Uniform flow in open channel

Uniform flow is flow in which depth, area, velocity and discharge are the same at every section along the channel. The following conditions must be satisfied.

1. Prismatic channel: the cross-section (shape and size) is constant along the length.
2. Constant bed slope: the bed slope $S_0$ does not change.
3. Constant roughness: the boundary roughness (Manning's $n$) is the same all along.
4. Steady flow: the discharge does not change with time, and is constant along the channel (no lateral inflow or outflow).
5. Sufficient length: the channel is long and straight enough that the effect of entrance and exit controls dies out.
6. No control or obstruction in the reach (no weir, gate, drop, or bend) and no backwater.
7. Equilibrium of forces: the gravity force component along the slope equals the boundary shear resistance, $\gamma A L S_0=\tau_0 P L$. This gives

$$S_0=S_w=S_f\quad\text{(bed, water surface and energy line parallel)}$$

Strictly, true uniform flow is rare in nature. It is assumed for design of canals and for long reaches of rivers.

For a given area and side slope, the most efficient (least wetted perimeter) trapezoid is used for each case. Then the cost per metre of channel is compared in terms of the unit cost of lining.

Given data

$Q=15$ m³/s, $z=1$ (1H:1V), $S_0=1/1000$, $C=45$ (unlined) or $70$ (lined). Let cost of lining per m² $=1$ unit; excavation cost per m³ $=3$ units.

Most efficient trapezoidal section ($z=1$)

$$b=2y(\sqrt{1+z^2}-z)=2y(\sqrt2-1)=0.8284\,y,\qquad R=\frac y2$$ $$A=(b+zy)y=1.8284\,y^2,\qquad P=b+2y\sqrt{1+z^2}=3.6569\,y$$

Chezy's equation $Q=CA\sqrt{RS_0}$

$$15=C\,(1.8284\,y^2)\sqrt{\frac y2\times0.001}$$

Cost per metre length (assuming the excavated section equals the flow section)

• Unlined: excavation only $=3\times A=3\times9.798=29.39$ units.
• Lined: excavation $=3\times6.881=20.64$ units, plus lining $=1\times P=7.09$ units; total $=27.74$ units.

Answer: lined channel costs 27.74 units/m against 29.39 units/m for the unlined channel, so the concrete-lined channel (y = 1.94 m, b = 1.61 m) is more economical (cheaper by about 5.6%).

Condition for maximum discharge

From Manning/Chezy, $Q=\dfrac{1}{n}AR^{2/3}S^{1/2}$. For a given area $A$, slope and roughness, discharge is maximum when $R=A/P$ is maximum, i.e. when the wetted perimeter $P$ is minimum.

Part 1: Proportions for given area $A$ and side slope $z$ (H:V)

Trapezoid with bed width $b$, depth $y$:

$$A=(b+zy)y\;\Rightarrow\;b=\frac{A}{y}-zy$$ $$P=b+2y\sqrt{1+z^2}=\frac{A}{y}-zy+2y\sqrt{1+z^2}$$

For minimum $P$ at constant $A$ and $z$:

$$\frac{dP}{dy}=-\frac{A}{y^2}-z+2\sqrt{1+z^2}=0\;\Rightarrow\;\frac{A}{y^2}=2\sqrt{1+z^2}-z$$

Since $A/y^2=b/y+z$:

$$\boxed{b=2y\left(\sqrt{1+z^2}-z\right)}$$

(Second derivative $d^2P/dy^2=2A/y^3>0$, so this is a minimum.)

Hydraulic radius:

$$A=y^2\left(2\sqrt{1+z^2}-z\right),\quad P=\frac{A}{y}-zy+2y\sqrt{1+z^2}=2y\left(2\sqrt{1+z^2}-z\right)$$ $$R=\frac{A}{P}=\frac{y}{2}\qquad\text{(for every most efficient trapezoid)}$$

Also, the condition can be written as $b+2zy=2y\sqrt{1+z^2}$, i.e. top width $=$ sum of the two sloping sides, so a semicircle of radius $y$ can be inscribed in the section.

