PIPELINE TO SPACE CARBON NANOTUBES

cyclonebuster · Oct 25, 2008

<div id="post_message_41024">A carbon nano tube pipeline from the Equator to geo. orbit could pump massive amounts of H2 and O2 to space to be stored as rocket fuel. The pipeline would only have to be 1/4 inch diameter. Rachmaninoff, It would be in a Geo. orbit like the space elevator!! Here are some numbers worked out on the Idea. I think even more powerfull pumps could work thus lowering the number of them needed by ten fold. 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/vchembo...ensitygas.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 wich 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.</div>

cyclonebuster · Oct 25, 2008

<div id="post_message_41024">A carbon nano tube pipeline from the Equator to geo. orbit could pump massive amounts of H2 and O2 to space to be stored as rocket fuel. The pipeline would only have to be 1/4 inch diameter. Rachmaninoff, It would be in a Geo. orbit like the space elevator!! Here are some numbers worked out on the Idea. I think even more powerfull pumps could work thus lowering the number of them needed by ten fold. 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/vchembo...ensitygas.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 wich 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.</div>

3488 · Oct 26, 2008

Hi cyclonebuster,I moved this thread to SB&T as in this forum, you are likely to receive better quality answers to your opening post. This is a very specialist topic, that really falls outside of general Astronomy & perhaps the Space Science normally related to in SS&A, the science primarily of natural bodies as opposed to manmade constructs or speculative technologies.Andrew Brown.  <div class="Discussion_UserSignature"> "I suddenly noticed an anomaly to the left of Io, just off the rim of that world. It was extremely large with respect to the overall size of Io and crescent shaped. It seemed unbelievable that something that big had not been visible before". Linda Morabito on discovering that the Jupiter moon Io was volcanically active. Friday 9th March 1979.http://www.launchphotography.com/ http://anthmartian.googlepages.com/thisislandearthhttp://web.me.com/meridianijournal </div>

webtaz99 · Oct 27, 2008

Things that make you go, "Hmmmmm". <div class="Discussion_UserSignature"> </div>

cyclonebuster · Nov 10, 2008

Replying to: <DIV CLASS='Discussion_PostQuote'>Things that make you go, "Hmmmmm". Posted by webtaz99</DIV>It appears that diamonds would also be up for the task with this newest technique. See link below:http://www.sciencedaily.com:80/releases/2008/10/081027174541.htm

DrRocket · Nov 10, 2008

Replying to: <DIV CLASS='Discussion_PostQuote'>A carbon nano tube pipeline from the Equator to geo. orbit could pump massive amounts of H2 and O2 to space to be stored as rocket fuel. The pipeline would only have to be 1/4 inch diameter. Rachmaninoff,It would be in a Geo. orbit like the space elevator!!Here are some numbers worked out on the Idea. I think even more powerfull pumps could work thus lowering the number of them needed by ten fold.Calculation of the amount of pumps neededAs 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)^2Where:Quote rho is the density of the gas on the high pressure side of the pumpP is the pressure of the gas on the high pressure side of the pumpg is the force of gravity at the pumpG is the universal gravitational constantM_E is the mass of the earthR is the radius of the earthr is the distance from the center of the earthh is altitudeN 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-005k=1 seems the most practical. From: http://www.elmhurst.edu/~chm/vchembo...ensitygas.htmlMethane DataHere are some densities

ensities of Common Elements and Compounds(Substance Density kg/m^3)Quote Hydrogen gas 0.000089e3Helium gas 0.00018e3Air 0.00128e3Carbon Dioxide 0.001977e3Water 1.00e3Methane 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 wich 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/kNow to get it for others we can do this trickQuote For hydrogen:1000*(density of hydrogen/density of air)= 1000*0.000089e3/0.00128e3=69.5313For 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. Posted by cyclonebuster</DIV>A 1/4 tube that long is an EXTREMELY long and thin column subject to Euler buckling.  It won't work. <div class="Discussion_UserSignature"> </div>

neilsox · Dec 28, 2008

Replying to: <DIV CLASS='Discussion_PostQuote'>A 1/4 tube that long is an EXTREMELY long and thin column subject to Euler buckling.  It won't work. Posted by DrRocket</DIV>A very massive counter weigh is needed slightly beyond GEO altitude or a small counter weight  at about 144,000 kilometers altitude. They are thinking 91,000 kilometers = 60,000 miles for the space elevator.Will a one meter tube 400,000 meters long also Euler buckle, if kept under moderate tension?  Considerable energy is needed to keep the tube from flopping back to Earth since it is still subject to about 0.9g = It is moving much slower than orbital speed. This slow speed = 1100 miles per hour makes transfering the gas to a much faster space craft difficult.Even if the CNT meets space elevator specs, the oxygen column is likely too heavy to an altitude of 400 kilometers. The enroute pumps may be too heavy even for the hydrogen tube. The jet steam blowing on the tubes may also be a problem at jet stream altitude even if the tubes are much longer than 400 kilometers. I think the jet stream rarely crosses the equator, but the base can be up to 2000 miles from the equator with only moderate loss of performance.  Neil

