REACTOR HEAT GENERATION
The energy released from fission appears as kinetic energy of fission fragments, fission neutrons and gamma and
beta radiation. Approximately 193 Mev is released directly as a result of fission, and consists of that energy
produced promptly at the time of fission, and a later delayed component resulting from delayed neutrons and decay
of radioactive fission products. The prompt component is associated with the kinetic energy of fission products,
prompt fission neutrons and prompt fission gammas. An additional 7 Mev per fission is produced by the capture of
excess fission neutrons and the subsequent gamma and beta decay of activation products. Thermal energy (heat) is
produced as these particles interact with, and transfer their kinetic energy to, the lattice atoms of the fuel and other
reactor materials. Of the 200 Mev available from a fission reaction, approximately 10 Mev is due to the kinetic
energy of neutrinos associated with beta decay and is unrecoverable.
Fission fragments and beta particles have very short ranges in reactor materials and their kinetic energy can be
considered absorbed at their point of origin. Approximately 160 Mev is carried as kinetic energy of the fission
fragments alone, and as such all of this energy would be deposited locally in the fuel. Gamma radiation on the other
hand has a relatively long range as compared to the dimensions of the reactor fuel and therefore deposits its energy
both in the fuel, the moderator and reactor structure. Due to the large mass of fuel in the core however, the majority
of the gamma energy is deposited in the fuel, though the point of interaction may be far from the point of origin. In
thermal reactors, the energy released by the fission neutrons is deposited primarily in the moderator as a result of the
thermalization process. In Light Water Reactors where the moderator and the coolant are the same, this results in
approximately 5 Mev per fission deposited directly into the coolant by neutron thermalization alone. The amount of
heat produced in various reactor components is a function of the reactor materials and configuration. This is
particularly true of the activation component. The following values however provide guidelines when more precise
information is unavailable.
ENERGY DISTRIBUTION
(a) Fuel
(b) Moderator
(c) Reactor structure
(d) Neutrinos
180 Mev/fission
8 Mev/fission
2 Mev/fission
10 Mev/fission
It is of interest to note, that of the approximately 190 Mev per fission that is recoverable, only 180 Mev or
approximately 95 % is produced in the fuel itself, with the remainder coming from outside the fuel. Of additional
interest is the heat produced by the decay of the radioactive fission fragments, their daughters and activation
products produced by excess fission neutrons. At the beginning of core life this energy is unavailable. However,
these radioactive products reach equilibrium after a short period of time, and at equilibrium constitute
approximately 7% of the reactor power.
Example:
Determine the energy released in the fission reaction
92 U
235
+ 0 n1 → 56 Ba137 + 36 Kr 97 +2 0 n1
A mass balance on the reactants gives
235.0439 + 100867
.
→ 136.9061 + 96.9212 + 2 × 100867
.
236.0526 → 235.8446
42
Δm = 236.0526 − 2358446
.
= 0.2080 amu
The energy associated with the mass deficit is then
Mev
Mev
= 193.6
amu
fission
0.2080 amu × 931
As the overwhelming majority of the energy deposited in the fuel is due to the short ranged fission fragments and
beta particles, we approximate the heat generation rate in the reactor fuel as a constant times the fission rate
v
v
q ′′′(r ) = GΣ f φ (r )
Mev
cm3 − sec
where:
v
q ′′′(r ) = Volumetric heat generation rate (Energy/vol-time)
G
= Fission energy absorbed by the fuel per fission (≅180 Mev/fission)
Both the cross section and the flux are in general functions of position, energy and time, i.e.
v
v
v
q ′′′ (r , E , t ) = GΣ f (r , E , t )φ (r , E , t )
(1)
where the flux and volumetric heat generation rate are now differential quantities with respect to energy. If we
assume steady-state operation, the heat generation rate is independent of time and the volumetric heat generation
rate as a function of position is
v
q ′′′(r ) = G
∫
∞
v
v
Σ f (r , E )φ (r , E )dE
(2)
0
The energy dependence of the flux is usually treated by rewriting the integral as the sum of integrals over energy
bands or groups
v
q′′′(r ) = G
NG
∑ ∫ Σ (rv, E)φ (rv, E)dE
E g −1
f
g =1
(3)
Eg
and defining a group averaged flux and cross section such that the fission rate over the group is preserved, i.e.
