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Radiation Shielding and Reactor Criticality Fall 2012 By Yaohang Li, Ph.D.

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Presentation on theme: "Radiation Shielding and Reactor Criticality Fall 2012 By Yaohang Li, Ph.D."— Presentation transcript:

1 Radiation Shielding and Reactor Criticality Fall 2012 By Yaohang Li, Ph.D.

2 Review Last Class –Test of Randomness –Chi-Square Test –KS Test This Class –Monte Carlo Application in Nuclear Physics Radiation Shielding Reactor Criticality Simulation of Collisions –Assignment #3 Markov Chain Monte Carlo

3 Monte Carlo Method in Nuclear Physics Flux of uncharged particles through a medium –Uncharged particles paths between collisions are straight lines do not influence one another –independence –allow us to take the behavior of a relatively small sample of particles to represent the whole –Randomness derive the Monte Carlo methods directly from the physical processes

4 Problem Definition Particle (Photon or Neutron) –energy E –instantaneously at the point r –traveling in the direction of the unit vector  Traveling of the Particle –At each point of its straight path it has a chance of colliding with an atom of the medium No collision with an atom of the medium –continue to travel in the same direction  with same energy E A probability of  c  s that the particle will collide with an atom of the medium –  s: a particle traverses a small length of its straight line –  c : cross section »depends on the nature of surrounding medium »energy E

5 Cross Section Determining  c –The medium remains homogeneous within each of a small number of distinct regions over each region,  c is a constant  c change abruptly on passing from one region to the next –Example Uranium rods immersed in water –  c a function of E in the rods –  c another function of E in the water

6 Collision Collision Probability – cdf of the distance that the particle travels before collision F c (s) = 1 – exp(-  c s) Three situations of collision –Absorption the particle is absorbed into the medium –Scatter the particle leaves the point of collision in a new direction with a new energy with probability  (E i ) –fission (only arises when the original particle is a neutron) several other neutrons, known as secondary neutrons, leaves the point of collision with various energies and directions Probability of the three situations –Governed by the physical law –Known distribution from Monte Carlo point of view

7 Shielding and Criticality Problems The Shielding Problem –When a thick shield of absorbing material is exposed to  - radiation (photons), of specified energy and angle of incidence, what is the intensity and energy-distribution of the radiation that penetrates the shield? The Criticality Problem –When a pulse of neutron is injected into a reactor assembly, will it cause a multiplying chain reaction or will it be absorbed, and in particular, what is the size of the assembly at which the reaction is just able to sustain itself?

8 Elementary Approach –Exact realization of the physical model Not very efficient –Tracking of simulated particles from collision to collision Starting with a particle (E, , r) Generate a number s with the exponential distribution –F c (s) = 1 – exp(-  c s) If the straight-line path from r to (r+s  ) does not intersect any boundary (between regions) –the particle has a collision Otherwise –proceed as far as the first boundary –if this is the outer boundary, the particle escapes from the system Repeat the procedure

9 Improvements of the Elementary Approach Problem –There may be too many or too few particles –Consider a reactor containing a very fissile component Every neutron entering this region may give rise to a very large number coming out –Give us more tracks than we have time to follow Solution –“Russian Roulette” Pick out one of the particles –discard it with probability p –otherwise allow this particle to continue but multiply its weight (initially unity) by (1-p) -1 The number of particles is reduced to manageable size –“Splitting” To increase the sample sizes –a particle of weight w may be replaced by any number k of identical particles of weights w1, …, wk »w1+…+wk=w

10 Special Methods for the Shielding Problem Outstanding feature of the shielding problem –The proportion of photons that penetrate the shield is very small, say one in 10 6. –To estimate an accuracy of 10% require the number of 10 8 paths. Hit or miss Quite inefficient Solution –Semi-analytic method –Allows the same random paths to be used for shields of other thickness –Simplification Only think about three coordinates –Energy E –Angle between the direction of motion and the normal to the stab –Distance z from the incident face of the slab

