Manning Equation
Open-channel flow has a free surface exposed to atmosphere. For uniform flow, the bed slope, water-surface slope, and energy slope are commonly taken equal.
$$Q=\frac{1}{n}AR^{2/3}S^{1/2}$$$$R=\frac{A}{P}$$
Critical Flow and Froude Number
Critical flow separates subcritical and supercritical conditions. For a rectangular channel:
$$Fr=\frac{V}{\sqrt{gy}}$$$$y_c=\left(\frac{q^2}{g}\right)^{1/3}$$
Hydraulic Jump and Non-uniform Flow
A hydraulic jump is a rapidly varied flow where high-velocity shallow flow changes into deeper slower flow with energy loss.
$$\frac{y_2}{y_1}=\frac{1}{2}\left(\sqrt{1+8Fr_1^2}-1\right)$$
Critical Depth in a Rectangular Channel
A rectangular channel carries 3.6 m3/s and is 2.0 m wide. Find critical depth.
$$q=\frac{Q}{b}=\frac{3.6}{2.0}=1.8\text{ m}^2/\text{s}$$$$y_c=\left(\frac{1.8^2}{9.81}\right)^{1/3}=0.691\text{ m}$$
Answer: $y_c=0.691\text{ m}$.
Manning Rectangular Channel Discharge
A rectangular channel is 3.0 m wide and carries water 1.2 m deep on a slope of 0.0016. If $n=0.015$, find the discharge for uniform flow.
$$A=by=3.0(1.2)=3.60\text{ m}^2$$$$P=b+2y=3.0+2(1.2)=5.40\text{ m}$$$$R=\frac{A}{P}=\frac{3.60}{5.40}=0.667\text{ m}$$$$Q=\frac{1}{n}AR^{2/3}S^{1/2}=\frac{1}{0.015}(3.60)(0.667)^{2/3}(0.0016)^{1/2}$$$$Q=7.34\text{ m}^3/\text{s}$$
Answer: $Q=7.34\text{ m}^3/\text{s}$.
Most Efficient Rectangular Section
A rectangular channel must carry 5.0 m3/s with $n=0.014$ and slope $S=0.001$. For the most efficient rectangular section, find the normal depth and width.
For the best rectangular section, $b=2y$, $A=2y^2$, and $R=y/2$.
$$Q=\frac{1}{n}(2y^2)\left(\frac{y}{2}\right)^{2/3}S^{1/2}$$$$5.0=\frac{1}{0.014}(2y^2)\left(\frac{y}{2}\right)^{2/3}(0.001)^{1/2}$$$$y=1.31\text{ m},\quad b=2y=2.62\text{ m}$$
Answer: $y=1.31\text{ m}$ and $b=2.62\text{ m}$.
Hydraulic Jump Sequent Depth and Energy Loss
In a rectangular channel, supercritical flow has depth $y_1=0.40\text{ m}$ and velocity $V_1=6.0\text{ m/s}$. Find the sequent depth and energy loss across the hydraulic jump.
$$Fr_1=\frac{V_1}{\sqrt{gy_1}}=\frac{6.0}{\sqrt{9.81(0.40)}}=3.03$$$$\frac{y_2}{y_1}=\frac{1}{2}\left(\sqrt{1+8Fr_1^2}-1\right)=\frac{1}{2}\left(\sqrt{1+8(3.03)^2}-1\right)=3.82$$$$y_2=3.82(0.40)=1.53\text{ m}$$$$\Delta E=\frac{(y_2-y_1)^3}{4y_1y_2}=\frac{(1.53-0.40)^3}{4(0.40)(1.53)}=0.588\text{ m}$$
Answer: $y_2=1.53\text{ m}$ and energy loss is 0.588 m.
Problem: Trapezoidal Channel — Manning Uniform Flow
A trapezoidal channel has a bottom width of 2.0 m and side slopes of 1.5H:1V (1.5 horizontal to 1 vertical). The channel is lined with concrete (Manning's n = 0.013) and laid on a slope of S = 0.0009. Find the normal discharge when the water depth is 1.20 m.
Compute the geometric elements for a trapezoidal section with b = 2.0 m, z = 1.5 (side slope), y = 1.20 m:
$$A = (b + zy)y = (2.0 + 1.5 \times 1.20)(1.20) = (2.0 + 1.80)(1.20) = 3.80(1.20) = 4.56 \text{ m}^2$$
$$P = b + 2y\sqrt{1 + z^2} = 2.0 + 2(1.20)\sqrt{1 + (1.5)^2} = 2.0 + 2.40\sqrt{3.25}$$
$$\sqrt{3.25} = 1.803, \quad P = 2.0 + 2.40(1.803) = 2.0 + 4.327 = 6.327 \text{ m}$$
$$R = \frac{A}{P} = \frac{4.56}{6.327} = 0.7208 \text{ m}$$
$$Q = \frac{1}{n}AR^{2/3}S^{1/2} = \frac{1}{0.013}(4.56)(0.7208)^{2/3}(0.0009)^{1/2}$$
$$(0.7208)^{2/3}: \quad (0.7208)^{1/3} = 0.8963, \quad (0.8963)^2 = 0.8034$$
$$Q = 76.92(4.56)(0.8034)(0.03) = 76.92(4.56)(0.02410) = 8.455 \text{ m}^3/\text{s}$$
Answer: Normal discharge is Q ≈ 8.46 m³/s. Trapezoidal channels are widely used in irrigation because the sloped sides are stable in soil and the section is close to the most efficient trapezoidal shape (half-hexagon) when $z = 1/\sqrt{3} \approx 0.577$.
