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# Rectangular Orifice

Discharge through a rectangular orifice

## Discharge Through A Small Rectangular Orifice

An orifice is considered to be small, if the head of water above the orifice if over 5 times the height of the orifice.

In a small rectangular orifice, the velocity of water in the entire cross-section of the jet is approximately constant, so the discharge can be approximately with the relation,
$Q&space;=&space;C_{d}&space;\times&space;a&space;\times&space;\sqrt&space;{2gh}$
or
$Q&space;=&space;C_{d}&space;\times&space;(b\times&space;d)&space;\times&space;\sqrt&space;{2gh}$

where,
• $\inline&space;C_{d}$ = Coefficient of discharge for the orifice
• a = Cross sectional area of the orifice
• h = Height of the liquid above the center of the orifice
• b = Width of the orifice
• d = Depth of the orifice

## Discharge Through A Large Rectangular Orifice

With a large rectangular orifice, the velocity of the liquid particles is not constant, because there is a considerable variation of effective pressure head over the height of an orifice: Velocity of liquid varies with the available pressure head of the liquid.

Now consider a large rectangular orifice, in one side of of the tank, discharging water as shown in figure. Let,
• $\inline&space;H_{1}$ = Height of liquid above the top of the orifice
• $\inline&space;H_{2}$ = Height of liquid above the bottom of the orifice
• $\inline&space;b$ = Breadth of the orifice
• $\inline&space;C_{d}$ = Coefficient of discharge

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

$\inline&space;\therefore$ Area of the strip = $\inline&space;b.dh$

We know that the theoretical velocity of water through the strip = $\inline&space;\sqrt&space;{2gh}$ and the discharge through the strip, $\inline&space;dq$ = $\inline&space;C_{d}\times$ Area $\inline&space;\times$ Theoretical velocity
$\therefore&space;dq&space;=&space;C_{d}.b.dh.\sqrt&space;{2gh}$

Total discharge through the whole orifice may be found out by integrating the above equation between the limits $\inline&space;H_{1}$ and $\inline&space;H_{2}$, i.e.
$Q&space;=&space;\int_{H_{1}}^{H_{2}}C_{d}.b.dh.\sqrt&space;{2gh}$
$\Rightarrow&space;Q&space;=&space;C_{d}.b.\sqrt&space;{2g}\int_{H_{1}}^{H_{2}}\sqrt&space;h&space;dh$
$\Rightarrow&space;Q&space;=&space;\frac{2}{3}&space;C_{d}.b.\sqrt&space;{2g}(H_{2}^\frac{3}{2}&space;-&space;H_{1}^\frac{3}{2})$

Example:
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##### Example - Discharge through a large rectangular orifice
Problem
A large rectangular orifice of 1.5m wide and 0.5m deep is discharging water from a tank. If the water level in the tank is 3m above the top edge of the orifice, find the discharge through the orifice. Take coefficient of discharge for the orifice as 0.6.
Workings
Given,
• b = 1.5m
• d = 0.5m
• $\inline&space;H_{1}$ = 3m
• $\inline&space;C_{d}$ = 0.6

$\inline&space;\therefore$ Height of water above the bottom of the orifice, $\inline&space;H_{2}$ = 3+0.5 = 3.5m

So the discharge through the orifice,
$Q&space;=&space;\frac{2}{3}&space;C_{d}.&space;b&space;\sqrt&space;{2g}&space;(H_{2}^\frac{3}{2}&space;-&space;H_{1}^\frac{3}{2})$
$\Rightarrow&space;Q&space;=&space;\frac{2}{3}&space;\times&space;0.6&space;\times&space;1.5&space;\times&space;\sqrt&space;{2\times&space;9.81}&space;[(3.5)^\frac{3}{2}&space;-&space;(3)^\frac{3}{2}]$
$\therefore&space;Q&space;=&space;0.6\times&space;4.429&space;\times&space;(6.548-5.196)&space;=&space;3.59&space;m^3/s$
Solution
Discharge through the orifice, Q = 3.59 m3 /s