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Stepped Notch

Discharge over a Stepped Notch

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Key Facts

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

= Moment of Inertia.
$\omega$ = Angular velocity
$\Omega$ = 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 stepped notch is a combination of rectangular notches as shown in figure. It is thus obvious that the discharge over such a notch will be the sum of the discharges over the different rectangular notches.

Consider a stepped notch as shown in figure. For the purpose of analysis, let us split up the notch into two rectangular notches 1 and 2. The total discharge over the notch will be the sum of the discharges over the two rectangular notches.

Let,

H₁ = Height of the liquid above the sill of notch 1
b₁ = Breadth of notch 1
H₂, b₂ = Corresponding values for notch 2, and
C_d = Coefficient of discharge for both the notches

From the geometry of the notch, we get,

$Q_1 = \frac{2}{3}C_d.b_1\sqrt {2g}\times H_1^{\frac{3}{2}}$

and the discharge over the notch 2,

$Q_2 = \frac{2}{3}C_d.b_2\sqrt {2g}\times (H_2^{\frac{3}{2}} - H_1^{\frac{3}{2}})$

Now the total discharge over the notch,

Example:

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Example - Discharge over a Stepped Notch

Problem

Find the discharge in m³ /s over a stepped notch shown in figure.

The level of water coincides with the top of the notch. Take C_d for all sections 0.6. All dimensions are in mm.

Workings

Given,

C_d = 0.6

We know that the total discharge over the stepped notch,

$Q = \frac{2}{3}C_d.b_1\sqrt {2g}\times H_1^{\frac{3}{2}} + \frac{2}{3}C_d.b_2\sqrt {2g}\times (H_2^{\frac{3}{2}} - H_1^{\frac{3}{2}}) + \frac{2}{3}C_d.b_3\sqrt {2g}\times (H_3^{\frac{3}{2}} - H_2^{\frac{3}{2}})$

$\Rightarrow Q = \frac{2}{3}\times 0.6\times 1.2\sqrt {2\times 9.81}\times (0.4)^{\frac{3}{2}} + \frac{2}{3}\times 0.6\times 0.6\sqrt {2\times 9.81}\times (0.6^{\frac{3}{2}} - 0.4^{\frac{3}{2}}) + \frac{2}{3}\times 0.6\times 0.15\sqrt {2\times 9.81}\times (0.65^{\frac{3}{2}} - 0.6^{\frac{3}{2}})$

$\therefore Q = 0.538 + 0.225 + 0.016 = 0.779\;m^3/s$

Solution

Discharge over the notch = 0.779 m³ /s