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No need for insults. If people don't understand, that reflects on your clarity of explanation. It's a continuous change from:

high vacuum, low HHO generation energy, high flywheel energy loss

to

low vacuum, high HHO generation energy, low flywheel energy loss

You can choose to operate near one end of that range but whereever it is, you're getting more of something and less of something else.


Either way I think this discussion is all a bit pointless without using numbers or laws of physics. Sure you might get some overall improvement in some area over some other engine. But it's a long way from 30% to 101%. Without 101% this engine design is no use at all. Don't think people haven't been trying every possible configuration of pistons and combustion chambers for the past century or two, as well has electrolysis which is as old as the hills. What you see in use only represents the tiny minority of inventions that ended up being the best at their particular job.




Yea of course people will use waste heat for heating nearby things when they can, that doesn't take an invention, it's just obvious. Like the cabin heater in a car. There's even a domestic appliance which burns gas to generate electricity and uses the waste heat to provide hot water and room heating. You sell the electricity back to the grid and end up spending a bit less on power. I heard those big old computers were connected to their building's heating system to reuse the waste heat. Once I even tried to design a device to store heat from my car's engine while driving, and transfer it into my house when I got home. Turned out to be just too inconvenient. But might have saved a little bit of money.

.
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Quote:
high vacuum, low HHO generation energy, high flywheel energy loss

to

low vacuum, high HHO generation energy, low flywheel energy loss


you cant explain it that easily.
but it can be explained if we walk through every image.
and include all the energies.

the forces provided to the flywheel.
the forces that drain from the flywheel.

but first we need to find the force to the piston from
the HHO explosion.

everything else is simple or should be.

we already know the Hydrogen has 140 mj energy per kg
and
gasoline has 40 mj per kg

so we can take a stardard small gas engine and determine how long
it will run on a tank of gas , then divide the volume of the tank of gas by the time it will run and the rpm's and power strokes of the engine.

to find an amount of explosive force to compare with HHO.

we can even use the same exact piston size and stroke of the small gas engine.

this way we will know how much force would be provided to the piston of a gas engine , and since the compression that is caused
by the gas engine is a even steven opperation the compression
differences between the two engines dont really matter.

I will see what I can find.


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I have a feeling I can see where all our discussions are heading.

You suggest a simple idea.
I show it doesn't work.
You suggest a more complicated modification to it.
I show it doesn't work.
You suggest an even more complicated modification to it.
It's too hard to me to figure out if it works or not.
You walk away believing it would work.

Is that where this is going? Is that perhaps the reason so many people in the world believe perpetual motion is possible?

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no way , if we can show it wont work then thats all it takes

otherwise we show it does work !!!

I found some information on a commercially sold product
that will help in our determination.

the generator complete

the generator and engine

the engine itself for specs.
the engine

11 HP rated
5500 watt rated
6.6 gallon tank capacity
5.3 hrs. @ rated load (5500 watts)

Torque (ft./lbs.) 18.50 at 2,500 RPM
Max. RPM 3,600
Bore x Stroke (in.) 3.5 x 2.5

I suppose the rated HP , and rated 5500 watts utilize the maximum 3600 rpm.

------------------------------------------------------------------------------------------

1 gallon = 3.785 liters
1 (cu in) = 0.016387064 liters
1 pound = 0.45359237 kilograms

6.6 gallons = 6.6 x 231 cu in = 1524.6 cu inch

1524.6 x 0.016387064 = 24.9837177744 liters

so 6.6 U.S. gallons = 24.9837177744 liters

gasoline weights from 5.8 to 6.5 lbs per gallon.

we can use 6 lbs per gallon.
we can use 2.721 kg per gallon

we have 6.6 gallons of gasoline x 6 pounds per gallon = 39.6 pounds per 6.6 gallons
we have 39.6 pounds of gasoline x 0.45359237 kilograms = 17.962 kilograms per 6.6 gallons

in 5.3 hours the gas engine uses 6.6 gallons
in 5.3 hours the gas engine uses 24.981 liters

so in 1 hour
6.6 gallons / 5.3 hours = 1.245 gallons
24.981 / 5.3 = 4.713 liters

in 1 hour the gas engine uses 1.245 gallons
in 1 hour the gas engine uses 4.713 liters

so in 1 minute
1.245 / 60 = 0.02075 gallons
4.713 / 60 = 0.07855 liters

in 1 minute the gas engine uses 0.02075 gallons
in 1 minute the gas engine uses 0.07855 liters

so in 1 second
0.02075 / 60 = 0.00034583 gallons
0.07855 / 60 = 0.00130916 liters

in 1 second the gas engine uses 0.00034583 gallons
in 1 second the gas engine uses 0.00130916 liters

