Как выбрать гостиницу для кошек
14 декабря, 2021
As the preceding section on fission emphasized, each fission reaction of an U235 atom leads to the release of an average of 2.5 neutrons. Hence, to sustain a continuous fission reaction (i. e., a chain reaction), these neutrons should be able to initiate the fission of at least another fissile atom. Note the schematic of a chain reaction involving U235 atoms in Figure 1.1. A majority of the neutrons (~99.25%) produced due to the fission reaction of U235, known as prompt neutrons, are released instantaneously (within 10-14 s). But there are about 0.75% neutrons that are released over
Figure 1.1 A schematic chain reaction of U235 fissile atoms in progress (for the sake of simplicity, it is assumed that two neutrons are released due to the fission of one U235 atom and fission fragments created in each fission are also not shown). |
a longer period (over ~20 s) and these neutrons are called delayed neutrons. These delayed neutrons play a very important role in controlling the fission chain reaction.
There is always a competition for neutrons between various processes, namely, (i) fission reaction of fissile atom nuclei, (ii) nonfission capture of neutrons by uranium and other reactions, (iii) nonfission capture of neutrons by other components in the reactor core, and (iv) leakage of neutrons from the core. The reaction can be termed as a chain reaction when the number of neutrons consumed in the processes (ii)-(iv) is at least equal to or less than that consumed in the process (i). Thus, neutron economy plays a very important role in the design of a nuclear reactor. The need for a favorable neutron economy necessitates certain conditions to be met by a chain reacting system. For a given geometry, there is a certain minimum size of a chain reacting system, called the critical size (in terms of volume), for which the production of neutrons by fission just balances the loss of neutrons by leakage and so on, and the chain reaction can be sustained independently. The mass corresponding to the critical size is called critical mass. Dependent on the relative generation of fission neutrons and their loss, the reactor is said to be in different stages: subcritical (neutron loss more than the production), critical (balance between the neutron production and loss, к = 1), or supercritical (the neutron production is more than the loss, к > 1). The multiplication factor к is often used to express the criticality condition of a reactor. This factor is basically the net number of neutrons per initial neutron.
A lot of space in crystals is empty and not occupied by lattice atoms(recall the definition of atomic packing factor). However, understanding interstitial sites in all the common lattice structures is an important step to understanding more complex crystal structures. Furthermore, interstitial impurity/alloying elements and self-interstitial atoms generally locate themselves in such sites. Figure 2.15a illustrates the octahedral position at the center of an FCC unit cell. Similar octahedral locations can be imagined at the center of each cube edge indicated by an “open” circle. However, to depict fully these octahedral positions just as shown in the center of the cube, one needs to construct additional neighboring unit cells. Tetrahedral interstitial positions (indicated by the “open” circles) of the FCC unit cell are shown in Figure 2.15b. Note that all the tetrahedral positions are entirely within the unit cell unlike the octahedral positions. Figures 2.16 and 2.17 show similar octahedral and tetrahedral sites in both BCC and HCP unit cells.
Figure 2.16 BCC unit cell. (a) Octahedral positions. (b) Tetrahedral positions (the “filled” circles represent the lattice atoms and the “open” circles interstitial positions).
2.1.7
A simple model for calculating the atomic displacements is due to Kinchin and Pease [3], known as the Kinchin-Pease model. Before discussing this model, let us consider a case of collision between a high-energy neutron of mass M1 and a lattice atom of mass M2 (Figure 3.5). At certain energy ranges, neutron-nucleus interaction can be described by an elastic scattering process, where both kinetic energy and momentum of the particles are conserved before and after the collision. In the majority of the events like this, binary (two-body) collision is an appropriate approximation. A neutron has no electrical charge (i. e., neutral particle), and hence does not get perturbed by electrical fields exerted by the nucleus or electrons.
Let E1 be the energy of the neutron and due to the collision, the atom is scattered with energy T transferred from the neutron. There exists a threshold energy Ed for the atom to be displaced that is related to the binding energy of the atom. Thus, the struck atom will be displaced only if the transferred energy T is at least equal to or greater than this threshold value. As we have noted before, the displacement threshold energy is very small on the order of ~25 eV, while the transferred energy might be in keV or MeV range. Thus, there is a very high probability that the struck atom will get displaced and become a PKA. This PKA with reasonably high energy will thus interact with other atoms and displace them from their positions leading to secondary, tertiary, and other higher order knock-on atoms.
