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# PHYSICS OF NUCLEAR REACTORS MANE 4480

RPI

GPA 3.56

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This 7 page Class Notes was uploaded by Hugh Wilkinson on Monday October 19, 2015. The Class Notes belongs to MANE 4480 at Rensselaer Polytechnic Institute taught by Yaron Danon in Fall. Since its upload, it has received 25 views. For similar materials see /class/224911/mane-4480-rensselaer-polytechnic-institute in Mechanical and Aerospace Engineering at Rensselaer Polytechnic Institute.

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Date Created: 10/19/15

Chapter 3 I hermalization 31 DOPPIer Effect Consider a neutron of mass m that interacts With a nuclear of mass M that is at rest The center of mass energy is given by 1 mM 1 where M is the reduced mass and z is the neutron energy in the lab system If the target nuclear is moving then the center of mass energy is given by 1 EC Eva 32 Where 2 is the relative speed in the lab system For example 7 111 1 111 q lb Elm F Figure 31 An illustration of the relative speeds of the neutron m and the nucleus M is shown above The motion of the target nucleus is affected by the effective temperature T of the target We want to calculate the effect of the temperature change on the reaction cross section JCEn l Where En in the neutron energy in the lab systm The reaction rate can be calculated by Where I is the beam intensity N is the atom density in cm3 and 6En T is the average doppler broadened cross section We describe the nuclei density by NVdV 67 which is the humber Of target rlriclei movirlg with Velocities betweerl V arld V I dv The reactiorl rate carl be writterl as IN5CEJn T fanNCV6ECCZV 34 Where n is the number dehsity Of neutrorls ir L the beam The beam irlterlsity carl be writterl as cm I 7 n zquot 35 Where an is the heutrorl Speed 11 1 the lab system We can solve for the doppler broadened cross section 5EJquot 139 N n f vTGCECNVdV 36 In order to evaluate this equatiorl we heed to express EC and pr 11 1 terms Of the lab erlergy Of the heutrorl arld target rlriclei The relative velocity is Thus E t 7 62 t 7 IV 7 V 392 t 7 62 I 7 V2 7 V V 3 8 C 7 H 7 7 H 7 H H H or 12 M m 2Equot EC En ET 7 g 7 Z 39 m I M m I M m C Where En arld ET are the lab system ehergies Of the heutron and the rlucleus respectively arld VZ is the comporlerlt Of the rlricleris velocity alorlg the directiorl Of the heutrorl 11 quotT 5 39 7 Z Figure 32 The illustratiorl above gives the velocities Of the rlricleus arld the heutrorl For heavy rlriclei M m z 1 and x O 310 m I M m I M C We get 2E 12 EC 7 En 7 g 7quot VZ 311 m gt C Sirlce EC arld vr deperld Ohly Oh the Z comp Oheht Of the velocity we carl Simplify the irlte gr atiorl fNVdV NVx V3 VZ de dVy dVZ fNVZdVZ 312 68 The equation for the doppler broadened cross section becomes 5031 T fmascwwzmvz 313 where vn is the neutron velocity 311 The Free Gas Model The simplest approximation is to assume that the target nuclei behave as gas In this case the energy and velocity distribution is given by the Maxwellian distribution M 12 MVZZ NVZNW exp W 314 where k 3617 10 5 is Boltzmann39s constant T is the temperature of the gas and M is the atomic mass of the gas atoms lnserting equation 314 into equation 313 and using the Breit Wigner formula for the cross section I E0 r2 Ty E0 1 07021 00 7 l 00 315 F EC 4 EC 1302 r2 F 133 2EEEE0gt2 1 we get 7 1 0 Fy E0 1 M 12 MVZZ 07Equot T N vnfmm VIE Cumwf 1N2nkT exp 2kT W1 33916 F Rearranging we get Mv2 7 03901quot M 12 0 1 E0 equotP kTZ 07Ean 1 TIM fwv nxlE C AEEEwYJrldVZ 317 F The term 13 can be simplified by noting that v Z EC and vn 2 13 318 y m U 213C m EcmmM N E vn l y 213 V13 mM N V13 33919 69 Thus provided M gtgt 111 The doppler broadened cross section is now MV2 7 00Fy E0 M szoo exP kTZ 07EnT r JFK 27TH foo 2EFEO21dVZ 320 F This can be rewritten by making some substitutions 2 2 r xfEn Eo yfEc Eo 525 321 where TD is known as the