Part 2: Side slope varied (best of all trapezoids)

Now keep $A$ and $y$ constant and vary $z$:

$$P=\frac{A}{y}-zy+2y\sqrt{1+z^2}$$ $$\frac{dP}{dz}=-y+\frac{2yz}{\sqrt{1+z^2}}=0\;\Rightarrow\;\sqrt{1+z^2}=2z\;\Rightarrow\;z=\frac{1}{\sqrt3}=0.577$$

So the side slope angle is $\tan\theta=1/z=\sqrt3$, $\theta=60^\circ$. Then

$$b=2y\left(\sqrt{1+\tfrac13}-\tfrac{1}{\sqrt3}\right)=2y\left(\frac{2}{\sqrt3}-\frac{1}{\sqrt3}\right)=\frac{2y}{\sqrt3}$$

The sloping length is $y\sqrt{1+z^2}=\dfrac{2y}{\sqrt3}=b$. Hence bed and both sloping sides are equal, with $60^\circ$ between sides: half of a regular hexagon.

     b
  _________
 /\       /\      60 deg side slopes
/60\     /60\     each side = b = 2y/sqrt(3)
 \  \___/  /

Hydraulic radius from Part 1 is $R=y/2$, which is the largest possible for any trapezoidal section of given area, so the half-hexagon is the most efficient of all trapezoidal sections.

Result: $b=2y(\sqrt{1+z^2}-z)$, $R=y/2$; for the best section $z=1/\sqrt3$ (half hexagon).

Geometry of a partly full circular channel

Let $\theta$ be the angle (radians) subtended at the centre by the wetted arc, $d$ the diameter and $y$ the depth.

$$A=\frac{d^2}{8}(\theta-\sin\theta),\qquad P=\frac{d\,\theta}{2},\qquad R=\frac{A}{P}=\frac d4\left(1-\frac{\sin\theta}{\theta}\right),\qquad y=\frac d2\left(1-\cos\frac\theta2\right)$$

    ___
  /  |  \      y = depth
 |   |th |     th = angle at centre
  \  | _/      subtended by wetted arc
    ---

Depth for maximum velocity

Chezy: $V=C\sqrt{RS_0}$. With $C$ and $S_0$ constant, $V$ is maximum when $R$ is maximum:

$$\frac{dR}{d\theta}=\frac d4\left[-\frac{\theta\cos\theta-\sin\theta}{\theta^2}\right]=0\;\Rightarrow\;\theta\cos\theta=\sin\theta\;\Rightarrow\;\tan\theta=\theta$$

Solving by trial, $\theta=4.4934$ rad $=257.45^\circ$ (the root between $\pi$ and $2\pi$).

$$y=\frac d2\left(1-\cos\frac{\theta}{2}\right)=\frac d2\left(1-\cos 128.73^\circ\right)=0.8128\,d$$

So the depth for maximum velocity is $y\approx0.81\,d$.

Discharge comparison with full flow

At maximum velocity: $A=\dfrac{d^2}{8}(\theta-\sin\theta)=0.6837\,d^2$ and $R=0.3043\,d$.

Running full: $A_f=\dfrac{\pi}{4}d^2=0.7854\,d^2$, $R_f=\dfrac d4=0.25\,d$.

$$\frac{V_{max}}{V_f}=\sqrt{\frac{R}{R_f}}=\sqrt{\frac{0.3043}{0.25}}=1.1033$$ $$\frac{Q_{V_{max}}}{Q_f}=\frac{A\sqrt R}{A_f\sqrt{R_f}}=\frac{0.6837\sqrt{0.3043}}{0.7854\sqrt{0.25}}=0.9604$$

Answer: depth for maximum velocity $=0.81\,d$; velocity is 1.103 times the full-flow velocity, and the discharge is 0.960 times (about 96.0% of) the full-pipe discharge. So the maximum velocity occurs at a discharge slightly less than the full-flow value.

Hydraulic exponent

For a given discharge or condition in a prismatic channel, the hydraulic exponent relates a section property to the depth. The second hydraulic exponent $N$ (for uniform flow) is defined by

$$K^2\propto y^{N},\qquad K=\frac1nAR^{2/3}=\frac{A^{5/3}}{nP^{2/3}}$$

where $K$ is the conveyance. (The first exponent $M$ is defined by $A^3/T\propto y^M$ for critical flow.)

