Foretony at 15:19, 13 April 2016

2016-04-13T15:19:16Z

← Older revision		Revision as of 15:19, 13 April 2016
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	~~The energy deposited by electrons scattered into a 3.48 diameter stainless steel beam pipe (1.65 mm thick) from a PbBi target as a function of a uniform Solenoidal magnetic field.~~

			Niowave Positron Project Progress for December 2015
	~~{\| border="1"~~
	~~{\| border="1"~~
	~~\| B-field (Tesla) \|\| Hot Spot (<math>MeV/e^-</math>)~~
	\|-
	~~\| 0.0 \|\| 0.35~~
	\|-
	~~\| 0.3 \|\| 0.35~~
	\|-
	~~\| 1.0 \|\| 0.35~~
	\|-
	~~\| 1.5 \|\| 0.22~~
	\|-
	~~\| 2.0 \|\| 0.10~~
	\|-
	~~\| 4.0 \|\| 0.002~~
	\|+
	\|}

			The energy deposited by electrons scattered into a 1.65 mm thick stainless steel beam pipe from the PbBi target as a function of a uniform solenoidal magnetic field was investigated for a 3.48 diameter stainless steel beam pipe. The simulation program GEANT4 was used to determine the amount of energy an electron lost in the stainless steel beam pipe. The beam pipe surface was divided into unit squares that were a centimeter in length and the energy deposited by a sample of 5 million electrons was summed for each unit square. The left side figure below shows a predicted peak energy deposition of 0.35 MeV/e^- in the 1.65 mm thick stainless steel beam pipe per 10 MeV electron impinging the 2 mm thick LBE target that has been sandwiched between two 0.25 mm thick stainless steel windows.

	~~To convert this~~ deposited ~~energy per incident electron on~~ the ~~target~~ to a heat load in the ~~pipe you need~~ to ~~divide by~~ the area of the ~~pipe.~~		The energy deposited in the beam pipe can be converted to an average heat load if a average beam current is assumed.
			Under the assumption of a 1mA (10^-3 C/s) average beam current, the deposited energy (E_{dep}) per cm^2 per electron can be converted to the deposited power (P) per area using the equation

	~~A histogram is filled with 1 cm bins along the Z axis. The surface area becomes~~ <math>1 cm \times 2 \pi 3.~~48/2 =~~ 10.~~933 cm^2~~</math>~~. The beam pipe diameter assumed is 3.48 cm.~~		<math>P= E_{dep} \left( \frac{\mbox{MeV}}{\mbox{cm}^2 \mbox{e}^-}\right) \times \left( \frac{ \mbox{e}^-}{1.6 \times 10^{-19}\mbox{C}} \right ) \times \left( \frac{1 \times 10^{-3} \mbox{C}}{\mbox{s}} \right ) \times \left( \frac{1.6 \times 10^{-13}\mbox{W} \cdot \mbox{ s}}{\mbox{MeV} }\right ).</math>

	~~When filling the histogram binned 1 cm in Z~~, ~~you should weight it by~~ the ~~amount of depositred~~ energy ~~divided by the circumference of the pipe and divided by the number~~ of ~~incident electrons on the target (5 million). The energy units are converted to~~ keV ~~by multiplying the numberator by 100 or in this case dividing by 5000 instead of 5 million.~~		In other words, the energy deposited in units of keV/cm^2/e by a 1 mA electron beam is equivalent to a heat load in the units of Watts per cm^2.


	~~TH1F *T00N=new TH1F("T00N","T00N",100,-1000.5,-0.5)~~

	~~Electrons->Draw("evt.EoutPosZ>>T00N","evt.DepE~~/~~10.088/5000")~~



	~~To convert From Mev~~/ e- to ~~kW/cm^2 assuming~~ a ~~current~~ of ~~1mA (10^-3 C/s) you~~

	~~<math>\left( \frac{\mbox{MeV}}{\mbox{~~cm}^2 ~~\mbox{e}^-}\right) \times \left( \frac{ \mbox{e}^-}{1.6 \times 10^{-19}\mbox{C}} \right ) \times \left( \frac{1 \times 10^{-3} \mbox{C}}{\mbox{s}} \right ) \times \left( \frac{1~~.~~6 \times 10^{-13}\mbox{W} \cdot \mbox{ s}}{\mbox{MeV} }\right )</math>~~

	<math>\left( \frac{\mbox{keV}}{\mbox{cm}^2 \mbox{e}^-}\right) = \left( \frac{\mbox{W} }{\mbox{cm}^2 } \right )</math>		<math>\left( \frac{\mbox{keV}}{\mbox{cm}^2 \mbox{e}^-}\right) = \left( \frac{\mbox{W} }{\mbox{cm}^2 } \right )</math>

			Based on this observation the peak heat load, as shown in the figure below on the right, for a 1 mA electron beam is 35 W/cm^2. The electrons will begin to intersect a 3.48 diameter beam pipe less when the solenoidal magnetic field strength increases beyond one Tesla. A solenoidal field between two and four Tesla would substantially reduce the electron heat load on the beam pipe and place the positrons in a tight helical orbit for transportation.


