8.4 Open Channel Flow, Manning's Equation, and Hydraulic Jump

8.4 Open Channel Flow, Manning's Equation, and Hydraulic Jump

Open channel hydraulics encompasses fluid conveyance where the flowing liquid possesses an unconfined free surface subjected to local atmospheric pressure. In civil engineering practice and the CELE licensure examination, open channel flow governs stormwater drainage, culverts, spillway chutes, irrigation canals, and natural river floodplains. Unlike pressurized pipe flow, where the cross-sectional area is fixed by the conduit geometry, open channel flow depth ($y$) varies dynamically with discharge, channel slope, boundary roughness, and upstream/downstream controls.

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1. Open Channel Geometry Fundamentals

The hydraulic response of an open channel depends on five fundamental geometric properties of its flow cross-section:

Cross-Sectional Geometry Formulations

• Rectangular Channel (width $b$, depth $y$): $A = b y, \quad P = b + 2y, \quad R_h = \frac{b y}{b + 2y}, \quad T = b, \quad D_h = y$
• Trapezoidal Channel (bottom width $b$, depth $y$, side slope $1\text{V}:z\text{H}$): $A = (b + z y) y, \quad P = b + 2y \sqrt{1 + z^2}, \quad T = b + 2 z y, \quad D_h = \frac{(b + zy)y}{b + 2zy}$
• Triangular Channel (side slope $1\text{V}:z\text{H}$, included vertex angle $2\theta$ where $z = \tan \theta$): $A = z y^2, \quad P = 2y \sqrt{1 + z^2}, \quad R_h = \frac{z y}{2\sqrt{1 + z^2}}, \quad T = 2 z y, \quad D_h = \frac{y}{2}$

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2. Uniform Steady Flow: The Chezy and Manning Equations

In steady uniform flow, gravity driving forces exactly balance boundary shear resistance. Consequently, the water depth remains constant at the normal depth ($y_n$), the velocity profile is invariant along the reach, and the bed slope ($S_0$), water surface slope ($S_w$), and energy grade line slope ($S_f$) are all strictly parallel ($S_0 = S_w = S_f = S$).

The Chezy Formula

Developed by Antoine de Chézy in 1769 from force equilibrium: $v = C \sqrt{R_h S}$ where $C$ is the Chezy resistance factor ($\text{m}^{1/2}/\text{s}$).

The Manning Equation (SI Metric)

Robert Manning (1889) established the empirical relation linking Chezy's $C$ to boundary roughness: $C = \frac{1}{n} R_h^{1/6}$. Substituting into Chezy's equation produces the Manning Equation in SI units: $v = \frac{1}{n} R_h^{2/3} S^{1/2}$ $Q = A v = \frac{1}{n} A R_h^{2/3} S^{1/2}$ where $Q$ is discharge ($\text{m}^3/\text{s}$), $A$ is flow area ($\text{m}^2$), $R_h$ is hydraulic radius ($\text{m}$), $S$ is longitudinal bed slope (dimensionless, $\text{m/m}$), and $n$ is Manning's roughness coefficient ($\text{s}/\text{m}^{1/3}$).

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3. Hydraulically Most Efficient (Most Economical) Cross-Sections

A channel cross-section is defined as hydraulically most efficient (most economical) when it conveys the maximum discharge $Q$ for a given flow area $A$, bed slope $S$, and roughness $n$. By inspecting Manning's formula ($Q \propto R_h^{2/3} = (A/P)^{2/3}$), maximizing $Q$ for a constant area $A$ requires minimizing the wetted perimeter $P$ ($dP/dy = 0$). Minimizing $P$ also minimizes excavation volume and concrete lining cost.

