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IZhO 2007, Теория — Задача 3. Nuclear fusion

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III International Zhautykov Olympiad — THEORETICAL COMPETITION, January 16, 2007 · 16 януари 2007 г.

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The time available for the theoretical competition is 4 hours. There are three questions.
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Условие

One of the most important problems of modern plasma physics is the so-called controlled nuclear fusion. The most promising way of achieving the controlled nuclear fusion is the TOKAMAK, that is the magnetic trap for confining hot plasma.

The most perspective fusion reaction is based on hydrogen isotopes that are deuterium (D) and tritium (T) and it is symbolically written as

D + T \to {}^{4}He + n \qquad (1)

whereas the reaction products are the helium isotope ${}^{4}He$ and neutron $n$ .

a) [1 point] Assuming the particle masses are known, determine, analytically and numerically, the energy $E_0$ , released in a single fusion reaction (1). [1 т.]

b) [0.5 points] In nuclear physics the energy is frequently counted in electron-volts (eV). One electron-volt is the energy that a single electron acquires passing an accelerating voltage of one volt. Calculate, analytically and numerically, the energy, found in question a), in electron-volts. [0,5 т.]

c) [1 point] Neglecting the initial energies of reacting particles, determine, analytically and numerically, the amounts of energy taken away by the helium nucleus $E_{He}$ and the neutron $E_n$ . [1 т.]

d) [1 point] In Tokamaks the deuterium-tritium mixture is heated up to a very high temperature, such that all the atoms are fully ionized (such a state is called a plasma) and their average kinetic energy is equal to $(3/2)k_BT$ . To make reaction (1) possible it is necessary to overcome the Coulomb repulsion of reacting nuclei and bring them close to a distance of $a = 10^{-14}\mathrm{m}$ where nuclei attraction forces prevail. Estimate, analytically and numerically, the temperature $T$ at which reaction (1) becomes possible for the majority of the colliding particles. [1 т.]

Fortunately, at thermal equilibrium the energies of particles are not even: some particles can have energy larger than $(3/2)k_BT$ while some might have less than $(3/2)k_BT$ . Due to the presence of energetic particles the fusion reaction can take place even at temperatures lower than determined in d). The number of fusion reactions taking place per unit of time per unit of volume is given by the expression $n_Dn_T\sigma(\mathrm{v})\mathrm{v}$ , where $n_D$ and $n_T$ are the deuterium and tritium concentrations, respectively, $\sigma(\mathrm{v})$ is the so-called cross section of the reaction, and $\mathrm{v}$ refers to the relative velocity of the particles participating in the reaction.

e) [3 points] Determine, analytically and numerically, the average reaction rate $< \sigma(\mathrm{v})\mathrm{v} >$ at the temperature $T_0 = 10^8\mathrm{K}$ , assuming that the effective velocity distribution is Maxwellian with the reduced mass $\mu = m_Dm_T/(m_D + m_T)$ . The dependence of $\sigma(\mathrm{v})$ is graphically represented in the Fig.3 below. [3 т.]

Graph of σ/σ0 (vertical axis, 0 to 5) against E, keV (horizontal axis, 0 to 200) on a grid: the curve rises from about 0 near 10 keV to a maximum of about 5.1 near 65–70 keV, then decreases to about 1 at 200 keV. — Fig.3. The cross section versus the relative energy of particles in the reaction expressed in $\mathrm{keV} = 10^3\mathrm{eV}$ . $\sigma_0 = 10^{-28}\mathrm{m}^2$

f) [3 points] The matter heated to such a high temperature is in the state of plasma and the atoms are fully ionized. On the average, the total density of nuclei equals the density of electrons to satisfy the neutrality charge. To heat up the matter to the optimum temperature $T_0 = 10^8\mathrm{K}$ it is necessary to put in per unit of time a certain amount of energy, which should not exceed the amount released in fusion reactions. The energy of helium nucleus can totally be used for power production while only $\eta = 0.3$ of the neutron energy can be utilized. The rate of loss of thermal energy is given by $P_{loss} = E/\tau$ , where $E$ is the energy density and $\tau$ is the time of plasma retention. Assuming that the particle concentrations are $n_D = n_T = n/2$ , obtain, analytically and numerically, the inequality that guarantees the positive production of energy in a controlled nuclear fusion and relates the plasma concentration $n$ and retention time $\tau$ . [3 т.]

g) [2.5 points] The plasma with temperature $T_0 = 10^8\mathrm{K}$ has a quite high pressure for a simple wall to sustain. A magnetic field, generated by a superconducting electromagnets of Tokamaks, is used to confine such matter. Magnetic field is used to compensate the thermal pressure. In the presence of magnetic field, the additional pressure due to magnetic field is $P = \gamma\mu_0^{\alpha}B^{\beta}$ , where $\gamma = 1/2$ , $\mu_0$ is the magnetic constant and $B$ is the magnetic field. Magnetic field is completely screened inside the plasma, so the additional pressure is absent there. Using dimensional analysis, find $\alpha$ and $\beta$ and determine, analytically and numerically, the magnetic field needed to confine the plasma with $n_D = n_T = n = 10^{20}\mathrm{m}^{-3}$ . [2,5 т.]

