Fabrication of uranium is guided by factors like its reaction tendency with air and hydrogen and anisotropic properties. Nevertheless, a host of fabrication techniques including rolling, forging, casting, extrusion, swaging, wire and tube drawing, machining, and powder metallurgy can be used. Here, we discuss some aspects of rolling and powder metallurgy.

In theory, uranium can be rolled in any of the three phase regimes (alpha, beta, or gamma). However, since the beta phase is relatively hard, it needs almost two-three times more force when rolling in the temperature range of 660-770 °C compared to rolling at 650 °C. On the other hand, the gamma phase has adequate ductility and is easily susceptible to warping and sagging effects. Hence, most rolling is done in the alpha-phase regime. The recrystallization temperature for heavily cold worked uranium is about 450 °C. For hot rolling, special protection sheaths like nickel along with graphite barrier or other barrier materials used in between the sheath and uranium sheet are used to prevent oxidation of the surface during rolling.

Uranium powder is generally obtained via hydriding method. Uranium powders are highly pyrophoric. That is why special cautions should be taken while using powder metallurgy technique. Paraffin wax or oil is used to protect the powder sur­faces during powder processing. As uranium is considered somewhat toxic, ura­nium powders should be handled under controlled atmosphere like in a “glove box.” Powder products are generally made by the following techniques:

i) Cold compaction with sintering in the high-temperature gamma-phase regime (1095-1120 ° C).

ii) Cold compaction, sintering, repressing, and final annealing either in the alpha — or gamma-phase regime.

iii) Hot pressing in the high-alpha-phase regime.

Postirradiation annealing of irradiated RPV steels near 450-490 °C for several hours (70-150 h) could lead to complete recovery of the transition tempera­ture, even after very high neutron fluences (>102°ncm~2). Thus, this is regarded as a potential technique to eliminate or minimize the radiation embrittlement effect. Figure 6.35 [27] illustrates the point in a ferritic steel containing 1.6 wt% Ni, 0.007 wt% P, and 0.06 wt% Cu. It is important to note

that Ni, Cu, and P are known to promote radiation embrittlement. Even in this case, thermal annealing at 490 °C can recover the DBTT and the upper shelf energy. In this particular case, the upper shelf energy appears to go above that of the unirradiated alloys. It is thought that the phosphorus already was to some extent in segregated state before the irradiation and thermal annealing effect removes phosphorus segregation.

6.2.3

Thorium has only two allotropes, both of which are of cubic type. The alpha phase of thorium is FCC (4 atoms per unit cell) and retained up to ~1400 °C. The alpha phase has a lattice constant (a) of 5.086 A and density of 11.72 gcm~3, at ambient temperature (25 °C). Above ~1400 °C, thorium assumes BCC lattice structure (beta phase) and is stable up to its melting point of ~1750 °C. The lattice constant (a) of beta-thorium is 4.11 A at 1450 °C and density is 11.1 gcm~3. Hence, it can be noted that thorium has a higher melting point and lower density than uranium. Thermal conductivity of thorium is about 30% higher than that of uranium at 100 °C and about 8% greater at 650 °C. Figure 7.16 shows the variation of specific heat at con­stant pressure (CP) for (highly pure) thorium produced by iodide process.

In Chapter 1, we have discussed liquid metal fast breeder reactor (LMFBR). Both EBR-1 and EBR-2 used liquid metal coolants. Hence, it is important to know the effect of liquid metals on the nuclear components. Due to thermodynamic imbalance between structural metal and liquid metal (Na, K, Li, NaK, etc.), the structural metals/alloys exhibit liquid metal embrittlement as per the following:

i) Due to dissolution and precipitation (because of AT, . . . ), depending on the solubility of liquid in the solid.

ii) Particle migration/diffusion of solid metal into liquid and vice versa.

iii) Penetration of the liquid metal atoms into solid metal (structural) mainly through GBs leading to rupture.

Radiation has minor effect on liquid metal embrittlement, but in general enhan­ces due to decomposition of liquid metal or transmutation of atoms/nuclei.

Generally, austenitic stainless steels 304 and 316 SS are excellent candidates, but swelling could be a major limiting factor. Refractory metals tend to have high degree of resistance against liquid metals.

