p hi t s setting of various source part i multi-purpose particle and heavy ion transport code system...
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PHITSSetting of various source
Part I
Multi-Purpose Particle and Heavy Ion Transport code System
Title 1
Aug. 2015 revised
Goal of this lecture
Purpose 2
Transport simulation with various kinds of sources
Simulation with 60Co source placed in two positions
Source having continuous energy distribution
sourceA.inp
3Check Input File
Basic setupProjectile:
Geometry:Tally:
Geometry
150MeV proton (pencil beam with radius 1.0cm)Water cylinder (10cm radius and 20cm thickness) [t-track] fluence distribution[t-cross] proton energy spectrum coming into water
Water150MeVProton
track_xz.eps cross_eng.eps
Table of Contents
4
Table of contents
1. Source having energy distribution
A) Continuous energy distribution
B) Discrete energy distribution
2. Setup of multiple source
3. Summary
Source having energy distribution
Energy distribution 5
A source having energy distribution can be treated in PHITS as well as a single energy source
Proton beam
having energy distribution
How to set 1
Energy distribution 6
At [source] section, set s-type=4 (for cylinder), 5 (for rectangular), 10 ( for sphere )
Set e-type subsection (Unit of energy is MeV or angstrom)
[ S o u r c e ] totfact = 1.0 s-type = 4 proj = proton$ e0 = 150. r0 = 1.0 z0 = -10. z1 = -10. dir = 1.0 e-type = 1 ne = 2 0.0 4 50.0 1 100.0
[ S o u r c e ] totfact = 1.0 s-type = 1 proj = proton e0 = 150. r0 = 1.0 z0 = -10. z1 = -10. dir = 1.0
How to set 2
Energy distribution 7
3 ways to specify energy distribution ( switched by e-type ) e-type=1*: Continuous distribution with integral value e-type=21*: Continuous distribution with differential value (particle/MeV) e-type=8*: Discrete distribution
e-type = 1 ne = n e(1) w(1) e(2) w(2) ・ ・ ・ ・ ・ ・ e(n) w(n) e(n+1)
For continuous distribution (f.g. e-type=1), specify data of number of energy group (ne), energy bin (e(i)), and probability to generate particle (w(i))
Number of e(i) is n+1 in totalNumber of w(i) is n(Log scale is applied when ne is given in minus)
For discrete distribution (f.g. e-type=8), specify data of number of energy point (ne), energy point(e(i)), and probability to generate particle (w(i))
e-type = 8 ne = n e(1) w(1) e(2) w(2) ・ ・ ・ ・ ・ ・ e(n) w(n)
Numbers of e(i) and w(i) are n
* To change weight or give energy with angstrom, use the other e-type ( See Sec. 4.3.15 of the manual )
Exercise 1
8
Set proton beam of the spectrum with the energy bin of [0,50], [50,100], [100,150] in MeV and the beam intensity of 1:3:2 in ratio (See right figure)
[ S o u r c e ] totfact = 1.0 s-type = 1 proj = proton e0 = 150. r0 = 1.0 z0 = -10. z1 = -10. dir = 1.0 e-type =
sourceA.inp
Energy distribution
e-type = 1 ne = n e(1) w(1) e(2) w(2) ・ ・ ・ ・ ・ ・ e(n) w(n) e(n+1)
e-type=1 format
• Change s-type to 4• Add e-type subsection and set energy distribution• Comment out the line of e0=150
Answer 1
9
[ S o u r c e ] totfact = 1.0 s-type = 4 proj = proton$ e0 = 150. r0 = 1.0 z0 = -10. z1 = -10. dir = 1.0 e-type = 1 ne = 3 0.0 1 50.0 3 100.0 2 150.0
sourceA.inp
Energy distribution
cross_eng.eps
The ratio of intensity is 1:3:2(The ratio is given in integral value for e-type=1 )
Set proton beam of the spectrum with the energy bin of [0,50], [50,100], [100,150] in MeV and the beam intensity of 1:3:2 in ratio
[ S o u r c e ]・ ・ ・ ・ ・ ・ e-type = 1 ne = 3 0.0 1 50.0 3 100.0 2 150.0
Exercise 2
10
Change the energy bin of [100:150] to [100:200]
sourceA.inp
Energy distribution
Source intensity is given by integral value with energy for e-type=1
• Change the energy range of the third bin• Check the beam flux when the energy bin
range becomes 50, 50, and 100 MeV
[ S o u r c e ]・ ・ ・ ・ ・ ・ e-type = 1 ne = 3 0.0 1 50.0 3 100.0 2 200.0
Answer 2
11
sourceA.inp
Energy distribution
cross_eng.eps
The ratio of intensity integrated by energy of the three bin is 1:3:2(The ratio is 1:3:1 if it is given in per unit energy or differential value)
Change the energy bin of [100:150] to [100:200]
[ S o u r c e ]・ ・ ・ ・ ・ ・ e-type = 1 ne = 3 0.0 1 50.0 3 100.0 2 200.0
Exercise 3
12
Give the ratio of the intensity of 1:3:2 in differential value for the energy bin of [0:50], [50:100], [100:200]
• Use e-type=21
sourceA.inp
Energy distribution
The ratio is given in differential value for e-type=21(Choose this option to use derivative spectrum)
