The Space Hose

[buzz]

Hi !

As promised I would like to discuss with you in this thread the idea of a new kind of low cost Space Tower: the Space Hose.

In a nutshell it is about using a lightweight hose made from PE foil which is blown trough from the bottom and is using the frictional forces of the flowing air to produce continuous lift for supporting the weight of the hose.

It was designed as an alternative approach to solving the N-prize problem which is about putting a 9,99 gram satellite into space for 9 orbits and winning
£ 9999,99 when staying within the £ 999,99 budget. Because of the geostationary orbit a space tower offers it would mean keeping the tower upright for a total of 9 days.

You can find a brief presentation including most of the relative simple math in the attached PDF file.

I'm aware that this approach is not a very realistic one due to the huge stability problems when going for a single hose, but the math showed that it could be feasable to support a 100km hose and the needed raw material and energy consumption would be within the N-prize budget, hence I think it is worth sharing with you.

By using plain air at a reasonable blowing speed as the medium for continuously transfering the frictional force to the hose it overcomes most of the limitations of the existing inflatable space tower and the space fountain concept.A head diffusor is making the air blowing out sidewards on top with only a small downward momentum to support the payload and prevent tearing the hose.

And it was fun to find such a new solution and do the math to support it, so I think it is worth sharing the concept!

Have fun reading the slides and Input is welcome !

gutemine/buzz

PS: Sorry, for the bad graphics - I had to compress heavily to get below the 256k limit of the forum for attachments

[wildspace]

the weight of Air in the tube seems to me to be a problem . after all you are lifting a 50 km. pile of air it seems the energy required would be large about 1 kg. per square cm . could such a tube be lifted to a starting altitude by a balloon and inflated there

[buzz]

Thanks for your feeedback!

I'll try to calculate this for you:

The Air Pressure at the ground is 100.000 Pa (which is N/m²) - which is 10N/cm² and hence your 1kg/cm² are exact - but this is a pressure not an energy

And this pressure is nicely balanced by the surrounding air at the same pressure (or every time you stand up from your bed in the morning you would be in big troubles)

In General the Energy needed is cinetic energy plus hydrostatic energy.

cinetic energy:

E=mv²/2

Hydrostatic energy:

E=mgh

Hence for moving 1kg to 100km height at the suggested speed of 3,5m/s (= 12,6km/h) you need:

E=1*3,5²/2+1*9,81*100000=981006 J (Joule or kgm²/s²) - which is about 981kJoule

Air has a densitiy at sea level of 1,293kg/m³

The suggested hose has a volume of 0,05*100000=5017m³ The mass of the air in the hose would be therefore at sea level 5017*1,293=6556,89kg which is approximately 6,5 tons

BUT in reality it is much lower, because pressure and density go down in the atmosphere pretty fast when you go upwards - Pressure goes from 100000Pa to 100Pa only (1/1000) at 100km height and the same applies to the density.

Decrease It is about -50% every 5km to get a raw idea on the rate of decrease in density and pressure.
This approximately logarithmic density gradient can be integrated up to get the 'correct' mass of air in the hose.
If I remember the result right the total mass is only about 2 time the mass of the first 5km - meaning the real mass of the air in the hose should be about 655kg - not something that you want to carry on your shoulders, I agree. But it is balanced by the outside pressure oft he hose, so no need to worry too much.

So you would need an energy of 981*655=642555kJ or 642MJ This looks like a lot but 1kg of Jet Fuel has when burt a Thermal energy of 44MJ - so you would not need more fuel then fits in the tank of your car.

If you take 1 day time to blow up the hose (which means at the end that the entire air inside the hose is moving at the suggested speed up to the suggested height) this would be a power consumption of approximately 642555000/24/3600=7437J/s or Watt or 7,4kW - in the slides I used 15kW, because I calculated only the energy to keep it flowing not for erecting it and I applied also an efficiency factor for he fan and the calculated flow speed includes also the friction component to provide the lift force on the hose.

Even if the entire hose would be laying on the ground and you would have to blow the entire 6,5 tons of air the needed energy would be 10x higher, which is still something your pickup truck has under the bonnet.

gutemine

[buzz]

Hi !

Just a small update on the progress we had so far:

With the help of the physicsforums.com I found a bug in the slide on Blowing into vacuum.

