View source for Faraday Cup Temperature

= 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.

 

[http://wiki.iac.isu.edu/index.php/PhotoFission_with_Polarized_Photons_from_HRRL Go Back]