∫
E g −1
v
v
v
v
Σ f (r , E )φ (r , E )dE ≡ Σ f g (r )φ g (r )
(4)
Eg
where
v
∫
E g −1
v
φ g (r ) ≡ φ (r , E )dE
(5)
Eg
Note, this implies the appropriate group averaged cross section is given by
43
v
Σ f g (r ) ≡
∫
E g −1
v
v
Σ f (r , E )φ (r , E )dE
Eg
∫
=
E g −1
∫
E g −1
v
v
Σ f (r , E )φ (r , E )dE
Eg
v
φ (r , E )dE
v
φg (r )
(6)
Eg
The volumetric heat generation rate may then be written in terms of the group fluxes and group averaged cross
sections as
v
q′′′(r ) = G
NG
∑Σ
fg
v
v
(r )φ g (r )
(7)
g =1
From Equation 7, we can infer that to a very good approximation the spatial distribution of the heat generated in a
nuclear reactor is proportional to the spatial distribution of the fission rate. In thermal reactors, the fission rate is
dominated by neutrons in the thermal energy range, primarily due to the large macroscopic cross section associated
with thermal fission relative to the fission cross section at higher energies. The volumetric heat generation rate can
then be approximated as
v
q′′′(r ) = G
NG
∑Σ
fg
v
v
v
v
(r )φ g (r ) ≅ GΣ f th (r )φth (r )
(8)
g =1
which implies the heat generation rate is proportional to the thermal neutron flux distribution in the fuel.
44
HEAT GENERATED IN A FUEL ROD
Consider a long, thin fuel rod oriented vertically at some arbitrary point within a cylindrical core. As the neutron
flux is in general a function of space, the flux and therefore heat generation rate in any particular rod will be a
function of space.
Figure 1: Core Flux Distribution in the Vicinity of a Fuel Rod
Due to the relatively small cross sectional area of a fuel rod compared to that of the core, the flux in the vicinity of
the fuel rod may be considered constant in terms of the overall core radial behavior with the magnitude of the flux
governed by the rod position. The axial distribution in the rod however follows the axial distribution in the core.
At any location within the core, the thermal flux within an individual fuel rod is depressed radially due to the strong
neutron absorption in the fuel.
Thermal Flux
Fast Flux
Moderator
Fuel
Figure 2: Fast and Thermal Flux Distributions in a Fuel Rod
If we assume the flux within the fuel rod to be separable in the radial and axial directions then
φ (r , z ) = φ 0 Φ(r )Ψ ( z )
(1)
45
with the amplitude φ0 a function of core position. This implies the volumetric heat generation rate is also separable
in r and z such that
q′′′(r , z ) = q0′′′ Φ (r )Ψ ( z )
(2)
The total heat generated in the rod is the integral of the volumetric heat generation rate over the rod volume
q=
∫
q ′′′(r , z )dV
(3)
V
or for a cylindrical fuel element
∫
q = q0′′′
R
∫
H
Φ (r )2πrdr Ψ ( z )dz
0
(4)
0
Assume for sake of illustration that the flux in the fuel rod is uniform radially and cosine shaped axially with z = 0
the core mid plane, i.e.
⎛ πz ⎞
q ′′′(r , z ) = q0′′′cos⎜
(5)
⎟
⎝ He ⎠
where H e is the extrapolated core height. The total heat generated in the fuel rod would be
∫
H 2
⎛ πz ⎞
⎟ dz
cos⎜
⎝ He ⎠
(6)
⎛ 2 He ⎞ ⎛ πH ⎞
q = q0′′′πR 2 ⎜
⎟
⎟ sin⎜
⎝ π ⎠ ⎝ 2 He ⎠
(7)
q = q0′′′πR 2
−H 2
or
where R is the fuel (pellet) radius. If the extrapolation distances are small relative to the dimensions of the reactor,
then
q ≅ q0′′′ 2 R2 H
(8)
Relationships which are useful in describing the heat generated in reactor fuel elements include:
Linear Heat Rate: The linear heat rate q ′ is defined to be the heat generated per unit length in a fuel element.
Local linear heat rate can be related to the heat produced at a specific location within a fuel
element and the volumetric heat generation rate at that location through the following
relationships. The total heat produced within a differential length dz about z is
q′( z )dz =
∫ q′′′(r , z)dAdz
v
(9)
Ax
such that the local linear heat rate is
q′( z ) =
∫ q′′′(r , z)dA
v
(10)
Ax
46
The total heat generated within the fuel element is then
q=
∫
H
q ′( z )dz .
(11)
0
The average linear heat rate in a particular element is obtained by averaging the local linear heat
rate over the height.
q′ =
1
H
∫
H
q ′( z )dz =
0
q
.
