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Lamarsh Solution Chap7

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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 …show more content…

1 1

d  ( x) d 
1
x

   ( H )  
Sin(2 x / H )     ( H )   Cos(2 x / H )  dx dx 
H H

 H 2




d  ( x)
 0.005$ / cm  0.5cent / cm dx x  3H / 4

7.31
There is a decrease in T so let’s examine the effects of sign of temperature coefficients,

If  T  () decrease in T  decrease in k  reduces P
 gives further dec. in k  shut down(unstable)
If  T  () decrease in T  increase

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