OK so i decided to throw some math in here
kill me not
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:
http://en.wikipedia.org/wiki/Air-fuel_ratio1)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 liter3)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
.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}=
0,224With 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
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
