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# Triangular Notch

Discharge over a Triangular Notch

## Overview

A triangular notch is also called a V-notch. Consider a triangular notch, in one side of the tank, over which water is flowing as shown in figure.

Let,
• H = Height of the liquid above the apex of the notch
• θ = Angle of the notch
• Cd = Coefficient of discharge

From the geometry of the figure, we find that the width of the notch at the water surface,
$=&space;2H\tan&space;\frac{\theta}{2}$

$\inline&space;\therefore$ Area of the strip = $\inline&space;2(H-h)\tan&space;\frac{\theta}{2}.&space;dh$

We know that the theoretical velocity of water through the strip = $\inline&space;\sqrt&space;{2gh}$

and discharge over the notch,
$dq&space;=&space;C_d&space;\times&space;Area\;of\;strip&space;\times&space;Theoretical\;velocity$
$\Rightarrow&space;dq&space;=&space;C_d&space;\times&space;2(H-h)\tan&space;\frac{\theta}{2}.&space;dh\sqrt&space;{2gh}$

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

$Q&space;=&space;\int_{0}^{H}&space;C_d&space;\times&space;2(H-h)\tan&space;\frac{\theta}{2}.&space;dh\sqrt&space;{2gh}$

$\Rightarrow&space;Q&space;=&space;2C_d\sqrt&space;{2g}\times&space;\tan&space;\frac{\theta}{2}&space;\int_{0}^{H}&space;(H-h)\sqrt&space;h&space;dh$

$\Rightarrow&space;Q&space;=&space;2C_d\sqrt&space;{2g}\times&space;\tan&space;\frac{\theta}{2}\int_{0}^{H}&space;(Hh^\frac{1}{2}-h^\frac{3}{2})&space;dh$

$\therefore&space;Q&space;=&space;\frac{8}{15}C_d&space;\sqrt&space;{2g}&space;\tan&space;\frac{\theta}{2}\times&space;H^{\frac{5}{2}}$

A triangular notch gives more accurate results for low discharges than rectangular notch and the same triangular notch can measure a wide range of flows accurately.

Example:
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##### Example - Discharge through a triangular notch
Problem
A right-angled V-notch was used to measure the discharge of a centrifugal pump. If the depth of water at V-notch is 200mm, calculate the discharge over the notch in liters per minute. Assume coefficient of discharge as 0.62.
Workings
Given,
• $\inline&space;\theta&space;=&space;90^{\circ}$
• $\inline&space;H&space;=&space;200\;mm&space;=&space;0.2\;m$
• $\inline&space;C_d&space;=&space;0.62$

We know that the discharge over the triangular notch,
$Q&space;=&space;\frac{8}{15}C_d&space;\sqrt&space;{2g}&space;\tan&space;\frac{\theta}{2}\times&space;H^{\frac{5}{2}}$

$\Rightarrow&space;Q&space;=&space;\frac{8}{15}\times&space;0.62\times&space;\sqrt&space;{2\times&space;9.81}&space;\tan&space;45^{\circ}\times&space;(0.2)^{\frac{5}{2}}$

$\Rightarrow&space;Q&space;=&space;1.465\times&space;0.018&space;=&space;0.026&space;m^3/s$

$\therefore&space;Q&space;=&space;26\;liters/s&space;=&space;1560\;liters/min$
Solution
Discharge over the notch = 1560 liters/s