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Rectangular Orifice

Discharge through a rectangular orifice

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Contents

Discharge Through A Small Rectangular Orifice
Discharge Through A Large Rectangular Orifice
Page Comments

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.

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,

$Q = C_{d} \times a \times \sqrt {2gh}$

(2)

$Q = C_{d} \times (b\times d) \times \sqrt {2gh}$

(3)

where,

$\inline C_{d}$ = 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.

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Now consider a large rectangular orifice, in one side of of the tank, discharging water as shown in figure. Let,

$\inline H_{1}$ = Height of liquid above the top of the orifice
$\inline H_{2}$ = Height of liquid above the bottom of the orifice
$\inline b$ = Breadth of the orifice
$\inline C_{d}$ = Coefficient of discharge

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

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

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

$\therefore dq = C_{d}.b.dh.\sqrt {2gh}$

Total discharge through the whole orifice may be found out by integrating the above equation between the limits $\inline H_{1}$ and $\inline H_{2}$ , i.e.

$Q = \int_{H_{1}}^{H_{2}}C_{d}.b.dh.\sqrt {2gh}$

$\Rightarrow Q = C_{d}.b.\sqrt {2g}\int_{H_{1}}^{H_{2}}\sqrt h dh$

$\Rightarrow Q = \frac{2}{3} C_{d}.b.\sqrt {2g}(H_{2}^\frac{3}{2} - H_{1}^\frac{3}{2})$

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
$H_{1}$ = 3m
$C_{d}$ = 0.6

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

So the discharge through the orifice,

$Q = \frac{2}{3} C_{d}. b \sqrt {2g} (H_{2}^\frac{3}{2} - H_{1}^\frac{3}{2})$

$\Rightarrow Q = \frac{2}{3} \times 0.6 \times 1.5 \times \sqrt {2\times 9.81} [(3.5)^\frac{3}{2} - (3)^\frac{3}{2}]$

$\therefore Q = 0.6\times 4.429 \times (6.548-5.196) = 3.59 m^3/s$

Solution

Discharge through the orifice, Q = 3.59 m³ /s