As far as fuel types go, the dividing line is pre- and post
proton-proton fusion IMTU.
Hydrogen (protium) is so abundant and so energy dense (1g ~ 7
megawatt-days) that others don't make much sense unless damper tech is
unavailable.
As previously noted, the experience in handling liquid hydrogen in the
Traveller universe is on par with the real world experience of working
with iron and steel...
The usual candidate reactions are:
i. Deuterium-tritium (D-T)
ii. Deuterium-deuterium
iii. Helium-3-deuterium
iv. Helium-3-helium-3
v. Boron-11-proton
vi. Lithium-6-deuterium
vii. Lithium-6-proton
For the reactions listed,
D = deuteron, T = triton, n = neutron, p = proton
Reaction energies are expressed in mega electron volts [MeV]; 1 MeV =
1.6 x 10^(-13) joules.
i. D + T -> He-4 + n + 17.6MeV
2g deuterium and 3g tritium yields 1.7 x 10^12 joules or 19.6 megawatt-days.
This is the easiest fusion reaction to initiate (thermonuclear weapons)
and the subject of current power research.
Deuterium has a cosmic abundance of about 20 atoms per million hydrogen
atoms. Concentration occurs in gas giants (~26 for Jupiter), comets
(160-200) and water world oceans (160 average for Earth's oceans).
Tritium has a half life of 12.3 years, beta-decaying to helium-3.
It is produced by neutron bombardment of lithium-6 or neutron capture of
the heavy water moderator of fission reactors.
Neutron production has implications for embrittlement and activation of
the walls of the reactor.
ii. Two reactions of equal probability:
D + D -> T + p + 4.03MeV (50%)
D + D -> He-3 + n + 3.27MeV (50%)
2g deuterium yields 1.1 X 10^11 joules, or 30.6 megawatt-hours.
Deuterium-tritium and deuterium-helium-3 side reactions are possible.
iii. D + He-3 -> He-4 + p + 18.3MeV
2g deuterium and 3g helium-3 yields 1.76 x 10^12 joules, or 20.4
megawatt-days.
Helium-3 is deposited by the stellar wind and can be found in gas giant
atmospheres (1 in 10,000 helium atoms) and on the surface of vacuum
worlds (1-50 parts per billion by mass in the lunar regolith).
It is also concentrated (0.5-5 parts per million) in planetary
hydrocarbon (oil and gas) deposits due to the decay of radionuclides.
iv. He-3 + He-3 -> He-4 + 2p + 12.86MeV
3g helium-3 yields 9.09 x 10^11 joules, or 10.5 megawatt-days.
v. p + B-11 -> 3He-4 + 8.7MeV
1 gram hydrogen and 11 grams boron releases 8.38 x 10^11 joules, or 9.7
megawatt-days.
Boron-11 makes up 80% of naturally occurring boron and is stable.
Helium-boron fusion produces neutrons but is an unlikely side reaction
(~0.2%).
It is harder to start than deuterium-deuterium fusion (about 16x more
energy required).
vi. D + Li-6 -> 2He-4 + 22.4MeV
2 grams of D and 6 grams Li-6 releases 2.15 x 10^12 joules, or 25
megawatt-days.
Lithium-6 makes up about 6% of naturally occurring lithium and is stable.
Alternate (side) reactions:
D + Li-6 -> He-3 + He-4 + n + 2.56MeV
D + Li-6 -> Li-7 + p + 5 MeV
D + Li-6 -> Be-7 + n + 3.4MeV
vii. p + Li-6 -> He-4 + He-3 + 4MeV
1 gram hydrogen and 6 grams Li-6 releases 3.85 x 10^11 joules, or 4.46
megawatt-days (107 megawatt-hours).
The lithium reactions are harder to start than boron-proton fusion.
On conversion drives and spacecraft design:
Some figures follow.
Hopefully they will format properly with the list software.
Assumptions:
- 30 day endurance required.
- 75% efficiency in converting fuel mass to kinetic energy.
[1] [2] [3]
720 0.086 0.995
1440 0.173 0.98
2160 0.259 0.955
2880 0.346 0.92
3600 0.432 0.875
4320 0.519 0.82
5040 0.605 0.753
5760 0.692 0.675
6480 0.778 0.581
[1] G-hours for 30 days' endurance, corresponds to manoeuver drive
ratings 1 to 9.
[2] Terminal velocity as a fraction of speed of light for continuous
acceleration over 30 days.
[3] Payload mass fraction - proportion of final loaded mass that isn't
fuel for the M-drive.
Note that the standard design sequences work up to the third row of the
table above.
So you could have higher acceleration ratings but with reduced endurance
e.g. 6G drive with 1440 G-hours fuel is good for 1440/6 = 240 hours or
10 days of sustained acceleration.
Rob O'Connor
