USING ELECTROLYSIS TO TEST FOR POWER OUTPUT

Electrolysis is fine for removing unwanted body hair and putting chrome onto bumper hitches.

It's of absolutely no use for measuring power output.

Electrolysis counts electrons.  So you can tell how many electrons went through the electrolytic cell.

But Power is volts times amps.  El;ectrolysis only gives you the amps part, and only summed up over an interval of time.

To get an accurate power reading, you need to measure the instantaneous amps and volts, as quickly as possible.  
With electrolysis, you'd be hard pressed to weigh the electrodes more often than once a minute.  That's way too slow.

So if you want accurate power readings, forget about using electrolysis.  Use some actual commercial power meter, one that measures the instantaneous volts and amps.  

Dont try using a typical multimeter-- they're useless for this.  I have a Simpson 262, an otherwise wonderful meter, except it greatly exaggerates the voltage of pulses-- I've seen it read 400 volts when fed with a 12 volt narrow pulse.


A lot of the "over 100%" devices put out strange pulse waveforms that give wonderful power readings-- but only on meters that are not designed to measure funny pulses.

I agree with what you are saying but we have performed this test with /without the device. The theory is that because of Faradays law  The mass of a substance produced at an electrode during electrolysis is proportional to the number of moles of electrons transferred at that electrode:, that after using the device, more gas is produce, then that would mean that because of the device more electrons were "sent" to the electrolysis unit. Also, if you examine the website, you will see a bombcaloriemeter test that we performed, which proportionally produced approx. the same amount of work but in the form of heat.

Conservation of Energy isn't just a good idea.    IT'S THE LAW.

A quick glance at your block diagrams and schematics suggests a number of problems, inadequate instrumentation being one of them.

But the most obvious omission is the failure to account for resistive heating and energy loss in the interconnecting cables.

I may be willing to grant that your box doesn't have a power source.  But it certainly has an impedance.  
What it it?  And how does it compare to the impedance of the electrolysis cell?

An impedance matching resistor between the capacitor and the load cell can have change the overall efficiency by 50%.
These are simple calculations.  Any first year Electrical Engineering student can do them.

If you post the specs here, I do them for you, and several other people will check my work.

The question is:  Who are you trying to fool?  Yourself or someone else.

I suppose you can use electrolysis to measure the total charge in As that has been transferred. In conjunction with a record of the voltage(s) involved in the circuits it can be a measurement of the total work (J) during a specific time. The average power will be available in the figures. The power in a particular instant or time window  will be very difficult to determine with that method, as laid out above.
/RID

You know, the average guy, if he REALLY wanted to measure power, would use a POWER METER.  They're available most anywhere, even on eBay.  


They would not use some half-tushkis water and baking soda contraption.   At least not without running lots of tests of it against a known working power meter.

And your capacitor is not a very good source of charge-- if you look at the graphs in the spec sheets, most aluminum elect. capacitors have a very steep capacitance versus tempetature curve-- they can lose 30% of capacitance when cold.  They also have considerable leakage resistance and a bit of series resistance too, so the amount of charge you can get out depends on how fast you draw it out.


All you've shown so far is the ability to blow 150% more bubbles, under some uncontrolled conditions.  Not impressive.

If you're really sincere about this, please get a real power meter and a real power source.  


.

The web site reads like a science fair project done for a teacher who knows very little physics.
There are several points.
The biggest is that you imply non-conservation of energy. If true, that would be a scientific breakthrough of the first magnitude, thus your experiment has to be clearly done and exceedingly well documentated.
Point 2. You have confused the concepts of energy and power. Before anybody will pay any attention to your results you must be clear in your own mind about the difference.
Point 3. Your spelling is deficient. That is not too bad here on the internet with quick replies to questions. i make may spelling errors myself.  But on a poster board they stand out and make it difficult psycologically to pay attention to your results.

