sol_ch7_all

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,

2

2

T

T

d dM dM

dM,D O dM,Be dM,C

d,D O d,Be d,C

k f 1

2.065 from table 6.3 so f 0.484

We will use t t (1 f ) and t from table 7.1

t 4.3e 2; t 3.9e 3; t 0.017

Then,

t =0.022188sec;t =2.0124e-3sec;t 8.772e 3sec

7.5

One‐delayed‐neutron group reactivity equation;

p 1

p p

l where 0.0065; 0.1sec

1 l 1 l

pFor l 0.0sec

pFor l 0.0001sec

pFor l 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 lifetime increases.And it is strongl

recommended that before exam,study figure 7.1 .

7.8

tfT

f i

i

2e 4 from figure 7.2 so you can ignore jump in power(flux)

in this positive reactivity insertion situation

PP Pe then t=ln T 3.456hr

P

7.10

In eq 7.19

a T

a Ta T

T a T a T

prompt delayed

prompt neutrons:(1- )k in a critical reactor(from 7.21)

delayed neutrons:p C

kdC0 C p C k

dt p

s (1- )k k

Now you can compare their values

prompt

de

(1- )

layed


7.12

t0

t

0 T

t

9 0 800

P 1P(t) e in here then, and

T

PP(t) e in here take T=-80sec

1

PP 10 e t 25.24min.

1 ( 5)

7.14

,0 0 ,1 ,1 ,0 1 0 0

,1 1 ,1 ,1 1 1

F FF Fa1 a0

1 0 a1 a0F M F M

a1 a a0 a

F M

0 a0 a

1

k pf ,critical state k 1 k k pf pf f1

k pf ,original state k k pf f

f f and we know =0.95 and finally,

f 0.9511 1 (

f 0.95

F M

a0 a

)

7.16

t

T0 0

i i

i

20min 60sec/ mina)2P P e T 1731.6sec.

ln 2

b)From fig 7.2 rectivity is small so small reactivity assumption can be used as,

1 1T= t 0.0848(from table 7.3)=4.89e-5=4.89e-3%

1731.6

4also in dollars=

.89e-57.52e 3$ 0.752cents

0.0065(U235)

7.17

i i

i

8hr 60min 60sec 8hr 60min 60seca)100MW 1MWe 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 1T= t

6253.8

0.0324(from table 7.3)=5.18e-6

7.18

1

1

(300 100)sect t

0 0 100secT T

(t 3t t

0 0T T

1a)From fig 7.1 when 0 0 so T= T

b)Use prompt jump approximation,

P P 10wattsP(t)= e e e 82watts

0.0991 1

1

c)Use T=-80sec.

P P 82wattsP(t)= e e e

81 1 ( )

1

00)sec

80sec


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

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50

0.02

0.04

0.06

0.08

0.1

0.12

0.14

X: 3.9

Y: 0.1004

rod radius

reac

tivity

reactivity vs rod radius(a)

7.23

TT 1 2

a

T T

T T

a)For a slab this equation is solved you know as,

qx x(x) A sinh( ) A cosh( ) then to find the constants you must introduce

L L

2 boundary conditions

d d1 1 1B.C.1: 0 @ x=0 and B.C.2: @ x=(

dx dx d

1

T

a

TT

a

m/2)-a

Introducing B.C.1 you find A 0 and B.C.2

xcosh( )

q LA2=- 1d

sinh((m 2a) / 2L) cosh((m 2a) / 2L)L

So finally,

xcosh( )

q L(x) 1d

sinh((m 2a) / 2L) cosh((m 2a) / 2L)L

b)

Neutron current

T@(m/2)-a

@(m/2)-a

density at the blade surface,

d LJ D

ddxcoth((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...

