### Author Topic: Charles Law of thermodynamics  (Read 11149 times)

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##### Re: Charles Law of thermodynamics
« Reply #8 on: October 06, 2009, 20:29:43 pm »
What Logic said.
Its not about volume, It's density.

You still need to get HHO out of stonch by mixing it with other gasses. In turn will change its burn rate.

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##### Re: Charles Law of thermodynamics
« Reply #9 on: October 06, 2009, 21:34:22 pm »
What Logic said.
Its not about volume, It's density.

You still need to get HHO out of stonch by mixing it with other gasses. In turn will change its burn rate.

Ok. Then tell me what happens if you mix HHO with other gasses.
Could it not be the same as making the volume bigger of the HHO?
Is it not that you want to make more space between atoms?

Steve

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##### Re: Charles Law of thermodynamics
« Reply #10 on: October 06, 2009, 22:06:37 pm »
Steve did you even watch the New Zealand Videos?

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##### Re: Charles Law of thermodynamics
« Reply #11 on: October 06, 2009, 22:32:32 pm »
Steve did you even watch the New Zealand Videos?

Yes, i seen them. I have put them here on the forum myself.

Steve

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##### Re: Charles Law of thermodynamics
« Reply #12 on: October 07, 2009, 01:24:34 am »
Steve did you even watch the New Zealand Videos?

I did, Do you remember the Part where stan clearly stated he used 180 degree's out of phase for the fuel cell. How about where he Talks about his "Boiler Plate Configeration," Which is the Boiler that was used to run the buggy in the video when he was Stating all of this.

A alternator is 180 degree's out of phase, There is just 3 coils, we call this 3 phase. Its where 3 180 degree out of phase signals Overlap at 120 degree's.

In other words, if you Research just 1 coil from an stator and not 3, you will see it is 180 out of phase.

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##### Re: Charles Law of thermodynamics
« Reply #13 on: October 07, 2009, 13:38:29 pm »
Then you lose power.

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##### Re: Law of Gay Lussac
« Reply #14 on: October 07, 2009, 14:19:18 pm »
Ok, let me try to convince you here. If you have HHO, that means you have the ideal mix of H and O. We dont want more oxygen because the mix will not be ideal anymore. That mix of H and O is like 2.5 times more powerfull then a basic petrol/air mix.
If you use petrol drops you MUST add ambient air for couple of reasons.
1. need of oxygen for the burning process and volume
2. need of nitrogen for volume (it automatically adjusts burnrates too)

1 and 2 provide a volume with a certain burnrate/power
HHO has with less volume the same power with a faster burnrate
So what to do to match HHO to a current engine?
Well, we can mix the HHO with ambient air, which i did, and you are getting lots of nitrogen in the mix, which creates more volume and less HH per volume.
But what happens too, is that the ideal mix of H and O is gone because of the extra oxygen.

Thats why i think we should heat up HHO. Make the gas expand 2.5 times. You get 2.5 time more volume with less H and O, but still in the ideal mix for burning.

Steve

I understand all that and the ideal gass law etc Steve.
Let me try put it another way:
You fill a balloon with HHO.
The Balloon contains a fixed number of atoms of H & O.
Heating the balloon makes it expand/increase in volume.
Cooling the balloon makes it contract/decrease in volume.
But you still have exactly the same number of H & O atoms.
Those atoms oxidise/explode to H2O when you add a spark.
The size of the explosion/energy released is the same no matter the size of the balloon.
The engine turns maybe 20% of that explosion into usefull work.
20% of say 100KJ is 20KJ irrespective of revs, load or anything else.
The only way to get the engine to produce more power is to get more HHO into it.
The only way to fit more HHO into a fixed volume/the cylinder at a fixed pressure is to decrease the temperature.

Now you want to increase the volume of a fixed mass/weight of HHO so that you have more liters per minute to feed to the engine; so that you dont have to add Air.
That will work, but you will get no more power out of the engine, even if the revs increase, as the same mass of HHO is still being used.

I think Stan lowered the burn rate of his HHO by adding steam to the engine:
Steam and HHO go into cylinder on intake stroke.
Steam and HHO compressed in cylinder on compression stroke.
Compressing a gass will increase the temperature of the gass.
So feeding in steam at just the right temp  will mean that the compression stroke will increase the temperature of the steam to the point where thermal decomposition takes place at TDC/just the right time;;; with the help of the burning of the small amount of HHO...
Not all the steam will be pyrolysed to HHO and that will slow the burn rate down.
(Using air/nitrogen forms NOx in an endothermic reaction. (heat used up) Using steam causes expansion/power)

There are 2 ways to heat water with electricity:
The way we are used to is to heat a resistor with electricity and then use the hot resistor to heat the water.
The second way is to send ac into water:
The water has a resistance just like a kettle element has resistance and thus you get the resistive heating just as you would in a kettle.
But you are now also imparting kenitic energy to ions in the water.
These ions bump into other water molecules at very hight speed and the kenitic energy is translated into extra heat energy; over and above what you get from resistance alone.
I think Stans 180 out of phase heater was this simple AC to electrodes heating of the water.
The reason he used electricity and not waste exhaust heat alone is so that he could get precise control over the temperature of the water; so that it would pyrolyse at the right time/TDC.

This methode of heating water is said to be over unity.

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##### Re: Charles Law of thermodynamics
« Reply #15 on: October 07, 2009, 15:05:33 pm »
Then you lose power.

True and exactly what i want it to do....