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Old 11-04-2011, 07:37 PM   #1
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Default HHO Generator Hybrid Vehicle

I thought of this idea recently and I just mentioned it on another thread so thought I would start its own thread.

HHO or Browns gas is now perfectedly accepted as authentic. In fact it always has been because 'apparently' Jewellers have been using a similar system for years to repair jewellery.

Nowadays HHO generators for welding are available. They consume current but actually quite a surprisingly small amount. I've read various figures.

This idea is based on a Hybrid design but really inspired by my electric vehicle ideas. I wondered what would be wrong with having a Hydrogen (HHO) generator on board a standard combustion vehicle and powering it with batteries. At the end of the day you would need to recharge your batteries and refill with water.

This is not overunity or perpetual motion etc.

This is just running a car on gas (HHO) and accepting that the fuel you will be required to fill up with is electricity and water.

Other than that the car stays the same.

Remove the fuel tank in the boot, install an HHO welding generator with a water tank and a load of batteries. Get an inverter to convert the batteries to 220V and connect it all up. The output gas of the generator is fumigated in through the air filter or if you have carburettor's using the standard LPG flange that fits between the carb and the intake manifold.

This is standard LPG practice. Fumigating through the air filter is called single point injection in LPG speak. Using flanges between carbs and manifold means you use the carb butterfly's to regulate fuel air.

Taking a look at examples.

This company -
Guangzhou SANTI Electro-Mechanical Equipment Co., Ltd
Produce this generator -
Micro Flame Generator/Hho Generator Sanho-250
There are other manufacturers with similar models.

I got a Lotus 2.2 litre 4 cylinder engine. I know from recent experience I need 2 litres every 30 seconds at 1.5 bar for full performance.
Thats 4 litres a minute.

The generator mentioned above pumps 250 litres and hour. 4.1 litres a minute. So thats about right.

It weighs 33kgs and consumes 1000watts at 220Volts so current is 4 or 5 amps according to P=V I

Lets keep it simple. a 12V 100aH battery feeding through an inverter, like the ones you use for camping to run a portable TV off a car battery, will lose about 20% power as inverter is 80% efficient.

The battery will weigh about 15-20Kgs
So you end up with 220V and 80aH or 5 amps for 16 hours - although you wouldn't want to flatten it. You could use 2 batteries wired in parallel.

Install a water tank. The generator consumes 70c.c per hour

Fumigate the gas through the air filter or into the manifold.

Your adding about 50-60Kgs in weight but you've removed the fuel tank and pump etc.

Set up and drive the car like an LPG vehicle and accept that you need to recharge the batteries and re-fill with water when neccessary. I commute 2 hours a day so would need a tank to hold sufficient water for the week and hopefully recharge the battery weekly.

I'm not sure what BTU's this HHO represents so can't predict power out of it would be the same. If you need more to gas per minute then get bigger generator and more batteries.

Anyone know why this wouldn't work?
(I can think of vibration damaging the generator in the boot)

Guangzhou SANTI Electro-Mechanical Equipment Co., Ltd




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Old 12-04-2011, 08:16 PM   #2
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Some results.

On Tue, Apr 12, 2011 at 3:03 PM, wrote:


I've got a 1985 Lotus Excel 2.2 litre with an aluminum engine, twin Dellorto carburettors and of course no ECU. I believe the engine draws between 2 and 4 litres of petrol per minute to fuel it. I do most of the maintenance myself.

Could I use your punch 5.0 with dedicated car batteries to generate enough HHO gas to feed into the engine via LPG carburettor type mixers to run the car entirely on HHO?
I realise I would need sufficient batteries onboard to power it and I would need to recharge the batteries when they run flat and add water where neccessary but is it feasible to run a car entirely on HHO?

At 15 amps for 1 litre per minute, I would need to supply 60 amps to get 4 litres per minute. A 220 aH battery would therefore last 3.5 hours of driving before being completely flat. I could install 2 batteries in parallel to ensure enough power and install an onboard Electric Vehicle type battery pack recharger to top up the batteries.

