Nuclear Fuel Engineering
K. Idemitsu
• Introduction of nuclear fuels
• Performance of current nuclear fuels
• Heat generation rate and
temperature distribution in a fuel
• Restructuring of fuel during irradiation
• Behavior of fission products in irradiated fuels
and Small Test
• Introduction of nuclear fuels
Ca. 30 % of electricity is generated by nuclear energy in Japan.
Advantages of nuclear energy;
High energy density, low cost, no carbon dioxide release, etc.
Disadvantage;Radioactivity as by-product
Coal
Oil
Liq.Natur.Gas
Solar Heat
Ocean Temp. Diff.
Logshore current
Wavelet
Solar cell
Wind Power
Geothermal
Nuclear Power
Hydro Power
By fuel burning
By others(construction etc.)
Carbon release per unit electricity
g-C/kWh
• Introduction of nuclear fuels
Current nuclear fuel is made of UO2
or MOX(Mixed Oxide) :(U,Pu)O2
Nuclear Fuel Pellets
FUGEN
PWR
JOYO
MOMJU
BWR
• Introduction of nuclear fuels
Upper plug
Plenum spring
Plenum is the spece for fission gas released from fuel
pellets in the rod.
Fuel pellet
φ 10mm x h 10mm
BWR
φ 8mm x h 12mm
PWR
Fuel rod
Cladding
Zircaloy
Zr based alloy
BWR Zry-2 (Zr-1.5 Sn - 0.12 Fe - 0.05 Ni - 0.1 Cr)
PWR Zry-4 (Zr-1.5 Sn - 0.15 Fe - 0.00 Ni - 0.1 Cr)
Lower plug
• Introduction of nuclear fuels
Fuel Assembly
Fuel pellet
Spacer grid
4m
Fuel rod
• Introduction of nuclear fuels
Fuel Assembly
BWR
6 x 6, 7 x 7, 8 x 8
PWR
14 x 14, 15 x 15, 17 x 17
BWR
Reactor core
-4m x h 4m
• Introduction of nuclear fuels
Disadvantage;Radioactivity as by-product
For the safety there are 5 containers in a reactor.
Pressure vessel
3rd container
Container vessel
4th container
Building
5th container
Fuel pellet
1st container
Fuel rod
2nd container
• Performance of current nuclear fuels in Japan
Low leakage of RNs from fuel rods into coolant by failure
(PWR)
4000 50
(BWR)
20000
PWR
3000
15000
BWR
40
30
10000
2000
5000
1000
20
10
PWR
BWR
0
1970 1980 1990 2000
No. of Bundles in core
0
0
1970
1980 1990 2000
No. of Failures
• Performance of current nuclear fuels in USA
10
Failure cause
･Crud induced locallized corrosion
･Debris fretting
･Fablication failure
･Pellet cladding interaction
･Crudding/corrosion
8
6
4
2
0
1980
1985
1990
1995
• Performance of current nuclear fuels in Japan
Heat generation
Burn up ; how much energy produced per unit fuel
odometer of fuel
GWd/tU (30 to 50 GWd/tU in Japan)
%FIMA(Fission per initial metal atom)
Heat rate ; how much power produced per unit fuel
speedmeter of fuel
GW/m (30 kW/m in Japan)
Integral of thermal conductivity
Tcent
K dT
Tsurf
K : thermal conductivity of fuel
Tsurf : temperature of fuel surface
Tcent : temperature of fuel center
• Performance of current nuclear fuels in Japan
for example
Linear heat rate 30 kW/m
30kW/m x 4 m = 120 kW per fuel rod
120kW/rod x 63rods = 7.6MW per assembly
7.6MW/assembly x 400 assembly = 3 GWth per reactor
3 GWth x 0.33 =1 GWe
BWR 8 x 8
Fuel rod
Water rod
Burn up ; how much energy produced per unit fuel
%FIMA(Fission of initial metal atom)
When 1% of U fission, total energy will produce •••.
• 1t of U; 1,000,000 g/238 * 6.02 1023 = 2.53 1027
• fission ; 2.53 1025
• energy of fission; 200MeV = 3.2 10-11 J
• energy from 1t of U ; 2.53 1025 * 3.2 10-11 = 8.1 1014 J
= 8.1 1014 Ws /86400 s/d = 9.37 109 Wd
= 9.37 GWd
1 %FIMA = 9.37 GWd/tU
If you want to use GWd/t UO2 ,
you should use 270 instead of 238.
• Heat generation rate and
temperature distribution in a fuel
Simple calculation of T distribution in a fuel rod
Assumption
• uniform heat generation per unit volume ; q (W/m3)
• temp. independent thermal conductivity;K (W/m/°C)
Heat generated in inner cylinder; πr2q
dr
r
a
Heat flux through cylinder;
πr2q rq
=
(W/m2) = – K dT
dr
2
2πr
a rq
0 2
dr =
Tcent
Tsurf
a2q Linear Heat rate
K dT =
=
(W/m)
4
4π
• Heat generation rate and
temperature distribution in a fuel
a rq
r 2
T(r)
dr =
Tsurf
K dT = T (r) – Tsurf K
2
2
a2q r2q a2q
r
r
Linear
heat
rate
=
–
=
1– 2 =
1– 2
4
4
4
4π
a
a
Parabolic distribution
Tcent
Tsurf
Linear heat rate is determined
by center temperature
not by pellet diameter.
Tcent
Tsurf
K dT = Linear Heat rate = Tcent – Tsurf K
4π
Linear heat rate is determined
by center temperature
not by pellet diameter.
• How high the center temperature in a fuel pellet
K = 2 W/m/°C, Linear heat rate = 30 kW/m
Tcent
Tsurf
K dT = Linear Heat rate = Tcent – Tsurf K
4π
Tcent - Tsurf = 1200 °
Tcent = 1700 - 1800 °C
If linear heat rate = 50 kW/m,
Tcent = 2500 - 2600 °C
• Phase diagram of U-O system
Melting point of UO2 2840°C
Temperature (°C)
Why are fuel rods getting slim?
To keep 7.6MW per assembly
7.6MW/assembly ÷ 63 rod = 120kW/rod
120kW/rod ÷ 4 m = 30 kW/m
Linear heat rate 30 kW/m
BWR 6 x 6
BWR 7 x 7
BWR 8 x 8
Fuel rod
Water rod
Number of fuel rods
Linear heat rate (kW/m)
36
53
49
39
63
30
• Restructuring of fuel during irradiation
There is big temperature gradient in fuel pellet.
Central void
>50GWd/tU
Columnar grain
>1700°C
Equiaxed grain
>1400°C
Undisterbed region
Rim region
>70GWd/tU
• Behavior of fission products in irradiated fuels
1.0
94
136
Fission gas yields (%)
Isotopes
0.0001
100
mass number
Fission yields
Pu
0.30
84
0.85
0.50
85
0.15
0.13
86
Kr
1.40
0.80
Total Kr
2.8
1.7
Kr
0.001
239
0.40
Kr
0.01
U
83
Kr
0.1
235
131
Xe
3.2
3.8
132
Xe
4.7
5.3
134
Xe
6.6
7.5
136
Xe
5.9
6.6
20.4
23.2
Total Xe
• Behavior of fission products in irradiated fuels
M + O2 ⇔ MO2
K=
∆G f
aMO2
= exp –
aMPO2
RT
∆GO2 = RT ln PO2 :Oxygen Potenial
a
∆G f = RT ln PO2 + RT ln a M
MO2
Elemental
Oxide
Temperature (K)
Oxygen potential of fuel ∆GO2
Formation energy of oxides