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Submerged Orifice

Discharge through a submerged orifice

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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 Wholly Drowned Orifice

When the outlet side of an orifice is beneath the surface of liquid it is known as a wholly submerged orifice as shown in fig.1. In such orifices, the coefficient of contraction is equal to one.

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Consider a wholly drowned orifice discharging water as shown in fig.1.

Let,

$\inline H_{1}$ = Height of water (on the upstream side) above the top of the orifice
$\inline H_{2}$ = Height of water (on the upstream side) above the bottom of the orifice
$\inline h$ = Difference between the two water levels on either side of the orifice
$\inline C_{d}$ = Coefficient of discharge
$\inline C_{v}$ = Coefficient of velocity
$\inline C_{c}$ = Coefficient of contraction

$\inline \therefore$ Area of orifice = $\inline b(H_{2}-H_{1})$

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

$\inline \therefore$ Actual velocity of water = $\inline C_{v}\sqrt {2gh}$

From the relation of hydraulic coefficients we know that, $\inline C_{d} = C_{v}\times C_{c}$

Since coefficient of contraction is 1 in this case, therefore $\inline C_{d} = C_{v}$

$\inline \therefore$ Actual velocity of water = $\inline C_{d}\sqrt {2gh}$

Now the discharge through the orifice,

$\inline Q$ = Area of orifice $\inline \times$ Actual velocity

$\Rightarrow Q = b(H_{2}-H_{1}) \times C_{d}\sqrt {2gh}$

$\therefore Q = C_{d}.b(H_{2}-H_{1}) \times \sqrt {2gh}$

If depth of the drowned orifice (d) is given instead of $\inline H_{1}$ and $\inline H_{2}$ , then in such cases the discharge through the wholly drowned orifice is:

$Q = C_{d}.b.d.\sqrt {2gh}$

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Example:

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Example - Discharge through a wholly drowned orifice

Problem

A drowned orifice 1.5m wide and 0.5m deep is provided in one side of a tank. Find the discharge in liters/s through the orifice, if the difference of water levels on both the sides of the orifice be 4m. Take $C_{d}$ = 0.64.

Workings

Given,

$b$ = 1.5m
$d$ = 0.5m
$h$ = 4m
$C_{d}$ = 0.64

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

$= 0.64 \times 1.5 \times 0.5 \times \sqrt {2\times 9.84 \times 4}$

$= 0.48 \times 8.859$

$= 4.25m^3 /s = 4250 liters/s$

Solution

Discharge = 4250 liters/s