1.3.2 Microscopic cross-section

For neutrons of a given energy one or more of the foregoing nuclear reactions can occur. It is necessary to have a method of calculating how many neutrons are undergoing which reaction.

Clearly, the rate R at which neutrons undergo any reaction is dependent on the number of target nuclei and the number of bombarding neutrons. We can say that R is proportional to N0 where N is the number of nuclei/m3 for the target material and Ф is the intensity of the neutron radiation measured in neu — trons/sm:. о is called neutron flux and is the product of the number of neutrons/m3 and the average neu­tron speed і/, which may be in any direction, i. e., ф = nv neutrons/sm-. Let the constant of proportionality be a, so that

R = a N ф

This is usually written in the order R — Na ф

———— і————— —————————————————

and the value of о for each type of nuclear reaction is dependent on the target material and the neutron speed, a may be regarded as a measure of the like­lihood, or the probability in the non-mathematical sense, of a given reaction occurring. Alternatively, because о has the unit of m2, it is also thought of as the ‘effective’ area presented to the incident neutron by the target nucleus. Hence a is called the micro­scopic cross-section for the neutron reaction but it must be emphasised that it is not the physical area of the nucleus (a can have values orders of magnitude greater or less than the physical area). Generally, but not always, microscopic cross-section values are in the range 10"26 to 10"30 m2 and so it is convenient to have a new unit for cross-sections. This is called the barn and is defined by 1 barn = 10“28 m2.

Since neutrons can have more than one type of re­action the total microscopic cross-section at, the like­lihood of the neutrons interacting with the nuclei, can be expressed as the sum of a number of partial cross-sections.

Thus:

at — <7S + aa

= Oe + Cq + a^ + CTf

where the subscripts refer to total, scattering, absorp­tion, elastic, inelastic, capture and fission respectively. To indicate the usefulness of these ideas Fig 1.6 shows examples of how the microscopic cross-section depends on the material, the neutron energy and of course the particular nuclear reaction. In referring to the figure it is convenient to regard the neutron energy as being in three parts:

• Low energy, slow neutrons — energies up to a

few eV.

• Intermediate energy — few eV to 100 keV,

say.

• High energy, fast neutrons — energies greater

than 100 keV.

From Fig 1.6, then, the following may be inferred:

• Boron 10 capture cross-section, Fig 1.6 (a).

For slow neutrons the capture cross-section for B-10 is very large (thousands of barns) and decreases progressively over the slow to intermediate energy range. In fact ac, using logarithmic scale, decreases linearly with increasing neutron speed. This linearity is not unusual and such materials are known as ‘/v absorbers’. For the lighter dements the /v dependence of ac may persist to several hundred eV but only to a few eV for the heavier elements. The extremely large values of ac for B-10 explains why boron, which contains 19.6% B-10, is incor-

porated for neutron absorption purposes in control rods, in the coolant of water reactors and in fuel storage ponds. [2] energy range but there is a series of peaks in the intermediate energy range where ac has extremely large values, in the thousands of barns. These peaks are called resonance capture peaks and are associated with discrete energy values of the ura­nium nucleus. It will be seen later that the resonance capture of neutrons has important implications in reactor design.

• Uranium 235 and 238 fission cross-sections, Fig

1.5 (c).

Resonance cross-section peaks are not unusual and are seen in the fission cross-section values for U-235. Much more important here however is to note the large of values for slow neutrons, several hundreds of barns, in contrast to fractions of a barn for very fast neutrons. Figure 1.6 (c) also gives the fission cross-section fff for U-238. This shows that fission of U-238 is possible but only by high energy neu­trons and even then is not very likely. There is a threshold value of 1.1 MeV below which fission of U-238 will not occur. An explanation for this is given in Section 3 of this chapter.

Graphs similar to those given in Fig 1.6 are more or less readily available in the literature for a wide range of materials and applicable to neutron and other nuclear reactions.

