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

  1. 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:
     
  2. jcsd
  3. mheslep

    mheslep 3,572
    Gold Member

    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.
     
  4. 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 travelling arc (Jacob’s ladder)?

    I still have not found how many amps are required but I will keep searching. :)
     
  5. I get half that; 150KV. How did you arrive at 300?
     
  6. 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.
     
  7. Doh! My mistake. 150KV per conductor. 300KV total, of course. But can one really ignore electron and ion drift?
     
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