LAMARSH SOLUTIONS CHAPTER-5 PART-1

5.4

a)

max 20

(sin(0.06285 )) ' cos(0.06285) nlim ( ) 5 13 5 13 0.06285 =3.14e12

(r)' 1 cm secr

rr e x e x x

(We used the L’Hospital rule.)

b)

2

sin(0.06285) 0.0628 cos(0.0628 ) sin(0.0628 )( ) 0.8 (5 13 ) 0.8 5 13 ( )

r

xr r rJ r Dgrad D x e x x e x

r r r

and this is a vector , you know!

c)

24 (50) 1.579 15escN R J e neutrons escape from the system.

5.10

Diffusion equation ,

2 ( )a sr

D D

For an infinite system leakage is zero . so

2 0 and then,

2

and J=-Dgrad 0 ,(derivative of constant)a

sL s

D

5.11

a)

The one dimensional diffusion eq. for a slab is,

2 2

2 2 2

1a

SD S

x x L D

2

( ) particular+homogenous sinh( ) cosh( )L S x x

x A BD L L

Because the system has mirror symmetry around x=0 plane,derivative of at x=0 must be zero,

0 cosh( ) sinh( )@ 0 A=0A x B x

x soL L L L

then,

2

( ) cosh( )x L S

x BL D

And the second boundary condition must be satisfied,which is ( ) 0a d and then,

( ) 0

cosh( )

a

s

a d Ba d

L

, finally we find that,where

2

a

L S S

D

cosh( )

( ) (1

cosh( )a

xs Lx

a d

L

b)

The equation of continuity;

[rate of change in number of neutrons in V] = [rate of production in number of neutrons in V] – [rate

of absorption in number of neutrons in V] - [rate of leakage in number of neutrons from V]

Production = Absorption + Leakage

# Produced = S x 2a

# Absorption = 0

[ / ]2 ( ) 2 ( )

cosh(( ) / )

a

a

LSinh a Lx dx S a

a d L

# Leakage = [ / ]

2 ( ) 2 2cosh(( ) / )

d sinh a LJ a D SL

dx a d L

[ / ] [ / ]2 2 ( ) 2

cosh(( ) / ) cosh(( ) / )

LSinh a L sinh a LS a S a SL True

a d L a d L

5.12

a)

We can write for a finite sphere the flux equation as,

( / ) ( / )( )

Sinh r L Cosh r Lr A B

r r

Boundary conditions;

(i) 2

0lim4 ( )r

r J r S

and

(ii) ( ) 0R d R

From first boundary condition: 4

SB

D ; From second boundary cond: [ / ]A B Coth R L

And inserting these constants into ( )r we find

cosh( / )sinh( / ) sinh( / )cosh( / ) sinh( / )( ) ( )

4 sinh( / ) 4 [ / ]

S r L R L r L R L S R r Lr

Dr R L DrSinh R L

b)

2 ( )( ) 4

4 (( ) / ) (( ) / )surface

S R d SLeakage A J R R

RLSinh R d L LSinh R d L

c)

( )

Leakage R dP

R dSLSinh

L

5.15

n( ) sin( ) J ( ) ( )nn n n

A rr r D r

r R r

,here D1=2.2cm,D2=1.7cm,D3=1.05cm

32

n

1

4 J ( ) 2.0173 18 n/secleak

n

N R R e