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Revision 1 December 2014 Fission Instructor Guide Reviewed by: Cassandra Bitler Project Manager, OGF 11/14/2014 Date Approved by: Robert Coovert Manager, INPO Learning Development 11/14/2014 Date Approved by: Kevin Kowalik Chairperson, Industry OGF Working Group 11/14/2014 Date NOTE: Signature also satisfies approval of associated student guide and PowerPoint presentation. GENERAL DISTRIBUTION GENERAL DISTRIBUTION: Copyright © 2014 by the National Academy for Nuclear Training. Not for sale or for commercial use. This document may be used or reproduced by Academy members and participants. Not for public distribution, delivery to, or reproduction by any third party without the prior agreement of the Academy. All other rights reserved. NOTICE: This information was prepared in connection with work sponsored by the Institute of Nuclear Power Operations (INPO). Neither INPO, INPO members, INPO participants, nor any person acting on behalf of them (a) makes any warranty or representation, expressed or implied, with respect to the accuracy, completeness, or usefulness of the information contained in this document, or that the use of any information, apparatus, method, or process disclosed in this document may not infringe on privately owned rights, or (b) assumes any liabilities with respect to the use of, or for damages resulting from the use of any information, apparatus, method, or process disclosed in this document. ii Table of Contents INTRODUCTION................................................................................................................. 1 TLO 1 NEUTRON INTERACTIONS WITH MATTER .............................................................. 2 Overview ..................................................................................................................... 2 ELO 1.1 Neutron Scattering Interactions .................................................................... 3 ELO 1.2 Neutron Elastic Scattering Interactions ........................................................ 5 ELO 1.3 Neutron Absorption Reactions ..................................................................... 6 TLO 1 Summary ......................................................................................................... 7 TLO 2 FISSION PROCESS .................................................................................................. 8 Overview ..................................................................................................................... 8 ELO 2.1 Excitation Energy ......................................................................................... 9 ELO 2.2 Fission Process Definition .......................................................................... 10 ELO 2.3 Fissile Material ........................................................................................... 12 ELO 2.4 Fission Materials ........................................................................................ 16 ELO 2.5 Binding Energy Per Nucleon Curve ........................................................... 17 TLO 2 Summary ....................................................................................................... 19 TLO 3 PRODUCTION OF HEAT FROM FISSION ................................................................. 20 Overview ................................................................................................................... 20 ELO 3.1 Energy Release Per Fission ........................................................................ 21 ELO 3.2 Fission Fragment Yield .............................................................................. 23 ELO 3.3 Calculate Fission Energy: Binding Energy and Conservation of Mass ..... 25 ELO 3.4 Fission Heat Production ............................................................................. 29 TLO 3 Summary ....................................................................................................... 30 TLO 4 INTRINSIC AND INSTALLED NEUTRON SOURCES ................................................. 31 Overview ................................................................................................................... 31 ELO 4.1 Source Neutrons ......................................................................................... 31 ELO 4.2 Intrinsic Source Neutrons ........................................................................... 32 ELO 4.4 Intrinsic Source Neutrons Over Core Life ................................................. 39 TLO 4 ........................................................................................................................ 41 Summary ................................................................................................................... 41 TLO 5 RELATIONSHIP BETWEEN NEUTRON FLUX, MICROSCOPIC, AND MACROSCOPIC CROSS-SECTIONS ........................................................................................................... 43 Overview ................................................................................................................... 43 ELO 5.1 Neutron Reaction Terms ............................................................................ 44 ELO 5.2 Neutron Energy Terms ............................................................................... 53 ELO 5.3 Neutron Energies versus Cross-Sections .................................................... 54 ELO 5.4 Temperature Affects to Macroscopic Cross-Sections ................................ 57 ELO 5.5 Neutron Flux Distribution .......................................................................... 60 ELO 5.6 Reaction Rate ............................................................................................. 62 ELO 5.7 Neutron Flux and Reactor Power ............................................................... 64 TLO 5 Summary ....................................................................................................... 67 FISSION SUMMARY ......................................................................................................... 68 iii This page is intentionally blank. iv Fission Revision History Revision Date Version Number Purpose for Revision Performed By 11/14/2014 0 New Module OGF Team 12/10/2014 1 Added signature of OGF Working Group Chair OGF Team Duration ο· 6 hours and 55 minutes Logistics Ensure that the presentation space is properly equipped with the following: ο· ο· ο· ο· ο· Projector Internet access, if needed Whiteboard or equivalent Space for notes, parking lot, mockups, or materials Sufficient space for all students Ensure that the following course materials are prepared and staged: ο· ο· ο· ο· All student materials Instructor materials Media, photos, and illustrations Props, lab equipment, or simulator time, as applicable Ensure that all students have fulfilled the course prerequisites, if applicable. Instructor preparation ο· Review the course material prior to beginning the class. ο· Review the NRC exam bank and as many new exams as are available prior to the class to ensure that you are prepared to address those items. Introduction This module focuses on neutron interactions, including fission, building on the atomic structure module. To provide complete knowledge of fission, it is important to understand all of the various types and probabilities of neutron interactions that exist. This module will cover neutron scattering and absorption reactions, materials used for nuclear fuel, the production of heat from fission, neutron sources for shutdown and startup conditions, and Rev 1 Logistics ο§ Use PowerPoint slides 1β3 and the instructor guide (IG) to introduce the Fission module. 1 the various terms used for describing and measuring neutron intensity, reaction probabilities, and reaction rates. You will then apply this background knowledge to understand the calculations for determining reactor thermal power output. Objectives At the completion of this training session, the trainee will demonstrate mastery of this topic by passing a written exam with a grade of 80 percent or higher on the following Terminal Learning Objectives (TLOs): 1. Describe neutron interactions with matter. 2. Describe the process of nuclear fission and the types of material that can undergo fission. 3. Explain the production of heat from fission. 4. Describe intrinsic and installed neutron sources and their contribution to source neutron strength over core life. 5. Explain the relationship between neutron flux, microscopic and macroscopic cross-sections; and their effect on neutron reaction rates. TLO 1 Neutron Interactions with Matter Overview Duration ο§ 5 minutes Logistics ο§ Use PowerPoint slides 4β5 and the IG to introduce TLO 1. Neutrons can cause many different types of interactions. The neutron may simply scatter off the nucleus, which can result in a reduction of the neutron's energy, or be absorbed within the nucleus and removed from the system. When neutrons are absorbed into the nucleus, the nucleus becomes excited. This results in emission of gamma rays and/or particles, or it can result in fission. Objectives Upon completion of this lesson, you will be able to do the following: 1. Describe the following neutron scattering interactions, including conservation principles: a. Elastic scattering b. Inelastic scattering 2. Define the following: a. Resonance elastic scattering b. Potential elastic scattering 3. Describe the following reactions where a neutron is absorbed in a nucleus: 2 Rev 1 a. Radiative capture b. Particle ejection c. Fission ELO 1.1 Neutron Scattering Interactions Introduction A neutron scattering reaction occurs after by a neutron striking a nucleus and emits a single neutron. The net effect of the reaction is that the free neutron has merely bounced off, or scattered as it interacts with the nucleus. In some cases, the initial and final neutrons are not the same. There are two categories of scattering reactions, elastic, and inelastic scattering. These collisions are of great importance for the process of making thermal or low energy neutrons available for a thermal reactor. Neutron Scattering Interactions Elastic Scattering Elastic scattering, also referred to as the billiard ball effect, is most probable with light nuclides. In an elastic scattering reaction, the neutron does not contribute to nuclear excitation of the target nucleus because of a conservation of momentum. Energy completely transfers from the neutron to the target nucleus, conserving the kinetic energy of the system. The target nucleus gains an amount of kinetic energy equal to the kinetic energy the neutron loses. This kinetic energy transfer is dependent on target size and angle. The figure below illustrates an elastic scattering reaction. Figure: Elastic Scattering The following mathematically illustrates the conservation of momentum and kinetic energy during an elastic scattering reaction: Conservation of momentum (ππ£) (ππ π£π,π ) + (π π π£π,π ) = (ππ π£π,π ) + (π π π£π,π ) 1 Conservation of kinetic energy (2 ππ£ 2 ) Rev 1 3 Duration ο§ 10 minutes Logistics ο§ Use PowerPoint slides 6β12 and the IG to present ELO 1.1. Inform ο§ The importance of these scattering reactions will be shown later in the course. 1 1 1 1 2 2 2 2 ( ππ π£π,π ) + ( π π π£π,π ) = ( ππ π£π,π ) + ( π π π£π,π ) 2 2 2 2 Where: mn = mass of the neutron mT = mass of the target nucleus vn,i = initial neutron velocity vn,f = final neutron velocity vT,i = initial target velocity vT,f = final target velocity Inelastic Scattering In inelastic scattering, the target nucleus absorbs the incident neutron by forming a compound nucleus and transfers the kinetic energy into nuclear excitation. The compound nucleus emits a neutron of lower kinetic energy than the incident neutron, leaving the nucleus in an excited state. A gamma emission occurs if the excitation state is not at a preferred level, dropping excess energy to reach ground state. Inelastic scattering is most probable with heavy nuclides and neutrons above 1 MeV in energy. The figure below illustrates inelastic scattering. Figure: Inelastic Scattering Some amount of kinetic energy transfers into excitation energy of the target nucleus. The total kinetic energy of the outgoing neutron and nucleus is less than the kinetic energy of the incoming neutron. Therefore, there is no conservation of kinetic energy; however, momentum is conserved. 4 Rev 1 Knowledge Check True/False: The difference between elastic and inelastic scattering where neutrons are concerned is elastic scattering involves no energy being transferred into excitation energy of the target nucleus. Inelastic scattering involves a transfer of kinetic energy into excitation energy of the target nucleus. A. True B. False ELO 1.2 Neutron Elastic Scattering Interactions Introduction Elastic scattering of neutrons by nuclei can occur in two ways: ο· ο· Resonance elastic scattering Potential elastic scattering Resonance Elastic Scattering Resonance elastic scattering, also referred to as compound elastic scattering is the least likely type of scattering event to occur. The nucleus absorbs the incident neutron, temporarily forming a compound nucleus. The nucleus also emits a neutron, returning the nucleus to its ground state and conserving total kinetic energy. This type of elastic scattering is dependent upon the initial kinetic energy possessed by the neutron. Potential Elastic Scattering Potential elastic scattering is the most likely type of scattering event to occur. The interaction between neutrons in potential elastic scattering referred to as the billiard ball effect, resembles two hard billiard balls bouncing off each other. Potential elastic scattering takes place with incident neutrons of up to 1 MeV. The incident neutron does not touch the nucleus nor does a compound nucleus form in this type of scattering. Instead, the incident neutron, as it comes close to the target nucleus, scatters because of the short-range nuclear forces of the nucleus. The result is no energy transfer into nuclear excitation, conservation of momentum, and conservation of kinetic energy of the system. The target nucleus gains the amount of kinetic energy that the incident neutron loses. Rev 1 5 Duration ο§ 10 minutes Logistics ο§ Use PowerPoint slides 13β16 and the IG to present ELO 1.2. Inform ο§ Ensure that students can describe the differences of these types of elastic scatterings. Knowledge Check Which of the following statements accurately describes resonance elastic scattering of a neutron? A. A nucleus absorbs a neutron, and a gamma ray is ejected. B. A nucleus absorbs a neutron and breaks into two smaller particles. C. An approaching neutron is repelled by the target nucleus. D. A nucleus absorbs a neutron and a neutron is reemitted. ELO 1.3 Neutron Absorption Reactions Duration ο§ 10 minutes Logistics ο§ Use PowerPoint slides 17β21 and the IG to present ELO 1.3. Inform ο§ Ensure that students can describe the differences of these types of elastic scatterings. Introduction Most absorption reactions result in the loss of a neutron, the production of charged particles, gamma ray, or fission fragments and the release of neutrons and energy. Three types of absorption reaction are: ο· ο· ο· Radiative capture Particle ejection Fission Radiative Capture In radiative capture, the incident neutron interacts with the target nucleus forming a compound nucleus. The compound nucleus, with the additional neutron, then decays to its ground state via gamma emission. The equation below shows an example of radiative capture with uranium-238. 