We can now investigate the temperature distribution which can occur across a fuel rod due to conduction when the rod is placed in the core and commences heat generation. In the previous two examples of heat conduction through solids we have specified the sur­face temperatures and assumed that the heat to be conducted is supplied from some external source, e. g., steam or water. In a reactor fuel element however, heat is being generated from within the element and the surface temperature will be determined by the rate at which heat can be extracted by the coolant. (It is important to note that this principle illustrates one of the basic methods of controlling reactor tem­peratures.) It is necessary to be able to determine the temperature distribution through the element in terms of:

• The rod surface temperature.

• The geometry,

• The rate of heat generation.

The heat conduction equations can be used to obtain a good approximation of this temperature distribution across a cylindrical rod (Fig 1.19). In this case the surface temperature is tf; the geometry is the radius

Fig. 1.19 Temperature distribution across a cylindrical
fuel rod

r; and the rate of heat generation is proportional to the volume Qv. Taking a unit length of the rod in the direction of the rod axis at a surface of radius x:

Rate of heat generated 7rx2Qy

Rate of heat being conducted through the area is 2x x к dt/dx. For a steady state condition: rate of heat generated = rate of heat conducted.

7ГХ2 Qi/ = — 2тгх к dt/dx t = — Qp x2/4k + C

At the rod surface t = tf and x = r,

therefore t = Qy/4k (r2 — x2) + tf (1.3)

the variation in temperature across the rod is thus parabolic as shown in Fig 1.20.

The maximum temperature differential is

tF — tf = Qi>r2/4k (1.4)

where tF is the temperature at the centre of the rod.

In practice the heat generation/unit volume varies across the rod but the error in assuming it to be constant is small. This expression for the maximum temperature differential illustrates that if this value is to be kept to a minimum with a given rate of heat

t

X

I

under these conditions (typical of magnox and AGR) the flow can be divided into three very approximate regions:

• A thin boundary layer in which the fluid can be regarded as stationary and in contact with the solid surface.

• A thin transfer layer between this boundary layer and the main body of the fluid. The flow conditions in this layer are assumed to be smooth and parallel to the solid surface.

generation the designer should aim for a small dia­meter rod using a material with a high thermal con­ductivity. The thermal conductivity of the AGR oxide fuel for instance is appreciably less than that of the magnox metal fuel and must be compensated for by a smaller diameter rod. In practice this ideal is modified by other factors such as reactor physics considerations and the availability of suitable fuel rod materials.

• A turbulent region usually occupied by the main body of the fluid in which continuous mixing and energy losses occur due to changes in flow direc­tions and local velocities. This continuous mixing produces an approximately constant temperature throughout the region which is equal to the mean coolant temperature (tc).

Although the mechanism of heat transfer under these conditions is complex it can be simplified by assum­ing that the transfer through the boundary and trans­fer layers (whose combined thickness is x) is due to conduction only. Thus from Equation (1.1)

Consider an infinite mass of natural uranium, re­presenting the simplest conceptual design for a nu­clear reactor. Assume nj fast neutrons (i. e., with energy of fission neutrons — 2 MeV) are introduced into the reactor. To determine the value of k® for the reactor it is necessary to calculate the number of neutrons in the generation following the absorption in the uranium of all the original generation of ni fast neutrons.

In considering the possible fates of the nj neutrons it is important to recall that natural uranium consists overwhelmingly of U-238 (U-238 : U-235 = 138 : 1). There are three possibilities.

4.3.1 Fission

Figure 1.6 (c) shows that fast neutrons can induce fission in U-238 provided they have energy in excess of the threshold 1.1 MeV, the fission cross-section being much the same value as for U-235. Thus the fission events taking place in the reactor can be re­garded as being predominantly of U-238 nuclei and giving rise to the next generation of fast neutrons, ni say.

4.3.2 Capture

Some of the n і neutrons will undergo direct capture in the natural uranium, i. e., non-fission absorptions in the U-238 and U-235.

4.3.3 Scatter

Some of the n) neutrons will be scattered but only those undergoing inelastic scattering are significant.

