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Fusion Propulsion
Fusion Propulsion
{| class="wikitable"
| colspan="16" rowspan="1" |Table 1 - Significant fusion reactions
|-
| colspan="4" rowspan="1" |Reaction
|→
| colspan="9" rowspan="1" |Products
|Total
Energy release
|Atomic mass of products
|-
|
| colspan="3" rowspan="1" |
|
|
|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
|-
| colspan="16" rowspan="1" |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
|}
[[Category:Propulsion]]
[[Category:Propulsion]]

Revision as of 14:52, 2 September 2021

Fusion Propulsion

Table 1 - Significant fusion reactions
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

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