1 238 239 β 239 0 π+ πβ( π) β π+ πΎ 0 92 92 92 0 Particle Ejection In particle ejection, an incident neutron interacts with the target nucleus, forming a compound nucleus. The new compound nucleus excites to a high enough energy level for it to eject a new particle with the incident neutron remaining in the nucleus. The nucleus may or may not exist in an excited state depending upon the mass-energy balance of the reaction after ejection of the new particle. The equation below shows a common example of this, specifically the boron-10 neutron reaction resulting in a lithium nucleus and alpha particle. 6 Rev 1 1 10 11 β 7 4 π + π΅ β ( π΅) β πΏπ + πΌ 0 5 5 3 2 Fission Fission is a very important absorption reaction. An incident neutron interacts with the target nucleus, the nucleus absorbs the incident neutron, and the nucleus splits into two smaller nuclei, called fission fragments. This reaction releases multiple neutrons and a considerable amount of energy in addition to the fission fragments. A fission reaction typically produces two fission fragments, 2 to 3 neutrons, and energy (in the form of kinetic energy and gamma rays). The following shows the fission of a uranium-235 atom. 1 236 β 140 93 1 235 π+ πβ( π) β πΆπ + π π + 3 ( π) 0 92 55 37 0 92 Knowledge Check What type of neutron interaction has occurred when a nucleus absorbs a neutron and ejects proton? A. Fission B. Fusion C. Particle ejection D. Radiative capture TLO 1 Summary Neutron Interactions ο· Interactions where a neutron scatters off a target nucleus are either elastic or inelastic. ο· Elastic scattering - there is conservation of both kinetic energy and momentum, and no energy is transferred into excitation energy of the target nucleus. This is most probable with light nuclides. - Resonance elastic scattering involves absorption of the neutron by the target nucleus, forming a compound nucleus, followed by the re-emission of a neutron. - In potential elastic scattering, the neutron does not touch the nucleus and does not form a compound nucleus. The neutron is scattered by the short-range nuclear forces when it approaches close to the nucleus. ο· Inelastic scattering - some amount of kinetic energy transfers into excitation energy of the target nucleus. The total kinetic energy of the outgoing neutron and nucleus is less than the kinetic energy of the Rev 1 7 Duration ο§ 10 minutes Logistics ο§ Use PowerPoint slides 22β24 and the IG to review TLO 1 material. Use directed and nondirected questions to students, check for understanding of ELO content, and review any material where student understanding of ELOs is inadequate. incoming neutron. Therefore, there is no conservation of kinetic energy; however, there is conservation of momentum. ο· Radiative capture is the absorption of a neutron by the target nucleus, resulting in an excited nucleus that subsequently (typically within a small fraction of a second) releases its excitation energy in the form of a gamma ray. ο· Particle ejection occurs when a target nucleus, absorbs a neutron, resulting in an excited compound nucleus. The compound nucleus immediately ejects a particle (for example, alpha, or proton). ο· Fission - an incident neutron adds sufficient energy to the target nucleus that the target nucleus splits apart, releasing two fission fragments, several neutrons, and energy. Now that you have completed this lesson, you should be able to: 1. Describe the following neutron scattering interactions, including conservation principles: a. Elastic scattering b. Inelastic scattering 2. Define the following: a. Resonance elastic scattering b. Potential elastic scattering 3. Describe the following reactions where a neutron is absorbed in a nucleus: a. Radiative capture b. Particle ejection c. Fission TLO 2 Fission Process Overview Duration ο§ 5 minutes Logistics ο§ Use PowerPoint slides 25β26 and the IG to introduce TLO 2. Nuclear fission is the splitting apart of a heavy nuclide into two fission products with the release of energy and additional neutrons. The release of neutrons causes additional fissions to occur, causing a self-sustaining fission rate capable of producing sufficient heat for power production. Objectives Upon completion of this lesson, you will be able to do the following: 1. Define the following terms: a. Excitation energy (Eexc) 8 Rev 1 b. Critical energy (Ecrit) 2. Explain the fission process using the liquid drop model of a nucleus. 3. Define the following terms: a. Fissile material b. Fissionable material c. Fertile material d. Thermal neutrons 4. Explain binding energy per nucleon. 5. Explain the shape of the Binding Energy per nucleon versus mass number curve including its significance to fission energy release. ELO 2.1 Excitation Energy Introduction Excitation energy is the energy above ground state of a nucleus. Critical energy is the required excitation energy for fission to occur. Excitation Energy Excitation energy (Eexc) is the measure of how far the energy level of a nucleus is above its ground state. The last topic discussed the many neutron reactions that can cause an increase in the excitation energy of a nucleus. Critical Energy The excitation energy must be above a specific minimum value for that nuclide for fission to occur. The critical energy (Ecrit) is the minimum excitation energy required for fission to occur for a particular nuclide. The reaction excites the target nucleus by an amount equal to binding energy of the neutron plus the neutron's kinetic energy when an incident neutron strikes a target nucleus. If the binding energy is less than the required critical energy for the nucleus, additional energy is required to cause the nucleus to fission. This energy could be in the form of kinetic energy from the incident neutron. For this reason, neutrons of low kinetic energy cannot cause fission with some types of isotopes used in nuclear reactor fuels. The discussion of binding energy in this course covers this concept in more detail. Rev 1 9 Duration ο§ 10 minutes Logistics ο§ Use PowerPoint slides 27β29 and the IG to present ELO 2.1. Inform ο§ Explain that the required amount of excitation energy to cause fission is dependent on the fuel isotope. That explains why not all isotopes of uranium make good fuels. Knowledge Check True or False: Excitation must be at least equal to critical energy for fission to occur. A. True B. False ELO 2.2 Fission Process Definition Duration ο§ 10 minutes Logistics ο§ Use PowerPoint slides 30β37 and the IG to present ELO 2.2. Inform ο§ Review the stages of the liquid drop model. Introduction In a fission reaction, the incident neutron interacts with the target nucleus, and the target nucleus absorbs the incident neutron. This creates a compound nucleus that is excited at such a high energy level (Eexc > Ecrit) that the compound nucleus splits (fissions) into two large fragments plus some neutrons. In addition, the fission process releases a large amount of energy in the form of radiation and fragment kinetic energy. Fission Definition As previously discussed, the attractive nuclear force between nucleons holds the nucleus together. In doing so, it is resisting the opposing electrostatic forces within the nucleus. Characteristics of the attractive nuclear force are: ο· Very short range attractive force, with essentially no strength beyond nuclear scale dimensions (β10-13 cm) ο· Stronger than the repulsive electrostatic forces within the nucleus ο· The force is independent of nucleon pairing. Attractive forces between pairs of neutrons, pairs of protons or neutron proton pairs are identical. ο· Saturable, that is, a nucleon can attract only a few of its nearest neighbors. Fission Process Example One theory of fission considers the fission of a nucleus similar to the splitting of a liquid drop. Molecular forces hold a liquid drop together, making the drop spherical in shape and resist deformation. Nuclear forces hold the nucleus together in the same manner. The next figure illustrates the liquid drop model. 10 Rev 1 Figure: Liquid Drop Model Liquid Drop Model Steps The steps illustrated above for the liquid drop model are as follows: The undisturbed nucleus in the ground state is undistorted, and its attractive nuclear forces are greater than the repulsive electrostatic forces between the protons within the nucleus. The nucleus becomes an excited compound nucleus when the incident neutron strikes the target nucleus and is absorbed. This compound nucleus temporarily contains all the charge and mass involved in the reaction, and exists in an excited state. The excitation energy added to the compound nucleus is equal to the binding energy contributed by the incident neutron plus its kinetic energy. The excitation energy may cause the nucleus to oscillate and distort in shape. If the excitation energy is greater than an isotope's specific critical energy, the oscillations may cause the compound nucleus to become dumbbellshaped as it starts to come apart. As this occurs, the attractive nuclear forces (short-range) in the neck area reduce due to saturation, while the repulsive electrostatic forces (long-range) remain almost as strong. Less force holds the nucleus together. When the repulsive electrostatic forces exceed the attractive nuclear forces, nuclear fission occurs - the nucleus breaks apart into two fission fragments and releases neutrons and energy. Rev 1 11 Knowledge Check In the Liquid Drop Model of fission, the nucleus absorbs a neutron, becomes distorted into a dumbbell shape, and splits into two nuclei. Which of the following forces is responsible for the nucleus splitting? A. Liquid drop force B. Gravitational force C. Electrostatic force D. Fission force ELO 2.3 Fissile Material Duration ο§ 20 minutes Logistics ο§ Use PowerPoint slides 38β47 and the IG to present ELO 2.3. Inform ο§ An earlier module covered binding energy. The next ELO covers binding energy in more detail. Do not get into detail in this ELO. ο§ A later section will cover neutron energy levels. Introduction There are two categories of nuclear fuel materials: those that can fission with thermal neutrons and those that require neutrons of higher energies for fission to be possible. It is possible to convert some non-fissionable materials into materials capable of fission. This section will explain the following terms: ο· ο· ο· ο· Fissile material Fissionable material Fertile material Thermal neutrons Fissile Material A fissile material consists of nuclides that will fission with incident neutrons of any energy level. They are the desired nuclides for use in thermal nuclear reactors. The advantage of fissile materials is that they can fission with neutrons possessing zero kinetic energy (thermal neutrons). Thermal neutrons have very low kinetic energy levels because they are in approximate equilibrium with the thermal energy of surrounding materials. They add essentially no kinetic energy to the reaction, however add enough BE to reach the critical energy, Ecrit. Fissile materials will fission after absorbing a thermal neutron. It is possible for fissile materials to fission with thermal neutrons because the change in binding energy supplied by the neutron addition alone is sufficient to exceed the critical energy. Examples of fissile materials are uranium-235, uranium-233, and plutonium-239. 12 Rev 1 Fissionable Material A fissionable material is composed of nuclides for which fission is possible. All fissile nuclides fall into this category as well as those nuclides that fission only from high-energy neutrons. The change in binding energy that occurs from neutron absorption causes an excitation energy level insufficient to reach critical energy for fission in fissionable materials. Therefore, the incident neutron must supply additional excitation energy by adding kinetic energy to the reaction. Experimental observation shows the reason for this difference between fissile and fissionable materials. A nucleus with an even number of neutrons or protons is more stable than a nucleus with an odd number. Adding a neutron to an odd numbered nucleus (changing it to an even numbered nucleus) produces higher binding energy than adding a neutron to an even numbered nucleus. The nuclides thorium-232, uranium-238, and plutonium-240 are fissionable materials that require high-energy neutrons to cause fission. The table below lists the critical energy (Ecrit) and the binding energy change for an added neutron (BEn) to target nuclei of interest. The change in binding energy plus the kinetic energy must equal or exceed the critical energy (ΞBE + KE β₯ Ecrit) for fission to be possible. Target Nucleus Critical Energy Ecrit Binding Energy of Last Neutron BEn BEn - Ecrit 232 90Th 7.5 MeV 5.4 MeV -2.1 MeV 238 92U 7.0 MeV 5.5MeV -1.5 MeV 235 92U 6.5 MeV 6.8 MeV +0.3 MeV 233 92U 6.0 MeV 7.0 MeV +1.0 MeV 234 94Pu 5.0 MeV 6.6 MeV +1.6 MeV As seen in the table above, uranium-235 can fission with thermal neutrons because its BEn is greater than its critical energy. This makes U-235 a fissile material. Uranium-238, on the other hand, has a critical energy of 7.0 MeV, which is greater than the 5.5 MeV BEn added by the neutron. Therefore, a fission Rev 1 13 neutron must have at least 1.5 MeV to cause fission of U-238. This makes uranium-238 a fissionable material. Fertile Materials All of the neutron absorption reactions that do not result in fission lead to the production of new nuclides through the process known as transmutation. These nuclides can be transmuted again or undergo radioactive decay to produce nuclides known as transmutation products. Nuclear reactions can produce transmutation products because several fissile nuclides do not exist in nature. Fertile materials are materials (nuclides) that can undergo transmutation to become fissile materials. The figure below shows the chain of transmutation mechanisms from which two fertile