Open channel flow

This chapter covers the following:

• Open channel section properties and flow section factors
• Specific energy and critical depth
• Hydraulic elements for partial flow in circular sewers
• Specific momentum and hydraulic jump
• Best hydraulic efficient sections

FE exam tip: The equations in this chapter - section factors, Manning’s equation, critical depth, the momentum function, and conjugate depth - are all provided in the FE Reference Handbook (Water Resources and Environmental Engineering section). The tested skill is locating and applying the right equation quickly, not memorizing it.

All worked examples here use SI units. If a problem mixes SI and USCS values, convert everything to one system before calculating, and carry full precision through multi-step problems (like a hydraulic jump followed by an energy-loss calculation) rather than rounding intermediate depths.

Open channel section properties and flow section factors

Open channel flow is driven by gravity, so the channel geometry strongly affects how much water can be conveyed and how the flow behaves. We describe that geometry using section properties (area, wetted perimeter, hydraulic radius, etc.). We then combine those properties into section factors, which make discharge and force calculations more direct.

These ideas show up repeatedly in:

• uniform-flow calculations (for example, Manning’s equation)
• comparisons of channel efficiency
• checks on how changing depth or shape changes conveyance capacity

Section properties

Rectangle

• Area:

$$A=by$$

• Wetted perimeter:

$$P=b+2y$$

• Hydraulic radius:

$$R=\frac{by}{b+2y}$$

• Top width:

$$T=b$$

• Hydraulic depth:

$$D=y$$

• Section factor:

$$Z=b{y}^{1.5}$$

Example: For a rectangular channel with $b=4m$ and $y=2m$: $A=8m^2$, $P=8m$, $R=1m$, $T=4m$, $D=2m$, $Z=11.31m^{2.5}$.

Trapezoid

• Area:

$$A=(b+zy)y$$

• Wetted perimeter:

$$P=b+2y\sqrt{1+z^2}$$

• Hydraulic radius:

$$R=\frac{(b+zy)y}{b+2y\sqrt{1+z^2}}$$

• Top width:

$$T=b+2zy$$

• Hydraulic depth:

$$D=\frac{(b+zy)y}{b+2zy}$$

• Section factor:

$$Z=\frac{[(b+zy)y{]}^{1.5}}{\sqrt{b+2zy}}$$

Example: For $b=3m$, $z=1$, and $y=2m$: $A=10m^2$, $P\approx 8.65m$, $R\approx 1.16m$, $T=7m$, $D=1.43m$, $Z\approx 11.95m^{2.5}$.

Triangle

• Area:

$$A=z{y}^2$$

• Wetted perimeter:

$$P=2y\sqrt{1+z^2}$$

• Hydraulic radius:

$$R=\frac{z{y}^2}{2y\sqrt{1+z^2}}=\frac{zy}{2\sqrt{1+z^2}}$$

• Top width:

$$T=2zy$$

• Hydraulic depth:

$$D=\frac{1}{2}y$$

• Section factor:

$$Z=\frac{\sqrt{2}}{2}z{y}^{2.5}$$

Example: For $z=1$ and $y=2m$: $A=4m^2$, $P\approx 5.66m$, $R\approx 0.71m$, $T=4m$, $D=1m$, $Z\approx 4.0m^{2.5}$.

Flow section factors

Flow section factors package the geometric properties into forms that are convenient for common open-channel calculations. The uniform-flow factors below are $A{R}^{2/3}{S}^{1/2}$ written out for each shape, so Manning’s equation gives $Q=\frac{1}{n}\times$ (factor) in SI units. For critical flow, the Handbook condition $\frac{Q^2}{g}=\frac{A^3}{T}$ is the same as $Z_c=\frac{Q}{\sqrt{g}}$, because $Z=A\sqrt{A/T}$. So critical depth is found by computing $Z_c$ from the discharge and then solving the shape’s $Z$ expression (from the section properties above) for $y$.

Rectangle

• Uniform flow section factor:

$$\left[\frac{(by{)}^{5/3}{S}^{1/2}}{(b+2y{)}^{2/3}}\right]$$

• Critical depth:

$$y_c={\left(\frac{Z}{b}\right)}^{2/3}$$

Example: A rectangular channel with $b=4m$ carries $Q=12m^3/s$: $Z_c=12/\sqrt{9.81}=3.83m^{2.5}$, so $y_c=(3.83/4{)}^{2/3}\approx 0.97m$ - the same answer the unit-discharge form $y_c=(q^2/g{)}^{1/3}$ gives for this channel below.

Trapezoid

• Uniform flow section factor:

$$\left[\frac{[(b+zy)y{]}^{5/3}{S}^{1/2}}{(b+2y\sqrt{1+z^2}{)}^{2/3}}\right]$$

• Critical depth (no closed form; solve the Handbook’s critical-flow condition for $y_c$ by trial):

$$\frac{A^3}{T}=\frac{[(b+z{y}_c)y_c{]}^3}{b+2z{y}_c}=\frac{Q^2}{g}$$

Example: For $b=3m$, $z=1$, and $Q=20m^3/s$, the target is $Q^2/g=40.8m^5$. Trying $y=1.40m$: $A=6.16m^2$, $T=5.80m$, $A^3/T=40.3$ (slightly low). Trying $y=1.41m$: $A=6.22m^2$, $T=5.82m$, $A^3/T=41.3$ (slightly high). So $y_c\approx 1.40m$.