In the collision process between the incoming high-energy particle and the atom being displaced, if the knocked-on atom flies away with energy, T the incoming neutron will have energy E1 — T — Ed, at a scattering angle of в. In general, however, the energy consumed in the scattering process is assumed negligible and thus the scattered neutron will have energy equal to E1 — T. Of course, the neutron
Figure 3.5 Various characteristics of an elastic collision event are shown in (a) the laboratory coordinates and (b) in the center-of-mass coordinates. |
with energy [E1 — T] and the PKA with energy Tproduce other knock-on atoms. A simple calculation can show the likelihood of neutron-nucleus interaction. If we consider a neutron scattering cross section (a) as 1 b (i. e., 10—24 cm) and the number density of atoms (N) in a material as 0.85 x 1023 cm—3 (typical of alpha-Fe), the distance between collisions, that is, the mean free path of the neutron (calculated by 1/Na) will be several centimeters (in this case, ~11.8 cm).
Analytical derivation based on the assumption of the kinetic energy and momentum conservation for an isotropic elastic scattering event between a neutron and an atom nucleus gives the following relations:
where Л — (4M1M2 )/(M1 + M2)2 and En is a generalized form of average incident neutron energy and is essentially equal to E10.
From Eq. (3.8), it is clear that the value of Tdepends on the scattering angle, the energy of the incident particle and the masses of the collided particles.
We can see that the maximum energy that can be transferred is given for the angle of scattering of 180°. In other words, the minimum possible value of cos(e), that is, —1, gives the transferred energy the highest value (Tmax), which is then given by the following relation:
T LE
max — L n
Of course, the minimum energy (Tmin) that can be transferred would be zero given a в value of0°.
Table 3.2 A summary of maximum percentage energy transfer that takes place from a 2 MeV neutron to different nuclides in the event of an isotropic elastic scattering (mass of neutron = 1.008665 amu).
|
However, the neutron-nucleus elastic scattering event could lead to glancing collisions. In that case, the use of average recoil energy (Tavg) is more appropriate:
Tavg = AEn • (3.10)
Table 3.2 lists the maximum percentage energy that can be transferred to different nuclides in the event of an elastic collision between a fast neutron (kinetic energy of 2 MeV) and the nuclides. One important observation from this table could be that the neutron transfers more energy to lighter nuclei. Thus, nuclides with low mass numbers generally act as good moderators. As we have discussed in brief, the materials often used are light water (H2O), heavy water (D2O), and graphite (C).
Even though earlier in the chapter we assumed that neutrons are scattered through every possible direction (Isotropic Scattering), in reality they tend to scatter preferentially in the forward directions, and hence the scattering behavior is anisotropic. That is why Eqs (3.9) and (3.10) may need to be corrected using a correction factor (f). The value off varies with the nuclide type; for example, for Be it is 0.56, whereas for C and Cu they are 0.84 and 0.6, respectively.
A beryllium (Be) reflector is exposed to a neutron of 1 MeV energy. Let us assume the masses of the neutron and the Be atom are 1 and 9.01 amu, respectively. (a) Calculate the maximum and average energies transferred from the neutron to the beryllium atom based on isotropic elastic scattering event. (b) Determine the average energy transferred if the anisotropic scattering occurs (f— 0.56).
Solution
From Eq. (3.8), we know that Tmax = AEn, where Л = (4M1M2)/ (M1 + M2 )2.
a) In order to find out Tmax, we need to first calculate Л. Given M1 = 1 amu and M2 = 9.01 amu,
(4)(1)(9.01) (1 + 9.01)2
Hence, Tmax = (0.36) x (1 MeV) = 0.36 MeV.
Tavg = -^ = 0.18 MeV.
b) For anisotropic elastic scattering event, Taniso(avg) = f (LEn/2) = (0.56 x 0.18 MeV) = ~0.1 MeV.
Now, let us try to get an expression of the displacement damage rate (Rd) defined as the number of displacements per unit volume (cm3) per second. Rd is basically proportional to the number of target atoms per cm3, N (i. e., number density of atoms), and the displacement cross section od(En) for neutrons with energy En, and the neutron flux W(En), and can generally be written as
Rd = N ■ Od(En)- W(En). (3.11)
The important unit of radiation damage, displacements per atom (or dpa), can now be defined as
for neutrons with energies varying from 0 to 1.
