Latest revision as of 05:49, 14 October 2010

Calculating the temperature of a Faraday Cup Rod

Number of particles per second and corresponding beam power

Assume electron beam parameters at FC location are:

Frequency:     f = 300 Hz
Peak current:  I = 3 Amps
Pulse width:   t = 50 ps
Beam energy:   E = 45 MeV

The number of electrons per second at FC location are:

[math] N = \frac {Q} {e} = \frac {f I ∆t} {e} = \frac {(300\ Hz)(3\ A) (50\ ps)} {1.6*10^{-19}\ C} = 0.28*10^{12} [/math]

The corresponding beam power for 45 MeV electron beam is:

[math] P = (0.28*10^{12}\ \frac{e}{sec}) (44\ MeV) (1.6*10^{-19} \frac{J}{eV}) = 1.9\ W[/math]

Temperature calculation

Now apply the Stefan-Boltzmann Law for one Faraday cup rod

[math] P = (0.924)\ (A)\ (\sigma)\ (T^4) [/math]

Here,

 A is the radiated area of the rode
 [math] \sigma = 5.67\cdot 10^{-8} \frac {W}{m^2 K^4} [/math] is the Stefan-Boltzmann constant.

Assume the all beam power goes to one FC rod and area of radiation is 2 cm^2 (two back sides) we can calculate the corresponding temperature:

[math] T^4 = \frac {P}{0.924\ A\ \sigma} = \frac {1.9\ W}{(0.924)(2\cdot 10^{-4}\ m^2)(5.67\cdot 10^{-8} \frac {W}{m^2 K^4})} = 0.1813\cdot10^{12}K^4[/math]

So

[math] T = 652\ K [/math]

Conclusion

Because the melting point of Aluminum is 933.5 K we are safety.

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 = Calculating the temperature of a Faraday Cup Rod =
-==Number of particles per second hitting one rod==
+==Number of particles per second and corresponding beam power==
-Assume electron beam parameters at Faraday Cup location are:
+Assume electron beam parameters at FC location are:
-  Frequency:  <math> f=300Hz </math>
+  Frequency:     f = 300 Hz
-  Peak current: <math> I=50 Amps </math>
+  Peak current:  I = 3 Amps
-  Pulse width: <math> ∆t= 50 ps = 5*10^{-11} seconds </math>
+  Pulse width:   t = 50 ps
-  Beam energy: <math> E=45 MeV </math>
+  Beam energy:   E = 45 MeV
-By <math> Q=It </math>, we have <math> N*e=f*I*∆t </math>
+The number of electrons per second at FC location are:
-Where <math> N </math> is the number of electrons that hit the target per second, <math> e </math> is electron charge and <math> f </math>, <math> I </math> and <math> ∆t </math> are given above.
+ <math> N = \frac {Q} {e} = \frac {f I ∆t} {e} = \frac {(300\ Hz)(3\ A) (50\ ps)} {1.6*10^{-19}\ C} = 0.28*10^{12} </math>
-<math> N = \frac {f I ∆t} {e} = \frac {(300 Hz)(50 A) (5*10^{-11} ps)} {1.6*10^{-19} C} = 4.6875*10^{12} </math>
+The corresponding beam power for 45 MeV electron beam is:
-So, we have around <math> 4.6875*10^{12}</math> electrons per second or <math> 15.625*10^9 </math> electrons per pulse.
+ <math> P = (0.28*10^{12}\ \frac{e}{sec}) (44\ MeV) (1.6*10^{-19} \frac{J}{eV}) = 1.9\ W</math>
-==Calculating the stopping power due to collision of one 44 MeV electron in Aluminum==
+==Temperature calculation==
-From NIST ([http://physics.nist.gov/PhysRefData/Star/Text/ESTAR.html] see link here) the stopping power for one electron with energy of 44 MeV in Aluminum is <math> 1.78 MeV cm^2/g </math>.
+Now apply the Stefan-Boltzmann Law for one Faraday cup rod
-<math> 1 mil = \frac {1} {1000} inch * 2.54 \frac {cm} {inch} = 0.00254 cm </math>
+ <math> P = (0.924)\ (A)\ (\sigma)\ (T^4) </math>
-The effective length of 1/2 mil Al:
+Here,
-<math> (2.70 \frac {g}{cm^3})(0.00127 cm) = 0.003429 \frac {g}{cm^2} </math>
+  A is the radiated area of the rode
+  <math> \sigma = 5.67\cdot 10^{-8} \frac {W}{m^2 K^4} </math> is the Stefan-Boltzmann constant.
-The total stopping power due to collisions on Al per incident electron:
+Assume the all beam power goes to one FC rod and area of radiation is 2 cm^2 (two back sides) we can calculate the corresponding temperature:
-<math> (1.78 MeV \frac {cm^2}{g})(0.003429 \frac {g}{cm^2}) = 0.0061 \frac {MeV}{electron} </math>
+ <math> T^4 = \frac {P}{0.924\ A\ \sigma} = \frac {1.9\ W}{(0.924)(2\cdot 10^{-4}\ m^2)(5.67\cdot 10^{-8} \frac {W}{m^2 K^4})} = 0.1813\cdot10^{12}K^4</math>
-The energy deposited per pulse:
+So
-<math> (0.0061 \frac {MeV}{electron})(15.625*10^9 \frac {electrons}{pulse}) = 95.3125*10^6 \frac {MeV}{pulse} </math>
+ <math> T = 652\ K </math>
-The energy deposited per second:
-<math> (95.3125*10^6 \frac {MeV}{pulse})(300 \frac {pulses}{second}) = 28.6*10^9 \frac {MeV}{second} </math>
-==Calculating the temperature increase==
-The power deposited in 1/2 mil Al is:
-<math> P = (28.6*10^{15} \frac {eV}{second})(1.6*10^{-19} \frac {C}{eV}) = 4.575*10^{-3} W </math>
-Stefan-Boltzmann Law (Wien Approximation) says <math> P = (0.924)(Area)(\sigma)(T^4) </math>
-Solving for Temperature and taking into account the two sides of the converter we get:
-<math> T^4 = \frac {P}{(0.924)(2A)(\sigma)} </math>
-where <math> \sigma </math> is the Stefan-Boltzmann constant, <math> \sigma = 5.67*10^{-8} \frac {W}{m^2 K^4} </math>.  Assume a beam spot diameter on the converter surface of 5mm, or an area of <math> A = 19.62 mm^2 = 19.62*10^{-6} m^2 </math>.
-Plugging in the numbers we see that the temperature will increase <math> 217.2 K </math>.  Now, adding in the temperature of the converter at room temperature we get :
-<math> T = 300 + 217.2 = 517.2 K</math>
-The melting temperature of Aluminum is <math> 933.5 K </math>.
 ==Conclusion==
-An Aluminum converter that is 1/2 mil thick being struck by a 44 MeV electron beam with a 50 picosecond pulse width, 300 Hz rep rate, and 50 Amp peak current is found to be safe from melting.
+Because the melting point of Aluminum is 933.5 K we are safety.
 [http://wiki.iac.isu.edu/index.php/PhotoFission_with_Polarized_Photons_from_HRRL Go Back]

Difference between revisions of "Faraday Cup Temperature"

Latest revision as of 05:49, 14 October 2010

Contents

Calculating the temperature of a Faraday Cup Rod

Number of particles per second and corresponding beam power

Temperature calculation

Conclusion

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