Difference between revisions of "Faraday Cup Temperature"

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= Calculating the temperature of a Faraday Cup Rod =
 
= Calculating the temperature of a Faraday Cup Rod =
  
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==Number of particles per second hitting one rod==
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==Number of particles per second and corresponding beam power==
  
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Assume electron beam parameters at Faraday Cup location are:
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Assume electron beam parameters at FC location are:
  
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  Frequency:    f=300 Hz
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  Frequency:    f = 300 Hz
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  Peak current:  I=3 Amps
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  Peak current:  I = 3 Amps
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  Pulse width:  t= 50 ps
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  Pulse width:  t = 50 ps
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  Beam energy:  E=45 MeV
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  Beam energy:  E = 45 MeV
  
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The number of electrons that hit the FC per second is:
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The number of electrons per second at FC location are:
  
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<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}\frac {e^-}(sec)} </math>
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<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>
  
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So, we have around <math> 4.6875*10^{12}</math> electrons per second or <math> 15.625*10^9 </math> electrons per pulse.
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The corresponding beam power for 45 MeV electron beam is:
  
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==Calculating the stopping power due to collision of one 44 MeV electron in Aluminum==
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<math> P = (0.28*10^{12}\ \frac{e}{sec}) (44\ MeV) (1.6*10^{-19} \frac{J}{eV}) = 1.9\ W</math>
  
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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>.
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==Temperature calculation==
  
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<math> 1 mil = \frac {1} {1000} inch * 2.54 \frac {cm} {inch} = 0.00254 cm </math>
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Now apply the Stefan-Boltzmann Law for one Faraday cup rod
  
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The effective length of 1/2 mil Al: 
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<math> P = (0.924)\ (A)\ (\sigma)\ (T^4) </math>
  
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<math> (2.70 \frac {g}{cm^3})(0.00127 cm) = 0.003429 \frac {g}{cm^2} </math>
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Here,
  
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The total stopping power due to collisions on Al per incident electron:
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  A is the radiated area of the rode
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  <math> \sigma = 5.67\cdot 10^{-8} \frac {W}{m^2 K^4} </math> is the Stefan-Boltzmann constant. 
  
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<math> (1.78 MeV \frac {cm^2}{g})(0.003429 \frac {g}{cm^2}) = 0.0061 \frac {MeV}{electron} </math>
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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:
  
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The energy deposited per pulse:
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<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>
  
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<math> (0.0061 \frac {MeV}{electron})(15.625*10^9 \frac {electrons}{pulse}) = 95.3125*10^6 \frac {MeV}{pulse} </math>
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So
  
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The energy deposited per second:
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<math> T = 652\ K </math>
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<math> (95.3125*10^6 \frac {MeV}{pulse})(300 \frac {pulses}{second}) = 28.6*10^9 \frac {MeV}{second} </math>
 
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==Calculating the temperature increase==
 
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The power deposited in 1/2 mil Al is:
 
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<math> P = (28.6*10^{15} \frac {eV}{second})(1.6*10^{-19} \frac {C}{eV}) = 4.575*10^{-3} W </math>
 
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Stefan-Boltzmann Law (Wien Approximation) says <math> P = (0.924)(Area)(\sigma)(T^4) </math>
 
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Solving for Temperature and taking into account the two sides of the converter we get:
 
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<math> T^4 = \frac {P}{(0.924)(2A)(\sigma)} </math>
 
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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>.
 
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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 :
 
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<math> T = 300 + 217.2 = 517.2 K</math>
 
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The melting temperature of Aluminum is <math> 933.5 K </math>.
 
  
 
==Conclusion==
 
==Conclusion==
  
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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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Because the melting point of Aluminum is 933.5 K we are safety.
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[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.


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