MetalicaRap:Physics Principles

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Revision as of 05:45, 3 October 2012 by Rapatan (talk | contribs) (Principles)
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Problem A ; How long will it take to vaporize smallest cylinder of Titanium

Normally the flattening of blobs through repeated surface spot melting, but during negative machining some metal will be vaporised.

Speed calculation based on this scenario;

100KV at 11.5mA Beam power 1.15kW/<math>mm^2</math> or 115 kW/<math>cm^2</math>. 5-15µ deep beam energy beam spot dia. 7µ.

85% energy transfer from beam to metal powder

According to my tables the heat of evaporation of Ti is 4x104 J/cm3 (8.953 MJ/kg times 4.5x103 kg/m3)


Treat as cylinder. 15µ*7µ*7µ x<math>10^{-12} cm^3</math> or 600µ<math>^3</math>. 600 x<math>10^{-12}</math><math>cm^3</math> at 115k W*0.85%/<math>cm^2</math> in to 40µ<math>^2</math> spot same as <math>40*10^{-8}</math> <math>cm^2</math>.

Is 40*<math>10^{-8}cm^2</math>*115kW*0.85%= 0.04 W in to metal volume ,

How many joules to vaporize this volume; <math>4*10^{4}J/cm^3</math> for <math>600*10^{-12}cm^3</math>

<math>600*10^{-12}cm^3</math>* <math>4*10^{4}J/cm^3</math>=<math>2.4*10^{-5}</math> Joules

Getting energy of 0.04 joule in a second so it will take <math>2.4*10^{-5}</math>/0.04W= <math>60*10^{-5}</math> Seconds or 600µS to vaporize cylinder,

This is so little heat conduction not a problem so this will remain cold process.

Problem B: Pressure drop from evaporated material


Taking the above scenario (100KV at 11.5mA Beam power 1.15kW/<math>mm^2</math> or 115 kW/<math>cm^2</math>. 5-15µ deep beam energy beam spot dia. 7µ. 85% energy efficiency)

We get 40*<math>10^{-8}cm^2</math>*115kW*0.85%= 0.04 W energy from beam in to the metal target.


Vaporizing Titanium at with .04 W into metal volume evaporating at the price of <math>5*10^4 J/cm^3</math> evaporates <math>\frac{0.04W}{5*10^4 J/cm^3}=8*10^{-7} cm^3</math>

With the density of metallic Ti 4.5 g/cm3 the above volume amounts to 3.6x10-6 g. And with a molar mass of Ti 204.37 g/mol it is 1.76x10-8 mol.

Now: ONE mole of a gas takes a volume of 24 Liter at a pressure of 1 atmosphere or 101,325 Pa = 1013.25 millibar.

Relative to this an amount of 1.76x10-8 mol vaporized titanium in a 1000 L (1 m3) volume will per second contribute to the pressure with the following value;

<math>\Delta P /S =\frac{1.76*10^{-8} mol/s}{1 mol}*\frac{24L}{1000L}*1013.25 mBar =4.43*10^{-7} mBar/s </math>

Thus the pump has to remove 4.3x10-7 millibars of pressure per second. Maybe this is not at all a problem. I don’t know about pumping capacities.

I just wanted to put attention to it. (loss of vacuum significant enough to lead to loss of beam quality?)

Problem C;Electron beam control via power supply effectiveness calculation.

Cal At switching frequency of 0.2Mhz ; the 416 2500pF output capacitors charged to 150V will contribute 1x10 -16 Joules after switch off [1] , for 1KW beam giving in metal for half a switching cycle 850W * 2.6 x10 -6 Secs = 2 x 10 -3 Joules . this would melt an extra 55µm cube of Titanium metal. This can be reduced by switching the beam off on unit cube earlier and by reducing printing speeds through lower beam energy.

Problem D;Tentative problem of shortage of supply of Indium and/or Selenium for solar cell manufacturing CIGS calculation.

Cal This method is very poor potentially misleading, but until we get something better here it is; taking average parts per million occurrence rate of Indium in crust and mass of crust, given 2µ m thick layer of indium necessary on CIGS solar cell.

What is the theoretical maximum power generation capacity of solar per year if all the indium recoverable on land was used in CIGS Solar cells?

What area would those solar cells cover?

Indium 0.049 parts per million, given mass of crust is 1.9 x10 22 , total indium in crust is 9.3 x10 14 Kg . Mine-able deposits are on land factor 0.3 ; 10% of land area factor 0.1 ; within a mile of surface factor 0.025 ; So Min able deposits 9.3 x10 14 Kg x 0.3 x0.1 x 0.25 = 6.9 x10 11 Kg , density of Indium is 7.31g per cm3 , So indium volume per Kg is 136cm3 , 1 cm3 covers 5000 cm2 in a 2µ thick layer. So area per Kg 136cm3 x 5000=680 000 cm2 or 68 m2 of CIGS solar per Kg of Indium, given a 15W CIGS panel is 0.63 m long 0.298 m wide so panel area is 0.18 m2 power per m2 is therefore 83watts per m2 of CIGS, Total theoretical power is 6.9 x10 11 Kg x 68 m2 per Kg x 83 watts per m2 = 3.93 x10 15 Watts. For poor climate 1800 peak sunshine hours a year, Total annual energy 3.93 x10 15 Watts x 60seconds x 60 minuits x 1800 hrs peak sunshine per year = 2.55 x10 22 Joules per year or 25 500 exa Joules

Current 2012 World annual energy usage from all sources is 600 exa Joules. So with CIGS producing 25 500 exa Joules it could easily ( 42 times over ) provide the required energy.

CIGS solar area of 6.9 x10 11 Kg x 68 m2 per Kg = 4.692 x10 13 m2 or 46 920 000 Km2 to produce 25 500 exa joules,

given total required energy is 600 exa joules so required area for CIGS is 1 117 000 km2, given total earth land area is148 940 000 , CIGS area would be 0.7% of total land area.

(others have quoted a lower figure of TerraWatts or x10 12 Watts ie 1000 times less than this) ( if the lowest quote in the terra watts range is used CIGS solar could provide 25 exa Joules of 600 exa joules ie 5% of energy demand).


[2] Maths formula Wiki authoring Help.