Chapter9_Thermal_performance dec09

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Chapter 9. Thermal Performance
9.1. Introduction ............................................................................................................. 2
9.2. The Fission Heat Generation .............................................................................. 2
9.2.1. Energy Release from fission ............................................................................... 2
9.2.2. Fuel Burnup ......................................................................................................... 3
9.3. Fission Heat Removal ........................................................................................... 4
9.3.1. Heat Conduction in the Fuel .............................................................................. 4
9.3.2. Heat Transfer Resistances .................................................................................. 7
Gap Closure ................................................................................................................ 8
9.4. Axial Temperature Profile .................................................................................. 10
9.5. Thermal Conductivity .......................................................................................... 12
9.5.1. Thermal Conductivity in Porous Oxide .......................................................... 12
9.5.2.Thermal Conductivity Variation with Temperature ..................................... 14
9.6. Thermal Margins and Operational Limits ...................................................... 14
9.6.1. The LOCA ......................................................................................................... 15
9.6.2. Departure from nucleate boiling ..................................................................... 15
9.6.3. Cladding Temperature Limits ......................................................................... 16
9.6.4 Stored Energy ..................................................................................................... 17
Problems ........................................................................................................................ 17
References ..................................................................................................................... 20
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2/6/2016
9.1. Introduction
One of the most important factors in designing nuclear power plants is the calculation of
the power produced in the core and its removal by the coolant. To that end, coolant is
circulated through the core and heat flows from the fuel rods to the coolant, which
experiences a temperature rise as it passes through the core, as explained in Chapter 1.
Since the heat is generated inside the fuel rods, temperature gradients are established
inside the rods that enable the heat to flow outwards from the rods to the coolant, which
results in a temperature profile within the rods. This temperature profile has a strong
influence on the mechanical properties, on the microstructure evolution under irradiation
and on corrosion processes occurring in the fuel and the cladding. Thus, a sound
understanding of the factors governing the temperature distribution within a reactor fuel
element is essential to predicting its performance during reactor exposure.
Both the high temperatures and the steep temperature gradients are important for
predicting fuel element performance. The temperature controls processes such as graingrowth, densification, fission product diffusion in the fuel and radiation damage
accumulation, and corrosion rates in the fuel cladding. The temperature gradients in the
fuel cause pore migration and formation of the central void, cause thermal stresses or fuel
cracking, and pellet-cladding interaction. In the cladding, temperature gradients cause
hydride rim formation and overall hydrogen redistribution.
The topic of this chapter is the thermal performance of the fuel rod and the core. This
includes a calculation of the temperature distribution in the fuel rod and in the core, and
its change with reactor exposure. The thermal margins and accident conditions are also
briefly discussed.
9.2. Fission Heat Generation
Nuclear energy is generated in light-water reactors by the fission of uranium induced by
absorption of neutrons. The fission of the uranium atom splits it into two smaller parts
(the fission products), and releases two to three energetic neutrons (average energy ~ 2
MeV), as well as other particles such as betas, gammas and neutrinos. The neutrons
released in the fission reaction cause fissions in other nuclei, a process called fission
chain reaction.
9.2.1. Energy Release from fission
The overall energy release in a single fission reaction is 200 MeV, about 90% of which is
deposited in the fuel pellet. Although every nuclide in the transuranic region can be
fissioned if enough energy is imparted to it, the fission cross-section for certain fissile
nuclides (U-233, U-235, Pu-239, Pu-241) is greatly increased if the neutron energy is
reduced to values close to the thermal energy (~ 0.025 eV at reactor temperature). In
thermal reactors (such as the light water reactors considered in this book), this energy
reduction is achieved by passing the neutron flux through a moderator (water) which
causes the neutrons to lose energy by successive collisions until they are in thermal
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equilibrium with their surroundings. To ensure that criticality (a fission chain reaction at
a constant rate) can occur with thermal neutrons, natural uranium is enriched to a larger
percentage of the isotope U-235 (3-5% in current reactors).
In parallel with neutron-induced fission, neutron absorption can occur in uranium,
causing transuranic (heavier than uranium) elements to be formed. Of chief importance
among these is Pu-239, which forms directly by neutron absorption in U-238 followed by
beta decay. This creates another fissile isotope which contributes to the fission rate. In
parallel with the depletion of U-235, Pu-239 is created from U-238 so that at the end of
the fuel residence time in the reactor, up to one third of the reactor energy is produced by
plutonium fission. Not all the fissile material (U and Pu) is used by the time the fuel is
removed from the reactor, so it is also possible to reprocess the spent fuel to extract U
and Pu to fabricate mixed-oxide fuel (MOX), which consists of (U,Pu)O2.
Thus, there are various sources of heat in a reactor core: fast fission of U-238, thermal
fission of U-235 and Pu-239, and the radioactive decay of fission products. The last term
is important in loss-of-coolant accident scenarios, since shutting down the fission chain
reaction does not eliminate this heat source and cooling needs to be provided to avoid
fuel damage (see chapter 28). However, during normal operation the most significant heat
generation term is thermal fission in U-235 and Pu-239 (the fast fission rate is a few
percent that of thermal fission).
The first step in calculating the temperature profile is to calculate the rate of heat
production by nuclear reactions, mostly nuclear fission. If we consider a fuel element at
the beginning of life, containing enriched uranium and take thermal fission in U-235 to
be the dominant heat production mechanism, the fission rate is given by
F ( fission / cm3s)   f   N f  f   qNU f 
(9.1)
wheref is the macroscopic thermal fission cross section (cm-1),  is the thermal neutron
flux (n.cm-2.s-1), Nf is the fissile atom density (fissile atom/cm3), f the thermal fission
cross section (barn), q is the enrichment (fissile atom density/uranium density) , and NU
is the uranium density (atom/cm3). The theoretical atom density in uranium dioxide is
2.5x 1022 atom/cm3, while the thermal fission cross-section for U-235 is ~ 550 barn.
9.2.2. Fuel Burnup
The term burnup indicates the degree of usage of the fuel although it clearly has nothing
to do with burning but rather with fissioning. There are three common measures of the
integrated amount of irradiation to which the fuel has been subjected. The first is the
fission density given by
t
F   F (t )dt  Ft [ fission / cm3 ]
0
for a constant fission rate. The second is the fractional burnup  given by
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(9.2)