H
(12)
The core averaged linear heat rate can be obtained from the core thermal output, and the number
of fuel elements in the core by recognizing that the total heat generated in the reactor fuel is
simply the sum of the heat generated in the individual fuel elements. If qi is the heat generated
in an arbitrary fuel element, then
n
γ f Q& =
n
∑ ∑ q′ H
qi =
i
i =1
(13)
i =1
where : Q& = Core thermal output from all sources
γ f = Fraction of core thermal energy generated in the fuel
n
= Total number of fuel elements
If we define the core averaged linear heat rate as
then
n
∑ q′
i
(14)
γ f Q& = qc′ nH
(15)
1
qc′ ≡
n
i =1
or
qc′ =
Q& γ
f
nH
(16)
It will be shown later, that linear heat rate can be related to fuel temperature. As a
result, maximum linear heat rate is usually set by fuel melt or other maximum
temperature considerations. For a given fuel height, the maximum linear heat rate
dictates the number of fuel elements in a core.
Heat Flux:
The heat flux is the heat transfer rate per unit surface area. While the heat flux can be
referenced to any surface, in reactors the heat flux is most often referenced to the outer clad
surface, i.e. the clad/coolant interface. Assuming all heat transfer in a fuel element is in the
47
radial direction, at steady-state the local heat flux can be related to local linear heat rate and
volumetric heat generation rate by the simple energy balance
∫ q′′′(r, z)dAdz = q′(z)dz
q′′( z ) Pw dz =
(17)
Ax
or
q′′( z ) Pw =
∫ q′′′(r, z)dA = q′( z)
(18)
Ax
where Pw is the heated peremiter.
We can then relate the heat flux and total heat generated in a fuel element by
∫ q′′( z)P dz .
H
q=
w
(19)
0
The average heat flux in a particular fuel element is then
q′′ =
1
H
∫
H
q′′( z )dz =
0
q
q
=
.
Pw H As
(20)
As with the core averaged linear heat rate, the core averaged heat flux can be obtained from the
core thermal output, and the number of fuel elements in the core by
qc′′ =
Q& γ f
nAs
=
Q& γ
f
nPw H
(21)
As will be shown later, maximum heat flux is usually set by critical heat flux (DNB,
Dryout) considerations. For a given fuel height and number of fuel elements, the
maximum heat flux dictates the cross sectional dimensions (radius, thickness, etc.) of
the fuel.
48
HEAT GENERATED IN A REACTOR CORE
We have seen, that the total heat generated in the reactor fuel is simply the sum of the heat generated in the
individual fuel elements
∑
n
γ f Q& =
(1)
qi
i =1
where again the magnitude of qi is a function of the fuel element’s location in the core. This is equivalent to
integrating the fission rate over the entire core, i.e.
γ f Q& =
∫
v v
GΣ f (r )φ (r )dV
(2)
Vcore
Due to the complex spatial distribution of the flux and the cross section, this integral can only be evaluated under
special conditions. Let’s consider one such special case, where we assume the fission cross section is a constant in
the fuel, and zero if outside the fuel.
v v
r ∈ rf
⎧Σ fo
⎪
v
Σ f (r ) = ⎨
⎪ 0
⎩
(3)
v v
r ∉ rf
v
where r f denotes locations within the fuel. We define an equivalent homogeneous cross section for the entire core
( Σ ′f = σ f N ′ ) such that the total number of fuel atoms is conserved. The total heat generated in the fuel can then
be written as
γ f Q& =
∫
v
GΣ ′f φ (r )dV
(4)
Vcore
If N o is the fuel number density, then the total number of fuel atoms is
N oV fuel = N ′Vcore
N ′ = No
(5)
V fuel
Vcore
The equivalent homogeneous macroscopic cross section can then be written in terms of our original fuel
macroscopic cross section as
Σ ′f = σ f N ′ = σ f N o
V fuel
Vcore
= Σ fo
V fuel
Vcore
(6)
such that the total heat generated in the fuel is
49
γ f Q& =
∫
⎛ V fuel ⎞ v
GΣ fo ⎜
⎟φ (r )dV
⎝ Vcore ⎠
(7)
Vcore
If we further assume, that the heat generation rate in the moderator and structural materials is proportional to the
core wide fission rate, then
Q& =
∫
⎛ V fuel ⎞ v
G ′Σ fo ⎜
⎟φ (r )dV
⎝ Vcore ⎠
(8)
Vcore
where G ′ contains contributions to the heat generation rate from all sources. Evaluation of Equation 8 still requires
specification of the core wide flux distribution. If we assume the local variations in the flux due to fuel elements is
small compared to the total flux, then we can treat the core as approximately homogeneous and use Diffusion
v
Theory or some other suitable neutron flux model to generate φ (r ). One group Diffusion Theory gives the
following simple flux shapes for idealized reactor geometries.
