Helium for flying-Buoyancy

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1. Apr 13, 2016

Behrouz

Hello everyone,
Following Buoyancy formula, how much helium is required to help someone with 80 kg weight against gravity?
While the Buoyancy depends on volume, what happens if we compress helium?
Regards,
Behrouz

2. Apr 13, 2016

Merlin3189

The difference in density between air and He is about 1.08 kg/m3 so you can lift about 1kg per m3 of He. To lift a man of 80 kg you need more than 74 m3 of He plus some to lift the container.

If you compress the He to, say, twice the density, then the difference in density is now 0.91 kg/m3, so you now need at least 88 m3

Your best bet for minimum volume is not to use any gas at all - a vacuum. Then you get about 1.25 kg/m3 of buoyancy and need only 64 m3 of vacuum to lift him. The snag is the container may need to be much stronger to hold the vacuum than to hold a gas at atmospheric pressure.

3. Apr 13, 2016

Merlin3189

The buoyant force arises from the displacement of the surrounding fluid - in this case air. So the important point is the volume of air displaced. If you displace a smaller volume, you get less buoyant force: displace a bigger volume, get more buoyant force.

The only role for the He is to keep the air out. As I said, ideally you'd have a container of nothing, displacing 64 m3 of air, giving you 80 kg of buoyancy. But that container would have to be very strong to keep the air out and so very heavy. Then you'd need more volume to support the container. If you fill the container with something at the same pressure as air, then the container does not have to be strong - just airtight, like a balloon. But whatever you fill the container with weighs something and again you have to increase the volume to get extra buoyancy to support that. So you fill it with the lightest thing you can, Hydrogen gas. He gas is second best*.

If you compress the He, you just make it more dense and so heavier for each m3, then you need more volume to support the extra weight of He. You also might need a stronger balloon, which would be heavier!

The only advantage I can think of for compressing the He, might be to preserve buoyancy as the He leaks out. If the balloon were relatively inelastic, as the He leaked* the pressure inside would fall and the density would fall. If the resultant shrinking of the balloon was small enough, the net buoyancy could stay the same or even rise, until the He pressure fell to atmospheric, when the balloon would start to be squashed.

Incidentally, if there were no atmosphere, it would not matter how much Helium or Hydrogen you pumped into a balloon. It would just get heavier and however big the volume, generate no buoyant force. Hydrogen and Helium do not provide lift or buoyant force: they just push the air out of the way and that is what provides the buoyancy.

Edit: * One reason for using He is that it leaks a bit slower than H.

4. Apr 13, 2016

rbelli1

And doesn't explode.

BoB

5. Apr 15, 2016

Behrouz

Hello Merlin,
That was great. thank you so much for that.
You're right. I've started modelling the container I have in mind and I'll do the finite element analysis on that.
I'll keep you all posted.
Thanks again.

6. Apr 15, 2016

There have been a couple of people who made the news by going aloft using plenty of large (4 ft. diameter) helium balloons. The calculations show it requires about 100 or more such balloons to support a 200 pound payload. "Lawn Chair Larry" is the most well known of these, (about 15 years ago in the Los Angeles, California area), and Kent Couch travelled about 200 miles in the Seattle, Washington area with a similar apparatus.

7. Apr 16, 2016

Behrouz

Thanks Charles. great story. I checked "Lawn Chair Larry".

8. Apr 16, 2016

Very good. I would have posted a "link", but the story I googled didn't have an easy share feature. Anyway, yes, a fun thing to google. It's quite entertaining.

9. Apr 18, 2016

PeterVermont

10. Jun 29, 2016

Behrouz

Hello merlin,

I have decided to test the idea on a vacuum container and I'm going to try some different materials (first by computer mechanical analysis of pressure vessels).
May I ask the formula you used which resulted in displacing 64 m3 of air, giving you 80 kg of buoyancy? The reason I'm asking is that I'm looking for parameters in that and the way I can can manipulate them.

Thanks again.
Regards,
Behrouz

11. Jun 30, 2016

Merlin3189

I didn't really use formulae explicitly, just the relationships between the variables. I suppose the formulae you would use are
mass = density x volume [ m=ρv ]
buoyant force = displaced mass of fluid x g [F = mg ]
net buoyant force = gross buoyant force - weight of container - weight of contents

My earlier numbers were just rough and I may have used densities at STP (0oC), so these are at NTP (20oC, 760mm Hg)
density of air = 1.2 kg/m3
density of H = 0.084 kg/m3
density of He = 0.17 kg/m3
density of vacuum = 0 kg/m3

Mass of air displaced = gross volume x density of air (gives gross buoyant force)
Mass of gas (or vacuum) in container = internal volume x density of gas (or vacuum)
Net buoyant force = (mass of air displaced - mass of gas in container - mass of container) x g
(or omit g if you work in kg force)

So a gross volume of 64m3 of air has a mass = 64m3 x 1.2 kg/m3 = 76.8 kg, a bit less than I said, because I had used STP density of 1.29 kg/m3 before (giving 82.6kg.)
Then, as I said, you need to subtract the mass of the container, to get the net buoyant force.
A sphere to contain 64 m3 would be 2.48m radius and have a surface area of 77.4 m2.
So, to get any net buoyant force, it would need a mass of less than about 1kg/m2. For aluminium that means less than 1/3 mm thick, or for polycarbonate less than 0.8mm thick. But it must withstand atmospheric pressure of 100kg/m2

PeterVermont's links go into these issues in more detail, including the calculations of the strengths of containers (which is not so simple.)

