[ASK] Is This Debit Question Solvable?

[Monoxdifly]

A water container whose volume is 400 liters has 3 vertical holes. The first hole is located at the height which shows 300 liters. The second hole is located at the height which shows 150 liters. The third hole is parallel to the bottom. If the water in the container is drained by opening the holes one by one from the top to the bottom (second hole isn't opened until the first hole can no longer drain the water, third hole also isn't opened until the first and second holes can no longer drain the water), and the water container is empty in 10 hours, what are the debit (speed of draining water) of each holes?

 

[MarkFL]

In physics, we typically find after making certain reasonable assumptions (container is open to the atmosphere, area of draining hole is small compared to the area of the surface of the water), that Bernoulli's equations tells us that the velocity $v$ of the water emerging from the hole is given by:

$$v=\sqrt{2gh}$$

where $h$ is the height of the water above the hole.

Thus, the rate at which the water will drain from any of the holes will not be constant.

I would say not knowing the geometry of the container (among other things), would make this problem unsolvable. There simply isn't enough information given. :D

 

[I like Serena]

Let's assume a container with constant horizontal intersection area $A$.
And let's assume holes that each have the same area $a$.

Let's look only at the volume $V$ above a hole when that is open.
Let $h$ be the height of the water surface above the hole.
And let the initial volume be $V_0$.

Then (with MarkFL's formula):
$$V=Ah \quad\Rightarrow\quad h=\frac V A$$
$$v=\sqrt{2gh} = \sqrt{\frac{2gV}{A}} $$
$$\d V t = -av = -a\sqrt{\frac{2gV}{A}} = -B\sqrt V \quad \text{where }B=a\sqrt{\frac{2g}{A}} \tag 1$$
Solving the ODE gives us:
$$t=\frac 2B(\sqrt{V_0} - \sqrt{V}) \tag 2$$
$$V=\frac 14(2\sqrt{V_0}-Bt)^2 \quad\Rightarrow\quad \d V t=-\frac B2(2\sqrt{V_0}-Bt) \tag 3$$
We can deduce $B$ from (2) in combination with the total time of 10 hours (after adding the 3 contributions to each other).
Substituting in (1) or (3) gives us the drain speeds as function of $V$ respectively $t$. (Thinking)