E5 Relevant Equations of Steady Open Channel Flow
The Chezy equation for uniform flow can be written:
The Chezy Equation
$\displaystyle Q$  $\displaystyle=AC\sqrt{RS_{o}}$  
$\displaystyle=ACR^{1/2}S_{o}^{1/2}$  (Eq1) 
Using the Manning’s n and Chezy C relationship $C=\frac{R^{1/6}}{n}$ we obtain the Manning’s equation for uniform flow:
The Manning’s Equation
$\displaystyle Q$  $\displaystyle=\frac{1}{n}AR^{1/6}\sqrt{RS_{o}}$  
$\displaystyle=\frac{1}{n}AR^{2/3}S_{o}^{1/2}$  (Eq2) 
Where $R=A/P$ and these are calculated depending on the shape of the channel crosssection e.g. Rectangular, Trapezoidal, etc. Table 1, below, may help.
Wide Channels
Note that when a channel is said to be wide then it is assumed to be approximately rectangular and that the width $B$, is much greater than depth $y$, i.e. $B\gg y$. This results in:
$\displaystyle R$  $\displaystyle=\frac{A}{P}=\frac{By}{B+2y}\approx\frac{By}{B}$  
$\displaystyle R$  $\displaystyle\approx y$ 
Also, for wide channels we often specify flow per unit width i.e. $q$ in units of $m^{3}/s/m$.
The Manning’s Equation  wide channel
In terms of $q$
This has the consequence that a normal depth calculation becomes explicit i.e.
The Chezy Equation  wide channel
In terms of $q$
And the explicit normal depth calculation becomes
In terms of $q$
Froude Number $Fr$
$\displaystyle Fr$  $\displaystyle=\frac{V}{\sqrt{gD_{m}}}=\frac{V}{\sqrt{g\frac{A}{B}}}$  (Eq11)  
$\displaystyle=\frac{Q}{A\sqrt{g\frac{A}{B}}}$  (Eq12)  
$\displaystyle=\left(\frac{Q^{2}B}{gA^{3}}\right)^{1/2}$  (Eq13) 
and
For a rectangular channel $A=By$ and $q=Q/B$, so
$\displaystyle Fr$  $\displaystyle=\frac{V}{\sqrt{gy}}$  (Eq15)  
$\displaystyle=\frac{Q}{By\sqrt{gy}}$  (Eq16) 
and
For a wide channel
Critical depth, $y_{c}$
At critical depth $y_{c}$, $Fr=1$, so equation Eq13 becomes
and similarly equation Eq14 become
Either of these can be solved, for $y_{c}$. You must use the appropriate expressions for geometry for $A$ and $B$.
For a rectangular channel, $A=By$ and $B=b$ so
$\displaystyle 1$  $\displaystyle=\frac{Q}{By\sqrt{gy_{c}}}$  
$\displaystyle 1$  $\displaystyle=\frac{Q^{2}}{B^{2}gy_{c}^{3}}$  (Eq21) 
And the explicit expression for critical depth $y_{c}$ is obtained:
In terms of $q$
Critical Slope, $So_{c}$
Critical slope, $So_{c}$ is the slope $S_{o}$ in the uniform flow equation (Manning’s or Chezy) at which depth is equal to critical depth, $y_{c}$.
Rearranging the Manning’s equation Eq2 and combining with the Froude number equated to 1, equation Eq13 or equation Eq14 (and using the subscript c to indicate parameters evaluated at critical depth) results in:
Critical Slope  wide channel
For a wide rectangular channel then as $P\approx B$ and thus $R=y$ this results in:
Bernoulli and Specific Energy
Where $\alpha=1$ in most cases, but can be calculated using:
where $V\;(=Q/A)$ is the mean velocity.
Conservation of Specific Energy We nearly always assume that specific energy is conserved so from point 1 to point 2 downstream in a flow, where point 2 may be raised a small height $\Delta z$, then
Note that for a converging or narrowing channel the with changes and $B_{1}\neq B_{2}$ and thus $q_{1}\neq q_{2}$. In terms of $q$ the specific energy equation is written between points 1 and 2 as.
