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Rectangular notch

Discharge over a rectangular notch

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Key Facts

Gyroscopic Couple: The rate of change of angular momentum ( $\inline \tau$ ) = $\inline I\omega\Omega$ (In the limit).

$\inline I$ = Moment of Inertia.
$\inline \omega$ = Angular velocity
$\inline \Omega$ = Angular velocity of precession.

Blaise Pascal (1623-1662) was a French mathematician, physicist, inventor, writer and Catholic philosopher.

Leonhard Euler (1707-1783) was a pioneering Swiss mathematician and physicist.

Theory

Consider a rectangular notch in one side of a tank over which water is flowing as shown in figure.

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Let,

H = Height of water above sill of notch
b = Width or length of the notch
C_d = Coefficient of discharge

Let us consider a horizontal strip of water of thickness dh at a depth of h from the water level as shown in figure.

$\inline \therefore$ Area of the strip

$= b.dh$

(2)

We know know that the theoretical velocity of water through the strip,

$= \sqrt {2gh}$

(3)

Discharge through the strip,

$dq = C_d \times Area\;of\;strip\;\times\;Theoretical\;velocity$

$\Rightarrow dq = C_d.bdh \sqrt {2gh}$

The total discharge over the whole notch, may be found out by integrating the above equation within the limits 0 and H.

$Q = \int_{0}^{H} C_d.b.dh\sqrt {2gh}$

$\Rightarrow Q = C_d.b\sqrt {2g}\int_{0}^{H} h^{\frac{1}{2}}.dh$

$\therefore Q = \frac{2}{3}C_d.b\sqrt {2g}(H)^{\frac{3}{2}}$

Example:

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Example - Discharge over a rectangular notch

Problem

A rectangular notch 0.5m wide has constant head of 400 mm. Find the discharge over the notch in liters per second, if the coefficient of discharge for the notch is 0.62.

Workings

Given,

b = 0.5 m
H = 400 mm = 0.4 m
C_d = 0.62

We know that discharge over the rectangular notch,

$Q = \frac{2}{3}C_d.b\sqrt {2g}(H)^{\frac{3}{2}}\;m^3/s$

$\Rightarrow Q = \frac{2}{3}\times 0.62\times 0.5 \sqrt {2\times9.81}(0.4)^{\frac{3}{2}}\;m^3/s$

$\Rightarrow Q = 0.915\times 0.253 = 0.231\;m^3/s = 231\;liters/s$

Solution

Discharge over the notch = 231 liters/s