Reactors 2 Solutions

B. Rouben UN0802 Reactor Exercises 2

1. A resea research rch reactor reactor is in the shape of a parallelep parallelepiped iped with with a square base base of  side 5.2 m and a height 6.8 6.8 m. The reactor reactor is filled uniformly with with a fuel of  -1 one-group properties  νΣ   νΣ ff   !.!!"2 cm #and  ν  2.$5% and Σ a  !.!!"! cm -1.   The reactor reactor operates operates steadily steadily at a fission fission power power of 15 &'. The a(erage a(erage -1 (alue of )nergy per fission E ff   2!! &e*+ and 1 e* 1.6,1!  . /0eglect the   etrapolation distance. #a% 'hat is the (alue of the diffusion coefficient3 #b% 'hat are the a(erage and maimum (alues of the neutron flu3 #c% 'hat is the maimum maimum (alue of the neutron flu3 #d% At what rate is the fuel consumed consumed in the entire reactor #in nuclides.s nuclides.s -1% and at the centre of the reactor #in nuclides.cm -4.s-1%3 olution on net page.

olution

( a) We %no that   $ 2

∴ # =

(ν Σ − Σ )   f  

 $

a

2

 ν Σ  f        − Σ a  % eff         =  #

(1) and  % eff  

=1

(" reactor  operates  steadily" )

(2) 2

2

We can  get   $   from the dim ensions :  $

∴ # =

0.0072 − 0.0070 9.434 * 10 −5

2

2

3.1416    3.1416   = 9.434 * 10 −5 = 2 *      +      520     680  

−2

= 2.12 cm

(b) The max imum   flu! is at  the centre φ ( 0,0,0)

=  A =

3.87 P  VE  R Σ  f  

(  from  "amarsh,  for  a  parallelepiped )

3.87 * 15 MW 

=

 0.0072 cm 2.45  

520 cm * 680 cm * 200  MeV  *  2

cm

2

−1

       

= 3.357 *1012

cm

−2

. s

−1

 "et  the average   flu! be φ . We can   find  it  as   follos : Total   Poer  =  E  R *  Average  Fission  Rate * Volume

∴φ  =

15  MW 

 0.0072 cm 2.45  

200 MeV  * 

−1

  2 2    * 520 cm * 680 cm  

=  E  R Σ  f   φ V  = 8.675 *1011

cm −2 . s −1

(c) Sorry, this is a repetition of    part  of   (b)! ( d ) The   fuel  is consumed  by absorption, not    just    fission!

∴ Total    fuel  consumptio n in

reactor  =

&  Fuel  consumptio n at  centre

Σ aφ V  = 0.0070 cm −1 * 8.675 * 1011 cm − 2 . s −1 * 520cm 2 * 680cm

= 1.117 * 1018  s −1 of   reactor  = Σ aφ ( 0,0,0 ) = 0.0070 cm −1 * 3.357 *1012 = 2.350 * 1010 cm −3 . s −1

cm − 2 . s −1

2. A homogeneous+ bare cylindrical reactor with diameter 6.5 m and height 5.8 m is critical+ and is operated at a fission power of 8!! &'. The material of  which it is composed has #in 1 energy group% ν  = 2.48+ Σ f  !.!!$2 cm -1+ Σ a  !.!! cm-1   [Note: neglect the extrapolation distance.] a% b% c% d%

7alculate the diffusion length. 'hat is the ratio of leaage to absorption3 'hat is the ratio of leaage t absorption3 7alculate the ratio of the flu on the cylinder ais at a distance of 1.8 m from the centre of the reactor to the flu on the cylinder ais at a distance of !.2 m from the centre.

e% 'hat is the a(erage fission rate per cm 4 in the reactor3 olution on net page.

olution To calculate  "2 , e need  to %no  #. We do it  this ay : Re actor  is critical  ⇒  Material   $uc%ling  = 'eometrical   $uc%ling  2

2

2

2

 π     2.405   =  3.1416   +  2.405   'eometrical   $uc%ling  =    +            &       R     580     325   ν Σ − Σ ∴ Material  $uc%ling  ≡  f   a = 0.8410 *10 − 4 cm −2 ⇒  # =  "2

=

 #

Σa

 # 2.38 * 0.0042 − 0.0099

=

0.8410 * 10 1.142 2 0.0099

cm

−4

cm − 2

= 0.8410 *10 − 4

cm = 1.142 cm

= 115.3 cm 2

 #iffusion  "ength  " = 10.74 cm

olution  Ratio of   lea%age to absorption =

 #$ 2

=

115.3 cm * 0.8410

Σa

0.0099

cm −2

cm −1

= 0.0097

olution

 π  (  , here  (  is measured   from     &   

The a!ial   flu! is  proportional  to cos the midplane of   the cylinder 

∴ Ratio of    flu! on a!is 1 m  from the centre to  flu! at  1.5 m  from centre =

cos( π  * 180 / 580) cos( π  * 20 / 580)

= 0.564

f% 'hat is the a(erage fission rate per cm 4 of the reactor3 olution  Fission rate = =

 Poer  rate 200 MeV 

=

Total   Poer  200 MeV  * Volume 800  MW 

200  MeV  * 3.1416 * ( 325

11

cm)

2

= 1.299 * 10

* 580

cm

 fissions.cm −3 . s −1

cm − 2