General expression

Take logarithms and differentiate with respect to $y$:

$$\ln K=\frac53\ln A-\frac23\ln P-\ln n$$ $$\frac{y}{K}\frac{dK}{dy}=\frac53\frac yA\frac{dA}{dy}-\frac23\frac yP\frac{dP}{dy}=\frac N2$$

(since $K\propto y^{N/2}$, $\dfrac yK\dfrac{dK}{dy}=\dfrac N2$). So

$$N=\frac{10}{3}\frac{y}{A}\frac{dA}{dy}-\frac43\frac yP\frac{dP}{dy}$$

Rectangular section (wide channel, $b\gg y$)

$A=by$, $\dfrac{dA}{dy}=b$, so $\dfrac yA\dfrac{dA}{dy}=\dfrac{y\,b}{by}=1$.

$P=b+2y\approx b$, $\dfrac{dP}{dy}=2$, so $\dfrac yP\dfrac{dP}{dy}=\dfrac{2y}{b+2y}\to0$.

$$N=\frac{10}{3}(1)-\frac43(0)=\boxed{\frac{10}{3}}$$

Directly: for a wide rectangle $R\approx y$, so $K=\dfrac{b}{n}y\cdot y^{2/3}=\dfrac bn y^{5/3}$, and $K^2\propto y^{10/3}$ ✓.

Hydraulic exponent

The second hydraulic exponent $N$ (for uniform flow) is defined by $K^2\propto y^N$, where the conveyance is

$$K=\frac1nAR^{2/3}=\frac{A^{5/3}}{nP^{2/3}}$$

(The first exponent $M$ is defined by $A^3/T\propto y^M$ for critical flow.)

General expression

$$\frac{y}{K}\frac{dK}{dy}=\frac53\frac yA\frac{dA}{dy}-\frac23\frac yP\frac{dP}{dy}=\frac N2$$

Triangular section (side slope $z$)

$$A=zy^2,\qquad \frac{dA}{dy}=2zy,\qquad \frac yA\frac{dA}{dy}=\frac{y\cdot2zy}{zy^2}=2$$ $$P=2y\sqrt{1+z^2},\qquad \frac{dP}{dy}=2\sqrt{1+z^2},\qquad \frac yP\frac{dP}{dy}=1$$ $$\frac N2=\frac53(2)-\frac23(1)=\frac{10}{3}-\frac23=\frac83$$ $$\boxed{N=\frac{16}{3}}$$

Directly: $R=\dfrac{zy}{2\sqrt{1+z^2}}\propto y$, so $K\propto y^2\cdot y^{2/3}=y^{8/3}$ and $K^2\propto y^{16/3}$ ✓. This is exact for a triangle for all depths because $A\propto y^2$ and $P\propto y$.

Definition

The momentum correction factor for a section of total area $A$, discharge $Q$ and mean velocity $V=Q/A$ is

$$\beta=\frac{\int v^2\,dA}{V^2A}$$

For a compound channel divided into subsections, each with mean velocity $V_i$, area $A_i$ (velocity taken uniform in each subsection):

$$\beta=\frac{\sum V_i^2A_i}{V^2A}$$

Proof

1. Total discharge $Q=\sum Q_i$, $A=\sum A_i$ and $V=\dfrac{Q}{A}$.
2. For each subsection, conveyance $K_i$ is defined by $Q_i=K_i\sqrt{S_f}$. All subsections have the same energy slope $S_f$ (same water surface and energy line). Therefore

$$Q=\sqrt{S_f}\sum K_i,\qquad V_i=\frac{Q_i}{A_i}=\frac{K_i\sqrt{S_f}}{A_i}$$

3. Substituting,

$$\sum V_i^2A_i=\sum\frac{K_i^2S_f}{A_i^2}A_i=S_f\sum\frac{K_i^2}{A_i}$$ $$V^2A=\frac{Q^2}{A}=\frac{S_f\left(\sum K_i\right)^2}{\sum A_i}$$

4. Dividing,

$$\beta=\frac{S_f\sum\dfrac{K_i^2}{A_i}}{\dfrac{S_f\left(\sum K_i\right)^2}{\sum A_i}}$$ $$\boxed{\beta=\frac{\sum\left(\dfrac{K_i^2}{A_i}\right)\left(\sum A_i\right)}{\left(\sum K_i\right)^2}}$$

Hence proved. For a single uniform section, $\beta=1$; for compound sections $\beta>1$ because the velocities in the main channel and flood plains differ greatly.

