8.3 Solutions
475
Area of the unit cell A = πr
2
1
Now,
Σ am
Σ au
=
N m
N u
×
σ am
σ au
=
ρ m /A m
ρ u /A u
×
σ am
σ au
=
1.62/12
18.7/238
×
4.5 × 10
−3
7.68
= 1.01 × 10
−3
V m
V u
=
18
2
− π (1.5)
2
π (1.5) 2
= 44.8
φ m
φ u
= 1.6
Hence, f =
1
1 + (1.01 × 10 −3 ) × 44.8 × 1.6
= 0.933
8.80 The equation for a critical reactor is
∇
2
φ + B
2
φ = 0
( 1 )
Where φ is the neutron flux and B
2 is the buckling.
For spherical geometry, Eq. (1) becomes
d
2
φ
dr 2 +
2
r
dφ
dr
+ B
2
φ = 0
( 2 )
which has the solution
φ =
A
r
sin(πr/R)
( 3 )
where A is the constant of integration and R is the radius of the bare reactor
dφ
dr
= −
A
r 2 sin
πr
R
+
π A
Rr
cos
π
r
R
(4)
d
2
φ
dr 2 =
2A
r 3 sin
π
r
R
−
π A
Rr
2
cos
πr
R
−
Aπ
2
R 2 r
sin
π
r
R
(5)
Therefore (2) becomes
−
Aπ
2
dr 2 sin
πr
R
+
B
2 A
r
sin
π
r
R
= 0
Therefore, B
2
=
π
2
R 2
Or the critical radius,
R =
π
B
(6)
B
2
=
k ∞ − 1
M 2 =
1.54 − 1
250
= 2.16 × 10
−3 cm
−2
B = 0.04647
R =
π
0.04647
= 67 cm