# Height of water fountain due to pressure difference

This was a question on a fluid dynamics exam: the pressure $$p_A$$ on a plane is lower than the atmospheric pressure. We fill a water bottle with a straw on the ground and open it in the plane. Water will come out due to the pressure difference. Calculate the maximum height of the water fountain above the straw.

What I think happens is that the only place where water can come out is at the straw, so we have a force upwards due to the pressure difference $$p_{bottle}-p_A$$. Water will only move upwards if this force can counteract the gravitational force $$\rho_{water} g$$ (I think we can negate the fact that we are high up and the gravitational force is weaker). Then I tried to calculate it as if it was a point mass thrown up with a force $$p_{water}-p_A$$. But to be honest I'm not even able to calculate the initial velocity (I'm notoriously bad at physics).

Any help or insight will be appreciated.

• I think the gravitational force should be $m_{water} g$. – Abhay Hegde Jan 20 '19 at 17:04
• It is not really given what the mass is or how much water is in the bottle so I was thinking to just use unit force. (We have done this multiple times in the course, but I am not sure if it is applicable here as well though) – Jarne Renders Jan 20 '19 at 17:14
• If the water were incompressible, how much would its volume increase when you released the pressure? – Chet Miller Jan 20 '19 at 19:31
• I'm not exactly sure, I found on the internet that by Boyle's law the pressure is inversely related to the volume. So if the pressure is released, it drops with $p_{bottle}-p_A$, which means the volume increases by $p_{bottle}-p_A$ times some constant? – Jarne Renders Jan 20 '19 at 19:58

Use Bernoulli's Principle: the energy equation for incompressible, inviscid fluid flow:

$$p_{bottle}+\frac12 \rho v_{bottle}^2+\rho gh_0=p_A+\frac12 \rho v_1^2+\rho gh_1$$

Because the bottle is much wider than the straw and with the continuum equation: $$A_{bottle}v_{bottle}=A_{straw}v_{straw}$$

Where the $$A$$ are cross-sections and with: $$\frac{A_{straw}}{A_{bottle}}<< 1$$

So that $$v_{bottle}\approx 0$$.

Now as the water flows out of the straw at $$v_{straw}$$ it will decelerate due to gravity, until $$v_1=0$$ and it has reached the height $$h_1$$ (the water then starts falling back to Earth).

Plug it all in and we have: $$p_{bottle}+\rho gh_0\approx p_A+\rho gh_1$$

Or:

$$\Delta h\approx \frac{p_{bottle}-p_A}{\rho g}$$

In essence what happens here is that pressure energy is converted to potential energy.

• The Bernoulli equation applies to steady state flow. This is a transient flow, so the Bernoulli equation does not apply. Furthermore, since liquid is close to being incompressible, once the pressure on fluid is released, the liquid volume will not change, and essentially nothing will happen. The fluid pressure will essentially drop instantaneously to the new pressure of the surroundings. Aside from that, there will be no fountain. – Chet Miller Jan 20 '19 at 23:46
• @ChesterMiller: I think you should formulate your comment as an answer. – Gert Jan 21 '19 at 2:34

In my judgment, it is invalid to apply the Bernoulli equation to this problem because the Bernoulli equation applies only to steady state flows (or to flows that are nearly steady state), and this problem involves neither. However, water is very nearly incompressible, particularly for a change in pressure that is relatively small (like in the present case). So the volume of water will not change significantly when the pressure on it is reduced. That means that, for all practical purposes, there will be no fountain effect. All that will happen will be that the pressure in the water within the bottle will drop to the new pressure, essentially instantaneously.

• The problem is that there is not enough detail given. Your answer assumes that the bottle is completely filled of water which is perfectly reasonable. Another possibility is that there is air and water in the bottle in which case there will be a noticeable movement of water in the straw as the air expands but who knows the pressure of the air when a steady state is reached? – Farcher Jan 21 '19 at 14:10
• @farcher These are valid comments. I assumed that when the question said “filled,” it meant completely filled with water. If there were actually air in the head space, this would change things, but that would have to be specified along with the volume of water and the straw diameter. – Chet Miller Jan 21 '19 at 14:39
• My comment, as was my +1, was there to support your answer but also to show that to answer the question fully without any extra information is difficult. – Farcher Jan 21 '19 at 15:21