SHOWING SOLUTIONS
Friday, 10 May 2019
14:00 – 15:30
EXAMINATION FOR THE DEGREES OF M.SCI. (HONOURS) AND
M.SC.
[ PHYS5038 ]
Nuclear Power Reactors
Candidates should answer Question 1 (16 marks)
and either Question 2A or Question 2B (24 marks each)
Answer each question in a separate booklet
Candidates are reminded that devices able to store or display text or images
may not be used in examinations without prior arrangement.
Approximate marks are indicated in brackets as a guide for candidates.
PHYS5038 Nuclear Power Reactors
Fundamental constants
name symbol value
speed of light c 2.998× 108 m s−1
permeability of free space µ0 4pi × 10−7 H m−1
permittivity of free space 0 8.854× 10−12 F m−1
electronic charge e 1.602× 10−19 C
Avogadro’s number N0 6.022× 1023 mol−1
electron rest mass me 9.110× 10−31 kg
proton rest mass mp 1.673× 10−27 kg
neutron rest mass mn 1.675× 10−27 kg
Faraday’s constant F 9.649× 10−4 C mol−1
Planck’s constant h 6.626× 10−34 J s
fine structure constant α 7.297× 10−3
electron charge to mass ratio e/me 1.759× 1011 C kg−1
quantum/charge ratio h/e 4.136× 10−15 J s C−1
electron Compton wavelength λe 2.426× 10−12 m
proton Compton wavelength λp 1.321× 10−15 m
Rydberg constant R 1.097× 107 m−1
Bohr radius a0 5.292× 10−11 m
Bohr magneton µB 9.274× 10−24 J T−1
nuclear magneton µN 5.051× 10−27 J T−1
proton magnetic moment µp 1.411× 10−26 J T−1
universal gas constant R 8.314 J K−1 mol−1
normal volume of ideal gas – 2.241× 10−2 m3 mol−1
Boltzmann constant kB 1.381× 10−23 J K−1
First radiation constant 2pihc2 c1 3.742× 10−16 W m2
Second Radiation constant hc/kB c2 1.439× 10−2 m K
Wien displacement constant b 2.898× 10−3 m K
Stefan-Boltzmann constant σ 5.670× 10−8 W m−2 K−4
gravitational constant G 6.673× 10−11 m3 kg−1 s−2
impedance of free space Z0 3.767× 102 Ω
Derived units
quantity dimensions∗ derived unit
energy ML2T−2 J
force MLT−2 N
frequency T−1 Hz
gravitational field strength LT−2 N kg−1
gravitational potential L2T−2 J kg−1
power ML2T−3 W
entropy ML2T−2 J K−1
heat ML2T−2 J
capacitance M−1L−2T 4I2 F
charge IT C
current I A
electric dipole moment LTI C m
electric displacement L−2TI C m−2
electric polarisation L−2TI C m−2
electric field strength MLT−3I−1 V m−1
electric (displacement) flux TI C
electric potential ML2T−3I−1 V
inductance ML2T−2I−2 H
magnetic dipole moment L2I A m2
magnetic field strength L−1I A m−1
magnetic flux ML2T−2I−1 Wb
magnetic induction MT−2I−1 T
magnetisation L−1I A m−1
permeability MLT−2I−2 H m−1
permittivity M−1L−3T 4I2 F m−1
resistance ML2T−3I−2 Ω
resistivity ML3T−3I−2 Ω m
∗ M = mass, L = length, T = time, I = current
SHOWING SOLUTIONS
1 (a) In the early days of commercial nuclear power development, some coun-
tries adopted a once-through fuel cycle strategy while others opted for a
reprocessing-recycling strategy. Give an example of an advantage and a dis-
advantage associated with each strategy. [4]
Solution:
[A bit of bookwork.]
For the once-through cycle, one could include:
• advantages: [1]
proliferation resistant; lower occupational dose to radiation workers; less
expensive.
