TF IsotopeTracers4Cracks

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.

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

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]

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.

Energy distribution

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.

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)

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)

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.

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)

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.

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