Upload
abdul-manan
View
897
Download
5
Embed Size (px)
Citation preview
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