RRY115 - Fusion energy
Notation:
Blackboard: Goes on the blackboard
Ask: a question for the class
Normal: Communicated orally
(Extra: If people ask; extra instructions; etc.)
(Board instructions: what I should draw, etc.)
TODO: I need to look up what this means before exercise
Exercise 1 2016– ignition and gain
Quick recap and notation
Nuclear fusion The fusion of nuclei into heavier nuclei. For light nuclei, the process
releases energy q = (mf − mi )c2 > 0. Man-made fusion is routinely achieved in
nuclear physics experiments. (Experiments: GSI (Darmstadt), ISOLDE (CERN), etc)
Confinement More problematic is to confine the energy released to make it avaliable
for initiating more fusion reactions. (If many nuclear engineers, can make parallels to fission, where neutrons
need to be slowed down to be available for further reactions. In fusion, we need many, fast particles =¿ ionized gas.)
Ignited plasma A plasma which can sustain fusion reactions: the energy from the fusion
reactions cancels energy losses.
Sub-ignited plasma Plasma can sustain itself with additional input power. Can still
have a positive gain.
Gain Eout /Ein
1.1 – Fusion car
Some fusion engineering students decide to travel by car from Gothenburg to Paris for
their holiday break, a distance of 1200 km. The students are driving a new fusion-powered
automobile which burns fully catalyzed D-D fuel. If 0.0153 % of the atoms (out of the
isotopes of hydrogen) in sea water are deuterium atoms how many liters of sea water are
required for the drive? Assume that approximately 2 MJ of energy is spent per km.
Burning of gasoline releases approximately 40 MJ/kg of energy which corresponds to
30 MJ/l. For comparison, how many liters of gasoline would be required if the car was
running on gasoline?
Solution
How are we to solve this?
Step-by-step:
• How much energy is needed to travel 1200 km
1
RRY115 - Fusion energy
• How much energy can be extracted from 1 l of fuel
Subproblem 1 1200 · 2 = 2400 MJ
Subproblem 2 This is the tricky problem.
We want energy from a fusion reactions. Most energetically favorable is to fuse
light elements, relevant reactions can be found on p. 20, table 3.1 in fusion compendium.
Which reactions are relevant for our problem?
(Anything which converts p ↔ n has very small cross-section due to weak interaction: p-p irrelevant.
Anything with a neutron reactant is irrelavant (and unlisted), since neutrons not
confined and have basically no time to react.
Anything between 2 reaction products: too low density to happen
Anything with proton reactant: low cross-section. p-T: would form He4 [very stable
=⇒ high energy release], but energy release is higher than binding energy of p,n of He4,
so would either go back to p+T, go to n+He3 [endothermic]. He4 + γ basically does not
happen when strong decay is possible. Same compond He4 nucleus is formed in D-D, and
decays for same reason.
p-D: Needs a second particle in final state to conserve energy+momentum, thus
need a photon, and thus has lower cross-section than purely strong interactions.)
Relevant reactions are thus:
D + D → He3 + n + 3.27 MeV
D + D → T + p + 4.03 MeV
(50%)
(50%)
D + He3 → α + p + 18.3 MeV
D + T → α + n + 17.6 MeV
(1)
(2)
(3)
(4)
We will consider two scenarios:
1. A basic scenario, where only the initial D-D reaction takes place. This is a reactor
operating at a temperature ideal for D-D reactions.
2. An advanced scenario: where all the T and He3 react with D.
In the basic scenario, each reaction, consuming two deuterium, produces, on average
(3.27 + 4.03)/2 MeV = 3.65 MeV, thus 1.825 MeV/D.
In the advanced scenario, each reaction chain 3D → He4 + n + p + 21.6 MeV,
consumes three deuterium for21.6 MeV/3D = 7.21 MeV/D, so that four times the energy
can be extracted from the same amount of deuterium.
Finally, we need to know how many deuterium nuclei there are in 1 l of water.
Using the average weight of a water molecule, we first calculate the number of water
molecules, which gives the number of deuterium atoms by ND = 0.000153NH = 2NH2 O .
The density of water is approximately 1 kg/l. The average weight of a water molecule is
2
RRY115 - Fusion energy
Pn
Pα
Pout,plasma
α
n
2 · (0.000153 · 2u + (1 − 0.000153) · u) + 16u Thus the number of water molecules per liter
is
1 kg/l
NH2 O /l =
≈ 3.35 · 1025 l−1
(0.000153 · 4 + (1 − 0.000153) · 2 + 16)u
so
ND2 /l = 2 · 0.000153NH2 O = 10.2 · 1022 l−1
Assuming we fuse every deuterium, 1 l of water gives us, upon convering MeV to MJ,
1 eV ≈ 1.6 × 10−19 J. 2990MJ/l for the basic, and 11800MJ/l for the advanced scenario.
