Carbon Nano Tube Pipeline to Space: A Revolutionary Idea

[errorist]

Any gas coild be used for an Ion drive engine.It does not have to be xenon.Btw,xenon could also be pumped up there.One other advantage it would have. It would have less of a chance of being impacted by a micro metiorite because it is so much smaller.
The sky hook has a million times the mass thus a million times the cost. This thing is but 1/4 inch Inside diameter.

 

[LunchBox]

The sky hook has a million times the mass thus a million times the cost. This thing is but 1/4 inch Inside diameter.

So all the pumping is going to be done from the surface of the Earth? Run a quick pressure analysis and see what kind of pressures that is going to require near the bottom. Then see just how many CNT layers you need to contain that via thin wall pressure vessel. Your OD may surprise you.

Cheers...

 

[errorist]

Calculation of the amount of pumps needed
As previously mentioned the additional amount of pumps needed per length is proportion to the force of gravity. We can neglect centripetal acceleration because it is to small to make much difference. I claim this can be written mathematically as:

dN/dh=(1/k)*/(P*g/rho)/ (P_o*g_o/rho_o)^2

Where:
Quote
rho is the density of the gas on the high pressure side of the pump
P is the pressure of the gas on the high pressure side of the pump
g is the force of gravity at the pump
G is the universal gravitational constant
M_E is the mass of the earth
R is the radius of the earth
r is the distance from the center of the earth
h is altitude
N is the number of pumps.
k is the number of attenuation constants between pumps. The fraction of gas remaining is given by e^(-k)
k=1 gives 0.3679, k=5 gives 0.0067, k=10 gives 4.5400e-005
k=1 seems the most practical.


From: http://www.elmhurst.edu/~chm/vchembook/123Adensitygas.html
Methane Data
Here are some densities:
Densities of Common Elements and Compounds
(Substance Density kg/m^3)
Quote
Hydrogen gas 0.000089e3
Helium gas 0.00018e3
Air 0.00128e3
Carbon Dioxide 0.001977e3
Water 1.00e3
Methane 0.0006557e3


The calculations will be done for air. Notice that methane is lighter then air.
To find the number of pumps needed we integrate the above expression from the radius of the Earth to GEO.
N=(1/k)int(P*g/rho),/(P_o*g_o/rho_o)^2, h=0…36e9)

Not that (P_o*g_o/rho_o) is the distance over which the pressure drops by 1/e.
Quote
=1000 Pa * 9.8 m/s^2/0.00128e3 kg/m^2=7.6563e+003 m for air.

In the calculation below we will use 6.92105e3 instead of 7.6563e+003 so are integral agrees exatly at the first pump which will be at 7.6563e+003.

Which is about 7.7 km. If the pressure at the high pressure side of each side is the same the expression for N becomes
Quote
N=(1/(k*6.92105e+003))int(g/g_o, h=0…36e6)= (1/k)int(G M_E/(h+R)^2/g_o, h=0…36e6) /0.3013
= (1/k*6.92105e+003)*int(6.67e-11 * 5.98e24/h^2/9.9, h=6e6…42e6)
=(1/k)* (-6.67e-11 * 5.98e24)/(9.9*6.92105e+003)*((1/42e6)-(1/6.38e6))=1000/k


Now to get it for others we can do this trick
Quote
For hydrogen:
1000*(density of hydrogen/density of air)
= 1000*0.000089e3/0.00128e3=69.5313
For methane:
773.8850*0.0006557e3/0.00128e3=512.66


So that is 70 pumps for hydrogen, 513 pumps for methane and 1000 pumps for air. The pumps start out being spaced 7 km apart and get further apart as the altitude increases. Also note that when the pumps get further apart the tubes must get wider to keep the viscous forces the same.

 

[LunchBox]

So it can't be any smaller than a skyhook, because you have "stations" hanging onto it at regular intervals. These are effectively the same as "cars" on a skyhook.

Your point that it could be smaller is now moot.

Cheers...

 

[errorist]

The pumps are even smaller and are part of the structure and the mass is still a million times less than the skyhook.

 

[FredGarvin]

That's the second time you have gone about this line of reasoning and I guess it will be a second time telling you that those calculations are incomplete to say the least. The only thing those calculations are doing is accounting for gravity. Try looking into frictional and compressibility effects.

Someone please close this thread.

 

[LunchBox]

Someone please close this thread.

I second that.

Cheers...

 

[errorist]

"Try looking into frictional and compressibility effects."

I'll look at these factors later but for now I can see for sure these pumps could also overcome these losses.

 

[LunchBox]

The pumps are even smaller and are part of the structure and the mass is still a million times less than the skyhook

What is going to power these pumps?

 

[errorist]

Perhaps,microwaves??

 

[errorist]

If you had a lot of pumps a RF wave guide would probably be the way to go to power them.

 

[brad50]

Why come to the surface harvest the atmosphere then pump to low Earth orbit, geo orbit is a anchor point. They will not get hydrogen oxygen is 1/2 of the fuel. Make a high thrust orbital tug gets its oxygen from are gas station hydrogen would come the old way. This tug would come down to suborbital levels dock with a suborbital craft boost to low orbit . Then transfer to proper orbit or a ion tug.We would not have to wait till cabin nano tubes can reach earth. Any one know how strong they are now, how far they can reach, is there any oxygen there how thick.

 

[russ_watters]

Welcome to PF. This thread is far too old to be resurrected.