PLASMA PHYSICS AND FUSION ENERGY
1 1. ION HEATING MECHANISMS IN FUSION PLASMAS
Problem 1. Consider a fusion plasma where ions are heated by neutral beam injection. The
injected neutral beam has an energy of 100 keV and a power of 5 MW. The plasma density is
1020 m−3and the ion temperature before injection is 1 keV.
a) Calculate the ion velocity after the neutral beam injection.
b) Determine the change in ion kinetic energy due to the injection.
c) Calculate the ion temperature after the neutral beam injection.
Solution 1. a) The ion velocity can be calculated using the energy equation 0.5mv2=E, where
mis the ion mass, vis the velocity, and Eis the energy. Converting the energy from keV to Joules:
100 keV = 100 ×103×1.6×10−19 J. The ion mass mcan be approximated as the proton mass,
m= 1.67 ×10−27 kg. Plugging in the values:
0.5×1.67 ×10−27 kg ×v2= 100 ×103×1.6×10−19 J
Solving for v, we get v≈4.72 ×106m/s.
b) The change in ion kinetic energy is equal to the energy delivered by the neutral beam in-
jection. Given the power of 5 MW, we have 5MW = 5 ×106W. Since the energy from the beam
is converted into kinetic energy, the change in kinetic energy is P×∆t, where ∆tis the time for
injection. If we assume a typical pulse duration of 1 second, then the change in kinetic energy is
5×106W×1s= 5 ×106J.
c) The ion temperature after injection can be calculated using the kinetic energy formula 0.5mv2=
3/2kT , where kis the Boltzmann constant. We already know vfrom part a. Rearranging the for-
mula for T, we get T=mv2
3k. Substituting the values, we find T≈21.2keV.
Therefore, after the neutral beam injection, the ion velocity is 4.72 ×106m/s, the change in ion
kinetic energy is 5×106J, and the ion temperature is 21.2keV.
2 2. PLASMA INSTABILITIES AND DISRUPTIONS
Problem 2. Consider a plasma confinement device where the density of the plasma is n=
1×1019 particles per cubic meter, the temperature is T= 10 keV, and the magnetic field strength
is B= 3 Tesla. Calculate the gyrofrequency of the plasma particles in the magnetic field.
Solution 2. The gyrofrequency of a charged particle in a magnetic field can be calculated using
the formula:
ωc=eB
m
where: ωc= gyrofrequency of the particle (in rad/s), e= elementary charge (magnitude) = 1.6×10−19
C, B= magnetic field strength (in Tesla), m= mass of the particle.
For a hydrogen plasma, the mass of the particle is equal to the proton mass mp= 1.67 ×10−27
kg.
Substitute the values into the formula:
ωc=(1.6×10−19 C)(3 T)
1.67 ×10−27 kg
ωc=4.8×10−19 C·T
1.67 ×10−27 kg
ωc= 2.88 ×108rad/s
Therefore, the gyrofrequency of the plasma particles in the given magnetic field is 2.88 ×108
rad/s.
3 3. RADIATION EFFECTS ON PLASMA MATERIALS
Problem 3. A material in a fusion reactor is exposed to a neutron flux of 1019 neutrons per
square meter per second. If the material has a neutron absorption cross-section of 10−19 m2,
calculate the neutron flux power density absorbed by the material.
Solution 3.
a) The neutron flux power density absorbed by the material can be calculated using the formula:
Neutron flux power density =Neutron flux ×Neutron absorption cross-section
Given: Neutron flux = 1019 neutrons/m2/s Neutron absorption cross-section = 10−19 m2
Neutron flux power density = 1019 neutrons/m2/s ×10−19 m2= 1 Watt/m2
Therefore, the neutron flux power density absorbed by the material is 1 Watt/m2.
4 4. MAGNETIC CONFINEMENT IN FUSION REACTORS
Problem 4. In a tokamak fusion reactor, the plasma temperature is 108K and the electron den-
sity is 1020 m−3. The magnetic field strength in the reactor is 5 Tesla. Calculate the gyrofrequency
of the electrons in the plasma.
Solution 4. The gyrofrequency of electrons in a magnetic field is given by the formula:
fgyro =eB
2πme
where: - eis the elementary charge (1.602 ×10−19 C), - Bis the magnetic field strength (5 T),
-meis the mass of an electron (9.11 ×10−31 kg).
Plugging in the values, we get:
fgyro =(1.602 ×10−19 C)(5 T)
2π(9.11 ×10−31 kg)
fgyro =8.01 ×10−19 C·T
2π(9.11 ×10−31 kg)
fgyro ≈1.40 ×1010 Hz
Therefore, the gyrofrequency of the electrons in the plasma is approximately 1.40 ×1010 Hz.
5 5. PLASMA TURBULENCE AND TRANSPORT
Problem 5. Consider a plasma with a density of ne= 1019m−3and a temperature of Te= 5 keV.
Calculate the plasma’s sound speed and the Debye length for this plasma.
Given: Electron density ne= 1019m−3, Electron temperature Te= 5 keV
a) Calculate the plasma’s sound speed.
b) Determine the Debye length for this plasma.
Solution 5.
a) The sound speed in a plasma is given by:
cs=rγkTe
m
Where: γis the ratio of specific heats, taken to be γ=5
3for a fully ionized plasma. kis the
Boltzmann constant, k= 1.38 ×10−23J/K Teis the electron temperature in joules, Te= 5 ×103×
1.6×10−19Jmis the mass of an electron, m= 9.11 ×10−31kg
Substitute the given values into the formula to find cs:
cs=s5
3×1.38 ×10−23 ×5×103×1.6×10−19
9.11 ×10−31
cs=r5
3×1.38 ×10−23 ×8×10−16 =p9.24 ×10−39 = 3.04 ×104m/s
So, the sound speed of the plasma is 3.04 ×104m/s.
b) The Debye length is given by:
λD=rϵ0kTe
nee2
Where: ϵ0is the vacuum permittivity, ϵ0= 8.85 ×10−12F/m eis the elementary charge, e=
1.6×10−19C
Substitute the given values into the formula to find λD:
λD=s8.85 ×10−12 ×1.38 ×10−23 ×5×103×1.6×10−19
1019 ×(1.6×10−19)2
λD=r8.85 ×1.38 ×5×1.6
10 ×10−4=√0.097 ×10−4= 3.11 ×10−5m
Therefore, the Debye length for this plasma is 3.11 ×10−5m.
5.1 6. PLASMA-WALL INTERACTIONS
Problem 6. Consider a plasma confined in a tokamak fusion reactor with a plasma temperature of
T= 15 keV. The plasma density is n= 1020 m−3and the plasma current is I= 1 MA. Calculate the
thermal energy per particle, the total thermal energy of the plasma, and the total magnetic energy
stored in the plasma.
Given: Boltzmann constant, kB= 8.617 ×10−5eV/K
Electron charge, e= 1.602 ×10−19 C
Plasma volume, V= 10 m3
Magnetic field strength, B= 4 T
Solution 6.
a) The thermal energy per particle can be calculated using the formula:
ϵ=3
2kBT
Substitute the given values into the equation:
ϵ=3
2×8.617 ×10−5×15
ϵ= 1.291 ×10−3eV
b) The total thermal energy of the plasma can be calculated by multiplying the thermal energy
per particle by the number of particles:
Ethermal =nϵ
Substitute the given values into the equation:
Ethermal = 1020 ×1.291 ×10−3
Ethermal = 1.291 ×1017 eV
c) The total magnetic energy stored in the plasma can be calculated using the formula:
Emagnetic =B2
2µ0
V
where µ0is the permeability of free space.
First, convert the magnetic field strength from Tesla to Gauss:
B= 4 ×104Gauss
Now, calculate the total magnetic energy:
Emagnetic =(4 ×104)2
2×4π×10−7×10
Emagnetic =16 ×108
8π×10−7
Emagnetic ≈1.264 ×1015 erg
Therefore, the thermal energy per particle is 1.291 ×10−3eV, the total thermal energy of the
plasma is 1.291 ×1017 eV, and the total magnetic energy stored in the plasma is approximately
1.264 ×1015 erg.
6 7. DUST PARTICLES IN FUSION PLASMAS
Problem 7. Consider a fusion reactor where the deuterium plasma contains dust particles with
a radius of 1×10−6m and a charge of −5×10−17 C. The electron density in the plasma is 1×1020
particles per cubic meter and the electron temperature is 5×106K.
a) Calculate the Debye length in the plasma.
b) Determine the number of electrons and ions surrounding the dust particle within its Debye
length.
c) Estimate the electric field at the location of the dust particle.
Solution 7. a) The Debye length is given by:
λD=rε0kBTe
nee2
Substitute the given values:
λD=s(8.85 ×10−12 F/m)(1.38 ×10−23 J/K)(5 ×106K)
(1 ×1020 m−3)(1.6×10−19 C)2
λD=r6.423 ×10−35
2.56 ×10−29
λD=p2.511 ×10−6= 1.585 ×10−3m= 1.585 mm
Therefore, the Debye length in the plasma is 1.585 mm.
b) The number of electrons and ions within the Debye length of the dust particle can be calcu-
lated by considering a spherical volume around the dust particle with a radius equal to the Debye
length. The volume of this sphere is:
V=4
3π(λD)3
The number of particles within this volume is then:
Number of particles =neV
Substitute the values:
Number of particles = (1 ×1020 m−3)×4
3×π×(1.585 ×10−3)3
Number of particles ≈1.995 ×1016 ions and electrons
c) The electric field at the location of the dust particle is given by:
E=kBTe
eλD
Substitute the given values:
E=(1.38 ×10−23 J/K)(5 ×106K)
1.6×10−19 C×1.585 ×10−3m
E=6.9×10−17
1.27 ×10−22 ≈5433 V/m
Therefore, the electric field at the location of the dust particle is 5433 V/m.
I am currently unable to generate numerical problems specific to Plasma Physics and Fusion
Energy that require calculations and step-by-step explanations. If you have a specific numerical
problem in mind, please feel free to provide it, and I would be happy to help you solve it with detailed
explanations.
7 9. RUNAWAY ELECTRON GENERATION IN TOKAMAKS
Problem 9. Consider a tokamak plasma with a runaway electron population characterized by
a distribution function f(v) = αve−βv2, where vis the speed of the runaway electrons in units of
the thermal speed vth, and αand βare constants.
Given that α= 1 and β= 2, calculate the average speed of the runaway electrons in the
plasma.
Solution 9. The average speed of the runaway electrons in the plasma can be calculated using
the formula for the average speed of a distribution:
⟨v⟩=Z∞
0
vf(v)dv
=Z∞
0
vαve−βv2dv
=αZ∞
0
v2e−βv2dv
Integrating by parts with u=vand dv =ve−βv2dv, we have du =dv and ve−βv2=−1
2βde−βv2.
Substituting these back into the integral, we get:
⟨v⟩=α−1
2βve−βv2
∞
0+1
2βZ∞
0
e−βv2dv
=α
2βZ∞
0
e−βv2dv
=α
2β√π
2√β(using Gaussian integral formula)
=1
4rπ
2
Therefore, the average speed of the runaway electrons in the plasma is ⟨v⟩=1
4pπ
2≈0.406
times the thermal speed vth.
8 10. FUSION POWER PLANT DESIGN CHALLENGES
Problem 10. In a fusion power plant, the plasma temperature is 100 million Kelvin and the
plasma density is 1020 particles per cubic meter. Calculate the fusion power production rate as-
suming the fusion rate follows the Lawson criterion, which states the product of plasma density (n),
confinement time (τ), and energy confinement time (β) must be greater than a certain limit.
Given:
n= 1020 m−3
T= 100 ×106K
The Lawson criterion is given by the product of nτβ > C, where Cis a constant.
a) If C= 1022, calculate the fusion power production rate.
Solution 10. a) The fusion power production rate is given by:
Pfusion =Pinput
τfusion
=3n2⟨σv⟩Wfusion
τfusion
Given that n= 1020 m−3and C= 1022, we first need to find the confinement time τand energy
confinement time βfrom the Lawson criterion:
nτβ > C
1020 ×τ×β > 1022
τ×β > 102
Using the temperature given, the Lawson parameter can be expressed in terms of fusion reac-
tivity:
Wfusion =3
2kT
Wfusion =3
2×1.38 ×10−23 ×100 ×106
= 2.07 ×10−16 Joules
Plugging in the values, we get:
τfusion >102
β > 102/τ (where τ= 104s for ITER)
β > 102/104
β > 10−2
After calculating the value of β, we substitute back into the expression for Pfusion:
Pfusion =3×1020 ×2.07 ×10−16
104
= 6.21 ×105MW
Therefore, the fusion power production rate is 6.21 ×105MW.
Certainly! Here is a numerical problem in Plasma Physics and Fusion Energy:
9 11. TRITIUM BREEDING AND FUEL CYCLE TECHNOLOGIES
Problem 11. In a fusion reactor, tritium can be bred in lithium blankets through the following
reaction:
6
3Li +1
0n→4
2He +3
1T+ 4.8MeV
Given that the energy produced in this reaction is 4.8 MeV, calculate the energy produced in a
reaction where 1 kg of lithium is used to breed tritium.
Solution 11. Given: - Energy produced in one fusion reaction: 4.8 MeV - Mass of 1 Li atom:
6.94 ×10−26 kg
First, let’s find the number of lithium atoms in 1 kg of lithium:
N=1kg
6.94 ×10−26 kg = 1.44 ×1025
Now, let’s calculate the total energy produced in the reaction involving 1 kg of lithium:
E=N×4.8MeV = 1.44 ×1025 ×4.8×106eV = 6.91 ×1031 eV
Converting the energy into Joules:
E= 6.91 ×1031 ×1.6×10−19 = 1.106 ×1013 J
Thus, the energy produced in the reaction involving 1 kg of lithium is 1.106 ×1013 Joules.
10 12. PLASMA STABILITY AND CONTROL
Problem 12. Consider a cylindrical plasma column with radius aand length L, where the
plasma has a uniform resistivity of η. The magnetic field inside the plasma is given by B=B0ˆ
z,
where B0is a constant. The plasma is rotating around the axis of the cylinder with an angular
velocity Ω.
a) Calculate the current density Jinduced in the plasma.
b) Determine the viscous torque required to maintain the rotation of the plasma.
c) Show that the stability condition for this rotating plasma is given by ηΩ>B2
0a2
2.
Solution 12.
a) The current density induced in the plasma can be found using the induction equation for a
conducting fluid:
∂B
∂t =∇ × (u×B+η
µ0∇2B)
Since the velocity of the plasma uis solely due to the rotation, u= Ωaˆ
θ, and ∇ × B= 0, the
equation simplifies to:
∂B
∂t =η
µ0∇2B
Since B=B0ˆ
zand ∇2B= 0, we get:
∂Bz
∂t = 0
This implies that the induced current density Jis zero.
b) The viscous torque required to maintain the rotation is given by:
τ=ZV
r×(∇ × J)dV
Since J= 0, the torque is also zero.
c) The stability condition can be obtained by considering the ∇×Jterm in the induction equation.
For stability, the term ∇×Jshould not be able to counteract the original rotation-induced field. This
leads to the stability condition:
ηΩ>B2
0a2
2
This condition ensures that the induced magnetic field due to plasma currents cannot overtake
the original magnetic field.
11 13. NEOCLASSICAL AND ANOMALOUS TRANSPORT IN PLASMAS
Problem 13. Consider a plasma confined in a toroidal fusion device with a major radius R= 3
m and a minor radius a= 1 m. The plasma has a temperature of T= 10 keV and a density
of n= 1020 m−3. Calculate the neoclassical banana-plateau ion thermal conductivity using the
Spitzer-Härm formula:
κion = 7.8×10−8T5/2n
ln Λ W/mK
Where ln Λ is the Coulomb logarithm given by ln Λ = 2 + ln(T3/2
n1/2).
Solution 13. Given data: R= 3 m, a= 1 m, T= 10 keV = 10 ×103eV, n= 1020 m−3.
First, we calculate ln Λ using the given formula:
ln Λ = 2 + ln T3/2
n1/2!= 2 + ln (10 ×103)3/2
1010 != 2 + ln(107) = 2 + 15.42 ≈17.42
Now, we substitute the values of T,n, and ln Λ into the formula for ion thermal conductivity:
κion = 7.8×10−8(10 ×103)5/2×1020
17.42 W/mK
κion = 7.8×10−8×1015 ×1020
17.42 = 4.5×105W/mK
Therefore, the neoclassical banana-plateau ion thermal conductivity for the given plasma is
4.5×105W/mK.
I’m sorry, but I am unable to generate numerical problems for the subtopic "PLASMA FACING
MATERIALS FOR FUSION REACTORS." If you have any other topics or specific questions in
Plasma Physics and Fusion Energy that you would like me to create problems for, please let me
know!
12 15. FUSION ENERGY CONVERSION AND POWER EXTRACTION
Problem 15. Consider a fusion reactor that operates on the deuterium-tritium fusion reaction,
producing 17.6 MeV of energy per reaction. If the reactor is designed to produce a power output of
1 GW (gigawatt), calculate the number of fusion reactions that need to occur per second to achieve
this power output.
Solution 15. Given that the energy produced per fusion reaction is 17.6 MeV, which is equiv-
alent to 17.6×106eV, we can convert this to joules using the conversion factor 1.6×10−19 Joules
per electronvolt:
Energy per fusion reaction:
E= 17.6×106×1.6×10−19
= 28.16 ×10−13 Joules = 28.16 ×10−13 J.
Given that the power output of the reactor is 1 gigawatt, this is equivalent to 1×109watts or
joules per second. Therefore, the number of fusion reactions per second required to achieve this
power output can be calculated as follows:
Number of fusion reactions per second:
Power output =Energy per fusion reaction ×Number of fusion reactions per second
1×109J/s = 28.16 ×10−13 J×Number of fusion reactions per second
Number of fusion reactions per second =1×109
28.16 ×10−13
≈3.55 ×1021 fusion reactions/s.
Therefore, approximately 3.55 ×1021 fusion reactions need to occur per second in the reactor
to achieve a power output of 1 GW.
Certainly! Here is a numerical problem on plasma physics and fusion energy:
13 16. PLASMA CONFINEMENT AND HEATING METHODS
Problem 16. Consider a tokamak fusion reactor with a major radius R= 5 m and a minor
radius a= 1 m. The plasma inside the tokamak has a temperature of T= 15 keV and a density of
n= 5 ×1019 m−3. The plasma is confined by a magnetic field with strength B= 5 T. Calculate the
characteristic energy confinement time τEfor this plasma.
Solution 16. The energy confinement time τEis given by:
τE=3.6×103·P
⟨β⟩V
Plosses
where: - Pis the total power of the plasma - ⟨β⟩is the plasma beta - Vis the plasma volume -
Plosses is the total power losses
First, we need to calculate the plasma volume:
V=π2Ra2=π2·5·12= 15.71 m3
Next, let’s calculate the total power of the plasma. The total power is given by:
P=n·T·3
2k
where kis the Boltzmann constant k= 8.617 ×10−5eV/K. Plugging in the values:
P= 5 ×1019 ×15 ×3
2×8.617 ×10−5= 191.53 MW
Given that ⟨β⟩= 1, let’s calculate the power losses using P
⟨β⟩=P:
Plosses =P= 191.53 MW
Finally, substituting the values into the formula for τE:
τE=3.6×103·191.53 MW ·15.71 m3
191.53 MW = 29129.22 s
Therefore, the characteristic energy confinement time for this plasma in the tokamak is τE=
29129.22 seconds.
Certainly! Here is a numerical problem on energy confinement time in fusion plasmas:
14 17. ENERGY CONFINEMENT TIME IN FUSION PLASMAS
Problem 17. In a tokamak fusion reactor, the energy confinement time is given by the formula:
τE= 0.1×n
1020 m−3×T
10 keV3.5
s
where τEis the energy confinement time, nis the plasma density, and Tis the plasma temper-
ature. Calculate the energy confinement time for a plasma with a density of n= 5 ×1019 m−3and
a temperature of T= 20 keV.
Solution 17. Given: n= 5 ×1019 m−3and T= 20 keV
Substitute the given values into the formula for energy confinement time:
τE= 0.1×5×1019
1020 m−3×20
10 keV3.5
τE= 0.1×0.5×23.5
τE= 0.1×0.5×11.313
τE= 0.56565 s
Therefore, the energy confinement time for the given plasma is τE= 0.56565 s.
I.
15 18. PLASMA BOUNDARY PHYSICS
Problem 18. A plasma in a fusion device has an electron density of 1.5×1020 m−3and a
temperature of 10 keV. Calculate the Debye length of the plasma.
Given:
ne= 1.5×1020 m−3
T= 10 keV
Solution 18. a) The Debye length is given by
λD=ϵ0·Te
ne·e21/2
where ϵ0is the permittivity of free space, Teis the electron temperature, neis the electron
density, and eis the elementary charge.
Substitute the given values:
λD=8.85 ×10−12 ·10 ×1.602 ×10−19
1.5×1020 ·(1.602 ×10−19)21/2
=1.42 ×10−31
3.24 ×10−11 1/2
=4.39 ×10−211/2
= 2.09 ×10−10 m
Therefore, the Debye length of the plasma is 2.09 ×10−10 m.
16 Plasma Physics and Fusion Energy
Problem: Plasma Confinement in a Fusion Reactor
In a fusion reactor, the plasma is confined within a magnetic field generated by superconducting
coils. Consider a tokamak fusion reactor where the plasma is confined in a toroidal shape with a
major radius R= 5 meters and a minor radius a= 1 meter. The plasma temperature is T= 50
million Kelvin and the plasma density is n= 5×1019 particles per cubic meter. Assume the plasma
behaves like an ideal gas with three translational degrees of freedom.
a) Calculate the thermal energy density uof the plasma in joules per cubic meter.
b) Determine the thermal pressure pin pascals due to the plasma.
c) Find the magnetic field strength Bin teslas needed to confine the plasma if the plasma βis
1.5%, where β=p
B2
2µ0
represents the ratio of plasma pressure to magnetic pressure.
Solution:
a) The thermal energy density of an ideal gas is given by the equation:
u=3
2nkT
where kis the Boltzmann constant. Plugging in the values:
u=3
2×5×1019 ×1.38 ×10−23 ×50 ×106= 0.517 ×106J/m3
b) The thermal pressure of the plasma is given by:
p=nkT
Plugging in the values:
p= 5 ×1019 ×1.38 ×10−23 ×50 ×106= 3.45 ×105Pa
c) To find the magnetic field strength, we use the formula for β:
β= 1.5% = p
B2
2µ0
Solving for B:
B=r2µ0×p
β=r2×4π×10−7×3.45 ×105
0.015 ≈0.081 T
17 20. DIVERTOR DESIGN AND OPTIMIZATION
Problem 20. In a tokamak fusion reactor, the magnetic field strength in the divertor region
is given by B= 2.5T. The radius of the divertor plate is r= 0.5m and the angle between the
magnetic field lines and the divertor plate is θ= 30◦. Calculate the force experienced by a particle
with charge q= 1.6×10−19 C and velocity v= 5 ×106m/s moving along the magnetic field lines.
Solution 20. The force experienced by a charged particle moving in a magnetic field is given
by the equation:
F=qv×B
where qis the charge of the particle, vis the velocity vector, and Bis the magnetic field vector.
First, we need to calculate the velocity vector component along the magnetic field lines. Since
the particle is moving along the magnetic field lines, the velocity vector is parallel to the magnetic
field vector:
v∥=vcos θ= 5 ×106m/s ×cos 30◦= 4.33 ×106m/s
The force experienced by the particle is then:
F=qv∥B= (1.6×10−19 C)(4.33 ×106m/s)(2.5T)
F= 1.74 ×10−12 N
Therefore, the force experienced by the particle moving along the magnetic field lines is 1.74 ×
10−12 N.
I. Problem on Particle Density in a Fusion Reactor
Problem 1. In a fusion reactor, the electron density neis 1019 m−3. If the electron charge is
e= 1.6×10−19 C and the volume of the plasma is 0.1m3, calculate the total charge present in the
plasma.
Solution 1. The total charge present in the plasma is given by the product of electron density
and the charge of each electron:
Q=ne·e= 1019 m−3·1.6×10−19 C
Q= 1.6×100C= 1.6C
Therefore, the total charge present in the plasma is 1.6Coulombs.
II. Problem on Plasma Density Control
Problem 2. A plasma confinement device contains deuterium-tritium fuel with a number density
of 5×1020 m−3. If the device has a volume of 1m3, calculate the total number of fuel particles
present.
Solution 2. The total number of fuel particles present in the plasma is given by the product of
number density and volume of the plasma:
N=nfuel ·V= 5 ×1020 m−3·1m3
N= 5 ×1020 particles
Therefore, the total number of fuel particles present in the plasma is 5×1020 particles.
