Cyclic Fusion Reactor. Colliding Beams.

[chrismb]

Certainly I want more qualified criticism.

So I have given it. What of the electron energy budget? How many 80MeV electrons do you need to create for each fusion to take place?

[Joseph Chikva]

Certainly I want more qualified criticism.

So I have given it. What of the electron energy budget? How many 80MeV electrons do you need to create for each fusion to take place?

I am ready to discuss 33MeV electrons for the following example:
Deuterium+Tritium Reaction
• Deuterium – 300 keV
• Tritium – 200 keV
• Center-of-mass collision energy – 20.2 keV
• Reaction cross section – 0.4 barn
• Ions’ relative velocity – 1.8*10^6 m/s
• Electron – 33 MeV
• Relativistic factor of electrons – 65.6
For the frame of reference connected with Deuterium beam this corresponds to electron’s relativistic factor equal to 66.7 and for Tritium – 66.3)
For condition that ions in their frame of reference should be in negative potential well we need the less electron number dencity:
-for Deuterium - 1/4449
-for Tritium - 1/4395
Deuterium nuclei has velocity =0.018c and Tritium =0.012c and electron 0.9999....c

[chrismb]

The question is the energy budget - how many electrons do you need to generate for each fusion reaction?

[Joseph Chikva]

The question is the energy budget - how many electrons do you need to generate for each fusion reaction?

And for effective pinch the required current of electron beam should be more than only ~1/52.6 of current of Tritium beam and ~1/80 of Deuterium.
And from these reasonings we can accept the ratio between electron and Tritium and Deuterium currents equal to 1/50 and 1/75 correspondently.

The raw estimation of required energy that should to be put into the beams specified on a single fusion event (initial energy consumption of a single fusion event)
In case if only 80% of nuclei will react initially we should spend per one fusion event:

(300keV+1/75*33MeV+200keV+1/50*33MeV)/0.8=2MeV

[chrismb]

[]

[Joseph Chikva]

The question is the energy budget OF THE ELECTRONS - how many ELECTRONS do you need to generate for each fusion reaction THAT TAKES PLACE?

Have you a problem with arithmetic?
(1/50+1/75)*33MeV=1.1MeV

[chrismb]

No, sorry, it was comprehending your meaning, so I deleted the message after I comprehended it and decided I needed to re-phrase it, but you seem to be on line permanently and answered immediately before I deleted. I was expecting A ratio; electrons per fusion.

What is the argument you put forward for this ratio? What is the loss rate at the end of the device? Remember, 'current' and 'total population' are different for charges at different velocities.

[Joseph Chikva]

No, sorry, it was comprehending your meaning, so I deleted the message after I comprehended it, but you seem to be on line permanently and answered immediately before I deleted. I was expecting A ratio; electrons per fusion.

No problem.

[Joseph Chikva]

What is the argument you put forward for this ratio? What is the loss rate at the end of the device? Remember, 'current' and 'total population' are different for charges at different velocities.

Argument is simple: mentioned above number densities correspond to mentioned electron currents which should then be summarized.
Energy losses via braking radiation and also we should compensate the alignment of ions velocities. And in the same electric field Deuterium nuclei have higher rate of acceleration.

[Joseph Chikva]

For example we can build reactor into each impulse of which we will put the following energy:
• 3MJ for deuterium
• 2MJ for tritium
• 11MJ for electrons
And repeatability 10Hz
This is rather costly. Energy storage capacitors are very expensive. For example General Atomic produces high quality capacitors for these application and 2MJ costs about 20-25 millions USD.

[chrismb]

Argument is simple: mentioned above number densities correspond to mentioned electron currents which should then be summarized.
Energy losses via braking radiation and also we should compensate the alignment of ions velocities. And in the same electric field Deuterium nuclei have higher rate of acceleration.

But Argument doesn't provide all details!

You would need to know the residence time of an electron in your device to relate current with density. Very fast streams will give lower density for a given current. Slower streams have higher density for given current. How fast - and for how long (i.e. energy confinement time of electrons?).

What does analysis say for electron energy confinement time versus power output?

[Joseph Chikva]

Argument is simple: mentioned above number densities correspond to mentioned electron currents which should then be summarized.
Energy losses via braking radiation and also we should compensate the alignment of ions velocities. And in the same electric field Deuterium nuclei have higher rate of acceleration.

But Argument doesn't provide all details!

You would need to know the residence time of an electron in your device to relate current with density. Very fast streams will give lower density for a given current. Slower streams have higher density for given current. How fast - and for how long (i.e. energy confinement time of electrons?).

What does analysis say for electron energy confinement time versus power output?

Electron and ions or in other words non-neutral plasma have the same confinement time. And besides that they are in potential well of each other they also confined by external fields of betatron type device.
Or you are asking the feasible confinement time?
Betatron by its method of creating of circular electric field very similar to TOKAMAK as on both devices plasma torus and circulated beam considered as secondary winding of transformer.
We will have lower circulated current and much lower temperature. In TOKAMAKs confinement time has seconds order.
And we need about tens milliseconds as projected repeatability of device is 10Hz.

[hanelyp]

Electron and ions or in other words non-neutral plasma have the same confinement time.

Dubious claim, especially when electric potential wells are involved. Can you back it up?

Electric potential and cyclotron radius (related to mass/charge ratio) both have an impact on particle confinement time.

[Joseph Chikva]

Electron and ions or in other words non-neutral plasma have the same confinement time.

Dubious claim, especially when electric potential wells are involved. Can you back it up?

In case when electric potential wells are not involved confinement time equal to zero.

Electric potential and cyclotron radius (related to mass/charge ratio) both have an impact on particle confinement time.

Electric potential of what?

If impact
Cyclotron radius how? What do you think increasing radius we increase or decrease confinement time?

Feasible confinement time depends on development of instabilities. And not on nothing else. As the offered system cinematically (single particle’s behavior) is absolutely stable.
Strictly speaking the instabilities task can not be solved analytically. But on base of common experience and taking into account that the offered confinement concept is very similar to TOKAMAK and also the temperature in device is much lower than in TOKAMAK, we can wait that the feasible confinement time will not be less than in at least in TOKAMAKs (seconds order) and we need only milliseconds.

High relativistic factor of electrons should also add the stability as it has been proved by G.I.Budker.

[Joseph Chikva]

I just calculated energy consumption for the following parameters:
• Deuterium - 450keV
• Tritium - 300keV
• Electron - 40.6MeV
And energy consumption specified per a single fusion event is about 1.8MeV

Taking into account that collision energy in center-of-mass frame for this case is 30keV (30keV vs. 20.2keV) and so ratio between fusion and scattering sections is more attractive, we should also wait less further energy losses that for previous case estimated as 0.7MeV per each event.