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 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 resistivity ML3T 3I 2 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: μˉ235 = 2 3A = 0.00284 μˉ238 = 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 μˉ) 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