To travel from Gothenburg to Paris, we thus need 0.8 l and 0.2 l of water, respectively.
This can be compared with burning gasoline, which releases 30 MJ/l, which means
that for a drive of 1200 km consuming 2 MJ/km we need
2 · 1200 MJ
= 80 l
30 MJ/l
of gasoline.
Gain factor and alpha particle heating
How does the energy balance look in a steady-state ignited fusion reactor? Ask them to
discuss: input power, output power and power sources.
(Circle with power created inside (wiggly arrows from inside) and going out. Alpha power!
Add straight arrows out due to neutrons.)
Once ignited, we have an infinite gain. Reactors are profitable even without infinite gain.
(Increase transport losses and add input power)
The neutron and α particle are both born in a fusion reaction. How do they share their
energy?
D + T → α + n + 17.6 MeV
3
(5)
RRY115 - Fusion energy
Pn
Pα
Pin,plasma
Pout,plasma
α
n
The released energy can be derived from mass difference, and is distributed as kinetic
energy between α and n. Tα + Tn = mα vα2 /2 + mn vn2 /2 = Ef . We also have momentum
mn
conservation. In the center of mass frame: mα vα + mn vn = 0 =⇒ vα = − m
vn so
α
1
mn
5
2
Tα = mα mn vn /2 = 4 Tn and Ef = 4 Tn so 1/5 goes to the α and 4/5 of the energy goes to
the neutron.
Sub-ignited plasma: Plasma can sustain itself with additional input power. Can
still have a positive gain.
To quantify how sub-ignited a plasma is, one can define the ratio of total heating
and heating due to fusion born alpha particles.
fα =
Sα
Sα + Sh
where Sα is the heating power (density) due to fusion born alpha particles, and Sh is all
external heating power.
The gain can be quantified with the Q-factor
Q=
Pout − Pin
Pin
where Pout is the output power, and Pin the input power.
Problem
We want to derive a relation between the gain factor and the fraction of alpha heating:
fα =
Q
.
5+Q
4
RRY115 - Fusion energy
Solution
How to solve?
Step-by-step:
• Relate output to input power
• Calculate how energy from α particles contribute to both
The output power is given by
Pout = (Sn + Sout,plasma ) · V,
where V is the plasma volume and
Sn =
En
Sα = 4Sα
Eα
is the neutron power density, which is 4 times the alpha power density. Splasma out represents any losses from the plasma – which can be further split up, but since we are in a
steady state, we know that it will be equal to the total power heating the plasma. The
input power is the external heating power
Pin = Sh · V.
Note that the volume factors V in Pout and Pin cancel.
Using power balance
Sα + Sh = Sout,plasma
the expression for Q can be rewritten as
Q=
Sn + Sout,plasma − Sh
Sn + Sα
Sα
Pout − Pin
=
=
=5 .
Pin
Sh
Sh
Sh
From the definition of fα =
Sα
Sα +Sh
we get
Sh
1 − fα
=
Sα
fα
Q=5
⇒
Sα
fα
=5
Sh
1 − fα
fα =
Q
.
5+Q
5
⇒
RRY115 - Fusion energy
1
Expression in terms of triple product
The triple product is given by pressure times the energy confinement time
pτE ,
which is a measure of how much energy is confined in the plasma, weight with how long
the energy is confined. We want to express the Q factor in terms of this.
We can decompose the losses from the plasma in Bremsstrahlung radiation losses
and losses due to transport mechanisms in the plasma.
Sout,plasma = SB + SL
In steady state:
Sα
.
fα
The transport losses are difficult to model, but it is usually expressed in terms of the
confinement time.
p
SL = Kk
τE
Sout,plasma = SB + SL = Sα + Sh =
(The radiation losses can be written as
SB = KB
p2
T 3/2
Usually in fusion experiments,
transport losses dominate radiation losses, SL SB , and thus
KB = 0.052 and Kk = 0.15, for pressure in 105 Pa, T in keV and τE in seconds. )
S L = Kk
Sα
p
≈
τE
fα
When fα = 1, we have ideal ignition, with triple product
K k p2
(pτ )I =
Sα
Am I allowed to just set fα to one? No other terms in our expressions must change: we
have the same Sα and p, and τE responds to changes in heating power to maintain steady
state. and thus
pτE
.
pτE = fα (pτE )I =⇒ fα =
(pτE )I
so that
Q=5
fα
5pτE
=⇒ Q =
.
1 − fα
(pτE )I − pτE
Engineering gain factor
The Q factor tells us how energy we put into the plasma translates into energy out from
the plasma. The engineering gain factor tells us how much electrical power we take from
the grid, and how much electrical power we produce.
6
RRY115 - Fusion energy
ηout
E
Pout
Pn
Pα
Pout,plasma
ηout
Pin,plasma
ηe
PinE
α
QE =
n
PE − PE
net electric power out
= out E in ,
electric power in
Pin
where PinE is the actual electric power required to drive the external heating sources and
E
Pout
is the electric power output.