number of fissions
F

initial number of uranium atoms NU
(9.3)
Finally, the burnup can be given as the number of megawatts days of thermal energy
released by a fuel containing one metric ton of uranium. The 200 MeV released in each
fission corresponds to 0.95 MW.day per gram fissioned. Thus
 [MWd / MTU ]    0.95 106  9.5 105  [%]
(9.4)
At the end of life, fuel burnup used to be about 3%, and thus approximately equal to
30,000 MWd/MtU Average fuel discharge burnups have been increasing for the past two
decades, now reaching over 50,000 MWd/ton for pressurized water reactors and
approaching that value for boiling water reactors [1]. The Nuclear Regulatory
Commission has established a limit of 62,500 MWd/ton. Current efforts exist to certify
fuel for higher burnups, as discussed in Chapter 28.
9.3. Fission Heat Removal
The thermal design of a fuel element is constrained by various limits. It is necessary to
stay well below the fuel and the cladding melting temperatures at all times. It is also
desirable to increase as much as possible the coolant outlet temperature from the core to
maximize thermal efficiency. It is undesirable to have large temperature gradients in the
fuel elements, as that leads to a variety of issues, as mentioned above. These constraints
would dictate a fuel with high thermal conductivity and high melting point. This is
achievable with metallic fuels, whose thermal conductivities are much higher than that of
UO2. However, while metallic fuels exhibit serious dimensional instabilities under
irradiation (Ch.27), uranium dioxide turns out to be a remarkably stable fuel matrix: it
can undergo thermal cycling, can withstand severe amounts of radiation damage and can
accommodate within its lattice the fission products and transuranic atoms produced
during irradiation. The melting temperature of uranium dioxide is in the temperature
range 2800-2865°C, as shown in the phase diagram in Chapter 2.
9.3.1. Heat Conduction in the Fuel
The removal of heat from the cylindrical fuel element occurs in the radial direction,
through a series of heat resistances, by conduction and convection. The heat transfer
geometry is shown in Figure 9.1.
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TFC
TFS TCI TCS TW
R+tgap+ tC
R+tgap
R
Figure 9.1: The heat transfer geometry in a nuclear fuel rod
Heat is generated in the fuel pellet radius R and flows radially through the fuel, the pelletcladding gap, and the cladding itself to reach the coolant. Because the axial temperature
variation is relatively small, and there is little azimuthal temperature variation, the steadystate heat conduction equation is written as:
1 d 
dT 
 r F
  q '''  0
r dr 
dr 
(9.5)
where T is the temperature, r the radial position in the pellet, F is the fuel thermal
conductivity and q ''' is the volumetric heat generation rate (W/cm3) given by
q '''  FEF  3.2 1011 F [W / cm3 ]
(9.6)
where EF is the energy deposited in the fuel pellet with each fission (about 180 MeV).
Equation (9.5) assumes a heat generation rate that is independent of r and of T (this is
not strictly true, because of self-shielding, and other flux gradient effects.) If the variation
of q ''' with r is known, an average volumetric heat generation rate q ''' can be calculated.
One useful quantity is the linear heat generation rate (W/cm) obtained by
q '   R 2 q '''
(9.7)
Equation (9.5) can be solved if two boundary conditions are specified. These are:
T ( R)  TFS
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(9.8)
dT
dr
0
(9.9)
r 0
where TFS is the fuel surface temperature .
Integrating equation (9.5) twice we obtain
T (r )  TFS  
q ''' r 2
 C1 ln r  C2
4 F
(9.10)
Applying the boundary conditions:
T (r )  TFS 
q ''' R 2  r 2 
q'
1  2  
4 F  R  4 F
 r2 
1  2 
 R 
(9.11)
or
T (r )  TFS
r2
 1 2
TFC  TFS
R
(9.12)
where TFS is the fuel pellet surface temperature and TFC is the fuel centerline
temperature. Thus a parabolic temperature profile is established whenever heat
generation in the volume is homogenous. The equivalent forms of equation (9.11) for
plate and sphere geometry are respectively:
T (r )  TFS 
q ''' L2  x 2 
1   for plate
2 F  L2 
q ''' R 2 
r2 
T (r )  TFS 
1


