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Author Topic: Water Arc Explosions (aka arc electrolysis and H2O-arc)  (Read 1517 times)

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Picking nits here... but the Watt-hour is a measure of energy, not power, and 100 watt-hours is the same as 360,000 Joules (100 watt-hours x 60 minutes/hour x 60 seconds/minute = 360,000 watt-seconds or Joules.)

And you have raised 2.1 liters of water by 42 degrees C, so that should have required
4.18 Joules per Degree Per milliliter (or gram), or 2100 ml x 4.18 J x 42 C  = 368,676 Joules -- an apparent OU result.

But not by much, COP=1.024 . Small errors in the initial temperature or volume measurements or in the input power or time would cancel out the slight OU figure.
 
Still, it's a nice result.    O0

(please check my math, somebody!)
 


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Picking nits here... but the Watt-hour is a measure of energy, not power, and 100 watt-hours is the same as 360,000 Joules (100 watt-hours x 60 minutes/hour x 60 seconds/minute = 360,000 watt-seconds or Joules.)

And you have raised 2.1 liters of water by 42 degrees C, so that should have required
4.18 Joules per Degree Per milliliter (or gram), or 2100 ml x 4.18 J x 42 C  = 368,676 Joules -- an apparent OU result.

But not by much, COP=1.024 . Small errors in the initial temperature or volume measurements or in the input power or time would cancel out the slight OU figure.
 
Still, it's a nice result.    O0

(please check my math, somebody!)
 

There are 3 other things that must be taken into account here TK.

1- the glass jar was also heated to the same temperature as the water,and so,the heating of that volume of glass must also be taken into consideration.

2-The glass jar was in no way insulated,and so,some of the heat energy would have been dissipated into the surrounding environment.

3- As the watt meter was before the FWBR,any energy dissipated by the FWBR would also have to be taken into consideration.

Against-

The watt meter was showing a power factor of .63-->what dose this mean for the calculated !energy! used?.
With a power factor of .63,would the watt meter have shown a larger or smaller amount of energy used?-or dose it calculate for the indicated power factor ?.

We have two things heating the water in this setup-as far as i can tell-and that is
1-resistive heating
2-the HHO produced during the process is also ignited,and the heat produced also heats the water.

A much more controlled experiment would have to be done,to gain any sort of accurate measurements with this one.


Brad
   

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All true and it's good of you to remind me. I thought of something else on the OU side too: there is likely some water boiling to steam in the arc channel as well, and the enthalpy of vaporization represents another big "bill" that is paid by the input power, if that steam escapes. If the steam re-condenses inside the water though, perhaps that balances it out.


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would be nice to read your musings on this [a possible experiment]
Silver melted with gas flame in a metal container (or with one of these portable inductive cooking plates). The liquid gravity fed* to a reaction chamber where the two streams of the molten silver meet.  Add some hydrogen + lithium or potassium.
Current from a MIG welder applied to the silver streams.  See how much the reaction chamber heats up.

If molten silvers scares you then the experiment shown in this video at time index 22:00 looks simpler.


Now, to be a devil's advocate:
I have never seen a convincing proof from BLP of the energy magnitude of that bright light.  How do we know that the bright light is not just an underunity silver welding arc ?
They have never shown that graphite sphere absorb that bright light and heat up to incandescence (although that's excusable because heated graphite burns up in air. That's why they plan to surround it all with pressurized Argon).

Thermolysed H2O or HO or HHO as a catalyst in a graphite sphere is implausible because oxygen + graphite burn up when heated.


* Gravity feed does not facilitate silver recirculation. For that the MHD Lorentz pump is a good solution. That pump requires DC current passing through the molten silver in a tube, orthogonal to magnetic flux from two NdFeB magnets.
« Last Edit: 2017-04-09, 15:37:26 by verpies »
   
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Good points, all.
Brad in particular wrote: "it was called !cold fusion!-although i believe that to be incorrect."  TOTALLY AGREE.

Now, re: measuring the input power or energy:

Brad:
Quote
3- As the watt meter was before the FWBR,any energy dissipated by the FWBR would also have to be taken into consideration.

