# increase in water bottle pressure caused by squeezing

For an experiment I wondered how mass flow rate would be affected by pressure in a water bottle entirely filled with water (no air present). Now the initial pressure in the bottle would be rho X height X g, but I am thinking of increasing the pressure by having a G-clamp on the side of the bottle. How much could you e.g. measure the force needed to reduce the initial radius of the water bottle by 1cm? Until now I thought it might have something to do with Third law, but I don't entirely would know how to find an exact value out of it. Is there a formula or something similarly relevant and useful?

Edit

Well so basically I have a water bottle filled with water and I have a g-clamp which will be tightened so that the pressure inside the bottle becomes even higher. What I am wondering is how much force is/needs to be applied to the G-clamp in order to apply a quantity of pressure on the bottle.

• If you could edit your question to make it clearer that would be great. It's a bit hard to understand what you're asking. Mar 29, 2017 at 19:31
• Well so basically I have a water bottle filled with water and I have a g-clamp which will be tightened so that the pressure inside the bottle becomes even higher. What I am wondering is how much force is/needs to be applied to the G-clamp in order to apply a quantity of pressure on the bottle... Mar 29, 2017 at 19:34
• That's not a simple question. The answer will depend in complicated ways on the shape of the bottle and on its material properties. The easiest way to answer the question is the empirical method: Do an experiment with a load cell to measure the clamping force, and a pressure gauge to measure the fluid pressure. Other than that, I think your next best shot at it would be to use Finite Element Analysis (en.wikipedia.org/wiki/Finite_element_method) Mar 29, 2017 at 22:05
• What exactly are you trying to do? Your opening sentence suggests that you are trying to find out how pressure in the bottle affects flow through the nozzle when the cap is removed. If you explain your goal, we can all avoid attempts to solve unnecessary intermediate steps. ... This seems to be a duplicate of your earlier question Flow of pressure in water bottle. Mar 29, 2017 at 22:14

There are two forces acting on the gripping edges of your G-clamp: the pressure difference between the inside water and the outside air, and the membrane stress of the bottle itself. Here is a sketch: $\alpha$ is the angle between bottle and G-clamp edge: When the bottle does not bend at the G-clamp edge, $\alpha=90°$. The G-clamp edge is assumed to be circular with diameter $d$. (This can be easily generalized if required.)

The force on the G-clamp $F_{\mathrm{clamp}}$ can be calculated by $$F_{\mathrm{clamp}}=\frac{\pi}{4}d^2 \, \Delta p + \sin(\alpha) \, \sigma\, \pi\,d\,t$$ with pressure difference between water and air $\Delta p$, membrane stress inside the bottle $\sigma$ and bottle wall thickness $t$.

The membrane stress $\sigma$ can be calculated using the formula for thin-walled pressure vessels: $$\sigma=\frac{\Delta p \, R}{2 \, t}$$ with $R$ being the local bending radius at the clamp (or very close to it).

In total: $$F_{\mathrm{clamp}}= \left( 1 + 2 \, \sin(\alpha) \, \frac Rd \right) \frac{\pi}{4}d^2 \, \Delta p$$

The difficult task is to determine $\alpha$ and $R$. However, if you can ensure $\alpha=90°$, i.e. the bottle does not bend at the G-clamp edge, then $\sin \alpha = 0$ and the clamp force is directly converted to pressure inside the bottle. This can be achieved by placing the bottle between two boards larger than the bottle.

Note: The hydrostatic pressure distribution is only important for very large bottles. The equation in your question ($\Delta p=\rho\,g\,z$ with $\Delta p$ being the pressure difference to $z=0$, vertical position $z$ pointing downward and fluid density $\rho$) is correct. For $\rho_{\mathrm{Water}}\approx 1000\,\mathrm{\frac{kg}{m^3}}$ and $g_{\mathrm{Earth}}\approx9,81\,\mathrm{\frac{N}{kg}}$, $\Delta p=1\,\mathrm{bar}=100\,000\,\mathrm{\frac{N}{m^2}}$ is reached for $z\approx 10.2\,\mathrm{m}$. If you feel it is important for your problem, you can superpose the hydrostatic pressure with the pressure calculated above.