Difference between revisions of "Faraday Cup Temperature"

From New IAC Wiki
Jump to navigation Jump to search
 
(10 intermediate revisions by the same user not shown)
Line 5: Line 5:
 
Assume electron beam parameters at FC location are:
 
Assume electron beam parameters at FC location are:
  
−
  Frequency:    f=300 Hz
+
  Frequency:    f = 300 Hz
−
  Peak current:  I=3 Amps
+
  Peak current:  I = 3 Amps
−
  Pulse width:  t= 50 ps
+
  Pulse width:  t = 50 ps
−
  Beam energy:  E=45 MeV
+
  Beam energy:  E = 45 MeV
  
 
The number of electrons per second at FC location are:
 
The number of electrons per second at FC location are:
Line 17: Line 17:
  
 
  <math> P = (0.28*10^{12}\ \frac{e}{sec}) (44\ MeV) (1.6*10^{-19} \frac{J}{eV}) = 1.9\ W</math>
 
  <math> P = (0.28*10^{12}\ \frac{e}{sec}) (44\ MeV) (1.6*10^{-19} \frac{J}{eV}) = 1.9\ W</math>
−
 
  
 
==Temperature calculation==
 
==Temperature calculation==
Line 27: Line 26:
 
Here,
 
Here,
  
−
   A is the radiated area of the rode.radiatedanradiated
+
   A is the radiated area of the rode
−
   <math> \sigma = 5.67*10^{-8} \frac {W}{m^2 K^4} </math> is the Stefan-Boltzmann constant.   
+
   <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:
 
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*10^{-4}\ m^2)(5.67*10^{-8} \frac {W}{m^2\ K^4})} = 0.18\ K^4</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>
  
 
So
 
So
  
−
  <math> T = 424\ K </math>
+
  <math> T = 652\ K </math>
  
 
==Conclusion==
 
==Conclusion==
  
−
Because the melting point of Aluminum is <math> 933.5 K </math> we are safety.
+
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]
 
[http://wiki.iac.isu.edu/index.php/PhotoFission_with_Polarized_Photons_from_HRRL Go Back]

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.


Go Back

Retrieved from "https://wiki.iac.isu.edu/index.php?title=Faraday_Cup_Temperature&oldid=53787"