• disadvantages: [1]
does not represent a sustainable use of resources (disposal of material with
huge energetic value); more waste material, heat and radiotoxicity in
geological repository; waste is longer-lived.
For reprocessing-recycling (the opposite of the above):
• advantages: [1]
extraction of energetic value of plutonium; shorter-lived waste as a result,
and less of it.
• disadvantages: [1]
has led to the build-up of stockpiles of plutonium; multi-recycling is very
challenging without fast reactor deployment.
(b) A loss of coolant in a nuclear power plant can lead to serious core dam-
age and/or a containment breach. Describe four possible consequences of an
uncontrolled increase in reactor core temperature. [4]
Solution:
[A bit of bookwork.]
Anything sensible here ([1] for each), including:
• The potential for a prompt-supercritical reactivity insertion and subsequent
nuclear explosion;
• Hydrogen gas production from the heated fuel rod cladding and subsequent
chemical explosion;
• Steam pressure build-up in water-cooled reactors and subsequent steam
explosion;
• Moderator fires in graphite-moderated reactors as a result of an uncontrolled
Wigner release;
PHYS5038 Nuclear Power Reactors 3/12 Q1 continued over. . .
Q1 continued
SHOWING SOLUTIONS
• Melting of the fuel assemblies leading to a breach in the reactor pressure
vessel (and subsequently any/all of the above).
(c) How many elastic collisions would a fast neutron need to undergo in order
to thermalise in light water and in heavy water? What else must be considered
when identifying which of the two materials is the more effective moderator? [4]
Solution:
[A simple bit of problem solving, similar to problems in the problem sheet and
lectures.]
The average logarithmic energy loss per collision for light and heavy water is:
ξ18 =
2
A+ 2/3
= 0.107
ξ20 =
2
A+ 2/3
= 0.0968 [1]
The neutron lethargy u is related to the initial (Ei) and final (Ef ) kinetic energy
such that:
Ei = 2× 106 eV
Ef = 0.025 eV
u = ln
(
Ei
Ef
)
= 18.20 [1]
The number of collisions is therefore:
N18coll =
u
ξ18
= 170.1
N20coll =
u
ξ20
= 188.0 [1]
We must also consider which material has the highest elastic scattering cross
section. [1]
(d) An important consideration in reactor physics is that many isotopes are
1/v neutron absorbers, which means that their microscopic absorption cross
section scales with increasing kinetic energy as the reciprocal of the neutron
velocity. If σa for such an isotope decreases by a factor of four, what is the
implication for the neutron kinetic energy, neutron flux, macroscopic cross
section for absorption and the corresponding reaction rate? [4]
Solution:
[This is an unseen question.]
The neutron velocity must have increased by a factor of 4, so:
PHYS5038 Nuclear Power Reactors 4/12 Q1 continued over. . .
Q1 continued
SHOWING SOLUTIONS
• The kinetic energy increases by a factor of 16 (since E ∝ v2). [1]
• The neutron flux increases by a factor of 4 (since φ ∝ v). [1]
• The macroscopic cross section decreases by a factor of 4 (since Σa ∝ σa). [1]
• The reaction rate remains unchanged (since Ra = Σa φ). [1]
Efast = 2 MeV, Ethermal = 0.025 eV
PHYS5038 Nuclear Power Reactors 5/12 Paper continued over. . .
SHOWING SOLUTIONS
2A The European pressurised water reactor (EPR) generates 4.3 GW of thermal
power and can be loaded with mixed oxide fuel (MOX). Its core can be approx-
imated as a homogeneous mix with density 8 g cm−3 of light water (20 wt. %)
and MOX fuel (80 wt. %).
(a) Assuming a plutonium enrichment of 10 % and isotopic composition of
75 % Pu-239 and 25 % Pu-240, show that the target densities for the fuel
isotopes are N238 = 1.46× 1022 cm−3, N239 = 1.21× 1021 cm−3 and
N240 = 4.01× 1020 cm−3. [4]
Solution:
[A simple bit of problem solving, similar to problems in the problem sheet and
lectures.]
