reactiv measurement exmple
DESCRIPTION
Reactivity measurementsTRANSCRIPT
Problem An experiment is set up to measure the reactivity worth of a control rod by the source multiplication method. The reactor being used is cubical with the bottom parallel to the ground, and has a side of 4m (The extrapolation length can be ignored.). An external neutron source is introduced in the (subcritical) reactor, and a detector placed at the center of the reactor indicates 100. Next, a control rod with a reactivity worth of 10 mk is inserted in the core. The new reading is 50. Subsequently, the first control rod is removed and a second rod, of unknown worth, is inserted. The new detector reading is also 50. However, it is later determined that the operator making the measurements has inadvertently moved the detector before the last reading, so that it was 1m above the center of the reactor when the last reading was taken. What is the reactivity worth of the second rod?
Solution Figure
In a cubical reactor of side L, the shape of the flux is, regardless of the actual reactivity or of the presence of the external neutron source or control rod:
)(coscoscos)0,0,0()( rAzL
yL
xL
r
For any detector position, we can write that the ratio between the flux at that position and the core integrated flux is the same, regardless of the reactivity of the reactor, because the shape is the same. For a fixed position 0r
we have:
constantˆ
)( 0
r
At the same time,
nvˆ So:
reactivity oft independen )()(
constantv
)(ˆ
)(0
000 rn
r
n
rr
The overall steady-state point kinetics equations for a subcritical reactor with external neutron source is:
6,...10
06
1
iCn
SCn
iii
iii
For the second equation, it follows that:
nnCnC ii
iii
iii
6
16
1
Substituting this into the first equation, we obtain:
SnSnSnn 00
Now, for any fixed position r0, we can write:
Srnrn
n
rr )()(
)()( 00
00
For the first measurement where the reactor has an initial negative reactivity 0 , we have:
000
0 )()(
Srr
For the second measurement where we insert rod 1, we have:
1000
1 )()(
Srr
For the third measurement (rod 2) we have:
2011
2 )()(
Srr
By dividing second measurement by the first, we obtain:
10
0
0
10
0
0
00
01
)(
)(
)(
)(
Sr
Sr
r
r
Solving for 0 we obtain in sequence:
10
0
00
01
)(
)(
r
r
010
00
1
00
00
1
)(
)(
)(
)(
r
r
r
r
10
00
1
00
00
1
)(
)(1
)(
)(
r
r
r
r
)(
)(1
11
)(
)(
)(
)(
00
01
1
00
01
10
00
1
0
r
rr
r
r
r
1)(
)(
01
00
10
r
r
If we divide the third measurement by the second, we obtain:
20
0
0
1
0
20
0
1
00
12
)(
)(
)(
)(
)(
)(
r
r
Sr
Sr
r
r
Equivalent to:
20
0
1
0
00
12
)(
)(
)(
)(
r
r
r
r
Solving for 2 we obtain in sequence:
20
0
1
0
00
12
)(
)(
)(
)(
r
r
r
r
)(
)(
)(
)(
1
0
00
02
020
r
r
r
r
0
1
0
00
12
02
)(
)(
)(
)(
r
r
r
r
1)(
)(
)(
)(
0
1
12
00
02 r
r
r
r
Now, substituting the value found previously for 0 we obtain:
1
)(
)(
)(
)(
1)(
)( 0
1
12
00
01
00
12 r
r
r
r
r
r
This is very similar with what we obtained for the regular source multiplication method, except for the presence of the
factor )(
)(
0
1
r
r
.
W can determine gamma by using the shape of the flux.
VVVV
dVr
r
dVrA
rA
dVrA
rA
dVr
rr
)(
)(
)(
)(
)(
)(
)(
)()( 0000
0
Similarly, we find:
V
dVr
rr
)(
)()( 1
1
It follows that:
)(
)(
)(
)(
)(
)(
)(
)(
0
1
0
1
0
1
r
r
dVr
r
dVr
r
r
r
V
V
For our case:
2
1
4cos
04
cos04
cos04
cos
14
cos04
cos04
cos
)0,0,0(
)1,0,0(
m
The unknown reactivity worth is then:
mkmk
r
r
r
r
r
r14.41
2
1
50
100
150
10010
1)(
)(
)(
)(
1)(
)( 0
1
02
00
01
00
12
Abbreviated solution In an open books/notes exam you could have used the following approach: Use directly the formula for the unknown reactivity:
1
)(
)(
1)(
)( 12
00
01
00
12 r
r
r
r
Express
2)(
4cos
1)(
14
cos04
cos04
cos
04
cos04
cos04
cos)(
)(
)()()(
12
12
12
12
02
12
02
rr
rr
rrr
Substitute the expression into the formula for the reactivity:
mkmk
r
r
r
r14.41
2
1
50
100
150
10010
12)(
)(
1)(
)(1
20
0
01
00
12