When potential voltage is applied, electrons flow from tool to workpiece. These electrons will collide with dielectric medium(water) resulting in ionisation of dielectric molecules.
H2O + e- = H+ + OH-
2H+ = H2 + e-
2OH- = H2O + 0.5 O2 + 2e-
These electrons released will move towards workpiece for removal. Please explain me how tool wear occurs?
Hello,
though some electron flow happens, most of the erosion process is not due to this (which would have to be considered 'electro-chemical etching'). Instead, most of the erosion in EDM is achieved via thermal effects - evaporating the 'meterial'. This evaporation happens on both sides of the spark gap, thus eroding not only the workpiece but also the tool.
First, there is no potential voltage because a voltage is the dif- ference between two potentials. In this way, they can assign a potential to each electrode and maintain its gap voltage.
According to their statement, the electrons would emerge from the tool, that is not true, because in the "normal case" the workpiece is cathodic. To explain it very simply, the "light" electrons tend to hit the tool (anodically), forming a space charge cloud that forms the larger-area anode flux point. Only when the positive ions of the dielectric hit the cathodic workpiece does the discharge channel form with a much smaller cathode base point. They can then observe the thermal behavior in the two electrodes (different current densities). The wear is thus generated primarily during ignition because the tool electrode is already heated and "removed" before, without the discharge channel.
@H.P.Schulze sir, I have seen some literature that takes workpiece as cathode and some that takes vice versa. However, during experiments, MRR is high if workpiece is made anode. Morever, considering equal K.E. of eletcrons and ions, electrons have mass 4 order less than ions, so their velocity is very high as compared to ions (equal K.E.). Electrons hit workpiece and tries to remove an atom. This is in atomic level. On macro scale, its seen that material removal takes place by melting and vaporization.
The explanation of the ablation processes at anode and cathode are fundamentally no different, whether they have anodic or cathodic tool poles. The "normal case" with pulse durations greater than 2 μs (depending on the material) requires an anodic polarity of the tool electrode because the primary wear decreases before and during the discharge channel generation as a percentage of the heating phase. In the heating phase, the removal of the workpiece is dominated by the higher current density at the cathode root point. At pulse durations of less than 1 μs, the described effect of the effect on the percentage is significantly larger, so that it makes sense to change the polarity.
Heating phase has a 70:30 ratio on anode and cathode (literature). Heating in EDM will be dominated by I^2*R*t. I think heating will also be more in anode. And considering similar melting point materials, anode will erode more (Shouldn't the workpiece be anode???). If we consider pulse duration less than 1 μs (achieved by RC circuit), material removal phenomena is both by melting and vaporization (literature). Fundamental of material removal is almost the same I believe.
In WEDM workpiece is anode and wire is cathode. Electron flow ( towards anode) is greater in first few microsec. and is responsible for material removal. After first few microsec (say 4) the movement of ions towards cathode is dominant causing erosion of cathode. Thus, in WEDM, ions are causing tool wear (cathode).
For WEDM (and also EDM) completely correct, only the times are not correct.
Again, I restrict it to a needle pulse, e.g. With 200 ns rise and 300 ns waste. The mentioned phase of the base point formation takes place in a period of 200 ns and also with the EDM it is not much higher. An exception is when I make the voltage rise very flat, which is useful for finish editing.
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