CHAPTER 4 Reactor Statics. Table of Contents

Transcription

1 CHAPTER 4 Reactor Statics Prepared by Dr. Benjamin Rouben, & Consulting, Adjunct Proessor, McMaster University & University o Ontario Institute o Technology (UOIT) and Dr. Eleodor Nichita, Associate Proessor, UOIT Summary: This chapter is devoted to the calculation o the neutron lux in a nuclear reactor under special steady-state conditions in which all parameters, including neutron lux, are constant in time. The main calculation method explored in this chapter is the neutron-diusion equation. Analytical solutions are derived or simple neutron-diusion problems in one neutron energy group in systems o simple geometry. Two-group diusion theory and the approximate representation o the diusion equation using inite dierences applied to a discrete spatial mesh are introduced. The rudimentary reactor-physics design o CANDU reactors is presented. The two-step approach to neutronics calculations is presented: multi-group lattice transport calculations, ollowed by ull-core, ew-group diusion calculations. Finally, the chapter covers uel-property evolution with uel burnup and speciic eatures o CANDU neutronics resulting rom on-line reuelling. Table o Contents Introduction Overview Learning Outcomes... 3 Neutron Diusion Theory Time-Independent Neutron Transport Fick s Law and Time-Independent Neutron Diusion Diusion Boundary Condition with Vacuum at a Plane Boundary Energy Discretization: The Multi-Group Diusion Equation... 3 One-Group Diusion in a Uniorm Non-Multiplying Medium Plane Source Point Source Flux Curvature in Source-Sink Problems and Neutron In-Leakage One-Group Diusion in a Uniorm Multiplying Medium with No External Source Uniorm Ininite Reactor in One Energy Group Uniorm Finite Reactors in One Energy Group Uniorm Finite Reactors in Various Geometries Flux Curvature and Neutron Out-Leakage Reactors in Two Neutron-Energy Groups with No External Source The Neutron Cycle and the Four-Factor Formula The Two-Energy-Group Model Neutron Diusion Equation in Two Energy Groups Uniorm Ininite Medium in Two Energy Groups Uniorm Finite Reactors in Two Energy Groups UNENE, all rights reserved. For educational use only, no assumed liability. Reactor Statics September 04

2 The Essential CANDU 6 Solution or Neutron Flux in a Non-Uniorm Reactor Finite-Dierence Form o the Neutron-Diusion Equation Iterative Solution o the Neutron-Diusion Equation Energy Dependence o Neutron Flux The Fission-Neutron Energy Range The Thermal Energy Range The Slowing-Down Energy Range Neutron Flux over the Full Energy Range Basic Design o Standard CANDU Reactors CANDU Reactor Physics Computational Scheme Lattice Calculation Reactivity-Device Calculation Full-Core Calculation Evolution o Lattice Properties Summary o Relationship to Other Chapters Problems Appendix A: Reactor Statics Acknowledgements List o Figures Figure Negative lux curvature in homogeneous reactor... 9 Figure Ininite-slab reactor... Figure 3 Flux with physical and unphysical values o buckling... 3 Figure 4 Ininite-cylinder reactor... 4 Figure 5 Ordinary Bessel unctions o irst and second kind... 5 Figure 6 Rectangular-parallelepiped reactor... 6 Figure 7 Finite-cylinder reactor... 8 Figure 8 Energies o ission neutrons... 3 Figure 9 Sketch o cross section versus neutron energy... 3 Figure 0 Neutron cycle in thermal reactor Figure Face view o very simple reactor model Figure Flux in a cell and in immediate neighbours Figure 3 Linear treatment o lux in central cell and one neighbour Figure 4 Flux in cell at reactor boundary... 4 Figure 5 Energy distribution o ission neutrons Figure 6 Variation o lux with energy i absorption is smooth Figure 7 Variation o lux with energy in the presence o resonances Figure 8 Sketch o lux variation over ull energy range Figure 9 CANDU basic lattice cell... 5 Figure 0 Supercell or calculating device incremental cross sections... 5 Figure Simple model with superimposed reactivity device Figure Evolution o uel isotopic densities Figure 3 Ininite-lattice reactivity vs. irradiation Figure 4 Finite-reactor reactivity vs. irradiation Figure 5 Averaging reactivity with daily reuelling UNENE, all rights reserved. For educational use only, no assumed liability. Reactor Statics September 04

3 Reactor Statics 3 Introduction. Overview This chapter is devoted to the calculation o the neutron lux in a nuclear reactor under special steady-state conditions in which all parameters, including neutron lux, are constant in time. The position-, energy-, and angle-dependent steady-state neutron-transport equation is derived by writing the detailed neutron-balance equation. This is done in Chapter 3 o the book. However, it is also done here or completeness. The steady-state diusion equation is subsequently derived using a linear approximation o the angular dependence o the neutron lux. Multi-group neutron-energy discretization is also introduced. Analytical solutions are derived or simple neutron diusion problems. First, the one-group diusion equation is solved or a uniorm non-multiplying medium in simple geometries. Subsequently, the one-group diusion equation is solved or a uniorm multiplying medium (a homogeneous nuclear reactor ) in simple geometries. The importance o neutron leakage and the concepts o criticality and the neutron cycle are introduced. O course, real reactors are almost never homogeneous, and rarely can neutron energies be accurately represented by a single energy group. However, two energy groups are oten suicient to represent neutron diusion in a thermal reactor because most o the issions are induced by thermal neutrons. This chapter thereore proceeds to introduce two-group diusion theory and the approximate representation o the diusion equation using inite dierences applied to a discrete spatial mesh. The latter enables the treatment o non-uniorm reactor cores. Following this treatment o the basic theory o neutron transport and diusion, the rudimentary reactor-physics design o CANDU reactors is presented, and the associated neutron energy spectrum is discussed. Subsequently, more general core modelling concepts are presented, including the two-step approach to neutronics calculations: multi-group lattice transport calculations, ollowed by ull-core, ew-group diusion calculations. The inal section in the chapter concentrates on uel-property evolution with uel burnup and speciic eatures o CANDU neutronics resulting rom on-line reuelling.. Learning Outcomes The goal o this chapter is or the student to understand: The neutron transport resulting rom the detailed neutron balance; The neutron-diusion equation resulting rom the linear approximation o the angular dependence o the neutron lux; Analytical solutions o the diusion equation in simple geometries or both multiplying and non-multiplying media; In this document, the term neutron lux is used to denote the quantity = n, where n is neutron density and is neutron speed. The units o lux are cm - s - or m - s -. Some newer texts use the term lux density or luence rate. All three terms reer to the same physical quantity. UNENE, all rights reserved. For educational use only, no assumed liability. Reactor Statics September 04

4 4 The Essential CANDU The concepts o neutron cycle, criticality, and buckling; Energy discretization o the diusion equation using neutron energy groups and spatial discretization using inite dierences; The two steps o neutronics calculations: lattice-cell transport calculations and ullcore diusion calculations; and Basic design elements o CANDU reactors and their implications or the physical attributes o such reactors. Neutron Diusion Theory. Time-Independent Neutron Transport The time-independent neutron-transport equation is an equation in six variables: three variables r or the position o the neutron in the reactor, two variables ˆ or the direction o neutron motion, and one variable E or neutron energy (we could also use neutron speed v instead o E, because there is a one-to-one relationship between v and E). Because we are dealing with time independence (i.e., reactor statics), the neutron-transport equation expresses the act that the number o neutrons at any position, moving in any direction, and with any energy, is unchanging over time. In other words, i we consider a dierential volume in the six-dimensional variable space, i.e., drdˆ de about r, ˆ, E, the rate o neutrons entering the dierential volume must be equal to the rate o neutrons exiting the dierential volume. To write the equation, we must thereore consider all phenomena by which neutrons enter or exit the dierential volume or are created or destroyed within the volume. The phenomena which we must consider are: Neutron production by nuclear ission; Neutron scattering, which can change the neutron s energy and/or its direction o motion; Neutron absorption; Neutron spatial leakage into or out o the dierential volume. [Note: Another process or neutron production is the (n, n) nuclear reaction. reaction produces many ewer neutrons than ission and is neglected here.] However, this The rates at which neutrons enter or exit the dierential volume by virtue o these various phenomena, expressed per unit dierential volume per second, are given one by one in the ollowing equations. The rate o birth o neutrons by ission is given by: where r, E' E 4 neutrons o energy E ; E ' ˆ ' ˆ r, E ' r, ', E ' dˆ ' de ', () is the neutron-yield cross section at position r or issions induced by UNENE, all rights reserved. For educational use only, no assumed liability. Reactor Statics September 04

5 Reactor Statics 5 r, ˆ ', E' is the angular lux o neutrons at position, o energy E, and moving in direction ' ; and (E) is the raction o all ission neutrons born with energy E. Note: In this chapter, or simplicity, a single ission spectrum (E) is used or issions induced in all nuclides, whereas in reality the ission spectrum should be taken as dierent or dierent nuclides. This ully correct treatment is what is done in Chapter 3. Note: In the above equation, delayed neutrons are not reerred to separately. The neutron production rate includes both prompt and delayed neutrons. This is acceptable because it can be shown that in (true) steady state, accounting separately or delayed-neutron production reduces in any case to the above expression. The rate o production o neutrons rom an external source (not related to issions in the reactor uel) is given by: S r, E, () S r, E is the external source strength or neutrons o energy E at position r. where Note: We will assume that the source is isotropic. The rate o neutron gain rom neutrons entering the dierential volume by scattering rom other neutron directions o motion or other neutron energies is given by: where r, E' E, ˆ ' ˆ s energy E and rom direction r, E ' E, ˆ ' ˆ r, ˆ ', E ' d ˆ ' de ', s E ' ' is the cross section or scattering neutrons rom energy E to ' to direction ˆ. The rate o neutron loss rom absorption and rom neutrons exiting the dierential volume by scattering is given by:,, ˆ t r E r, E, (4) r E, the total cross section at position r or neutrons o energy E, where t, where r E a, s r, E r, E E ', ˆ ˆ ' dˆ ' de ', a is the absorption cross section. The net rate o neutron spatial leakage (i.e., diusion) out o the dierential volume is given by: J r, ˆ, E, (5) where J r, ˆ, E energy E. is the angular current at position r o neutrons moving in direction ˆ Note: Because J r,, E r,, E ˆ ˆ ˆ UNENE, all rights reserved. For educational use only, no assumed liability. Reactor Statics September 04 (3) with, the neutron leakage [Eq. (5)] can also be written as: ˆ r, ˆ, E. (6) Taking into account all the above rates, the neutron balance can then be expressed in the time-