Note that the neutrons with En < Ed/L do not generate any displaced atoms since Tshould be >Ed (i. e., the maximum transferable energy given by LEn > Ed to produce a knock-on atom).
The displacement cross section is a function of the sum of the number of atomic displacements, n(T), produced by PKAs with energies T from Ed (the minimum needed) to Tmax(= LEn). The interaction probability is given by the differential energy transfer cross section, od(En, T)dE, for producing a PKA with energy (T, dT) due to the interaction with a neutron of energy En.
Therefore, we can write
Г LEn
Od(En)= On(En, T)■ n(T)dT. (3.13)
Ed
Note that the minimum energy of the PKA to produce displacements is Ed (Л = 1) since scattering between like atoms and the maximum energy of PKA is Tmax = LEn.
The Kinchin-Pease model is the simplest of all radiation damage models. It gives an estimate of the number of displacements produced by a PKA. The model is based on certain assumptions:
1) The cascade is created due to a sequence of binary (two-body) elastic collisions and the potential is based on the hard sphere model approximation.
2) Atomic displacements occur only when T> Ed (i. e., no quantum effects).
3) No energy is passed to the lattice during the collision phase.
4) Energy loss by electronic stopping is given by a cutoff energy known as electronic cutoff energy (Ec). If the PKA energy is more than Ec, no additional displacements occur until electron energy losses reduce the PKA energy to Ec. For all energies less than Ec, electronic stopping is ignored and only atomic collisions take place.
5) Atomic arrangement is random (i. e., the effect of crystal structure is neglected).
6)
No annihilation of defects is assumed.
For the derivation of the K-P model, readers are referred to Refs [4, 5]. Here, we present the end result of the K-P model derivation:
According to Eq. (3.14), the number of displaced atoms can be plotted as a function of PKA energy, as shown in Figure 3.6. The relation (3.14a) is self-explanatory from the very definition of Ed. That is, if T is less than Ed, we expect no atom displacement (i. e., n(T) = 0). However, for the relation (3.14b), when Tis greater than Ed but less than 2Ed, there could be two possibilities. One scenario could be that the struck atom does not get energy Ed from the PKA and stays in the same place. Second scenario could be that the struck atom gets more than Ed energy to be dislodged from the lattice site. But the PKA now with less than Ed energy falls back to
the original site of the lattice atom. So, in either case, only one displacement takes place. When the PKA energy is greater than 2Ed but less than Ec, the number of displacements increases monotonically with PKA energy until it reaches the electronic cutoff energy (Ec). Beyond that, the number of displacements does not change. Even though in the simple K-P model it was assumed that all energy losses of the PKA go toward elastic collisions, with the increasing PKA energy, a greater fraction of energy will be lost into electronic excitation and ionization. In this case, we need to recognize that PKAs are basically fast ions. So, the number of displacements calculated at higher PKA energy would not be all consumed in creating the displacement damage. There are several modifications to the simple K-P model depending on the relaxation of the various assumptions.
Now getting back to the expression of Rd developed in Eq. (3.11) and combining with Eq. (3.13), we get the following expression:
From the theory of isotropic elastic scattering and hard sphere approximation, we can get a relation between double differential energy transfer cross section and scattering cross section:
where elastic collision cross section weakly depends on the neutron energy (En). Thus, it can be shown that
and
where Ф is the total integrated neutron flux given by
Ф = / W(En)dEn.
J0
|
|
■ Example 3.3
Consider 0.5 MeV neutrons (ael = 3 b) with a flux of 5 x 1015ncm-2s-1 interacting with alpha-iron (atomic mass number 56, lattice constant 0.287 nm) target. Find Rd.
Solution
Given are En = 0.5 MeV, ael = 3 b, Ed = 24 eV, Ф = 5 x 1015 n cm-2 s-1, and a0 = 0.287 nm.
From Eq. (3.17), we have
We have values of all the terms in the above equation except N and Л.
For Fe, we can calculate the number density of atoms (N) in the target material. There are two ways to do it.
