PHITS
Multi-Purpose Particle and Heavy Ion Transport code System
Melt a snowman by proton beam
Mar. 2016 revised
title
1
Purpose of this exercise
Let us consider with current technology how realistic
beam weapons in animations are by performing
numerical exercise of proton beam melting a snowman
Lecture on
• Geometry setup
• Change of source
• Concept of normalization
Courtesy of D. Satoh
Purpose
2
snowman.inp
Basic setup
Projectile： 100MeV proton (pencil beam with radius 1.0cm)
Geometry： Water sphere of 5cm radius at the origin
Tally： [t-track] flux distribution
[t-deposit] absorbed dose (Gy/source) in water sphere
Definition of volume of sphere is necessary at [volume] section
…
x:
y:
p:
h:
#
Serial Num. of Region
Dose [Gy/source]
xlin ylog afac(0.8) form(0.9)
x
n
n
y(all
),l3
n
num
reg
volume
dose
r.err
1
1
5.2360E+02
2.9853E-11 0.0032
deposit.out
track.eps
2.9853 x 10-11 (Gy/source)
Check Input File
3
Sections in snowman.inp
snowman.inp
[Title]
Title of the simulation
[Parameters]
Define history number etc.
[Source]
Define source
[Material]
Define materials
[Surface]
Define surfaces
[Cell]
Define cells
[Volume]
Define volume of cells
[T-Track]
Draw particle trajectory
[T-Deposit]
Calculate deposition energy
Sections in Input file
II. Source
I. Geometry
III. Tally
(Detector)
4
Flow chart of this exercise
1. Set geometry of a snowman
2. Set beam condition
3. Determine beam current and
power to melt a snowman
Procedure
5
Setup geometry of a snowman
\phits\utility\rotate3dshow
Geometry to construct
in this exercise
Direction
• A simple structure of big and small ice balls with aluminum plate
• Material of the ice balls are 1g/cm3 water (without temperature option*)
• The aluminum plate is placed on the small ice ball without gap
*Temperature option influence only motion of low energy neutron
6
Step 1: Construct a big ice ball
Set a big ice ball around original 10cm sphere (radius 20cm at center z=0cm)
Hint
• Geometry check can be done with icntl = 8
• A spherical surface centering the origin is defined by ”so [radius]”
• Define the region of the big ice ball not to overlap with the region of
the original sphere of 5cm radius (avoid double defined region)
Step 1
7
Step 2: Construct a small ice ball
Set a small ice ball on top of the big ice ball (radius 15 cm at center z=-25cm)
Hint
• A spherical surface centering on z axis is defined by
”sz [center z position] [radius]”
• Exclude region of big ice ball from small ice ball or vise versa,
otherwise double defined region is created by two ice balls
Step 2
8
Step3: Set aluminum plate
1. Define aluminum (27Al) at [material] section
2. Construct an aluminum plate of 10cm radius and 4cm thickness
(-40cm < z < -36cm)
Hint
•
•
•
•
A cylindrical surface parallel to z axis is defined by “cz [radius]”
A plane perpendicular to z axis is defined by ”pz [z position]”
Density of aluminum is 2.7g/cm3 (negative value for mass density)
Exclude region of the aluminum plate from that of the small ice ball
Step 3
9
Step 4: Set proton beam condition
Tune a proton beam energy so that the absorbed dose at central sphere is
maximized (Tune the energy at the level of 1MeV)
Hint
•
•
•
•
Transport calculation can be executed with icntl = 0
Beam energy is given by e0 parameter at [source] section
Absorbed dose at central sphere can be checked by deposit.out
Proton absorbed dose is maximized at the Bragg peak
Step 4
10
Step 5: Tune proton beam current
Calculate absorbed dose (Gy) in 1 second deposited by
10nA beam used in general proton therapy
Hint
• Default output of PHITS is normalized to per particle emitted from source
Maxcas and maxbch control statistical uncertainty, and are not related to norm
• Set totfact at [source] section to change the normalization
f.g. Absorbed dose for 100 protons is given by setting totfact=100.0
• 1A denotes status that 1C current is conducting in 1 second
• The electric charge of a proton is 1.6x10-19C
Step 5
11
Answer 5
Check the scale
change with totfact
1. Number of emitted protons for 1 ampere in 1 second is
1.0 / 1.6E-19 = 6.25E18 (particles)
2. Number of emitted protons for 10nA in 1 second is
6.25E18 x 10 x 1E-9 = 6.25E10 (particles)
Absorbed dose with totfact = 6.25E10 is 1.2740 (Gy) (for 294MeV proton beam)
Answer to Step 5
12
Step6: Calculate beam current to melt
central sphere in the snow man
1. Assume the snowman is made of ice of -10C
2. Calculate the proton beam current (in ampere) and the power (in
Watt) necessary to heat up the ice to 0 C in 1 second
Hint (assumption for simplicity)
• Specific heat of ice is 0.5 (cal/g/K) = 2.1 (J/g/K)
• Latent heat of ice (heat necessary for phase transition from ice to
water) is 333.5（J/g）
• 1Gy = 1（J/kg） = 0.001（J/g）
• Beam power (in MW) = Particle energy (in MeV)  Beam current (in A)
By the way…
The maximum power of the biggest power accelerator
facilities in the world is the order of 1MW
Step 6
13
Answer 6
You can solve the problem by answering the following questions step by step
1. How much is the absorbed dose in central sphere by 294 MeV proton
beam with 10nA in 1 second ?
2. How much heat is needed to heat up the ice 10K and melt the ice ?
3. How much current is required to give that heat in 1 second ?
4. How much is the power at this beam current ?
About 1MW beam power is required to melt a snowman
(Details are given in answer\answer-snowman-en.ppt)
• Melting a snowman is just what we can do with current technology
(Of course, with longer time, we can melt metals)
• Beam weapons in animation which destroy robots instantly are
far beyond current technology !
Answer to Step 6
14
Summary
• Geometry of a snowman was constructed and
PHITS simulation was conducted with tuned
proton beam energy to melt it effectively
• Normally, tally outputs of PHITS are
normalized to per particle
• Tally outputs have to be re-normalized by
using totfact parameter to simulate actual
conditions
Summary
15