Elements, atoms and more. Contents. Atoms. Binding energy per nucleon. Nuclear Reactors. Atom: cloud of electrons around a nucleus

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1 Delft University of Technology Nuclear Reactors Jan Leen Kloosterman, Reactor Institute Delft, TU-Delft Challenge the future Contents Elements, atoms and more Introductory physics Reactor physics Time dependence Core thermal hydraulics Fuel and reactor design Fast reactors Safety of reactors Generation-IV Electrons Protons Neutrons Atoms 3 4 Atoms 0-0 m 0-4 m Binding energy per nucleon Atom: cloud of electrons around a nucleus the nucleus consists of Z protons and N neutrons Z is the atomic number, determines the chemical element (always an integer) A = N+Z is the number of nucleons in the nucleus; also called the atomic weight the neutral atom has Z electrons surrounding a positively charged nucleus 5 6

2 Binding energy per nucleon Stable nuclides: pairing effect 30x8.5=955 MeV E90 MeV 35x7.5= 76 MeV Stable Nuclides Z (Magic numbers and double magic-number nuclides are in bold) (to be continued) Sc 0 Ca Ca Ca Ca Ca Ca 9 K K K 8 Ar Ar Ar. 7 Cl Cl 6 S S S S P 4 Si Si. Si. 3 Al Mg Mg Mg... Na Ne Ne Ne 9 F... 8 O O O 7 N N 6 C... C. 5 B.... B 4 Be Li Li He He..... P D N Effect of Pairing Nucleons Mass defect: Zm Nm m A p n Z Pairing of nucleons is energetically more favourable and gives more stable nuclides Z N No. of stable nuclides even even 66 even odd 57 odd even 53 odd odd Coulomb barrier for fission E a Inducing fission Adding a neutron (easily enters the nucleus) releases E b ( 36 U)- E b ( 35 U)=6.6 MeV E b ( 39 U)- E b ( 38 U)=5. MeV All odd heavy nuclides fission easily (thermal neutrons) All even heavy nuclides need energy threshold in MeV range Most important nuclides: 33 U, 35 U, 38 U, 3 Th, 39 Pu, 40 Pu, 4 Pu 36 U: E a =5.3 MeV 39 U: E a =5.5 MeV Fissile Fissionable 9 0 Fissile/fissionable nuclides 0 4 Uranium isotopes Fissile Fission cross section f (barn) U-33 U-35 Pu-39 U-38 Fissionable Not fissile Fissile Good fuel Energy (ev) 99,3% 0,7%

3 Fission product yields Sr-90 LW R 0 U-35 Pu-39 0 Pu I-3 Cs-37 Another view at fission yields Yield (%) Mass number A Java app 3 4 Karlsruher nuclide chart Atomic number Z Gamma decay Alpha decay Positron decay Beta decay 5 6 Neutron number N Spontaneous fission Fission yields (integrated) 7 Primordial nuclides 8 3

4 Exercise: Lifetime of the Earth Solution: Lifetime of the earth For both U-35 and U-38 holds: Nuclide T a Yield % ½ U U U Calculate the lifetime of the earth (or more precise: the lifetime of the material the earth is made of ) 9 ln N t N0expt N0exp t T/ N T N exp t N T N0 exp t Assume N N (equal amounts of U-35 and U-38 formed) T ln From this you can calculate T 60 a (6 billion years) 0 0 Neutron induced nuclear reactions A X n Z Scattering Nuclear reactions Absorption Elastic n A X Z Inelastic A X n Z Capture A X Z H- capture cross section Potential scattering Compound nucleus formation A A Z Z Fission X Y n Compound nucleus formation Shape of a single resonance p-wave scattering s-wave scattering v E E 0 Line spacing smaller for higher E / E 0 ( Ec) 0 ; y ( E c E0) Ec y 3 4 4

5 U-38 capture cross section Resolved region: reduced height for increasing E U-38 capture cross section (groups) /v slope Unresolved resonance region 5 6 U-35 capture and fission xsecs U-38 capture and fission xsecs fission ratio capture 7 8 Nuclear Data sets The fission process Radio-active 35 Un X Y n00 MeV CH O CO H O 8eV

6 Distribution of energy Exercise: Energy from gram U UnX Y n00 MeV Calculate the energy contained in gram of Uraniu m. 3 3 Solution: Energy from gram U-35 Fossils equivalent to gram of U35 Gasoline Coal 35 UnX Y n00 MV e gram of uranium contains atoms A When completely fissioned, the energy production equals N A E =8.0 J E =0.95 MWd Rule of thumb: gram fissioned gives MWd of energy 500 liter 3000 kg Exercise: Velocity of neutron and FP 35 Un X Y n00 MeV The average energy of a fission neutron is MeV, the kinetic energy of the fission products is 80% of total. Calculate the velocity of a fission neutron. m n kg Calculate the velocity of a fission product with A 0 In reality there is a light and heavy FP. How is the energy distributed over two products? Solution: Velocity of neutron and FP E mv 6 9 E 0.60 vn 7 m.6750 n 7 0m/s 6 9 E vfp.0 m/s 7 mfp Two fission fragments with mass m and M M m Conservation of momentum gives mv MV MV m mv m M Ratio of kinetic energies is = MV MV m So the lightest FP has highest kinetic energy!

