If one graphed the volume per time of air being expelled from a common elastic balloon out of a hole of constant size relative to the balloon's surface area, would the curve of the graph be exponential or polynomial? I have neither the means to perform the experiment nor the brains to theorize, but I'm very interested.

  • 1
    $\begingroup$ What with turbulence and reaction at the nozzle and a high likely hood that the flow can vary over multiple orders of magnitude in Reynolds number I expect that the only easy answer is "a complicated one". It may be possible to find BOTE approximation some parts of the process. $\endgroup$ – dmckee --- ex-moderator kitten Sep 16 '11 at 20:47
  • $\begingroup$ There was a question on pressure vs. diameter of balloons here on this site. So You just need to calculate the volume expelled depending on pressure.[ :=)] What do You mean with "hole of constant size relative to the balloon's surface area," ? is the hole constant in diameter or does it vary with the ballons diameter? $\endgroup$ – Georg Sep 16 '11 at 20:49
  • $\begingroup$ The ratios stay the same - hole@time1/surfacearea@time1 = hole@time2/surfacearea@time2, so yes, the hole would decrease/increase in size proportionately to the balloon. $\endgroup$ – Daniel Sep 16 '11 at 20:51
  • $\begingroup$ Do You think that that is the "natural" behaviour of a balloon nozzle? In fact balloon nozzles often have a kind of "fart" effect, because Bernoulli enchanted them to do so. $\endgroup$ – Georg Sep 16 '11 at 20:57
  • $\begingroup$ Then imagine a large balloon with a large hole, which would eliminate much of the other air disturbances. I'm trying to concentrate on simple air flow, even if that's not exactly how a real balloon would do it. $\endgroup$ – Daniel Sep 16 '11 at 21:05

The outflow velocity is given by the law of energy conservation. If you call the pressure inside the balloon p, the velocity is given by Bernoulli's equation, it comes out as the square root of the pressure difference. The pressure difference is approximately constant at first, with some little effects due to reaching the strain limit, ignore these effects. At first, the graph is constant. (As Georg pointed out, this is incorrect. The balloon gets significantly thinner long before the strain limit is reached. I didn't consider the thinning, but it leads to a pressure reduction for large balloons. The balloon has a certain amount of stress per unit mass, which is increasing linearly as you blow it up, but decreasing in any area as the reduction amount of mass in that area, in terms of the radius, as r_0^2/r^2. The flow rate rises for a while as 1/r before falling, but the analysis of the reduction region below, which is what you were asking about, is fine)

When the balloon begins to reduce its pressure, you would get an exponential if the pressure difference p goes down according to

$$ {dp\over dt} \propto -p $$

But the rate at which gas leaves is the square-root of the pressure difference, so it is more like

$$ {dp\over dt} \propto -\sqrt{p}$$

$$ p \propto (t-t_0)^2$$

Which crashes down to zero at a finite time, quadratically. This is not exactly right either, because it's the volume that is decreasing as the square root of the pressure, and the pressure is due to extra surface, or extra $r^2$, so in terms of r

$$ p \propto r^2 - r_0^2$$ $$ {dV\over dt} \propto - \sqrt{p}$$ $$ r^2 {dr\over dt} \propto - \sqrt{r^2 - r_0^2}$$

which, when you look at $r$ near $r_0$, has the same square root behavior. So $r$ would hit $r_0$ quadratically, as $(t-t_0)^2$, right before a certain time $t_0$.

But after reaching $r_0$ there is still a finite pressure, just from the gravity of the top of the balloon pushing down on the remaining air inside the balloon. So, in this part, the pressure is constant again, at a much lower value. This makes the balloon hit zero linearly, and the previous answer is modified by the extra pressure to smooth out the quadratic behavior to fit with the linear one at the end.

  • $\begingroup$ I was expecting it to be exponential. That's interesting. $\endgroup$ – Daniel Sep 17 '11 at 15:52
  • $\begingroup$ You are too proud to follow hints? If You had looked for the theme on this forum, You would have found this link, where Fig. 5.7 shows that You are wrong. colorado.edu/engineering/CAS/courses.d/Structures.d/… $\endgroup$ – Georg Sep 17 '11 at 17:45
  • $\begingroup$ There is nothing on the link which is contradicting what I said, although my answer was wrong--- the dependence is (t-t_f)^2. $\endgroup$ – Ron Maimon Sep 17 '11 at 18:29
  • $\begingroup$ @drɱ65 δ: I screwed up in an idiotic way. I just fixed it. There are small extra effects due to strain limits, which can be ignored for the regime you care about. $\endgroup$ – Ron Maimon Sep 17 '11 at 19:02
  • 1
    $\begingroup$ @xpda: it does--- the (t-t_0)^2 I give is for the linear regime, so it's all in this extra-pressure regime. I didn't realize the pressure goes down again past this region, so the curve isn't constant at large R. I suppose this is what Georg was saying. $\endgroup$ – Ron Maimon Sep 18 '11 at 2:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.