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Author Topic: HHO as a car fuel calculations  (Read 14755 times)

Offline boostmr2

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HHO as a car fuel calculations
« on: April 06, 2008, 12:09:22 AM »
Any of you math geeks out want to give this an attempt?

In the average automobile, lets say a 2-litre 4 cylinder motor, it is good to have a 14 to 1 air fuel ratio.

Air is only needed for the oxygen it contains, which is approximately 20%.

This means the oxygen to fuel ratio is more like 2.8 to 1, lets say 3 to 1 just for theory.

Would this be the same ratio we should expect for burning pure hydrogen as the fuel?  For each hydrogen molecule we need 3 oxygen for a clean burn?

Just a thought, I get lost when it comes to chemistry on a molecular level.  Hydrogen may require a completely different air/fuel ratio than petrol because of it's different properties.  But if someone could figure out what this is, then it's very possible to calculate exactly how many litres/minute HHO (or just H) need to be produced to keep a vehicle running at lets say a max of 8k rpm's consistently.  This seems to be the biggest problem right now.

The question comes up because most vehicle manufacturer's are leaning towards a fuel tank setup, not a fuel-cell setup for hydrogen vehicles.  I believe it is completely plausible to produce enough Hydrogen or HHO onboard the vehicle, and pressurize it, just like petrol is pressurized, and inject it into a cylinder for burn.  Since Hydrogen is much more volatile as well, it should cause the gases (air) to expand just as fast as petrol would but using less gas to produce the same power.  Plus its cleaner burning!  Any ideas?

Offline z.monkey

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Re: HHO as a car fuel calculations
« Reply #1 on: April 07, 2008, 10:47:17 PM »

That's a good question.  It may be up to experimentation.  I also think that the HHO generation may need to be throttled, meaning the HHO production needs to be variable and proportional to the position of the accelerator pedal, or the engine RPM.  One clue may be at NASA.  The Space Shuttle burns LOX and LH2 in the main engines.  If I am not mistaken the ration is 1:2.7.  The two tanks in the external tank are the clue.  The LH2 tank is 53,488 cubic feet and the LOX tank is 19,744 cubic feet.  The tanks are designed to be expended at the same time.  So hey, thanks NASA, that is your tax dollars doing something good for a change.  So we need a better control system for the Electrolyzer, and maybe a better Electrolyzer.  I know how to make a better control system.  Make it processor controlled.  Check out this other thread.,4293.0.html

So far this idea has been more or less ignored in this forum.  By using a processor the control circuit can take external input such as throttle position, RPM, road speed, and environmental variables to get the Electrolyzer producing HHO in the correct ratio to the air intake.  This way we don't starve the engine or burn it up.  Modern gasoline engine management systems do this.  There are sensors all over the motor, and the engine management computer is continually adjusting the mixture to provide optimum performance.  You can really tell this too.  We take it for granted.

Here's to a cleaner, more powerful tomorrow...

Offline Creativity

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Re: HHO as a car fuel calculations
« Reply #2 on: April 08, 2008, 12:00:14 AM »
burning H in O gives  water, reaction goes as follows:

2 H2+O2->2 H2O
so the ratio is 2:1 in volume.H and O are vary unstable as single atoms and gets really fast to the form of H2 and O2.So u need to base your calculations on O2 and H2.Both are gases and both occupy the same volume per Moll.You get out of the electolyser the exact ratio that you need for full burn.So air oxygen is the excess oxygen not needed in the case you burn only HHO from the elecrolyser.Some excess oxygen can ensure full burn of HHO.

If u want to calculate situation for any specific engine we just need the fuel type it is currently using and the kg of fuel per hour per kW burnt in this engine.Based on this u can easily calculate the amount of HHO needed to generate the same amount of work as other fuel.

i can do it for u tomorrow,

Offline z.monkey

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Re: HHO as a car fuel calculations
« Reply #3 on: April 08, 2008, 12:15:27 PM »
Uh, Creativity,
H2O2 is Hydrogen Peroxide...
That's rocket fuel...

