Post Reply 
 
Thread Rating:
  • 0 Votes - 0 Average
  • 1
  • 2
  • 3
  • 4
  • 5
Optimal Fuel ratio Calculation
Author Message
M98Ranger Offline
Member
***

Posts: 32
Joined: Dec 2008
Reputation: 0
Post: #1
Optimal Fuel ratio Calculation
I just spent about 40 minutes crunching some numbers and found something out that all of you might be interested in knowing.

I am first going to tell you my conclussion and then I will give you some calculations to chew on. I have the file written down in mathematica and (assuming you get to me before January 7 when classes start for me) I would be happy to give you all of my work so you can verify the accuracy of it.

What I found was that based on a 14.7 to 1 air to fuel ratio with Octane (gasoline, C8H18) as your fuel the unburned hydrogen that you come up with (assuming a few things ....like perfect combustion and other things) is around 8.4 kmols (in 1 revolution assuming 3 liters per minute Brown gas production). This is good information because now I can see that in order to utilize all of the hydrogen in the mix (assuming near complete combustion), one would have to increase the air fuel mixture to 19.81 to 1 if you were running at 2000 rpm.

That means that anything less than a 19.8 ratio will give you unburned hydrogen or more likely unburned hydrocarbon by virtue of the fact that hydrogen is likely to be the first thing that burns as it is closer to the spark. That is what the numbers say anyway. There are a lot of other factors that go into combustion other than just conservation of energy. The other part of this is that for every unit of O2 that you have in the air, you have 3.67 units of N2. Some N2 exits as NO (NOX) but most of it essentially comes out as it goes in (except hotter)....therefore N2 is effectively steeling a great amount of the combustion heat in every cycle. If we take that knowledge to the next level then as we increase the air to fuel ratio 1 unit we are increasing the N2 content by 3.67 Units, which can absorb quite a bit of energy.

So not only are we fighting the fact that the engine is using some of the gas in our engines to power the alternator to get us the hydrogen, but as we lean out the fuel ratio, Nitrogen is taking energy in multiples of 3.67*(Nitrogen Latent Energy at flame Temperature). That coupled with water in the HHO production could take a considerable amount more energy out the tail pipe than I realized.

Water in the chamber can be controled through judicious use of water traps / bubbler / what ever you want to call them. The dryer the gas the better obviously. Plus if there is sulfur mixed into your octane (which is often the case) then at dewpoint temperature for atmospheric pressure you are going to get sulfuric acid eating away the innards of your exhaust....so water is not a good thing to put in your engine. The big point though, is that N2 is probably going to be your controlling factor in so far as where your "break even" point is going to be found (your high efficiency point, if you want to call it that).

If you want I could put out my combustion equation for Octane plus Air in a 14.7 to 1 ratio, and then compare that to a 3 liter per minute Browns gas production with a 14.7 to 1 ratio (it has 8.4 kmols per revolution of unused hydrogen piping out the exhaust). And then a 19.81 to 1 air/gas ratio and the accompanying increase in N2 (N2 that has been heated by combustion increases approximately 28 moles per revolution).

PS: I am just now getting done at 12am this morning so please give me some slack if some things are not making sense. I have ADD so evening is not the best time for me to do these sorts of things. But its exciting to me, so I thought I would share it with you all. Thanks for your time.
01-05-2009 11:02 PM
Find all posts by this user Quote this message in a reply
benny Offline
Member
***

Posts: 332
Joined: Sep 2008
Reputation: 5
Post: #2
RE: Optimal Fuel ratio Calculation
(01-05-2009 11:02 PM)M98Ranger Wrote:  I just spent about 40 minutes crunching some numbers and found something out that all of you might be interested in knowing.

I am first going to tell you my conclussion and then I will give you some calculations to chew on. I have the file written down in mathematica and (assuming you get to me before January 7 when classes start for me) I would be happy to give you all of my work so you can verify the accuracy of it.