Against-

The watt meter was showing a power factor of .63-->what dose this mean for the calculated !energy! used?.
With a power factor of .63,would the watt meter have shown a larger or smaller amount of energy used?-or dose it calculate for the indicated power factor ?.
A much more controlled experiment would have to be done,to gain any sort of accurate measurements with this one.


Brad

Why not measure the DC voltage and current AFTER the FWBR?  I realize there are fluctuations, but using an analog meter should give a decent mean value for the current, and the voltage here should be fairly constant.  Then, to a first approximation, the input power is simply

Pin = V * I  after the FWBR.
 
QUESTION:  Is there a better way to measure the Input Power (after the FWBR, in such an interesting experiment as Brad's)?



   

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Hi Steve.

I recently bought a DC watt meter, seems to work reasonably well but it only has a maximum of 60 V at 100 A capability.

With some fiddling Brad could do a bit of " power factor correction " by adding a capacitor across the input, alternatively a bit of mathematics could ascertain the actual input power. We were taught how to draw " Vector diagrams " but it's so long ago now that knowledge has been " overwritten " ! :)

Cheers Graham.


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Hi Steve.

I recently bought a DC watt meter, seems to work reasonably well but it only has a maximum of 60 V at 100 A capability.

With some fiddling Brad could do a bit of " power factor correction " by adding a capacitor across the input, alternatively a bit of mathematics could ascertain the actual input power. We were taught how to draw " Vector diagrams " but it's so long ago now that knowledge has been " overwritten " ! :)

Cheers Graham.

Adding a cap across the input-i dont think will work,as the resistance of the load(cell) changes,as the water heats up--and by a large margin.
EG-at the start of the test(water cold),we are drawing around 1600 watts !according to the watt meter!,and then as the water nears the 75*C mark,we are only drawing around half that-800 watts,and as we near 100*C,it drops down to around 300 watts.

This is not an inductive load-(or not one we know of),but a resistive load,where the arc is playing hell on the resistive value's of that load.
Then there is the explosions and recombination of the HHO back to water--who knows what kind of havoc that is playing on the watt meter.

Placing a large value cap on the output of the FWBR,may make things a little better.

Can anyone explain what the power factor value dose in the way of power being used?
Will the watt meter account for the power factor value,and calculate the energy being used(watt hours)using the shown power factor?,as the power factor seemed to be quite stable at .63 throughout the test.

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Hi Brad.

It's now over 40 years ago since technical college, but I'm pretty sure, being a resistive variable load won't stop a PF correction capacitor from doing its thing. Perhaps one of our erudite members will connect a capacitor and correct me!

Cheers Graham.

PS. Brad, didn't you use a duff Platinum heater plug for your tests?


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Hi Brad.

It's now over 40 years ago since technical college, but I'm pretty sure, being a resistive variable load won't stop a PF correction capacitor from doing its thing. Perhaps one of our erudite members will connect a capacitor and correct me!

Cheers Graham.

PS. Brad, didn't you use a duff Platinum heater plug for your tests?

The anode was S/S,and the cathode was an iridium element from a glow plug.

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It's now over 40 years ago since technical college, but I'm pretty sure, being a resistive variable load won't stop a PF correction capacitor from doing its thing.
Perhaps one of our erudite members will connect a capacitor and correct me!
If the cap is after the FWBR and power is measured after the cap then the PF will tend to unity as the capacitance increases.
The PF measured before the FWBR, it will get worse (away from unity), because the current will flow into the FWBR-->cap only at the peaks of the AC sine wave.
   

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The anode was S/S,and the cathode was an iridium element from a glow plug.

Brad

Thanks Brad, I was obviously thinking about thermocouples!


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If the cap is after the FWBR and power is measured after the cap then the PF will tend to unity as the capacitance increases.
The PF measured before the FWBR, it will get worse (away from unity), because the current will flow into the FWBR-->cap only at the peaks of the AC sine wave.

Hi Verpies.

I'm unsure but I think Brad was using a wattmeter on the primary of the transformer. If he was using a Microwave oven transformer they are known to be " mighty " rotten on power factor.

I have looked at many of my " decent " transformers they have a remarkably good PF , usually in the 0.9 Lag area.