6 6 The Essential CANDU independent neutron-transport equation as ollows: E S r, E, ', ˆ ', ' ˆ ' ', ', ˆ ' ˆ, ˆ ', ' ˆ r E r E d de r E E r E d ' de ' 4 4 E' ˆ ' E' ˆ ',,,,, 0. t s ˆ r E r E J r ˆ E (7) Note again that or simplicity a single ission spectrum (E) has been used here. See Chapter 3 or the ully correct treatment. The second term o the equation can be simpliied because the ission cross section does not depend on ˆ ', so that the integral over ˆ ' in the second term reduces to the angle-integrated lux, and Eq. (7) becomes: E S r, E, ', ' ', ', ˆ ' ˆ, ˆ ', ' ˆ r E r E de r E E r E d ' de ' 4 4 E' E' ˆ ',,,,, 0, t s ˆ r E r E J r ˆ E where r E' r, ˆ ', E' at position r., dˆ ' ˆ ' UNENE, all rights reserved. For educational use only, no assumed liability. Reactor Statics September 04 (7) is the total (angle-integrated) lux o neutrons with energy E The neutron-transport equation (7) or (7) is an exact statement o the general steady-state neutron-balance problem. However, it can immediately be seen that it is very complex: in addition to its dependence on six independent variables, it is an integrodierential equation. Because o its complexity, this equation cannot be solved analytically except or problems in the very simplest geometries, and real problems require numerical solution by computer. Note that when there is no external source S, the equation appears to be a linear homogeneous equation, which does not generally have a solution (except a trivial zero solution or the lux) or arbitrary values o the nuclear properties. In this case, a solution can be ound by modiying the nuclear properties. Mathematically, this can be done by modiying the yield cross section by dividing it by a quantity, k e, which can be selected to ensure a non-trivial solution. This quantity k e is called the multiplication constant. Two general categories o codes exist to solve the neutron-transport equation: deterministic codes (which solve the equation directly by numerical means) and Monte Carlo codes (where stochastic methods are used to model a very large number typically millions or even hundreds o millions o neutron births and their travel and event histories, rom which the multiplication constant and lux and power distributions can be evaluated using appropriate statistics o these histories). Although the application o either type o transport computer code to ull-core reactor models requires very signiicant computer resources and execution time to achieve a high degree o accuracy, both methods, especially Monte Carlo codes, have seen much greater application in whole-reactor analysis in the last decade or so. While core-wide pin-power reconstruction and time-dependent kinetics calculations are still beyond reach, static eigenvalue calculations and global lux and power distributions can now be carried out routinely using Monte Carlo codes. However, detailed discussion o either type o transport code will not be covered in the present work. The traditional way o attacking neutronics problems in reactors has been to solve the transport equation numerically in relatively small regions (such as a basic lattice cell or a small collection

7 Reactor Statics 7 o cells) to compute region-averaged properties, and then to use these to solve the ull-core reactor problem with a simpliied version o the transport equation, the neutron-diusion equation. This computational scheme is discussed in greater detail in Section 9.. Fick s Law and Time-Independent Neutron Diusion To derive a simpliied version o the neutron-transport equation, we note that, because the ission and total cross sections do not depend on the neutron direction o motion ˆ (i.e., nuclei do not care rom which direction neutrons interact with them ), and also because the most important quantity in reactor physics is the ission rate (which determines power production), it may be very eective to try to obtain an equation which is independent o ˆ and which r, E [also called the integral lux]. involves only the angle-integrated lux To achieve this, we can attempt to remove the neutron direction o motion by simply integrating Eq. (7) over ˆ. Let us see what this gives, term by term. The irst two terms in (7) do not contain ˆ, and because the integral o ˆ over a sphere is 4, their integrals, which we can call T and T, can be written directly as: T S r E (8), T E r, E ' r, E ' de '. E ' Similarly, integrating the ourth term gives a result that depends on the integral lux only:,, ˆ, ˆ T r E r E d r, E r, E. 4 ˆ t Integrating the third term gives a more complex relation: ˆ ˆ T r, E ' E, ' r, ˆ ', E ' dˆ ' de ' dˆ. 3 ˆ E ' ˆ ' s We can, however, simpliy T 3 by noting that, because no absolute direction in space is special, the scattering cross section s cannot depend on the absolute directions ˆ and ˆ ', but only on the scattering angle between the directions o the incoming and scattered neutrons, or even more speciically on the cosine o that angle, i.e., on ˆ ' ˆ. Using this act, we can then simpliy the integral in Eq. () by integrating over ˆ irst (actually over, because s does not depend on the azimuthal angle o scattering, and integrating over that azimuthal angle simply gives ). The result is then a product o two integrals:, ', ˆ ' ˆ T r E E d' r, E ' E, d de ' where we have deined 3 E ' ˆ ' E ' s r, E ' r, E ' E de ', s r, E ' E r, E ' E, d. s s t (9) (0) () () (3) UNENE, all rights reserved. For educational use only, no assumed liability. Reactor Statics September 04

8 8 The Essential CANDU Equation () is a useul result because it depends on the integral lux r, E' For the ith term, we get: where 5 ˆ ˆ ˆ T J r, E, d r, E, ˆ d ˆ ˆ ˆ J r, E, J r E J r E dˆ,,, ˆ is the current o neutrons o energy E at position r. Unortunately, this result is not at all o the orm we would like, i.e., it is not at all expressible in terms o the integral lux, because it is clear that integrating a unction o the vector quantity J r, E, ˆ over ˆ will obviously give in general a result totally unrelated to the integral lux r, E! In summary, integrating Eq. (7) over ˆ gives: S r, E E r, E ' r, E ' de ' r, E ' E r, E ' de ' E ' E ' s t r, E r, E J r, E 0, but the last term on the let-hand side (the leakage term) has thwarted our eorts to achieve an equation in the integral lux only. However, one approximation which is oten used in diusion problems (diusion o one material through another) can help us here. This approximation, called Fick s Law, applies in lowabsorption media i the angular lux varies at most linearly with angle and i the neutron source is isotropic [already assumed in writing Eq. (7)]. Under these conditions it states that the current J r, E is in the direction in which the integral lux r, E decreases most rapidly, i.e., it is proportional to the negative gradient o r, E: J r, E D r, E r, E, (7) where the quantity D r, E is called the diusion coeicient and can be written as /(3 tr ), with tr being the neutron transport cross section. The reader is reerred to Appendix A (Reactor Statics) o this chapter or the derivation o Fick s Law. Here, we will continue to derive the inal orm o the neutron-diusion equation. Equation (7), when substituted into Eq. (6), inally gives an equation in the integral lux only, the time-independent neutron-diusion equation: S r, E E r, E ' r, E ' de ' s r, E ' E r, E ' de ' E ' E '. (8) r, E r, E D r, E r, E 0 t only. (4) (5) (6) UNENE, all rights reserved. For educational use only, no assumed liability. Reactor Statics September 04

9 Reactor Statics 9 The time-independent neutron-diusion equation is much simpler than the neutron-transport equation and can be used to solve or the lux in specially prepared ull-core reactor models. Note that the out-leakage term (the last term on the let-hand side) has a positive sign, even though this term nominally represents a loss o neutrons; the + sign arises as a result o the sign in relationship (7) between the current vector and the lux gradient. Beore the diusion equation can be used, it is important to understand that Fick s Law is only an approximation; it is valid only in low-neutron-absorption media, when the integral lux does not vary too quickly and when angular lux varies weakly with angle (at most linearly). Thereore, Fick s Law is not an especially good approximation in the vicinity o strong absorbers, where the spatial lux variation is very large, or, or the same reason, near interaces between regions with large variations in nuclear properties or near external suraces..3 Diusion Boundary Condition with Vacuum at a Plane Boundary To solve the diusion equation, which is a second-order partial dierential equation, throughout the reactor volume, we need to deine boundary conditions at the surace o the reactor. The vacuum boundary conditions in transport theory are quite clear: the angular current (or angular lux) at the boundary must be zero or any direction pointing to the inside o the reactor (assuming that the reactor has no re-entrant surace): r, ˆ, E 0 (9) or r at the boundary and any ˆ such that ˆ eˆ 0, where ê is the unit outgoing normal at r. This condition cannot be applied in diusion theory, which depends on the integral lux only, because we have lost all the directional inormation o the angular lux. In diusion theory, we need boundary conditions, at most, on the integral lux and its gradient. The conditions can be generalized to demand that the total rate o incoming neutrons be zero. We will now proceed to derive the boundary conditions used in diusion theory. We irst derive a general relationship between the angular lux and the current using the approximation that the angular lux is linear in angle. Linearity in angle means that the angular lux can be written as a linear unction o the x-, y-, and z-components o the angle ˆ : ˆ a b b b, (0) x x y y z z where a, b x, b y, b z are constants which can be determined in terms o the integral lux and current. Let us irst consider the integral lux ˆ d ˆ. From Eq. (0), a dˆ b dˆ b dˆ b dˆ. () x x y y z z The irst integral in Eq. () has value 4, whereas the others have value 0 (it is clear that in integrating a single component o the angle over all solid angles, the + and components cancel out). This means that = 4a, which implies that UNENE, all rights reserved. For educational use only, no assumed liability. Reactor Statics September 04

10 0 The Essential CANDU Now let us consider the current: From Eq. (0), J a. () 4 ˆ ˆ d ˆ. J a ˆ dˆ b ˆ dˆ b ˆ dˆ b ˆ dˆ. x x y y z z It is clear that the irst integral in Eq. (4) is zero, or the same reason that the + and components cancel out. Consider the other integrals in Eq. (4), or instance, ˆ d ˆ. This is a x vector integral, and only the x-component will survive because or d ˆ 0, or the same reason, cancellation. On the other hand, the x-component gives ˆ x d, which can easily be shown to be equal to 4 4/3 [by symmetry, it must be equal to ˆ 3 z d d ]. 3 3 Thereore, we can conclude that i.e., (and similarly or y and z). 4 J b ˆ ˆ ˆ xi by j bzk, 3 Now, substituting Eqs. () and (6) into Eq. (0), b x (3) (4) UNENE, all rights reserved. For educational use only, no assumed liability. Reactor Statics September 04 x y z (5) 3 J x (6) 4 ˆ 3 3 J ˆ xx J y y J zz J Let us now apply Eq. (7) at a plane boundary with vacuum. Assume that the boundary plane is perpendicular to the z-axis and the polar axis is the outward normal to the boundary, i.e., the unit vector kˆ in the positive z-direction. The total rate o incoming neutrons is: J ˆ kd ˆ ˆ, where the integral over the hal-space - means integrating over the entire azimuthal angle (i.e., over ), but only over hal the polar angle, i.e., over rom / to (or over cos rom - to 0). Using Eq. (7), the total rate o incoming neutrons is: Because ˆ kˆ z ˆ ˆ kd ˆ ˆ 3 ˆ ˆ ˆ ˆ ˆ. 4 4 J kd (7) (8) (9), then, in the integral o the second term, only the part involving z will

11 Reactor Statics survive (or the same reason stated beore), and: 4 Total rate o incoming neutrons = * * d * * J k Applying Fick s Law and setting the total rate o incoming neutrons to 0 gives: D ˆ D d k 0, i.e., 0 4 dz 4 which can be written as d, dz D boundary, d extr ˆ * 0 d 4 J kˆ. (30) which leads to where tr is the transport cross section. d d extr extr D, (3) 3 tr D, (3) 3 Actually, a more advanced analysis gives a slightly higher, more correct value or d extr : d extr D. 3 Equations (30)-(33) result in a zero net incoming current at the physical boundary. The geometric interpretation o Eq. (30) is that the relative neutron lux near the plane boundary has a slope o -/d extr, i.e., the lux would extrapolate linearly to 0 at a distance d extr beyond the boundary. This is oten stated as the lux goes to 0 at an extrapolation distance d extr beyond the boundary. Such an interpretation is not literally correct: the lux cannot go to zero in a vacuum, because there are no absorbers to remove the neutrons; the lux only appears to be heading to the zero value at the extrapolation point. In act, instead o using the slope version (30), the boundary condition is oten applied by extending the reactor model to the extrapolation point and demanding a zero lux at that point. Now, typical values or the diusion coeicients D in CANDU reactors are cm, and thereore the extrapolation distance d extr is o the order o 3 cm. For systems o large dimension, e.g., large power reactors which are several metres in size, the extrapolation distance is sometimes neglected i a slightly approximate answer is suicient. [Note: Equation (33) applies in principle to plane boundaries only. Slightly dierent ormulas or the extrapolation distance would apply to curved boundaries; however, the dierence is small unless the radius o curvature o the boundary is o the same order o magnitude as d extr.] tr tr tr (33) UNENE, all rights reserved. For educational use only, no assumed liability. Reactor Statics September 04