In this case the neutrons will emerge from the inelastic scattering event with energy less than the threshold value of 1.1 MeV necessary for U-238 fission and will subsequently be captured in the resonance capture peaks of the U-238. In this context therefore inelastic scattering may be regarded as leading in­directly to capture.

Elastic scatter events may be ignored because the neutrons are effectively unchanged by the collision and are still identified as being of the original nj neutron generation.

4.3.4 koo for natural uranium and fast neutrons

The initial n і neutrons will therefore either cause fission or be captured (directly or indirectly via in­elastic collision). The fraction of ni causing fission may be calculated using the cross-section values for the nuclear reactions and hence the next generation of neutrons, ndetermined by

Of n і

П 2 — ———————- 1%

<7f + + a,

where v is the average number of neutrons released per fission = 2.55 for 2 MeV impinging neutrons. Using the values for cross-sections given in Table 1.2:

n2 = [0.29п[/(0.29 + 0.04 + 2.47)] 2.55 — 0.26n,

Therefore k® = 0.26 < 1. Hence natural uranium, no matter what the geometry, cannot of itself sustain a chain reaction.

4.4 To achieve к OO > 1

The value of k® is determined by the balance be­tween neutron production in the reactor fuel and neutron loss by absorption in the reactor materials and by leakage out of the reactor for a finite system.

Production : Absorption + Leakage.

In Section 4.3 of this chapter, natural uranium was exposed to fast neutrons. It was found that к® < 1 and thus this is not a viable nuclear reactor design. There are two ways by which k« may be increased:

• Change the properties of the fuel.

• Change the properties of the impinging neutrons.

Knowing the mass of the neutron and proton one would expect to be able to calculate the mass of a nucleus, knowing its constituent^ For example, the nuclide Au-197 (gold) has 118 neutrons and 79 protons. The nucleus would be expected to have a mass of:

(118 x 1.008665 u) + (79 x 1.007277 u)

= 198.597353 u

Direct measurement however gives a value of 196.9232 u: there is a ‘mass defect1 of 1.6741 u. The energy equivalent to the mass defect, 1559 MeV, was released at the time of formation of the gold nucleus.

All nuclei (with the exception of hydrogen, H-I) have some measure of mass defect — an individual nucleon has less mass when it is part of a nucleus than when it is in isolation. The following expression allows comparison between the mass defect of a nucleon in different nuclides:

Z/rip + N/«n — Л/

Mass defect per nucleon = ——————————-

A

where /tip and tnn are the proton and neutron masses and Л/ the observed mass of the nucleus. As an example, for gold:

, 1.6741

Mass defect per nucleon = ————— u

197

= 0.0085 u s 7.9 MeV

Figure 1.3 (a) shows how the mass per nucleon varies with the atomic number Z. It may be seen that the nuclides with the largest mass defect per nucleon are those with medium atomic numbers, rather less deficit for the heavier nuclei and substantially less for the lighter elements.

The energy equivalent to the mass defect per nucleon is the energy required to overcome the nuclear forces and ‘force’ a nucleon out of the nucleus — the nucleon having a slightly larger mass after its successful removal. The energy equivalent to the mass defect of a nucleus is known as the binding energy: the energy required to dismantle the nucleus into its individual constituent nucleons or, alternatively, the energy re­leased when the nucleons come together to form the nucleus. Figure 1.3 (b) gives the binding energy per nucleon for the different atomic numbers and is the mirror image of the mass defect in Fig 1.3 (a) but in the energy units MeV.

Figure 1.3 (b) shows that if the nucleus of a heavy element is split into two nuclei of medium atomic numbers each nucleon will have less mass than pre­viously and the equivalent amount of energy will be spontaneously released. To illustrate this assume for simplicity that uranium 236 divides into exactly two halves (an unlikely event in practice), giving two palladium 118. From Fig 1.3 (b) we have:

Binding energy of 236 n = 236 x 8.5 MeV 92 U

Binding energy of 2 x 118n = 118 x 7.5 x 2 MeV
46 p

Energy released = 236 MeV

Most of the energy released appears as kinetic energy of the two palladium nuclei. The above process is of course fission and is the source of energy for all

(a) Variation of mass per nucleon with atomic number

[bj Variation of binding energy per nucleon with atomic number

Fig. 1.3 Variation of mass and binding energy with
atomic number

present day nuclear reactors. Fission is discussed in detail in Section 3 of this chapter.

