Difference between revisions of "Radiators Temperature"
Line 1: | Line 1: | ||
+ | =Calculating Radiators Equilibrium Temperature= | ||
==1.Calculating number of particles per second == | ==1.Calculating number of particles per second == | ||
Line 39: | Line 40: | ||
Stopping power can be found from nuclear data tables <math> (dE/dx)_{ave} </math> and thickness is 0.001 times of radiation length. From Particle Data group we got radiation length and average total stopping powers around 15MeV for electrons in these materials from National Institute of Standards and Technology | Stopping power can be found from nuclear data tables <math> (dE/dx)_{ave} </math> and thickness is 0.001 times of radiation length. From Particle Data group we got radiation length and average total stopping powers around 15MeV for electrons in these materials from National Institute of Standards and Technology | ||
− | =Table of Radiation Lengths= | + | ===Table of Radiation Lengths=== |
Note:These data is from Particle Data group,Link: [http://pdg.lbl.gov/AtomicNuclearProperties/]. | Note:These data is from Particle Data group,Link: [http://pdg.lbl.gov/AtomicNuclearProperties/]. | ||
Line 58: | Line 59: | ||
|}<br> | |}<br> | ||
− | =Table of energy calculations= | + | ===Table of energy calculations=== |
For the thickness of 0.001 Radiation Length (0.0001RL) of radiators. | For the thickness of 0.001 Radiation Length (0.0001RL) of radiators. | ||
Note: <math>(dE/dx)_{coll}</math> is from National Institute of Standards and Technology. Link: [[http://physics.nist.gov/PhysRefData/Star/Text/ESTAR.html]]) | Note: <math>(dE/dx)_{coll}</math> is from National Institute of Standards and Technology. Link: [[http://physics.nist.gov/PhysRefData/Star/Text/ESTAR.html]]) |
Revision as of 17:06, 1 June 2008
Calculating Radiators Equilibrium Temperature
1.Calculating number of particles per second
We have electron beam of:
Frequency: f=1000Hz
Peak current: I=10mAmp=0.01 Amp
Pulse width: ∆t= 50 ns=5*10-8 seconds
So, how many electrons we have in each second?
By Q=It, we have
N*e=f*I*∆t
Where Ne is the total electron numbers hits target per second, e is electron charge and f, I and ∆t are given above. So
N= f*I*∆t/e=1000*0.01*5*10-8/(1.6*10-19)=3.12075*1012
So, we have around 3.12075*1012 electrons hit radiator per second.
2.Calculating Energy deposited per second
If we find the energy deposited by each electron and multiply to the total number of electrons in each second, we will find the total energy per second deposited in radiator.
To find energy deposited by each electron, we need to use formula
Where is is energy deposited by one electron, is mean energy loss (also stopping power) by collision of electron and t 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.
Stopping power can be found from nuclear data tables
and thickness is 0.001 times of radiation length. From Particle Data group we got radiation length and average total stopping powers around 15MeV for electrons in these materials from National Institute of Standards and TechnologyTable of Radiation Lengths
Note:These data is from Particle Data group,Link: [1].
Elements | Radiation Lengths |
Al | 24.01 |
W | 6.76 |
Ti | 16.16 |
Fe | 13.84 |
Table of energy calculations
For the thickness of 0.001 Radiation Length (0.0001RL) of radiators. Note: [2]])
is from National Institute of Standards and Technology. Link: [Elements | t( | )(MeV) | (MeV/s) | (J/s) | |
Al | 1.676 | 0.02401 | 0.00402076 | ||
W | 1.247 | 0.00676 | 0.00842972 | ||
Ti | 1.555 | 0.01616 | 0.0251288 | ||
Fe | 1.529 | 0.01384 | 0.02116136 |
In above table,we took the total numbers of electrons per second and multiply it to Energy deposited by one electron,get total energy deposited per second (which is power).