Niowave 12-2015

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.

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]

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");