nuclides, thorium-232 and uranium-238, produce uranium-233 and plutonium-239, respectively. Figure: Transmutation Mechanisms Breeding and Conversion If a reactor contains fertile material in addition to its fissile fuel, depletion of original fuel produces new fuel in a process called conversion. Breeder Reactors Glossary Breeder reactors are reactors specifically designed to produce fissionable fuel. In such reactors, the amount of fissionable fuel produced is greater than the amount of fuel depleted. The process is called conversion if less fuel is produced than used, and the reactor is termed a converter. Commercial Pressurized Water Reactors (PWRs) would be converters. Thermal Neutrons Thermal neutrons are neutrons with very low kinetic energy levels (essentially zero) because they are in equilibrium with the thermal energy of surrounding materials. In reality, thermal neutrons undergo scattering 14 Rev 1 collisions, gaining and losing equal amounts of energy in successive collisions. They add essentially no kinetic energy to a neutron absorption reaction. The goal is to thermalize as many neutrons as possible in a commercial PWR, as a thermal reactor. Tables that summarize data for thermal neutrons' cross-sections at 68°F (20°C) allow comparisons under similar temperatures. At 68°F thermal neutron energy is 0.025 eV with a velocity of 2.2 X 105 cm/sec. As temperature increases, thermal neutron energy and velocity increase by the relationships shown below: 1 π 2 πΈπ = (0.025 ππ) ( ) ππ 1 ππ π 2 ππ = (2.2 × 105 )( ) π ππ ππ Where: Ep = Most probable energy (eV) Vp = Most probable velocity (cm/sec) To = Table temperature of 68°F (528°R) T = New temperature in °R (°F + 460°) Knowledge Check What is the difference between a fissionable material and a fissile material? Rev 1 A. There is no difference between a fissionable and a fissile material, except for the number of protons and neutrons located in the nuclei of the particular materials. B. A fissionable material can become fissile by capturing a neutron with zero kinetic energy, whereas a fissile material can become fissionable by absorbing a neutron that has some kinetic energy. C. Fissile materials require a neutron with some kinetic energy in order to fission, whereas fissionable materials will fission with a neutron that has zero kinetic energy. D. Fissionable materials require a neutron with some 15 kinetic energy in order to fission, whereas fissile materials will fission with a neutron that has zero kinetic energy. ELO 2.4 Fission Materials Duration ο§ 5 minutes Logistics ο§ Use PowerPoint slides 48β50 and the IG to present ELO 2.4. Inform ο§ Review total binding energy calculation coming from the delta mass defect. ο§ Review binding energy added as a result of the addition of one neutron (last lesson). ο§ Discuss the BE/A calculation. Introduction Binding energy per nucleon (BE/A) equals the average energy required to remove a nucleon from a specific nucleus. The total binding energy related to the mass defect and the binding energy related to the addition of a single neutron to a nucleus determines a thermal neutron's ability to cause fission. The next two sections go into more detail about BE/A. Binding Energy per Nucleon Binding energy per nucleon (BE/A) is equal to the average energy required to remove a single nucleon from a specific nucleus. It is determined by dividing the total binding energy of a nuclide by the total number of nucleons in its nucleus. The example below illustrates this calculation. Example: Given that the total binding energy (BE) for a U-238 nucleus is 1,804.3 MeV, calculate the binding energy per nucleon (BE/A) for U-238. Solution: π΅πΈ 1,804.3 πππ = = 7.6 πππ π΄ 238 Knowledge Check True or False: Binding energy per nucleon is independent of the specific nuclide. 16 A. True B. False Rev 1 ELO 2.5 Binding Energy Per Nucleon Curve Introduction This lesson describes the relationship of binding energy per nucleon as the mass number changes. As the number of nucleons in a nucleus increases, the total binding energy also increases. Many scientific experiments have determined that the rate of increase is not linear. This non-linear relationship results in a variation of the binding energy per nucleon for nuclides of different mass numbers. Binding Energy per Nucleon Curve Example Plotting the average BE/A against the atomic mass numbers (A) shows the variation in the binding energy per nucleon, as shown in the figure below. Figure: Binding Energy per Nucleon versus Mass Number The figure above shows that as the atomic mass number (A) increases, BE/A increases, until A reaches about 60, and BE/A decreases as A increases above 60. The BE/A curve reaches a maximum value of 8.79 MeV at A = 56 and decreases to about 7.6 MeV for A = 238. No stable nuclei exist with A greater than 209. The general properties of nuclear forces determine the general shape of the BE/A curve. Very short-range attractive forces existing between nucleons hold the nucleus together and long range repulsive electrostatic (coulomb) forces existing between positively charged protons in the nucleus are forcing nucleons apart. As the atomic number (number of protons) and the atomic mass number of a nucleus increase, the repulsive electrostatic forces within the nucleus Rev 1 17 Duration ο§ 15 minutes Logistics ο§ Use PowerPoint slides 51β54 and the IG to present ELO 2.5. Inform ο§ Explain the BE/A curve and point out fission vs. fusion. ο§ The next TLO discusses delta binding energy. increase due to the greater number of protons. Increasing the proportion of neutrons in the nucleus overcomes this increased repulsion. The slope of the nuclide stability curve illustrates the effect of increasing repulsive forces. The increase in the neutron-to-proton ratio only partially compensates for the increasing proton-proton repulsive force in heavier naturally occurring elements. Less energy is required, on average, to remove a nucleon from the nucleus because repulsive forces are increasing. Therefore, the BE/A decreases. In the case of fissile materials (heaviest nuclei), only a small distortion from a spherical shape (a small energy addition) is required for the relatively large electrostatic forces attempting to force the two halves of the nucleus apart to overcome the attractive nuclear forces holding the two halves together. Consequently, the heaviest nuclei are more easily fissionable than lighter nuclei. The BE/A of a nucleus is also an indication of its stability. Generally, the higher the BE/A, the more stable the nuclide is. The increase in the BE/A as the atomic mass number decreases from 260 to 60 is the main reason for energy liberation in the fission process as the fuel isotope is split into two fragments of lower A values. The fission arrow in the figure above denotes this area. A later section will discuss delta binding energy. However, the increase in the BE/A as the atomic mass number increases from 1 to 60 is the reason that energy release is possible in a fusion event that combines rather than splits atoms. The fusion arrow in the figure above denotes this area. Knowledge Check True or False. Generally, less stable nuclides have a higher BE/A than the more stable ones. 18 A. True B. False Rev 1 TLO 2 Summary Nuclear Fission ο· Nuclear fission liquid drop model of a nucleus - In the ground state, the nucleus is nearly spherical in shape. - After absorbing a neutron, the nucleus will be in an excited state, start to oscillate, and become distorted. - If distortion causes the nucleus to become dumbbell-shaped, the repulsive electrostatic forces may overcome the short-range attractive nuclear forces causing the nucleus to split in two. ο· Excitation energy is the amount of energy a nucleus has above its ground state. ο· Critical energy is the minimum excitation energy that a nucleus must have before it can fission. ο· Fissile material is material for which fission is possible with neutrons that have zero kinetic energy. ο· Fissionable material is material for which fission caused by neutron absorption is possible, but the binding energy from the additional neutron is not sufficient for excitation above critical energy. Therefore, the neutron must have additional kinetic energy to make fission possible. ο· Fertile material is fissionable material that can undergo transmutation to become fissile material. ο· Thermal neutrons have very low kinetic energy levels (essentially zero) because they are in energy equilibrium with the thermal motion of surrounding materials. ο· Transmutation is the process of neutron absorption and subsequent decay, which changes one nuclide to another nuclide. ο· Conversion is the process of transmuting fertile material into fissile material where the amount of fissile material produced is less than the fissile material consumed. ο· Breeding is the process of transmuting fertile material into fissile material where the amount of fissile material produced is more than the amount of fissile material consumed. ο· The curve of binding energy per nucleon increases quickly through the light nuclides and reaches a maximum at a mass number of about 56. The curve decreases slowly for mass numbers greater than 60. ο· The heaviest nuclei are easily fissionable because they require only a small distortion from the spherical shape allowing the electrostatic forces to overcome the attractive nuclear forces, causing the nucleus to split. ο· Fission of uranium-235 versus fission of uranium-238: - Uranium-235 fissions with thermal neutrons because the binding energy released by the absorption of a neutron is greater than the critical energy. - The binding energy released by uranium-238 absorbing a neutron is less than the critical energy, so the neutron must have additional kinetic energy for fission to be possible. Now that you have completed this lesson, you should be able to: Rev 1 19 Duration ο§ 10 minutes Logistics ο§ Use PowerPoint slides 55β59 and the IG to review TLO 2 material. Inform ο§ Use directed and nondirected questions to students, check for understanding of ELO content, and review any material where student understanding of ELOs is inadequate. 1. Define the following terms: a. Excitation energy (Eexc) b. Critical energy (Ecrit) 2. Explain the fission process using the liquid drop model of a nucleus. 3. Define the following terms: a. Fissile material b. Fissionable material c. Fertile material d. Thermal neutrons 4. Explain binding energy per nucleon. 5. Explain the shape of the Binding Energy per nucleon versus mass number curve including its significance to fission energy release. TLO 3 Production of Heat from Fission Overview Duration ο§ 5 minutes Logistics ο§ Use PowerPoint slides 60β61 and the IG to introduce TLO 3. Fission of heavy nuclides converts a small amount of mass into a large amount of energy. There are two ways to calculate the energy released by fission: you can base computations on the change in mass that occurs during the fission or by the difference in binding energy per nucleon between the fissile nuclide and the fission products. This TLO discusses the process of fission and heat production. Objectives Upon completion of this lesson, you will be able to do the following: 1. Describe the average total amount of energy released per fission event including: a. Energy released immediately from fission b. Delayed fission energy 2. Describe which fission products nuclides are most likely to result from fission. 3. Describe the energy released from fission by the following methods: a. Change in binding energy b. Conservation of mass-energy c. Decay energy 4. Describe how heat is produced as a result of fission. 20 Rev 1 ELO 3.1 Energy Release Per Fission Introduction Total energy released per fission varies from one fission event to the next, depending on what fission products the reaction forms, but the average total energy released from uranium-235 fission with a thermal neutron is approximately 200 MeV. The majority of this energy (approximately 83 percent) is from the kinetic energy of the fission fragments. Energy Release per Fission Details The table below shows the average energy distribution for the energy released in a U-235 thermal fission, approximately 200 MeV. Instantaneous Energy Value Kinetic Energy of Fission Fragments 165 MeV Kinetic Energy of Fission Neutrons 5 MeV Instantaneous Gamma Rays 7 MeV Capture Gamma Ray Energy 10 MeV Delayed Energy Value Kinetic Energy of Beta Particles 7 MeV Decay Gamma Rays 6 MeV Neutrinos** 10 MeV** Total Energy Released 200 MeV **Not included in total Some fission neutrons undergo radiative capture reactions, producing gamma ray emissions as their compound nuclei drop to ground state. This provides the additional 10 MeV of instantaneous energy from Capture Gamma Ray energy shown in the table above. The total energy released per fission does not include the delayed energy from neutrinos, because neutrinos do not interact with materials inside the reactor. Rev 1 21 Duration ο§ 10 minutes Logistics ο§ Use PowerPoint slides 62β65 and the IG to present ELO 3.1. Inform ο§ Review the breakdown of energy release per fission. Delayed Fission Energy (Decay Heat) About seven percent (13 MeV) releases sometime after the instant of fission of the 200 MeV released per fission. Fissions mostly cease with the reactor shut down, but heat energy, referred to as decay heat, continues to release from the reactor because of the decay of fission products. After reactor shutdown, decay heat production tapers off but remains a significant source of heat for a very long time. Systems exist to remove this decay heat after reactor shutdown in order to prevent damage to the reactor core. The Three Mile Island event is an example of what can happen if this cooling source is unavailable following a shutdown. The reactor operations module discusses decay heat in detail. Isotope specifies, including the particular isotopes that are undergoing fission, and their classification as fissionable or fissile material have a small effect on the amount of energy released. For instance, the fission of U-235 by a slow neutron yields nearly identical energy to that of a U-238 fission by a fast neutron. Knowledge Check Which of the following statements correctly describes the amount of energy released from a single fission event? 22 A. Approximately 200 MeV are released during fission. 13 MeV are released instantaneously, and 187 MeV are released later (delayed). B. Approximately 200 MeV are released during fission. 187 MeV are released instantaneously, and 13 MeV are released later (delayed). C. Approximately 200 eV are released during fission. 13 eV are released instantaneously, and 187 eV are released later (delayed). D. Approximately 200 eV are released during fission. 