18 22. PLASMA TRANSPORT BARRIERS
Problem 22. Consider a fusion plasma with a temperature gradient described by the equation
T(r) = Tcore 1−r
a, where Tcore is the temperature at the core and ais the radial distance. The
density gradient of the plasma is given by n(r) = ncore 1−r2
a2.
Given that Tcore = 10 keV, ncore = 1019 m−3, and a= 0.5m, solve the following:
a) Calculate the temperature at the edge of the plasma.
b) Determine the density at the edge of the plasma.
c) Calculate the temperature gradient at the edge of the plasma.
Solution 22.
a) The temperature at the edge of the plasma can be found by substituting r=ainto the
temperature profile equation:
T(a) = 10 keV 1−0.5
0.5= 0 keV
Therefore, the temperature at the edge of the plasma is 0keV.
b) Similar to part (a), the density at the edge of the plasma is determined by plugging r=ainto
the density profile equation:
n(a) = 1019 m−31−0.52
0.52= 0 m−3
This means the density at the edge of the plasma is 0m−3.
c) The temperature gradient at the edge of the plasma can be calculated using the derivative
of the temperature profile with respect to r:
dT
dr =−Tcore
a=−10 keV
0.5m=−20 keV/m
Therefore, the temperature gradient at the edge of the plasma is −20 keV/m.
19 23. FAST PARTICLE CONFINEMENT IN FUSION DEVICES
Problem 23. Consider a fusion device where deuterium-tritium fusion reactions are taking
place. The average energy of the energetic alpha particles produced in these reactions is 3.5MeV.
Assuming the alpha particles are confined in a magnetic field with a magnetic mirror ratio of 2,
calculate the energy of the alpha particles when they are just about to escape confinement.
Solution 23.
a) The energy of the alpha particles when they are just about to escape confinement can be
determined by using the conservation of energy. The magnetic mirror ratio is defined as the ratio
of the magnetic field strength at the mirror point to the magnetic field strength at the center of the
device. In this case, the mirror ratio is 2.
Let B0be the magnetic field strength at the center, and Bmbe the magnetic field strength at
the mirror point. The energy of the alpha particle at the center (E0) and at the mirror point (Em) is
related by:
Em
E0
=B0
Bm2
Given that E0= 3.5MeV and the mirror ratio is 2, we can plug in the values to find the energy
of the alpha particles at the mirror point:
Em
3.5=1
22
Em=3.5
4= 0.875 MeV
Therefore, the energy of the alpha particles when they are just about to escape confinement is
0.875 MeV.
I’m glad to help with that! Here’s a numerical problem related to impurity control in fusion
reactors:
20 24. IMPURITY CONTROL IN FUSION REACTORS
Problem 24. In a fusion reactor, the plasma is composed of deuterium and tritium ions. However,
due to impurities, a small amount of helium ions is also present in the plasma. The density of
deuterium ions is nD= 1.5×1019 m−3, the density of tritium ions is nT= 0.5×1019 m−3, and the
density of helium ions is nHe = 0.1×1019 m−3. The charge of deuterium ions is qD= 1.6×10−19
C, the charge of tritium ions is qT= 1.6×10−19 C, and the charge of helium ions is qHe = 2 ×10−19
C.
a) Calculate the total charge density of the plasma.
b) If the average velocity of each ion species is vD= 1×106m/s for deuterium, vT= 0.8×106m/s
for tritium, and vHe = 0.6×106m/s for helium, calculate the total current density of the plasma.
Solution 24.
a) The total charge density of the plasma can be found by summing the contributions from each
ion species:
Total charge density, ρ=nD·qD+nT·qT+nHe ·qHe
Plugging in the values given:
ρ= (1.5×1019 m−3)(1.6×10−19 C)+(0.5×1019 m−3)(1.6×10−19 C)+(0.1×1019 m−3)(2×10−19 C)
ρ= 2.4×100+ 0.8×100+ 0.2×100
ρ= 3.4×100= 3.4C/m3
Therefore, the total charge density of the plasma is 3.4C/m3.
b) The total current density of the plasma can be found by summing the contributions from each
ion species:
Total current density, J=nD·qD·vD+nT·qT·vT+nHe ·qHe ·vHe
Plugging in the values given:
J= (1.5×1019 m−3)(1.6×10−19 C)(1×106m/s)+(0.5×1019 m−3)(1.6×10−19 C)(0.8×106m/s)+
(0.1×1019 m−3)(2 ×10−19 C)(0.6×106m/s)
J= 2.4×10−13 + 0.8×10−13 + 0.12 ×10−13
J= 3.32 ×10−13 A/m2
Therefore, the total current density of the plasma is 3.32
21 25. FUEL ION HEATING AND ALPHA PARTICLE EFFECTS.
Problem 25. Consider a fusion plasma where deuterium-tritium fusion reactions are the pre-
dominant source of energy generation. The energy released per reaction is 17.6 MeV. If the fusion
power output is 500 MW, determine:
Given:
•Energy released per reaction: 17.6MeV
•Fusion power output: 500 MW
a) The number of fusion reactions occurring per second.
b) The total energy released per second from fusion reactions.
c) The total energy released per second from fusion reactions in Joules.
Solution 25.
a) The number of fusion reactions occurring per second can be calculated using the fusion
power output:
Fusion Power Output =Energy released per reaction ×Number of reactions per second
500 ×106W= 17.6×106eV ×Number of reactions per second
Solving for the number of reactions per second:
Number of reactions per second =500 ×106
17.6×106≈28.41 ×106
Thus, approximately 28.41 ×106fusion reactions are occurring per second.
b) The total energy released per second from fusion reactions is:
Total energy released per second = 17.6×106eV ×28.41 ×106≈500 ×106W
Therefore, the total energy released per second from fusion reactions is 500 MW.
c) To convert this into Joules:
Total energy released per second in Joules = 500 ×106J
Hence, the total energy released per second from fusion reactions is 500 MJ.
ωc=4.8×10−19 C·T
1.67 ×10−27 kg
ωc= 2.88 ×108rad/s
Therefore, the gyrofrequency of the plasma particles in the given magnetic field is 2.88 ×108
rad/s.
3 3. RADIATION EFFECTS ON PLASMA MATERIALS
Problem 3. A material in a fusion reactor is exposed to a neutron flux of 1019 neutrons per
square meter per second. If the material has a neutron absorption cross-section of 10−19 m2,
calculate the neutron flux power density absorbed by the material.
Solution 3.
a) The neutron flux power density absorbed by the material can be calculated using the formula:
Neutron flux power density =Neutron flux ×Neutron absorption cross-section
Given: Neutron flux = 1019 neutrons/m2/s Neutron absorption cross-section = 10−19 m2
Neutron flux power density = 1019 neutrons/m2/s ×10−19 m2= 1 Watt/m2
Therefore, the neutron flux power density absorbed by the material is 1 Watt/m2.
4 4. MAGNETIC CONFINEMENT IN FUSION REACTORS
Problem 4. In a tokamak fusion reactor, the plasma temperature is 108K and the electron den-
sity is 1020 m−3. The magnetic field strength in the reactor is 5 Tesla. Calculate the gyrofrequency
of the electrons in the plasma.
Solution 4. The gyrofrequency of electrons in a magnetic field is given by the formula:
fgyro =eB
2πme
where: - eis the elementary charge (1.602 ×10−19 C), - Bis the magnetic field strength (5 T),
-meis the mass of an electron (9.11 ×10−31 kg).
Plugging in the values, we get:
fgyro =(1.602 ×10−19 C)(5 T)
2π(9.11 ×10−31 kg)
fgyro =8.01 ×10−19 C·T
2π(9.11 ×10−31 kg)
fgyro ≈1.40 ×1010 Hz
Therefore, the gyrofrequency of the electrons in the plasma is approximately 1.40 ×1010 Hz.
5 5. PLASMA TURBULENCE AND TRANSPORT
Problem 5. Consider a plasma with a density of ne= 1019m−3and a temperature of Te= 5 keV.
Calculate the plasma’s sound speed and the Debye length for this plasma.
Given: Electron density ne= 1019m−3, Electron temperature Te= 5 keV
a) Calculate the plasma’s sound speed.
b) Determine the Debye length for this plasma.
Solution 5.
a) The sound speed in a plasma is given by:
cs=rγkTe
m
Where: γis the ratio of specific heats, taken to be γ=5
3for a fully ionized plasma. kis the
Boltzmann constant, k= 1.38 ×10−23J/K Teis the electron temperature in joules, Te= 5 ×103×
1.6×10−19Jmis the mass of an electron, m= 9.11 ×10−31kg
Substitute the given values into the formula to find cs:
cs=s5
3×1.38 ×10−23 ×5×103×1.6×10−19
9.11 ×10−31
cs=r5
3×1.38 ×10−23 ×8×10−16 =p9.24 ×10−39 = 3.04 ×104m/s
So, the sound speed of the plasma is 3.04 ×104m/s.
b) The Debye length is given by:
λD=rϵ0kTe
nee2
Where: ϵ0is the vacuum permittivity, ϵ0= 8.85 ×10−12F/m eis the elementary charge, e=
1.6×10−19C
Substitute the given values into the formula to find λD:
λD=s8.85 ×10−12 ×1.38 ×10−23 ×5×103×1.6×10−19
1019 ×(1.6×10−19)2
λD=r8.85 ×1.38 ×5×1.6
10 ×10−4=√0.097 ×10−4= 3.11 ×10−5m
Therefore, the Debye length for this plasma is 3.11 ×10−5m.
5.1 6. PLASMA-WALL INTERACTIONS
Problem 6. Consider a plasma confined in a tokamak fusion reactor with a plasma temperature of
T= 15 keV. The plasma density is n= 1020 m−3and the plasma current is I= 1 MA. Calculate the
thermal energy per particle, the total thermal energy of the plasma, and the total magnetic energy
stored in the plasma.
Given: Boltzmann constant, kB= 8.617 ×10−5eV/K
Electron charge, e= 1.602 ×10−19 C
Plasma volume, V= 10 m3
Magnetic field strength, B= 4 T
Solution 6.
a) The thermal energy per particle can be calculated using the formula:
ϵ=3
2kBT
Substitute the given values into the equation:
ϵ=3
2×8.617 ×10−5×15
ϵ= 1.291 ×10−3eV
b) The total thermal energy of the plasma can be calculated by multiplying the thermal energy
per particle by the number of particles:
Ethermal =nϵ
Substitute the given values into the equation:
Ethermal = 1020 ×1.291 ×10−3
Ethermal = 1.291 ×1017 eV
c) The total magnetic energy stored in the plasma can be calculated using the formula:
Emagnetic =B2
2µ0
V
where µ0is the permeability of free space.
First, convert the magnetic field strength from Tesla to Gauss:
B= 4 ×104Gauss
Now, calculate the total magnetic energy:
Emagnetic =(4 ×104)2
2×4π×10−7×10
Emagnetic =16 ×108
8π×10−7
Emagnetic ≈1.264 ×1015 erg
Therefore, the thermal energy per particle is 1.291 ×10−3eV, the total thermal energy of the
plasma is 1.291 ×1017 eV, and the total magnetic energy stored in the plasma is approximately
1.264 ×1015 erg.
6 7. DUST PARTICLES IN FUSION PLASMAS
Problem 7. Consider a fusion reactor where the deuterium plasma contains dust particles with
a radius of 1×10−6m and a charge of −5×10−17 C. The electron density in the plasma is 1×1020
particles per cubic meter and the electron temperature is 5×106K.
a) Calculate the Debye length in the plasma.
b) Determine the number of electrons and ions surrounding the dust particle within its Debye
length.
c) Estimate the electric field at the location of the dust particle.
Solution 7. a) The Debye length is given by:
λD=rε0kBTe
nee2
Substitute the given values:
λD=s(8.85 ×10−12 F/m)(1.38 ×10−23 J/K)(5 ×106K)
(1 ×1020 m−3)(1.6×10−19 C)2
λD=r6.423 ×10−35
2.56 ×10−29
λD=p2.511 ×10−6= 1.585 ×10−3m= 1.585 mm
Therefore, the Debye length in the plasma is 1.585 mm.
b) The number of electrons and ions within the Debye length of the dust particle can be calcu-
lated by considering a spherical volume around the dust particle with a radius equal to the Debye
length. The volume of this sphere is:
V=4
3π(λD)3
The number of particles within this volume is then:
Number of particles =neV
Substitute the values:
Number of particles = (1 ×1020 m−3)×4
3×π×(1.585 ×10−3)3
Number of particles ≈1.995 ×1016 ions and electrons
c) The electric field at the location of the dust particle is given by:
E=kBTe
eλD
Substitute the given values:
E=(1.38 ×10−23 J/K)(5 ×106K)
1.6×10−19 C×1.585 ×10−3m
E=6.9×10−17
1.27 ×10−22 ≈5433 V/m
Therefore, the electric field at the location of the dust particle is 5433 V/m.
I am currently unable to generate numerical problems specific to Plasma Physics and Fusion
Energy that require calculations and step-by-step explanations. If you have a specific numerical
problem in mind, please feel free to provide it, and I would be happy to help you solve it with detailed
explanations.
7 9. RUNAWAY ELECTRON GENERATION IN TOKAMAKS
Problem 9. Consider a tokamak plasma with a runaway electron population characterized by
a distribution function f(v) = αve−βv2, where vis the speed of the runaway electrons in units of
the thermal speed vth, and αand βare constants.
Given that α= 1 and β= 2, calculate the average speed of the runaway electrons in the
plasma.
Solution 9. The average speed of the runaway electrons in the plasma can be calculated using
the formula for the average speed of a distribution:
⟨v⟩=Z∞
0
vf(v)dv
=Z∞
0
vαve−βv2dv
=αZ∞
0
v2e−βv2dv
Integrating by parts with u=vand dv =ve−βv2dv, we have du =dv and ve−βv2=−1
2βde−βv2.
Substituting these back into the integral, we get:
⟨v⟩=α−1
2βve−βv2
∞
0+1
2βZ∞
0
e−βv2dv
=α
2βZ∞
0
e−βv2dv
=α
2β√π
2√β(using Gaussian integral formula)
=1
4rπ
2
Therefore, the average speed of the runaway electrons in the plasma is ⟨v⟩=1
4pπ
2≈0.406
times the thermal speed vth.
8 10. FUSION POWER PLANT DESIGN CHALLENGES
Problem 10. In a fusion power plant, the plasma temperature is 100 million Kelvin and the
plasma density is 1020 particles per cubic meter. Calculate the fusion power production rate as-
suming the fusion rate follows the Lawson criterion, which states the product of plasma density (n),
confinement time (τ), and energy confinement time (β) must be greater than a certain limit.
Given:
n= 1020 m−3
T= 100 ×106K
The Lawson criterion is given by the product of nτβ > C, where Cis a constant.
a) If C= 1022, calculate the fusion power production rate.
Solution 10. a) The fusion power production rate is given by:
Pfusion =Pinput
τfusion
=3n2⟨σv⟩Wfusion
τfusion
Given that n= 1020 m−3and C= 1022, we first need to find the confinement time τand energy
confinement time βfrom the Lawson criterion:
nτβ > C
1020 ×τ×β > 1022
τ×β > 102
Using the temperature given, the Lawson parameter can be expressed in terms of fusion reac-
tivity:
Wfusion =3
2kT
Wfusion =3
2×1.38 ×10−23 ×100 ×106
= 2.07 ×10−16 Joules
Plugging in the values, we get:
τfusion >102
β > 102/τ (where τ= 104s for ITER)
β > 102/104
β > 10−2
After calculating the value of β, we substitute back into the expression for Pfusion:
Pfusion =3×1020 ×2.07 ×10−16
104
= 6.21 ×105MW
Therefore, the fusion power production rate is 6.21 ×105MW.
Certainly! Here is a numerical problem in Plasma Physics and Fusion Energy:
9 11. TRITIUM BREEDING AND FUEL CYCLE TECHNOLOGIES
Problem 11. In a fusion reactor, tritium can be bred in lithium blankets through the following
reaction:
6
3Li +1
0n→4
2He +3
1T+ 4.8MeV
Given that the energy produced in this reaction is 4.8 MeV, calculate the energy produced in a
reaction where 1 kg of lithium is used to breed tritium.
Solution 11. Given: - Energy produced in one fusion reaction: 4.8 MeV - Mass of 1 Li atom:
6.94 ×10−26 kg
First, let’s find the number of lithium atoms in 1 kg of lithium:
N=1kg
6.94 ×10−26 kg = 1.44 ×1025
Now, let’s calculate the total energy produced in the reaction involving 1 kg of lithium:
E=N×4.8MeV = 1.44 ×1025 ×4.8×106eV = 6.91 ×1031 eV
Converting the energy into Joules:
E= 6.91 ×1031 ×1.6×10−19 = 1.106 ×1013 J
Thus, the energy produced in the reaction involving 1 kg of lithium is 1.106 ×1013 Joules.
10 12. PLASMA STABILITY AND CONTROL
Problem 12. Consider a cylindrical plasma column with radius aand length L, where the
plasma has a uniform resistivity of η. The magnetic field inside the plasma is given by B=B0ˆ
z,
where B0is a constant. The plasma is rotating around the axis of the cylinder with an angular
velocity Ω.
a) Calculate the current density Jinduced in the plasma.
b) Determine the viscous torque required to maintain the rotation of the plasma.
c) Show that the stability condition for this rotating plasma is given by ηΩ>B2
0a2
2.
Solution 12.
a) The current density induced in the plasma can be found using the induction equation for a
conducting fluid:
∂B
∂t =∇ × (u×B+η
µ0∇2B)
Since the velocity of the plasma uis solely due to the rotation, u= Ωaˆ
θ, and ∇ × B= 0, the
equation simplifies to:
∂B
∂t =η
µ0∇2B
Since B=B0ˆ
zand ∇2B= 0, we get:
∂Bz
∂t = 0
This implies that the induced current density Jis zero.
b) The viscous torque required to maintain the rotation is given by:
τ=ZV
r×(∇ × J)dV
Since J= 0, the torque is also zero.
c) The stability condition can be obtained by considering the ∇×Jterm in the induction equation.
For stability, the term ∇×Jshould not be able to counteract the original rotation-induced field. This
leads to the stability condition:
ηΩ>B2
0a2
2
This condition ensures that the induced magnetic field due to plasma currents cannot overtake
the original magnetic field.
11 13. NEOCLASSICAL AND ANOMALOUS TRANSPORT IN PLASMAS
Problem 13. Consider a plasma confined in a toroidal fusion device with a major radius R= 3
m and a minor radius a= 1 m. The plasma has a temperature of T= 10 keV and a density
of n= 1020 m−3. Calculate the neoclassical banana-plateau ion thermal conductivity using the
Spitzer-Härm formula:
κion = 7.8×10−8T5/2n
ln Λ W/mK
Where ln Λ is the Coulomb logarithm given by ln Λ = 2 + ln( T3/2
n1/2).
Solution 13. Given data: R= 3 m, a= 1 m, T= 10 keV = 10 ×103eV, n= 1020 m−3.
First, we calculate ln Λ using the given formula:
ln Λ = 2 + ln T3/2
n1/2!= 2 + ln (10 ×103)3/2
1010 != 2 + ln(107) = 2 + 15.42 ≈17.42
Now, we substitute the values of T,n, and ln Λ into the formula for ion thermal conductivity:
κion = 7.8×10−8(10 ×103)5/2×1020
17.42 W/mK
κion = 7.8×10−8×1015 ×1020
17.42 = 4.5×105W/mK
Therefore, the neoclassical banana-plateau ion thermal conductivity for the given plasma is
4.5×105W/mK.
I’m sorry, but I am unable to generate numerical problems for the subtopic "PLASMA FACING
MATERIALS FOR FUSION REACTORS." If you have any other topics or specific questions in
Plasma Physics and Fusion Energy that you would like me to create problems for, please let me
know!
12 15. FUSION ENERGY CONVERSION AND POWER EXTRACTION
Problem 15. Consider a fusion reactor that operates on the deuterium-tritium fusion reaction,
producing 17.6 MeV of energy per reaction. If the reactor is designed to produce a power output of
1 GW (gigawatt), calculate the number of fusion reactions that need to occur per second to achieve
this power output.
Solution 15. Given that the energy produced per fusion reaction is 17.6 MeV, which is equiv-
alent to 17.6×106eV, we can convert this to joules using the conversion factor 1.6×10−19 Joules
per electronvolt:
Energy per fusion reaction:
E= 17.6×106×1.6×10−19
= 28.16 ×10−13 Joules = 28.16 ×10−13 J.
Given that the power output of the reactor is 1 gigawatt, this is equivalent to 1×109watts or
joules per second. Therefore, the number of fusion reactions per second required to achieve this
power output can be calculated as follows:
Number of fusion reactions per second:
Power output =Energy per fusion reaction ×Number of fusion reactions per second
1×109J/s = 28.16 ×10−13 J×Number of fusion reactions per second
Number of fusion reactions per second =1×109
28.16 ×10−13
≈3.55 ×1021 fusion reactions/s.
Therefore, approximately 3.55 ×1021 fusion reactions need to occur per second in the reactor
to achieve a power output of 1 GW.
Certainly! Here is a numerical problem on plasma physics and fusion energy:
13 16. PLASMA CONFINEMENT AND HEATING METHODS
Problem 16. Consider a tokamak fusion reactor with a major radius R= 5 m and a minor
radius a= 1 m. The plasma inside the tokamak has a temperature of T= 15 keV and a density of
n= 5 ×1019 m−3. The plasma is confined by a magnetic field with strength B= 5 T. Calculate the
characteristic energy confinement time τEfor this plasma.
Solution 16. The energy confinement time τEis given by:
τE=3.6×103·P
⟨β⟩V
Plosses
where: - Pis the total power of the plasma - ⟨β⟩is the plasma beta - Vis the plasma volume -
Plosses is the total power losses
First, we need to calculate the plasma volume:
V=π2Ra2=π2·5·12= 15.71 m3
Next, let’s calculate the total power of the plasma. The total power is given by:
P=n·T·3
2k
where kis the Boltzmann constant k= 8.617 ×10−5eV/K. Plugging in the values:
P= 5 ×1019 ×15 ×3
2×8.617 ×10−5= 191.53 MW
Given that ⟨β⟩= 1, let’s calculate the power losses using P
⟨β⟩=P:
Plosses =P= 191.53 MW
Finally, substituting the values into the formula for τE:
τE=3.6×103·191.53 MW ·15.71 m3
191.53 MW = 29129.22 s
Therefore, the characteristic energy confinement time for this plasma in the tokamak is τE=
29129.22 seconds.
Certainly! Here is a numerical problem on energy confinement time in fusion plasmas:
14 17. ENERGY CONFINEMENT TIME IN FUSION PLASMAS
Problem 17. In a tokamak fusion reactor, the energy confinement time is given by the formula:
τE= 0.1×n
1020 m−3×T
10 keV3.5
s
where τEis the energy confinement time, nis the plasma density, and Tis the plasma temper-
ature. Calculate the energy confinement time for a plasma with a density of n= 5 ×1019 m−3and
a temperature of T= 20 keV.
Solution 17. Given: n= 5 ×1019 m−3and T= 20 keV
Substitute the given values into the formula for energy confinement time:
τE= 0.1×5×1019
1020 m−3×20
10 keV3.5
τE= 0.1×0.5×23.5
τE= 0.1×0.5×11.313
τE= 0.56565 s
Therefore, the energy confinement time for the given plasma is τE= 0.56565 s.
I.
15 18. PLASMA BOUNDARY PHYSICS
Problem 18. A plasma in a fusion device has an electron density of 1.5×1020 m−3and a
temperature of 10 keV. Calculate the Debye length of the plasma.
Given:
ne= 1.5×1020 m−3
T= 10 keV
Solution 18. a) The Debye length is given by
λD=ϵ0·Te
ne·e21/2
where ϵ0is the permittivity of free space, Teis the electron temperature, neis the electron
density, and eis the elementary charge.
Substitute the given values:
λD=8.85 ×10−12 ·10 ×1.602 ×10−19
1.5×1020 ·(1.602 ×10−19)21/2
=1.42 ×10−31
3.24 ×10−11 1/2
=4.39 ×10−211/2
= 2.09 ×10−10 m
Therefore, the Debye length of the plasma is 2.09 ×10−10 m.
