radiation shielding and reactor criticality fall 2012
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Radiation Shielding and Reactor Criticality Fall 2012. By Yaohang Li, Ph.D. 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 - PowerPoint PPT PresentationTRANSCRIPT
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Radiation Shielding and Reactor Criticality
Fall 2012
By Yaohang Li, Ph.D.
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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
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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
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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 cs 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
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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
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Collision• Collision Probability
– cdf of the distance that the particle travels before collision• Fc(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 (Ei)
– 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
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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?
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Elementary Approach• 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– Fc(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
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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
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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
106.
– To estimate an accuracy of 10% require the number of 108 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
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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– Pi()
• 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)
• ci=cosi; i=c(Ei); i=[1-(Ei)]c(Ei)
– P0()=exp(- i/c0)
• the probability that the particle passes through z= before suffering any scatterings
– Pi+1()
• A particle crosses z= between its (i+1)th and (i+2)th scatterings
n
nn
EEEhh
,...,,
,...,,
10
10
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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)=Pi(’)
• ii: the particle suffered the (i+1)th scattering between the planes z= ’ and z= ’+d’
– P(ii)= i d’/|ci|
• iii: after scattering, the particle now travels with energy E i+1 in direction i+1
– P(iii)= exp(- i+1(- ’) /ci+1)
– Then
– The probability of penetrating the shield is
0
111 ||
'}/)'(exp{)'()(
i
iiiii c
dcPP
))((0
ii tPE
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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
))((0
ii tPE ))((
0
N
ii tPE
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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 pc
– Proportional to capture cross section
• Scatter with probability ps
– Proportional to scattering cross section
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Scattering
•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
X
Z
v v'
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Solid Angle
• 2D– measured by unit angles
(radians)
– full circle subtends 2
• 3D– measured by unit solid angles
(steradians)
– full sphere subtends 4
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Probability of Scattering
•Scattering equally in all directions– probability p(,)dd=d/4
•Definition of the Solid Angle
– then d = sindd– we can get p(,) = sin/4
•Probability density for and
S
dd sin
2
0
sin2
1),()( dpp
0 2
1),()( dpp
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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
1)( dxxp
x
rdxxpxP ')'()(
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Randomizing the Angles
=2r is uniformly distributed between 0 and 2
– Then we can get cos = 1-2r
– cos is uniformly distributed between -1 and +1
0
sin2
1xdxr
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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
/)/1()( lelp
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Neutron Transport Algorithm (1)
• Input parameters– thickness of the plate t
– capture probability pc
– scattering probability ps
– mean free path
• Initial value z=0
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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<0, add one to the number of reflected neutrons. 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
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An Improved Method
• Instead of Considering A Neutron– Consider a set of neutrons
– ps portion of neutrons are scattered
• All scattered neutrons will move to a new direction
– pc portion of neutrons are captured
• A better convergence rate
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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
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What I want you to do?
• Review Slides• Review basic probability/statistics concepts• Work on your Assignment 3