Definition

The energy (kinetic energy) correction factor for a section of area $A$ and mean velocity $V=Q/A$ is

$$\alpha=\frac{\int v^3\,dA}{V^3A}$$

For a compound channel with subsections of area $A_i$ and mean velocity $V_i$:

$$\alpha=\frac{\sum V_i^3A_i}{V^3A}$$

Proof

1. $Q=\sum Q_i$, $A=\sum A_i$, $V=Q/A$.
2. The conveyance of each part is $K_i$, with $Q_i=K_i\sqrt{S_f}$ (all subsections share the same energy slope $S_f$). Then

$$Q=\sqrt{S_f}\sum K_i,\qquad V_i=\frac{Q_i}{A_i}=\frac{K_i\sqrt{S_f}}{A_i}$$

3. Numerator:

$$\sum V_i^3A_i=\sum\frac{K_i^3S_f^{3/2}}{A_i^3}A_i=S_f^{3/2}\sum\frac{K_i^3}{A_i^2}$$

4. Denominator:

$$V^3A=\frac{Q^3}{A^2}=\frac{S_f^{3/2}\left(\sum K_i\right)^3}{\left(\sum A_i\right)^2}$$

5. Dividing,

$$\boxed{\alpha=\frac{\sum\left(\dfrac{K_i^3}{A_i^2}\right)\left(\sum A_i\right)^2}{\left(\sum K_i\right)^3}}$$

Hence proved. For one uniform section $\alpha=1$; for a compound section $\alpha>1$ (typically 1.1 to 2 or more), and it must be used in $E=y+\alpha V^2/2g$.

The bed slope is not given in the question. It is assumed to be $S_0=1$ in 1000 $=0.001$ and the working is shown so that another slope can be substituted (depth varies as $S_0^{-3/16}$).

Given data

$z_1=0.5$, $z_2=1.5$ (H:V), $Q=12$ m³/s, $n=0.014$.

Condition for most efficient section

Least wetted perimeter for a given area gives (as for a symmetric trapezoid) the condition that the top width equals the sum of the sloping sides, $R=y/2$:

$$A=by+\frac{(z_1+z_2)}{2}y^2=by+y^2$$ $$P=b+y\left(\sqrt{1+z_1^2}+\sqrt{1+z_2^2}\right)=b+y(1.1180+1.8028)=b+2.9208\,y$$

Minimising $P=\dfrac{A}{y}-\dfrac{z_1+z_2}{2}y+s\,y$ with $dP/dy=0$ gives

$$b=y\left(\sqrt{1+z_1^2}+\sqrt{1+z_2^2}-(z_1+z_2)\right)=(2.9208-2)\,y=0.9208\,y$$

Then

$$A=1.9208\,y^2,\qquad P=3.8416\,y,\qquad R=\frac{A}{P}=\frac y2$$

Normal depth from Manning's equation

$$Q=\frac1nAR^{2/3}S_0^{1/2}\;\Rightarrow\;12=\frac{1}{0.014}(1.9208\,y^2)\left(\frac y2\right)^{2/3}(0.001)^{1/2}$$ $$y^{8/3}=4.3905\;\Rightarrow\;y=1.742\ \text{m}$$ $$b=0.9208\times1.742=1.604\ \text{m},\qquad A=5.826\ \text{m}^2,\qquad P=6.690\ \text{m}$$

Top width $T=b+(z_1+z_2)y=5.087$ m.

Critical depth

Critical flow condition: $\dfrac{Q^2}{g}=\dfrac{A_c^3}{T_c}$ with $A_c=by_c+y_c^2$, $T_c=b+2y_c$:

$$\frac{12^2}{9.81}=14.679=\frac{(1.604\,y_c+y_c^2)^3}{1.604+2y_c}$$

Solving by trial: $y_c=1.349$ m. Check: $A_c=3.982$ m², $T_c=4.301$ m, $V_c=3.014$ m/s, $Fr=1$.

Answer (for $S_0=0.001$): most efficient section has $b=1.60$ m, $y=1.74$ m, side slopes 0.5H:1V and 1.5H:1V, and the critical depth is $y_c=1.35$ m. Since $y_n>y_c$, the flow is subcritical.

Given data

$D=1.5$ m, $r=0.75$ m, flow depth $y=1.125$ m. The depth is more than the radius ($y/D=0.75$), so the water surface is above the centre.

     ____  <- water surface (chord)
   /  a   \     a = angle at centre by surface
  |   *    |    * = centre
   \______/

(i) Angle $\alpha$

The centre is $y-r=1.125-0.75=0.375$ m below the water surface.

$$\cos\frac{\alpha}{2}=\frac{y-r}{r}=\frac{0.375}{0.75}=0.5\;\Rightarrow\;\frac\alpha2=60^\circ,\quad \alpha=120^\circ$$

Angle subtended at the centre by the wetted arc (the part below the surface) is $\theta=360^\circ-120^\circ=240^\circ=4.1888$ rad. This $\theta$ is used for area and perimeter.

