# PROBLEM OF CONTROLLED NUCLEAR FUSION

PROBLEM OF CONTROLLED NUCLEAR FUSION

Mechanical engineer

Contact to the author: garaf@ Inbox.ru

Work on the project of nuclear fusion is conducted around the world for nearly sixty years and is still not complete. The problem is that the implementation of the nuclear fusion reaction to a device in which a period of time necessary to keep the core in light of the atoms specified concentration with an energy of 100 keV. According to the Lawson criterion to apparatus for nuclear fusion was energetically favorable necessary that the concentration of particles in the product during its retention is not less than a certain value. Since the high-temperature plasma is not highly stable, this criterion is not satisfied.

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I propose to reflect on another circuit holding high-energy particles. The gist of it is this. Inside the vacuum chamber position 1 (see Figure 1) is placed (suspended) a negatively charged spherical kernel (hereinafter kernel) Pos.2 pos.3 via conductor which serves to hold the core within the chamber and for applying charge of the nucleus and retraction . At some distance from the core at an angle towards the core of about 90 degrees from a pre pos.4 accelerated particles are thrown from one species involved in the reaction with an initial velocity which is insufficient to hold it in orbit around pos.6 nucleus. In the reverse direction from the source device 5 take a form other particles participating in the reaction, ie, toward each other. As the velocity of the particles is not sufficient for a centrifugal particle retention in orbit, the particles under the influence of Coulomb attraction will start the spiral orbit closer to the nucleus. According to the law of conservation of angular momentum of the particle velocity as it approaches the nucleus will increase. When reaching a certain radius, and the centrifugal forces of attraction are balanced, and the orbit of the particles is nearly circular. You can pick up such a face-off conditions opposite of moving particles that orbit them practically coincide. In the case of inelastic collision of oppositely directed particle reaction occurs in their synthesis, which is what we need. In the case of elastic collisions scatter particles in opposite directions. Here there is a very important issue, not whether we are their losses. Suppose, for simplicity, the particles that we are trying to push the same (for example, the nuclei of deuterium atoms). In this case, it follows from the theory of impact upon collision, they will exchange their values velocities is not essential. But may change significantly their direction. From physics we know that the field of Coulomb forces (as well as gravity) particles may leave this field when the kinetic energy of the particle is equal to or exceeds the potential energy of attraction.

The condition of a retention of particles on a fixed circular orbit is the equality of forces of attraction and centrifugal force. Speed Hold on a circular orbit at 1.41 times less than the rate of overcoming the force of gravity. In this case, after the collision of the particle velocity and the distance to the kernel has not changed much, therefore, the particles can not come off from the core. But they will change their orbits. Since the direction of the velocity vectors of the particles after the collision can be arbitrary, the particle orbits are elliptical and will be rotated relative to the plane of its initial orbital plane. In an elliptical orbit has two extreme points: perihelion — the point of closest approach to the center of gravity (the center of the nucleus) and aphelion — the point of maximum distance from the center of attraction. For us is unacceptable orbit whose perihelion distance from the center of gravity is less than or equal to the radius of the kernel. We define the conditions under which this can happen and what is the probability of particle loss as a result of their falling into the nucleus. Suppose the particle rotated to a collision with another particle around the nucleus in a circular orbit poses. 1 at a speed VZ (See Figure 2). After the collision, the particle in question

• change in the velocity vector Ve1 or Ve2, Item 2 and the orbit becomes elliptical reaching perihelion in the nuclear surface. It should be noted that the orbits of the particles after the collision velocities Ve1 or Ve2, the same, so further discussion will communicate with the velocity Ve1. According to the law of conservation of angular momentum:

m • Ve1• H = m • VBishop• r, (1)

where m — mass of the particle,

H — Shoulder velocity vector Ve1,

VBishop — The speed of the particle at perihelion,

r — radius of the nucleus.

As

H = R • sin? ,

? — The angle between the line connecting the centers of nuclei and particles, and the vector Ve1.

Equation (1) can be transformed to:

VBishop= Ve1• (R? R) • sin?. (2)

In accordance with the law of conservation of energy the sum of kinetic and potential components of the total energy of the particle as it moves in the potential field remains the same:

?• m • VBishop2— k • (qQ? r) =? • m • Ve12— k • (qQ? R),

where k = 1 / (4?0) — The coefficient of proportionality?0 = 8,85 • 10-12F / m — dielectric constant,

q — the electric charge of the particle,

Q — electric charge of the nucleus,

After the conversion, taking into account (2)

?• m • Ve12[R2?
r2• sin2?-1] = k • qQ [1? R-1? R] (3)

Find the speed of the particle confinement conditions in a circular orbit:

m • VZ 2? R = k • (qQ? R2)

From the above reasoning

VZ ?
Ve1

And from here

m • Ve12 ? m • VZ 2 = K • (qQ? R)

Paste this expression into (3) and after conversion obtain

sin? = (r / R) • [(2R-r)? r]?

From the analysis of the last formula, we can conclude that the smaller the ratio r / R is less than that angle? And, consequently, the loss of particles due to drop them into the nucleus. Therefore, to reduce loss of particles necessary to core radius as small as possible. But on the other hand, to the nucleus to keep the largest number of particles, the nuclear charge is necessary to provide the highest, as the total charge of all held-core particles can not exceed the value of the charge on the nucleus itself. It follows that the kernel must have a maximum charge density.

To the fall of the particles in the nucleus, they are not lost in vain, the core can be coated with a layer of material in contact, which will stand out other useful charged particles that would be involved in the synthesis or a nuclear reaction occurs.

In order to reduce the loss of particles from falling into the nucleus can apply another method of supplying the particles into the chamber. It is to first apply one type of particle. Going on a close orbit and moving in the same direction of the particle due to the magnetic fields generated due to their motion, are attracted to each, other forming a dense torus-shaped harness. Then, the particles start to apply in reverse. The advantage of this method is that when "firing" tight rope probability of inelastic collisions of particles increases and decreases the probability of falling particles in the nucleus. Is then fed into chamber particles fed for the first time, then provide the particles in the opposite direction and so forth.

The proposed scheme enables retention of high particle synthesis reaction process control, as feed particles, and due to changes in the charge of the nucleus (with decreasing magnitude of the charge on the nucleus particles orbit radius increases and therefore their concentration varies and vice versa). If the amount of charge will fluctuate, the particle orbit will also vary, causing an intersection of the orbits of the particles in different directions and thus increase the probability of mutual collision. Moreover, because particles of elliptical orbits relativistic mass change particles during their movement along the orbit will precess, ie their orbits will rotate in the direction of rotation of the particles. This effect increases the probability of intersection of the orbits of the particles.

Garafutdinov AA