What kind of curve would describe the rate of air coming out of a balloon?

Question

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.

Answer 1

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.