| FUSION REVISITED Let's look at a possible design for a fusion rocket motor. The engine is hemispherical consisting of a heavy metal (uranium, tungsten, irridium, nickel, chromium, ceramic oxides of these, titanium dioxide, possibly in layers to reflect various wavelengths of x-rays and gamma rays. TiO2 and Cr could reflect visible light and some UV.) or combination of metals. I estimate 80% reflectivity. It's 100 meters in diameter. Coils of superconducting cable surround the outer surface and there are loops beyond the open side of the hemisphere to channel the plasma. Ruby lasers with huge crystals grown in outer space micro-gravity furnaces bounce beams off dielectric mirrors (99% reflectivity) and are focused onto pellets of fusion fuel with pure silica lenses. Would diamond lenses or some other substance be superior? The pellets might consist of selenium surrounding lithium deuteride or supercold helium 3. Laser implosion microbombs have been made with selenium outer layers if memory serves me correctly. If lithium deuteride is used as fuel there will be 17.1 MeV neutrons released. U238 will absorb 14 MeV neutrons and undergo fission, so if we use lithium deuteride we might not want to use a uranium radiation reflector and a beryllium neutron reflector probably won't help because beryllium absorbs neutrons in the 5 to 15 MeV range and breaks down into helium four and nuetrons. Other metals of the reflector could be activated by neutrons. Helium 3 would be the best fuel because it doesn't release neutrons when it fuses and the charged particles could be channeled into the exhaust by the magnetic loops. Let's make the ship's mass 10,000 tons. The engine might amass 15,000 tons. We will energize 1,000 tons of hydrogen to 20,000 kps. This will give us a delta V of 784 kps or 1.75 million miles per hour! That would take us into the outer solar system fast. We will brake with mag-sails. If we travel at 875,000 mph, brake with mag-sails, return at 875kmph and brake with sails, we can make an astounding two way trip. One way flight time to Uranus would be 81 days! The energy required will be: 0.5(1,000,000 kg.)(20,000,000 m/s)^2 = 2E20j If we thrust for 100 hours, we generate 5.55E14 watts With deuterium and helium 3 a proton with 3.7 MeV and a helium 4 nucleus with 14.7 MeV form that have a total of 18.4 MeV of energy but about 20% is lost as Bremmstralung radiation as particles accelerate other particles in the magnetic field. If the reflector absorbs 20% of this energy ( some will be transmitted by the reflector and will not heat it) while the rest is focused onto the pellet and plasma (which carries away most of the energy) and some is reflected into space, delivering an amount of thrust I am incapable of calculating ( radiation pressure in implosion fusion bombs is said to be very high), then we can calculate the amount absorbed by the reflector: 4pi(50^2)/2= 15,700 square meters. (0.20)( 0.20*5.55E14)= 2.22E13 (2.22E13)/ 15,700 = 1.414E9 watts/ square meter or 1414 megajoule/sq. m. Since a megajoule equals 238.8 kcals, 337.7 thousand kcals fall on each square meter. Since uranium has a specific heat capacity of 0.0278 kcal/ deg kg. we find that if the shield is uranium one tenth of a meter thick or 1.91 tons (1910 kg.) then the reflector will heat up by: 423,000 kcals/ 0.0278 kcals/ deg. (1910) = 6359 deg. C in one second without cooling. This reflector would amass 29,990 tons. Lets increase the acceleration time to 500 hours or 21 days- 1,800,000 seconds. We then find that 1.11E14 watts are generated and (0.20)(0.20*1.11E14)/15,700= 282.8 Mj/m^2 Our shield will heat up to 1272 C in one second without cooling. If we halve the reflector mass to 0.955 tons per square meter then the temp will rise by 2544 C. in one second without cooling and the reflector mass is about 15,000 tons. Tungsten which is about as dense as uranium has a melting point of 3390 C. and uranium only 1130 C. but uranium dioxide melts at 2800 C. If we reduce the reflector mass to 10,000 tons or 637 kg. per square meter. and double the acceleration time to 1000 hours the uranium reflector with no cooling heats up by 1907 C and a tungsten reflector would increase in temperature 1648 C. in one second w/o cooling. I've been assuming the metals would conduct heat rapidly when bombarded by radiation and heat through thoroughly. The smooth polished surfaces would actually reflect the radiation. Plasma will be kept off the reflector by magnetic fields. Uranium dioxide does not conduct heat as well as the metal. The engine with a 10,000 ton reflector allows 5000 tons for coil loops, lasers, cooling systems, etc. The reflector would be laced with drilled passages for coolant to circulate through. Helium comes to mind but some susbstance like ammonia or a CFC or HFC might be better because of their high specific heats and density. CFCs and HFCs are almost chemically inert so they wouldn't corrode the reflector metals (?). The heat the coolant absorbs could run turbogenerators for electrical power to pump the lasers perhaps. I've also been assuming that the reflector absorbs 20% of the energy falling on it and the other 80% is bounced back at the plasma. I may be wrong. Some wavelengths will be reflected better and some, like gamma rays, will pass right through the reflector. This is only a simple analysis. If my assumptions are correct, then we are boiling water in a paper box with a torch, and that can be done. Dare we trust this machine? If computers sense a cooling system glitch they could shut down the drive instantly. The engine could also run in pulsed mode-one second for igniting pellets and one second for cool down. Ships to the outer system would travel in convoys of perhaps three ships. Landers and other cargo would be sent to the target planet years ahead of time. Experience with the fusion engine on unmanned craft would demonstrate its reliability or lack thereof. Also the reflector would radiate its heat into space and this might give it more than a couple of seconds to melt if computers didn't shut the drive down in a cooling system failure event. It seems like a huge device at 100 meters diameter but it is attached to a 10,000 tons ship-the mass of a naval destroyer or the Eiffel Tower. The ship would be made of materials far lighter than steel and iron. It would be composed of graphite, C60 fullerenes perhaps, polymers, magnesium, beryllium and aluminum. Ten thousand tons of these will make for a ship far larger than a destroyer!!! This ship would be more like the Battlestar Galactica than the Enterprise! Taking 1000 hours or 42 days to reach 875,000 mph (392 kps) would involve an acceleration of only 0.108 m/s. The ship would travel about 700 million kilometers in this time. It would have roughly another 2 billion kilometers to go to reach Uranus traveling in an almost straight line. That would take another 60 days not counting deceleration, so it's safe to say we could reach the outer system in less than six months aboard a ship as great as a sea going luxury liner. The specific power of the engine would be 3700 kW/kg. (2E20)/(3600*1000 hrs)= 5.55E13 5.55E13/15,000,000 kg.