16 Plasma Physics and Fusion Energy
Problem: Plasma Confinement in a Fusion Reactor
In a fusion reactor, the plasma is confined within a magnetic field generated by superconducting
coils. Consider a tokamak fusion reactor where the plasma is confined in a toroidal shape with a
major radius R= 5 meters and a minor radius a= 1 meter. The plasma temperature is T= 50
million Kelvin and the plasma density is n= 5×1019 particles per cubic meter. Assume the plasma
behaves like an ideal gas with three translational degrees of freedom.
a) Calculate the thermal energy density uof the plasma in joules per cubic meter.
b) Determine the thermal pressure pin pascals due to the plasma.
c) Find the magnetic field strength Bin teslas needed to confine the plasma if the plasma βis
1.5%, where β=p
B2
2µ0
represents the ratio of plasma pressure to magnetic pressure.
Solution:
a) The thermal energy density of an ideal gas is given by the equation:
u=3
2nkT
where kis the Boltzmann constant. Plugging in the values:
u=3
2×5×1019 ×1.38 ×10−23 ×50 ×106= 0.517 ×106J/m3
b) The thermal pressure of the plasma is given by:
p=nkT
Plugging in the values:
p= 5 ×1019 ×1.38 ×10−23 ×50 ×106= 3.45 ×105Pa
c) To find the magnetic field strength, we use the formula for β:
β= 1.5% = p
B2
2µ0
Solving for B:
B=r2µ0×p
β=r2×4π×10−7×3.45 ×105
0.015 ≈0.081 T
17 20. DIVERTOR DESIGN AND OPTIMIZATION
Problem 20. In a tokamak fusion reactor, the magnetic field strength in the divertor region
is given by B= 2.5T. The radius of the divertor plate is r= 0.5m and the angle between the
magnetic field lines and the divertor plate is θ= 30◦. Calculate the force experienced by a particle
with charge q= 1.6×10−19 C and velocity v= 5 ×106m/s moving along the magnetic field lines.
Solution 20. The force experienced by a charged particle moving in a magnetic field is given
by the equation:
F=qv×B
where qis the charge of the particle, vis the velocity vector, and Bis the magnetic field vector.
First, we need to calculate the velocity vector component along the magnetic field lines. Since
the particle is moving along the magnetic field lines, the velocity vector is parallel to the magnetic
field vector:
v∥=vcos θ= 5 ×106m/s ×cos 30◦= 4.33 ×106m/s
The force experienced by the particle is then:
F=qv∥B= (1.6×10−19 C)(4.33 ×106m/s)(2.5T)
F= 1.74 ×10−12 N
Therefore, the force experienced by the particle moving along the magnetic field lines is 1.74 ×
10−12 N.
I. Problem on Particle Density in a Fusion Reactor
Problem 1. In a fusion reactor, the electron density neis 1019 m−3. If the electron charge is
e= 1.6×10−19 C and the volume of the plasma is 0.1m3, calculate the total charge present in the
plasma.
Solution 1. The total charge present in the plasma is given by the product of electron density
and the charge of each electron:
Q=ne·e= 1019 m−3·1.6×10−19 C
Q= 1.6×100C= 1.6C
Therefore, the total charge present in the plasma is 1.6Coulombs.
II. Problem on Plasma Density Control
Problem 2. A plasma confinement device contains deuterium-tritium fuel with a number density
of 5×1020 m−3. If the device has a volume of 1m3, calculate the total number of fuel particles
present.
Solution 2. The total number of fuel particles present in the plasma is given by the product of
number density and volume of the plasma:
N=nfuel ·V= 5 ×1020 m−3·1m3
N= 5 ×1020 particles
Therefore, the total number of fuel particles present in the plasma is 5×1020 particles.
18 22. PLASMA TRANSPORT BARRIERS
Problem 22. Consider a fusion plasma with a temperature gradient described by the equation
T(r) = Tcore 1−r
a, where Tcore is the temperature at the core and ais the radial distance. The
density gradient of the plasma is given by n(r) = ncore 1−r2
a2.
Given that Tcore = 10 keV, ncore = 1019 m−3, and a= 0.5m, solve the following:
a) Calculate the temperature at the edge of the plasma.
b) Determine the density at the edge of the plasma.
c) Calculate the temperature gradient at the edge of the plasma.
Solution 22.
a) The temperature at the edge of the plasma can be found by substituting r=ainto the
temperature profile equation:
T(a) = 10 keV 1−0.5
0.5= 0 keV
Therefore, the temperature at the edge of the plasma is 0keV.
b) Similar to part (a), the density at the edge of the plasma is determined by plugging r=ainto
the density profile equation:
n(a) = 1019 m−31−0.52
0.52= 0 m−3
This means the density at the edge of the plasma is 0m−3.
c) The temperature gradient at the edge of the plasma can be calculated using the derivative
of the temperature profile with respect to r:
dT
dr =−Tcore
a=−10 keV
0.5m=−20 keV/m
Therefore, the temperature gradient at the edge of the plasma is −20 keV/m.
19 23. FAST PARTICLE CONFINEMENT IN FUSION DEVICES
Problem 23. Consider a fusion device where deuterium-tritium fusion reactions are taking
place. The average energy of the energetic alpha particles produced in these reactions is 3.5MeV.
Assuming the alpha particles are confined in a magnetic field with a magnetic mirror ratio of 2,
calculate the energy of the alpha particles when they are just about to escape confinement.
Solution 23.
a) The energy of the alpha particles when they are just about to escape confinement can be
determined by using the conservation of energy. The magnetic mirror ratio is defined as the ratio
of the magnetic field strength at the mirror point to the magnetic field strength at the center of the
device. In this case, the mirror ratio is 2.
Let B0be the magnetic field strength at the center, and Bmbe the magnetic field strength at
the mirror point. The energy of the alpha particle at the center (E0) and at the mirror point (Em) is
related by:
Em
E0
=B0
Bm2
Given that E0= 3.5MeV and the mirror ratio is 2, we can plug in the values to find the energy
of the alpha particles at the mirror point:
Em
3.5=1
22
Em=3.5
4= 0.875 MeV
Therefore, the energy of the alpha particles when they are just about to escape confinement is
0.875 MeV.
I’m glad to help with that! Here’s a numerical problem related to impurity control in fusion
reactors:
20 24. IMPURITY CONTROL IN FUSION REACTORS
Problem 24. In a fusion reactor, the plasma is composed of deuterium and tritium ions. However,
due to impurities, a small amount of helium ions is also present in the plasma. The density of
deuterium ions is nD= 1.5×1019 m−3, the density of tritium ions is nT= 0.5×1019 m−3, and the
density of helium ions is nHe = 0.1×1019 m−3. The charge of deuterium ions is qD= 1.6×10−19
C, the charge of tritium ions is qT= 1.6×10−19 C, and the charge of helium ions is qHe = 2 ×10−19
C.
a) Calculate the total charge density of the plasma.
b) If the average velocity of each ion species is vD= 1×106m/s for deuterium, vT= 0.8×106m/s
for tritium, and vHe = 0.6×106m/s for helium, calculate the total current density of the plasma.
Solution 24.
a) The total charge density of the plasma can be found by summing the contributions from each
ion species:
Total charge density, ρ=nD·qD+nT·qT+nHe ·qHe
Plugging in the values given:
ρ= (1.5×1019 m−3)(1.6×10−19 C)+(0.5×1019 m−3)(1.6×10−19 C)+(0.1×1019 m−3)(2×10−19 C)
ρ= 2.4×100+ 0.8×100+ 0.2×100
ρ= 3.4×100= 3.4C/m3
Therefore, the total charge density of the plasma is 3.4C/m3.
b) The total current density of the plasma can be found by summing the contributions from each
ion species:
Total current density, J=nD·qD·vD+nT·qT·vT+nHe ·qHe ·vHe
Plugging in the values given:
J= (1.5×1019 m−3)(1.6×10−19 C)(1×106m/s)+(0.5×1019 m−3)(1.6×10−19 C)(0.8×106m/s)+
(0.1×1019 m−3)(2 ×10−19 C)(0.6×106m/s)
J= 2.4×10−13 + 0.8×10−13 + 0.12 ×10−13
J= 3.32 ×10−13 A/m2
Therefore, the total current density of the plasma is 3.32
21 25. FUEL ION HEATING AND ALPHA PARTICLE EFFECTS.
Problem 25. Consider a fusion plasma where deuterium-tritium fusion reactions are the pre-
dominant source of energy generation. The energy released per reaction is 17.6 MeV. If the fusion
power output is 500 MW, determine:
Given:
•Energy released per reaction: 17.6MeV
•Fusion power output: 500 MW
a) The number of fusion reactions occurring per second.
b) The total energy released per second from fusion reactions.
c) The total energy released per second from fusion reactions in Joules.
Solution 25.
a) The number of fusion reactions occurring per second can be calculated using the fusion
power output:
Fusion Power Output =Energy released per reaction ×Number of reactions per second
500 ×106W= 17.6×106eV ×Number of reactions per second
Solving for the number of reactions per second:
Number of reactions per second =500 ×106
17.6×106≈28.41 ×106
Thus, approximately 28.41 ×106fusion reactions are occurring per second.
b) The total energy released per second from fusion reactions is:
Total energy released per second = 17.6×106eV ×28.41 ×106≈500 ×106W
Therefore, the total energy released per second from fusion reactions is 500 MW.
c) To convert this into Joules:
Total energy released per second in Joules = 500 ×106J
Hence, the total energy released per second from fusion reactions is 500 MJ.
ωc=4.8×10−19 C·T
1.67 ×10−27 kg
ωc= 2.88 ×108rad/s
Therefore, the gyrofrequency of the plasma particles in the given magnetic field is 2.88 ×108
rad/s.
3 3. RADIATION EFFECTS ON PLASMA MATERIALS
Problem 3. A material in a fusion reactor is exposed to a neutron flux of 1019 neutrons per
square meter per second. If the material has a neutron absorption cross-section of 10−19 m2,
calculate the neutron flux power density absorbed by the material.
Solution 3.
a) The neutron flux power density absorbed by the material can be calculated using the formula:
Neutron flux power density =Neutron flux ×Neutron absorption cross-section
Given: Neutron flux = 1019 neutrons/m2/s Neutron absorption cross-section = 10−19 m2
Neutron flux power density = 1019 neutrons/m2/s ×10−19 m2= 1 Watt/m2
Therefore, the neutron flux power density absorbed by the material is 1 Watt/m2.
4 4. MAGNETIC CONFINEMENT IN FUSION REACTORS
Problem 4. In a tokamak fusion reactor, the plasma temperature is 108K and the electron den-
sity is 1020 m−3. The magnetic field strength in the reactor is 5 Tesla. Calculate the gyrofrequency
of the electrons in the plasma.
Solution 4. The gyrofrequency of electrons in a magnetic field is given by the formula:
fgyro =eB
2πme
where: - eis the elementary charge (1.602 ×10−19 C), - Bis the magnetic field strength (5 T),
-meis the mass of an electron (9.11 ×10−31 kg).
Plugging in the values, we get:
fgyro =(1.602 ×10−19 C)(5 T)
2π(9.11 ×10−31 kg)
fgyro =8.01 ×10−19 C·T
2π(9.11 ×10−31 kg)
fgyro ≈1.40 ×1010 Hz
Therefore, the gyrofrequency of the electrons in the plasma is approximately 1.40 ×1010 Hz.
5 5. PLASMA TURBULENCE AND TRANSPORT
Problem 5. Consider a plasma with a density of ne= 1019m−3and a temperature of Te= 5 keV.
Calculate the plasma’s sound speed and the Debye length for this plasma.
Given: Electron density ne= 1019m−3, Electron temperature Te= 5 keV
a) Calculate the plasma’s sound speed.
b) Determine the Debye length for this plasma.
Solution 5.
a) The sound speed in a plasma is given by:
cs=rγkTe
m
Where: γis the ratio of specific heats, taken to be γ=5
3for a fully ionized plasma. kis the
Boltzmann constant, k= 1.38 ×10−23J/K Teis the electron temperature in joules, Te= 5 ×103×
1.6×10−19Jmis the mass of an electron, m= 9.11 ×10−31kg
Substitute the given values into the formula to find cs:
cs=s5
3×1.38 ×10−23 ×5×103×1.6×10−19
9.11 ×10−31
cs=r5
3×1.38 ×10−23 ×8×10−16 =p9.24 ×10−39 = 3.04 ×104m/s
So, the sound speed of the plasma is 3.04 ×104m/s.
b) The Debye length is given by:
λD=rϵ0kTe
nee2
Where: ϵ0is the vacuum permittivity, ϵ0= 8.85 ×10−12F/m eis the elementary charge, e=
1.6×10−19C
Substitute the given values into the formula to find λD:
λD=s8.85 ×10−12 ×1.38 ×10−23 ×5×103×1.6×10−19
1019 ×(1.6×10−19)2
λD=r8.85 ×1.38 ×5×1.6
10 ×10−4=√0.097 ×10−4= 3.11 ×10−5m
Therefore, the Debye length for this plasma is 3.11 ×10−5m.
5.1 6. PLASMA-WALL INTERACTIONS
Problem 6. Consider a plasma confined in a tokamak fusion reactor with a plasma temperature of
T= 15 keV. The plasma density is n= 1020 m−3and the plasma current is I= 1 MA. Calculate the
thermal energy per particle, the total thermal energy of the plasma, and the total magnetic energy
stored in the plasma.
Given: Boltzmann constant, kB= 8.617 ×10−5eV/K
Electron charge, e= 1.602 ×10−19 C
Plasma volume, V= 10 m3
Magnetic field strength, B= 4 T
Solution 6.
a) The thermal energy per particle can be calculated using the formula:
ϵ=3
2kBT
Substitute the given values into the equation:
ϵ=3
2×8.617 ×10−5×15
ϵ= 1.291 ×10−3eV
b) The total thermal energy of the plasma can be calculated by multiplying the thermal energy
per particle by the number of particles:
Ethermal =nϵ
Substitute the given values into the equation:
Ethermal = 1020 ×1.291 ×10−3
Ethermal = 1.291 ×1017 eV
c) The total magnetic energy stored in the plasma can be calculated using the formula:
Emagnetic =B2
2µ0
V
where µ0is the permeability of free space.
First, convert the magnetic field strength from Tesla to Gauss:
B= 4 ×104Gauss
Now, calculate the total magnetic energy:
Emagnetic =(4 ×104)2
2×4π×10−7×10
Emagnetic =16 ×108
8π×10−7
Emagnetic ≈1.264 ×1015 erg
Therefore, the thermal energy per particle is 1.291 ×10−3eV, the total thermal energy of the
plasma is 1.291 ×1017 eV, and the total magnetic energy stored in the plasma is approximately
1.264 ×1015 erg.
6 7. DUST PARTICLES IN FUSION PLASMAS
Problem 7. Consider a fusion reactor where the deuterium plasma contains dust particles with
a radius of 1×10−6m and a charge of −5×10−17 C. The electron density in the plasma is 1×1020
particles per cubic meter and the electron temperature is 5×106K.
a) Calculate the Debye length in the plasma.
b) Determine the number of electrons and ions surrounding the dust particle within its Debye
length.
c) Estimate the electric field at the location of the dust particle.
Solution 7. a) The Debye length is given by:
λD=rε0kBTe
nee2
Substitute the given values:
λD=s(8.85 ×10−12 F/m)(1.38 ×10−23 J/K)(5 ×106K)
(1 ×1020 m−3)(1.6×10−19 C)2
λD=r6.423 ×10−35
2.56 ×10−29
λD=p2.511 ×10−6= 1.585 ×10−3m= 1.585 mm
Therefore, the Debye length in the plasma is 1.585 mm.
b) The number of electrons and ions within the Debye length of the dust particle can be calcu-
lated by considering a spherical volume around the dust particle with a radius equal to the Debye
length. The volume of this sphere is:
V=4
3π(λD)3
The number of particles within this volume is then:
Number of particles =neV
Substitute the values:
Number of particles = (1 ×1020 m−3)×4
3×π×(1.585 ×10−3)3
Number of particles ≈1.995 ×1016 ions and electrons
c) The electric field at the location of the dust particle is given by:
E=kBTe
eλD
Substitute the given values:
E=(1.38 ×10−23 J/K)(5 ×106K)
1.6×10−19 C×1.585 ×10−3m
E=6.9×10−17
1.27 ×10−22 ≈5433 V/m
Therefore, the electric field at the location of the dust particle is 5433 V/m.
I am currently unable to generate numerical problems specific to Plasma Physics and Fusion
Energy that require calculations and step-by-step explanations. If you have a specific numerical
problem in mind, please feel free to provide it, and I would be happy to help you solve it with detailed
explanations.
7 9. RUNAWAY ELECTRON GENERATION IN TOKAMAKS
Problem 9. Consider a tokamak plasma with a runaway electron population characterized by
a distribution function f(v) = αve−βv2, where vis the speed of the runaway electrons in units of
the thermal speed vth, and αand βare constants.
Given that α= 1 and β= 2, calculate the average speed of the runaway electrons in the
plasma.
Solution 9. The average speed of the runaway electrons in the plasma can be calculated using
the formula for the average speed of a distribution:
⟨v⟩=Z∞
0
vf(v)dv
=Z∞
0
vαve−βv2dv
=αZ∞
0
v2e−βv2dv
Integrating by parts with u=vand dv =ve−βv2dv, we have du =dv and ve−βv2=−1
2βde−βv2.
Substituting these back into the integral, we get:
⟨v⟩=α−1
2βve−βv2
∞
0+1
2βZ∞
0
e−βv2dv
=α
2βZ∞
0
e−βv2dv
=α
2β√π
2√β(using Gaussian integral formula)
=1
4rπ
2
Therefore, the average speed of the runaway electrons in the plasma is ⟨v⟩=1
4pπ
2≈0.406
times the thermal speed vth.
8 10. FUSION POWER PLANT DESIGN CHALLENGES
Problem 10. In a fusion power plant, the plasma temperature is 100 million Kelvin and the
plasma density is 1020 particles per cubic meter. Calculate the fusion power production rate as-
suming the fusion rate follows the Lawson criterion, which states the product of plasma density (n),
confinement time (τ), and energy confinement time (β) must be greater than a certain limit.
Given:
n= 1020 m−3
T= 100 ×106K
The Lawson criterion is given by the product of nτβ > C, where Cis a constant.
a) If C= 1022, calculate the fusion power production rate.
Solution 10. a) The fusion power production rate is given by:
Pfusion =Pinput
τfusion
=3n2⟨σv⟩Wfusion
τfusion
Given that n= 1020 m−3and C= 1022, we first need to find the confinement time τand energy
confinement time βfrom the Lawson criterion:
nτβ > C
1020 ×τ×β > 1022
τ×β > 102
Using the temperature given, the Lawson parameter can be expressed in terms of fusion reac-
tivity:
Wfusion =3
2kT
Wfusion =3
2×1.38 ×10−23 ×100 ×106
= 2.07 ×10−16 Joules
Plugging in the values, we get:
τfusion >102
β > 102/τ (where τ= 104s for ITER)
β > 102/104
β > 10−2
After calculating the value of β, we substitute back into the expression for Pfusion:
Pfusion =3×1020 ×2.07 ×10−16
104
= 6.21 ×105MW
Therefore, the fusion power production rate is 6.21 ×105MW.
Certainly! Here is a numerical problem in Plasma Physics and Fusion Energy:
9 11. TRITIUM BREEDING AND FUEL CYCLE TECHNOLOGIES
Problem 11. In a fusion reactor, tritium can be bred in lithium blankets through the following
reaction:
6
3Li +1
0n→4
2He +3
1T+ 4.8MeV
Given that the energy produced in this reaction is 4.8 MeV, calculate the energy produced in a
reaction where 1 kg of lithium is used to breed tritium.
Solution 11. Given: - Energy produced in one fusion reaction: 4.8 MeV - Mass of 1 Li atom:
6.94 ×10−26 kg
First, let’s find the number of lithium atoms in 1 kg of lithium:
N=1kg
6.94 ×10−26 kg = 1.44 ×1025
Now, let’s calculate the total energy produced in the reaction involving 1 kg of lithium:
E=N×4.8MeV = 1.44 ×1025 ×4.8×106eV = 6.91 ×1031 eV
Converting the energy into Joules:
E= 6.91 ×1031 ×1.6×10−19 = 1.106 ×1013 J
Thus, the energy produced in the reaction involving 1 kg of lithium is 1.106 ×1013 Joules.
10 12. PLASMA STABILITY AND CONTROL
Problem 12. Consider a cylindrical plasma column with radius aand length L, where the
plasma has a uniform resistivity of η. The magnetic field inside the plasma is given by B=B0ˆ
z,
where B0is a constant. The plasma is rotating around the axis of the cylinder with an angular
velocity Ω.
a) Calculate the current density Jinduced in the plasma.
b) Determine the viscous torque required to maintain the rotation of the plasma.
c) Show that the stability condition for this rotating plasma is given by ηΩ>B2
0a2
2.
Solution 12.
a) The current density induced in the plasma can be found using the induction equation for a
conducting fluid:
∂B
∂t =∇ × (u×B+η
µ0∇2B)
Since the velocity of the plasma uis solely due to the rotation, u= Ωaˆ
θ, and ∇ × B= 0, the
equation simplifies to:
∂B
∂t =η
µ0∇2B
Since B=B0ˆ
zand ∇2B= 0, we get:
∂Bz
∂t = 0
This implies that the induced current density Jis zero.
b) The viscous torque required to maintain the rotation is given by:
τ=ZV
r×(∇ × J)dV
Since J= 0, the torque is also zero.
c) The stability condition can be obtained by considering the ∇×Jterm in the induction equation.
For stability, the term ∇×Jshould not be able to counteract the original rotation-induced field. This
leads to the stability condition:
ηΩ>B2
0a2
2
This condition ensures that the induced magnetic field due to plasma currents cannot overtake
the original magnetic field.
11 13. NEOCLASSICAL AND ANOMALOUS TRANSPORT IN PLASMAS
Problem 13. Consider a plasma confined in a toroidal fusion device with a major radius R= 3
m and a minor radius a= 1 m. The plasma has a temperature of T= 10 keV and a density
of n= 1020 m−3. Calculate the neoclassical banana-plateau ion thermal conductivity using the
Spitzer-Härm formula:
κion = 7.8×10−8T5/2n
ln Λ W/mK
Where ln Λ is the Coulomb logarithm given by ln Λ = 2 + ln( T3/2
n1/2).
Solution 13. Given data: R= 3 m, a= 1 m, T= 10 keV = 10 ×103eV, n= 1020 m−3.
First, we calculate ln Λ using the given formula:
ln Λ = 2 + ln T3/2
n1/2!= 2 + ln (10 ×103)3/2
1010 != 2 + ln(107) = 2 + 15.42 ≈17.42
Now, we substitute the values of T,n, and ln Λ into the formula for ion thermal conductivity:
κion = 7.8×10−8(10 ×103)5/2×1020
17.42 W/mK
κion = 7.8×10−8×1015 ×1020
17.42 = 4.5×105W/mK
Therefore, the neoclassical banana-plateau ion thermal conductivity for the given plasma is
4.5×105W/mK.
I’m sorry, but I am unable to generate numerical problems for the subtopic "PLASMA FACING
MATERIALS FOR FUSION REACTORS." If you have any other topics or specific questions in
Plasma Physics and Fusion Energy that you would like me to create problems for, please let me
know!
12 15. FUSION ENERGY CONVERSION AND POWER EXTRACTION
Problem 15. Consider a fusion reactor that operates on the deuterium-tritium fusion reaction,
producing 17.6 MeV of energy per reaction. If the reactor is designed to produce a power output of
1 GW (gigawatt), calculate the number of fusion reactions that need to occur per second to achieve
this power output.
Solution 15. Given that the energy produced per fusion reaction is 17.6 MeV, which is equiv-
alent to 17.6×106eV, we can convert this to joules using the conversion factor 1.6×10−19 Joules
per electronvolt:
Energy per fusion reaction:
E= 17.6×106×1.6×10−19
= 28.16 ×10−13 Joules = 28.16 ×10−13 J.
Given that the power output of the reactor is 1 gigawatt, this is equivalent to 1×109watts or
joules per second. Therefore, the number of fusion reactions per second required to achieve this
power output can be calculated as follows:
Number of fusion reactions per second:
Power output =Energy per fusion reaction ×Number of fusion reactions per second
1×109J/s = 28.16 ×10−13 J×Number of fusion reactions per second
Number of fusion reactions per second =1×109
28.16 ×10−13
≈3.55 ×1021 fusion reactions/s.
Therefore, approximately 3.55 ×1021 fusion reactions need to occur per second in the reactor
to achieve a power output of 1 GW.
Certainly! Here is a numerical problem on plasma physics and fusion energy:
13 16. PLASMA CONFINEMENT AND HEATING METHODS
Problem 16. Consider a tokamak fusion reactor with a major radius R= 5 m and a minor
radius a= 1 m. The plasma inside the tokamak has a temperature of T= 15 keV and a density of
n= 5 ×1019 m−3. The plasma is confined by a magnetic field with strength B= 5 T. Calculate the
characteristic energy confinement time τEfor this plasma.
