Maximum density?
Moderator: Alyrium Denryle
Maximum density?
I'm trying to figure out how to calculate the maximum density that one can achieve before things are so compressed that critical mass is reached and fusion or fission is achieved. I know it has to do with the Coloumb barrier, but that's all I remember from my schooling. I don't even recall what to search for beyond that. If anyoen could help I'd appreciate it.
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Winston Blake
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Lawson criterion might be something to start with.
Robert Gilruth to Max Faget on the Apollo program: “Max, we’re going to go back there one day, and when we do, they’re going to find out how tough it is.”
Fusion primarily - I'm thinking fuel tanks in spacecraft and I'll start by asumign that they will use H2 or something similar for propellent.drachefly wrote:'Fusion or fission"? The question is very very different for the two.
Which are you more concerned with?
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Fuel tanks? Oh, wow, that's a completely different question. You mean the max density to which the fluid can be compressed in large quantities for an extended period with a pressure small enough that a reasonable material could hold it in?
Hmm..
On spacecraft, actually, volume is cheap - space is big - but mass is expensive. So the designers won't worry too much about volume. So, I'd go with the density of liquid deuterium. According to wikipedia, the density of solid deuterium is 195 kg/m^3 at the triple point (18 kelvins, I don't know the pressure but it shouldn't be outrageous), and that should be similar to the density of the liquid.
Hmm..
On spacecraft, actually, volume is cheap - space is big - but mass is expensive. So the designers won't worry too much about volume. So, I'd go with the density of liquid deuterium. According to wikipedia, the density of solid deuterium is 195 kg/m^3 at the triple point (18 kelvins, I don't know the pressure but it shouldn't be outrageous), and that should be similar to the density of the liquid.
I'm doing scifi universes - materials holding it are a null isue, I'm worried about fuel limitations - how fast will I consume it? So I need max density to fit it in the ship
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Winston Blake
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Based on this data:
Lawson criterion for D-D fusion is 30x D-T
Triple product Lawson criterion for D-T fusion is 1e21 Kev s*m^-3
Since I haven't studied any of this stuff this is probably wrong, but given that the 'triple product' Lawson criterion seems to be directly proportional to the normal one, I can guess that for deuterium the triple product must be greater than (30 * 1e21 Kev s*m^-3) for ignition, or (30*(1e21Kev / (1.5k)) (K*s*m^-3)) assuming an ideal gas of monatomic deuterium.
Then the electron density has to be that figure divided by temperature and confinement time. Given that I have no idea how to find the parameters that define confinement time (energy content and power loss), I'll take a wild guess of 1s and let somebody correct this later. I'll take temperature as 298K.
Given that the molar mass of monatomic deuterium is apparently 2.01g, I assume that 'electron density' can be converted into mass density simply by multiplying it by (2.01g/Avogadro's number), i.e. one electron per atom.
So the calculation ends up being (copy into Google):
(2.01g/avogadro's number ) *(30*(1e21Kev / (1.5k)) (s*m^-3))/((1e0s) * 298K) = 2.6 kg/m^3.
This is obviously wrong, I'd expect it to be huge simply because D-D fusion at room temperature must be extremely difficult. Hopefully any people who know their stuff and have seen this thread but haven't had the time to help out can fix my method (particularly the confinement time). OTOH, there's probably a book full of tables where this stuff can be looked up
.
An idea I just had is that, since the sun is mostly hydrogen, the pressure at its surface must be roughly that required to fuse hydrogen. It's 6000K and 1e23 particles/m^3 at the surface, so going by my above assumption about converting particle density to mass density, that's about 1.7e-9 kg/m^3. Maybe somebody can use this to get an estimate for hydrogen.
Lawson criterion for D-D fusion is 30x D-T
Triple product Lawson criterion for D-T fusion is 1e21 Kev s*m^-3
Since I haven't studied any of this stuff this is probably wrong, but given that the 'triple product' Lawson criterion seems to be directly proportional to the normal one, I can guess that for deuterium the triple product must be greater than (30 * 1e21 Kev s*m^-3) for ignition, or (30*(1e21Kev / (1.5k)) (K*s*m^-3)) assuming an ideal gas of monatomic deuterium.
Then the electron density has to be that figure divided by temperature and confinement time. Given that I have no idea how to find the parameters that define confinement time (energy content and power loss), I'll take a wild guess of 1s and let somebody correct this later. I'll take temperature as 298K.
Given that the molar mass of monatomic deuterium is apparently 2.01g, I assume that 'electron density' can be converted into mass density simply by multiplying it by (2.01g/Avogadro's number), i.e. one electron per atom.
So the calculation ends up being (copy into Google):
(2.01g/avogadro's number ) *(30*(1e21Kev / (1.5k)) (s*m^-3))/((1e0s) * 298K) = 2.6 kg/m^3.
This is obviously wrong, I'd expect it to be huge simply because D-D fusion at room temperature must be extremely difficult. Hopefully any people who know their stuff and have seen this thread but haven't had the time to help out can fix my method (particularly the confinement time). OTOH, there's probably a book full of tables where this stuff can be looked up
.An idea I just had is that, since the sun is mostly hydrogen, the pressure at its surface must be roughly that required to fuse hydrogen. It's 6000K and 1e23 particles/m^3 at the surface, so going by my above assumption about converting particle density to mass density, that's about 1.7e-9 kg/m^3. Maybe somebody can use this to get an estimate for hydrogen.
Robert Gilruth to Max Faget on the Apollo program: “Max, we’re going to go back there one day, and when we do, they’re going to find out how tough it is.”
Winston:
Don't suppose they'll store it at the freezing point of water, we have better storage systems on the freaking space shuttle. Bring it down to 4 kelvins and it'd get MUCH denser, being a liquid. Even if it were a gas, PV = NKT would help you out by a factor of about 100.
Ender:
Brown dwarfs also have a mean pressure that makes our strongest pressure vessels look puny. Like, it's nearly pressurized enough to start fusion in its core. Liquids are nearly incompressible, the space savings from compressing it would be insignificant compared to the hazards in the event of failure. Like...
case 1: near room pressure liquid. Danger if leaks: could burn, but it has little incentive to leak.
case 2: have compressed the liquid to 99% of its normal volume. Danger if leaks: explosion, just from release of pressure.
If anything ever ever goes wrong with those fuel tanks, they'll take option 1.
Don't suppose they'll store it at the freezing point of water, we have better storage systems on the freaking space shuttle. Bring it down to 4 kelvins and it'd get MUCH denser, being a liquid. Even if it were a gas, PV = NKT would help you out by a factor of about 100.
Ender:
Brown dwarfs also have a mean pressure that makes our strongest pressure vessels look puny. Like, it's nearly pressurized enough to start fusion in its core. Liquids are nearly incompressible, the space savings from compressing it would be insignificant compared to the hazards in the event of failure. Like...
case 1: near room pressure liquid. Danger if leaks: could burn, but it has little incentive to leak.
case 2: have compressed the liquid to 99% of its normal volume. Danger if leaks: explosion, just from release of pressure.
If anything ever ever goes wrong with those fuel tanks, they'll take option 1.