 for sphere
6 F  R 2 
(9.13)
where L is the plate half-thickness and R is the radius of the sphere.
From equation (9.12) it follows that the linear power at a given axial position is given by
q '  (TFC  TFS )4 F
(9.14)
This heat flux flows from the fuel through the fuel-cladding gap and through the cladding
and into the coolant. Heat transfer through the cladding is achieved by conduction.
Although heat transfer through the gap occurs through a gas, no flow or convective
effects exist. As a result the process can be thought of as one of conduction with
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 gap  hgap t gap . where is the gap thickness and hgap the convective heat transfer
coefficient in the gap.
Solving the heat transfer equations for the cladding we find that the temperature drop
through these two hollow cylinders is given by
TFS  TCI  q '
TCI  TCS  q '
ln[( R  t gap ) / R ]
2 k gap
ln[( R  t gap  tC ) /( R  t gap )]
2 C
(9.15)
(9.16)
where TCI is the cladding inner radius temperature, TCS is the cladding outer radius
temperature,  C is the cladding thermal conductivity and tC is the cladding wall thickness.
Heat transfer from the cladding surface to the coolant is achieved by convection.
q'
(TCS  TW ) 
(9.17)
h
where TW is the bulk coolant temperature and h is the convective heat transfer coefficient
between the cladding wall and the coolant, given for example by the Dittus-Boelter
relation [2].
If t gap  R then
q '  (TFC  TFS )4 F 
2 k gap R(TFS  TCI )
t gap

2 C (TCI  TCS )
 h(TCS  TW ) (9.18)
ln[( R  tC ) / R]
9.3.2. Heat Transfer Resistances
Using the well-known electric analogy, we can write that the temperature difference T
(analogous to voltage), established through a material with a certain thermal resistance R
(analogous to resistivity) creates a heat flux q’ (analogous to electric current) given by
q' 
T