Offline Creativity

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Re: HHO as a car fuel calculations
« Reply #4 on: April 08, 2008, 03:10:34 PM »
:) where i wrote H2O2 ? it is just that hydrogen as well as oxygen forms a two atomic molecule when in gas state.So u have H2,O2 and by the way the same for nitrogen N2 .A lot of gases have this form in normal conditions.So when u have electrolysis you get H2 above one electrode and O2 above the other one,when u mix them u have H2,O2 gas mixture.H2O2 is liquid and it is a molecule of 4 atoms.

check yourself:

Offline Creativity

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Re: HHO as a car fuel calculations
« Reply #5 on: April 14, 2008, 12:57:34 AM »
OK so i decided to throw some math in here :) kill me not  ;) ;D use it if u need :)

Let us calculate how much an engine needs to run.Goal is to calculate an amount of air and fuel used.I will calculate the air needs in liters as we are interested here in volumes to compare it later to the volumes of H2 or H2,O2 gases needed to run an engine.
First of all who still has no knowledge of AFR (air fuel ratio) please read:

1)Now our good running engine uses around 14,7 g of air per 1g of gasoline (heptan,octane mixture),that comes out of AFR calculation of perfect burning.I will make an example how to calculate the AFR for a given fuel:

C7H16 + 11O2 ->7C02 + 8H20

heptane has a moll mass = 7*12+16*1=100 [g/mol]
we are using air as an oxidant so we have also nitrogen present and it is 79 % of volume of air.It will give us around 41 molecules of N2 in this situation.So our air mole mass=11*16*2 + 41*14*2=1.500 [g/mol]
 now 1.500/100 = 15 this is our AFR for heptane.

2)just how much is this 14,7 g of air in volume?

Air has density of 1,2 g/liter.Volume is then =12,25 liter

3)let assume that our engine uses 10l of gasoline to drive our car 100km/h in 1hour.So an engine of around 2-2,5 liters of displacement,rated with around 150HP( i assume ,but i feel it can be also quite good calculated out of engine efficiency and gasoline caloric value).
How much air volume will it use?

Density of gasoline 737,22 [g/liter] shows us that we used 7.372g of it for our trip.
So air used=14,7*7.372=108.368 g.
It is then =108.368/1,2= 90.307 liters of air.

4)How much of air was used per minute?

90.307/60=1505 liters

5)how much fuel per minute was used?

7.372/60=123 g

6) and when the fuel was evaporated how much space it occupied?

Gasoline expands around 220 times in volume when evaporated(comes from comparison of density ratio of liquid gasoline with the gaseous form) , so we used 2200liters of evaporated gasoline.
It is around 2200/60=37 [liters/minute].

OK let as got some attention to our big question of using H2,O2.From a SAE technical paper i have a graphic describing the effects of adding H2 into the internal combustion engine runned on CNG.Graphic speaks about the possibility of further shift of lean mixture AFR limit when H2 was feed to the fuel.Unfortunately i can`t put this graphic here because it is copy righted.But i think copyright allows me to say what is the effect observed :) Lean limit of pure gasoline is around AFR =25:1 according to this source.Adding of 8% of H2 into the combustion shifts AFR to lean limit of 26 :1.So it is a shift of 4%.Further addition of around 20% of H2 shifts AFR to 27,5:1 resulting in relative 9% shift.So it makes some sence to add H2 to the engine.

This effect was described for addition of pure H2,however i believe if H2,O2 was used effect could be slightly different.Result of electrolysis in form of a H2,O2 mix has 33% of volume made of O2.I could think of two situations:
a)extra oxygen would lean the charge even more and counteract the H2 effect
b)oxygen would assist the burning of the fuel by adding extra heat to burn even leaner charge.

Now let me try to calculate the H2 burning in air.I will try to make analogical calculations as for heptane,just to see how much of the H2 would we need to run our can in a situation as above.

2H2 + O2 ->2H2O

AFR (2*16+3,76*2*14)/(2*2)=34,25 :1

it is 34,25/1,2=28,5 liters of air per 1g of H2.

This time we need to know how much energy was released in gasoline engine,because we want to make our hydrogen car to run at least as well as our gasoline car.In the first example we used 7.372g of gasoline for our trip,this would translate to 327MJ produced.I won`t take into consideration the efficiency of the engine,i just assume it won`t change substantially under pure H2.I make this assumption to make first approximation of amount of H2 required to run an engine.Later we can add efficiency to the calculation.
327MJ translates to 327[MJ]/130[MJ/Kg]= 2.517 g of hydrogen used to make the same work.
Now volumes:

0,09 g/liter is the density of H2 gas.

