Talk:Air–fuel ratio

Can someone put the air fuel ratio into laymans terms? If the ratio is 14.7 to 1 and air weighs .0807 lbs per cubic foot, and gasoline weighs 46.24 lbs per cubic foot, then what is the % of fuel to air, presuming the fuel is gasoline? I think is should be about 5% fuel to air. Hhilliard (talk) 14:18, 12 May 2008 (UTC)

AFR

AFR is measured in volume not weight. For every cubic feet of gasoline that enters into the combustion chamber as vapor or mist (as near as possible)approxamatly 14.7 cubic feet of air is needed to obtain a Stiochimetric mixture. This is measured. in CFM (cubic feet per minute). —Preceding unsigned comment added by 134.134.136.3 (talk) 17:51, 5 August 2008 (UTC)

Stoichiometric Ratio

Someone might want to look into stoichiometric AF ratio a little more. I understand it to be 14.6 and not 14.7, as per Heywood

marine engine air fuel ratio on large container vessels - probable design faults attributable to stochiometric air requirement caused extensive damages like scuffiing , sudden severe wear and assosciated commercial losses to due to its effect on ship's schedule is becoming more and more common.

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Indeed, I came to this subject expecting to see the 14.7 number proven as stoichiometric for some example, perhaps octane. After hours of calculations and searching I have become convinced that a 14.7 AFR for octane does indeed not produce stoichiometricity. Having a look at the aricles provided by the SAE I am suspicious that this number may have been derived using instrumentation to make measurements of the ehxaust gas and intake values of standard engine and fuel combination.

Having the calculations available in this section would seem to me to be critical to the understanding.

Bill S

Proposed Merger

It’s been added to Wikipedia:Proposed mergers to gain a little more attention and help it merge faster. AeroKnight 18:18, 24 September 2006 (UTC)

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"air/fuel mixture is approximately 14.7 times the mass of air to fuel"

Sentence is ambiguouly structured. After reading this part of the sentence it is not clear to me what that means. The mixture has more mass than air to fuel? Is it 14.7 : 1 or 1 : 14.7?

Stoich Redirect?

How prevelant is the "stoich" abbreviation for Air-fuel ratio? I've always known "stoich" as a shortened versio of stoichiometry. Google seems to agree, if you search for "stoich", it brings up "See results for: stoichiometry". If no one has any reasonable objections, I will change the redirect in a few days. Brien Clark 07:41, 3 January 2007 (UTC)

Equivalence Ratio - Atomic mass of O = 8?

In the calculation of the equivalence ratio, we ca read that the mass of O2 is 2 times 8... BUT the atomic mass of O is 16g/mol so it would be 2*16 instead of 2*8. Morever, it is repeated several times while for the other masses (H and C) it's the good value.

Thanks for explaining me why it is not 16 and if it was a mistake, would you correct it?

Regards, Yannick

You are correct. It should be 16 and not 8. I will fix it. Myth ( Talk) 18:33, 26 January 2007

Please Comment on this

Why does a lean mixture run hot?? If fuel is added to improve cooling, the release of unburned hydrocarbons (soot) into the air would be very large. It would overwhelm the catalytic converter.

I do not buy the argument that fuel vaporization contributes to engine cooling, if there is not a corresponding increase in hydrocarbon emission.

This is a common question, and one to which I have not yet found a satisfactory answer. Could you add a few sentences on this subject to your entry?

Thanks,

Dave 70.231.147.222 09:26, 27 April 2007 (UTC)

You can not find a satisfactory answer because that statement is false. The hottest combustion temperatures occur at approximately the stoichiometric ratio, and drop off as you go rich OR lean from there.

Vessbot (talk) 09:37, 27 September 2008 (UTC)

Mixtures a little lean of stoichiometric cause the engine to run hotter than stoichiometric. (Very lean mixtures run cool, eventually running so cool that combustion is no longer supported and the engine stops.) It is my understanding that lean mixtures burn more slowly than richer mixtures so the heat is generated later in the power stroke, causing the engine to run hotter even though power output is less than at stoichiometric. Very lean mixtures burn so much slower that eventually the fuel-air mixture is not yet fully combusted when the exhaust valve opens, possibly leading to back-firing.