Triangle

• Uniform flow section factor:

$$\frac{z^{5/3}{y}^{8/3}{S}^{1/2}}{(4(1+z^2){)}^{1/3}}$$

• Critical depth:

$$y_c={\left(\frac{\sqrt{2}Z}{z}\right)}^{0.4}$$

Example: A triangular channel with $z=1$ carries $Q=5m^3/s$: $Z_c=5/\sqrt{9.81}=1.60m^{2.5}$, so $y_c=(\sqrt{2}\cdot 1.60/1{)}^{0.4}\approx 1.39m$.

Manning’s equation for open channel flow

Manning’s equation relates discharge to channel geometry, roughness, and slope:

$$Q=\frac{1}{n}A{R}^{2/3}{S}^{1/2}$$

Where:

• $n$ = Manning’s roughness coefficient
• $A$ = Area of flow
• $R$ = Hydraulic radius
• $S$ = Slope of the energy grade line

Example: For $b=3m$, $y=2m$, $n=0.015$, $S=0.001$: $A=6m^2$, $P=7m$, $R=0.857m$, $Q\approx 11.4m^3/s$.

Specific energy and critical depth

Specific energy measures the energy of the flow relative to the channel bottom at a particular cross-section. For a fixed discharge, the specific energy varies with depth, and there is a minimum value. The depth at that minimum is the critical depth, which separates:

• subcritical flow (deeper, slower)
• supercritical flow (shallower, faster)

This relationship is used to identify control sections and to analyze transitions where the flow regime may change.

Specific energy (E)

Specific energy is the energy relative to the channel bottom at a given cross-section:

$$E=y+\frac{v^2}{2g}$$

Where:

• $y$ = Flow depth ($m$)
• $v$ = Flow velocity ($m/s$)

Example: For $y=1.5m$, $v=2m/s$: $E=1.5+\frac{4}{19.62}\approx 1.71m$.

Critical depth ($y_c$) for rectangular channels

For a rectangular channel, critical depth can be written in terms of unit discharge:

$$y_c={\left(\frac{q^2}{g}\right)}^{1/3}$$

Where:

• $q=\frac{Q}{B}$ = Unit discharge ($m^2/s$)
• $B$ = Channel width ($m$)

Example: For $Q=12m^3/s$, $B=4m$: $q=3m^2/s$, $y_c=(9/9.81{)}^{1/3}\approx 0.97m$.

Froude number at critical depth

At critical flow, the Froude number equals 1:

$$F_r=\frac{v}{\sqrt{gy}}=1$$

Example: For $v=3m/s$, $y=0.92m$: $F_r=3/\sqrt{9.81\cdot 0.92}\approx 1$ (critical).

Pitfall: A given specific energy (below the channel’s minimum) generally corresponds to two possible depths - one subcritical ($F_r<1$, deeper and slower) and one supercritical ($F_r>1$, shallower and faster). Check the Froude number to determine which root matches the physical flow regime described in the problem. Also remember that a trapezoidal channel has no closed-form critical depth: solve $A^3/T=Q^2/g$ by trial, as in the trapezoid example above.

Hydraulic elements for partial flow in circular sewers

A circular pipe flowing partially full behaves like an open channel: the water surface is exposed to the atmosphere, and the hydraulic properties depend on depth. As depth changes, the flow area and wetted perimeter change nonlinearly, so designers often work with ratios relative to full-flow conditions.

In circular sewer pipes, flow often occurs at partial depth. Important hydraulic ratios are:

• Depth ratio: $\frac{y}{D}$
• Area ratio: $\frac{A}{A_f}$
• Wetted perimeter ratio: $\frac{P}{P_f}$
• Hydraulic radius ratio: $\frac{R}{R_f}$
• Velocity ratio: $\frac{v}{v_f}$
• Discharge ratio: $\frac{Q}{Q_f}$

Where:

• $D$ = Diameter of pipe
• Subscript $f$ = full-flow condition

Example: For $D=1.0m$, $y=0.5m$: $y/D=0.5$. From the standard constant-$n$ hydraulic-elements chart, $R/R_f\approx 1.0$, $v/v_f\approx 1.0$, and $Q/Q_f\approx 0.50$.

The curves do not peak at full flow. With a constant Manning $n$, velocity is greatest near $y/D=0.81$ (about $1.14v_f$) and discharge near $y/D=0.94$ (about $1.08Q_f$), both above their full-flow values: close to the crown the wetted perimeter grows faster than the flow area, so the hydraulic radius falls.

Sewage flow ratio curves

These are empirical curves used in circular sewer design to determine:

• Flow depth ratio $\frac{y}{D}$
• Velocity ratio $\frac{v}{v_f}$
• Discharge ratio $\frac{Q}{Q_f}$

Example problem

Given: Circular pipe with $D=1.2\text{m}$, flow depth $y=0.6\text{m}$, slope $S=0.001$, and $n=0.013$. Find the discharge.