12 The Essential CANDU.4 Energy Discretization: The Multi-Group Diusion Equation Neutron energy E is o course a continuous variable. However, the diusion equation (8) with continuous E is not usually solved as is. Instead, the equation is rewritten in multi-group orm by subdividing the energy range into a number G o intervals, called energy groups and labelled with g rom to G (g = corresponding to the interval with the highest energies and g = G to the interval with the lowest energies). In each group g, the continuous lux r, E is replaced by an average group lux, r. g The nuclear cross sections in each group are assumed to have been appropriately averaged over the corresponding energy intervals. In multi-group notation, the cross sections and variables are written as ollows: S r, E S r g E g r, E, g r s r, E ' E s, g ' g r t r, E t, g r,. D r E D r The multi-group diusion equation then takes the orm: G g g, g ' g ' g' S r r r G s, g ' g g ' t, g g g g g ' g r r r r D r r 0, g,..., G. 3 One-Group Diusion in a Uniorm Non-Multiplying Medium We will start the analysis o systems with simple systems and increase complexity gradually. Assume that all neutrons are lumped into a single energy group, i.e., G =. There is then no need or the subscript g. With a single energy group, the notion o scattering across groups has no meaning, and the incoming scattering term in s cancels the outgoing scattering term inherent in the total cross section t (another way o saying this is that the out-scattering and in-scattering terms cancel one another i there is only one group). This means that we can drop the second term in the diusion equation and write a instead o t in the third term. In this section, we will deal with neutron diusion in non-multiplying media, i.e., in media where the ission cross section is zero and the neutron lux is driven by an external neutron source. This type o problem is sometimes called a source-sink problem. We will assume that the medium is uniorm outside the source, and also that it is ininite in size. Uniorm properties also mean that the diusion coeicient can be taken outside the divergence operator. With these assumptions, the diusion equation becomes: S r a r D r 0, (34) or S r r D r 0. a UNENE, all rights reserved. For educational use only, no assumed liability. Reactor Statics September 04

13 Reactor Statics 3 3. Plane Source We take the source as an ininite plane source, which we can place, without loss o generality, in the y-z plane. In this case, the lux is a unction o x only, (x), and the Laplacian can be written as: The diusion equation outside the source, i.e., or any x dierent rom 0, becomes: x d dx We can simpliy this equation by dividing by D. I we deine. d D a x 0. (35) dx D L, (36) called the diusion area (and L called the diusion length), the equation becomes: d dx x a x 0. (37) L For x > 0, Eq. (37) has mathematical solutions exp(x/l) and exp(-x/l), which give a general solution: x/ L / x Ae Ce However, the term e x/l goes to as x and thereore cannot be part o a physically acceptable solution or x > 0. The solution or x > 0 must then be x Ce x / L. By let-right symmetry, we can see that the ull solution or any x must be x x L. x / L Ce, (38) with C being a constant which we can determine rom the boundary condition at x = 0. I S is the source strength per unit area o the plane, then the number o neutrons crossing outwards per unit area in the positive x-direction must tend to S / as x 0. Thereore, S lim J x x0 d x S lim D x0 dx CD x/ L S lim e x0 L CD S SL, i. e., C. L D The neutron lux outside the source is then inally: UNENE, all rights reserved. For educational use only, no assumed liability. Reactor Statics September 04

14 4 The Essential CANDU x SL D x / L e. (39) It is interesting to try to interpret the physical meaning o the diusion area L. Let us calculate the mean square distance that a neutron travels in the x-direction rom the source (at x = 0) to its absorption point. We can do this by averaging x with the absorption rate a as a weighting unction: x 0 x dx 0 a a dx With the orm (39) or the lux, we can evaluate the integrals and show that x L, i.e., we can interpret L as one-hal the square o the average distance (in one dimension) between the neutron s birth point and its absorption point. 3. Point Source Let us now take the source as a single point source, assumed isotropic. Without loss o generality, we can place the source at the origin o co-ordinates, r 0. To solve or the lux in this geometry, we need to write the Laplacian in spherical coordinates. In view o the spherical symmetry o the problem, there is no dependence on angle (whether polar or azimuthal), the lux is a unction o radial distance r only, i.e., (r), and the Laplacian is then: d d d d r. r dr dr dr r dr The diusion equation outside the source, i.e., or all points except the origin, is then: D r a r 0, d d i. e., D a r 0, dr r dr D r a r 0, d d i. e., D a r 0, dr r dr d r d r. (40) (4) (4) r 0. (43) dr r dr L r To solve this equation, we can write in the orm r. Then, in terms o, Eq. (43) r becomes: UNENE, all rights reserved. For educational use only, no assumed liability. Reactor Statics September 04

15 Reactor Statics 5. d r dr r r r, d r d r d r d r r dr dr dr dr r r r r 3 Substituting these orms into Eq. (43) results in the simpler orm: d dr r r 0. (44) L Equation (44) has mathematical solutions exp(r/l) and exp(-r/l), which give a general solution: e r / L r Ae C e r/ L r/ L e r A C r e r r/ L r/ L Now, the term as r and thereore cannot be part o a physically acceptable r solution. The solution must then be with C being a constant which remains to be determined. continuity condition at the origin.. r/ L e r C, (45) r. To determine C, we can use the I S is the source strength, then the number o neutrons crossing the surace o a small sphere outwards must tend to S as the sphere s radius tends to 0, and thereore: lim 4 r0 r J r S d lim 4 r0 dr r D S r e r/ L e r/ L lim 4 L r0 r S 4 CD S, i. e., C. 4 D The neutron lux outside the source is then, inally: r CD S r r/ L S e. (46) 4 D r Again, it is interesting to interpret the physical meaning o the diusion area L. Let us calculate the mean square distance that a neutron travels outwards rom the source (at r = 0) to its UNENE, all rights reserved. For educational use only, no assumed liability. Reactor Statics September 04

16 6 The Essential CANDU absorption point. We can do this by averaging r with the absorption rate a as a weighting unction: 0 r r 0 4 r dr. a 4 a r dr With the orm (46) or the lux, we can evaluate the integrals and show that r 6 L, (47) i.e., we can interpret L as one-sixth o the square o the average distance outwards between the neutron s birth point and its absorption point. 3.3 Flux Curvature in Source-Sink Problems and Neutron In-Leakage Let us consider the out-leakage term (call it Leak out ) in the above examples. Leak out leakage out per dierential physical volume dr J Dr r. Applying this to the lux rom a plane source (Eq. 39) gives, or x > 0: SL S Leak D e e D L x/ L x/ L out We can also show that Leak out < 0 or x < 0. I we apply the ormula to the lux rom a point source (Eq. 46): r/ L r/ L S e d d S e Leakout D D, which we can show 4 D r dr r dr 4 D r S 4 L e r r/ L 0. We see that in all cases, Leak out < 0, i.e., the leakage is inwards (in-leakage) at any point outside the source. This can actually be seen in the general case rom S r r D r 0, a stated earlier as Eq. (34), which or any point outside the source gives D r r 0. In other words, we can say that any point outside an external source in a non-multiplying medium sees a positive lux curvature and neutron in-leakage, i.e., any dierential volume is a net sink or absorber o neutrons: Positive lux curvature neutron in-leakage. 4 One-Group Diusion in a Uniorm Multiplying Medium with No External Source We will now switch to the analysis o reactor systems with no external source, meaning that all neutrons are produced by neutron-induced nuclear ission. In this section, as in the previous one, we will start by assuming that all neutrons are lumped into a single energy group, i.e., G = UNENE, all rights reserved. For educational use only, no assumed liability. Reactor Statics September a

17 Reactor Statics 7. There is then, again, no need or the subscript g. Also, as explained in Section 3, we can drop the term in s and replace t by a. The neutron-diusion equation in this case is: r r D r 0. a A very important point to note is that without the external source, this is a linear homogeneous equation in the lux (i the properties are truly constant and independent o the lux). This means that i we ind one solution o the equation, then any multiple is also a solution. Thereore, the absolute value o the lux cannot possibly be deduced rom the diusion equation (incidentally, not rom the transport equation either). This is totally dierent rom problems with external sources, which drive the absolute value o the lux. 4. Uniorm Ininite Reactor in One Energy Group Let us irst assume a uniorm reactor, ininite in size. This assumption makes all points in space equivalent, which means that the neutron lux would also have to be constant throughout space. Moreover, there can be no leakage rom one point to another; thereore, the leakage term can be dropped rom the equation. The equation in the ininite uniorm multiplying medium is then: (48) 0. (49) This equation is interesting. The only solution is a trivial solution, i.e., a null lux, = 0, unless a. a (50) In other words, unless the composition o the medium is exactly balanced so that Eq. (50) is satisied, the uniorm ininite reactor cannot really operate in steady state (except in a trivial zero-lux situation). We can call Eq. (50) the criticality condition or a uniorm ininite reactor. What happens i the criticality condition in Eq. (50) is not satisied? Then there is no non-trivial solution, but is this all that we can say? Actually, we can ensure that there is always a non-trivial solution i we modiy Eq. (49) by tuning the neutron-yield cross section by dividing it by a new modiying actor which we call k, as in: k 0. (5) A non-trivial solution o Eq. (5) can always be guaranteed by selecting the value o k as: k a. (5) What this means is that i the composition o the uniorm ininite reactor is modiied rom the original composition by dividing the neutron-yield cross section by k deined as in Eq. (5), the modiied uniorm ininite reactor can then be operated in steady state with a non-zero lux; this modiied reactor is now critical. What is the value o the lux? We can see rom Eq. (5) that with the modiied neutron-yield cross section, the lux can have any value; in other words, the critical uniorm ininite reactor can operate at any lux (and thereore power) value! This is a direct consequence o the (apparent) homogeneity o the equation (i.e., the equation is o the a UNENE, all rights reserved. For educational use only, no assumed liability. Reactor Statics September 04

18 8 The Essential CANDU orm F = 0, with F an arbitrary operator independent o the lux). Comparing Eqs. (50) and (5), the criticality condition or a uniorm ininite reactor can be seen to be: k or criticality. The deviation o k rom tells us how ar the reactor with the original composition (, ) is rom criticality. k is called the ininite reactor multiplication constant and can a have the ollowing values: a k < : the ininite reactor is said to be subcritical; k = : the ininite reactor is critical; k > : the ininite reactor is said to be supercritical. The physical interpretation o k can be stated as ollows: k a a neutron production rate. neutron loss rate A quantity related to k, but centred at 0 instead o at, is the reactivity,, deined as: Thereore: k < < 0: the ininite reactor is subcritical k = = 0: the ininite reactor is critical k > > 0: the ininite reactor is supercritical. (53) (54). (55) k Real-lie reactors are designed to be operated at critical or very close to critical. Thereore, in most situations, we would expect k to be very close to and to be very close to 0. As a result, while (and k ) are absolute, dimensionless numbers, a new ractional unit o milli-k (or mk) is deined or reactivity, where: mk (56) For example, = + mk (or mk) means that = (or 0.00). Another reactivity unit, pcm 0.0 mk, is commonly used in Europe. 4. Uniorm Finite Reactors in One Energy Group We will now analyze uniorm reactors o inite size (at least in some dimensions). The nuclear properties are assumed uniorm throughout the reactor, but not all points in the reactor are equivalent spatially (some are urther rom or closer to the reactor boundary), so that the lux is a unction o space, r, and also the leakage term must remain in the diusion equation: r r D r 0, presented earlier as Eq. (48). This can be rewritten as: D r r r. a a (57) UNENE, all rights reserved. For educational use only, no assumed liability. Reactor Statics September 04

