Heating a Plasma to 100 Million Kelvin Gunnar Froberg¨ and Thomas Nygren

C3. TOKAMAK HEATING Heating a Plasma to 100 Million Kelvin Gunnar Froberg¨ and Thomas Nygren

Abstract—In this work techniques for heating the fusion way such that the high temperatures involved does not destroy reactor ITER to thermonuclear temperatures, over 100 million the walls. The plasma is heated in several ways: injection kelvin, is investigated. The temperature is numerically computed of high energy deuterium, resistive heating, radio waves and for different heating configurations. The heat leakage is modeled to occur only via diffusion. The diffusion is assumed to be a finally, when the temperature is high enough: thermonuclear combination of Bohm and gyro-Bohm diffusion. Basic conditions fusion. for a fusion reactor has been studied. The power needed for the different heat sources for the plasma to ignite is computed. Plots of the temperature profiles are included in the results together II.IMPLEMENTATION with plots showing the Q-value dependency on the power and A tokamak is a type of fusion reactor where the vessel the major radius. is shaped like a toroid. In order to confine the plasma and minimize the contact with the wall strong magnetic fields are I.INTRODUCTION applied. Note that the plasma temperature must be maintained HE International Energy Outlook 2013 predicts that the and the walls must be kept cool to prevent meltdown of the T world energy consumption will grow by 56 percent be- reactor walls. The fuel is assumed to consist of an equal tween 2010 and 2040[1]. The increase in the electric consump- mix of completely ionized deuterium and tritium with the ion tion will have to be produced somehow and as coal and oil temperature being similar to the electron temperature. On a will be increasingly expensive and is widely considered to be side-note, a completely ionized gas is the same thing as a bad for the environment, an alternative is needed. That could plasma. be weather dependent renewables, gas, fission and hopefully To achieve confinement, the orbits of the plasma particles fusion. must stay within the torus to avoid contact with the walls. This Gas releases carbon dioxide, many renewable sources are is achieved as the particles gyrate around the magnetic fields weather dependent and works intermittently and fission pro- lines, making the trajectories helical. The poloidal magnetic duces long lived highly radioactive waste. As an alternative field is a result of the toroidal movement of the charged research is being done on fusion with the goal to create a way particles in the plasma and the toroidal field is produced by to produce clean and sustainable electricity. coils surrounding the rector vessel. The plasma current is Fusion of hydrogen isotopes releases energy due to the induced via other coils. difference in mass between the fused particles and the particles Ignition is the state of the reactor when the heating from going in to the reaction. The reaction that is the easiest the fusion reactions is enough to maintain the temperature to achieve is the one between deuterium and tritium which without auxiliary heating. The Lawson criteria is a simple produces an alpha particle and a neutron. The neutron will criteria for estimating the conditions for ignition in a fusion have a high speed. The neutrons from the fusion reactions reactor. The criterion is a product of the number density n, gets absorbed by the lithium blanket in the reactor wall which energy confinement time τE, and the plasma temperature T in turn then heats water in a secondary cooling circuit. The (here measured in eV). For fusion of deuterium and tritium temperature difference between the hot water and the cooling the following version of the Lawson criterion holds. water creates an energy flow, in this case in the form of steam. n τ T ≥ 3 · 1021 (1) The steam flows through a turbine and therefore turning it. The e E mechanical energy in the turbine is then converted to electrical The Lawson criteria can be derived as follows: The confine- energy in a generator connected to the turbine. The neutrons ment time τE is defined as the energy density in the plasma is captured by a section of the wall containing lithium which divided by the power loss Ploss will create tritium that will later be used in the fusion reaction. W W The first fusion reactor to produce net energy is called ITER τE = ⇒ Ploss = (2) and is being built in southern France. It is planned to be ready Ploss τE to run experiments in November 2020[2] and begin fusion in The energy is defined as 2027. This reactor is build for the purpose of testing, it will not produce power to the grid. Some of the objectives with W = 3nT e the projects are: 3nT e ⇒ Ploss = (3) 1) To create an ignited plasma τE 2) To get a Q, defined by (6), value greater than 10. The heating from the alpha particles in the fusion reactions 3) To maintain fusion for eight minutes. has to exceed the power losses ITER is a tokamak type reactor which means that the plasma is magnetically confined in a torus. The plasma is confined in a fEα ≥ Ploss C3. TOKAMAK HEATING

1 2 Where f = 4 n hσvi is the number of fusion reactions occurring per volume and time and Eα is the kinetic energy of the alpha particle from the fusion. This gives us the inequality a 1 2 3nT e

n hσviEα ≥ M

4 τE a

g n

e R0

i 12T e c

⇒ nτE ≥ (4) a E hσvi x α i s By miltiplying both sides with T we get

12T 2e τEnT ≥ (5) Eαhσvi

By approximating hσvi ∝ T 2 we get that the right hand side is a constant which according to [3, Eq. 1.5.5 p. 11] is Figure 1. Geometrical description of a torus seen from above 21 τEnT > 3 · 10 .