0,09*2.517=27.967 liters of H2 gas

As in our previous example all of it was used in 1h time for our trip.It is 27.967/60 = 466 liters of gaseous H2 per minute!.
Norm amount to produce from on-board electrolysis alone,no?But that is not the end now.Just compare the numbers:

Our car on gasoline used 90.307 liters of air + 10 liters of not yet evaporated fuel.This all went for the trip.I use 10l of gasoline because when it was feed to the engine most of its evaporation process took place inside of the cylinder.
Our car would need to use 27.967 liters of H2 + 2.517*28,5 liters of air in total to burn it for the trip.IT is 99.701/90.317= 1,1 times more volume of gases would have to go through the engine(so volumetric efficiency would have to improve).We simply suck not enough oxygen from the air to burn our H2.Our engine would have to be supercharged or rpm would have to go higher for the same power demand.

Situation would be different if we used 2H2,O2 mixture(as from electrolyser).In that case AFR would be a bit more tricky to calculate.We have here an extra oxygen feed and exactly as much as we need,In this case oxygen from the air is not needed at all and a result will be that we run lean mixture.Problem gets not easier when we feed more H2,O2.In that case we get more power and hot running engine,feeding even more will result in big explosions and only way i can see would be inject water to cool down the engine and slow down the burning.Maybe dumping of some oxygen from the electrolyser outside of the engine could help,just to achieve a stoichiometric mixture for H2 without adding more of H2 and ruining the engine.

Coming back to our car trip  ;D.We still want to produce the same energy as in pure gasoline or H2 case.This time we have extra oxygen available so volumetric efficiency won`t suffer,no supercharging or higher revving will be needed.We can then stick to gasoline base case  90.317 liters of charge introduced to the engine during the trip.The reason is that we can manipulate the amount of oxygen supplied to burn H2.We have an excess of oxygen so we will dump all the oxygen that could cause lean burn.We use as in a pure H2 case  27.967 liters of H2.The  90.317 - 27.967=62.350 liters will have to contain all the oxygen we need.Only air won`t support enough oxygen (as shown in pure H2 usage case),supplying of all the O2 from electrolyse,next to ordinary air will give us too much oxygen.Wear have to find how much of the O2 we need to burn stoichiometric(the best).
 coming back to :

2H2 + O2 ->2H2O

What we see here is we use 1 volume of O2 for every 2 volumes of H2 to have a nice burn.We have 27.967 liters of H2 so we need the half of it in volume of O2.

1-{[62.350 -27.967/2])/62.350}= 0,224

With above formule i calculated what part of the total air and O2 supplied has to be the oxygen.In easy words we need upgrade air to have 22,4% of oxygen.The rest of the oxygen we don`t need anymore.With this i offer you ,the one who had a long road to read through all this calculations  ;D  8) my respect :).As a reward i can bring u step closer to the solution of an on-board hydrogen production.

I am an author of an idea as follows:
As u see oxygen release from the electrolysis is not what we want.We can make a small amount of O2 (2,5% of air volume sucked to the engine) but it costs us a big penalty of energy used to release this oxygen from OH bond.Sure we can Strip it and get this one extra hydrogen,that is what you all do in electrolysis.I say it is not the way.Use this energy to strip another water molecule of the only one Ht.As well as combustion engine,fuel cell can also use oxygen from the air.What i see, is usage of the low energy electrolysis with minimal OH bond braking rate.This cell will produce almost only hydrogen that can be feed to the fuelcell to produce electricity with air oxygen.This system has a chance to become overunity.The secret of the not stripping of the OH bond will stay for a moment here.I have to finalise my long research on it first.

all the best,
Bartosz Paszkowski
« Last Edit: April 15, 2008, 02:35:14 PM by Creativity »

Offline 22350

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Re: HHO as a car fuel calculations
« Reply #6 on: June 03, 2008, 07:54:58 AM »
:) where i wrote H2O2 ? it is just that hydrogen as well as oxygen forms a two atomic molecule when in gas state.So u have H2,O2 and by the way the same for nitrogen N2 .A lot of gases have this form in normal conditions.So when u have electrolysis you get H2 above one electrode and O2 above the other one,when u mix them u have H2,O2 gas mixture.H2O2 is liquid and it is a molecule of 4 atoms.

check yourself:
amazing.  if you look at dingels fuel cell, he has a smaller tube, which he says is to extract the oxygen separately from the hydrogen.  I wonder if he found the right mixture of the two gases, with a modified air intake.