19 Reactor Statics 9 The let-hand side o Eq. (57) is the leakage out o the point r. I we integrate Eq. (57) over the volume V o the reactor: D r dr r dr. (58) V a The let-hand side o Eq. (58) is the total leakage out o the reactor. This o course must be positive: neutrons can only go out o the reactor, not into it, because there are no sources o neutrons outside which can eed neutrons into the reactor. Thereore, i Eq. (58) is true as is, the right-hand side must be positive, and because the integral o the lux must also be positive, so must the quantity ). We can thereore write Eq. (57) as: where we have deined ( a a r D i e r B r V r.., g, B g (59) a. (60) D The quantity B g, as it appears in Eq. (59), is called the geometrical buckling o the reactor. Its physical meaning can be understood by rewriting Eq. (60) as: r Bg. (6) r In other words, the geometrical buckling is the negative relative curvature o the neutron lux. Because we have established that B is deinitely positive, this means that the lux curvature is g negative (see Figure ), and in act, in view o Eq. (59), the relative curvature is uniorm (and negative) in a homogeneous inite reactor. Note that this statement is strictly true or homogeneous reactors only. In real reactors, the relative lux curvature (and the buckling) can vary and even change sign. Figure Negative lux curvature in homogeneous reactor Because the neutron lux curvature/geometrical buckling must bend the lux to bring it to 0 UNENE, all rights reserved. For educational use only, no assumed liability. Reactor Statics September 04

20 0 The Essential CANDU almost at the reactor boundary (actually, at an extrapolation distance d extr beyond the physical boundary), it is clear that the lux curvature will be large or reactors o small dimensions and small i the reactor dimensions are large. Thereore, B in Eq. (59) is purely a geometrical quantity. Returning now to Eq. (57) and using the deinition o geometrical buckling, the diusion equation or homogeneous reactors can be rewritten as: r a r DBg r 0. (6) This equation has the same characteristics as Eq. (49). The only solution is a null lux, r 0, unless a DBg. g (63) In other words, unless the composition o the reactor is exactly balanced so that Eq. (63) is satisied, the uniorm reactor cannot really operate in steady state (except in a trivial zero-lux situation). We can call Eq. (63) the criticality condition or a uniorm inite reactor. What happens i the criticality condition Eq. (63) is not satisied? Then there is no non-trivial solution. However, just as we did or the ininite medium, we can ensure that there is always a non-trivial solution i we modiy Eq. (6) by tuning the neutron-yield cross section by dividing it by a similar parameter, called k e, as in: k e 0. r r DB r a A non-trivial solution o Eq. (64) can always be guaranteed by selecting the value o k e as: which can also be written as: k g (64), e (65) a DBg B g a ke. (66) D This means that i the composition o the uniorm reactor is modiied rom the original composition by dividing the neutron-yield cross section by k e deined as in Eq. (65), the modiied uniorm reactor can then operate in steady state (i.e., as a critical reactor) with a non-zero lux. Again, as in the ininite medium, the lux in the modiied critical reactor can have any value, that is, the critical uniorm reactor can operate at any lux (and thereore power) value! In summary, the criticality condition or a uniorm inite reactor is: or k e DB a g (67) UNENE, all rights reserved. For educational use only, no assumed liability. Reactor Statics September 04

21 Reactor Statics B g a a k. D D L Equation (67) is clearly a generalization o the criticality condition Eq. (53) or the ininite medium, where the buckling was 0, i.e., B 0 (lat lux). The deviation o k e rom tells us how ar the reactor with the original composition (, a, D) is rom criticality. g Equation (68) is intriguing. The let-hand side, B, is a geometrical quantity, as already noted. The right-hand side, on the other hand, is a unction o the nuclear properties only and is not a geometrical quantity. It is nonetheless called a buckling, the material buckling: B m g a (68) k. (69) L In view o this, the criticality condition or a uniorm reactor in one neutron-energy group can be expressed as: Geometrical Buckling Material Buckling, Bg Bm. (70) The physical interpretation o k e can be obtained rom: r neutron production rate ke. (7) DB r DB r neutron loss rate (by absorption and leakage) a g a g This ratio o production to loss is the same at any point in a homogeneous reactor and is o course then also the same as the ratio o the reactor-integrated production and loss. Incidentally, Eq. (64), a linear equation with a boundary condition (zero lux at the extrapolation distance beyond the boundary), is mathematically an eigenvector problem, which can best be seen by rewriting the equation in the orm: DB r r, (7) where is the eigenvalue o the problem. a g (73) Eigenvalues o eigenvector problems can take on only distinct, non-continuous values. This tells us that k e cannot have just any value, but it can have only a certain number o distinct values. Later, we will see that the physical k e is the largest o these distinct values. 4.3 Uniorm Finite Reactors in Various Geometries Equations (67) and (68) relate the reactor multiplication constant to the reactor properties and to geometrical buckling, but we do not yet have a value or the buckling or or the distribution o the neutron lux in the reactor. In this section, we will solve or the lux distribution and or the value o the reactor multiplication constant k e or reactors o various geometries (and we k e UNENE, all rights reserved. For educational use only, no assumed liability. Reactor Statics September 04

22 The Essential CANDU can thereore also derive the value o the buckling). We will do this by showing how to solve the diusion equation in the ollowing orm [rewritten rom Eq. (59)]: r B r 0. (74) 4.3. Ininite slab reactor Let us consider a reactor in the shape o a slab o physical width a in the x-direction and ininite in the y- and z-directions. The reactor is centred at x = 0; see Figure. This is a one-dimensional problem, meaning that is a unction o x only. The Laplacian reduces to d dx, and the diusion equation (74) is now: d dx x g x B 0. g (75) This equation has mathematical solutions (x) sin(b g x) and (x) cos(b g x). However, by letright symmetry, sin(b g x) is not a viable solution (because the lux would be negative in hal the space). Thereore, the only physical solution must have the orm (x) = Acos(B g x), where A is a constant. Figure Ininite-slab reactor a However, we must also satisy the boundary condition or the problem, which is ex 0, where a ex is the extrapolated width and x = a ex / is the point at which the lux appears to go aex to zero. Thereore, we must have A cos Bg 0, which means that the values o B g are n limited to Bg, where n = any odd integer. a ex While any odd integer value o n gives a mathematical solution o Eq. (75), only n = is a physically acceptable solution, because higher values o n would give cosine unctions which would UNENE, all rights reserved. For educational use only, no assumed liability. Reactor Statics September 04

23 Reactor Statics 3 become negative or some values o x beore returning to 0 at a ex (see Figure 3). Thereore, the inal solution or the lux distribution in an ininite slab reactor is and the buckling is x aex x Acos, (76) B g a ex. (77) Figure 3 Flux with physical and unphysical values o buckling Note: The value o buckling which gives the only physical solution is the smallest o the mathematically possible values, and the corresponding value o k e or the physical solution is the largest o the mathematically possible values (i.e., the physical eigenvalue is the smallest o the possible values). Equation (76) gives the lux distribution in the reactor. However, the amplitude A, which gives the absolute value o the lux, cannot be obtained rom the diusion equation, as noted at the beginning o Section 4. To ind A, an extraneous condition has to be imposed, or instance the actual value o the lux at a point, or the total power o the reactor, or the power in some region o the reactor. For example, i we assume a value P or the power per unit area o the y-z plane and note that E is the energy released per ission (typically 00 MeV 3.*0 - joules), we have: UNENE, all rights reserved. For educational use only, no assumed liability. Reactor Statics September 04

24 4 The Essential CANDU a/ a/ x P E xdx E A cos dx a a/ a/ ex a/ a ex x a ex a i.e., P E Asin A E sin aex aex P A, a aexe sin aex and the absolute lux in the slab is a/ P x a aex aexe sin a ex x cos. I the extrapolation distance is ignored, i.e., a ex = a, this reduces to: P x a x cos. ae (78) (79) 4.3. Ininite cylindrical reactor We now consider a reactor in the shape o a uniorm cylinder o physical radius R and o ininite length in the axial (z) direction, as shown in Figure 4. The problem is independent o the azimuthal angle, and the lux is a unction o the r variable only. Figure 4 Ininite-cylinder reactor The Laplacian can now be written as d d r, r dr dr and the diusion equation (74) becomes: UNENE, all rights reserved. For educational use only, no assumed liability. Reactor Statics September 04

25 Reactor Statics 5 r d d r B g r 0 r dr dr d r d r.e., g 0. i B r dr r dr This dierential equation is actually well known to mathematicians: it is called Bessel s equation o order 0, and its mathematical solutions are the ordinary Bessel unctions o the irst and second kind, J 0 (Br) and Y 0 (Br) respectively; see Figure 5. (80) Figure 5 Ordinary Bessel unctions o irst and second kind From Figure 5, we can see that Y 0 (Br) goes to - as r 0 and is thereore not acceptable as a physical solution or the neutron lux. The only solution is then: r AJ Bgr (8) 0. The lux must go to 0 at the extrapolated radial boundary R ex, i.e., we must have: J 0 BgRex 0. (8) Figure 5 shows that J 0 (r) has several zeroes, called r i : the irst is at r.405, and the second at r 5.6. However, because the neutron lux cannot have regions o negative values, the only physically acceptable value or B g is As a result, the lux in the ininite cylinder is given by: B g.405. (83) R ex and the buckling or the ininite cylinder is.405r, Rex r AJ 0 (84) UNENE, all rights reserved. For educational use only, no assumed liability. Reactor Statics September 04

26 6 The Essential CANDU.405 Bg. Rex I we use the reactor power P per unit axial height to determine A and neglect the extrapolation distance, the ollowing value is obtained: A P, E R (85) (86) and the absolute lux is 0.738P.405r. r J 0 E R R (87) Parallelepiped reactor We now look at a ully inite reactor geometry (i.e., the reactor is inite in every dimension, and is uniorm): the parallelepiped o sides a, b, c (see Figure 6). In this case, we use the Cartesian r x, y, z. The diusion equation becomes: co-ordinate system, and the lux is written as,,,,,, d x y z d x y z d x y z B,, 0. g x y z (88) dx dy dz Figure 6 Rectangular-parallelepiped reactor To solve this equation, we try a separable orm or the neutron lux, i.e., x y z x g yh z,,, (89) where, g, and h are unctions to be determined. Substituting orm (89) into Eq. (88), we get: d x d g y d h z g( y) h( z) ( x) h( z) ( x) g( y) B g x g( y) h( z) 0. (90) dx dy dz Dividing the equation by (x)g(y)h(z): d x d g y d h z B. g (9) x dx g y dy h z dz The let-hand side o Eq. (9) is a sum o three terms which are unctions only o x, y, and z UNENE, all rights reserved. For educational use only, no assumed liability. Reactor Statics September 04

27 Reactor Statics 7 respectively. The right-hand side is a constant. The only way in which this can happen is i each o the three terms on the let-hand side is a constant on its own, i.e., i we can write: d x d g y d h z B,,, x B y B z (9) x dx g y dy h z dz where B x, By, Bz are constants and Each o the equations in (9), e.g., B B B B (93) x y z g. d x d x B x B x x 0, (94) x dx dx is exactly the same equation as or the slab reactor, with the same solution where x cos B x, (95) B x x. (96) a The ull solution or the neutron lux in the parallelepiped reactor is thereore: ex The quantities x y z x, y, z Acos cos cos. aex bex cex (97) Bx, By, Bz, aex bex cex are called the partial bucklings in the three directions, and the total buckling is (98) B g a b c ex ex ex I we normalize the lux to the total ission power P o the reactor and neglect the extrapolation distance, we can show that the normalization constant, A, is given by A 3 P 8abcE.. (99) (00) Finite-cylinder reactor In the case o a inite-cylinder uniorm reactor o radius R and height H (see Figure 7), we use the cylindrical co-ordinate system to write the lux as r r, z. The Laplacian can be written as: UNENE, all rights reserved. For educational use only, no assumed liability. Reactor Statics September 04