My daily commute is about an hour each way, generally at 50 - 60 mph which would probably only need 2 litres of fuel per minute.

I see it a bit like an electric HHO hybrid but would need to recharge the batteries daily and add water - this is still preferable and cheaper than a tank of petrol.

I presume rust may be a problem but if I replace the Piston Rings, Cylinder Liners, Exhaust manifold and anything associated with the valves with components that won't rust (it has stainless steel exhaust - the block and head is aluminum) then could pure HHO power the car?

I am very tempted to buy your kit and test it out, but do you know in advance if this would or wouldn't work?

Would rusting in the engine be a problem?

Many thanks for any help towards answering this question.
No this would not work you would need a LOT more hho than your calculations include. You could get some killer gas milage though using the punch 5.0 to supliment the burn.

Punch HHO

HHO Kits Direct -
Hi I've got a 1985 Lotus Excel 2.2 litre with an aluminum engine, twin Dellorto carburettors and of course no ECU. I believe the engine draws between 2 and 4 litres of petrol per minute to fuel it. I do most of the maintenance myself.

Could I use a HHO generator with dedicated car batteries to generate enough HHO gas to feed into the engine via LPG carburettor type mixers to run the car entirely on HHO?
I realise I would need sufficient batteries onboard to power it and I would need to recharge the batteries when they run flat and add water where neccessary but is it feasible to run a car entirely on HHO?
It is not possible yet. There are those of us working on that, but it is not efficient enough to produce substantial amounts to run only hho.
But there are some enthusiasts that are upgrading to high output alternators that use less horsepower than normal ones, so they have an extra 100 amps available to run a dual generator system...
cool, so with two generators they can get even better mileage gains I presume?
especially if you are running a manual carb set up where you can adjust the air fuel and even timing if need be...
Yep, I've got twin Dellorto carbs on my Lotus Excel - I guess I would need to change the jets?
no- I wouldn't do that. Think of the hho as a catalyst, not a replacement
But to improve mileage would I not need to change anything or just add the gas and go easy on the accelerator pedal?
Most of the time just add gas and go easy on the pedal. We have had some people enjoy the extra power- so mileage gains were not huge, but they didn't care.
Ok, but to get mileage gains - how would I do it on a carburettor car?
As I said- you can adjust air fuel screw on carbs to lean out a bit, and adjust driving. There will be more power because of it.
Ahh right - I get it yes. On my car I've got air bleed and the Idle adjustment screws. The idle adjust affects the mix all through the rev range.
I am not a tech support guy- you would need to talk to Bob for details beyond this
Cool thanks for the help, I think I understand a bit more though so thanks again
Good luck!
So even the guys selling it say No!
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Old 13-04-2011, 07:56 PM   #3
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I've been giving this more thought and done a bit of reading.

Consider a 1 litre 2 stroke engine.
Air fuel mixtures for combustion engines should be 14.7% (call it 15% for easier maths)

At 1000rpm there are 500 intake strokes. 2 strokes make 1 revolution.

Thus we need to supply 15% of 500 litres per min

=75 litres per min.

14V HHO cells typically comsume 15 amps to produce 1 litre of fuel per min. 30 amps for 2 litres. (1 cell)

You would need 38 HHO cells all drawing 30 amps and producing 2 litres per min to power a 1 litre 2 stroke at 1000rpm

Consider a 2 litre 4 stroke. (I'm having trouble getting my head around this one but I think the following is correct.)

