Lamarsh Solutions

Chapter-3 Part-1

3.2

Flux is independent of angle =1 2I I =2e10+1e10 =3e10

2

n

cm secx

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;

2e10 030

1e10

1 2

1 2

J=J +J

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

J=(2e10+1e10cos(30))i+1e10sin(30)j

3.4

(a)

2

3

n3e8

I ncm x secn= 136422 4 cm/sec cme

(b) This reaction can be shown as,

27 φ x x V=RAl ,a.k.a “reaction rate”.Now calculate the necessary inputs,

2

th

1

2.699 x 0.6022 24 0.23 24

26.9815

0.01385

N x e x e cm

cm

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

0.01

2.699

m grV

so

0.013 8 0.01385 1.539 4 Al27 atoms/sec

2.699

grR e x x e

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

28

28 (1 )t

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

max

1.539 4 / sec1.539 4 / sec 4.159 7

3.7 10

e dise dis e Ci

e

3.7

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

( )p x dx

/ 2

/ 2

( ) t t t

a

x a a

t

a

p x dx e dx e e

3.11

Atom densities of components;

100

i ai

i

NN

M

C Cr Ni Fe

C a Cr a Ni a Fe aN N N N we know the percentages as

.08 %

19 %

10 %

70.92 %

C wt

Cr wt

Ni wt

Fe wt

a

a

a

a

0.0034

3.1

4.43

2.55

forC b

forCr b

forNi b

forFe b

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

6 11.07 10100

C CC aC a a

i

NN cm

M

10.0536100

Cr CrCr aCr a a

Cr

NN cm

M

10.0357100

Ni NiNi aNi a a

Ni

NN cm

M

10.155100

Fe FeFe aFe a a

Fe

NN cm

M

10.243cm

3.14

First let’s look at how the problem 2.63 will be solved,

1 1 1 1 70.5 21.3 5.5 2.7( ) 239.4548

100 100 239.05 240.05 241.05 242.06

iPu

i i Pu

MM M M

2271.45365PuOM and

2270.0496UOM

Now we should find the percents by weight the amounts of Pu isotopes,U and O in the fuel

239.4548/ 0.3 0.264636

239.4548 2 15.99

238.0508/ 0.7 0.617055

238.0508 2 15.99

/ 1 (0.264636 0.617055) 0.11839

Puw ox

Uw ox

Ow o

And using the well known equation to find the atoms ,

239

240

241

242

0.264636 0.705( ) 0.6022 24 4.6998 20

239.054

0.264636 0.213( ) 0.6022 24 1.4135 20

240.054

0.264636 0.055 ( ) 0.6022 24 0.3636 20

241.057

0.264636 0.027 ( ) 0

242.0587

xN Pu x e e atoms

xN Pu x e e atoms

xN Pu x e e atoms

xN Pu x

242

.6022 24 0.1777 20

0.264636 0.027( ) 0.6022 24 0.1777 20

242.0587

e e atoms

xN Pu x e e atoms

238 0.617055( ) 0.6022 24 15.6 20

238.0508

0.11839( ) 0.6022 24 44.367 20

15.99

N U x e e atoms

N O x e e atoms

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

N/1e20

atoms

N/1e20*10.6

atoms

a (barns) f (barns) a ( 1cm ) f ( 1cm )

O 44.367 470.29 0.00027 0 0.000012697 0

U28 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

Total 6.0506543 4.087941

16.05065a cm and 14.08794f cm

3.16 will be included to the second part

3.31

Absorption rate:

0 0

0

( ) ( ) ( ) ( )a a aF n E E E dE E

13 2

0 1.5 10 / secneutrons cm

24 30.03343 10 /N atoms cm and 0.664a b from Table II.3.

13 11 30.664 0.03343 1.5 10 3.33 10 / secaF atoms cm

3.35

0 0 0( ) ( )Cd In Ag

a a a aF E 0( )Cd Cd Cd In In In Ag Ag

a a a a aN g N g N

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

w/o Weight a (b) ag

N (atoms)

Cd 5 112.4 2450 2.5589 42.6788 10

In 15 114.82 193.5 1.1011 47.8671 10

Ag 80 107.87 63.6 ---- 444.6612 10

13 310.66 10 / secaF neutrons cm