Difference between revisions of "TF IsotopeTracers4Cracks"

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=Photon Attenuation in Shale=
+
=Definition of Shale=
−
 
 
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The attenuation length for a 1.8 MeV photon is 0.16 /cm and 0.895 is 0.22/cm through shale.
 
−
 
 
−
[[File:XCOM_attenuation4Shale.pdf]]
 
−
 
 
−
 
 
  
−
=Yittrium in Shale=
+
Shale was assumed to have the chemical composition of
  
−
A simulation of the penetration of the 0.895 and 1.8 MeV photons from Yittrium through shale.
+
<math>\mbox{Al}_2\mbox{Si}_2\mbox{O}_5(\mbox{OH})_4</math>
−
 
 
−
GEANT4 create a point source or 895 keV photon and another with 1800 keV photons iostropically distributed.
 
  
−
A 4" thick piece of shale was placed between the source and the detector.
+
and a density of 2.6 g/<math>\mbox{cm}^3</math>
  
−
shale was defined as:
 
 
<pre>
 
<pre>
  
Line 31: Line 22:
  
 
</pre>
 
</pre>
 +
 +
=Photon Attenuation in Shale=
 +
 +
According to the XCOM database, the attenuation length for a 1.8 MeV photon is 0.16 /cm and 0.895 is 0.22/cm through shale.
 +
 +
[[File:XCOM_attenuation4Shale.pdf]]
 +
 +
 +
Directing a 1.8 MeV beam of photons in GEANT4 towards various thicknesses of shale produced the graph below.
 +
 +
[[File:GEANT4AttShale_3212012.png| 200 px]]
 +
 +
XCOM predicts an attenuation coefficient of 0.0441 cm^2/g which becomes 0.12/cm when you multiply by the shale density of 2.6 g/cm^3.
 +
The fit to GEANT4's predictions above produces a value of 0.1/cm when you plot the number of photons that pass through the shale and still have an energy of 1.8 MeV. 
 +
 +
If I use a point source isotropically emitting 1800 keV photons I observe
 +
 +
[[File:GEANT4AttShalePntSrc_322013.png| 200 px]]
 +
 +
which gives a the value of 0.1/cm as well.
 +
 +
=Yittrium in Shale=
 +
 +
A simulation of the penetration of the 0.895 and 1.8 MeV photons from Yittrium through shale.
 +
 +
GEANT4 create a point source or 895 keV photon and another with 1800 keV photons iostropically distributed.
 +
 +
A 4" thick piece of shale was placed between the source and the detector.
  
  
Line 42: Line 61:
  