(9.19)
The resistances are all in series, such that the total thermal resistance  in going from
the centerline of the fuel to the coolant is given by the sum of the individual resistances.
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   i 
1
4 F

t gap
2 gap

1
 R  tC 
ln 


2 C  R  2 ( R  tC ) h
1
(9.20)
For the case of the fuel rod when the gap is closed (tgap=0 and thus TFS=TCI) then
equation (9.19) is written:
TFC  TW
(9.21)
t gap
1
1
1
 R  tC 


ln

4 F 2 gap R 2 C  R  2 ( R  tC )h
where TW is the cooling water temperature, h is the convection coefficient from the clad
to the coolant, and tC and C are the cladding thickness and thermal conductivity.
q' 
Gap Closure
The change in gap width upon fuel and cladding expansion is given by
hot
cold
tgap  tgap
 t gap
 Rc  R
(9.22)
where Rc is the change in cladding radius and R is the change in fuel radius with
temperature. It can be shown (see problem 9.1) that the radial swelling strain of the fuel
when a parabolic profile is present is given by
R
  F (T  T fab )
(9.23)
R
where T is the average fuel temperature and Tfab is the fuel fabrication temperature and
F is the coefficient of thermal expansion of UO2 . The corresponding swelling increase
for the cladding is approximately equal to
 FR 
Rc
  c (Tc  T fab )
Rc
Thus the change in gap width upon heating is given by
hot
cold
tgap  tgap
 tgap
 Rcc (Tc  Tfab )  RF F (T  Tfab )
If an additional strain from fuel fission gas swelling occurs, it can be inserted in the
equation above.
Example 9.1: Calculation of a Fuel Rod Temperature Profile
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(9.24)
(9.25)
We can now evaluate the temperature profile for a typical fuel rod. The properties listed
in table 9.1 can be used for this purpose;
Table 9.1. Thermal properties of fuel materials
Material
Cp (J/gK)
 (g/cm3)
UO2
10.98
0.330
Zircaloy
6.5
0.35
Steel
8.0
0.5
(W/mK)
(K-1)
3
17
17
1.45x10-5
5-10x10-6
9.6x10-6
For this calculation we assume
TW=310°C
q’=15000 W/m
h=20000 W/m2K
Fuel pellet radius R=0.5 cm
Fuel cladding thickness tC= 0.06 cm
Gap width tgap= 0.03 cm
from this we calculate the temperature drops in each resistance:
From the coolant bulk through the coolant film to the cladding wall= 21.3 °C
Through the cladding thickness = 15.9 °C
In the fuel-cladding gap= 4.7 °C
From the outer skin to the centerline of the fuel pellet=1250 °C
Given the above, the fuel centerline temperature is 1602 °C.
This is shown as the blue curve in Figure 9.2. It is clear that the greatest contributor to the
increase in centerline temperature is the temperature rise in the fuel, but the gap can have
a significant impact as well. This is illustrated by keeping all parameters the same, while
substituting xenon for helium. As burnup increases, xenon gas is released into the fuelcladding gap degrading its thermal conductivity. If the xenon conductivity of 0.007565
W/mK is used instead of that of helium, we obtain a temperature drop in the gap of 94.7
°C (an increase of almost 100 °C), leading to a centerline temperature of 1691 °C (red
curve). All other factors being equal, every temperature rise in the coolant will cause a
corresponding temperature rise in the centerline temperature.
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2000
Coolant
Fuel rod
1800
Gap
1600
Fuel cladding
Temperature (C)
1400
1200
1000
800
600
Bulk Coolant
Temperature
400
200
0
0.00
0.20
0.40
0.60
0.80
1.00
Radial Distance (cm)
Figure 9.2 Temperature Profile in a fuel rod (see example 9.1)
9.4. Axial Temperature Profile
Although the most significant temperature variation is in the radial direction it is also
important to calculate the axial temperature variation. The axial power distribution can be
written as
 z 
(9.26)
q '( z )  qo 'cos 

 H core 
where qo’ is the linear generation rate at the centerline of the core, and Hcore= zoutlet-zinlet
is the height of the core.
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In addition to a radial temperature profile within the fuel rods, an axial temperature
profile also exists, created by the combination of an axial fission rate profile and by the
changing temperature of the coolant as it passes through the core, starting out cold and
leaving hot. This temperature profile can be evaluated using a heat balance on the
coolant. The overall rise in temperature of the coolant is sufficient to remove the heat
generated by nuclear fission.
Fig.9.3:
For the element of coolant shown in Figure 9.3, the enthalpy of the fluid entering the
element is
QC pW (TW  Tref )
(9.27)
where Q  m / N rods is the mass flow rate associated with one fuel rod, C pW is the
specific heat of water, and Tref is a reference water temperature. Upon exiting the control
volume, the enthalpy increase of the water is
dTW
dz
(9.28)
dz
At steady-state this increase is equal to the energy produced in the fuel section of the
control volume
dT
QC pW W dz  q ' dz
(9.29)
dz
Assuming no radial flux variation, an energy balance in the coolant gives
QC pW
QC pW (Toutlet  Tinlet ) 
zoutlet