(ii) Area and wetted perimeter

$$A=\frac{r^2}{2}(\theta-\sin\theta)=\frac{0.75^2}{2}(4.1888-\sin240^\circ)=\frac{0.5625}{2}(4.1888+0.8660)=1.4217\ \text{m}^2$$ $$P=r\theta=0.75\times4.1888=3.1416\ \text{m}$$

Hydraulic radius

$$R=\frac{A}{P}=\frac{1.4217}{3.1416}=0.4525\ \text{m}$$

Top width

$$T=2r\sin\frac\alpha2=2\times0.75\times\sin60^\circ=1.2990\ \text{m}$$

Section factor for uniform flow

$$AR^{2/3}=1.4217\times(0.4525)^{2/3}=0.8380\ \text{m}^{8/3}$$

(The section factor for critical flow, $Z=\sqrt{A^3/T}=1.4873$ m$^{2.5}$, can be computed with the same $A$ and $T$.)

Answer: $\alpha=120^\circ$, $R=0.453$ m, $T=1.299$ m, $AR^{2/3}=0.838$ m$^{8/3}$.

Most economical triangular section

Let side slope be $z$ (H:V), depth $y$:

$$A=zy^2,\qquad P=2y\sqrt{1+z^2}$$

For a given area, $y^2=A/z$, so

$$P^2=4y^2(1+z^2)=\frac{4A(1+z^2)}{z}$$

Minimise $P$ with respect to $z$:

$$\frac{d}{dz}\left(\frac{1+z^2}{z}\right)=\frac{2z\cdot z-(1+z^2)}{z^2}=0\;\Rightarrow\;z^2=1\;\Rightarrow\;z=1$$

So the best triangular section has side slopes $45^\circ$ (right-angled vertex at the bottom):

$$A=y^2,\qquad P=2\sqrt2\,y,\qquad R_t=\frac{y^2}{2\sqrt2\,y}=\frac{y}{2\sqrt2}$$

Most economical rectangular section

$b=2y$: $A=2y^2$, $P=4y$, so

$$R_r=\frac{2y^2}{4y}=\frac y2$$

Comparison (at the same depth $y$)

$$\frac{R_t}{R_r}=\frac{y/(2\sqrt2)}{y/2}=\frac{1}{\sqrt2}$$ $$\boxed{R_t=\frac{1}{\sqrt2}R_r}$$

Hence the hydraulic radius of the most economical triangular section is $1/\sqrt2$ times that of the most economical rectangular section of the same depth.

Horton-Einstein method: each part of the section has the same mean velocity as the whole section, and the equivalent roughness is

$$n_e=\left[\frac{\sum P_in_i^{3/2}}{P}\right]^{2/3}$$

where $P_i$ is the wetted perimeter of the part having roughness $n_i$, and $P=\sum P_i$ is the total wetted perimeter.

Assumptions from the figure

Main channel: bed width 20 m, depth 4 m, side slopes 1V:4H. Flood plain on each side: horizontal part 5 m wide at the level of the top of the main channel, with an outer bank 1V:2H. Total depth $=5$ m, so the water depth on the flood plain is $5-4=1$ m. The vertical interface lines between the main channel and flood plain are not counted as wetted perimeter.

 flood plain   ~~~~~~~~~~~~~~~~~~~~~  flood plain
  \ 5 m    (1 m deep)      5 m /
   \_______      ________/
           \    /  4 m deep
            \__/  20 m

Wetted perimeters

• Main channel (n = 0.021): slope length each side $=4\sqrt{1+4^2}=4\times4.1231=16.492$ m

$$P_m=20+2\times16.492=52.985\ \text{m}$$

• Each flood plain (n = 0.039): horizontal part $5$ m + outer slope length $1\times\sqrt{1+2^2}=2.236$ m

$$P_f=5+2.236=7.236\ \text{m};\quad\text{both}=2P_f=14.472\ \text{m}$$

• Total: $P=52.985+14.472=67.457$ m.