Problem: Most Efficient Trapezoidal Section Design
Design the most efficient (best hydraulic) trapezoidal channel to carry Q = 10 m³/s with Manning's n = 0.014 on a slope S = 0.0004. The most efficient trapezoidal section is a half-hexagon: side slopes z = 1/√3 ≈ 0.5774, and the relationship b = 2y(√(1+z²) − z) holds. Determine the required bottom width b and normal depth y.
For the most efficient trapezoidal section: $z = 1/\sqrt{3} = 0.5774$, $R = y/2$, and $b = 2y(\sqrt{1+z^2} - z) = 2y(\sqrt{1+1/3} - 1/\sqrt{3}) = 2y(2/\sqrt{3} - 1/\sqrt{3}) = 2y/\sqrt{3}$.
$$A = (b + zy)y = \left(\frac{2y}{\sqrt{3}} + \frac{y}{\sqrt{3}}\right)y = \frac{3y}{\sqrt{3}} \cdot y = \sqrt{3}\,y^2$$
$$R = \frac{y}{2}$$
$$Q = \frac{1}{n}(\sqrt{3}\,y^2)\left(\frac{y}{2}\right)^{2/3}S^{1/2}$$
$$10 = \frac{1}{0.014}(\sqrt{3}\,y^2)\left(\frac{y}{2}\right)^{2/3}(0.0004)^{1/2}$$
$$10 = 71.43(1.732\,y^2)(0.6300\,y^{2/3})(0.02)$$
$$10 = 71.43(1.732)(0.6300)(0.02)\,y^{8/3}$$
$$10 = 1.5607\,y^{8/3} \Rightarrow y^{8/3} = 6.407$$
$$y = (6.407)^{3/8} = (6.407)^{0.375}$$
Compute: $\ln(6.407) = 1.857$, $0.375(1.857) = 0.696$, $e^{0.696} = 2.006$ m.
$$y \approx 2.006 \text{ m}, \quad b = \frac{2y}{\sqrt{3}} = \frac{2(2.006)}{1.732} = 2.316 \text{ m}$$
Answer: Normal depth y ≈ 2.01 m and bottom width b ≈ 2.32 m. The most efficient trapezoidal section minimizes wetted perimeter for a given area, reducing excavation and lining costs while maximizing conveyance.
Problem: Froude Number and Flow Classification
A 4.0 m wide rectangular channel carries a discharge of 12 m³/s at a uniform depth of 1.80 m. Determine: (a) the Froude number and classify the flow, (b) the critical depth, (c) the critical velocity, and (d) the minimum specific energy (critical specific energy).
$$V = \frac{Q}{A} = \frac{12}{4.0(1.80)} = \frac{12}{7.20} = 1.667 \text{ m/s}$$
$$Fr = \frac{V}{\sqrt{gy}} = \frac{1.667}{\sqrt{9.81(1.80)}} = \frac{1.667}{\sqrt{17.66}} = \frac{1.667}{4.202} = 0.397$$
$Fr < 1$: flow is subcritical (tranquil).
$$q = \frac{Q}{b} = \frac{12}{4.0} = 3.0 \text{ m}^2/\text{s}$$
$$y_c = \left(\frac{q^2}{g}\right)^{1/3} = \left(\frac{9.0}{9.81}\right)^{1/3} = (0.9174)^{1/3} = 0.9716 \text{ m}$$
$$V_c = \sqrt{g y_c} = \sqrt{9.81(0.9716)} = \sqrt{9.531} = 3.087 \text{ m/s}$$
$$E_c = \frac{3}{2}y_c = \frac{3}{2}(0.9716) = 1.457 \text{ m}$$
Answer: (a) Fr = 0.397 — subcritical flow. (b) Critical depth = 0.972 m. (c) Critical velocity = 3.09 m/s. (d) Minimum specific energy = 1.457 m. The actual flow depth of 1.80 m is above critical depth (0.972 m), confirming subcritical conditions consistent with Fr < 1.
