Difference between revisions of "Faraday Cup Temperature"
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Assume the whole power of beam is absorbed by just one Faraday Cup rod. For 45 MeV electron beam this power is: | Assume the whole power of beam is absorbed by just one Faraday Cup rod. For 45 MeV electron beam this power is: | ||
− | <math> P = 0.28*10^{12}\frac{e}{sec} \dot 44\ MeV \dot 1.6*10^{-19} \frac{J}{eV}</math> | + | <math> P = (0.28*10^{12}\ \frac{e}{sec}) \dot (44\ MeV) \dot (1.6*10^{-19} \frac{J}{eV})</math> |
The energy deposited per pulse: | The energy deposited per pulse: |
Revision as of 04:52, 14 October 2010
Calculating the temperature of a Faraday Cup Rod
Number of particles per second hitting one rod
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 is:
Power deposited in 1 FC rod
Assume the whole power of beam is absorbed by just one Faraday Cup rod. For 45 MeV electron beam this power is:
The energy deposited per pulse:
The energy deposited per second:
Calculating the temperature increase
The power deposited in 1/2 mil Al is:
Stefan-Boltzmann Law (Wien Approximation) says
Solving for Temperature and taking into account the two sides of the converter we get:
where
is the Stefan-Boltzmann constant, . Assume a beam spot diameter on the converter surface of 5mm, or an area of .Plugging in the numbers we see that the temperature will increase
. Now, adding in the temperature of the converter at room temperature we get :
The melting temperature of Aluminum is
.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.