I'm annoyed really, I seem to have forgotten most of my training, now having to rely on others for help. But around a 5 microfarad capacitor on the input would bring the transformer to near unity.

Cheers Graham.


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Hi Verpies.

I'm unsure but I think Brad was using a wattmeter on the primary of the transformer. If he was using a Microwave oven transformer they are known to be " mighty " rotten on power factor.

I have looked at many of my " decent " transformers they have a remarkably good PF , usually in the 0.9 Lag area.

I'm annoyed really, I seem to have forgotten most of my training, now having to rely on others for help. But around a 5 microfarad capacitor on the input would bring the transformer to near unity.

Cheers Graham.

Transformer  ???

There was no transformer.
The 240 volt AC mains went straight into the FWBR--no transformer.


Brad
   

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Transformer  ???

There was no transformer.
The 240 volt AC mains went straight into the FWBR--no transformer.


Brad

Well,  whadayno!!     :)

So what's causing that rubbish power factor? Is your wattmeter capable of discerning both, leading and lagging?

Cheers Graham.


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It's wrong to even consider a power factor with non-sinusoidal current.
For that an I vs. V scopeshot is needed.
   
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Well,  whadayno!!     :)

So what's causing that rubbish power factor? Is your wattmeter capable of discerning both, leading and lagging?

Cheers Graham.

Well,it's going to be a !yuck! situation for the watt meter-i would think.
Lots of explosions and current pulses going on in that jar.

@ verpies
Cant use my scope,as it just trips out the RCD--common ground issues
Cant use my DMMs either,as they just go mental trying to get some form of steady reading--it's like watching an epileptic staring into a strobe light.  :D


Brad
   

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I know.  If you have a 1:1 isolation transformer then use it. It will allow the scope to be used safely.
   
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  Using a scope then taking V(t) * I(t) = P(t) would work, then integrate the power vs time to get Energy (over a certain time interval).

HOWEVER - I think there must be a way to measure the voltage and current AFTER the FWBR without using a scope... even if the resistance is variable.

Graham wrote:  "I recently bought a DC watt meter, seems to work reasonably well but it only has a maximum of 60 V at 100 A capability."

Would that work in this case - if the max voltage coming out of the FWBR was say, 55V?  and not flat-DC... 

Or is there a "DC watt meter" than can handle higher voltages (and currents if needed) -- that would work with the output of a FWBR? 
I think 120V max would be sufficient; and 55V might work for arc electrolysis also. 

Another approach, is to take the DC from a set of batteries.  Another - use a super cap for the input power, then calculate the energy used  straightforwardly  (E = 1/2 C (Vend^2 - Vinitial^2)
   

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I know.  If you have a 1:1 isolation transformer then use it. It will allow the scope to be used safely.

Hi Verpies.

Your post reflected my musing whilst driving back from the supermarket.

Brad could also, if available, use a couple of transformers step down and back up to get isolation.

How would the wattmeter be affected looking at the input in this scenario?

Cheers Graham.


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How would the wattmeter be affected looking at the input in this scenario?
If the transformer/s is/are tightly coupled then not at all.
   
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  So to measure the input energy going into some type of H2O-arc device, one could use:

1 - output from a FWBR (fed by the grid) going into the device - measure V using a true-RMS meter and I using an analog meter.  A good estimate of the Power-in is thus:  Pin = V * I.

2.  A super-cap and simply measure the initial and final voltage, then
Ein = 1/2 C (Vinit^2 - Vfinal^2)

Eout -- I would measure using calorimetry.  (Sure there will be losses, which can be minimized - but if Eout / Ein is sufficiently large, then a few losses won't really matter.)
 Any comments on these methods - especially comments on better methods - would be appreciated.

   

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A quick sketch of the HHO bubble switch idea.

Bubble size must be controlled but that is not hard to do.

https://en.wikipedia.org/wiki/Breakdown_voltage


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I was musing on the hho bubble ignition idea and wondered if it was necessary to complete the circuit ? Because of the large dielectric breakdown differential between the liquid water and the hho gas I pondered that the gas bubble itself must be raised to the electrostatic energy level needed to breakdown the thin liquid water of the boundary layer liquid coating on the electrodes. Would this create hotspots within the bubble and cause a combustion event without the necessity of closing the circuit, and creating a spark path, by breaking down both electrodes water layers ? Any thoughts on this gentlemen ??

