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Variable Cross Section

Time of emptying a tank of variable cross-section through an 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.


Previously the time of emptying of geometrical tanks (i.e., rectangular, hemispherical and circular) was discussed. But, sometimes, we come across tanks, which have variable cross-section. In such cases, there are two variables instead of one, as in the case of tanks of uniform cross-section. Since a single relation cannot be derived for different cross-sections, it is therefore essential that such problems should be solved from the first principles i.e., from the equation.

dt = \frac{-A.dh}{C_{d}.a\sqrt {2gh}}

This can be best understood from the following examples.

Example - Time of emptying a tank of variable cross-section
A rectangular tank of 20m\times12m at the top and 10m\times6m at the bottom is 3m deep as shown in figure.
There is an orifice of 450mm diameter at the bottom of the tank. Determine the time taken to empty the tank completely, if coefficient of discharge is 0.64.
  • Top length = 20m
  • Top width = 12m
  • Bottom length = 10m
  • Bottom width = 6m
  • Depth of water = 3m
  • Diameter of orifice, d = 450mm = 0.45m
  • C_{d} = 0.64

We know that the area of the orifice,
a = \frac{\pi}{4}\times d^2 = \frac{\pi}{4}\times (0.45)^2 = 0.159m^2

First of all, let us consider a small strip of water of thickness (dh) at a height (h) from the bottom of the tank. From the geometry of the figure, we find that the length of the strip of water,
l = 10 + \frac{10h}{3}

and the breadth of the strip of water,
b = 6 + \frac{6h}{3} = 6 + 2h

So, the area of the strip,
A = l\times b =(10 + \frac{10h}{3})(6 + 2h) = 6.67 h^2 + 40h + 60

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

The total time required to empty the tank may be found by integrating the above equation between the limits 3 and 0 (because initial head of water is 3m and final head of water is 0m).

T = \int_{3}^{0} \frac{-A.dh}{C_{d}.a.\sqrt {2gh}}

\;\;\;\;\;= \frac{-1}{C_{d}.a.\sqrt {2g}}\int_{3}^{0}A.h^{-\frac{1}{2}}.dh

\;\;\;\;\;= \frac{1}{C_{d}.a.\sqrt {2g}}\int_{0}^{3}(6.67 h^2 + 40h + 60).h^{-\frac{1}{2}}.dh

\;\;\;\;\;= \frac{1}{C_{d}.a.\sqrt {2g}}\int_{0}^{3}(6.67 h^{\frac{3}{2}} + 40h^{\frac{1}{2}} + 60h^{-\frac{1}{2}}).dh

\;\;\;\;\;= \frac{1}{0.64\times 0.159\times \sqrt {2\times 9.81}}\times [2.668\times (3)^{\frac{5}{2}}+26.67\times (3)^{\frac{3}{2}}+120\times (3)^{\frac{1}{2}}]

\;\;\;\;\;= 860\;s = 14\;min\;20\;s

Time taken to empty the tank = 14 min 20 s