Difference between revisions of "Fusion Propulsion"
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| − | Fusion Propulsion | + | Fusion Propulsion may be used for Mars transfer in the future, once the technology is functional. Mars has a higher amount of deuterium in its water than Earth, so this basic fusion fuel might be easier to obtain on Mars. |
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| − | | colspan="16" rowspan="1" |Table 1 - Significant fusion reactions | + | | colspan="16" rowspan="1" |Table 1 - Significant fusion reactions that could be used for a fusion engine |
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| colspan="4" rowspan="1" |Reaction | | colspan="4" rowspan="1" |Reaction | ||
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|Total | |Total | ||
Energy release | Energy release | ||
| − | |Atomic mass of products | + | |Atomic mass |
| + | of products | ||
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|ratio | |ratio | ||
| − | |MeV/ fusion | + | |MeV/ |
| + | fusion | ||
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| − | |MeV/ fusion | + | |MeV/ |
| + | fusion | ||
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| − | |MeV/ fusion | + | |MeV/ |
| + | fusion | ||
|GJ/kg | |GJ/kg | ||
|N | |N | ||
Revision as of 05:55, 2 September 2021
Fusion Propulsion may be used for Mars transfer in the future, once the technology is functional. Mars has a higher amount of deuterium in its water than Earth, so this basic fusion fuel might be easier to obtain on Mars.
| Table 1 - Significant fusion reactions that could be used for a fusion engine | |||||||||||||||
| Reaction | → | Products | Total
Energy release |
Atomic mass
of products | |||||||||||
| ratio | MeV/
fusion |
MeV/
fusion |
MeV/
fusion |
GJ/kg | N | ||||||||||
| 1 | D | + | T | → | He4 | 100% | 3.49 | + | n0 | 14.1 | 340 000 | 5 | |||
| 2a | D | + | D | → | He3 | 50% | 0.82 | + | n0 | 2.45 | 79 000 | 4 | |||
| 2b | + | D | → | T | 50% | 01.01 | + | p+ | 03.02 | 97 000 | 4 | ||||
| 3 | D | + | He3 | → | He4 | 100% | 3.6 | + | p+ | 14.7 | 353 000 | 5 | |||
| 4 | T | + | T | → | He4 | 100% | + | 2x n0 | 11.3 | 182 000 | 6 | ||||
| 5 | He3 | + | He3 | → | He4 | 100% | + | 2x p+ | 12.9 | 207 000 | 6 | ||||
| 6a | He3 | + | T | → | He4 | + | n+ | 12.1 | + | p+ | 195 000 | 6 | |||
| 6b | + | T | → | He4 | 4.8 | + | D | 9.5 | 230 000 | 6 | |||||
| 6c | + | T | → | He4 | 0.5 | + | n0 | 1.9 | + | p+ | 11.9 | 230 000 | 6 | ||
| 7a | D | + | Li6 | → | He4 | + | He4 | 22.4 | 270 000 | 8 | |||||
| 7b | + | Li6 | → | He3 | + | He4 | + | n0 | 2.56 | 31 000 | 8 | ||||
| 7c | + | Li6 | → | Li7 | + | p+ | 5.0 | 60 000 | 8 | ||||||
| 7d | + | Li6 | → | Be7 | + | n0 | 3.4 | 41 000 | 8 | ||||||
| 8 | p+ | + | Li6 | → | He4 | 1.7 | + | He3 | 2.3 | 55 000 | 7 | ||||
| 9 | n | + | Li6 | → | He4 | + | T | 2.3 | 7 | ||||||
| 10 | He3 | + | Li6 | → | He4 | + | He4 | + | p+ | 16.9 | 181 000 | 9 | |||
| 11 | p+ | + | B11 | → | 3x He4 | 8.7 | 70 000 | 12 | |||||||
| 12 | 6D | → | 2x He4 | 8.92 | + | 2n0 | 16.55 | + | 2p+ | 17.72 | 348 000 | 12 | |||
| Total energy release = sum of energies (MeV) x 1.6e-19 J/eV / mass of reaction products
Mass of reaction products = N x atomic mass unit (1.66e-27 kg) In a fusion reaction, the portion of the fuel that actually reacts is called the burnup fraction. As a first order of magnitude evaluation, we can establish that a fusion drive with a burnup fraction of 10%, using 1 kg of deuterium+he3 per second will produce 353 000 GJ/kg x 1kg/s x 0,1 = 35 300 GW of power (Line 4, table 1). To find the thrust from the fusion reaction, F=ṁ*2Pt*ηn . For an efficiency (ηn) of 80%, the force is 5 600 KN or 576 tonnes. To find the exhaust velocity Ve=F/ṁ= 56 500 000 N / 1 kg/s = 5 650 000 m/s | |||||||||||||||