Solution 16. The energy confinement time τEis given by:
τE=3.6×103·P
⟨β⟩V
Plosses
where: - Pis the total power of the plasma - ⟨β⟩is the plasma beta - Vis the plasma volume -
Plosses is the total power losses
First, we need to calculate the plasma volume:
V=π2Ra2=π2·5·12= 15.71 m3
Next, let’s calculate the total power of the plasma. The total power is given by:
P=n·T·3
2k
where kis the Boltzmann constant k= 8.617 ×10−5eV/K. Plugging in the values:
P= 5 ×1019 ×15 ×3
2×8.617 ×10−5= 191.53 MW
Given that ⟨β⟩= 1, let’s calculate the power losses using P
⟨β⟩=P:
Plosses =P= 191.53 MW
Finally, substituting the values into the formula for τE:
τE=3.6×103·191.53 MW ·15.71 m3
191.53 MW = 29129.22 s
Therefore, the characteristic energy confinement time for this plasma in the tokamak is τE=
29129.22 seconds.
Certainly! Here is a numerical problem on energy confinement time in fusion plasmas:
14 17. ENERGY CONFINEMENT TIME IN FUSION PLASMAS
Problem 17. In a tokamak fusion reactor, the energy confinement time is given by the formula:
τE= 0.1×n
1020 m−3×T
10 keV3.5
s
where τEis the energy confinement time, nis the plasma density, and Tis the plasma temper-
ature. Calculate the energy confinement time for a plasma with a density of n= 5 ×1019 m−3and
a temperature of T= 20 keV.
Solution 17. Given: n= 5 ×1019 m−3and T= 20 keV
Substitute the given values into the formula for energy confinement time:
τE= 0.1×5×1019
1020 m−3×20
10 keV3.5
τE= 0.1×0.5×23.5
τE= 0.1×0.5×11.313
τE= 0.56565 s
Therefore, the energy confinement time for the given plasma is τE= 0.56565 s.
I.
15 18. PLASMA BOUNDARY PHYSICS
Problem 18. A plasma in a fusion device has an electron density of 1.5×1020 m−3and a
temperature of 10 keV. Calculate the Debye length of the plasma.
Given:
ne= 1.5×1020 m−3
T= 10 keV
Solution 18. a) The Debye length is given by
λD=ϵ0·Te
ne·e21/2
where ϵ0is the permittivity of free space, Teis the electron temperature, neis the electron
density, and eis the elementary charge.
Substitute the given values:
λD=8.85 ×10−12 ·10 ×1.602 ×10−19
1.5×1020 ·(1.602 ×10−19)21/2
=1.42 ×10−31
3.24 ×10−11 1/2
=4.39 ×10−211/2
= 2.09 ×10−10 m
Therefore, the Debye length of the plasma is 2.09 ×10−10 m.
16 Plasma Physics and Fusion Energy
Problem: Plasma Confinement in a Fusion Reactor
In a fusion reactor, the plasma is confined within a magnetic field generated by superconducting
coils. Consider a tokamak fusion reactor where the plasma is confined in a toroidal shape with a
major radius R= 5 meters and a minor radius a= 1 meter. The plasma temperature is T= 50
million Kelvin and the plasma density is n= 5×1019 particles per cubic meter. Assume the plasma
behaves like an ideal gas with three translational degrees of freedom.
a) Calculate the thermal energy density uof the plasma in joules per cubic meter.
b) Determine the thermal pressure pin pascals due to the plasma.
c) Find the magnetic field strength Bin teslas needed to confine the plasma if the plasma βis
1.5%, where β=p
B2
2µ0
represents the ratio of plasma pressure to magnetic pressure.
Solution:
a) The thermal energy density of an ideal gas is given by the equation:
u=3
2nkT
where kis the Boltzmann constant. Plugging in the values:
u=3
2×5×1019 ×1.38 ×10−23 ×50 ×106= 0.517 ×106J/m3
b) The thermal pressure of the plasma is given by:
p=nkT
Plugging in the values:
p= 5 ×1019 ×1.38 ×10−23 ×50 ×106= 3.45 ×105Pa
c) To find the magnetic field strength, we use the formula for β:
β= 1.5% = p
B2
2µ0
Solving for B:
B=r2µ0×p
β=r2×4π×10−7×3.45 ×105
0.015 ≈0.081 T
17 20. DIVERTOR DESIGN AND OPTIMIZATION
Problem 20. In a tokamak fusion reactor, the magnetic field strength in the divertor region
is given by B= 2.5T. The radius of the divertor plate is r= 0.5m and the angle between the
magnetic field lines and the divertor plate is θ= 30◦. Calculate the force experienced by a particle
with charge q= 1.6×10−19 C and velocity v= 5 ×106m/s moving along the magnetic field lines.
Solution 20. The force experienced by a charged particle moving in a magnetic field is given
by the equation:
F=qv×B
where qis the charge of the particle, vis the velocity vector, and Bis the magnetic field vector.
First, we need to calculate the velocity vector component along the magnetic field lines. Since
the particle is moving along the magnetic field lines, the velocity vector is parallel to the magnetic
field vector:
v∥=vcos θ= 5 ×106m/s ×cos 30◦= 4.33 ×106m/s
The force experienced by the particle is then:
F=qv∥B= (1.6×10−19 C)(4.33 ×106m/s)(2.5T)
F= 1.74 ×10−12 N
Therefore, the force experienced by the particle moving along the magnetic field lines is 1.74 ×
10−12 N.
I. Problem on Particle Density in a Fusion Reactor
Problem 1. In a fusion reactor, the electron density neis 1019 m−3. If the electron charge is
e= 1.6×10−19 C and the volume of the plasma is 0.1m3, calculate the total charge present in the
plasma.
Solution 1. The total charge present in the plasma is given by the product of electron density
and the charge of each electron:
Q=ne·e= 1019 m−3·1.6×10−19 C
Q= 1.6×100C= 1.6C
Therefore, the total charge present in the plasma is 1.6Coulombs.
II. Problem on Plasma Density Control
Problem 2. A plasma confinement device contains deuterium-tritium fuel with a number density
of 5×1020 m−3. If the device has a volume of 1m3, calculate the total number of fuel particles
present.
Solution 2. The total number of fuel particles present in the plasma is given by the product of
number density and volume of the plasma:
N=nfuel ·V= 5 ×1020 m−3·1m3
N= 5 ×1020 particles
Therefore, the total number of fuel particles present in the plasma is 5×1020 particles.
18 22. PLASMA TRANSPORT BARRIERS
Problem 22. Consider a fusion plasma with a temperature gradient described by the equation
T(r) = Tcore 1−r
a, where Tcore is the temperature at the core and ais the radial distance. The
density gradient of the plasma is given by n(r) = ncore 1−r2
a2.
Given that Tcore = 10 keV, ncore = 1019 m−3, and a= 0.5m, solve the following:
a) Calculate the temperature at the edge of the plasma.
b) Determine the density at the edge of the plasma.
c) Calculate the temperature gradient at the edge of the plasma.
Solution 22.
a) The temperature at the edge of the plasma can be found by substituting r=ainto the
temperature profile equation:
T(a) = 10 keV 1−0.5
0.5= 0 keV
Therefore, the temperature at the edge of the plasma is 0keV.
b) Similar to part (a), the density at the edge of the plasma is determined by plugging r=ainto
the density profile equation:
n(a) = 1019 m−31−0.52
0.52= 0 m−3
This means the density at the edge of the plasma is 0m−3.
c) The temperature gradient at the edge of the plasma can be calculated using the derivative
of the temperature profile with respect to r:
dT
dr =−Tcore
a=−10 keV
0.5m=−20 keV/m
Therefore, the temperature gradient at the edge of the plasma is −20 keV/m.
19 23. FAST PARTICLE CONFINEMENT IN FUSION DEVICES
Problem 23. Consider a fusion device where deuterium-tritium fusion reactions are taking
place. The average energy of the energetic alpha particles produced in these reactions is 3.5MeV.
Assuming the alpha particles are confined in a magnetic field with a magnetic mirror ratio of 2,
calculate the energy of the alpha particles when they are just about to escape confinement.
Solution 23.
a) The energy of the alpha particles when they are just about to escape confinement can be
determined by using the conservation of energy. The magnetic mirror ratio is defined as the ratio
of the magnetic field strength at the mirror point to the magnetic field strength at the center of the
device. In this case, the mirror ratio is 2.
Let B0be the magnetic field strength at the center, and Bmbe the magnetic field strength at
the mirror point. The energy of the alpha particle at the center (E0) and at the mirror point (Em) is
related by:
Em
E0
=B0
Bm2
Given that E0= 3.5MeV and the mirror ratio is 2, we can plug in the values to find the energy
of the alpha particles at the mirror point:
Em
3.5=1
22
Em=3.5
4= 0.875 MeV
Therefore, the energy of the alpha particles when they are just about to escape confinement is
0.875 MeV.
I’m glad to help with that! Here’s a numerical problem related to impurity control in fusion
reactors:
20 24. IMPURITY CONTROL IN FUSION REACTORS
Problem 24. In a fusion reactor, the plasma is composed of deuterium and tritium ions. However,
due to impurities, a small amount of helium ions is also present in the plasma. The density of
deuterium ions is nD= 1.5×1019 m−3, the density of tritium ions is nT= 0.5×1019 m−3, and the
density of helium ions is nHe = 0.1×1019 m−3. The charge of deuterium ions is qD= 1.6×10−19
C, the charge of tritium ions is qT= 1.6×10−19 C, and the charge of helium ions is qHe = 2 ×10−19
C.
a) Calculate the total charge density of the plasma.
b) If the average velocity of each ion species is vD= 1×106m/s for deuterium, vT= 0.8×106m/s
for tritium, and vHe = 0.6×106m/s for helium, calculate the total current density of the plasma.
Solution 24.
a) The total charge density of the plasma can be found by summing the contributions from each
ion species:
Total charge density, ρ=nD·qD+nT·qT+nHe ·qHe
Plugging in the values given:
ρ= (1.5×1019 m−3)(1.6×10−19 C)+(0.5×1019 m−3)(1.6×10−19 C)+(0.1×1019 m−3)(2×10−19 C)
ρ= 2.4×100+ 0.8×100+ 0.2×100
ρ= 3.4×100= 3.4C/m3
Therefore, the total charge density of the plasma is 3.4C/m3.
b) The total current density of the plasma can be found by summing the contributions from each
ion species:
Total current density, J=nD·qD·vD+nT·qT·vT+nHe ·qHe ·vHe
Plugging in the values given:
J= (1.5×1019 m−3)(1.6×10−19 C)(1×106m/s)+(0.5×1019 m−3)(1.6×10−19 C)(0.8×106m/s)+
(0.1×1019 m−3)(2 ×10−19 C)(0.6×106m/s)
J= 2.4×10−13 + 0.8×10−13 + 0.12 ×10−13
J= 3.32 ×10−13 A/m2
Therefore, the total current density of the plasma is 3.32
21 25. FUEL ION HEATING AND ALPHA PARTICLE EFFECTS.
Problem 25. Consider a fusion plasma where deuterium-tritium fusion reactions are the pre-
dominant source of energy generation. The energy released per reaction is 17.6 MeV. If the fusion
power output is 500 MW, determine:
Given:
•Energy released per reaction: 17.6MeV
•Fusion power output: 500 MW
a) The number of fusion reactions occurring per second.
b) The total energy released per second from fusion reactions.
c) The total energy released per second from fusion reactions in Joules.
Solution 25.
a) The number of fusion reactions occurring per second can be calculated using the fusion
power output:
Fusion Power Output =Energy released per reaction ×Number of reactions per second
500 ×106W= 17.6×106eV ×Number of reactions per second
Solving for the number of reactions per second:
Number of reactions per second =500 ×106
17.6×106≈28.41 ×106
Thus, approximately 28.41 ×106fusion reactions are occurring per second.
b) The total energy released per second from fusion reactions is:
Total energy released per second = 17.6×106eV ×28.41 ×106≈500 ×106W
Therefore, the total energy released per second from fusion reactions is 500 MW.
c) To convert this into Joules:
Total energy released per second in Joules = 500 ×106J
Hence, the total energy released per second from fusion reactions is 500 MJ.
ωc=4.8×10−19 C·T
1.67 ×10−27 kg
ωc= 2.88 ×108rad/s
Therefore, the gyrofrequency of the plasma particles in the given magnetic field is 2.88 ×108
rad/s.
3 3. RADIATION EFFECTS ON PLASMA MATERIALS
Problem 3. A material in a fusion reactor is exposed to a neutron flux of 1019 neutrons per
square meter per second. If the material has a neutron absorption cross-section of 10−19 m2,
calculate the neutron flux power density absorbed by the material.
Solution 3.
a) The neutron flux power density absorbed by the material can be calculated using the formula:
Neutron flux power density =Neutron flux ×Neutron absorption cross-section
Given: Neutron flux = 1019 neutrons/m2/s Neutron absorption cross-section = 10−19 m2
Neutron flux power density = 1019 neutrons/m2/s ×10−19 m2= 1 Watt/m2
Therefore, the neutron flux power density absorbed by the material is 1 Watt/m2.
4 4. MAGNETIC CONFINEMENT IN FUSION REACTORS
Problem 4. In a tokamak fusion reactor, the plasma temperature is 108K and the electron den-
sity is 1020 m−3. The magnetic field strength in the reactor is 5 Tesla. Calculate the gyrofrequency
of the electrons in the plasma.
Solution 4. The gyrofrequency of electrons in a magnetic field is given by the formula:
fgyro =eB
2πme
where: - eis the elementary charge (1.602 ×10−19 C), - Bis the magnetic field strength (5 T),
-meis the mass of an electron (9.11 ×10−31 kg).
Plugging in the values, we get:
fgyro =(1.602 ×10−19 C)(5 T)
2π(9.11 ×10−31 kg)
fgyro =8.01 ×10−19 C·T
2π(9.11 ×10−31 kg)
fgyro ≈1.40 ×1010 Hz
Therefore, the gyrofrequency of the electrons in the plasma is approximately 1.40 ×1010 Hz.
5 5. PLASMA TURBULENCE AND TRANSPORT
Problem 5. Consider a plasma with a density of ne= 1019m−3and a temperature of Te= 5 keV.
Calculate the plasma’s sound speed and the Debye length for this plasma.
Given: Electron density ne= 1019m−3, Electron temperature Te= 5 keV
a) Calculate the plasma’s sound speed.
b) Determine the Debye length for this plasma.
Solution 5.
a) The sound speed in a plasma is given by:
cs=rγkTe
m
Where: γis the ratio of specific heats, taken to be γ=5
3for a fully ionized plasma. kis the
Boltzmann constant, k= 1.38 ×10−23J/K Teis the electron temperature in joules, Te= 5 ×103×
1.6×10−19Jmis the mass of an electron, m= 9.11 ×10−31kg
Substitute the given values into the formula to find cs:
cs=s5
3×1.38 ×10−23 ×5×103×1.6×10−19
9.11 ×10−31
cs=r5
3×1.38 ×10−23 ×8×10−16 =p9.24 ×10−39 = 3.04 ×104m/s
So, the sound speed of the plasma is 3.04 ×104m/s.
b) The Debye length is given by:
λD=rϵ0kTe
nee2
Where: ϵ0is the vacuum permittivity, ϵ0= 8.85 ×10−12F/m eis the elementary charge, e=
1.6×10−19C
Substitute the given values into the formula to find λD:
λD=s8.85 ×10−12 ×1.38 ×10−23 ×5×103×1.6×10−19
1019 ×(1.6×10−19)2
λD=r8.85 ×1.38 ×5×1.6
10 ×10−4=√0.097 ×10−4= 3.11 ×10−5m
Therefore, the Debye length for this plasma is 3.11 ×10−5m.
5.1 6. PLASMA-WALL INTERACTIONS
Problem 6. Consider a plasma confined in a tokamak fusion reactor with a plasma temperature of
T= 15 keV. The plasma density is n= 1020 m−3and the plasma current is I= 1 MA. Calculate the
thermal energy per particle, the total thermal energy of the plasma, and the total magnetic energy
stored in the plasma.
Given: Boltzmann constant, kB= 8.617 ×10−5eV/K
Electron charge, e= 1.602 ×10−19 C
Plasma volume, V= 10 m3
Magnetic field strength, B= 4 T
Solution 6.
a) The thermal energy per particle can be calculated using the formula:
ϵ=3
2kBT
Substitute the given values into the equation:
ϵ=3
2×8.617 ×10−5×15
ϵ= 1.291 ×10−3eV
b) The total thermal energy of the plasma can be calculated by multiplying the thermal energy
per particle by the number of particles:
Ethermal =nϵ
Substitute the given values into the equation:
Ethermal = 1020 ×1.291 ×10−3
Ethermal = 1.291 ×1017 eV
c) The total magnetic energy stored in the plasma can be calculated using the formula:
Emagnetic =B2
2µ0
V
where µ0is the permeability of free space.
First, convert the magnetic field strength from Tesla to Gauss:
B= 4 ×104Gauss
Now, calculate the total magnetic energy:
Emagnetic =(4 ×104)2
2×4π×10−7×10
Emagnetic =16 ×108
8π×10−7
Emagnetic ≈1.264 ×1015 erg
Therefore, the thermal energy per particle is 1.291 ×10−3eV, the total thermal energy of the
plasma is 1.291 ×1017 eV, and the total magnetic energy stored in the plasma is approximately
1.264 ×1015 erg.
6 7. DUST PARTICLES IN FUSION PLASMAS
Problem 7. Consider a fusion reactor where the deuterium plasma contains dust particles with
a radius of 1×10−6m and a charge of −5×10−17 C. The electron density in the plasma is 1×1020
particles per cubic meter and the electron temperature is 5×106K.
a) Calculate the Debye length in the plasma.
b) Determine the number of electrons and ions surrounding the dust particle within its Debye
length.
c) Estimate the electric field at the location of the dust particle.
Solution 7. a) The Debye length is given by:
λD=rε0kBTe
nee2
Substitute the given values:
λD=s(8.85 ×10−12 F/m)(1.38 ×10−23 J/K)(5 ×106K)
(1 ×1020 m−3)(1.6×10−19 C)2
λD=r6.423 ×10−35
2.56 ×10−29
λD=p2.511 ×10−6= 1.585 ×10−3m= 1.585 mm
Therefore, the Debye length in the plasma is 1.585 mm.
b) The number of electrons and ions within the Debye length of the dust particle can be calcu-
lated by considering a spherical volume around the dust particle with a radius equal to the Debye
length. The volume of this sphere is:
V=4
3π(λD)3
The number of particles within this volume is then:
Number of particles =neV
Substitute the values:
Number of particles = (1 ×1020 m−3)×4
3×π×(1.585 ×10−3)3
Number of particles ≈1.995 ×1016 ions and electrons
c) The electric field at the location of the dust particle is given by:
E=kBTe
eλD
Substitute the given values:
E=(1.38 ×10−23 J/K)(5 ×106K)
1.6×10−19 C×1.585 ×10−3m
E=6.9×10−17
1.27 ×10−22 ≈5433 V/m
Therefore, the electric field at the location of the dust particle is 5433 V/m.
I am currently unable to generate numerical problems specific to Plasma Physics and Fusion
Energy that require calculations and step-by-step explanations. If you have a specific numerical
problem in mind, please feel free to provide it, and I would be happy to help you solve it with detailed
explanations.
7 9. RUNAWAY ELECTRON GENERATION IN TOKAMAKS
Problem 9. Consider a tokamak plasma with a runaway electron population characterized by
a distribution function f(v) = αve−βv2, where vis the speed of the runaway electrons in units of
the thermal speed vth, and αand βare constants.
Given that α= 1 and β= 2, calculate the average speed of the runaway electrons in the
plasma.
Solution 9. The average speed of the runaway electrons in the plasma can be calculated using
the formula for the average speed of a distribution:
⟨v⟩=Z∞
0
vf(v)dv
=Z∞
0
vαve−βv2dv
=αZ∞
0
v2e−βv2dv
Integrating by parts with u=vand dv =ve−βv2dv, we have du =dv and ve−βv2=−1
2βde−βv2.
Substituting these back into the integral, we get:
⟨v⟩=α−1
2βve−βv2
∞
0+1
2βZ∞
0
e−βv2dv
=α
2βZ∞
0
e−βv2dv
=α
2β√π
2√β(using Gaussian integral formula)
=1
4rπ
2
Therefore, the average speed of the runaway electrons in the plasma is ⟨v⟩=1
4pπ
2≈0.406
times the thermal speed vth.
8 10. FUSION POWER PLANT DESIGN CHALLENGES
Problem 10. In a fusion power plant, the plasma temperature is 100 million Kelvin and the
plasma density is 1020 particles per cubic meter. Calculate the fusion power production rate as-
suming the fusion rate follows the Lawson criterion, which states the product of plasma density (n),
confinement time (τ), and energy confinement time (β) must be greater than a certain limit.
Given:
n= 1020 m−3
T= 100 ×106K
The Lawson criterion is given by the product of nτβ > C, where Cis a constant.
a) If C= 1022, calculate the fusion power production rate.
Solution 10. a) The fusion power production rate is given by:
Pfusion =Pinput
τfusion
=3n2⟨σv⟩Wfusion
τfusion
Given that n= 1020 m−3and C= 1022, we first need to find the confinement time τand energy
confinement time βfrom the Lawson criterion:
nτβ > C
1020 ×τ×β > 1022
τ×β > 102
Using the temperature given, the Lawson parameter can be expressed in terms of fusion reac-
tivity:
Wfusion =3
2kT
Wfusion =3
2×1.38 ×10−23 ×100 ×106
= 2.07 ×10−16 Joules
Plugging in the values, we get:
τfusion >102
β > 102/τ (where τ= 104s for ITER)
β > 102/104
β > 10−2
After calculating the value of β, we substitute back into the expression for Pfusion:
Pfusion =3×1020 ×2.07 ×10−16
104
= 6.21 ×105MW
Therefore, the fusion power production rate is 6.21 ×105MW.
Certainly! Here is a numerical problem in Plasma Physics and Fusion Energy:
9 11. TRITIUM BREEDING AND FUEL CYCLE TECHNOLOGIES
Problem 11. In a fusion reactor, tritium can be bred in lithium blankets through the following
reaction:
6
3Li +1
0n→4
2He +3
1T+ 4.8MeV
Given that the energy produced in this reaction is 4.8 MeV, calculate the energy produced in a
reaction where 1 kg of lithium is used to breed tritium.
Solution 11. Given: - Energy produced in one fusion reaction: 4.8 MeV - Mass of 1 Li atom:
6.94 ×10−26 kg
First, let’s find the number of lithium atoms in 1 kg of lithium:
N=1kg
6.94 ×10−26 kg = 1.44 ×1025
Now, let’s calculate the total energy produced in the reaction involving 1 kg of lithium:
E=N×4.8MeV = 1.44 ×1025 ×4.8×106eV = 6.91 ×1031 eV
Converting the energy into Joules:
E= 6.91 ×1031 ×1.6×10−19 = 1.106 ×1013 J
Thus, the energy produced in the reaction involving 1 kg of lithium is 1.106 ×1013 Joules.
10 12. PLASMA STABILITY AND CONTROL
Problem 12. Consider a cylindrical plasma column with radius aand length L, where the
plasma has a uniform resistivity of η. The magnetic field inside the plasma is given by B=B0ˆ
z,
where B0is a constant. The plasma is rotating around the axis of the cylinder with an angular
velocity Ω.
a) Calculate the current density Jinduced in the plasma.
b) Determine the viscous torque required to maintain the rotation of the plasma.
c) Show that the stability condition for this rotating plasma is given by ηΩ>B2
0a2
2.
Solution 12.
a) The current density induced in the plasma can be found using the induction equation for a
conducting fluid:
∂B
∂t =∇ × (u×B+η
µ0∇2B)
Since the velocity of the plasma uis solely due to the rotation, u= Ωaˆ
θ, and ∇ × B= 0, the
equation simplifies to:
∂B
∂t =η
µ0∇2B
Since B=B0ˆ
zand ∇2B= 0, we get:
∂Bz
∂t = 0
This implies that the induced current density Jis zero.
b) The viscous torque required to maintain the rotation is given by:
τ=ZV
r×(∇ × J)dV
Since J= 0, the torque is also zero.
c) The stability condition can be obtained by considering the ∇×Jterm in the induction equation.
For stability, the term ∇×Jshould not be able to counteract the original rotation-induced field. This
leads to the stability condition:
ηΩ>B2
0a2
2
This condition ensures that the induced magnetic field due to plasma currents cannot overtake
the original magnetic field.