1. Most Economical Rectangular Channel

For a rectangle of area $A = b y$, wetted perimeter is $P = b + 2y = \frac{A}{y} + 2y$. $\frac{dP}{dy} = - \frac{A}{y^2} + 2 = 0 \implies A = 2y^2 \implies b y = 2y^2 \implies \boxed{b = 2y}$ $\boxed{R_h = \frac{A}{P} = \frac{2y^2}{2y + 2y} = \frac{y}{2}}$ The optimal rectangular channel has a width equal to twice the water depth, and its hydraulic radius equals half the depth.

2. Most Economical Trapezoidal Channel (The Semi-Hexagon)

For a trapezoidal channel with variable bottom width $b$ and side slope $z$:

• Setting $\frac{\partial P}{\partial y} = 0$ and $\frac{\partial P}{\partial z} = 0$ yields a semi-regular hexagon:
  • Side slope angle with horizontal: $\theta = 60^\circ \implies z = \frac{1}{\tan 60^\circ} = \frac{1}{\sqrt{3}} \approx 0.57735$
  • Bottom width: $b = \frac{2}{\sqrt{3}} y \approx 1.1547 y$
  • Top width: $T = 2 b = \frac{4}{\sqrt{3}} y \approx 2.3094 y$
  • Flow Area: $A = \sqrt{3} y^2 \approx 1.73205 y^2$
  • Wetted perimeter: $P = 2\sqrt{3} y \approx 3.4641 y$
  • Hydraulic radius: $\boxed{R_h = \frac{y}{2}}$
• Inscribed Circle Criterion: Any most economical polygonal channel of any number of sides forms a polygon circumscribed about a semicircle whose center lies on the free water surface and whose radius is $r = y$.

3. Circular Conduits Flowing Partially Full

Due to boundary perimeter friction near the crown, circular pipes flowing full do not convey the maximum flow or velocity:

• Maximum Velocity Depth: Occurs at depth $y = 0.810 d_0$ ($v_{\max} = 1.14 v_{\text{full}}$), corresponding to a subtended central angle $\theta \approx 257^\circ$.
• Maximum Discharge Depth: Occurs at depth $y = 0.938 d_0$ ($Q_{\max} = 1.08 Q_{\text{full}}$), corresponding to a subtended central angle $\theta \approx 302^\circ$.

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4. Specific Energy and Critical Flow

Specific Energy ($E$) is the total mechanical energy head measured relative to the channel invert (channel bottom datum): $E = y + \frac{v^2}{2g} = y + \frac{Q^2}{2g A^2}$

Loading diagram...

The Specific Energy Curve and Alternate Depths

Plotting specific energy $E$ versus depth $y$ for a fixed discharge $Q$ reveals two asymptotes ($y = 0$ and $E = y$). For any specific energy value $E > E_{\min}$, there exist two possible alternate depths:

1. A small depth with high velocity (supercritical flow).
2. A large depth with low velocity (subcritical flow).

Critical Flow Condition

Critical depth occurs at the point of minimum specific energy ($dE/dy = 0$): $\frac{dE}{dy} = 1 - \frac{Q^2}{g A^3} \frac{dA}{dy} = 0$ Since the rate of change of area with depth is the top surface width ($\frac{dA}{dy} = T$): $\boxed{\frac{Q^2 T}{g A^3} = 1 \iff \frac{v^2}{g (A/T)} = 1 \iff \frac{v}{\sqrt{g D_h}} = 1}$ This is the universal critical flow criterion for an open channel of arbitrary cross-sectional shape.