6.4.4

Stress is defined as the force acting on a plane, and in the case of a complex geome­try under external force, it is convenient to define stress as

F

a — lim —. (A.1)

A!0 A

We thus note that to designate stress (a) we need to define the direction along which the force (F) acts on an area (A) designated with a normal to the area. Force F is a vector with magnitude and direction and is known as a first-rank tensor usu­ally referred to as F; (with i being x, y, or z). Similarly, an area A; refers to the area whose normal is along the i-direction. On the other hand, a stress needs to be des­ignated by two directions, one each for the force and area, respectively, and is referred to as a second-rank tensor (Sj acting on an area whose normal is along the “i” axis with a force along the “j” direction). When the plane normal and the force direction are same, they are referred to as “normal” stresses such as axx, ayy, and azz. Tensile stresses are regarded as positive and compressive stresses are regarded as negative. i / j results in a shear stress such as a у with force acting on a plane whose normal is “i” with “j" being the force direction; in this case, the shear stresses are considered to be “negative” if one of the indices (i or j) is negative (opposite to one another) and “positive” if both i and j are along positive or negative direction.

In general, stress tensor (aij) may have nine different components:

axx axy axz

ayx ayy ayz I. (A.2)

azx azy azz J

Since a body under force is regarded as rigid or with “no net moment,” Sj = Sji or stress is a “symmetrical” tensor, thereby having six terms with three normal (diago­nal) and three shear (nondiagonal) terms,

axx axy axz

axy ayy ayz I ; (A.3a)

axz ayz azz

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.

Figure A.1 Stress tensor designations.

which is often written as

Ox txy Txz

Oy tyz. (A.3b)

In Eq. (A.3b), the diagonal terms are referred by single index such as oy, which is the same as Oyy, while shear stresses are usually referred to as “t”. These are illustrated in Figure A.1.

For any given situation (a solid under load or force), the various stress component (Oj) values depend on the coordinate system selected. Values for stress components given for a set of coordinate system (xyz) can be converted to a differ­ent set of coordinates (Xy’z’) through tensor transformations by knowing the angles between xx, yy, and zz :

oj = aikaflOU, (A.4)

where ay are direction cosines, for example, axy = cos(0xy), where 0xy is the angle between X and y coordinates. It is important to note that trace of the stress tensor (sum of diagonal terms) remains independent of the coordinate system. That means

A.1

Plane Stress

In special cases such as thin plates/sheets where stresses along the thickness direc­tion are zero and thin-walled cylinders (with outer diameter >10 x thickness)

A.3 Hydrostatic and Deviatoric Stresses | 361

where stresses along tube wall thickness are negligible, they are called “plane stresses” meaning that stresses in one of the three directions are negligible so that the stress tensor is two dimensional with three different components (oxx, oyy,

and — syx).

A.2

As we know from our fundamental physics classes, heat transfer modes are of three types — conduction, convection, and radiation. The first two processes are of

importance in nuclear reactor materials selection. Most important example in nuclear reactor is the choice of fuel and cladding materials. The safety and effi­ciency of the reactor depends on how efficiently the heat generated inside the fuel can be removed. Hence, thermal conductivity is an important property. Otherwise, the fuel will melt such as in a “loss of coolant accident” (LOCA) scenario. Similarly, heat transfer properties are also important for various balance-of-plant features, such as heat exchangers, condensers, and other ancillary equipment (such as steam generator in a PWR system).

1.9.1.4 Availability and Cost

This is the last but not the least general consideration in the materials selection process. Availability and cost are the economic consideration that may trump tech­nical considerations. If a material is not available in the form or at an allowable price, the prudent engineering decision would be to find an alternative material with similar properties or a different form of material or make changes in design to allow different characteristics. Cost-benefit analysis must be at the heart of that process. In this regard, the fair question becomes, “If it were your money, would you refuse to buy the item because it costs too much?”