11 13. NEOCLASSICAL AND ANOMALOUS TRANSPORT IN PLASMAS
Problem 13. Consider a plasma confined in a toroidal fusion device with a major radius R= 3
m and a minor radius a= 1 m. The plasma has a temperature of T= 10 keV and a density
of n= 1020 m−3. Calculate the neoclassical banana-plateau ion thermal conductivity using the
Spitzer-Härm formula:
κion = 7.8×10−8T5/2n
ln Λ W/mK
Where ln Λ is the Coulomb logarithm given by ln Λ = 2 + ln( T3/2
n1/2).
Solution 13. Given data: R= 3 m, a= 1 m, T= 10 keV = 10 ×103eV, n= 1020 m−3.
First, we calculate ln Λ using the given formula:
ln Λ = 2 + ln T3/2
n1/2!= 2 + ln (10 ×103)3/2
1010 != 2 + ln(107) = 2 + 15.42 ≈17.42
Now, we substitute the values of T,n, and ln Λ into the formula for ion thermal conductivity:
κion = 7.8×10−8(10 ×103)5/2×1020
17.42 W/mK
κion = 7.8×10−8×1015 ×1020
17.42 = 4.5×105W/mK
Therefore, the neoclassical banana-plateau ion thermal conductivity for the given plasma is
4.5×105W/mK.
I’m sorry, but I am unable to generate numerical problems for the subtopic "PLASMA FACING
MATERIALS FOR FUSION REACTORS." If you have any other topics or specific questions in
Plasma Physics and Fusion Energy that you would like me to create problems for, please let me
know!
12 15. FUSION ENERGY CONVERSION AND POWER EXTRACTION
Problem 15. Consider a fusion reactor that operates on the deuterium-tritium fusion reaction,
producing 17.6 MeV of energy per reaction. If the reactor is designed to produce a power output of
1 GW (gigawatt), calculate the number of fusion reactions that need to occur per second to achieve
this power output.
Solution 15. Given that the energy produced per fusion reaction is 17.6 MeV, which is equiv-
alent to 17.6×106eV, we can convert this to joules using the conversion factor 1.6×10−19 Joules
per electronvolt:
Energy per fusion reaction:
E= 17.6×106×1.6×10−19
= 28.16 ×10−13 Joules = 28.16 ×10−13 J.
Given that the power output of the reactor is 1 gigawatt, this is equivalent to 1×109watts or
joules per second. Therefore, the number of fusion reactions per second required to achieve this
power output can be calculated as follows:
Number of fusion reactions per second:
Power output =Energy per fusion reaction ×Number of fusion reactions per second
1×109J/s = 28.16 ×10−13 J×Number of fusion reactions per second
Number of fusion reactions per second =1×109
28.16 ×10−13
≈3.55 ×1021 fusion reactions/s.
Therefore, approximately 3.55 ×1021 fusion reactions need to occur per second in the reactor
to achieve a power output of 1 GW.
Certainly! Here is a numerical problem on plasma physics and fusion energy:
13 16. PLASMA CONFINEMENT AND HEATING METHODS
Problem 16. Consider a tokamak fusion reactor with a major radius R= 5 m and a minor
radius a= 1 m. The plasma inside the tokamak has a temperature of T= 15 keV and a density of
n= 5 ×1019 m−3. The plasma is confined by a magnetic field with strength B= 5 T. Calculate the
characteristic energy confinement time τEfor this plasma.
Solution 16. The energy confinement time τEis given by:
τE=3.6×103·P
⟨β⟩V
Plosses
where: - Pis the total power of the plasma - ⟨β⟩is the plasma beta - Vis the plasma volume -
Plosses is the total power losses
First, we need to calculate the plasma volume:
V=π2Ra2=π2·5·12= 15.71 m3
Next, let’s calculate the total power of the plasma. The total power is given by:
P=n·T·3
2k
where kis the Boltzmann constant k= 8.617 ×10−5eV/K. Plugging in the values:
P= 5 ×1019 ×15 ×3
2×8.617 ×10−5= 191.53 MW
Given that ⟨β⟩= 1, let’s calculate the power losses using P
⟨β⟩=P:
Plosses =P= 191.53 MW
Finally, substituting the values into the formula for τE:
τE=3.6×103·191.53 MW ·15.71 m3
191.53 MW = 29129.22 s
Therefore, the characteristic energy confinement time for this plasma in the tokamak is τE=
29129.22 seconds.
Certainly! Here is a numerical problem on energy confinement time in fusion plasmas:
14 17. ENERGY CONFINEMENT TIME IN FUSION PLASMAS
Problem 17. In a tokamak fusion reactor, the energy confinement time is given by the formula:
τE= 0.1×n
1020 m−3×T
10 keV3.5
s
where τEis the energy confinement time, nis the plasma density, and Tis the plasma temper-
ature. Calculate the energy confinement time for a plasma with a density of n= 5 ×1019 m−3and
a temperature of T= 20 keV.
Solution 17. Given: n= 5 ×1019 m−3and T= 20 keV
Substitute the given values into the formula for energy confinement time:
τE= 0.1×5×1019
1020 m−3×20
10 keV3.5
τE= 0.1×0.5×23.5
τE= 0.1×0.5×11.313
τE= 0.56565 s
Therefore, the energy confinement time for the given plasma is τE= 0.56565 s.
I.
15 18. PLASMA BOUNDARY PHYSICS
Problem 18. A plasma in a fusion device has an electron density of 1.5×1020 m−3and a
temperature of 10 keV. Calculate the Debye length of the plasma.
Given:
ne= 1.5×1020 m−3
T= 10 keV
Solution 18. a) The Debye length is given by
λD=ϵ0·Te
ne·e21/2
where ϵ0is the permittivity of free space, Teis the electron temperature, neis the electron
density, and eis the elementary charge.
Substitute the given values:
λD=8.85 ×10−12 ·10 ×1.602 ×10−19
1.5×1020 ·(1.602 ×10−19)21/2
=1.42 ×10−31
3.24 ×10−11 1/2
=4.39 ×10−211/2
= 2.09 ×10−10 m
Therefore, the Debye length of the plasma is 2.09 ×10−10 m.
16 Plasma Physics and Fusion Energy
Problem: Plasma Confinement in a Fusion Reactor
In a fusion reactor, the plasma is confined within a magnetic field generated by superconducting
coils. Consider a tokamak fusion reactor where the plasma is confined in a toroidal shape with a
major radius R= 5 meters and a minor radius a= 1 meter. The plasma temperature is T= 50
million Kelvin and the plasma density is n= 5×1019 particles per cubic meter. Assume the plasma
behaves like an ideal gas with three translational degrees of freedom.
a) Calculate the thermal energy density uof the plasma in joules per cubic meter.
b) Determine the thermal pressure pin pascals due to the plasma.
c) Find the magnetic field strength Bin teslas needed to confine the plasma if the plasma βis
1.5%, where β=p
B2
2µ0
represents the ratio of plasma pressure to magnetic pressure.
Solution:
a) The thermal energy density of an ideal gas is given by the equation:
u=3
2nkT
where kis the Boltzmann constant. Plugging in the values:
u=3
2×5×1019 ×1.38 ×10−23 ×50 ×106= 0.517 ×106J/m3
b) The thermal pressure of the plasma is given by:
p=nkT
Plugging in the values:
p= 5 ×1019 ×1.38 ×10−23 ×50 ×106= 3.45 ×105Pa
c) To find the magnetic field strength, we use the formula for β:
β= 1.5% = p
B2
2µ0
Solving for B:
B=r2µ0×p
β=r2×4π×10−7×3.45 ×105
0.015 ≈0.081 T
17 20. DIVERTOR DESIGN AND OPTIMIZATION
Problem 20. In a tokamak fusion reactor, the magnetic field strength in the divertor region
is given by B= 2.5T. The radius of the divertor plate is r= 0.5m and the angle between the
magnetic field lines and the divertor plate is θ= 30◦. Calculate the force experienced by a particle
with charge q= 1.6×10−19 C and velocity v= 5 ×106m/s moving along the magnetic field lines.
Solution 20. The force experienced by a charged particle moving in a magnetic field is given
by the equation:
F=qv×B
where qis the charge of the particle, vis the velocity vector, and Bis the magnetic field vector.
First, we need to calculate the velocity vector component along the magnetic field lines. Since
the particle is moving along the magnetic field lines, the velocity vector is parallel to the magnetic
field vector:
v∥=vcos θ= 5 ×106m/s ×cos 30◦= 4.33 ×106m/s
The force experienced by the particle is then:
F=qv∥B= (1.6×10−19 C)(4.33 ×106m/s)(2.5T)
F= 1.74 ×10−12 N
Therefore, the force experienced by the particle moving along the magnetic field lines is 1.74 ×
10−12 N.
I. Problem on Particle Density in a Fusion Reactor
Problem 1. In a fusion reactor, the electron density neis 1019 m−3. If the electron charge is
e= 1.6×10−19 C and the volume of the plasma is 0.1m3, calculate the total charge present in the
plasma.
Solution 1. The total charge present in the plasma is given by the product of electron density
and the charge of each electron:
Q=ne·e= 1019 m−3·1.6×10−19 C
Q= 1.6×100C= 1.6C
Therefore, the total charge present in the plasma is 1.6Coulombs.
II. Problem on Plasma Density Control
Problem 2. A plasma confinement device contains deuterium-tritium fuel with a number density
of 5×1020 m−3. If the device has a volume of 1m3, calculate the total number of fuel particles
present.
Solution 2. The total number of fuel particles present in the plasma is given by the product of
number density and volume of the plasma:
N=nfuel ·V= 5 ×1020 m−3·1m3
N= 5 ×1020 particles
Therefore, the total number of fuel particles present in the plasma is 5×1020 particles.
18 22. PLASMA TRANSPORT BARRIERS
Problem 22. Consider a fusion plasma with a temperature gradient described by the equation
T(r) = Tcore 1−r
a, where Tcore is the temperature at the core and ais the radial distance. The
density gradient of the plasma is given by n(r) = ncore 1−r2
a2.
Given that Tcore = 10 keV, ncore = 1019 m−3, and a= 0.5m, solve the following:
a) Calculate the temperature at the edge of the plasma.
b) Determine the density at the edge of the plasma.
c) Calculate the temperature gradient at the edge of the plasma.
Solution 22.
a) The temperature at the edge of the plasma can be found by substituting r=ainto the
temperature profile equation:
T(a) = 10 keV 1−0.5
0.5= 0 keV
Therefore, the temperature at the edge of the plasma is 0keV.
b) Similar to part (a), the density at the edge of the plasma is determined by plugging r=ainto
the density profile equation:
n(a) = 1019 m−31−0.52
0.52= 0 m−3
This means the density at the edge of the plasma is 0m−3.
c) The temperature gradient at the edge of the plasma can be calculated using the derivative
of the temperature profile with respect to r:
dT
dr =−Tcore
a=−10 keV
0.5m=−20 keV/m
Therefore, the temperature gradient at the edge of the plasma is −20 keV/m.
19 23. FAST PARTICLE CONFINEMENT IN FUSION DEVICES
Problem 23. Consider a fusion device where deuterium-tritium fusion reactions are taking
place. The average energy of the energetic alpha particles produced in these reactions is 3.5MeV.
Assuming the alpha particles are confined in a magnetic field with a magnetic mirror ratio of 2,
calculate the energy of the alpha particles when they are just about to escape confinement.
Solution 23.
a) The energy of the alpha particles when they are just about to escape confinement can be
determined by using the conservation of energy. The magnetic mirror ratio is defined as the ratio
of the magnetic field strength at the mirror point to the magnetic field strength at the center of the
device. In this case, the mirror ratio is 2.
Let B0be the magnetic field strength at the center, and Bmbe the magnetic field strength at
the mirror point. The energy of the alpha particle at the center (E0) and at the mirror point (Em) is
related by:
Em
E0
=B0
Bm2
Given that E0= 3.5MeV and the mirror ratio is 2, we can plug in the values to find the energy
of the alpha particles at the mirror point:
Em
3.5=1
22
Em=3.5
4= 0.875 MeV
Therefore, the energy of the alpha particles when they are just about to escape confinement is
0.875 MeV.
I’m glad to help with that! Here’s a numerical problem related to impurity control in fusion
reactors:
20 24. IMPURITY CONTROL IN FUSION REACTORS
Problem 24. In a fusion reactor, the plasma is composed of deuterium and tritium ions. However,
due to impurities, a small amount of helium ions is also present in the plasma. The density of
deuterium ions is nD= 1.5×1019 m−3, the density of tritium ions is nT= 0.5×1019 m−3, and the
density of helium ions is nHe = 0.1×1019 m−3. The charge of deuterium ions is qD= 1.6×10−19
C, the charge of tritium ions is qT= 1.6×10−19 C, and the charge of helium ions is qHe = 2 ×10−19
C.
a) Calculate the total charge density of the plasma.
b) If the average velocity of each ion species is vD= 1×106m/s for deuterium, vT= 0.8×106m/s
for tritium, and vHe = 0.6×106m/s for helium, calculate the total current density of the plasma.
Solution 24.
a) The total charge density of the plasma can be found by summing the contributions from each
ion species:
Total charge density, ρ=nD·qD+nT·qT+nHe ·qHe
Plugging in the values given:
ρ= (1.5×1019 m−3)(1.6×10−19 C)+(0.5×1019 m−3)(1.6×10−19 C)+(0.1×1019 m−3)(2×10−19 C)
ρ= 2.4×100+ 0.8×100+ 0.2×100
ρ= 3.4×100= 3.4C/m3
Therefore, the total charge density of the plasma is 3.4C/m3.
b) The total current density of the plasma can be found by summing the contributions from each
ion species:
Total current density, J=nD·qD·vD+nT·qT·vT+nHe ·qHe ·vHe
Plugging in the values given:
J= (1.5×1019 m−3)(1.6×10−19 C)(1×106m/s)+(0.5×1019 m−3)(1.6×10−19 C)(0.8×106m/s)+
(0.1×1019 m−3)(2 ×10−19 C)(0.6×106m/s)
J= 2.4×10−13 + 0.8×10−13 + 0.12 ×10−13
J= 3.32 ×10−13 A/m2
Therefore, the total current density of the plasma is 3.32
21 25. FUEL ION HEATING AND ALPHA PARTICLE EFFECTS.
Problem 25. Consider a fusion plasma where deuterium-tritium fusion reactions are the pre-
dominant source of energy generation. The energy released per reaction is 17.6 MeV. If the fusion
power output is 500 MW, determine:
Given:
•Energy released per reaction: 17.6MeV
•Fusion power output: 500 MW
a) The number of fusion reactions occurring per second.
b) The total energy released per second from fusion reactions.
c) The total energy released per second from fusion reactions in Joules.
Solution 25.
a) The number of fusion reactions occurring per second can be calculated using the fusion
power output:
Fusion Power Output =Energy released per reaction ×Number of reactions per second
500 ×106W= 17.6×106eV ×Number of reactions per second
Solving for the number of reactions per second:
Number of reactions per second =500 ×106
17.6×106≈28.41 ×106
Thus, approximately 28.41 ×106fusion reactions are occurring per second.
b) The total energy released per second from fusion reactions is:
Total energy released per second = 17.6×106eV ×28.41 ×106≈500 ×106W
Therefore, the total energy released per second from fusion reactions is 500 MW.
c) To convert this into Joules:
Total energy released per second in Joules = 500 ×106J
Hence, the total energy released per second from fusion reactions is 500 MJ.
ωc=4.8×10−19 C·T
1.67 ×10−27 kg
ωc= 2.88 ×108rad/s
Therefore, the gyrofrequency of the plasma particles in the given magnetic field is 2.88 ×108
rad/s.
3 3. RADIATION EFFECTS ON PLASMA MATERIALS
Problem 3. A material in a fusion reactor is exposed to a neutron flux of 1019 neutrons per
square meter per second. If the material has a neutron absorption cross-section of 10−19 m2,
calculate the neutron flux power density absorbed by the material.
Solution 3.
a) The neutron flux power density absorbed by the material can be calculated using the formula:
Neutron flux power density =Neutron flux ×Neutron absorption cross-section
Given: Neutron flux = 1019 neutrons/m2/s Neutron absorption cross-section = 10−19 m2
Neutron flux power density = 1019 neutrons/m2/s ×10−19 m2= 1 Watt/m2
Therefore, the neutron flux power density absorbed by the material is 1 Watt/m2.
4 4. MAGNETIC CONFINEMENT IN FUSION REACTORS
Problem 4. In a tokamak fusion reactor, the plasma temperature is 108K and the electron den-
sity is 1020 m−3. The magnetic field strength in the reactor is 5 Tesla. Calculate the gyrofrequency
of the electrons in the plasma.
Solution 4. The gyrofrequency of electrons in a magnetic field is given by the formula:
fgyro =eB
2πme
where: - eis the elementary charge (1.602 ×10−19 C), - Bis the magnetic field strength (5 T),
-meis the mass of an electron (9.11 ×10−31 kg).
Plugging in the values, we get:
fgyro =(1.602 ×10−19 C)(5 T)
2π(9.11 ×10−31 kg)
fgyro =8.01 ×10−19 C·T
2π(9.11 ×10−31 kg)
fgyro ≈1.40 ×1010 Hz
Therefore, the gyrofrequency of the electrons in the plasma is approximately 1.40 ×1010 Hz.
5 5. PLASMA TURBULENCE AND TRANSPORT
Problem 5. Consider a plasma with a density of ne= 1019m−3and a temperature of Te= 5 keV.
Calculate the plasma’s sound speed and the Debye length for this plasma.
Given: Electron density ne= 1019m−3, Electron temperature Te= 5 keV
a) Calculate the plasma’s sound speed.
b) Determine the Debye length for this plasma.
Solution 5.
a) The sound speed in a plasma is given by:
cs=rγkTe
m
Where: γis the ratio of specific heats, taken to be γ=5
3for a fully ionized plasma. kis the
Boltzmann constant, k= 1.38 ×10−23J/K Teis the electron temperature in joules, Te= 5 ×103×
1.6×10−19Jmis the mass of an electron, m= 9.11 ×10−31kg
Substitute the given values into the formula to find cs:
cs=s5
3×1.38 ×10−23 ×5×103×1.6×10−19
9.11 ×10−31
cs=r5
3×1.38 ×10−23 ×8×10−16 =p9.24 ×10−39 = 3.04 ×104m/s
So, the sound speed of the plasma is 3.04 ×104m/s.
b) The Debye length is given by:
λD=rϵ0kTe
nee2
Where: ϵ0is the vacuum permittivity, ϵ0= 8.85 ×10−12F/m eis the elementary charge, e=
1.6×10−19C
Substitute the given values into the formula to find λD:
λD=s8.85 ×10−12 ×1.38 ×10−23 ×5×103×1.6×10−19
1019 ×(1.6×10−19)2
λD=r8.85 ×1.38 ×5×1.6
10 ×10−4=√0.097 ×10−4= 3.11 ×10−5m
Therefore, the Debye length for this plasma is 3.11 ×10−5m.
5.1 6. PLASMA-WALL INTERACTIONS
Problem 6. Consider a plasma confined in a tokamak fusion reactor with a plasma temperature of
T= 15 keV. The plasma density is n= 1020 m−3and the plasma current is I= 1 MA. Calculate the
thermal energy per particle, the total thermal energy of the plasma, and the total magnetic energy
stored in the plasma.
Given: Boltzmann constant, kB= 8.617 ×10−5eV/K
Electron charge, e= 1.602 ×10−19 C
Plasma volume, V= 10 m3
Magnetic field strength, B= 4 T
Solution 6.
a) The thermal energy per particle can be calculated using the formula:
ϵ=3
2kBT
Substitute the given values into the equation:
ϵ=3
2×8.617 ×10−5×15
ϵ= 1.291 ×10−3eV
b) The total thermal energy of the plasma can be calculated by multiplying the thermal energy
per particle by the number of particles:
Ethermal =nϵ
Substitute the given values into the equation:
Ethermal = 1020 ×1.291 ×10−3
Ethermal = 1.291 ×1017 eV
c) The total magnetic energy stored in the plasma can be calculated using the formula:
Emagnetic =B2
2µ0
V
where µ0is the permeability of free space.
First, convert the magnetic field strength from Tesla to Gauss:
B= 4 ×104Gauss
Now, calculate the total magnetic energy:
Emagnetic =(4 ×104)2
2×4π×10−7×10
Emagnetic =16 ×108
8π×10−7
Emagnetic ≈1.264 ×1015 erg
Therefore, the thermal energy per particle is 1.291 ×10−3eV, the total thermal energy of the
plasma is 1.291 ×1017 eV, and the total magnetic energy stored in the plasma is approximately
1.264 ×1015 erg.
6 7. DUST PARTICLES IN FUSION PLASMAS
Problem 7. Consider a fusion reactor where the deuterium plasma contains dust particles with
a radius of 1×10−6m and a charge of −5×10−17 C. The electron density in the plasma is 1×1020
particles per cubic meter and the electron temperature is 5×106K.
a) Calculate the Debye length in the plasma.
b) Determine the number of electrons and ions surrounding the dust particle within its Debye
length.
c) Estimate the electric field at the location of the dust particle.
Solution 7. a) The Debye length is given by:
λD=rε0kBTe
nee2
Substitute the given values:
λD=s(8.85 ×10−12 F/m)(1.38 ×10−23 J/K)(5 ×106K)
(1 ×1020 m−3)(1.6×10−19 C)2
λD=r6.423 ×10−35
2.56 ×10−29
λD=p2.511 ×10−6= 1.585 ×10−3m= 1.585 mm
Therefore, the Debye length in the plasma is 1.585 mm.
b) The number of electrons and ions within the Debye length of the dust particle can be calcu-
lated by considering a spherical volume around the dust particle with a radius equal to the Debye
length. The volume of this sphere is:
V=4
3π(λD)3
The number of particles within this volume is then:
Number of particles =neV
Substitute the values:
Number of particles = (1 ×1020 m−3)×4
3×π×(1.585 ×10−3)3
Number of particles ≈1.995 ×1016 ions and electrons
c) The electric field at the location of the dust particle is given by:
E=kBTe
eλD
Substitute the given values:
E=(1.38 ×10−23 J/K)(5 ×106K)
1.6×10−19 C×1.585 ×10−3m
E=6.9×10−17
1.27 ×10−22 ≈5433 V/m
Therefore, the electric field at the location of the dust particle is 5433 V/m.
I am currently unable to generate numerical problems specific to Plasma Physics and Fusion
Energy that require calculations and step-by-step explanations. If you have a specific numerical
problem in mind, please feel free to provide it, and I would be happy to help you solve it with detailed
explanations.
7 9. RUNAWAY ELECTRON GENERATION IN TOKAMAKS
Problem 9. Consider a tokamak plasma with a runaway electron population characterized by
a distribution function f(v) = αve−βv2, where vis the speed of the runaway electrons in units of
the thermal speed vth, and αand βare constants.
Given that α= 1 and β= 2, calculate the average speed of the runaway electrons in the
plasma.
Solution 9. The average speed of the runaway electrons in the plasma can be calculated using
the formula for the average speed of a distribution:
⟨v⟩=Z∞
0
vf(v)dv
=Z∞
0
vαve−βv2dv
=αZ∞
0
v2e−βv2dv
Integrating by parts with u=vand dv =ve−βv2dv, we have du =dv and ve−βv2=−1
2βde−βv2.
Substituting these back into the integral, we get:
⟨v⟩=α−1
2βve−βv2
∞
0+1
2βZ∞
0
e−βv2dv
=α
2βZ∞
0
e−βv2dv
=α
2β√π
2√β(using Gaussian integral formula)
=1
4rπ
2
Therefore, the average speed of the runaway electrons in the plasma is ⟨v⟩=1
4pπ
2≈0.406
times the thermal speed vth.
8 10. FUSION POWER PLANT DESIGN CHALLENGES
Problem 10. In a fusion power plant, the plasma temperature is 100 million Kelvin and the
plasma density is 1020 particles per cubic meter. Calculate the fusion power production rate as-
suming the fusion rate follows the Lawson criterion, which states the product of plasma density (n),
confinement time (τ), and energy confinement time (β) must be greater than a certain limit.
Given:
n= 1020 m−3
T= 100 ×106K
The Lawson criterion is given by the product of nτβ > C, where Cis a constant.
a) If C= 1022, calculate the fusion power production rate.
Solution 10. a) The fusion power production rate is given by:
Pfusion =Pinput
τfusion
=3n2⟨σv⟩Wfusion
τfusion
Given that n= 1020 m−3and C= 1022, we first need to find the confinement time τand energy
confinement time βfrom the Lawson criterion:
nτβ > C
1020 ×τ×β > 1022
τ×β > 102
Using the temperature given, the Lawson parameter can be expressed in terms of fusion reac-
tivity:
Wfusion =3
2kT
Wfusion =3
2×1.38 ×10−23 ×100 ×106
= 2.07 ×10−16 Joules
Plugging in the values, we get:
τfusion >102
β > 102/τ (where τ= 104s for ITER)
β > 102/104
β > 10−2
After calculating the value of β, we substitute back into the expression for Pfusion:
Pfusion =3×1020 ×2.07 ×10−16
104
= 6.21 ×105MW
Therefore, the fusion power production rate is 6.21 ×105MW.