Equivalent roughness

$$n_e=\left[\frac{52.985\,(0.021)^{1.5}+14.472\,(0.039)^{1.5}}{67.457}\right]^{2/3}$$ $$(0.021)^{1.5}=0.003043,\qquad (0.039)^{1.5}=0.007702$$ $$n_e=\left[\frac{0.2727}{67.457}\right]^{2/3}=(0.004043)^{2/3}=0.0254$$

Answer: equivalent Manning's roughness $n_e\approx0.0254$ (between 0.021 and 0.039, weighted towards the larger perimeter of the main channel).

Plotting the section

Take distance as $x$ and depth as measured below the water surface (the depth is $0$ at both banks, so the water surface is level with the ground at $x=0$ and $x=10$ m). Top width $T=10$ m, maximum depth $5$ m at $x=5$ m.

 0    2    4    6    8   10  (m)
 ~~~~~~~~~~~~~~~~~~~~~~~~~~  water surface
 \                        /
  \__ 3 m ___      ___   /
          5 m  \  /   /
                \/

(points: (0,0), (1,3), (3,3), (5,5), (7,3), (10,0))

Area and wetted perimeter by segments

Area of each strip = width × mean depth; length = $\sqrt{\Delta x^2+\Delta y^2}$.

(i) Area

$A=28.00$ m².

(ii) Equivalent hydraulic radius

$$R=\frac{A}{P}=\frac{28.00}{15.062}=1.8590\ \text{m}$$

(iii) Conveyance factor

$$K=\frac{1}{n}AR^{2/3}=\frac{1}{0.02}\times28.00\times(1.8590)^{2/3}=2116.7\ \text{m}^3/\text{s}$$

(so that $Q=K\sqrt{S_0}$; the section factor is $AR^{2/3}=42.333$ m$^{8/3}$.)

(iv) Hydraulic depth

$$D_h=\frac{A}{T}=\frac{28.00}{10}=2.80\ \text{m}$$

(v) Section factor for uniform flow

$$AR^{2/3}=28.00\times(1.8590)^{2/3}=42.333\ \text{m}^{8/3}$$

Answer: $A=28.0$ m², $K=2117$, $R=1.859$ m, $D_h=2.8$ m, $AR^{2/3}=42.33$ m$^{8/3}$.

Manning's equation

Manning's equation is an empirical formula for mean velocity in open channels, for uniform (turbulent, rough) flow:

$$V=\frac{1}{n}R^{2/3}S^{1/2}\qquad\Rightarrow\qquad Q=\frac{1}{n}AR^{2/3}S^{1/2}\quad(\text{SI units})$$

It is related to Chezy's formula $V=C\sqrt{RS}$ by $C=R^{1/6}/n$.

The slope $S$ and the energy diagram

$S$ is the energy slope $S_f$, the loss of total energy head per unit length:

$$S=S_f=-\frac{dH}{dx},\qquad H=z+y+\frac{\alpha V^2}{2g}$$

On the energy diagram it is the slope of the energy grade line (EGL), measured relative to the horizontal. It is not necessarily the bed slope.

 EGL -----\
 WSL  ------\  \  S = slope of EGL (S_f)
 Bed  -------\---\--

For uniform flow, $S_f=S_w=S_0$, since depth and velocity do not change, so the EGL, water surface and bed are parallel. Then $S$ may be taken as the bed slope. For non-uniform flow, $S_f$ is used at each section (the local friction slope), found from the Manning equation: $S_f=\dfrac{n^2V^2}{R^{4/3}}$.

Why uniform flow is assumed in most applications

1. Most canals, flumes and drains are prismatic with constant slope and roughness, and long reaches reach equilibrium (normal depth).
2. The bed slope can be measured, but the energy slope cannot be measured directly. With uniform flow, $S_f=S_0$ so only the bed slope is needed.
3. In gradually varied flow, depth changes slowly, so the uniform-flow formula is valid for local friction loss at each section.
4. It gives a simple design method: for a given $Q$, $n$ and $S_0$, find the normal depth.
5. Manning's equation was itself derived from data of (nearly) uniform flow.

Hence, Manning's equation gives acceptable accuracy when the flow is turbulent, steady and nearly uniform.

