E field distribution in spark discharge

1. Jun 1, 2012

fnsaceleanu

Hi,

I just joined this forums.
I'm currently doing my master thesis in mechanical engineering, however, my topic is related to plasma physics.
As part of my results, I need to simulate an audio frequency electric discharge in a spark plug in air, in Ansys HFSS software.

The discharge is at 35 kHz, and an oscilloscope obtains the voltage between electrode and ground plate, and current through the ground plate, every nanosecond.

Is this info enough to estimate the dielectric constant and loss tangent of the air plasma, every nanosecond?
I'm interested to know how the electric field varies in space, from the data obtained through the oscilloscope.

Any help/suggestion is appreciated.
Thanks!

Florin

2. Jun 6, 2012

Jano L.

Phys. Rev. 83,3 (1951)

1951 Finkelburg Segal The Potential Field in and around a Gas Discharge

Jano

3. Jun 6, 2012

eman3

Florin,
I'm not really qualified to answer this, so take my thoughts with a grain of salt. That said:
Dielectric constant should be related to the rate of change of current and voltage between electrode and ground plate prior to the arc discharge. At that point I think the spark plug would be acting as a very low-value air-dielectric capacitor. If you know how fast this "capacitor" charges given a particular applied voltage, and you know the dimensions of the "capacitor" (the size of the electrode and ground plate, and distance between them) you should be able to determine the dielectric constant.

I don't know what a loss tangent is, so I can't help you there...

4. Jun 8, 2012

fnsaceleanu

and Eman, could you show me the formula that relates the dielectric constant to the rate of change of current and potential between electrode and ground? (I have the current rate of change and potential difference)

I went another way around it, and assumed that plasma conductivity, omega, (E field divided by curreny density J) is simply maximum E field divided by maximum measured current through the wire. Then I found the dielectric constant from conductivity value and the AC frequency.

But as you said, there's probably a better way to calculate that.
Thanks!

5. Jun 8, 2012

fnsaceleanu

Ok, I can see how to relate the permittivity to the rate of change of voltage and current.. formula for a capacitor, where the capacitance is a function of the permittivity and geometry.
Is this what you meant?

Also, the loss tangent is the angle between the real permittivity and its imaginary part.
In plasmas, the permittivity has an imaginary part that is related to absorption/reflection of the electromagnetic waves.

So now the question is how to use this complex permittivity... I'm guessing that's why discharge gaps are modelled as resistor/capacitor, so you have a real and a fluctuating loss.

6. Jun 8, 2012

Bobbywhy

Just a comment: conventionally the audio frequency range is from 20 Hz to 20 kHz. 35kHz would be “ultrasonic” because we humans cannot hear it.

7. Jun 15, 2012

eman3

I'm not really sure what you're asking, nor am I really sure I'm the person to give you a definitive answer... but this seems like it might help:

http://www.avx.com/docs/Catalogs/cbasic.pdf [Broken]

Equation III relates the current to the capacitance and dV/dt, and capacitance contains the dielectric constant.

This discussion also seems relevant: