Variable Cross Section

A rectangular tank of 20m$\times$12m at the top and 10m$\times$6m 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.

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).

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

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

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

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

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

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

Time taken to empty the tank = 14 min 20 s

A tank in the form of frustum of a cone is 3m high and filled with water. It has a diameter of 2.4m at the top and 1.2m at the bottom. There is an orifice in the bottom of the tank, whose coefficient of discharge is 0.6. Find the size of the orifice which can empty the tank in 6 minutes.

Given,

• Depth of water = 3m
• Diameter at the top = 2.4m
• Diameter at the bottom = 1.2m
• Time of emptying = 6 min = 360 s
• $C_{d}$ = 0.6

Let, $d$ = Diameter of the orifice in meters

First of all, 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 diameter of the strip of water, $= 1.2 + \frac{1.2h}{3} = (1.2 + 0.4h) m$

$\therefore Area, A = \frac{\pi}{4}(1.2+0.4h)^2 m^2$

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

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).

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

$\Rightarrow 360 = \frac{-1}{C_{d}.a.\sqrt {2g}}\int_{3}^{0}\frac{\pi}{4}(1.2+0.4h)^2 \times h^{-\frac{1}{2}}.dh$

$\Rightarrow 360 = \frac{-\pi}{4C_{d}.a.\sqrt {2g}}\int_{3}^{0}(1.44+0.16h^2+0.96h)\times h^{-\frac{1}{2}}.dh$

$\Rightarrow 360 = \frac{0.296}{a}[2.88\times (3)^{\frac{1}{2}}+0.064\times (3)^{\frac{5}{2}}+0.64\times (3)^{\frac{3}{2}}]$

$\Rightarrow 360 = \frac{2.756}{a}$

$\therefore a = \frac{2.756}{360} = 0.0077 m^2$

$\therefore$ Diameter of orifice, $d = \frac{\sqrt 4a}{\pi} = \frac{\sqrt {4\times 0.0077}}{\pi} = 0.099 m = 99 mm$

Diameter of orifice = 99 mm