7 How to extract energy from Uranium? Neutron yield from fission Geothermal energy: 40 MeV per nuclide Fission energy: 00 MeV per nuclide energy of incoming neutron (MeV) Nuclide and energy dependent! Reactor Physics Fission neutron spectrum Fission spectrum (MeV - ) E E Eexp met T,3 MeV 3 T T 0 E E' E' de' 3 T MeV Energy (MeV) Moderation of neutrons Energy transfer in collision Moderation U-35 Pu-39 U Grazing shot: energy transfer almost zero Fission cross section (barn) Fission spectrum Frontal collision: energy transfer large Frontal collision: energy transfer small Energy (ev)

8 Energy transfer in collisions Energy transfer in collisions Elastic scattering Conservation energy and momentum Collision with light nuclide gives larger energy transfer than with heavy nuclide Hydrogen is lightest element largest E-transfer p E E' for E E' E E pe E' M mass nucleus A A m mass neutron A Area= E E Energy transfer in collisions Energy transfer in collisions Average energy: E' E' p E E' de' E Average loss fraction: E E E A E A Independent of energy! Average logaritmic energy loss per collision: E E ln p E E ' de ' E ' E A 3 This can be used to easily calculate the number of collisions needed to moderate a neutron from energy EH to energy EL EH ln E n L Example: moderation from Mev ev Element A n H D C Na U Moderator quality A good moderator has the following properties: s large large small a These properties can be combined into two parameters: Moderator power s s Moderator quality - Moderator Moderator power cm Moderator qua H O.35 7 DO Be C a lity

9 Neutron spectrum in the epi-thermal range If s E is constant, then it follows for the neutron spectrum in the epi-thermal range: E C E Elastic scatter cross section (cm - ) water graphite Energy (ev) 49 Scattering in the thermal range Thermal scattering kernel P(E E') *kT 0*kT Hydrogen E'/E Lamarsh, Introduction to Nuclear Reactor Theory kt Scattering in the thermal range Scattering in the thermal range Thermal scattering kernel P(E E') *kT 0*kT E'/E kt Graphite Thermal energy range upscattering of neutronen Result is a Maxwell neutron distribution: no E M E Eexp 3/ kt kt and corresponding neutron flux density: n0 M E 3/ kt m Average neutron energy: / E Eexp kt 3 E EM EdE kt 0 Thermal Maxwell spectrum M (E) E kt=0.05 ev 93 K Energy (ev) 5 kt Thermal neutron density Neutron spectrum in a reactor Thermal Maxwell spectrum M (E) K 600 K 900 K 00 K The neutron spectrum contains three contributions: Fission spectrum Epi-thermal /E spectrum Thermal Maxwell spectrum Energy (ev)

10 neutron U-35 Description neutronbalance Moderator U-38 U-35 U-39 Moderator Np-39 U-38 Pu-39 June 8, Pu Neutron balance Multiplication factor, k Fission Capture Leakage Fission # neutrons in generation n+ k # neutrons in generation n k=: critical, stationary situation, stable k>: supercritical, increasing population, unstable k<: subcritical, decreasing population, unstable Scattering Neutron cycle in a thermal reactor absorption in fuel START thermal neutron p fpp f t absorption in non-fuel f fast fission f p PP t fast leakage resonance absorption thermal leakage P f p P f 59 Pf p Pt Six-factor formula: definitions fast fission factor number of fast neutrons = number of fast neutrons produced by thermal fission P f = fast non-leakage probability = fraction of fast neutrons not leaking from the core p = resonance escape probability = fraction of neutrons passing the resonance region P t = thermal non-leakage probability = fraction of thermal neutrons not leaking from the core f = thermal utilization factor number of neutrons absorbed in the fuel = total number of thermal neutrons absorbed = neutron reproduction factor average number of neutrons per fission = number of neutrons absorbed in t he fuel June 8,