187 eV are released instantaneously, and 13 eV are released later (delayed). Rev 1 ELO 3.2 Fission Fragment Yield Introduction Fissions do not produce identical results on each occurrence. In fact, both the number of neutrons and the resultant fission fragments vary. Scientific experiments have developed a yield curve of fission product probabilities. Most Probable Fission Fragments Resultant fission fragments have masses that vary widely. The figure below shows the percent yield of various atomic mass numbers. The most probable pair of fission fragments for a thermal neutron fission of uranium235 have masses of about 95 and 140. NOTE: The vertical axis of the fission yield curve is a logarithmic scale further amplifying the higher probability for mass numbers of 95 and 140. Rubidium-93 and cesium-140 are very likely to result from fission. Note Note The vertical axis of the fission yield curve is a logarithmic scale further amplifying the higher probability for mass numbers of 95 and 140. Rubidium-93 and cesium-140 are very likely to result from fission. Figure: Uranium-235 Fission Yield for Fast and Thermal Neutrons versus Mass Number Rev 1 23 Duration ο§ 10 minutes Logistics ο§ Use PowerPoint slides 66β69 and the IG to present ELO 3.2. Inform ο§ Review the fission yield curve. Stress the most likely mass numbers. As shown in the above figure, the fission fragment yield varies considerably. An example of one fission fragment yield is: 235 π+π β 236 πβ β 140 ππ + 94 ππ + 2π Notice in the example above that two neutrons result from this fission. Normally, one fission event releases two or three fast neutrons. The figure below shows average number of neutrons released per fission event for various fuels. Isotope Average Neutrons Released per Fission (v) U-233 2.492 U-235 2.418 Pu-239 2.871 Pu-241 2.927 Knowledge Check List the two most likely fission fragments resulting from fission of uranium-235. A. 24 Cesium, Rubidium Rev 1 ELO 3.3 Calculate Fission Energy: Binding Energy and Conservation of Mass Introduction Nuclear fission releases enormous quantities of energy; to compute these energies, start with the fission reaction equation. Consider first a typical fission reaction as shown below to begin these calculations. 1 236 β 140 93 1 235 π+ πβ( π) β πΆπ + π π + 3 ( π) 0 92 55 37 0 92 When the compound nucleus splits in the equation above, it breaks into two fission fragments, cesium-140, rubidium-93, and three neutrons. Both fission products will decay by multiple Ξ²- emissions (not shown) because of the high neutron-to-proton ratios they possess. Change in Binding Energy From the binding energy per nucleon curve (BE/A) we can determine the amount of energy released by a "typical" fission by plotting the uranium235 fission reaction on the curve below and calculating the change in binding energy (ΞBE) between the reactants on the left-hand side of the fission equation and the products on the right-hand side. Figure: Change in Binding Energy for Typical Fission The figure above shows that the binding energy per nucleon for the products (rubidium-93 and cesium-140) is greater than that for the reactant (uranium-235). The system has become more stable by releasing energy equal to the increase in total binding energy of the system when there is an increase in Rev 1 25 Duration ο§ 30 minutes Logistics ο§ Use PowerPoint slides 70β79 and the IG to present ELO 3.3. Inform ο§ Explain and work through the example calculations. Demonstrate ο§ Demonstrate methods used for calculating the energy released from fission. the total binding energy of a system. The energy liberated is equal to the increase in the total binding energy of the system in a fission process. Example: We find the total binding energy for a nucleus by multiplying the binding energy per nucleon by the number of nucleons, as shown in the table below. Nuclide BE per Nucleon (BE/A) Mass Number (A) Binding Energy (BE/A) x (A) 93 37Rb 8.7 MeV 93 809 MeV 140 55Cs 8.4 MeV 140 1,176 MeV 235 92U 7.6 MeV 235 1,786 MeV Note that the βAβ symbols in the above table are not the same and do not cancel out in the binding energy calculation of the last column. The energy released will be equivalent to the difference in binding energy (BE) between the reactants and the products. βπ΅πΈ = π΅πΈπππππ’ππ‘π β π΅πΈπππππ‘πππ‘π = (π΅πΈπ πβ93 + π΅πΈπΆπ β140 ) β (π΅πΈπβ235 ) = (809 πππ + 1,176 πππ) β (1,786 πππ) = 199 πππ Conservation of Mass-Energy During the fission process a decrease in the system mass occurs. The loss of mass equals the energy liberated. The conservation of mass-energy method considers this relationship between mass and energy. This method is more accurate than the ΞBE method since it considers all mass changes immediately occurring from fission. Therefore, measurements that require more precision should use this method. The instantaneous energy (Einst) is the energy released immediately after the fission process and is equal to the energy equivalent of the mass lost. This calculation method includes: ο· Add the nucleon masses of the fuel isotope including the incident neutron mass (reactants). ο· Add the nucleon masses of the fission products and released neutrons (products). ο· Subtract the mass of the products from the reactants. ο· Multiply the mass calculated by 931.5 MeV to convert to energy. 26 Rev 1 The table below facilitates this calculation. Mass of the Reactants 235 π 92 1 π 0 Totals 235.043924 amu 1.008665 amu Mass of the Products 93 π π 37 140 πΆπ 55 1 3 π 0 236.052589 amu 92.91699 amu 139.90910 amu 3.02599 amu 235.85208 amu Mass Difference = πππ π ππ π‘βπ π ππππ‘πππ‘π β πππ π ππ π‘βπ πππππ’ππ‘π = 236.052589 πππ’ β 235.85208 πππ’ = 0.200509 πππ’ Energy Equivalent = πππ π × 931.5 πππ/πππ’ = 0.200509 πππ’ × 931.5 πππ/πππ’ = 186.8 πππ Estimation of Decay Energy Decay energy (EDecay) is the additional energy added to the instant fission reaction when the fission fragments decay by Ξ²- emission. The equations below show the decay chains for rubidium-93 and cesium-140. π½β π½β π½β π½β 93 93 93 93 93 π π β ππ β π β ππ β ππ 37 38 39 40 41 π½β π½β π½β 140 140 140 140 πΆπ β π΅π β πΏπ β πΆπ 55 56 57 58 The energy released during the decay for each chain will be equivalent to the mass difference between the original fission product and the sum of the final stable nuclide and the beta particles emitted. The total decay energy is the sum of the energies of the two chains. The example below shows the steps to calculate the total decay energy. Example The energy released the decay chain of rubidium-93 is: Rev 1 27 931.5 πππ πΈπ·ππππ¦ = [ππ πβ93 β (πππβ93 + 4 ππππππ‘πππ )] ( ) πππ’ = [92.91699 πππ’ β (92.90638 πππ’ + 4 (0.0005486 πππ’))] ( 931.5 πππ ) πππ’ 931.5 πππ = 0.008416 πππ’ ( ) πππ’ = 7.84 πππ The following steps show the calculation of energy released in the decay chain of cesium-140. 931.5 πππ πΈπ·ππππ¦ = [ππΆπ β140 β (ππΆπβ140 + 3 ππππππ‘πππ )] ( ) πππ’ = [139.90910 πππ’ 931.5 πππ β (139.90543 πππ’ + 3 (0.0005486 πππ’))] ( ) πππ’ 931.5 πππ = 0.000202 πππ’ ( ) πππ’ = 1.89 πππ The total decay energy is the sum of the energies of the two chains, or 9.73 MeV. Knowledge Check The change in binding energy per nucleon between the fissile nuclide and the fission products, is used to calculate _________. 28 A. fission fragment speed B. decay heat C. energy released by fission D. increase in mass Rev 1 ELO 3.4 Fission Heat Production Introduction The majority of the energy liberated in the fission process releases immediately after the fission occurs. This energy appears as kinetic energy of the fission fragments and neutrons, and instantaneous gamma rays. The remaining energy releases over time after the fission occurs and appears as kinetic energy of the decay products. Fission Heat Production All of the energy released in fission, with the exception of the neutrino energy, transforms into heat via ionization and scattering. Fission fragments, with a positive charge and kinetic energy, cause ionization directly as they remove orbital electrons from the surrounding atoms through this ionization process. Kinetic energy transfers to the surrounding atoms of the fuel material, increasing temperature. Beta particles and gamma rays also give up energy through ionization, and fission neutrons interact and lose their energy through scattering. Knowledge Check Which of the following is NOT a method by which heat is produced from fission? Rev 1 A. Fission fragments causing direct ionizations resulting in an increase in temperature. B. Beta particles and gamma rays causing ionizations resulting in increased temperature. C. Neutrons interacting and losing their energy through scattering, resulting in increased temperature. D. Neutrinos interacting and losing their energy through scattering and ionizations resulting in increased temperatures. 29 Duration ο§ 5 minutes Logistics ο§ Use PowerPoint slides 80β82 and the IG to present ELO 3.4. Inform ο§ Explain the relationship of KE and temperature. TLO 3 Summary Duration ο§ 10 minutes Logistics ο§ Use PowerPoint slides 83β84 and the IG to review TLO 3 material. Use directed and nondirected questions to students, check for understanding of ELO content, and review any material where student understanding of ELOs is inadequate. Fission Energy Release ο· A typical fission event releases approximately 200 MeV of energy. - Approximately 187 MeV of energy releases instantaneously. - Approximately 13 MeV appears as delayed energy. ο· Fission products generally decay by Ξ²- emission. ο· Fission product yields are near mass numbers of 95 and 140. ο· The energy released by fission equals the difference in mass between the reactants before fission and the fission fragments and fission neutrons after fission, multiplied by 931.5 MeV per amu. ο· Another method to determine the energy released by fission considers the change in binding energy per nucleon between the fissile nuclide and the fission products. ο· Heat is produced from fission by: - Fission fragments cause direct ionizations resulting in an increase in temperature. - Beta particles and gamma rays cause ionizations resulting in increased temperature. - Neutrons interact and lose their energy through scattering, resulting in increased temperature. ο· Decay heat is a result of the decay of fission products. During reactor operation, decay heat accounts for approximately 7 percent of the heat produced from fission. In a shutdown reactor, decay heat becomes a significant source of heat and requires removal to prevent core damage (Three Mile Island, as an example). Objectives Now that you have completed this lesson, you should be able to: 1. Describe the average total amount of energy released per fission event including: a. Energy released immediately from fission b. Delayed fission energy 2. Describe which fission products nuclides are most likely to result from fission. 3. Describe the energy released from fission by the following methods: a. Change in binding energy b. Conservation of mass-energy c. Decay energy 4. Describe how heat is produced as a result of fission. 30 Rev 1 TLO 4 Intrinsic and Installed Neutron Sources Overview Neutrons from a variety of sources are always present in a reactor core, even when the reactor is shut down. Some neutrons result from naturally occurring (intrinsic) neutron sources, while others are from fabricated (installed) neutron sources designed for the reactor. Neutrons produced by sources other than normal neutron-induced fission are source neutrons. Duration ο§ 5 minutes Logistics ο§ Use PowerPoint slides 85β86 and the IG to introduce TLO 4. Objectives Upon completion of this lesson, you will be able to do the following: 1. Describe the purpose and importance of source neutrons. 2. Describe, including examples, each of the following types of intrinsic neutron sources: a. b. c. d. Spontaneous fission Photo-neutron reactions Alpha-neutron reactions Transuranic elements 3. Describe the purpose and type of installed neutron sources. 4. Describe the primary source of intrinsic source neutrons in the reactor for the following conditions: a. At beginning and end of core life b. Immediately following a reactor shutdown c. Several weeks after reactor shutdown ELO 4.1 Source Neutrons Introduction In addition to fission, other reactions also produce neutrons. The term source neutrons refers to these non-fission produced neutrons. Source neutrons are important because they help to monitor the fission process during a nuclear reactor startup. Source Neutrons Source neutrons ensure that the neutron population during shutdown conditions remains high enough to allow the operators visible indication of the source range nuclear instrumentation neutron level. This is important during reactor shutdown and startup conditions to confirm instrument operability and to monitor the reactor's neutron population changes. There are two classes of source neutrons: intrinsic and installed neutron sources. The following sections discuss these two classes. Rev 1 31 Duration ο§ 5 minutes Logistics ο§ Use PowerPoint slides 87β88 and the IG to present ELO 4.1. Inform ο§ Elaborate on the need for source neutrons. ο§ Subcritical multiplication is covered later in the course. Knowledge Check Source neutrons are important because they: A. Extend the neutron lifetime allowing a nuclear chain reaction to occur. B. Allow for visible indication of neutron level in a shutdown nuclear reactor. C. Shorten the neutron generation time allowing for operational control of a nuclear reactor. D. Contribute a larger percentage of the thermal neutron population than the fast neutron population in an operating nuclear reactor. ELO 4.2 Intrinsic Source Neutrons Duration ο§ 20 minutes Logistics ο§ Use PowerPoint slides 89β101 and the IG to present ELO 4.2. Inform ο§ Ensure students understand the various intrinsic neutron sources. Introduction Intrinsic neutron sources are nuclei that yield neutron-producing reactions, and that occur in reactor core fuel related materials. This section discusses the following types of neutron reactions capable of producing source neutrons and transuranic contributions: ο· ο· ο· Spontaneous fission Photo-neutron reactions Alpha-neutron reactions Intrinsic Neutron Sources A limited number of neutrons will be present, even in a reactor core with no operation history. This is due to spontaneous fission of some heavy nuclides present in the fuel. Uranium-238, uranium-235, and plutonium239 are examples of these heavy nuclides. Later in core life, the transuranic elements provide more spontaneous fission neutrons. Photo-neutron reactions provide a significant source of neutrons in a reactor that operated at power. The interaction of a gamma ray and a deuterium nucleus produce these neutrons. Interactions between alpha particles and various isotopes in the reactor core also result in source neutron production. Alpha particles from the decay of heavy nuclides interact with oxygen-18 and boron-11 in the core to produce neutrons. Transuranic elements building in the core also produce source neutrons from these reactions. 