CivilSolve

Fluid Mechanics & Hydraulics

Critical Depth & Froude Number

Critical depth from the Fr = 1 condition Q²T/(gA³) = 1 for rectangular, trapezoidal, triangular, or circular sections (rectangular closed form (q²/g)^(1/3) shown as a check), plus — when an actual depth is given — the Froude number Fr = V/√(g·A/T) using the hydraulic depth, with sub/supercritical classification. Use for: 'find the critical depth', flow-regime classification, choke and control-section checks. Not for: normal depth (use manning-normal-depth) or pressurized pipes.

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Worked examples

These are the solver's own reference problems — the answers come from published sources or independent hand computation, never from the solver itself. The solution below is the live solver output for each.

Example 1

SI rectangular: b = 3 m, Q = 12 m³/s

Given

section
shape rectangularbottom width 3 m
flowrate
12 m^3/s

Assumptions

  • Critical flow is defined by minimum specific energy at fixed discharge — equivalently a Froude number of one.
  • Hydrostatic pressure distribution and uniform velocity over the section.

Solution steps

  1. Critical-flow condition

    At critical depth the Froude number equals one, which for the rectangular channel means the discharge, top width, and area satisfy the condition below — the depth enters through both A(y) and T(y).

    Q²·T/(g·A³) = 1

    Q = 12 m^3/s

  2. Solve for the critical depth

    A bracketed root search on the critical-flow residual converges to the depth where the condition holds exactly.

    Q²·T(yc)/(g·A(yc)³) − 1 = 0

    y_c = 1.177 m; A = 3.53 m^2, T = 3 m, V_c = 3.4 m/s

    = 1.177 m

  3. Rectangular closed-form check

    For a rectangular channel the condition collapses to a closed form in the unit discharge q = Q/b — it must (and does) reproduce the root-found depth.

    yc = (q²/g)^(1/3) with q = Q/b

    q = 4 m²/s per metre of width → y_c = 1.177 m

    = 1.177 m

Results

Critical depth yc

1.177m

Velocity at critical depth Vc

3.4m/s

Where this answer was checked
source
Rectangular critical-depth closed form yc = (q²/g)^(1/3) (NCEES FE Reference Handbook) — q = Q/b = 4 m²/s
verified by
hand-recomputed
derivation
yc = (4²/9.80665)^(1/3) = (1.631546)^(1/3) = 1.177244 m. Vc = Q/(b·yc) = 12/3.531732 = 3.397767 m/s (= √(g·yc), the rectangular identity).

Example 2

US rectangular: b = 10 ft, Q = 200 cfs

Given

section
shape rectangularbottom width 10 ft
flowrate
200 cfs

Assumptions

  • Critical flow is defined by minimum specific energy at fixed discharge — equivalently a Froude number of one.
  • Hydrostatic pressure distribution and uniform velocity over the section.

Solution steps

  1. Critical-flow condition

    At critical depth the Froude number equals one, which for the rectangular channel means the discharge, top width, and area satisfy the condition below — the depth enters through both A(y) and T(y).

    Q²·T/(g·A³) = 1

    Q = 200 cfs

  2. Solve for the critical depth

    A bracketed root search on the critical-flow residual converges to the depth where the condition holds exactly.

    Q²·T(yc)/(g·A(yc)³) − 1 = 0

    y_c = 2.317 ft; A = 23.2 ft^2, T = 10 ft, V_c = 8.63 ft/s

    = 2.317 ft

  3. Rectangular closed-form check

    For a rectangular channel the condition collapses to a closed form in the unit discharge q = Q/b — it must (and does) reproduce the root-found depth.

    yc = (q²/g)^(1/3) with q = Q/b

    q = 1.858 m²/s per metre of width → y_c = 2.317 ft

    = 2.317 ft

Results

Critical depth yc

2.317ft

Velocity at critical depth Vc

8.63ft/s

Where this answer was checked
source
Rectangular critical-depth closed form in US units, yc = (q²/g)^(1/3) — q = 20 cfs/ft
verified by
hand-recomputed
derivation
US route: yc = (20²/32.174)^(1/3) = (12.4324)^(1/3) = 2.31660 ft. SI cross-check: q = 1.858061 m²/s → yc = (q²/9.80665)^(1/3) = 0.706100 m = 2.316602 ft; Vc = 8.6333 ft/s.

Example 3

SI trapezoidal: b = 2 m, 1.5H:1V — Q chosen so yc = 0.75 m

Given

section
shape trapezoidalbottom width 2 mside slope h 1.5
flowrate
5.450456 m^3/s

Assumptions

  • Critical flow is defined by minimum specific energy at fixed discharge — equivalently a Froude number of one.
  • Hydrostatic pressure distribution and uniform velocity over the section.

Solution steps

  1. Critical-flow condition

    At critical depth the Froude number equals one, which for the trapezoidal channel means the discharge, top width, and area satisfy the condition below — the depth enters through both A(y) and T(y).

    Q²·T/(g·A³) = 1

    Q = 5.45 m^3/s

  2. Solve for the critical depth

    A bracketed root search on the critical-flow residual converges to the depth where the condition holds exactly.

    Q²·T(yc)/(g·A(yc)³) − 1 = 0

    y_c = 0.75 m; A = 2.34 m^2, T = 4.25 m, V_c = 2.33 m/s

    = 0.75 m

Results

Critical depth yc

0.75m

Velocity at critical depth Vc

2.33m/s

Where this answer was checked
source
Critical condition Q = √(g·A³/T) forward-computed at y = 0.75 m, then inverted — trapezoidal geometry at critical flow
verified by
hand-recomputed
derivation
At y = 0.75: A = 0.75·(2 + 1.5·0.75) = 2.34375 m²; T = 2 + 2·1.5·0.75 = 4.25 m. Q = √(9.80665·2.34375³/4.25) = √(9.80665·12.87415/4.25) = √29.70642 = 5.450456 m³/s. Vc = Q/A = 2.325528 m/s.