Problem: Circular Pipe Flowing Partially Full
A 1200 mm diameter concrete sewer pipe (Manning's n = 0.013) is laid on a slope of 0.0008. The pipe flows at a depth of 900 mm (75% full by depth). Find the discharge and flow velocity. Note: for a circular pipe, the geometric properties at partial depth require the central half-angle θ (in radians) where $\cos\theta = 1 - 2d/D$ and $d$ is the depth.
D = 1.2 m, d = 0.9 m, d/D = 0.75. Find the central half-angle θ:
$$\cos\theta = 1 - \frac{2d}{D} = 1 - 2(0.75) = 1 - 1.50 = -0.50$$
$$\theta = \arccos(-0.50) = 120° = \frac{2\pi}{3} \text{ rad} = 2.094 \text{ rad}$$
$$\text{Full angle subtended} = 2\theta = 4.189 \text{ rad}$$
$$A = \frac{D^2}{8}(2\theta - \sin 2\theta) = \frac{(1.2)^2}{8}(4.189 - \sin 240°)$$
$$\sin 240° = -\sin 60° = -0.8660, \quad 2\theta - \sin 2\theta = 4.189-(-0.866)= 5.055$$
$$A = \frac{1.44}{8}(5.055) = 0.180(5.055) = 0.910 \text{ m}^2$$
$$P = D\theta = 1.2(2.094) = 2.513 \text{ m} \quad(\text{arc length for } 2\theta)$$
Wait — wetted perimeter uses full angle $2\theta$: $P = D \cdot \theta_{total} = 1.2 \times 4.189/1 = ?$ Let me use $P = D\theta_{half} \times 2 = 1.2(2.094) = 2.513$ m where $\theta_{half}$ is the half-angle from top. Actually: $P = R_0 \cdot 2\theta$ where $R_0 = D/2 = 0.6$ m: $P = 0.6(4.189) = 2.513$ m.
$$R = \frac{A}{P} = \frac{0.910}{2.513} = 0.362 \text{ m}$$
$$Q = \frac{1}{n}AR^{2/3}S^{1/2} = \frac{1}{0.013}(0.910)(0.362)^{2/3}(0.0008)^{1/2}$$
$$(0.362)^{2/3}: (0.362)^{1/3} = 0.7136, \quad (0.7136)^2 = 0.5092$$
$$Q = 76.92(0.910)(0.5092)(0.02828) = 76.92(0.01311) = 1.008 \text{ m}^3/\text{s}$$
$$V = \frac{Q}{A} = \frac{1.008}{0.910} = 1.108 \text{ m/s}$$
Answer: Discharge Q ≈ 1.01 m³/s and velocity V ≈ 1.11 m/s. Interestingly, a circular pipe flowing about 93% full by depth carries slightly more discharge than when completely full — at 75% full, both Q and V are below their maximum but significantly above the full-pipe values divided by area ratio.
Problem: Hydraulic Jump — Power Dissipated and Location
Water flows at Q = 18 m³/s in a 6-m wide rectangular channel. The flow upstream of a sluice gate has depth y₁ = 0.50 m. A hydraulic jump forms downstream. Determine: (a) the upstream velocity and Froude number, (b) the sequent depth y₂, (c) the energy loss in the jump, and (d) the power dissipated (kW) if the channel width is uniform.
$$V_1 = \frac{Q}{by_1} = \frac{18}{6(0.50)} = 6.0 \text{ m/s}$$
$$Fr_1 = \frac{V_1}{\sqrt{gy_1}} = \frac{6.0}{\sqrt{9.81(0.50)}} = \frac{6.0}{2.214} = 2.710$$
$Fr_1 > 1$: supercritical flow — jump will occur.
$$\frac{y_2}{y_1} = \frac{1}{2}\left(\sqrt{1 + 8Fr_1^2} - 1\right) = \frac{1}{2}\left(\sqrt{1 + 8(2.710)^2} - 1\right)$$
$$= \frac{1}{2}\left(\sqrt{1 + 58.73} - 1\right) = \frac{1}{2}(\sqrt{59.73} - 1) = \frac{1}{2}(7.729 - 1) = 3.364$$
$$y_2 = 3.364(0.50) = 1.682 \text{ m}$$
$$\Delta E = \frac{(y_2 - y_1)^3}{4y_1 y_2} = \frac{(1.682 - 0.50)^3}{4(0.50)(1.682)} = \frac{(1.182)^3}{3.364}$$
$$= \frac{1.651}{3.364} = 0.491 \text{ m}$$
$$P_{dissipated} = \gamma Q \Delta E = 9.81(18)(0.491) = 86.73 \text{ kW}$$
Answer: (a) V₁ = 6.0 m/s, Fr₁ = 2.71 (supercritical). (b) Sequent depth y₂ = 1.68 m. (c) Energy loss ΔE = 0.491 m. (d) Power dissipated = 86.7 kW. The hydraulic jump converts kinetic energy to heat and turbulence — this energy dissipation is deliberately exploited in stilling basins downstream of spillways and sluice gates to prevent scour.