1000 Solved Problems in Modern Physics

8.3 Solutions 475

Area of the unit cellA=πr 12

Now,

Σam

Σau

=

Nm

Nu

×

σam

σau

=

ρm/Am

ρu/Au

×

σam

σau

=

1. 62 / 12

18. 7 / 238

×

4. 5 × 10 −^3

7. 68

= 1. 01 × 10 −^3

Vm

Vu

=

182 −π(1.5)^2

π(1.5)^2

= 44. 8

φm

φu

= 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)

Whereφis the neutron flux andB^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)

whereAis the constant of integration andRis the radius of the bare reactor

dφ

dr

=−

A

r^2

sin

(πr

R

)

+

πA

Rr

cos

(

π

r

R

)

(4)

d^2 φ

dr^2

=

2 A

r^3

sin

(

π

r

R

)

−

πA

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