Each cylinder = 500cc

There is 1 intake stroke in the 4 stroke process

So 1000rpm = 250 intake strokes of 0.5 litres per cylinder

= 125 litres per min

15% of 125 litres = 18.75 litres per minute of fuel, per cylinder

4 Cylinders = 4 x 18.75 = 75 litres per minutre, per 1000 rpm

Again, you would need 38 HHO cells drawing 30 amps and producing 2 litres per cell to fuel the car at 1000rpm
38 HHO cells x 30 amps = 1140

My Lotus redlines at 7000rpm. Therefore to fuel it to max power I would need

7 x 38 HHO cells = 266 HHO cells
7 x 1140 amps = 7980 amps

You can buy batteries with 220aH ratings. You would need 37 batteries to power the system for 1 hour!
Batteries of those capacities weigh 25Kgs = 925Kgs!

Sticking with the 1 hour, if I redlined it at 130mph for 1 hour I would have travelled 130 miles - generally speaking an electric car with that many onboard batteries would do way more miles.

Running a car on HHO - ain't gonna happen with current 14V HHO cells.
You could wire the batteries in series for 240V pack and use thatat low current (as per the OP) but even still, how many of those welding HHO generators can you fit into a car?

My mistake was the 2 litres of fuel from the fuel pump in my OP. That 2 litres of fuel or 2 litres of LPG is then atomised to represent 15% of the air fuel mixure - petrol and LPG are both stored as liquids. 2 litres of gas from the HHO cell, then mixed into the volume of air passing through the engine would be vastly underpowering the car. If your supplying gas then you need to constantly supply 15% of the volume of air fuel. As it can be seen above - this is way more than 2 litres per min.

(What I can't get my head around is, there are 4 cylinders all on the same crank, each one drawing 1 intake stroke every 4 strokes and don't forget it takes 2 strokes to complete 1 revolution of the crank (piston up and down once = 1 rpm). So how many intake strokes per min per 1000 rpm are there for all 4 cylinders?)

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Old 20-04-2011, 12:40 AM   #4
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Some good info.

I'm surprised at the lack of chemical knowledge around here.

The answer is easy to find and I'll show you all the math behind it:

Water has a density of 1g/mL, so 1L = 1000g.

Water has a molar mass of 18.01g. Therefore, the number of moles of water in one liter is 1000/18.01 = 55.52 moles.

One mole of water decomposes into one oxygen and two hydrogen atoms. As such, the decomposition of one mole of water yields one mole of hydrogen gas and one-half mole of oxygen gas through the following equation:

(H2O) --> (H2) + 1/2(O2) (remember that the gases are diatomic)

Subsequently, 55.52 moles of water decompose into 55.52 moles of hydrogen gas and 27.76 moles of oxygen gas.

According to the Ideal Gas Law, at standard temperature and pressure (1 atmosphere, 293.15 Kelvin), one mole of any gas will occupy 22.4 liters of volume.

One liter of water has produced 55.52 moles of H2 and 27.76 moles of O2 for a combined total of 83.28 moles of gas. Each mole of gas occupies 22.4 liters, so 22.4*83.28 = 1865.5 liters of gas.

And thus we arrive at the answer: One liter of water makes 1865.5 liters of gas.

__________________________________________________ ______________

As far as running an engine on nothing but the gas mix, generation problems aside:

The lower flammable limit of hydrogen in air is 4% by volume. Therefore, it would be theoretically possible to run an engine on a blend of oxyhydrogen and air which would amount to 4% hydrogen delivered at the manifold.

Air is a mixture of many gases, mainly nitrogen and oxygen. The average molar mass of air is 28.97. This is the average weight of 22.4 liters of air at STP.

To achieve 4% hydrogen by volume in air by adding a mix of 2:1 H2:O2 takes a bit of simple math. The equation goes like this:

(% gas A)(molar mass of gas A) + (% gas B)(molar mass of gas B) = (100%)(molar mass of mixed gases)

Therefore, we can solve for %HHO needed as long as we know the molar mass of the target gas. The target gas in this case is 4% by volume hydrogen and the rest air. Let us determine the molar mass:

1 mole of air is 28.97g and occupies 22.4 liters. 4% by volume of 22.4 = 0.896. 22.4-0.896 = 21.504. Therefore, the target gas has 21.504 liters, or (.96 moles) of air and .896 (or 0.04 moles) of hydrogen. Cumulatively, the molar mass is (.96*28.97) + (.04*2) = 27.89 grams per mole.