Fusion is the process by which the nuclei of two light elements combine to form a single nucleus, again resulting in the total mass being reduced and the equivalent energy released. Fusion reactions are the source of the energy of the sun where hydrogen

1.4.2 Radioactive nuclei In the radioactive decay of an unstable nucleus the proton/neutron combination may undergo one of the following changes:

+ °/3 + 1 Positron

Argon

Fig. 1.4 Graph showing region of nuclei stability

	Read More 26.08.2015   Modern Power Station Practice Fast fission factor ‘e’ The fn 1 thermal neutrons in the fuel give rise to 17 fn 1 fast neutrons by thermal fission. But, as shown in Fig 1.6 (c) and discussed in Section 3.2.1 of this chapter, fast neutrons (> 1.1 MeV) may induce U-238 to fission. It is not a very likely event because, firstly, the cross-section for fisson is low (~ 0.3 barns) and, secondly, fission is possible only by fast neutrons which have not undergone a scattering event. Scat­tering will reduce the fast neutron energy to below the fission threshold of 1.1 MeV. U-235 ENRICHMENT,% Fig. 1.14 Variation of thermal fission factor with enrichment Nevertheless, U-238 fission is possible and must be included. The gain in the number of neutrons by fast fission is given by the fast fission factor: number of fast neutrons arising from _ thermal and fast fission e = —————————————————— number of fast neutrons arising from thermal fission only Because of the condition that fission is possible only for a first collision between fast neutrons and U-238 the value of e may be regarded as being mainly deter­mined by the geometry of the fuel design. The value of 1.03 used here is probably greater than would ap­ply to most practical reactor designs. The total fast neutrons, then, is given by eTjfnj [1.03 x 1160 = 1190 fast neutrons]. Read More 26.08.2015   Modern Power Station Practice Macroscopic cross-section When dealing with matter in bulk, the reaction rate Ыаф is often written as Еф, where £ = Ncr and is known as the macroscopic cross-section. E (capital sigma) may be regarded as the total collision area presented by the target nuclei per unit volume of material. As it has dimensions (area/volume), m_1, £ may also be interpreted as the probability per metre of track length that a neutron will interact with the material. Now the average distance that a neutron travels without interacting is known as the mean free path, X. It may readily be shown that the mean free path is the reciprocal of the macroscopic cross-section, i. e., X = l/Ee. The subscripts used for a are also used for X and £ as appropriate: Xt, Xa, X$ … £[, Eat ••• 1.3.3 Thermal neutrons Neutrons in thermal equilibrium with the surrounding material are referred to as thermal neutrons. It may be shown, using the methods of the kinetic theory of gases, that the mean kinetic energy of thermal neu­trons in surroundings at a temperature of TK is given by: 1/2 mV2 = 3/2 kT where m is the mass of the neutron у is the mean neutron energy к is the Boltzmann’s constant (1.38 x 10-‘ J/K) Nuclear data are often given for the ‘standard’ temperature of 293 К (20°C). Calculation for thermal neutrons in surroundings at 293 К gives a mean ve­locity of 2200 m/s and corresponding kinetic energy of 0.025 eV. Thus for uranium 235 and for thermal neutrons at the ‘standard’ 2200 m/s we may write: <Т( = О a + = <7C + fff + <?e + Uj 690 = 101 + 579 + 10 + 0 The values for a are taken from Table 1.2. Sometimes the term ‘slow neutron’ is used as if synonymous with ‘thermal neutron’ — this is not so. Also, thermal neutrons are sometimes regarded as being necessarily at 20°C — again this is not so. For example, neutrons in equilibrium with their sur­roundings at, say, 250°C or 500°C would be thermal neutrons with corresponding speed and energy values of (2940 m/s)/(0.045 eV) or (3570 m/s)/(0.067 eV) respectively. Read More 26.08.2015   Modern Power Station Practice Heat transfer from a solid to a fluid We have applied the laws of heat transfer by con­duction to solids only and they can be used with a reasonable degree of accuracy where the thermal con­ductivity of the material is known and is constant. For instance through a solid uranium or plutonium rod fuel element and its associated metallic sheath. However, the purpose of the fuel element is to gen­erate heat which can then be transferred from the sur­face of the element to a fluid which, in turn, either directly powers the turbine (BWR) or transfers the heat to a secondary circuit of the power system (mag­nox, AGR and PWR). The fluid used for the magnox and AGR reactor systems is CO2 which has a relatively poor thermal conductivity and must be circulated at high velocity to give the ‘turbulent’ conditions necessary for high rates of heat transfer. Figure 1.21 illustrates that QT/A = q = (k/x)(ts — tc) In practice the ratio (k/x) is difficult to measure or calculate and it is replaced by an empirical constant h: QT/A = q = h(ts — tc) (1.5) The factor h is known as ‘the surface heat transfer