q '( z )dz
(9.30)
zinlet
where Q is the core mass flow rate in (kg/s), Cp is the coolant specific heat (J/kg°C),
Tinlet and Toutlet are the inlet and outlet coolant temperatures. Inserting equation (9.26) into
equation (9.30) and integrating to a height z instead of to the top of the core will yield the
temperature of the coolant at height z.
qo ' H core    z  
sin 
  1
 QC pW   H core  
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TW ( z )  TWinlet 
(9.31)
As expected the highest temperature occurs at the outlet.
9.5. Thermal Conductivity
The heat conduction equations derived above apply only when the thermal conductivity is
constant (does not depend on temperature, chemistry or fuel microstructure). This section
discusses the thermal conductivity of the fuel rod constituent materials, (especially the
fuel), their variation with temperature, composition and burnup and the methods used to
take these variations into account. Because most of these variations occur in the fuel, the
next sections discuss fuel thermal conductivity. In the as-fabricated fuel the thermal
conductivity depends on many parameters, as discussed in the following sub-sections.
For the cladding the thermal conductivity is most strongly affected by the formation of
the outer wall oxide (as discussed in chapter 22).
The thermal conductivity of the fuel, gap and cladding is essential to determining the
temperature distribution and transient thermal response of the fuel rod. The thermal
conductivity of the fuel also determines the amount of stored heat in the fuel which is a
consequence of the large gradients that have to be established order for the heat to flow
out at steady state.
9.5.1. Thermal Conductivity in Porous Oxide
When uranium dioxide is fabricated by sintering (see chapter 16) it is possible to control
the sintering conditions so that the pores initially present are eliminated to varying
degrees during the sintering process, resulting in a solid with less than the ideal
theoretical density of 10.95 g/cm3. The presence of the pores allows for free space to
accommodate fission gases, thus reducing swelling. However, the effect of those pores is
to diminish the overall thermal conductivity of the material. In addition, during
irradiation porosity develops in the form of bubbles and voids whose distribution varies
with radius, linear power and with burnup.
It is then necessary to evaluate the thermal conductivity of the composite material, which
is done in the following
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Figure 9.4. Geometry for analysis of effect of pores on thermal conductivity [3]
Consider a homogenous distribution of cube-shaped pores. We can then define a “unit
cell of volume associated with each pore, as shown in Figure 9.4. If we consider that heat
flow occurs along the y direction, then two pathways exist for the heat to transverse the
cube: through the “pore tube”, or around it. Clearly when going around the pore, the
thermal conductivity will be that of the dense UO2 while going through the tube the
thermal conductivity will be a composite of pore and uranium dioxide. The overall
conductivity is then
 F  (1  f P )UO  f P tube
(9.32)
where fP is the fraction of cross sectional area perpendicular to heat flow occupied by the
pore. Each tube consists of a square pillar conductity of dense UO2 and of a pore. The
fraction occupied by the pore in the tube is ftube. Then the effective conductivity of the
tube is
f
(1  f Ptube )
1
(9.33)
 Ptube 
 tube  pore
UO2
By substituting (9.33) into (9.32) we obtain
2



2
1  [(1  ftube ) / ftube ]( pore / UO2 ) 
 ( /  ) 
 F  (1  f P )UO2 1  pore UO2 
ftube


 F  (1  f P )UO 
1  ( pore / UO2 )
(9.34)
(9.35)
Note that the porosity pF is given by
pF  f p f tube
(9.36)
f p  pF 2/ 3
(9.37)
For cubic pores
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ftube  pF 1/ 3