This makes sense because we know that adding hydrogen to air will decrease its molar mass because it is much less dense than air.

Anyway, now to find the amount of HHO needed to make air 4% hydrogen. We come back to the equation:

(%HHO)(molar mass of HHO) + (100-%HHO)(molar mass of air) = (100%)(molar mass of mixed gases)

And insert the values

(%HHO)(18.01) + (100-%HHO)(28.97) = (100%)(27.89)

and then solve for %HHO, which comes to 9.9% after a bit of algebra.

Therefore, you need to add at least 9.9% HHO to air to make it combustible.

__________________________________________________ ______________

So what does this mean? How about an example:

In a 2-liter engine at 1000RPM, you need a fuel/air intake of 1000 liters per minute. (500 2-liter intakes in 1000 rotations in one minute) 9.9%vol of the gas has to be HHO in order for combustion to take place, so 9.9% of 1000L = 99 liters.

Your gas generator must produce, at minimum, 99 liters of HHO per minute. This guarantees, however, that the fuel/air entering the engine is combustible. This makes no statement as to whether the combusting mixture will release enough energy to keep the engine spinning, much less move a vehicle. That calls for a whole other set of equations.


Assuming you people will want "That next set of equations," I've compiled them here:

We'll have to work with a practical example in this case. Let us examine again the 2-liter engine spinning at 1000 RPM. How much energy can it output running on only 4% hydrogen? This is an easy calorific equation which can be readily calculated.

We know that when hydrogen reacts to form water, a certain amount of energy is released. We also know that of the 1000 liters of air entering the engine each minute, 40L, or 4% of it, is hydrogen gas. 40L of hydrogen gas, according to the Ideal Gas Law, is equivelant to 1.786 moles.

Two moles of hydrogen react with one mole of oxygen to produce one mole of water and some heat. The heat of combustion of hydrogen and subsequent formation of water is 285.83 kilojoules per mole of hydrogen reacted. Therefore, 1.786 moles of hydrogen per minute will generate 510.49 kilojoules of energy per minute. This can be converted to horsepower since one horsepower is equivalent to 44.74 kJ/min.

As it turns out, 510.49 kilojoules per minute equates to 11.04 horsepower. This is, of course, assuming that all of the heat energy released is converted to mechanical energy. We know this to be very untrue - the efficiency of a modern internal combustion engine is around 19%. This means that just about 2.09 horsepower would be mechanical - not even enough to keep the engine spinning - while the rest would be lost as heat.

So how much HHO would it take to produce something like 50 horsepower with a 19% efficient engine? We simply run the equations backward: 50hp = 2237 kJ/min divided by 19% = 11774 kJ/min. One mole hydrogen is equivalent to 285.83kJ, so that makes 41.19 moles of hydrogen needed. In liters, 41.19 moles is 922.7 liters of hydrogen per minute.

Assuming you'd be feeding the engine pure HHO at this point, we know that half a mole of oxygen is needed for every mole of hydrogen, so 922.7 + 461.4 = 1384.1 liters of HHO per minute, or 23 liters per second. As a fun fact, if the engine displacement is 2L, it would have to spin at 1384.1 RPM while producing this 50hp.

On a last note, let's compare this to gasoline. Gasoline has a listed average energy density of 34,200,000 joules/liter when reacted with air. One liter of hydrogen (or 1/22.4th of a mole) contains (1/22.4)*285,830 = 12,760 joules per liter when reacted with air. Take 2/3 of the 12,760 to account for the oxygen in the mix and you end up with 8,507 joules per liter of HHO.

