Difference between revisions of "Converter Temperture"

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Line 9: Line 9:
 
Frequency:  f=1000Hz
 
Frequency:  f=1000Hz
  
Peak current:  I=10mAmp=0.01 Amp
+
Peak current:  I=10~mAmp=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?
 
So, how many electrons we have in each second?
Line 17: Line 17:
 
By Q=It, we have
 
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.
+
                            <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
 
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.
+
 
 +
                            <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==
 
==2.Calculating Energy deposited per second==
Line 32: 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}*t </math>
+
                                                     <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>(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> {(\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.
Line 50: Line 57:
 
|Elements
 
|Elements
 
|Radiation Lengths <math> (g/cm^{2} )</math>
 
|Radiation Lengths <math> (g/cm^{2} )</math>
|-
 
|Al || 24.01
 
 
|-
 
|-
 
|W || 6.76
 
|W || 6.76
 
|-
 
|-
|Ti || 16.16
 
|-
 
|Fe || 13.84
 
 
|}<br>
 
|}<br>
  
Line 73: Line 75:
 
|<math>E_{dep/s}</math> (MeV/s)
 
|<math>E_{dep/s}</math> (MeV/s)
 
|<math>E_{dep/s}</math> (J/s)
 
|<math>E_{dep/s}</math> (J/s)
|-
 
|Al ||1.676 || 0.02401 ||0.00402076 ||<math>1.26*10^{11}</math> ||<math>2.01*10^{-2}</math>
 
 
|-
 
|-
 
|W  ||1.247 ||0.00676 ||0.00842972 ||<math>2.63*10^{10}</math>  ||<math>4.21*10^{-3}</math>  
 
|W  ||1.247 ||0.00676 ||0.00842972 ||<math>2.63*10^{10}</math>  ||<math>4.21*10^{-3}</math>  
|-
 
|Ti ||1.555 ||0.01616 || 0.0251288||<math>7.84*10^{10} </math> ||<math>1.26*10^{-2}</math>
 
|-
 
|Fe ||1.529 ||0.01384 ||0.02116136 ||<math>6.60*10^{10}</math>  ||<math>1.06*10^{-2}</math>
 
 
|-
 
|-
 
|}
 
|}
Line 111: Line 107:
 
|Stefan-Boltzmann Constant
 
|Stefan-Boltzmann Constant
 
|<math>T_{equ}</math> (K)
 
|<math>T_{equ}</math> (K)
|-
 
|Al ||0.002  || 0.00000628 ||<math>5.6704*10^{-8}</math> ||487.5
 
 
|-
 
|-
 
|W  ||0.002  ||0.00000628 ||<math>5.6704*10^{-8}</math> ||329.8  
 
|W  ||0.002  ||0.00000628 ||<math>5.6704*10^{-8}</math> ||329.8  
|-
 
|Ti ||0.002  ||0.00000628 ||<math>5.6704*10^{-8}</math>||433.4
 
|-
 
|Fe ||0.002  ||0.00000628 ||<math>5.6704*10^{-8}</math>||415.2
 
 
|-
 
|-
 
|}
 
|}
Line 126: Line 116:
 
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.
 
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.
  
===Al===
 
Equilibrium temperature for Al, unit is K. (Melting point of Al is 933K)
 
 
{| border="1" cellpadding="20" cellspacing="0"
 
 
|-
 
|
 
|2(mm)
 
|4(mm)
 
|6(mm)
 
|8(mm)
 
|10(mm)
 
|-
 
|-
 
|0.001RlK || 487.5 K||344.7 K || 281.5 K|| 234.8 K|| 218.0 K
 
|-
 
|0.005Rl K|| 729.1 K||515.5 K||420.9 K||364.5 K|| 326.0 K
 
|-
 
|0.01Rl K || 867.0 K|| 613.1 K|| 550.6 K||433.5 K||387.7 K
 
|-
 
|}
 
 
[[Image:Equilibrium Temperature for Al  .jpg]]
 
  
 
===W===
 
===W===
Line 173: Line 140:
  
 
[[Image:Equilibrium Temperature for W.jpg]]
 
[[Image:Equilibrium Temperature for W.jpg]]
 
===Ti===
 
Equilibrium temperature for Ti, unit is K. (Melting point of Ti is 1941K)
 
 
{| border="1" cellpadding="20" cellspacing="0"
 
 
|-
 
|
 
|2(mm)
 
|4(mm)
 
|6(mm)
 
|8(mm)
 
|10(mm)
 
|-
 
|-
 
|0.001Rl || 433.4 K||306.5  K|| 250.2 K||216.7 K||193.8 K
 
|-
 
|0.005Rl || 648.1 K|| 458.3 K|| 374.2 K|| 324.0 K|| 289.8 K
 
|-
 
|0.01Rl  ||770.7  K||545.0  K|| 445.0 K||385.4 K|| 344.7 K
 
|-
 
|}
 
 
[[Image:Equilibrium Temperature for Ti.jpg]]
 
 
===Fe===
 
Equilibrium temperature for Fe, unit is K. (Melting point of Fe is 1811 K)
 
 
{| border="1" cellpadding="20" cellspacing="0"
 
 
|-
 
|
 
|2(mm)
 
|4(mm)
 
|6(mm)
 
|8(mm)
 
|10(mm)
 
|-
 
|-
 
|0.001Rl || 415.2 K|| 293.6 K||239.7 K||207.6 K|| 185.7 K
 
|-
 
|0.005Rl || 620.8 K||439.0  K|| 358.4 K||310.4 K|| 277.6 K
 
|-
 
|0.01Rl  ||738.3  K|| 522.1 K||426.3 K||369.2 K||330.2 K
 
|-
 
|}
 
 
[[Image:Equilibrium Temperature for Fe.jpg]]
 
 
 
  
 
==5. Results for worst cases ( 80mA peak current,1000Hz frequency, 50ns pulse width ) ==
 
==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.
 