 
=Insert a single crack=
 
=Insert a single crack=
 +
 +
A simulation was performed that predicted the transmission of photons through 10 centimeters of shale.  The shale is in the form of a cylinder with a hole through the central axis.
 +
 +
The source was a uniform points source.  One set of simulations used a 1.8 MeV photon and another used a 0.895 photon.
 +
 +
 +
[[File:TF_Crack_Yit_04092013.png| 200 px]]
 +
 +
==Energy distribution==
 +
 +
[[File:TF_Crack_Edist_895.png| 200 px]][[File:TF_Crack_Edist_1800.png| 200 px]]
 +
 +
==Position Distribution==
 +
 +
The Shale is a cylinder and the detector is a rectangle.  A cut is used to look only at photons within the cylinder.
 +
 +
 +
Units are Centimeters.
 +
 +
[[File:TF_Crack_PositnDist.png| 200 px]][[File:TF_Crack_PositnDistCuts.png| 200 px]]
 +
 +
==Rates==
 +
 +
A uniform point source sent 10^7 photons of a specific energy in all directions.
 +
 +
1 Curie = <math>3.7 \times 10^{10}</math> decays/sec
 +
 +
===Surface rates -vs- Crack size===
 +
 +
The crack size decreases as you move from left to right in the table.
 +
 +
 +
A projection is made along the X-axis  with the cuts
 +
 +
#radius < 100 cm
 +
#Energy > Source energy (0.7 or 1.7 MeV)
 +
 +
 +
The result below is from a '''1800 keV''' uniform point source and a 10 cm thick shale wall with a cylindrical hole through the center (X=0:Y=0)
 +
 +
{| border="1"  |cellpadding="20" cellspacing="0
 +
|-
 +
|[[File:TF_100mmCrack_XvsY.png| 200 px]] || [[File:TF_25mmCrack_XvsY.png| 200 px]]|| [[File:TF_10mmCrack_XvsY.png| 200 px]]|| [[File:TF_5mmCrack_XvsY.png| 200 px]]|| [[File:TF_1mmCrack_XvsY.png| 200 px]]
 +
|-
 +
|100 mm hole|| 25 mm hole||10 mm ||5 mm|| 1 mm
 +
|-
 +
|}
 +
 +
 +
The result below is from a '''895 keV''' uniform point source and a 10 cm thick shale wall with a cylindrical hole through the center (X=0:Y=0)
 +
 +
{| border="1"  |cellpadding="20" cellspacing="0
 +
|-
 +
|[[File:TF895_100mmCrack_XvsY.png| 200 px]] || [[File:TF895_25mmCrack_XvsY.png| 200 px]]|| [[File:TF895_10mmCrack_XvsY.png| 200 px]]|| [[File:TF895_5mmCrack_XvsY.png| 200 px]]|| [[File:TF895_1mmCrack_XvsY.png| 200 px]]
 +
|-
 +
|100 mm hole|| 25 mm hole||10 mm ||5 mm|| 1 mm
 +
|-
 +
|}
 +
 +
'''Now change the detector size from a 5 x 5 cm area to a 0.5 x 0.5 cm area.'''
 +
 +
{| border="1"  |cellpadding="20" cellspacing="0
 +
|-
 +
|[[File:TF_100mmCrack_XvsY5x5.png| 200 px]] || [[File:TF_25mmCrack_XvsY5x5.png| 200 px]]|| [[File:TF_10mmCrack_XvsY5x5.png| 200 px]]|| [[File:TF_5mmCrack_XvsY5x5.png| 200 px]]|| [[File:TF_1mmCrack_XvsY5x5.png| 200 px]]
 +
|-
 +
|100 mm hole|| 25 mm hole||10 mm ||5 mm|| 1 mm
 +
|-
 +
|}
 +
 +
 +
The result below is from a '''895 keV''' uniform point source and a 10 cm thick shale wall with a cylindrical hole through the center (X=0:Y=0)
 +
 +
{| border="1"  |cellpadding="20" cellspacing="0
 +
|-
 +
|[[File:TF895_100mmCrack_XvsY5x5.png| 200 px]] || [[File:TF895_25mmCrack_XvsY5x5.png| 200 px]]|| [[File:TF895_10mmCrack_XvsY5x5.png| 200 px]]|| [[File:TF895_5mmCrack_XvsY5x5.png| 200 px]]|| [[File:TF895_1mmCrack_XvsY5x5.png| 200 px]]
 +
|-
 +
|100 mm hole|| 25 mm hole||10 mm ||5 mm|| 1 mm
 +
|-
 +
|}
 +
 +
==== Projections====
 +
 +
As observed above, the 1 mm radius hole may be observed if you use a detector that is 5 x 5 mm.
 +
 +
 +
 +
Notice that the 895 keV photons show a more distinct peak for the 1 mm radius holes than the 1800 keV photons.  The bins are 2cm in size to represent a 2 cm x 2cm area detector.  A cut was made restricting the Y-position to +/- 1 cm.
 +
 +
 +
 +
 +
{| border="1"  |cellpadding="20" cellspacing="0
 +
|-
 +
|[[File:41113EnergyDep_100mmHole_2x2cmdetectr.png| 200 px]] || [[File:41113EnergyDep_25mmHole_2x2cmdetectr.png| 200 px]]|| [[File:41113EnergyDep_10mmHole_2x2cmdetectr.png| 200 px]]|| [[File:41113EnergyDep_5mmHole_2x2cmdetectr.png| 200 px]]|| [[File:41113EnergyDep_1mmHole_2x2cmdetectr.png| 200 px]]
 +
|-
 +
|100 mm hole|| 25 mm hole||10 mm ||5 mm|| 1 mm
 +
|-
 +
|}
 +
 +
===Peak ratio -vs- distance from center===
 +
 +
The analysis below compares the number of photons at two energies of 0.895 and 1.8 MeV that penetrating the shale as a function of the distance from the radial center.  A 2 x 2 cm detector is used.
 +
 +
 +
sample root command
 +
PhotonTrack->Draw("evt.Energy","sqrt(evt.PosX*evt.PosX+evt.PosY*evt.Poroot [27]  abs(evt.PosY)<1 &&evt.PosX>0 && evt.PosX< 4 && evt.Energy>1.7");
 +
 +
 +
{| border="1"  |cellpadding="20" cellspacing="0
 +
|-
 +
| rowspan="3"  | Distance (cm) ||colspan="16" style="text-align: center;" | Energy
 +
|-
 +
| colspan="3" style="text-align: center;" | 100 mm hole|| colspan="3" style="text-align: center;" | 25 mm hole||colspan="3" style="text-align: center;" | 10 mm ||colspan="3" style="text-align: center;" | 5 mm|| colspan="3" style="text-align: center;" | 1 mm
 +
|-
 +
| 0.895 MeV ||1.8 || Ratio||0.895  || 1.8 ||Ratio||0.895 ||1.8  ||Ratio||0.895 ||1.8  ||Ratio||0.895 ||1.8  ||Ratio
 +
|-
 +
|2 ||144 ||285 || 0.51 ||
 +
|}
 +
 +
 +
  
