Author Topic: spark timing for implosion?  (Read 8487 times)

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spark timing for implosion?
« on: July 15, 2010, 18:14:07 pm »
Hi,
i ask myself why the  spark timing is adjusted to create a spark at the top center position.
We know that if water is converted to H2 and O2 the volume of the gas is in the 1:1600 range. That means when the gas is burned and converts back to water it implodes, because the volume of 1600 is converted to water with volume 1.
Wouldn't the spark be best timed at the bottom center position and the resulting vacuum will pull the piston up?


Experiments regarding the implosion of HHO:
As you can see when the reaction of the gas occurs water is sucked into the reactor chamber by a vacuum. I hope you can understand what he is doing, he is explaining everything, but in german.






This is assuming that no air is mixed with the gas obviously, which would counter the implosion effect by expansion of the heated air.
« Last Edit: July 15, 2010, 20:01:55 pm by haithar »

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Re: Adjusting the burn rate
« Reply #1 on: July 15, 2010, 20:01:28 pm »
whoops i don't know why i called it adjusting the burn rate, will change it asap  :-X

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Re: spark timing for implosion?
« Reply #2 on: July 15, 2010, 22:06:25 pm »
The question is: is the energy liberated through implosion stronger than the heat generated with combustion, which could expand a second gas (air) and move a piston like it usually happens with petrol.
Or could both effects be used together for a higher efficiency..

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Re: spark timing for implosion?
« Reply #3 on: July 15, 2010, 22:15:24 pm »
Man, There is no implosion!!!


2H2 + O2 will not become liquid H20 after it recombine. It will become superheated steam. Not liquid.  Only superheated steam.


Man the amount of 2H2+O2 that you need to inject inside the cylinder to have an equivalent to gasoline explosion represents only 6% of the cylinder in volume. So NO IMPLOSION. PLEASE!!!


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Re: spark timing for implosion?
« Reply #4 on: July 15, 2010, 22:57:52 pm »
how would you explain the experiments in the videos then?




also where do you have the 6% from? i'm interested in that, because i may be running some motor tests in the not so far future.

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Re: spark timing for implosion?
« Reply #5 on: July 15, 2010, 23:35:14 pm »
Hi haitar


I have calculated the energy content of the gasoline and compared to that of water...


But would be easier than that because meyer already gave us the amount of water that should be injected. And comparatively is quite right!


Take 7,4ul thus 0,0000074 liters and multiply it by 1800 witch is the amount of gas that this amount of water will expand into when transformed into 2h2+02. Thus you have 0,01332 liters of gas per injection or 13,32ml. So lets take 250ml per cylinder you have % =13,3*100/250 =5,3%


Actually meyers car was 1600cc thus 400ml per cylinder witch would lead you to need only 3,33% (h2+02) not only H2


Hope You understood what i mean... ( the 1800 value is not correct so you should expect an increment in % also because the temperature would increment a few ok)


I didn't watched yet to the video but i'm going to... 


Best regards

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Re: spark timing for implosion?
« Reply #6 on: July 16, 2010, 00:12:37 am »
I think he's right, it would be steam, because it releases energy, exothermic reaction, it would only turn to water if you removed this heat... but when you heat things they expand, so it would go counter to your implosion suction... I don't understand the german in the videos.


Anyway, test the theory and see if you could make something like that work
I do agree that the exothermic energy released will most likely heat the created water-molecules so much that they will (probably) convert to steam.
One would have to calculate to get the temperature and state of matter the water/steam is in, this is important to know the density of the created steam. On the other hand it's possible that the enthalpy of the gas is given off to the outside and the steam is quickly converted into water again.


The video shows the following: A HHO generator is plugged into the top connection of the reactor chamber, the bottom connection is connected to the water reservoir seen. there are two valves above and beneath the reactor which can be closed and opened.
The reactor has a cable which creates a spark to ignite the HHO mix.


In the first video they ignite the HHO (top valve is open) and see what happens: The water is is sucked into the reaction chamber, the result of underpressure in the chamber after the reaction. They try different versions with valves closed, open, ... In the third video they test different hose diameters and watch how long it takes for the water to go into the reaction chamber.

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Re: spark timing for implosion?
« Reply #7 on: July 16, 2010, 03:41:18 am »
Hello,


Is easy to know if water will become liquid again or if it will become steam, however there is still a problem related to the relation between temperature/pressure/state      but  ... Lets calculate it together...


Knowing that 1 mole of water totalize 1 mole of hydrogen H2 + 1/2 mole of oxygen O2 and that it means 18g of water, and that water have a specific heat that makes that 1g water raise _1__ degrees Celsius ° applying to it __1__ calories, and also knowing that water to change state (liquid to gas) needs yet another + amount of energy.   40kj per gram
 
1° how much energy 1 mole of hydrogen H2 + 1/2 mole of oxygen O2 reforming into water release? What is the heat of forming of water?
H2(g)  +   1/2 O2 (g)  H2O (g)
The heat of formation of water vapor is -241.8 kJ. (Why is this negative? - heat isreleased!)


2° Conversion kJ x cal

241.9 kJ = 57766.7 cal

So reforming 1 mole of water witch is 18g releases 57766.7 calories



3° 57766.7 cal divided by 18 grams should raise the temperature of water of 3.209° Celcius 



Therefore as water was previously gas before exploding i really think that it will become only superheated steam... 
 
However lets calculate further ...




« Last Edit: July 16, 2010, 04:16:26 am by sebosfato »