This shows that every liter of HHO is equivalent to just 0.25mL or 0.00845 fl. oz. of gasoline with respect to energy content. Conversely, one gallon of gasoline is worth 15,140 liters of HHO.

Electronut, as much as I respect your mathematical prowess, you are missing a fundamental point in your calculations.

When you say 1L of HHO gas produces only 12670 J, you are forgetting that 1L of WATER Produces approximately ****1866 LITERS **** of HHO Gas !!!!

This shows that the energy in 1L of water is at least 23642220 J

Not Bad for stuff that falls from the sky.

All the mathematical comprehension in the world cant help you if haven't got your numbers straight in the first place.

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Old 20-04-2011, 12:55 AM   #5
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Some more good info posted as comments to a negative article.

July 15, 2008 at 8:29 pm

There are some technical terms to show how HHO can be used as a fuel “supplement”, and by using it as that, mileage increases can be seen.
Lets explore them:

Its a know scientific fact that you cant get more energy out of something than you put in. For example, a 1.5v battery cannot produce more than 1.5v.

Correct? Good.

Same goes for anything.

Energy cannot be created or destroyed, however, it CAN be converted from one form to another, ie, from chemical to mechanical, to kinectic.(as an automobile does).

The automobile takes the stored energy in the form of gasoline, diesel, natural gas, etc and converts that stored energy to mechanical energy via the engine.

Part of that mechanical energy is converted to electrical energy via the alternator, and part of that energy is converted to kinectic energy via the transmission to move the car forward, and part of that energy is converted to heat (wasted energy).

So, the stored energy potential in a tank of gas is converted to several other forms of energy to move your car.

This is an automobile in its energy form.

Now the automobiles alternator(generator) doesnt just charge the battery, it produces all the electricity to run the car and all electronics contained. For example, you can start your car, and completly remove the battery and the engine will still run until it runs out of fuel.

(I dont recommend this though; todays cars electronics are very sensitive and *could* be damaged.)

The battery’s job is to start the car, and run lights and other stuff only if the engine is not running. Your battery will die if you leave your lights on after your turn your engine off, but it will not die while the car is running with the lights on.

Now the alternator only produces what the car and its electronics demand at any given moment. It is CAPABLE of producing in the range of 80A on some vehicles and heavy duty ones can produce up to 120A. Putting an extra load on the alternator by running an HHO generator @ 20A will have almost NO negative effect of reduced gas mileage. Have you noticed any decrease in mileage just because you run with your headlights on? They draw 20-30A when on. The air conditioner doesn’t draw power from the alternator. It creates “drag” on the engine by engaging a mechanical clutch connected to a pump. Yes, the ac compressor is just a pump, it siphons mechanical energy, not electrical. Only the fan that blows the air out of your ducts draws any electrical energy.

Try this.

Start your car.

Turn on your lights. No RPM change. Now turn on the wipers. Still no change. Turn on the radio. No change. Turn on the fan(not the AC, just the fan). Again no change in engine load that would require the engine to work harder. Turn on every electrical thing in your car and all you are doing is drawing on a RESERVE of electrical energy already being produced by the alternator.
OK, on to the next point.

An HHO generator “mostly” draws 20A (yes, some do draw more, others less) but from what I’ve seen the average is 20A. This 20A draw on the alternator is almost like turning on your headlights, fan and radio at once.

Now the 20A that the generator is using will produce varying amounts of HHO, some claiming up to 4 Litres per minute. Lets be reasonalble and call it more like it really is and say *most* will produce only about 1.5 ltrs/min.

Thats 1500ml/min. Divide that by 60sec. and you get a rate of 25ml/sec.

Electrical energy is measured by WATTS. To get watts, you multiply the Amps by the volts, ie, the alternator is wired to output around 13 to 14 volts.

so the mathematical equation for this is 20A x 14v = 280Watts. Thats the energy output of the *standard* HHO generator. Lets be generous by saying that the cell will be 90% efficient, losing only 10% in heat.