1.9.2

In addition to the slip, another method by which the crystal can plastically deform is “twinning.” When the arrangement of atoms on one side of a boundary is a mir­ror image of that on the other side due to a homogeneous shear action, the bound­ary is known as a twin boundary (Figure 2.38). Twin boundaries appear in pairs where the orientation difference introduced by one boundary is nullified by the other. Twin boundaries are formed during annealing (called annealing twins) or during deformation (deformation twins). Annealing twins generally appear as

straight edged band, whereas deformation twins appear with a lenticular shape with jagged edges. The energy of twin boundaries is in the range of 0.01­0.05 J m-2, which is clearly less than those of the high-angle grain boundaries.

With a large enough applied stress to a crystal containing dislocations, some dislo­cations can move and create plastic deformation. Thus, work is done on the crystal by the dislocation as it moves. The dislocation also experiences an opposing force against its movement. One can then evaluate the force on a dislocation due to the applied stress using the principle of virtual work. The work done by the applied stress when the dislocation moves completely through the crystal is equated to the work done by the resisting force f — Fl, where l is the dislocation length and F is the force per unit length), as shown in Figure 4.14. The resulting displacement due to the force (rl1l2) is b and thus the work done by the applied stress (t) is rl1l2b. The dislocation moves through the crystal a distance І2 doing work against the resist­ance force f (-Fl1) given by Fl1l2. Thus, the force (F) per unit length on the dislocation is given by the following relation:

where t is the shear stress resolved in the glide plane along the direction of the Burgers vector (b). Since the force is acting along x-direction, it is indeed F = tbi.

In the general case, in three — dimensions with a dislocation and Burgers vector (b = a[hkl]) exposed to a stress tensor Oj, one can use the Peach-Koehler formula to calculate the force vector per unit length:

F = Gxt = (o, j ■ b)xt. (4.15b)

Equation (4.15b) is very useful in evaluating the forces on dislocations due to complex stress states.

determine the shear stress (t) needed by a dislocation line to maintain a radius of curvature R (with the center at O). The angle subtended by a very small arc segment of length ds is given by dy = ds/R, which is a very small value in radians. The outward force on the dislocation segment is (tb)(ds). On the other hand, the oppos­ing line tension force is 2T sin (dy/2). For small values of dy, this force becomes Tdy. The dislocation line will be in equilibrium if the outward force and the inward force are equal, that is,

T dy = (tb)(ds)

or

Tdy_ T bds bR

Using Eq. (4.16), we can write

aGb2 aGb bR ~ ~R~

This is thus the stress required to keep the dislocation curved with a radius of curvature R. Alternatively, a dislocation pinned between two points tends to become curved with a radius R under the applied stress t given by

aGb R =

t

This implies that in the absence of any external stress, the dislocation wants to be as straight as possible with the minimum possible length.

■ Example 4.5

A stress

acts on an edge dislocation with b along [110] and line vector along [112]

in Ni (FCC) (G = 110 GPa, a = 3.6 A).

a) What is the magnitude of the Burger’s vector in millimeter and how is it represented?

b) What are the x, y, z components of the unit line vector?

c) Calculate the force per unit length (N mm-1) on the dislocation due to the applied stress state? (Hint: Use Peach-Koehler formula.)

d) What is the force (per unit length) along the Burger’s vector (i. e., perpen­dicular to the dislocation line)?

e) If the dislocation is pinned between two points separated by 1000 A, determine the radius of curvature.

Solution a) Representation: b = (a/2) [110 and magnitude:

|b| _a_ 3.6 x 1Q-7 1 1 72 1.414

so that

fi(1QQ — 25) — j(-2QQ — 25) + k(1QQ + 5Q)}

Thus,

F = — P f75i + 225j + 15Qk}.

26

d) Force along the Burgers vector: Fb = F ■ b, where 011 1

Q

so that

implying that the force acts along the — b direction.

e)

Recall:

4.2.5

For many alloys, there is a range of temperatures over which the ductile-brittle transition occurs. This leads to the difficulty in specifying ductile-brittle transition temperature (DBTT).

a) DBTT is often described as the temperature at which CVN assumes certain value (40 J or 30 ft lb).

b) In some cases, DBTT corresponds to some given fracture appearance (50% fibrous appearance).