Some responces
 "But the most obvious omission is the failure to account for resistive heating and energy loss in the interconnecting cables."
The reason why this is not mentioned is that we used the same exact componets for both test,meaning that the device had to deal with the same impedences as the control experiment only difference is that 1 test had the device inline. If any thing, because of the internal resistances of the device, there should have been less useful work performed.
 
Nothing else was added or taken away to influence the test in ether direction.

Maybe I should rephrase my original question- Here we go. If you were me, how would you test the device for total power output to find if it is actualy producing additional power.

You musta counted wrong when adding up the squares for your calorimiter curves.

I took your data, plotted it in Excel, and had it compute the area under the curves:

Results:  4160  units without device,    4164 with the device.

That's under 0.1%.  I certainly made bigger errors than that trying to read the data off the chart.

minutes      without      with            area w/o      area with

5      90      90            90      90
6      122      93            122      93
7      130      101            130      101
8      125      105            125      105
9      118      110            118      110
10      115      112            575      560
15      98      105            490      525
20      90      94            900      940
30      83      85            830      850
40      78      79            780      790
50                              
                        without                    with
                  Tot areas:      4160      4164

The energy stored in a capacitor is   ½ C V²

       C = 0.1 farad    and   V = 20 volts  gives you  20 joules of stored energy.

============================
From   http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/electrol.html

      The energy required to seperate 1 mole of water is  286 kJ.

============================
From   http://www.ausetute.com.au/moledefs.html

       One mole of hydroghen gas takes up 24.5 liters at room temperature.

============================
So if you have a perfect electrolyzing machine you should get

      (20 J) (24.5 liters)
       --------------------   =    1.71  ml  of  hydrogen gas                   ==>   How much are you getting?
              286 kJ

============================

Try this test:

Charge up you capacitor to 10 volts instead of 20.  This gives you a stored energy of 5 joules.

Run this test four times in a row without your scavanger to give you a total of 20 joules.   How does the volume of gas produced compare to the 20 volt tests with and without the scavenger?

My prediction:  Bigger than either.

============================

What does it mean?  20 volts is not the most efficient voltage for electrolysis.

============================

What do I think your scavenger is doing?  

   Adding a series resistance between the capacitor and the cell lowers the voltage on cell and increases the discharge time.

   Sure you get more gas production.  But all you've done is taken a horribly inefficent process and made it less so.

We've told you already but you don't like what you're hearing.

If you really want the right answer, you'll have to get better equipment, like a an actual calibrated RMS-measuring wattmeter.
Electrolysis can't get you where you want to go.

Side note: The meter test has already been performed by a  Electical engineering instructer at a local tech. college, who also has full knowlege of the devices operation,. After testing and calculating the results, the meter also showed overunity. The instructer disreguarded the results blaming equipment error.

These over-unity guys need to restrict themselves to simple demonstrations so they will stop confusing themselves.  It seems like all their experiments are designed to be just complex enough to confuse the amateur observer.  They typically use a power meter that for one reason or another can't measure true RMS accurately enough.

What they need to do is set up something that perpetually lights up a light bulb.  That's simple.  That would impress.  10W would be fine.  Instead, they always have some contraption that requires external power and they have to persuade and convince you that something is happening.

Almost forgot, the meter belong to the college and during the test, plotted the voltage and current at 50,000 points per second.

whiskers1:  Why don't you just hook the output of your 'device' to its own input?  Won't it be a perpetual energy source then?  

Looking at your apparatus and write up:  

   It looks like the gas collector covers both electrodes.  
   If that is correct, you should get 1.71 ml of H2 and 0.86 of O2 (2.57 ml total) for each 20 joule input.

   You also say that your run each test 15 times.   That should give you a total volume of 38.5 ml for 600 joules of power input.

   What are the units on your gas volume scale?

> After testing and calculating the results, the meter also showed overunity.

Why does your story keep changing every minute?   Not a good sign.

It's also not a good sign when you show two curves and state one has much more area than the other, while a simple eyeballing indicates otherwise.  A simple calculation proves otherwise also.

