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Fuel Calculations HHO Hydrogen

Please NOTE 

1 .A Running car is using alot of air volume from ambient natural air this ar must be ionized and the O2 Content Enhanced by stripping off electrons.  

2. Also we Meter mix EGR valve air from exhaust standard on all vehicles

3 We use Precise injection from MegaSquirt Secure Supplies 

4 Use a Split Gas Cell from Secure Supplies 


Calculating HHO requirement:
PLEASE NOTE this is for THE AIR FUEL GASEOUS RATIO IN THE CYLINDER

3.2 Combustive Properties of Hydrogen
The properties of hydrogen are detailed in Section 1. The properties that contribute to its use as a combustible fuel are its:
• wide range of flammability
• low ignition energy
• small quenching distance
• high autoignition temperature
• high flame speed at stoichiometric ratios
• high diffusivity
• very low density

Wide Range of Flammability

Hydrogen has a wide flammability range in comparison with all other fuels. As a result, hydrogen can be combusted in an internal combustion engine over a wide range of fuel-air mix-tures. A significant advantage of this is that hydrogen can run on a lean mixture. A lean mixture is one in which the amount of fuel is less than the theoretical, stoichiometric or chemically ideal amount needed for combustion with a given amount of air. This is why it is fairly easy to get an engine to start on hydrogen.
Generally, fuel economy is greater and the combustion reac-tion is more complete when a vehicle is run on a lean mix-ture. Additionally, the final combustion temperature is generally lower, reducing the amount of pollutants, such as nitrogen oxides, emitted in the exhaust. There is a limit to how lean the engine can be run, as lean operation can sig-nificantly reduce the power output due to a reduction in the volumetric heating value of the air/fuel mixture.

Low Ignition Energy
Hydrogen has very low ignition energy. The amount of energy needed to ignite hydrogen is about one order of magnitude less than that required for gasoline. This enables hydrogen engines to ignite lean mixtures and ensures prompt ignition.
Unfortunately, the low ignition energy means that hot gases and hot spots on the cylinder can serve as sources of igni-tion, creating problems of premature ignition and flashback. Preventing this is one of the challenges associated with run-ning an engine on hydrogen. The wide flammability range of
hydrogen means that almost any mixture can be ignited by a hot spot.

Small Quenching Distance
Hydrogen has a small quenching distance, smaller than gasoline. Consequently, hydrogen flames travel closer to the cylinder wall than other fuels before they extinguish. Thus, it is more difficult to quench a hydrogen flame than a gasoline flame. The smaller quenching distance can also increase the tendency for backfire since the flame from a hydrogen-air mixture more readily passes a nearly closed intake valve, than a hydrocarbon-air flame.

High Autoignition Temperature
Hydrogen has a relatively high autoignition temperature. This has important implications when a hydrogen-air mix-ture is compressed. In fact, the autoignition temperature is an important factor in determining what compression ratio an engine can use, since the temperature rise during com-pression is related to the compression ratio.The temperature may not exceed hydrogen’s autoignition temperature without causing premature ignition. Thus, the absolute final temperature limits the compression ratio. The high autoignition temperature of hydrogen allows larger compression ratios to be used in a hydrogen engine than in a hydrocarbon engine.
This higher compression ratio is important because it is re-lated to the thermal efficiency of the system as presented in Section 3.7. On the other hand, hydrogen is difficult to ignite in a compression ignition or diesel configuration, because the temperatures needed for those types of ignition are rela-tively high.

High Flame Speed
Hydrogen has high flame speed at stoichiometric ratios. Un-der these conditions, the hydrogen flame speed is nearly anorder of magnitude higher (faster) than that of gasoline. This means that hydrogen engines can more closely approach the thermodynamically ideal engine cycle. At leaner mixtures, however, the flame velocity decreases significantly.

High Diffusivity
Hydrogen has very high diffusivity. This ability to disperse in air is considerably greater than gasoline and is advanta-geous for two main reasons. Firstly, it facilitates the forma-tion of a uniform mixture of fuel and air. Secondly, if a hydrogen leak develops, the hydrogen disperses rapidly. Thus, unsafe conditions can either be avoided or minimized.

Low Density
Hydrogen has very low density. This results in two problems when used in an internal combustion engine. Firstly, a very large volume is necessary to store enough hydrogen to give a vehicle an adequate driving range. Secondly, the energy den-sity of a hydrogen-air mixture, and hence the power output, is reduced.

3.3 Air/Fuel Ratio
The theoretical or stoichiometric combustion of hydrogen and oxygen is given as:
2H2 + O2
= 2H2O
Moles of H2 for complete combustion
= 2 moles
Moles of O2 for complete combustion
= 1 mole
Because air is used as the oxidizer instead oxygen, the nitro-gen in the air needs to be included in the calculation:
Moles of N2 in air
= Moles of O2 x (79% N2 in air / 21% O2 in air)
= 1 mole of O2 x (79% N2 in air / 21% O2 in air)
= 3.762 moles N2
Number of moles of air
= Moles of O2 + moles of N2
= 1 + 3.762
= 4.762 moles of air
Weight of O2
= 1 mole of O2 x 32 g/mole
= 32 g
Weight of N2
= 3.762 moles of N2 x 28 g/mole
= 105.33 g
Weight of air
= weight of O2 + weight of N (1)
= 32g + 105.33 g
= 137.33 g
Weight of H2
= 2 moles of H2 x 2 g/mole
= 4 g
Stoichiometric air/fuel (A/F) ratio for hydrogen and air is:
A/F based on mass:
= mass of air/mass of fuel
= 137.33 g / 4 g
= 34.33:1
A/F based on volume:
= volume (moles) of air/volume (moles) of fuel
= 4.762 / 2
= 2.4:1
The percent of the combustion chamber occupied by hydro-gen for a stoichiometric mixture:
% H2
= volume (moles) of H2/total volume (2)
= volume H2/(volume air + volume of H2)
= 2 / (4.762 + 2)

Stoichiometric air/fuel (A/F) ratio for hydrogen and air is:= 29.6% 

As these calculations show, the stoichiometric or chemically correct A/F ratio for the complete combustion of hydrogen in air is about 34:1 by mass. This means that for complete combustion, 34 pounds of air are required for every pound of hydrogen. This is much higher than the 14.7:1 A/F ratio re-quired for gasoline.