doppler width and is given by 12 4E0kT D A 322 The doppler broadened cross section is then given by 001quot E 12 mm r V Mac 323 where 1pCx is given by 1 2 2 c f BXPHC x w d 324 mac 2W7 1W2 y lt gt The dependence of the cross section on temperature is shown in figure 33 for the 238U resonance at 667 eV 312 High Temperatures The 1p function has some similarity to a Gaussian distribution with a width of and the exponential term centered at x y At high temperatures 1 D is large and the C Gaussian distribution is broad In this case the term 1 dominates The expression can be approximated by 2 W exp High T 325 D 00F 570311 T Zia D Thus at large T the cross section has a Gaussian shape with a standard deviation PD and the cross section shape does not depend on the resonace parmaters 70 20000 Has a h V a soon u E Lt 0Irrilrtfv f zm pgplgmpgi b 0003050709 who Eu Figure 33 Doppler broadening of the 667 eV resonance in 2380 313 10 Cross Section At low incident neutron energies the cross section is due to sum of resonances tails and behaves like or 10 0 ugh do we get uuo rnruuvE on H4 N in this case the cross section shape does not depend on the target nudei distribution 00 and the cross section remains or 10 314 Scattering Asimiiar procedure can he applied to the scattering cross section We get duo000000300 on Physics ofNuclear Reactors Spring 2012 MAN E4480 Homework 1 1 Well collimated thermal 00253 eV and fast 2 MeV neutron beams are incident on two U02 fuel plates with enrichment of 35 w and density of 10 gcm3 The thickness of each plate is 04 cm Answer V a What are the U235 U238 and O densities in unit of atomsbarn b What is the probability that the thermal and fast neutrons will collide with the fuel plates 0 What is the probability that the thermal neutrons will cause ssion in the fuel d What is the probability that the fast neutrons will cause ssion in the fuel e What is the faction of thermal neutron that gets captured in the rst plate f What is the faction of thermal neutron that gets captured in the second plate The neutron cross sections can be obtained from httpwwwnndcbnlgpvexforendfOOig Cross section retrieval example 1 2 3 On the right deselect all libraries with the exception of ENDFBVl0 To retrieve the 23 5 U total cross section use the following of search Options Target U 235 Reaction ntot Quantity sig Plot the data and click on the right link See plotted data Scroll to nd the required energy and cross section value repeat for the other materials and reactions of interest 2 A reactor is shielded by a concrete wall density235 gcm3 with a thickness of 2 m The composition of the concrete is given in Table I For the purpose of shielding calculation assume a well collimated thermal 00253 eV neutron beam with intensity similar to the reactor neutron ux of 1012 ncmZsec is hitting a 1 cm2 area of the concrete slab Calculate a The mean free path of thermal neutron in concrete b The number of neutron transmitted through the shield with no interaction c The number of capture reactions with hydrogen occurring in the shield d The hydrogen capture gamma energy write the nuclear reaction and use mass difference for the calculation Table I Concretecoposition for qustion 2 1 Element eght F ractiot 39 30 0500 0315 Al 2048 Na 0017 Ca I 0083 Fe 0012 N K 0019 Use Table III of appendix A on page 606 for the thermal cross sections Physics ofNuclear Reactors Spring 2012 MANE4480 3 Calculate the mean free path mfp of thermal 00253 eV neutrons in heavy water What is the atomic ratio of B1 0 boron10 nd x in DZOBX that should be introduced to the heavy water to reduce its mfp to that of light water The neutron total cross section can be obtained from httpwwwnndcbnlgovexforendf00isp Select the ENDFBVIIO database and use the following search options Target HO Reaction ntot Quantity sig Repeat for the other materials of interest Use a density of 11 gcm3 for heavy water

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