Fluid Mechanics & Hydraulics
Normal (uniform-flow) depth from Manning's equation Q = (1/n)·A·R^(2/3)·√S for rectangular, trapezoidal, triangular, or circular (partly full) sections, with velocity, area, hydraulic radius, top width, and Froude number at that depth. Handles the circular-pipe capacity edge: flows above the open-channel maximum refuse as surcharge, and near-full flows with two normal depths report the lower with a warning listing both. Use for: 'find the normal depth', channel/culvert uniform flow, 'does the pipe surcharge'. Not for: gradually varied profiles, critical depth (use critical-depth-froude), or pressurized pipes.
Direct solving is free and needs no account. Have a word problem instead? Submit it as text.
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, n = 0.015, S = 0.002 — Q chosen so yn = 1.2 m
Given
- section
- shape rectangularbottom width 3 m
- manning n
- 0.015
- slope
- 0.002
- flowrate
- 8.19091 m^3/s
Assumptions
- Uniform (normal) flow: the depth is constant along the channel, with the bed slope balancing friction; Manning's equation in SI form with k = 1.
- The channel is prismatic with constant slope and roughness.
Solution steps
Manning's equation
Normal depth is the depth at which the rectangular channel conveys the design discharge. The geometry terms A and R both depend on y, so y is found as the root of Q − (1/n)·A·R^(2/3)·√S = 0.
Q = (1/n)·A·R^(2/3)·√S (SI units, k = 1)
Q = 8.19 m^3/s; n = 0.015; S = 0.002
Solve for the normal depth
A bracketed root search on the Manning residual converges to the depth; the geometry below is evaluated there.
Q − (1/n)·A(y)·R(y)^(2/3)·√S = 0
y_n = 1.2 m; A = 3.6 m^2, P = 5.4 m, R = 0.667 m, T = 3 m
= 1.2 m
Velocity and Froude number at normal depth
The Froude number uses the HYDRAULIC depth Dh = A/T (not the flow depth): Fr below one is subcritical (tranquil), above one supercritical (rapid).
V = Q/A; Fr = V/√(g·A/T)
V = 2.28 m/s; D_h = A/T = 1.2 m; Fr = 0.663 → subcritical (tranquil) at normal depth
= 2.28 m/s
Results
Mean velocity at normal depth
2.28m/s
Hydraulic radius R at yn
0.667m
Froude number at yn (Dh = A/T)
0.663
Where this answer was checked
- source
- Manning's equation forward-computed at y = 1.2 m, then inverted — Q = (1/n)·A·R^(2/3)·√S with rectangular geometry
- verified by
- hand-recomputed
- derivation
- At y = 1.2: A = 3·1.2 = 3.6 m²; P = 3 + 2·1.2 = 5.4 m; R = 0.666667 m; R^(2/3) = 0.763143. Q = (1/0.015)·3.6·0.763143·√0.002 = 66.6667·3.6·0.763143·0.0447214 = 8.19091 m³/s. V = Q/A = 2.27525 m/s; Dh = A/T = 1.2 m; Fr = 2.27525/√(9.80665·1.2) = 0.66325.
Example 2
US trapezoidal: b = 10 ft, 2H:1V, n = 0.013, S = 0.001 — Q chosen so yn = 3 ft
Given
- section
- shape trapezoidalbottom width 10 ftside slope h 2
- manning n
- 0.013
- slope
- 0.001
- flowrate
- 279.98 cfs
Assumptions
- Uniform (normal) flow: the depth is constant along the channel, with the bed slope balancing friction; Manning's equation in SI form with k = 1.
- The channel is prismatic with constant slope and roughness.
Solution steps
Manning's equation
Normal depth is the depth at which the trapezoidal channel conveys the design discharge. The geometry terms A and R both depend on y, so y is found as the root of Q − (1/n)·A·R^(2/3)·√S = 0.
Q = (1/n)·A·R^(2/3)·√S (SI units, k = 1)
Q = 280 cfs; n = 0.013; S = 0.001
Solve for the normal depth
A bracketed root search on the Manning residual converges to the depth; the geometry below is evaluated there.
Q − (1/n)·A(y)·R(y)^(2/3)·√S = 0
y_n = 3 ft; A = 48 ft^2, P = 23.4 ft, R = 2.05 ft, T = 22 ft
= 3 ft
Velocity and Froude number at normal depth
The Froude number uses the HYDRAULIC depth Dh = A/T (not the flow depth): Fr below one is subcritical (tranquil), above one supercritical (rapid).
