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,

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

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

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

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

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

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

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

Now the discharge through the orifice,

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

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

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