28 8 The Essential CANDU and the diusion equation becomes d d d dr r dr dz, (0),,, d r z d r z d r z B, 0. g r z (0) dr r dr dz Figure 7 Finite-cylinder reactor To solve this equation, we again try a separable orm or the neutron lux, i.e., r z r g z,, (03) where and g are unctions to be determined. Substituting Eq. (03) into Eq. (0), we get: Let us divide this equation by (r)g(z): d r d r d g z g z r B g r g z 0. (04) dr r dr dz d r d r d g z B. g (05) r dr r r dr g z dz The let-hand side o Eq. (05) is the sum o a unction o r and a unction o z. The right-hand side is a constant. The only way in which this can happen is i the parts in r and z are each individually equal to a constant, i.e., i we can write: d r d r d g z B,, r B z (06) r dr r r dr g z dz where B B r, g are constants and B B B (07) r g g. The equations in (06) have been seen beore; they are the same equations as or the ininite cylinder and the slab reactor respectively and thereore have the same solutions: r J 0 Brr, g z cos Bzz, (08) UNENE, all rights reserved. For educational use only, no assumed liability. Reactor Statics September 04

29 Reactor Statics 9 where B r.405 and Bz, (09) R H ex where R ex and H ex are the extrapolated radius and the extrapolated axial dimension o the reactor. The ull solution or the neutron lux in the inite-cylinder reactor is thereore: ex.405r z, cos, Rex Hex r z AJ 0 B r z Rex.405, B Hex are called the radial and axial bucklings respectively, and the total buckling is (0) ().405 Bg. Rex Hex () I we normalize the lux to the total ission power P o the reactor and neglect the extrapolation distance, we can show that the normalization constant is: A 3.63P R HE. (3) Spherical reactor The last geometry we will look at is a spherical uniorm reactor o radius R. We use the spherical co-ordinate system, and because there is spherical symmetry, the lux can be written as r r. The Laplacian can be written as: and the diusion equation becomes r dr Let us try to represent (r) in the orm where the unction r d r dr d, dr r r d d r B g r 0. dr r, (4) (5) r (6) r is to be determined. Equation (5) then reduces to: UNENE, all rights reserved. For educational use only, no assumed liability. Reactor Statics September 04

30 30 The Essential CANDU which inally reduces to d d r r r r Bg 0 r dr r r dr r d d r r r dr dr r i.e., 0 r r Bg d r d r d r r r dr dr dr r i.e, r B 0, g d The general solution o this equation is: which gives dr r r B 0. g (7) r A Bgr C Bgr sin cos, Bgr cos Bgr sin r A C. r r (8) However, the cosine term goes to as r 0, which is physically not acceptable, whereas the sine term is acceptable because it has a inite limit as r 0 (as can be veriied using L Hôpital s rule). Thereore, the inal solution or the spherical reactor is sinbgr r A, r where, or the same reason as in the other geometries (to guarantee no negative lux), B g must take the lowest allowable value, i.e., so that inally r B g, (9) R ex r sin Rex A. (0) R I we normalize the lux by imposing a value P or the total ission power o the reactor and neglect the extrapolation distance, we ind that: P 4 A 4 AR E P R E 4.4 Flux Curvature and Neutron Out-Leakage. () Let us consider the out-leakage term (call it Leak out ) in the above examples: Leak out outleak- UNENE, all rights reserved. For educational use only, no assumed liability. Reactor Statics September 04

31 Reactor Statics 3 age per dierential physical volume dr J D r r uniorm slab reactor (Eq. 76) gives, or x > 0:. Applying this to the lux in a x x cos cos 0. Leakout D A DA aex aex aex0 We can similarly show or all other uniorm reactors that Leak out > 0. This can actually be seen in the general case rom r a r D r 0 (presented earlier as Eq. (48)), which or any point in the uniorm reactor gives D r r DB r 0. This a means that we can say that any point in a uniorm reactor sees neutron out-leakage (i.e., any dierential volume is a source o neutrons). Note again that the above inding applies to uniorm reactors, not to all reactors in general. Real reactors may have regions which are net sinks o neutrons, where the lux curvature is positive and where there is a net in-leakage o neutrons. The general rule that we can iner rom the above and also rom what we learned in source-sink problems is that: Positive lux curvature Net neutron in-leakage, while Negative lux curvature Net neutron out-leakage. 5 Reactors in Two Neutron-Energy Groups with No External Source 5. The Neutron Cycle and the Four-Factor Formula Neutrons born in ission have high energy (see the sketch o their energy distribution in Figure 8). Figure 8 Energies o ission neutrons UNENE, all rights reserved. For educational use only, no assumed liability. Reactor Statics September 04

32 3 The Essential CANDU It can be seen that the peak in their energy distribution is ~0.73 MeV. A -MeV neutron has a speed o ~3,800 km/s: ission neutrons are ast neutrons. There are essentially no ission neutrons born with thermal energies ~ ev; the reerence thermal energy is 0.05 ev, speed,00 m/s, T = 93.6 K. Figure 9 Sketch o cross section versus neutron energy However, the probability o neutrons inducing ission in issile nuclei (such as 35 U) is orders o magnitude larger or thermal neutrons (with energies o a small raction o ev) than or ast neutrons (see Figure 9). This is why thermal reactors (such as CANDU reactors) use a neutron moderator (such as heavy water in CANDU) to slow neutrons down. This leads to the neutron cycle in thermal reactors: neutrons are born ast (i.e., with relatively high energy, typically in the range o MeV) and are encouraged to slow down to the thermal range, where they induce more issions. O course, ast neutrons do induce a small number o issions; in act, only ast neutrons can ission non-issile nuclides (e.g., 38 U), because there is usually a minimum energy threshold or such ission to occur. Moreover, while ollowing the cycle, neutrons can also escape (leak out o the reactor) or be captured in non-productive absorptions at any energy, in particular in the resonance range (between ~ ev and ~0 5 ev), where many strong resonance-absorption peaks (mostly leading to non-productive captures) exist. To reduce resonance capture, most power reactors are designed with the uel lumped into (mostly) cylindrical elements in uel-rod assemblies (inside uel channels in the case o CANDU) surrounded by a moderator. Note also that some up-scattering (increase in neutron energy on collision with a nucleus) can occur in the neutron cycle; this is much less than the rate o down-scattering (moderation) o neutrons, and the up-scattering cross section is sometimes neglected. The neutron cycle in a thermal reactor and with no up-scattering (i.e., scattering rom a low-energy group to a high-energy group neglected) is illustrated in Figure 0. UNENE, all rights reserved. For educational use only, no assumed liability. Reactor Statics September 04

33 Reactor Statics 33 Figure 0 Neutron cycle in thermal reactor We can derive an equation or the ininite-lattice multiplication constant k by ollowing the various phases o the neutron cycle in the neutron-generation view, as ollows: Imagine that N neutrons are born in thermal issions in one generation (each thermal ission produces about.5 neutrons on average). There will also be some neutrons born in ast issions in the same generation. Deine as the ratio o the total number o ission neutrons born in both ast and thermal issions to the number born in thermal issions. is called the ast-ission ratio ( > ). Thereore the total number o ast neutrons is now N. These neutrons start to slow down, but some are captured in non-productive resonances. I we deine p as the resonance-escape probability (p < ), the number o neutrons which survive to thermal energies is now Np. These thermal neutrons can be absorbed in uel or in non-uel components o the medium. I we deine as the ratio o thermal neutrons absorbed in uel nuclides to the total number o thermal neutrons absorbed in this same generation, then the number o neutrons absorbed in uel in the current generation is Np; is called the uel utilization ( ). Some o the neutrons absorbed in uel may induce more issions, and some may not. Deine as the average number o ission neutrons released per thermal absorption event in the uel. Then the total number o ission neutrons born (in this new generation!) rom these thermal issions is now Np. is called the reproduction actor ( > ). We have now gone around the cycle one ull time, and we can compare the number o neutrons in the same phase in successive generations: the ratio in successive generations is k, given by: N p k p. N () UNENE, all rights reserved. For educational use only, no assumed liability. Reactor Statics September 04

34 34 The Essential CANDU 5. The Two-Energy-Group Model The neutron energy range in the neutron cycle is very wide, eight or nine orders o magnitude (rom ~0 MeV to ~0.0 ev). From this it is obvious that the one-neutron-energy-group diusion treatment, while very instructive, cannot be very accurate. Because a very large majority o issions are induced by thermal neutrons and the ission rate is the most important rate to calculate correctly (because it is essentially the heat-production rate), a two-energy-group treatment is oten suiciently accurate in diusion calculations. The nuclear properties in two groups must o course be obtained by proper averaging o the detailed many-group properties determined by a transport-theory calculation. Next, we will analyze the two-energy-group model. 5.3 Neutron Diusion Equation in Two Energy Groups Two luxes With two energy groups, we will need: r and r, with index or the ast (sometimes called the slowingdown) group and index or the thermal group; a r and a r ; r and luxes respectively; Two absorption cross sections, Two neutron-yield cross sections, r, acting on the ast and thermal Down-scattering and up-scattering cross sections, r and r Two diusion coeicients, D r and D r. respectively; The diusion equation or two energy groups is actually two equations, one or each group. There must be neutron balance in each group. As previously indicated, all neutrons are born in the ast group: r r r r r r a r r r r D r r 0, r r r r a 5.4 Uniorm Ininite Medium in Two Energy Groups r r D r r 0. (3) (4) As we did in one energy group, let us irst assume a uniorm reactor, ininite in size. Again, all points in space are equivalent, which means that the neutron luxes are constant through space and there is no leakage rom one point to another; thereore the leakage terms can be dropped rom the equations. The diusion equations (3) and (4) in the ininite uniorm multiplying medium then become: 0 a 0, a which can be written as a linear homogeneous system o two equations: UNENE, all rights reserved. For educational use only, no assumed liability. Reactor Statics September 04

35 Reactor Statics 35 a (5) 0 a 0. (6) Now, this system can have a non-trivial lux solution i and only i the determinant o the lux coeicients is zero, i.e.: a a (7) 0. Satisying the criticality condition, Eq. (7), requires a very ine balance between the nuclear properties. What i that balance is not achieved and Eq. (7) is not satisied? Then there is no non-trivial solution, i.e., we do not have a real operating ininite-medium reactor. However, we can ensure that there is always a non-trivial solution i we tune the neutron-yield cross sections by dividing them by a parameter, k (to be determined). The diusion equations then become: and the criticality condition is: k a 0 k a 0, (8) (9) k a a 0. k From Eq. (30), we can determine the value that k must have or criticality: As beore, we can deine the reactivity as k (30) a. (3) a a a a. k The dierence o k rom unity (or o rom 0) tells us how ar rom critical the original uniorm medium is. When the ast-ission and up-scattering cross sections are neglected, the orm obtained or k is simpler and instructive: The second actor in Eq. (3), k, noupscattering & no ast ission a. (3), a a gives the probability that ast neutrons will down-scatter relative to their probability o being absorbed or down-scattered; i.e., it is the probability o ast neutrons surviving to thermal UNENE, all rights reserved. For educational use only, no assumed liability. Reactor Statics September 04

36 36 The Essential CANDU energies, which is simply the resonance-escape probability p in Eq. (). The irst actor in Eq. (3) is the number o ission neutrons produced per thermal absorption, i.e., it is the reproduction actor. Another important quantity to determine in the two-group model, which has no meaning in the one-group treatment, is the ratio o group luxes. In the uniorm ininite medium, this can be obtained most simply rom Eq. (9):. (33) a Note that the criticality condition (30) ensures that the same value would be obtained rom Eq. (8). 5.5 Uniorm Finite Reactors in Two Energy Groups We now analyze uniorm inite reactors. The diusion equations are now: r r r r r D r a 0 r r r D r a 0. (34) (35) We can try to ind a solution or the lux which is separable in space and energy, i.e., where the two-row lux vector r r r can be written as a group-dependent amplitude times a group-independent lux shape. I we can solve the equation with such a solution, then it is a good solution: r A r. r (36) A r Substituting Eq. (36) into Eq. (35), we get: and i we divide by : A a A A r D A r r 0, r r A aa A (37). (38) D A The right-hand side o Eq. (38) is a single number, independent o space, which we can write as B. [Note: We would have reached a similar conclusion i we had substituted Eq. (36) into Eq. (34).] Equation (38) is analogous to Eq. (6), i.e., B is a group-independent geometric buckling that is applicable to both the ast and the thermal groups: r B r, g,. (39) g Because this is exactly the same equation as we ound with one energy group, the lux distribution in two groups is exactly the same as in one energy group: i.e., a product o cosines in a g UNENE, all rights reserved. For educational use only, no assumed liability. Reactor Statics September 04