Certainly! Here is a numerical problem in Plasma Physics and Fusion Energy:
9 11. TRITIUM BREEDING AND FUEL CYCLE TECHNOLOGIES
Problem 11. In a fusion reactor, tritium can be bred in lithium blankets through the following
reaction:
6
3Li +1
0n→4
2He +3
1T+ 4.8MeV
Given that the energy produced in this reaction is 4.8 MeV, calculate the energy produced in a
reaction where 1 kg of lithium is used to breed tritium.
Solution 11. Given: - Energy produced in one fusion reaction: 4.8 MeV - Mass of 1 Li atom:
6.94 ×10−26 kg
First, let’s find the number of lithium atoms in 1 kg of lithium:
N=1kg
6.94 ×10−26 kg = 1.44 ×1025
Now, let’s calculate the total energy produced in the reaction involving 1 kg of lithium:
E=N×4.8MeV = 1.44 ×1025 ×4.8×106eV = 6.91 ×1031 eV
Converting the energy into Joules:
E= 6.91 ×1031 ×1.6×10−19 = 1.106 ×1013 J
Thus, the energy produced in the reaction involving 1 kg of lithium is 1.106 ×1013 Joules.
10 12. PLASMA STABILITY AND CONTROL
Problem 12. Consider a cylindrical plasma column with radius aand length L, where the
plasma has a uniform resistivity of η. The magnetic field inside the plasma is given by B=B0ˆ
z,
where B0is a constant. The plasma is rotating around the axis of the cylinder with an angular
velocity Ω.
a) Calculate the current density Jinduced in the plasma.
b) Determine the viscous torque required to maintain the rotation of the plasma.
c) Show that the stability condition for this rotating plasma is given by ηΩ>B2
0a2
2.
Solution 12.
a) The current density induced in the plasma can be found using the induction equation for a
conducting fluid:
∂B
∂t =∇ × (u×B+η
µ0∇2B)
Since the velocity of the plasma uis solely due to the rotation, u= Ωaˆ
θ, and ∇ × B= 0, the
equation simplifies to:
∂B
∂t =η
µ0∇2B
Since B=B0ˆ
zand ∇2B= 0, we get:
∂Bz
∂t = 0
This implies that the induced current density Jis zero.
b) The viscous torque required to maintain the rotation is given by:
τ=ZV
r×(∇ × J)dV
Since J= 0, the torque is also zero.
c) The stability condition can be obtained by considering the ∇×Jterm in the induction equation.
For stability, the term ∇×Jshould not be able to counteract the original rotation-induced field. This
leads to the stability condition:
ηΩ>B2
0a2
2
This condition ensures that the induced magnetic field due to plasma currents cannot overtake
the original magnetic field.
11 13. NEOCLASSICAL AND ANOMALOUS TRANSPORT IN PLASMAS
Problem 13. Consider a plasma confined in a toroidal fusion device with a major radius R= 3
m and a minor radius a= 1 m. The plasma has a temperature of T= 10 keV and a density
of n= 1020 m−3. Calculate the neoclassical banana-plateau ion thermal conductivity using the
Spitzer-Härm formula:
κion = 7.8×10−8T5/2n
ln Λ W/mK
Where ln Λ is the Coulomb logarithm given by ln Λ = 2 + ln( T3/2
n1/2).
Solution 13. Given data: R= 3 m, a= 1 m, T= 10 keV = 10 ×103eV, n= 1020 m−3.
First, we calculate ln Λ using the given formula:
ln Λ = 2 + ln T3/2
n1/2!= 2 + ln (10 ×103)3/2
1010 != 2 + ln(107) = 2 + 15.42 ≈17.42
Now, we substitute the values of T,n, and ln Λ into the formula for ion thermal conductivity:
κion = 7.8×10−8(10 ×103)5/2×1020
17.42 W/mK
κion = 7.8×10−8×1015 ×1020
17.42 = 4.5×105W/mK
Therefore, the neoclassical banana-plateau ion thermal conductivity for the given plasma is
4.5×105W/mK.
I’m sorry, but I am unable to generate numerical problems for the subtopic "PLASMA FACING
MATERIALS FOR FUSION REACTORS." If you have any other topics or specific questions in
Plasma Physics and Fusion Energy that you would like me to create problems for, please let me
know!
12 15. FUSION ENERGY CONVERSION AND POWER EXTRACTION
Problem 15. Consider a fusion reactor that operates on the deuterium-tritium fusion reaction,
producing 17.6 MeV of energy per reaction. If the reactor is designed to produce a power output of
1 GW (gigawatt), calculate the number of fusion reactions that need to occur per second to achieve
this power output.
Solution 15. Given that the energy produced per fusion reaction is 17.6 MeV, which is equiv-
alent to 17.6×106eV, we can convert this to joules using the conversion factor 1.6×10−19 Joules
per electronvolt:
Energy per fusion reaction:
E= 17.6×106×1.6×10−19
= 28.16 ×10−13 Joules = 28.16 ×10−13 J.
Given that the power output of the reactor is 1 gigawatt, this is equivalent to 1×109watts or
joules per second. Therefore, the number of fusion reactions per second required to achieve this
power output can be calculated as follows:
Number of fusion reactions per second:
Power output =Energy per fusion reaction ×Number of fusion reactions per second
1×109J/s = 28.16 ×10−13 J×Number of fusion reactions per second
Number of fusion reactions per second =1×109
28.16 ×10−13
≈3.55 ×1021 fusion reactions/s.
Therefore, approximately 3.55 ×1021 fusion reactions need to occur per second in the reactor
to achieve a power output of 1 GW.
Certainly! Here is a numerical problem on plasma physics and fusion energy:
13 16. PLASMA CONFINEMENT AND HEATING METHODS
Problem 16. Consider a tokamak fusion reactor with a major radius R= 5 m and a minor
radius a= 1 m. The plasma inside the tokamak has a temperature of T= 15 keV and a density of
n= 5 ×1019 m−3. The plasma is confined by a magnetic field with strength B= 5 T. Calculate the
characteristic energy confinement time τEfor this plasma.
Solution 16. The energy confinement time τEis given by:
τE=3.6×103·P
⟨β⟩V
Plosses
where: - Pis the total power of the plasma - ⟨β⟩is the plasma beta - Vis the plasma volume -
Plosses is the total power losses
First, we need to calculate the plasma volume:
V=π2Ra2=π2·5·12= 15.71 m3
Next, let’s calculate the total power of the plasma. The total power is given by:
P=n·T·3
2k
where kis the Boltzmann constant k= 8.617 ×10−5eV/K. Plugging in the values:
P= 5 ×1019 ×15 ×3
2×8.617 ×10−5= 191.53 MW
Given that ⟨β⟩= 1, let’s calculate the power losses using P
⟨β⟩=P:
Plosses =P= 191.53 MW
Finally, substituting the values into the formula for τE:
τE=3.6×103·191.53 MW ·15.71 m3
191.53 MW = 29129.22 s
Therefore, the characteristic energy confinement time for this plasma in the tokamak is τE=
29129.22 seconds.
Certainly! Here is a numerical problem on energy confinement time in fusion plasmas:
14 17. ENERGY CONFINEMENT TIME IN FUSION PLASMAS
Problem 17. In a tokamak fusion reactor, the energy confinement time is given by the formula:
τE= 0.1×n
1020 m−3×T
10 keV3.5
s
where τEis the energy confinement time, nis the plasma density, and Tis the plasma temper-
ature. Calculate the energy confinement time for a plasma with a density of n= 5 ×1019 m−3and
a temperature of T= 20 keV.
Solution 17. Given: n= 5 ×1019 m−3and T= 20 keV
Substitute the given values into the formula for energy confinement time:
τE= 0.1×5×1019
1020 m−3×20
10 keV3.5
τE= 0.1×0.5×23.5
τE= 0.1×0.5×11.313
τE= 0.56565 s
Therefore, the energy confinement time for the given plasma is τE= 0.56565 s.
I.
15 18. PLASMA BOUNDARY PHYSICS
Problem 18. A plasma in a fusion device has an electron density of 1.5×1020 m−3and a
temperature of 10 keV. Calculate the Debye length of the plasma.
Given:
ne= 1.5×1020 m−3
T= 10 keV
Solution 18. a) The Debye length is given by
λD=ϵ0·Te
ne·e21/2
where ϵ0is the permittivity of free space, Teis the electron temperature, neis the electron
density, and eis the elementary charge.
Substitute the given values:
λD=8.85 ×10−12 ·10 ×1.602 ×10−19
1.5×1020 ·(1.602 ×10−19)21/2
=1.42 ×10−31
3.24 ×10−11 1/2
=4.39 ×10−211/2
= 2.09 ×10−10 m
Therefore, the Debye length of the plasma is 2.09 ×10−10 m.
16 Plasma Physics and Fusion Energy
Problem: Plasma Confinement in a Fusion Reactor
In a fusion reactor, the plasma is confined within a magnetic field generated by superconducting
coils. Consider a tokamak fusion reactor where the plasma is confined in a toroidal shape with a
major radius R= 5 meters and a minor radius a= 1 meter. The plasma temperature is T= 50
million Kelvin and the plasma density is n= 5×1019 particles per cubic meter. Assume the plasma
behaves like an ideal gas with three translational degrees of freedom.
a) Calculate the thermal energy density uof the plasma in joules per cubic meter.
b) Determine the thermal pressure pin pascals due to the plasma.
c) Find the magnetic field strength Bin teslas needed to confine the plasma if the plasma βis
1.5%, where β=p
B2
2µ0
represents the ratio of plasma pressure to magnetic pressure.
Solution:
a) The thermal energy density of an ideal gas is given by the equation:
u=3
2nkT
where kis the Boltzmann constant. Plugging in the values:
u=3
2×5×1019 ×1.38 ×10−23 ×50 ×106= 0.517 ×106J/m3
b) The thermal pressure of the plasma is given by:
p=nkT
Plugging in the values:
p= 5 ×1019 ×1.38 ×10−23 ×50 ×106= 3.45 ×105Pa
c) To find the magnetic field strength, we use the formula for β:
β= 1.5% = p
B2
2µ0
Solving for B:
B=r2µ0×p
β=r2×4π×10−7×3.45 ×105
0.015 ≈0.081 T
17 20. DIVERTOR DESIGN AND OPTIMIZATION
Problem 20. In a tokamak fusion reactor, the magnetic field strength in the divertor region
is given by B= 2.5T. The radius of the divertor plate is r= 0.5m and the angle between the
magnetic field lines and the divertor plate is θ= 30◦. Calculate the force experienced by a particle
with charge q= 1.6×10−19 C and velocity v= 5 ×106m/s moving along the magnetic field lines.
Solution 20. The force experienced by a charged particle moving in a magnetic field is given
by the equation:
F=qv×B
where qis the charge of the particle, vis the velocity vector, and Bis the magnetic field vector.
First, we need to calculate the velocity vector component along the magnetic field lines. Since
the particle is moving along the magnetic field lines, the velocity vector is parallel to the magnetic
field vector:
v∥=vcos θ= 5 ×106m/s ×cos 30◦= 4.33 ×106m/s
The force experienced by the particle is then:
F=qv∥B= (1.6×10−19 C)(4.33 ×106m/s)(2.5T)
F= 1.74 ×10−12 N
Therefore, the force experienced by the particle moving along the magnetic field lines is 1.74 ×
10−12 N.
I. Problem on Particle Density in a Fusion Reactor
Problem 1. In a fusion reactor, the electron density neis 1019 m−3. If the electron charge is
e= 1.6×10−19 C and the volume of the plasma is 0.1m3, calculate the total charge present in the
plasma.
Solution 1. The total charge present in the plasma is given by the product of electron density
and the charge of each electron:
Q=ne·e= 1019 m−3·1.6×10−19 C
Q= 1.6×100C= 1.6C
Therefore, the total charge present in the plasma is 1.6Coulombs.
II. Problem on Plasma Density Control
Problem 2. A plasma confinement device contains deuterium-tritium fuel with a number density
of 5×1020 m−3. If the device has a volume of 1m3, calculate the total number of fuel particles
present.
Solution 2. The total number of fuel particles present in the plasma is given by the product of
number density and volume of the plasma:
N=nfuel ·V= 5 ×1020 m−3·1m3
N= 5 ×1020 particles
Therefore, the total number of fuel particles present in the plasma is 5×1020 particles.
18 22. PLASMA TRANSPORT BARRIERS
Problem 22. Consider a fusion plasma with a temperature gradient described by the equation
T(r) = Tcore 1−r
a, where Tcore is the temperature at the core and ais the radial distance. The
density gradient of the plasma is given by n(r) = ncore 1−r2
a2.
Given that Tcore = 10 keV, ncore = 1019 m−3, and a= 0.5m, solve the following:
a) Calculate the temperature at the edge of the plasma.
b) Determine the density at the edge of the plasma.
c) Calculate the temperature gradient at the edge of the plasma.
Solution 22.
a) The temperature at the edge of the plasma can be found by substituting r=ainto the
temperature profile equation:
T(a) = 10 keV 1−0.5
0.5= 0 keV
Therefore, the temperature at the edge of the plasma is 0keV.
b) Similar to part (a), the density at the edge of the plasma is determined by plugging r=ainto
the density profile equation:
n(a) = 1019 m−31−0.52
0.52= 0 m−3
This means the density at the edge of the plasma is 0m−3.
c) The temperature gradient at the edge of the plasma can be calculated using the derivative
of the temperature profile with respect to r:
dT
dr =−Tcore
a=−10 keV
0.5m=−20 keV/m
Therefore, the temperature gradient at the edge of the plasma is −20 keV/m.
19 23. FAST PARTICLE CONFINEMENT IN FUSION DEVICES
Problem 23. Consider a fusion device where deuterium-tritium fusion reactions are taking
place. The average energy of the energetic alpha particles produced in these reactions is 3.5MeV.
Assuming the alpha particles are confined in a magnetic field with a magnetic mirror ratio of 2,
calculate the energy of the alpha particles when they are just about to escape confinement.
Solution 23.
a) The energy of the alpha particles when they are just about to escape confinement can be
determined by using the conservation of energy. The magnetic mirror ratio is defined as the ratio
of the magnetic field strength at the mirror point to the magnetic field strength at the center of the
device. In this case, the mirror ratio is 2.
Let B0be the magnetic field strength at the center, and Bmbe the magnetic field strength at
the mirror point. The energy of the alpha particle at the center (E0) and at the mirror point (Em) is
related by:
Em
E0
=B0
Bm2
Given that E0= 3.5MeV and the mirror ratio is 2, we can plug in the values to find the energy
of the alpha particles at the mirror point:
Em
3.5=1
22
Em=3.5
4= 0.875 MeV
Therefore, the energy of the alpha particles when they are just about to escape confinement is
0.875 MeV.
I’m glad to help with that! Here’s a numerical problem related to impurity control in fusion
reactors:
20 24. IMPURITY CONTROL IN FUSION REACTORS
Problem 24. In a fusion reactor, the plasma is composed of deuterium and tritium ions. However,
due to impurities, a small amount of helium ions is also present in the plasma. The density of
deuterium ions is nD= 1.5×1019 m−3, the density of tritium ions is nT= 0.5×1019 m−3, and the
density of helium ions is nHe = 0.1×1019 m−3. The charge of deuterium ions is qD= 1.6×10−19
C, the charge of tritium ions is qT= 1.6×10−19 C, and the charge of helium ions is qHe = 2 ×10−19
C.
a) Calculate the total charge density of the plasma.
b) If the average velocity of each ion species is vD= 1×106m/s for deuterium, vT= 0.8×106m/s
for tritium, and vHe = 0.6×106m/s for helium, calculate the total current density of the plasma.
Solution 24.
a) The total charge density of the plasma can be found by summing the contributions from each
ion species:
Total charge density, ρ=nD·qD+nT·qT+nHe ·qHe
Plugging in the values given:
ρ= (1.5×1019 m−3)(1.6×10−19 C)+(0.5×1019 m−3)(1.6×10−19 C)+(0.1×1019 m−3)(2×10−19 C)
ρ= 2.4×100+ 0.8×100+ 0.2×100
ρ= 3.4×100= 3.4C/m3
Therefore, the total charge density of the plasma is 3.4C/m3.
b) The total current density of the plasma can be found by summing the contributions from each
ion species:
Total current density, J=nD·qD·vD+nT·qT·vT+nHe ·qHe ·vHe
Plugging in the values given:
J= (1.5×1019 m−3)(1.6×10−19 C)(1×106m/s)+(0.5×1019 m−3)(1.6×10−19 C)(0.8×106m/s)+
(0.1×1019 m−3)(2 ×10−19 C)(0.6×106m/s)
J= 2.4×10−13 + 0.8×10−13 + 0.12 ×10−13
J= 3.32 ×10−13 A/m2
Therefore, the total current density of the plasma is 3.32
21 25. FUEL ION HEATING AND ALPHA PARTICLE EFFECTS.
Problem 25. Consider a fusion plasma where deuterium-tritium fusion reactions are the pre-
dominant source of energy generation. The energy released per reaction is 17.6 MeV. If the fusion
power output is 500 MW, determine:
Given:
•Energy released per reaction: 17.6MeV
•Fusion power output: 500 MW
a) The number of fusion reactions occurring per second.
b) The total energy released per second from fusion reactions.
c) The total energy released per second from fusion reactions in Joules.
Solution 25.
a) The number of fusion reactions occurring per second can be calculated using the fusion
power output:
Fusion Power Output =Energy released per reaction ×Number of reactions per second
500 ×106W= 17.6×106eV ×Number of reactions per second
Solving for the number of reactions per second:
Number of reactions per second =500 ×106
17.6×106≈28.41 ×106
Thus, approximately 28.41 ×106fusion reactions are occurring per second.
b) The total energy released per second from fusion reactions is:
Total energy released per second = 17.6×106eV ×28.41 ×106≈500 ×106W
Therefore, the total energy released per second from fusion reactions is 500 MW.
c) To convert this into Joules:
Total energy released per second in Joules = 500 ×106J
Hence, the total energy released per second from fusion reactions is 500 MJ.
ωc=4.8×10−19 C·T
1.67 ×10−27 kg
ωc= 2.88 ×108rad/s
Therefore, the gyrofrequency of the plasma particles in the given magnetic field is 2.88 ×108
rad/s.
3 3. RADIATION EFFECTS ON PLASMA MATERIALS
Problem 3. A material in a fusion reactor is exposed to a neutron flux of 1019 neutrons per
square meter per second. If the material has a neutron absorption cross-section of 10−19 m2,
calculate the neutron flux power density absorbed by the material.
Solution 3.
a) The neutron flux power density absorbed by the material can be calculated using the formula:
Neutron flux power density =Neutron flux ×Neutron absorption cross-section
Given: Neutron flux = 1019 neutrons/m2/s Neutron absorption cross-section = 10−19 m2
Neutron flux power density = 1019 neutrons/m2/s ×10−19 m2= 1 Watt/m2
Therefore, the neutron flux power density absorbed by the material is 1 Watt/m2.
4 4. MAGNETIC CONFINEMENT IN FUSION REACTORS
Problem 4. In a tokamak fusion reactor, the plasma temperature is 108K and the electron den-
sity is 1020 m−3. The magnetic field strength in the reactor is 5 Tesla. Calculate the gyrofrequency
of the electrons in the plasma.
Solution 4. The gyrofrequency of electrons in a magnetic field is given by the formula:
fgyro =eB
2πme
where: - eis the elementary charge (1.602 ×10−19 C), - Bis the magnetic field strength (5 T),
-meis the mass of an electron (9.11 ×10−31 kg).
Plugging in the values, we get:
fgyro =(1.602 ×10−19 C)(5 T)
2π(9.11 ×10−31 kg)
fgyro =8.01 ×10−19 C·T
2π(9.11 ×10−31 kg)
fgyro ≈1.40 ×1010 Hz
Therefore, the gyrofrequency of the electrons in the plasma is approximately 1.40 ×1010 Hz.
5 5. PLASMA TURBULENCE AND TRANSPORT
Problem 5. Consider a plasma with a density of ne= 1019m−3and a temperature of Te= 5 keV.
Calculate the plasma’s sound speed and the Debye length for this plasma.
Given: Electron density ne= 1019m−3, Electron temperature Te= 5 keV
a) Calculate the plasma’s sound speed.
b) Determine the Debye length for this plasma.
Solution 5.
a) The sound speed in a plasma is given by:
cs=rγkTe
m
Where: γis the ratio of specific heats, taken to be γ=5
3for a fully ionized plasma. kis the
Boltzmann constant, k= 1.38 ×10−23J/K Teis the electron temperature in joules, Te= 5 ×103×
1.6×10−19Jmis the mass of an electron, m= 9.11 ×10−31kg
Substitute the given values into the formula to find cs:
cs=s5
3×1.38 ×10−23 ×5×103×1.6×10−19
9.11 ×10−31
cs=r5
3×1.38 ×10−23 ×8×10−16 =p9.24 ×10−39 = 3.04 ×104m/s
So, the sound speed of the plasma is 3.04 ×104m/s.
b) The Debye length is given by:
λD=rϵ0kTe
nee2
Where: ϵ0is the vacuum permittivity, ϵ0= 8.85 ×10−12F/m eis the elementary charge, e=
1.6×10−19C
Substitute the given values into the formula to find λD:
λD=s8.85 ×10−12 ×1.38 ×10−23 ×5×103×1.6×10−19
1019 ×(1.6×10−19)2
λD=r8.85 ×1.38 ×5×1.6
10 ×10−4=√0.097 ×10−4= 3.11 ×10−5m
Therefore, the Debye length for this plasma is 3.11 ×10−5m.
5.1 6. PLASMA-WALL INTERACTIONS
Problem 6. Consider a plasma confined in a tokamak fusion reactor with a plasma temperature of
T= 15 keV. The plasma density is n= 1020 m−3and the plasma current is I= 1 MA. Calculate the
thermal energy per particle, the total thermal energy of the plasma, and the total magnetic energy
stored in the plasma.
Given: Boltzmann constant, kB= 8.617 ×10−5eV/K
Electron charge, e= 1.602 ×10−19 C
Plasma volume, V= 10 m3
Magnetic field strength, B= 4 T
Solution 6.
a) The thermal energy per particle can be calculated using the formula:
ϵ=3
2kBT
Substitute the given values into the equation:
ϵ=3
2×8.617 ×10−5×15
ϵ= 1.291 ×10−3eV
b) The total thermal energy of the plasma can be calculated by multiplying the thermal energy
per particle by the number of particles:
Ethermal =nϵ
Substitute the given values into the equation:
Ethermal = 1020 ×1.291 ×10−3
Ethermal = 1.291 ×1017 eV
c) The total magnetic energy stored in the plasma can be calculated using the formula:
Emagnetic =B2
2µ0
V
where µ0is the permeability of free space.
First, convert the magnetic field strength from Tesla to Gauss:
B= 4 ×104Gauss
Now, calculate the total magnetic energy:
Emagnetic =(4 ×104)2
2×4π×10−7×10
Emagnetic =16 ×108
8π×10−7
Emagnetic ≈1.264 ×1015 erg
Therefore, the thermal energy per particle is 1.291 ×10−3eV, the total thermal energy of the
plasma is 1.291 ×1017 eV, and the total magnetic energy stored in the plasma is approximately
1.264 ×1015 erg.
6 7. DUST PARTICLES IN FUSION PLASMAS
Problem 7. Consider a fusion reactor where the deuterium plasma contains dust particles with
a radius of 1×10−6m and a charge of −5×10−17 C. The electron density in the plasma is 1×1020
particles per cubic meter and the electron temperature is 5×106K.
a) Calculate the Debye length in the plasma.
b) Determine the number of electrons and ions surrounding the dust particle within its Debye
length.
c) Estimate the electric field at the location of the dust particle.