Flux Distributions in Ideal Geometries
⎛ πx ⎞
⎟
⎝ ae ⎠
φ0 cos⎜
Infinite Slab
⎛ πx ⎞
⎛ πy ⎞
⎛ πz ⎞
⎟ cos⎜ ⎟ cos⎜ ⎟
⎝ ae ⎠
⎝ be ⎠
⎝ ce ⎠
φ0 cos⎜
Parallelepiped
⎛ πr ⎞
sin⎜ ⎟
πr Re ⎝ Re ⎠
φ0
Sphere
⎛ 2.405r ⎞ ⎛ πz ⎞
⎟ cos⎜
⎟
⎝ Re ⎠ ⎝ He ⎠
φ0J 0 ⎜
Finite Cylinder
Table 1 (All dimensions are extrapolated)
Flux shapes in actual reactor systems almost never follow the simple functional forms given in Table 1. To account
for power variations due to non ideal geometries and/or uncertainties due to manufacturing tolerances and physical
changes during operation, the concept of a Hot Spot or Power Peaking Factor is introduced. The Power Peaking
Factor is defined such that
Fq =
Maximum Core Heat Flux
.
Core Averaged Heat Flux
(9)
This implies that the maximum local heat flux at any point in the core is
qmax
′′ = Fq qc′′
(10)
and since the core averaged heat flux is proportional to the linear heat rate
qmax
′ = Fq qc′ .
(11)
50
The simple flux shapes given in Table 1 can be used to estimate the power peaking factor as illustrated in the
following example.
Example:
Compute the power peaking factor for a cylindrical reactor having a neutron flux distribution
⎛ 2.405r ⎞ ⎛ πz ⎞
⎟ cos⎜
⎟
⎝ Re ⎠ ⎝ He ⎠
φ0 J 0 ⎜
where z = 0 is at the core midplane.
SOLUTION
The local heat flux in a power reactor is proportional to the local fission rate and therefore the local flux, i.e.
⎛ 2.405r ⎞ ⎛ πz ⎞
⎟ cos⎜
⎟
q ′′(r , z ) ∝ φ0J 0 ⎜
⎝ Re ⎠ ⎝ He ⎠
⎛ 2.405r ⎞ ⎛ πz ⎞
⎟ cos⎜
⎟
q ′′(r , z ) = C1J 0 ⎜
⎝ Re ⎠ ⎝ He ⎠
For this flux distribution, the maximum heat flux occurs at r = 0, and z = 0, such that qmax
′′ = C1 . The core averaged
heat flux is obtained by averaging the heat flux over the core volume
q ′′ =
1
Vcore
∫
⎛ 2.405r ⎞ ⎛ πz ⎞
⎟ cos⎜
⎟ dV
C1J 0 ⎜
⎝ Re ⎠ ⎝ He ⎠
Vcore
q ′′ =
1
2
πRcore
H
q ′′ =
1
2
H
πRcore
∫ ∫
H /2
− H /2
Rcore
0
⎛ 2.405r ⎞ ⎛ πz ⎞
⎟ cos⎜
⎟ 2πrdrdz
C1J 0 ⎜
⎝ Re ⎠ ⎝ He ⎠
⎛ 4 R R H ⎞ ⎛ 2.405Rcore ⎞ ⎛ πH ⎞
C1 ⎜ core e e ⎟ J 1 ⎜
⎟ sin⎜
⎟
⎝ 2.405 ⎠ ⎝
Re
⎠ ⎝ 2 He ⎠
If the reactor dimensions are large compared to the extrapolation distances Re ≅ Rcore , He ≅ H and the above
reduces to
q ′′ =
1
π
⎛ 4 ⎞
⎟ J ( 2.405)
C1 ⎜
⎝ 2.405⎠ 1
The power peaking factor is then
Fq =
C1
(2.405)π
=
= 3.64
4J 1 ( 2.405)
⎛ 4 ⎞
⎟ J 1 (2.405)
C1 ⎜
π ⎝ 2.405⎠
q max
′′
=
1
q ′′
51
The fuel in power reactors is normally loaded in such a way as to reduce the power peaking factor, with typical
values around 2.3.