Yes they do heat up.

But really, how efficient is an HHO Cell.

there is a formula, and that formula is this:

MMW = ML per minute per watt

ml per min / Watts = MMW

Let take a *standard* cell and run the numbers.

1500 ml / 280w = 5.35714285714(MMW)

Now we have a solid number ( this is for you math geeks, like me ). We can round it out to 5.4 MMW

Now we have to calculate amount of air/fuel that the engine uses. Most modern cars ( 1996 to present) has a preset Air/Fuel ratio programmed into their computers. That is 14.7 parts of air to 1 part of fuel. 14.7:1

To figure out how much air/fuel per minute the engine uses, use this formula:
RPM x Engine displacement. I’ll use mine. A 2.2L engine that idles at 600 RPM. So 2.2 x 600 = 1320 litres of a/f per minute. In seconds that is this number:
2.2 x (600/60) = 22 litres of a/f mix per second. then divide that number by 4 (cylinders) = 5.5L a/f mix. Breaking it down more: 5.5l(or 5500ml) / 14.7(air
to fuel ratio) = 374.5ml of air and 1ml of gas for a total of 375.5ml total a/f mix.

Now take your ml/min and break it down to ml/sec: 1500/60=25ml/sec
We now have our math worked out.

So we have in our (one) combustion chamber 25ml of HHO and 375ml of a/f mix

Will these numbers help burn the a/f mix more efficiently? Probably. With the addition of an electronic fuel injection device, and the more efficient burn of the a/f mix, i can see the potential increase in mpg.


Lets look at the electronic fuel injection device:

The concept is that a more efficient burn in the cylinder will result in less unburned gas passing thru the exhaust. The exhaust system has a sensor in it to measure the amount of oxygen in the exhaust stream and compare it to the ambient(outside) air. There will always be a difference in the two and that difference causes the 02 sensor to generate a voltage that ranges from 0.1v to 1.0v. That voltage is sent to the computer to be compared to a set value. The computer then uses that voltage to adjust the fuel injectors “injector pulse width” ( the time the injector stay open) and try to maintain a 14.7:1 a/f ratio.

Since the 02 content is constantly changing due to varying conditions ( driver demand ), the 02 sensor generates a varying voltage: the higher the voltage the computer sees that as a “rich” condition, meaning less 0xygen and too much gas, so it leans out the mix. On the other end of the scale, the computer interprets the low voltage as a “lean” condition and it adjusts the injector pulse width to put more gas into the mix.

The Efie adds voltage to the existing voltage the 02 sensor generated and sends the total to the computer, in effect, fooling the computer into see a “rich” condition, so the computer “leans” out the fuel mix by shortening the injector pulse width.

So now you have a “leaner” air/fuel ratio in the cylinder. For example, the normal idle will bring in 375 ml a/f mix/sec but with the computer leaning out the mix, that number drops. The HHO now has a better chance to increase the burn characteristics of the a/f mix.

So this suggest the mpg WILL increase.

How much? It will vary greatly due to so many factors. Altitude, load(weight), speed, and driver demand.

But the net effect would seem to substantiate an mpg increase.

So there you have it in very technical terms as to why I think this is a viable thing to explore.
• Craig Spitler
December 16, 2010 at 1:42 am

I have skeptically watched friends install these aftermarket devices. Surprisingly they have reported 10 to 20% improvements in mileage, which is nothing to sniff at! My own background is in physics, and at first I could not understand how increasing the electrical HP demand on the motor could result in improved mileage, but after much research I think I have the answer: it’s called fuel hybridization. What’s at work is increasing the burn efficiency in the combustion chamber.

