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I'm working with stratospheric balloons (latex ones) and I want to put a valve on it so it can float for a longer time. I'm trying to define which valve I should use, which demands I estimate the flow os gas it can address. Also, it would be great to be able to model the problem mathematically, so I can design an altitude control system.

Thinking about this, I got some doubts related to the physics of the latex balloon membrane and the gas inside of it. I'm trying to understand what exactly makes the gas inside a balloon come out when we open a valve placed on its "mouth".

I know the pressure inside the ballon must be greater than the external in order to the gas flow. But when the system is in equilibrium, the internal pressure must be equal to the external, so the ballon doesn't expand/contract. Or ALMOST that. I think most people thinking about this just ignore or deem negligible the force the membrane of the balloon exerts in the gas inside of it. I came to the following conclusion:

$$p_i = p_o + p_b$$

where:

$p_o$ = external pressure

$p_i$ = internal pressure

$p_b$ = pressure exerted by the balloon membrane into the gas in its interior


This justifies why the internal pressure would be greater than the external and the system would, still, remain in equilibrium. If this is correct, I can conclude that what will make the gas come out is pb, at any altitude. So, if I have pb, I could use Bernoulli's principle to calculate the velocity with which the gas come out and, from this, find which is the flow rate I need to model the control problem and guide me on the valve specification.

I have two concerns with this approach:

  1. Is it correct? Physically speaking, this is what really happens? Can someone give me more insights on the dynamics of mechanical tension on the membrane and gas pressure inside the balloon?

  2. I can't find a way to calculate the pressure the membrane would exert on the gas (pb). I know that I can calculate the tensions that appears in the membrane surface accordingly with the internal pressure (with pressure vessels theory). I have a feeling that I can derive the force with which the membrane compress the gas from this tensions, and it seems to me it would be only a geometrical/calculus problem, but I'm struggling getting it forward. I have idealized a simplified model in which I apply Hook's Law, but I'm stuck with infinitesimal areas. Any one can guide me here?

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This sounds plausible to me. I think your odd of calculating pb are small. As $p_b$ would be only dependent on the pressure difference, you might be able to perform some experiments on ground level, by varying $p_i$. –  Bernhard Apr 20 '13 at 6:25
    
Hummm, good to know I'm not doing nothing stupid. I got to this problem by thinking how the balloon will behave when it is at high altitudes, under very low external pressure. I want to know if the gas will flow out of the balloon in this conditions (to know if the valve will work well). I'm not sure if experiments on ground level would be sufficient to predict the behavior at low pressure (and temperature)... –  john1034 Apr 20 '13 at 11:46
    
@john1034: Float for a longer time? Do you mean it should stop ascending, or do you mean it should not burst? I'm only asking because balloons basically ascend until they burst. You could halt the ascent by letting out some of the gas. –  Mike Dunlavey Apr 20 '13 at 13:26
    
If you had a GPS receiver on the balloon (which most typically do) you could use that altitude and its rates to control your valve. –  OSE Apr 20 '13 at 14:34
    
Usually, the common latex balloon will burst before 2 hours of flight. I would like to extend the time it stays at 20-30Km to at least 5 hours (10 hours would be great). It doesn't matter if the ballon stays afloat or burst, as long as it keeps in this range for at least 5 hours. I don't need rigid altitude control, too. My biggest problem here is to determine how much helium flow my valve could handle as altitude (and external pressure and temperature) changes, so I can project a controller to handle the problem. –  john1034 Apr 20 '13 at 14:41

2 Answers 2

  1. Yes.
  2. You need to solve two problems.

First, given the natural radius of the spherical membrane $R_0$ (the radius with no tension in the membrane) and the current membrane radius $R$, membrane's modulus of elasticity $E$ and Poisson's ratio $\mu$, calculate tension stress in the membrane. If you consider an infinitesimally small square (with the side of $\delta l_0$) of the spherical membrane under no tension with thickness $d_0$, currently it will be a square with a side of $\delta l=\delta l_0\frac{R}{R_0}$. The tension stress $\sigma$ will expand the square in two directions. Calculation of tension stress from strain (using elasticity modulus and Poisson's ratio, and assuming the membrane material is isotropic) is a standard task, see e.g. https://en.wikipedia.org/wiki/Poisson%27s_ratio , although you may wish to find a better source. The exact value of the Poisson's ratio is not very important.

Second, given tension stress, calculate differential pressure in equilibrium. To do this, consider the condition of equilibrium for a half of the spherical membrane: $(p_i-p_o)\pi R^2=2\pi R d\sigma$ (assuming the membrane is thin: $d_0<<R_0$).