The key here is parsing the jargon to realise that U-238 as the matrix is present in
the fuel at 90% by weight, with Pu-239 at 7.5% and Pu-240 at 2.5%. [1]
That means:
N238 =
w238 ρNA
A
=
(0.8 · 0.9) · 8 · 6.02× 1023
238
= 1.46× 1022 cm−3 [1]
N239 =
w239 ρNA
A
=
(0.8 · 0.075) · 8 · 6.02× 1023
239
= 1.21× 1021 cm−3 [1]
N240 =
w240 ρNA
A
=
(0.8 · 0.025) · 8 · 6.02× 1023
240
= 4.01× 1020 cm−3 [1]
(b) Show that the thermal fission rate is 2.26 × 1013 cm−3 s−1, the thermal
radiative capture rate is 1.21 × 1013 cm−3 s−1 and the neutron yield factor is
1.86. Calculate the specific burn-up for this fuel. [8]
Solution:
[A simple bit of problem solving, similar to problems in the problem sheet and
lectures. The last part is unseen and might prove a little trickier.]
Recognising that R = φΣ, the thermal fission and capture rates are:
Rf = φN σf = 0 +
(2.5× 1013 · 1.21× 1021 · 747× 10−24) +
0
= 2.26× 1013 cm−3 s−1 [1]
Rγ = φN σγ = (2.5× 1013 · 1.46× 1022 · 2.68× 10−24) +
(2.5× 1013 · 1.21× 1021 · 272× 10−24) +
(2.5× 1013 · 4.01× 1020 · 289× 10−24)
= 1.21× 1013 cm−3 s−1 [2]
PHYS5038 Nuclear Power Reactors 6/12 Q2A continued over. . .
Q2A continued
SHOWING SOLUTIONS
Since the capture-to-fission ratio is just α = Σγ/Σf and the number of neutrons
released per fission for Pu-239 is 2.87:
α =
Σγ
Σf
=
Rγ
Rf
=
1.21
2.26
= 0.54
η =
ν
1 + α
=
2.87
1.54
= 1.86 [2]
The specific burn-up is a measure of the useful energy extracted from the fuel and
is given by P t/m. We therefore need to calculate m, the mass of fuel in the reactor
core from the volume:
V =
P
Rf f
=
4.3× 109
2.26× 1013 · 200 · 1.6× 10−13 = 5.94× 10
6 cm3 [1]
mfuel = 0.8 ρ V = 0.8 · 8 · 5.94× 106 = 38.00× 106 g = 38.00 tHM [1]
burn− up = P t
m
=
4.3 · 365
38.00
= 41 GWd/tHM [1]
(c) Calculate the thermal utilisation factor for this reactor core. If the reso-
nance escape probability is 0.9 and the fast fission factor is 1.05, calculate k∞.
Discuss your result for the latter in the context of fuel design and/or reactor
operation. [8]
Solution:
[Unseen, but similar to a problems in the lecture notes and problem sheet. The
last part requires a bit of reasoning.]
We begin by calculating the target densities in the light water moderator:
N1 =
w1 ρNA
A
=
(0.2 · 218 ) · 8 · 6.02× 1023
1
= 1.07× 1023 cm−3 [1]
N16 =
w16 ρNA
A
=
(0.2 · 1618 ) · 8 · 6.02× 1023
16
= 5.35× 1022 cm−3 [1]
This then allows us to calculate the absorption cross section in the moderator:
ΣaM = N1 σγ1 +N16 σγ16
= 1.07× 1023 · 0.33× 10−24 + 5.35× 1022 · 0.0002× 10−24
= 0.035 cm−1 [1]
We can easily get the fuel equivalent from part(b):
ΣaF =
Rf
φ
+
Rγ
φ
=
2.26
2.5
+
1.21
2.5
= 1.39 cm−1 [1]
PHYS5038 Nuclear Power Reactors 7/12 Q2A continued over. . .
