Air pressure

[Zweiblumen]

Maybe I'm just too lazy to Google or something, how ever I'm hoping that someone can point me in the right direction. What I want to calculate is how the pressure inside a balloon changes with altitude. Lets say that we have a balloon that is inflated to 1 psi at sealevel, what would the pressure be at 40.000 feet ?

Yours
Zweiblumen

[Redlemon]

I think the pressure inside the balloon remains the same; however, as the balloon increases in altitude, the pressure outside of the balloon drops, resulting in the balloon increasing in size. Oh, except for the counter-effect from the drop in temperature. I don't know which would be more significant.

Remember the perv(n)ert!

[Redlemon]

Oh, and a balloon that's "inflated to 1 psi at sealevel"? Standard pressure at sea level is 14.7 psi. Your balloon is not inflated.

[telekinetic]

There are two ways to model a balloon. The simplest case is that the membrane is infinitely deformable, in which case the volume of the balloon adjusts until the pressure inside and outside the balloon are equal.

Let's will deal with that first:

PV=nRT is exactly correct, as red lemon implied (P is absolute pressure, V is volume, T is absolute temperature). We're keeping n and R constant, so we can reorder to P1V1/T1=P2V2/T2.

Now we need to make some assumptions about initial and final temperature and pressure. Due to the vast multitude of starting conditions, we'll use the ISO standard temperature and pressure, or STP: 59 degrees F and 14.7PSI. Let's assume you've filled the balloon to 1 liter. (I know I'm making a mess of the units, but it realy doesn't matter, since they all cancel--this might as well be 'one lungful')

The ever-handy Wolfram|Alpha gives us the conditions at 40,000 feet as 5.5inHg pressure (which google handily converts to 2.7PSI) and -70 degrees F. We obviously don't know the volume yet, so let's use the ideal gas law to solve for it

Before we can plug our numbers in, we need to get our temperature in absolute units. Since we're already in english units, let's just convert to Rankine by adding 459. This gives us:

(14.7*1)/(59+459)=(2.7*V2)/(-70+459)

(14.7 * 1 * 389)/(518*2.7) = V2

4.09 = V2

Therefore, your balloon, if brought from sea level in 60 degree weather up to 40,000 feet, would have swelled to 4 times its size. Do note that we canceled the gas constants out of this, so this works equally well for helium.


The second model brings the physical properties of the balloon itself into play. Assuming we are talking about a latex balloon, and not some non-deformable material like, say, a giant mylar balloon filled to less than 1/4th of its capacity, thus leaving plenty of room for the air to 'grow', then the latex exerts some nonzero force inwards on the air in the balloon, increasing the pressure slightly. If you have read this far into the wall of text that is way more than I intended to write, please make a bad science pun (possibly about hot air?) in your reply. Anyways, to determine the effect on your balloon, you would need to know what that pressure increase was at sea level compared to what it would be at your increased volume. However, as long as your balloon is big enough to not pop at altitude, it is pretty safe to assume that this effect will be negligible....if it's not, just start with a larger balloon.

Originally Posted by Zweiblumen

Maybe I'm just too lazy to Google or something, how ever I'm hoping that someone can point me in the right direction. What I want to calculate is how the pressure inside a balloon changes with altitude. Lets say that we have a balloon that is inflated to 1 psi at sealevel, what would the pressure be at 40.000 feet ?

Yours
Zweiblumen

To strictly answer your question, as I've shown above, in a deformable vessel, it is safe to assume the pressure doesn't change, and saying you are filling a balloon with 1psi has no meaning unless you know the physical properties of the balloon walls (Young's modulus of elasticity, thickness, and internal surface area would probably be enough to figure it out)

If you instead had, say, a steel canister filled with some volume of air, just raising the altitude isn't going to change the absolute pressure in the vessel--only a temperature change could do that. If you wanted to find the relative pressure change (that is, the forces on the walls of the vessel--0psi when you screw the lid on at sea level), you'd adjust the internal pressure for the temperature change, and then just do simple subtraction of the absolute pressure outside (we already found that to be 2.7PSI) from the absolute pressure inside. Both are going down, since it is getting colder, but I suspect the pressure drops faster than the temperature, so you should see an increase relative pressure.

