NPTEL – Chemical Engineering – Nuclear Reactor Technology
Breeding
K.S. Rajan
Professor, School of Chemical & Biotechnology
SASTRA University
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NPTEL – Chemical Engineering – Nuclear Reactor Technology
Table of Contents
1 NEED FOR BREEDING .................................................................................................................. 3 1.1 COMPARISON OF THERMAL AND FAST SPECTRUM FOR BREEDING .................................................... 3 2 REFERENCES/ADDITIONAL READING ................................................................................... 7 Joint Initiative of IITs and IISc – Funded by MHRD
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NPTEL – Chemical Engineering – Nuclear Reactor Technology
In this lecture we shall discuss briefly about breeding in nuclear reactors
At the end of this lecture, the learners will be able to
(i)
understand the need for breeding in nuclear reactors
(ii)
identify materials and conditions for breeding
(iii)
classify breeders
1 Need for Breeding
Per-capita energy consumption of a nation is an indication of growth and prosperity of
a nation. As all of us are aware that energy consumption throughout the world is
increasing continuously due to several reasons that include population increase, better
standard of living in terms of comfort and industrialization. Hence any mode of
energy generation meant to serve industry and society needs to be sustainable. The
problem of rapid depletion of oil and coal reserves is well known. One method of
achieving sustainability is to improve the efficiency of energy generation/conversion
such that the energy generated from unit mass of fuel is maximized. Another method
is the re-use of material in one form or the other in order to minimize the amount of
fresh fuel inventory required. Hence inline with these principles, power generation
from nuclear fuel must also be made sustainable for long-term use.
The natural uranium contain very little amount of fissile isotope (0.7 %) while the rest
is U-238. Hence in typical pressurized heavy nuclear reactors, to utilize one kg of
fissile isotope for power generation, nearly 100 kg of natural uranium is used. During
fission reaction by neutron bombardment, the fissile material is destroyed or
consumed. If fissile material could be produced from fertile material available
abundantly in the core, inventory of fresh fuel required can be reduced. The
conversion of fertile material to fissile material due to neutron irradiation in the core
can be achieved by one of the nuclear reactions called nuclear transmutation. This is
called breeding in which fissile isotopes are produced from fertile isotopes.
Nuclear reactors that produce more fissile isotope than they consume are called
breeders or breeder reactors. Depending upon the neutron energy utilized for
breeding, breeders are classified as fast breeders or thermal breeders.
1.1 Comparison of thermal and fast spectrum for breeding
Fast Breeders or Fast Breeder Reactors (FBR) are the common breeder reactors
utilized in the world. For successful breeding, at least one neutron must be exclusively
available for interaction with fertile isotope for every neutron absorbed by fuel, after
taking into account of neutron absorption in core and structural materials. For
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NPTEL – Chemical Engineering – Nuclear Reactor Technology
example, for every fast neutron absorbed by Pu-239, 2.45 neutrons are produced. If
one neutron is utilized for sustenance of chain reaction and on an average 0.45
neutron is lost, the remaining one neutron is available for conversion of fertile to
fissile isotope. With U-235, an average of 2.1 neutrons are produced for every neutron
absorbed. After accounting for neutrons required for sustenance of chain reaction and
for losses, on an average less than one neutron is available which is insufficient for
breeding. In other words, the number of neutrons produced per neutron absorbed must
be greater than 2 by atleast a certain amount to facilitate breeding.
Now let us analyze the scenario in thermal spectrum. The number of neutrons
produced per neutron absorbed in Pu-239 and U-235 due to bombardment of thermal
neutrons is 2.04 and 2.06 respectively, clearly indicating that thermal breeders with
these fissile isotopes are impractical. Hence most of the breeder reactors are fast
reactors utilizing Pu-239 as the fissile isotope. Hence such breeder reactors are called
fast breeder reactors. Fast breeder technology for power generation is confined only to
few countries like France, Russia, China, India, Japan and South Korea. It may be
noted that thermal reactors are used in several countries for power generation while
only the above countries have active fast breeder programme.
One of the less-explored or lesser known fissile material is U-233. This isotope does
not occur in nature and hence produced artificially in reactors using Th-232 as fertile
isotope. In Indian fast reactor, Th-232 is used in the blanket, from which U-233 is
produced. A fast reactor operating with U-233 as the fissile material will produce 2.31
neutrons per every neutron absorbed and hence with appropriate core design aimed at
minimizing neutron loss, a fast breeder reactor operating with U-233 as fissile
material can also be built.
Breeding is possible even with thermal neutrons if U-233 is used as fissile material, as
2.26 neutrons are produced per neutron absorbed. A breeder reactor utilizing thermal
neutrons is called thermal breeder. India is developing an Advanced Heavy Water
Reactor that will act as thermal breeder utilizing U-233 as fissile isotope and Th-232
as fertile isotope. Availability of large reserves of thorium in India (world’s 1/3 of
total thorium reserves) is a prime reason for India’s interest in breeders utilizing U233.
