## Neutron Flux Density

**criticality of the reactor core**. In other words, we do not know anything about the

**power level**of the reactor core. If we want to know the

**reaction rate**or

**thermal power**of the reactor core, it is necessary to know

**how many neutrons**are traveling through the material.

It is convenient to consider the** neutron density**, that is the number of neutrons existing in one cubic centimeter. The neutron density is represented by the symbol **n** with units of neutrons/cm^{3}. In reactor physics, the** neutron flux** is more likely used, because it expresses better the** total path length** covered by **all neutrons**. The total distance these neutrons can travel each second is determined by their velocity and therefore the **neutron flux density** value is calculated as the **neutron density (n)** multiplied by **neutron velocity (v)**.

**Ф = n.v**

where:

**Ф – neutron flux (neutrons.cm ^{-2}.s^{-1})**

**n – neutron density (neutrons.cm**

^{-3})**v – neutron velocity (cm.s**

^{-1})**The neutron flux**, which is the number of neutrons crossing through some arbitrary cross-sectional unit area in

**all directions**per unit time, is a

**scalar quantity**. Therefore it is also known as the

**scalar flux**. The expression

**Ф(E).dE**is the total distance traveled during one second by all neutrons with energies between E and dE located in 1 cm

^{3}.

The connection to the **reaction rate**, respectively the reactor power, is obvious. Knowledge of the **neutron flux** (the **total path length** of all the neutrons in a cubic centimeter in a second) and the** macroscopic cross sections** (the probability of having an interaction **per centimeter path length**) allows us to compute the rate of interactions (e.g. rate of fission reactions). **The reaction rate** (the number of interactions taking place in that cubic centimeter in one second) is then given by multiplying them together:

where:

**Ф – neutron flux (neutrons.cm ^{-2}.s^{-1})**

**σ – microscopic cross section (cm**

^{2})**N – atomic number density (atoms.cm**

^{-3})We have to distinguish between the** neutron flux** and the **neutron intensity**. Although both physical quantities have the** same units**, namely, neutrons.cm^{-2}.s^{-1}, their physical interpretations are different. In contrast to the neutron flux, the neutron intensity is the number of neutrons crossing through some arbitrary cross-sectional unit area** in a single direction** per unit time (a surface is perpendicular to the direction of the beam). The **neutron intensity** is a **vector quantity**.

**thermal reactor**contains about

**100 tons**of uranium with an average enrichment of

**2%**(do not confuse it with the enrichment of the

**fresh fuel**). If the reactor power is 3000MW

_{th}, determine the

**reaction rate**and the

**average core thermal flux**.

**Solution:**

Multiplying the reaction rate per unit volume (RR = Ф . Σ) by the total volume of the core (V) gives us the** total number of reactions** occurring in the reactor core per unit time. But we also know the amount of energy released per one fission reaction to be about **200 MeV/fission**. Now, it is possible to determine the **rate of energy release** (power) due to the fission reaction. It is given by following equation:

**P = Ф . Σ _{f} . E_{r} . V = Ф . N_{U235} . σ_{f}^{235} . E_{r} . V**

where:

**P – reactor power (MeV.s ^{-1})**

**Ф – neutron flux (neutrons.cm**

^{-2}.s^{-1})**σ – microscopic cross section (cm**

^{2})**N – atomic number density (atoms.cm**

^{-3})**Er – the average recoverable energy per fission (MeV / fission)**

**V – total volume of the core (m**

^{3})The amount of fissile ^{235}U per the volume of the reactor core.

m_{235} [g/core] = 100 [metric tons] x 0.02 [g of ^{235}U / g of U] . 10^{6} [g/metric ton]

= **2 x 10 ^{6} grams of ^{235}U** per the volume of the reactor core

The atomic number density of ^{235}U in the volume of the reactor core:

N_{235} . V = m_{235} . N_{A} / M_{235}

= 2 x 10^{6} [g 235 / core] x 6.022 x 10^{23} [atoms/mol] / 235 [g/mol]

= **5.13 x 10 ^{27} atoms / core**

The microscopic fission cross-section of

^{235}U (for thermal neutrons):

**σ _{f}^{235} = 585 barns**

The average recoverable energy per ^{235}U fission:

**E _{r} = 200.7 MeV/fission**

**Mixed oxide fuel**, commonly referred to as

**MOX fuel**, is nuclear fuel that contains more than one oxide of fissile material. MOX fuels usually consist of plutonium blended with natural uranium. For simplicity and for purposes of this example, it will be assumed, that the reactor core contains only MOX fuel (100% MOX) and that the

**averaged percentage of plutonium**(averaged over all the used fuel assemblies –

**not all FAs are fresh FAs**) in MOX fuel

**is equal to 4%**. Note that, the initial percentage of plutonium in

**fresh MOX fuel is around ~7%**.

Moreover, it will be assumed, the recycled plutonium contains only the fissile ^{239}Pu. In reality, MOX fuel always contain significant amounts of higher isotopes – ^{240}Pu, ^{241}Pu and ^{242}Pu. The presence of these amounts **will be neglected in this example**.