Since hydrogen is a gaseous fuel at ambient conditions it displaces more of the combustion chamber than a liquid fuel. Consequently less of the combustion chamber can be occupied by air. At stoichiometric conditions, hydrogen dis-places about 30% of the combustion chamber, compared to about 1 to 2% for gasoline. Figure 3-3 compares combustion chamber volumes and energy content for gasoline and hy-drogen fueled engine

Figure 3-3 Combustion Chamber Volumetric and Energy Comparison for Gasoline and Hydrogen Fueled Engines ATTACHED 

Depending the method used to meter the hydrogen to the engine, the power output compared to a gasoline engine can be anywhere from 85% (intake manifold injection) to 120% (high pressure injection).
Because of hydrogen’s wide range of flammability, hydrogen engines can run on A/F ratios of anywhere from 34:1 (stoichiometric) to 180:1. The A/F ratio can also be ex-pressed in terms of equivalence ratio, denoted by phi (Φ). Phi is equal to the stoichiometric A/F ratio divided by the actual A/F ratio. For a stoichiometric mixture, 

the actual A/F ratio
is equal to the stoichiometric A/F ratio and thus the phi equals unity (one). For lean A/F ratios, phi will be a value less than one. For example, a phi of 0.5 means that there is only enough fuel available in the mixture to oxidize with half of the air available. Another way of saying this is that there is twice as much air available for combustion than is theo-retically required.


This means that for complete combustion,  

SEE CHART ATTACHED  

1 Pound Gasoline  is 3.8 liters
1 Pound Hydrogen is 6.408 liters


Air Fuel Ratio 
 ( we have to Account for Density Temp and Pressure)

14.7 pounds of air are required for every pound of gasoline
        Liters of air are required for every 1 liter of Gasoline  (convert)

34 pounds of air are required for every pound of hydrogen.
        Liters of air are required for every 1 liter  of Hydrogen (convert)


Example  engine: 305 cc Generator

PLEASE NOTE this is for THE AIR FUEL GASEOUS RATIO IN THE CYLINDER


305 cc at 3600 rpm would draw 1800 x 305 cc = 549 LPM of whatever your running on.

Feeding it with 8 LPM of HHO would leave 541 LPM to be made up with air, 
water mist, recycled exhaust gases or anything else that might help.

 If you convert those figures to a mixture percentage you get a 1.48% mix. 

You only need approximately 1 - 2% mix to run an Internal combustion engine
so your production should be bang on for that engine. Hope this helps.

That last calculation I didn't scratch down was:
4.43cc / 305cc  or 1.45% mix

So both of our calculations are correct.  

Now that's obviously assuming you get a full gulp on intake and that the engine is running at full speed producing a 60 cycle AC output on the generator.  

At lower RPM the cell has more time to produce gas and the engine gets a richer mixture to work with, so starting and idling the engine should be relatively easy.

What is interesting is the tiny amount of pure HHO diluted in that big cylinder can actually produce a big enough bang to overcome friction and still do useful work. 

Pretty sure about 1% being the minimum requirement? 
 The good thing is I see no way to over-rev the engine and blow it up.


The stoichiometric  Air-Fuel mixture according to WIKIPEDIA is :

Fuel         By mass [4]      By volume [5]        Percent fuel by mass
Natural gas   17.2 : 1   9.7  : 1                            5.8%
Propane (LP)   15.67 : 1   23.9 : 1                     6.45%
Ethanol            9 : 1           —                          11.1%
Methanol           6.47 : 1     —                          15.6%
Hydrogen           34.3 : 1   2.39 : 1                  2.9%
Diesel                14.5 : 1   0.094 : 1                6.8%
Gasoline           14.7 : 1       —                        6.8%

PLEASE NOTE this is for THE AIR FUEL GASEOUS RATIO IN THE CYLINDER

Note that : an engine can run very lean AFR with hydrogen i-e : 80:1

PLEASE NOTE this is for THE AIR FUEL GASEOUS RATIO IN THE CYLINDER

Simple calculations for LPM are based on air to fuel ratio: e.g:

1 Liter engine,  takes in 1/4 Liter of air per revolution(Because its 4 stroke),

so @ 1000 RPM , air=250LPM,  if Air-Fuel ratio=25:1, then Fuel=10 Lpm
                                             if Air-Fuel ratio=50:1, then Fuel=5 Lpm
                                              if Air-Fuel ratio=80:1 then Fuel=3.1 Lpm

@ 2000 Rpm , Air=500 Lpm , if Air-Fuel ratio=25:1, then Fuel=20 Lpm
                                             if Air-Fuel ratio=50:1, then Fuel=10 Lpm
                                              if Air-Fuel ration=80:1 then Fuel=6.2 Lpm

so a generic FORMULA can be:

PLEASE NOTE this is for THE AIR FUEL GASEOUS RATIO IN THE CYLINDER

FUEL(Lpm)= Engine Capacity/4 * RPM * AFR

then e.g: Engine capacity = 1 Liter, RPM=1000, AFR= 25:1 =1/25

then FUEL(LPM) = 1/4 * 1000 * 1/25 =   10 LPM

in case of Stan Meyer as he stated he ran car with very lean mix it can be:

if Engine capacity = 1.6L , RPM=3000, AFR =60:1
then , FUEL= 1.6/4 * 3000 * 1/60 = 20 LPM 

PLEASE NOTE this is for THE AIR FUEL GASEOUS RATIO IN THE CYLINDER

from this.
1 Liter engine,  takes in 1/2 Liter of air per revolution(Because its 4 stroke).
-i think 1/2 of rpm cause 4 stroke that has 2 cycles(revolution).meter show 2X rpm but drain air X rpm.

