# Time to empty conical tank to half of its original height

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?

• Hi Rob, I have added mathjax to your post, to increase the chance that you might get an answer. Will you please check it, as I might have got things like the square root wrong. If you have to, just leave the mathjax out of problem terms. Best of luck with it. The mathjax reference page is at meta.math.stackexchange.com/questions/5020/… thanks – user108787 Aug 28 '16 at 15:48

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.

• your use of $d(y)$, $d$ and $dy$ is slightly confusing, can you fix this? – nluigi Aug 29 '16 at 18:27
• Hi Gert Thanks for your reply, bit It seems that you haven't completely taken into account the small orifice at the bottom of the conical tank. I mean in the surface area of water in the tank, part of your answer and the integration. I get diameter at any height d (y) = [(d/h) * y - (d/n) * (y/h)], which simplifies to d(y) = [d (1/n + y/h - y/nh )]. It's just that I'm not quite sure how to get the answer $(4 \sqrt2 - 1) n^2 \sqrt h / 12.4 \sqrt g$ from there. – Rob Wilkinson Dec 26 '17 at 1:51
• @RobWilkinson - You would do this by redefining the coordinate system. The lowest tip of the cone would become $y=0$. Where the orifice is would be $y_2$ and the starting position of the fluid would be $y_1$. Now integrate the last formula between $y_1$ and $y_2$. Thanks. – Gert Dec 26 '17 at 13:55
• @Gert - From this website scribd.com/document/274250309/FLUID-MECHANICS-NOTES-pdf on pg 54 -57, it has theory about time to empty a conical tank. From this if the diameter varied from d at the top to d/n for the orifice at the bottom, then I get, diameter of surface area at any height d (y) = [(dy)/(h)-(dy)/(nh)+(d)/(n)]^2 and that gives dt = [(dy)/(h)-(dy)/(nh)+(d)/(n)]^2 / [(0.62 * d^2 * sqrt 2g)/ (n^2)] . So after integrating this formula between the limits of y = h and y = 1/2h wouldn't you get a more accurate time for emptying a conical tank to one half it's original height. – Rob Wilkinson Mar 22 '18 at 15:11
• @Gert - Thanks for your reply Gert. I've rearranged your formula to dt={(n^2)(y^1.5) dy / h^2} / (0.62 * sqrt * 2g). I have then integrated between the limits of y = h and y = 1/2h, or as you have put it y1 = h and y2 = 1/2h. So now I have managed to get the answer T = $(4 \sqrt2 - 1) n^2 \sqrt h / 12.4 \sqrt g$ . However can please show me how you got d(y) = d * y / h . – Rob Wilkinson Mar 22 '18 at 15:21