High-gain nuclear fusion shown achievable through simulation
40 replies, posted
[QUOTE=Pierrewithahat;35285011]Yeah matter is essentially just "condensed" energy in a sense, that's why you can directly convert matter into energy through fusion and fission. Also we'd never be able to get that much usable energy cause a sizeable portion of the energy released in fusion and fission is released as heat and we have major issues trying to harness heat.[/QUOTE]
Actually heat is the only way reactors can generate electricity. Fusion might be hugely complex but eventually it boils down to using the released energy to heat water and use the steam to power turbines.
Yeah I meant waste heat though, and even then we can't directly use the heat itself we're using mechanical forces of a dynamo/turbine being activated by water/vapor/steam pressure not the actual heat itself which needs to be removed from the system by coolant.
[QUOTE=pebkac;35285983]But here he's not referring to the average atomic mass you get on the periodic table, it's the actual mass of a certain isotope. The atomic mass unit u is defined to be 1/12th of mass of carbon 12. Only carbon 12 has the mass of exactly 12u, for other elements it's not exactly the same as number of nucleons * u due to binding energies involved.[/QUOTE]
He's using the values of the periodic table of hydrogen and helium, which won't make sense because of two reasons:
a) the isotopic distribution I spoke off
b) Hydrogen doesn't have neutrons, but protons don't weigh the same as neutrons, so you can't just say that the mass of 1 He-4 is the same as 4 H-1 because a helium doesn't weigh 4 protons mass worth.
[QUOTE=Kendra;35286989]He's using the values of the periodic table of hydrogen and helium, which won't make sense because of two reasons:
a) the isotopic distribution I spoke off
b) Hydrogen doesn't have neutrons, but protons don't weigh the same as neutrons, so you can't just say that the mass of 1 He-4 is the same as 4 H-1 because a helium doesn't weigh 4 protons mass worth.[/QUOTE]
It still works on the same principle
The mass of 2 deuterium nuclei is 4.02820u
The mass of helium-3 and a neutron is 4.024272u
There is a mass difference which we can convert to energy in MeV which leads to 3.24 MeV
[QUOTE=Kendra;35286989]He's using the values of the periodic table of hydrogen and helium, which won't make sense because of two reasons:
a) the isotopic distribution I spoke off
b) Hydrogen doesn't have neutrons, but protons don't weigh the same as neutrons, so you can't just say that the mass of 1 He-4 is the same as 4 H-1 because a helium doesn't weigh 4 protons mass worth.[/QUOTE]
You're right about that, I just skimmed through his post quickly and it seemed to be about right at a first glance. But still, with hydrogen and helium the mass of most common isotopes is practically the same as the relative atomic mass (due to scarcity of other isotopes):
[url]http://en.wikipedia.org/wiki/Hydrogen_atom[/url]
[QUOTE]Isotope mass 1.007825 u[/QUOTE]
[url]http://en.wikipedia.org/wiki/Helium_4[/url]
[QUOTE]Isotope mass 4.002602 u[/QUOTE]
Now I'm not sure what process hydrogen 1 would have to undergo to fuse into helium 4 but I still think he gave a decent enough approximation.
Wow, I just saw a Modern Marvels episode on this!
[IMG]http://upload.wikimedia.org/wikipedia/commons/thumb/7/78/FusionintheSun.svg/421px-FusionintheSun.svg.png[/IMG]
So apparently you end up with two neutrinos and two positrons apart from the photons and helium. I don't know much about nuclear physics apart from what they're teaching me in high school, but I assume the positrons are going to annihilate with electrons at some point and the energy taken away by neutrinos is essentially "lost" due to them barely interacting with matter at all. Would anyone more knowledgeable care to elaborate how much energy that is, if it's at all significant?
[QUOTE=Glorbo;35278761]Nuclear fusion is much safer than fission. It's very delicate, so if you somehow cut the reaction the reactor would quickly decelerate and just stop working, and the worst thing you can get is a small, localized explosion in the reactor chamber (which can be prevented with enough shielding). Compare that with fission, which can accelerate very quickly if left uncooled or unwatched. There isn't even going to be any nuclear fallout or radiation because the materials used aren't radioactive.[/QUOTE]
And don't forget the fact that in a normal fission reactor enough fuel is stored IN the reactor to keep it running for a long while. When a (typical) meltdown happens that fuel continues it's reaction and builds up massive heat and becomes uncontrollable.
But in a fusion reactor they constantly feeds the reaction with fuel. So if shit goes down, just cut the fuel and the reactor stops.
[I]They're waiting for you, Gordon. In the test chamber.[/I]
[QUOTE=Pierrewithahat;35286651]Yeah I meant waste heat though, and even then we can't directly use the heat itself we're using mechanical forces of a dynamo/turbine being activated by water/vapor/steam pressure not the actual heat itself which needs to be removed from the system by coolant.[/QUOTE]
Don't forget the X-rays and other types of radiation we can't really use for power generation.
[editline]25th March 2012[/editline]
[QUOTE=Fear_Fox;35288454]And don't forget the fact that in a normal fission reactor enough fuel is stored IN the reactor to keep it running for a long while. When a (typical) meltdown happens that fuel continues it's reaction and builds up massive heat and becomes uncontrollable.
But in a fusion reactor they constantly feeds the reaction with fuel. So if shit goes down, just cut the fuel and the reactor stops.[/QUOTE]
But after years of use, the free neutrons will irradiate the reactor walls and damage them over time. The reactor it's self would be more radioactive than normal fission waste, but only for 50 years.
[QUOTE=OvB;35280891]Remember who we're dealing with though:
[media]http://www.youtube.com/watch?v=XOI-Va5aU3U[/media][/QUOTE]
I can't stop laughing at 0:30
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