----------------------------------------------------------------------
3600 rpm / 60 seconds = 60 revolutions per second.
a 4 stroke engine has 2 strokes per revolution
and only 1 of the 4 strokes is a intake stroke.

which equates to 15 intake strokes per second at 3600 rpm.

in 1 second the gas engine uses 0.00034583 gallons
in 1 second the gas engine uses 0.00130916 liters

which means that each intake stroke consumes

0.00034583 gallons / 15 = 0.0000230553 gallons per intake stroke
0.00130916 liters / 15 = 0.0000872773 liters per intake stroke

0.0000230553 gallons x 6 pounds per gallon = 0.0001383318 pounds
0.0001383318 pounds = 0.00006275 kg

gasoline has 44.4 Mj/kg energy content
Hydrogen has 142.0 Mj/kg energy content

44.4 Mj/kg x 0.00006275 kg = 0.0027861 Mj per stroke
44,000,000 j x 0.00006275 kg = 2786.1 j

since the power stroke comes from the energy content in the gasoline
the available energy from each power stroke is 0.0027861 Mj
or 2786.1 j
at 15 strokes per second the amount of energy avilable to the power stroke
is 2786.1 j x 15 = 41791.5 j or 41.7 kW

----------------------------------------------------------------------

wow now thats a lot of wasted energy wouldnt you say.
getting only 5.5 kW from a fuel that has 41.7 kW of energy in it!!!!!!!

what a waste , but anyway we now have several things to work with
we can just apply them to the HHO engine and see what the reults would be.

-----------------------------------------------------------------------

so give me some time and I will try to see what I can come up with


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Hehe you really like to do things the hard way! How's this -

On one tank of fuel:

Output energy = 5.3hrs * 3600 s/hr * 5500W = 105MJ
Input energy = 6.6gallons * 2.72kg/gallon * 44.4MJ/kg = 800MJ
Efficiency = 105/800 = 13%

Same as your answer of
5.5kW / 41.7kW = 13%

smile

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well I thought it would be nice to have the exact or a very close volume of gasoline fuel used per power stroke this way an equal volume of HHO can be used for comparison.
although an equal volume of HHO is 3 times more powerfull than gasoline.

which will most likely allow for a reduction in HHO volume per power stroke compared to the power strokes in a gasoline engine in a working model.

because the gasoline engine has a higher rpm to achieve the output energy.
I can use the same volume of HHO fuel per unit of time to give momentum to the flywheel and use a much lower rpm , because of the time required to produce the HHO.

for instance in 1 second the gas engine uses 0.00034583 gallons of gasoline.

I might design a HHO engine that only has 1 power stroke every second vs the 15 power strokes every second of the gasoline engine.
this way I can have 15 times the HHO fuel per power stroke.
or maybe every 5 seconds.
this way I can have 75 times the fuel that is used in the gasoline engine over a 5 second period only I will use it at 5 second intervals to slow down the engine.

kind of like a slow motion engine that can be easily understood.
but that will have more output energy than it has input
energy.

I have been busy on other things lately but thought I would comment on that.



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I don't think you can cram more fuel in than will fit there, without pressurizing it. If HHO has 3 times the energy density of petrol/air mixture at the same pressure then all you can do it put the same volume in and get 3 times the energy out. Of course you're also putting 3 times the energy in so it's not necessarily improving efficiency.

No need to slow it down to match the HHO generator. Just use a faster HHO generator, or a bunch of slow ones.

It still works the same if you consider it as a continuous process that just consumes x liters of fuel per second, without breaking it down into individual strokes. You could actually make it a more continuous process by having more cylinders and/or some buffer reservoir for the fuel.