37 Reactor Statics 37 parallelepiped reactor, an axial cosine times a radial Bessel unction in a cylindrical reactor, and a sine-over-r unction in a spherical reactor. The only new quantity is the ratio o the group luxes! Let us now return to the criticality conditions or a inite reactor. Substituting Eq. (39) into Eqs. (34) and (35) gives a linear homogeneous system, and thereore we can divide the yield cross sections by k e as usual to ensure a non-trivial solution: k r ( D B ) r 0. a DB r r 0 e ke a (40) (4) The criticality condition or inite reactors in two energy groups is ound by equating the determinant o this system to zero, which yields: k e a DB a DB a DB a DB. (4) I we consider the simpler case obtained by neglecting up-scattering and ast ission, as we did ollowing Eq. (3), we get: k e, noupscattering & no ast ission a DB a DB a DB (43). a DB a D B The ratio o Eq. (43) to Eq. (3) gives the eect o leakage: a D B a D B k k D B D B e a a a a a a. (44) The irst actor is the ratio o (ast absorption + down-scattering) to (ast absorption + downscattering + ast leakage). Thereore, the irst actor represents the ast non-leakage probability, which we can call P NL. In the same way, the second actor represents the thermal non-leakage probability, P NL. Thereore, Eq. (44) is equivalent to and the total non-leakage probability is k k P P (45) e NL NL, P P P (46) NL NL NL. I we use the our-actor ormula or k, then Eq. (46) becomes a six-actor ormula or k e : k p P P (47) e NL NL. UNENE, all rights reserved. For educational use only, no assumed liability. Reactor Statics September 04

38 38 The Essential CANDU 6 Solution or Neutron Flux in a Non-Uniorm Reactor For uniorm (homogeneous) reactors o various geometries, we were able to solve the diusion equation analytically and ind closed orms or the neutron-lux distribution. However, real reactors are not homogeneous. The steady-state neutron-diusion equation must then be solved numerically. This section shows one method or doing this in the two-neutron-energy group case. We start with the two-group neutron-diusion-equation system, Eqs. (3) and (4), which we rewrite with the required k e added as a divisor to the ission cross sections (6.): ( r r r r ) / ke r r r r a r r r r a r r D r r 0 r r D r r 0. (48) (49) All cross sections in a lattice cell are homogenized (averaged spatially within the cell) and condensed to two groups using a multi-group transport code or each lattice cell, but in a real reactor, they must also incorporate the eects o reactivity devices superimposed upon the basic lattice. 6. Finite-Dierence Form o the Neutron-Diusion Equation The simplest method or solving the neutron-diusion equation in a non-homogeneous reactor starts by developing a inite-dierence orm o the equations. Figure Face view o very simple reactor model Consider a two-dimensional view o a reactor model, where each spatial dimension has been subdivided into intervals (see Figure ). The model is then a collection o homogeneous cells (which may generally all have dierent nuclear properties). These cells can be basic lattice cells or subdivisions o these cells. The inite-dierence method then consists o solving or the lux distribution at a single point in each cell, the centre o the cell in the mesh-centred ormulation, which we will use here. UNENE, all rights reserved. For educational use only, no assumed liability. Reactor Statics September 04

39 Reactor Statics 39 Let us look at one o these cells (which we label with a superscript C or Central) and its six nearest neighbours in the three directions, labelled with a superscript n = to 6 ( to 4 in x-y geometry). We get the inite-dierence orm o the diusion equation i we integrate Eqs. (48) and (49) over the volume o cell C: C C r r C C C C r a r D r dr 0 (50) C ke C r C C C a r D r dr 0. (5) C where we have dropped the dependence on r or the properties because the cells are homogeneous. There are two types o integrals in Eqs. (50) and (5). The irst type does not involve the divergence operator " ". In these integrals, the (homogeneous) cross sections can be taken out o the integral sign, or example: C C r r dr C a a. C For the integral on the right-hand side, the approximation is made that the integral is equal to C C C C the value o the lux at the centre o the cell multiplied by the volume V x y z o the C C C cell ( x, y, z being the dimensions o the cell in x, y, and z). Then: C r C C C C a a V. (5) The second type o integral has the divergence operator in the integrand. This type o integral simply represents the leakage out o cell C to its neighbours. By Gauss s theorem, the volume integral over the divergence is equal to a surace integral, or example, C C D r dr D r nds ˆ J nds ˆ, (53) C S S where S is the surace o the cell, nˆ is the outer normal to the surace, and by Fick s Law, J is the net outward current at a point on the surace o the cell. Figure Flux in a cell and in immediate neighbours UNENE, all rights reserved. For educational use only, no assumed liability. Reactor Statics September 04

40 40 The Essential CANDU Figure 3 Linear treatment o lux in central cell and one neighbour Because the cell has six aces, the surace integral is evaluated over the six interaces between cell C and its six neighbours, n = -6. I we take as an example the ace towards the cell with the lowest x-value (n = ) or cell C, we can evaluate the surace integral o the current in the ast group over that ace (which we call ace C) as ollows. First, denote the group- lux at the C centre o the ace by (see Figure ). Then let us assume that the lux is linear between the centre o the cell and the centre o the ace (see Figure 3). Then the group- current rom cell C to its neighbouring cell, at the centre o the ace, is: C C C C d C, C. C J D D dx Similarly, the group- current rom cell to cell C is: x (54) J D, C C. x However, the current must be continuous at the ace, i.e., J., C J, C (55) (56) Then by equating Eqs. (54) and (55) and solving or the interace lux, we ind that: D x D x. C C C C C C D x D x We can then substitute this into Eq. (55) and calculate the current at the centre o the ace: J D D C C C C C, C D C C C D x D x x (57) UNENE, all rights reserved. For educational use only, no assumed liability. Reactor Statics September 04. (58)

41 Reactor Statics 4 I we assume that the average current over the ace is equal to the current at the centre o the ace, the total current over ace C is: C D D y z J nds ˆ A, C C C C C D x D x C C C (59) where A C is a coupling coeicient: D D y z C C C C C C D x D x A. (60) A very similar process can o course be perormed or all six aces between cell C and its neighbours. The total outward current in Eq. (53) is then: 6 Cn C n n J nds ˆ A. (6) S Using all the above results, the inite-dierence neutron-diusion Eqs. (50) and (5) then become: C C C C C C C C C C C Cn C n V V V A ke n 6 a 0 (6) 6 a V V A 0. (63) C C C C C C C Cn C n n These equations couple cell C to its closest neighbours. These and similar equations or all the other cells in the model make up a coupled system o linear homogeneous equations or the luxes at the centres o the cells. I there are N cells in the model, the system is composed o N equations. To solve this coupled system, we need also the boundary conditions at the model edges, which we look at now. The edge cells have neighbour cells only towards the interior o the model. In directions outward rom the model, the diusion boundary condition with vacuum is that the lux goes to zero at the extrapolation distance beyond the boundary. However, we can express the boundary condition in the same orm as Eqs. (6) and (63) by creating a dummy neighbour cell o width d extr, where d extr is the extrapolation length =.3 D C [see Eq. (33)] and orcing the lux to be zero at the extrapolation distance (see Figure 4). UNENE, all rights reserved. For educational use only, no assumed liability. Reactor Statics September 04

42 4 The Essential CANDU Figure 4 Flux in cell at reactor boundary 6. Iterative Solution o the Neutron-Diusion Equation The inite-dierence neutron-diusion-equation system described above is a coupled set o N homogeneous equations, where N is the number o real cells in the model. However, actually we seem to have (N + ) unknowns: two values o lux or each o the N cells, plus the value o k e. In act, however, there are only N unknowns because the overall lux normalization is arbitrary in the homogeneous diusion equation and only relative luxes can be determined. The absolute lux normalization actor must be determined using an extraneous condition such as the total reactor ission power. In real-lie reactors, the number o cells N is very large (typically tens o thousands). Consequently, the system o equations can be solved only by iterative techniques. The irst step is that a lux guess must be selected; this can be as simple as a lat lux distribution, i.e., the same value in all cells. A irst guess must also be made or k e ; or instance, we can use the guess k e =. Once a lux guess has been established, the lux iterations can begin. At each iteration, the computer program progresses rom cell to cell in the x-direction, the y-direction, and inally the z-direction, with each cell considered as the central cell C. The two equations (6) and (63) C are solved simultaneously to obtain the luxes in cell C, C and, using the latest values o the luxes in the neighbouring cells. As the sweep progresses through the cells, all the luxes are updated and used as the latest lux values when the cells are called upon as neighbours. During this sweep, the value o k e is kept constant. However, when a ull sweep over all reactor cells (a ull iteration ) is completed, the value o k e is updated. It can be recalculated rom the deinition o the reactor multiplication constant: UNENE, all rights reserved. For educational use only, no assumed liability. Reactor Statics September 04

43 Reactor Statics 43 k e Neutron Production Neutron Absorption Neutron Leakage [ r r r r ] dr r, a r r a r r J r J r dr r where the latest luxes rom the latest iteration are used and the integrals are evaluated as sums over the cells. One iteration is not suicient to obtain a sel-consistent solution o the entire system o initedierence equations. We must repeat the iterations until the luxes converge, i.e., until the relative dierence in lux in each cell rom one iteration to the next is very small, smaller than an accepted tolerance, typically ~0-5. A convergence criterion is also needed or k e ; typically, a dierence o 0.00 or 0.0 mk rom one iteration to the next is used. Once convergence has been reached, we have the sought-ater solution or the lux shape, i.e., we have the unnormalized (relative) lux distribution. To ind the absolute values or the lux, we can normalize the lux distribution to the total reactor ission power, or example. 7 Energy Dependence o Neutron Flux In Section 4, we studied the spatial dependence o the neutron lux in one energy group in uniorm reactors. In Sections 5 (uniorm reactors) and 6 (non-uniorm reactors), we studied how to determine the neutron lux both in space and in two energy groups. But what can be said about the variation o the neutron lux with energy when treated as a continuous variable (not just in two groups)? We examine this in the present section. When we consider the range o neutron energies rom ~ (or a ew) MeV down to a raction o ev, we can subdivide the lux energy spectrum into three broad regions (note that the region boundaries cannot be considered sharp ): Fission-neutron energies (>~0.5 MeV) Slowing-down (or moderation) range (~0.5 MeV - ~0.65 ev) Thermal range (< ~0.65 ev) We will look at the variation o neutron lux with energy in these three regions in turn. 7. The Fission-Neutron Energy Range This range consists o energies above ~0.5 MeV, up to several MeV, with a maximum lux magnitude around 0.7 MeV. The energy distribution ( spectrum ) o neutrons in this range is determined by experimental measurement. It is ound to be well approximated by:.036 E E 0.453e sinh.9 E, (64) where E is in MeV (see Figure 5). [Note: In Chapter 3, the Watt spectrum is used.] Note that this is a distribution o the number o neutrons, but the lux can be obtained simply by multiplying by the neutron speed (not to be conused with, the ission neutron yield). UNENE, all rights reserved. For educational use only, no assumed liability. Reactor Statics September 04

