## Reaction Rate

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})The reaction rate for various types of interactions is found from the appropriate cross-section type:

**Σ**_{t}. Ф – total reaction rate**Σ**_{a}. Ф – absorption reaction rate**Σ**_{c}. Ф – radiative capture reaction rate**Σ**_{f}. Ф – fission reaction rate

## Reaction Rate and Reactor Power Calculation

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 = RR . E _{r} . V = 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})## Example – Reaction Rate and Reactor Power

A typical **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:**

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**