11 Six-factor formula: numbers k k P P pf eff f t η = neutron reproduction factor (.65) ε = fast fission factor (.0) p = resonance escape probability (0.87) f = thermal utilization factor (0.7) P f = fast non-leakage probability (0.97) P t = thermal non-leakage probability (0.99) Fast and thermal non-leakage factors k keff kpf Pt Bg L with: L measure for the total distance neutrons travel until absorption as a thermal neutron. This distance includes the moderation process and the distance the thermal neutron travels until absorption. Material L t (cm) L f (cm) L (cm) H O D O Be C The four and six factor formulas Multiplication factor for infinitely large medium: k pf Exercise: Fuel enrichment f N(35) f(35) Fuel N(35)[ (35) (35)] N(38) (38) a f c c Effective multiplication factor: k k P P eff f t Good approximation for reactors with Natural uranium as a fuel Graphite as a moderator 63 Calculate the useful enrichment N5 Exercise : e by plotting as function of e N N Data for thermal neutrons: f c c barn barn barn Solution: Fuel enrichment 35 e f N35 with e e e N N a c Centrifuge cascades Enrichment above 5% is expensive and does not give much gain What is the limit value of η? e % 65 66

12 The k as function of the moderator to fuel ratio k p f Critical mass k f pp P f t k k Undermoderated Overmoderated LWR Relatively, a small number of neutrons leak out of the system Relatively, a large number of neutrons leak out of the system At k =, the critical mass has been achieved Neutron economy: reflector Feedback mechanisms U-35 Moderator Moderator feedback Doppler feedback U-38 U U-39 ) Stable system (important for control) ) June 8, 0 Loss of cooling capacity shuts down the reactor 70 3) Np-39 Loss of moderator shuts downd the reactor 70 Doppler feedback mechanism Doppler feedback mechanism barn Capture probability in resonance is E cap E 00 E E tot E 05 E Due to the vibration of the nucleus, the effective resonance broadens. The area remains virtually constant Capture probability in broadened resonance is Resonance 00 barn E cap E 50 E E tot E 55 E Broadened resonance 50 barn Scatter 5 barn 5 E E E E 7 7

13 Time dependence Multiplication factor production by fission k loss by absorption and leakage k=: critical, stationary situation, stable k>: supercritical, increasing population, unstable k<: subcritical, decreasing population, unstable N(t) k> N(0) k= k< 73 t Life time of neutrons Time from birth of the neutron until absorption Prompt life time dominated by moderation and diffusion sec for thermal reactors, 0-6 sec for fast reactors birth death Infinite system: l p L a v v a Exercise: Controllability of fission chain reactions Time-dependence of neutron flux: dn k nt dt l p 4 If k.00 and lp 0 sec typical values for LWR, calculate the time needed for the neutron density to double Solution: Controllability of fission chain reactions Time-dependence of neutron flux: dn k k nt nt n0exp t dt l p l p k k l 4 If.00; p 0 sec 0 lp k ln t ln t 0.07 sec lp 0 Cannot be controlled! 6 For fast reactor even worse typically 0 sec l p Delayed neutrons Coming from instable fission products Br Kr Kr n * % s Prompt neutron Small fraction of total neutron yield U-35: 0.65% d Delayed neutron

14 Time dependence with del. neutrons delayed neutrons k delayed neutrons Average neutron life time including delayed neutrons l l l t 6 6 i i i i i l t e p i p d U-35 =0.685% : 0.09 sec and 3 s c d U-35 U-35 Doubling time in exercise would increase to mi n! MATLAB: nudel=0.0668; nu=.43; beta=[0.038, 0.3, 0.88, 0.407, 0.8, 0.06]*nudel/nu; thalf=[54.5,.84, 6.0,.3, 0.496, 0.79]; lambda=log()./thalf; sum(beta./lambda); 79 - prompt neutrons k(-) prompt neutrons As long as k(-)< delayed neutrons are needed to sustain the fission chain reaction and the reactor period will be dominated 80 by the delayed neutron decay time. U-35 neutron Core thermal hydraulics of LWRs Moderator U-38 U-35 Pu-39 Moderator U-38 Pu Fuel assembly of a PWR Zircalloy r UO a b H Cosine shaped power profile z ( z) max cos H ''' ''' z q ( z) qmax cos H Axial temperature distribution center line outer radius H

15 Radial temperature distribution (Departure from) nucleate boiling kt r q''' r k thermal conductivity T temperature F q ''' power density Integration over fuel pin gives: q ' 75 W/cm LWR TF q' ave 4 k 95 W/cm LMFBR Independent of radius! LWR: T 400 K, T 00K F cl Fuel pin Nucleate Boiling Departure from Nucleate Boiling Thin vapour layer Maximum heat flux: burnout Fuel assembly of a PWR Two pellets sufficient to generate all electricity for European family per year Pressurized Water Reactor Exercise: Mass flow condenser circuit A nuclear reactor with thermal power of 500 MWth discharge waste heat in a river. If the temperature increase of the river water is 0 K, calculate the mass flow of the condensor circuit heat capacity water is c 4, J. g K