	{\| border="1"		{\| border="1"
	\| [[File:BeamPipeDepEmev-vs-B.png \|~~200px~~]] \|\| [[File:BeamPipeDepPower-vs-B.png \|~~200px]]\|\| [[File:BeamPipeDepPower-vs-lowB.png \|200px~~]]		\| [[File:BeamPipeDepEmev-vs-B.png \|400px]] \|\| [[File:BeamPipeDepPower-vs-B.png \|400px]]
	\|+ Energy deposited (MeV) along a 1 m long beam pipe of stainless steel 1.65 mm thick.		\|+ Energy deposited (MeV) along a 1 m long beam pipe of stainless steel 1.65 mm thick.
	\|}		\|}

	~~With SS windows~~
	Positrons->Draw("sqrt(evt.BeamPosPosXevt.BeamPosPosX+evt.BeamPosPosYevt.BeamPosPosY)","evt.BeamPosMomZ>0 && evt.BeamPosPosZ>-500 && sqrt(evt.BeamPosPosXevt.BeamPosPosX+evt.BeamPosPosYevt.BeamPosPosY)<97.4/2");

Foretony: Created page with "The energy deposited by electrons scattered into a 3.48 diameter stainless steel beam pipe (1.65 mm thick) from a PbBi target as a function of a uniform Solenoidal magnetic fiel…"

2016-04-12T20:50:25Z

Created page with "The energy deposited by electrons scattered into a 3.48 diameter stainless steel beam pipe (1.65 mm thick) from a PbBi target as a function of a uniform Solenoidal magnetic fiel…"

New page

The energy deposited by electrons scattered into a 3.48 diameter stainless steel beam pipe (1.65 mm thick) from a PbBi target as a function of a uniform Solenoidal magnetic field.

{| border="1"
{| border="1"
| B-field (Tesla) || Hot Spot (<math>MeV/e^-</math>)
|-
| 0.0 || 0.35
|-
| 0.3 || 0.35
|-
| 1.0 || 0.35
|-
| 1.5 || 0.22
|-
| 2.0 || 0.10
|-
| 4.0 || 0.002
|+
|}

To convert this deposited energy per incident electron on the target to a heat load in the pipe you need to divide by the area of the pipe.

A histogram is filled with 1 cm bins along the Z axis. The surface area becomes <math>1 cm \times 2 \pi 3.48/2 = 10.933 cm^2</math>. The beam pipe diameter assumed is 3.48 cm.

When filling the histogram binned 1 cm in Z, you should weight it by the amount of depositred energy divided by the circumference of the pipe and divided by the number of incident electrons on the target (5 million). The energy units are converted to keV by multiplying the numberator by 100 or in this case dividing by 5000 instead of 5 million.

TH1F *T00N=new TH1F("T00N","T00N",100,-1000.5,-0.5)

Electrons->Draw("evt.EoutPosZ>>T00N","evt.DepE/10.088/5000")

To convert From Mev/ e- to kW/cm^2 assuming a current of 1mA (10^-3 C/s) you

<math>\left( \frac{\mbox{MeV}}{\mbox{cm}^2 \mbox{e}^-}\right) \times \left( \frac{ \mbox{e}^-}{1.6 \times 10^{-19}\mbox{C}} \right ) \times \left( \frac{1 \times 10^{-3} \mbox{C}}{\mbox{s}} \right ) \times \left( \frac{1.6 \times 10^{-13}\mbox{W} \cdot \mbox{ s}}{\mbox{MeV} }\right )</math>

<math>\left( \frac{\mbox{keV}}{\mbox{cm}^2 \mbox{e}^-}\right) = \left( \frac{\mbox{W} }{\mbox{cm}^2 } \right )</math>

{| border="1"
| [[File:BeamPipeDepEmev-vs-B.png |200px]] || [[File:BeamPipeDepPower-vs-B.png |200px]]|| [[File:BeamPipeDepPower-vs-lowB.png |200px]]
|+ Energy deposited (MeV) along a 1 m long beam pipe of stainless steel 1.65 mm thick.
|}

With SS windows
Positrons->Draw("sqrt(evt.BeamPosPosX*evt.BeamPosPosX+evt.BeamPosPosY*evt.BeamPosPosY)","evt.BeamPosMomZ>0 && evt.BeamPosPosZ>-500 && sqrt(evt.BeamPosPosX*evt.BeamPosPosX+evt.BeamPosPosY*evt.BeamPosPosY)<97.4/2");

Niowave 12-2015 - Revision history

Foretony at 15:19, 13 April 2016

Foretony: Created page with "The energy deposited by electrons scattered into a 3.48 diameter stainless steel beam pipe (1.65 mm thick) from a PbBi target as a function of a uniform Solenoidal magnetic fiel…"