# Lamarsh Solution Chap7

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; 0.1sec1

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…*

T 665C ( 938K ) I (T ) I (T0 )(1 13.31* 1 ) 1.1264I (T0 )

I (T ) 0.09211.1264 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) 2 105 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