---What is

the area of the blades in the cell:

From fig 7.9,assume unit depth into the page so the cross sectional area of one of four blades,

A=(l-a) 1

Divide by the total number of neutrons thermalizing per seco

R 2

nd in the cell

---What is the volume of the cell:

From fig 7.9,assume unit depth into the page so

V=(m-2a) (m 2a) 1

So as in page 358

4(l a) 1f

d(m 2a)coth((m 2a) / 2L)

L

7.25

You should find the B-10 average atom density in the reactor

3

3 1

B aB3

Total mass of B-10=50rods 500g=25 10 g

25e3N 0.6022e24 1.39e27atoms

10.8

Atom density averaged over whole reactor volume,

1.39e27N 2.9e21 atoms/cm 2.9e21 0.27b 7.8e 4cm

4(48.5)

3

use eq.7.62 the

w

7.8e 4n find, 0.0938 9.4%

0.00833 0.000019

7.27

100H cm and 0.5$@ x H

a) For 3 / 4 75 x H cm

1( ) ( ) (2 / ) (3 / 4) 0.4545$

2

so the positive reactivity insertion is

-0.4545$-(-0.5$)=0.04545$

xx H Sin x H H

H

b) The rate of reactivity per cm can be found by differentiating the reactivity equation over

the distance.

( ) 1( ) (2 / )

2

d x d xH Sin x H

dx dx H

1 1( ) (2 / )H Cos x H

H H

( )0.005$ / 0.5 /

3 / 4

d xcm cent cm

dx x H

7.31

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

T

T

If ( )

decrease in T decrease in k reduces P

gives further dec. in k shut down(unstable)

If ( )

decrease in T increase in k increase in P

inc. in T and finally reactor returns to its original state

!(stable)

7.33

exp F F

M sM M

N V Ip

V

I: Resonance Integral

sM : Scattering Cross-Section of Moderator

M : Constant

2 1.5 0.75a a (rod radius)

11 0

(300 )( ) (300 )(1 ( ))

2

I KdII T I K T T

dT T

( ) lnsM M M

F F

VI T p

N V

0 0( ) ( ) ln0.912 0.0921T T I T I T k k where sM M M

F F

Vk

N V

For slightly enriched uranium dioxide reactor take 310.5 / g cm (See Chapter 6).

1 /A C a where 461 10A and 22.68 10C (Table 7.4) 1 0.009503

0 1 0665 ( 938 ) ( ) ( )(1 13.31* ) 1.1264 ( ) T C K I T I T I T

( ) 0.0921 1.1264 0.1037 I T k k

@665

1exp ( ) exp 0.1037 0.9014

o C

kp I T

k k

7.34

0

0

5 0 1

70 21

550 287

(287 21) 2 10 5.32 3 0.532% 0.81$

where =0.0065

T

F C

F C

dC e

dT T

7.37

First you should solve problem 7.36 to find the fraction of expelled water,

0

0

0

23 3

vessel water 0

3 0 3

v

0

0

575F 301 C6 C increase in T

585F 307 C

DV 6.5m V v 3.25m

4

vT v 3.25m 3e 3 6 C 5.85e 2m

v

v0.018

v

Then find f after expelling,

,0 0 ,1 ,1 ,0 1 0 0

,1 1 ,1 ,1 1 1

F FF Fa1 a0

1 0 a1 a0F M F M

a1 a a0 a

F M

0 a0 a

1

k pf ,critical state k 1 k k pf pf f1

k pf ,original state k k pf f

f f and we know =0.95 and finally,

f 0.9511 1 (

f 0.95

F M

a0 a

F F

a a0 F M F M

a a a a

M

a0 F

a 0

0

0

0

T 0

)

f f (1 )

1in here f 0.682 so 1

f

so

1f 0.956

11 0.0982 ( 1)

f

f-fThen = 0.287

f

0.287Finally, (f ) 0.0478per C

T 6 C


7.39

The reactivity equivalent of equilibrium xenon is to be;

I X T

X Tp

where 13 20.770 10 / secX cm and 0.00237X and 0.0639I

2.42 and 1p

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5-0.03

-0.025

-0.02

-0.015

-0.01

-0.005

0

X: 4.8

Y: -0.02695

thermal flux x 1e14

reactivity

Note the convergence …..