 F  (1  pF2/ 3 )UO 1 
2
For the case where the  pore is <<  UO2 then

(9.38)
( pore / UO2 ) 

p1/F 3

 F  (1  pF2/3 )UO
2
(9.39)
(9.40)
9.5.2.Thermal Conductivity Variation with Temperature
The integration of equation (9.5) was performed assuming that the thermal conductivity
is independent of the radial position r. In fact the thermal conductivity of UO2 varies
significantly with temperature [3] which in turn varies markedly with radial position.
For example an empirical fit of data has given the following equation for thermal
conductivity of UO2
 F  0.0130 
1
[W / cm. C ]
(0.038  0.45 pF )T
(9.41)
To take into account the dependence of thermal conductivity with temperature it would
then be necessary to solve the heat conduction equations with  UO2 being f(T). This would
normally be done by numerical methods.
9.6. Thermal Margins and Operational Limits
A nuclear reactor is different from conventional power plants in that because of the need
to avoid fuel damage, the operational constraints are on the maximum temperature in the
core, rather than on the average temperature. Reactor operational conditions depend both
on the maximum allowable fuel rod temperature (centerline temperature) and the
maximum allowable cladding temperature. Because the neutron flux and coolant
temperature vary axially and radially through the core, so do the fuel rod temperatures.
To base the normal operation limits on the highest maximum temperature they have to be
calculated for the hottest channel.
The power profiles that give rise to the temperature profile can be characterized using the
power peaking factor
PF 
 (r ) q '(r )

 (r ) q '(r )
(9.42)
It is necessary to set thermal limits and to ensure that the reactor does not exceed those
during normal operation or accident conditions. In light-water reactors the accidents of
greatest concern (those associated with the greatest potential for fuel damage) are those in
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which the temperature of the fuel is raised beyond acceptable limits, either (i) through the
production of excessive power (as in a reactivity initiated accident, Chapter 28) or (ii)
having too little coolant, as in a loss of coolant accident (LOCA).
9.6.1. The Loss-of-Coolant Accident (LOCA)
The loss of coolant accident is a design-basis accident in light water reactors. The
postulated accident required for analysis is a guillotine break in a main primary coolant
pipe, allowing coolant to freely discharge out of the primary system into the containment
building. In that event, although the control rods shut down the reactor, power continues
to be produced by the decay heat of the fission products. At steady-state, this amounts to
approximately 10% of the full-power thermal output of the reactor, which for a typical
1000MWe reactor amounts to ~ 300 MW. In addition, some energy is stored in the core
from the temperature gradients necessary to ensure heat flow. The combination of decay
heat and stored heat distribution can cause the cladding temperature to rise above
acceptable limits if not enough emergency cooling is supplied. In the case of loss of
cooling, the cladding temperature can rise above acceptable limits. Thus, it is necessary
to ensure that the reactor continues to be cooled. The Emergency Core Cooling System
(ECCS) then injects water into the core, allowing the cladding temperature to remain low
and avoiding fuel failure.
Figure 9.5. Cladding to water heat flux as a function of temperature
If for some reason there is a failure to deliver adequate core cooling, there is a mismatch
between power produced and heat removed, the cladding temperature rises, as does the
temperature of water/steam mixture that forms. The heat flux in such a situation is shown
in Figure 9.5. As the temperature of the water/steam mixture rises, the heat flux initially
also rises, due to the high degree of nucleate boiling that occurs, as a higher rate of
bubble generation causes improved mixing and more efficient heat transfer. Thus in the
first region the heat transfer coefficient increases as the increased mixing produced by the
bubbles causes more efficient heat transfer.
9.6.2. Departure from nucleate boiling
As the temperature of the cladding increases the bubble density increases until the
bubbles coalesce and a continuous film is formed. Once that occurs (point c), the heat
flux is severely decreased as the heat transfer coefficient of steam is much lower than that
Light Water Reactor Materials © Donald Olander and Arthur Motta
2/6/2016
of water. This departure from nucleate boiling (DNB) occurs at a critical heat flux qc’ and
can have severe consequences for the fuel. To avoid reaching the critical heat flux,
thermal margins are established. According to equation (9.42), the peaking factor in the
hottest channel needs to be such that the heat flux in the hottest channel is lower than the
critical heat flux by a safety margin in normal operation, as illustrated in Figure 9.6.
Figure 9.6. Illustration of DNB limits
The Departure from Nucleate Boiling Ration (DNBR) is then a useful measure of how
close this dangerous situation is approached.
9.6.3. Cladding Temperature Limits
Once DNB has occurred cladding temperature s quickly increases. This temperature rise,
in turn exponentially increases the rate of chemical reaction between the Zircaloy
cladding and the steam. As the reaction proceeds, the heat of reaction causes further
acceleration of the reaction rate, while the hydrogen produced and released to the
containment increases the risk of explosion.
Since DNB would only be expected to occur in highly improbable accident scenarios, it
is not necessary to show that fuel failure does not occur. Rather, it is only necessary to
demonstrate in that case that the consequences are not excessive, i.e. that fuel does not
get damaged to the point that a coolable geometry cannot be maintained. The cladding
temperature limit was established to avoid excessive reaction of the Zircaloy cladding
with water. It has been shown that a high temperature excursion and consequent reaction
with steam and oxygen absorption by the cladding causes significant post-quench
cladding embrittlement [4]. As explained in Chapter 28, such limits are on maximum
cladding temperature
TC  1204 C
and on the equivalent cladding reacted
Light Water Reactor Materials © Donald Olander and Arthur Motta
2/6/2016
(9.43)
ECR  17%
(9.44)
To avoid this scenario, a cladding temperature limit has been established of 1204°C
(2200°F) such that this runaway reaction process cannot occur.
9.6.4 Stored Energy
One of the consequences of a low fuel thermal conductivity is that large temperature
gradients are needed in the fuel pellet to drive the heat flux out from the fuel to the
coolant. The large temperature gradient means that there is a significant amount of stored
energy in the fuel. This is one of the reasons why cooling must continue to be provided
even in the case of a loss of coolant accident. In the event there was no cooling at the
cladding surface the heat redistributes across the fuel pin until the temperatures are even
across the rod. This redistribution would likely cause the cladding temperature to rise
above acceptable limits.
The stored energy per unit length of fuel rod is
Estored  C
R
F
p
 2 rdr (T  T )
S
(9.45)
0
where  C pF is the volumetric heat capacity of the fuel (J/m3C)
For a parabolic profile with constant thermal conductivity, we obtain the stored energy by
inserting equation (9.11) into equation (9.45) to obtain
q '  r2 
1 