1:80 for hydrogen 
-but if HHO that must less than maybe 1:100(1%) or 0.5:100(0.5%). if include ionize air+laser intake must less than that maybe 0.1:100(0.1)% .this mean 4.8 LPM(0.1%) that enough to run 100 km/h(1600cc).


Keep in mind the A/F ratio for petroleum products is based on doped fuel molecules, most of which do not burn in the combustion chamber.  

HHO is already at stochiometric ratio; when you add air which is mostly nitrogen, you have a partially lean mixture, but a mixture that has a controlled burn rate to coincide with the needed burn rate inside an internal combustion engine.  

Even better is to add a plasma spark and water mist to the mixture; doing

 So should reduce the LpM HHO drastically to something manageable by even brute force electrolysis. 


PLEASE NOTE this is for THE AIR FUEL GASEOUS RATIO IN THE CYLINDER

=================================
NOTE 

A four-stroke engine (also known as four-cycle) is an <a>internal combustion engine</a> in which the <a>piston</a> completes four separate strokes which comprise a single thermodynamic cycle. A stroke refers to the full travel of the piston along the cylinder, in either direction. While risqué slang among some automotive enthusiasts names these respectively the "suck," "squeeze," "bang" and "blow" strokes.<a>[1]</a> they are more commonly termed
INTAKE: this stroke of the piston begins at top dead center. The piston descends from the top of the cylinder to the bottom of the cylinder, increasing the volume of the cylinder. A mixture of fuel and air is forced by atmospheric (or greater) pressure into the cylinder through the intake port.
COMPRESSION: with both intake and exhaust valves closed, the piston returns to the top of the cylinder compressing the air or fuel-air mixture into the cylinder head.
POWER: this is the start of the second revolution of the cycle. While the piston is close to Top Dead Centre, the compressed air–fuel mixture in a gasoline engine is ignited, by a <a>spark plug</a> in gasoline engines, or which ignites due to the heat generated by compression in a diesel engine. The resulting pressure from the <a>combustion</a> of the compressed fuel-air mixture forces the piston back down toward bottom dead centre.
EXHAUST: during the exhaust stroke, the piston once again returns to top dead centre while the exhaust valve is open. This action expels the spent fuel-air mixture through the exhaust valve(s).
-i think 1/2 of rpm cause 4 stroke that has 2 cycles(revolution).
meter show 2X rpm but drain air X rpm.

sorry my mistake, it is 1/2 of Rpm.

but still let us assume that we use 1600 cc engine.

AFR 80:1                         (not very lean mixture)
Engine capacity=1600cc    (approx 40+ hp )
RPM=2500                       (approx 100km/h)

then HHO required= 1.6/2*2500*1/80=  25LPM

PLEASE NOTE this is for THE AIR FUEL GASEOUS RATIO IN THE CYLINDER

NON MEYERS
and using brute force , a 5mmw cell will consume approx 6.6Hp (approx 5000Watt),
still  its running a 40+ Hp engine while consuming 6.6hp 


when your driving constant 100km/h the engine do not use its full power.
 The 40+hp is only the maximum power which the engine is able to produce with gasoline. 

This is a quite other calculation  SEE Part 2 

 So you need for (perhaps) 15hp (don´t know exactly) 25LPM.

Meyers System of Electrolysis using the cell and water in it as a capacitor makes a 185% efficiency gain 
and is factual and Patented and now open Source. 

Stanley Meyer Water fuel Cell is a Water Capacitor and Highly Efficient electrolyzer. 

Use Secure Supplies Step Charge PWM and Switch and Also Use a Chock with your transformers. 

========================================================
Calculating Vehicle Weights       PART 2


Air and rolling resistance

Essentially two forces slow a vehicle , namely the air and rolling resistance . This article explains the relationships and shows how much effort has to be made in tuning measures in order to increase the speed limit.
The rolling resistance is calculated as follows :

Froll = Cr × m × g

in which

Cr is a rolling resistance coefficient, which is in our case about 0.015,
m is the mass of the vehicle , including passengers (ie about 1900 kg for the 8 series )
and g is the acceleration due to gravity 9.81 m/s2 is .

The rolling resistance of a BMW 8 Series is thus 0.015 × 1900 kg x 9.81 m/s2 = 280 N

This size is , as you can see , not dependent on the speed , but on the car's weight. So it will become more and more insignificant with increasing speed. Nevertheless, the engine still has to overcome a force of 280 N = 28.5 kg , to keep the 8 moves .

The air resistance is calculated as follows:

FLuft A / 2 × Cw × D × v2 =

in which

A is the face of the vehicle in m2,
Cw is the drag coefficient ,
D is the density of air , ie 1.29 kg/m3 , and
v the driven speed in m / sec .

So here are the speed comes into play .

 Since when applying the equation only on a car , many values to constants , this sake of clarity, are now summarized : The BMW 8 Series has a frontal area of 2:07 m2 on . This is compensated by the very low drag coefficient of 0.29 of 2.07 m2 × 0:29 = 0.6 m2 are left to air resistance area of . Here you can see what effect the drag coefficient to drag. The smaller it is, the smaller the air appears the car to which they must move around - the car is streamlined . The 850CSi has a Cd value of 0.31 , but has a different face ( which is not known to me - deeper , different mirrors ) .

Half the drag area must now be multiplied nor with the specific density of our atmosphere. The then obtained ( 0.6 m2 × 1.29 kg/m3) / 2 = 0387 kg / m.