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Quote:
Of course you're also putting 3 times the energy in so it's not necessarily improving efficiency.


well we can consider at least 1 of those three times can supply the power to generate the HHO in the vacuum created by the explosion of the 3 times of energy normaly used in a gasoline engine.

as the piston strokes out , and as the hydrogen implodes.
but mostly as the piston strokes out via the 3 times energy transfered to the fast moving heavy flywheel.

and the additional efficientcy of having only 1 stroke
vs the 4 strokes of a normal gasoline engine kind of makes it seem not so inefficient at all.

unless your just a simple common nay sayer.

a 1 gallon non pressurized HHO explosion

that should send a piston down a cylinder wouldnt you think?



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Originally Posted By: paul
well we can consider at least 1 of those three times can supply the power to generate the HHO in the vacuum


So far you haven't shown this.
As far as I'm concerned I consider in a perfect system without any energy losses due to friction or thermal radiation, the energy gained by this powerful explosion matches exactly the energy necessary to do the electrolysis.

Since there are many losses (one of the biggest ones will be the electrolysis it self), you have basically devised a combustion engine which is less efficient then an electric engine of the same parameters.

You are the one claiming your engine would be remarkable efficient, but so far you didn't show anything more then your claims.

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Quote:
the energy gained by this powerful explosion matches exactly the energy necessary to do the electrolysis.


how much energy is necessary to do the electrolysis?

you havent shown this , yet you jump to the conclusion that it will take just as much energy as the explosion delivers.

unless you can show that the explosion delivers less energy than the energy required to make the HHO then
your point is moot / invalid / worthless.

after all we already know that a gasoline powered generator can produce 5.5 kw using 1/3 the energy
that is contained in a equal amount of Hydrogen.

so a hydrogen powered generator using the same amount of fuel should produce 3 times that energy even if used in a
in-efficient internal combustion engine that has been designed to use gasoline.

I should be able to power a 16.5 kW generator when using
HHO in the in-efficient internal combustion engine.

so I should be able to use 11 kW to produce the HHO
and still have the 5.5 kW from the generator to use as a power source if I use a in-efficient internal combustion engine .

to charge up my electric car and to power my home.

so naturally I would expect that using HHO in an engine designed to use gasoline would result in a automatic
87% in-efficient engine opperation.

but if I use the HHO in an efficient engine that has been designed to use HHO I could get a whole lot more as the gasoline engine waste apx 87% of the energy in the gasoline fuel it uses.




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Originally Posted By: paul

how much energy is necessary to do the electrolysis?

Havn't we already agreed that HHO generation is something like 80% efficient in commercial plants and theoretically up to 97%? It doesn't matter if the gas produced has 3 times the energy density of petrol, or 1000 times, it's still going to need 3 or 1000 times as much electrical energy to get an equivalent amount of fuel.


Quote:

but if I use the HHO in an efficient engine that has been designed to use HHO I could get a whole lot more as the gasoline engine waste apx 87% of the energy in the gasoline fuel it uses.


Yes you probably would get some improvement above 13%, but even if it was 3 times as efficient, it still isn't producing as much energy as the HHO generator requires. You could use a fuel cell with what, 80% efficiency? But that too means the power output of the fuel cell will be less than the power needed to produce the HHO at 97% efficiency.




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1 liter of petrol, when burnt with air produces 3 times the energy as 1 liter of liquid hydrogen.

Uh oh!!

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Quote:
So sure the engine might produce 3 times the power, but it also requires 3 times the power in the HHO to do that, which requires 3 times the electrical input to the HHO generator. So there's still no improvement.


kallog , there wouldnt be a need to supply 3 times the power to the HHO generator.

the HHO itself has 3 times the power of gasoline.

so the engine produces 3 times the power
and the HHO already has 3 times the power
but it only takes 1 times the electrical input.

knowing that HHO already has 3 times the power and knowing that a gasoline engine only has a 13% efficiency
theres room to think that a HHO engine would be much more efficient than a gasoline engine.


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Quote:
1 liter of petrol, when burnt with air produces 3 times the energy as 1 liter of liquid hydrogen.

Uh oh!!


uh oh!!! would you have a link to that info , so that I
can read it myself?

