Author Topic: Demonstration Cell - Tubular Cluster Array  (Read 16871 times)

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Re: Demonstration Cell
« Reply #8 on: September 16, 2009, 15:23:31 pm »
If you use Permanent Magnets then you can not control the output of the alternator, and thus not control the gas production. It will be a fixed value. Unless of course you did something else as well, like duty the output.

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Re: Demonstration Cell
« Reply #9 on: September 16, 2009, 22:41:10 pm »
I have built a Tubular Cluster Array from Stan's drawing, with four 3" tubes.
I have built a power source, consisting of a Variac, Diode Bridge, fuses, a 1/2 horse power motor @ 1725 rpm, pulley a bit greater than 2:1, and a de-regulated Delco Remy Alternator.



Very cool when can we expect you yourtube?
Thanks,
BW

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Re: Demonstration Cell
« Reply #10 on: September 16, 2009, 23:24:09 pm »
How much electrical energy does it take to mechanically turn that alternator shaft with your motor @ 1700 rpm ?

At least the power consumed by your cell circuit + the losses , I know that ...

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Re: Demonstration Cell
« Reply #11 on: September 16, 2009, 23:58:37 pm »
Burnt Wire, I'll make some videos when I get the cell finished and producing better. I've pretty much run through all the basic experiments I can do right now, I need to make some modifications now.



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Re: Demonstration Cell
« Reply #12 on: September 17, 2009, 18:26:13 pm »
How much electrical energy does it take to mechanically turn that alternator shaft with your motor @ 1700 rpm ?

At least the power consumed by your cell circuit + the losses , I know that ...

One HP is equal to 746 watts, You can google that! That equation is not perfect.

Your question was based on Rpm's and not wattage, The Rpm is not as important as friction, the reason the equation is not perfect. If the alternator is being driving at 1700 Rpm's with No load, you only have Friction, If the alternator is at 1700 Rpm's and it is producing 500 watts, then you can usually use math on the system, Transformers have losses.

If the Equation is not perfect, and you know 1 hp is 746 watts, you know the driver motor is 120 volts and consumes 10 amps then 120 * 10, 1200 watts. So based on that 746 "1 HP" Divided into 1200 watts is 1.6 HP. Thats an example thats a bit misleading But you get the ideal..

Tip! Your cars alternator was clocked at Producing 12.5 volts 15 amps. How much HP Was the alternator taking from the car? You was using 187 watts, so a small portion of 1 HP ;)

1 watt = 0.00134102209 HP

so, 187 watts * 0.00134102209 = 0.25077113083 HP

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Re: Demonstration Cell
« Reply #13 on: September 19, 2009, 20:57:21 pm »
Sixth Run
Tap water
~12 volts on variac
Testing the nozzle again, sustaining a flame, maintains about 1/2 psi with flame, probably due to back pressure of the bubbler.
nozzle got hot and melted out of the nylon hose... no new results, but it's neat.
Been looking into tiny drill-bits and alumina tubes, should get something ordered soon...
Few other things that need to be done... have a multi-meter, so I'll be measuring a few things to get more familiar with the system.

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Re: Demonstration Cell
« Reply #14 on: October 04, 2009, 21:53:19 pm »
Ok Hydro, I've been thinking, and this is what I'm going to try next chance I get, and I'm also posting it here hoping someone else with a similar set up will try it too. (probably mid week I will get a chance)

Procedure:
1. Fill cell right to the top with water.
2. Siphon out 500 mL of water to create 500 mL cavity.
3. Start Cell, record psi after 1 minute. If cell maxes at 5 psi before 1 minute, then try 30 seconds...
4. Release gas pressure into inverted bottle full of water, let it rise up to release pressure, measure expanded volume now that psi is at 0.

This will provide a relationship between psi/min and cc/min. However we would not be considering the fact that hydrogen gas is less dense at atmospheric pressure, therefore in this test we will have some error that may or may not be significant, but it should still give a result that we can compare directly with Stan's result.

Stan says 7000 cc/min, my cell is 7% the size of his, assuming gas production is directly proportional to cell size, I might expect 490 cc/min.

I will perform this test with Rain Water. If I feel like it would be useful, I will repeat it again with Tap Water, Distilled Water, and maybe Well Water, and Salt Water...

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Re: Demonstration Cell
« Reply #15 on: October 04, 2009, 23:33:20 pm »
You're going to be very disappointed, It took me 12 tubes to make 600 cc's a min, This was at around 25 volts 30 amps.

Clearly from the information I just provided, You can Pinpoint that I used 6 diodes because the voltage across my 12 tubes was 25 volts. If i would have only used 2 diodes to full wave then the voltage would have been much lower.

Keep in mind, stan full waved at a lower voltage, Do you understand what I am pointing out, Or do you understand you can Full wave 2 different ways? If this seams a but fuzzy to you, I can provide a normal transformer schematic showing 2 ways to do this. That would clear that part up. Also note his voltage was 12, and not 12,000, His amperage was not small. When I hit resonance using 6 tubes, I had about 8 to 10 volts across the 6, about 15 amps, I produced about 275 cc's of gas per min with that setup using only about 120 watts, Which is why at resonance this did not bother the 2k watt motor. Thats The best setup to see on a scope.