What I found was that based on a 14.7 to 1 air to fuel ratio with Octane (gasoline, C8H18) as your fuel the unburned hydrogen that you come up with (assuming a few things ....like perfect combustion and other things) is around 8.4 kmols (in 1 revolution assuming 3 liters per minute Brown gas production). This is good information because now I can see that in order to utilize all of the hydrogen in the mix (assuming near complete combustion), one would have to increase the air fuel mixture to 19.81 to 1 if you were running at 2000 rpm.

That means that anything less than a 19.8 ratio will give you unburned hydrogen or more likely unburned hydrocarbon by virtue of the fact that hydrogen is likely to be the first thing that burns as it is closer to the spark. That is what the numbers say anyway. There are a lot of other factors that go into combustion other than just conservation of energy. The other part of this is that for every unit of O2 that you have in the air, you have 3.67 units of N2. Some N2 exits as NO (NOX) but most of it essentially comes out as it goes in (except hotter)....therefore N2 is effectively steeling a great amount of the combustion heat in every cycle. If we take that knowledge to the next level then as we increase the air to fuel ratio 1 unit we are increasing the N2 content by 3.67 Units, which can absorb quite a bit of energy.

So not only are we fighting the fact that the engine is using some of the gas in our engines to power the alternator to get us the hydrogen, but as we lean out the fuel ratio, Nitrogen is taking energy in multiples of 3.67*(Nitrogen Latent Energy at flame Temperature). That coupled with water in the HHO production could take a considerable amount more energy out the tail pipe than I realized.

Water in the chamber can be controled through judicious use of water traps / bubbler / what ever you want to call them. The dryer the gas the better obviously. Plus if there is sulfur mixed into your octane (which is often the case) then at dewpoint temperature for atmospheric pressure you are going to get sulfuric acid eating away the innards of your exhaust....so water is not a good thing to put in your engine. The big point though, is that N2 is probably going to be your controlling factor in so far as where your "break even" point is going to be found (your high efficiency point, if you want to call it that).

If you want I could put out my combustion equation for Octane plus Air in a 14.7 to 1 ratio, and then compare that to a 3 liter per minute Browns gas production with a 14.7 to 1 ratio (it has 8.4 kmols per revolution of unused hydrogen piping out the exhaust). And then a 19.81 to 1 air/gas ratio and the accompanying increase in N2 (N2 that has been heated by combustion increases approximately 28 moles per revolution).

PS: I am just now getting done at 12am this morning so please give me some slack if some things are not making sense. I have ADD so evening is not the best time for me to do these sorts of things. But its exciting to me, so I thought I would share it with you all. Thanks for your time.

Isn't the idea of igniting the fuel mix in the cylinder to produce 'instant' heat and thereby cause rapid gas (air) expansion in the cylinder. Does it matter if the gas in the cylinder is air or N2, or any of the other gases produced in the fuel burn/explosion? If heated, they all expand.
As far as my understanding of the process goes, what added HHO does is give a better, faster, more complete burn of all the fuel mixture, thereby increasing engine, or fuel use , efficiency.

If you could harness all the waste heat, dumped via the exhaust pipe, for re-use in the engine then that would be something.
Think steam engine? Double use of steam. High pressure steam cylinder, followed by low pressure steam cylinder.
(This post was last modified: 01-06-2009 03:24 AM by benny.)
01-06-2009 03:23 AM
Find all posts by this user Quote this message in a reply
M98Ranger Offline
Member
***

Posts: 32
Joined: Dec 2008
Reputation: 0
Post: #3
RE: Optimal Fuel ratio Calculation
Benny,

