LAMARSH SOLUTIONS CHAPTER-7 PART-1

7.1
Look at example 7.1 in the textbook,only the moderator materials are different
Since the reactor is critical,

k   T f  1
T  2.065 from table 6.3 so f  0.484
We will use t d  t dM (1  f ) and t dM from table 7.1 t dM,D2O  4.3e  2; t dM,Be  3.9e  3; t dM,C  0.017
Then,
t d,D2O =0.022188sec;t d,Be =2.0124e-3sec;t d,C  8.772e  3sec

7.5
One‐delayed‐neutron group reactivity equation;



lp
1  lp





where   0.0065;   0.1sec1
1  lp   

For lp  0.0sec

For lp  0.0001sec

For lp  0.001sec

Note:In this question examine the figure 7.2 and see that to give a constant period value ,say
1 sec,you should give much more reactivity as p.neutron lifet ime increases.And it is …show more content…

1  A  C / a where A  61104 and C  2.68 102 (Table 7.4)  1  0.009503
T  665C ( 938K )  I (T )  I (T0 )(1  13.31* 1 )  1.1264I (T0 )
 I (T )  0.09211.1264  k  0.1037k

1

k

p@ 665o C  exp   I (T )   exp   0.1037   0.9014
k

k


7.34

70 F  210C
550 F  287 0C d  
T 

   (287  21)  2  105 0C dT T where  =0.0065

1

 5.32e  3  0.532%  0.81$

7.37
First you should solve problem 7.36 to find the fraction of expelled water,

575F  301 0 C
585F  307 C
0

Vvessel

 6 0 C increase in T

D 2

 6.5m3  Vwater  v 0  3.25m3
4

v
  v T  v  3.25m3  3e  3  6 0 C  5.85e  2m3 v0 

v
 0.018 v0 Then find f after expelling,

k  ,0  pf 0 ,critical state k  ,1  pf1 ,original state



k  ,1  1 k  ,1



k  ,1  k  ,0 k  ,1



pf1  pf 0 f  1 0
pf1
f1

 a1F
a 0 F f0  and we know  a1F =0.95  a 0 F and finally,
F
M
F
M
 a1   a
a 0  a

f1 

f0
1 0.95 a 0 F   a M
1  1
(
) f1 0.95  a 0 F   a M f0 

a F
a F  a M

f

in here f 0  0.682 so

a F
 a F  (1  ) a M
a M 1
 1
a F f0

so f 1
1
1  0.0982  (  1) f0  0.956

f-f 0
 0.287 f  0.287
Finally,  T (f ) 
 0  0.0478per 0 C
T
6C

Then =

LAMARSH SOLUTIONS CHAPTER-7 PART-4

7.39
The reactivity equivalent of equilibrium xenon is to be;

 

 I   X T where X  0.770 1013 / cm2  sec and  X  0.00237 and  I  0.0639
 p X  T

  2.42 and p    1

0

-0.005