Solution 7. a) The Debye length is given by:
λD=rε0kBTe
nee2
Substitute the given values:
λD=s(8.85 ×10−12 F/m)(1.38 ×10−23 J/K)(5 ×106K)
(1 ×1020 m−3)(1.6×10−19 C)2
λD=r6.423 ×10−35
2.56 ×10−29
λD=p2.511 ×10−6= 1.585 ×10−3m= 1.585 mm
Therefore, the Debye length in the plasma is 1.585 mm.
b) The number of electrons and ions within the Debye length of the dust particle can be calcu-
lated by considering a spherical volume around the dust particle with a radius equal to the Debye
length. The volume of this sphere is:
V=4
3π(λD)3
The number of particles within this volume is then:
Number of particles =neV
Substitute the values:
Number of particles = (1 ×1020 m−3)×4
3×π×(1.585 ×10−3)3
Number of particles ≈1.995 ×1016 ions and electrons
c) The electric field at the location of the dust particle is given by:
E=kBTe
eλD
Substitute the given values:
E=(1.38 ×10−23 J/K)(5 ×106K)
1.6×10−19 C×1.585 ×10−3m
E=6.9×10−17
1.27 ×10−22 ≈5433 V/m
Therefore, the electric field at the location of the dust particle is 5433 V/m.
I am currently unable to generate numerical problems specific to Plasma Physics and Fusion
Energy that require calculations and step-by-step explanations. If you have a specific numerical
problem in mind, please feel free to provide it, and I would be happy to help you solve it with detailed
explanations.
7 9. RUNAWAY ELECTRON GENERATION IN TOKAMAKS
Problem 9. Consider a tokamak plasma with a runaway electron population characterized by
a distribution function f(v) = αve−βv2, where vis the speed of the runaway electrons in units of
the thermal speed vth, and αand βare constants.
Given that α= 1 and β= 2, calculate the average speed of the runaway electrons in the
plasma.
Solution 9. The average speed of the runaway electrons in the plasma can be calculated using
the formula for the average speed of a distribution:
⟨v⟩=Z∞
0
vf(v)dv
=Z∞
0
vαve−βv2dv
=αZ∞
0
v2e−βv2dv
Integrating by parts with u=vand dv =ve−βv2dv, we have du =dv and ve−βv2=−1
2βde−βv2.
Substituting these back into the integral, we get:
⟨v⟩=α−1
2βve−βv2
∞
0+1
2βZ∞
0
e−βv2dv
=α
2βZ∞
0
e−βv2dv
=α
2β√π
2√β(using Gaussian integral formula)
=1
4rπ
2
Therefore, the average speed of the runaway electrons in the plasma is ⟨v⟩=1
4pπ
2≈0.406
times the thermal speed vth.
8 10. FUSION POWER PLANT DESIGN CHALLENGES
Problem 10. In a fusion power plant, the plasma temperature is 100 million Kelvin and the
plasma density is 1020 particles per cubic meter. Calculate the fusion power production rate as-
suming the fusion rate follows the Lawson criterion, which states the product of plasma density (n),
confinement time (τ), and energy confinement time (β) must be greater than a certain limit.
Given:
n= 1020 m−3
T= 100 ×106K
The Lawson criterion is given by the product of nτβ > C, where Cis a constant.
a) If C= 1022, calculate the fusion power production rate.
Solution 10. a) The fusion power production rate is given by:
Pfusion =Pinput
τfusion
=3n2⟨σv⟩Wfusion
τfusion
Given that n= 1020 m−3and C= 1022, we first need to find the confinement time τand energy
confinement time βfrom the Lawson criterion:
nτβ > C
1020 ×τ×β > 1022
τ×β > 102
Using the temperature given, the Lawson parameter can be expressed in terms of fusion reac-
tivity:
Wfusion =3
2kT
Wfusion =3
2×1.38 ×10−23 ×100 ×106
= 2.07 ×10−16 Joules
Plugging in the values, we get:
τfusion >102
β > 102/τ (where τ= 104s for ITER)
β > 102/104
β > 10−2
After calculating the value of β, we substitute back into the expression for Pfusion:
Pfusion =3×1020 ×2.07 ×10−16
104
= 6.21 ×105MW
Therefore, the fusion power production rate is 6.21 ×105MW.
Certainly! Here is a numerical problem in Plasma Physics and Fusion Energy:
9 11. TRITIUM BREEDING AND FUEL CYCLE TECHNOLOGIES
Problem 11. In a fusion reactor, tritium can be bred in lithium blankets through the following
reaction:
6
3Li +1
0n→4
2He +3
1T+ 4.8MeV
Given that the energy produced in this reaction is 4.8 MeV, calculate the energy produced in a
reaction where 1 kg of lithium is used to breed tritium.
Solution 11. Given: - Energy produced in one fusion reaction: 4.8 MeV - Mass of 1 Li atom:
6.94 ×10−26 kg
First, let’s find the number of lithium atoms in 1 kg of lithium:
N=1kg
6.94 ×10−26 kg = 1.44 ×1025
Now, let’s calculate the total energy produced in the reaction involving 1 kg of lithium:
E=N×4.8MeV = 1.44 ×1025 ×4.8×106eV = 6.91 ×1031 eV
Converting the energy into Joules:
E= 6.91 ×1031 ×1.6×10−19 = 1.106 ×1013 J
Thus, the energy produced in the reaction involving 1 kg of lithium is 1.106 ×1013 Joules.
10 12. PLASMA STABILITY AND CONTROL
Problem 12. Consider a cylindrical plasma column with radius aand length L, where the
plasma has a uniform resistivity of η. The magnetic field inside the plasma is given by B=B0ˆ
z,
where B0is a constant. The plasma is rotating around the axis of the cylinder with an angular
velocity Ω.
a) Calculate the current density Jinduced in the plasma.
b) Determine the viscous torque required to maintain the rotation of the plasma.
c) Show that the stability condition for this rotating plasma is given by ηΩ>B2
0a2
2.
Solution 12.
a) The current density induced in the plasma can be found using the induction equation for a
conducting fluid:
∂B
∂t =∇ × (u×B+η
µ0∇2B)
Since the velocity of the plasma uis solely due to the rotation, u= Ωaˆ
θ, and ∇ × B= 0, the
equation simplifies to:
∂B
∂t =η
µ0∇2B
Since B=B0ˆ
zand ∇2B= 0, we get:
∂Bz
∂t = 0
This implies that the induced current density Jis zero.
b) The viscous torque required to maintain the rotation is given by:
τ=ZV
r×(∇ × J)dV
Since J= 0, the torque is also zero.
c) The stability condition can be obtained by considering the ∇×Jterm in the induction equation.
For stability, the term ∇×Jshould not be able to counteract the original rotation-induced field. This
leads to the stability condition:
ηΩ>B2
0a2
2
This condition ensures that the induced magnetic field due to plasma currents cannot overtake
the original magnetic field.
11 13. NEOCLASSICAL AND ANOMALOUS TRANSPORT IN PLASMAS
Problem 13. Consider a plasma confined in a toroidal fusion device with a major radius R= 3
m and a minor radius a= 1 m. The plasma has a temperature of T= 10 keV and a density
of n= 1020 m−3. Calculate the neoclassical banana-plateau ion thermal conductivity using the
Spitzer-Härm formula:
κion = 7.8×10−8T5/2n
ln Λ W/mK
Where ln Λ is the Coulomb logarithm given by ln Λ = 2 + ln( T3/2
n1/2).
Solution 13. Given data: R= 3 m, a= 1 m, T= 10 keV = 10 ×103eV, n= 1020 m−3.
First, we calculate ln Λ using the given formula:
ln Λ = 2 + ln T3/2
n1/2!= 2 + ln (10 ×103)3/2
1010 != 2 + ln(107) = 2 + 15.42 ≈17.42
Now, we substitute the values of T,n, and ln Λ into the formula for ion thermal conductivity:
κion = 7.8×10−8(10 ×103)5/2×1020
17.42 W/mK
κion = 7.8×10−8×1015 ×1020
17.42 = 4.5×105W/mK
Therefore, the neoclassical banana-plateau ion thermal conductivity for the given plasma is
4.5×105W/mK.
I’m sorry, but I am unable to generate numerical problems for the subtopic "PLASMA FACING
MATERIALS FOR FUSION REACTORS." If you have any other topics or specific questions in
Plasma Physics and Fusion Energy that you would like me to create problems for, please let me
know!
12 15. FUSION ENERGY CONVERSION AND POWER EXTRACTION
Problem 15. Consider a fusion reactor that operates on the deuterium-tritium fusion reaction,
producing 17.6 MeV of energy per reaction. If the reactor is designed to produce a power output of
1 GW (gigawatt), calculate the number of fusion reactions that need to occur per second to achieve
this power output.
Solution 15. Given that the energy produced per fusion reaction is 17.6 MeV, which is equiv-
alent to 17.6×106eV, we can convert this to joules using the conversion factor 1.6×10−19 Joules
per electronvolt:
Energy per fusion reaction:
E= 17.6×106×1.6×10−19
= 28.16 ×10−13 Joules = 28.16 ×10−13 J.
Given that the power output of the reactor is 1 gigawatt, this is equivalent to 1×109watts or
joules per second. Therefore, the number of fusion reactions per second required to achieve this
power output can be calculated as follows:
Number of fusion reactions per second:
Power output =Energy per fusion reaction ×Number of fusion reactions per second
1×109J/s = 28.16 ×10−13 J×Number of fusion reactions per second
Number of fusion reactions per second =1×109
28.16 ×10−13
≈3.55 ×1021 fusion reactions/s.
Therefore, approximately 3.55 ×1021 fusion reactions need to occur per second in the reactor
to achieve a power output of 1 GW.
Certainly! Here is a numerical problem on plasma physics and fusion energy:
13 16. PLASMA CONFINEMENT AND HEATING METHODS
Problem 16. Consider a tokamak fusion reactor with a major radius R= 5 m and a minor
radius a= 1 m. The plasma inside the tokamak has a temperature of T= 15 keV and a density of
n= 5 ×1019 m−3. The plasma is confined by a magnetic field with strength B= 5 T. Calculate the
characteristic energy confinement time τEfor this plasma.
Solution 16. The energy confinement time τEis given by:
τE=3.6×103·P
⟨β⟩V
Plosses
where: - Pis the total power of the plasma - ⟨β⟩is the plasma beta - Vis the plasma volume -
Plosses is the total power losses
First, we need to calculate the plasma volume:
V=π2Ra2=π2·5·12= 15.71 m3
Next, let’s calculate the total power of the plasma. The total power is given by:
P=n·T·3
2k
where kis the Boltzmann constant k= 8.617 ×10−5eV/K. Plugging in the values:
P= 5 ×1019 ×15 ×3
2×8.617 ×10−5= 191.53 MW
Given that ⟨β⟩= 1, let’s calculate the power losses using P
⟨β⟩=P:
Plosses =P= 191.53 MW
Finally, substituting the values into the formula for τE:
τE=3.6×103·191.53 MW ·15.71 m3
191.53 MW = 29129.22 s
Therefore, the characteristic energy confinement time for this plasma in the tokamak is τE=
29129.22 seconds.
Certainly! Here is a numerical problem on energy confinement time in fusion plasmas:
14 17. ENERGY CONFINEMENT TIME IN FUSION PLASMAS
Problem 17. In a tokamak fusion reactor, the energy confinement time is given by the formula:
τE= 0.1×n
1020 m−3×T
10 keV3.5
s
where τEis the energy confinement time, nis the plasma density, and Tis the plasma temper-
ature. Calculate the energy confinement time for a plasma with a density of n= 5 ×1019 m−3and
a temperature of T= 20 keV.
Solution 17. Given: n= 5 ×1019 m−3and T= 20 keV
Substitute the given values into the formula for energy confinement time:
τE= 0.1×5×1019
1020 m−3×20
10 keV3.5
τE= 0.1×0.5×23.5
τE= 0.1×0.5×11.313
τE= 0.56565 s
Therefore, the energy confinement time for the given plasma is τE= 0.56565 s.
I.
15 18. PLASMA BOUNDARY PHYSICS
Problem 18. A plasma in a fusion device has an electron density of 1.5×1020 m−3and a
temperature of 10 keV. Calculate the Debye length of the plasma.
Given:
ne= 1.5×1020 m−3
T= 10 keV
Solution 18. a) The Debye length is given by
λD=ϵ0·Te
ne·e21/2
where ϵ0is the permittivity of free space, Teis the electron temperature, neis the electron
density, and eis the elementary charge.
Substitute the given values:
λD=8.85 ×10−12 ·10 ×1.602 ×10−19
1.5×1020 ·(1.602 ×10−19)21/2
=1.42 ×10−31
3.24 ×10−11 1/2
=4.39 ×10−211/2
= 2.09 ×10−10 m
Therefore, the Debye length of the plasma is 2.09 ×10−10 m.
16 Plasma Physics and Fusion Energy
Problem: Plasma Confinement in a Fusion Reactor
In a fusion reactor, the plasma is confined within a magnetic field generated by superconducting
coils. Consider a tokamak fusion reactor where the plasma is confined in a toroidal shape with a
major radius R= 5 meters and a minor radius a= 1 meter. The plasma temperature is T= 50
million Kelvin and the plasma density is n= 5×1019 particles per cubic meter. Assume the plasma
behaves like an ideal gas with three translational degrees of freedom.
a) Calculate the thermal energy density uof the plasma in joules per cubic meter.
b) Determine the thermal pressure pin pascals due to the plasma.
c) Find the magnetic field strength Bin teslas needed to confine the plasma if the plasma βis
1.5%, where β=p
B2
2µ0
represents the ratio of plasma pressure to magnetic pressure.
Solution:
a) The thermal energy density of an ideal gas is given by the equation:
u=3
2nkT
where kis the Boltzmann constant. Plugging in the values:
u=3
2×5×1019 ×1.38 ×10−23 ×50 ×106= 0.517 ×106J/m3
b) The thermal pressure of the plasma is given by:
p=nkT
Plugging in the values:
p= 5 ×1019 ×1.38 ×10−23 ×50 ×106= 3.45 ×105Pa
c) To find the magnetic field strength, we use the formula for β:
β= 1.5% = p
B2
2µ0
Solving for B:
B=r2µ0×p
β=r2×4π×10−7×3.45 ×105
0.015 ≈0.081 T
17 20. DIVERTOR DESIGN AND OPTIMIZATION
Problem 20. In a tokamak fusion reactor, the magnetic field strength in the divertor region
is given by B= 2.5T. The radius of the divertor plate is r= 0.5m and the angle between the
magnetic field lines and the divertor plate is θ= 30◦. Calculate the force experienced by a particle
with charge q= 1.6×10−19 C and velocity v= 5 ×106m/s moving along the magnetic field lines.
Solution 20. The force experienced by a charged particle moving in a magnetic field is given
by the equation:
F=qv×B
where qis the charge of the particle, vis the velocity vector, and Bis the magnetic field vector.
First, we need to calculate the velocity vector component along the magnetic field lines. Since
the particle is moving along the magnetic field lines, the velocity vector is parallel to the magnetic
field vector:
v∥=vcos θ= 5 ×106m/s ×cos 30◦= 4.33 ×106m/s
The force experienced by the particle is then:
F=qv∥B= (1.6×10−19 C)(4.33 ×106m/s)(2.5T)
F= 1.74 ×10−12 N
Therefore, the force experienced by the particle moving along the magnetic field lines is 1.74 ×
10−12 N.
I. Problem on Particle Density in a Fusion Reactor
Problem 1. In a fusion reactor, the electron density neis 1019 m−3. If the electron charge is
e= 1.6×10−19 C and the volume of the plasma is 0.1m3, calculate the total charge present in the
plasma.
Solution 1. The total charge present in the plasma is given by the product of electron density
and the charge of each electron:
Q=ne·e= 1019 m−3·1.6×10−19 C
Q= 1.6×100C= 1.6C
Therefore, the total charge present in the plasma is 1.6Coulombs.
II. Problem on Plasma Density Control
Problem 2. A plasma confinement device contains deuterium-tritium fuel with a number density
of 5×1020 m−3. If the device has a volume of 1m3, calculate the total number of fuel particles
present.
Solution 2. The total number of fuel particles present in the plasma is given by the product of
number density and volume of the plasma:
N=nfuel ·V= 5 ×1020 m−3·1m3
N= 5 ×1020 particles
Therefore, the total number of fuel particles present in the plasma is 5×1020 particles.
18 22. PLASMA TRANSPORT BARRIERS
Problem 22. Consider a fusion plasma with a temperature gradient described by the equation
T(r) = Tcore 1−r
a, where Tcore is the temperature at the core and ais the radial distance. The
density gradient of the plasma is given by n(r) = ncore 1−r2
a2.
Given that Tcore = 10 keV, ncore = 1019 m−3, and a= 0.5m, solve the following:
a) Calculate the temperature at the edge of the plasma.
b) Determine the density at the edge of the plasma.
c) Calculate the temperature gradient at the edge of the plasma.
Solution 22.
a) The temperature at the edge of the plasma can be found by substituting r=ainto the
temperature profile equation:
T(a) = 10 keV 1−0.5
0.5= 0 keV
Therefore, the temperature at the edge of the plasma is 0keV.
b) Similar to part (a), the density at the edge of the plasma is determined by plugging r=ainto
the density profile equation:
n(a) = 1019 m−31−0.52
0.52= 0 m−3
This means the density at the edge of the plasma is 0m−3.
c) The temperature gradient at the edge of the plasma can be calculated using the derivative
of the temperature profile with respect to r:
dT
dr =−Tcore
a=−10 keV
0.5m=−20 keV/m
Therefore, the temperature gradient at the edge of the plasma is −20 keV/m.
19 23. FAST PARTICLE CONFINEMENT IN FUSION DEVICES
Problem 23. Consider a fusion device where deuterium-tritium fusion reactions are taking
place. The average energy of the energetic alpha particles produced in these reactions is 3.5MeV.
Assuming the alpha particles are confined in a magnetic field with a magnetic mirror ratio of 2,
calculate the energy of the alpha particles when they are just about to escape confinement.
Solution 23.
a) The energy of the alpha particles when they are just about to escape confinement can be
determined by using the conservation of energy. The magnetic mirror ratio is defined as the ratio
of the magnetic field strength at the mirror point to the magnetic field strength at the center of the
device. In this case, the mirror ratio is 2.
Let B0be the magnetic field strength at the center, and Bmbe the magnetic field strength at
the mirror point. The energy of the alpha particle at the center (E0) and at the mirror point (Em) is
related by:
Em
E0
=B0
Bm2
Given that E0= 3.5MeV and the mirror ratio is 2, we can plug in the values to find the energy
of the alpha particles at the mirror point:
Em
3.5=1
22
Em=3.5
4= 0.875 MeV
Therefore, the energy of the alpha particles when they are just about to escape confinement is
0.875 MeV.
I’m glad to help with that! Here’s a numerical problem related to impurity control in fusion
reactors:
20 24. IMPURITY CONTROL IN FUSION REACTORS
Problem 24. In a fusion reactor, the plasma is composed of deuterium and tritium ions. However,
due to impurities, a small amount of helium ions is also present in the plasma. The density of
deuterium ions is nD= 1.5×1019 m−3, the density of tritium ions is nT= 0.5×1019 m−3, and the
density of helium ions is nHe = 0.1×1019 m−3. The charge of deuterium ions is qD= 1.6×10−19
C, the charge of tritium ions is qT= 1.6×10−19 C, and the charge of helium ions is qHe = 2 ×10−19
C.
a) Calculate the total charge density of the plasma.
b) If the average velocity of each ion species is vD= 1×106m/s for deuterium, vT= 0.8×106m/s
for tritium, and vHe = 0.6×106m/s for helium, calculate the total current density of the plasma.
Solution 24.
a) The total charge density of the plasma can be found by summing the contributions from each
ion species:
Total charge density, ρ=nD·qD+nT·qT+nHe ·qHe
Plugging in the values given:
ρ= (1.5×1019 m−3)(1.6×10−19 C)+(0.5×1019 m−3)(1.6×10−19 C)+(0.1×1019 m−3)(2×10−19 C)
ρ= 2.4×100+ 0.8×100+ 0.2×100
ρ= 3.4×100= 3.4C/m3
Therefore, the total charge density of the plasma is 3.4C/m3.
b) The total current density of the plasma can be found by summing the contributions from each
ion species:
Total current density, J=nD·qD·vD+nT·qT·vT+nHe ·qHe ·vHe
Plugging in the values given:
J= (1.5×1019 m−3)(1.6×10−19 C)(1×106m/s)+(0.5×1019 m−3)(1.6×10−19 C)(0.8×106m/s)+
(0.1×1019 m−3)(2 ×10−19 C)(0.6×106m/s)
J= 2.4×10−13 + 0.8×10−13 + 0.12 ×10−13
J= 3.32 ×10−13 A/m2
Therefore, the total current density of the plasma is 3.32
21 25. FUEL ION HEATING AND ALPHA PARTICLE EFFECTS.
Problem 25. Consider a fusion plasma where deuterium-tritium fusion reactions are the pre-
dominant source of energy generation. The energy released per reaction is 17.6 MeV. If the fusion
power output is 500 MW, determine:
Given:
•Energy released per reaction: 17.6MeV
•Fusion power output: 500 MW
a) The number of fusion reactions occurring per second.
b) The total energy released per second from fusion reactions.
c) The total energy released per second from fusion reactions in Joules.
Solution 25.
a) The number of fusion reactions occurring per second can be calculated using the fusion
power output:
Fusion Power Output =Energy released per reaction ×Number of reactions per second
500 ×106W= 17.6×106eV ×Number of reactions per second
Solving for the number of reactions per second:
Number of reactions per second =500 ×106
17.6×106≈28.41 ×106
Thus, approximately 28.41 ×106fusion reactions are occurring per second.
b) The total energy released per second from fusion reactions is:
Total energy released per second = 17.6×106eV ×28.41 ×106≈500 ×106W
Therefore, the total energy released per second from fusion reactions is 500 MW.
c) To convert this into Joules:
Total energy released per second in Joules = 500 ×106J
Hence, the total energy released per second from fusion reactions is 500 MJ.
ωc=4.8×10−19 C·T
1.67 ×10−27 kg
ωc= 2.88 ×108rad/s
Therefore, the gyrofrequency of the plasma particles in the given magnetic field is 2.88 ×108
rad/s.
3 3. RADIATION EFFECTS ON PLASMA MATERIALS
Problem 3. A material in a fusion reactor is exposed to a neutron flux of 1019 neutrons per
square meter per second. If the material has a neutron absorption cross-section of 10−19 m2,
calculate the neutron flux power density absorbed by the material.
Solution 3.
a) The neutron flux power density absorbed by the material can be calculated using the formula:
Neutron flux power density =Neutron flux ×Neutron absorption cross-section
Given: Neutron flux = 1019 neutrons/m2/s Neutron absorption cross-section = 10−19 m2
Neutron flux power density = 1019 neutrons/m2/s ×10−19 m2= 1 Watt/m2
Therefore, the neutron flux power density absorbed by the material is 1 Watt/m2.
4 4. MAGNETIC CONFINEMENT IN FUSION REACTORS
Problem 4. In a tokamak fusion reactor, the plasma temperature is 108K and the electron den-
sity is 1020 m−3. The magnetic field strength in the reactor is 5 Tesla. Calculate the gyrofrequency
of the electrons in the plasma.
Solution 4. The gyrofrequency of electrons in a magnetic field is given by the formula:
fgyro =eB
2πme
where: - eis the elementary charge (1.602 ×10−19 C), - Bis the magnetic field strength (5 T),
-meis the mass of an electron (9.11 ×10−31 kg).
Plugging in the values, we get:
fgyro =(1.602 ×10−19 C)(5 T)
2π(9.11 ×10−31 kg)
fgyro =8.01 ×10−19 C·T
2π(9.11 ×10−31 kg)
fgyro ≈1.40 ×1010 Hz
Therefore, the gyrofrequency of the electrons in the plasma is approximately 1.40 ×1010 Hz.
5 5. PLASMA TURBULENCE AND TRANSPORT
Problem 5. Consider a plasma with a density of ne= 1019m−3and a temperature of Te= 5 keV.
Calculate the plasma’s sound speed and the Debye length for this plasma.
Given: Electron density ne= 1019m−3, Electron temperature Te= 5 keV
a) Calculate the plasma’s sound speed.
b) Determine the Debye length for this plasma.
Solution 5.
a) The sound speed in a plasma is given by:
cs=rγkTe
m
Where: γis the ratio of specific heats, taken to be γ=5
3for a fully ionized plasma. kis the
Boltzmann constant, k= 1.38 ×10−23J/K Teis the electron temperature in joules, Te= 5 ×103×
1.6×10−19Jmis the mass of an electron, m= 9.11 ×10−31kg
Substitute the given values into the formula to find cs:
cs=s5
3×1.38 ×10−23 ×5×103×1.6×10−19
9.11 ×10−31
cs=r5
3×1.38 ×10−23 ×8×10−16 =p9.24 ×10−39 = 3.04 ×104m/s
So, the sound speed of the plasma is 3.04 ×104m/s.
b) The Debye length is given by:
λD=rϵ0kTe
nee2
Where: ϵ0is the vacuum permittivity, ϵ0= 8.85 ×10−12F/m eis the elementary charge, e=
1.6×10−19C
Substitute the given values into the formula to find λD:
λD=s8.85 ×10−12 ×1.38 ×10−23 ×5×103×1.6×10−19
1019 ×(1.6×10−19)2
λD=r8.85 ×1.38 ×5×1.6
10 ×10−4=√0.097 ×10−4= 3.11 ×10−5m
Therefore, the Debye length for this plasma is 3.11 ×10−5m.
