LAMARSH SOLUTIONS CHAPTER-6 PART/2

IMPORTANT TABLES:

*TABLE 3.2 NON-1/V FACTORS

*TABLE 3.4 THERMAL DATA (0.0253ev) FOR THE FISSILE NUCLIDES

*TABLE 5.2 THERMAL NEUTRON DIFFUSION PARAMETERS OF COMMON MODERATORS AT 20C

*TABLE 5.3 FAST GROUP CONSTANTS FOR VARIOUS MODERATORS

*TABLE 6.1 NOMINAL ONE-GROUP CONSTANTS FOR A FAST REACTOR

*TABLE 6.2 BUCKLINGS AND FLUXES FOR CRITICAL BARE REACTORS(ASSUME D IS SMALL)

*TABLE 6.3 VALUES OF T

*APPENDIXES:TABLE II.2 AND TABLEII.3

6.20

Assume d<<R

2 2( )BR

=3.95e-3 cm-2

a)

0

0

( )

( ) ( )

aM aM FFF M M

aF M aF aF M

E MMm Z m Z m

M g T E M

, eq(6.88) , where

2 2

2

1

1

TM

T TM

B MZ

B

,eq(6.83), 2 2 2480 102 582T T TM L cm

And in here, 0 0( ) 0.0092 , ( ) 681 , ( ) 0.978, 2.065aM aF aF TE b E b g T so inserting into

eq(6.88) we’ll find , 1.79 3F Mm e m

Then the total mass of Be in the reactor as follows,

31.85 /M g cm , 3 5 345.24 10

3V R cm

1.85 969 1.73M Fm V kg m kg [ANS]

b)

Thermal flux is given by,

R2

1( ) ( ), from table 6.2 in here P=50kW=5e4joules/sec ; E 3.2 11

4T

R f

r Pr A Sin A e joule

r R R E

F 0 from 6.85 N = = (T) ( ) V V 2

F A F Af f f f fF f

F F

m N m NN g E

M M

;and the values,

-1

0(T)=0.976; ( ) 582 =4.26e-3cmfF f fg E b and the constant A is found as follows,

2

5 4 13.66 13 ( ) 3.66 13 ( )

4 50 3.2 11 4.26 3T

e rA e r e Sin

e e r R

[ANS]

c)

4: 2.3 10 n/cm2.sec in here D=0.5 for Be

4r R R f

d D PLeakage D e

dr R E

[ANS]

d)

cons.rate=1.05(1+ ) P g/day=1.05(1+0.169) 50e-3MW=0.0613 g/day [ANS]

6.26

235 2 2 2

0.84 2.13 1.7892 ; is from table5.2

(300 ) 2.055 from tab.6.3 and 721

M

T T T T

D cm d D cm D

C M L cm

a)

2 -2

2 2

2 2

For a critical reactor; 1 =1.012e-4 cm 1

where 3 ( ) and 3 / 2 537.32

T

kB

M B

B a B a a d cma

[ANS]

b)

max 25

3

25 25 0 25

25

u

3.87 [from table 6.2] 1113.5 24 / 3

( ) (573 )2

1113.5 24235 4.3466 5 gr U-235 then found the mass of the natural uranium as,

0.6022 24

4.3466 5m 60 tons nat.

0.0072

R f f

PN e atoms cm

a E N E g K

em e

e

e

Note:U in here 0.0072 is the percent of U235 in natural U

6.37

a)

2 22.80 (from equal volumes approach) 1.58b L L b cm ,which is the radius of equivalent

radius [ANS]

b)

2

2 20.290F

M

V R L

V b L R L

[ANS]

c)

F ML 1.55 ; L 2.85 ; thus,

0.75 0.75 1.58x= =0.48 ; y= =0.26 ; z= 0.55

1.55 2.85 2.85

Then as x,y and z are less than 0.75 you can use eq(6.117) and eq(6.118) with a good accuracy,

After finding E and F the

cm cm

1

0 2, . 25 ,25 0 25 28 ,28 0 28

1

0 2, , 0 , 0

n finally we'll use eq.6.114 in where,

T( ) ( ( ) ( ) ( ) ( ) ) ( ) and

2 T

T( ) ( ( ) ( ) ( ) ( ) ) ( )

2 T

in here you can use the table II.3

a nat U a a a a

a water H a H a H O a O a O

T N E g T N E g T

T N E g T N E g T

1 1

235 238

,at 0.0253ev, 0.3677 & 0.0222 with a good accuracy or

you can calculate yourself the macroscopic cross sections, inserting the following above

680 & 2.73 & 0.664 and the ot

F M

a a

M

a a a

cm cm

b b b

3 3her constants as 19.1 / & 1 / so finally

0.797

F Mg cm g cm

f

[ANS]

d)

2 24 3

1.46 from table(6.6) where

4.83 10 10 / from table II.3

F F

M sM M

F

N V I

VM sMp e

N atoms cm

I A C a where 2.8& 38.3A C from Table6.5

0.883P [ANS]

e)Using fig. 6.10 and 1.5 cm rods on that figure

1.04 [ANS]

f)