However, the most conservative estimate would be at which the fracture surface becomes fully 100% fibrous. Structures constructed from alloys that exhibit the ductile-brittle transition temperature should be used at temperatures above the transition temperature to avoid catastrophic failure. Materials with FCC crystal structures (such as aluminum — and copper-based alloys and austenitic stainless steels) do not exhibit DBTT. BCC and HCP alloys mainly experience this transi­tion, which is mainly due to the highly temperature-sensitive yield stress at lower temperatures. For these materials, the transition temperature is sensitive to both the alloy composition and microstructure. It is to note that decreasing the average grain size of the material decreases DBTT. Figure 5.21 shows the effect of carbon content on the impact energy in plain carbon steels with various carbon concen­trations. (0.01-0.67 wt%). Impact energy could also vary depending on the mechanical anisotropy.

5.1.5

The swelling behavior in a material depends on composition. An example is shown in Figure 6.17 for nickel-based materials at 425 °C. Nickel with just 0.4% impurities shows greater resistance to swelling than the high-purity nickel, and Inconel-600 (73Ni-17Cr-8Fe, in wt%) does not show any swelling (actually shows densifica — tion!) during irradiation. The swelling resistance observed in Inconel is due to the fine coherent precipitates present in this type of material. In coherent precipitates, the precipitate-matrix interface is continuously bonded. Coherent precipitates (var­iable bias sink) act as recombination sites for vacancies and interstitials, thus reduc­ing swelling. However, precipitates need to be used as direct control of swelling due to instability of the structure in radiation environment.

FERRITIC ALLOYS

FLUENCE (1022 n/cm2, E > 0.1 MeV)

Figure 6.18 Comparison between the swelling behavior of a ferritic/martensitic steel with austenitic stainless steels (316 type) as a function of fast neutron fluence at 420 °C Ref. [16].

It has generally been a common knowledge found out from the EBR-II irradiation testing that ferritic/martensitic (F-M) alloys show greater resistance to void swelling compared to austenitic steels. Swelling behavior of six com­mercial ferritic/martensitic steels is compared with a 316-type austenitic stain­less steel as a function of neutron fluence at ~420 °C in Figure 6.18. At the peak swelling temperature (400-420 °C), only <2% swelling was noted in F-M alloys like HT-9 (12Cr-1MoWV) and T-91 (9Cr-1Mo-V-Nb) steels even after irradiation dose of 200 dpa. The complex microstructure of the ferri — tic/martensitic steels involving lath boundaries, various types of precipitates, high dislocation density, and so on provides numerous defect sink sites that help in limiting void swelling. On the other hand, in austenitic stainless steels, presence of Cr, Ni, and other elements in higher amounts leads to the forma­tion of helium through transmutation reactions and promote void swelling. However, recent studies have shown that F-M steels also swell at a greater rate at very high radiation doses; the phenomenon is generally not observed in conventional irradiation experiments as the incubation dose required is rela­tively large compared to that of the austenitic stainless steels.

Void swelling characteristics in a 16Cr-4Al-Y2O3 oxide dispersion- strength­ened (ODS) steel were studied by Kimura et al. [17]. They also compared the results with those of a non-ODS, reduced activation martensitic steel (JLF-1, 9Cr-WVTa alloy). The ODS alloys contain a high number density of nanometric (<5 nm) Y-Ti based oxide precipitates. They used dual ion irradiation at 500 °C (773 K) at a higher dose rate to accumulate extensive displacement damage. The

and (b) afterthedual ion irradiation up to 60dpa, respectively. (c and d) JLF-1 steel before and after the same irradiation condition, respectively [17].

volumetric swelling has been plotted against the displacement damage levels in Figure 6.19a. Interestingly, the ODS steel showed not much void swelling and was almost independent of the dose level. On the other hand, the JLF-1 steel exhibited comparatively higher volumetric swelling levels. They also presented evidence from TEM studies of both alloys irradiated up to 60 dpa, as shown in Figure 6.19b. The helium bubbles formed were much finer and numerous in the ODS alloy compared to the JLF-1 that showed larger bubbles. This micro­structural evidence supports the hypothesis that helium bubbles are created at the Y-Ti-O nanoprecipitates, and thus help keep a reduced level of volumetric swelling in the ODS alloys.