The force that opposes the drag the 8s, can now be calculated by the newly made combining all constant values for the car : FLuft = 0387 kg / m × v2 . The fact that the driven speed square flows into the equation , at high speeds, extreme forces are expected to :

0 km / h : 0 N = 0 kg
50 km / h : 75 N = 8 kg + rolling resistance ( 280 N , 29 kg) = 37 kg
100 km / h : 299 N = 30 kg + rolling resistance ( 280 N , 29 kg) = 59 kg
150 km / h : 672 N = 69 kg + rolling resistance ( 280 N , 29 kg) = 98 kg
200 km / h : 1194 N = 122 kg + rolling resistance ( 280 N , 29 kg) = 151 kg
250 km / h : 1866 N = 190 kg + rolling resistance ( 280 N , 29 kg) = 219 kg
300 km / h : 2688 N = 274 kg + rolling resistance ( 280 N , 29 kg) = 303 kg
350 km / h : 3658 N = 373 kg + rolling resistance ( 280 N , 29 kg) = 402 kg
400 km / h : 4778 N = 488 kg + rolling resistance ( 280 N , 29 kg) = 517 kg
As nice as this table may be, continue to really show she does a . What is missing is the power in Watts that is required in order to achieve these speeds . It is calculated as follows:

P = ( Froll + FLuft ) × v
= ( Cr × m × g + A / 2 × Cw × D × v2) × v
= Cr × m × g × v + A / 2 × Cw × D × v3
After multiplying the clip can be seen that the required power rises again with the third power . This means that you need eight times as much power for twice the speed and three times even 27 times as much ! In a table that looks like this:

Speed total resistance required performance
50 km / h 355 N 5 kW = 7 PS
100 km / h 579 N 16 kW = 22 hp
150 km / h 952 N 40 kW = 54 hp
200 km / h 1474 N 82 kW = 111 PS
250 km / h 2146 N 149 kW = 202 hp
300 km / h 247 kW = 2968 N 336 PS
350 km / h 3938 N 383 kW = 520 hp
400 km / h 5058 N 562 kW = 764 hp
One can see how from 250 km / h Lestung needed fast in incredible heights . Now it becomes clear why Bugatti Veyron 16.4 in need 1000 hp to the targeted 400 km / h to crack. The above values apply only to the BMW 8 series or vehicles with identical aerodynamics .

So we are quite ready but still can not , because the calculated horsepower must not touch the engine, but at the wheels ! This means that the engine power must be higher to compensate for the loss of gearbox and drive train. This loss is on rear-drive vehicles, approximately 17 % ( front-wheel drive 15%), so that one obtains the following values for the motor power at the end :

Speed performance to
the wheels of motor performance
50 km / h 7 hp 8 hp
100 km / h 22 hp 25 hp
150 km / h 54 PS 64 PS
200 km / h 111 hp 130 hp
250 km / h 202 hp 237 hp
260 km / h 226 hp 264 hp
270 km / h 250 hp 293 hp
280 km / h 277 hp 324 hp
290 km / h 306 hp 358 hp
300 km / h 336 hp 393 hp
310 km / h 368 hp 431 hp
350 km / h 520 hp 609 hp
400 km / h 764 hp 893 hp
The values for the loss in the drive train are not measured values , but a kind of estimate which I have formed in finding such data on the internet. Although he is a bit at the top, but given the engine power of a Alpina B12 5.7 coupes as well as its speed and its own experience, he is not too high.

To make matters even more now added that the transmission must be carefully so that the top speed is achieved even on the maximum output. The 380 hp of a 850CSi never bring to just under 300 km / h , because they rest one at 5300 rev / min and 250 km / h . Beyond that power falls off again , which of course reduces the possible top speed . If you keep the standard gearbox you will at some engine tuning not around to raise the speed limit. Which brings us to the next point .

Due to the growth of the power hunger with the cube of the speed , extensive measures are needed to achieve noticeable change in top . To only ten percent to be faster , third is more of engine power required ( 1.13 = 1.33) . Conversely, it is achieved by the maximum achievable chip-tuning in naturally aspirated performance improvement from just under 10% only a maximum speed increase of 3% ( cube root of 1.1). In previously possible 290 km / h with catapults the 299 km / h at least very close to the magic 300 , but weaker vehicles come with 160 km / h before only to 165 after that. Questionable whether this is worth it .

Finally, once the general formula , with the required engine power can be calculated at a predetermined speed:

Pmotor = (( A / 2 x D x Cd x V3 ) + ( Cr × m × g × v )) × 1.17

with

A: frontal area in m2
Cw : drag coefficient (0.29 for the 8 series )
D: density of air (1.29 kg/m3)
Cr : rolling resistance coefficient ( approximately 0015 , no units)
m: mass of the vehicle in kilograms (about 1900kg for the 8 series )
g : acceleration due to gravity ( 9.81 m/sec2 )
v : driving speed in m / sec ( = km / h / 3.6)
1.17 correction factor to compensate for the energy loss in the drive train ( 1.15 for front-wheel drive )
Pmotor : Motor power in W ( 736 W = 1 hp)
All Substituting constant for the 8 values and sums , so you will receive:

Pmotor = ( 0387 kg / m × 280 v3 + N × v) × 1:17


Further References 

www.securesupplies.biz

http://www.e31.net/luftwiderstand.html

Staley Meyer Patents 


I would like to hear suggest edit or corrections , I am simply trying to put information togther
which advances the knowledge this document has attachments also make sure you see the picture attachements 

Reference Documents https://onedrive.live.com/redir?resid=A5F9DB5B451D72AC%213464

Warm Regards 

Daniel 

Skype Email 

Suggestions or corrections.
danieldonatelli@hotmail.com 

Mb + 66 83 647 3443


If for Some reason you <a>cynical </a> and disagree with this , 
please Email  

danieldonatelli@hotmail.com

Detail the Corrections
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Post has attachment
Fuel Calculations HHO Hydrogen

Please NOTE 

1 .A Running car is using alot of air volume from ambient natural air this ar must be ionized and the O2 Content Enhanced by stripping off electrons.  

