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 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 small so small reactivity assumption can be used as, 1
1
T=  i t i  
 0.0848(from table 7.3)=4.89e-5=4.89e-3%
i
1731.6
4.89e-5
also in dollars=
 7.52e  3$  0.752cents
0.0065(U235)
t
T

a)2P0  P0e  T 

7.17
8hr  60 min 60sec
8hr  60 min 60sec
T
 6253.8sec(very large)
T
ln100
b)We will make small reactivity insertion approximation using the insight given by figure 7.2 for U-235 so,
1
1
T=  i t i  
 0.0324(from table 7.3)=5.18e-6
i
6253.8
a)100MW  1MWe

7.18

a)From fig 7.1 when   0 1  0 so T=

1
T
1

b)Use prompt jump approximation,
t
t
P0 T
P0 T 10watts (300100)sec
P(t)=
e
e
e 100sec  82watts

0.099

1
1

1
c)Use T=-80sec.
300)sec
t
t
P0 T
P0 T 82watts  (t 80sec
P(t)=
e
e
e

8

1
1  ( )

1

LAMARSH SOLUTIONS CHAPTER-7 PART-3

7.20
Insert 7.56 into 7.57 and plot reactivity vs rod radius
Using eq. 7.57 and 7.56 we plotted and found the radius value for 10% reactivity=3.9 cm reactivity vs rod radius(a)
0.14

0.12
X: 3.9
Y: 0.1004

reactivity

0.1

0.08

0.06

0.04

0.02

0

0

0.5

1

1.5

2

2.5
rod radius

3

3.5

4

4.5

5

7.23

a)For a slab this equation is solved you know as,
x
xq
T (x)  A1 sinh( )  A 2 cosh( )  T then to find the constants you must introduce L
L a
2 boundary conditions
1 d T
1 d T
1
B.C.1:
 0 @ x=0 and B.C.2:
  @ x=(m/2)-a
T dx
T dx
d
Introducing B.C.1 you find A1  0 and B.C.2
x


cosh( )


q
L
A2=- T 1 

d
a 
sinh((m  2a) / 2L)  cosh((m  2a) / 2L) 
L

So finally,
x


cosh( )


qT
L
T (x)  1 

d
a 
sinh((m  2a) / 2L)  cosh((m  2a) / 2L) 
L

b)
Neutron current density at the blade surface,
d
L
J @(m/2)-a   D T

d
dx @(m/2)-a
 coth((m  2a) / 2L)
L
Let 's follow the instructions in the question
Multiply the n.current density by the area of the blades in the cell......
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