exoscientist · Dec 29, 2008

Replying to: <DIV CLASS='Discussion_PostQuote'>A 1/4 tube that long is an EXTREMELY long and thin column subject to Euler buckling.  It won't work. Posted by DrRocket</DIV>  Euler buckling applies to a tall structure in compression. He is considering here a pipe supported at the top so is instead in tension. In this case buckling would not apply.  Bob Clark   <div class="Discussion_UserSignature"> </div>

exoscientist · Dec 29, 2008

Replying to: <DIV CLASS='Discussion_PostQuote'>A carbon nano tube pipeline from the Equator to geo. orbit could pump massive amounts of H2 and O2 to space to be stored as rocket fuel. The pipeline would only have to be 1/4 inch diameter. Rachmaninoff,It would be in a Geo. orbit like the space elevator!!Here are some numbers worked out on the Idea. I think even more powerfull pumps could work thus lowering the number of them needed by ten fold.Calculation of the amount of pumps neededAs 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)^2Where:Quote rho is the density of the gas on the high pressure side of the pumpP is the pressure of the gas on the high pressure side of the pumpg is the force of gravity at the pumpG is the universal gravitational constantM_E is the mass of the earthR is the radius of the earthr is the distance from the center of the earthh is altitudeN 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-005k=1 seems the most practical. From: http://www.elmhurst.edu/~chm/vchembo...ensitygas.htmlMethane DataHere are some densities

ensities of Common Elements and Compounds(Substance Density kg/m^3)Quote Hydrogen gas 0.000089e3Helium gas 0.00018e3Air 0.00128e3Carbon Dioxide 0.001977e3Water 1.00e3Methane 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 wich 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/kNow to get it for others we can do this trickQuote For hydrogen:1000*(density of hydrogen/density of air)= 1000*0.000089e3/0.00128e3=69.5313For 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. Posted by cyclonebuster</DIV>  I like your suggestion but your calculations are hard to read in text format. Perhaps you could write it in MS Word or in HTML and attach it or save as an image and embed it in your post.   Bob Clark   <div class="Discussion_UserSignature"> </div>

exoscientist · Dec 29, 2008

I proposed a similar idea in this post to sci.astro but only to LEO:From: "Robert Clark" <rgregorycl...@...> Newsgroups: sci.astro,sci.physics,sci.mech.fluids,sci.engr.mech,sci.space.policy Subject: "Rockets not carrying fuel" and the space tower. http://groups.google.com/group/sci.astro/msg/db628394d9994d3e?hl=en&dmode=source  I was using liquid water as the fluid only because it is incompressible so the pressure required is easier to calculate. Very likely you would actually want to use a gas since the pressure required would be less. Your calculations probably could tell the pressure required to just get to LEO. Note I was envisioning just going to LEO because the tower would be self-supporting as it would be supported by high pressure fluid being directed downward at regular distances along its length. However, a key question I have is could the gas that is being vented be recycled? If you had this gas being just vented to the air that would be quite wasteful, and perhaps dangerous if the gas is hydrogen. Could you have a wider shroud around the pipeline that caught this vented gas so that it fell back down to the ground to be used again? The question about this I have is this gas now moving downward and being constrained by the shroud would bump into it with a downward momentum and might impart a downward momentum to the pipe and shroud and cancel out the upward force you get from the venting.   Bob Clark <div class="Discussion_UserSignature"> </div>

exoscientist · Feb 20, 2009

This invention shows the concept behind it is valid:Fluid Motion: JetLev-Flyer H2O-Propelled Jet Pack By Chuck Squatriglia February 02, 2009 | 5:02:03 PM http://blog.wired.com/cars/2009/02/fluid-motion--.html James Bond-style jetpack powered by high-pressure water invented. A German entrepreneur, Hermann Ramke, has invented a James Bond-like jetpack powered by high-pressure water, called the JetLev-Flyer. Last Updated: 10:35AM GMT 17 Feb 2009 http://www.telegraph.co.uk/news/newstopics/howaboutthat/4640262/James-Bond-style-jetpack-powered-by-high-pressure-water-invented.html      Bob Clarkhttp://www.telegraph.co.uk/news/new...-powered-by-high-pressure-water-invented.html  <div class="Discussion_UserSignature"> </div>

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PIPELINE TO SPACE CARBON NANOTUBES

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