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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.

  • \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:

Example - Discharge through a wholly drowned orifice
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 C_{d} = 0.64.
  • \inline b = 1.5m
  • \inline d = 0.5m
  • \inline h = 4m
  • \inline 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
Discharge = 4250 liters/s

Last Modified: 10 Jan 12 @ 17:32     Page Rendered: 2022-03-14 17:46:32