2. Also we Meter mix EGR valve air from exhaust standard on all vehicles

3 We use Precise injection from MegaSquirt Secure Supplies 

4 Use a Split Gas Cell from Secure Supplies 


Calculating HHO requirement:
PLEASE NOTE this is for THE AIR FUEL GASEOUS RATIO IN THE CYLINDER

3.2 Combustive Properties of Hydrogen
The properties of hydrogen are detailed in Section 1. The properties that contribute to its use as a combustible fuel are its:
• wide range of flammability
• low ignition energy
• small quenching distance
• high autoignition temperature
• high flame speed at stoichiometric ratios
• high diffusivity
• very low density

Wide Range of Flammability

Hydrogen has a wide flammability range in comparison with all other fuels. As a result, hydrogen can be combusted in an internal combustion engine over a wide range of fuel-air mix-tures. A significant advantage of this is that hydrogen can run on a lean mixture. A lean mixture is one in which the amount of fuel is less than the theoretical, stoichiometric or chemically ideal amount needed for combustion with a given amount of air. This is why it is fairly easy to get an engine to start on hydrogen.
Generally, fuel economy is greater and the combustion reac-tion is more complete when a vehicle is run on a lean mix-ture. Additionally, the final combustion temperature is generally lower, reducing the amount of pollutants, such as nitrogen oxides, emitted in the exhaust. There is a limit to how lean the engine can be run, as lean operation can sig-nificantly reduce the power output due to a reduction in the volumetric heating value of the air/fuel mixture.

Low Ignition Energy
Hydrogen has very low ignition energy. The amount of energy needed to ignite hydrogen is about one order of magnitude less than that required for gasoline. This enables hydrogen engines to ignite lean mixtures and ensures prompt ignition.
Unfortunately, the low ignition energy means that hot gases and hot spots on the cylinder can serve as sources of igni-tion, creating problems of premature ignition and flashback. Preventing this is one of the challenges associated with run-ning an engine on hydrogen. The wide flammability range of
hydrogen means that almost any mixture can be ignited by a hot spot.

Small Quenching Distance
Hydrogen has a small quenching distance, smaller than gasoline. Consequently, hydrogen flames travel closer to the cylinder wall than other fuels before they extinguish. Thus, it is more difficult to quench a hydrogen flame than a gasoline flame. The smaller quenching distance can also increase the tendency for backfire since the flame from a hydrogen-air mixture more readily passes a nearly closed intake valve, than a hydrocarbon-air flame.

High Autoignition Temperature
Hydrogen has a relatively high autoignition temperature. This has important implications when a hydrogen-air mix-ture is compressed. In fact, the autoignition temperature is an important factor in determining what compression ratio an engine can use, since the temperature rise during com-pression is related to the compression ratio.The temperature may not exceed hydrogen’s autoignition temperature without causing premature ignition. Thus, the absolute final temperature limits the compression ratio. The high autoignition temperature of hydrogen allows larger compression ratios to be used in a hydrogen engine than in a hydrocarbon engine.
This higher compression ratio is important because it is re-lated to the thermal efficiency of the system as presented in Section 3.7. On the other hand, hydrogen is difficult to ignite in a compression ignition or diesel configuration, because the temperatures needed for those types of ignition are rela-tively high.

High Flame Speed
Hydrogen has high flame speed at stoichiometric ratios. Un-der these conditions, the hydrogen flame speed is nearly anorder of magnitude higher (faster) than that of gasoline. This means that hydrogen engines can more closely approach the thermodynamically ideal engine cycle. At leaner mixtures, however, the flame velocity decreases significantly.

High Diffusivity
Hydrogen has very high diffusivity. This ability to disperse in air is considerably greater than gasoline and is advanta-geous for two main reasons. Firstly, it facilitates the forma-tion of a uniform mixture of fuel and air. Secondly, if a hydrogen leak develops, the hydrogen disperses rapidly. Thus, unsafe conditions can either be avoided or minimized.

Low Density
Hydrogen has very low density. This results in two problems when used in an internal combustion engine. Firstly, a very large volume is necessary to store enough hydrogen to give a vehicle an adequate driving range. Secondly, the energy den-sity of a hydrogen-air mixture, and hence the power output, is reduced.

3.3 Air/Fuel Ratio
The theoretical or stoichiometric combustion of hydrogen and oxygen is given as:
2H2 + O2
= 2H2O
Moles of H2 for complete combustion
= 2 moles
Moles of O2 for complete combustion
= 1 mole
Because air is used as the oxidizer instead oxygen, the nitro-gen in the air needs to be included in the calculation:
Moles of N2 in air
= Moles of O2 x (79% N2 in air / 21% O2 in air)
= 1 mole of O2 x (79% N2 in air / 21% O2 in air)
= 3.762 moles N2
Number of moles of air
= Moles of O2 + moles of N2
= 1 + 3.762
= 4.762 moles of air
Weight of O2
= 1 mole of O2 x 32 g/mole
= 32 g
Weight of N2
= 3.762 moles of N2 x 28 g/mole
= 105.33 g
Weight of air
= weight of O2 + weight of N (1)
= 32g + 105.33 g
= 137.33 g
Weight of H2
= 2 moles of H2 x 2 g/mole
= 4 g
Stoichiometric air/fuel (A/F) ratio for hydrogen and air is:
A/F based on mass:
= mass of air/mass of fuel
= 137.33 g / 4 g
= 34.33:1
A/F based on volume:
= volume (moles) of air/volume (moles) of fuel
= 4.762 / 2
= 2.4:1
The percent of the combustion chamber occupied by hydro-gen for a stoichiometric mixture:
% H2
= volume (moles) of H2/total volume (2)
= volume H2/(volume air + volume of H2)
= 2 / (4.762 + 2)

Stoichiometric air/fuel (A/F) ratio for hydrogen and air is:= 29.6% 

As these calculations show, the stoichiometric or chemically correct A/F ratio for the complete combustion of hydrogen in air is about 34:1 by mass. This means that for complete combustion, 34 pounds of air are required for every pound of hydrogen. This is much higher than the 14.7:1 A/F ratio re-quired for gasoline.