 
[[TF_IsotopeTracers#Tracers_for_Cracks]]
 
[[TF_IsotopeTracers#Tracers_for_Cracks]]

Latest revision as of 18:37, 17 April 2013

Definition of Shale

Shale was assumed to have the chemical composition of

[math]\mbox{Al}_2\mbox{Si}_2\mbox{O}_5(\mbox{OH})_4[/math]

and a density of 2.6 g/[math]\mbox{cm}^3[/math] 


G4Element* O = new G4Element("Oxygen"  , "O", z=8., a= 16.00*g/mole);
G4Element* Al = new G4Element("Aluminum"  , "Al", z=13., a= 26.98*g/mole);
G4Element* Si = new G4Element("Silicon"  , "Si", z=14., a= 28.085*g/mole);
G4Element* H = new G4Element("Hydrogen"  , "H", z=1., a= 1.008*g/mole);

G4Material* Shale = new G4Material("Shale", density= 2.6*g/cm3, nel=4);
Shale->AddElement(Al, 15*perCent);
Shale->AddElement(Si, 15*perCent);
Shale->AddElement(O, 38*perCent);
Shale->AddElement(H, 32*perCent);

Photon Attenuation in Shale

According to the XCOM database, the attenuation length for a 1.8 MeV photon is 0.16 /cm and 0.895 is 0.22/cm through shale.

File:XCOM attenuation4Shale.pdf


Directing a 1.8 MeV beam of photons in GEANT4 towards various thicknesses of shale produced the graph below.

GEANT4AttShale 3212012.png

XCOM predicts an attenuation coefficient of 0.0441 cm^2/g which becomes 0.12/cm when you multiply by the shale density of 2.6 g/cm^3. The fit to GEANT4's predictions above produces a value of 0.1/cm when you plot the number of photons that pass through the shale and still have an energy of 1.8 MeV.

If I use a point source isotropically emitting 1800 keV photons I observe

GEANT4AttShalePntSrc 322013.png

which gives a the value of 0.1/cm as well.

Yittrium in Shale

A simulation of the penetration of the 0.895 and 1.8 MeV photons from Yittrium through shale.

GEANT4 create a point source or 895 keV photon and another with 1800 keV photons iostropically distributed.

A 4" thick piece of shale was placed between the source and the detector.


Two ratios were constructed.

[math]R_p = \frac{\mbox{counts under the 1800 keV peak}}{\mbox{counts under the 895 keV peak}}[/math]

[math]R_I = \frac{\mbox{number of photons from 1800 keV source penetrating shale}}{\mbox{number of photons from 895 keV source penetrating shale}}[/math]

Y88ShaleNoCrack 1.png

Insert a single crack

A simulation was performed that predicted the transmission of photons through 10 centimeters of shale. The shale is in the form of a cylinder with a hole through the central axis.