Q2A continued
SHOWING SOLUTIONS
And then the thermal utilisation factor is:
f =
ΣaF
ΣaF + ΣaM
=
1.39
1.39 + 0.035
= 0.98 [1]
We can then use the four-factor formula for k∞:
k∞ = η f p = 1.86 · 0.98 · 0.9 · 1.05
= 1.72 [1]
This is a large value of k∞ meaning that a large amount of reactivity control will
be needed. This could include: burnable poisons incorporated into fuel assemblies
(such as Gd-157), more soluble poisons in the water coolant (boric acid) or routine
operation with a larger than normal negative reactivity insertion associated with
the control rod. [1]
(d) The EPR has a cylindrical core, which means that the neutron flux is
described by:
φ(r, z) =
(
3.63P
f Σf V
)
J0
(
2.405 r
R
)
cos
(pi z
H
)
,
where R and H are the radius and height of the reactor core and all other
symbols have their usual meanings. At what values of r and z is the flux
maximum and what value does it take? If Σtr = 2.37 cm
−1, what is the flux
at z = H + 0.35 cm? [4]
Solution:
[This is completely unseen.]
Recognising that φav = P/(f Σf V ) or otherwise, the maximum flux for a
cylindrical reactor (at r = 0 and z = 0) is just:
φmax = 3.63φav = 9.08× 1013 cm−2 s−1 [2]
The flux drops to 0 at z = H + 0.71λtr and since:
λtr =
1
Σtr
= 0.43 cm
0.71λtr = 0.31 < 0.35 cm,
the flux is 0 at H + 0.35 cm. [2]
PHYS5038 Nuclear Power Reactors 8/12 Q2A continued over. . .
Q2A continued
SHOWING SOLUTIONS
H-1: σγ = 0.33 b
O-16: σγ = 0.0002 b
U-238: σγ = 2.68 b
Pu-239: σγ = 272 b, σf = 747 b, ν = 2.87
Pu-240: σγ = 289 b
1 barn = 1× 10−24 cm2
Average EPR neutron flux φav = 2.5× 1013 cm−2 s−1
Energy released per fission f = 200 MeV
PHYS5038 Nuclear Power Reactors 9/12 Paper continued over. . .
SHOWING SOLUTIONS
2B A hypothetical fast reactor prototype is fuelled with UOX (uranium dioxide)
fuel at a U-235 enrichment of 15% and a density of 10.9 g cm−3. It has a
spherical core with radius of 0.3 m.
(a) Show that the fast macroscopic fission, capture and scattering cross sec-
tions for this fuel are 0.0124 cm−1, 0.0020 cm−1 and 0.132 cm−1, respectively. [5]
Solution:
[A simple bit of problem solving, similar to problems in the problem sheet and
lectures.]
N235 =
w235 ρNA
A
=
0.15 · 10.9 · 6.02× 1023
235
= 4.19× 1021 cm−3 [1]
N238 =
w238 ρNA
A
=
0.85 · 10.9 · 6.02× 1023
238
= 2.34× 1022 cm−3 [1]
The macroscopic fission cross section is
Σf235 = N235 σf235 = 4.19× 1021 · 1.22× 10−24 = 0.00511 cm−1
Σf238 = N238 σf238 = 2.34× 1022 · 0.31× 10−24 = 0.00725 cm−1
Σf = 0.00511 + 0.00725 = 0.0124 cm
−1 [1]
The macroscopic capture cross section is
Σγ235 = N235 σγ235 = 4.19× 1021 · 0.087× 10−24 = 0.000365 cm−1
Σγ238 = N238 σγ238 = 2.34× 1022 · 0.070× 10−24 = 0.00164 cm−1
Σγ = 0.000365 + 0.00164 = 0.0020 cm
−1 [1]
The macroscopic scattering cross section is
Σs235 = N235 σs235 = 4.19× 1021 · 4.45× 10−24 = 0.019 cm−1
Σs238 = N228 σs238 = 2.34× 1022 · 4.83× 10−24 = 0.113 cm−1
Σs = 0.019 + 0.113 = 0.132 cm
−1 [1]
(b) Calculate the fast removal cross section. Why is this parameter not used
in diffusion calculations for a fast reactor? [5]
PHYS5038 Nuclear Power Reactors 10/12 Q2B continued over. . .
