Niowave 11-2015

A 10 MeV electron beam with a radius of 0.5 cm was incident on a 2 mm thick PbBi target. The target is positioned at Z = -902 mm. Niowave Positron Project Progress for November 2015

The next step in the positron project, after determining the PbBi thickness for maximum positron production, was to estimate the heat that would be deposited from the primary electron beam on the beam pipe. Electrons scattered by the PbBi target will intersect a beam pipe used to transport the desired positrons. Electrons traversing the beam pipe can deposit energy that will heat the beam pipe. The goal of this simulation was to estimate the amount of the deposited energy to determine if there is sufficient cooling to prevent a beam pipe breach. Below is a preliminary design of the beam pipe dimensions.

Element	dimension
Inner beam pipe radius	1.74 cm
Inner beam pipe thickness	0.165 cm
water jacket thickness	0.457 cm
outer beam pipe radius	2.362 cm
outer beam pipe thickness	0.165 cm
Solenoid inner radius	2.527 cm
Solenoid outer radius	4.406 cm

A comparison between the prediction from MCNPX and GEANT4 was once again performed. As shown in the figure below, GEANT4 predicts a peak energy deposition of 0.35 MeV/e while MCNPX predicts about 0.38 MeV/e. GEANT4 predicts a total of [math]3.08\times 10^7[/math] MeV (the integral adds up the energy in each 1cm bin) of energy will be deposited in a 1m long beam pipe surrounding a 2 mm thick PbBi target located at Z=-902 mm when 20 million electrons impinge the target.

Energy deposited (MeV) in a 1 m long 3.48 cm diameter beam pipe surrounding a 2 mm target located at Z=-900 mm.

The energy deposited by electrons scattered into a 1.65 mm thick stainless steel beam pipe from the LBE 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.

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

[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]

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.

add units to the histograms below

Energy deposited (MeV) along a 1 m long beam pipe of stainless steel 1.65 mm thick.

if you assume a 1 mA beam of electrons then this becomes

[math]\left ( \frac{3.5 \; \mbox{W} }{ \mbox{cm}^2 } \right) [/math]

I converted the above histogram to deposited power by 1000 mA, divide by the number of incident electrons, divide by the circumference of the beam pipe, convert the number of electrons to Coulombs, and use a unit conversion from MeV to W-s per MeV.

[math]\left(Counts \frac{\mbox{MeV}}{\mbox{cm}}\right) \times \left( \frac{1}{2 \times 10^{7} \mbox{e}^-} \right ) \times \left( \frac{1}{ \pi \times 3.48 \mbox{cm}} \right ) \times \left( \frac{1. \times 10^{-3}\mbox{ C}}{\mbox{s} }\right )\times \left( \frac{\mbox{e}^- }{1.6 \times 10^{-19}\mbox{ C}}\right ) \left( \frac{1.6 \times 10^{-13}\mbox{W} \cdot \mbox{ s}}{\mbox{MeV} }\right ) \times [/math]

If you use the above factors to weight the histogram, then the figure below shows that GEANT4 predicts a power deposition density of [math]= 4 \frac{W}{cm^2}[/math], 1 cm downstream of the target. Back scattered electrons appear to create the hottest spot of [math]= 15 \frac{W}{cm^2}[/math] about 1cm upstream of the target.

Power Deposition Zoomed in and 902 mm offset applied	Power deposition over the 1 m long beam pipe

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");

Positron Collection rates with 60 cm long Solenoid

Sample Positron Production Events

When the solenoid is 1.5 Tesla, a 10 MeV electron produces a 6.5 MeV photon that pair produces a 4.4 MeV positron and a 1 MeV electron	Same Event but this time the solenoid is 0.3 Tesla and the positron hits the beam pipe, annihilates and makes two 511 keV photons

Sample Brem event producing no positrons

When the solenoid is set to 1.5 Tesla, a 10 MeV electron produces three photons less than 1 MeV in the target, two of them compton scatter in the beam pipe	The same event but this time the electron produces only 1 photon than ionizes in the target

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");

B-field (Tesla)		34.8 mm diameter pipe	47.5	60.2	72.9	97.4
0.0	0.35	1,2,4,4,5	2,3,4,4,6	4,4,6,7,9	6,8,9,10,11	16,14,15,16,17
0.1		225,236,250,246,249=241[math] \pm[/math] 10	282,282,293,294,306=291[math] \pm[/math] 10	373,366,370,364,373=369[math] \pm[/math] 4	451,437,440,438,451=443[math] \pm[/math] 7	602,584,563,558,570=575[math] \pm[/math] 18
0.3	0.35	626,619,596,619,611 =614[math] \pm[/math] 11	720,726,706,730,717=720[math] \pm[/math] 9	871,864,840,841,834 =850[math] \pm[/math] 16	987,968,939,943,952 =958[math] \pm[/math] 20	1118,1106,1069,1067,1080=1088[math] \pm[/math] 23
0.6		929,935,949,969,961=949[math] \pm[/math] 17	1022,1031,1046,1059,1052 =1042[math] \pm[/math] 15	1120,1130,1152,1154,1146 =1140[math] \pm[/math] 15	1168,1184,1210,1221,1206 =1198[math] \pm[/math] 21	1212, 1218,1240,1254,1242=1233[math] \pm[/math] 18
1.0	0.35	1117,1085,1083,1061,1085=1086[math] \pm[/math] 20	1188,1154,1140,1111,1134=1145[math] \pm[/math] 28	1225,1190,1178,1149,1172 =1183[math] \pm[/math] 28	1243.1208,1195,1164,1184=1199[math] \pm[/math] 30	1252,1219,1206,1178,1200=1211[math] \pm[/math] 27
1.5	0.22
2.0	0.10	1198,1210,1215,1223,1176=1204[math] \pm[/math] 18	1216,1227,1235,1241,1196 =1223[math] \pm[/math] 18	1237,1243,1252,1257,1214=1241[math] \pm[/math] 17	1249,1252,1262,1266,1225 =1251[math] \pm[/math] 16	1257,1262,1270,1276,1234=1260[math] \pm[/math] 16
4.0	0.002

Positron Rates -vs- Solenoid Field for 2mm thick PbBi target and several Beam pipe diameters