44 44 The Essential CANDU 7. The Thermal Energy Range Figure 5 Energy distribution o ission neutrons Next, let us look at the thermal energy range, with neutron energies less than ~0.65 ev. Here, the neutrons are in thermal-equilibrium balance with the ambient medium at temperature T. Analogously to the atoms in an ideal gas at thermal equilibrium, the neutron population has a Maxwellian (or approximately Maxwellian) distribution. In terms o lux: E E E (65) kt e. kt Room temperature is by convention taken as T = 93.6 K = 0.4 o C, which gives kt = ev. The energy value E = kt is the most probable neutron energy in a Maxwellian lux distribution, and the corresponding thermal neutron speed is: 7.3 The Slowing-Down Energy Range v kt 0.053eV 00 m / s. (66) The slowing-down energy range is intermediate between the ission-energy range and the thermal-energy range. This is the most complex range because o its very broad size and the highly complex resonance scheme presented to neutrons by heavy nuclides, e.g., 38 U. Thereore, especially when considering neutron absorption in the resonance range in uel, neutron slowing-down can be very diicult to calculate and is now mostly done by numerical computation using complex lattice codes. However, under certain approximations, it is possible to derive an analytic orm or the energy distribution o neutrons as they slow down in the moderator through collisions with light nuclei Slowing down in a hydrogen moderator The simplest case is slowing down in hydrogen. Using analysis o the kinematics o neutron UNENE, all rights reserved. For educational use only, no assumed liability. Reactor Statics September 04

45 Reactor Statics 45 collisions with hydrogen nuclei, the slowing-down lux (E) in hydrogen can be derived. It can be shown that below the lower boundary E s o the ission-neutron energy range and neglecting neutron absorption relative to scattering (a airly good approximation), the slowing-down lux is inversely proportional to energy E: S E, (67) E where S is the ission source and s is the scattering cross section (assumed to be independent o energy). This provides an important, simple, basic ormula or the slowing-down spectrum, even i it is somewhat o an approximation. Another way o interpreting this relationship is that the product E(E) is nearly constant with energy below E s Slowing down in a non-hydrogenous moderator The analysis o the slowing-down spectrum in a non-hydrogenous moderator is more complex, but under similar approximations, the same orm o the slowing-down lux can be ound: E, s S E s (68) This has the same orm as in hydrogen, with an additional actor o in the denominator. is the average lethargy gain per collision, where the lethargy u is deined as: Reer to Chapter 3 or greater detail. E s u ln. E (69) Relaxing the no-absorption approximation I the absorption cross section is no longer neglected, but is assumed to vary smoothly with energy (i.e., in the absence o resonances), the /E lux spectrum can be modiied to: E ' ' E S s a E de ' exp. Et E (70) t E E ' E The exponential actor in Eq. (70) represents the probability that the neutron survives slowing down to energy E; i.e., it is the resonance-escape probability to energy E, which we can denote as p(e). From the general orm o the slowing-down lux, the ollowing simpliied statements about the product E(E) can be deduced: I absorption is neglected, and under the assumption that the scattering cross section does not depend on energy, E(E) is constant (lat) with E. I absorption is included, and assuming a smooth variation o the absorption cross section with E, then E(E) will decrease smoothly or decreasing E. These statements are shown in graphical orm in Figure 6. UNENE, all rights reserved. For educational use only, no assumed liability. Reactor Statics September 04

46 46 The Essential CANDU Figure 6 Variation o lux with energy i absorption is smooth Slowing down in the presence o resonances Let us now consider absorption resonances. At a resonance energy, the neutron lux decreases signiicantly because o the very high absorption cross section at that energy. As a result o the dip in lux, the absorption rate in the resonance (which is proportional to the lux) is reduced relative to the absorption rate with the background lux value at energies above and below the resonance, i.e., the product a is smaller than i the lux were unaected. This is called resonance sel-shielding. On account o resonances, then, the slowing-down lux will be distorted relative to the smooth curve in the previous subsection, as shown in Figure 7. Figure 7 Variation o lux with energy in the presence o resonances 7.4 Neutron Flux over the Full Energy Range With the results in the previous subsections, we are able to piece together the neutron lux over the energy range rom ission energies to the thermal range, using: the ission spectrum at energies above about 500 kev; UNENE, all rights reserved. For educational use only, no assumed liability. Reactor Statics September 04

47 Reactor Statics 47 the slowing-down spectrum to about 0.65 ev. [Note that in the thermal energy range, neutrons can gain as well as lose energy in collisions. To be consistent with the approximation o no up-scattering in the derivation o the slowing-down spectrum, the boundary between thermal and epithermal energies should be selected suiciently high to ensure negligible up-scattering rom the thermal region to the epithermal region. This is one reason or the typical choice o 0.65 ev, which is about 5 times the most probable energy o 0.05 ev at room temperature, as the lower energy boundary o the epithermal range.] the Maxwellian spectrum at thermal energies. [Note that it is not a perect Maxwellian, being distorted somewhat by neutron absorption]. The piecing together o the neutron-lux portions in the various energy regions is shown in the sketch in Figure 8. Figure 8 Sketch o lux variation over ull energy range 8 Basic Design o Standard CANDU Reactors In this section, we look at the basic eatures o the standard CANDU reactor design. The CANDU design is based on the ollowing design principles and the rationale or each: Use natural uranium (NU) as uel. The reason or this choice is that Canada has substantial, high-grade, domestic uranium mineral resources and, or a reactor cooled and moderated by heavy water (D O), it is not necessary to use enriched uranium uel. Developing a uranium-enrichment industry is extremely expensive, primarily because o the sophisticated technology involved as well as the costs associated with the stringent materials saeguards, physical security, and criticality saety requirements associated with enriched nuclear materials. Moreover, the production cost o uranium-isotope enrichment is also high using current technology, in part due to the energy requirements. Consequently, natural-uranium UNENE, all rights reserved. For educational use only, no assumed liability. Reactor Statics September 04

48 48 The Essential CANDU uel is much cheaper and simpler to abricate than enriched-uranium uel. Use heavy water (D O) as neutron moderator and reactor coolant. The ission chain reaction cannot be sel-sustaining with light water (H O) as moderator and NU as uel. The reason is that normal hydrogen (H) has a airly large absorption cross section or thermal neutrons. Hence, light water robs neutrons rom the chain reaction, and the uel must be enriched to restore a sel-sustaining balance between neutron production and capture. On the other hand, deuterium (D), or heavy hydrogen, has a very small neutron-absorption cross section and enables the use o NU uel: heavy water provides high neutron economy. Although D O is very expensive to produce due to the low abundance o D relative to H in nature, the large mass dierence between D and H makes separation relatively easy to achieve using chemical technology processes, and, unlike enriched uel, the D O retains its avourable nuclear properties or the ull lie o the reactor. Use a pressure-tube rather than a pressure-vessel design. The coolant takes heat away rom the uel (to create steam and produce electricity in a turbine generator) and becomes very hot. I we want to keep the coolant liquid (or to limit boiling to a small amount), we must pressurize it (to about 00 atmospheres, ~0 MPa). Thereore, a pressure boundary must be provided. In light-water reactors (LWRs) a large, thick, steel pressure vessel is used that surrounds the entire reactor core. The irst prototype CANDU reactor [Nuclear Power Demonstration (NPD), the irst reactor to produce electricity in Canada (about 0 MWe)], was initially conceived with a pressure-vessel design. However, using D O as the moderator results in a larger vessel than or an LWR, because the smaller scattering cross section o D O (compared to that o H O) requires a larger moderator volume. Concerns were expressed about the huge size o the pressure vessels which would be required or larger CANDU reactors, because such vessels would be diicult to abricate and very expensive. The design o NPD, and that o all later CANDU reactors, was thereore changed to a pressure-tube design, in which the pressure boundary occurs in a small, relatively thin, uel channel that surrounds each uel assembly module and its associated coolant. A pressuretube design was made possible only by emerging research on zirconium, which showed that zirconium had a very small neutron-absorption cross section and would be a very good candidate metal or the pressure tubes. Using steel, with its much larger absorption cross section, would not permit a working pressure-tube design or a reactor using natural-uranium uel. The pressure-tube design has the additional saety eature that the break o one pressure tube is more manageable than the break o a large pressure vessel. Some additional CANDU design eatures and some consequences o the design decisions are explored here: The natural-uranium lattice has low excess reactivity because the uel is not enriched. This is a saety advantage because it limits the amount o reactivity increase available in an accident. It is also another rationale or adopting a pressure-tube design which allows on-power reuelling o the reactor to sustain the core reactivity without shutting down the reactor. Otherwise, batch reuelling o a CANDU reactor in a pressure vessel would be required every ew months, which would lower its capacity actor on account o the lost operating time during reuelling and during the associated, time-consuming, shutdown/cool-down and start-up/heat-up transients. CANDU NU uel achieves a discharge burnup in the ~7,500 ~9,500 MW.d/Mg(U) range, depending on the speciic reactor design. This is much lower than the dis- UNENE, all rights reserved. For educational use only, no assumed liability. Reactor Statics September 04

49 Reactor Statics 49 charge burnup in LWRs (~30,000 50,000 MW.d/Mg(U)). However, considering uel enrichment or LWRs (about 3.5% 5% 35 U) and the depleted-uranium tailings which are discarded, CANDU reactors are more eicient than LWRs and extract ~5% more energy per Mg o mined natural uranium. The pressure-tube design means that the main reactor vessel (the calandria) is not pressurized. The moderator is insulated rom the hot coolant by placing a calandria tube concentric with each pressure tube, with an insulating gas in the intermediate (annulus) space. The moderator can be maintained relatively cool at ~70 C by circulating it in its own heat-exchanger circuit, which means that more o the neutrons will have lower thermal energies than in an LWR and consequently will more readily induce ission in issile nuclides, e.g., 35 U and 39 Pu. CANDU reactivity devices are located interstitially between pressure tubes and are thereore in a benign (low-temperature and unpressurized) environment, which is a saety advantage. CANDU reactivity devices are not subject to being ejected rom the reactor by coolant pressure. Heavy water is used as coolant instead o light water in all operating CANDU reactors or additional neutron economy. Light water was used as coolant in the Gentilly- prototype, and organic coolant was used in the Whiteshell Reactor (WR-) engineering test unit, both o which are no longer in operation. Light water is also used as coolant in the proposed Advanced CANDU Reactor (ACR) design, to reduce capital costs associated with heavy water, and also in the Supercritical-Water-Cooled Reactor (SCWR) design. All operating CANDU reactors have horizontal pressure tubes (the Gentilly- prototype and the WR- had vertical pressure tubes). This orientation promotes symmetry because coolant can be circulated in opposite directions in alternate tubes (i.e., using bi-directional coolant low), making average neutronics conditions essentially identical at the two ends o the reactor (unlike the situation in LWRs). With vertical pressure tubes, the gradient in coolant temperature and density as the coolant picks up heat and moves upward makes the neutronic and thermal-hydraulic conditions asymmetric. Horizontal pressure tubes also acilitate bi-directional reuelling (i.e., uelling in opposite directions in adjacent channels), which urther promotes symmetry. When the uel is manuactured as short (~50-cm-long) uel bundles which can easily be pushed through the pressure tube rom either end, the associated uelling machines will also be short, less cumbersome, and require less radiation shielding. With vertical tubes, the entire uel string (consisting o a vertical stack o perhaps or 3 bundles) would have to be removed rom the top o the reactor, so that a very long uel assembly would have to be used (together with a long and heavy transer lask), or short uel bundles would have to be connected to a common stringer. Whereas the loss o moderator shuts any (thermal) reactor down, the loss o coolant in a CANDU reactor does not have the same eect because the coolant is separated rom the moderator. Instead, the loss o coolant in operating CANDU reactors generates a positive reactivity insertion (called the coolant void reactivity, or CVR); this is due to a number o spectral eects which change mostly the ast-ission actor and the resonance-escape probability. As a consequence, a loss-o-coolant accident in CANDU results in a positive power pulse, which must be quickly turned around by UNENE, all rights reserved. For educational use only, no assumed liability. Reactor Statics September 04