32 Rev 1 Spontaneous Fission A limited number of neutrons will always be present, even in a reactor core that has never been operated, due to spontaneous fission of some heavy nuclides that are present in the fuel. Uranium-238, uranium-235, and plutonium-239 undergo spontaneous fission to a limited extent. Uranium238, for example, yields almost 60 neutrons per hour per gram through spontaneous fission. The table below shows a comparison of the rate at which different heavy nuclides produce neutrons by spontaneous fission. t½ (Ξ±-decay) Nuclide t½ (Fission) Neutrons/sec/gram 235 92U 1.8 x 1017 years 6.8 x 108 years 8.0 x 10-4 238 92U 8.0 x 1015 years 4.5 x 109 years 1.6 x 10-2 239 94Pu 5.5 x 105 years 2.4 x 104 years 3.0 x 10-2 240 94Pu 1.2 x 1011 years 6.6 x 103 years 1.0 x 103 252 98Cf 66.0 years 2.65 years 2.3 x 1012 The four isotopes in a reactor that contribute most to the source neutron population via spontaneous fission are: ο· ο· ο· ο· Uranium-235 Uranium-238 Curium-242 Curium-244 Uranium-235 and uranium-238 from the core fuel load contribute approximately 1 x 106 neutrons per second (n/sec) to the source neutron population. Curium (a transuranic element) isotopes produced during reactor power operation are also a source of neutrons via spontaneous fission. Prior to a new core reactor startup, uranium spontaneous sources are significant contributors to the source neutron population. Curium-242 becomes a major producer of source neutrons via spontaneous fission after power operation and up to approximately 20,000 MWd/T of core irradiation. Beyond 20,000 MWd/T, curium-244 becomes a more predominant source neutron producer from spontaneous fission. One ton of spent nuclear fuel will contain on the order of 20 grams of curium. Rev 1 33 Photo-Neutron Reactions Source neutrons from photo-neutron reactions become significant in a reactor with an operational history at power. The equation below describes the interaction between a gamma ray and a deuterium nucleus. This reaction is referred to as a photo-neutron reaction because it is initiated by electromagnetic (gamma photon) radiation and results in the production of a neutron - photo-neutron. The high-energy gammas for this reaction come from the decay of fission products. 1 2 1 πΎ+ π»β π»+ π 0 1 1 Once the reactor has been operated for a short time there is an abundant supply of high-energy gammas (2.22 MeV or greater). Commercial PWRs have some deuterium present in the moderator/coolant because the atom percentage of naturally occurring deuterium is 0.015 percent. This is sufficient deuterium for production of photo-neutrons following power operation. However, after shutdown, gamma ray emitters decay off and the quantity of gamma rays decreases with time; therefore, the photo-neutron production rate also decreases with reactor shutdown time. Alpha Neutron Reactions Interactions between alpha particles and various isotopes in the reactor core also result in the production of source neutrons. Alpha particles occur from the decay of heavy elements in the fuel. They interact with naturally occurring oxygen-18 and boron-11, from the soluble boron added to the reactor coolant system, to produce neutrons. Transuranic elements produced during power operation also provide source neutrons via alphaneutron reaction. Oxygen-18 Oxygen-18 is naturally occurring in small quantities (0.2 percent) in the reactor coolant system and in uranium oxide fuel (U3O8). Uranium-235, uranium-238, curium-242, and curium-244 all undergo alpha decay. The following table lists the half-lives (alpha decay) for uranium and plutonium. The heaviest of these elements have long half-lives; therefore, they will produce only limited alpha particles. t½ (Ξ±-decay) Nuclide t½ (Fission) 235 92U 1.8 x 1017 years 6.8 x 108 years 8.0 x 10-4 238 92U 8.0 x 1015 years 4.5 x 109 years 1.6 x 10-2 239 94Pu 5.5 x 105 years 2.4 x 104 years 3.0 x 10-2 34 Neutrons/sec/gram Rev 1 t½ (Ξ±-decay) Nuclide t½ (Fission) Neutrons/sec/gram 240 94Pu 1.2 x 1011 years 6.6 x 103 years 1.0 x 103 252 98Cf 66.0 years 2.65 years 2.3 x 1012 This reaction does not contribute significantly to the source neutron population because uranium does not produce many alpha particles and the abundance of oxygen-18 in the reactor core is low. The equation below describes this reaction. 4 18 21 1 π»π + π β ππ + π 8 10 0 2 Boron-11 Another intrinsic neutron source is a reaction involving natural boron and fuel. Consider reactor coolant with added soluble boron. The soluble boron contains boron-11 (80.1 percent of natural boron). Boron-11 and available alpha particles from the heavy elements react to produce source neutrons. The equation below describes this reaction. 4 11 14 1 π»π + π΅ β π + π 5 0 2 7 The boron-11 must be in very close proximity to the fuel for this reaction to be possible because of the short travel length of alpha particles. Transuranic Elements A transuranic element is one that is beyond uranium (atomic number greater than 92). These elements build up in the core during reactor power operations due to various fuel isotopes undergoing neutron capture reactions. The equation below describes the example of uranium-238 undergoing neutron capture to produce plutonium-239. 238 1 239 π+ πβ π 92 0 92 Transuranic elements contribute a significant number of source neutrons in highly exposed reactor fuel. Either spontaneous fission or alpha-neutron reactions produce these types of neutrons. Isotopes providing the major contribution to these source neutron-producing reactions are the transuranic elements curium and americium. The following reactions illustrate the production of curium and alpha particles. Curium-242 production: Rev 1 35 239 4 242 1 ππ’ + π»π β πΆπ + π 94 96 0 2 Alpha production (163-day half-life): 242 238 4 πΆπ β ππ’ + π»π 96 94 2 One gram of curium-244 produces about 5 x 105 neutrons per second due to alpha-neutron reactions and about 1.2 x 107 neutrons per second due to spontaneous fission. ο· One gram of curium-242 produces about 2.5 x 107 neutrons per second from alpha-neutron reactions and about 2.3 x 107 neutrons per second from spontaneous fission. ο· One gram of americum-241 produces approximately 4 x 103 neutrons per second from alpha-neutron reactions. This isotope does not undergo spontaneous fission. ο· Transuranic neutron sources produce about 100 neutrons per second for every gram of fuel in the core or approximately 1 x 107 neutrons per second core wide in a typical nuclear reactor core. Knowledge Check Select the three major contributors (types of reactions) to the intrinsic neutron source in a nuclear reactor. 36 A. Photo neutron B. Alpha neutron C. Spontaneous fission D. Oxygen-18 neutron Rev 1 Knowledge Check Some heavy nuclides will produce neutrons directly from a reaction in the reactor referred to as _______________ _____________. A. photo neutron B. alpha neutron C. spontaneous fission D. oxygen-18 alpha ELO 4.3 Installed Neutron Sources Introduction Many reactors have additional neutron sources installed to ensure a reliable and sufficient number of source neutrons because intrinsic neutron sources can be relatively weak or dependent upon the recent power history of the reactor. This may be especially important with all new fuel in the core. Installed Neutron Sources These installed neutron sources ensure that shutdown neutron levels are high enough for the nuclear instruments to detect at all times. This provides the operators with valid indication of the reactor's status. Installed neutron sources are assemblies in the reactor for the sole purpose of producing source neutrons. Californium-252 The artificial nuclide californium-252 is a strong source of neutrons that emits neutrons at the rate of about 2 x 1012 neutrons per second per gram from spontaneous fission. It is not widely used as an installed neutron source in commercial PWRs because of its high cost and short half-life (2.65 years). Beryllium Sources Alpha-Neutron Beryllium Source Many installed neutron sources use an alpha-neutron reaction with beryllium. These sources are composed of metallic beryllium (100 percent beryllium-9) with an alpha particle emitter, such as a compound of radium, polonium, or plutonium. The reaction that occurs is below. 4 13 β 12 1 9 π΅π + πΌ β ( πΆ) β πΆ + π 6 6 0 4 2 Rev 1 37 Duration ο§ 10 minutes Logistics ο§ Use PowerPoint slides 102β107 and the IG to present ELO 4.3. Inform ο§ Ensure students understand the various installed neutron sources. The beryllium is intimately (homogeneously) mixed with the alpha emitter and is usually enclosed in a stainless steel capsule. Photo-neutron Beryllium Source Another installed neutron source that is commonly used is the photoneutron reaction with beryllium. Beryllium is a good source because its stable isotope beryllium-9 has a weakly attached neutron with a binding energy of only 1.66 MeV. Thus, a gamma ray with greater energy than 1.66 MeV can cause neutrons to eject because of the photo-neutron reaction shown below. 1 9 8 πΎ + π΅π β π΅π + π 0 4 4 Antimony-Beryllium Source The most common installed source is antimony-beryllium (Sb-Be). Many startup neutron sources use antimony and beryllium because after activation with neutrons, the radioactive antimony becomes an emitter of high-energy gammas. The reactions below show the reactions resulting in gamma emissions. 123 1 124 ππ + π β ππ + πΎ 51 0 51 π½ β 124 124 0 ππ β ππ + π+πΎ 51 52 β1 The activated antinomy also decays with a 60-day half-life to produce a gamma ray of sufficient energy to interact with the beryllium to produce a neutron. 1 9 8 πΎ + π΅π β π΅π + π 0 4 4 Antimony-beryllium (Sb-Be) photo-neutron sources of this type are constructed differently from the alpha-neutron types. One design incorporates a capsule of irradiated antimony enclosed in a beryllium sleeve. Stainless steel cladding then encases the entire assembly. A PWR may have several neutron sources of this type installed. 38 Rev 1 Knowledge Check Which of the following would be used to provide a source neutron population for a nuclear reactor core that has never been operated (newly installed core)? A. Photo-neutron source B. Transuranic source C. Startup source D. Installed source ELO 4.4 Intrinsic Source Neutrons Over Core Life Introduction Installed neutron sources provide a steady source of neutrons throughout core life, although over long periods their strength does decay. Intrinsic neutrons, however, do change in importance over core life. Duration ο§ 10 minutes Logistics ο§ Use PowerPoint slides 108β113 and the IG to present ELO 4.4. Inform ο§ Ensure that students understand how using all new fuel or reused fuel in a core reload will make a difference in relative source strength of intrinsic sources. ο§ Some reactors do remove their installed sources, finding intrinsic sources adequate. Intrinsic Source Neutrons over Core Life Photo-neutron source strength changes significantly over core life. Initially, the core contains no fission products, and therefore there are no high-energy gamma rays available to initiate photo-neutron reactions resulting in source neutron production. In addition, deuterium concentration, which is low in the beginning of life, limits source neutron production. Both of these factors change and increase source neutron production as core life increases. At times past approximately core middle-of-life (MOL), photo-neutron reactions are the largest contributor to the source neutron population immediately following reactor shutdown. There are variations depending on amounts of new or reused fuel present. Photo-neutron reactions contribute large amounts of source neutrons for several days following shutdown from power operations. Transuranic Source Strength versus Core Life The transuranic alpha-neutron intrinsic neutron source tends to contribute the greatest number of source neutrons (than other intrinsic sources) in a reactor core past middle-of-life (MOL) conditions, after several weeks of reactor shutdown from power operations. The photo-neutron source strength decays off after a couple of weeks. Whether the photo-neutron or transuranic source provides the greatest source neutron strength will depend on time in core life and the fuel mix new or partially reused. Once-burned and twice-burned fuel will possess Rev 1 39 transuranic alpha-neutron sources as the reactor core's dominant intrinsic neutron source. Intrinsic Neutron Source Strength at Core Beginning of Life At the beginning of core life (BOL) for a reactor, the relative strengths of the intrinsic sources should be as follows (strongest to weakest): Immediately following reactor shutdown from power: ο· ο· ο· Photo-neutron sources Spontaneous fission sources Alpha-neutron (transuranic) sources Several weeks following reactor shutdown from power: ο· ο· ο· Spontaneous fission sources Photo-neutron sources Alpha-neutron (transuranic) sources Note Note The alpha neutron (transuranic source) does not produce a significant number of source neutrons until later in core life or if once or twice burned fuel is used. Intrinsic Neutron Source Strength at Core End of Life As a reactor core approaches end-of-life (EOL) conditions, the relative strengths of the intrinsic sources would be as follows (strongest to weakest): Immediately following reactor shutdown from power: ο· ο· ο· Photo-neutron sources Alpha-neutron (transuranic) sources Spontaneous fission sources Several weeks following reactor shutdown from power: ο· ο· ο· 40 Alpha-neutron (transuranic) sources Spontaneous fission sources Photo-neutron sources Rev 1 Knowledge Check Which of the following reactions contributes the most to the source neutron population in a nuclear reactor core that is early in core life, shortly after the reactor is shut down? A. Photo-neutron source B. Transuranic source C. Spontaneous fission source D. Californium source TLO 4 Summary Source Neutrons ο· Source neutrons help initiate the fission process during a nuclear reactor startup. - They also ensure that the neutron population remains high enough to allow a visible indication of neutron level during shutdown and startup conditions. ο· Intrinsic neutron sources are sources of neutrons from materials that are in the reactor for other purposes such as fuel, burnable poison, or moderator. ο· Installed neutron sources are materials or components placed in the reactor specifically to produce source neutrons. ο· Intrinsic neutron sources are: - Spontaneous