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Drowned Orifice

Discharge through a drowned 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 Partially Drowned Orifice

If the outlet side of the orifice is partly under water, it is known as a partly drowned or partly submerged orifice as shown in fig.2.

The discharge through a partially drowned orifice is obtained by treating the lower portion as a drowned orifice and the upper portion as an orifice running free, and then by adding the two discharges thus obtained.

We know the discharge through the free portion,

and discharge through the drowned orifice,

Now total discharge, Q = Q_{1}+Q_{2}

Example - Discharge through a partially drowned orifice
An orifice in one side of a large tank is rectangular in shape, 2 meters broad and 1 meter deep. The water level on one side of the orifice is 4 meters above its top edge. The water level on the other side of the orifice is 0.5 meter below its top edge as shown in fig.


Calculate the discharge through the orifice per second if C_{d} = 0.63
  • b = 2m
  • d = 1m
  • H_{1} = 4m
  • H_{2} = 4+1 = 5m
  • H = 4+0.5 = 4.5m
  • C_{d} = 0.63

Since the orifice is partially drowned, therefore let us split up the orifice into two portion will be treated as a free orifice and the lower portion as a drowned orifice.

The discharge through the free portion of the orifice,
Q_{1} = \frac{2}{3} C_{d}. b \sqrt {2g} (H_{2}^\frac{3}{2} - H_{1}^\frac{3}{2})
\Rightarrow Q_{1} = \frac{2}{3} \times 0.63 \times \sqrt {2\times 9.81} [(4.5)^\frac{3}{2} - (4)^\frac{3}{2}]
\therefore Q_{1} = 1.86 \times (9.546 - 8.0) = 2.88 m^3 /s

The discharge through the drowned portion of the orifice,
Q_{2} = C_{d}.b (H_{2}-H)\times \sqrt {2gh}
\Rightarrow Q_{2} = 0.63\times 2(5-4.5)\times \sqrt {2\times 9.81\times 4.5}
\therefore Q_{2} = 0.63\times 9.396 = 5.92 m^3 /s

\therefore Total discharge Q = Q_{1}+Q_{2} = 2.88 + 5.92 = 8.8m3 /s
Total discharge = 8.8m3 /s