# Converter Temperture

# 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:

secondsSo, how many electrons we have in each second?

By Q=It, we have

` `

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, we have around 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 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 Technology### Table of Radiation Lengths

Note:These data is from Particle Data group,Link: [1].

Elements | Radiation Lengths |

W | 6.76 |

### 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) | |

W | 1.247 | 0.00676 | 0.00842972 |

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).

` `

## 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

= 2Aσ

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

` `

so

` T =[`

/(2Aσ)]^{1/4}

### Table of equilibrium temperatures

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

Elements | d (m) | 2A( | )Stefan-Boltzmann Constant | (K) |

W | 0.002 | 0.00000628 | 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 |

## 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 |

## 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

(Θ₀ = ), (Θ₀= ) and (Θ₀ = ) 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 |

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