Definitions

• First hydraulic exponent $M$ (for critical flow): $Z^2=\dfrac{A^3}{T}\propto y^{M}$. It is used in the critical flow condition $\dfrac{Q^2}{g}=\dfrac{A^3}{T}$.
• Second hydraulic exponent $N$ (for uniform flow): the conveyance $K=\dfrac1nAR^{2/3}\propto y^{N/2}$, i.e. $K^2\propto y^N$.

Wide rectangular channel (width $b$, depth $y$)

• $A=by$, $T=b$, and $R\approx y$ since $b\gg y$.

$$\frac{A^3}{T}=\frac{b^3y^3}{b}=b^2y^3\;\Rightarrow\;\boxed{M=3}$$ $$K=\frac{1}{n}(by)\,y^{2/3}=\frac bn\,y^{5/3}\;\Rightarrow\;K^2\propto y^{10/3}\;\Rightarrow\;\boxed{N=\frac{10}{3}=3.33}$$

Triangular channel (side slope $z$)

• $A=zy^2$, $T=2zy$, $P=2y\sqrt{1+z^2}$, so $R=\dfrac{zy}{2\sqrt{1+z^2}}\propto y$.

$$\frac{A^3}{T}=\frac{z^3y^6}{2zy}=\frac{z^2}{2}y^5\;\Rightarrow\;\boxed{M=5}$$ $$K=\frac1n(zy^2)\,(c\,y)^{2/3}\propto y^{2+2/3}=y^{8/3}\;\Rightarrow\;K^2\propto y^{16/3}\;\Rightarrow\;\boxed{N=\frac{16}{3}=5.33}$$

The figure with dimensions and Manning's coefficients was not available. The method is given in full and applied to an assumed section with stated values, so that the numbers can be replaced by those of the figure.

Part 1: Boundary shear stress equation

Consider a uniform flow in a length $L$ of a prismatic channel of area $A$, wetted perimeter $P$ and bed slope $S_0$ (angle $\theta$).

  ----------------------------  WSL
  \  W sin(th) ->  /         
   \ ___ control volume / tau0 on boundary

Forces along the flow direction (pressure forces at the two ends are equal since depth is constant):

• Gravity component: $W\sin\theta=\gamma AL\sin\theta\approx\gamma AL\,S_0$
• Boundary resistance: $\tau_0\,P\,L$

For uniform flow (no acceleration):

$$\gamma AL\,S_0=\tau_0PL\;\Rightarrow\;\boxed{\tau_0=\gamma RS_0}$$

where $R=A/P$. This is the average shear stress on the boundary.

Part 2: Compound channel (method)

1. Divide the section by vertical lines into subsections (main channel and flood plains); the dividing lines are not counted in wetted perimeters.
2. For each subsection $i$: $A_i$, $P_i$, $R_i=A_i/P_i$ and $Q_i=\dfrac{1}{n_i}A_iR_i^{2/3}S_0^{1/2}$.
3. Total discharge $Q=\sum Q_i$.
4. Effective (equivalent) Manning coefficient from the whole section, treated as a single section:

$$n_e=\frac{A\,R^{2/3}S_0^{1/2}}{Q},\quad A=\sum A_i,\quad P=\sum P_i,\quad R=\frac AP$$

Illustration with assumed data

Main channel: bed 6 m wide, 2 m deep, $n=0.025$. Flood plains: 8 m wide each side, water depth 1 m over the plain, $n=0.040$, bank vertical. $S_0=0.002$. (Assumed.)

Answer (for the assumed section): $Q=44.67$ m³/s and $n_e=0.0295$. For the actual figure, repeat steps 1-4 with the given dimensions and $n$ values.

The normal depth $y_n$ satisfies Manning's equation, which cannot be solved directly for $y$. It is found numerically by the bisection method on

$$f(y)=AR^{2/3}-\frac{nQ}{\sqrt{S_0}}=0,\qquad A=(b+zy)y,\quad P=b+2y\sqrt{1+z^2},\quad R=\frac AP$$

$f(y)$ increases with $y$, so the root is unique.

Algorithm

1. Read $b$, $z$, $n$, $Q$, $S_0$.
2. Compute the target value $K_0=nQ/\sqrt{S_0}$.
3. Set lower limit $y_{lo}=0.0001$ and upper limit $y_{hi}=1$; while $f(y_{hi})<0$, double $y_{hi}$.
4. Repeat: $y_{mid}=(y_{lo}+y_{hi})/2$. If $f(y_{mid})>0$ set $y_{hi}=y_{mid}$, else $y_{lo}=y_{mid}$.