In order to achieve high core temperatures, the wall needs to be far away from the center for it to be able to withstand the temperatures. The temperature proﬁle is independent of b the minor radius of the containment vessel. The temperature decreases with the distance to the center. This means that the core temperature can be increased if the wall is further from the center. a ITER aims at getting the fusion energy gain factor Q = 10 Where Q is the fusion power divided by the auxiliary heating.

P Q = fusion (6) Pauxiliary

This will require a temperatures in excess of 100 million Kelvin. This study looks at ITER, tokamaks and fusion in a very simpliﬁed way and give phenomenological explanations Figure 2. Minor and major axis of an ellipse of some of the the requirements of the reactor.

B. Derivation of the temperature equation A. Geometry The temperature of the plasma can be calculated from the The geometry of the tokamak is a torus with major radius energy conservation[4, Eq 4.1 page 61] R0 and a minor radius a. Figure 1 shows the geometry of 3 ∂p the torus from above, the bright circle is a crosssection. It + ∇ · pv + p∇ · v + ∇ · q = S (8) 2 ∂t is somewhat complicated to use toroidal coordinates so from here on we will consider the toroid to be a periodic cylinder Where p is the pressure, t is the time, v is the velocity, q is with a period of 2πR0 which is an equivalent problem. the energy flux and S is the power density from sources and Consider a torus shaped shell with a thickness δr and a sinks. The flows are in this case negligible, thus minor radius r in the tube, we then get the circumference O of the magnetic axis, the surface area A of the shell and a ∇ · pv = 0, p∇ · v = 0 small volume element dV . This simplifies equation (8) to 2 2 O = 2πR0,A = π(r + δr) − πr = 2πrδr 3 ∂p + ∇ · q = S 2 ∂t 2 dV = AO = 2πrδrO = 4π R0rδr (7) Stationarity gives ∇ · q = S Since ITER will have a plasma that is toroidal with an (9) elongation κ, the elongation will be used when integrating b or over the volume. The elongation κ = a is defined as the major radius divided by the major axis of an ellipse, see Figure 2. ∇ · qdV = q · ndAˆ = SdV (10) ˚Ω ‹∂Ω ˚Ω C3. TOKAMAK HEATING

This tells us that the power flux through the surface, ∂Ω is After some algebra we get that equal to the source inside the volume Ω. The divergence in s cylindrical coordinates can be written as T √ √ ρ = ( md + mt) (21) 1 ∂ 2eB2 ∇ · q = rq (11) tor r ∂r r md and mt is the mass of deuterium respectively tritium. In this work we assume that the energy flux qr given by Fick’s first law q = −Dgb∇U (12) D. Ohmic heating Dgb is given by (16) and U is the energy density. Ohmic heating is caused by the collisions of electrons and ∂U qr = −Dgb ions in the plasma. The ramping up of the externally applied ∂r magnetic field induces a current in the plasma. The resistivity where of the plasma creates friction, creating a force balance with 3 U = (n T e + n T e)[4, p. 61] the electric field from the change in the magnetic field. 2 e e i i The power density per unit volume of Ohmic heating SO is With the density ne = ni and the temperature(in eV) Ti = Te = T we get S = ηj2 (22) ∂T O qr = −3neeDgb (13) ∂r Where j is the current density and η is the resistivity combining (9), (11) and (13) (2m )1/2e2 ln(Λ) 1 ∂ ∂T S(r, T ) η = e [7, Eq. 11.30] (23) ⇒ rD = − (14) 2 3/2 gb 12 (πTe) r ∂r ∂r 3nee 0

The term S is a sum of all the heat sources and sinks: Where me is the electron mass, e is the elementary charge, 0 is the vacuum permittivity and T is the electron temperature S = S + S + S + S + S + S (15) e α B NBI ECRH ICRH O in eV. The numerical value is Those terms are explained in theirs own subsections. S is the α 5 · 10−5 ln(Λ) fusion source, the energy from the alpha particle, SB is the η = 3/2 (24) sink from Bremsstrahlung. Te

C. Bohm diffusion ln(Λ) ≈ 20[Eq. 37][4] In this work we use a combined Bohm/gyro-Bohm diffusion (22) can be derived as follows. Consider an inﬁnitesimal work coefﬁcient[5] on a charged particle dW = F dr. The power on N particles Bohm Gyro−Bohm Dgb = a1χ + a2χ (16) becomes