fission - Photo-neutron reactions - Alpha-neutron reactions ο· Spontaneous fission - results in production of fission fragments and neutrons - Uranium-235 - Uranium-238 - Curium-242 - Curium-244 ο· Photo-neutron reaction: 1 2 1 πΎ+ π»β π»+ π 0 1 1 ο· After approximately MOL, the photo-neutron reaction is largest contributor of intrinsic neutron sources. ο· Alpha-neutron reactions: 4 18 21 1 π»π + π β ππ + π 8 10 0 2 4 11 14 1 π»π + π΅ β π + π 5 0 2 7 Rev 1 41 Duration ο§ 10 minutes Logistics ο§ Use PowerPoint slides 114β121 and the IG to review TLO 4 material. Use directed and nondirected questions to students, check for understanding of ELO content, and review any material where student understanding of ELOs is inadequate. ο· Transuranic elements - contribute a significant number of source neutrons in highly exposed reactor fuel. Either spontaneous fission or alpha-neutron reactions produce these types of neutrons. ο· Intrinsic neutron source strength at core beginning of life - at BOL the relative strengths of the intrinsic sources are (strongest to weakest): - Immediately following reactor shutdown from power: o Photo-neutron sources o Spontaneous fission sources o Alpha-neutron (transuranic) sources - Several weeks following reactor shutdown from power: o Spontaneous fission sources o Photo-neutron sources o Alpha-neutron (transuranic) sources - Note that the alpha neutron (transuranic source) does not produce a significant number of source neutrons until later in core life or if once or twice burned fuel is used. ο· Intrinsic neutron source strength at core end of life - At EOL conditions, the relative strengths of the intrinsic sources are (strongest to weakest): - Immediately following reactor shutdown from power: o Photo-neutron sources o Alpha-neutron (transuranic) sources o Spontaneous fission sources - Several weeks following reactor shutdown from power: o Alpha-neutron (transuranic) sources o Spontaneous fission sources o Photo-neutron sources ο· Examples of installed neutron sources are: - Spontaneous fission of californium-252 results in fission fragments and free neutrons - Beryllium alpha-neutron reaction (alpha from the decay of plutonium, polonium, or radium): 4 13 β 12 1 9 π΅π + πΌ β ( πΆ) β πΆ + π 6 6 0 4 2 - Beryllium photo-neutron reaction (high-energy gamma from decay of antimony-124): 1 9 8 πΎ + π΅π β π΅π + π 0 4 4 - Antimony-beryllium source β this is the most common installed source. The activated antinomy also decays with a 60-day halflife to produce a gamma ray of sufficient energy to interact with the beryllium to produce a neutron. 123 1 124 ππ + π β ππ + πΎ 51 0 51 π½ β 124 124 0 ππ β ππ + π+πΎ 51 52 β1 42 Rev 1 Objectives Now that you have completed this lesson, you should be able to: 1. Describe the purpose and importance of source neutrons. 2. Describe, including examples, each of the following types of intrinsic neutron sources: a. b. c. d. Spontaneous fission Photo-neutron reactions Alpha-neutron reactions Transuranic elements 3. Describe the purpose and type of installed neutron sources. 4. Describe the primary source of intrinsic source neutrons in the reactor for the following conditions: a. At beginning and end of core life b. Immediately following a reactor shutdown c. Several weeks after reactor shutdown TLO 5 Relationship between Neutron Flux, Microscopic, and Macroscopic Cross-Sections Overview Fission neutrons are born at an average energy of about 2 MeV and interact with reactor core materials in various absorption and scattering reactions. Scattering reactions are useful for thermalizing neutrons. Thermal neutrons may be absorbed by fissile nuclei to produce fission or absorbed in fertile material resulting in the production of fissionable fuel. Additionally, some neutrons are absorbed in structural components, reactor coolant, and other non-fuel materials resulting in the removal of neutrons from the fission process. To determine these neutron interaction rates, it is necessary to identify the number of neutrons available and the probability interaction. To assist in quantifying neutron availability and reaction probabilities, we use terms including neutron flux, microscopic and macroscopic cross-section. The complexity of designing a reactor requires predicting of the various reaction rates, both in specific portions of the core and averages throughout the core. One example is the calculation of the reactor's thermal output, knowing the fission rate and core volume. Rev 1 43 Duration ο§ 5 minutes Logistics ο§ Use PowerPoint slides 122β125 and the IG to introduce TLO 5. Objectives Upon completion of this lesson, you will be able to do the following: 1. Explain the following terms, including any mathematical relationships: a. Atomic density b. Neutron flux c. Fast neutron flux d. Thermal neutron flux e. Microscopic cross-section f. Barn g. Macroscopic cross-section h. Mean free path 2. Define the following neutron classes: a. Fast b. Intermediate c. Slow 3. Describe how the absorption and scattering cross-section of typical nuclides varies with neutron energies in the 1 eV region, and the resonance absorption region. 4. Describe the macroscopic cross-section and mean free path at various temperatures. 5. Describe radial and axial neutron flux distribution. 6. Describe how changes in neutron flux and macroscopic cross-section affect reaction rates. 7. Describe the relationship between neutron flux and reactor power. ELO 5.1 Neutron Reaction Terms Duration ο§ 40 minutes Logistics ο§ Use PowerPoint slides 126β151 and the IG to present ELO 5.1. Inform ο§ Ensure all students have a clear understanding of these terms. ο§ Stress that students will be working with these terms in the future. Introduction The following terms are important for understanding the theoretical concepts of nuclear reactors and are foundational for future lessons: ο· ο· ο· ο· ο· ο· Atomic density Neutron flux Fast neutron flux Thermal neutron flux Microscopic cross-section Mean free path Atomic Density An important property of a material is its atomic density. The atomic density is the number of atoms of a given type per unit volume of the material. Use the following equation to calculate the atomic density of a substance. 44 Rev 1 π= πππ΄ π Where: N = Atomic density (atoms/cm3) Ο = density (g/cm3) NA = Avogadroβs number (6.022 x 1023 atoms/mole) M = gram atomic weight or gram molecular weight Example: A block of aluminum has a density of 2.699 g/cm3. If the gram atomic weight of aluminum is 26.9815 g, calculate the atomic density of the aluminum. Solution: π= πππ΄ π π 23 ππ‘πππ ) 3 (6.022 × 10 ππππ ππ = π 26.9815 ππππ ππ‘πππ = 6.024 × 1022 ππ3 2.699 Neutron Flux Macroscopic cross-sections for neutron reactions with a specific material determine the probability of one neutron undergoing a specific reaction per centimeter of its travel through that material. It is necessary to determine how many neutrons travel through the material, and the distance (in centimeters) the neutrons travel each second, to determine how many reactions actually occur in a field of neutrons. These calculations are based on the number of neutrons existing in one cubic centimeter at any one instant and the total distance they travel each second while in that cubic centimeter. The number of neutrons existing in a cm3 of material at any instant is the neutron density, represented by the symbol n with units of neutrons/cm3. The neutrons' velocities determine the total distance these neutrons travel each second. Neutron flux is the number of neutrons passing through the unit area (cm2) per unit time. Neutron flux (Ξ¦) is the total path length covered by all neutrons in one cubic centimeter during one second. Its units are neutrons per square centimeter per second (n/cm2/sec). The equation below shows Rev 1 45 the relationship between neutron flux, neutron density, and neutron velocity. Ξ¦ = ππ£ Where: Ξ¦ = neutron flux (neutron/cm2-sec) n = neutron density (neutrons/cm3) v = neutron velocity (cm/sec) The term neutron flux refers to parallel beams of neutrons traveling in a single direction. The intensity (I) of a neutron beam is the product of the neutron density and the average neutron velocity. The directional beam intensity is equal to the number of neutrons per unit area and time (neutrons/cm2-sec) falling on a surface perpendicular to the direction of the beam. Neutron flux is modeled as many neutron beams traveling in various directions in a nuclear reactor. The neutron flux becomes the scalar sum of these directional flux intensities (added as numbers and not vectors), expressed as follows: π· = πΌ1 + πΌ2 + πΌ3 + πΌπ . All of the directional beams contribute to the total reaction rate since the atoms in a reactor do not prefer neutrons coming from any particular direction. In reality, at any given point within a reactor, neutrons are traveling in all directions. There are two classes of neutron flux: thermal neutron flux and fast neutron flux in a nuclear reactor. The next section defines these neutron energies. Thermal Neutron Flux Thermal neutron flux (Ξ¦th) is the number of thermal neutrons crossing a unit area in the reactor in a given amount of time. The units are neutrons (thermal) per square centimeter per second (n/cm2/sec) as with total neutron flux. Remember that neutron flux is omni-directional, meaning neutrons can enter a particular square centimeter of reactor material from any direction. Therefore, Ξ¦th also equals the total distance of all thermal neutrons diffused (moved) in a particular unit volume in one second. Fast Neutron Flux Fast neutron flux (Ξ¦f) is number of fast neutrons crossing a unit area in a given amount of time. The units are neutrons (fast) per square centimeter per second (n/cm2/sec) as with thermal neutron flux. Fast neutron flux also equals the distance that fast neutrons diffuse (move) in a particular unit volume in one second. The next section defines fast neutrons. 46 Rev 1 Cross-Sections The probability of a neutron interacting with a nucleus for a particular reaction depends on the nucleus involved and, importantly, the energy of the neutron. Most materials are more likely to absorb a thermal neutron than a fast neutron. Additionally, the probability of the neutron interaction varies with the type of reaction involved. Note Note Cross-section depends on the characteristics of the nucleus, not on the size of nucleus. Microscopic Cross-Section (Ο) The probability of a particular reaction occurring between a neutron and a nucleus is the microscopic cross-section (Ο) of the nucleus. This crosssection varies with the energy of the neutron, and is the effective area the nucleus presents to the neutron for the particular reaction. Barn is the unit of measure for this area. A larger effective area yields a greater probability for reaction. Total Microscopic Cross-Section (ΟT) A neutron interacts with an atom in two basic ways: scattering or absorption reaction. The probability of a neutron being absorbed is the microscopic cross-section for absorption (Οa). The probability of a neutron scattering is the microscopic cross-section for scattering (Οs). The sum of these two microscopic cross-sections is the total microscopic cross-section (ΟT). ππ = ππ + ππ Microscopic Cross-Section for Scattering (Οs) Both the absorption and the scattering microscopic cross-sections have two components. For instance, the scattering cross-section is the sum of the elastic scattering cross-section (Οse) and the inelastic scattering cross-section (Οsi). ππ = ππ π + ππ π Microscopic Cross-Section for Absorption (Οa) The microscopic absorption cross-section (Οa) includes all reactions except scattering. However, for most purposes it has two categories, fission (Οf) and capture (Οc). ππ = ππ + ππ Rev 1 47 The Chart of Nuclides uses two different conventions to represent microscopic cross-section for capture, depending on the type of particle ejected from the nucleus after the capture reaction. ο· ο· If a gamma ray results from the capture, Οs is used If an alpha particle results from the capture, ΟΞ± is used A later section discusses fission and radiative capture of neutrons in detail. Barns Microscopic cross-sections are expressed in units of area - usually square centimeters. A square centimeter is very large compared to the effective area of a nucleus. The old story goes that a physicist once referred to the measure of a square centimeter as being "as big as a barn" when viewed on a nuclear level. The name has persisted, so now microscopic cross-sections have units of barns. The conversion to cm2 is below. 1 ππππ = 10β24 ππ2 Consider boron-10, for example. This nuclide has a relatively low mass with an absorption cross-section of 3,838 barns resulting in an ejected alpha particle. A boron-10 nucleon presents a very large effective area for neutron interaction. Lead-208 is a nuclide of relatively high mass and only has an absorption cross-section of 8 microbarns (8 x 10-6 barns) for the same reaction. Lead-208 presents a very small effective area for neutron interaction. Macroscopic Cross-Section (Ξ£) Whether or not a neutron interacts with a certain volume of material depends not only on its microscopic cross-section but also on the number of nuclei within that volume (atomic density). Macroscopic cross-section (Ξ£) is the probability of a given reaction occurring per unit travel of the neutron. Macroscopic cross-section is relates to microscopic cross-section (Ο) by the following relationship. Ξ£ = ππ Where: Ξ£ = macroscopic cross-section (cm-1) N = atomic density (atoms/cm3) Ο = microscopic cross-section (cm2) The number of target nuclei per unit volume increases and the probability of interaction increases if atomic density increases. 