4 F  R 2 
0
from where we obtain the stored energy per unit length of fuel rod (J/m)
Estored  C
R
F
p
 2 rdr
(9.46)
 C pF q ' R 2
(9.47)
8 F
Thus the coolant needs to provide enough heat transfer capability to remove at least this
much heat, in addition to the heat generated by fission product decay which is about 10%
of the thermally rated power at the moment of shutdown.
Estored 
Problems
9.1. Verify that the burnup when expressed in megawatt.day/metric ton is equal to 9.5 x
105 the fractional burnup.
9.2
The temperature dependence of the thermal conductivity of UO2 is given
theoretically by:
Light Water Reactor Materials © Donald Olander and Arthur Motta
2/6/2016
k= ko/(1+BT)
where ko and B are constants and T is the temperature in K. Using this equation, derive
the equation giving the temperature difference between the fuel centerline and the fuel
surface. The fuel surface temperature is Ts and the rod linear power is P.
9.3 In calculating the temperature distribution in the cladding, it is possible to neglect the
curvature and approximate the geometry as a slab.
(a) Derive the equation for the temperature difference between the inside and
outside surfaces for cladding of thickness tc and inner radius Rci on a fuel rod operating
at linear power q’. The cladding thermal conductivity is c.
(b) Show mathematically how the result of (a) reduces to for the case of thin
cladding(tc<<Rci).
(c) The cladding of PWR fuel is 0.7 mm thick and 8 mm in diameter. What is the
fractional error in the temperature difference due to the use of slab geometry?
9.4. Demonstrate equation (9.21)
9.5. Prove that the fractional thermal expansion of the outside diameter of a solid cylinder
of radius R with thermal expansion coefficient  in a parabolic temperature distribution is
 T , where T is the volume-averaged temperature rise of the solid.
9.6. Derive equation (9.13)
9.7 Duplex fuel consists of two radial zones with different enrichments of the uranium.
Such pellets reduce the fuel centerline temperature below that of conventional uniformlyenriched fuel operating at the same linear power. In a particular fuel design, the center of
the pellet to a radius ri consists of natural uranium and the outer annulus ri<r<R contains
4% enriched fuel. This fuel design is to be compared to conventional single-enrichment
fuel containing 3.2% 235U, neglecting neutron-flux depression in the pellet and assuming
a temperature-independent fuel thermal conductivity.
(a) What is the value of ri that gives the same average power density in the duplex
and homogeneous fuels?
(b) What is the ratio of the temperature difference between the fuel centerline and
surface for the two designs at the same linear power?
9.8 Solutions to transient heat-conduction problems often are expressed in terms of a
dimensionless time t/d2 and a dimensionless position r/d, where  =k/Cp is the thermal
diffusivity, d is a characteristic dimension of the body, and t and r are time and position,
Light Water Reactor Materials © Donald Olander and Arthur Motta
2/6/2016
respectively. The ratio d2/ is a characteristic time for the problem. When the imposed
time scale of the transient is much longer than the characteristic time, the quasi-steadystate form of the heat conduction equation is sufficient for the analysis. If the transient
occurs in times shorter or comparable to the characteristic time, the full time-dependent
heat conduction equation must be solved.
(a) A PWR is shut down from full power in 1/2 minute. Considering typical