This is the most common form.
Relationship between critical depth and specific energy
Specific Force
Momentum force $F$
Momentum force (from Newton’s 2nd law), is given by
Where $beta$ is a momentum correction factor, often taken as equal to 1, but should the flow across a section, or along a reach be very non uniform then this expression can be used:
Pressure Force $P$
Pressure force $P$ is the force due to changes in pressure, given by:
For a rectangular channel
Hydraulic Jump
For a hydraulic jump in a rectangular channel the following relate the the depths on the upstream $y_{1}$, and downstream $y_{2}$, sides of the jump. These depths are know as conjugate depths and are depths of equal specific force
And
The equations can be manipulated to give this expression for energy loss in a jump
Gradually Varied Flow Equation
$S_{f}$ is calculated from the uniform flow equations (Chezy or Manning’s) where it is equated to the slope of the channel $S_{o}$. From Eq1 and Eq2 we get:
And for a wide channel
and
Direct step method
(Where the method to obtain the mean value is flexible, see below.)
Standard step method
(Where the method to obtain the mean value is flexible, see below.)
Numerical Integration There are a great many of numerical methods for integrating firstorder numerical method, such as the Gradually Varied Flow equation. Here three are identified that are used in this course.
For distance from depth integrations we have:

i)
As in the firstorder Euler method (where the subscript ${}_{o}$ indicates at the known, or initial, point), i.e. $y_{0}$:
$\Delta x=\Delta y\left(\frac{1Fr^{2}}{S_{o}S_{f}}\right)_{o}$ (Eq50) 
ii)
At the averaged depth $y_{1/2}=(y_{0}+y_{1})/2$
$\Delta x=\Delta y\left(\frac{1Fr^{2}}{S_{o}S_{f}}\right)_{mean}=\Delta y% \left(\frac{1Fr^{2}_{y{{}_{1/2}}}}{S_{o}S_{f_{y_{1/2}}}}\right)=\Delta y% \left(\frac{1Fr^{2}}{S_{o}S_{f}}\right)_{{1/2}}$ (Eq51) 
iii)
or the whole function averaged between the initial and subsequent point:
$\displaystyle\Delta x=\Delta y\left(\frac{1Fr^{2}}{S_{o}S_{f}}\right)_{mean}$ $\displaystyle=\Delta y\left.\left[\left(\frac{1Fr_{0}^{2}}{S_{o}S_{f_{y_{0}}% }}\right)+\left(\frac{1Fr_{1}^{2}}{S_{o}S_{f_{y_{1}}}}\right)\right]\middle/% 2\right.$ $\displaystyle=\Delta y\left.\left[\left(\frac{1Fr}{S_{o}S_{f}}\right)_{0}+% \left(\frac{1Fr}{S_{o}S_{f}}\right)_{1}\right]\middle/2\right.$ (Eq52)
For the depth from distance formulation we solve
This is usually applied using an Euler approach similar to i) above.
Common Channel Section geometric Formulae
All of the above expressions must be adapted for the appropriate channel geometry. Table 1 of three very common channel shapes may help:
Rectangle  Trapezoid  Circle  
($\phi$ in radians)  
Area, $A$  $by$  $(b+sy)y$  $\frac{1}{8}(\phi\sin\phi)D^{2}$ 
Wetted perimeter, $P$  $b+2y$  $b+2y\sqrt{1+s^{2}}$  $\frac{1}{2}\phi D$ 
Top width, $B$  $b$  $b+2sy$  $(\sin(\phi/2))D$ 
Hydraulic Radius, $R=\frac{A}{P}$  $\frac{by}{b+2y}$  $\frac{(b+sy)y}{b+2y\sqrt{1+x^{2}}}$  $\frac{1}{4}\left(1\frac{\sin\phi}{\phi}\right)D$ 
Hydraulic mean depth, $D_{m}=\frac{A}{B}$  $y$  $\frac{(b+sy)y}{b+2sy}$  $\frac{1}{8}\left(\frac{\phi\sin\phi}{\sin(\phi/2)}\right)D$ 