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Two orifices

Time of emptying a tank with two orifices
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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.


Previously, the cases of time of emptying the tank through one orifice at its bottom were discussed. But, sometimes, the tank has two (or even more) orifices. There may be two conditions.

  • When the orifices are at the same level
  • When the orifices are at different levels
Orifices At Same Level
When the orifices are at same level, they will be discharging the water under the same head. In that case both of the orifices will behave as a single orifice of area equal to sum of the two orifices.
Orifices At Different Levels
When the two orifices are at different levels, the water will flow through both the orifices under their respective heads so long as the water level reaches the upper orifice. After this, the water will flow through lower orifice only.

In such cases, the problem can be divided into two parts, i.e., first up to the upper orifice and then upto the bottom orifice.

The following examples will help more to solve this type of problems.


Example - Time of emptying a tank with two orifices
A vertical circular tank of 600mm diameter and 2.5m height is full of water. It contains two orifices each of 1300 mm2 area, one at the bottom of the tank and the other at a height of 1.25m above the bottom as shown in figure.


Determine the time required to empty the tank. Take coefficient of discharge for both of the orifices as 6.2.
  • Diameter of the tank = 600 mm = 0.6 m
  • H = 2.5 m
  • Area of each orifice, a = 1300 mm2
  • C_{d} = 0.62

For the sake of simplicity, let us divide the example into two parts, i.e., first up to the center of the top orifice, and then up to the bottom orifice.

First of all, consider the first part of the example. In that case, the water is flowing through both the orifices.

Now consider an instant, when the height of water above the center of the top orifice be h meters. At that instant, the height of water above the bottom orifice will be (h + 1.25) meters.

We know that the surface area of the tank,
a = \frac{\pi}{4}\times d^2 = \frac{\pi}{4}\times 0.6^2 = 0.2827m^2

Now let us use the general equation for the time to empty a tank,
\therefore dt = \frac{-A.dh}{C_{d}.a.\sqrt {2gh}}

\;= \frac{-A.dh}{C_{d}.a.\sqrt {2g(h+1.25)}+C_{d}.a.\sqrt {2gh}}

\;= \frac{-A.dh}{C_{d}.a.\sqrt {2g}[\sqrt {(h+1.25)}+\sqrt h]}

The total time (T_{1}) required to bring the water level up to the center of the top orifice may be found by integrating the above equation between the limits 1.25m and 0. Therefore

T_{1} = \int_{1.25}^{0} \frac{-A.dh}{C_{d}.2a.\sqrt {2g}[\sqrt {(h+1.25)}+\sqrt h]}

\Rightarrow T_{1} = \frac{A}{C_{d}.2a.\sqrt {2g}} \int_{0}^{1.25} \frac{dh}{\sqrt {(h+1.25)}+\sqrt h}

Multiplying the numerator and denominator by [\sqrt (h+1.25)-\sqrt h]

T_{1} = \frac{0.2827}{0.62\times 2\times 0.0013\times \sqrt {2\times 9.81}} \int_{0}^{1.25} \frac{\sqrt {(h+1.25)}-\sqrt h}{1.25}dh

\therefore T_{1} = 31.7\times 0.771 = 24.4 s

Now consider the flow of water below the center of the top orifice. A little consideration will show that now the water will be flowing through the bottom orifice only. We know that the time required to empty the tank,

T_{2} = \frac{2A\sqrt H_{1}}{C_{d}.a.\sqrt {2g}} = \frac{2\times 0.2827\times \sqrt 1.25}{0.62\times 0.0013\times \sqrt 19.62} = 201.5s

Total time,
T = T_{1}+T_{2} = 24.4+177.1 = 201.5s = 3\;min\;21\;s
Total time required = 3 min 21.5 s