11 The Semi-Analytic Method (I) A random history –for a particle which undergoes a suitably large number n of scatterings in the medium The semi-analytic method –P i (  ) The probability that a particle has a history hi and also crosses the plane z=  between its ith and (i+1)th scatterings –Abbreviation (α is the absorption probability) c i =cos  i ;  i =  c (E i );  i =[1-  (E i )]  c (E i ) –P 0 (  )=exp(-  i  /c 0 ) the probability that the particle passes through z=  before suffering any scatterings –P i+1 (  ) A particle crosses z=  between its (i+1)th and (i+2)th scatterings

12 The Semi-Analytic Method (II) The semi-analytic method –the (i+1)th scattering occurred on some a plane z=  ’ where 0<  ’< . –Compound event i: immediately prior to the (i+1)th scattering the particle crossed z=  ’ –P(i)=P i (  ’) ii: the particle suffered the (i+1)th scattering between the planes z=  ’ and z=  ’+d  ’ –P(ii)=  i d  ’/|c i | iii: after scattering, the particle now travels with energy E i+1 in direction  i+1 –P(iii)= exp(-  i+1 (  -  ’) /c i+1 ) –Then –The probability of penetrating the shield is

13 Probability of Penetration Replace with Approximate unbiased estimator of penetration probability N = 25, 12, 9, 6 is efficient for shields of water, iron, tin, and lead

14 Neutron Transport Transmission of Neutrons –Bulk matter Plate –thickness t –infinite in the x and y directions –z axis is normal to the plate –Neutron at any point in the plate Capture with probability p c –Proportional to capture cross section Scatter with probability p s –Proportional to scattering cross section

15 Scattering –polar angle  –azimuthal angle  we are not interested in how far the neutron moves in x or y direction, the value of  is irrelevant

16 Solid Angle 2D –measured by unit angles (radians) –full circle subtends 2  3D –measured by unit solid angles (steradians) –full sphere subtends 4 

17 Probability of Scattering Scattering equally in all directions –probability p( ,  )d  d  =d  /4  Definition of the Solid Angle –then d  = sin  d  d  –we can get p( ,  ) = sin  /4  Probability density for  and 

18 Non-uniform Random Sample Generation Revisit Probability Density p(x) Then –r is a uniform random number Inverse Function Method –use r to represent x

19 Randomizing the Angles  –  =2  r –  is uniformly distributed between 0 and 2   –Then we can get cos  = 1-2r –cos  is uniformly distributed between -1 and +1

20 Path Length Path length –distance traveled between subsequent scattering events –obtained from the exponential probability density function –l=- lnr is the mean free path or the cross section constant  c

21 Neutron Transport Algorithm (1) Input parameters –thickness of the plate t –capture probability p c –scattering probability p s –mean free path Initial value z=0

22 Neutron Transport Algorithm (II) 1.Determine if the neutron is captured or scattered. If it is captured, then add one to the number of captured neutrons, and go to step 5 2.If the neutron is scattered, compute cos  by cos  = 1-2r and l by l=- lnr. Change the z coordinate of the neutron by lcos  3.If z t, add one to the number of transmitted neutrons. In either case, skip to step 5 below. 4.Repeat steps 1-3 until the fate of the neutron has been determined. 5.Repeat steps 1-4 with additional incident neutrons until sufficient data has been obtained

23 An Improved Method Instead of Considering A Neutron –Consider a set of neutrons –p s portion of neutrons are scattered All scattered neutrons will move to a new direction –p c portion of neutrons are captured A better convergence rate

24 Summary Nuclear Simulation –Radiation Shielding –Reactor Criticality –Particle Assumption Cross Section Collision –Elementary Method –Improvements for the Elementary Method Russian Roulette Splitting –Special methods for the shielding problem Semi-Analytic Method –Neutron Transport Problem –Nonuniform Distribution Samples

25 What I want you to do? Review Slides Review basic probability/statistics concepts Work on your Assignment 3


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