Difference between revisions of "HRRL Positron Rotating W Target"

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<math>M_{\mbox{Al}}= \rho \times V = 2.7 g/cm^3 (0.2 cm)\times \pi</math>
 
<math>M_{\mbox{Al}}= \rho \times V = 2.7 g/cm^3 (0.2 cm)\times \pi</math>
  
−
<math>M_{\mbox{Al}}= \rho \times V = 2.7 g/cm^3 (0.2 cm) \times \pi ( 9" 2.54 cm/")^2 =2.7 g/cm^3 0.2 cm \pi 523 cm^2 = 886 g</math>
+
<math>M_{\mbox{Al}}= \rho \times V = 2.7 g/cm^3 (0.2 cm) \times \pi ( 9 in 2.54 cm/in)^2 =2.7 g/cm^3 0.2 cm \pi 523 cm^2 = 886 g</math>
  
−
<math>I_{\mbox{Al disk}} =886 g \times 523 cm^2 = 463423 g cm^2 \times (1 oz/28 g) (1 "/2.54 cm) (1"/2.54 cm) = 2565 oz-in^2
+
<math>I_{\mbox{Al disk}} =886 g \times 523 cm^2 = 463423 g cm^2 \times (1 oz/28 g) (1 in/2.54 cm) (1in/2.54 cm) = 2565 oz-in^2
 
</math>
 
</math>
  

Revision as of 03:52, 8 April 2011

Beam Heat Load

Accelerator Settings
beam energy 10 MeV
Rep Rate: 300 Hz
I_peak: 200 mA/pulse
pulse width: 0.1 [math]\mu[/math]s (100 ns)
Vacuum: [math]10^{-8}[/math] Torr
[math]200 \frac{\mbox{mA}}{\mbox{pulse}} \times 300 \frac{\mbox{pulses}}{\mbox{sec}} \times 0.1 \mu \mbox{s}= 60 \mu A[/math]

Time on for one second is 30 us. and Time off is 0.999999 s The average current 0.999999 * 0 + 0.000001*300*200mA = 60[math]\mu[/math]A.

P = (10 MeV) [math]\times[/math] (60 [math]\mu[/math]A) = 600 Watts.

Target Chamber design

There is an 18" space in Z for the target chamber. If you want to rotate a disk on the 5" wide motor you will need at least a a 10" x 10" area.

So we could try as a first design a chamber which is 18" x 10" x 10". We need the following

  • 2 beam flanges flush with the walls of the chamber.
  • 2 flanges to allow access for hands to work on the target motor inside. It should able to fit hands through.
  • a flange on top which will allow a 10" diameter target disk to come out.
  • a flange on the bottom for the ion pump.

Gate Valve - Target distance

The distance from the front of the gate valve to the most beam upstream part of the target chamber is

88.43 cm

The gate valve closes in 0.7 sec.


So water would need to travel at a speed of 88.43 cm/0.7 sec = 1.26 m/sec

Properties of Tungsten

  • Melting Point = 3695 K.
  • Heat Capacity = [math](25~ ^{o}C)~ 24.27 ~ J ~ mol^{-1} ~ K^{-1}[/math]


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

Tungsten Temperature as a function of heat load

  • IAC beamline pressure = [math]10^{-8} [/math] Torr
  • 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

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. Number of particle per second is: 


                           [math]  N = \frac {f \times I \times \Delta t}{e} [/math]


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 a electron comes from two parts: 1) the emission of electromagnetic radiation arising from scattering in the electric field of a nucleus (bremsstrahlung); 2) Collisional energy loss when passing through matter. Bremsstrahlung will not contribute to the temperature, because 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: Tungsten Stopping Power.