= 3700 kW/kg. Since P= e*b*A* ((T^4)-(Tinf^4)) P= heat loss in watts e= emissivity of surface material b= 5.6076E-8 watts/(m^2*deg. K^4) A=surface area in square meters T=material temp in deg. K Tinf= temp of interstellar vacuum= 4 def. K approx. Since carbon has an emissivity of 0.95 and we can put a carbon coating on the outside of the reflector, and the tungsten reflector could reach 1921 K in one second if it is at 0 C. initially: We find that P= 0.95* 5.6076E-8 * 15,700 square meters * ((1921^4)-(4^4)) = 1.139E10 watts With the 1000 hour acceleration we had 2E20/(3600*1000)=5.55E13 and 5.55E13*(0.20)*(0.20)=2.22E12 watts absorbed by the reflector. At 7177 K it would reach steady state with equal amounts of energy coming in and equal amounts radiated. Of course, carbon sublimates at 3600 C and tungsten and uranium at lower temps. What if we extended the carbon backing on a tungsten heat sink to make a cone shaped "nozzle"? And tripled the surface area? Carbon is a good conductor of heat as far as ceramics go and will draw heat from the metallic reflector and radiate it away if it isn't too thick. We reach steady state at 5453 K. If we increase the area ten fold with our low mass graphite radiator, we reach steady state at 4036 K. At twenty times as much area we are down to 3394 K. or 3121 C. and niether our carbon or our tungsten will melt. This would be a passive cooling system with no working parts to fail. Heat pipes in the tungsten heat sink could conduct heat. Uranium would be molten but perhaps it could be contained between layers of tungsten. Further refinements are certainly possible. By using lithium deuteride or pellets of deuterium and tritium made by bombarding lithium with neutrons spalled from heavy metal targets blasted by protons from solar powered particle accelerators on the Moon, we could reduce the Bremmstralung radiation to 1/140th of the fusion power but much of the energy would be released as neutrons. This would preclude the use of uranium and require the use of chromium, tungsten, nickel, etc. for the reflector. The neutrons could be absorbed by a lithium jacket to breed more tritium and the hot lithium could run generators that power the lasers and pulsed inductive thrusters that have no electrodes or grids to burn out like magnetoplasmic or electrostatic ion drives do. The reflector/heat sink would not have to be so massive if Bremmstralung rays are not nearly as intense. The lithium jacket and waste heat radiators might be lighter than the heavy reflector/heat sink of the D+3He engine. Deuterium+tritium reactions are easier to ignite than D+3He For D+3He reactors it might be possible to develop better, lighter radiation reflectors. Just by increasing reflectivity from 80% to 90% the reflector mass can be halved. There might be less Bremmstralung loss also in a rocket engine and transmitted gamma rays would not heat the reflector. Another strategy would be to simply let the radiation leak out and make no effort to contain it! The magnetic loops would be shielded and the engine made of material that is mostly transparent to the continuous spectrum of Bremmstralung radiation. Perhaps pure silica would do. Fusion power would be cut to 80% but this is no great loss. |
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| ABOVE) Better fusion reactor. It focuses radiation onto the plasma better. Graphite has good thermal conductivity as far as ceramics go and high melting point. It has very high emmisivity. Heat pipes and active cooling systems are also used. Further refinements are possible. IE, tungsten coated with graphite. Tungsten has m.p. almost as high as carbon and much better thermal conductivity (W 0.31 kcal*10E-3/deg cm s versus graphite 0.012), but graphite has much higher emissivity, thus tungsten could conduct heat away from the radiation reflector rapidly and if coated with thin layer of graphite it could radiate waste heat into the vacuum better. Lasers are the key to fusion rockets. Magnetic containment fusion (TOKAMAKs, SPHEROMAKs) are too massive. |
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| ABOVE) Stern view of finned reactor. |
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| Rather than try to contain the radiation it may be better to just let it radiate away into space. Most of the energy of the fusion reaction will be carried away by the plasma. The superconducting loops will need some shielding+reflecting material on the inside and will be cooled by liquid nitrogen. If the loops are transparent to the x-rays and gamma rays there will be little heating effect. |
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| Deuterium+tritium fusion looses very little energy to Bremmstralung radiation so not much reflector mass is needed if any. Most of the fusion reaction energy is carried away by neutrons. These could be captured by a lithium jacket and used to generate power for pulsed inductive thrusters that add thrust along with the plasma of alpha particles from the engine. This system is more complex but DT fusion is easier to achieve and total system mass compared to engine with heavy reflector/heat sink and radiator discussed above may be lower. |
| Learn more about fusion http://fti.neep.wisc.edu/neep602/lecture27.html |
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