(Point at the previous figure with input and output powers, and add small converter
stations with efficency ηe and ηout )
ηe tells us how much energy we emit towards the plasma vessel, and how much is converted
into useless heat in our heating system.
We actually need another coefficient, ηa , which tells us how much of the emitted
energy is absorbed by the plasma.
(Draw some of input power being reflected and some passing through the plasma.)
The reason why we need two coefficients to characterize how efficiently we put power
into the plasma is that the power transmitted or reflected from the plasma 1 − ηa , does
not become USELESS heat, but will heat the walls of our vessel and ultimately produce
power in our turbines.
Problem
Derive the relation between the engineering gain factor and pτE :
QE = 2
pτE − 0.37(pτE )I
.
(pτE )I − pτE
7
RRY115 - Fusion energy
ηout
ηout
E
Pout
Pn
Pα
Pin,plasma
ηe
Pout,plasma
ηout
PinE
α
n
Solution
Basically, we need wallplug efficiencies for any process that inserts power into the plasma,
and efficiencies for any power that leaves the plasma. We will make very simple assumptions and take one, constant efficiency coefficient for all outputs.
By our assumed input efficiencies
Pin = ηe ηa PinE ,
Pout is a bit more tricky: first, we should add the unabsorbed heating power to the
plasma power output
Pout = Pout,plasma + Pn + (1 − ηa )ηe PinE .
Secondly, in a tritium breeding reactor, the neutrons will not only give away their kinetic
energy, but once slowed down, they will react with Lithium to breed tritium, in a reaction
which releases additional energy. (Extra arrows next to the n arrows)
6
Li + n → 4 He + T + 4.8 MeV.
In
steady state, each tritium must be replaced, and thus every fusion reacton gives a neutron
which undergoes the above reaction, for an extra ELi = 4.8 MeV per neutron. Since α,
neutron and lithium power all scale like (energy released) × (number of reactions), we get
( There is an additional reaction with 7 Li which absorbs energy 7 Li + n → 4 He + T + n − 2.5 MeV, but it is a lot less likely to occur.)
Pn + PLi =
En + ELi
Pα = 5.4Pα .
Eα
8
RRY115 - Fusion energy
E
Pout
ηout
ηout
Pn
PLi
Pα
Pin,plasma
ηe
Pout,plasma
ηout
PinE
α
n
Finally, power balance gives
Pout,plasma = Pα + Pin = Pα + ηe ηa PinE
Thus
Pout = Pout,plasma +Pn +(1−ηa )ηe PinE +PLi = ηe ηa PinE +(1−ηa )ηe PinE +6.4Pα = ηe PinE +6.4Pα
and the output electrical power becomes
E
Pout
= ηout Pout = ηout ηe PinE + ηout 6.4Pα
and thus
E
Pout
− PinE = (ηout ηe − 1)PinE + ηout 6.4Pα
so
QE = (ηout ηe − 1) + ηout 6.4
Pα
.
PinE
Lastly, if we neglect Bremsstrahlung, we can write Pα /Pin =
fα
1−fα
which implies that
Pα
ηe ηa fα
=
E
1 − fα
Pin
and thus
QE = (ηout ηe − 1) + 6.4ηout ηe ηa
or, using fα =
fα
1 − fα
pτE
(pτE )I
QE =
(6.4ηout ηe ηa − (ηout ηe − 1))pτE + (ηout ηe − 1)(pτE )I
.
(pτE )I − pτE
9
RRY115 - Fusion energy
Assuming ηout ≈ 0.4, ηe ≈ ηa ≈ 0.7 finally gives
QE =
pτE − 0.37(pτE )I
1.98pτE − 0.72(pτE )I
≈2
(pτE )I − pτE
(pτE )I − pτE
Engineering and physics gain factor relation
Finally, we are in a position to derive a simple relation between the phyiscs and engineering
gain factor
Problem
Derive the relation between the physics and engineering gain factors:
Q = 4(QE + 0.72).
Solution
From 4.5.2 we know that
QE =
1.97pτE − 0.72 (pτE )I
1.25pτE − 0.72 ((pτE )I − pτE )
1.25pτE
=
=
− 0.72
(pτE )I − pτE
(pτE )I − pτE
(pτE )I − pτE
and we recall from earlier that
Q=5
pτE
.
(pτE )I − pτE
Combing these expressions we see that
QE =
1.25
Q − 0.72
5
which can be rearranged into
Q=
5
(QE + 0.72) = 4 (QE + 0.72) .
1.25
Thus we see that to achieve QE > 0, we need Q > 2.88, and the goal for ITER,
Q = 10 =⇒ QE = 1.78, could in principle produce net power to the grid.
10