−4 −5 −2 −2 dW dr a = 1.6·10 +8·10 a = 1.75·10 +3.5·10 P = N = NF = NF v (25) Where we use 1 2 , 2 2 dt dt and 2 2πa2B T The Coulomb force F = qE and the electric ﬁeld is E = ηj. χBohm = tor (17) µ0R0Iplasma Btor This gives and P = qvηjN (26) T ρ χGyro−Bohm = (18) aBtor Where v is the difference is velocity between the ions and the Btor = 5.3 T[6] and the gyro radius ρ is electrons. From the deﬁnition of power density we get v ρ = th (19) P Nqvηj ω S = = (27) g O V V Where ω = eBtor is the cyclotron frequency and v = g m th N q 2eT √ With V = n we get m is the thermal velocity. Since ρ ∝ m and the plasma is a combination of deuterium and tritium we construct an P S = = nqvηj (28) average mass, m, as O V √ √ √ m + m m = d t and with the current density nqv = j we end up with the 2 desired √ √ 2 ( md + mt) 2 ⇒ m = (20) SO = ηj 4 C3. TOKAMAK HEATING

xa a+r injection of fast neutral particles, in this case deuterium, into dNNeu (r) N0 − = − e λNBI the plasma chamber. Since neutral particles are not affected dx λNBI by magnetic fields they are able to penetrate into the plasma. To account for the finite width of the beam the source is Inside the neutrals undergo charge-exchange in which they corrected by a factor √ r , where ε is the width of the beam, r2+ε2 lose an electron and thus becomes confined by the magnetic which also prevents the diverging behavior in the formula for field. small r In order to calculate the NBI power source SNBI we assume E N a−r a+r In 0 − λ − λ SNBI (r) = √ e NBI + e NBI N0 particles each having a mass m is inserted each second. 2 2 2 4κπ R0 r + ε λNBI The total power is P (35) mv2 r 2P P = N0 ⇒ v = (29) 2 N0m Every second there is a probability of ionization via charge-exchange. Charge-exchange is a collision between a charged and a neutral particle where charge is transferred between the particles in the collision while the energy of the particles stays the same.

The probability of ionization per second will be denoted as p. Assuming that the particles have the same velocities, v, before and after the collisions we define the penetration depth, λ , as NBI v λ = (30) NBI p NBI injector The penetration depth is the distance into the plasma where the number of particles flowing through have decreased to −1 Figure 3. Geometrical description of NBI injector as seen from above N0e . We assume that the fast neutrons will be inserted orthogonal to the torus, towards the symmetry axes shown in As NBI sends through highly energetic particles with an Figure 3. Call this direction xˆ. exponentially decaying profile, some energy will be deposited into the opposite side of the beam injector, in this case the The difference in the number of non-ionized deuterium center. Too much of this energy will damage the reactor and particles N in dx is so the maximum shine trough allowed for NBI is set to five dN −1 Neu = N percent of the power. dx λNBI x F. Fusion reactions − λ ⇒ NNeu(x) = N0e NBI (31) The kind of nuclear fusion that will be used in ITER is 2 3 4 1 The change rate of ions correlated to the change in numbers 1D + 1T −−→ 2He(3.5MeV) + 0n(14.1MeV).[4, Eq. 2.17] of neutral particles: Because the neutron is neutral, it will not become a part of the plasma and instead go straight to the reactor wall. This means dNNeu dNIon = − δr that only the alpha particles will contribute to the plasma dr heating. Each inserted neutral deuterium particle has an energy of EIn According to [4, Eq. 4.5 p. 63] the power density Sα is dP = E dN (32) given by Beam In Ion 2 Eαn hσvi dP Sα = (36) S = Beam 4 NBI dV where hσvi is given by With (7) we get r ξ −3ξ −dNNeu hσvi = C1 · θ e [ , ] EIn δr 2 3 8 Eq 12 dr mrc T SNBI = 2 = 4κπ R0rδr T θ = [8, Eq 13] EIn dNNeu 1 − T (C2+T (C4+TC6)) − 2 (33) 1+T (C3+T (C5+TC7)) 4κπ R0r dr 1/3 B2 EIn dNNeu dNNeu ξ = G [8, Eq 14] SNBI (r) = − 2 + 4θ 4κπ R0r dx x a (34) The constants are given in Table I C3. TOKAMAK HEATING

