Niowave 11-2015
GEANT4 predicts that scattered electrons, photons, and positrons (mostly scattered electrons) deposit
According to the above figure, GEANT4 predicts a total of
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. The peak energy deposition is 0.3 MeV/eIf this energy were uniformly distributed along the 5 mm thick beam pipe having a diameter of 3.48 cm then I would see
if you assume a 1 mA beam of electrons then this becomes
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.
If you use the above factors to weight the histogram, then the figure below shows that GEANT4 predicts a power deposition density of
, 1 cm downstream of the target. Back scattered electrons appear to create the hottest spot of 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 ( | )
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
. 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
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
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 | 10282,282,293,294,306=291 | 10373,366,370,364,373=369 | 4451,437,440,438,451=443 | 7602,584,563,558,570=575 | 18|
0.3 | 0.35 | 626,619,596,619,611 =614 | 11720,726,706,730,717=720 | 9871,864,840,841,834 =850 | 16987,968,939,943,952 =958 | 201118,1106,1069,1067,1080=1088 | 23
0.6 | 929,935,949,969,961=949 | 171022,1031,1046,1059,1052 =1042 | 151120,1130,1152,1154,1146 =1140 | 151168,1184,1210,1221,1206 =1198 | 211212, 1218,1240,1254,1242=1233 | 18|
1.0 | 0.35 | 1117,1085,1083,1061,1085=1086 | 201188,1154,1140,1111,1134=1145 | 281225,1190,1178,1149,1172 =1183 | 281243.1208,1195,1164,1184=1199 | 301252,1219,1206,1178,1200=1211 | 27
1.5 | 0.22 | |||||
2.0 | 0.10 | 1198,1210,1215,1223,1176=1204 | 181216,1227,1235,1241,1196 =1223 | 181237,1243,1252,1257,1214=1241 | 171249,1252,1262,1266,1225 =1251 | 161257,1262,1270,1276,1234=1260 | 16
4.0 | 0.002 |