Gasoline is a long carbon chain that burns readily, but not so completely in the time it has to do its job. By introducing hydrogen gas, it facilitates a faster and cleaner burn during the power stroke, which results in increased engine efficiency. Some people forego the messy HHO generator and inject propane instead, which also increases the burn efficiency, although not as much as hydrogen would. I was talking with an older gentleman, that told me how in the old days, they would “spike” diesel water pumps on the farm with a trickle of propane, to double the run time. He thought that was just common knowledge…
Ted Sharp
January 6, 2011 at 4:18 am

This old engineer agrees with physicist Mr. Craig Spitler who has it right. The Hydrogen flame front velocity within the combustion chamber of an IC engine is several thousand times that of its vaporized petrochemical based fuel.

Accordingly, the intermixed Hydrogen acts as an accelerant of the total flame front throughout the cylinder causing more complete / efficient combustion of the petrochemical fuel. The Hydrogen does not of itself contribute a practical increase in energy release in the combustion cycle. However, it has been demonstrated Hydrogen can be used by itself as an IC engine fuel, but in far greater volume (hence cost )than when used only as an accelerant.

In summary, Hydrogen is not a cost effective substitute for petrochemical fuels, but it is a cost effective way to increase fuel efficiency of IC engines. An increase in gasoline and Diesel engine fuel efficiency in the order of 20% to 40% has been, and is daily being, demonstrated in practical applications.

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Old 20-04-2011, 12:59 AM   #6
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Lots of accurate info regarding Meyer's Fuell Cell replication and explanation of it and Brown's gas otherwise known as HHO.

PDF file -


Welcome to Panacea University and the faculty section of the main Panacea-BOCAF website. Here you will find, via free access, the most current versions of educational semi-textbooks and reports of experiments having to do with alternative engineering that are not currently taught in the accepted universities. Our aims are to protect and provide studies and archives of information dealing with, for instance, free energy technology, suppressed energy technology, mileage boosting, lowering emissions, alternative fuels, interesting motor modifications and studies in rotors and magnets—sometimes called experimental magnetic motors or experimental magnetic generators.

Link - http://www.panaceauniversity.org/
I don't know anything about Panacea University, but I downloaded and read the PDF - there is some good info in there about HHO and Meyers cell.
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Old 20-04-2011, 01:19 AM   #7
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My conclusion.

If a 500cc cylinder is passing 125litres per 1000rpm then to mix as an additive - since it takes about 9% to be used as fuel and petrol takes 14.7% you would need to add in the region of 2& to 5% HHO to have an appreciable effect. At 1000rpm, at 6000rpm we would need 12 litres per min, per cylinder. That’s about 50 litres per min at 6000rpm which equals 25 generators and 25x30amps - 750amps.

Otherwise, with 1 generator, you would likely only see any difference at tickover which would diminish through the rev range to 2000 rpm, at which point 1 generator cell on a 4 cylinder car would be contributing 0.25% fuel additive to each cylinder.

Meyer’s cell replication require high quality steel, meticulous cleaning / anealing and 2 to 3 months of conditioning before anything useable can be got from it. Pulsed current is obviously a must, not 35 amps direct from the alternator. In fact Meyer’s device claimed to use little current and mainly voltage potential.

The central claim – one that I subscribe to and think is probably valid, is the idea of resonant frequency of water, and applying that frequency to disassociate the molecules. However it seems the resonant frequency is dependent on mass. Thus it seems likely the exact frequency required will be dependant on the size of the fuel cell.

It would also ‘guess’ that the spacing between the plates should also be linked to wavelength.

The Ravi cell PDF also talks about acoustic resonance - the electrical input resonates the tubes like tuning forks which vibrate the water

With all this in mind, I find it hard to see how some of the very simple HHO cell setups with steady DC current, in a unit haphazardly put together with low quality components, without months of conditioning can possibly have any affect on MPG other than the charge cooling process described in my other thread regarding water or steam injection.

I’m certainly not discrediting the HHO idea completely. I know there is real science behind it.

Its just the ability of some HHO cells to deliver HHO Gas in any useful amount, without meticulous building and engineering to incorporate the system into a modern ECU car that I now question.

