P- Substitute
A = by = 7 × 6.9 = 48.30 ft²; P = b + 2y = 20.80 ft; R = A/P = 2.322 ftQ = (K/n) A R^(2/3) S^(1/2) = (1.486/0.025)(48.30)(2.322)^(2/3)(0.002)^(1/2) = 225 ft³/sec
- Result
- 225 ft³/sec, 3 significant figures
- Check
- the mean velocity is Q/A = 4.66 ft/sec; doubling n would halve Q.
- Why the others are wrong
- B: counted only one side wall in the wetted perimeter
- C: is the velocity; it was not multiplied by the area
- D: used the depth y for the hydraulic radius
Problem 3 · FE, Open channel flow, Manning
A rectangular channel is 2.7 m wide and flows 2.9 m deep at uniform flow. Manning's n = 0.017 and the bed slope is 0.004. The discharge is most nearly:
Handbook: Civil Engineering, Manning's Equation, FE Reference Handbook 10.6
Show the worked solution
- Answer
- B (27.6 m³/s)
- Given
- b = 2.7 m, y = 2.9 m, n = 0.017, S = 0.004, si =
- Find
- discharge Q (m³/s)
- Handbook
- Civil Engineering, Hydraulics, Manning's Equation, page 303
- Equation
Q = (K/n) A R_H^(2/3) S^(1/2), K = 1.0 (SI), R_H = A/P- Substitute
A = by = 2.7 × 2.9 = 7.830 m²; P = b + 2y = 8.500 m; R = A/P = 0.9212 mQ = (K/n) A R^(2/3) S^(1/2) = (1.0/0.017)(7.830)(0.9212)^(2/3)(0.004)^(1/2) = 27.6 m³/s
- Result
- 27.6 m³/s, 3 significant figures
- Check
- the mean velocity is Q/A = 3.52 m/s; doubling n would halve Q.
- Why the others are wrong
- A: counted only one side wall in the wetted perimeter
- C: is the velocity; it was not multiplied by the area
- D: used K = 1.486 (USCS) instead of 1.0
Problem 4 · FE, Open channel flow, Manning
A rectangular channel is 3.0 m wide and flows 1.1 m deep at uniform flow. Manning's n = 0.013 and the bed slope is 0.004. The discharge is most nearly:
Handbook: Civil Engineering, Manning's Equation, FE Reference Handbook 10.6
Show the worked solution
- Answer
- D (11.9 m³/s)
- Given
- b = 3.0 m, y = 1.1 m, n = 0.013, S = 0.004, si =
- Find
- discharge Q (m³/s)
- Handbook
- Civil Engineering, Hydraulics, Manning's Equation, page 303
- Equation
Q = (K/n) A R_H^(2/3) S^(1/2), K = 1.0 (SI), R_H = A/P- Substitute
A = by = 3.0 × 1.1 = 3.300 m²; P = b + 2y = 5.200 m; R = A/P = 0.6346 mQ = (K/n) A R^(2/3) S^(1/2) = (1.0/0.013)(3.300)(0.6346)^(2/3)(0.004)^(1/2) = 11.9 m³/s
- Result
- 11.9 m³/s, 3 significant figures
- Check
- the mean velocity is Q/A = 3.59 m/s; doubling n would halve Q.
- Why the others are wrong
- A: used the depth y for the hydraulic radius
- B: counted only one side wall in the wetted perimeter
- C: used K = 1.486 (USCS) instead of 1.0
Problem 5 · FE, Open channel flow, Manning
A rectangular channel 15 ft wide carries 200 ft³/sec. The critical depth is most nearly:
Handbook: Civil Engineering, Specific Energy and Critical Depth, FE Reference Handbook 10.6
Show the worked solution
- Answer
- B (1.77 ft)
- Given
- Q = 200 ft³/sec, b = 15 ft, g = 32.174 ft/sec² (handbook), si =
- Find
- critical depth yc (ft)
- Handbook
- Civil Engineering, Hydraulics, Specific Energy and Critical Depth, page 301
- Equation
yc = (q²/g)^(1/3) for a rectangular channel, q = Q/b- Substitute
q = Q/b = 200/15 = 13.33 ft²/secyc = (q²/g)^(1/3) = (13.33²/32.174)^(1/3) = 1.77 ft
- Result
- 1.77 ft, 3 significant figures
- Check
- at yc the Froude number is 1: V = q/yc = 7.542 equals √(g·yc) = 7.542.
- Why the others are wrong
- A: is the minimum specific energy (1.5 yc), not the critical depth
- C: used the total discharge instead of the discharge per unit width
- D: took the square root instead of the cube root
Problem 6 · FE, Open channel flow, Manning
Water flows at 1,440 ft³/sec in a rectangular channel 16 ft wide at a depth of 4.7 ft. The Froude number is most nearly:
Handbook: Civil Engineering, Froude Number, FE Reference Handbook 10.6
Show the worked solution
- Answer
- D (1.56)
- Given
- Q = 1,440 ft³/sec, b = 16 ft, y = 4.7 ft, g = 32.174 ft/sec² (handbook), si =
- Find
- Froude number
- Handbook
- Civil Engineering, Hydraulics, Froude number, page 302
- Equation
Fr = V/√(g·yh), yh = A/T (= y for a rectangular channel)- Substitute
V = Q/(by) = 1,440/(16 × 4.7) = 19.15 ft/secFr = V/√(g·yh) with yh = y for a rectangle: 19.15/√(32.174 × 4.7) = 1.56
- Result
- 1.56, 3 significant figures
- Check
- Fr > 1, so the flow is supercritical.