I think you are absolutely correct in thinking that the N2 / H2O will expand considerably. Additionally, all of the gases (which can be assumed to ideal) expand in proportion with temperature increase. The negative aspect comes in when you think about the fact that N2 is just a place holder so to speak, it is an inert gas that does pretty much nothing (except turn into NOX which is a polluting agent). N2 does not net any energy output, therefore it is just taking up space and absorbing energy bringing the ultimate adiabatic flame temperature down. This can be visualized easier by realizing just how much of the product and reactant consists of N2. If N2 was not there then the other contents would utilize the net energy gain from combustion by increasing in temperature and pressure quite a bit higher than they normally would have reached had the N2 been there. H2O is a bit different in that H2O is a product of H2 combusting. So it is there regardless. However, added water in the form of humidity (introduced either through the air, or through electrolysis), also is an energy carrier that keeps the adiabatic flame temperature down for everything else. Higher flame temperature is important in that it increases pressure, but more than that, it is conducive to better combustion. H2O does expand and do a little work but the amount of work it does compared to the kilojoules of energy it consumes per mol is quite miniscule. That is the short answer.

Also, I totally agree that you can make the engine more efficient by for instance putting heat exchangers in strategic locations and insulating everything and then redirecting the heat back to do useful work....(kind of like a feedwater heater in a rankine cycle "electricity generation"). The mitigating factor in all such design as I understand it is always whether or not the added design and material would justify the gains you would get by it. On an individual basis (ie a person like you or me who might have time to design something like that) it might be way worth it. But as far as a manufacturer doing it, he might have to charge too much for the system for the small amount of gains in efficiency he finds.

Because I start school tomorrow and I don't want to give you a half a**ed answer I am going to do part of a Combustion of a Gaseous Fuel with moist air problem that is found in my text. You won't be able to verify the numbers I get off of tables (unless you know how to go look them up on the internet which you can if you like), but the concept is the important thing anyway. By the way, I had the same question as you a while back about whether water helps or hurts. Here goes. I will note the important parts. It won't hurt my feelings is all I am saying.

The composition of gas to be combusted is 72% CH4, 9% H2, 14% N2, 2% O2 and 3% CO2. The gas is burned with the stoichiometric amount of air. The air is 20 Celsius, 1atm and 80% relative humidity (that is 80% of the water that the air can hold at 20 Celsius is contained in vapor form in the air). My explanations are not for you (benny), but for others that may not quite understand. Also, the total pressure is 1atm (101.4kpa).

Assumptions:
-Fuel is completely consumed in the combustion process
-There is no free O2 left in the product
-combustion gases are ideal gases

(*Important*)The saturation pressure of water is 2.3392 kpa @ 20 Celsius. That means that any less than 2.3kpa and the water turns to superheated steam, any more and it becomes subcooled liquid.

(*Important*)In this problem the moisture in the air does not react with anything ie in other words the moisture in the air is "inert". It goes in and it comes out unchanged. I am going to find the partial pressure contribution of humidity to the total pressure (which we said in the problem statement was 1atm (101.4kpa). Then, I will calculate the force in pounds that the partial pressure of the air exerts on the piston to give you an idea of the contribution. I could follow up with the partial contributions of all the other products and possibly the energy of fussion / latent energy absorbtion that each product sucks up in the combustion process, but I think you will see that the water contributes a very small amount to the overall final product pressure. Additionally, if I decided to do the numbers you would see that the H2O (vapor in the air, as opposed to the H2O product) absorbs 100s of kilojoules per kilomol of substance. That is energy better utilized going elsewhere. Here goes, (skip the balance equation and solve if you want)

(*Important*) , (0.80*(2.3392kpa)=1.871kpa) And the total pressure of all the products is 101.4kpa (1atm).....that means that the water vapor makes up a total of 1.871/101.4=1.84517% of the total pressure in the combustion chamber.