[ S o u r c e ]・ ・ ・ ・ ・ ・ e-type = 21 ne = 3 0.0 1 50.0 3 100.0 2 200.0
Answer 3
13
sourceA.inp
Energy distribution
The ratio of the intensity is 1:3:2 in differential value(The ratio is 1:3:4 in integral value)
cross_eng.eps
Give the ratio of the intensity of 1:3:2 in differential value for the energy bin of [0:50], [50:100], [100:200]
Table of Contents
14
Table of contents
1. Source having energy distribution
A) Continuous energy distribution
B) Discrete energy distribution
2. Setup of multiple source
3. Summary
Source having discrete energy
Energy distribution 15
A source emitting several discrete energy radiations such as 60Co and 134Cs can be treated in PHITS
60Co source
60Co emits two gamma-rays of the energy of 1.173 and 1.333 MeV after the beta decay60Co
60Ni
1.173MeV) 100%
1.333MeV) 100%
[ S o u r c e ] totfact = 1.0 s-type = 4 proj = proton$ e0 = 150. r0 = 1.0・ ・ ・ ・ ・ ・ dir = 1.0e-type = 21 ne = 3 0.0 1 50.0 3 100.0 2 200.0
Exercise 4
16
Simulate 60Co source
• Change the source to photon
• Set isotropic point source [Change the parameters of radius (r0) and direction (dir)]
• Set the normalization factor (totfact) to 2 ( 60Co emits two gamma-ray per decay)
=> Tally output becomes amount per Bq
• Use e-type=8 and set the photon energies of 1.173MeV and 1.333MeV with the ratio of 1:1
• Set [t-cross] to tally photon fluence from 0 to 2 MeV with 10keV resolution (200 groups) [change emax, ne, part]
sourceA.inp
Energy distribution
e-type = 8 ne = n e(1) w(1) ・ ・ ・ ・ ・ ・ e(n) w(n)
e-type=8 format
[ S o u r c e ] totfact = 2.0 s-type = 4 proj = photon$ e0 = 150. r0 = 0.0 z0 = -10. z1 = -10. dir = all e-type = 8 ne = 2 1.173 1 1.333 1
Answer 4
17
sourceA.inp
Energy distribution
cross_eng.eps
[ T - C r o s s ]・ ・ ・ ・ ・ ・ emin = 0.0emax = 2.0ne = 200unit = 1axis = engfile = cross_eng.outoutput = fluxpart = photonepsout = 1 track_xz.eps
60Co source
Simulate 60Co source
Table of Contents
18
Table of contents
1. Source having energy distribution
A) Continuous energy distribution
B) Discrete energy distribution
2. Setup of multiple source
3. Summary
Setup of multiple source
Multi source 19
Multiple source with different radiation type, position, or energy distribution can be treated in PHITS
60Co source 60Co source
60Co sources placed at right and left of target with the intensity ration of 2:1
How to set
20
At [source] section, set multiple source subsection starting with ”<source>=relative intensity”
Set totfact to normalize the total source intensity
Multi source
[ S o u r c e ] totfact = 2.0 <source> = 2.0 s-type = 1 proj = proton・ ・ ・ ・ ・ ・
<source> = 1.0 s-type = 4 proj = neutron・ ・ ・ ・ ・ ・
<source> = 3.0 s-type = 8 proj = photon・ ・ ・ ・ ・ ・
Normalization factorIf it is positive, particles are produced with the ratio of the relative intensityIf it is negative, same number of particles are produced with weight of the normalized intensity
Triple source
Relative ratio of intensity for each source(In this case, 2:1:3 from the top)
Exercise 5
21
Set 60Co sources at the left and right of the cylindrical water (z=-10, 40cm) with intensity ratio of 2:1• Add the <source> line to create two subsections
• Set two point sources at the position of z=-10 and 40cm (Change z0 and z1 to set a point source)
• Set the intensity to produce photon with the ratio of 2:1 from the left (z=-10cm) and the right (z=40cm) respectively
60Co source 60Co source
Multi source
z axis
z=-10cm z=40cm
Answer 5
22Multi source
sourceA.inp
track_xz.eps
60Co source(With intensity ratio of right:left = 2:1)
[ S o u r c e ] totfact = 2.0 <source> = 2.0 s-type = 4 proj = photon$ e0 = 150. r0 = 0.0 z0 = -10. z1 = -10.・ ・ ・ ・ ・ ・ <source> = 1.0 s-type = 4 proj = photon r0 = 0.0 z0 = 40. z1 = 40. dir = all e-type = 8 ne = 2 1.173 1 1.333 1
Set 60Co sources at the left and right of the cylindrical water (z=-10, 40cm) with intensity ratio of 2:1
Table of Contents
23
Table of contents
1. Source having energy distribution
A) Continuous energy distribution
B) Discrete energy distribution
2. Setup of multiple source
3. Summary
Summary
Summary 24
• The source of continuous and discrete energy distribution can be treated by choosing s-type and e-type at the [source] section
• Simulation with multiple source can be conducted by the setup of <source> subsections
Refer to “setting of various source B” (phits-lec-sourceB-jp.ppt) for the setup of source using “Dump data”
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