Example:
A nuclear reactor is to be constructed to produce 3411 Mwt, 97.4 % of which is produced in the fuel. The fuel
elements are in the form of cylindrical rods. The power distribution predicted by a neutronics analysis indicates a
core power peak-to-average ratio of 2.5. Accident analyses place the maximum allowable linear heat rate at 13.58
kW/ft at any point in the core and requires the maximum core heat flux not exceed 474,500 Btu/hr-ft2 at any point.
a) If the fuel rods are to be 12 feet long, how many fuel rods are required?
b) What diameter are the rods?
c) In light water reactors, the fuel rods are typically arranged in a square lattice. Assuming a rod pitch (center-tocenter spacing) of 0.496 inches, what is the effective diameter of the core?
SOLUTION
a) The number of rods in the core are related to the total core power and the linear heat rate through the relationship
q′
γ f Q& = nqc′ H = n max H
Fq
The number of rods in the core is then
Fq γ f Q&
n=
Hq max
′
=
(2.5)(0.974)(3411 × 103 )
= 50,968
(12)(1358
. )
b) The rod diameter is dictated by the surface heat flux. The relationship between the surface heat flux and the
linear heat rate is
′ = qmax
′′ Pw = q′max
′ πD
qmax
such that the rod diameter is
D=
qmax
′
πqmax
′′
=
(13.58 )(3413)
= 0. 03109 ft = 0.373 inches
π( 474, 500 )
c) For a rod pitch of S = 0.496 inches, the total core area is
Acore = nS 2 = (50, 968 )( 0. 496 / 12)2 = 87. 076 ft 2
The effective core diameter is then
π
De2
= 87. 076 ft 2 ⇒ De =
4
( 4 )(87. 076 )
π
= 10.53 ft
52
HEAT GENERATION DURING SHUTDOWN
In a reactor shutdown, the reactor power does not immediately go to zero, but falls off rapidly with a rate
governed by the longest lived delayed neutron precursor. This may be easily shown if we assume that reactor
power can be described by the point kinetics equations
dP ( ρ − β ) Pk
=
+
dt
l
∑λ C
i
(1)
i
i
dCi β i Pk
=
− λi Ci
dt
l
(2)
or taking advantage of the definition of reactivity
dP (k − 1 − βk ) P
=
+
dt
l
∑λ C
i
(3)
i
i
dCi β i Pk
=
− λi Ci
dt
l
(4)
If we further assume that following control rod insertion, neutron multiplication is very small ( k ≅ 0 ) then the
solution for the power may be approximated by
dP P
+ =
dt l
∑λ C
i
(5)
i
i
dCi
= −λi Ci
dt
(6)
which has solution
P (t ) = P (0)e − t / l +
∑ (λ C−(λ0)) [e
i
i i
1
l
− λi t
− e−t / l
]
(7)
i
Note, l is the prompt neutron lifetime and is extremely short ( ≅ 10 −4 seconds) such that exponential terms
containing l die out quickly leaving only those asssociated with the delayed neutron precursors. As time
progresses, the short lived delayed neutron precursors also die out such that eventually only terms associated with
the longest lived delayed neutron precursor remain and control the rate at which the reactor power decays. In
addition to this residual fission energy, the reactor also continues to generate heat due to the decay of fission
fragments and activation products built up during operation. The magnitude of this heat source and the rate at
which it decays depends partly on the operating history of the core. In particular, shutdown power (Ps) depends on
the operating power level (Po), the operating time (to) at which the reactor operated at power level Po, and the
shutdown time (ts). In reality, the decay heat source is the many beta and gamma transitions of the excited nuclei
formed as fission fragments or neutron capture products. To account for all, or even most of these decay chains is
impractical at best for routine estimates of the decay heat rate. As a result, empirical fits have been developed
which relate the ratio of the decay power or shutdown power of the reactor to the operating power in terms of the
operating and shutdown times. The spatial distribution of decay heat can be assumed to follow the operating power
distribution. To obtain an idea of the rate at which decay heat is built up in a reactor core the following table is
provided.
Ps/PoVersus Operating Time
53
Operating Time (sec)
Ps/Po
0
60
3,600
36,000
360,000
3,600,000
infinite
0
0.0328
0.0567
0.0635
0.0665
0.0682
0.0699
Table 1
The rate of decay of shutdown power is seen by the following table for an infinite operating time. These tables
contain information from the ANS Standard for computing decay heat.
Ps/Po Versus Shutdown Time
Shutdown Time (sec)
Ps/Po
1
10
100
1,000
10,000
100,000
0.0625
0.0500
0.0331
0.0185
0.0097
0.0048
Table 2
Note: The decay heat rate approaches its equilibrium value very quickly, and as a result
the decay heat rate is a strong function of the reactor operating history with the most
recent operating conditions being the most influential.
