# 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 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:pC

in a critical reactor(from 7.21)

k

dC

0 C a T pC 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 0.95a 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 (300100)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......

Please join StudyMode to read the full document