For a sharp$-$crested weir with crest level Ps above the base and head of water $H$ above the crest

level, the Bernoulli equation can be applied along a streamline between sections $1$ and $2$ :

Fig $4\mathrm{.}1$ Flow over a weir $V_2=[2g(h+\frac{V_1^2}{2g}){]}^{\frac{1}{2}}$

The flow rate through the strip dh is

$dQ=V_{2}dA=V_2(bdh)=V_2[2(H-h)tan()]dh$

or$,$

$dQ=2\sqrt[\phantom{2}]{2g}tan()(H-h)[h+\frac{V_1^2}{2g}{]}^{\frac{1}{2}}dh$

If the approach velocity head $\frac{V_1^2}{2}g$ is neglected,

$Q_{\text{i}\text{d}\text{e}\text{a}\text{l}}={\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}dQ=2\sqrt[\phantom{2}]{2g}tan(){\displaystyle \underset{0}{\overset{H}{}}}(H-h)h^{0\mathrm{.}5}dh$

This is the ideal flow rate, as it is based on several simplifying assumptions. We define a Weir

Coefficient $Q_{\text{A}\text{c}\text{t}\text{u}\text{a}\text{l}}=C_W{Q}_{\text{I}\text{d}\text{e}\text{a}\text{l}}=C_W\frac{8}{15}tan()\sqrt[\phantom{2}]{2g}{H}^{2\mathrm{.}5}$sW$2$:

The discharge from a $90V-$notch weir $($shown in Fig $12)$ fills a $10$ Litre vessel in $27$

seconds and produces a stage of $36mm$ upstream of the weir crest. If the value of

$P_s=125mm$ and the width of the channel is $250mm,$ what is the mean approach

velocity head?

The answer is $4\mathrm{.}32$ micro$-$m$.$