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# Submerged Orifice

Discharge through a submerged orifice

## 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.

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

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

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

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

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

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

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

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

Now the discharge through the orifice,

$\inline&space;Q$ = Area of orifice $\inline&space;\times$ Actual velocity

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

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

If depth of the drowned orifice (d) is given instead of $\inline&space;H_{1}$ and $\inline&space;H_{2}$, then in such cases the discharge through the wholly drowned orifice is:
$Q&space;=&space;C_{d}.b.d.\sqrt&space;{2gh}$

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 $\inline&space;C_{d}$ = 0.64.
Workings
Given,
• $\inline&space;b$ = 1.5m
• $\inline&space;d$ = 0.5m
• $\inline&space;h$ = 4m
• $\inline&space;C_{d}$ = 0.64

$\therefore&space;Q&space;=&space;C_{d}.b.d&space;\sqrt&space;{2gh}$
$=&space;0.64&space;\times&space;1.5&space;\times&space;0.5&space;\times&space;\sqrt&space;{2\times&space;9.84&space;\times&space;4}$
$=&space;0.48&space;\times&space;8.859$
$=&space;4.25m^3&space;/s&space;=&space;4250&space;liters/s$
Solution
Discharge = 4250 liters/s