I would presume that you are ignighting the hydrogen in a low or slow oxygen environment.
since the air only has a 21% oxygen content the liquid hydrogen would need to WAIT untill more oxygen
is sucked in before more hydrogen would ignight causing a slow burn or explosion.

uh ohhh !!!

I would think that when people performed test to determine the energy content in the two fuels
they would have thought about this.
and they probabbly did , thats why hydrogen shows as 140 MJ and gasoline shows as 44 MJ per kg





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Yep, you're using MJ/kg. I was using MJ/liter. By coincidence the situation is reversed, but it empahsises the meaninglessness of energy density on efficiency.

Wikipedia "energy density".

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You're just twisting words into knots, let's use numbers.

1. Engine produces 15W of mechanical power
2. Alternator converts that to 14W of electricity
3. HHO generator converts that to 14W of HHO
4. Engine burns that to produce 15W of mechanical power

Step 2 assumes 93% efficient alternator, but you can have 99.99% if you want.

In step 3 I generously allowed a 100% efficient HHO generator.

Step 4 requires >100% efficiency.




Originally Posted By: paul

and the HHO already has 3 times the power
but it only takes 1 times the electrical input.


The first line is correct, but you just invented the 2nd line. It directly contradicts the HHO generator efficiencies that we sorted out on long ago.

How does the HHO generator "know" that it's making HHO instead of petrol? It can't even make petrol. Why should it s performance in any way depend on petrol's energy density? Will it mysteriously get even better if we imagine how much energy a kg of plutonium has?

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Originally Posted By: paul

how much energy is necessary to do the electrolysis?
you havent shown this , yet you jump to the conclusion that it will take just as much energy as the explosion delivers.


"Extraordinary claims require extraordinary evidence"
YOU are the one claiming to have found new laws of physics.
While this is not impossible (others have done this before) it is an extraordinary claim, so YOU are the one to provide the evidence.


Originally Posted By: paul
unless you can show that the explosion delivers less energy than the energy required to make the HHO


The electrolysis of one mol of water produces a mol of hydrogen gas and a half-mol of oxygen gas.
The process must provide the energy for the dissociation plus the energy to expand the produced gases.
To achieve this under athmospheric pressure you will need 285.83 kJ of energy.

This overall amount of energy 285.83 kJ / 1.5 mol of HHO is STORED in the HHO. You can get exactly this amount back by recombination of the molecules.
Not magically more.

In a combustion engine, which is far away from 100% efficiency you won't even get the 285.83 kJ back as electrical energy, but a lot of losses by thermal radiation.

Reality is far more complex and above my current understanding, but maybe this page gives you some useful insights:

http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/electrol.html#c2


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the molar mass of 1 molecule of water = 1 H + 1 H + 16 O = 18 gmol-1

mass of 1 litre of water = 1 kg

Michael Farraday has shown that 1 liter of HHO can be produced
in 1 hour using 2.34 watts or 2.34 watt hours.

Quote:
Michael Faraday was an exceptional and highly respected
researcher who investigated the electric current needed to convert
water into hydrogen gas and oxygen gas by electrolysis. His results
are accepted by pretty much every scientist everywhere. While he
expressed the results of his work in terms which would be meaningless
to the average person, his result is that an electrical input of
2.34 watts produces one litre of hydroxy gas in one hour.


and you are correct

The enthalpy of combustion for hydrogen is 286 kJ/mol

but 2.34 watts delivered over a 1 hour time period is only 8424 J

it would take 1800 hours to convert 1 liter of water into HHO
using 2.34 watts
but durring the 1 hour we can convert 1/1800 liters of water
into HHO using 8424 J

1 kg / 1800 = .000555 grams of water
--------------------------------------------------------

now for the energy we can get back

The molar mass of water is 18.01528 gmol-1

Avogadro's constant is 6.0221415 1023

So the mass of one molecule is

18.01528 gmol-1 6.0221415 1023 = 2.991507 10-23g

Which is 2.991507 10-26kg

or: 0.00000000000000000000000002991507kg

so we determine the number of water molecules we have converted
into HHO durring the 1 hour by dividing the mass of 1 molecule
of water by the mass that was converted in the 1 hour.

