Use Bernoulli's equation to derive Torricelli's Law (check any website for this) for the velocity out of the hole; $ v = \sqrt{2 g h(t)} $, where g is gravity and $ h(t) $ is the height of the fluid in the tank at any time.
Write a balance on the mass of the fluid in the tank as:
$$ \text{in - out + gen = accumulation} $$
$$ \rho Q_{in} - \rho Q_{out} = \frac{d(\rho V)}{dt} $$
where the generation term is zero, $\rho$ is the fluid density (constant here) and $Q_{in}$ and $Q_{out}$ is the flow rate in and out of the tank, respectively. $Q_{in}$ is zero so we get:
$$ \frac{dV}{dt} = -Q_{out} $$
The flow out is $v A = \sqrt{2 g h(t)} A$, where $A$ is the area of the hole which is calculated by knowing the diameter of the circular hole; given in the problem statement as 1 inch.
The volume of the tank, $V = a_t h(t)$, is the height, $h(t)$ times the area, $a_{t}$.
Putting it all together, we get the separable first order differential equation for the height of the fluid in the tank versus time:
$$ \frac{dh}{dt} = -\frac{A}{a_t}\sqrt{2g}\sqrt{h}$$
Prepare it for Integration
$$ \frac{dh}{\sqrt{h}} = -\frac{A}{a_t}\sqrt{2g}{dt}$$
Integrate the equation. The upper bound for $dh$ is $h(t)$. The lower bound is $h(0)=H$ . For $dt$ we integrate from $t$ to $0$:
$$ 2 (\sqrt{h(t)} - \sqrt{H}) =-\frac{A}{ a_t}\sqrt{2g} t$$
Solve for $h(t)$
$$ h(t)=[\sqrt{H} -\frac{A}{2 a_t}\sqrt{2g} t]^2 $$
To find the time when the tank empties, set $h$ equal to zero and solve for $t$:
$$ T= \sqrt{\frac{H}{2g}} \frac{2 a_t}{A}$$
Plug that back in the previous equation to clarify time dependence on t:
$$ h(t)=H[1 - t/T]^2 $$
The times to empty the same tank for two different starting heights, $H_1$ and $H_2$ is:
$$ \frac{T_1}{T_2} = \sqrt{\frac{H_1}{H_2}} $$
So, finally, if $H_2 = 2 H_1$ as in the problem statement, then the time to empty the tank is not double, but $\sqrt{2}$ times longer.
Make sense?
Paul Safier
Best Answer
Yes, at least if you ignore the little droplets that leave the bottom of the upper bucket a little damp. There will also be some water in the tube, but it won't be much higher than the level of the water in the lower bucket. (There might be some capillary action that raises it a little, but probably not much.)
I suspect the reason you ask this is because you wonder if the pressure at the top of the tube will be lower than the pressure at the bottom of the tube -- because the bottom will be immersed under a bucketful of water, but the top of the tube will have no pressure. And indeed, that pressure difference will be there. So it might seem as if the water should want to drain from high pressure to low, and thus up the tube.
The reason it doesn't do that is because of the weight of the water in the tube itself. Once the draining is done, the pressure just inside the bottom of the tube will equal the pressure just outside it in the lower bucket, which means that the water won't move any more. The source of the pressure in the tube is exactly the same as the source of pressure in the bucket: the weight of the water above.