Full flow area and radius

$$A_f=\frac{\pi D^2}{4}=\frac{\pi \cdot (1.2{)}^2}{4}=1.131{\text{m}}^2$$

$$R_f=\frac{D}{4}=\frac{1.2}{4}=0.3\text{m}$$

Full flow discharge

$$Q_f=\frac{1}{n}{A}_f{R}_f^{2/3}{S}^{1/2}$$

$$Q_f=\frac{1}{0.013}\cdot 1.131\cdot (0.3{)}^{2/3}\cdot (0.001{)}^{1/2}\approx 1.23{\text{m}}^3/\text{s}$$

Partial flow (using approximate flow ratio)

Since $y/D=0.5$, $Q/Q_f\approx 0.50$ (matching the standard constant-$n$ hydraulic-elements chart):

$$Q=0.50\cdot 1.23\approx 0.62{\text{m}}^3/\text{s}$$

Specific momentum and hydraulic jump

Specific momentum (also called the momentum function or specific force) is used for rapidly varied flow, where depth changes over a short distance. A hydraulic jump is the classic example: supercritical flow transitions abruptly to subcritical flow, dissipating energy.

Using the momentum principle, you can determine:

• conjugate (sequent) depths
• jump characteristics and forces
• design parameters for stilling basins and energy dissipators

Specific momentum (M)

The momentum function or specific force for a rectangular channel:

$$M=\frac{Q^2}{gA}+A\cdot \bar{y}$$

Where:

• $\bar{y}$ = Depth from the water surface to the centroid of the flow area ($\bar{y}=y/2$ for a rectangular channel)

Example: For $Q=10m^3/s$, $b=3m$, $y=2m$: $A=6m^2$, $\bar{y}=1m$, $M=\frac{100}{9.81\cdot 6}+6\cdot 1\approx 7.7m^3$.

Hydraulic jump (conjugate depths)

For a rectangular channel:

$$y_2=\frac{1}{2}{y}_1\left(\sqrt{1+8F_1^2}-1\right)$$

Where:

• $y_1$ = Depth before jump
• $y_2$ = Depth after jump
• $F_1$ = Froude number before jump

Example: For $y_1=0.5m$, $F_1=3$: $y_2=0.25(\sqrt{73}-1)\approx 1.89m$.

Energy loss in hydraulic jump

$$\Delta E=\frac{(y_2-y_1{)}^3}{4y_1{y}_2}$$

Example: Using $y_1=0.5m$, $y_2=1.89m$: $\Delta E\approx 0.71m$.

Best hydraulic efficient sections

A best hydraulic section conveys a given discharge with the minimum wetted perimeter for a given area. Minimizing wetted perimeter increases hydraulic radius, which generally improves conveyance and reduces energy loss.

These proportions are commonly used in preliminary design for canals and lined channels where efficiency and economy matter.

Trapezoid (half of a hexagon)

Area

$$A=\sqrt{3}{y}^2$$

Wetted perimeter

$$P=2\sqrt{3}y$$

Hydraulic radius

$$R=\frac{1}{2}y$$

Top width

$$T=\frac{4}{3}\sqrt{3}y$$

Hydraulic depth

$$D=\frac{3}{4}y$$

Section factor

$$Z=1.5y^{2.5}$$

Rectangle (half of a square)

Area

$$A=2y^2$$

Wetted perimeter

$$P=4y$$

Hydraulic radius

$$R=\frac{1}{2}y$$

Top width

$$T=2y$$

Hydraulic depth

$$D=y$$

Section factor

$$Z=2y^{2.5}$$

Triangle (half of a square)

Area

$$A=y^2$$

Wetted perimeter

$$P=2\sqrt{2}y$$

Hydraulic radius

$$R=\frac{1}{4}\sqrt{2}y$$

Top width

$$T=2y$$

Hydraulic depth

$$D=\frac{1}{2}y$$

Section factor

$$Z=\frac{\sqrt{2}}{2}{y}^{2.5}$$

Semicircle

Area

$$A=\frac{\pi }{2}{y}^2$$

Wetted perimeter

$$P=\pi y$$

Hydraulic radius

$$R=\frac{1}{2}y$$

Top width

$$T=2y$$

Hydraulic depth

$$D=\frac{\pi }{4}y$$

Section factor

$$Z=\frac{{\pi }^{3/2}}{4}{y}^{2.5}\approx 1.39y^{2.5}$$

Example: comparing the best-hydraulic sections at $y=2m$

Shape	$A$ ($m^2$)	$P$ ($m$)	$R$ ($m$)	$T$ ($m$)	$D$ ($m$)	$Z$ ($m^{2.5}$)
Trapezoid	6.93	6.93	1.0	4.62	1.5	8.49
Rectangle	8.0	8.0	1.0	4.0	2.0	11.3
Triangle	4.0	5.66	0.707	4.0	1.0	4.0
Semicircle	6.28	6.28	1.0	4.0	1.57	7.87