Foretony at 15:19, 13 April 2016

2016-04-13T15:19:16Z

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…"

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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.

{| 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");

@@ Line 1: / Line 1: @@
+Niowave Positron Project Progress for December 2015
-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.
-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.
 <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"
-| [[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.
 |}

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…"