MetalicaRap:Physics Principles

Principles

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)

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 B: 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?)

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.

Discussion

[2] Maths formula Wiki authoring Help.