12. Jun 30, 2016

The formula for the density of air comes from the ideal gas law $PV=nRT$ where n=number of moles and $R =.08206$ liter-atmosphere/(moles-degrees Kelvin) is the universal gas constant. This gives $n/V=P/(RT)$ and density $\delta =M.W. (n/V) =M.W. (P/(RT))$. (M.W.=molecular weight). To get in units of grams/liter, 1 mole=6.02 E+23 molecules and each proton has mass of 1.67 E-24 grams. The M.W. of He=4, the M.W. of $H_2$ (hydrogen, it's diatomic)=2. M.W. of Nitrogen $N_2$=28 , and M.W. of $O_2$=32. Air is 78% $N_2$ and 21 $O_2$, so that for air, you can use M.W.=30 (approximately). T degrees Kelvin=T centigrade+273. These calculations should give you the density numbers given for S.T.P. or for any other pressure and temperature. Also 1 gram/liter=1 kg/m^3 since 1000 liters=1 m^3. Computing for air at S.T.P. $\delta_{air S.T.P.}=30*(6.02 E+23)*(1.67 E-24)/((.08206)(273))=1.3$ grams/liter=1.3 kg/m^3.

Last edited: Jun 30, 2016
13. Jun 30, 2016

Baluncore

Vacuum is difficult for structural reasons. Neutral pressure is much easier and does not leak as fast as a positive pressure balloon. But there must still be a pressure against the upper part of the envelope to provide the lifting force.

If the lifting gas inside the envelope could be heated then you would get more buoyancy with less volume. Heating could be economic if the envelope was a “greenhouse” in the sun, or was made from very light weight closed cell foam with a metallised surface to reduce heat loss.

Another alternative would be to replace air with water. Given the right shaped pool or bath, an 80kg man would have neutral buoyancy in only a few litres of water.

14. Jun 30, 2016

cjl

I think it's the other way around - I think hydrogen is less prone to leakage than helium. Hydrogen is a larger atom, and it forms diatomic molecules, while helium is a monatomic gas.

15. Jul 1, 2016

Merlin3189

Thank you for pointing this out. I made a mistaken off the cuff comment on this because I did not think about He being monatomic and I thought I had heard about H2 gas being unusually difficult to contain (though I could well be wrong about that.)
However, on looking into this a bit more I have found an article on Permeability of Balloon Fabrics to H2 and He which experimentally determined greater permeability to hydrogen, despite noting that there are theoretical grounds for thinking the reverse should be the case. (This is not the work I refered to above and I can't find any reference to that for the moment. So, as I said, that comment may be unfounded.) If they are correct, then my comment was still a mistake, just serendipitously accurate!

I would query that. Net upward force would be obtained if the (external) pressure against the lower part were greater than the (external) pressure against the upper part. In principle the vacuum balloon would create maximum buoyancy, even though there is no (or very little) internal pressure on the upper surface. The problem with the vacuum is only with the strength (& thence weight) of the container. The internal pressure of H2 or He merely serves to reduce the required structural strength (& thence weight.)

An interesting comment because there are several toy solar airships - large black plastic balloons - which are just that.

When the thread started I thought we just had to explain why putting more He into the container did not give more buoyancy. Then the vacuum appeared and it became a matter of, well that would be ideal, except for the weight of the container. Now I find that it may even be feasible with modern (or post-modern) materials. So the world never ceases to surprise.

16. Jul 1, 2016

cjl

Interesting! I was going purely off molecular size (and He is unquestionably smaller), but I hadn't considered that other factors might be at play. I'll have to look into it when I have a bit of free time, since I'm sure there are some fascinating interactions going on there.

17. Jul 8, 2016

JBA

18. Jul 9, 2016

Baluncore

For hydrostatic reasons, the (external) pressure against the lower part will always be greater than the (external) pressure against the upper part. I think you have neglected the essential hydrostatic gradients in your analysis.

I was referring to the situation where the envelope was not under tension due to excessively high internal pressure. In that situation the neutral pressure balloon has two advantages, it leaks less and it does not raise the density of the lifting gas by unnecessary compression.

A hot air balloon has a big open hole at the bottom where the burner circulates and heats the internal parcel of air. A neutral pressure lifting-gas filled balloon can also be modelled as having a small hole at the bottom. Both systems will be at equilibrium when the differential pressure across the hole is neutral = zero.

For the sake of the following discussion I maintain the concept of a neutral pressure difference across a small hole in the envelope at the bottom of the balloon.

If there is buoyant lifting force then it has to come from somewhere. That is where it gets interesting and where it can be shown that it is the upper surface in the neutral pressure balloon that does the lifting.

It is clear that the buoyancy comes from the lower density of the internal gas compared with the external atmosphere. But understanding the mechanism is more difficult, so we start with the observation that hydrostatic pressure falls with height above the Earth's surface. The rate of pressure fall with height is proportional to the density of the gas.

The hydrostatic pressure of the atmosphere outside the balloon therefore falls faster with height than in the lower density lifting gas within the envelope. That explains why the internal pressure at the top of the balloon is greater than the external atmospheric pressure. The balloon envelope differential pressure rises with height, from zero at the hole at the bottom, to a maximum at the top.

The buoyant lifting force on the envelope is transferred to a suspended payload by tension in the envelope. The shape of the envelope results from a complex interaction of the differential pressure and envelope tension forces. All of which explains why a partly filled or neutral pressure balloon looks like a sphere of ice cream on an ice cream cone.

Now there comes a question. What lifting advantage is there in having a greater than neutral pressure inside the envelope ?

Last edited: Jul 9, 2016