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Trapezoidal Notch

Discharge over a Trapezoidal 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.

Overview

A trapezoidal notch is a combination of a rectangular notch and two triangular notches as shown in figure. It is, thus obvious that the discharge over such a notch will be the sum of the discharge over the rectangular and triangular notches.

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Consider a trapezoidal notch ABCD as shown in figure. For the purpose of analysis, split up the notch into a rectangular notch BCFE and two triangular notches ABE and DCF. The discharge over these two triangular notches is equivalent to the discharge over a single triangular notch of angle θ.

Let,

$\inline H$ = Height of the liquid above the sill of the notch
$\inline C_{d1}$ = Coefficient of discharge for the rectangular portion
$\inline C_{d2}$ = Coefficient of discharge for the triangular portion
$\inline b$ = Breadth of the rectangular portion of the notch
$\inline \frac{\theta}{2}$ = Angle, which the sides make with the vertical

$\inline \therefore$ Discharge over the trapezoidal notch,

$\inline Q$ = Discharge over the rectangular notch + Discharge over the triangular notch

$\therefore Q = \frac{2}{3}C_d.b\sqrt {2g}(H)^{\frac{3}{2}} + \frac{8}{15}C_d \sqrt {2g} \tan \frac{\theta}{2}\times H^{\frac{5}{2}}$

Example:

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Example - Discharge over the trapezoidal notch

Problem

A trapezoidal notch notch of 1.2m wide at the top and 450mm at the bottom is 300mm high. Find the discharge through the notch, if the head of water is 225mm. Take coefficient of discharge as 0.6.

Workings

Given,

Width of the notch = 1.2m
b = 450mm = 0.45m
Height of the notch = 300mm = 0.3m
H = 225mm = 0.225m
C_d = 0.6

From the geometry of the notch, we get,

$\tan \frac{\theta}{2} = \frac{1200 - 450}{2}\times \frac{1}{300} = \frac{750}{600} = 1.25$

and the discharge over trapezoidal notch,

$Q = \frac{2}{3}C_d.b\sqrt {2g}(H)^{\frac{3}{2}} + \frac{8}{15}C_d \sqrt {2g} \tan \frac{\theta}{2}\times H^{\frac{5}{2}}$

$= \frac{2}{3}\times 0.6\times 0.45\sqrt {2\times 9.81}\times (0.225)^\frac{3}{2} + \frac{8}{15}\times 0.6 \sqrt {2\times 9.81}\times 1.25\times (0.225)^\frac{5}{2}\;m^3 /s$

$= 0.085 + 0.043 = 0.128\;m^3 /s = 128\;liters/s$

Solution

Discharge through the notch = 128 liters/s