48 Rev 1 Microscopic Calculation Most materials are composed of several elements. Most elements are composed of several isotopes; therefore, most materials involve many cross-sections, one for each isotope involved. Each macroscopic crosssection is determined to include all the isotopes for a material, which are added together as follows: π΄ = π1 π1 + π2 π2 + π3 π3 +. . . . . . .. ππ ππ Where: Nn = the number of nuclei per cm3 of the nth element Οn = the microscopic cross-section of the nth element Example: Find the macroscopic thermal neutron absorption cross-section for iron, which has a density of 7.86 g/cm3. The iron microscopic cross-section for absorption is 2.56 barns and the gram atomic weight is 55.847 g. Solution: Step 1: Using the equation for atomic density, calculate the atomic density of iron. π= πππ΄ π 7.68 = π ππ‘πππ (6.022 × 1023 ) ππππ ππ3 π 55.847 ππππ = 8.48 × 1022 ππ‘πππ ππ3 Step 2: Using the atomic density and given microscopic cross-section, calculate the macroscopic cross-section. Ξ£π = πππ = 8.48 × 1022 ππ‘πππ 1 × 10β24 ππ2 (2.56 πππππ ) ( ) ππ3 1 ππππ = 0.217 ππβ1 Rev 1 49 Microscopic versus Macroscopic Cross-Section The difference between the microscopic and macroscopic cross-sections is extremely important and deserves restatement. The microscopic crosssection (Ο) represents the effective target area that a single nucleus presents to a bombarding particle (neutron). The units are in barns or cm2. The macroscopic cross-section (Ξ£) represents the total effective target area of the nuclei contained in 1 cm3 of the material. The units are 1/cm or cm-1. Mean Free Path A neutron has a certain probability of undergoing a particular interaction in one centimeter of travel (Ξ£) in a material. The inverse of this probability describes how far the neutron will travel (average) before undergoing an interaction. This average distance of travel by a neutron before interaction is the mean free path for interaction and represented by the symbol Ξ». The equation below expresses the mathematical relationship between the mean free path (Ξ») and the macroscopic cross-section (Ξ£): π= 1 Ξ£ Mean Free Path Calculation The macroscopic cross-section must be determined to calculate the mean fee path. The following example will demonstrate a mean free calculation for a material composed of both aluminum and silicon. Example An alloy is composed of 95 percent aluminum and 5 percent silicon (by weight). The density of the alloy is 2.66 g/cm3. The table below shows the properties of aluminum and silicon. Element Gram Atomic Weight Οa (barns) Οs (barns) Aluminum 26.9815 0.23 1.49 Silicon 28.0855 0.16 2.20 Perform the following: Calculate the atomic densities for aluminum and silicon. Determine absorption and scattering macroscopic cross-sections for thermal neutrons. 50 Rev 1 Calculate the mean free paths for absorption and scattering. Solution: Step 1: The density of aluminum is 95 percent of the total density. Using the formula for atomic density, calculate the atomic densities for aluminum and silicon. ππ΄π = ππ΄π ππ΄ ππ΄π π ππ‘πππ ) (6.022 × 1023 ) 3 ππππ ππ = π 26.9815 ππππ ππ‘πππ = 5.64 × 1022 ππ3 0.95 (2.66 πππ = πππ ππ΄ πππ π ππ‘πππ ) (6.022 × 1023 ) 3 ππππ ππ = π 28.0855 ππππ ππ‘πππ = 2.85 × 1021 ππ3 0.05 (2.66 Step 2: Calculate the absorption and scattering macroscopic cross-sections from the microscopic cross-sections and the combined atomic densities (95 percent aluminum and 5 percent silicon). Ξ£π = ππ΄π ππ,π΄π + πππ ππ,ππ ππ‘πππ ) (0.23 × 10β24 ππ2 ) ππ3 ππ‘πππ + (2.85 × 1021 ) (0.16 × 10β24 ππ2 ) ππ3 = (5.64 × 1022 = 0.0134 ππβ1 Ξ£π = ππ΄π ππ ,π΄π + πππ ππ ,ππ Rev 1 51 ππ‘πππ ) (1.49 × 10β24 ππ2 ) ππ3 ππ‘πππ + (2.85 × 1021 ) (2.20 × 10β24 ππ2 ) ππ3 = (5.64 × 1022 = 0.0903 ππβ1 Step 3: Determine mean free paths from the calculated absorption and scattering macroscopic cross-sections. ππ = = 1 Ξ£π 1 0.01345 ππβ1 = 74.3 ππ ππ = = 1 Ξ£π 1 0.0903 ππβ1 = 11.1 ππ Conclusions: ο· A neutron must travel an average of 74.3 cm to interact by absorption in this alloy. ο· A neutron must travel an average of 11.1 cm to interact by scattering in this alloy. Knowledge Check _______________ is the total path length traveled by all neutrons in one cubic centimeter of material during one second. 52 A. Mean free path B. Neutron flux C. Gamma flux D. Atomic density Rev 1 ELO 5.2 Neutron Energy Terms Introduction There are three classes of neutron energy levels: fast, intermediate and slow. Neutrons that are in energy equilibrium with their surrounding are thermal neutrons. They are the most important for thermal reactors. Neutron Energies Terms The figure below illustrates the relative energy levels of neutrons and their flux distribution in a typical thermal reactor. Figure: Neutron Energy Definitions of the three classes of neutron energy levels are as follows: ο· Fast neutrons - Fission neutrons are born as fast neutrons. They are categorized with energy levels greater than 0.1 MeV (105 eV) ο· Intermediate neutrons - neutrons with energy levels between 1 eV and 0.1 MeV ο· Slow neutrons - neutrons with energy levels less than 1 eV ο· Thermal Neutrons β neutrons with an energy level of 0.025 eV (2.2 x 105 cm/sec. velocity) at 68° F. Velocity and energy increase with temperature. Rev 1 53 Duration ο§ 5 minutes Logistics ο§ Use PowerPoint slides 152β154 and the IG to present ELO 5.2. Inform ο§ Review the neutron distribution curve. Knowledge Check Thermal neutrons are classified in the intermediate neutron energy range. A. True B. False ELO 5.3 Neutron Energies versus Cross-Sections Duration ο§ 15 minutes Logistics ο§ Use PowerPoint slides 155β165 and the IG to present ELO 5.3. Inform ο§ Explain the differences between scattering and absorption crosssections. Introduction Neutron energies affect both neutron absorption and neutron scattering cross-sections. A higher energy neutron has a lower probability of interaction in general. Resonance peaks at certain specific energies possess very high cross-sections. Neutron Energies versus Cross-Sections Neutron absorption cross-sections have three unique regions of resonance probability related to neutron energy. These are the 1/v region, the resonance region, and the fast neutron region. Each region has a different relationship to changing neutron energy. Neutron scattering cross-sections are somewhat different. Resonance elastic scattering and inelastic scattering cross-sections do have resonance peaks, but they are smaller than absorption peaks. Neutron energy has little effect on potential elastic scattering cross-sections. Scattering reactions are good since they do not lose neutrons but cause them to thermalize for use as fission neutrons. Neutron Scattering Cross-Section versus Incident Neutron Energy The elastic scattering cross-sections are generally small, typically 5 barns to 10 barns, with the exception of hydrogen. This is close to the magnitude of the actual geometric cross-sectional area expected for atomic nuclei. The cross-section is essentially constant and independent of neutron energy up to around 1 MeV in potential elastic scattering. Resonance elastic scattering and inelastic scattering exhibit resonance peaks similar to those associated with absorption cross-sections. The resonances occur at lower energies for heavy nuclei than for light nuclei. In general, the variations in scattering cross-sections are very small when compared to the variations that occur in absorption cross-sections. Neutron scattering cross-sections are most important in the moderator where neutrons are thermalized in a thermal reactor. Elastic scatterings are most probable in light nuclei, Οs = Οselastic since water molecules (the moderator) are light nuclei. The following figure illustrates the relative 54 Rev 1 consistency of Οs throughout the neutron energy ranging from 0.025 eV to 0.1 MeV. The standard elastic scattering cross-section table is based on a value of 0.025 eV at 68 degrees F. Figure: Elastic Scattering Cross-Section versus Neutron Energy Neutron Absorption Cross-Section versus Incident Neutron Energy The variation of absorption cross-sections with neutron energy is complex. The absorption cross-sections are small, ranging from a fraction of a barn to a few barns for slow and thermal neutrons for many elements. For a considerable number of nuclides of moderately high to high mass numbers the variation of absorption cross-sections with incident neutron energy reveals three distinct regions on an absorption cross-section versus neutron energy curve. These regions are the 1/v region, resonance peaks region and fast neutron region. The figure below illustrates these regions. Figure: Typical Neutron Absorption Cross-Section versus Neutron Energy Rev 1 55 1/v Region In the 1/v region, the cross-section decreases steadily with increasing neutron energy; this low energy region includes thermal neutrons (< 1 eV). In this region the absorption cross-section, which is often high, is inversely proportional to the velocity (v). This region is frequently referred to as the "1/v region" because of the inverse relationship to velocity or energy (velocity and energy are directly proportional, but both are inversely proportional to cross-section). Resonance Region Beyond the 1/v region, there is a resonance region where the cross-sections rise sharply to high value peaks. These are resonance peaks with neutrons in the intermediate energy (epithermal neutrons) range. These energies are resonance peaks and are a result of the affinity of the nucleus for neutrons whose energies closely match its discrete, quantum (preferred) energy levels. These peaks occur where the neutron's binding energy plus its kinetic energy are exactly equal to the amount of energy required to raise a compound nucleus from its ground state to a quantum level energy state. Resonance absorption occurs when these peaks match energy required to reach a quantum energy state. The following example problem further illustrates the absorption in the resonance region. Example: Calculate the kinetic energy a neutron must possess to undergo resonant absorption in uranium-235 at this resonance energy level assuming that uranium-235 has a nuclear quantum energy level at 6.8 MeV above its ground state. Solution: π΅πΈ = [πππ π ( 235 π) + πππ π (πππ’π‘πππ) β πππ π ( 235 π)] × 931 π΅πΈ = (235.043925 + 1.008665 β 236.045563) × 931 π΅πΈ = (0.007025 πππ’) × 931 πππ πππ’ πππ πππ’ πππ = 6.54 πππ πππ’ 6.8 πππ β 6.54 πππ = 0.26 πππ The difference between the binding energy and the quantum energy level equals the amount of kinetic energy the neutron must possess. The typical heavy nucleus will have many closely spaced resonances starting in the low energy (eV) range. This is because heavy nuclei are complex and have more possible configurations and corresponding energy states. Light nuclei, 56 Rev 1 being less complex, have fewer possible energy states and fewer resonances distributed at higher energy levels. Fast Neutron Region The absorption cross-section steadily decreases as the energy of the neutron increases for higher neutron energies. This is the fast neutron region. The absorption cross-sections are usually less than 10 barns in this region. This explains the low probability for fast fissioning of neutrons in this region. Knowledge Check At low neutron energies (<1 eV) the absorption crosssection for a material is ___________ proportional to the neutron velocity. (Fill in the blank). A. inversely ELO 5.4 Temperature Affects to Macroscopic Cross-Sections Duration ο§ 20 minutes Logistics ο§ Use PowerPoint slides 166β173 and the IG to present ELO 5.4. Inform ο§ Explain and work through the example calculations. Demonstrate ο§ Demonstrate methods used for calculating macroscopic crosssections and mean free paths at other than tableprovided temperatures. Introduction As neutron energy varies, microscopic absorption cross-sections vary significantly. The microscopic cross-section values given on most charts and tables are at an ambient temperature of temperature of 68°F. This equates to a neutron velocity of 2,200 eV of energy. You must correct the absorption cross-section for the new temperature value if a material is at higher temperatures. Calculate the macroscopic cross-section and mean free path following the determination of the temperature corrected microscopic cross-section. Note Note Rev 1 Temperature corrections apply to any cross-section involving absorption (for example, Οa, Οc, Οf); however, they do not apply to scattering cross-sections, as neutron energies up to 1 MeV have little effect on the predominant elastic scatters occurring in the moderator. (ELO 5.3) 57 Temperature Effects on Microscopic Cross-Sections The following calculations illustrate how macroscopic cross-section and mean free path vary with temperatures. Step 1. Action Obtain the standard (at 68°F) table value of the desired microscopic cross-section. 2. Convert temperatures to °R or °K. °π = °πΉ + 460 °πΎ = °πΆ + 273 3. Calculate new microscopic cross-section using: 1 ππ 2 π = ππ ( ) π Where: Ο = microscopic cross-section corrected for temperature Οo = microscopic cross-section at reference temperature (68°F or 20°C) To = reference temperature (68°F) in degrees Rankine (°R) or Kelvin (°K) T = temperature for which corrected value is being calculated Temperature Effects to Macroscopic Cross-Sections Demonstration Example: What is the fission macroscopic cross-section and mean free path for uranium-235 for thermal neutrons at 500°F? Given: Uranium-235 has a Οf of 583 barns at 68°F. Atomic density of uranium-235 equals 7.03 x 1020 atoms/cm3 at 500°F 58 Rev 1 Solution: Steps 1, 2, and 3: 1 ππ 2 ππ = ππ,π ( ) π 1 68β + 460 2 = 583 πππππ ( ) 500β + 460 = 432 πππππ Step 4: Ξ£ = ππ ππ = 7.03 × 1020 ππ‘πππ ππ2 β24 × 432 π΅ππππ × (1 × 10 ) ππ3 π΅πππ ππ = 0.304 ππβ1 Step 5: π= π = 1 Ξ£ 1 0.304 ππβ1 π = 3.29 ππ Note Note Rev 1 As temperature increases, the microscopic cross-section for fission decreases (less probability for fission). The fission macroscopic cross-section dropped from 0.41 to 0.30 cm-1 and the fission mean free path increased from 2.44 cm to 3.29 cm. 59 Knowledge Check If an isotope has an absorption microscopic crosssection of 5,000 barns at 68°F, what happens to its absorption microscopic cross-section if temperature is increased to 1,000°F? A. Increases B. Decreases C. No change D. Need more data to answer ELO 5.5 Neutron Flux Distribution Duration ο§ 15 minutes Logistics ο§ Use PowerPoint slides 174β180 and the IG to present ELO 5.5. Inform ο§ Ensure students understand radial and axial neutron flux distribution, as well as the differences between the two. ο§ Explain the importance of reactor safety. Introduction Neutron flux distribution is very important; both operator and reactor engineering sections monitor it. This topic is important for safe operation of the reactor. This section provides initial information on neutron flux distribution for the operator. Neutron Flux Distribution Neutrons interact with all materials in a reactor, absorptions, scattering, etc. The type of material found in particular locations of the core affects the neutron flux in that area. For example, fast neutron flux in materials with large scattering cross-sections will cause fast neutrons to slow down to lower energy levels more quickly. This reduces the fast neutron flux. At same time, intermediate and thermal neutron flux increases as a larger percentage of fast neutrons slow down to these energy levels. Similarly, thermal neutron flux in a material with large absorption crosssections decreases as a larger number of thermal neutrons are absorbed in material thereby reducing the thermal neutron flux levels. Axial and Radial Neutron Flux Commercial PWR nuclear reactor cores are cylindrical in shape. Axial and radial flux, describe neutron flux profiles from top to bottom (axial) and across (radial) the core. Axial flux is the side view of the core (top to bottom) and radial flux is the top view of the core (side to side). The next figure graphically illustrates axial and radial flux (assuming a uniformly distributed fuel with no zoning at the BOL). 