Author Topic: Resonant Cell  (Read 14868 times)

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Resonant Cell
« on: November 19, 2009, 02:05:05 am »
In this post I will go into the resonant cell design.I will describe the set up that Stan used and how it all worked.

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Re: Resonant Cell
« Reply #1 on: November 19, 2009, 02:28:51 am »
This section will go into the details of the cell in the attached file at the end of this post.

Inside the cell was 11 tube sets.They were in a circular pattern,just like the demo cell.All of the tube sets were sunken down into the body of the case.The tubes were only three inches in length.The ouside tube was shorter than the inside,in order to keep them at the proper spacing.The water entered from the bottom of the tubes through a passage that fed all the tubes together.Water flows up through the tubes and into the water chamber above the tubes.Then there is a passage in the center of the cell that the water flows back down through,and out of the cell back to the tank.

More will follow.
Don

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Re: Resonant Cell
« Reply #2 on: November 19, 2009, 08:57:41 am »
Don this gives me the impression that the system is not under pressure when in operation, what do you think?
Where is the pump situated, is it pumping before the inlets of the tubes, or is it sucking from the outlets ?

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Re: Resonant Cell
« Reply #3 on: November 19, 2009, 22:10:28 pm »
More details about the cell chamber.

The top of the chamber has a pressure relief valve set by spring pressure.There is also a 0-50 psi transducer to monitor said pressures.Also the gas output port is there as well.The outlet port has the quenching tube hooked to it.This tube has five sets of six holes in it.These holes are between .010-.015 inches each.It's about 5/16 inch dia.

At the base is where the water inlet and outlet are located.This cell is run under pressure.The water pump is the same as a late model fuel injection type of pump.Albiet with stainless internals.

On the inside wall there was five or six screws that went through to the outside in a single row going from the bottom up.These I would take to be a water level sensor.

So thats all there is to the cell itself.

The  most important thing here is the tubes are completely covered in delrin.

Don

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Re: Resonant Cell
« Reply #4 on: November 20, 2009, 12:19:02 pm »
More details about the cell chamber.

The top of the chamber has a pressure relief valve set by spring pressure.There is also a 0-50 psi transducer to monitor said pressures.Also the gas output port is there as well.The outlet port has the quenching tube hooked to it.This tube has five sets of six holes in it.These holes are between .010-.015 inches each.It's about 5/16 inch dia.

At the base is where the water inlet and outlet are located.This cell is run under pressure.The water pump is the same as a late model fuel injection type of pump.Albiet with stainless internals.

On the inside wall there was five or six screws that went through to the outside in a single row going from the bottom up.These I would take to be a water level sensor.

So thats all there is to the cell itself.

The  most important thing here is the tubes are completely covered in delrin.

Don

Thanks Don.

The question is still how much flow of gas can one tube set handle. Brian and I made many tests with tubes. There is a limit on the flow of gas.
How much gas should one tube deliver?

Steve



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Re: Resonant Cell
« Reply #5 on: November 20, 2009, 22:37:18 pm »
I don't think that anybody has even come close to finding out how much gas a tube can flow.You may have pumped alot of amps through a cee,and seen alot of gas and think that's all it will do,but I don't think your close.
If you pump water up through your tubes like Stan's cell,than you will be able to really make gas.That's only going to happen with resonace.Your not going to see it with straight electrolysis.

I don't have any idea on how much gas you can get.But if Stan's writings are for real,than he states 5 gallons of water per hour with one set of 3 inch tubes.
Don

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Re: Resonant Cell
« Reply #6 on: November 21, 2009, 01:00:04 am »
Lets do the math!   :D

3 inch tube set, 1/8 inch gap, 5 gallons per hour, as per the control and driver patent

I know 1/16th inch gap is mentioned in other documents, but 5 gallons per hour is only mentioned with the 1/8th inch gap, as far as i know.

5 gallons per hour is 5.26 ml per second
1 US gallon = 3 785.41178 ml
((3 785.41178 x 5) / 60) / 60 = 5.25751636

5.26 ml of water is split per second
5.26 ml *1800 = 9460 ml of fuel gas per second (Hydrogen, Oxygen, Dissolved Air)
= 2.50 gallons of gas per second

What is the volume of the water gap?

V = A * L
L = 3 inch
A = Pi(1 inch)*2 - Pi(.5 inch)&2 = 2.356 inch^2

V = 2.356*3 = 7.0685 cubic inch

7.0685 (cubic inch) = 115.831962 ml

Basically, we have to have a flow rate equal (actually, GREATER than) to the gas production, which is 2.5 gallons at zero pressure, what pressure does this operate at? up to 50 psi?

Say at max pressure, 50 psi, and using the density of the gas as the same as air, you have 4.407 times the density, so at 50 psi we could have the flow rate 4.407 times less than 2.5 gallons per second, so that is 2.4/4.407 = 0.567 gallons per second, or ...

Water flow rate into tube set must be = 2147 ml per second or, 2.2 liter per second, or 129 liter per minute.

At pressures lower than 50 psi, you would need a higher flow rate than this... at 20 psi you would need... 4 liters per second, or 240 liters per minute

How fast is the water moving? (at 20 psi)
Flow rate = Area * Velocity
4000 ml = 0.0015199 square meters * Velocity

Velocity = 2.63 meters per second

That is not exceptionally fast...

so you would need to move about a gallon per second, to each tube set, now I assume the cell used in that patent was a single tube set

This cell Don is talking about with 11 tube sets (10 used) might have a 1/16th inch gap, i don't know. But I imagine he was not using a pump that big, and that he was not getting the same production.


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Re: Resonant Cell
« Reply #7 on: November 21, 2009, 02:31:51 am »
Donald,your math is wrong on the gap volume.
A=Pi*R^2
A= Area
Pi=3.14159...
R^2=Radius squared
V=A*L
V=volume
L=length
outside tube id = .750 inches (.375 inch radius)
inside tube od = .500 inches (.250 inch radius)
.375 * .375 = .140625 inches
.250 * .250 = .0625 inches
3.14159...* .140626 = .441786467 square inches, outside tube id
3.14159...* .0625 = .196349541 square inches, inside tube od
.441789609 * 3 = 1.325359401 cubic inches
.196349541 * 3 = .589048623 cubic inches
1.325359401 - .589048623 = .736310778 cubic inches area

Those are the numbers I get
Don
« Last Edit: November 21, 2009, 14:57:15 pm by Dynodon »