Difference between revisions of "TF IsotopeTracers4Cracks"

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=Definition of Shale=
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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>
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 +
<pre>
 +
 +
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);
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G4Element* Si = new G4Element("Silicon"  , "Si", z=14., a= 28.085*g/mole);
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G4Element* H = new G4Element("Hydrogen"  , "H", z=1., a= 1.008*g/mole);
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G4Material* Shale = new G4Material("Shale", density= 2.6*g/cm3, nel=4);
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Shale->AddElement(Al, 15*perCent);
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Shale->AddElement(Si, 15*perCent);
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Shale->AddElement(O, 38*perCent);
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Shale->AddElement(H, 32*perCent);
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 +
</pre>
  
 
=Photon Attenuation in Shale=
 
=Photon Attenuation in Shale=
  
−
The attenuation length for a 1.8 MeV photon is 0.16 /cm and 0.895 is 0.22/cm through shale.
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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]]
 
[[File:XCOM_attenuation4Shale.pdf]]
Line 12: Line 35:
  
 
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.
 
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.  A least squares fit gave the parameters
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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.   
  
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slope=-0.118033 +/- 0.0423836
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If I use a point source isotropically emitting 1800 keV photons I observe
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y-intercept=13.4456 +/- 0.102718
 
  
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but the curve from these parameters was too low to believe. It is not worth the time right now to figure out what is wrong.  The fit by hand which gives 0.1/cm is enough to convince one that GEANT4 is correctly attenuating the photons.
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[[File:GEANT4AttShalePntSrc_322013.png| 200 px]]
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 +
which gives a the value of 0.1/cm as well.
  
 
=Yittrium in Shale=
 
=Yittrium in Shale=
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A 4" thick piece of shale was placed between the source and the detector.
 
A 4" thick piece of shale was placed between the source and the detector.
−
 
−
shale was defined as:
 
−
<pre>
 
−
 
−
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);
 
−
 
−
</pre>
 
  
  
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=Insert a single crack=
 
=Insert a single crack=
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 +
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.
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 +
 +
[[File:TF_Crack_Yit_04092013.png| 200 px]]
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==Energy distribution==
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[[File:TF_Crack_Edist_895.png| 200 px]][[File:TF_Crack_Edist_1800.png| 200 px]]
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==Position Distribution==
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 +
The Shale is a cylinder and the detector is a rectangle.  A cut is used to look only at photons within the cylinder.
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Units are Centimeters.
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[[File:TF_Crack_PositnDist.png| 200 px]][[File:TF_Crack_PositnDistCuts.png| 200 px]]
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==Rates==
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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
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===Surface rates -vs- Crack size===
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The crack size decreases as you move from left to right in the table.
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A projection is made along the X-axis  with the cuts
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 +
#radius < 100 cm
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#Energy > Source energy (0.7 or 1.7 MeV)
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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)
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{| border="1"  |cellpadding="20" cellspacing="0
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|-
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|[[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]]
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|-
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|100 mm hole|| 25 mm hole||10 mm ||5 mm|| 1 mm
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|-
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|}
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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)
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{| border="1"  |cellpadding="20" cellspacing="0
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|-
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|[[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]]
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|-
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|100 mm hole|| 25 mm hole||10 mm ||5 mm|| 1 mm
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|-
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|}
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'''Now change the detector size from a 5 x 5 cm area to a 0.5 x 0.5 cm area.'''
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{| border="1"  |cellpadding="20" cellspacing="0
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|-
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|[[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]]
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|-
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|100 mm hole|| 25 mm hole||10 mm ||5 mm|| 1 mm
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|-
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|}
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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
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|-
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|[[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]]
 +
|-
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|100 mm hole|| 25 mm hole||10 mm ||5 mm|| 1 mm
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|-
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|}
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==== Projections====
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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.
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{| border="1"  |cellpadding="20" cellspacing="0
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|-
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|[[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]]
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|-
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|100 mm hole|| 25 mm hole||10 mm ||5 mm|| 1 mm
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|-
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|}
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===Peak ratio -vs- distance from center===
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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.
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sample root command
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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");
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{| border="1"  |cellpadding="20" cellspacing="0
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|-
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| rowspan="3"  | Distance (cm) ||colspan="16" style="text-align: center;" | Energy
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|-
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| 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
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|-
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| 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
 +
|-
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|2 ||144 ||285 || 0.51 ||
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|}
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[[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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