(*Important*)Pressure is force divided by area as you know. and therefore the amount of force that is exerted on the piston can be determined by dividing out the area of the piston face. Lets say that the piston face has a radius of 6 centimeters (just throwing a number out there). Then the area of the piston face is 3.1459*(6cm)^2 and if we convert the answer to meters^2 then (3.1459*6*6)/100^2=0.0113097m^2 then ForceofWater=PartialPressureOfWater*AreaofPiston=(1870Pa)*(0.0113097m^2)= 21.14592 Newtons of force....1 newton = 0.224808943 pounds force, therefore 21.14592 Newtons converted to pounds would be 21.14592*0.224808943=4.75453 Pounds. I hope that helps.
(This post was last modified: 01-06-2009 02:08 PM by M98Ranger.)
01-06-2009 11:41 AM
Find all posts by this user Quote this message in a reply
Gary Offline
Member
***

Posts: 1,542
Joined: Jun 2008
Reputation: 4
Post: #4
RE: Optimal Fuel ratio Calculation
I don't get what you mean when you say (vapor in the air, as opposed to the H2O product) ; I'm wondering just what H20 product, the vapor in HHO or are you intimating water or mist injection?
That is a subject some of us are trying to figure out, and one guy is convinced that HHO along with H20 injection is going to be highly beneficial - but he gets that I think from Stan Meyers 3 step concept of HHO, H20, and resonant frequency input.
I just want to know all I can about water, mist, and steam injection, if you can extrapolate that by a guess. And I do refer to using the naturally aspirated engine, not turbo or anything.
01-06-2009 08:20 PM
Find all posts by this user Quote this message in a reply
M98Ranger Offline
Member
***

Posts: 32
Joined: Dec 2008
Reputation: 0
Post: #5
RE: Optimal Fuel ratio Calculation
(01-06-2009 08:20 PM)Gary Wrote:  I don't get what you mean when you say (vapor in the air, as opposed to the H2O product) ; I'm wondering just what H20 product, the vapor in HHO or are you intimating water or mist injection?
That is a subject some of us are trying to figure out, and one guy is convinced that HHO along with H20 injection is going to be highly beneficial - but he gets that I think from Stan Meyers 3 step concept of HHO, H20, and resonant frequency input.
I just want to know all I can about water, mist, and steam injection, if you can extrapolate that by a guess. And I do refer to using the naturally aspirated engine, not turbo or anything.

Sorry. What I meant was that I am reffering to "the vapor in the air" (ie the vapor due to humidity) rather than "the vapor due to the combustion". There is H2O vapor going into the chamber with the air.... and then there is the vapor that is made as a ("product of the combustion"). Sorry I didn't explain myself well.

Okay as for the question "I'm wondering just what H20 product, the vapor in HHO or are you intimating water or mist injection?" The problem that I am reffering to has humidity (vapor in the air) coming into the combustion chamber. For all intensive purposes if you think about it, the source of the vapor that gets into the combustion chamber with the gases doesn't really matter. You could think of it as vapor due to electrolysis or vapor due to to the humidity. The one thing that the water vapor in the combustion chamber wouldn't be due to is water created by virtue of the combustion of H2 with O2 in the chamber. The reason why it wouldn't be in the chamber is because the old air/gas/combustion products all get cycled out before a fresh mix of air/fuel gets pulled into the chamber.

I don't beleive that it would be good to inject "mist" into the chamber. For one thing, at a given temperature and pressure the air will only hold so much vapor. SO in order to get it to hold more vapor you have to put heat energy into it somehow to raise the temperature. The other "mist" that is injected into the chamber is going to coagulate on the walls of the chamber awaiting the heat of fusion required to turn it into steam. So you have two energy sinks if you inject mist into the chamber. One is the water (liquid form) and the other is the water (vapor form). Where is the energy going to come from? From the fuel/Oxygen/Brown's gas. At what cost? The cost is going to be a lowering of the overall flame temperature. What is that going to do? That is going to cause there to be more unburned hydrocarbon going out the exhaust. And that is just the beginning of the implications. If you follow it out I could foresee that possibly the oxygen in the air is unburned causing the O2 sensor to send less voltage to the ECM which will cause the fuel mix to be increased, which would increase the unburned hydrocarbons coming out the exhaust and diminish fuel mileage gains (if there were any). Perhaps not, but you can follow my reasoning.