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.
 
===Al===
 
Equilibrium temperature for Al, unit is K. (Melting point of Al is 933K)
 
 
{| border="1" cellpadding="20" cellspacing="0"
 
 
|-
 
|
 
|1(mm)
 
|2(mm)
 
|3(mm)
 
|4(mm)
 
|-
 
|-
 
|0.00005Rl || 548 K||387 K || 317 K||274 K
 
|-
 
|0.0001Rl  || 652 K||461 K|| 376 K||326 K
 
|-
 
|0.0005Rl  || 975 K|| 689 K||562 K|| 487 K
 
|-
 
|}
 
 
[[Image:Al_2.jpg|down|800px]]
 
  
  
Line 274: Line 168:
  
 
[[Image:W_2.jpg|down|800px]]
 
[[Image:W_2.jpg|down|800px]]
 
 
 
===Ti===
 
Equilibrium temperature for Ti, unit is K. (Melting point of Ti is 1941K)
 
 
{| border="1" cellpadding="20" cellspacing="0"
 
 
|-
 
|
 
|1(mm)
 
|2(mm)
 
|3(mm)
 
|4(mm)
 
|-
 
|-
 
|0.00005Rl || 487 K||334 K || 281 K|| 244 K
 
|-
 
|0.0001Rl || 597 K||410 K||335 K|| 290 K
 
|-
 
|0.0005Rl || 1226 K|| 867 K|| 500 K|| 433 K
 
|-
 
|}
 
 
[[Image:Ti_2.jpg|down|800px]]
 
 
 
 
===Fe===
 
Equilibrium temperature for Fe, unit is K. (Melting point of Fe is 1811 K)
 
 
{| border="1" cellpadding="20" cellspacing="0"
 
 
|-
 
|
 
|1(mm)
 
|2(mm)
 
|3(mm)
 
|4(mm)
 
|-
 
|-
 
|0.00005Rl || 467 K||330 K || 270 K|| 233 K
 
|-
 
|0.0001Rl || 555 K||393 K||321 K||278 K
 
|-
 
|0.0005Rl ||830  K|| 587 K|| 479 K||415 K
 
|-
 
|}
 
 
[[Image:Fe_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) ==
 
==6. More results for worst cases ( 80mA peak current,1000Hz frequency, 50ns pulse width for Θ₀ = 0.2,0.4,0.6 dgrees) ==
Line 330: Line 173:
 
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.
 
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.
  
===Al===
 
Equilibrium temperature for Al, unit is K. (Melting point of Al is 933K)
 
 
{| border="1" cellpadding="20" cellspacing="0"
 
 
|-
 
|
 
|2(mm)
 
|3(mm)
 
|4(mm)
 
|-
 
|-
 
|Θ₀=0.2˚, t=3.9E-5 RL=3.5μm|| 364 K|| 298 K|| 258 K
 
|-
 
|Θ₀=0.4˚,t=1.35E-4RL=12μm  || 497 K|| 406 K|| 351 K
 
|-
 
|Θ₀=0.6˚,t=2.8E-4RL=25μm  || 596 K|| 487 K|| 422 K
 
|-
 
|}
 
 
[[Image:Al_3.jpg|down|800px]]
 
  
 
===W===
 
===W===
Line 375: Line 197:
 
[[Image:W_3.jpg|down|800px]]
 
[[Image:W_3.jpg|down|800px]]
  
===Ti===
 
Equilibrium temperature for Ti, unit is K. (Melting point of Ti is 1941K)
 
 
 
{| border="1" cellpadding="20" cellspacing="0"
 
 
|-
 
|
 
|2(mm)
 
|3(mm)
 
|4(mm)
 
|-
 
|-
 
| Θ₀=0.2˚, t=3.9E-5 RL=1.4μm|| 329 K|| 264 K|| 229 K
 
|-
 
| Θ₀=0.4˚,t=1.35E-4RL=4.8μm || 442 K|| 360 K|| 312 K
 
|-
 
| Θ₀=0.6˚,t=2.8E-4RL=10μm  || 530 K|| 432 K|| 375 K
 
|-
 
|}
 
 
[[Image:Ti_3.jpg|down|800px]]
 
 
===Fe===
 
Equilibrium temperature for Fe, unit is K. (Melting point of Fe is 1811 K)
 
 
 
{| border="1" cellpadding="20" cellspacing="0"
 
 
|-
 
|
 
|2(mm)
 
|3(mm)
 
|4(mm)
 
|-
 
|-
 
| Θ₀=0.2˚, t=3.9E-5 RL=0.7μm|| 310 K|| 253 K|| 219 K
 
|-
 
| Θ₀=0.4˚,t=1.35E-4RL=2.5μm || 423 K|| 346 K|| 299 K
 
|-
 
| Θ₀=0.6˚,t=2.8E-4RL=5μm    || 508 K|| 415 K|| 359 K
 
|-
 
|}
 
 
[[Image:Fe_3.jpg|down|800px]]
 
  
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+
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Latest revision as of 03:32, 8 June 2010

Positrons


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: [1].

Elements Radiation Lengths [math] (g/cm^{2} )[/math]
W 6.76


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: [[2]])

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

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)

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

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)

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

down

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)

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


down


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