54
Example:
An empirical relationship for determining decay heat rates is given by
P
= 0.1 ( t s + 10) −0.2 − ( t o + t s + 10) −0.2 + 0.87 ( t o + t s + 2 × 107 ) −0.2 − 0. 87 ( t s + 2 × 107 ) −0.2
P0
where to is the time (seconds) the reactor operated at power P0 and ts is the time (seconds) since reactor shutdown.
a) Compute and plot the decay heat rate as a function of operating and shutdown times for operating times of
one day, one week, one month and one year.
b) Determine the operating time required for the decay heat rate to reach 95% of its equilibrium value.
c) A reactor operates for the first 6 months of a one year cycle at 50% power, and the remaining six months at
100 % power. Compare the decay heat rate at the end of the year to that which would be obtained had the
reactor operated for the entire year at 100 % power.
SOLUTION
a) The decay heat rate as a function of operating and shutdown time is given in the following graphs
0.06
0.05
Decay Heat Rate
0.04
Ptot 0.03
k
0.02
0.01
0
0
5
10
15
20
25
30
35
40
45
50
t
k
Time (hours)
Decay heat rate for operation and shut down times of one day
55
0.06
0.05
Decay Heat Rate
0.04
Ptot 0.03
k
0.02
0.01
0
0
2
4
6
8
10
12
14
16
t
k
Time (days)
Decay heat rate for operation and shut down times of one week
0.06
0.05
Decay Heat Rate
0.04
Ptot 0.03
k
0.02
0.01
0
0
10
20
30
40
50
60
70
t
k
Time (days)
Decay heat rate for operation and shut down times of one month
56
0.06
0.05
Decay Heat Rate
0.04
Ptot 0.03
k
0.02
0.01
0
0
100
200
300
400
500
600
700
800
t
k
Time (days)
Decay heat rate for operation and shut down times of one year
b) The equilibrium decay heat rate (P∞) is 6 % of the operating power level for this correlation. The time to reach
95 % of this value (P = 5.7 %) is found by iteratively solving the above decay heat equation for to, with ts = 0.
The resulting operating time is 15.12 days.
c) For the first six months, the decay heat rate is due only to continuous operation at 50 % power. For the second
six months, the decay heat rate is the sum of the decay heat rates from operation at 50 % power for the total
operating time and the decay heat rate from operating at 50 % power for the second six month period. For the
decay heat rate written as
⎧0
⎪
P ( Po , t o , t s ) = ⎨
⎪ P × 01
. (t s + 10) −0.2 − (t o + t s + 10) −0.2 + 0.87(t o + t s + 2 × 107 ) −0.2 − 0.87(t s + 2 × 107 ) −0.2
⎩ o
[
to < 0
]
to > 0
this may be expressed as
Pop ( to ) = P(.5, to , 0 ) + P(.5, to − 6 months , 0 )
The reference decay heat rate is
Pref ( to ) = P(1, to , 0 ) .
The resulting decay heat rates for one year of operation are illustrated below.
57
0.06
0.05
Decay Heat Rate
0.04
Pop
i
0.03
Pref
i
0.02
0.01
0
0
50
100
150
200
250
300
350
400
t
i
Time (days)
The initial steep rise in the decay heat rate upon initiation of reactor operation, and corresponding steep decline
in the decay heat rate upon reactor shutdown is due to the buildup and decay of the short lived fission products.
It is obvious from these graphs, that the short time behavior of the decay heat is dominated by the most recent
operating conditions.
Example:
A power reactor operates at 3400 Mwt for one year. Determine the decay heat rate 5 minutes following reactor
shutdown.
SOLUTION
We again use the empirical equation for decay heat
P
= 0.1 ( t s + 10) −0.2 − ( t o + t s + 10) −0.2 + 0.87 ( t o + t s + 2 × 107 ) −0.2 − 0. 87 ( t s + 2 × 107 ) −0.2
P0
t o = 1 year = 316
. × 107 seconds
t s = 5 minutes = 300 seconds
From the given equation for decay heat
Ps
= 0.028 ⇒ Ps = (3400)(0.028) = 95 Mwt
Po
58
It should be obvious from this level of decay heat, that core cooling is necessary even when the reactor is shutdown.
In power reactors, the lack of sustained cooling can result in severe structural damage, including core melt.
