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#11
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alternate working fluids for nuclear thermal rockets?
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#12
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alternate working fluids for nuclear thermal rockets?
In article ,
Gordon D. Pusch wrote: CO2, with a molecular weight of 44, gives performance grossly inferior to a LOX/kerosene chemical rocket... ...Although, didn't Zubrin once float the scheme of using carbon _monoxide_ as a propellant for a "Martian Indigenous Fueled Nuclear Thermal Rocket" ??? I think he was proposing CO2 for that, but it's been a while since I read that paper, and I could be remembering it incorrectly. That was somewhat of a special case: you don't need great rocket performance for ballistic hops from place to place on the Martian surface, and there is a huge advantage in being able to refuel from the atmosphere. You still have the materials problems, not to mention the wee issue -- which as I recall, Zubrin neglected -- of conducting surface operations in the immediate vicinity of a lightly-shielded highly-radioactive reactor. But it does have its good points. -- MOST launched 30 June; first light, 29 July; 5arcsec | Henry Spencer pointing, 10 Sept; first science, early Oct; all well. | |
#13
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alternate working fluids for nuclear thermal rockets?
In article , Gordon D. Pusch wrote:
(Henry Spencer) writes: CO2, with a molecular weight of 44, gives performance grossly inferior to a LOX/kerosene chemical rocket. Moreover, at high temperatures it shows some tendency to break down to CO and O2. This improves performance a little, but that free oxygen is very hard on an orthodox NTR core and chamber. (There are materials that can stand it, but they're very different from the ones in LH2 or ammonia NTRs, and the technology for oxidizing-propellant NTRs is poorly developed.) ...Although, didn't Zubrin once float the scheme of using carbon _monoxide_ as a propellant for a "Martian Indigenous Fueled Nuclear Thermal Rocket" ??? There's a section about CO2-fuelled NIMFs (as he calls them) in /Case for Mars/, fwiw, in the somewhat more free-ranging half of the book g. His comment was that by just grabbing and compressing raw CO2, and heating it for a NTR, you could get a low but acceptable thrust. And, well, when you're flying on air you forgive a lot of problems with efficiency. He doesn't say anything about using CO as a propellant there, though. There's mention of a CO-O2 rocket soemwhere, and idle mention as an aside about using residual CO from a process as a fuel, but nothing about nuclear-thermal in the book. -- -Andrew Gray |
#14
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alternate working fluids for nuclear thermal rockets?
In article ,
James Nicoll wrote: The original context for this was a thread on "ultimate rockets", rockets where the dominent cost was fuel rather than labour. Obviously that doesn't describe the current state of affairs. Also, with an NTR system, the dominant consumables cost almost certainly will be fission fuel, not propellant. This is not your friendly next-door nuclear power plant :-) that needs refueling only quite rarely; NTRs are much more aggressive systems with much higher power densities and much higher burn rates. (The Phoebus NTR, the original design goal of the Rover program -- essentially a nuclear J-2 replacement -- and nearly flight-ready when development stopped, was the most powerful nuclear reactor of any kind ever built.) -- MOST launched 30 June; first light, 29 July; 5arcsec | Henry Spencer pointing, 10 Sept; first science, early Oct; all well. | |
#15
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alternate working fluids for nuclear thermal rockets?
In article ,
Henry Spencer wrote: In article , James Nicoll wrote: The original context for this was a thread on "ultimate rockets", rockets where the dominent cost was fuel rather than labour. Obviously that doesn't describe the current state of affairs. Also, with an NTR system, the dominant consumables cost almost certainly will be fission fuel, not propellant. This is not your friendly next-door nuclear power plant :-) that needs refueling only quite rarely; NTRs are much more aggressive systems with much higher power densities and much higher burn rates. (The Phoebus NTR, the original design goal of the Rover program -- essentially a nuclear J-2 replacement -- and nearly flight-ready when development stopped, was the most powerful nuclear reactor of any kind ever built.) At the risk of sounding ignorant, how would I estimate the U consumption? I see U-235 runs about $30,000.00/kg. Each kg is good for what, 8.1x10^13 J? Or very roughly 2x10^7 kilowatt-hours? That seems pretty cheap. -- It's amazing how the waterdrops form: a ball of water with an air bubble inside it and inside of that one more bubble of water. It looks so beautiful [...]. I realized something: the world is interesting for the man who can be surprised. -Valentin Lebedev- |
#16
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alternate working fluids for nuclear thermal rockets?
In article ,
James Nicoll wrote: Also, with an NTR system, the dominant consumables cost almost certainly will be fission fuel, not propellant. This is not your friendly next-door nuclear power plant :-) that needs refueling only quite rarely... At the risk of sounding ignorant, how would I estimate the U consumption? Mmm, the key variable is the energy efficiency -- how much of the fission energy release actually goes into jet kinetic energy -- and *that* I don't know offhand. There'll be some lost energy in molecular vibration etc., and a fair bit in escaping neutrons and other radiation. Let's guess 50%... I see U-235 runs about $30,000.00/kg. Each kg is good for what, 8.1x10^13 J? The energy value is at least roughly right, although I don't have the right references handy; the price I can't vouch for. :-) Okay, let's see. Phoebus was about 200 klb thrust -- as I said, it was a J-2 replacement -- so that's about 1 MN. Isp was in the general vicinity of 800 s, that's about 8 km/s. So the jet power was 0.5*1M*8k = 4 GW. At 50%, we need an 8 GW reactor. So 1000 s of operation, a reasonably long burn, requires 8 TJ = 0.1 kg = about $3000 at that price. Hmm, that's not so bad. Of course, you probably won't get anywhere near 100% burnup efficiency, so you probably have to multiply that by a factor of several. Still, if that price is right, that's no big deal. -- MOST launched 30 June; first light, 29 July; 5arcsec | Henry Spencer pointing, 10 Sept; first science, early Oct; all well. | |
#17
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alternate working fluids for nuclear thermal rockets?