50 50 The Essential CANDU shutdown-system action to avoid overheating the uel. In act, the postulated largeloss-o-coolant accident (LLOCA) scenario is the reason or the adoption o redundancy in CANDU shutdown capability design. CANDU reactors have two ast shutdown systems, which are physically and logically independent o one another and each ully capable o shutting the reactor down rom any credible coniguration. Shutdown system consists o cadmium shut-o rods which all under gravity (initially spring-assisted) into the reactor rom above. Shutdown system consists o the injection o a solution o neutron-absorbing gadolinium (a high neutron absorber) under high pressure through nozzles directly into the moderator. The lattice pitch (distance between the centres o neighbouring tubes) in all operating CANDU reactors is cm. This is not the optimum value in the sense o maximizing the lattice reactivity (thereore minimizing the reuelling rate and maximizing the average uel discharge burnup), which is closer to 34 cm or so. However, the larger volume o D O moderator would result in a higher capital cost; the shorter pitch was selected to minimize the levellized unit-energy cost. A shorter lattice pitch o 4 cm was selected or the ACR (to reduce moderator cost, among other reasons), but the reduced moderation would not allow the chain reaction to be sel-sustaining with natural-uranium uel, and the ACR would need to use enriched uel. The workhorse control devices in all operating CANDU reactors are liquid zone controllers. These control 4 compartments in which amounts o light water (used or its much higher neutron-absorption cross section than heavy water) can be varied uniormly across all compartments to control reactivity or to shape the power distribution dierentially. 9 CANDU Reactor Physics Computational Scheme Because o the strong heterogeneity o reactor lattices (see the example o the CANDU basic lattice cell in Figure 9) and because nuclear uel is a strong neutron absorber, the diusion equation cannot be used in reactors based on the detailed lattice geometry. For cases where it is not desired to apply neutron transport in ull-core reactor models, a two-stage or three-stage process has been developed to enable calculation o the neutron lux and the power distribution using the diusion equation. This process is illustrated here or CANDU reactors. UNENE, all rights reserved. For educational use only, no assumed liability. Reactor Statics September 04

51 Reactor Statics 5 Figure 9 CANDU basic lattice cell The computational scheme or CANDU neutronics consists o three stages. Separate computer programs have been developed to perorm the calculations corresponding to each stage. The three stages are: lattice-cell calculation reactivity-device (supercell) calculation inite-core calculation. 9. Lattice Calculation The irst stage involves solving or the lux distribution in the basic lattice cell, which consists o the uel, coolant, pressure and calandria tubes, and moderator, but no reactivity devices (see Figure 9). This calculation is perormed using a deterministic multi-group neutron transport code such as WIMS-AECL. Then the transport code averages the nuclear properties (or absorption, moderation, ission, etc.) over the cell according to the calculated reaction rates in each sub-region o the lattice cells to determine the eective neutronic properties o the whole lattice cell (i.e., ew (typically )-group macroscopic cross sections and diusion coeicients). This averaging (homogenization) o the properties o the basic lattice cell dilutes the strong absorption o the uel with the much weaker absorption in the moderator. With these homogenized properties, Fick s Law becomes a good approximation over most o the reactor model, and the diusion equation can be used to calculate the ull-core neutron-lux and power distributions. The lattice properties govern the neutron-multiplying behaviour o the reactor lattice, and thereore they must be obtained or all ages o the reactor uel, i.e., the transport code must perorm a depletion calculation to evolve the lattice-cell neutronic properties to relect the changes in the nuclide composition o the uel with irradiation/burnup. The lattice properties must also relect any modiied conigurations o the lattice. For example, i the coolant is assumed voided in a hypothetical loss-o-coolant accident, or the temperature o the material in a sub-region is assumed to change, then the lattice properties must change accordingly. UNENE, all rights reserved. For educational use only, no assumed liability. Reactor Statics September 04

52 5 The Essential CANDU 9. Reactivity-Device Calculation This type o calculation determines the eect o a reactivity device on the nuclear properties in its vicinity. This eect is expressed in the orm o incremental cross sections which are added to the (unperturbed) homogenized properties o neighbouring lattice cells in a modelled volume around the device, called a supercell ; see Figure 0. For instance, shuto-rod incremental cross sections dictate the local eicacy o the shuto rods in absorbing neutrons and shutting down the ission chain reaction. The incremental cross sections o a device are obtained by calculating the dierences in the supercell homogenized properties between the case when the device is inserted into the supercell and the case when it is withdrawn rom it. The incremental cross sections are added to the lattice-cell cross sections in the modelled volume around the device s position in the reactor core, when that particular device is inserted. Figure 0 Supercell or calculating device incremental cross sections Because the supercell also contains highly absorbing material, transport theory must again be used. CANDU reactivity devices are perpendicular to the uel channels. Thereore, realistic supercell models are three-dimensional, and as a result supercell calculations are best done using the DRAGON transport code, which can perorm calculations in three dimensions. 9.3 Full-Core Calculation In the inal stage, a ull-core reactor model is assembled. The ull-core model must incorporate, or each cell in the reactor, the homogenized unperturbed-lattice-cell cross sections which apply there, together with the incremental cross sections o nearby reactivity devices in their appropriate locations. The model thereore consists o homogenized basic-lattice cells on which are superimposed homogenized subcells representing reactivity devices. One o many such device model areas is shown in Figure ; in the white subcells, the nuclear cross sections are obtained by summing the basic-cell properties and the appropriate device incremental properties. UNENE, all rights reserved. For educational use only, no assumed liability. Reactor Statics September 04

53 Reactor Statics 53 A ull, 3-D reactor model constructed in this way is used to calculate the three-dimensional lux and power distributions in the core. Because homogenized properties are used, ew-group diusion theory can be applied to the ull-core reactor model. Figure Simple model with superimposed reactivity device 0 Evolution o Lattice Properties When nuclear uel is burned in a reactor, changes occur in the material composition o the uel. These changes are generally called uel (isotopic) depletion. In this section, we look at how the properties o the CANDU lattice evolve with uel irradiation (or burnup). The ollowing changes occur cumulatively in CANDU nuclear uel with time: The 35U depletes (i.e., its concentration, which starts at 0.7 atom% or resh natural uranium, decreases). Fission products accumulate; most o these are radioactive, and many have a signiicant neutron-absorption cross section. 39 Pu is produced by neutron absorption in 38U and two subsequent beta decays: 38 U n 39 U* 39 Np 39 Np 39 Pu. 35 Pu participates strongly in the ission chain reaction (because it is issile, like 35U), but while it continues to be created at about the same rate rom 38U, its net rate o increase slows. Further neutron absorptions lead rom 39Pu to 40Pu (non-issile), and then to 4Pu (issile). Other higher actinides are also ormed (e.g., curium, americium). The total issile raction in the uel (35U + 39Pu + 4Pu) decreases monotonically. The evolution o the concentration o these three nuclides is shown in Figure. A typical graph o reactivity o the ininite CANDU lattice versus uel irradiation, obtained rom a cell calculation using a transport code, is shown in Figure 3. The ollowing points are o note: UNENE, all rights reserved. For educational use only, no assumed liability. Reactor Statics September 04

54 54 The Essential CANDU The resh-uel ininite lattice (where the uel has not yet received any irradiation) has a high reactivity (~78 mk when the 35Xe and other saturating ission products have built up). To achieve a steady state in the ininite lattice with resh uel, a corresponding amount o negative reactivity must be added to the lattice [e.g., by dissolving a neutron poison (i.e., a material with a large neutron absorption cross section) in the moderator] to suppress the initial supercriticality. The reactivity starts to decrease immediately on account o 35U depletion. It then starts to increase or a while, on account o production o 39Pu, which is slightly more eective than 35U. Note the slight delay due to the 39Np ~-day hallie. However the rate o increase o reactivity slows (because the net rate o plutonium production decreases), and the reactivity proceeds through a maximum, called the plutonium peak, with increasing burnup (note that this is not a peak in 39Pu concentration, but in lattice-cell reactivity!). Following the plutonium peak, the reactivity decreases monotonically on account o the continuing depletion o 35U and the continuing accumulation o ission products. The ininite lattice reaches zero reactivity at an irradiation corresponding to a burnup o ~6,700 MW.d/Mg(U). A homogeneous ininite lattice with uel beyond that burnup would be subcritical. Figure Evolution o uel isotopic densities UNENE, all rights reserved. For educational use only, no assumed liability. Reactor Statics September 04

55 Reactor Statics 55 Figure 3 Ininite-lattice reactivity vs. irradiation However, remember that the ininite lattice does not experience reactivity losses due to neutron leakage; moreover, it does not account or neutron absorption in all the reactivity devices within the core. Consequently, a homogeneous reactor with all uel at the same irradiation would reach zero reactivity at a much lower burnup than the ininite lattice. In the CANDU-6 reactor, leakage is about -30 mk, and the average in-core reactivity-device load is ~-8 mk. Thereore, an estimate o the reactivity e o a inite homogeneous CANDU-6 reactor can be obtained by subtracting 48 mk rom the reactivity o the ininite lattice; see Figure 4. Figure 4 Finite-reactor reactivity vs. irradiation In a homogeneous reactor, a signiicant amount o additional negative reactivity (~30 mk) is required to suppress the initial supercriticality. A homogeneous CANDU-6 would reach zero reactivity at an irradiation corresponding to a burnup o ~4,000 MW.d/Mg(U). A homogeneous CANDU-6 with uel beyond that burnup would be subcritical [note that the argument is hypothetical in any case because a homogeneous inite reactor would not remain homogeneous with burnup]. Thereore, i the CANDU-6 were to be batch-reuelled, the discharge burnup would be only ~4,000 MW.d/Mg(U). UNENE, all rights reserved. For educational use only, no assumed liability. Reactor Statics September 04

56 56 The Essential CANDU However, CANDU reactors are reuelled on-power, and thereore there is always (except near start o lie) a mixture o resh uel and uel with high irradiation. The uel with high irradiation has negative local reactivity, but this is compensated or by the positive local reactivity o lowirradiation uel. The proper mixture o uel in this inhomogeneous reactor (obtained by the proper rate o daily reuelling) maintains the reactor critical day-to-day. Physically, the older uel does the job o reducing the high reactivity o the young uel, a job which moderator poison does in the batch-reuelled reactor. The dierence is that whereas the poison is just a parasitic absorber, the older uel does provide issions and thereore additional energy. The mixture o new and old uel makes it possible to drive the discharge burnup to a much higher value than that deduced rom the homogeneous reactivity curve. We can guess (or calculate) approximately how ar we can drive the exit burnup by determining what value gives equal positive and negative areas under the reactivity curve [this tells us where the average e would be 0]; see Figure 5. From the igure, we can see that positive and negative areas are equal when the average exit (or discharge) burnup to which we can take the uel with daily reuelling is ~7,500 MW.d/Mg(U). This is almost twice the discharge-to-burnup value attainable in the batch-reuelled reactor and represents quite a beneit provided by on-power reuelling! Figure 5 Averaging reactivity with daily reuelling Summary o Relationship to Other Chapters The Neutron Physics chapter explains the interactions that neutrons have with matter and deines all the basic quantities needed in the study o the ission chain reaction and the analysis o reactors. The present chapter covers mostly the time-independent neutron-diusion equation and the analysis o steady-state (time-independent) neutron distributions, either in the presence o external sources or in ission reactors. It has not covered time-dependent phenomena, which are let or the ollowing chapter. The Reactor Kinetics chapter covers the time-dependent neutron-diusion equations and UNENE, all rights reserved. For educational use only, no assumed liability. Reactor Statics September 04