10.2 Open Channel Flow

9.2 Open Channel Flow

Open channel flow is defined as the flow of a liquid with a free surface exposed to atmospheric pressure. Unlike closed-conduit pressurized flow, the primary driving force in open channel systems is gravity, with the fluid flowing downhill along a channel slope. This section covers uniform flow, specific energy, critical flow, and the design and analysis of common hydraulic control structures.

Geometric Properties of Channels

To analyze flow in open channels, engineers must define several geometric parameters based on the shape of the cross-section:

• Flow depth ($y$): The vertical distance from the lowest point of the channel bottom to the free surface.
• Flow area ($A$): The cross-sectional area of the flow perpendicular to the flow direction.
• Wetted perimeter ($P$): The length of the channel boundary in contact with the liquid.
• Hydraulic radius ($R_h$): The ratio of flow area to wetted perimeter ($R_h = A/P$).
• Top width ($T$): The width of the channel cross-section at the free surface.
• Hydraulic depth ($D$): The ratio of flow area to top width ($D = A/T$). For rectangular channels, the hydraulic depth is equal to the flow depth ($D = y$).

Common Cross-Sections and Geometry Formulas

For a rectangular channel of width $b$ and depth $y$:

• $A = b y$
• $P = b + 2y$
• $R_h = \frac{b y}{b + 2y}$
• $T = b$
• $D = y$

For a trapezoidal channel of bottom width $b$, depth $y$, and side slopes $z:1$ (horizontal:vertical):

• $A = (b + z y) y$
• $P = b + 2 y \sqrt{1 + z^2}$
• $R_h = \frac{(b + z y) y}{b + 2 y \sqrt{1 + z^2}}$
• $T = b + 2 z y$
• $D = \frac{(b+zy)y}{b+2zy}$

---

Uniform Flow and Manning's Equation

Uniform flow occurs when the depth, flow area, velocity, and discharge at every section of the channel are constant. In this state, the energy grade line, water surface, and channel bottom are all parallel, meaning the energy slope ($S_f$), water surface slope ($S_w$), and channel bottom slope ($S_0$) are equal ($S_0 = S_w = S_f = S$). The depth of flow in a uniform state is called the normal depth ($y_n$).

Manning's Equation

The most common empirical formula used to analyze uniform flow is Manning's equation. In US Customary units, the average velocity and volumetric flow rate ($Q = V A$) are: $V = \frac{1.486}{n} R_h^{2/3} S^{1/2}$ $Q = \frac{1.486}{n} A R_h^{2/3} S^{1/2}$

In SI units, the velocity and flow rate are: $V = \frac{1}{n} R_h^{2/3} S^{1/2}$ $Q = \frac{1}{n} A R_h^{2/3} S^{1/2}$ where:

• $n$ is the Manning roughness coefficient (dimensionless coefficient, representing boundary friction).
• $R_h$ is the hydraulic radius (ft or m).
• $S$ is the channel slope (dimensionless, e.g., ft/ft).
• $A$ is the flow area ($\text{ft}^2$ or $\text{m}^2$).
• $Q$ is the discharge (cfs or $\text{m}^3\text{/s}$).

Solving for normal depth ($y_n$) when $Q$, $n$, $S$, and the cross-section geometry are known is an implicit mathematical problem. It typically requires trial-and-error, graphical charts, or numerical solver tools.

---

Specific Energy and Critical Flow

Specific energy ($E$) is the energy of flow at any section of the channel measured relative to the channel bottom as the datum: $E = y + \frac{V^2}{2g} = y + \frac{Q^2}{2g A^2}$ For a constant discharge $Q$, the specific energy is a function of depth $y$ only.

Specific Energy Diagram

When plotting $E$ against $y$ for a constant $Q$:

1. The curve has two branches. The upper branch represents high depth and low velocity (subcritical flow). The lower branch represents low depth and high velocity (supercritical flow).
2. The two branches meet at a point of minimum specific energy ($E_{min}$). The depth corresponding to this minimum energy is called the critical depth ($y_c$).
3. For any energy level $E > E_{min}$, there are two possible depths, known as alternate depths (one subcritical and one supercritical).

Mathematical Definition of Critical Flow

Critical flow represents the boundary between subcritical and supercritical flow regimes. The general criterion for critical flow in any channel shape is: $\frac{Q^2 T}{g A^3} = 1$ At critical depth, the Froude number is exactly equal to 1.

For a rectangular channel of width $b$, the flow per unit width (unit discharge) is defined as $q = Q/b$. The critical depth simplifies to: $y_c = \sqrt[3]{\frac{q^2}{g}} = \sqrt[3]{\frac{Q^2}{g b^2}}$ The minimum specific energy in a rectangular channel is: $E_{min} = 1.5 y_c$

Froude Number

The dimensionless Froude number ($Fr$) represents the ratio of inertial forces to gravitational forces: $Fr = \frac{V}{\sqrt{g D}}$ where $D = A/T$ is the hydraulic depth.

• If $Fr < 1$: The flow is subcritical. Gravitational forces dominate, the velocity is less than the wave celerity, and downstream conditions can affect the upstream flow.
• If $Fr = 1$: The flow is critical.
• If $Fr > 1$: The flow is supercritical. Inertial forces dominate, the velocity is greater than the wave celerity, and downstream disturbances cannot travel upstream.

---

Hydraulic Structures

Engineers use hydraulic structures to control flow depth, measure discharge, or dissipate energy.