60 Rev 1 Figure: Axial and Radial Neutron Flux in a Reactor Core The axial and radial neutron flux distribution across a nuclear reactor is a spatial representation of neutron flux level throughout the core. Fission produces neutrons, the moderator slows them down, the core materials and fuel within the core boundaries capture neutrons, and the neutrons undergo fission, leading to more of these same processes (the cycle repeats). All of these neutron processes affect the neutron flux distribution in the core. Neutron flux distribution is generally highest in the center of the core and drops off toward the core boundaries (top and bottom, sides). Neutrons produced near any edge (top, bottom, sides) of the core are more likely to leak out of the core and not cause fission. This reduces thermal neutron flux in the outer boundaries of the core. Neutrons produced toward the center of the core have less leakage and a greater probability of thermalizing and causing more fission events; therefore, neutron flux levels are higher toward the center of the core. Self-Shielding The neutron flux level may be lower in some localized areas than in others because of self-shielding. For example, the interior of a fuel pin receives a lower average neutron flux level than the outer surfaces since an appreciable fraction of the neutrons will have been absorbed in the outer layers of the fuel, thereby reducing neutron availability to the fuel pellet interior. The concept of self-shielding is also important relating to neutron poisons in the reactor core; a later section of the course will cover self-shielding in detail. Rev 1 61 Knowledge Check The neutron flux that varies from the top to bottom of the core is called the ________ flux and the neutron flux that varies from across the top is ________. A. axial; radial ELO 5.6 Reaction Rate Duration ο§ 20 minutes Logistics ο§ Use PowerPoint slides 181β185 and the IG to present ELO 5.6. Demonstrate ο§ Demonstrate methods used for calculating reaction rate. Introduction Two factors are required in order to calculate the number of interactions taking place in a cubic centimeter in one second. These two factors are the total path length of all the neutrons in a cubic centimeter per second (neutron flux [Ξ¦]), and the probability of an interaction per centimeter path length (macroscopic cross-section [Ξ£]). Multiply these two factors together to get the number of interactions taking place in that cubic centimeter in one second. The resulting value is the reaction rate, denoted by the symbol R. The type of reaction rate calculated will depend on the macroscopic crosssection used in the calculation. Normally, the reaction rate is of greatest interest is the fission reaction rate. Calculate the Reaction Rate Step Action 1. Obtain the average thermal neutron flux in the reactor. 2. Obtain the fuel macroscopic cross-section for the particular material and reaction of interest. Determine the microscopic cross-section and atomic density for material and calculate the macroscopic cross-section if the macroscopic cross is unknown using: Ξ£ = ππ Where: Ξ£ = macroscopic cross-section (cm-1) N = atomic density (atoms/cm3) Ο = microscopic cross-section (cm2) 3. Calculate the reaction rate using the following formula: π = ΣΦ 62 Rev 1 Step Action Where: R = reaction rate (reactions/cm3-sec) Ξ£ = macroscopic cross-section (cm-1) Ξ¦ = neutron flux (neutrons/cm2-sec) Note This reaction rate is only calculating the reactions occurring in one cubic centimeter. This unit value requires multiplication by the core volume to determine the reactions in the entire core. The next session will discuss the conversion. Note Reaction Rate Example: A reactor has a macroscopic fission cross-section of 0.1 cm-1, and thermal neutron flux of 1013 neutrons/cm2-sec, what is the fission rate in that cubic centimeter? Solution: π π = ΦΣf = (1 × 1013 = 1 × 1012 πππ’π‘ππππ ) (0.1 ππβ1 ) ππ2 β π ππ πππ π ππππ ππ3 β π ππ Neutron Flux and Macroscopic Cross-Section Versus Reaction Rate Increasing either the neutron flux or the macroscopic cross-section increases the reaction rate. Macroscopic cross-section decreases with fuel burnup and decreasing the reaction rate. An operator can increase neutron flux to compensate and maintain the reaction rate. Rev 1 63 Knowledge Check Calculate the reaction rate (fission rate) in a one cubic centimeter section of a reactor that has a macroscopic fission cross-section of 0.2 cm-1, and a thermal neutron flux of 1014 neutrons/cm2-sec. A. R = 2.0 × 1014 B. R = 0.2 × 1013 neutrons cm2 βsec neutrons cm2 βsec C. R = 2.0 × 1013 neutrons D. π = 20 × 1013 πππ’π‘ππππ cm3 βsec ππ3 βπ ππ ELO 5.7 Neutron Flux and Reactor Power Duration ο§ 15 minutes Logistics ο§ Use PowerPoint slides 186β192 and the IG to present ELO 5.7. Inform ο§ Explain and work through the example calculation. Demonstrate ο§ Demonstrate methods used for calculating reactor power from the reaction rate. Introduction Multiplying the reaction rate per unit volume by the total volume of the core equals the total number of reactions occurring in the core per unit time. It is possible to calculate the rate of energy release (power) due to a certain reaction given the amount of energy involved in each reaction. Neutron Flux and Reactor Power Step-by-Step Tables The number of fissions to produce one watt of power requires the following conversion factors in a reactor where average energy per fission is 200 MeV: ο· ο· ο· 1 fission = 200 MeV 1 MeV = 1.602 x 10-6 ergs 1 erg = 1 x 10-7 watt-sec 1 πππ 1 πππ 1 πππ π πππ 1 π€ππ‘π‘ ( )( )( ) β7 β6 1 × 10 π€ππ‘π‘β π ππ 1.602 × 10 πππ 200 πππ πππ π πππ = 3.12 × 1010 π πππππ This is equivalent to stating that 3.12 x 1010 fissions release 1 watt-second of energy. You can use this equation to calculate the power released in a reactor. Multiply the reaction rate by the volume of the reactor to obtain the total fission rate for the entire reactor. Divide the total fission rate by the number of fissions per watt-sec to obtain the power released by fission in the reactor in watts. Solve for reactor thermal power (watts or Megawatts) using the following steps based on this equation. 64 Rev 1 Step 1. Action Obtain the following reactor data: ο· ο· ο· Volume of core (cm3) Reaction rate Fission macroscopic cross-section and average thermal neutron flux or, π = ΣΦ 2. Calculate reactor power using the following equation, substitute reaction rate if known: π= Ξ¦π‘β Ξ£π π πππ π ππππ 3.12 × 1010 π€ππ‘π‘β π ππ Where: P = power (watts) Ξ¦th = thermal neutron flux (neutrons/cm2 - sec) Ξ£f = macroscopic cross-section for fission (cm-1) V = volume of core (cm3) Neutron Flux and Reactor Power Demonstration Example: Calculate reactor power given the following: ππππ’ππ ππ ππππ (ππ3 ) = 20,000 ππ3 πππ π ππππ ο· π ππππ‘πππ πππ‘π = 1 × 1015 ππ3 βπ ππ ο· Solution: Step 1: ππππ’ππ ππ ππππ (ππ3 ) = 20,000 ππ3 πππ π ππππ ο· π ππππ‘πππ πππ‘π = 1 × 1015 ππ3 βπ ππ ο· Step 2: Rev 1 65 π= Ξ¦π‘β Ξ£π π πππ π ππππ 3.12 × 1010 π€ππ‘π‘β π ππ Substitute: Reaction rate for thermal neutron flux multiplied by the macroscopic cross-section for fission π = 1 × 1015 × 20,000 3.12 × 1010 π = 6.41 × 108 π€ππ‘π‘π = 641 πππππ€ππ‘π‘π π‘βπππππ Relationship between Neutron Flux and Reactor Power The volume of the reactor is constant in an operating reactor. The number (density) of fuel atoms is also relatively constant over a relatively short period (days and weeks). With a constant atomic density and microscopic cross-section, the macroscopic cross-section is also constant. By examining the equation for power, we see that reactor power and the neutron flux are directly proportional if the reactor volume and macroscopic cross-section are constant. However, if the fission macroscopic crosssection decreases from the reactor fuel, the neutron flux must increase to maintain power constant. This will deplete (macroscopic cross-section = atomic density time microscopic cross-section) and atomic density will decrease as a result. Knowledge Check With a reaction rate of 2 x 1013 neutrons/cm3-sec what is the reactor power level? Assume the entire volume of the core is 10,000 cm3. A. π= B. 66 πππ’π‘ππππ (10,000 ππ3 ) ππ3 β π ππ πππ π ππππ 3.12 × 1010 π€ππ‘π‘β π ππ 2 × 1013 π= C. Ξ¦π‘β Ξ£π π πππ π ππππ 3.12 × 1010 π€ππ‘π‘β π ππ π = 6.41 × 106 π€ππ‘π‘π Rev 1 TLO 5 Summary Neutron Reaction Rates Summary ο· Atomic density (N) is the number of atoms of a given type per unit volume of material. ο· Microscopic cross-section (Ο) is the probability of a given reaction occurring between a single neutron and a single nucleus. - Microscopic cross-sections are measured in units of barns, where 1 barn = 10-24 cm2. ο· Macroscopic cross-section (Ξ£) is the probability of a given reaction occurring per unit length of travel of the neutron. - The units for macroscopic cross-section are cm-1. - Calculate the macroscopic cross-section for a material using the equation below: π΄ = ππ - Calculate the macroscopic cross-section for a mixture of materials using the equation below: π΄ = π1 π1 + π2 π2 + π3 π3 +. . .. ππ ππ ο· ο· ο· ο· ο· ο· ο· Rev 1 The absorption cross-section for a material usually has three distinct regions. - At low neutron energies (<1 eV) the cross-section is inversely proportional to the neutron velocity. - Resonance peaks exist at intermediate (epithermal) energy levels. Resonance absorption occurs when the sum of the kinetic energy of the neutron and its binding energy is equal to amount of energy required to raise a compound nucleus from its ground state to a quantum level energy state. - For higher neutron energies, the absorption cross-section steadily decreases as the neutron energy increases. The mean free path (Ξ») is the average distance that a neutron travels in a material between interactions. - The mean free path equals 1/Ξ£. Neutron flux (Ξ¦) is the total path length traveled by all neutrons in one cubic centimeter of material during one second. - Thermal neutron flux (Ξ¦th) is number of thermal neutrons crossing a unit area in the reactor in a given amount of time: n/cm2/sec. - Fast neutron flux (Ξ¦f) is number of fast neutrons crossing a unit area in a given amount of time. The neutron flux distribution is generally highest in the center of the core and drops off toward the core boundaries (top and bottom, sides). The magnitude of axial flux is a side view of the core flux (top to bottom). The magnitude of radial flux is a top view of the reactor core flux (side to side). Self-shielding is where the local neutron flux is depressed within a material due to neutron absorption near the surface of the material. 67 Duration ο§ 10 minutes Logistics ο§ Use PowerPoint slides 193β197 and the IG to review TLO 5 material. Inform ο§ Use directed and nondirected questions to students, check for understanding of ELO content, and review any material where student understanding of ELOs is inadequate. ο· The reaction rate is the number of interactions of a particular type occurring in a cubic centimeter of material in a second. ο· Calculate the reaction rate using the equation below. π = π·π΄ ο· Fission reaction rate multiplied by the core volume is directly proportional to reactor power. Objectives Now that you have completed this lesson, you should be able to: 1. Explain the following terms, including any mathematical relationships: a. Atomic density b. Neutron flux c. Fast neutron flux d. Thermal neutron flux e. Microscopic cross-section f. Barn g. Macroscopic cross-section h. Mean free path 2. Define the following neutron classes: a. Fast b. Intermediate c. Slow 3. Describe how the absorption and scattering cross-section of typical nuclides varies with neutron energies in the 1 eV region, and the resonance absorption region. 4. Describe the macroscopic cross-section and mean free path at various temperatures. 5. Describe radial and axial neutron flux distribution. 6. Describe how changes in neutron flux and macroscopic crosssection affect reaction rates. 7. Describe the relationship between neutron flux and reactor power. Fission Summary Duration ο§ 5 minutes Logistics ο§ Use PowerPoint slides 198β202 to review the Fission module. This module focused on the fission process and energy released through fission. TLO 1 introduced neutron interactions, covering resonance and potential elastic scattering where kinetic energy is conserved, and inelastic scattering where there is momentum conservation. TLO 1 also covered radiative capture, where a target nucleus absorbs a neutron to become an excited nucleus, releasing excitation energy in the form of a gamma ray. Alternatively, the excited nucleus may eject an alpha or proton particle. 68 Rev 1 Finally, if the nucleus has gained enough energy, it may split apart, releasing two fission fragments, neutrons, and energy. TLO 2 focused on the processes included in nuclear fission, definitions of key terms involved in the fission process and classification of different materials with respect to their fission capability. TLO 3 covered the energy released during a fission event, including methods of calculating energy released in fission, and the heat generated in the fission process, as well as the heat generated during decay of fission products. TLO 4 focused on neutron sources, both those that are intrinsic, and those installed specifically to produce source neutrons. This TLO described the variation of source neutrons over the life of the core, including which sources provide the highest source of neutrons at varying points in the core life. TLO 5 presented a summary of neutron reaction rates, discussing microscopic and macroscopic cross-sections and calculation methods, as well as neutron flux and its relationship to the above cross-sections, and the effect these parameters have on neutron reaction rates. Now that you have completed this module, you should be able to demonstrate mastery of this topic by passing a written exam with a grade of 80 percent or higher on the following TLOs: 1. Describe neutron interactions with matter. 2. Describe the process of nuclear fission and the types of material that can undergo fission. 3. Explain the production of heat from fission. 4. Describe intrinsic and installed neutron sources and their contribution to source neutron strength over core life. 5. Explain the relationship between neutron flux, microscopic and macroscopic cross-sections, and their effect on neutron reaction rates. Rev 1 69