A typical** thermal reactor** contains about **100 tons of fuel** (HM – heavy metal). If the reactor power is 3000MWth, determine the** reaction rate** and the **average core thermal flux**.

**Solution:**

Multiplying the reaction rate per unit volume (RR = Ф . Σ) by the total volume of the core (V) gives us the **total number of reactions** occurring in the reactor core per unit time. But we also know the amount of energy released per one fission reaction is about **207 MeV/fission**. Now, it is possible to determine the** rate of energy release** (power) due to the fission reaction, it is given by following equation:

**P = Ф . Σ _{f} . E_{r} . V = Ф . N_{Pu239} . σ_{f}^{239} . E_{r} . V**

where:

**P – reactor power (MeV.s ^{-1})**

**Ф – neutron flux (neutrons.cm**

^{-2}.s^{-1})**σ – microscopic cross section (cm**

^{2})**N – atomic number density (atoms.cm**

^{-3})**Er – the average recoverable energy per fission (MeV / fission)**

**V – total volume of the core (m**

^{3})The amount of fissile ^{239}Pu per the volume of the reactor core.

m_{239} [g/core] = 100 [metric tons] x 0.02 [g of ^{239}Pu / g of fuel] . 10^{6} [g/metric ton]

= **4 x 10 ^{6} grams of ^{239}Pu** per the volume of the reactor core

The atomic number density of ^{239}Pu in the volume of the reactor core:

N_{239} . V = m_{239} . N_{A} / M_{239}

= 4 x 10^{6} [g 239 / core] x 6.022 x 10^{23} [atoms/mol] / 239 [g/mol]

= **10.1 x 10 ^{27} atoms / core**

The microscopic fission cross-section of

^{235}U (for thermal neutrons):

**σ _{f}^{239} = 750 barns**

The average recoverable energy per ^{239}Pu fission:

**E _{r} = 207 MeV/fission**

## Neutron Flux – Uranium vs. MOX

**uranium loaded reactor core**was calculated, was

**3.11 x 10**. In comparison with this value, the average neutron flux in

^{13 }neutrons.cm^{-2}.s^{-1}**100% MOX fueled core**is about

**2.6 times lower**(

**1.2 x 10**), while the reaction rate remains almost the same. This fact is of importance in the reactor core design and in the design of reactivity control. It is primarily caused by:

^{13 }neutrons.cm^{-2}.s^{-1}**higher fission cross-section of**. The fission cross-section is about 750 barns in comparison with 585 barns for^{239}Pu^{235}U.**higher energy release per one fission event**. In order to generate the same amount energy a MOX core do not require such the neutron flux as a uranium fueled core.**larger fissile loading.**The main reason is in the larger fissile loading. In MOX fuels, there is relatively high buildup of^{240}Pu and^{242}Pu. Due to the relatively lower fission-to-capture ratio, there is higher accumulation of these isotopes, which are parasitic absorbers and that results in a reactivity penalty. In general, the average**regeneration factor η**is lower forfuel than for^{239}Pu^{235}U

The relatively lower average neutron flux is MOX cores has following consequences on reactor core design:

- Because of the lower neutron flux and the larger thermal absorption cross section for
, reactivity worth of control rods, chemical shim (PWRs) and burnable absorbers is less with MOX fuel.^{239}Pu - The high fission cross-section of
and the lower neutron flux lead to^{239}Pu**greater power peaking**in fuel rods that are located near water gaps or when MOX fuel is loaded with uranium fuel together.

## Neutron Flux and Fuel Burnup

**over a relatively short period**of time (days or weeks), the atomic number density of the fuel atoms remains relatively constant. Therefore in this short period, also the

**average neutron flux remains constant**, when reactor is operated at a constant power level. On the other hand, the atomic number densities of fissile isotopes over a period of months decrease due to the fuel burnup and therefore also the macroscopic cross-sections decrease. This results is slow

**increase in the neutron flux**in order to keep the desired power level.

**Nuclear and Reactor Physics:**

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