Helium lifting ability

According to this pressure curve,

latex balloon pops at ~1.5 psi above ambient.

R, Tom Q.

According to this pressure curve,

there's a definite pressure spike when the balloon starts to inflate.

For those that can't reach the site (Jeff, et. al.). The graph shows a sharp rise in pressure when inflation begins, pressure drops sharply, than rises relatively slowly until the balloon pops.

Also, the site shows an experiment where two balloons, one slightly inflated & one almost fully inflated, are connected by a tube which is pinched shut. When the tube is allowed to open, the air goes from the less inflated balloon to the more fully inflated one.

Of course, the site could be full of s**t, but it _seems_ right.

R, Tom Q.

I thought that MAYBE I actually still had the paper those guys wrote--but no. Memory sez they did an aluminum geodetic frame covered with sheet aluminum triangles, all sealed with wax. I don't remember what stopped them--the sheets or the frame collapsing, or all the joints leaking.

David

"D> >A high school classmate made a year long project of building a vacuum balloon.

Momentum?

-c.

The only way that there would *not* be a pressure gradient across the height of the balloon is if the material outside the balloon has zero weight. This would happen only if there's no mass (i.e. a vacuum), or if there's no apparent gravity (i.e. inside the orbiting space station). If the material around the balloon does have weight, any weight, then there must be a pressure gradient equal and opposite to the weight - or the air wouldn't be at rest.

There's also a pressure gradient inside the balloon. Gravity is the same inside and outside the balloon, but the density of helium is less than air, so the pressure gradient is less inside than outside.

Dave

Yes. Which by definition means that there is also a pressure gradient between the top and bottom of the balloon, albeit a very minute one. The essential bit though is only that there be a density gradient. The fact that this is caused primarily by pressure is not relevant. It could also be caused by temperature. The balloon rises to a point at which it displaces exactly its own mass. If it were to rise higher it would displace less than its mass and so sink again. The reverse would happen if it were to sink from the equilibrium point.

I'm not sure what you are taking the term "pressure gradient" to be but I think you are misunderstanding it. It simply means there is a static density gradient and the original point might have been better made using this term. There is no way a balloon filled with air can float in air of the same density unless the material of the balloon is also less dense than air.

Exactly. But in incompressible fluids objects generally either sink to the bottom or float to the top. In compressible ones they can find an equilibrium point from which they rise no higher. Strictly speaking even the tiny amount of compressibility in liquids means that there is a small density gradient and a possible equilibrium position for a submerged object but it's a very unstable equilibrium and requires that the density of the object falls somewhere within the tiny range of densities of the liquid between the top and bottom.

Dave Baker - Puma Race Engines

"How's life Norm?" "Not for the squeamish, Coach" (Cheers, 1982)

OK, replace pressure gradient with density gradient if you will.

But this is only apparent if the helium balloon is ballasted so it is neither rising or sinking - if it is floating free between the floor and ceiling. The effect is small, much smaller than what makes the balloon provide lift in the first place.

The density gradient effect can be seen with one of those 'cartesian diver' toys.

Jim

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The pressure differential is only half of the equation. It has to be balanced against the weight of the balloon. In the case of a helium-filled balloon that rises, the net force resulting from the pressure differential is greater than the weight of the balloon, so it pushes the balloon up. For an air-filled balloon, the weight of the balloon is greater than the resultant buoyant force, so the balloon sinks.

The amount of force involved is not as tiny as you think. At sea level at 59 deg F, the pressure gradient is about 0.000531 psi / ft, or

0.0765 psf / ft. To simplify calculations, let's assume we have a balloon in the shape of a cube 1 ft on each side. The difference in the upward pressure acting on the bottom square foot and the downward pressure acting on the upper square foot is therefore 0.0765 lb, or 1.22 oz, acting in an upward direction. (Note that these numbers are the same as the weight of one cubic foot of air under these conditions

-- this isn't a coincidence!) This force is greater than the combined weight of a cubic foot of helium (at near ambient pressure) and a balloon; it's less than the combined weight of a cubic foot of air and a balloon. (An empty balloon this size would probably weigh 1/3 to 1/2 oz, based on the weight of a smaller balloon I had lying around.)

The weight of the fluid displaced is simply an easily computed surrogate for the force due to the pressure differential. Both values turn out to be exactly the same (regardless of whether the fluid is compressible or not). But physically speaking, the balloon moves because of the forces acting on it, and the force acting on it due to the surrounding atmosphere is transmitted through pressure (which is the result of individual gas molecules impinging on the balloon).

Bert

Very good description Bert.

John

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On Mon, 21 Jul 2003 02:05:39 GMT, Jeff Wisnia wrote something ......and in reply I say!:

It will go past a maximum point, because as the balloon stretches, it becomes weaker and therefore requires less PSI to expand, but inside will always be higher than outside pressure.

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