Lamarsh Solution Chap7

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 strongl recommended that before exam,study figure 7.1 .

7.8

  2e  4 from figure 7.2 so you can ignore jump in power(flux) in this positive reactivity insertion situation t P
Pf  Pi e T then t=ln f  T  3.456hr
Pi

7.10
In eq 7.19

prompt neutrons:(1-)k   a T delayed neutrons:pC


in a critical reactor(from 7.21)

k   dC  0  C   a T  pC   k   a T dt p

 s T  (1-)k   a T   k   a T
  

  prompt delayed

Now you can compare their values prompt (1-)

delayed

LAMARSH SOLUTIONS CHAPTER-7 PART-2

7.12


P0  t

1
P(t)  e in here
  then, and  


T
t
P0 T
P(t)  e in here take T=-80sec

1

t

P0
P0 10  e 80  t  25.24 min .
1  (5)
9

7.14

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 f1  F f0  and we know  a1F =0.95 a 0 F and finally,
M
F
M
 a1   a
a 0  a f0 1 0.95a 0 F  a M
1  1
(
) f1 0.95  a 0 F  a M
7.16
20 min 60sec/ min
 1731.6sec. ln 2
b)From fig 7.2 rectivity is