NPTEL – Chemical Engineering – Nuclear Reactor Technology
Principles of Heat Generation in Thermal
Reactors
K.S. Rajan
Professor, School of Chemical & Biotechnology
SASTRA University
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NPTEL – Chemical Engineering – Nuclear Reactor Technology
Table of Contents
1 PRINCIPLE OF HEAT GENERATION ........................................................................................ 3 1.1 REACTOR POWER ....................................................................................................................................... 5 2 REFERENCES/ADDITIONAL READING ................................................................................... 8 Joint Initiative of IITs and IISc – Funded by MHRD
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NPTEL – Chemical Engineering – Nuclear Reactor Technology
This lecture will focus on the basic principles of heat generation in thermal reactors.
The components contributing to the total heat generated will also be discussed for
light water and heavy water reactors.
At the end of this lecture, the learners will be able to
(i)
(ii)
(iii)
List the components contributing to total energy release
Understand the distribution of energy released in various components of
reactor
Relate reactor power with neutron flux and cross section
1 Principle of Heat Generation
The principle modes of heat generation in a nuclear reactor are the reactions (nuclear
reactions) that neutrons undergo with nuclei of various materials used in the reactor.
Nuclear fission is exothermic with the energy released due to fission attributed to
splitting of heavy nuclei into fragments. The energy released is proportional to the
difference in mass of the reactants (neutron & nucleus) and that of products (fission
fragments). On an average, 207 MeV (1 MeV = 1.61 x10-13 J) of energy is released
per fission event in U-235, of which around 200 MeV can be recovered as heat. This
energy is sum of the kinetic energies & decay energy of the fission fragments, kinetic
energy of new neutrons and the energy of gamma radiation. This energy can be
further classified into energy release due to fission and neutron capture.
The recoverable energy released due to fission (200 MeV) can be further classified
into instantaneous energy and delayed energy. The components of instantaneous
energy release are the kinetic energy of fission fragments (168 MeV), kinetic energy
of new neutrons (5 MeV) and gamma radiation (7 MeV). The contributions for the
delayed energy come from β-decay (8 MeV) and γ-decay (7 MeV) of fission
fragments.
The energy released due to neutron capture is the result of non-fission reactions
between excess neutrons (unutilized for chain reaction) and their β-decay and γ-decay
(5 MeV).
Figure 1 illustrates the energy contribution from a nuclear reactor using U-235 as the
fissile isotope. The contributions of various nuclear reactions towards the recoverable
energy are also shown.
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NPTEL – Chemical Engineering – Nuclear Reactor Technology
Fig 1. Contribution of different nuclear reactions towards recoverable heat energy: (a) energy in MeV; (b)
energy contribution in % of total recoverable energy
Now let’s discuss the distribution of this energy among various components of the
reactor.
The fission fragments have very short range (< 0.25 mm). The β-radiation too has
short range (< 1 mm). Hence their energy is released within the fuel element itself.
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NPTEL – Chemical Engineering – Nuclear Reactor Technology
The energy released during thermalization of high-energy neutrons rests in the
moderator. The energy of the gamma radiation is released in the fuel as well as in
structural elements. Table 1 shows the distribution of energy in various components
of the reactor.
Table 1. Distribution of fission energy in various components of the thermal reactor
Energy Source
Location of its release
Kinetic energy of fission fragments
Fuel
Kinetic energy of new neutrons
Moderator
γ-radiation (instantaneous)
Fuel and structural components
γ-radiation (delayed)
Fuel and structural components
β-decay of fission products (delayed)
Fuel
Neutron capture (delayed)
Fuel and structural components
For a light-water reactor, 92 % of the total energy released stays in the fuel while
about 3 % is released in the moderator. The remaining (5 %) is released in the
structural elements.
In a CANDU-type reactor, about 94 % of the total energy is released in fuel, 5% is
released in moderator, while the rest is released in pressure tubes, calandria, coolant
and shielding.
1.1 Reactor Power
Let us look at the factors that influence the power of a reactor. It may be recalled that
200 MeV of (recoverable) energy is released for every fission event (Ef). Hence, the
reactor power is the product of energy released per fission (Ef) and the number of
fission events.
Energy released (J/s) = Energy released per fission (Ef)* Number of fissions events
per second (fission/s)
(1)
Neutron flux (φ), defined as the number of neutrons per unit fission cross section per
unit time is a key parameter in influencing the number of fission events.
Number of fissions events per second (fission/s)= Neutron flux (neutron/cm2s) * Total
fission cross section (cm2).