Critical Flow in Rectangular Channels

Defining unit discharge $q = \frac{Q}{b}$ (discharge per meter of channel width, $\text{m}^3/(\text{s}\cdot\text{m})$): $A = b y_c, \quad T = b \implies \frac{Q^2 b}{g (b y_c)^3} = \frac{q^2}{g y_c^3} = 1$ $\boxed{y_c = \sqrt[3]{\frac{q^2}{g}} = \left(\frac{q^2}{g}\right)^{1/3}}$ $v_c = \sqrt{g y_c}, \quad \frac{v_c^2}{2g} = \frac{y_c}{2}$ $\boxed{E_{\min} = y_c + \frac{v_c^2}{2g} = y_c + 0.5 y_c = 1.5 y_c = \frac{3}{2} y_c}$

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5. The Froude Number and Flow Regimes

The dimensionless Froude Number ($Fr$) represents the ratio of inertial forces to gravitational forces, and physically corresponds to the ratio of flow velocity $v$ to the celerity (speed) of an elementary gravity surface wave $c = \sqrt{g D_h}$: $Fr = \frac{v}{\sqrt{g D_h}} = \frac{v}{\sqrt{g (A/T)}}$

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6. The Hydraulic Jump in Rectangular Channels

A hydraulic jump is a rapid, turbulent, irreversible open-channel phenomenon wherein flow transitions abruptly from an unstable supercritical state ($Fr_1 > 1, y_1 < y_c$) to a stable subcritical state ($Fr_2 < 1, y_2 > y_c$), dissipating tremendous kinetic energy through roller vortices.

Momentum Conservation and Bélanger's Sequent Depths

Because energy is lost in the jump, Bernoulli's equation cannot predict the downstream depth. Instead, applying the Linear Momentum Equation across the jump control volume (neglecting boundary shear over the short jump length) yields equality of the Specific Force (Momentum Function $M$): $M = \frac{Q^2}{g A} + A \bar{y} = \text{constant}$ For a rectangular channel of width $b$, integrating hydrostatic pressure and momentum fluxes produces the Bélanger Sequent (Conjugate) Depth Equation: $\boxed{y_2 = \frac{y_1}{2} \left(\sqrt{1 + 8 Fr_1^2} - 1\right)}$ Conversely, expressing upstream depth in terms of downstream Froude number: $y_1 = \frac{y_2}{2} \left(\sqrt{1 + 8 Fr_2^2} - 1\right)$

Hydraulic Jump Characteristics

1. Head Loss Across Jump ($\Delta E$): The total energy head dissipated by turbulent rollers: $\Delta E = E_1 - E_2 = \frac{(y_2 - y_1)^3}{4 y_1 y_2}$
2. Height of Jump ($h_j$): $h_j = y_2 - y_1$.
3. Length of Jump ($L_j$): Empirically, $L_j \approx 5\text{ to }7 (y_2 - y_1)$ for $Fr_1$ between $4.5$ and $9.0$.
4. Power Dissipated in Jump ($P$): $P = \gamma Q \Delta E \quad (\text{kW for } Q \text{ in m}^3/\text{s}, \gamma \text{ in kN/m}^3, \Delta E \text{ in m})$
5. Jump Classification by Initial Froude Number ($Fr_1$):
   • $1.0 < Fr_1 \le 1.7$: Undular Jump (slight standing waves on surface, negligible energy loss).
   • $1.7 < Fr_1 \le 2.5$: Weak Jump (small surface rollers, $\Delta E / E_1 < 15\%$).
   • $2.5 < Fr_1 \le 4.5$: Oscillating Jump (jet oscillates from bottom to surface, produces destructive waves downstream).
   • $4.5 < Fr_1 \le 9.0$: Steady / Well-Established Jump (best operating range for stilling basins, $45\%\text{–}70\%$ energy dissipation).
   • $Fr_1 > 9.0$: Strong / Chutting Jump (rough, violent action, up to $85\%$ energy dissipation).

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7. Worked Example: Comprehensive Open Channel & Hydraulic Jump Analysis

Problem Statement: A rectangular concrete spillway apron has a channel width of $b = 6.00\text{ m}$. Water issues from beneath a high-head sluice gate at a uniform depth of $y_1 = 0.600\text{ m}$ with an initial velocity of $v_1 = 12.00\text{ m/s}$. The flow then enters a horizontal stilling basin and forms a hydraulic jump. Taking $\gamma = 9.81\text{ kN/m}^3$ and $g = 9.81\text{ m/s}^2$, determine:

1. The total discharge $Q$ and unit discharge $q$.
2. The initial Froude number $Fr_1$ and initial specific energy $E_1$.
3. The critical depth $y_c$.
4. The sequent (conjugate) depth $y_2$ downstream of the jump.
5. The energy head loss $\Delta E$ across the jump.
6. The power dissipated by the jump in kilowatts.

