How many volts/amps for an electric arc between points?

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Discussion Overview

The discussion revolves around calculating the voltage and current required to maintain a continuous electric arc between two conductive points within a hollow insulating cylinder. The context includes theoretical considerations of electric arc formation, ionization of air, and the effects of temperature on voltage requirements.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant seeks guidance on calculating the necessary voltage and current for an electric arc between two sharp metal contacts in a cylinder.
  • Another participant states that the breakdown voltage for air is approximately 3 million volts per meter, suggesting that around 300,000 volts is needed to initiate the arc over a 10 cm distance, noting that this value can vary with environmental conditions.
  • A participant expresses an intention to use high voltage generators to achieve the required voltage, while questioning how temperature might affect the voltage needed for discharge in a closed system.
  • There is a disagreement regarding the required voltage, with one participant asserting a figure of 150,000 volts for 10 cm, while another clarifies that this is the voltage per conductor, leading to a total of 300,000 volts.
  • Concerns are raised about the effects of ozone and nitrous oxide buildup on the contacts, as well as the potential influence of electron and ion drift on the calculations.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the exact voltage required for the electric arc, with differing views on the breakdown voltage and the implications of environmental factors. The discussion remains unresolved regarding the precise calculations and effects of temperature.

Contextual Notes

Participants acknowledge that the breakdown voltage can vary with pressure and moisture content, and there are uncertainties regarding the effects of temperature on voltage requirements and the influence of ion drift.

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Can someone help me find these calculations or give me a point in the right directions?

If I have a hollow insulating cylinder (has a diameter of 5 cm and a length of 14 cm) with two (conductive) sharp metal point contacts at each end (measuring 2cm each leaving 10 cm exactly between the points). Inside the cylinder is normal atmosspheric pressure of standard air.

How do I calculate how many volts and amps I require to have a continuous electric arc between the contacts?

Thanks for your help and all answers are welcome.

:smile:
 
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Thats a rather complicated question. To get any current flow at all you must first ionize, or breakdown the air. The breakdown voltage for air is roughly 3 million volts per meter - that varies with pressure and moisture content - so you need about 300,000 volts to start the arc. After that things get more complicated.
 
First of all thank you mheslep for the speedy reply, I appreciate it.

I could use an array of extremely high voltage generators to produce 300,000 to 330,000 volts and a continuous 10cm electric arc (assuming a standard pressure and temperature air mix), but this would still be a process of trial and error.

One more question is:

Assuming the cylinder is closed to outside influence and the walls are a suitable insulator to temperature, would the voltage required to generate the discharge should drop when the temperature increases on a steady curve, just like the rising temp of a high voltage traveling arc (Jacob’s ladder)?

I still have not found how many amps are required but I will keep searching. :)
 
mheslep said:
Thats a rather complicated question. To get any current flow at all you must first ionize, or breakdown the air. The breakdown voltage for air is roughly 3 million volts per meter - that varies with pressure and moisture content - so you need about 300,000 volts to start the arc. After that things get more complicated.

I get half that; 150KV. How did you arrive at 300?
 
3 × 10^6 V/m

I have found several sources like hypertextboox qouteing the average votage for dielectric breakdown of air around 3 × 10^6 V/m I took this as a base line and divided 3,000,000 by 100 then mutliplied it by 10 to arive at 300,000 volts for every 10cm.

Exclueding the build up of ozone and nitrous oxide on the sharp metal contacts causing corrision of course.

Where did you get 150,000 volts for 10cm?

as always thanks for the reply.
 
Doh! My mistake. 150KV per conductor. 300KV total, of course. But can one really ignore electron and ion drift?
 

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