I'm preparing research about high voltage electric distribution for my university but I don't understand why the arc only happen when current is flowing.

I made my internship in an electric distribution company and they told me you can't use disconnectors when a circuit is on load; you have to use high voltage circuit breakers first and they will extinguish arc and then you can use disconnectors to cut off voltage.

First I thought that happens because of capacitive & inductive loads but if that's true it shouldn't happen when load is ohmic but basically short circuits are ohmic load (phase to earth short circuit) and they cause arc too.

Also electric company's voltage values : 31.5 kV and 154kV . Load current 100-300A , short circuit current 400-1000A and they use 10 ohm resistor for earth.

They use SF6 gassed circuit breakers and I remember something about "arc extinguish when phase angle is 180" (V=0) and something about current voltage phase difference.

  • \$\begingroup\$ Your last, added note should help point a direction. This is AC, not DC, and twice each full cycle the voltage and current (not necessarily at the same time) goes through zero. The SF6 gas (used to be oil) acts to quench when the current reaches its zero -- if it quenched early (as vacuum interrupters may), then there is a high risk of over-voltage events. There cannot be an arc if the current is zero, since the arc is created by the flow of charges (the definition of "non-zero current.") I'm sure I'm probably misunderstanding your real question, though. \$\endgroup\$ – jonk Sep 6 '19 at 3:35
  • \$\begingroup\$ I thought same thing too but voltage-current difference only happen if load is reactive but short circuits are active power so they shouldn't cause voltage-current difference . Also normally main grid shouldn't have reactive power , like this why factories use capacitors to stabilize their reactive power otherwise they will break whole grid. But you must right about vacuum interrupters can cause high risk of over voltage because I read same thing in catalog for vacuum interrupters \$\endgroup\$ – Mordecai Sep 6 '19 at 3:43
  • \$\begingroup\$ I believe there's always some imbalance. Particularly during load shedding events. \$\endgroup\$ – jonk Sep 6 '19 at 4:05
  • \$\begingroup\$ at what short-circuit magnitude does the energy stored in the grid inductance become a problem? the inductance causes a phaseshift. \$\endgroup\$ – analogsystemsrf Sep 6 '19 at 8:15

Arcing when an electric switch is operated is a very complex subject. Switching off a high load lowers the contact pressure and separates 2 or more metal contact pieces. During this events, the resistance increases. This can cause very hot spots, gas plasma resp. accelerated electrons in the air layer between the minimal distance between the 2 contacts. Dependent on the voltage and material, the resulting UV radiation may free more electrons from one or both contact metals via the photoelectric effect. Those electrons could also free electrons and ions if they hit the positive metal dependent on free waylength between hitting the gas molecules (air pressure) and kinetic energy resp. voltage. The positive ions (e.g. oxygen, palladium) and the electrons each are shielding the negative and positive poles to some extent. The plasma may generate high frequency oscillations. Tunneling is also involved.

All these points can cause the current to continue flowing through the hot plasma. So high direct current is not easy to interrupt even with pure ohmic load.

Direct current is much more difficult to interrupt if any inductivity is involved, and in reality there is no circuit without inductivity: transformers, motors, coils of net filters, but also pure wires are inductivities. Compensation with capacities is never intended to 100% degree because of resonance effects.

Since the current through an inductivity is steady, it can not be interrupted at once but must be allowed to continue to flow even after the begin of the separation of the metal contacts in the switch. i(t-) = i(t+) means that the same current which is flowing just before switching off will also flow just after separating the metal contacts.

This is why every circuit diagram with a voltage source, an inductivity and a switch (with or without a resistor) all in series is a contradiction eo ipso, but it can be found in some (text) books/scripts.

To switch a circuit when no current is flowing is much more easy since no hot zones and no plasma will be present.

Edit: Didn't see SF6 is used. This gas decreases the mentioned problems to some extent.

  • \$\begingroup\$ So high current is the problem , is it would be same if power DC instead of AC ? \$\endgroup\$ – Mordecai Sep 6 '19 at 8:47
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    \$\begingroup\$ In general, DC is much more difficult to switch off. This is why the rating printed on switches is lower for DC (voltage and/or current) then AC. \$\endgroup\$ – xeeka Sep 6 '19 at 13:18
  • \$\begingroup\$ Okay now I understand also I'm really thankful for answers ,it cleared everything but I want to make sure I understand it correctly : When I cut off switch ,resistance slowly raises so small amount of current still flows and this causing air (or whatever inside) to turn into plasma then because of plasma switch lose resistance so current raise more ... But even I cut the current without any arc then it will cause high risk of over voltage so best option is let the arc happen then use SF6 to extinguish arc ? It's like the first law of thermodynamics , I have to convert energy into something \$\endgroup\$ – Mordecai Sep 6 '19 at 18:14

Short circuts in general are NOT resistive! In fact most of them are very highly reactive. Consider isolated generator supplying active power to the load, suppose (bolted i. e. 0 impedance) fault occurs somewhere on the transmission line, then load will be shot circuited and no power will be delivered to it. Generator will be feeding fault through transmission line, which are mostly reactive and so are generator windings. In fact short-circuit studies often neglects the resistance and approximates fault-feeding network as purely reactive. You're true in saying that if there were no energy storing elements (L & C), there'd be absolutely no arcing. In presence of L however, in language of elementary circuit theory, if fault occurs at t=0 then, iL(0-)=iL(0+), and due to such constraint, arcing takes place (this is not complete reason but serves the purpose). The physics are well explained in other answer!

NOTE : 'Arc' (after it happens) IS purely resistive (but remaining network isn't), this means arc current is proportional to "voltage across arc". This might've confused you thinking that short-circuts are resistive.

  • \$\begingroup\$ Even without any L and C, there could be arcing due to the Joule heat and plasma at the contact zone. In general, the arc current is not proportional to the arc voltage because of the non - linearity of the plasma. \$\endgroup\$ – xeeka Sep 6 '19 at 13:22

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