Almost all the reactors in the United States and a majority in the world are thermal reactors wherein thermal neutrons cause the bulk of the fission reactions. If one starts to think about designing a prototype reactor, the several design elements need to be flawlessly integrated. Figure 1.6 shows such a sche­matic for a primitive thermal reactor. The tubes containing fuels are generally made of metallic alloys (also known as fuel cladding). The radioactive fuels (such as uranium) could be in metallic, alloy, or ceramic forms. The fuel clad­ding serves many purposes: it provides mechanical support to the fuel, keeps the fission products from leaving the fuel element, and protects the fuels from corrosion from the coolant. The fuel elements are arranged in a distinct regular pattern (square, hexagon, etc., dictated by neutronics and other factors) with the moderator. Moderator slows down the neutron to sustain the fission reaction with thermal neutrons. The fuel-moderator assembly is surrounded by a

Figure 1.6 A schematic of a simple reactor design. (adapted from C. O. Smith, Nuclear Reactor Materials, Addison-Wesley, Reading, MA, 1967)

reflector. The purpose of a reflector is to direct all neutrons generated toward the core so that neutron leakage can be controlled, thus improving the neutron economy. On the outside, the reactor is lined by shielding materials that absorb neutrons and gamma rays that escape the core and reduce the radiation inten­sity to a tolerable level so that people near the reactor are not exposed to these radiations. The control rod (usually an assembly) helps control the chain reaction by absorbing neutrons, maintaining the steady state of operation. Hence, the control materials are neutron-absorbing materials (boron, hafnium, and so forth), and are generally fabricated in the form of rods (in some cases, plates). A reactor is typically equipped with two types of control rods — regulat­ing rods for routine control reasons and safety rods (to permit shutdown in the case of emergency). Even though coolant is not shown in Figure 1.6, it is an important component of a reactor. As a huge amount of heat is generated in the fuel elements, the heat needs to be removed continuously in an efficient manner in order to maintain a safe, steady-state reactor operation. This means an efficient coolant is needed. The coolant can be a gas or liquid (such as light or heavy water, carbon dioxide, liquid metals, and molten salts). However, it is important to remember that the presence of any coolant tends to adversely affect the neutron economy. Hence, the balance between the reduction in the neutron economy and the efficiency of heat removal needs to be carefully considered.

1.8.2

This structure is named after the mineral fluorite, or CaF2. In this solid, the valency of the cation is twice that of the anion, that is, MX2. This structure can be described in two ways. Figure 2.22 shows the view of a CaF2-type unit cell where the FCC cation sublattice is interlaced with a simple cubic anion sublattice. That means all the tetrahedral interstitial positions of the FCC cation sublattice are filled with anions. Thus, each fluorite unit cell contains four cations and eight anions or four CaF2 molecules. This type of structure has the following relation between the lattice constant and ionic radii: a = (4Д/3) (R+ + R~), which can be derived from purely geometrical aspects noting that the F ion is at a/4, a/4, a/4 position, thus, R+ + R~ becomes the body diagonal of a cube with side equal to a/4. Some examples of the fluorite-type crystal structure are ZrO2, HfO2, UO2, ThO2, PuO2, and CeO2. UO2 is the most widely used nuclear fuel in most com­mercial power reactors. One of the beneficial effects of using UO2 is its high melting point (2860 °C), thus giving it excellent stability. UO2 structure has more empty space than a UC structure due to the simple cubic structure of the interior F~ sublattice. Even though this feature aids in providing more space for the fission products to accumulate inside the structure, it also reduces the fissile atom density. Notably, when the cations and anions exchange their positions in the fluorite structure (i. e., becomes M2X type), the resulting structure is called an “antifluorite” structure. Examples of such crystal structure are alkali metal oxides, that is, Li2O, Na2O, K2O, and Rb2O.

■ Example 2.3

Calculate the density of uranium dioxide (UO2) from the first principles given: atomic weights of uranium and oxygen are 238 and 16, respectively. The atomic radius data are U4+ = 0.105 nm and O2~ = 0.132 nm.

Solution

The theoretical density of UO2 can be obtained from the mass of the atoms in the unit cell divided by the volume of the unit cell.

The lattice constant of a UO2 unit cell (a) is given by (4/^/3) (R+ + R~) = (4/P3) (0.105 + 0.132) nm = 0.547 nm.

The volume of the UO2 unit cell (a3) is given by (0.547 nm)3 = 0.164 nm3 = 0.164 x (10~9)3 m3 = 0.164 x 10~27m3.

UO2 has a fluorite-type crystal structure, that is, the effective number of U4+ ions is 4 and of O2~ ions is 8.