The reason rich mixtures burn cooler than stoichiometric is explained in the following sentences in Air-fuel ratio: Rich mixtures produce cooler combustion gases than does a stoichiometric mixture, primarily due to the excessive amount of carbon which oxidises to form carbon monoxide, rather than carbon dioxide. The chemical reaction oxidizing carbon to form carbon monoxide releases significantly less heat than the similar reaction to form carbon dioxide. (Carbon monoxide retains significant potential chemical energy. It is itself a fuel whereas carbon dioxide is not.) Dolphin51 (talk) 12:26, 28 September 2008 (UTC)

non-rocket scientist, D in chemistry viewpoint

198.176.208.75 (talk) 17:49, 7 October 2008 (UTC) I would like the article to explain practical internal combustion engine air-fuel ratio better in terms of flame temperature. In blasting class for mining engineering, evaluating explosive mixtures for maximum flame temperature at "oxygen balance" (=ideal or stoichiometric conditions, I guess?) in order to impart the most energy to the rock being blasted was the goal. I know that in gasoline engines, going from a rich air-fuel mixture toward oxygen balance, by leaning, increases engine temperature and reduces CO and unburned hydrocarbons in the exhaust (obviously there are many other factors). I am also pretty sure that real internal combustion engines cannot run for very long at oxygen balance before they overheat and fail, hence part of the need for all the secondary means of reducing CO and other pollutants. Conversely, diesel engines run hotter by increasing the amount fuel or decreasing air-fuel ratio, in fact leaning a diesel engine results in cooler temperatures until it stops running. Aslo, diesel engines can be used underground without killing people because they produce almost no CO, compared to gasoline engines. Does all this mean that gasoline engines generally run on the fuel rich side of oxygen balance, and diesel engines generally run on the fuel poor side of oxygen balance - when they are running correctly anyway? Please expand the article in this practical direction or re-educate me as necessary here, if possible. Thanks.

Proposed changes

I plan of making some changes to the article. Before I do that I would like to know your suggestions on them.

Equivalence ratio section

Update the equivalence ratio section by making the oxidizer to be more specific, in this case air. The article should be talking about air and fuel mixtures, but equivalence ratio is a more broader concept, since it does not restrict the oxidizer definition to air or oxygen. It would be better to have a separate article for it. -- Myth (Talk) 20:01, 4 July 2007 (UTC)

Fuels

Most of the article assumes that the fuel is a hydrocarbon. While this maybe true for most applications, I believe the definition does not have anything to do with it. It would be better if the article also reflects this where ever possible be more general in the definition of the fuel. -- Myth (Talk) 20:01, 4 July 2007 (UTC)

Other terms used section

Should AFR be again mentioned here? -- Myth (Talk) 20:01, 4 July 2007 (UTC)

Effect on temperature for rich/lean mixtures

Someone recently added the explanation as to why rich / lean mixtures have lower temperature than stoichiometric mixtures. While the explanation is correct, it does not belong here. A more appropriate section for such explanation will be the adiabatic flame temperature article. As such this explanation should be removed from this article and moved to the flame temperature article. -- Myth (Talk) 20:01, 4 July 2007 (UTC)

I disagree strongly that the information as to why rich and lean mixtures have lower combustion temperatures does not belong on this page. On 27 April, on this page, Dave wrote "Please comment on this." He asked how rich mixtures can run cool without depositing excessive soot throughout the exhaust system. He said "This is a common question." In the 2 months since then no-one has posted any information in response to Dave's questions. Clearly this is a common question, but one that doesn't have a common answer. Any discussion about stoichiometric mixture must be accompanied by some information about why this mixture burns hotter and produces more power than mixture strengths either side of stoichiometric. Dolphin51 12:56, 6 July 2007 (UTC)

First of all, rich mixture can achieve higher temperature than stoichiometric mixture (This is because CO₂ has higher specific heat capacity than CO). So even though less energy is released in case of rich mixture, it is easier to raise temperature of the mixture. Furthermore, this largely depends on the fuel we are using. Here we assumed it is a hydrocarbon.

Secondly, a "common question" is a relative term. The reason why I proposed to move it the adiabatic flame temperature article is because it more appropriate for discussing temperature changes and would be easier to explain it in detail there.