Dielectric Breakdown and its Influence on Ignition

http://oai.dtic.mil/oai/oai?verb=getRecord&metadataPrefix=html&identifier=ADA264925

Abstract : Electrical conduction, dielectric breakdown and the consequent ignition of small pellets, 0.2 to 5 mm in thickness and 5 to 7 mm in diameter, of MTV compositions SR886B, SR886E and SR886E have been investigated. The investigations were made using d.c fields and high-speed photography at framing rates of up to one million frames per second was employed to follow the breakdown and ignition behaviour. It was found that under suitable conditions, the dielectric breakdown of the samples led to their ignition and the delay time between the breakdown and ignition could be as small as 20 - 25 microseconds. Furthermore, a special circuit was designed and constructed for determining the minimum ignition energies of samples of different thicknesses. It was found that for samples of thickness of approx. 0.2 mm the minimum electrostatic energy which could cause ignition was as low as approx. 1 mJ. Moreover, the minimum ignition energy was found to increase with the sample thickness. A model, based on our previous model for the electrical ignition of silver azide, has been proposed. According to this, the dielectric breakdown leads to the formation of high conductivity filaments in which Joule heating-results in the formation of high temperature hot spots. Ignition initiates at these hot spots. An essential feature of this model is that the dielectric breakdown is a prerequisite for electrical ignition.


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Further musings:

http://aip.scitation.org/doi/abs/10.1063/1.338806

Abstract

The fast propagation (v≊106 ms−1) of an ionized and thermalized channel (ne≊1018 cm−3, T≊2.8×104 K) is studied in ambient air at atmospheric pressure, using gliding discharges produced over a charged dielectric slab. For surface voltages of about 100 kV, 1‐m‐long gliding sparks follow a straight line without any preionization of the gas. In this way, the discharges can be investigated with full diagnostics, including measurements of the current and of the propagation velocity, recording of the light emission (electronic image converter, spectroscopy, Lichtenberg figures), holographic interferometry of the spark channels, and detection of transient electric fields by capacitive probes. The various measurements are synchronized from optical fiber devices located close to the sparks. The study shows that the thermalized spark channel is produced by the three following stages: (a) a predischarge stage where the electron temperature Te≊2 eV is much greater than the gas temperature T0<1500 K; (b) a transient arc stage, lasting about 10 ns where energy is transferred from the electric field to the gas ionization; and (c) a heating stage, lasting from 25 to 60 ns, where the electric field has been largely reduced.

The hho bubble would be at atmospheric pressure but is not ambient air, its oxygen and hydrogen and so will combust the final end product being water, pressure pulse, and a release of heat (with a volume reduction after event concludes).

As Mike commented about TK's video the spark appeared to follow the path of least resistance around the water droplet boundary layer. Would it do the same thing on the inside of the hho bubble ? The shortest distance is straight across between the two electrodes, so it will be interesting to find out what the actual behaviour is.

If the high voltage breaks through the first electrodes liquid water layer, would it then raise the electrostatic field strength of the gas hho bubble to a value that is high enough to break the second electrodes liquid water layer in order to complete the circuit and create an ionising spark path ?

If the high voltage must be raised to a value that is equal to the dielectric breakdown of the two liquid surface boundary layers on the electrode + the hho gas gap straight line distance how does the second electrode know that the value has been reached at the first electrode ?

Would the information travel around the hho gas bubble and liquid water interface (bubble surface) while the spark propogation is straight line shortest distance between the two electrodes ?

This experiment raises a lot of questions! Please feel free to chip in if anyone has any ideas on what would happen. Once the data has been gathered and the experiment observed in super slo mo the opportunity for musing in advance of evidence is gone.. and that can be where a lot of the fun is.

 O0

Dielectric strength

https://en.wikipedia.org/wiki/Dielectric_strength

In physics, the term dielectric strength has the following meanings:

    Of an insulating material, the maximum electric field that a pure material can withstand under ideal conditions without breaking down (i.e., without experiencing failure of its insulating properties).