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] [math] t~( g~cm^{-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)})\times(Number~of~electrons~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} =  A \sigma 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 

                                        [math] T = [  \frac{P_{dep}}{A\sigma} ]^{1/4} = [  \frac{N*E_{(dep~one)}}{A\sigma} ]^{1/4} [/math]



Vendors

Tungsten Disks

http://www.alibaba.com/product-gs/346924116/Tungsten_Disk_With_Screw.html


http://www.zlxtech.com.cn/products.asp?productcode=7401000404

http://www.cleveland-tungsten.com/


If the motor has a radius of 2.5" and I want to shield the motor from the beam using 2" of Pb and the beam is about 1" in diameter then I want a Tungsten disk that is

5.5" in radius = 13.75 cm in radius ~ 30 cm in diameter

Then density of Tungsten is

19.25 g/cm^3


19.25 g/cm^3 \times \pi (30 cm)^2 = 54,428 g/cm

= 8.7 kg if it is 0.16 cm thick

The moment of inertia for a disk about its center is MR^2/2

So I = 8.7kg \times (0.3 m)^2/2 = 0.4 kg m^2

I should look into a ring of tungsten attached to a less dense material.

VX-U42 torque is about 50 in-lbs = 5.65 N m

Max acceleration would be

[math]\alpha = \frac{5.65 N m}{0.4 kg m^2} = 14.4 rad/sec \times \frac{1}{2 \pi} = 2.3 cycles/sec[/math]

Brushless Motor

Target chamber: MDC vacuum products can send a prefab one you just need to weld up.

jbrooks@mdcvacuum.com

MDC does not have a fast closing gate vavle, they take a few seconds to close. Model GV1500

rick@empiremagnetics.com:

The VX grade motors, with dry lubes have been used in deep space (Wake Shield Module) where I'm told they were trying to get 10^-12 torr.


We need 10^{-8 }Torr

Vacuum lab grade motor U42.

http://www.empiremagnetics.com/prod_vac/prod_vac_vx_stepper_frame42.htm


Torque

The MRI rotary Unions require 2 in-oz of torque each

The U42 has a minimum static torque of 375 in-oz (1 Nm/8.851 oz-in) = 42.4 Nm

The Momentum of inertia of a thin disk about its axial axis is

[math]MR^2/2[/math]

Moment of Inertia of a ring:

[math]MR^2[/math]

The target will be a disk of Aluminum with an outer ring of tungsten.

[math]I_{\mbox{Al disk}} =M_{\mbox{Al}} R^2[/math]


[math]M_{\mbox{Al}}= \rho \times V = 2.7 g/cm^3 (0.2 cm)\times \pi[/math]

[math]M_{\mbox{Al}}= \rho \times V = 2.7 g/cm^3 (0.2 cm) \times \pi ( 9 in 2.54 cm/in)^2 =2.7 g/cm^3 0.2 cm \pi 523 cm^2 = 886 g[/math]

[math]I_{\mbox{Al disk}} =886 g \times 523 cm^2 = 463423 g cm^2 \times (1 oz/28 g) (1 in/2.54 cm) (1in/2.54 cm) = 2565 oz-in^2 [/math]

[math]Torque = 42.4 Nm = I \alpha[/math]

\alpha = 42.4 Nm/(0.04 kg m^2))=1.9 rad/sec^2

Rotary coupler/Union

ORNL rotating tungsten target

Full Name: Thomas J. McManamy Email Address: mcmanamytj@ornl.gov Phone Number: 865-576-0039 Fax Number: 865-241-6909 Postal Address: OAK RIDGE NATIONAL LABORATORY PO BOX 2008 MS6476 OAK RIDGE TN 37831-6476

http://www.ornl.gov/info/reporter/no116/nov09_dw.htm#revolution

http://www.dsti.com/products/?gclid=CPWEvP6ogagCFRs5gwod6SSXtA

http://www.gat-mbh.de/index.php?page=rotovac-en&group=produkte-en:rotating_unions-en

http://www.tengxuan.net/old/product.asp?Catalog=8

http://www.rotarysystems.com/?gclid=CID4r6z22acCFQM6gwodoQVQ9w 

   http://rotarysystems.com/series-006

Target specifications

Torque

Rotational Speed

Mechanical design

An Aluminum disk will be machined to have several Tungsten targets attached.