5. Stop when $y_{hi}-y_{lo}<10^{-6}$ (or 100 iterations).
6. Print $y_n=(y_{lo}+y_{hi})/2$.

Flow chart

        [Start]
           |
   [Read b,z,n,Q,S0]
           |
   [ylo=0.0001, yhi=1]
           |
     /-----------\ yes
   < f(yhi) < 0 ? >-----> [yhi = 2*yhi] --+
     \-----------/                         |
           | no  <-------------------------+
           v
   [ymid = (ylo+yhi)/2]<---------+
           |                     |
     /-----------\ yes           |
   < f(ymid) > 0 >--> [yhi=ymid] |
     \-----------/               |
           | no                  |
      [ylo=ymid]                 |
           |                     |
     /---------------\ no        |
   < yhi-ylo < 1e-6 ? >----------+
     \---------------/
           | yes
   [Print yn = (ylo+yhi)/2]
           |
        [Stop]

C program

#include <stdio.h>
#include <math.h>

/* f(y) = A*R^(2/3) - n*Q/sqrt(S): zero at the normal depth */
double f(double y, double b, double z, double n, double Q, double S)
{
    double A = (b + z * y) * y;                 /* flow area        */
    double P = b + 2.0 * y * sqrt(1.0 + z * z); /* wetted perimeter */
    double R = A / P;                           /* hydraulic radius */
    return A * pow(R, 2.0 / 3.0) - n * Q / sqrt(S);
}

int main(void)
{
    double b, z, n, Q, S, lo, hi, mid;
    int i;

    printf("Enter b (m), z (H:1V), n, Q (m3/s), S0: ");
    scanf("%lf %lf %lf %lf %lf", &b, &z, &n, &Q, &S);

    lo = 0.0001;
    hi = 1.0;
    while (f(hi, b, z, n, Q, S) < 0.0)   /* bracket the root */
        hi *= 2.0;

    for (i = 0; i < 100; i++) {          /* bisection */
        mid = 0.5 * (lo + hi);
        if (f(mid, b, z, n, Q, S) > 0.0)
            hi = mid;
        else
            lo = mid;
        if (hi - lo < 1e-6)
            break;
    }
    printf("Normal depth yn = %.4f m\n", 0.5 * (lo + hi));
    return 0;
}

Sample run

Input: b = 5, z = 1.5, n = 0.015, Q = 20, S0 = 0.0004

Output: Normal depth yn = 1.7563 m

(The result was checked independently: $A=13.409$ m², $R=1.183$ m, $AR^{2/3}=15.000$ and $nQ/\sqrt{S_0}=15.000$.)

Minimum lining means minimum wetted perimeter for the given flow, which is the most efficient hydraulic section. The channel has one vertical side ($z_1=0$) and one side slope $z_2=2$ (2H:1V).

Given data

$Q=10$ m³/s, $S_0=0.0001$, $n=0.015$, bed $b$, depth $h$.

Geometry

$$A=bh+\frac{z_1+z_2}{2}h^2=bh+h^2$$ $$P=b+h\left(\sqrt{1+0^2}+\sqrt{1+2^2}\right)=b+h(1+2.2361)=b+3.2361\,h$$

Condition for minimum $P$ at fixed $A$

$$P=\frac{A}{h}-h+3.2361\,h\;\Rightarrow\;\frac{dP}{dh}=-\frac{A}{h^2}+2.2361=0\;\Rightarrow\;\frac{A}{h^2}=2.2361$$ $$b=\frac{A}{h}-h=2.2361h-h=1.2361\,h$$

Then $A=2.2361h^2$, $P=4.4721h$ and $R=\dfrac{A}{P}=\dfrac h2$.

Manning's equation

$$Q=\frac1nAR^{2/3}S_0^{1/2}\;\Rightarrow\;10=\frac{1}{0.015}(2.2361h^2)\left(\frac h2\right)^{2/3}(0.0001)^{1/2}$$ $$h^{8/3}=\frac{10\times0.015}{2.2361\times0.5^{2/3}\times0.01}=10.6486\;\Rightarrow\;h=2.428\ \text{m}$$ $$b=1.2361\times2.428=3.001\ \text{m}$$

Check: $A=13.181$ m², $P=10.858$ m, $R=1.214$ m, $V=0.759$ m/s and $Q=\dfrac{1}{0.015}(13.181)(1.214)^{2/3}(0.01)=10.00$ m³/s.

Answer: depth $h=2.43$ m and bed width $b=3.00$ m (wetted perimeter $=10.86$ m is the minimum lining per metre).