The atomic weight of the uranium is 238. This means the mass of each U4+ ion is 238 amu or (238 x 1.66 x 10~27)kg (since 1 amu = 1.66 x 10~27 kg). There are four U4+ ions of a combined mass of (4 x 238 x 1.66 x 10~27) kg = 1.58 x 10~24kg.

Similarly, the mass of eight oxygen ions in the unit cell is given by 0.213 x 10~24kg.

The total mass of the UO2 unit cell is (1.58 + 0.212) x 10~24kg = 1.792 x 10~24kg.

The density of UO2 is then given by (1.792 x 10~24kg/0.164 x 10-27 m3) « 10 926kg m~3 « 10.93 g cm~3.

A dislocation experiences an opposing force (the basic level of lattice friction) when it tries to move through an otherwise perfect crystal (i. e., without any other defect acting as obstacles). The corresponding stress needed to move a dislocation in a particular direction in the crystal is known as Peierls-Nabarro (P-N) stress. The P-N stress is a direct consequence of the periodic force field present in crystal lattice and is very sensitive to any changes in the individual atom positions. That is, it is a function of the dislocation core structure, and hence developing a single analytical expression is difficult. However, the analysis forwarded by Peierls (1940) and Nabarro (1947) still gives us some important qualitative understanding that is of definite value. The P-N stress (tp_N) is given by the following relation:

where w is the dislocation core width, b is the distance between atoms in the slip direction, that is, the Burgers vector of the dislocation involved, G is the shear mod­ulus, and v is the Poisson’s ratio of the material. It can be shown that for screw dislocations, w is close to the interplanar spacing between slip planes (d); whereas

for edge dislocations, w is given by (d/(1 — n)). Dislocation core widths generally seem to vary between b and 5b (sometimes on the order of 10b for ductile metals), depending again on the interatomic potential and crystal structure. The related energy barrier is called Peierls energy.

Despite its limitations, the concept of P-N stress explains some important quali­tative aspects of plastic deformation in various crystalline materials. From Eq. (4.2), one can see that a wider dislocation core (i. e., with larger w) leads to lower value of P-N stress. This situation arises in ductile metals where the dislocation core is highly distorted and is not localized. However, ceramic materials (with covalent and/or ionic bonds) that show a very low or no ductility at lower temperatures have dislocations of narrower width. This also means that the P-N stress in ceramic materials is correspondingly high. Also, the presence of electrostatic forces in ceramic materials makes the dislocation movement difficult leading to lower plas­ticity. However, some ductility in ceramic materials can be obtained by increasing temperature that provides for thermal activation surmounting the relevant Peierls barrier. The relative higher P-N stress in BCC metals compared to FCC metals can also be explained using Eq. (4.2). In FCC metals, slip occurs on the close-packed planes that are greater distance apart, that is, d is higher). In the close-packed struc­ture, the magnitude of the Burgers vector (b) is smaller. That means d > b. It makes the magnitude of tP—N smaller. The opposite thing happens in BCC crystals that contain loosely packed slip planes with close separation from each other, that is, d < b. Again, from Eq. (4.2), we can see that tP—N will be higher in BCC and like typical not-so close-packed crystal lattices.

4.1.3

Load-bearing structures in engineering applications need some attributes in response to some particular forms of mechanical loading conditions. There are mechanical properties like strength and ductility that can be evaluated from simple uniaxial ten­sile testing, whereas dynamic mechanical properties like fatigue could only be eval­uated from cyclic loading testing. Here, we will assume that the readers have been familiar with a basic engineering mechanics course. In this chapter, tensile, impact, fracture toughness, and fatigue and creep properties will be briefly discussed. The American Society for Testing of Materials (ASTM) has developed specific standards for each test procedure. These will be referred to here from time to time. It is impor­tant to understand them before we start deliberations on the effects of radiation on these properties in Chapter 6.

Before we go into the details of the various sections of the mechanical properties, we first describe the modes and types of deformation. There are, in general, four ways that a load may be applied: tension, compression, shear, and torsion, as depicted in Figure 5.1; we will not be concerned here of torsion. Force applied per unit area is known as stress (s or t) and the sample displacement per unit length is the strain (e or y); tension is regarded as positive while compression negative.

An Introduction to Nuclear Materials: Fundamentals and Applications, First Edition.

K. Linga Murty and Indrajit Charit.

© 2013 Wiley-VCH Verlag GmbH & Co. KGaA. Published 2013 by Wiley-VCH Verlag GmbH & Co. KGaA.

