P HI T S Setting of various source Part I Multi-Purpose Particle and Heavy Ion Transport code System...

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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