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## Q. 6.P.2

The rate of discharge of water from a tank is measured by means of a notch, for which the flowrate is directly proportional to the height of liquid above the bottom of the notch.
Calculate and plot the profile of the notch if the flowrate is 0.1 m³/s when the liquid level is 150 mm above the bottom of the notch.

## Verified Solution

The velocity of fluid discharged as a height h above the bottom of the notch is:

$u=\sqrt{(2 g h)}$

The velocity therefore varies from zero at the bottom of the notch to a maximum value at the free surface.
For a horizontal element of fluid of width 2w and depth dh at a height h above the bottom of the notch, the discharge rate of fluid is given by:

$d Q=\sqrt{ }(2 g h) 2 w d h$

If the discharge rate is linearly related to the height of the liquid over the notch, H, w will be a function of h and it may be supposed that:

$w=k h^n$

where k is a constant.

Substituting for w in the equation for dQ and integrating to give the discharge rate over the notch Q then:

$Q=2 \sqrt{(2 g)} k \int_0^H h^n h^{0.5} d h$

$=2 \sqrt{(2 g)} k \int_0^H h^{n+0.5} d h$

$=2 \sqrt{(2 g)} k[1 /(n+1.5)] H^{(n+1.5)}$

Since it is required that $Q \propto H$:

$n+1.5=1$

and:                              $n=-0.5$
Thus:                           $Q=2 \sqrt{(2 g)} k H$

Since $Q=0.1 m ^3 / s$ when H = 0.15 m:

$k=(0.1 / 0.15)\left[1 /(2 \sqrt{(2 g)}]=0.0753 m ^{1.5}\right.$

Thus, with w and h in m:         $w=0.0753 h^{-0.5}$
and, with w and h in mm:        $w=2374 h^{-0.5}$

and using this equation, the profile is plotted as shown in $\underline{\underline{\text { Figure 6 a}}}$ 