Lamarsh Solutions Ch-3 Part1 2o6c23

Lamarsh Solutions Chapter-3 Part-1 3.2 Flux is independent of angle =   I1  I 2 =2e10+1e10 =3e10

n cm x sec 2

Current is dependent on angle ,it defines a vector.If we take the coordinate system as shown below we can find the current vector at that point Coordinate system:

+I

+j And the current vectors intersect as;

300

2e10

1e10

J=J1 +J 2 J1 =2e10i ; J 2 =1e10cos(30)i+1e10sin(30)j J=(2e10+1e10cos(30))i+1e10sin(30)j 3.4

n I cm x sec  1364 n n=   22e4 cm/sec cm3 3e8

(a)

2

(b) This reaction can be shown as,

 Al 27  φ x  x V=R ,a.k.a “reaction rate”.Now calculate the necessary inputs, 2.699 x0.6022e24 x0.23e  24cm 2 26.9815   0.01385cm1   N x  th 

in here 2.699 is the density of Al and 26.9.. is the mol. weight of Al

V

m





0.01gr 2.699

so R  3e8 x0.01385 x

0.01gr  1.539e4 Al27 atoms/sec 2.699

(c) Maximum activity can be found using,you ,

 Al 28  R(1  e t ) 28

and here for the max. act. you set t to infinity and find MAX ACT=R

 max  1.539e4dis / sec 

1.539e4dis / sec  4.159e  7Ci 3.7e10

3.7 The probability that a neutron will have its first collision in dx;

p( x)dx a

  p( x)dx   t et x dx  e t a  e t a / 2 a/2

3.11 Atom densities of components;

Ni 

i  N a 100M i

  NC aC  NCr aCr  N Ni aNi  N Fe aFe we know the percentages as

C  .08wt % Cr  19 wt % Ni  10wt % Fe  70.92wt % forC forCr forNi forFe

 a  0.0034b  a  3.1b  a  4.43b  a  2.55b

You can find these cross sections from table II.3 at the end Lamarsh

NC aC 

C  N a 100M i

NCr aCr 

N Ni aNi  N Fe aFe 

 aC  1.07 106 cm1

Cr  N a 100M Cr

Ni  N a 100M Ni

Fe  N a 100M Fe

 aCr  0.0536cm1

 aNi  0.0357cm1  aFe  0.155cm1

   0.243cm1 3.14 First let’s look at how the problem 2.63 will be solved,

i 1 1 1 1 70.5 21.3 5.5 2.7    (    )  M Pu  239.4548  M 100 i M i M Pu 100 239.05 240.05 241.05 242.06 M PuO2  271.45365 and MUO2  270.0496 Now we should find the percents by weight the amounts of Pu isotopes,U and O in the fuel

239.4548  0.264636 239.4548  2 x15.99 238.0508 Uw / o  0.7  0.617055 238.0508  2 x15.99 Ow / o  1  (0.264636  0.617055)  0.11839 Puw / o  0.3

And using the well known equation to find the atoms ,

0.264636 x0.705 x0.6022e24  4.6998e20atoms 239.054 0.264636 x0.213 N ( Pu 240 )  x0.6022e24  1.4135e20atoms 240.054 0.264636 x0.055 N ( Pu 241 )  x0.6022e24  0.3636e20atoms 241.057 0.264636 x0.027 N ( Pu 242 )  x0.6022e24  0.1777e20atoms 242.0587 0.264636 x0.027 N ( Pu 242 )  x0.6022e24  0.1777e20atoms 242.0587 N ( Pu 239 ) 

0.617055 x0.6022e24  15.6e20atoms 238.0508 0.11839 N (O)  x0.6022e24  44.367e20atoms 15.99 N (U 238 ) 

And now back to 3.14 ,we can find the asked cross sections using the tables for  a and  f which are thermal and the equation   N Examine the table below

 a (barns)

 f (barns)

 a ( cm1 )

470.29

0.00027

0

0.000012697 0

15.6

165.36

2.7

0

0.0446

0

Pu239

4.6998

49.81788

1011.3

742.5

5.03808

3.69897

Pu240

1.4135

14.9831

289.5

0.03

0.43376

0.0000449

Pu241

0.3636

3.85416

1377

1009

0.530717

0.38888

Pu242

0.1777

1.88362

18.5

0.2

0.0034846

0.0000376

6.0506543

4.087941

N/1e20 atoms

N/1e20*10.6

O

44.367

U28

atoms

Total

a  6.05065cm1

and

 f  4.08794cm1

3.16 will be included to the second part 3.31 Absorption rate: 

Fa   n( E ) ( E ) a ( E )dE   a ( E0 )0 0

0  1.5 1013 neutrons / cm2  sec N  0.03343 1024 atoms / cm3 and   0.664b from Table II.3. a  Fa  0.664  0.03343 1.5 1013  3.33 1011 atoms / cm3  sec 3.35

 f ( cm1 )

In Ag Cd Cd Cd F  a ( E0 )0  0 (Cd ga   aIn N In gaIn   aAg N Ag ) a   a   a )  0 ( a N

For one gram control rod sample at T=400C; w/o

Weight

 a (b)

ga

N (atoms)

Cd

5

112.4

2450

2.5589

2.6788 104

In

15

114.82

193.5

1.1011

7.8671104

Ag

80

107.87

63.6

----

44.6612 104

 Fa  10.66 1013 neutrons / cm3  sec