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Calculating Energy in Arc (Difficult problem)

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LAN_403

HV Enthusiast

Tue Dec 23 2008, 04:35PM

Registered Member #15 Joined: Thu Feb 02 2006, 01:11PM
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Posts: 3068

I'm presently faced with a difficult problem, in which some of you guy's expertise may be able to help out with.

Basically, the problem is determining how much energy is dissipated in an arc event. Very little information is available regarding this, and what information is available, usually conflicts with other information on the same topic.

For an example problem, say you have 10uF of capacitance which is charged at 50kV. You initiate an arc discharge (assume 100% discharge of capacitor) through a 100 ohm resistor. Now how much energy is dissipated in the arc? Assume the arc path is very short.

The energy of the capacitor itself is 12.5kJ. However, if you go on available documentation, the actual arc is very small in voltage (about 100V) which maintains a constant voltage over much of the discharge, so by energy calculations, the actual energy discharge is about 50mJ and the majority of the energy dissipated in the resistors.

However, this doesn't make much sense to me, as I've seen what 12.5kJ can do, even when discharged through a 100 ohm resistor.

Any thoughts?

rp181

Tue Dec 23 2008, 04:42PM

Registered Member #1062 Joined: Tue Oct 16 2007, 02:01AM
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It would be hard to measure accurately with all the variables. Maybe you could use dimensional properties to measure the energy, like the arc volume/density, light emitted and sound.

Steve Conner

Tue Dec 23 2008, 05:41PM

Registered Member #30 Joined: Fri Feb 03 2006, 10:52AM
Location: Glasgow, Scotland
Posts: 6706

Use a high-speed DSO to sample the arc volt drop and arc current, multiply them together sample-by-sample, and integrate. Some good DSOs will do this and display a trace of energy, if not, you can do it in Excel.

Arc drops start to increase again at really high currents. For instance, when the current density gets above what the electrodes can emit, an extra voltage drop appears, the energy of which goes into blowing the electrodes into tiny pieces. When energies get really high, the plasma tries to expand at supersonic speeds and is prevented from expanding freely by its own shockwave, which increases the drop still more, and so on.

Antonio

Wed Dec 24 2008, 08:02PM

Registered Member #834 Joined: Tue Jun 12 2007, 10:57PM
Location: Brazil
Posts: 644

A capacitive discharge arc is oscillatory. The current goes back and forth over the nonlinear resistor untill all the energy is dissipated. There is inductance in the discharge path too, so it's incorrect to assume that the capacitor voltage is applied to a small (nonlinear) resistor.

HV Enthusiast

Thu Dec 25 2008, 07:58PM

Registered Member #15 Joined: Thu Feb 02 2006, 01:11PM
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Antonio wrote ...

A capacitive discharge arc is oscillatory. The current goes back and forth over the nonlinear resistor untill all the energy is dissipated. There is inductance in the discharge path too, so it's incorrect to assume that the capacitor voltage is applied to a small (nonlinear) resistor.

Not with a series limiting resistor. The resistor critically damps almost all oscillations. In fact, both my voltage and current discharges are nice exponential decaying waveforms - no oscillations period.
Of course, without the series resistor, you definetely get some serious oscillation action.

I did some testing two days ago and got good results. Basically measured the voltage and current, and then used the Math function on the DSO to multiple the current / voltage. From there i can easily determine the energy from area under the curve. Seems to match the equations i have.

davidb

Mon Dec 29 2008, 08:23PM

Registered Member #1890 Joined: Mon Dec 29 2008, 08:17PM
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Measure the temperature rise of the resistors.

Calibrate by shorting the arc and doing a small discharge through just the resistors to find the temperature rise from that energy. Then after you fire an arc, you can calculate probably a pretty good estimate of the energy dissapated in the resistors.

Dr. Dark Current

Mon Dec 29 2008, 08:27PM

Registered Member #152 Joined: Sun Feb 12 2006, 03:36PM
Location: Czech Rep.
Posts: 3384

davidb wrote ...

Measure the temperature rise of the resistors.

Calibrate by shorting the arc and doing a small discharge through just the resistors to find the temperature rise from that energy. Then after you fire an arc, you can calculate probably a pretty good estimate of the energy dissapated in the resistors.

I think this would not work... Quote from the OP:

wrote ...
the actual energy discharge is about 50mJ and the majority of the energy dissipated in the resistors.

So the energy difference with shorted and open gap is about 0.00004%

tesla500

Tue Dec 30 2008, 08:40AM

Registered Member #347 Joined: Sat Mar 25 2006, 08:26AM
Location: Vancouver, Canada
Posts: 106

Dr. GigaVolt wrote ...

I did some testing two days ago and got good results. Basically measured the voltage and current, and then used the Math function on the DSO to multiple the current / voltage. From there i can easily determine the energy from area under the curve. Seems to match the equations i have.

Since the energy dissipated is defined as the integral of voltage*current in the arc, isn't this measurement the best way to go? Why try to calculate that which you can measure precisely?

David

badastronaut

Thu Jan 01 2009, 06:05PM

Registered Member #222 Joined: Mon Feb 20 2006, 05:49PM
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Posts: 96

Have you also taken into consideration the energy dissipated inside the capacitor?

Assuming the equivalent average resistance of the arc weighted by current squared is constant, the amount of energy disspated in the arc is related to resistance of the resistor. The arc and resistor form a voltage divider; if the resistance of the resistor drops, then the amount of energy dissipated in the arc must go up assuming the arc resistance is constant.

Dr. Dark Current

Thu Jan 01 2009, 07:07PM

Registered Member #152 Joined: Sun Feb 12 2006, 03:36PM
Location: Czech Rep.
Posts: 3384

badastronaut wrote ...

...assuming the arc resistance is constant.

Wrong assumption

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