Since hydrogen is a gaseous fuel at ambient conditions it displaces more of the combustion chamber than a liquid fuel. Consequently less of the combustion chamber can be occupied by air. At stoichiometric conditions, hydrogen dis-places about 30% of the combustion chamber, compared to about 1 to 2% for gasoline. Figure 3-3 compares combustion chamber volumes and energy content for gasoline and hy-drogen fueled engine

Figure 3-3 Combustion Chamber Volumetric and Energy Comparison for Gasoline and Hydrogen Fueled Engines ATTACHED 

Depending the method used to meter the hydrogen to the engine, the power output compared to a gasoline engine can be anywhere from 85% (intake manifold injection) to 120% (high pressure injection).
Because of hydrogen’s wide range of flammability, hydrogen engines can run on A/F ratios of anywhere from 34:1 (stoichiometric) to 180:1. The A/F ratio can also be ex-pressed in terms of equivalence ratio, denoted by phi (Φ). Phi is equal to the stoichiometric A/F ratio divided by the actual A/F ratio. For a stoichiometric mixture, 

the actual A/F ratio
is equal to the stoichiometric A/F ratio and thus the phi equals unity (one). For lean A/F ratios, phi will be a value less than one. For example, a phi of 0.5 means that there is only enough fuel available in the mixture to oxidize with half of the air available. Another way of saying this is that there is twice as much air available for combustion than is theo-retically required.


This means that for complete combustion,  

SEE CHART ATTACHED  

1 Pound Gasoline  is 3.8 liters
1 Pound Hydrogen is 6.408 liters


Air Fuel Ratio 
 ( we have to Account for Density Temp and Pressure)

14.7 pounds of air are required for every pound of gasoline
        Liters of air are required for every 1 liter of Gasoline  (convert)

34 pounds of air are required for every pound of hydrogen.
        Liters of air are required for every 1 liter  of Hydrogen (convert)


Example  engine: 305 cc Generator

PLEASE NOTE this is for THE AIR FUEL GASEOUS RATIO IN THE CYLINDER


305 cc at 3600 rpm would draw 1800 x 305 cc = 549 LPM of whatever your running on.

Feeding it with 8 LPM of HHO would leave 541 LPM to be made up with air, 
water mist, recycled exhaust gases or anything else that might help.

 If you convert those figures to a mixture percentage you get a 1.48% mix. 

You only need approximately 1 - 2% mix to run an Internal combustion engine
so your production should be bang on for that engine. Hope this helps.

That last calculation I didn't scratch down was:
4.43cc / 305cc  or 1.45% mix

So both of our calculations are correct.  

Now that's obviously assuming you get a full gulp on intake and that the engine is running at full speed producing a 60 cycle AC output on the generator.  

At lower RPM the cell has more time to produce gas and the engine gets a richer mixture to work with, so starting and idling the engine should be relatively easy.

What is interesting is the tiny amount of pure HHO diluted in that big cylinder can actually produce a big enough bang to overcome friction and still do useful work. 

Pretty sure about 1% being the minimum requirement? 
 The good thing is I see no way to over-rev the engine and blow it up.


The stoichiometric  Air-Fuel mixture according to WIKIPEDIA is :

Fuel         By mass [4]      By volume [5]        Percent fuel by mass
Natural gas   17.2 : 1   9.7  : 1                            5.8%
Propane (LP)   15.67 : 1   23.9 : 1                     6.45%
Ethanol            9 : 1           —                          11.1%
Methanol           6.47 : 1     —                          15.6%
Hydrogen           34.3 : 1   2.39 : 1                  2.9%
Diesel                14.5 : 1   0.094 : 1                6.8%
Gasoline           14.7 : 1       —                        6.8%

PLEASE NOTE this is for THE AIR FUEL GASEOUS RATIO IN THE CYLINDER

Note that : an engine can run very lean AFR with hydrogen i-e : 80:1

PLEASE NOTE this is for THE AIR FUEL GASEOUS RATIO IN THE CYLINDER

Simple calculations for LPM are based on air to fuel ratio: e.g:

1 Liter engine,  takes in 1/4 Liter of air per revolution(Because its 4 stroke),

so @ 1000 RPM , air=250LPM,  if Air-Fuel ratio=25:1, then Fuel=10 Lpm
                                             if Air-Fuel ratio=50:1, then Fuel=5 Lpm
                                              if Air-Fuel ratio=80:1 then Fuel=3.1 Lpm

@ 2000 Rpm , Air=500 Lpm , if Air-Fuel ratio=25:1, then Fuel=20 Lpm
                                             if Air-Fuel ratio=50:1, then Fuel=10 Lpm
                                              if Air-Fuel ration=80:1 then Fuel=6.2 Lpm

so a generic FORMULA can be:

PLEASE NOTE this is for THE AIR FUEL GASEOUS RATIO IN THE CYLINDER

FUEL(Lpm)= Engine Capacity/4 * RPM * AFR

then e.g: Engine capacity = 1 Liter, RPM=1000, AFR= 25:1 =1/25

then FUEL(LPM) = 1/4 * 1000 * 1/25 =   10 LPM

in case of Stan Meyer as he stated he ran car with very lean mix it can be:

if Engine capacity = 1.6L , RPM=3000, AFR =60:1
then , FUEL= 1.6/4 * 3000 * 1/60 = 20 LPM 

PLEASE NOTE this is for THE AIR FUEL GASEOUS RATIO IN THE CYLINDER

from this.
1 Liter engine,  takes in 1/2 Liter of air per revolution(Because its 4 stroke).
-i think 1/2 of rpm cause 4 stroke that has 2 cycles(revolution).meter show 2X rpm but drain air X rpm.