Offline jandre680

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Re: HHO as a car fuel calculations
« Reply #7 on: November 18, 2008, 03:38:38 AM »
Estimating Volume of Hydrogen Gas required to run an Internal Combustion Engine 100% on Hydrogen.

This site has a calculator for making that determination.  The numbers are higher than required because H2 burs more efficiently and the engine timing will need to be adjusted to TDC.

Offline Creativity

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Re: HHO as a car fuel calculations
« Reply #8 on: November 18, 2008, 01:58:31 PM »
that sustains my previous calculations that u need an order of hundreds of liters per minute.
when digging on their website i found an interesting article :

About blind setting of ignition to TDC i disagree.If u touch ur engine settings at least know why and how it affect the rest.None of the fuels has an infinite small time of burn(infinite time of flame front propagation).Let us say it takes 10milisec from ignition to peak pressure in the combustion chamber.This time will vary slightly due to varying conditions of charge preparation(different rpm's,different throttle position).We can consider it to be constant ,i just want to make a simple example that can be further worked out if desired.
What is desired, is to have this peak pressure slightly after TDC(measured experimentally for best performance/efficiency considerations)Advancing of ignition allows us to controll when will peak pressure occur according to piston movement.So we advance let say 15deg at 2000rpm and 25 deg at 5000rpm.We advance more at higher rpm's because piston moves faster and TDC is reached faster,so charge needs to be ignited earlier to reach peak pressure slightly after TDC.
The drawback of advancing of ignitions is lost work.Piston has to work against rising pressure of combustion until it reaches TDC.That's where use of faster burning fuel makes savings(winning this lost work back).that's not the only place through the cycle where u can win..

Hydrogen propagates flame front much faster,so we will get peak pressure much faster.Let us say time from ignition until peak pressure will be 0.1 milisec(100 times faster than for gasoline).Now it will translate to ignition advancement of 0.15deg before TDC at 2000rpm and 0.25deg for 5000rpm.Leaving the factory settings of ignition for pure hydrogen would be a disaster(kick back of an engine),however here also experiments would show what the right setting is desired in function of rpm's and load.

There is a thing not to forget!!! remember ur engine igniter has an automatic advancement/retarding circuit(either mechanical or on ignition map).
Even if u manage to retard timing u would have to make sure ur advancement curve is adjusted!!! for example retarding ignition to 0.5 deg before TDC at 2000rpm,rest untouched ,would leave u with 10.5 deg AFTER TDC at 5000rpm...far from optimum.So any change in base ignition timing would have to go hand in hand with adjusted maps or modified mechanical advancer and retarder.

Loose from ignition.Feeding ur engine with gaseous fuel reduces volumetric efficiency greatly.It means u can feed less air and get less power from ur engine this way.(analogy to CNG and LPG alternative fuels)

Offline CrazyEwok

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Re: HHO as a car fuel calculations
« Reply #9 on: December 11, 2008, 02:33:33 AM »
You all are missing one really big thing... Your Engine management system has a variable calculator built in... The only problem is its not designed for calculating with the power of H as the fuel. IMHO the best way would be if we can get an electrolizer to not only separate the H and O from the water but collect them separately. Then we can pump the right amount of H and O into the "fuel" lines to go to the injectors ( the lines might have to be replaced so that they are thinner thus reducing the actual amount of gas under the same pressure that the system is used to). Change your injectors to Vapour/gas injectors and let the cpu do the rest. The 2 hardest things will be pressurizing the lines to the appropriate pressure (a pressure switch on the fuel lines maybe used i guess to regulate the reaction chambers production  ::)) and making sure that the mix is right in the H and O. Asside from this we can use atmosphere as gap pressure and ensure that there is more than enough O to complete the burn hopefully reducing the chance of hydrolization...

if we can get the correct amount of flow in the fuel lines the CPU should advertantly do the rest for us by controlling input and amount through control of the injectors...

Hope that idea helps.