V = Q/A; Fr = V/√(g·A/T)
V = 5.83 ft/s; D_h = A/T = 2.18 ft; Fr = 0.696 → subcritical (tranquil) at normal depth
= 5.83 ft/s
US-customary form (footnote)
The familiar US form carries k = 1.486 ft^(1/3)/s — that constant is nothing but the unit conversion (1/0.3048)^(1/3); this solver computes in SI where k = 1 exactly, so both forms give the same depth.
Q = (1.486/n)·A·R^(2/3)·√S (Q cfs, A ft², R ft)
Results
Mean velocity at normal depth
5.83ft/s
Hydraulic radius R at yn
2.05ft
Froude number at yn (Dh = A/T)
0.696
Where this answer was checked
- source
- Manning's equation in the US 1.486 form, forward-computed at y = 3 ft — Q = (1.486/n)·A·R^(2/3)·√S with trapezoidal geometry
- verified by
- hand-recomputed
- derivation
- At y = 3 ft: A = 3·(10 + 2·3) = 48 ft²; P = 10 + 2·3·√5 = 23.4164 ft; R = 2.04984 ft; R^(2/3) = 1.61374. Q = (1.486/0.013)·48·1.61374·√0.001 = 114.308·48·1.61374·0.0316228 = 279.98 cfs. V = 279.98/48 = 5.833 ft/s. (The 1.486 handbook constant vs the exact (1/0.3048)^(1/3) = 1.48592 shifts Q by 0.005% — far inside tolerance.)
Example 3
SI circular half-full: D = 1 m, n = 0.013, S = 0.002 — Q chosen so yn = 0.5 m
Given
- section
- shape circulardiameter 1 m
- manning n
- 0.013
- slope
- 0.002
- flowrate
- 0.536115 m^3/s
Assumptions
- Uniform (normal) flow: the depth is constant along the channel, with the bed slope balancing friction; Manning's equation in SI form with k = 1.
- The channel is prismatic with constant slope and roughness.
Solution steps
Manning's equation
Normal depth is the depth at which the circular pipe flowing partly full conveys the design discharge. The geometry terms A and R both depend on y, so y is found as the root of Q − (1/n)·A·R^(2/3)·√S = 0.
Q = (1/n)·A·R^(2/3)·√S (SI units, k = 1)
Q = 0.536 m^3/s; n = 0.013; S = 0.002
Pipe capacity check
A circular section's conveyance peaks at about ninety-four percent depth, ABOVE the just-full value — so a pipe carries slightly more as an open channel near that depth than completely full.
Qmax = Q at the peak-conveyance depth y* (greater than Qfull)
Q_full = 1.072 m^3/s; Q_max = 1.153 m^3/s at y = 0.938 m
Solve for the normal depth
A bracketed root search on the Manning residual converges to the depth; the geometry below is evaluated there.
Q − (1/n)·A(y)·R(y)^(2/3)·√S = 0
y_n = 0.5 m; A = 0.393 m^2, P = 1.57 m, R = 0.25 m, T = 1 m
= 0.5 m
Velocity and Froude number at normal depth
The Froude number uses the HYDRAULIC depth Dh = A/T (not the flow depth): Fr below one is subcritical (tranquil), above one supercritical (rapid).
V = Q/A; Fr = V/√(g·A/T)
V = 1.37 m/s; D_h = A/T = 0.393 m; Fr = 0.696 → subcritical (tranquil) at normal depth
= 1.37 m/s
Results
Mean velocity at normal depth
1.37m/s
Flow area A at yn
0.393m^2
Hydraulic radius R at yn
0.25m
Froude number at yn (Dh = A/T)
0.696
Where this answer was checked
- source
- Manning's equation forward-computed at half depth (θ = π), then inverted — half-full circular geometry: A = πD²/8, P = πD/2, R = D/4
- verified by
- hand-recomputed
- derivation
- At y = D/2: A = π/8 = 0.392699 m²; P = π/2 = 1.570796 m; R = 0.25 m; R^(2/3) = 0.396850. Q = (1/0.013)·0.392699·0.396850·√0.002 = 76.9231·0.392699·0.396850·0.0447214 = 0.536115 m³/s. V = 1.36521 m/s; T = D = 1 m; Dh = 0.392699 m; Fr = 1.36521/√(9.80665·0.392699) = 0.69568.