- Why the others are wrong
- A: used √(2gy) instead of √(gy)
- B: used g = 9.807 in a USCS problem
- C: is Fr², not Fr
Problem 7 · FE, Open channel flow, Manning
A rectangular channel is 2.4 m wide and flows 2.2 m deep at uniform flow. Manning's n = 0.03 and the bed slope is 0.0005. The discharge is most nearly:
Handbook: Civil Engineering, Manning's Equation, FE Reference Handbook 10.6
Show the worked solution
- Answer
- D (3.32 m³/s)
- Given
- b = 2.4 m, y = 2.2 m, n = 0.03, S = 0.0005, si =
- Find
- discharge Q (m³/s)
- Handbook
- Civil Engineering, Hydraulics, Manning's Equation, page 303
- Equation
Q = (K/n) A R_H^(2/3) S^(1/2), K = 1.0 (SI), R_H = A/P- Substitute
A = by = 2.4 × 2.2 = 5.280 m²; P = b + 2y = 6.800 m; R = A/P = 0.7765 mQ = (K/n) A R^(2/3) S^(1/2) = (1.0/0.03)(5.280)(0.7765)^(2/3)(0.0005)^(1/2) = 3.32 m³/s
- Result
- 3.32 m³/s, 3 significant figures
- Check
- the mean velocity is Q/A = 0.630 m/s; doubling n would halve Q.
- Why the others are wrong
- A: used the depth y for the hydraulic radius
- B: used K = 1.486 (USCS) instead of 1.0
- C: counted only one side wall in the wetted perimeter
Problem 8 · FE, Open channel flow, Manning
Water flows at 297 ft³/sec in a rectangular channel 15 ft wide at a depth of 4.6 ft. The Froude number is most nearly:
Handbook: Civil Engineering, Froude Number, FE Reference Handbook 10.6
Show the worked solution
- Answer
- B (0.354)
- Given
- Q = 297 ft³/sec, b = 15 ft, y = 4.6 ft, g = 32.174 ft/sec² (handbook), si =
- Find
- Froude number
- Handbook
- Civil Engineering, Hydraulics, Froude number, page 302
- Equation
Fr = V/√(g·yh), yh = A/T (= y for a rectangular channel)- Substitute
V = Q/(by) = 297/(15 × 4.6) = 4.304 ft/secFr = V/√(g·yh) with yh = y for a rectangle: 4.304/√(32.174 × 4.6) = 0.354
- Result
- 0.354, 3 significant figures
- Check
- Fr < 1, so the flow is subcritical.
- Why the others are wrong
- A: used g = 9.807 in a USCS problem
- C: is Fr², not Fr
- D: used √(2gy) instead of √(gy)
Problem 9 · FE, Open channel flow, Manning
Water flows at 1,760 ft³/sec in a rectangular channel 9 ft wide at a depth of 6.8 ft. The Froude number is most nearly:
Handbook: Civil Engineering, Froude Number, FE Reference Handbook 10.6
Show the worked solution
- Answer
- A (1.94)
- Given
- Q = 1,760 ft³/sec, b = 9 ft, y = 6.8 ft, g = 32.174 ft/sec² (handbook), si =
- Find
- Froude number
- Handbook
- Civil Engineering, Hydraulics, Froude number, page 302
- Equation
Fr = V/√(g·yh), yh = A/T (= y for a rectangular channel)- Substitute
V = Q/(by) = 1,760/(9 × 6.8) = 28.76 ft/secFr = V/√(g·yh) with yh = y for a rectangle: 28.76/√(32.174 × 6.8) = 1.94
- Result
- 1.94, 3 significant figures
- Check
- Fr > 1, so the flow is supercritical.
- Why the others are wrong
- B: used √(2gy) instead of √(gy)
- C: divided by gy without the square root
- D: is Fr², not Fr
Problem 10 · FE, Open channel flow, Manning
A rectangular channel 10 ft wide carries 445 ft³/sec. The critical depth is most nearly:
Handbook: Civil Engineering, Specific Energy and Critical Depth, FE Reference Handbook 10.6
Show the worked solution
- Answer
- C (3.95 ft)
- Given
- Q = 445 ft³/sec, b = 10 ft, g = 32.174 ft/sec² (handbook), si =
- Find
- critical depth yc (ft)
- Handbook
- Civil Engineering, Hydraulics, Specific Energy and Critical Depth, page 301
- Equation
yc = (q²/g)^(1/3) for a rectangular channel, q = Q/b- Substitute
q = Q/b = 445/10 = 44.50 ft²/secyc = (q²/g)^(1/3) = (44.50²/32.174)^(1/3) = 3.95 ft
- Result
- 3.95 ft, 3 significant figures
- Check
- at yc the Froude number is 1: V = q/yc = 11.27 equals √(g·yc) = 11.27.
- Why the others are wrong
- A: used 2g instead of g
- B: used the total discharge instead of the discharge per unit width
- D: did not square the unit discharge
A new problem posts every day inside the lab, with the full worked solution the same evening.
Frequently asked questions
Where is open channel flow, Manning in the FE Reference Handbook?
Look in the Civil Engineering, Froude Number 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.