Converter Temperture - Revision history

Oborn: /* 2.Calculating Energy deposited per second */

2010-06-08T03:32:00Z

2.Calculating Energy deposited per second

2010-06-08T03:31:32Z

2.Calculating Energy deposited per second

2010-06-08T03:30:52Z

2.Calculating Energy deposited per second

2010-06-08T03:30:27Z

2.Calculating Energy deposited per second

2010-06-08T03:29:57Z

2.Calculating Energy deposited per second

2010-06-08T03:29:10Z

2.Calculating Energy deposited per second

2010-06-08T03:28:13Z

1.Calculating number of particles per second

2010-06-08T03:27:29Z

1.Calculating number of particles per second

2010-06-08T03:26:39Z

1.Calculating number of particles per second

2010-06-08T03:26:31Z

1.Calculating number of particles per second

← Older revision		Revision as of 03:32, 8 June 2010
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	To find energy deposited by each electron, we need to use formula		To find energy deposited by each electron, we need to use formula

−	<math> E_{dep~one}=(\frac{dE}{dx})_{coll}\times t </math>	+	<math> E_{dep~one}={(\frac{dE}{dx})}_{coll}\times t </math>

← Older revision		Revision as of 03:31, 8 June 2010
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−	Where is <math> E_{dep~one} </math> is energy deposited by one electron, <math> (dE/dx)_{coll} </math> is mean energy loss (also stopping power) by collision of electron and <math> t </math> is thickness of the radiator.	+	Where is <math> E_{dep~one} </math> is energy deposited by one electron, <math> {(\frac{dE}{dx})}_{coll} </math> is mean energy loss (also stopping power) by collision of electron and <math> t </math> is thickness of the radiator.

	Actually, energy loss of electron comes from two parts: the emission of electromagnetic radiation arising from scattering in the electric field of a nucleus (bremsstrahlung) and collisional energy loss when passing through matter. But bremsstrahlung will not contribute to the temperature, since it is radiation.		Actually, energy loss of electron comes from two parts: the emission of electromagnetic radiation arising from scattering in the electric field of a nucleus (bremsstrahlung) and collisional energy loss when passing through matter. But bremsstrahlung will not contribute to the temperature, since it is radiation.

← Older revision		Revision as of 03:30, 8 June 2010
Line 39:		Line 39:
	To find energy deposited by each electron, we need to use formula		To find energy deposited by each electron, we need to use formula

−	<math> E_{dep~one}=(dE/dx)_{coll}\times t </math>	+	<math> E_{dep~one}=(\frac{dE}{dx})_{coll}\times t </math>

← Older revision		Revision as of 03:30, 8 June 2010
Line 42:		Line 42:


−	Where is <math> E_{dep~one}</math> is energy deposited by one electron, <math>(dE/dx)_{coll}</math> is mean energy loss (also stopping power) by collision of electron and <math>t</math> is thickness of the radiator.	+	Where is <math> E_{dep~one} </math> is energy deposited by one electron, <math> (dE/dx)_{coll} </math> is mean energy loss (also stopping power) by collision of electron and <math> t </math> is thickness of the radiator.

	Actually, energy loss of electron comes from two parts: the emission of electromagnetic radiation arising from scattering in the electric field of a nucleus (bremsstrahlung) and collisional energy loss when passing through matter. But bremsstrahlung will not contribute to the temperature, since it is radiation.		Actually, energy loss of electron comes from two parts: the emission of electromagnetic radiation arising from scattering in the electric field of a nucleus (bremsstrahlung) and collisional energy loss when passing through matter. But bremsstrahlung will not contribute to the temperature, since it is radiation.

← Older revision		Revision as of 03:29, 8 June 2010
Line 42:		Line 42:


−	Where is <math> E_{dep one}</math> is energy deposited by one electron, <math>(dE/dx)_{coll}</math> is mean energy loss (also stopping power) by collision of electron and t is thickness of the radiator.	+	Where is <math> E_{dep~one}</math> is energy deposited by one electron, <math>(dE/dx)_{coll}</math> is mean energy loss (also stopping power) by collision of electron and <math>t</math> is thickness of the radiator.

	Actually, energy loss of electron comes from two parts: the emission of electromagnetic radiation arising from scattering in the electric field of a nucleus (bremsstrahlung) and collisional energy loss when passing through matter. But bremsstrahlung will not contribute to the temperature, since it is radiation.		Actually, energy loss of electron comes from two parts: the emission of electromagnetic radiation arising from scattering in the electric field of a nucleus (bremsstrahlung) and collisional energy loss when passing through matter. But bremsstrahlung will not contribute to the temperature, since it is radiation.

@@ Line 9: / Line 9: @@
 Frequency:  f=1000Hz
-Peak current:  I=10mAmp=0.01 Amp
+Peak current:  I=10~mAmp=0.01 Amp
-Pulse width: <math> \Delta t= 50 ns=5 \times10^{-8}</math> seconds
+Pulse width: <math> \Delta t= 50~ns=5 \times10^{-8}</math> seconds
 So, how many electrons we have in each second?
@@ Line 21: / Line 21: @@
-Where Ne is the total electron numbers hits target per second, e is electron charge and f, I and ∆t are given above.
+Where N is the total electron numbers hits target per second, e is electron charge and f, I and ∆t are given above.
 So

@@ Line 11: / Line 11: @@
 Peak current:  I=10mAmp=0.01 Amp
-Pulse width: </math \Delta t= 50 ns=5 \times10^{-8}</math seconds
+Pulse width: <math \Delta t= 50 ns=5 \times10^{-8}</math seconds
 So, how many electrons we have in each second?

@@ Line 11: / Line 11: @@
 Peak current:  I=10mAmp=0.01 Amp
-Pulse width: ∆t= 50 ns=5*10-8 seconds
+Pulse width: </math \Delta t= 50 ns=5 \times10^{-8}</math seconds
 So, how many electrons we have in each second?
@@ Line 31: / Line 31: @@
-So, we have around 3.12075*1012 electrons hit radiator per second.
+So, we have around <math> 3.12075 \times 10^{12 }</math> electrons hit radiator per second.
 ==2.Calculating Energy deposited per second==