# Breakdown voltage of humid air in uniform electric field

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1. Sep 10, 2017

### Ali_A

Could any one please let me know if breakthrough spark can happen, or, in general, what will happen in the below system? Note 1: the dielectric layers have a much higher breakthrough voltage than humid air. Note 2: the intensity of the electric field that develops in the flow of the humid air is higher than the humid air breakthrough voltage.

Last edited by a moderator: Sep 14, 2017 at 10:40 AM
2. Sep 10, 2017

### davenn

initial thoughts

for a start, you will need to find out the breakdown voltage for the dielectric material used taking into account its thickness ( x2)
the distance between the dielectric layers so that you can then find out the breakdown voltage for layer of humid air of that thickness
will the humidity of that air vary ?, of so then so will the breakdown voltage for it

3. Sep 11, 2017

### Ali_A

Hi Davenn,

I do not know if I really understand your point. But let's assume that the breakdown voltage of dielectric layers are high enough to avoid spark through them. And, the humidity is constant.

I would like just to know if how or whether arcing can occur in the humid air flow if the intensity of the electric field that develops in the flow of the humid air is higher than tits breakdown voltage?

4. Sep 12, 2017

### Rive

I think that's the point where it's wrong. You take this as if no current would flow at all.

Although no spark will happen (spark requires a minimal current), the humid air will be partially ionized, and will produce a (really small) current enough to keep the voltage between the surfaces of the dielectric layers around the breakdown voltage of the air flow.

At least I think so.

5. Sep 13, 2017 at 10:23 AM

### Ali_A

Thank you Rive!
Your explanation sounds logical and the point (that a current would flow) is what I am looking for. So, if we would have a current as you said, finally the voltage between the dielectric layers will discharge and the current will be vanished, what do you think? What would the final state of the system be?

6. Sep 14, 2017 at 4:18 AM

### Rive

I don't know. You should consider this setup as a circuit which is built by resistors at the GOhm scale.
This caliber requires really special knowledge and many thing depends on the exact technology. Even a simple fingerprint can change everything.

For example, in a typical real-world application such setups tends to gather dust//dust bunnies, which greatly changes the parameters.
Or if the air flow is fast enough then it'll blow out slightly ionized air. I don't know how should that be calculated.

7. Sep 14, 2017 at 7:37 PM

### Ali_A

Thank you!

Hopefully I can get the funding to build the setup and let you know the result in future. In the meanwhile I have to study a lot as this field is competently new to me.

8. Sep 15, 2017 at 9:22 AM

I am confused by your comment " I would like just to know if how or whether arcing can occur in the humid air flow if the intensity of the electric field that develops in the flow of the humid air is higher than it's breakdown voltage?"

How do you define "breakdown voltage"?

I.e. "Although air is normally an excellent insulator, when stressed by a sufficiently high voltage (an electric field strength of about 3 x 106 V/m or 3 kV/mm), air can begin to break down, becoming partially conductive."

Basically you are asking if the breakdown voltage (Voltage where arcing occurs) is it's breakdown voltage?

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In your diagram, you have 2 dielectric materials, the Air counts as one. The breakdown voltage is the where the energy stored in the material, proportional to its dielectric constant, exceeded it dielectric strength. Even though no conventional current flows, this is essentially a voltage divider.

For a static system most of this can be calculated easily. The trick here, IMO, will be the affects of the FLOW. As turbulence has peculiar impacts on moisture distribution, heat, and pressure -- all of which will affect both Dielectric constant and strength. AND lastly - turbulence is chaotic - so not predictable.

9. Sep 15, 2017 at 5:11 PM

### Tom.G

A somewhat related demonstration that I saw at a trade show many years ago was an arc in free air with no apparent voltage source. The arc was created with a high powered LASER focused to a small region. The electric field of the light was high enough to actually break down the air and show up as an arc!

I wouldn't have believed it if I hadn't seen it. That must have been a LOT of power. The LASER itself was about 4 to 6 feet long and its enclosure parhaps 16 inches square... sitting on a substantial steel stand.