Alternate Approach
An alternate approach to the purely empirical correlations given above, is to assume the numerous components
of the decay heat source can be lumped into a relatively small number of “groups”, similar to the approach taken
with delayed neutrons. If q j is the concentration of decay heat group j, then q j is assumed to satisfy the simple
balance equation
dq jV f
dt
= E j Σ f φ (t )V f − λ j q jV f
1424
3 1
424
3
fission
rate
loss rate
142
4
43
4
(8)
production rate
where E j is the yield fraction and λ j the decay constant for decay heat group j. As was shown previously, the heat
production rate is proportional to the fission rate, such that
dGq jV f
dt
= E j GΣ f φ (t )V f − λ j Gq jV f
14243
123
(9)
γj
Po (t )
where G is energy per fission and Po (t ) the total reactor power. The balance equation for decay heat group j can
then be written
dγ j
dt
= E j Po (t ) − λ jγ j
(10)
and the total decay heat source is
∑λ γ (t)
P(t ) =
j
(11)
j
j
In principle, Equation 10 can be solved for any operating history. Consider the special case of an infinite operating
time, such that the system has reached equilibrium and
dγ j
dt
=0
Then from Equation 10,
γ j (∞) =
E j Po
If the system is then shut down, then for any shutdown time t s , γ
dγ j
dt
(12)
λj
= −λ j γ j
j
is the solution of
(13)
subject to the initial condition
59
γ j (t s = 0 ) = γ j ( ∞ ) =
E j Po
(14)
λj
Solution of Equation 13 gives
γ j (t s ) =
E j Po
λj
exp(−λ j t s )
(15)
and the decay heat source is
P (t s )
≡ Γd (t s ) =
Po
∑ E exp(−λ t )
j
j s
j
Typical values for the yields and decay constants are given in Table 3 below.
Group
Ej
λ j (sec-1)
1
2
3
4
5
6
7
8
9
10
11
0.00299
0.00825
0.01550
0.01935
0.01165
0.00645
0.00231
0.00164
0.00085
0.00043
0.00057
1.772
0.5774
6.743 x 10-2
6.214 x 10-3
4.739 x 10-4
4.810 x 10-5
5.344 x 10-6
5.726 x 10-7
1.036 x 10-7
2.959 x 10-8
7.585 x 10-10
Table 3 Decay Heat Group Constants (From RETRAN code manual)
60
HEAT GENERATION IN REACTOR STRUCTURE
Gamma and neutron radiation emanating from the reactor core interacts with and is absorbed by structural
materials such as core barrels, pressure vessels, etc. The absorbed radiation is converted into heat which must be
removed. As core barrels and reactor pressure vessels are relatively thin compared to their diameter, they can be
accurately approximated as slabs. Consider a monodirectional, monoenergetic gamma flux incident upon a slab
wall as illustrated below.
x=0
x=L
Figure 1: Gamma Radiation Incident on a Slab Wall
The interaction rate within the slab is given by μφ( x ) where μ is the total attenuation coefficient. Photon
interactions inevitably lead to the production of short range electrons through Compton scattering, pair production,
and the photoelectric effect. We can again assume the deposition of the electron energy occurs locally such that the
heat generated by photons is proportional to the photon interaction rate. We therefore write the heat generation rate
as
q ′′′( x ) = E μ a φ( x )
(1)
where μ a is the energy absorption coefficient for the slab material at photon energy E. Values of attenuation
coefficient and energy absorption coefficient are given in Table 1. The energy absorption coefficient accounts for
the fraction of the incident photon energy carried by the electrons after an interaction. For the simple example of a
monodirectional beam, the uncollided gamma flux within the slab is
φo ( x ) = φ 0 exp( − μx )
(2)
To account for the contribution of scattered photons we introduce the concept of a Buildup Factor B( E , μx ) where
the Buildup Factor is defined as
B ( E , μx ) ≡
Total Energy Absorbed at x from Scattered and Unscattered Photons of Incident Energy E
Energy Absorbed at x from Unscattered Photons of Incident Energy E
(3)
The volumetric heat generation rate for this example would then be
q ′′′( x ) = E μ a φ 0 exp( − μx ) B( E , μx ) .
(4)
The Buildup Factor is an empirical fit to data obtained from detailed radiation transport calculations and is available
in most standard shielding texts.
