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Imagine such situation:
We have an empty balloon. We put a long, narrow tube into this balloon and we fill it with water.
Then, the volume (and surface) of the balloon will start to increase. The pressure in the balloon will increase, too.
I conducted this reserach. I messured the height of the water in the tube (starting at the center of the balloon) and then the circumference of the balloon. Then, I calculeted diameter and pressure in the balloon

$d = L/ \pi $ (m)
$p = 10^5 + 1000 gh$ (Pa)

Then, I plotted the relation between diameter and pressure
Graph

What role in this project did surface tension play? How can I consult it to the math?

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  • $\begingroup$ Excuse me. I have already corrected my error. I made an error in Excel. $\endgroup$ – ILoveChess Apr 9 '16 at 11:57
  • $\begingroup$ Yeah, it is presented on the graph $\endgroup$ – ILoveChess Apr 9 '16 at 12:44
  • $\begingroup$ You asked a similar question before. Did Formula there given for duplicate not okay? $\endgroup$ – Anubhav Goel Apr 9 '16 at 12:50
  • $\begingroup$ That formula is not entirely clear to me. $\endgroup$ – ILoveChess Apr 9 '16 at 12:54
  • $\begingroup$ T= ∆p D. Can you elaborate what's not clear? $\endgroup$ – Anubhav Goel Apr 9 '16 at 12:55
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Very interesting experiment you did!

1. The effect of surface tension due to the water/rubber and rubber/air interfaces is negligible.

For a radius of $R = 10$ cm, and taking a typical surface tension of $\gamma = 70$ mN/m, the increase of pressure in the balloon because of the interfaces is typiclally of the order of the Laplace pressure $\Delta p = 2 \gamma / R = 1.4$ Pa. This is totally negligible compared to the hydrostatic pressure in the ballon and the elastic energy of the membrane (see below).

2. There is another "surface tension" associated to the elastic energy of the membrane of the balloon, because of the elasticity of rubber.

The problem is that it is not an energy that is proportional to the increase of area of the membrane by stretching. Also, modeling the rubber of the balloon as a linear elastic material is not a good approximation at all. A common model is the Mooney-Rivlin material. Using this material, the formula is rather complex [1]: $$\Delta p(r) = 2 s_1 \frac{d_0}{r_0} \left[ \frac{r_0}{r} - \left( \frac{r_0}{r} \right)^7 \right] \left[ 1 - \frac{s_1}{s_{-1}} \left( \frac{r}{r_0} \right)^2 \right]$$ with $s_1$ and $s_{-1}$ the Mooney-Rivlin constants of the rubber, $d_0$ the thickness of the unstretched membrane, and $r_0$ the radius of the unstretched balloon.

The link below also gives some values for a typical balloon:

$$s_1 = 3 \text{ bar, } s_{-1} = -0.3 \text{ bar and } \frac{d_0}{r_0} = 0.5 \times 10^{-2}$$

Figure 1 in the link is exactly what you want to plot. It shows that the elastic pressure increase is typically of the order of 2 kPa with the parameters in the article, which is similar to what you found. You can see that for a given pressure you apply, you can have up to three possible equilibrium radii of the balloon, so if you want to relate this graph to your measurements, you will need to remember exactly the evolution of the pressure and radius with time as you add water.

Finally, be careful that this formula is valid for a spherical balloon, filled with air. With water, the pressure may not be the same everywhere in the balloon. What you measured is the pressure where the tube enters the balloon.

[1] https://www.researchgate.net/profile/Henning_Struchtrup/publication/245382031_Inflating_a_Rubber_Balloon/links/0c960529e161c41bc8000000.pdf

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