View source for HRRL Positron Rotating W Target

The first version of a positron converter target will be designed to distribute the heat load by rotating the tungsten target.

Calculate for 1 mm and 2 mm thick Tungsten


Look for Tungsten disks to attach to brushless motor and fit into beam pipe
= Properties of Tungsten=

Melting Point = 3695 K.

Heat Capacity = <math>(25~ ^{o}C)~ 24.27 J \dot mol^{-1} \dot K^{-1}</math>


Heat loss due to radiation ( \sigma T^4)

=Tungsten Temperature as a function of heat load=

The Tungsten heats up when an MeV energy electron impinges its surface.  The imperfect vacuum inside the beam pipe does allow some radiative cooling.


Conduction and Radiation
 
;IAC beamline pressure = <math>10^{-8} </math>
: Torr

=Calculating Radiators Equilibrium Temperature=
==1.Calculating number of particles per second ==

We have electron beam of:

Frequency:  f=1000Hz

Peak current:  I=10~mAmp=0.01 Amp

Pulse width: <math> \Delta t= 50~ns=5 \times10^{-8}</math> seconds

So, how many electrons we have in each second?

By Q=It, we have


                            <math> N \times e=f \times I \times \Delta t </math>


Where N is the total electron numbers hits target per second, e is electron charge and f, I and ∆t are given above.
So



                            <math>  N = \frac {f \times I \times \Delta t}{e}  = \frac {1000 \times 0.01 \times 5\times10^{-8} }{1.6 \times 10^{-19}} =3.12075 \times 10^{12 }</math>




So, we have around <math> 3.12075 \times 10^{12 }</math> 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

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


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.

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===
Note:These data is from Particle Data group,Link: [http://pdg.lbl.gov/AtomicNuclearProperties/].

{| border="1" cellpadding="20" cellspacing="0"


|-
|Elements
|Radiation Lengths <math> (g/cm^{2} )</math>
|-
|W || 6.76
|-
|}<br>

===Table of energy calculations===
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]])
{| border="1" cellpadding="20" cellspacing="0"


|-
|Elements
|<math>(dE/dx)_{coll} (MeV \; cm^2/g)</math>
|t( <math> gcm^{-2}</math>)
|<math>E_{dep one}</math> (MeV)
|<math>E_{dep/s}</math> (MeV/s)
|<math>E_{dep/s}</math> (J/s)
|-
|W  ||1.247 ||0.00676 ||0.00842972 ||<math>2.63*10^{10}</math>  ||<math>4.21*10^{-3}</math> 
|-
|}

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).

                                   <math>P{dep}= E{dep/s} = ( E {dep by one})*(Number of electron per second)</math>

==3.Calculating equilibrium temperature using Stefan–Boltzmann law==
If we assume that there is no energy conduction and total energy is just radiated from two surfaces of the radiators which are as big as beam spot,in our case beam spot is 2mm in diameter.  According to Stefan–Boltzmann law, this total power radiated will be 
                                                         <math> P_{rad}</math>= 2Aσ<math>T^{4}</math>
Where T is radiating temperature P is the radiating power, A is surface area that beam incident and σ is Stefan–Boltzmann constant or Stefan's constant. To reach equilibrium temperature, Power deposited in and power radiated should be. So
                                                        <math> P_{dep}=P_{rad} </math>

so

                                               T =[<math>  P_{dep}</math>/(2Aσ)]^{1/4}


===Table of equilibrium temperatures=== 

For 2mm diameter spot size and 0.001 time Radiation Length thickness

{| border="1" cellpadding="20" cellspacing="0"

|-
|Elements
|d (m)
|2A( <math> m^{2}</math>)
|Stefan-Boltzmann Constant
|<math>T_{equ}</math> (K)
|-
|W  ||0.002  ||0.00000628 ||<math>5.6704*10^{-8}</math> ||329.8 
|-
|}

==4. Results for different Thicknesses and Spot sizes ==

We can calculate separately for Al, W and Ti for different thickness and different beam spot diameter. Following tables are temperature calculation for Al, W  Ti and Fe in thickness of 0.001, 0.005 and 0.01 times of Radiation Length, in beam spot of diameter of 2, 4, 6, 8, 10 mm.


===W===
Equilibrium temperature for W, unit is K. (Melting point of W is 3695K)

{| border="1" cellpadding="20" cellspacing="0"

|-
|
|2(mm)
|4(mm)
|6(mm)
|8(mm)
|10(mm)
|-
|-
|0.001Rl || 239.8 K||233.2  K||190.4 K||164.9 K|| 147.5 K
|- 
|0.005Rl ||493.2 K|| 348.7 K||248.7 K||246.6 K|| 220.6 K
|- 
|0.01Rl  ||586.5  K||414.7 K|| 338.6 K|| 293.3 K|| 262.3 K
|-
|}

[[Image:Equilibrium Temperature for W.jpg]]

==5. Results for worst cases ( 80mA peak current,1000Hz frequency, 50ns pulse width ) ==

Now we calculate temperature for Al, W, Ti and Fe for different thickness and different beam spot diameter in worst case . Following tables are temperature calculation for Al, W Ti and Fe in thickness of 0.00001, 0.0005 and 0.0001 times of Radiation Length, in beam spot of diameter of 2, 3, 4 mm.


===W===
Equilibrium temperature for W, unit is K. (Melting point of Al is 3695K)

{| border="1" cellpadding="20" cellspacing="0"

|-
|
|1(mm)
|2(mm)
|3(mm)
|4(mm)
|-
|-
|0.00005Rl ||370  K||262 K || 214 K|| 158 K
|- 
|0.0001Rl ||441  K||312 K||254 K||220 K
|- 
|0.0005Rl  ||660  K|| 446 K|| 381 K||330 K
|-
|}

[[Image:W_2.jpg|down|800px]]

==6. More results for worst cases ( 80mA peak current,1000Hz frequency, 50ns pulse width for Θ₀ = 0.2,0.4,0.6 dgrees) ==

Now we calculate temperature for Al, W, Ti and Fe for different thickness and different beam spot diameter in worst case . Following tables are temperature calculation for Al, W Ti and Fe in thickness of <math>3.9*10^{-5}</math> (Θ₀ = <math>0.2^{o}</math>), <math>1.35*10^{-4}</math>(Θ₀= <math>0.4^{o}</math>) and <math>2.8*10^{-4}</math>(Θ₀ = <math>0.6^{o}</math>) times of Radiation Length, in beam spot of diameter of 2, 3, 4 mm.


===W===
Equilibrium temperature for W, unit is K. (Melting point of Al is 3695K)

{| border="1" cellpadding="20" cellspacing="0"

|-
|
|2(mm)
|3(mm)
|4(mm)
|-
|-
| Θ₀=0.2˚, t=3.9E-5 RL=0.15μm|| 246 K|| 201 K|| 174 K
|- 
| Θ₀=0.4˚,t=1.35E-4RL=0.5μm  || 336 K|| 275 K|| 237 K
|- 
| Θ₀=0.6˚,t=2.8E-4RL=1μm     || 404 K|| 329 K|| 285 K
|-
|}


[[Image:W_3.jpg|down|800px]]


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