Faraday Cup Temperature - Revision history

Oborn: /* Temperature calculation */

2010-10-14T05:49:54Z

Temperature calculation

2010-10-14T05:48:45Z

Number of particles per second and corresponding beam power

2010-10-14T05:45:55Z

Temperature calculation

2010-10-14T05:33:58Z

Conclusion

2010-10-14T05:33:37Z

Conclusion

2010-10-14T05:32:47Z

Temperature calculation

2010-10-14T05:32:01Z

2010-10-14T05:30:15Z

Temperature calculation

2010-10-14T05:29:00Z

Temperature calculation

2010-10-14T05:28:25Z

Temperature calculation

@@ Line 31: / Line 31: @@
 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})} = 0.18\cdot10^{12}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

@@ Line 5: / Line 5: @@
 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:
@@ 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>
 ==Temperature calculation==

@@ Line 36: / Line 36: @@
 So
-  <math> T = 424\ K </math>
+  <math> T = 652\ K </math>
 ==Conclusion==

@@ Line 40: / Line 40: @@
 ==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]

@@ Line 40: / Line 40: @@
 ==Conclusion==
-Because the melting point of Aluminum is <math> 933.5 K </math> we are safety.
+Because the melting point of Aluminum is <math> 933.5\ K </math> we are safety.
 [http://wiki.iac.isu.edu/index.php/PhotoFission_with_Polarized_Photons_from_HRRL Go Back]

@@ Line 27: / Line 27: @@
 Here,
-   A is the radiated area of the rode.radiatedanradiated
+   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.
@@ Line 37: / Line 37: @@
   <math> T = 424\ K </math>
 ==Conclusion==

@@ Line 28: / Line 28: @@
    A is the radiated area of the rode.radiatedanradiated
-   <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:
-  <math> T^4 = \frac {P}{0.924\ A\ \sigma} = \frac {1.9\ W}{(0.924)(2\cdot 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})} = 0.18\ K^4</math>
 So