7.42

For Xenon using eq.7.94

( )

6.39 2 and 2.37 3 (from table 7.5)

2.09 5 (from table 7.6)

You should make a correction to the thermal absorption cross

section as follows,

I X f T

X aX T

I X

X

X

where e e

e

0 0 0.5

a,X a,X

2

20(200 ) (20 ) ( )

2 200

(200 ) 0.886 1.236 2.65 6 1 24 0.316

(200 ) 9.17 19 9.17 5 finally,

0.06627 1 13

2.09 5 9.17 5 1 13

For Samarium using eq.7.94

Xe

a

aX

aX

f

f

P

aX

g C C

C e e

C e cm e b

eX

e e b e

S

where

0 0 0.5

a,S a,S

0.01071

20(200 ) (20 ) ( )

2 200

(200 ) 0.886 2.093 41 3 1 24 0.316

(200 ) 2.39 4 finally,

0.010712.39 4

P

S

a

aX

aX

f

g C C

C e e

C e b

Se b

Note:When finding fission cross sections you should find the atom density of

uranium 235 for this infinite thermal reactor.To do this ,refer to example 6.5 on

page 294 taking buckling zero and find a relation between moderator number

density and fuel density.

7.43

Using eq. 7.98

0.06627 1 13

2.42 1 13 0.773 13 p= =1

0.01071

2.42

Xe

Sm

e

e ewhere

7.44

First of all, we must write the rate equations for each element;

SmSm Sm Sm Sm Sm

a T f T

dNN N

dt

EuSm Sm Eu Eu Eu Eu

a T

dNN N N

dt

GdEu Eu Gd Gd

a T

dNN N

dt

a) For equilibrium reactivity;

( ) ( )i iX XN t N t dt

( )0

iXdN t

dt

and ignore

Sm Sm

a TN & Eu Eu

a TN

Inserted into all rate equations;

Sm

f TSm

SmN

Sm Sm Eu EuN N

Eu EuGd Gd

a

T

NN

Reactivity equation is found as below;

/Gd Sma f

p p

where 57 10Sm and 2.42 and 1p

52.893 10

b)

157Sm decays rapidly relative to 157Eu and half-life of the 157Sm is too small so,

0Sm

Sm Sm Sm

f T

dNN

dt

Sm Sm Sm

f TN

This equation is inserted into rate equation of 157Eu and 157Gd ;

EuSm Eu Eu

f T

dNN

dt

( )GdEu Eu Gd Gd

a T

dN tN N

dt

At shutdown 0 0&Eu GdN N are equal to equilibrium concentration for 157Eu and157Gd .

No fission & no absorption is observed.

From rate equation of 157Eu 0( )Eu

Sm

f TEu Eu t

Eu

EutN t N e e

From rate equation of 157Gd 0( ) (1 )

Sm

f TGd Gd

Eu

EutN t N e

From equilibrium of 157Gd 0

Sm

fGd

Gd

a

N

( ) (1 )

Sm Sm

f f TGd

Gd Eu

a

EutN t e

Maximum reactivity is reached at time goes to infinity!

max

1( ) ( )Gd Sm T

f Gd Eu

a

N t

/Gd

a f

p

where (1 )

GdGd Sm T a

a f Eu

(1 ) /Gd

Sm T a

Eu

where

51.162 10Eu s

54.386 10 0.675cents

7.47

a) For constant power;

( , ) ( , )R fF T

V

P E r t r t dV

So as N decreases ,flux should increase to keep power constant,

( )( ) ( ) (1)

( ) ( ), ( ) ( )

( ) ( ) (0) (0) constant

integrating (1) between 0,t we get,

( ) (0) (0) (0)

( ) (0)[1 (0) ]

)

( ) ( )

( )

FF aF T

R fF T fF F aF

F T F T

F F F aF T

F F aF T

R fF T

T

dN tN t t

dt

P E t t t N t

N t t N

N t N N t

N t N t

b

P E t t

Pt

E

1 1

( ) (0)[1 (0) ]

R fF F R fF F aF T

P

N t E N t

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