1) If the material is a crystalline solid and we know its crystal structure (here, alpha-Fe that is BCC), we first find out the volume (Vcell) of the unit cell of alpha-iron from its lattice parameter (a0 = 0.287 nm). We know that the effective number of atoms (q) in a BCC unit cell is 2. So, N = q/Vcell = 2/a03 = 2/(0.287 nm)3 = 0.85 x 1023 atoms per cm3.
2) Another way to calculate N can be applied to all state of matter. This has been explained in Chapter 1. This needs knowledge of density and atomic mass of the material.
For Л, we use
Hence,
(0.85 x 1023 cm-3)(0.069)(3 x 10-24 cm2) , , w 2
Rd =±—————— 4(24 eV) —————- “ ^0’5 x 106 eV (5 x 1015 n cm-2 s^1)
=~ 4.5 x 1017 displacements cm-3 s_1.
Now if we divide Rd just by N, we obtain ~5 x 10 6dpas x.
We can also find the dpa value by multiplying the above number by the days of neutron exposure (say, 30 days), we get a dpa of (~5 x 10-6 dpa s-1). (30 x 24 x 3600s) = ~13.0.
Furthermore, if we wish to calculate the number of displaced atoms per neutron, we can use the following expression:
ЛЕn (0.069)(0.5 x 106eV)
4Ed = 4(24 eV)
Summary
In this chapter, we focused our attention on understanding the concepts of primary radiation damage involving neutron-nucleus collision and subsequent PKA and lattice atoms interaction leading to the formation of displacement cascades. The displacement threshold concept is elucidated with a simple example. The value of displacement energy is generally about 25 eV. Then, the K-P model is introduced. This gives the number of atomic displacements produced by a PKA. Finally, the methods of calculating the damage displacement rate and dpa are described. The field of radiation damage is vast, but discussion on all these aspects is out of the scope of this chapter. Interested readers may refer to texts listed in Bibliography.
1.1 a) If a copper target (mass number 64) is exposed to monoenergetic neutrons
of flux 2 x 1015 n cm-2 s-1 for 6 days continuously, show that the number of atomic displacements per atom (dpa) is 2.856 (Ed = 22 eV, ael = 2 b).
b) What will be the enhancement in self-diffusion in copper at 600 °C if only vacancy survived for million displacements (due to in situ annealing)? Given: Coordination number of copper = 12, lattice constant = ~0.22 nm, vacancy formation energy = 20kcalmol-1, and vacancy migration energy = 18 kcal mol-1.
1.2 a) Using the simple Kinchin-Pease model, evaluate the number of atomic
displacements per atom (dpa) of iron due to a monoenergetic neutron (2MeV) flux of 3 x 1015ncm-2s-1 for 1 year (assume isotropic elastic scattering ael = 3 b and Ed = 40 eV. Iron is BCC with A = 56 and q = 7.9 g cm-3).
b) Calculate the dpa for neutrons of energy 0.25 MeV and compare with the above.
1.3 a) Calculate the average energy of primary knock-ons for Zr when struck with 2MeV neutrons (Zr: (HCP) A — 91, a — 3.23A, c — 5.147A, Ev — 30kcal mol_1, Ed — 59kcalmol_1, sel — 2 b, Ed — 25eV).
b) If in a zirconium target exposed to high-energy neutrons, PKAs were produced with 240 keV, calculate the number ofatoms displaced due to a PKA (assume Kinchin-Pease model).
c) Where do the displaced atoms go to and what is the primary defect created by these atomic displacements?
d) Some of these zirconium specimens were exposed to high-energy neutrons and dpa was calculated tobe2.5, while only one vacancy survived for billion displacements. Calculate the vacancy concentration in the irradiated sample at 500 °C.
e) Evaluate the %increase in lattice diffusivity ofthe irradiated material.
There are two basic mechanisms of precipitation strengthening: (a) particle shearing (particle cutting mechanism), and (b) dislocation bowing or bypassing mechanism (Orowan’s theory).
Corrosion couples can be categorized into three broad categories: (i) composition cells, (ii) concentration cells, and (iii) stress cells.