Hydraulic Jumps

A hydraulic jump is a rapid, turbulent transition from supercritical flow ($Fr_1 > 1$) to subcritical flow ($Fr_2 < 1$). This phenomenon is highly turbulent and is used to dissipate kinetic energy downstream of spillways, preventing erosion.

For a horizontal, rectangular channel, the depths before ($y_1$) and after ($y_2$) the jump are called sequent depths (or conjugate depths) and are related by the Belanger equation: $\frac{y_2}{y_1} = \frac{1}{2} \left( \sqrt{1 + 8 Fr_1^2} - 1 \right)$ The energy head loss ($h_L$) in the jump is calculated as: $h_L = E_1 - E_2 = \frac{(y_2 - y_1)^3}{4 y_1 y_2}$

Weirs

A weir is an obstruction placed in an open channel that forces water to flow over it, providing a reliable method for discharge measurement based on the head of water above the weir crest ($H$).

• Rectangular sharp-crested weir: The flow rate is computed as: $Q = C_w L H^{3/2}$ where:
  • $L$ is the length of the weir crest (ft or m).
  • $H$ is the head on the weir (ft or m), measured upstream of the drawdown zone.
  • $C_w$ is the weir discharge coefficient (typically around 3.33 in US Customary units).
• V-notch (triangular) weir: Excellent for measuring low flow rates. The equation is: $Q = C_d \frac{8}{15} \sqrt{2g} \tan\left(\frac{\theta}{2}\right) H^{5/2}$ For a standard 90-degree V-notch weir ($\theta = 90^\circ$) in US Customary units, this simplifies approximately to: $Q \approx 2.5 H^{2.5}$

Flumes

A flume is a specially shaped channel constriction (such as a Parshall flume) that forces flow through critical depth. By measuring the water level at a specific upstream point, engineers can determine the discharge using empirical calibration curves, which are less susceptible to sediment accumulation than weirs.

---

Comprehensive Example: Open Channel Analysis

Problem: A rectangular concrete channel ($n = 0.013$) has a width of 10 feet and a bottom slope of 0.002 ft/ft. The channel carries a constant discharge of 120 cfs. Determine:

1. The critical depth ($y_c$) and critical velocity ($V_c$).
2. The normal depth ($y_n$) under uniform flow conditions.
3. The Froude number ($Fr$) at normal depth, and classify the flow regime.

Solution Step 1: Critical Flow Calculations The unit discharge ($q$) is: $q = \frac{Q}{b} = \frac{120 \text{ cfs}}{10 \text{ ft}} = 12 \text{ cfs/ft}$ The critical depth ($y_c$) for a rectangular channel is: $y_c = \sqrt[3]{\frac{q^2}{g}} = \sqrt[3]{\frac{(12)^2}{32.2}} = \sqrt[3]{\frac{144}{32.2}} = \sqrt[3]{4.472} \approx 1.65 \text{ ft}$ The critical area is: $A_c = b y_c = 10 \text{ ft} \times 1.65 \text{ ft} = 16.5 \text{ ft}^2$ The critical velocity ($V_c$) is: $V_c = \frac{Q}{A_c} = \frac{120}{16.5} \approx 7.27 \text{ ft/s}$

Solution Step 2: Normal Depth (Uniform Flow) Calculation Using Manning's equation for flow rate in US Customary units: $Q = \frac{1.486}{n} A R_h^{2/3} S^{1/2}$ For a rectangular channel of depth $y$: $A = 10 y$ $P = 10 + 2y$ $R_h = \frac{A}{P} = \frac{10 y}{10 + 2y}$

Substitute these into Manning's equation: $120 = \frac{1.486}{0.013} (10 y) \left( \frac{10 y}{10 + 2y} \right)^{2/3} (0.002)^{1/2}$ Simplify the constant terms: $\frac{1.486}{0.013} \approx 114.308$ $(0.002)^{1/2} \approx 0.04472$ $120 = 114.308 \times 0.04472 \times 10 y \left( \frac{10 y}{10 + 2y} \right)^{2/3}$ $120 = 51.12 y \left( \frac{10 y}{10 + 2y} \right)^{2/3}$ Divide both sides by 51.12: $2.347 = y \left( \frac{10 y}{10 + 2y} \right)^{2/3}$

We solve this equation for $y$ (normal depth, $y_n$) using trial and error:

• Let $y = 1.90$: $R_h = \frac{10(1.90)}{10 + 2(1.90)} = \frac{19.0}{13.80} \approx 1.377$ $R_h^{2/3} = (1.377)^{2/3} \approx 1.237$ $\text{Value} = 1.90 \times 1.237 = 2.350 \text{ (extremely close to 2.347)}$

Thus, the normal depth is $y_n \approx 1.90 \text{ ft}$.

Solution Step 3: Froude Number and Flow Regime At normal depth $y_n = 1.90 \text{ ft}$: $A = 10 \times 1.90 = 19.0 \text{ ft}^2$ $V = \frac{Q}{A} = \frac{120}{19.0} \approx 6.32 \text{ ft/s}$ The hydraulic depth for a rectangular channel is $D = y_n = 1.90 \text{ ft}$. The Froude number is: $Fr = \frac{V}{\sqrt{g D}} = \frac{6.32}{\sqrt{32.2 \times 1.90}} = \frac{6.32}{\sqrt{61.18}} = \frac{6.32}{7.82} \approx 0.81$ Since $Fr = 0.81 < 1$ (and $y_n = 1.90 \text{ ft} > y_c = 1.65 \text{ ft}$), the uniform flow is subcritical.