The disk has a radius of about 9 inches. The circumference will be 2 \pi r \approx 45 inches. If the targets are 1" wide with 1" gaps, then you will have about 20 targets.


If you rotate the target at 1 Hz then in one second you can hit 20 targets. This means your rep rate can be 20 Hz.

If you rotate the target at 50 Hz then your rep rate can be 1 kHz.

But since you will have 20 gaps (empty target positions) then you could run at 2 kHz.

So a target rotating 25 Hz can accept a rep rate of 1 kHz.


The HRRL easily runs at 300 Hz. For a target with 20 foils you would run the motor at 300/40 = 7.5 Hz.

Heat load

You will need at least 2 holes in the Tungsten disk which you can use as an empty target. The Max pulse rate of the Linac is 1 kHz. So a 500 Hz motor would be the Max speed needed if you have 2 Windows.


How long can a 1.6 mm thick piece of Tungsten that has an area of 2.5" x 2.5" be in a 600 Watt heat source before melting.

Specific Heat Capacity = [math] 130 \frac{J}{kg K}[/math]

Thermal Conductivity = [math] 173\frac{W}{m K}=100\frac{BTU}{hr ft F}[/math]

The Melting point of Tungsten is 3600 Kelvin.

density = 19.25[math]\frac{g}{cm^3}[/math]

Volume = [math](.16 cm)(2.5 cm)^2 = 1 cm^3[/math]

Mass = [math]1 cm^2 \times 19.25 g/cm^3 = 19.25 g :\lt math\gt Q = C_m m \Delta T = 130 \frac{J}{kg K} \times 0.967 kg \times (3600-300 K) = 414843 J = 600 J/s *t [/math] Watt = J/s

[math]\Rightarrow t= 691 seconds.[/math]

If Zinc

C_m = 0.39 J/g/K Density = 7.13 g/cm^3 \Right arrow M = 7.13 \times 50.3 = 358 g, Melting point 693 K
Q = 0.39 \times 0.358 \times 393 = 55 Joules = 10^3 Watts * t \Rightarrow t = 0.055 s \Rightarrow 18 Hz.

Now You need to calculate how quickly water can absorb heat

Water properties
C_m = 4.12 J/g/K , Density = 1 g/cm^3 assuming aflow rate of 1 liter/sec \Rightarrow mass = 1 g, Boiling point water = 373.15


Tungsten's specific heat capacity is about 30 times larger than water. So for the same mass and power input water's temperature will change a factor of 30 more than tungsten.



In 1 second Tungsten's temperature change will be

\frac{600 J}{130 \times 0.967} = 5K


Copper coils which have water flowing through them are soldered to the Tungsten. The water is a heat sink transferring heat from the Tungsten to the water.

Heat transfer by conduction
[math]\frac{Q}{t} = \frac{\kappa A \Delta T}{d}[/math]
[math]\kappa \equiv[/math] Heat Conductivity

Q

Q: Does the W target has to be in the vacuum? 
   What if we have it in the air, at the ends of the beam line we have very thin vacuum window? 
   In photofission experiment we had 1 mil vacuum window. under 
   15 MeV e- beam, only 1/1000 electron will interact with the window, produce bremsstrahlung photon.

Q: Can we make positron target that is inner part of the vacuum box, then we rotate all the box. Heat will be transfered to box, 
   box can cooled by water from outside. 
   May be box does not have to rotation movement? I can do other moments like up-down and right-left.


  I think we can make a Tungsten target as a part of the beam line. That means outer part Tungsten target will be beam pine. 
Then we can use water cooling to cool our Tungsten target from outside. 
Before we had all the electrons hitting Faraday Cup, which did not melt the copper even when there was not cooling water 
circling around FC. 
Tungsten has higher melting point, plus we run cooling water around Tungsten target, then Tungsten shouldn't melt.


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 PositronsGo Back

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