The source was a uniform points source. One set of simulations used a 1.8 MeV photon and another used a 0.895 photon.


TF Crack Yit 04092013.png

Energy distribution

TF Crack Edist 895.pngTF Crack Edist 1800.png

Position Distribution

The Shale is a cylinder and the detector is a rectangle. A cut is used to look only at photons within the cylinder.


Units are Centimeters.

TF Crack PositnDist.pngTF Crack PositnDistCuts.png

Rates

A uniform point source sent 10^7 photons of a specific energy in all directions.

1 Curie = [math]3.7 \times 10^{10}[/math] decays/sec

Surface rates -vs- Crack size

The crack size decreases as you move from left to right in the table.


A projection is made along the X-axis with the cuts

  1. radius < 100 cm
  2. Energy > Source energy (0.7 or 1.7 MeV)


The result below is from a 1800 keV uniform point source and a 10 cm thick shale wall with a cylindrical hole through the center (X=0:Y=0)

TF 100mmCrack XvsY.png TF 25mmCrack XvsY.png TF 10mmCrack XvsY.png TF 5mmCrack XvsY.png TF 1mmCrack XvsY.png
100 mm hole 25 mm hole 10 mm 5 mm 1 mm


The result below is from a 895 keV uniform point source and a 10 cm thick shale wall with a cylindrical hole through the center (X=0:Y=0)

TF895 100mmCrack XvsY.png TF895 25mmCrack XvsY.png TF895 10mmCrack XvsY.png TF895 5mmCrack XvsY.png TF895 1mmCrack XvsY.png
100 mm hole 25 mm hole 10 mm 5 mm 1 mm

Now change the detector size from a 5 x 5 cm area to a 0.5 x 0.5 cm area.

TF 100mmCrack XvsY5x5.png TF 25mmCrack XvsY5x5.png TF 10mmCrack XvsY5x5.png TF 5mmCrack XvsY5x5.png TF 1mmCrack XvsY5x5.png
100 mm hole 25 mm hole 10 mm 5 mm 1 mm


The result below is from a 895 keV uniform point source and a 10 cm thick shale wall with a cylindrical hole through the center (X=0:Y=0)

TF895 100mmCrack XvsY5x5.png TF895 25mmCrack XvsY5x5.png TF895 10mmCrack XvsY5x5.png TF895 5mmCrack XvsY5x5.png TF895 1mmCrack XvsY5x5.png
100 mm hole 25 mm hole 10 mm 5 mm 1 mm

Projections

As observed above, the 1 mm radius hole may be observed if you use a detector that is 5 x 5 mm.


Notice that the 895 keV photons show a more distinct peak for the 1 mm radius holes than the 1800 keV photons. The bins are 2cm in size to represent a 2 cm x 2cm area detector. A cut was made restricting the Y-position to +/- 1 cm.



41113EnergyDep 100mmHole 2x2cmdetectr.png 41113EnergyDep 25mmHole 2x2cmdetectr.png 41113EnergyDep 10mmHole 2x2cmdetectr.png 41113EnergyDep 5mmHole 2x2cmdetectr.png 41113EnergyDep 1mmHole 2x2cmdetectr.png
100 mm hole 25 mm hole 10 mm 5 mm 1 mm

Peak ratio -vs- distance from center

The analysis below compares the number of photons at two energies of 0.895 and 1.8 MeV that penetrating the shale as a function of the distance from the radial center. A 2 x 2 cm detector is used.


sample root command 

PhotonTrack->Draw("evt.Energy","sqrt(evt.PosX*evt.PosX+evt.PosY*evt.Poroot [27]  abs(evt.PosY)<1 &&evt.PosX>0 && evt.PosX< 4 && evt.Energy>1.7");
Distance (cm) Energy
100 mm hole 25 mm hole 10 mm 5 mm 1 mm
0.895 MeV 1.8 Ratio 0.895 1.8 Ratio 0.895 1.8 Ratio 0.895 1.8 Ratio 0.895 1.8 Ratio
2 144 285 0.51



TF_IsotopeTracers#Tracers_for_Cracks

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