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Variable Cross Section

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

Overview

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:

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Example - Time of emptying a tank of variable cross-section
Problem
A rectangular tank of 20m\times12m at the top and 10m\times6m at the bottom is 3m deep as shown in figure.
23547/variable_1.png
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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.
Workings
Given,
  • 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

Solution
Time taken to empty the tank = 14 min 20 s