0
$\begingroup$

Let's hypothesize there is a flat ballon in space (vacuum, no gravity and no external pressure), which is connected to a gas pump. The gas and the material of the balloon doesn't interact. At a certain time, a valve opens the tank, and some gas inflates towards the flat balloon. The Ballon has a perfect spherical shape if inflated. The material of the balloon is particular, for example some combination of Mylar and Aluminium foil, so it will not burst in space (see ECHO I & II, as example). Let's suppose we know all about the material and the balloon geometry (initial radius, and infltated radius, thickness, elastic modulus, Poisson ratio, etc...) and the gas (density, attractive and repulsive forces between atoms, and so on...).


How can we model the inflation of the balloon in space? how much pressure is needed?


This question comes to me after reading about ECHO missions in wikipedia that: " During ground inflation tests, 18,000 kg (40,000 lb) of air were needed to fill the balloon, but while in orbit, several pounds of gas were all that was required to fill the sphere." [Thermal related problems are avoided for now :) ]

$\endgroup$
5
  • 1
    $\begingroup$ Are you asking what the relationship is between the pressure in the balloon and the amount that the balloon membrane stretches? $\endgroup$ Commented Oct 30, 2023 at 10:41
  • $\begingroup$ You will need to know how difficult it is to stretch the surface of the balloon. The external pressure is 0, so there will be no work required to inflate the balloon apart from overcoming surface tension and actually moving the gas inside the pump. $\endgroup$ Commented Oct 30, 2023 at 11:03
  • $\begingroup$ Voting to reopen - this really doesn't look like a homework question. $\endgroup$
    – gandalf61
    Commented Oct 30, 2023 at 12:55
  • $\begingroup$ @ChetMiller I suppose yes. Do you know where I should search for this data? $\endgroup$
    – nuwanda
    Commented Oct 31, 2023 at 11:23
  • $\begingroup$ See Post #24 of the following on-line reference: physicsforums.com/threads/… $\endgroup$ Commented Oct 31, 2023 at 13:31

2 Answers 2

0
$\begingroup$

Since the balloon is in a vacuum and the external pressure is (essentially) zero then any amount of gas, no matter how small, will expand to fill the whole balloon and will inflate it. This is why it takes only a few pounds of gas to inflate the balloon in space.

Since the balloon is inflated at a very low pressure, it will not be very rigid i.e. it will be easily deformed. But as long as it is not required to be rigid then it can be inflated with any amount of gas at all.

$\endgroup$
0
$\begingroup$

This depends on the material of the balloon. For instance, you could create a rigid sphere. Such a sphere would require zero air to hold it's shape (but would be much heavier than a balloon and would take up far too much space to launch).

The material of the balloon has some resistance to motion. It has folds and wrinkles that disappear only under some minimum force. If you know the force that is required to hold the envelope in your balloon shape, and you know the volume of that shape, then you can calculate the mass of air that would provide that pressure.

Short of finding this data on existing materials, you'd probably do an empirical measurement to find the necessary pressure difference to hold the shape. Then you could calculate the mass of air necessary to inflate.

$\endgroup$
2
  • $\begingroup$ So if I understand it well, we could use some formulas from the material of choice for calculating the needed pressure for having a "nice" shape without deformations or wrinkles. From the geometry the volume of the sphere. And so using the law of perfect gas, for example, we could calculate the the required number of moles of the gas? n=(pV)/(RT) $\endgroup$
    – nuwanda
    Commented Oct 31, 2023 at 11:20
  • $\begingroup$ Yes, that seems about right. Take the worst case temperature and leak rate for the mission and that'd be a good start point. $\endgroup$
    – BowlOfRed
    Commented Oct 31, 2023 at 15:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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