1:80 for hydrogen 
-but if HHO that must less than maybe 1:100(1%) or 0.5:100(0.5%). if include ionize air+laser intake must less than that maybe 0.1:100(0.1)% .this mean 4.8 LPM(0.1%) that enough to run 100 km/h(1600cc).


Keep in mind the A/F ratio for petroleum products is based on doped fuel molecules, most of which do not burn in the combustion chamber.  

HHO is already at stochiometric ratio; when you add air which is mostly nitrogen, you have a partially lean mixture, but a mixture that has a controlled burn rate to coincide with the needed burn rate inside an internal combustion engine.  

Even better is to add a plasma spark and water mist to the mixture; doing

 So should reduce the LpM HHO drastically to something manageable by even brute force electrolysis. 


PLEASE NOTE this is for THE AIR FUEL GASEOUS RATIO IN THE CYLINDER

=================================
NOTE 

A four-stroke engine (also known as four-cycle) is an <a>internal combustion engine</a> in which the <a>piston</a> completes four separate strokes which comprise a single thermodynamic cycle. A stroke refers to the full travel of the piston along the cylinder, in either direction. While risqué slang among some automotive enthusiasts names these respectively the "suck," "squeeze," "bang" and "blow" strokes.<a>[1]</a> they are more commonly termed
INTAKE: this stroke of the piston begins at top dead center. The piston descends from the top of the cylinder to the bottom of the cylinder, increasing the volume of the cylinder. A mixture of fuel and air is forced by atmospheric (or greater) pressure into the cylinder through the intake port.
COMPRESSION: with both intake and exhaust valves closed, the piston returns to the top of the cylinder compressing the air or fuel-air mixture into the cylinder head.
POWER: this is the start of the second revolution of the cycle. While the piston is close to Top Dead Centre, the compressed air–fuel mixture in a gasoline engine is ignited, by a <a>spark plug</a> in gasoline engines, or which ignites due to the heat generated by compression in a diesel engine. The resulting pressure from the <a>combustion</a> of the compressed fuel-air mixture forces the piston back down toward bottom dead centre.
EXHAUST: during the exhaust stroke, the piston once again returns to top dead centre while the exhaust valve is open. This action expels the spent fuel-air mixture through the exhaust valve(s).
-i think 1/2 of rpm cause 4 stroke that has 2 cycles(revolution).
meter show 2X rpm but drain air X rpm.

sorry my mistake, it is 1/2 of Rpm.

but still let us assume that we use 1600 cc engine.

AFR 80:1                         (not very lean mixture)
Engine capacity=1600cc    (approx 40+ hp )
RPM=2500                       (approx 100km/h)

then HHO required= 1.6/2*2500*1/80=  25LPM

PLEASE NOTE this is for THE AIR FUEL GASEOUS RATIO IN THE CYLINDER

NON MEYERS
and using brute force , a 5mmw cell will consume approx 6.6Hp (approx 5000Watt),
still  its running a 40+ Hp engine while consuming 6.6hp 


when your driving constant 100km/h the engine do not use its full power.
 The 40+hp is only the maximum power which the engine is able to produce with gasoline. 

This is a quite other calculation  SEE Part 2 

 So you need for (perhaps) 15hp (don´t know exactly) 25LPM.

Meyers System of Electrolysis using the cell and water in it as a capacitor makes a 185% efficiency gain 
and is factual and Patented and now open Source. 

Stanley Meyer Water fuel Cell is a Water Capacitor and Highly Efficient electrolyzer. 

Use Secure Supplies Step Charge PWM and Switch and Also Use a Chock with your transformers. 

========================================================
Calculating Vehicle Weights       PART 2


Air and rolling resistance

Essentially two forces slow a vehicle , namely the air and rolling resistance . This article explains the relationships and shows how much effort has to be made in tuning measures in order to increase the speed limit.
The rolling resistance is calculated as follows :

Froll = Cr × m × g

in which

Cr is a rolling resistance coefficient, which is in our case about 0.015,
m is the mass of the vehicle , including passengers (ie about 1900 kg for the 8 series )
and g is the acceleration due to gravity 9.81 m/s2 is .

The rolling resistance of a BMW 8 Series is thus 0.015 × 1900 kg x 9.81 m/s2 = 280 N

This size is , as you can see , not dependent on the speed , but on the car's weight. So it will become more and more insignificant with increasing speed. Nevertheless, the engine still has to overcome a force of 280 N = 28.5 kg , to keep the 8 moves .

The air resistance is calculated as follows:

FLuft A / 2 × Cw × D × v2 =

in which

A is the face of the vehicle in m2,
Cw is the drag coefficient ,
D is the density of air , ie 1.29 kg/m3 , and
v the driven speed in m / sec .

So here are the speed comes into play .

 Since when applying the equation only on a car , many values to constants , this sake of clarity, are now summarized : The BMW 8 Series has a frontal area of 2:07 m2 on . This is compensated by the very low drag coefficient of 0.29 of 2.07 m2 × 0:29 = 0.6 m2 are left to air resistance area of . Here you can see what effect the drag coefficient to drag. The smaller it is, the smaller the air appears the car to which they must move around - the car is streamlined . The 850CSi has a Cd value of 0.31 , but has a different face ( which is not known to me - deeper , different mirrors ) .

Half the drag area must now be multiplied nor with the specific density of our atmosphere. The then obtained ( 0.6 m2 × 1.29 kg/m3) / 2 = 0387 kg / m.

