Sol Ch6 Part1

LAMARSH SOLUTIONS CHAPTER-6 PART-1

6.1

Before solving this question,examine ex.6.1 in Lamarsh and for cross sections and other constants

use table 6.1;

0.03 32.33Pu Na

Pu Na Pu

and then;

,,

,

,

, ,

0.03 32.33

1 10.108 and 2.61

*1

1*

2.11 8 4 and finally k 2.222(sup )

Pu Na

Pu Na Pu

Na a Na Naa Na

NaPu a Pu

Pua Pu

Pu

a Pu a Na

fN Navgdro

MN

Navgdro

M

b e b f ercritical

6.2

a)

Sodium properties:

0f and 0.0008a b and 3.3tr b

Uranium-238 properties:

0.095f b and 0.255a b and 6.9tr b and 2.6

Uranium-235 properties:

1.4f b and 1.65a b and 2.6 and 6.9tr b and 2.6

318.6 /U g cm and 30.97 /Sodium g cm

3

235 18.6 25.6 /100 4.7616 /U g cm and 3

238 18.6 74.4 /100 13.8384 /U g cm

ANN

M

24240.6022 10

0.97 0.63 0.017 1023

NaN

2424

238

0.6022 1013.8384 0.37 0.01296 10

238UN

2424

235

0.6022 104.7616 0.37 0.00446 10

238UN

b)

235 238

235 238

235 238

235 238

0.0106630.9987

0.0106774

U U

aF U a U a

U U S

a U a U a S a

N Nf

N N N

c)

235 238

25 235 28 238

235 238

235 238

0.0074751.8227

0.010663

U U

f U f U f

U U

aF U a U a

N N

N N

d)

1.82033k f

6.6

max max

ave ave

P

P

2 3785.4 22243V R H ft lt

825 37.09 /ave aveP MW P kW lt

max . 1.5 37.09 55.64 /aveP P kW lt

6.9

2 21

1

k

B L

where 2 2( )BR

and 2

a

DL

R R d and 1

3 tr

D

and 2.13d D

1

LR d

k

We know the density of mixture:1g/cm3

3 30.97 / & 0.03 /100

ii s Pug cm g cm

24 30.023564 10 /SN atoms cm and 5 24 37.559 10 10 /PuN atoms cm

3.3S

tr b and 6.8Pu

tr b 10.07828 4.2585 9.0706tr cm D cm d cm

0.0008S

a b and 2.11Pu

a b 2 223877 154.5 430.07L cm L cm R cm

b)

4 24 11.85 1.2624 10 10Pu Pu

f fb cm

Maximum flux for a spherical reactor occurs at the center of sphere and assuming d is small

P=500MW,

14 2

max2 2 30

1 1( ) ( ) 3.8865 10 / sec

4 4 4limrR f R f R f

P r P r PSin Sin n cm

R E r R R E r R R E

c)

2 2

10.450 and 1 0.450 0.550

1L NLP P

B L

6.11

a) From table 6.2 using the result for parallelpiped result and inserting a,a,a instead of a,b,c for this

cube you can calculate everything as,

2 2 -23*( ) 5.1 5 cmB ea

b)

For the maximum flux insert x=0;y=0; and z=0 in the flux as,

max

3.87 3.87cos( )cos( )cos( ) 2.18 12 n/cm2/sec

R f R f

P x y z Pe

VE a a a VE

c)

Using eq. (6.46)

1 and P= and so 5.63 11 n/cm2/secav R f av

R f

PdV E dV e

V E V

d)

Using eq.3.58

consumption rate=1.05*(1+ )P g/day=24.55 g/day

6.3

R / 2

''' '''

max 0

0 / 2

R / 2

''' '''

max 0 max

0 / 2

''' ''

2.405q ( , ) q ( )cos( ) first integrate both sides with 2 and we

2.405 MWobtain 20MW=q ( )cos( )2 then find q 9.25 5

cm3

q ( , ) q

H

H

H

H

r zr z J rdrdz

R H

r zJ rdrdz e

R H

and r z

'

max 0 0

2.405 2.405*7 ( 22.7) MW( )cos( ) 9.25 5* ( )cos( ) 1.13 5

50 100 cm3

r zJ e J e

R H

NOTE!!!!!

' ' '

0 0 0 1

0 0

2 2

0 12 2

0

2.405 2.405 2.405( ) ( ) and using the relation ( ) ( )

2.405

2.405 2.405, ' ' ' ( ') ' *2.405* (2.405)

2.405 2.405

R R

R

r R r rJ rdr J dr J x x dx xJ x

R R R

r dr R Rsay x dx x J x dx J

R R