Nuclear fusion fuel question

Arrow · Post by **Arrow** » 2006-06-05 10:06pm

Based on previous threads I been involved with or started a while back, I decided I wanted to use nuclear fusion for my own scifi unverse, fueled by Lithium-7 Deuterium pellets. I've done some digging since them, and I can't find numbers for the density of this fuel, nor how much energy (expressed in joules/kilogram) is release by its reactions. Now, I envision these reactors to be able to run the full Deuterium Cycle of reactions (as shown here), which is 43.2 MeV (this appears to be the most energetic fusion reactor; please tell me if there's a better one). However, I don't have the background to know the equations and sources I need to get the answers I want.

So, can someone help me figure out the density of Li-7D and the energy released from its reaction?

Ariphaos · Post by **Ariphaos** » 2006-06-05 10:57pm

From here, lithium-6 deuteride has a density of 820 kg/m^3. L7 would be ~12.5% more, so roughly 920, I think.

A molar unit would be 9 grams, and the cycle you are referring to uses 6 parts deut to produce said 43.7 MeV.

One eV is 1.6022x10^-19 Joules.

So, per molar unit, we have 60200 * 1,600,000 * 43.7/6 = ~700 gigajoules per 9 grams.

I think I did that right, anyway O_o

Arrow · Post by **Arrow** » 2006-06-06 11:26am

So that would work out to around 77 terajoules per kilogram? Is that right? Mike's constants page list D-D fusion as 620 terajoules per kilogram (6.2*E14 J/K). Damn I wish my chem and physics books weren't buried in storage.

vargo · Post by **vargo** » 2006-06-06 04:56pm

check this vid out its talks about Water fuel

http://www.thatvideosite.com/view/2602.html

Ariphaos · Post by **Ariphaos** » 2006-06-06 06:43pm

Arrow wrote:So that would work out to around 77 terajoules per kilogram? Is that right? Mike's constants page list D-D fusion as 620 terajoules per kilogram (6.2*E14 J/K). Damn I wish my chem and physics books weren't buried in storage.

90 petajoules * .007 (roughly the Sun's mass-energy conversion rate) = ~630 terajoules. I always thought that was a standard hydrogen fusion reaction, however. this site seems to be working similarly to my math.

The majority is, since you're using Lithium-7 Deuteride (instead of Lithium 6), you're dividing that by 4.5 outright for your fusable mass. Also, it's not using the Lithium or standard hydrogen fusion, or various free neutrons that get produced, which will continue the fusion reaction at higher temperatures, so it's not the full reaction.

Arrow · Post by **Arrow** » 2006-06-06 08:03pm

What would the full reaction be in that case?

Ariphaos · Post by **Ariphaos** » 2006-06-07 01:36am

Arrow wrote:What would the full reaction be in that case?

http://hyperphysics.phy-astr.gsu.edu/hb ... rocyc.html

It's not entirely relevant, since it requires stellar densities for various reasons, and even then the average proton waits a billion years to convert to deuterium. However, if you were using Lithium 6 instead of 7 (Li-6 is rare though, granted), the Lithium would also be a net energy producer in the fusion reaction (see here).

However, since we end up with two free neutrons and two free protons at the end of the reaction, they may also participate in fusion, forming deuterium and messing around with others. It also compounds the neutron embrittlement of your reactor, however, so it's a little annoying.

Also, it's important to remember that some of the energy produced in the reactions we've described gets lost in neutrinos, though this varies per reaction.