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# Stepped Notch

Discharge over a Stepped Notch

## 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
• Cd = Coefficient of discharge for both the notches

From the geometry of the notch, we get,

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

and the discharge over the notch 2,

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

Now the total discharge over the notch,

$Q&space;=&space;Q_1&space;+&space;Q_2&space;+&space;........$

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 Cd for all sections 0.6. All dimensions are in mm.
Workings
Given,
• Cd = 0.6

We know that the total discharge over the stepped notch,

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

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

$\therefore&space;Q&space;=&space;0.538&space;+&space;0.225&space;+&space;0.016&space;=&space;0.779\;m^3/s$
Solution
Discharge over the notch = 0.779 m3 /s