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Ok, thank you all for your comments. That's how I solved it, but I'm not quite sure if my reasoning is correct. Any feedback would be great!

Searching for materials on pressure vessels theory, I found this: http://www.colorado.edu/engineering/CAS/courses.d/Structures.d/IAST.Lect05.d/IAST.Lect05.pdf

In the end, there's an example on inflating ballons where they derive an expression relating internal pressure with the expansion observed on the diameter.

$\beta =$$D_f\over{D_0}$

Where $D_f$ is the final diameter and $D_0$ is the initial diameter.

The expression relating $p_i$ and $\beta$ depends on $\nu$, the poisson constant of the material. I assume $\nu = 0.5$ as a good approximation for the rubber. With this value, I got the following expression:

$p_i =$$8.E.t_0.(\beta-1)+D_0.p_0.(2-\beta^3).\beta\over{D_0.\beta^4}$

Where $t_0$ is the thickness of the membrane of the balloon at the beginning, when the ballon has diameter $D_0$ and internal pressure $p_0$. $E$ is the Young's modulus for the rubber. This is a problem, as the rubber presents high non-linearity on its elasticity. Searching through the web, I decided to approximate this value to 1.0MPa (I found references saying latex has $E$ varying from 0.5 to 1.1 MPa).

Well, the derivation of the formula above assumes the external pressure is zero. Because of that, I assumed I can say $p_i$ really represents the differential pressure ($\Delta p=p_i - p_o$) between the internal and external pressures actuating on the membrane of the balloon.

I'm not sure this is correct, but I suppose it is, as it seems to me that the external pressure acts only as an offset to the state of the internal pressure, and the stress on the membrane of the balloon depends only on the strain caused by expansion of its diameter (which will be controlled by the external pressure but won't be affected in terms of value by it).

Said that, I stipulated values for the variables as follows:

$E = 1.0MPa$

$t_0 = 300um$

In the specifications of the balloon I'm going to use, they say that the diameter when it is barely inflated is 1.44m. I assumed this would be my $D_0$, which means $p_0 = 0$ (as it is barely inflated).

I wanted to know the capacity of flow at time near bursting. The specifications gave me the diameter at burst: 9.10m. I chose this as my $D_f$. Then, I got $\beta = 6.25$.

Solving the equation, I got $\Delta p= 5.73Pa$. That would be the pressure with which the membrane of the balloon would compress the helium inside. It can give me an estimate of the flow at this extreme conditions.

Well, with the differential pressure, I applied Bernoulli's principle to estimate the velocity of the gas if I opened a valve. I know the conditions in which the gas is can make Bernoulli simply don't work, but at least I can get an approximate value to think about.

From Bernoulli's principle, considering the potential energy associated with height negligible:

$p_i + 0.5\rho v_i^2 = p_o + 0.5\rho v_o^2$

Where $i$ subscript denote the variables on the internal side of the balloon and $o$ the external. $v_i$ is the velocity inside the balloon, relative to its membrane. I assume it is zero. Rearranging and substituting $p_i - p_o = \Delta p$:

$v_o = \sqrt{2\Delta p \over{\rho_{he}}}$

I know the volume at bursting is $4\over 3$$\pi (D_f/2)^3 = 394.6m^3$ and I know from initial conditions that the total mass of helium was approximately 0.49Kg (from initial volume of $3m^3$ and desity of $0.1634 Kg/m^3$). Assuming no leakage of helium through the whole flight, at burst time the density of helium will be:

$\rho_{he} = $$0.49\over{394.6}$$= 0.00124 kg/m^3$

With this value, I got $v_o = 96.1 m/s$, which is a very big value.

Assuming the neck of the balloon has a diameter of 3cm (it has) and that the gas can flow freely through it, I estimated the flow:

$A_{neck} = \pi.(0.015m)^2 = 7.07 x 10^{-4} m^2$

$q = A.v_o = 0.0679 m^3/s = 67.9 L/s$

That's a very HIGH flow. For the results on the velocity and flow, I assume I must have done something wrong. Maybe I just simplified things too much. Or maybe it is close to correct, and this would be good news, as I could reduce the diameter of the neck using a valve of my desire. Anyway, I recognize this problem is much more complex than depicted here. For example, this values would be true only for the time when I opened the valve. To see the whole dynamics, I would need to use differential equations as the diameter, volume and pressure changes with time, modifying everything else. I just wanted to have a feeling of how much flow I could get.

Anybody has any thoughts about this resolution? Can I trust in it, even barely? I have a feeling I may have done something very stupid. :/

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