The Line with the ideal gas formular between bottom and top should read:

Q on top =(100000*0,17/293)*183/100 = 106,18 m³/s

This means when blowing out trough the same diameter then on the bottom (0,05m²) would give only 2253,16m/sec theoretical blowout speed not the 5367m/sec on the slides.

This would be about 30% of the first cosmic speed needed for orbit which still seems to be pretty unrealistic. This means the Diameter of the Blowout nozzle would need to be changed by a factor of the squareoot of 3 to reach 7400m/sec in case this would be needed.

Probably I should also add a slide on the foil strenght, because this seems to be a hard part too:

In the slides I actually calculated already the pressure to support a PE foil hose that long (0,6 bar for the 4 micrometer foil, multiples of that if you use 12/25/50/100 my foil. This is pretty simple because if you assume worst case - all the weigth on the top then you can use simply the piston formular - weight*9,81 = gives the force in Netwon, then divide with the area of the hose (r²Pi) and you get the pressure - all hydrostatic pressure of the inside is balanced from outside.

300*9,81/0,05=58860N/m² which is less then 0,6 bar. Actually it should be much less, because most of the weight of the hose is not on top. AND it is a hose - so there is no 'piston' at all.

If you then take a 1m long part of the hose at the bottom the force on one half of the hose will be this pressure * height * diameter = 58860*1*0,25=14715N (approximately 1,5t of weight equivalent trying to tear the 1m hose piece)

This has to be held bei the 2m of foil (2 sides of the hose each 1m long) 4 micrometer thick

Stress in the foil will therefore be force/(2*height*thickness)
14715/(2*1*0,000004)=1839375000N/m²

Normaly strengt of a material is given in N/mm² so wie have to divide by 1.000.0000 - gives a stress in the foil of 1840N/mm² This is approximately 100x what normal PE foil can hold.

Dyneema, which is a kind of ultra long PE chain fabric now can hold 3000-4000N/mm² before tearing apart.

But this is actually the calculation for an inflatable space tower (and they suggest to use kevlar which has compareable strength)
If you have an open hose the math should be different. If you are open, there should be not a pressure surpluss beyond what the diffusor causes?

And also a few words on the windforces (maybe I should add a slide on this also - but we are still in the feedback collecting process):

I found the time to use the hopefully proper air resistance force formular:

A=0,25*15000=3750m² (the 15km we agreed that winds are counting - jetstream is gone beyond this - multiplied with the 10" diameter)
C=0,1 (drag coefficient)
v=13,89m/s² (a 50km/h "breeze")
F=airdensity*v²AC/2 = 1,293*13,89²*3750*0,1/2=46766N

actually this still is too high calculated because air density goes down to already 50% in 5km and 1/8 in 15km - but let us stay with this, because maybe the C value is too optimistic or the wind blows harder.

If I read the Dyneema Wiki entry right a single 8mm thick Dyneema string could hold this (58000N tearing strength, and 50kg/km) This string would have a weight of 750kg on these 15km. Probably your would use a bundle of smaller strings for better structural support.

We also found out at the Halfbakery that the hose probably under wind would form a kind of exponential or parabolic curve pointing to space - so that mainly only pull forces as described above would be stressing the hose. Which could be held by the suggested Dyneema strings as explained above. Then the foil only needs to hold the dynamic pressure of the wind which is not that bad. Let us calculate it again for a 50km/h wind = 13,88m/sec

dynamic pressure = airdensitiy * v²/2 = 1,2*13,88²/2=116Pa - so my assumed 100 Pa surplus should be suffient for up to 40km/h of wind to prevent the hose from collapsing - not bad, but a 200-300Pa surplus would be definitely better!

On the tpyical equatorial mountain suggested for a space tower the wind zone could be 5km shorter, so maybe 10km would be more realistic, and at the equator the winds are mainly blowing warm air upwards (which would be even positive for the energy bill of our hose) causing all the jet stream and trade winds at the areas above and below the equator.

I hope this summarizes all suggestions and additional things we found and checked so far, but we still have problems
understanding the gas dynamics in the hose !

Thanks to everybody for contributing !

gutemine

[buzz]

Hi !

Just a small update so that you don't think I have forgotten you or got completely lost:

I have spent the last few days to build a spreadsheet which calculates the standard atmosphere from 0 to 100km in 1km pieces, and then applied the formulars of the slides to these pieces to better understand what happens in the hose.