Additionally, as I alluded to in my first post, (I beleive) sulfur is known to be in most gasoline in small quantities. Therefore, if you get some Sulfur (dry), mixed with water vapor....then all you need to do is get near the tailpipe and at some temperature (maybe 19 celsius or so) the vapor turns to liquid and the SO4 combines with water to make sulfuric acid. Not good on the innards. Complications such as this (and other more poignant ones) are what made people switch from steam engine technology, (which is/was as I understand it) VERY high maintenance, to diesel engines.

Also, you have to realize the difference in the way that the energy is delivered in a steam engine vs. an gasoline engine. Steam engines heat the steam from a boiler. There is no boiler obviously in a gas engine. So when the steam (an inert that does not react with anything to give off heat) goes into the chamber, it takes up space that could be occupied by reactants such as Oxygen and hydrogen and gas which, when combusted give off a tremendous amount of energy. All the vapor does is absorb energy that would have brought the temperature up to a higher level which would equate to a higher efficiency of energy output.

It seems to me people are trying to mix the two technologies into one and it just doesn't work as far as I can see. I would have to see this paper that Stan Meyers wrote, "Stan Meyers 3 step concept of HHO, H20, and resonant frequency input" in order to ascertain what he is alluding to by connecting res. freq. H2O/HHO all together. I don't even have the slightest guess as of this moment.

I wish I had more time to read it and discuss this, but my class starts in less than 2 hours so I will have to cut it short. By the way, what I am saying is not necessarily hard and fast truth, I have been wrong before, so I am quite sure that there are holes in my reasoning. Taken as a whole though, I think my argument is sound though. Good luck everyone, keep up the good work! I will likely be back on here in four months or so.
(This post was last modified: 01-07-2009 04:50 AM by M98Ranger.)
01-07-2009 04:42 AM
Find all posts by this user Quote this message in a reply
Gary Offline
Member
***

Posts: 1,542
Joined: Jun 2008
Reputation: 4
Post: #6
RE: Optimal Fuel ratio Calculation
I don't know if anyone else if following this, but I'm not entirely in agreement yet. Water injection I suppose needs to be put into another thread and leave this one to focus more on A/F math, as intended. And if we don't even know who Stan Meyers is, the point is moot.
01-07-2009 06:15 AM
Find all posts by this user Quote this message in a reply
M98Ranger Offline
Member
***

Posts: 32
Joined: Dec 2008
Reputation: 0
Post: #7
RE: Optimal Fuel ratio Calculation
(01-07-2009 06:15 AM)Gary Wrote:  I don't know if anyone else if following this, but I'm not entirely in agreement yet. Water injection I suppose needs to be put into another thread and leave this one to focus more on A/F math, as intended. And if we don't even know who Stan Meyers is, the point is moot.

Could you post a link to "Stan Meyer's 3 step concept of HHO, H20, and resonant frequency input"?
01-07-2009 06:39 AM
Find all posts by this user Quote this message in a reply
benny Offline
Member
***

Posts: 332
Joined: Sep 2008
Reputation: 5
Post: #8
RE: Optimal Fuel ratio Calculation
M98Ranger

Quote:http://guns.connect.fi/innoplaza/energy/...index.html

It is known that a gram atom is equal to atomic mass of substance; a gram molecule is equal to molecular mass of substance. For example, the gram molecule of hydrogen in the water molecule is equal to two grams; the gram-atom of the oxygen atom is 16 grams. The gram molecule of water is equal to 18 grams. Hydrogen mass in a water molecule is 2x100/18=11.11%; oxygen mass is 16x100/18=88.89%; this ratio of hydrogen and oxygen is in one liter of water. It means that 111.11 grams of hydrogen and 888.89 grams of oxygen are in 1000 grams of water.

One liter of hydrogen weighs 0.09 g; one liter of oxygen weighs 1.47 g. It means that it is possible to produce 111.11/0.09=1234.44 liters of hydrogen and 888.89/1.47=604.69 liters of oxygen from one liter of water. It appears from this that one gram of water contains 1.23 liters of hydrogen. Energy consumption for production of 1000 liters of hydrogen is 4 kWh and for one liter 4 Wh. As it is possible to produce 1.234 liters of hydrogen from one gram of water, 1.234x4=4.94 Wh is spent for hydrogen production from one gram of water now.