A wet sponge in the airfilter would probably do the same thing.

Or a plastic bottle of water, with a lid with 2 pipes, one pipe, just above the water level in the bottle fed into the airfilter the other fed down into the bottom of the bottle and vented to atmosphere. Water bubbles through the water under the vacum from the engine and water vapour is fed into the engine for charge cooling. This has been written about on a few classic car sites. It can apparently gives a 5 or 6 percent increase in MPG.
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Old 07-04-2012, 05:42 PM   #8
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Post Physics and energy!

I am replying to a post in 2008 in order to clear up a common misunderstanding. In the energy arena, as in many others, it is important to be precise in one's wording.
It was said that ( and I quote ) " Its a know scientific fact that you cant get more energy out of something than you put in. For example, a 1.5v battery cannot produce more than 1.5v."

The first part of this is manifestly untrue, the second part has little to do with the first.

Please let me explain. As I said above,WORDING is everything. You CAN get more out of s system (or device) than you put in...much much more. The important word here is "you".

What the laws of thermodynamics actually say is that a system ( or device ) cannot output more energy that it receives as input.

The two statements are not the same in meaning.

Thermodynamics is not concerned with "you", it is only concerned with energy and the system that is outputting that energy. That is great because it enables "you" to take advantage of it.

A classic example is photovoltaic cells ( solar cells )

Here we also get a classic example of the difference between EFFICIENCY and C.O.P ( coefficient of performance)

Even today, photovoltaic cells that are available to the public are only about 17% efficient BUT.... COP is the useable energy output divided by the energy YOU have to put in. So a solar panel that outputs 100 Watts ( on a sunny day in England maybe !) does so with absolutely zero energy input from YOU. So the COP is 100 / 0 = infinity. This is free energy disregarding the outlay costs ( always )

It may waste 83% of the incident energy but the rest is yours for free as electricity!

Windmills are the same argument. ( they are more efficient though )

I realise this may seem like a pointless issue, but believe me it isn't. The lack of distinction has lead many people astray and caused countless arguments totally unnecessarily.

By the way the reason a 1.5 volt battery cannot output more than 1.5 volts has nothing to do with energy but I would rather not bore you with the details as it would get bogged down with esoteric stuff that will take too long.

While I am here, let me just make a couple of other points, if I may.

It has been said by many that it is NOT possible to run a car on water alone. This is also untrue. There are two ways of doing so, one being hydrolysis. There is at least one system that can run a car on Hydroxy gas alone, but is not particularly cheap or easy to build but does work. It will probably run an engine up to 2 Litres BUT only just and the real problem is coping with hard acceleration from standing. The other method is directly injecting water ( atomised of course ) into the cylinder and using a plasma arc to dissociate the hydrogen and oxygen just prior to the ignition spark This has been done.

One final point. It is a common misapprehension that a 100Ah lead acid battery will give you 1 amp for 100 hours ( probably true ) or 100 amps for 1 hour ( untrue ). Discharge should be limited to 20 hour rate so the the max current you should draw from said battery without causing damage to it and/ or seriously shortening it's life is 100 /20 = 5 Amps ! Clearly you can draw more than that without to much problem in reality but the figure should not be exceeded by too much or for too long. This the basis of one of the cost issues with solar systems - the sheer number of batteries you need to store serious energy to provide sustainable power for a home if you design it to discharge specs that won't damage them.

I hope I haven't upset anyone, and I hope this is of some use to someone. If somehow I screwed up and all this has been said before, I apologise. Please feel free to comment. If I can be of any help in answering any questions I will gladly do my best. I may well need to ask you guys something soon.
techknow is offline   Reply With Quote
Old 09-04-2012, 11:23 PM   #9
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Hey techknow,

welcome to the forum.

Nice post.

You've come to the right place!
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Old 29-01-2019, 05:35 AM   #10
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