Aluminum Converter - Revision history

Oborn at 20:52, 7 June 2010

2010-06-07T20:52:39Z

2010-06-07T20:50:31Z

2010-06-07T20:45:35Z

2010-06-07T20:33:35Z

2010-06-07T20:33:08Z

2010-06-07T20:26:29Z

2010-06-07T20:14:04Z

2010-06-07T19:59:24Z

2010-06-07T19:57:24Z

2010-06-07T19:53:29Z

@@ Line 31: / Line 31: @@
 The total stopping power due to collisions on Al per incident electron:
-<math> (1.78 MeV \frac {cm^2}{g})(0.003429 \frac {g}{cm^2}) = 0.0061 MeV per electron </math>
+<math> (1.78 MeV \frac {cm^2}{g})(0.003429 \frac {g}{cm^2}) = 0.0061 \frac {MeV}{electron} </math>
 The energy deposited per pulse:

@@ Line 21: / Line 21: @@
 ==Calculating the stopping power due to collision of one 44 MeV electron in Aluminum==
-From NIST ([http://physics.nist.gov/cgi-bin/Star/e_table.pl] 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>.
+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>.
 <math> 1 mil = \frac {1} {1000} inch * 2.54 \frac {cm} {inch} = 0.00254 cm </math>

← Older revision		Revision as of 20:45, 7 June 2010
Line 55:		Line 55:
	where <math> \sigma </math> is the Stefan-Boltzmann constant, <math> \sigma = 5.6710^{-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 = 19.6210^{-6} m^2 </math>.		where <math> \sigma </math> is the Stefan-Boltzmann constant, <math> \sigma = 5.6710^{-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 = 19.6210^{-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.

← Older revision		Revision as of 20:33, 7 June 2010
Line 53:		Line 53:
	<math> T^4 = \frac {P}{(0.924)(2A)(\sigma)} </math>		<math> T^4 = \frac {P}{(0.924)(2A)(\sigma)} </math>

−	where <math> \sigma </math> is the Stefan-Boltzmann constant, <math> \sigma = 5.6710^{-8} \frac {W}{m^2 K^4}. Assume a beam spot diameter on the converter surface of 5mm, or an area of <math> A = 19.62 mm^2 = 19.6210^{-6} m^2 </math>.	+	where <math> \sigma </math> is the Stefan-Boltzmann constant, <math> \sigma = 5.6710^{-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 = 19.6210^{-6} m^2 </math>.

@@ Line 45: / Line 45: @@
 The power deposited in 1/2 mil Al is:
-<math> 28.6*10^9 \frac {MeV}{second} = (28.6*10^{15} \frac {eV}{second})(1.6*10^{-19} \frac {C}{eV} = 4.575*10^{-3} W </math>
+<math> P = (28.6*10^{15} \frac {eV}{second})(1.6*10^{-19} \frac {C}{eV}) = 4.575*10^{-3} W </math>
-Stefan-Boltzmann Law (Wien Approximation) says:
+Stefan-Boltzmann Law (Wien Approximation) says <math> P = (0.924)(Area)(\sigma)(T^4) </math>
 Solving for Temperature and taking into account the two sides of the converter we get:
-<math> T = (\frac {P}{(0.924)(2A)(\sigma)})^{\frac {1}{4}} </math>
+<math> T^4 = \frac {P}{(0.924)(2A)(\sigma)} </math>
-Assume a beam spot diameter on the converter surface of 5mm, or an area of <math> A = 19.62 mm^2 </math>.
+where <math> \sigma </math> is the Stefan-Boltzmann constant, <math> \sigma = 5.67*10^{-8} \frac {W}{m^2 K^4}.  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>.

← Older revision		Revision as of 20:26, 7 June 2010
Line 39:		Line 39:
	The energy deposited per second:		The energy deposited per second:

−	<math> (95.312510^6 \frac {MeV}{pulse})(300 \frac {pulses}{second} = 28.610^9 \frac {MeV}{second} </math>	+	<math> (95.312510^6 \frac {MeV}{pulse})(300 \frac {pulses}{second}) = 28.610^9 \frac {MeV}{second} </math>

		+	==Calculating the temperature increase==
		+
		+	The power deposited in 1/2 mil Al is:
		+
		+	<math> 28.610^9 \frac {MeV}{second} = (28.610^{15} \frac {eV}{second})(1.610^{-19} \frac {C}{eV} = 4.57510^{-3} W </math>
		+
		+	Stefan-Boltzmann Law (Wien Approximation) says:
		+
		+	<math> P = (0.924)(Area)(\sigma)(T^4) </math>
		+
		+	Solving for Temperature and taking into account the two sides of the converter we get:
		+
		+	<math> T = (\frac {P}{(0.924)(2A)(\sigma)})^{\frac {1}{4}} </math>

← Older revision		Revision as of 20:14, 7 June 2010
Line 24:		Line 24:

	<math> 1 mil = \frac {1} {1000} inch * 2.54 \frac {cm} {inch} = 0.00254 cm </math>		<math> 1 mil = \frac {1} {1000} inch * 2.54 \frac {cm} {inch} = 0.00254 cm </math>
		+
		+	The effective length of 1/2 mil Al:
		+
		+	<math> (2.70 \frac {g}{cm^3})(0.00127 cm) = 0.003429 \frac {g}{cm^2} </math>
		+
		+	The total stopping power due to collisions on Al per incident electron:
		+
		+	<math> (1.78 MeV \frac {cm^2}{g})(0.003429 \frac {g}{cm^2}) = 0.0061 MeV per electron </math>
		+
		+	The energy deposited per pulse:
		+
		+	<math> (0.0061 \frac {MeV}{electron})(15.62510^9 \frac {electrons}{pulse}) = 95.312510^6 \frac {MeV}{pulse} </math>
		+
		+	The energy deposited per second:
		+
		+	<math> (95.312510^6 \frac {MeV}{pulse})(300 \frac {pulses}{second} = 28.610^9 \frac {MeV}{second} </math>
		+

@@ Line 15: / Line 15: @@
 Where <math> N </math> is the number of electrons that hit the target per second, <math> e </math> is electron charge and <math> f </math>, <math> I </math> and <math> ∆t </math> are given above.
-<math> N = f*I*∆t/e = \frac {(300 Hz)(50 A) (5*10^{-11} ps)} {1.6*10^{-19} C} = 4.6875*10^{12} </math>
+<math> N = \frac {f I ∆t} {e} = \frac {(300 Hz)(50 A) (5*10^{-11} ps)} {1.6*10^{-19} C} = 4.6875*10^{12} </math>
 So, we have around <math> 4.6875*10^{12}</math> electrons per second or <math> 15.625*10^9 </math> electrons per pulse.
@@ Line 28: / Line 28: @@
-Assume a beam spot diameter on the converter surface of 5mm, or an area of <math> A=19.62 mm^2 <\math>.
+Assume a beam spot diameter on the converter surface of 5mm, or an area of <math> A = 19.62 mm^2 </math>.