coefficient’; it is not constant and will vary with a large number of design and operating parameters, e. g., geometry, flow velocity, pressure, temperature. If the flow velocity in the reactor fuel channel is reduced to a value which produces no turbulence, the flow pattern is known as ‘streamline or viscous’, and there will be a continuous temperature gradient across the fluid. Equation (1.5) would still be valid pro­vided the temperature tc is the mean temperature of the fluid. For fluids with a higher thermal conductivity than the gaseous coolant used in magnox and AGR re­actors, e. g., water and liquid metals, a similar rela­tionship can also be applied, but in these cases the variation in temperature across the fluid will be dif­ferent to that applied to gases. Equation (1.5) is important to the designer and operator since it illustrates that for a given tempera­ture differential between the sheath and the coolant the rate of heat extraction from the fuel element is a function of: • The heat transfer surface area (A). • The surface heat transfer coefficient (h). ‘h’ is a function of many design and operating factors but in particular the degree of turbulence. The latter is most important with gaseous coolants where, in order to obtain an economic level of heat transfer, a high degree of turbulence is essential. Thus the design of the fuel element sheath geometry must provide: • A high degree of turbulence by presenting the gas flow with a ‘rough’ profile. • A high effective surface area without increasing the fuel element diameter (if the latter were increased Equation (1.4) shows that the temperature gradient across the fuel would increase thus reducing the surface temperature tf for a given maximum fuel temperature). These two effects are achieved by the ‘finning’ and ‘ribbing* geometry of the fuel element sheath. Figure 1.22 shows typical finning on the magnox fuel ele­ments and ribbing on the AGR fuel elements. Read More 26.08.2015   Modern Power Station Practice Fuel enrichment k® is less than unity in Section 4.3, largely because the inelastic collisions of the fast neutrons in the natural uranium fuel lead to the subsequent loss of the neutrons in the U-238 resonance capture peaks. The fast fission cross-section values of U-238 and U-235 are much the same and too low to counter the neutron losses. It follows that if the proportion of U-238 in the fuel is reduced, thus increasing the proportion of U-235, the effect is to replace a high neutron cap­ture material (U-238) by another material (U-235) which has much the same fast fission cross-section but less neutron capture characteristics. This is called fuel enrichment. As U-235 has no energy threshold for fission (in fact of increases with decreasing neutron energy) the net result of fuel enrichment is that the neutron produc­tion from fission increases and neutron absorption decreases. Calculation shows that U-235 enrichment of at least 20% is necessary to give k® > 1 for fast neutrons. This is the basis of ‘fast reactors’, the title fast reactor being derived from the fact that the neutrons inducing fission are largely energetic fast neutrons. Read More 26.08.2015   Modern Power Station Practice Half life The radioactive decay of unstable nuclei is a com­pletely random process and it is not possible to pre­dict when a particular nucleus is going to decay. When there are a large number of radioactive nuclei, however, one can statistically state that the proba­bility of decay per unit time is constant. That is, the number of nuclei likely to decay in an infinitesimal time interval is proportional to the number (N) pre­sent. Thus dN/dt = XN where X is the constant of proportionality and is known as the decay constant; its value depends on the particular radioactive isotope and is unique to that isotope. dN/dt is negative be­cause N is decreasing. Integration gives N = No exp (-Xt), the familiar exponential function where No is the number of radioactive nuclei of a particular isotope at time t = 0. The rate at which nuclei decay is often given in terms of the mean lifetime of a nucleus, r, or alter­natively in terms of the half value period, T±, often abbreviated to the half life, r is the inverse’ of the decay constant X and it follows from the exponential form of the decay that:
The a particle is identical to the helium nucleus 4He and the above may also be written:
4-
He
Thorium	a particle
Alpha emission tend to be from the heavier elements (Fig 1.4). • Spontaneous fission A further type of radioacti­vity is ‘spontaneous’ fission in which the nucleus splits into two roughly equal portions. Spontane­ous fission is confined to the heavy elements and in general is barely detectable in competition with the more prevalent a decay. Examples of dements which undergo spontaneous fissions are U-238 with about 26 fissions/gm h and Pu-240 with 106 fis — sions/gm h. Spontaneous fission is not to be con­fused with neutron induced fission, the basis of nuclear reactors, and which will be discussed later.

