Derive the dynamic equation for Gradually Varied Flow.
1 Answer

Derivation :

Following assumptions are made :

  1. Bed slope is small.

  2. Channel is rectangular and straight.

  3. Flow is steady and streamline are approximately parallel.

  4. Roughness is constant

Notations :

enter image description here

$\delta l$ = Length of channel of rectangular cross section

S = Bed slope

D = Depth at section A

$\bar v$ = Velocity at section A

$D+\delta D$ = Depth at section B

$\bar v +\delta \bar v$ = Velocity at section B

$i\delta l$ = loss in energy between section A and section B

$i$ = is the slope of the total energy line $s\delta l$ = Fall in bed level Applying bernoulies equation to A and B $\begin{aligned} s\delta l +D+\frac{\bar v^2}{2g} &=(D+\delta D)\frac{{\bar v+\delta \bar v}^2}{2g}+i\delta l \end{aligned}$ $ \space \space \qquad \qquad \begin{aligned}s\delta l =\delta D +\frac{\bar v\delta \bar v}{g}+i\delta l \end{aligned}$ $\space \qquad \qquad \begin{aligned} \delta D = s\delta l -\frac{\bar v\delta \bar v}{g}-i\delta l\end{aligned}$ $\qquad\qquad \begin{aligned} \frac{\delta D}{\delta l} =s-\frac{\bar v\delta \bar v}{g\delta l}-i ....(i) \end{aligned}$ Assuming a constant width of channel, for continuity of flow the discharge per unit width is constant from section to section and so, $\begin {aligned} \qquad \bar vD &=(\bar v+\delta \bar v)(D+\delta D)\end{aligned}$ $\qquad \begin {aligned}\bar vD &= \bar vD+ \bar v\delta D+ D\delta \bar v+\delta \bar v \delta D \end{aligned}$ $\qquad\space\space \begin {aligned} \delta \bar v &=-\frac{\bar v\delta D }{D} \end{aligned}$ $ \because \delta \bar v \delta D$ are negligible. Substituting in equation $(i)$ $\qquad \qquad \begin {aligned} \frac{\delta D}{\delta l} &=s-i+\frac{\bar v^2 \delta D}{gD\delta l}\end {aligned}$ $\qquad \qquad \begin {aligned}\frac{\delta D}{\delta l} &=\frac{s-i}{1-\bar v^2/gD} \end {aligned}$ This is the basic equation for non-uniform flow of gradually varied flow. Alternative forms $$\begin {aligned} \frac{\delta D}{\delta l}=\frac{s-i}{1-Fr^2}\ \because Fr=\frac {\bar v^2}{gD} \end{aligned}$$

Please log in to add an answer.