Step-by-Step Solution:

1. Compute discharge and unit discharge: $Q = A_1 v_1 = (b y_1) v_1 = (6.00\text{ m} \times 0.600\text{ m})(12.00\text{ m/s}) = 3.60\text{ m}^2 \times 12.00\text{ m/s} = 43.20\text{ m}^3/\text{s}$ $q = \frac{Q}{b} = \frac{43.20\text{ m}^3/\text{s}}{6.00\text{ m}} = 7.20\text{ m}^3/(\text{s}\cdot\text{m})$

2. Compute initial Froude number and specific energy: $Fr_1 = \frac{v_1}{\sqrt{g y_1}} = \frac{12.00\text{ m/s}}{\sqrt{(9.81\text{ m/s}^2)(0.600\text{ m})}} = \frac{12.00}{\sqrt{5.886}} = \frac{12.00}{2.4261} = 4.946$ Because $Fr_1 = 4.95 > 1$, the incoming flow is strongly supercritical. $E_1 = y_1 + \frac{v_1^2}{2g} = 0.600\text{ m} + \frac{(12.00\text{ m/s})^2}{2(9.81\text{ m/s}^2)} = 0.600 + \frac{144.0}{19.62} = 0.600 + 7.339 = 7.939\text{ m}$

3. Compute critical depth ($y_c$): $y_c = \sqrt[3]{\frac{q^2}{g}} = \sqrt[3]{\frac{(7.20)^2}{9.81}} = \sqrt[3]{\frac{51.84}{9.81}} = \sqrt[3]{5.2844} = 1.742\text{ m}$ (Notice that $y_1 = 0.600\text{ m} < y_c = 1.742\text{ m}$, confirming supercritical flow). $E_{\min} = 1.5 y_c = 1.5(1.742\text{ m}) = 2.613\text{ m}$

4. Calculate sequent depth ($y_2$) via the Bélanger Equation: $y_2 = \frac{y_1}{2} \left(\sqrt{1 + 8 Fr_1^2} - 1\right) = \frac{0.600}{2} \left(\sqrt{1 + 8(4.946)^2} - 1\right)$ $8 Fr_1^2 = 8(24.463) = 195.705$ $\sqrt{1 + 195.705} = \sqrt{196.705} = 14.025$ $y_2 = 0.300 \times (14.025 - 1) = 0.300 \times 13.025 = 3.908\text{ m} \approx 3.91\text{ m}$

5. Compute energy head loss ($\Delta E$): $\Delta E = \frac{(y_2 - y_1)^3}{4 y_1 y_2} = \frac{(3.908 - 0.600)^3}{4(0.600)(3.908)} = \frac{(3.308)^3}{9.379} = \frac{36.199}{9.379} = 3.860\text{ m}$ (Verification: $v_2 = \frac{Q}{b y_2} = \frac{43.20}{6.00 \times 3.908} = 1.842\text{ m/s}$. $E_2 = 3.908 + \frac{1.842^2}{19.62} = 3.908 + 0.173 = 4.081\text{ m}$. $\Delta E = E_1 - E_2 = 7.939 - 4.081 = 3.858\text{ m}$. Matches perfectly!)

6. Compute power dissipated by the jump: $P = \gamma Q \Delta E = (9.81\text{ kN/m}^3)(43.20\text{ m}^3/\text{s})(3.860\text{ m}) = 1,635.8\text{ kW} \approx 1.64\text{ MW}$

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8. CELE Exam Traps & Common Computational Errors