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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 (\tau) = I\omega\Omega (In the limit).
  • I = Moment of Inertia.
  • \omega = Angular velocity
  • \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.

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

  • 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}

If depth of the drowned orifice (d) is given instead of H_{1} and 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 C_{d} = 0.64.
  • b = 1.5m
  • d = 0.5m
  • h = 4m
  • C_{d} = 0.64

= 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