Numerical values:

Mass of deuterium $m_D = 3.34447\cdot 10^{-27}kg$ ,

Mass of tritium $m_T = 5.00732\cdot 10^{-27}kg$ ,

Mass of helium $m_{He} = 6.64432\cdot 10^{-27}kg$ ,

Mass of neutron $m_n = 1.67439\cdot 10^{-27}kg$ ,

Speed of light $c = 2.9979\cdot 10^{8}m/s$ ,

Elementary charge $e = 1.6022\cdot 10^{-19}C$ ,

Boltzmann constant $k_B = 1.3806\cdot 10^{-23}J/K$ ,

Electric constant $\varepsilon_0 = 8.854\cdot 10^{-12}\ F/m$ ,

Magnetic constant $\mu_0 = 1.257\cdot 10^{-6}\ H/m.$

Note:

The Maxwell distribution of the particle velocities with mass $\mu$ and temperature T has the following density function:

f(\mathrm{v}) = 4\pi\left(\frac{\mu}{2\pi k_BT}\right)^{3/2}\mathrm{v}^2\exp\left(-\frac{\mu\mathrm{v}^2}{2k_BT}\right)

Отговори

Решение

a) [1 point] In accordance with the Einstein formula the energy released in a single reaction is written as follows

E_0 = (m_D + m_T - m_{He} - m_n)c^2 = 2.973\cdot 10^{-12}\ J.

b) [0.5 points] The same energy is written in electron-volts as follows

E_0 = 2.973\cdot 10^{-12}/1.6022\cdot 10^{-19} = 1.856\cdot 10^{7}\ eV.

c) [1 point] If one can neglect the initial energies of reacting particles, the momentum and energy conservation laws immediately produce

E_n = \frac{m_{He}}{m_{He} + m_n} = 2.375\cdot 10^{-12}\ J = 1.482\cdot 10^{7}\ eV,

E_{He} = \frac{m_n}{m_{He} + m_n} = 5.984\cdot 10^{-13}J = 3.735\cdot 10^{6}\ eV.

d) [1 point] To overcome the Coulomb barrier nuclei should possess an appropriate thermal kinetic energy

\frac{3}{2}k_BT = \frac{e^2}{4\pi\varepsilon_0a}.

Thus

T = \frac{e^2}{6\pi\varepsilon_0ak_B} = 1.115\cdot 10^{9}\mathrm{K}.

e) [3 points] The reaction rate is given by an average over the Maxwellian distribution as follows

< \sigma(\mathrm{v})\mathrm{v} > = \int_0^{\infty}\mathrm{v}\sigma(\mathrm{v})f(\mathrm{v})d\mathrm{v}

with the Maxwellian distribution function

f(\mathrm{v}) = 4\pi\left(\frac{\mu}{2\pi k_BT}\right)^{3/2}\mathrm{v}^2\exp\left(-\frac{\mu\mathrm{v}^2}{2k_BT}\right).

Introducing the dimensionless value $x = \mu\mathrm{v}^2/2k_BT$ and substituting the latter expression into the previous one can obtain

< \sigma(\mathrm{v})\mathrm{v} > = \frac{2}{\sqrt{\pi}}\sigma_0\sqrt{\frac{2k_BT}{\mu}}\int_0^{\infty}x\exp(-x)\frac{\sigma}{\sigma_0}\left(x\frac{k_BT_0}{e10^3}\right)dx = \frac{2}{\sqrt{\pi}}\sigma_0\sqrt{\frac{2k_BT}{\mu}}\int_0^{\infty}f(x)dx.

The integrand function $f(x)$ is drawn with the aid of the cross section figure as follows

Graph of the integrand f(x) (vertical axis, 0 to 0.12) against x (horizontal axis, 0 to 20): zero up to about x = 1, a sharp peak of about 0.125 near x ≈ 3.5, then decaying to nearly zero by x ≈ 11 and staying at zero to x = 20.

Estimating the area under the line gives

< \sigma(\mathrm{v})\mathrm{v} > = 6.731\cdot 10^{-23}\frac{\mathrm{m}^3}{\mathrm{s}}.

f) [3 points] Plasma consists of electrons and ions and loses its thermal energy at time $\tau$ . The effective power of losses in the unit of volume is given by the expression

P_{losses} = \frac{3nk_BT}{\tau}.

This power should be less than that produced in fusion reactions (see e)

P_{fusion} = \frac{n^2}{4}< \sigma(\mathrm{v})\mathrm{v} >\left(E_{He} + \eta E_n\right).

Thus, $P_{fusion} > P_{losses}$

n\tau > \frac{12k_BT_0}{(E_{He} + \eta E_n)< \sigma(\mathrm{v})\mathrm{v} >} \approx 1.982\cdot 10^{20}\frac{\mathrm{s}}{\mathrm{m}^3}.

This is the so-called Lawson’s criterion.

g) [2.5 points] Dimensional analysis immediately produces $\alpha = -1$ and $\beta = 2$ . The thermal pressure of the plasma medium

P_{thermal} = 4nk_BT

should be compensated by the magnetic pressure from outside

P_{magnetic} = \frac{B^2}{2\mu_0}.

Thus, $P_{thermal} = P_{magnetic}$

B = 2\sqrt{2\mu_0nk_BT_0} = 1.178\,\mathrm{Ts}.

Оригинал в Архива: IZhO-2007-Theory_eng_sf.doc · официални решения: IZhO-2007-Theory_eng_sol_sf.doc