A simple fact worth pointing out:
First Light Fusion got its projectile to travel at 6.5KM per second and achieved some Fusion.
(1) Assume the mass of projectile is M, the kinetic energy required is 0.5*M*6.5*6.5 units.
= 21.125M units.
(2) If we use 2 projectiles in the Head-on Collision, the energy required is 2 times 0.5*M*3.25*3.25 = 10.5625M units.
(3) This means that the Input Energy required to achieve fusion in Head-on Collision is half that of Chasing Collision. The compression force is much higher as the pellet will not fly away.
(4) It is a trade-off of accurate timing of two projectiles hitting the pellet simultaneously (within one millionth of a second?). Such accuracy is within the capability of modern electronics.
I am more confident that we have a winner here.
Head-on Collision used half the energy
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Re: Head-on Collision used half the energy
Let us assume the projectile mass is 0.1kg, the kinetic energy of the two projectiles would be 0.1*6500*6500 Joules = 4,225,000 Joules.
The amount of mass to turn into energy will be 4,225,000/c/c (c= 299,792,458m/s)
= 7.0095e-10 kg
From the change in mass, determine the mass of the pellet.
a. Mass of Deuterium = 2.014 amu
b. Mass of tritium = 3.016 amu
c. Mass of Helium = 4.0026 amu
d. Mass of Neutron = 1.008665 amu
e. Mass before fusion - Mass after fusion = (2.014 + 3.016) - (4.0026 + 1.008665)
= 5.03 - 5.011265
= 0.018735 amu
f. Percentage change = (0.018735/5.03) x100% = 0.3710%
Amount of fuel = 7.0095e-10/0.3710% = 1.2671e-8 kg = 1.2671E-5 gm
This is extremely small! If we can use the same two projectiles to convert more fuel into energy, the Output Energy will easily be higher than Input Energy! This is a big jump in the Nuclear Fusion Field.
The amount of mass to turn into energy will be 4,225,000/c/c (c= 299,792,458m/s)
= 7.0095e-10 kg
From the change in mass, determine the mass of the pellet.
a. Mass of Deuterium = 2.014 amu
b. Mass of tritium = 3.016 amu
c. Mass of Helium = 4.0026 amu
d. Mass of Neutron = 1.008665 amu
e. Mass before fusion - Mass after fusion = (2.014 + 3.016) - (4.0026 + 1.008665)
= 5.03 - 5.011265
= 0.018735 amu
f. Percentage change = (0.018735/5.03) x100% = 0.3710%
Amount of fuel = 7.0095e-10/0.3710% = 1.2671e-8 kg = 1.2671E-5 gm
This is extremely small! If we can use the same two projectiles to convert more fuel into energy, the Output Energy will easily be higher than Input Energy! This is a big jump in the Nuclear Fusion Field.
Re: Head-on Collision used half the energy
(1) Assume (Data from Internet)each reaction (Fuel Pellet) will supply an average UK person per year. (4500 KWH)
a. https://www.bing.com/search?q=Electrici ... b7c15b006a
(2) Express that Energy in Joules. (1KWH =3600000 joules)
a. 4500 KWH = 3600000 * 4500 =16,200,000,000 Joules
(2) Change that from year to day (divide by 365)
=16,200,000,000/365 =4,383,561 Joules
Let us assume the projectile mass is 0.1kg, the kinetic energy of the two projectiles would be 0.1*6500*6500 Joules = 4,225,000 Joules. This translates to fuel mass of 1.2671E-5 gm
Similarly, 4,383,561 Joules translates to 1.2671E-5 * 4,383,561 / 4,225,000 =1.3146E-5 gm.
We need to have at least (1.2671E-5 + 1.3146E-5) = 2,5871E-5 grams of fuel take part in the fusion reaction. This should not be a problem.
a. https://www.bing.com/search?q=Electrici ... b7c15b006a
(2) Express that Energy in Joules. (1KWH =3600000 joules)
a. 4500 KWH = 3600000 * 4500 =16,200,000,000 Joules
(2) Change that from year to day (divide by 365)
=16,200,000,000/365 =4,383,561 Joules
Let us assume the projectile mass is 0.1kg, the kinetic energy of the two projectiles would be 0.1*6500*6500 Joules = 4,225,000 Joules. This translates to fuel mass of 1.2671E-5 gm
Similarly, 4,383,561 Joules translates to 1.2671E-5 * 4,383,561 / 4,225,000 =1.3146E-5 gm.
We need to have at least (1.2671E-5 + 1.3146E-5) = 2,5871E-5 grams of fuel take part in the fusion reaction. This should not be a problem.
Re: Head-on Collision used half the energy
Specific Heat of lithium = 3.6J/g k
c. Mass of lithium to absorb 4,383,561 joules and raise 1000 degrees Kelvin
= (4,383,561/3.6) /1000 grams = 1,212grams or 1.212 Kg.
This is well within the present technology. The Container need to contain a few Kilograms of Lithium to absorb the heat from the nuclear fusion and use that for the whole day.
The Pellet need to be changed only once a day. This can be done automatically.
c. Mass of lithium to absorb 4,383,561 joules and raise 1000 degrees Kelvin
= (4,383,561/3.6) /1000 grams = 1,212grams or 1.212 Kg.
This is well within the present technology. The Container need to contain a few Kilograms of Lithium to absorb the heat from the nuclear fusion and use that for the whole day.
The Pellet need to be changed only once a day. This can be done automatically.