Table 1. Number of neutrons produced per neutron absorbed for different isotopes in fast and thermal
spectrum
Neutron Energy
0.025 eV
> 1 MeV
Number of neutrons produced
per neutron absorbed (η)
U-235 Pu-239
U-233
2.06
2.04
2.26
2.10
2.45
2.31
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NPTEL – Chemical Engineering – Nuclear Reactor Technology
It is to be noted that the number of neutrons produced per neutron absorbed (η), also
called ‘reproduction factor’ shown in Table 1 is not a measured quantity. It is
determined from other measurable quantities as follows:
η0=υσf/σa = υσf/(σc+σa) = υ/(1+α)
(1)
In the above equation, σa, σf and σc refer to absorption, fission and capture cross
sections. ‘υ’ refers to the number of neutrons produced per fission. ‘η0’ is the number
of neutrons produced per neutron absorbed for a pure fissile material. When fuel is a
mixture of fertile and fissile material, reproduction factor (η) is
η=(Ν υ σf1+N2υ σf2)/(Ν σa1+ Ν σa2)
1
1
2
1
2
(2)
In Equation (2), the suffices 1 and 2 represent the isotopes 1 and 2 (say U-235 and U238); N1 and N2 are the number fractions of isotopes 1 and 2.
The following example shows calculation of reproduction factor for a typical
scenario.
Example – 1: Determine the reproduction factor in a fast reactor operating with
a pure fuel whose fission and absorption cross sections are 1.39 b and 1.64 b
respectively. The average number of neutrons produced per fission is 2.605
Solution:
Recall Eq. (1)
η0=υσf/σa = υσf/(σc+σa) = υ/(1+α)
σf = 1.39 b
σa= 1.64 b
υ = 2.605
Therefore, reproduction factor (η0) = 2.207
Example – 2: The following data pertain to a nuclear reactor utilizing natural
uranium (NU-235/NU-238=0.007) and operating in thermal spectrum:
U-235
U-238
σ f = 586 b σ f = 0 b
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NPTEL – Chemical Engineering – Nuclear Reactor Technology
σ a= 681 b
υ = 2.42
σ a= 2.7 b
υ =0
Determine the reproduction factor.
Solution:
Let ‘1’ and ‘2’ represent the isotopes U-235 and U-238 respectively. Using Eq. (2),
η=(Ν υ σf1+N2υ σf2)/(Ν σa1+ Ν σa2)
1
1
2
1
2
Dividing the numerator and denominator of right side of the above Equation by N2,
we get
η=(Ν υ σf1/ N2+υ σf2)/(Ν σa1/Ν + σa2)
1
1
2
1
2
η=(0.007∗2.42∗586)/(0.007∗681+2.7) = 1.33
Therefore, the reproduction factor under the above circumstances is 1.33
Example – 3: The following data correspond to U-233 in a thermal reactor.
σ f = 531 b
σ a= 576 b
υ = 2.49
Determine the reproduction factor if the fuel is pure.
Solution:
Recall Eq. (1),
η0=υσf/σa
Substituting the data in Eq. (1), reproduction factor is obtained as 2.30
Example - 4: If a fast reactor is loaded with U-235 and U-238 such that NU-235/NU238=0.25, comment on the possibility of breeding. Data on cross section and
average number of neutrons produced per fission are given below:
U-235
U-238
σ f = 1.4 b
σ f = 0.095 b
σ a= 1.65 b σ a= 0.255 b
υ = 2.60
υ = 2.60
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NPTEL – Chemical Engineering – Nuclear Reactor Technology
Solution:
Let ‘1’ and ‘2’ represent the isotopes U-235 and U-238 respectively. Using Eq. (2),
η=(Ν υ σf1+N2υ σf2)/(Ν σa1+ Ν σa2)
1
1
2
1
2
η=(Ν υ σf1/ N2+υ σf2)/(Ν σa1/Ν + σa2)
1
1
2
1
2
η=(0.25∗2.6∗1.4+2.6∗0.095)/(0.25∗1.65+0.255) = 1.73
The reproduction factor is 1.73 and hence it is not possible to achieve breeding with
this configuration.
Hence one may understand from Examples – 1 to 5, that the reproduction factor could
be varied by using fuels with different enrichment levels or composition and by use of
neutrons of appropriate energy (fast neutron or thermal neutron). In summary, fast
breeder reactors are practical with Pu-239 as the fissile material, while thermal
breeders can be built with U-233 as fuel.
2 References/Additional Reading
1. R.L. Murray, “Nuclear Energy: An Introduction to the Concepts, Systems, and
Applications of Nuclear Processes”, 5/e, Butterworth Heinemann, 2000
(Chapters 11 and 13)
2. A.E. Waltar, D.R. Todd, P.V. Tsvetkov (Eds.), “Fast Spectrum Reactors”,
Springer, 2012 (Chapter 1)
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