The fourth factor *p’ measures the efficiency of the design in ‘slowing down’ the fast neutrons to thermal energies. During the slowing down process many of the neutrons will diffuse back into the fuel and be captured in the resonance capture peaks of U-238, see Fig 1.6 (b). [In the numerical example of Fig 1,13, of the initial 1190 fast neutrons 180 neutrons are captured and 1010 thermalised.] The main interest, however, is in how many neutrons escape resonance capture; hence the resonance escape probability:

number of thermal neutrons emerging from
the slowing down process

15 ~ number of fast neutrons starting the slowing down process

Thus p^fni neutrons are successfully thermalised [0 85 x 1190 = 1010 thermal neutrons for p = 0.85].

The more fuel there is in the reactor relative to the moderator the more likely is resonance capture of the neutrons and the lower the value of p; ultimately p = 0 for 100% fuel. Conversely the more modera — [or there is the greater the value of p; ultimately p _ I for 100% moderator. Hence we may conclude that the value of p increases progressively from zero towards unity as more and more moderator is added to a given amount of fuel. This is in exact contrast to the thermal utilisation factor T, Section 6.2.1 of this chapter.

6.3 к * = peijf

In Fig 1.13 and the preceding sections the assump­tion was made of having an initial П| [1000] thermal neutrons in the moderator. These neutrons were all subsequently absorbed and replaced a neutron life cycle later by perjfni [1010] thermal neutrons in the moderator. Hence

peqfni
ni

к 00 = pei/f

This is known as the four factor formula In the numerical example here:

кос = 0.85 x 1.03 x 1.33 x 0.87 = 1.01

This represents a supercritical reactor in which the number of neutrons increase by a multiple of 1.01 per neutron life cycle. If the values of the four factors had been such that their product was unity then of course this would be an exactly critical reactor. A subcritical reactor implies a product less than unity.

The usefulness of the four factor formula is in giving an understanding of how the value of k® may change, or be changed, by for example altering the geometry or the operating characteristics of the re­actor. Thus the operator, by insertion or withdrawal of the control rods, changes T and hence k®. Again, the isotopic content of the fuel will change during irradiation as the U-235 is consumed and plutonium created by the neutron capture in U-238; V will change and hence k®.

Referring back to the reactor design considerations of Section 6.1 of this chapter we may now see that, for given moderator and fuel materials, k® is largely determined by the fuel enrichment and by the quan­tity of moderator relative to fuel. For a given amount of fuel T decreases and ‘p’ increases as the quantity of moderator is increased.

The task of the designer is to calculate the amount of moderator that will give maximum value for k® and this value can be increased further, if required, by enrichment. Assuming a simple rod design for the fuel the optimum lattice pitch and fuel rod diameter may then be easily determined.