[Zweiblumen]

Thank you all for your time.
Here is the real reason for the question, airliners request that the valve is removed when transporting a bike on a jet. The reason given is to prevent the tire/tube from exploding in-flight. An average mountainbike tire/tube is supposed to be inflated 55-85 psi (depends on type of tire) and narrower tires have higher rating (100+ psi). One can ride a bike down to 20-30 psi so I was trying to figure out the pressure "increase" at 40.000 feet for reference.
BTW Redlemon when measuring airpressure in tires (or balloons) it's the diffrence in pressure outside and inside that is beeing measured so a balloon inflated to 1 psi at sealevel has total 15.7 ps (1 + 14.7)

Yours
ZB

[Martian]

That post increased the relative pressure within my cranium, causing a rapid forceful decompression.

Blew my mind.

This is my substitute for a bad science pun, because I can't think of any.

[Redlemon]

Sorry, I'm working from a science perspective, I didn't know the nomenclature change.

I know that the cargo hold isn't kept at the same temperature and pressure as the passenger area, but I would think that it isn't the same as the outside. I mean, wouldn't that kill pets who are put in the cargo hold?

[telekinetic]

Originally Posted by Martian

That post increased the relative pressure within my cranium, causing a rapid forceful decompression.

Blew my mind.

This is my substitute for a bad science pun, because I can't think of any.

<3 That's one! Thanks for at least trying! :-p


By the way, Red Lemon, you're referring to Absolute pressure, where as the OP is talking about gauge pressure, which is only useful if the outside atmosphere isn't changing (which it is). You need to use absolute pressure to do any calculations....all the math needed is in my tl;dnr, you would just need to find out what the cargo hold conditions were at 40,000 feet. I guarantee it's not 2.7PSI and -70F!

[inBOIL]

Originally Posted by Redlemon

Sorry, I'm working from a science perspective, I didn't know the nomenclature change.

I know that the cargo hold isn't kept at the same temperature and pressure as the passenger area, but I would think that it isn't the same as the outside. I mean, wouldn't that kill pets who are put in the cargo hold?

I believe some sections of the cargo hold are pressurized, while others are not.

As far as the op, why not deflate the tires entirely and bring a portable hand pump with you?

[supersix2]

After doing some searching on the internet combined with my aerospace engineer powers, I found info that the cargo compartment is pressurized to the same pressure as the passenger compartment. Depending upon the flight altitude, the cabin is pressurized to a 5,000 ft to 8000 ft altitude equivalent. So your pressure in the cabin is between 10.9 - 12 psia. However, the cargo compartment isn't always thermally conditioned the way the passenger compartment is. As a result, the cargo compartment could be colder or hotter than the cabin depending upon the location inside the cargo compartment. It makes sense for them to be at the same pressure, otherwise the floor of the passenger compartment would have to be strong enough to withstand the differential pressure between the passenger compartment and the crew compartment. You would have to have reinforced pressure bulkheads between the two compartments and that would add unnecessary weight. There was an aviation mishap back a while back that involved a cargo hold door losing its seal causing the cargo compartment to depressurize. This caused the floor of the passenger compartment to buckle because of the delta pressure between the two volumes.

If your tires are inflated to 55 psig (~70 psia at sea level) then inside a cargo compartment, your tire pressure could increase as high as ~74 psia. That seems well within the operable range of the tire. I think they ask you to do this as a precaution since some tires may not be as well rated as others. However, if your tire can handle that high of a pressure there should be no problem transporting it with the valve closed. The thing you want to make sure is the units your tire is rated to psig (gauge pressure) or psia (absolute pressure). Since 55 psig and 55 psia are two completely different things. I think this also protects the airline from liability in case any damage occurred to your tires during flight (for any other reasons as well) so they can say "well you didn't follow our policy." So while your tire may be able to handle the pressure, it is probably just a better idea to follow the airline policy to avoid any unneeded fuss.