Just because "Dave" asked a question here and no one answered it for two months, does not make it necessary to include that explanation here. It will just make the article too cluttered with bits and pieces of information.

btw I am also not totally convinced that we should move the explanation, but I would like to discuss it further, before I make up my mind. Thanks. -- Myth (Talk) 18:13, 6 July 2007 (UTC)

Hi Myth. Thanks for your prompt response. My view is that, seeing the article has such a strong focus on the concept of stoichiometric mixture the article should present the complete picture. Stoichiometric mixture is more than just an academic concept in chemistry. It has an intensely practical application in that carburettors and fuel injection systems are designed to deliver charge that is close to the stoichiometric mixture, rather than very rich or very lean. Until recently this article revealed only that fuel injection systems aim to achieve stoichiometric mixture; and that stoichiometric mixtures burn very hot (presumably hotter than other mixtures.) These pieces of information beg the question "Why?" Good question. A reader thinking intuitively might imagine that the highest temperature (and the highest power) would be achieved with maximum rich mixture. You and I know that isn't so. My experience indicates that this really is a paradox. I have read a statement by a very competent practitioner who wrote that with rich mixtures, the exhaust gas contains unburnt fuel!

My view is that in any document where stoichiometric mixture is mentioned the explanation must carry on far enough to give the reader an understanding of the two major processes that cause combustion temperature and power output to vary with mixture strength, and to display a maximum that is neither fully rich nor fully lean - those two major processes are dilution by excess air, and the contribution made by carbon monoxide. You say that it is easier to explain the process in detail in the context of adiabatic flame temperature. The reader who is new to the concept of fuel-air ratio is not ready to understand the process to the level of detail that you are able to provide. He or she is looking for some information of the same level of difficulty as "air-fuel ratio", and that isn't rocket science. I agree the article is cluttered at present. Let's get rid of the extraneous information about n-heptane, iso-octane, alkanes, detergents, oxygenators, MTBE and ethanol. These things are not essential to an introductory article about air-fuel ratio (certainly not in the 3rd paragraph.) And on the matter of fuels, let's assume we are talking exclusively about hydrocarbons. So far the article is focussing on internal combustion engines and I'm not aware of any such engine that burns something other than hydrocarbons. Best regards Dolphin51 12:23, 7 July 2007 (UTC)

Well then it would be better to move the discussion about temperature to a separate section in the article. We can also include small details like why lean mixtures are preferred in IC engines especially because of better performance and lower NOx emissions.

Also it would be better not exclusively talk about hydrocarbons, reason being fuels like hydrogen are likely to be used extensively in future. Also the temperature variation with air-fuel ratio is valid for most of the fuels, so it is not necessary to restrict the article to hydrocarbons. -- Myth (Talk) 22:34, 7 July 2007 (UTC)

AFR for solid fuels

I suggest adding info or a link to info about the AFR for a typical grade of coal, or perhaps if there are significant differences, various grades of coal, coke, etc. GilesW (talk) 13:41, 10 March 2008 (UTC)

stock ECU information incorrect

This statement "Vehicle manufacturers do not provide a means of altering predefined fuel maps making such alterations impossible without replacing the stock ECU with a customizable system." is incorrect. I'm just deleting it for now.

Vehicles at least back into the 80's (Ford and GM included) have ECU's that can be remapped with a chip. Many more modern vehicles can be flashed through the OBD-II port or by other means. --167.102.224.45 (talk) 13:08, 18 March 2008 (UTC)

AFR measurments

AFR is measured in volume not weight. For every cubic feet of gasoline that enters into the combustion chamber as vapor or mist (as near as possible)approxamatly 14.7 cubic feet of air is needed to obtain a Stiochimetric mixture. This is measured. in CFM (cubic feet per minute). —Preceding unsigned comment added by 134.134.136.36 (talk) 18:03, 5 August 2008 (UTC)

This statement suggests an automobile gasoline tank of 100 liters (26 US gallons or 3.53 cubic feet) would consume only 51.9 cubic feet of air during combustion! If the automobile achieves 10 liters per 100 kilometres, it could travel 1000 km on its 100 liter tank of gas, and in traveling 1000 km it would require only 51.9 cubic feet of air. This could not possibly be correct!

Chemical equations are always based on mass rather than volume, because volume changes so significantly when pressure and temperature change. The correct statement is that the air-fuel ratio (eg 14.7) is a ratio of the masses of the air and the fuel, not a ratio of their volumes. Dolphin51 (talk) 02:38, 27 January 2009 (UTC)

Lambda?

The article describes a parameter called lambda, after the Greek letter $\lambda$. I don't doubt that there is a parameter that matches the description given here, and I'm sure the parameter is abbreviated $\lambda$, but what is the name of the parameter? Lambda is simply the Greek letter used as an abbreviation. Citing a source for this information would be a good move. Dolphin51 (talk) 11:59, 18 January 2009 (UTC)

Air Press v Stoich

Please explain why air pressure at sea level and stoichiometric are the same value? —Preceding unsigned comment added by 71.102.8.187 (talk) 17:04, 23 March 2009 (UTC)