properties, which components of the fuel rod(fuel and cladding) can be treated by quasisteady-state heat conduction during this transient?
(b) The first wall of a fusion reactor is a steel plate 1 cm thick. The back side of
the wall is maintained at constant temperature by a coolant. The front face is
suddenly exposed to a plasma disruption that increases the heat flux
instantaneously. Approximately how long is required for the increased heat
load on the front face be felt at the back face of the wall? The thermal
conductivity, heat capacity, and density of the particular steel are 0.23 W/cmK, 0.32 J/g-K, and 8.7 g/cm3, respectively.
9.9
A fuel rod operates at a linear power P with a radial power density distribution
H(r)=Ho+ar2, where Ho and a are constants. a is positive because the effect it describes is
the neutron flux depression in the fuel pellet. What is the difference between the
centerline-to-surface temperature difference for this rod compared to one operating at the
same linear power but with a uniform radial power density distribution?
9.10 A fast reactor fuel pin operates at a linear power of 700 W/cm at an axial location
where the coolant (liquid sodium) is 500oC. Under these conditions, a central portion of
the fuel is molten. The heat transfer properties of this fuel pin are:
kgap/tgap = 0.5 W/cm2-K
kc/tc = 9 W/cm2-K
hcoolant = 12 W/cm2-K
ksolid fuel = 0.03 W/cm-K
kliquid fuel = 0.05 W/cm-K
R = 0.35 cm
(a) What is the fuel surface temperature?
(b) To what fractional radius is the fuel molten?
(c) What is the fuel centerline temperature?
Light Water Reactor Materials © Donald Olander and Arthur Motta
2/6/2016
9.11 A key safety limit in light-water reactors is the stored energy in the fuel. This is the
thermal energy (relative to 25oC) contained in the temperature distributions in the fuel
pellet and in the cladding. In the event of a loss-of-coolant accident, flowing liquid water
is replaced by stagnant steam, which acts as an insulating blanket over the fuel rod. Even
though heat production by fission has been shut off, the original temperature profiles
relax to a constant temperature that is the same in both fuel and cladding. This final
temperature must not exceed a regulatory limit.
Before shutdown, the fuel operated at 500 W/cm linear power and the local water
coolant temperature was 300oC. The fuel pellet diameter is 1 cm and the cladding
thickness is 1 mm. The fuel-cladding gap is closed, so the temperature drop across the
gap is negligible. The external heat transfer coefficient in the water coolant is very large,
so there is no temperature drop here either.
(a) What are the fuel surface and centerline temperatures during operation?
(b) What is the final uniform fuel and cladding temperature after the adiabatic
relaxation of the original distributions?
Use the thermal properties given in the chapter.
References
[1]
[2]
[3]
[4]
R. Yang, O. Ozer, and H. Rosenbaum, "Current Challenges and Expectations of
High Performance Fuel for the Millenium," Light Water Reactor Fuel
Performance Meeting, Park City, Utah, ANS, 2000.
M. M. El-Wakil, Nuclear Heat Transport: American Nuclear Society, 1978.
D. R. Olander, Fundamental Aspects of Nuclear Reactor Fuel Elements: ERDA,
1976.
G. Hache and H. M. Chung, "The history of LOCA embrittlement criteria," 28th
Water Reactor Safety Information Meeting, Bethesda, USA, NRC, 2001, 205-237.
Light Water Reactor Materials © Donald Olander and Arthur Motta
2/6/2016
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