A hollow cone of height $h$ and diameter $d$ at the base is held with its axis vertical downwards and is filled with water. A small circular hole whose diameter is $1/n$ th that of the diameter of the base is made at the vertex. Assuming that the coefficient of discharge is equal to 0.62, show that the time taken for the depth of water to fall to one-half of its original value h cannot be less than
$(4 \sqrt2 – 1) n^2 \sqrt h / 12.4 \sqrt g$
How do you arrive at this formula?
So far I know that:
$dt = A dh / Q $and then you integrate to find time.
You have to express area in terms of water height above circular hole.
This a truncated cone and the area between two horizontal lines of the cone is constant and equal to $\pi/4 (d/n)^2$. We have to determine angle $\theta$.
Length of each opposite side (o) of triangle for the side slopes of cone is
$o = (d – d/n) / 2 $
Using $h$ and $o$, $tan \theta $ can be calculated: $\tan 0 = o/h$ where h is maximum height of water.
diameter $d$ at any height:$ d(h) = 2 \tan (0)(h) + (d/n)$
Muliplying by $\pi /4 $ gives area at any height.
But the challenge that I'm having here is that when you take $2tan (0)(h) + (d/n)$ and put $tan 0 = ((d – d/n) / 2) / h$ into $d(h)$, $h$ cancels out.
h relating to $tan \theta $ is the maximum height so $h = h$ for that and doesn't change, but $h$ in the $d(h)$ formula varies, so how do I handle this.
flow out of circular hole $Q = Cd a \sqrt {2gh}$
Does anyone have any ideas that can help me solve this thanks?
Best Answer
Suppose at some time $t$ the water level is $y$. With Torricelli's law and the discharge coefficient the outflow speed $v$ is:
$$v=C\sqrt{2\rho g y}$$
The volumetric throughput $Q_v(y)$ is at that point is:
$$Q_v(y)=vA=\frac{\pi}{4} (d/n)^2C\sqrt{2\rho g y}=\frac{\pi d^2C}{4n^2}\sqrt{2\rho g y}$$
In an infinitesimal amount of time $dt$ a volume $dV$ is dispensed:
$$dV=Q_v(y)dt=\frac{\pi d^2C}{4n^2}\sqrt{2\rho g y}dt$$
This causes the level in the tank to drop by $dy$ so that:
$$dV=-\frac{\pi}{4}\Big(\frac{d}{h}\Big)^2y^2dy$$
Because the diameter $d(y)$ of the liquid's surface is a function of $y$, so is the surface area: $d(y)=\frac{d}{h}y \implies A(y)=\frac{\pi}{4}\Big(\frac{d}{h}\Big)^2y^2$
The negative sign is needed because $dV > 0$ but $dy<0$.
So we have:
$$\frac{\pi d^2C}{4n^2}\sqrt{2\rho g y}dt=-\frac{\pi}{4}\Big(\frac{d}{h}\Big)^2y^2dy$$
Or:
$$C\sqrt{2\rho g}\sqrt{y}dt=-\frac{n^2}{h^2}y^2dy$$
Or:
$$C\sqrt{2\rho g}dt=-\frac{n^2}{h^2}y^{\frac{3}{2}}dy$$
To find the emptying time integrate between $y=h, t=0$ and $y=0, t$.
$$C\sqrt{2\rho g}\int_0^tdt=-\frac{n^2}{h^2}\int_h^0y^{\frac{3}{2}}dy$$
You can take it from there.