Tools · Hydraulics
Open channel flow and measurement
Guided step by step: section properties and Manning’s formula, Froude number, critical depth and specific energy; then measure discharge with weirs, notches and a current meter. Try each calculation yourself before the answer is shown.
Guided · step 1 of 10
Know the channel
A trapezoidal channel on a bed slope S₀ = 0.0005 with Manning’s n = 0.015. In uniform flow the depth stays the same along the channel (the normal depth), so the water surface is parallel to the bed and the energy line slope equals the bed slope.
b = 3 m, side slope z = 1.5 (z H : 1 V)
Depth of flow y = 1.2 m
Channel
Cross-section
Specific energy diagram
The curve is E = y + Q²/(2gA²) for Q = 7.314 m³/s. The point is your flow (E = 1.282 m). The nose C is critical flow, where E is least.
Full solution, all steps at once
- Know the channel
b = 3 m, side slope z = 1.5 (z H : 1 V)
Depth of flow y = 1.2 m
- Flow area AA = (b + zy)·y
A = (3 + 1.5 × 1.2) × 1.2 = 5.76 m²
- Wetted perimeter PP = b + 2y√1 + z²
P = 3 + 2 × 1.2 × √1 + 1.5² = 7.3267 m
- Hydraulic radius RR = AP
R = 5.767.3267 = 0.7862 m
- Mean velocity by Manning’s formulaV = (1/n)·R2/3·S₀1/2
V = (10.015) × 0.78622/3 × 0.00051/2
V = 66.667 × 0.8518 × 0.02236 = 1.2698 m/s
- Discharge QQ = A·V
Q = 5.76 × 1.2698 = 7.3141 m³/s (7314.1 L/s)
- Froude number: is the flow tranquil or rapid?D = ATFr = V√gD
T = 6.6 m, so D = 5.766.6 = 0.8727 m
Fr = 1.2698√9.81 × 0.8727 = 0.434
Fr < 1, so the flow is subcritical
- Critical depth ycQ²TgA³ = 1
Trying y = 0.7426 m: A = 3.0548 m², T = 5.2277 m
Q²TgA³ = 7.3141² × 5.22779.81 × 3.0548³ = 1 ✓
yc = 0.7426 m
- Specific energyE = y + V²2g
E = 1.2 + 1.2698²2 × 9.81 = 1.2 + 0.0822 = 1.2822 m
Emin (at yc) = 1.0347 m
Alternate depth with the same E: 0.4976 m