.000555 kg of water / 0.00000000000000000000000002991507 kg 1 molecule of water =
= 18552565602540531505933.4781882 water molecules

by multiplying the number of water molecules by the 286 kJ/mol
we get 5306033762326592010696974761.8252 J

we used 8424 J we got back 5306033762326592010696974761.8252 J

now that was me using your 286 kJ/mol and Farradays 2.34 watt hours per liter HHO

how bout that !!!

thats probably why they use hydrogen and oxygen to launch the space shuttle

instead of gasoline.



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Originally Posted By: paul

the molar mass of 1 molecule of water = 1 H + 1 H + 16 O = 18 g * mol^-1

Originally Posted By: paul

Michael Farraday has shown that 1 liter of HHO can be produced in 1 hour using 2.34 watts or 2.34 watt hours.
[...]
but 2.34 watts delivered over a 1 hour time period is only 8424 J


Let's see:

2.34 watt hours ~ 8424 Joule. So far ok.

Originally Posted By: paul

it would take 1800 hours to convert 1 liter of water into HHO using 2.34 watts


Didn't understand that one. Why 1800 hours?
I tried a different approach:

1 liter of an ideal gas, at standard temperature and pressure, will contain 0,0446 moles of molecules.
= 6.022 * 10^23 / 22.414 molecules.

The gas will be in molecular state, meaning H2 + O2 molecules in a 2:1 ratio.
= 0.272 * 10^23 molecules (H2 and O2) per liter
means: 0.091 * 10^23 of O2 and 0.181 * 10^23 of H2
or: 0,015 mole of O2 and 0,03 mole of H2

= they could recombine to 181 * 10^20 molecules of water.
= 181 * 10^20 * 3 * 10^-23g
= 0.543g

this is pretty much the same as your figures:

Originally Posted By: paul

0.555 g of water / 0.00000000000000000000000002991507 kg 1 molecule of water =
= 185... * 10^20 water molecules


How much energy can we get back of our one liter of HHO?

Originally Posted By: paul

Originally Posted By: Momos

The enthalpy of combustion for hydrogen is 286 kJ/mol




This figure is from http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/electrol.html,
while it is the energy input it is NOT the energy output (due to excess heat) and far less the usable energy in a combustion engine.

But nevertheless, lets see what we get:

286 kJ/mol of hydrogen
0.272 * 10^23 molecules (H2 and O2 in 2:1 ratio).
means: 0,091 * 10^23 of O2 and 0.181 * 10^23 of H2
= 0,03 mol of H2

0,03 * 286 kJ/mol = 8,58 kJ

Nearly the same as the 8424 J at the beginning (probably due to rounding and because my initial assumption about HHO as an ideal gas is wrong).

But why are you so far off?

Originally Posted By: paul

= 18552565602540531505933.4781882 water molecules
by multiplying the number of water molecules by the 286 kJ/mol
we get 5306033762326592010696974761.8252 J


Well: the energy is in Joule PER Mol.
You multiplied by the number of water molecules, instead you have to divide again by Avogadro's constant.
Doing this we get:
5306033762326592010696974761.8252 J / 6.0221415 1023
= 5306 * 10^24 J / 6.022 * 10^23
= 8810 J

There you are.
Still you could argue in this calculation you still have some energy gained (8424 J input and 8580 J respective 8810 J output). But we haven't consider the surrounding heat which is reducing the necessary input energy for the electrolysis (but also reduces the energy we can gain).
Furthermore this 8424 J of input energy are disputable, since I couldn't find a credible source for this figure.

And all this without even thinking about losses in alternator, excess heat, wheel, sub-optimal chamber design, etc. ...

[quote=paul]
thats probably why they use hydrogen and oxygen to launch the space shuttle instead of gasoline.
[quote]

Nobody has ever doubted hydrogen and oxygen are a great way to store energy. The potential energy in HHO might be 3 or 4 times larger as for gasoline. The same is true for tnt or plutonium.
STORING Energy is nice, but you can't get energy out of nothing.

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are you serious?
1000/18 = 55.55 mol
ok there are 55.55 mol in 1 liter of water.

so 55.55 mol / 1800 = .03 mol / liter HHO

286 kJ * .03 = 8826.27 J

your right there is more energy there , but its only
apx 400 J an hour.
not enought to worry about.

but it shows you can get back what you put in , and using a vacuum to generate the HHO will return even more.








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