Basically 1 litre of water produces 1794.89 litres of HHO

Making a couple assumptions for HHO use with ICE..

Constant HHO generation at 1 LPM
Constant engine speed : 3000 RPM
4 cylinder engine

Water produced per minute from HHO combustion = 1/1794.89 = 5.571e-4 litres per minute
Average water produced from HHO per rev = 5.571e-4 / 3000 = 1.8571e-7 litres
Average water produced per cylinder at assumed rates = 1.8571e-7 / 4 = 4.623e-8 litres
For higher HHO production rates simply multiply this result by the number of litres of HHO being produced.

Plugging these results into your equations will show that actual volume of water from HHO is so miniscule that it will have little or no effect on energy loss. Bear in mind that any HHO, when burned in an ICE produces superheated steam, from the conversion process of burning HHO alone, ergo is already at peak energy level, and as such would absorb little or no further energy from gas/petrol combustion.
If you also include ratio of HHO to fuel/air mix, for cylinder capacity, the water produced from HHO becomes even more insignificant when compared to water produced from actual gas/petrol combustion. Note that most water produced in the ICE combustion process is expelled from the exhaust system as steam, long before it can become the major cause of corrosion in an exhaust system

HHO as an additive is used as an enhancer for gas/petrol combustion as well as being a fuel in its own right. Results, although as yet somewhat inconsistent, do show that the addition of HHO does help with overall engine performance, and also helps reduce unwanted exhaust emissions..

The addition of water, not HHO, has been known for a long time to assist in the combustion process of ICE. Technique was used from the 1940s on aircraft to increase engine performance.
Water injection also helps reduce carbon deposits (de-coke) in engines.

Have a nice term at school.
(This post was last modified: 01-08-2009 04:29 AM by benny.)
01-08-2009 04:26 AM
Find all posts by this user Quote this message in a reply
Gary Offline
Member
***

Posts: 1,542
Joined: Jun 2008
Reputation: 4
Post: #9
RE: Optimal Fuel ratio Calculation
It's also used today in drag racing and other venues, is being experimented with towards FE. In my group, we may be using it for intentional cooling for running lean mixes.
There is no single document I know of (although I'm sure it exists; I just don't know which one) that explains Stan Meyers dune buggy that ran on water, but it's known that he approached it in that threefold way, for whatever reasons. There is simply far too much to study on ol' Stan for me to spend time with.
01-08-2009 06:38 AM
Find all posts by this user Quote this message in a reply
realtyroy Offline
Member
***

Posts: 158
Joined: Sep 2008
Reputation: 0
Post: #10
RE: Optimal Fuel ratio Calculation
(01-08-2009 06:38 AM)Gary Wrote:  It's also used today in drag racing and other venues, is being experimented with towards FE. In my group, we may be using it for intentional cooling for running lean mixes.
There is no single document I know of (although I'm sure it exists; I just don't know which one) that explains Stan Meyers dune buggy that ran on water, but it's known that he approached it in that threefold way, for whatever reasons. There is simply far too much to study on ol' Stan for me to spend time with.
I believe everything that Stan had before he died including the dune buggy was just auctioned off and will probably see things start to pop up with that info. Unless big oil got it?

Anyway that's what I heard.

RealtyRoy-HHO Scambuster
http://pmgen.com/hhoscambusters/

2002 Dodge Durango SLT 4.7
2001 Audi A6 2.8 Quattro
1964 Chev Impala SS 409 My baby!

A day is a day but the day I save gas is a good day!!
(This post was last modified: 01-08-2009 09:52 AM by realtyroy.)
01-08-2009 09:51 AM
Find all posts by this user Quote this message in a reply
Post Reply 


Forum Jump:


User(s) browsing this thread: 1 Guest(s)