Thus, from Figure 5.1, we note the following relations:

Note that the stress has units of N m~2 known as Pascal (Pa), and we generally talk about MPa (=106 Pa = Nmm~2), while the strain has no units and often referred to as %.

There are three types of deformation: elastic, anelastic, and plastic. Elastic defor­mation is instantaneous completely recoverable deformation where the body returns to the original shape once the external forces are removed. Anelastic defor­mation is time-dependent completely recoverable deformation, meaning that once the forces are removed, the body returns to the original shape in a time-dependent fashion. Plastic deformation is permanent deformation; i. e., it does not recover neither instantaneously nor as a function of time. During elastic deformation, force is directly proportional to the resulting strain as in Hooke’s law (force proportional

to displacement), and during uniaxial tension or compression, the proportionality constant is the Young’s modulus (E) also referred to as elastic modulus:

E = -. (5.2a)

e

Similarly, for shear loading,

—

G = -. (5.2b)

c

Here, for shear loading, G in Eq. (5.2b) is known as shear modulus or modulus of rigidity. In Section 5.1.1, we describe tensile properties, while shear character­istics will be mentioned as appropriate.

We note that when tensile stress is applied (Figure 5.1a), the diameter of the specimen decreases and the ratio of transverse contraction to longitudinal elonga­tion is given by the Poisson’s ratio (n), which is positive and has a value close to 0.3.

This implies that during elastic deformation, volume increases, which is known as elastic dilatation. For rectangular cross section in a tensile bar of an isotropic mate­rial with loading along the z-direction, the Poisson’s ratio is given by

Some materials such as Zr and Ti alloys in general exhibit anisotropy in which case the ratios of contractile width to axial elongation and contractile thickness to axial elongation may not be the same.

The volume change is given by the sum ofthe three orthogonal strains,

DV, r „ .

— = ex + ey + ez. (5.3a)

Under the application of hydrostatic pressure (P), the volume change is related to the bulk modulus (k), the inverse of which is known as compressibility (b):

The above relationships are usually derived in mechanical property courses and it is sufficient here to know that such interrelationships exist.

Example Problem

A cylindrical steel specimen (10 mm in diameter and 40 mm in length) is subjected to a stress of 100 MPa resulting in the length and diameter of the elastically deformed specimen of 40.019 and 9.9986 mm, respectively. Evaluate (i) elastic modulus, (ii) Poisson’s ratio, (iii) shear modulus, and (iv) change in volume.

184 I 5 Properties of Materials Solution

We first need to calculate the longitudinal (ej) and diametral (eD) strains:

Є = —- =————- 9 = 0.000475 = 0.0475% and eD =—-

‘ l 40 D D

= -0.00014 = -0.014%.

= 210 526 = 210.5 GPa.

2(1 + n) 2(1.295)

Change in volume can be found either by finding the final volume (pD2l/4) and subtracting from the initial volume or by adding the strains (el + 2eD) to be 1.953 x 10-4 or 0.0195%, an increase.

5.1.1

“Imagination is more important than knowledge. For knowledge is limited to all we now know and understand, while imagination embraces the entire world, and all there ever will be to know and understand.”

—Albert Einstein

We have learned about some aspects of the primary radiation damage in Chapter 3 and covered on the evaluation of damage in terms of displacements per atom (dpa). Radiation effects generally refer to the behavior (or properties) of materials in the aftermath of the radiation damage. As this book is on nuclear reactor materials, we will focus on the radiation effects caused by neutron irradiation. However, most of the radiation effects are universal in nature regardless of the type of radiation (i. e., proton, heavy ions, electrons) with specific different features. Let us make some general observations on the radiation effects: (i) defect concentration increases with fluence, (ii) nuclear transmutation occurs, (iii) chemical reactivity changes, (iv) diffusion increases, (v) new phases, both equilibrium and nonequilibrium, can take place, and (vi) impurities are produced. The changes of properties are, in gen­eral, proportional to neutron flux, irradiation time, and temperature. From the pre­vious chapters, it has been very clear that the essence of materials science as applied to nuclear field is in structure-property performance. Figure 6.1 illustrates this very aspect. Defects formed during primary radiation damage evolve to form clusters and subsequently extended defects and change the microstructural fea­tures in various ways. These microstructural changes, in turn, alter the materials properties. That is why materials undergo extensive irradiation testing before they could qualify to be fit for use in nuclear reactors. In this chapter, while we will dis­cuss general aspects of radiation effects, we will give examples from nonfuel (i. e., structural/fuel cladding, etc.) reactor materials. Some specific examples of radia­tion effects in fuels will be discussed in Chapter 7.

6.1