    For a specific configuration of dielectric material and electrodes, the minimum applied electric field (i.e., the applied voltage divided by electrode separation distance) that results in breakdown.

The theoretical dielectric strength of a material is an intrinsic property of the bulk material and is independent of the configuration of the material or the electrodes with which the field is applied. This "intrinsic dielectric strength" corresponds to what would be measured using pure materials under ideal laboratory conditions. At breakdown, the electric field frees bound electrons. If the applied electric field is sufficiently high, free electrons from background radiation may become accelerated to velocities that can liberate additional electrons during collisions with neutral atoms or molecules in a process called avalanche breakdown. Breakdown occurs quite abruptly (typically in nanoseconds), resulting in the formation of an electrically conductive path and a disruptive discharge through the material. For solid materials, a breakdown event severely degrades, or even destroys, its insulating capability.

Factors affecting apparent dielectric strength

    it decreases with increased sample thickness.[1] (see "defects" below)
    it decreases with increased operating temperature.
    it decreases with increased frequency.
    for gases (e.g. nitrogen, sulfur hexafluoride) it normally decreases with increased humidity.
    for air, dielectric strength increases slightly as humidity increases

Breakdown field strength

The field strength at which breakdown occurs depends on the respective geometries of the dielectric (insulator) and the electrodes with which the electric field is applied, as well as the rate of increase at which the electric field is applied. Because dielectric materials usually contain minute defects, the practical dielectric strength will be a fraction of the intrinsic dielectric strength of an ideal, defect-free, material. Dielectric films tend to exhibit greater dielectric strength than thicker samples of the same material. For instance, the dielectric strength of silicon dioxide films of a few hundred nm to a few μm thick is approximately 0.5GV/m.[2] However very thin layers (below, say, 100 nm) become partially conductive because of electron tunneling. Multiple layers of thin dielectric films are used where maximum practical dielectric strength is required, such as high voltage capacitors and pulse transformers. Since the dielectric strength of gases varies depending on the shape and configuration of the electrodes,[3] it is usually measured as a fraction of the dielectric strength of Nitrogen gas.


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Everyman Standing Order 02: Everyman is Responsible for Energy and Security.
Everyman Standing Order 03: Everyman knows Timing is Critical in any Movement.
   

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Here is a nice description of dielectric strength in polypropylene which mentions the creation of the conductive path. If the voltage was raised slowly over a period of time so that it was higher than the voltage necessary to breakdown the liquid layer on the first electrode (therefore the liquid must become conductive) it would then have to raise the voltage to a level equal to the breakdown on the second electrode, in order to complete the circuit.. this would provide a voltage 'gap' between the two electrodes that the hho bubble was occupying and should therefore electrostatically charge the hho bubble. If this voltage is higher than the voltage necessary to 'normally' create a conductive path through hho igniting it, but cannot complete the circuit for the spark to 'jump', how does the hho bubbles properties change ? if at all ? Would it hold a stable electrostatic charge that exceeds what would be required to normally ignite it due to a lack of being able to close the circuit ?

http://hypertextbook.com/facts/2009/CherryXu.shtml

In a dielectric, or an electric insulating material, electrons are bound to atoms and molecules and there is a high resistance to electric current, which means that the material has a zero or near zero electrical conductivity. When breakdown occurs, the present electric field frees the electrons. If the electric field is strong enough, the freed electrons may accelerate and liberate other electrons when they collide with neutral atoms or molecules. The liberation of electrons may turn the material into a conductor with its positive charge. In order to calculate how much the voltage an insulating material can hold before breaking down, the dielectric strength of the material must be found, which is measured in volts per unit length. The higher the electric strength, the more useful is the material as an insulator.

Dielectric strength is a property of an insulating material and it is defined as the ratio of the breakdown voltage to the material's thickness. Breakdown voltage is the maximum voltage a material can withstand before a conducting path forms through it.


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Everyman Standing Order 01: In the Face of Tyranny; Everybody Stands, Nobody Runs.
Everyman Standing Order 02: Everyman is Responsible for Energy and Security.
Everyman Standing Order 03: Everyman knows Timing is Critical in any Movement.
   
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