12.7 Open-Channel Flow, Manning's Equation, and Drag

12.7 Open-Channel Flow, Manning's Equation, and Drag

The NCEES Fluid Mechanics specification pairs these as one sub-topic: "Open-channel flow (e.g., Manning's equation, drag)." Open-channel flow differs fundamentally from pipe flow in that it has a free surface at atmospheric pressure, so the flow is driven by gravity acting on the channel slope rather than by a pressure difference.

Manning's Equation

$V = \frac{1}{n}R_h^{2/3}S^{1/2} \quad\text{(SI: m/s)}, \qquad Q = \frac{1}{n}A R_h^{2/3}S^{1/2}$

$V = \frac{1.486}{n}R_h^{2/3}S^{1/2} \quad\text{(US Customary: ft/s)}$

where $n$ is Manning's roughness coefficient (dimensional, and the same numerical value in both systems), $R_h$ is the hydraulic radius, and $S$ is the channel slope (dimensionless, often the bed slope).

The 1.486 factor is purely a unit conversion, equal to $3.2808^{1/3}$. Applying the SI form to feet under-predicts velocity by 33%, and forgetting it in US Customary work is one of the most common errors in this topic.

Hydraulic Radius

$R_h = \frac{A}{P}$

with $P$ the wetted perimeter — the length of channel boundary in contact with water, excluding the free surface. The free surface has no shear, so it contributes no resistance and is never counted.

A pipe flowing half full has the same hydraulic radius as one flowing full — both are $D/4$ — so Manning's equation gives them the same velocity. Since the half-full pipe has half the area, it carries half the discharge at the same velocity. And a partly full pipe actually carries its maximum discharge at about 94% of full depth, because the wetted perimeter grows faster than the area in the last few percent. This counterintuitive result is a favorite exam item.

Manning's Roughness Coefficients

Because $Q \propto 1/n$, a channel that silts up and vegetates from $n = 0.014$ to $n = 0.040$ loses 65% of its capacity at the same depth and slope — the usual explanation for a culvert that suddenly floods after years of service.

Worked Example: Trapezoidal Channel Discharge

A trapezoidal channel has a 3.0 m bottom width, 2:1 side slopes ($z = 2$), a flow depth of 1.2 m, $n = 0.025$, and a bed slope of 0.0008. Find the discharge.

$A = by + zy^2 = 3.0(1.2) + 2(1.2)^2 = 3.60 + 2.88 = 6.48\ \text{m}^2$

$P = b + 2y\sqrt{1+z^2} = 3.0 + 2(1.2)\sqrt{1+4} = 3.0 + 2.4(2.236) = 3.0 + 5.367 = 8.367\ \text{m}$

$R_h = \frac{6.48}{8.367} = 0.7746\ \text{m}$

$R_h^{2/3} = (0.7746)^{2/3} = e^{0.6667\ln 0.7746} = e^{0.6667(-0.2553)} = e^{-0.1702} = 0.8435$

$S^{1/2} = \sqrt{0.0008} = 0.02828$

$Q = \frac{1}{0.025}(6.48)(0.8435)(0.02828) = 40(6.48)(0.8435)(0.02828) = \boxed{6.18\ \text{m}^3/\text{s}}$

Mean velocity $V = Q/A = 6.18/6.48 = 0.954$ m/s. Froude check: $Fr = 0.954/\sqrt{9.81(1.2)} = 0.954/3.431 = 0.278$ — subcritical, as designed channels normally are.

Most Efficient Cross-Sections

For a fixed cross-sectional area, discharge is maximized when the wetted perimeter is minimized (maximizing $R_h$). This also minimizes lining cost.

Normal vs. Critical Depth

Comparing them classifies the channel slope: $y_n > y_c$ is a mild slope with subcritical uniform flow; $y_n < y_c$ is a steep slope with supercritical uniform flow.

$E = y + \frac{V^2}{2g} \qquad\text{(specific energy, measured from the channel bed)}$

Drag

$F_D = C_D\,\frac{1}{2}\rho V^2 A$

where $A$ is the reference area — usually the frontal (projected) area for bluff bodies, and the planform area for airfoils. Check which the problem intends, since it changes $C_D$ by a large factor.

Drag scales with $V^2$, so power to overcome it scales with $V^3$ ($P = F_D V$). Doubling a vehicle's speed quadruples the drag force and requires eight times the power. This cubic relationship is why highway fuel economy degrades so sharply with speed.

Terminal Velocity

At terminal velocity, weight minus buoyancy equals drag:

$V_t = \sqrt{\frac{2(\rho_s-\rho)Vol\,g}{C_D\rho A}}$

For a sphere of diameter $D$, substituting $Vol = \pi D^3/6$ and $A = \pi D^2/4$:

$V_t = \sqrt{\frac{4gD(\rho_s-\rho)}{3C_D\rho}}$

Worked Example: Drag Force and Power

A vehicle with frontal area 2.4 m² and $C_D = 0.31$ travels at 100 km/h through air ($\rho = 1.20$ kg/m³). Find the drag force and the power required, then compare with 130 km/h.

$V = \frac{100}{3.6} = 27.78\ \text{m/s}$

$F_D = 0.31\left(\tfrac{1}{2}\right)(1.20)(27.78)^2(2.4) = 0.31(0.60)(771.6)(2.4) = 344.4\ \text{N}$

$P = F_D V = 344.4(27.78) = 9{,}568\ \text{W} = \boxed{9.57\ \text{kW}}$

At 130 km/h ($V = 36.11$ m/s), the ratio is $(36.11/27.78) = 1.30$:

$F_D = 344.4(1.30)^2 = 582\ \text{N}, \qquad P = 9.57(1.30)^3 = \boxed{21.0\ \text{kW}}$

A 30% speed increase raises drag 69% and required power 120% — more than doubling it. That is the $V^3$ law made concrete.