>Almost forgot, the meter belong to the college and during the test, plotted the voltage and current at 50,000 points per second.

Okay, show us the data the compuer collected.

  If it's as you say, overunity, you've got an excellent shot at a Nobel Prize or two.

What makes you think that the area under your curves represent power? It does not!
As a matter of fact, the peak of your curves is a better indication of the amount of energy deposited in the calorimeter.
To explain the results of your work (which has a number of interesting aspects) you MUST learn the difference between power and energy.
I can help a bit, if you really want help, if you would post what you consider the definitions of power and of energy to be.

consider what your calorimeter curves would look like if your calorimeter were perfect (ie did not loose heat to the suroundings)

Ah yes, aburr is completely right, and I'm an idiot-- I just assumed the OP knew what he was talking about regarding the area under the curve.  it's the RISE in temperature that is proportional to the amount of heat pumped in.  the area has nothing to do with it..

And the calorimeter curves with the steep dropoffs indicates the calorimiter isnt too good at holding heat.  Ideally the curve would just flatten out if the insulation were perfect.

A few suggestions:
   
  (1) You're losing a lot of heat up the copper tubing. Coil up a bunch of the copper tubing and place it inside the calorimeter.

  (2)  Why be so frugal with the insulation?  A film canister isnt very big.  How about putting like many inches of foam insulation around the canister?  Better yet get some of that spray eurethane foam stuff so you can thoroughly cover it with unbroken insulation.  Or better yet, spray it in place  into a thermos bottle.  

  (3)  Use something calibrateable, like 100cc of water, as the heat storage medium, so you can get calibrated answers.  Right now, we have no idea what the heat capacity of the calorimeter is.

(4)  Go get some digital thermometers, $4.99 each opn eBay, which have a very small probe which won't drain much heat as that copper thingy.

(5) Using light bulbs as electrical to heat converters is a bit problematical-- too much juice and you blow the bulbs (you're putting 20 volts into 12v bulbs it seems).  And the bulbs are very non-linear in their resistance, which adds a bit more confusion into interpreting the results.  (They'll still be 100% efficient into converting electric power into heat, but they'll draw it in a non-linear way.)  Use something more docile, like a 10 ohm 10 watt resistor.

Also there's no substitute for size in a calorimeter-- the heat holding ability goes up as the cube of the linear dimensions, while the heat loss only goes up as the square.  So if you made your calorimeter ten times bigger in every dimension, it will hold 1000 times more heat per degree of rise, but only lose 100 times as much heat through its surface.  That's ten times better.

   Plus the heat loss will be yet again another 10 times less, as the rate of heat loss is proportional to the difference in indoor/outdoor temperature, which will be ten times less.    

Hope I got all those factors right this time :)

Various sources list the efficiency of water electrolysis at 30 to 35 percent.  Maybe 45 percent if you use a water-sulfur cycle at 800 degrees Celsius.

Recall that electrolysis happens at the electrode surfaces, not in the water between the electrodes.  That seems to suggest that any electrical resistance in the water is just going to contribute to heating the water, not to electrolysis.  So the water's resistance is going to be a considerable factor in the efficiency calculation.

Also I suspect the water can absorb some of the evolved hydrogen and oxygen, so that's another somewhat imponderable source of error.

Since the errors seem to be rather large and unpredictable, I'd suggest using a real power meter, or a better calorimiter.  
Perhaps something along the lines of a thermos bottle mostly full of water, with a 10 ohm resistor in the water.  And one of those very small thermal  sensors of a digital outdoor thermometer.   I suggest going with a bigger calorimiter as there seems to be nothing in the way of you pumping in as many loads of charge as it takes to get a considerable indicated temperature rise.

The questioner is hopelessly confused about the physics of the experiment.
My copy of the question shows an accepted answer.
Most of the responders has useful comments to make.

>>My copy of the question shows an accepted answer.

The asker accepted it after the ping:)