A composition cell can be established between two dissimilar metals. One example of it is the galvanized steel in which zinc coating is applied to carbon steel to protect it from rust and other environmental attacks. Here, zinc coating acts as a sacrificial anode, which means that the steel underneath is protected while zinc coating gets corroded. Thus, a metal higher on the electrochemical series can be used to reduce corrosion to a less active one. Note that from the relative positions of iron and zinc on the electrochemical series in Table 5.4, zinc is more anodic than iron. The unique aspect of sacrificial anode is that even if there is an accidental scratching on the zinc coating exposing the steel surface, the steel will still be protected until all zinc coating gets corroded. However, tin coating on iron or steel sheets is also used for protection in which case protection remains only up to the point when the tin coating acts as a barrier and is not compromised. However, if tin coating gets scratched or punctured, the corrosion effect will be concentrated in that punctured region (anode) with respect to remaining tin coating (cathode).
In the galvanized steel example, we found out benefits of the composition cell in corrosion protection. However, there are many other engineering examples of composition cells wherever dissimilar metals/alloys are made to work in contact and that situation is not desirable at all from an engineering point of view. A couple of examples include steel bolts in brass naval components and steel piping connected to a copper valve. Whenever a materials selection decision is made to use dissimilar metals/alloys in engineering applications, due care must be exercised. If not, composition cells will be set up and would cause undue galvanic corrosion problems. Galvanic cells can also be created at microlevel. Usually the single-phase metallic alloys are more corrosion resistant. As more second-phase particles are formed through precipitation, galvanic microcorrosion cells are set up between the particle and the matrix metal and lead to a poor corrosion resistance.
Gen-IV reactors are the futuristic reactors for which research and development efforts are currently in progress. These reactors will be more efficient, safer, longer lasting (60 years and beyond), proliferation-resistant, and economically viable compared to the present nuclear reactors. Six reactor designs were selected at the outset. They are summarized in Table 1.4 including information on the type (fast or thermal spectrum) of the reactor, coolants, and approximate core outlet temperatures. Two reactor concepts, sodium fast reactor (SFR) — along with the advanced burner reactor (ABR) concept under the erstwhile Advanced Fuel Cycle Initiative Program) — and very high-temperature reactor (VHTR) under the Next Generation Nuclear Plant (NGNP) are of the highest priority in the United States (Figure 1.15a and b, respectively). VHTR is the reactor concept employed for the Next Generation Nuclear Plant program where the heat generated will cogenerate electricity and hydrogen.
The demanding service conditions (higher neutron doses, exposure to higher temperatures, and corrosive environments) that the structural components will experience in these reactors would pose a significant challenge for the structural materials selection and qualification efforts. Because of the stringent requirements noted above, the materials employed in today’s commercial reactors are not suitable for use in Gen-IV reactors. For example, zirconium alloys (Zircaloy-2, Zircaloy-4, Zr-2.5Nb, M5) have been used routinely as fuel cladding and other reactor internals in both light and heavy water reactors because of their low neutron capture cross section and acceptable mechanical and corrosion resistance in high temperature
Table 1.4 A summary of Gen-IV reactor system designs.
water-cooled reactor (SCWR) |
Figure 1.15 Schematics of (a) a sodium fast reactor and (b) a very high-temperature reactor. (Courtesy: US Department of Energy Gen-IV Initiative) |
(probably never exceeding 350-380 °C under normal service conditions) aqueous environment. However, higher temperatures envisioned in Gen-IV reactors would limit the use of zirconium alloys because of increased susceptibility to hydrogen embrittlement due to severe hydride formation, allotropic phase changes at higher temperatures (a-fi phase), poor creep properties, and oxidation. It is instructive to note that some high-performance zirconium alloys may be of possible use in relatively low-temperature Gen-IV reactor design (such as SCWR). Furthermore, the out-of-core components (pressure vessel, piping, etc.) may need to be made from materials other than the low-alloy ferritic steels (e. g., A533B) currently employed primarily because similar components in Gen-IV reactors are expected to withstand much higher temperatures and neutron doses. Some of the fabrication difficulty involved in the VHTR construction demands mention here. For example, the pressure vessel for VHTR reactor as shown in Figure 1.16 is about double the size of the currently operating PWRs. Heavy component forgings will be needed in the construction of these huge pressure vessels.
Figure 1.16 A schematic size comparison between the reactor pressure vessels in a typical PWR and a conceptual VHTR. Courtesy: Nuclear News. |
|
||
|