2 Coefficient√ With T ∂T = 1 ∂T we get BG( keV) 34.3827 ∂r 2 ∂r 2 mrc (keV) 1124572 K dT 2 dT 2(r) 2y(r0) −9 0 0 C1 1.17302 · 10 y(r ) = r ⇒ = 0 C2 1.51361 · 10−2 2 dr dr r K C3 7.51886 · 10−2 Where −3 C4 4.60643 · 10 1.6 · 10−4 + 8 · 10−5 2πa2B 2 1 C5 1.35000 · 10−2 K = tor + −4 C6 −1.06750 · 10 2 µ0R0Iplasma Btor C7 1.36600 · 10−5 1.75 · 10−2 + 3.5 · 10−2 ρ Ti range keV 0.2 − 100 (39) (∆hσvi)max(%) 0.25 2 aBtor Table I Integration gives IMPROVEDFORMULASFORFUSIONCROSS-SECTIONSANDTHERMAL r 0 REACTIVITIES[8, TABLE VII] 2 2 2y(r ) 0 T (r) − T (r0) = 0 dr ˆr0 r K Inserting the expression for y(r0) from (38) G. Radio-frequency heating r 2 2 2 2 r1K dT ⇒ T (r) = T (r0) + − There are two types of radio frequency heating involved ˆ r0K 2 dr r0 r1 is this project. Ion Cyclotron Resonance Heating(ICRH) and r0 ! Electron Resonance Heating(ECRH). The waves get absorbed r00S(r00,T ) dr00 dr0 (40) when the wave frequency is the same as the cyclotron fre- ˆr1 3nee quency ωg of the ions and electrons respectively. Since Btor depends on the distance from the toroidal axis the cyclotron dT 2 With r0 = a and r1 = 0 we get that dr = 0 and to frequency will also depend on the distance. This means that 0 handle the divergent behavior when r approaches 0 we use there will be an area where the cyclotron frequency is the r = pr2 + (0.01 · a)2 in the denominator. same as source frequency at which the energy of the wave 0 ! will be absorbed. Since the wave frequency can be controlled, r 2 r r00S(r00,T ) T 2(r) = T 2− dr00 dr0 the absorption area can be chosen fairly freely. From [9] we a p 02 2 â r + (0.01 · a) K ˆ0 3nee identify that the absorption curve will take on a Gaussian (41) profile. In our equations we chose that the ECRH will be (41) is the equation to be solved numerically, see section III centered at a radius 0.3a and with a width 0.08a, while the ICRH will be centered at the magnetic axis with a width of III.METHODS 0.2a. The way this problem will be solved I Set the boundary temperature T a. Assume T a to be the H. Bremsstrahlung temperature everywhere and save it to T list(r). Create a minor radius Bremsstrahlung is the radiation process when a charged list for r of equal length starting at length(T list) and ending particle is accelerated by another charged particle through at minor radius. Coulomb collision. The power density radiated from II Interpolate T list to T. bremsstrahlung is III Integrate the inner integral using quad. IV Integrate the outer integral using quad. 3 2 1/2 SB = 5.35 · 10 Zeff n20 (T ) [4, Eq. 3.42 p. 56] (37) V Save T list(r). VI Estimate the error in Tk. Where Zeff ≈ 1 is the effective number of protons, n20 is VII If the absolute error is too large: go to II. 1020 the number density is units of m3 and T is the temperature in keV. The relative error in T is ||Tk+1 − Tk|| rel = I. The temperature equation on integral form ||Tk|| Equation (14) is rewritten here for numerical analysis. Let’s Where the index denotes the iteration number. first define The norm used is the L norm on a domain Ω ∂T rS(r, T ) 2 y(r) ≡ rDgb , f(r) ≡ − s ∂r 3nee 2 ||f||L2 = f(x) dV (42) Integrating (14) with respect to r gives the equation ˆΩ r 0 0 2 r dy r T 2(r) = T 2 − · ydr0 dr = f(r)dr a ˆ Kpr02 + (0.01 · a)2 ˆ dr ˆ a r1 r1 (outer integral) 0 0 r 00 00 r r S(r ,T ) 00 y(r0) − y(r ) = f(r)dr y = dr (inner integral) 1 (38) ˆ0 3ne ˆr1 C3. TOKAMAK HEATING

IV. RESULTS Fusion power gain factor vs major radius 55

1e8 Temperature over a cross-section 3.0 50 Iteration: 0 Iteration: 1 2.5 Iteration: 2 45 Iteration: 3 Iteration: 4 2.0 Iteration: 5 40 Q Iteration: 6 Iteration: 7 1.5

[K] Iteration: 8 35 T

1.0 30

0.5 25 4.0 4.5 5.0 5.5 6.0 6.5 7.0 7.5 8.0 R [m]

0.0 0.0 0.5 1.0 1.5 2.0 r [m] Figure 6. The fusion power gain factor versus the major radius of the tokamak, the plasma current density was held constant. Figure 4. The simulated plasma temperature over the minor radius of the torus.

1e9 p in red and P in blue vs r 1.0 q heating

Diffusion coefficient vs r 1.8 0.8 1.6

1.4 0.6