16 Delft University of Technology Solution: Mass flow condenser circuit Fuel assembly of a PWR The efficiency typically equals =33% Mass flow required equals: 9 P m 4 ton/sec Tc 04. Two pellets sufficient to generate all electricity for European family per year 9 9 Soluble boron during burnup Why may this concentration not be too high? Xe and Sm Burnable poison not fully depleted: loss of reactivity Important components in a PWR Pressure vessel PWR m 4 m 95 Challenge the future 6

17 Pressurizer PWR Steam generator PWR (Re)placement steam generator Exercise: Electricity from one pellet Dimension of one pellet: D8 mm; H 5mm; 0 g.cm UO -3 If the burnup of the fuel reaches 45 MWd/kgU, estimate the electricity that can be produced from one pellet Solution: Electricity from one pellet Boiling Water Reactor Dimension of one pellet: D8 mm; H 5mm V 0.75 cm 3 Volume pellet 0.75 cm -3 One pellet contains 7.5 g UO 6 g U Density UO 0 g.cm 3 A burnup of 45 MWd/kgU corresponds to 0.7 g uranium fissioned 0.7 g U 0.6 MWd th=600 kwhth 00 kwhe Two pellets almost sufficient to generate all electricity for a Dutch family per year How much coal is needed to generate the same amount of electricity? 0 0 7

18 BWR fuel assemblies Generations of BWR gen II gen III gen III+ ABWR ESBWR Reactor safety Reactor safety Three prevailing principles Reactivity control Fuel cooling (in and outside reactor pressure vessel) Safe confinement of radioactivity Feedback coefficients Decay heat production U-35 Moderator Moderator feedback Doppler feedback U-38 U-35 ) Stable system Pu-39 Source: ORNL ) Loss of coolant stops fission chain reaction 3) Loss of moderation stops fission chain reaction

19 European Pressurized Water Reactor Reactor building Turbine building Passively cooled core catcher Cooling water 4 veiligheidsgebouwen 4 x 00% 09 0 Containments nuclear power plant Inherently safe reactors Fuel rod Primary system (steel) Containments (x concrete+steel) High Temperature Reactor (HTR) HTR Fuel assembly 3 4 9

20 X-Ray image of HTR fuel pebble LOF incident in pebble-bed HTR 5 6 neutron Fast reactors U-35 Moderator U-38 U-35 Pu-39 Moderator U-38 Pu U-35 Snel neutron Pu-39 cross sections U-38 U-35 Pu-39 U-38 Pu-39 Pu

21 Pu-39 Snel neutron Uranium isotopes Pu-39 U-38 U-38 Pu-39 Plutonium gives more neutrons per fission! Not fissile Fissile Good fuel Pu-39 U-38 Also fuel!! Pu-39 U-38 Pu-39 99,3% 0,7% Pu-39 Coolants for fast reactors What makes up a good coolant for fast reactors? Stable (no decomposition due to radiation, etc) Compatible with fuel, cladding, heat exchangers, Low melting point, high boiling point Poor moderator, no buildup of activity Low cost, high availability Low pumping losses High heat transfer coefficients Liquid metals are very good coolants for FR Fuel pin design for fast reactors Two criteria for heat transfer from fuel pin to coolant: Maximum heat flux from fuel pin surface to coolant q < q DNB (300 W/cm in LWRs) Maximum linear power in fuel pin q < q max (660 W/cm in UO fuel elements) Reducing the fuel pin diameter and keeping constant the linear power: Reduces the fissile inventory of the reactor (good) Reduces the breeding gain of the reactor (bad/good) 3 4 Core design for breeding for many energies (even more excess neutrons) CR BR More fissile fuel can be produced than consumed Core design for breeding Inner core/ Driver core Breeding blanket Fast (spectrum) breeder: 38 U / 39 Pu mixture U(n,γ) U Np Pu β 39 β 39 3min.3d Neutrons leaking from the core zone will be captured in the breeding blanket and convert U-38 to Pu-39 5 June 8, 0 6 6

22 Fast reactors Nuclear fuel cycle in 050 LWR + HTR a a Pu+Am FR 300 a 7 8 Generations of nuclear reactors Six reactor types in Gen-IV Generation I Early Prototype Reactors - Shippingport - Dresden, Fermi I - Magnox Generation II Commercial Power Reactors - LWR-PWR, BWR - CANDU - VVER/RBMK Generation III Advanced LWRs - ABWR - System AP600 - EPR Evolutionary Designs Offering Improved Economics Gen I Gen II Gen III Generation IV - Highly Economical - Enhanced Safety - Minimal Waste - Proliferatio n Resistant Gen IV 9 30