Q2B continued
SHOWING SOLUTIONS
Solution:
[This is unseen. The first part is similar to problems seen in the problem sheets.]
The logarithmic energy loss per collision are:
ξ235 = 2/(235 + 2/3) = 0.0085
ξ238 = 2/(238 + 2/3) = 0.0084 [2]
Taking the average, the fast removal cross section is:
Σ1 =
ξΣs
ln (E0/Eth)
=
0.0085 · 0.132
ln (2× 106/0.025) = 0.0000617 cm [2]
The results clearly show that without a moderator, neutron absorption dominates
over thermalisation. This means that neutron diffusion in a fast reactor core ends
on an absorption reaction and that fast removal is therefore redundant. [1]
(c) Calculate the fast neutron diffusion length and geometric buckling factor
for this reactor. Show that the non-leakage probability is 43.8%. [10]
Solution:
[This is unseen, but similar to problems seen in the problem sheets.]
Start by calculating the average cosine of the elastic scattering angle:
µ¯235 =
2
3A
= 0.00284
µ¯238 =
2
3A
= 0.00280 [2]
We can simplify things by taking the average of these two and the values from
part(a):
Σa = 0.0124 + 0.0020 = 0.0144 cm
−1
Σtr = Σa + (1− µ¯) Σs
Σtr = 0.0144 + 0.997 · 0.132 = 0.146 cm−1 [3]
Now the diffusion coefficient and diffusion lengths are:
D =
1
3 Σtr
=
1
0.438
= 2.283 cm
L =
√
D
Σa
=
√
2.283
0.0144
= 12.59 cm [2]
PHYS5038 Nuclear Power Reactors 11/12 Q2B continued over. . .
Q2B continued
SHOWING SOLUTIONS
The geometric buckling factor for a spherical reactor is just:
B2g =
(
pi
Rextr
)2
=
(
pi
R+ 0.71λtr
)2
=
(
3.14
30 + 0.71 · (1/0.146)
)2
= 0.00811 cm−2 [2]
This means that the one-group non-leakage probability is:
PNL =
1
(1 +B2g L
2)
=
1
(1 + 0.00811 · 158.51) = 0.438
= 43.8 % [1]
(d) You now have enough information to calculate the infinite medium mul-
tiplication constant for this reactor using two different methods. How do the
results for each method compare? [4]
Solution:
[This is completely unseen.]
The first approach is obvious:
k∞ =
keff
PNL
=
1.0
0.438
= 2.228 [1]
The second approach is to calculate the neutron yield factor from the numbers
given in part (a) and realise that for fast reactor k∞ = η:
α =
Σγ
Σf
=
0.0020
0.0124
= 0.161
k∞ =
ν
1 + α
=
2.6
1.161
= 2.239 [2]
The results are reasonably close, even though the first calculation depends on the
core geometry. [1]
U-235 (fast): σγ = 0.087 b, σf = 1.22 b, σs = 4.45 b, ν = 2.60
U-238 (fast): σγ = 0.070 b, σf = 0.31 b, σs = 4.83 b, ν = 2.60
1 barn = 1× 10−24 cm2
Efast = 2 MeV, Ethermal = 0.025 eV
End of Paper
NOTE: SHOWING SOLUTIONS
PHYS5038 Nuclear Power Reactors 12/12 END