25 25 25 25

2825 2825 28

25

02825 25 f25

25

0 0

25 a25 28 a25

The ratio 138 for natural U.Using 2.42, 582 ,g (20 ) 0.976

681 ,g (20 ) 0.978, 2.70 ,g (20 ) 1.0017 gives,

1.32 and using t

f f

T

a aa a

f

a a

T

N

N

Nb C

N

b C b C

he four factor formula we can find k

0.9828 k fp

6.17

(-a) -100 0 100 (a)

a)

2 2

2 -2

( ) in here 2 where d=2.13D and D=0.84 for graphite finally

9.53 4 cm

B a a da

B e

b)

2

TM

2 2

2

F F 0 F

M M M

F

M

,

F,critical

3500 , 368 , 2.065

16.56 and the using the relation,

1

N N ( ) ( ) NZ= 3.34 5 so use,

N N N

N1.046 3 gr/cm3 and finally,

N

N

TM T

TM

T TM

aF aF aF

aM aM

F MF

M F

F cri

L

B MZ

B

g T Ee

Me

M

2.68 18 atoms/cm3tical AV

F

Ne

M

c)

2 2 2Z(1 ) and here f= 0.867 and so 463 2

Z+1T TM TL f L L cm

d)

k 1.79 T f

e)

max( ) cos( ) 5 12cos( ) and the current

d ( ) DJ=-D 5 12 sin( )

x xx e

a a

x xe

dx a a

f)

max

, ,25 25 0

25 0

1.57[ again from table 6.2]

( ) ( )2

where 200 2 2.13 0.84 203.57

( ) can be found from table II.2 in appendix

so you can find P from here!

R F critical f fF

f

PA

aE N E g T

a cm

E

6.27

assume d<<a

2

2

1

and k=1 for criticality

k

kBL

2 2

c c

2

R R2

R

in the power producing section,core,you should solve the equation

B 0

and in the reflector region there is no fission ;so solve then,

10

L

For criticality conditions:

a)

İnf. reflector core core İnf. reflector

-a/2 0 a/2

c 1 2

c2 c 1

R 3 4

R 4 R 3

c R

c Rc R

cos( ) sin( )

@ 0 B.C.1 0 cos( )

lim x must be finite so 0

then interface B.C.

a a[I] ( )= ( )

2 2

a a[II] -D @ D @ the

2 2

R R

R

x x

L L

x

L

A Bx A Bx

dx A A Bx

dx

A e A e

A A e

d dx x

dx dx

2

1 3

23c 1

c

n introduce these into the fluxes,

aI cos( ) =

2

aII D sin( )

2

acriticality condition gives:D tan( ) (*)

2

(*) must be satisfied by the reactor to be critical.

R

R

x

L

x

L

R

R

R

R

A B A e

ABA B D e

L

DIIB B

I L

b)

ref. core core ref.

-a/2-b -a/2 0 a/2 a/2+b

R R

c 1

x x

L L

R 3 4 R

R 3

R

c R

c

again A cos(Bx)

aA e A e from vacuum B.C. ( b) 0

2

a( b) x

a2so we can write =A sinh( ) note that @ x= b vacuum B.C. is satisfied.L 2

then interface B.C.

a a[I] ( )= ( )

2 2

[II] -D

c RR

1 3

R

3c 1 R

R R

Rc

R R

d da a@ x D @ x then introduce these into the fluxes,

dx 2 dx 2

a bI A cos(B ) = A sinh( )

2 L

Aa bII D BA sin(B ) D cosh( )

2 L L

DII a bcriticality condition gives:D B tan(B ) coth( ) (**)

I 2 L L

(**) must be s

atisfied by the reactor to be critical.

c)

R+b

R

c 1 2

c 2 c 1

R R RR 3 4 R R 3

c R

sin(Br) cos(Br)A A

r r

sin(Br)as r 0 must be finite so A 0 A

r

r r R b rsinh( ) cosh( ) sinh( )

L L LA A or using vacuum B.C. (R b) 0 A

r r r

then interface B.C.

[I] (R)= (R)

[II] -

c Rc R

R1 3

R Rc 1 3 R2 2

R

c

d dD @ r R D @ r R then introduce these into the fluxes,

dx dx

bsinh( )

Lsin(BR)I A = A

R R

b bcosh( ) sinh( )

L LBcos(BR) sin(BR)II D A ( ) A D ( )

R R L R R

II 1criticality condition gives:D (Bcot(BR) ) D

I R

RR

R

bcoth( )

L 1( ) (***)

L R

For the fluxes:

c 1

a / 2

R f 1 1

R fa / 2

c 1

R 22

R f 1 1

R f0

For infinite slab; A cos(Bx) so,

1.57PP=E A cos(Bx) dx A

aE

sin(Br)For the sphere; A so,

r

sin(Br) PBP=E A 4 r dr A

r 4 E (sin(BR) BR cos(BR))