(Henry Spencer) wrote in message ...
Okay, let's see. Phoebus was about 200 klb thrust -- as I said, it was a J-2 replacement -- so that's about 1 MN. Isp was in the general vicinity of 800 s, that's about 8 km/s. So the jet power was 0.5*1M*8k = 4 GW. At 50%, we need an 8 GW reactor. So 1000 s of operation, a reasonably long burn, requires 8 TJ = 0.1 kg = about $3000 at that price. The Nerva 12GW examined for the Hyperion sustainer produced about 1.3mlbf. Assuming I converted "589,670 kgf" to lbf correctly. http://www.astronautix.com/stages/hypainer.htm http://www.astronautix.com/engines/ner1mlbf.htm Hmm. There seems to be more jet power than thermal power in those nukes. Or my math is off. Or those entries are wrong. Mike Miller, Materials Engineer |
#18
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alternate working fluids for nuclear thermal rockets?
In article ,
Henry Spencer wrote: In article , James Nicoll wrote: Also, with an NTR system, the dominant consumables cost almost certainly will be fission fuel, not propellant. This is not your friendly next-door nuclear power plant :-) that needs refueling only quite rarely... At the risk of sounding ignorant, how would I estimate the U consumption? Mmm, the key variable is the energy efficiency -- how much of the fission energy release actually goes into jet kinetic energy -- and *that* I don't know offhand. There'll be some lost energy in molecular vibration etc., and a fair bit in escaping neutrons and other radiation. Let's guess 50%... I see U-235 runs about $30,000.00/kg. Each kg is good for what, 8.1x10^13 J? The energy value is at least roughly right, although I don't have the right references handy; the price I can't vouch for. :-) Okay, let's see. Phoebus was about 200 klb thrust -- as I said, it was a J-2 replacement -- so that's about 1 MN. Isp was in the general vicinity of 800 s, that's about 8 km/s. So the jet power was 0.5*1M*8k = 4 GW. At 50%, we need an 8 GW reactor. So 1000 s of operation, a reasonably long burn, requires 8 TJ = 0.1 kg = about $3000 at that price. Hmm, that's not so bad. Of course, you probably won't get anywhere near 100% burnup efficiency, so you probably have to multiply that by a factor of several. Still, if that price is right, that's no big deal. Yeah, I used the Ek of the jet and the mass ratio to guess at the amount of U-235 used and came to the conclusion that the cost of the H2 used would be much higher than for the U235, thus the look for cheaper working fluids. I looked up the price of U235 online and got prices from 30-50K per kilogram. I belatedly realise that since my older brother is manager of chemstores at the local university, he might well know what the going rate is right now. Now, to phrase the email so I don't sound like I am building a Device. t -- It's amazing how the waterdrops form: a ball of water with an air bubble inside it and inside of that one more bubble of water. It looks so beautiful [...]. I realized something: the world is interesting for the man who can be surprised. -Valentin Lebedev- |
#19
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alternate working fluids for nuclear thermal rockets?
In article ,
Henry Spencer wrote: In article , James Nicoll wrote: Also, with an NTR system, the dominant consumables cost almost certainly will be fission fuel, not propellant. This is not your friendly next-door nuclear power plant :-) that needs refueling only quite rarely... At the risk of sounding ignorant, how would I estimate the U consumption? Mmm, the key variable is the energy efficiency -- how much of the fission energy release actually goes into jet kinetic energy -- and *that* I don't know offhand. There'll be some lost energy in molecular vibration etc., and a fair bit in escaping neutrons and other radiation. Let's guess 50%... I see U-235 runs about $30,000.00/kg. Each kg is good for what, 8.1x10^13 J? The energy value is at least roughly right, although I don't have the right references handy; the price I can't vouch for. :-) It was on the basis of what may admittedly be crap assumptions [1] and a dodgy methodology that I came to the conclusion that Orion would have been insanely expensive. The number I got was around half a million dollars per kg of payload. 1: Including no drop in the price of U235, thrust charges that were mostly fission and no exciting new way to make small, cheap nuclear explosives. -- It's amazing how the waterdrops form: a ball of water with an air bubble inside it and inside of that one more bubble of water. It looks so beautiful [...]. I realized something: the world is interesting for the man who can be surprised. -Valentin Lebedev- |
#20
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alternate working fluids for nuclear thermal rockets?
In article ,
Mike Miller wrote: Okay, let's see. Phoebus was about 200 klb thrust... At 50%, we need an 8 GW reactor... The Nerva 12GW examined for the Hyperion sustainer produced about 1.3mlbf. Assuming I converted "589,670 kgf" to lbf correctly. "kgf" is a hideous abortion, but I think you got the conversion right. However, if you read http://www.astronautix.com/stages/hypainer.htm carefully, I think you'll find that the thrust number is for two engines. That would put it in about the right ballpark for "12GW" to be jet power. http://www.astronautix.com/engines/ner1mlbf.htm Now this one is just weird, since the name would strongly suggest a 1Mlb engine... I suspect confusion about units somewhere. -- MOST launched 30 June; first light, 29 July; 5arcsec | Henry Spencer pointing, 10 Sept; first science, early Oct; all well. | |
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