Physics problem: acceleration of water level in a container due to draining from an opening in its bottom

Problem statement

From an opening at the bottom of a tall container, water drains out. The cross-section area of the container is $S$, and the cross-section area of the draining stream of water is $\sigma$. The water level in the vessel moves downward with constant acceleration. Determine this acceleration.

Solution

I'll describe two methods to solve the problem: the first one is the one I used and it is based on Bernoulli's law (energy conservation) and the second one is using the Toricelli's formula and accompanying assumptions from the get go. The second one I would guess a more experimented physicist would use and be done with it (see the approach 2). If I were to think about the second approach right away, this post would not happen.

Approach 1

Let's denote the speed of the water at the surface of the container and at the draining hole as $v_0$ and $v_1$ respectively. Then we can write energy and mass conservation expressions for the two cross-sections as follows:

\begin{align*} \begin{cases} \frac{v_0^2}{2} + gh = \frac{v_1^2}{2} \\ v_1\sigma = v_0 S \end{cases} \end{align*}

where $h$ is the current water level in the container, ang $g$ is the acceleration due to gravity.

Now, if we eliminate $v_1$ from the energy conservation equation and apply a time derivative to the both sides of the equation we will obtain the following relation (we use an upper dot to denote a time derivative, $\dot{x}=\frac{dx}{dt}$):

\begin{equation*} 2 v_0 \dot{v_0} + 2g\dot{h}=\left(\frac{S}{\sigma}\right)^2 2 v_0 \dot{v_0} \end{equation*}

Let's denote the acceleration of the water level as $a=\dot{v_0}$, and also note that $v_0 = -\dot{h}$ (the minus sign here is added because the level $h$ is decreasing with time and I prefer to consider $v_0$ as speed value without direction). Using this notation, the previous equation can be written as:

\begin{equation*} 2 v_0 a - 2gv_0=\left(\frac{S}{\sigma}\right)^2 2 v_0 a \end{equation*}

Simplifying we get the following expression for the water level acceleration in the container:

\begin{equation*} a = \frac{g}{1 - \left(\frac{S}{\sigma}\right)^2} \end{equation*}

Now we could consider that the conditions given in the problem statement imply $S\gg\sigma \rightarrow S/\sigma \gg 1$, therefore we can approximate the above expression as:

\begin{equation*} a \approx -g\left(\frac{\sigma}{S}\right)^2 \end{equation*}

Note that $a = \dot{v_0} < 0$, meaning that the water level descent in the container is slowing down with time.

Approach 2

For this approach we recognize right away that the problem conditions allow the use of Toricelli's formula for the speed $v_1$ of draining water from a large container with the water level at $h$:

\begin{equation*} v_1=\sqrt{2gh} \end{equation*}

The draining water will cause the change to the water level as follows: \begin{align*} \dot{h}=-v_1\sigma/S=-\frac{\sigma}{S}\sqrt{2gh} \end{align*}

If we apply a time derivative to the above equation and substitute $\dot{h}$, we obtain:

\begin{align*} \ddot{h} & =-\frac{\sigma}{S}\sqrt{2g}\frac{1}{2\sqrt{h}}\dot{h} \\ & = g\left(\frac{\sigma}{S}\right)^2 \end{align*}

Since the water level acceleration is $a = \dot{v_0} = d_t \left( -\dot{h} \right) = -\ddot{h}$, we obtain the final expression using the above relation for the second derivative of the water level as:

\begin{align*} a = -g\left(\frac{\sigma}{S}\right)^2 \end{align*}

Note that in this case we did not have to make any approximations as they were already applied for the Toricelli's formula to be valid (mainly that the speed of the water level $v_0$ is much lower than the draining speed $v_1$).