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Rectangular Orifice

Discharge through a rectangular 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 Small Rectangular Orifice

An orifice is considered to be small, if the head of water above the orifice if over 5 times the height of the orifice.

In a small rectangular orifice, the velocity of water in the entire cross-section of the jet is approximately constant, so the discharge can be approximately with the relation, or

  • C_{d} = Coefficient of discharge for the orifice
  • a = Cross sectional area of the orifice
  • h = Height of the liquid above the center of the orifice
  • b = Width of the orifice
  • d = Depth of the orifice

Discharge Through A Large Rectangular Orifice

With a large rectangular orifice, the velocity of the liquid particles is not constant, because there is a considerable variation of effective pressure head over the height of an orifice: Velocity of liquid varies with the available pressure head of the liquid.

Now consider a large rectangular orifice, in one side of of the tank, discharging water as shown in figure. Let,
  • H_{1} = Height of liquid above the top of the orifice
  • H_{2} = Height of liquid above the bottom of the orifice
  • b = Breadth of the orifice
  • C_{d} = Coefficient of discharge

Consider a horizontal strip of thickness dh at a depth h from the water level as shown in figure.

\therefore Area of the strip = b.dh

We know that the theoretical velocity of water through the strip = \sqrt {2gh} and the discharge through the strip, dq = C_{d}\times Area \times Theoretical velocity
\therefore dq = C_{d}.b.dh.\sqrt {2gh}

Total discharge through the whole orifice may be found out by integrating the above equation between the limits H_{1} and H_{2}, i.e.
Q = \int_{H_{1}}^{H_{2}}C_{d}.b.dh.\sqrt {2gh}
\Rightarrow Q = C_{d}.b.\sqrt {2g}\int_{H_{1}}^{H_{2}}\sqrt h dh
\Rightarrow Q = \frac{2}{3} C_{d}.b.\sqrt {2g}(H_{2}^\frac{3}{2} - H_{1}^\frac{3}{2})

Example - Discharge through a large rectangular orifice
A large rectangular orifice of 1.5m wide and 0.5m deep is discharging water from a tank. If the water level in the tank is 3m above the top edge of the orifice, find the discharge through the orifice. Take coefficient of discharge for the orifice as 0.6.
  • b = 1.5m
  • d = 0.5m
  • H_{1} = 3m
  • C_{d} = 0.6

\therefore Height of water above the bottom of the orifice, H_{2} = 3+0.5 = 3.5m

So the discharge through the orifice,
Q = \frac{2}{3} C_{d}. b \sqrt {2g} (H_{2}^\frac{3}{2} - H_{1}^\frac{3}{2})
\Rightarrow Q = \frac{2}{3} \times 0.6 \times 1.5 \times \sqrt {2\times 9.81} [(3.5)^\frac{3}{2} - (3)^\frac{3}{2}]
\therefore Q = 0.6\times 4.429 \times (6.548-5.196) = 3.59 m^3/s
Discharge through the orifice, Q = 3.59 m3 /s