The force that opposes the drag the 8s, can now be calculated by the newly made combining all constant values for the car : FLuft = 0387 kg / m × v2 . The fact that the driven speed square flows into the equation , at high speeds, extreme forces are expected to :

0 km / h : 0 N = 0 kg
50 km / h : 75 N = 8 kg + rolling resistance ( 280 N , 29 kg) = 37 kg
100 km / h : 299 N = 30 kg + rolling resistance ( 280 N , 29 kg) = 59 kg
150 km / h : 672 N = 69 kg + rolling resistance ( 280 N , 29 kg) = 98 kg
200 km / h : 1194 N = 122 kg + rolling resistance ( 280 N , 29 kg) = 151 kg
250 km / h : 1866 N = 190 kg + rolling resistance ( 280 N , 29 kg) = 219 kg
300 km / h : 2688 N = 274 kg + rolling resistance ( 280 N , 29 kg) = 303 kg
350 km / h : 3658 N = 373 kg + rolling resistance ( 280 N , 29 kg) = 402 kg
400 km / h : 4778 N = 488 kg + rolling resistance ( 280 N , 29 kg) = 517 kg
As nice as this table may be, continue to really show she does a . What is missing is the power in Watts that is required in order to achieve these speeds . It is calculated as follows:

P = ( Froll + FLuft ) × v
= ( Cr × m × g + A / 2 × Cw × D × v2) × v
= Cr × m × g × v + A / 2 × Cw × D × v3
After multiplying the clip can be seen that the required power rises again with the third power . This means that you need eight times as much power for twice the speed and three times even 27 times as much ! In a table that looks like this:

Speed total resistance required performance
50 km / h 355 N 5 kW = 7 PS
100 km / h 579 N 16 kW = 22 hp
150 km / h 952 N 40 kW = 54 hp
200 km / h 1474 N 82 kW = 111 PS
250 km / h 2146 N 149 kW = 202 hp
300 km / h 247 kW = 2968 N 336 PS
350 km / h 3938 N 383 kW = 520 hp
400 km / h 5058 N 562 kW = 764 hp
One can see how from 250 km / h Lestung needed fast in incredible heights . Now it becomes clear why Bugatti Veyron 16.4 in need 1000 hp to the targeted 400 km / h to crack. The above values apply only to the BMW 8 series or vehicles with identical aerodynamics .

So we are quite ready but still can not , because the calculated horsepower must not touch the engine, but at the wheels ! This means that the engine power must be higher to compensate for the loss of gearbox and drive train. This loss is on rear-drive vehicles, approximately 17 % ( front-wheel drive 15%), so that one obtains the following values for the motor power at the end :

Speed performance to
the wheels of motor performance
50 km / h 7 hp 8 hp
100 km / h 22 hp 25 hp
150 km / h 54 PS 64 PS
200 km / h 111 hp 130 hp
250 km / h 202 hp 237 hp
260 km / h 226 hp 264 hp
270 km / h 250 hp 293 hp
280 km / h 277 hp 324 hp
290 km / h 306 hp 358 hp
300 km / h 336 hp 393 hp
310 km / h 368 hp 431 hp
350 km / h 520 hp 609 hp
400 km / h 764 hp 893 hp
The values for the loss in the drive train are not measured values , but a kind of estimate which I have formed in finding such data on the internet. Although he is a bit at the top, but given the engine power of a Alpina B12 5.7 coupes as well as its speed and its own experience, he is not too high.

To make matters even more now added that the transmission must be carefully so that the top speed is achieved even on the maximum output. The 380 hp of a 850CSi never bring to just under 300 km / h , because they rest one at 5300 rev / min and 250 km / h . Beyond that power falls off again , which of course reduces the possible top speed . If you keep the standard gearbox you will at some engine tuning not around to raise the speed limit. Which brings us to the next point .

Due to the growth of the power hunger with the cube of the speed , extensive measures are needed to achieve noticeable change in top . To only ten percent to be faster , third is more of engine power required ( 1.13 = 1.33) . Conversely, it is achieved by the maximum achievable chip-tuning in naturally aspirated performance improvement from just under 10% only a maximum speed increase of 3% ( cube root of 1.1). In previously possible 290 km / h with catapults the 299 km / h at least very close to the magic 300 , but weaker vehicles come with 160 km / h before only to 165 after that. Questionable whether this is worth it .

Finally, once the general formula , with the required engine power can be calculated at a predetermined speed:

Pmotor = (( A / 2 x D x Cd x V3 ) + ( Cr × m × g × v )) × 1.17

with

A: frontal area in m2
Cw : drag coefficient (0.29 for the 8 series )
D: density of air (1.29 kg/m3)
Cr : rolling resistance coefficient ( approximately 0015 , no units)
m: mass of the vehicle in kilograms (about 1900kg for the 8 series )
g : acceleration due to gravity ( 9.81 m/sec2 )
v : driving speed in m / sec ( = km / h / 3.6)
1.17 correction factor to compensate for the energy loss in the drive train ( 1.15 for front-wheel drive )
Pmotor : Motor power in W ( 736 W = 1 hp)
All Substituting constant for the 8 values and sums , so you will receive:

Pmotor = ( 0387 kg / m × 280 v3 + N × v) × 1:17


Further References 

www.securesupplies.biz

http://www.e31.net/luftwiderstand.html

Staley Meyer Patents 


I would like to hear suggest edit or corrections , I am simply trying to put information togther
which advances the knowledge this document has attachments also make sure you see the picture attachements 

Reference Documents https://onedrive.live.com/redir?resid=A5F9DB5B451D72AC%213464

Warm Regards 

Daniel 

Skype Email 

Suggestions or corrections.
danieldonatelli@hotmail.com 

Mb + 66 83 647 3443


If for Some reason you <a>cynical </a> and disagree with this , 
please Email  

danieldonatelli@hotmail.com

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