61
Photon Energy
(Mev)
0.5
1.0
1.5
2.0
3.0
5.0
10.0
Attenuation
Iron
Water
μ = 0.0966
μa = 0.0330
μ = 0.0706
μa = 0.0311
μ = 0.0574
μa = 0.0285
μ = 0.0493
μa = 0.0264
μ = 0.0396
μa = 0.0233
μ = 0.0301
μa = 0.0198
μ = 0.0219
μa = 0.0165
μ = 0.651
μa = 0.231
μ = 0.468
μa = 0.205
μ = 0.381
μa = 0.190
μ = 0.333
μa = 0.182
μ = 0.284
μa = 0.176
μ = 0.246
μa = 0.178
μ = 0.231
μa = 0.197
Coefficients (cm-1)
Lead
μ = 1.640
μa = 0.924
μ = 0.776
μa = 0.375
μ = 0.581
μa = 0.285
μ = 0.518
μa = 0.273
μ = 0.477
μa = 0.284
μ = 0.483
μa = 0.328
μ = 0.554
μa = 0.419
Concrete
μ = 0.2040
μa = 0.0700
μ = 0.1490
μa = 0.0650
μ = 0.1210
μa = 0.0600
μ = 0.1050
μa = 0.0560
μ = 0.0853
μa = 0.0508
μ = 0.0674
μa = 0.0456
μ = 0.0538
μa = 0.0416
Table 1: (Photon Attenuation and Energy Absorption Coefficients, from Todreas and Kazimi)
For energy distributions other than monoenergetic, the volumetric heat generation rate is somewhat more
complicated. We consider two cases: one where the incident photon flux consists of a finite number of discrete
photon energies, and the second where the incident photon flux is a continuous spectrum of energies.
Multiple Discrete Photon Energies
For the case of multiple discrete photon energies, the volumetric heat generation rate is the sum of the heat produced
by each incident photon, i.e.
q ′′′( x ) =
∑E μ
i
ai
φ 0i exp( − μ i x ) B( Ei , μ i x )
(5)
i
where the subscript i denotes the individual energies.
Continuous Spectrum of Photon Energies
If the incident photon flux is a continuous spectrum of energies, the volumetric heat generation rate is obtained by
integrating over all incident energies, i.e.
q ′′′( x ) =
∫
∞
E μa φ0 ( E ) exp( − μx ) B( E , μx )dE
(6)
0
where φ 0 ( E ) contains the incident photon spectrum.
62
HEAT GENERATION FROM RADIOISOTOPE SOURCES
Radioisotope power sources produce heat as a result of exothermic decay reactions. This heat is then converted
to electricity through thermoelectric devices or other similar means. Charged particles emitted during the decay
process have kinetic energies corresponding to the mass defect of the decay reaction. We again assume that the
kinetic energy of these particles is deposited locally at the point of decay, such that the heat generation rate is
proportional to the spatial distribution of the radioisotope within the power source. This is generally uniform. If
ΔE is the energy associated with the mass defect of the decay reaction, then the volumetric heat generation rate due
to radioactive decay is
v
v
q ′′′(r ) = ΔEλN (r )
(1)
= Decay constant of the radioisotope
where: λ
v
N (r ) = Radioisotope number density as a function of position
Example: (Adapted from Example 4-6, El-Wakil)
A radioisotope power source is fueled with 475 gm of Pu 238 C, 100 % enriched in Pu 238 . If the density of PuC is
12.5 gm/cm3, calculate the volumetric heat generation rate and total thermal output of the radioisotope source.
SOLUTION
Pu238 decays to U234 via alpha emission with an 86 year half life, i.e.
94 Pu
238
→92 U 234 + 2 He4 .
U234 has a half life of 2.47 x 105 years and relative to Pu238 can be considered stable. The mass defect is given by
Δm = 238. 0495 − 234. 0409 − 4. 0026
= 0. 006 amu
or in terms of energy
Mev
Δm
amu
= (931)( 0. 006 )
ΔE = 931
= 5.586 Mev/reaction
The decay constant is related to the half life through
λ=
0. 693 0. 693
0. 693
=
=
= 2.5519 × 10−10 sec−1 .
t 12
86 yr 2. 7156 × 109 sec
The Pu238 number density is given by
N=
ρPuC Av
M PuC
=
(12.5)( 6. 023 × 1023 )
= 3. 01 × 1022 nuclei/cm 3
238. 041 + 12. 01
such that the volumetric heat generation rate is
63
q ′′′ = ΔEλN = (5.586 )( 2.5519 × 10−10 )(3. 01 × 1022 )
= 4. 291 × 1013 Mev/cm 3− sec.
The total thermal output of the device is
Q = q′′′V = q ′′′
m
ρ
= ( 4. 291 × 1013 )( 475 12.5)
= 1. 631 × 1015 Mev sec
= 261 W
64