Question 5.16: A tank has the form of a frustum of a cone, with a diameter ...
A tank has the form of a frustum of a cone, with a diameter of 2.44 m at the top and 1.22 m at the bottom as shown in Fig. 5.38. The bottom contains a circular orifice whose coefficient of discharge is 0.60. What diameter of the orifice will empty the tank in 6 minutes if the full depth is 3.05 m?
Let the diameter of orifice be d_{0}, and at any instant t, the height of the liquid level above the orifice be h. Then during an infinitesimal time dt, discharge through the orifice is
q=C_{d} \frac{\pi d_{0}^{2}}{4} \sqrt{2 g h} d t
=0.60 \times \frac{1}{4} \pi d_{0}^{2} \sqrt{2 g h} d t
If the liquid level in the tank falls by an amount dh during this time, then from continuity,
-A_{h} d h=0.60 \frac{\pi d_{0}^{2}}{4} \sqrt{2 g h} d t (5.113)
where A_{h} is the area of the tank at height h.
From the geometry of the tank (Fig. 5.38),
\tan \alpha=\frac{(2.44-1.22)}{2 \times 3.05}=0.2
Therefore the diameter of the tank at height h = 1.22 + 2 ×0.2 h
Hence, A_{h}=(\pi / 4)(1.22+0.4 h)^{2}
Substituting the value of A_{h} in Eq. (5.113), we have
0.60 \times(1 / 4) \pi d_{0}^{2} \sqrt{2 \times 9.81 h} d t=-\pi / 4(1.22+0.4 h)^{2} d h
or d_{0}^{2} \int d t=\frac{1}{0.60 \times \sqrt{2 \times 9.81}} \int_{3.05}^{0}(1.22+0.4 h)^{2} h^{-1 / 2} d h
Since, the time of emptying \int d t=360 \text { seconds }
d_{0}^{2}=\frac{1}{0.60 \times \sqrt{2 \times 9.81} \times 360} \int_{3.05}^{0}\left(1.22+0.4 h^{2}\right) h^{-1 / 2} d h
Integrating and solving for d_{0}, we ge
d_{0}^{2}=0.010 m ^{2}
or d_{0}=0.1 m =100 mm