The goal of the demonstration is to illustrate the use of Bernoulli’s equation to compute the discharge velocity of a water pistol.

**Equipment**

- A water pistol or a syringe
- A tape measure
- Level
- OHP with a sheet with parallel lines drawn on it (helpful but not essential)
- Protractor or Set Square (optional)
- A few student volunteers

Leveling the vertical scale created by the OHP slide

Measuring the vertical scale

**Demonstration**

** **The demonstration is fairly simple in concept but it can be quite hard to maintain both a smooth flow rate and steady angle. Simply project the water vertically upward and measure the height to which it rises. This can be a little challenging as the height can be quite large depending on how hard you press the plunger. An alternative is to project it at some angle say 20^{o}-30^{o} from the horizontal. You will need the Protractor or Set Square for this. Firing at an angle will make the maximum height lower but possibly more challenging to measure. It might help to draw a series of horizontal lines on a board, or project a set of horizontal lines using an overhead projector. You can get the lines horizontal using the level. After you have completed the demonstration you will have the change in height from the discharge to the maximum height (H_{2}-H_{1}) and the angle of discharge (θ) measured from the horizontal.

**Analysis**

The analysis for the vertical case is the simplest. Start by writing Bernoulli’s equation from the release point (1) to the rise height (2)

P_{1}/γ+H_{1}+(u_{1}^{2})/2g=P_{2}/γ+H_{2}+(u_{2}^{2})/2g

The pressure everywhere is atmospheric and u_{2}=0 at the point of maximum vertical rise. Therefore, the discharge velocity is given by

u_{1}=(2g(H_{2}-H_{1}))^{1/2}

The angled release is a little more complex. Combining Pythagoras theorem and Bernoulli leads to

[P/γ+H+(u_{x}^{2}+u_{z}^{2})/2g]_{1}=[P/γ+H+(u_{x}^{2}+u_{z}^{2})/2g]_{2}

Where x and z denote the horizontal and vertical directions. Assuming there is negligible drag or energy loss then then the horizontal velocity will remain a constant. Therefore Bernoulli’s equation can be simplified to

u_{z1 }= u_{1} sinθ = (2g(H_{2}-H_{1}))^{1/2}

as the vertical velocity is zero at the maximum rise height. Therefore, the discharge velocity is given by

u_{1} = (2g(H_{2}-H_{1}))^{1/2}/ sinθ

Note that this equation can be applied to the vertical release case as sin90=1.

**Discussion**

There is a lot of uncertainty in the measurements so, as with so many other demonstrations, it is a good opportunity to talk about experimental errors and repeatability. It would be fairly straight forward to run this test for a few different release angles to test the relationship in the final equation. However, the main repeatability problem is likely to be getting the jet release velocity constant between tests.

An index of all the demonstrations posted on this blog can be found here. Don’t forget to follow @nbkaye on twitter for updates to this blog. If you have a demonstration that you use in class that you would like to share on this blog please email me (nbkaye@clemson.edu). I also welcome comments (through the comments section or via email) on improving the demonstrations.