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 one rode. Assume it's 2 cm^2
 [math] \sigma = 5.67*10^{-8} \frac {W}{m^2 K^4} [/math] is the Stefan-Boltzmann constant.  

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.

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