PHGN590
Introduction to Nuclear Reactor Physics
Fuchs Model the USGS TRIGA Pulse
J. A. McNeil
Physics Department
Colorado School of Mines
4/2009
Description of the Fuchs Model
The Fuchs model attempts to describe the response of a nuclear reactor to a supercritical reactivity intertion. We assume the time
scale of the pulse event is short compared to the delayed neutron precursor lifetimes; so this term is neglected. The time dependent one-group flux equation becomes:
eq1 = f '@tD ã Hr@tD - bL f@tD ê L
f£ @tD ã
H-b + r@tDL f@tD
L
To model the pulse, we assume the transient rod is withdrawn instantaneously giving a large positive reactivity, r0 > b. We next
assume that the time behavior of the reactivity is due to the rise in temperature in the fuel and is linear in the deviation of the
temperature from some reference value, T0.
r@t_D = r0 - a HT@tD - T0L
r0 - a H-T0 + T@tDL
where a is the temperature coefficient of reactivity. The temperature rise is due to the energy released by fission. The process is
much faster than the thermal transport time; so we treat the fuel rod as adiabatic. Therefore, if C is the heat capacity of the fuel
rod and g is the energy released per fission, we have
T@t_D = T0 +
1
C
Ÿ0 P@tD „ t
‡ P@tD „ t
t
0
t
T0 +
C
where the power is given by:
P@t_D = g Sf f@tD V
V g Sf f@tD
Putting this altogether gives the Fuchs equation (Hansen form):
2
Fuchs.nb
eq1
a Ÿ0 V g Sf f@tD „t
t
f @tD ã
-b + r0 -
C
£
f@tD
L
ü Solve the Fuchs equation using a functional transformation
Define the term in brackets as y[t]:
y@t_D = Hr0 - bL ê L -b + r0
L
a g Sf V
LC
V a g Sf Ÿ0 f@tD „ t
‡ f@tD „ t
t
0
t
-
CL
The initial conditions are:
Clear@yD
InitCond1 = y@0D ã Hr0 - bL ê L
a g Sf V
InitCond2 = y '@0D ã f0
LC
y@0D ã
-b + r0
L
y£ @0D ã -
V a g Sf f0
CL
Take the derivative, y'[t], solve for f[t], and plug it into the Fuchs equation to get the transformed Fuchs equation:
Clear@yD
eq2 = y ''@tD ã D@y@tD ^ 2, tD ê 2
y££ @tD ã y@tD y£ @tD
Integrate once gives
eq3 = y '@tD ã Hy@tD ^ 2 - k ^ 2L ê 2
y£ @tD ã
1
2
I-k2 + y@tD2 M
where k^2 is the integration constant which we solve using the initial condition #2.
ksoln =
r0 - b
H-b + r0L2
L2
Try direct solution
L
+
2
+
2 a g Sf V f0
LC
4 V2 a2 g2 Sf2 f02
C2 L2
2
Fuchs.nb
ysoln@t_D = y@tD ê. Flatten@DSolve@8eq3, y@0D ã Hr0 - bL ê L<, y@tD, tDD
LC
fsoln@t_D = ysoln'@tD
a g Sf V
k TanhB
1
2
-t k - 2 ArcTanhB
b - r0
kL
C k2 L SechA 12 I-t k - 2 ArcTanhA
F F
2
b-r0
EME
kL
2 V a g Sf
The total energy is the integral of the power:
Etotal = g Sf ‡ fsoln@tD „ t
¶
-¶
$Aborted
The total energy can be obtained from y[t]:
Energy@t_D = -b + r0
L
-
1201.11
LC
aV
ysoln@tD + Hr0 - bL ê L
C k L TanhA 12 I-t k - 2 ArcTanhA
Va
b-r0
EME
kL
3
4
Fuchs.nb
ü Plot the flux during the pulse
dvalue = 2.; Vrod = 394.78;
params = 8b Ø .007, a Ø 3. µ 10 ^ -5, C Ø 720, V Ø Vrod,
g Ø 190. µ 1.6 µ 10 ^ -13, Sf Ø .12, L Ø 40. µ 10 ^ -6, f0 Ø 3. µ 10 ^ 9<;
fplot@t_, d_D = HHfsoln@tD ê. k Ø ksolnL ê. r0 Ø d bL ê. params;
horizaxis =
StyleForm@"t HsL", FontFamily Ø "Tahoma", FontColor Ø Blue, FontWeight Ø Bold, FontSize Ø 12D;
vertaxis = StyleForm@" fHtLêf0 ", FontFamily Ø "Tahoma",
FontColor Ø Blue, FontWeight Ø Bold, FontSize Ø 12D;
plotname = StyleForm@
" Fuchs Model of Reactor Response to Supercritical Insertion of Reactivity",
FontFamily Ø "Tahoma", FontColor Ø Black, FontWeight Ø Bold, FontSize Ø 14D;
fscale = f0 ê. params; tplot = 0.3;
Plot@8fplot@t, dvalueD ê fscale<, 8t, 0, tplot<, PlotStyle Ø 88Red, Thickness@.005D<<,
Frame Ø True, GridLines Ø Automatic, PlotLabel Ø plotname,
FrameLabel Ø 8horizaxis, vertaxis<, ImageSize Ø 600, Background Ø LightYellow, PlotRange Ø AllD
Fuchs Model of Reactor Response to Supercritical Insertion of Reactivity
3.5 µ 106
3.0 µ 106
2.5 µ 106
fHtLêf0
2.0 µ 106
1.5 µ 106
1.0 µ 106
500 000
0
0.00
0.05
0.10
t HsL
0.15
0.20
0.25
0.30
Fuchs.nb
ü Find the peak location and evaluate the peak flux
tpeakGuess = .12;
dfdt@t_, d_D = Simplify@fsoln'@tDD;
dfdtplot@t_D = HHHdfdt@t, dD ê. k Ø ksolnL ê. r0 Ø d bL ê. paramsL ê. d Ø dvalue;
Plot@dfdtplot@tD, 8t, 0., 2 tpeakGuess<, PlotRange Ø AllD
tpeak = t ê. FindRoot@dfdtplot@tD, 8t, tpeakGuess<D;
fpeak = fplot@tpeak, dvalueD; fpeakRatio = fpeak ê f0 ê. params;
Plot@fplot@t, dvalueD - fpeak ê 2, 8t, tpeak - tpeak ê 2, tpeak + tpeak ê 2<D
thalf = t ê. FindRoot@fplot@t, dvalueD - fpeak ê 2, 8t, tpeak ê 2<D;
Etotal = Limit@HHEnergy@tD ê. k Ø ksolnL ê. r0 Ø dvalue bL ê. params, t Ø ¶D;
DT = Etotal V ê C ê. params;
Print@" The peak occurs at t = ", tpeak, " s"D
Print@"The peak flux is ", fpeak, " êcm^2ês"D
Print@"The peak flux ratio is ", fpeakRatioD
Print@"The pulse width is ", 2 Abs@tpeak - thalfD, " s"D
Print@"The total energy released is ", Etotal, " J"D
Print@"The temperature increase of the fuel is ", DT, "
6 µ 1017
4 µ 1017
2 µ 1017
0.05
0.10
0.15
0.20
-2 µ 1017
-4 µ 1017
-6 µ 1017
4 µ 1015
2 µ 1015
0.08
-2 µ 1015
-4 µ 1015
0.10
0.12
0.14
0.16
0.18
C"D
5
6
Fuchs.nb
The peak occurs at t = 0.120781 s
The peak flux is 1.02072 µ 1016 êcm^2ês
The peak flux ratio is 3.40241 µ 106
The pulse width is 0.0201457 s
The total energy released is 600.553 J
The temperature increase of the fuel is
329.287 C