(2)
Note: Please recall that the fission cross section (σf) is expressed per nucleus. Hence
the total fission cross section is the fission cross section per nucleus multiplied by the
number of fuel nuclei.
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NPTEL – Chemical Engineering – Nuclear Reactor Technology
Number of fissions events per second (fission/s)= Neutron flux (neutron/cm2s)*
Fission cross section per nucleus (cm2)*Number of fuel nuclei.
(3)
Therefore,
Energy released (J/s) = Energy released per fission (J)* Neutron flux (neutron/cm2s) *
Fission cross section per nucleus (cm2)*Number of fuel nuclei.
(4)
Specific power of the fuel, defined as the thermal energy released per unit mass of the
fuel, can be related to the energy released as follows:
Specific power of fuel (W/kg) = Energy released (J/s)/Mass of the fuel (kg)
(5)
Substituting Eq. (4) in Eq. (5), we get
Specific power of fuel (W/kg) = Energy released per fission (J)* Neutron flux
(neutron/cm2s) * Macroscopic fission cross section (cm2)*Number of fuel
nuclei/Mass of the fuel (kg).
Let P’ be the specific power of fuel, Ef be the energy released per fission in Joule, Nf
be the number of fuel nuclei per unit mass of the fuel then,
P’=EfφNfσf
(6)
Equation (6) is a simplified one assuming neutron flux to be independent of neutron
energy and radial position. Taking in to account the influence of radial position and
neutron energy on the neutron flux, we have
P’=EfNfφ(r,E)σf
(7)
Neutron flux corresponding to energy of thermal neutron is the thermal neutron flux,
φ(r). Hence the specific power of fuel as a function of radial position in the fuel is
given by
P’=EfNfφ(r)σf
(8)
Power density (P’’) is defined as the thermal energy released per unit volume of the
fuel.
P’’=EfNfφ(r)σfρf
(9)
If the average neutron flux is φ-, then Eq. (8) and (9) can be written as
P’=EfNfφ-σf
(10)
P’’=EfNfφ-σf ρf
(11)
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NPTEL – Chemical Engineering – Nuclear Reactor Technology
Example – 1: Determine the number of U-235 nuclei in 1 kg of natural UO2 fuel.
The molecular weight of Uranium dioxide is 270.
Solution: From the molecular weight of UO2 (270) and the atomic weight of Uranium
(238), one may calculate the mass of uranium in one kg of UO2 as follows:
Mass of Uranium in 1 kg of UO2 = 238/270 = 0.88 kg
Number of moles of Uranium in 1 kg of UO2= 0.88/238 = 3.697 gmole
Recalling the definition of one mole, there are 6.023 x 1023 (Avagadro number) atoms
or nuclei per mole of a substance.
Therefore, one kg of UO2 contains 2.227 x 1024 atoms.
Please recall that the natural uranium contains only 0.7 % (by mass) of U-235. Hence
the number of atoms of U-235 in one kg of UO2 is the product of atomic fraction of
U-235 and 2.227 x 1024
To convert mass % to atomic fraction, the following method is used
Atomic % of U-235 = (mass % of U-235/235)/(mass % of U-235/235+ mass % of U238/238)
Atomic % of U-235 = (0.7/235)/(0.7/235+99.3/238) = 0.007039.
Note: For natural uranium dioxide fuel, the mass % and atomic % of U-235 are same.
However, at higher levels of enrichment the mass % and atomic % may slightly
differ.
Therefore, one kg of natural UO2 contains ~ 2.227 x 1024 x 0.007= 1.559 x 1022 atoms
of U-235.
Example - 2: Determine the specific power and power density of a natural UO2
fuel in a heavy water reactor. The average neutron flux is 5x1013 cm-2s-1. The
fission cross section is 579 b. The density of UO2 is 18900 kg/m3.
Solution:
Writing Eq. (10) again, we have
P’=EfNfφ-σf
Ef = 200 MeV = 3.2 x 10-11 J; φ-=5x1013 cm-2s-1; σf= 579 b = 579*10-24 cm2
As seen in the example -1, the number of fissile nuclei per kg of natural uranium
dioxide is 1.559x1022.
Therefore, P’ = 14442 W/kg = 14.44 kW/kg
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P’’=P’ρf
P’’= 273 MW/m3
2 References/Additional Reading
1. Nuclear Energy: An Introduction to the Concepts, Systems, and Applications of
Nuclear Processes, 5/e, R.L. Murray, Butterworth Heinemann, 2000.
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