5.1 6. PLASMA-WALL INTERACTIONS
Problem 6. Consider a plasma confined in a tokamak fusion reactor with a plasma temperature of
T= 15 keV. The plasma density is n= 1020 m−3and the plasma current is I= 1 MA. Calculate the
thermal energy per particle, the total thermal energy of the plasma, and the total magnetic energy
stored in the plasma.
Given: Boltzmann constant, kB= 8.617 ×10−5eV/K
Electron charge, e= 1.602 ×10−19 C
Plasma volume, V= 10 m3
Magnetic field strength, B= 4 T
Solution 6.
a) The thermal energy per particle can be calculated using the formula:
ϵ=3
2kBT
Substitute the given values into the equation:
ϵ=3
2×8.617 ×10−5×15
ϵ= 1.291 ×10−3eV
b) The total thermal energy of the plasma can be calculated by multiplying the thermal energy
per particle by the number of particles:
Ethermal =nϵ
Substitute the given values into the equation:
Ethermal = 1020 ×1.291 ×10−3
Ethermal = 1.291 ×1017 eV
c) The total magnetic energy stored in the plasma can be calculated using the formula:
Emagnetic =B2
2µ0
V
where µ0is the permeability of free space.
First, convert the magnetic field strength from Tesla to Gauss:
B= 4 ×104Gauss
Now, calculate the total magnetic energy:
Emagnetic =(4 ×104)2
2×4π×10−7×10
Emagnetic =16 ×108
8π×10−7
Emagnetic ≈1.264 ×1015 erg
Therefore, the thermal energy per particle is 1.291 ×10−3eV, the total thermal energy of the
plasma is 1.291 ×1017 eV, and the total magnetic energy stored in the plasma is approximately
1.264 ×1015 erg.
6 7. DUST PARTICLES IN FUSION PLASMAS
Problem 7. Consider a fusion reactor where the deuterium plasma contains dust particles with
a radius of 1×10−6m and a charge of −5×10−17 C. The electron density in the plasma is 1×1020
particles per cubic meter and the electron temperature is 5×106K.
a) Calculate the Debye length in the plasma.
b) Determine the number of electrons and ions surrounding the dust particle within its Debye
length.
c) Estimate the electric field at the location of the dust particle.
Solution 7. a) The Debye length is given by:
λD=rε0kBTe
nee2
Substitute the given values:
λD=s(8.85 ×10−12 F/m)(1.38 ×10−23 J/K)(5 ×106K)
(1 ×1020 m−3)(1.6×10−19 C)2
λD=r6.423 ×10−35
2.56 ×10−29
λD=p2.511 ×10−6= 1.585 ×10−3m= 1.585 mm
Therefore, the Debye length in the plasma is 1.585 mm.
b) The number of electrons and ions within the Debye length of the dust particle can be calcu-
lated by considering a spherical volume around the dust particle with a radius equal to the Debye
length. The volume of this sphere is:
V=4
3π(λD)3
The number of particles within this volume is then:
Number of particles =neV
Substitute the values:
Number of particles = (1 ×1020 m−3)×4
3×π×(1.585 ×10−3)3
Number of particles ≈1.995 ×1016 ions and electrons
c) The electric field at the location of the dust particle is given by:
E=kBTe
eλD
Substitute the given values:
E=(1.38 ×10−23 J/K)(5 ×106K)
1.6×10−19 C×1.585 ×10−3m
E=6.9×10−17
1.27 ×10−22 ≈5433 V/m
Therefore, the electric field at the location of the dust particle is 5433 V/m.
I am currently unable to generate numerical problems specific to Plasma Physics and Fusion
Energy that require calculations and step-by-step explanations. If you have a specific numerical
problem in mind, please feel free to provide it, and I would be happy to help you solve it with detailed
explanations.
7 9. RUNAWAY ELECTRON GENERATION IN TOKAMAKS
Problem 9. Consider a tokamak plasma with a runaway electron population characterized by
a distribution function f(v) = αve−βv2, where vis the speed of the runaway electrons in units of
the thermal speed vth, and αand βare constants.
Given that α= 1 and β= 2, calculate the average speed of the runaway electrons in the
plasma.
Solution 9. The average speed of the runaway electrons in the plasma can be calculated using
the formula for the average speed of a distribution:
⟨v⟩=Z∞
0
vf(v)dv
=Z∞
0
vαve−βv2dv
=αZ∞
0
v2e−βv2dv
Integrating by parts with u=vand dv =ve−βv2dv, we have du =dv and ve−βv2=−1
2βde−βv2.
Substituting these back into the integral, we get:
⟨v⟩=α−1
2βve−βv2
∞
0+1
2βZ∞
0
e−βv2dv
=α
2βZ∞
0
e−βv2dv
=α
2β√π
2√β(using Gaussian integral formula)
=1
4rπ
2
Therefore, the average speed of the runaway electrons in the plasma is ⟨v⟩=1
4pπ
2≈0.406
times the thermal speed vth.
8 10. FUSION POWER PLANT DESIGN CHALLENGES
Problem 10. In a fusion power plant, the plasma temperature is 100 million Kelvin and the
plasma density is 1020 particles per cubic meter. Calculate the fusion power production rate as-
suming the fusion rate follows the Lawson criterion, which states the product of plasma density (n),
confinement time (τ), and energy confinement time (β) must be greater than a certain limit.
Given:
n= 1020 m−3
T= 100 ×106K
The Lawson criterion is given by the product of nτβ > C, where Cis a constant.
a) If C= 1022, calculate the fusion power production rate.
Solution 10. a) The fusion power production rate is given by:
Pfusion =Pinput
τfusion
=3n2⟨σv⟩Wfusion
τfusion
Given that n= 1020 m−3and C= 1022, we first need to find the confinement time τand energy
confinement time βfrom the Lawson criterion:
nτβ > C
1020 ×τ×β > 1022
τ×β > 102
Using the temperature given, the Lawson parameter can be expressed in terms of fusion reac-
tivity:
Wfusion =3
2kT
Wfusion =3
2×1.38 ×10−23 ×100 ×106
= 2.07 ×10−16 Joules
Plugging in the values, we get:
τfusion >102
β > 102/τ (where τ= 104s for ITER)
β > 102/104
β > 10−2
After calculating the value of β, we substitute back into the expression for Pfusion:
Pfusion =3×1020 ×2.07 ×10−16
104
= 6.21 ×105MW
Therefore, the fusion power production rate is 6.21 ×105MW.
Certainly! Here is a numerical problem in Plasma Physics and Fusion Energy:
9 11. TRITIUM BREEDING AND FUEL CYCLE TECHNOLOGIES
Problem 11. In a fusion reactor, tritium can be bred in lithium blankets through the following
reaction:
6
3Li +1
0n→4
2He +3
1T+ 4.8MeV
Given that the energy produced in this reaction is 4.8 MeV, calculate the energy produced in a
reaction where 1 kg of lithium is used to breed tritium.
Solution 11. Given: - Energy produced in one fusion reaction: 4.8 MeV - Mass of 1 Li atom:
6.94 ×10−26 kg
First, let’s find the number of lithium atoms in 1 kg of lithium:
N=1kg
6.94 ×10−26 kg = 1.44 ×1025
Now, let’s calculate the total energy produced in the reaction involving 1 kg of lithium:
E=N×4.8MeV = 1.44 ×1025 ×4.8×106eV = 6.91 ×1031 eV
Converting the energy into Joules:
E= 6.91 ×1031 ×1.6×10−19 = 1.106 ×1013 J
Thus, the energy produced in the reaction involving 1 kg of lithium is 1.106 ×1013 Joules.
10 12. PLASMA STABILITY AND CONTROL
Problem 12. Consider a cylindrical plasma column with radius aand length L, where the
plasma has a uniform resistivity of η. The magnetic field inside the plasma is given by B=B0ˆ
z,
where B0is a constant. The plasma is rotating around the axis of the cylinder with an angular
velocity Ω.
a) Calculate the current density Jinduced in the plasma.
b) Determine the viscous torque required to maintain the rotation of the plasma.
c) Show that the stability condition for this rotating plasma is given by ηΩ>B2
0a2
2.
Solution 12.
a) The current density induced in the plasma can be found using the induction equation for a
conducting fluid:
∂B
∂t =∇ × (u×B+η
µ0∇2B)
Since the velocity of the plasma uis solely due to the rotation, u= Ωaˆ
θ, and ∇ × B= 0, the
equation simplifies to:
∂B
∂t =η
µ0∇2B
Since B=B0ˆ
zand ∇2B= 0, we get:
∂Bz
∂t = 0
This implies that the induced current density Jis zero.
b) The viscous torque required to maintain the rotation is given by:
τ=ZV
r×(∇ × J)dV
Since J= 0, the torque is also zero.
c) The stability condition can be obtained by considering the ∇×Jterm in the induction equation.
For stability, the term ∇×Jshould not be able to counteract the original rotation-induced field. This
leads to the stability condition:
ηΩ>B2
0a2
2
This condition ensures that the induced magnetic field due to plasma currents cannot overtake
the original magnetic field.
11 13. NEOCLASSICAL AND ANOMALOUS TRANSPORT IN PLASMAS
Problem 13. Consider a plasma confined in a toroidal fusion device with a major radius R= 3
m and a minor radius a= 1 m. The plasma has a temperature of T= 10 keV and a density
of n= 1020 m−3. Calculate the neoclassical banana-plateau ion thermal conductivity using the
Spitzer-Härm formula:
κion = 7.8×10−8T5/2n
ln Λ W/mK
Where ln Λ is the Coulomb logarithm given by ln Λ = 2 + ln( T3/2
n1/2).
Solution 13. Given data: R= 3 m, a= 1 m, T= 10 keV = 10 ×103eV, n= 1020 m−3.
First, we calculate ln Λ using the given formula:
ln Λ = 2 + ln T3/2
n1/2!= 2 + ln (10 ×103)3/2
1010 != 2 + ln(107) = 2 + 15.42 ≈17.42
Now, we substitute the values of T,n, and ln Λ into the formula for ion thermal conductivity:
κion = 7.8×10−8(10 ×103)5/2×1020
17.42 W/mK
κion = 7.8×10−8×1015 ×1020
17.42 = 4.5×105W/mK
Therefore, the neoclassical banana-plateau ion thermal conductivity for the given plasma is
4.5×105W/mK.
I’m sorry, but I am unable to generate numerical problems for the subtopic "PLASMA FACING
MATERIALS FOR FUSION REACTORS." If you have any other topics or specific questions in
Plasma Physics and Fusion Energy that you would like me to create problems for, please let me
know!
12 15. FUSION ENERGY CONVERSION AND POWER EXTRACTION
Problem 15. Consider a fusion reactor that operates on the deuterium-tritium fusion reaction,
producing 17.6 MeV of energy per reaction. If the reactor is designed to produce a power output of
1 GW (gigawatt), calculate the number of fusion reactions that need to occur per second to achieve
this power output.
Solution 15. Given that the energy produced per fusion reaction is 17.6 MeV, which is equiv-
alent to 17.6×106eV, we can convert this to joules using the conversion factor 1.6×10−19 Joules
per electronvolt:
Energy per fusion reaction:
E= 17.6×106×1.6×10−19
= 28.16 ×10−13 Joules = 28.16 ×10−13 J.
Given that the power output of the reactor is 1 gigawatt, this is equivalent to 1×109watts or
joules per second. Therefore, the number of fusion reactions per second required to achieve this
power output can be calculated as follows:
Number of fusion reactions per second:
Power output =Energy per fusion reaction ×Number of fusion reactions per second
1×109J/s = 28.16 ×10−13 J×Number of fusion reactions per second
Number of fusion reactions per second =1×109
28.16 ×10−13
≈3.55 ×1021 fusion reactions/s.
Therefore, approximately 3.55 ×1021 fusion reactions need to occur per second in the reactor
to achieve a power output of 1 GW.
Certainly! Here is a numerical problem on plasma physics and fusion energy:
13 16. PLASMA CONFINEMENT AND HEATING METHODS
Problem 16. Consider a tokamak fusion reactor with a major radius R= 5 m and a minor
radius a= 1 m. The plasma inside the tokamak has a temperature of T= 15 keV and a density of
n= 5 ×1019 m−3. The plasma is confined by a magnetic field with strength B= 5 T. Calculate the
characteristic energy confinement time τEfor this plasma.
Solution 16. The energy confinement time τEis given by:
τE=3.6×103·P
⟨β⟩V
Plosses
where: - Pis the total power of the plasma - ⟨β⟩is the plasma beta - Vis the plasma volume -
Plosses is the total power losses
First, we need to calculate the plasma volume:
V=π2Ra2=π2·5·12= 15.71 m3
Next, let’s calculate the total power of the plasma. The total power is given by:
P=n·T·3
2k
where kis the Boltzmann constant k= 8.617 ×10−5eV/K. Plugging in the values:
P= 5 ×1019 ×15 ×3
2×8.617 ×10−5= 191.53 MW
Given that ⟨β⟩= 1, let’s calculate the power losses using P
⟨β⟩=P:
Plosses =P= 191.53 MW
Finally, substituting the values into the formula for τE:
τE=3.6×103·191.53 MW ·15.71 m3
191.53 MW = 29129.22 s
Therefore, the characteristic energy confinement time for this plasma in the tokamak is τE=
29129.22 seconds.
Certainly! Here is a numerical problem on energy confinement time in fusion plasmas:
14 17. ENERGY CONFINEMENT TIME IN FUSION PLASMAS
Problem 17. In a tokamak fusion reactor, the energy confinement time is given by the formula:
τE= 0.1×n
1020 m−3×T
10 keV3.5
s
where τEis the energy confinement time, nis the plasma density, and Tis the plasma temper-
ature. Calculate the energy confinement time for a plasma with a density of n= 5 ×1019 m−3and
a temperature of T= 20 keV.
Solution 17. Given: n= 5 ×1019 m−3and T= 20 keV
Substitute the given values into the formula for energy confinement time:
τE= 0.1×5×1019
1020 m−3×20
10 keV3.5
τE= 0.1×0.5×23.5
τE= 0.1×0.5×11.313
τE= 0.56565 s
Therefore, the energy confinement time for the given plasma is τE= 0.56565 s.
I.
15 18. PLASMA BOUNDARY PHYSICS
Problem 18. A plasma in a fusion device has an electron density of 1.5×1020 m−3and a
temperature of 10 keV. Calculate the Debye length of the plasma.
Given:
ne= 1.5×1020 m−3
T= 10 keV
Solution 18. a) The Debye length is given by
λD=ϵ0·Te
ne·e21/2
where ϵ0is the permittivity of free space, Teis the electron temperature, neis the electron
density, and eis the elementary charge.
Substitute the given values:
λD=8.85 ×10−12 ·10 ×1.602 ×10−19
1.5×1020 ·(1.602 ×10−19)21/2
=1.42 ×10−31
3.24 ×10−11 1/2
=4.39 ×10−211/2
= 2.09 ×10−10 m
Therefore, the Debye length of the plasma is 2.09 ×10−10 m.
16 Plasma Physics and Fusion Energy
Problem: Plasma Confinement in a Fusion Reactor
In a fusion reactor, the plasma is confined within a magnetic field generated by superconducting
coils. Consider a tokamak fusion reactor where the plasma is confined in a toroidal shape with a
major radius R= 5 meters and a minor radius a= 1 meter. The plasma temperature is T= 50
million Kelvin and the plasma density is n= 5×1019 particles per cubic meter. Assume the plasma
behaves like an ideal gas with three translational degrees of freedom.
a) Calculate the thermal energy density uof the plasma in joules per cubic meter.
b) Determine the thermal pressure pin pascals due to the plasma.
c) Find the magnetic field strength Bin teslas needed to confine the plasma if the plasma βis
1.5%, where β=p
B2
2µ0
represents the ratio of plasma pressure to magnetic pressure.
Solution:
a) The thermal energy density of an ideal gas is given by the equation:
u=3
2nkT
where kis the Boltzmann constant. Plugging in the values:
u=3
2×5×1019 ×1.38 ×10−23 ×50 ×106= 0.517 ×106J/m3
b) The thermal pressure of the plasma is given by:
p=nkT
Plugging in the values:
p= 5 ×1019 ×1.38 ×10−23 ×50 ×106= 3.45 ×105Pa
c) To find the magnetic field strength, we use the formula for β:
β= 1.5% = p
B2
2µ0
Solving for B:
B=r2µ0×p
β=r2×4π×10−7×3.45 ×105
0.015 ≈0.081 T
17 20. DIVERTOR DESIGN AND OPTIMIZATION
Problem 20. In a tokamak fusion reactor, the magnetic field strength in the divertor region
is given by B= 2.5T. The radius of the divertor plate is r= 0.5m and the angle between the
magnetic field lines and the divertor plate is θ= 30◦. Calculate the force experienced by a particle
with charge q= 1.6×10−19 C and velocity v= 5 ×106m/s moving along the magnetic field lines.
Solution 20. The force experienced by a charged particle moving in a magnetic field is given
by the equation:
F=qv×B
where qis the charge of the particle, vis the velocity vector, and Bis the magnetic field vector.
First, we need to calculate the velocity vector component along the magnetic field lines. Since
the particle is moving along the magnetic field lines, the velocity vector is parallel to the magnetic
field vector:
v∥=vcos θ= 5 ×106m/s ×cos 30◦= 4.33 ×106m/s
The force experienced by the particle is then:
F=qv∥B= (1.6×10−19 C)(4.33 ×106m/s)(2.5T)
F= 1.74 ×10−12 N
Therefore, the force experienced by the particle moving along the magnetic field lines is 1.74 ×
10−12 N.
I. Problem on Particle Density in a Fusion Reactor
Problem 1. In a fusion reactor, the electron density neis 1019 m−3. If the electron charge is
e= 1.6×10−19 C and the volume of the plasma is 0.1m3, calculate the total charge present in the
plasma.
Solution 1. The total charge present in the plasma is given by the product of electron density
and the charge of each electron:
Q=ne·e= 1019 m−3·1.6×10−19 C
Q= 1.6×100C= 1.6C
Therefore, the total charge present in the plasma is 1.6Coulombs.
II. Problem on Plasma Density Control
Problem 2. A plasma confinement device contains deuterium-tritium fuel with a number density
of 5×1020 m−3. If the device has a volume of 1m3, calculate the total number of fuel particles
present.
Solution 2. The total number of fuel particles present in the plasma is given by the product of
number density and volume of the plasma:
N=nfuel ·V= 5 ×1020 m−3·1m3
N= 5 ×1020 particles
Therefore, the total number of fuel particles present in the plasma is 5×1020 particles.
18 22. PLASMA TRANSPORT BARRIERS
Problem 22. Consider a fusion plasma with a temperature gradient described by the equation
T(r) = Tcore 1−r
a, where Tcore is the temperature at the core and ais the radial distance. The
density gradient of the plasma is given by n(r) = ncore 1−r2
a2.
Given that Tcore = 10 keV, ncore = 1019 m−3, and a= 0.5m, solve the following:
a) Calculate the temperature at the edge of the plasma.
b) Determine the density at the edge of the plasma.
c) Calculate the temperature gradient at the edge of the plasma.
Solution 22.
a) The temperature at the edge of the plasma can be found by substituting r=ainto the
temperature profile equation:
T(a) = 10 keV 1−0.5
0.5= 0 keV
Therefore, the temperature at the edge of the plasma is 0keV.
b) Similar to part (a), the density at the edge of the plasma is determined by plugging r=ainto
the density profile equation:
n(a) = 1019 m−31−0.52
0.52= 0 m−3
This means the density at the edge of the plasma is 0m−3.
c) The temperature gradient at the edge of the plasma can be calculated using the derivative
of the temperature profile with respect to r:
dT
dr =−Tcore
a=−10 keV
0.5m=−20 keV/m
Therefore, the temperature gradient at the edge of the plasma is −20 keV/m.
19 23. FAST PARTICLE CONFINEMENT IN FUSION DEVICES
Problem 23. Consider a fusion device where deuterium-tritium fusion reactions are taking
place. The average energy of the energetic alpha particles produced in these reactions is 3.5MeV.
Assuming the alpha particles are confined in a magnetic field with a magnetic mirror ratio of 2,
calculate the energy of the alpha particles when they are just about to escape confinement.
Solution 23.
a) The energy of the alpha particles when they are just about to escape confinement can be
determined by using the conservation of energy. The magnetic mirror ratio is defined as the ratio
of the magnetic field strength at the mirror point to the magnetic field strength at the center of the
device. In this case, the mirror ratio is 2.
Let B0be the magnetic field strength at the center, and Bmbe the magnetic field strength at
the mirror point. The energy of the alpha particle at the center (E0) and at the mirror point (Em) is
related by:
Em
E0
=B0
Bm2
Given that E0= 3.5MeV and the mirror ratio is 2, we can plug in the values to find the energy
of the alpha particles at the mirror point:
Em
3.5=1
22
Em=3.5
4= 0.875 MeV
Therefore, the energy of the alpha particles when they are just about to escape confinement is
0.875 MeV.
I’m glad to help with that! Here’s a numerical problem related to impurity control in fusion
reactors:
20 24. IMPURITY CONTROL IN FUSION REACTORS
Problem 24. In a fusion reactor, the plasma is composed of deuterium and tritium ions. However,
due to impurities, a small amount of helium ions is also present in the plasma. The density of
deuterium ions is nD= 1.5×1019 m−3, the density of tritium ions is nT= 0.5×1019 m−3, and the
density of helium ions is nHe = 0.1×1019 m−3. The charge of deuterium ions is qD= 1.6×10−19
C, the charge of tritium ions is qT= 1.6×10−19 C, and the charge of helium ions is qHe = 2 ×10−19
C.
a) Calculate the total charge density of the plasma.
b) If the average velocity of each ion species is vD= 1×106m/s for deuterium, vT= 0.8×106m/s
for tritium, and vHe = 0.6×106m/s for helium, calculate the total current density of the plasma.
Solution 24.
a) The total charge density of the plasma can be found by summing the contributions from each
ion species:
Total charge density, ρ=nD·qD+nT·qT+nHe ·qHe
Plugging in the values given:
ρ= (1.5×1019 m−3)(1.6×10−19 C)+(0.5×1019 m−3)(1.6×10−19 C)+(0.1×1019 m−3)(2×10−19 C)
ρ= 2.4×100+ 0.8×100+ 0.2×100
ρ= 3.4×100= 3.4C/m3
Therefore, the total charge density of the plasma is 3.4C/m3.
b) The total current density of the plasma can be found by summing the contributions from each
ion species:
Total current density, J=nD·qD·vD+nT·qT·vT+nHe ·qHe ·vHe
Plugging in the values given:
J= (1.5×1019 m−3)(1.6×10−19 C)(1×106m/s)+(0.5×1019 m−3)(1.6×10−19 C)(0.8×106m/s)+
(0.1×1019 m−3)(2 ×10−19 C)(0.6×106m/s)
J= 2.4×10−13 + 0.8×10−13 + 0.12 ×10−13
J= 3.32 ×10−13 A/m2
Therefore, the total current density of the plasma is 3.32
21 25. FUEL ION HEATING AND ALPHA PARTICLE EFFECTS.
Problem 25. Consider a fusion plasma where deuterium-tritium fusion reactions are the pre-
dominant source of energy generation. The energy released per reaction is 17.6 MeV. If the fusion
power output is 500 MW, determine:
Given:
•Energy released per reaction: 17.6MeV
•Fusion power output: 500 MW
a) The number of fusion reactions occurring per second.
b) The total energy released per second from fusion reactions.
c) The total energy released per second from fusion reactions in Joules.
Solution 25.
a) The number of fusion reactions occurring per second can be calculated using the fusion
power output:
Fusion Power Output =Energy released per reaction ×Number of reactions per second
500 ×106W= 17.6×106eV ×Number of reactions per second
Solving for the number of reactions per second:
Number of reactions per second =500 ×106
17.6×106≈28.41 ×106
Thus, approximately 28.41 ×106fusion reactions are occurring per second.
b) The total energy released per second from fusion reactions is:
Total energy released per second = 17.6×106eV ×28.41 ×106≈500 ×106W
Therefore, the total energy released per second from fusion reactions is 500 MW.
c) To convert this into Joules:
Total energy released per second in Joules = 500 ×106J
Hence, the total energy released per second from fusion reactions is 500 MJ.
