MetalicaRap:Physics Principles
Principles
Normally the flattening of blobs through repeated surface spot melting, but during negative machining some metal will be vaporised.
Problem; How long will it take to vaporize smallest cylinder of Titanium
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)
Cal
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: Pressure drop from evaporated material
Given;
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.
Cal
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?)
Discussion
Problem: Pressure from evaporated material
Not problem with tubo-pump but could be a problem with oil diffusion pump? what are typical oil diffusion pump pumping rates?. Calculation required!
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