Then yesterday (during my morning shower!) I actually found out why I had so big problems getting sensemaking results:

It was a typical border value problem. Actually I always tried to calculate from bottom to top, but it is much easier to do it top to bottom, because there you simply define the maximum blowout speed (for example speed of sound to prevent the hose going supersonic) and the needed surpressure required to hold the payload and provide sufficient pull power to keep the hose stable.

Then you calculate the pressure loss of the resulting flow 1km down, add it to the atmospheric pressure there and the requested surpressure and get with ideal gas law a new density there, and hence a new flow speed because of continuity law. This means a new Reynolds number and a new viscosity and Lambda which means you have all the starting values for the next 1km and so on.

The pressure and blowing speed at the bottom are then a simple result of this iteration down and not the other way around. Because you can change pressure and blowing speed at the bottom in a relatively wide range depending what pump/fan you use this is not really a problem, and much better then choosing them and then get weird results at the top and within the hose.

Then you calculate the speed of sound at all these points to check that nowhere the air flow is faster. When you then have the flow and pressure gradient of the entire hose you can calculate the tension forces in the hose and can check if the PE foil and/or Dyneema strength can hold it.

If you blow out at the speed of sound this also becomes a kind of event horizon, meaning the hose doesn't know/care what the diffusor afterwards does, if you add a de Laval nozzle to blow out supersonic, turn the air flow downwards to generate lift, etc.

Then you are done and have a Spreadsheet where you can start playing with different blowout speeds, diffusor pressures, different foil thickness, hose diameters,... to find the optimal hose.

I will polish the spreadsheet a little bit more so that everybody can use it and then probably tomorrow you can play with it. There are some quite interesting findings already from what I tried out.

So actually the formulars and the math was not that bad (and there was no real critics from you on this either), but the USAGE was simple a little bit dumb and I should have tried it the way I suggested already earlier instead of trying to enter 100km in a single formular which allows to get an idea if it would work, but produces only consfusion on how.

Thanks for your patience with me!

gutemine

[pyramids]

This topic surely deserves to receive a bump --- cool idea!

But the stability issue truly is daunting.

[pauldear]

nah....I've been unstable for ages and it hasn't hurt me.

But seriously, it's an interesting inversion of the normal situation. With rockets, the first concern is always getting enough oomph, and then stability emerges as a secondary (though daunting) problem.

In this case, if gutemine's got his physics right, the oomph becomes a trivial problem and stability comes to the fore. Still, if you can loft a structure like that so simply, you can afford (financially and intellectually) to throw a lot of resources at stability.
Paul

[pyramids]

Does anyone know about equatorial high altitude winds?

Regarding space elevators and similar things, I've somehow always assumed you'd have to design to deal with winds of jet stream proportion. But at least temporarily and at least at the surface, the tropics can be extremely calm in terms of winds. How is that further up and on the day to week timescale gutemine proposed? If we can somehow pick time and location to always remain in the single digit m/s range, one might yet reconsider active stabilization by blowing (leaking) out air from the tube, although gutemine's original slides had already discounted that idea. All those actuators would be expensive and potentially heavy, though...

[pauldear]

What sort of forces (taking into account the hose diameter, wind speeds and air density) are we talking about here? On the face of it, the problems with a space hose seem immense - but not quite as immense as the problems with a "conventional" space elevator.

(I can't think of anywhere else where one could talk about a "conventional" space elevator!)

Paul

[pyramids]

The space hose, an example of an unconventional space elevator. I love it, Paul!

It sounds "easy" because it does not require unavailable material strength.

Regarding aerodynamic forces, it will be difficult, however.
The formula is F = 0.5 * Cd * rho * v^2 * A
where Cd is a drag coefficient (for a tube it might be around 0.5, I guess),
rho is the air density (ca. 1.3 kg/m^3 at sea level, approx. halving every 5.5 km of additional altitude),
v is the airspeed and A is the cross-sectional area (seen by the moving air) of the object creating the aerodynamic drag force F.

Cd can be lowered significantly by a streamlined shape ("profile"), to maybe as low as 0.05, but then this shape must be oriented appropriately. There are profiles that create appropriate aligning forces, but this obviously will result in a compromise Cd (still, maybe 0.1 is attainable?). Keeping such shapes would require more than just the super-thin tube assumed in the original idea, so maybe we better stick to a circular profile!

Anyways, to plug in numbers: At sea level, even a 1 m/s wind (which will feel like absolute wind still conditions) will pull a 40cm diameter circular tube with about 0.1 N/m. At 10 m/s, it would be about 13 N/m.