# Max Power Output with Air-Fuel Ratio

• Automotive
• Mohankpvk
In summary, the conversation discusses the relationship between air fuel ratio and power output in an internal combustion engine. It is explained that a slightly richer air fuel ratio can produce higher power due to the presence of enough oxygen for all the fuel to be burnt completely. This results in a higher temperature, which is not desirable for maximum power output. The speaker clarifies that temperature does not drive the piston, but rather, pressure does. Therefore, more fuel in the mixture leads to higher pressure and thus, more power. However, the speaker also mentions that this can lead to high temperatures, which can be damaging to the engine components. The conversation also touches on the use of engine control modules and CD ignition to effectively use the fuel and produce more power.f

#### Mohankpvk

How can the slightly richer Air fuel ratio produce higher power while the temperature it produces is lesser than slightly leaner(slightly leaner than stoichiometric for best economy)?
FYI:I understood that ,at the stoichiometric ratio(precisely slightly leaner due to fast operation of engine) there will be enough oxygen for all the fuel to be burnt completely.So it results in higher temperature.
In case of a richer mixture, lesser than required oxygen combined with the cooling effect of the fuel(enthalpy of vapourisation), the temperature is lower.
But my doubt is,how can a lower temperature produce the best power(maximum)?
I thought, higher temperature=higher pressure=higher power.
I have attached a graph (I found on the internet).

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PV=rT
its all about Pressure, not temperature.
More pressure is more horsepower as this is what drives the piston. Temperature does not drive the piston. So you got more fuel (rich fuel air ratio) you got more power than the lean condition fuel air ratio. More fuel means more pressure ( all other things being the same).

Mohankpvk
PV=rT
its all about Pressure, not temperature.
More pressure is more horsepower as this is what drives the piston. Temperature does not drive the piston. So you got more fuel (rich fuel air ratio) you got more power than the lean condition fuel air ratio. More fuel means more pressure ( all other things being the same).
But by the gas equation(Gay Lussacs law), at constant volume the pressure is directly propotional to absolute temperature.This means higher the increase in temperature higher the increase in pressure.Then why isn't temperature high at the high pressure ratio(max power air fuel ratio)?

you are thinking that temperature causes pressure with constant volume. This is not the case. Higher pressure causes the higher temperature in a constant volume. The only way to get more pressure in a constant volume situation is to add more fuel to cause more pressure. Maximum power output has lower temperature than a lean burn scenario because the lean burn is to extend fuel mileage, not to produce maximum power. Two different outcomes intentionally designed for very specific purposes. No one wants high temperatures ( over 240° F) as this ruins engine oil, seals and not a desired result. Lean burn engines are designed to run at maximum limit of the engine components with fuel economy in mind. Higher than "normal " temperatures became the standard in the auto industry. The lean burn engines were running at the edge of detonation because of the limits of fuel octane ratings available. Add in greenie weenie ethanol and you have a hot running low energy engine with the best fuel mileage you can get for t he fuel available today.

Mohankpvk
The only way to get more pressure in a constant volume situation is to add more fuel to cause more pressure.
Thank you for answering.By the above statement,do you mean the pressure increases due addition of mass(fuel) or more fuel is burnt so more heat is released so higher pressure is achieved(constant volume heat addition)?

The only way to get more pressure is to add more fuel, with all other things being constant. per your post above, if you had a closed vessel of a specific volume, and heated it, the pressure would increase as you have stated. The PV=rT would be in effect but we are talking about the internal combustion engine. In the IC we change to volume of the combustion chamber from a reliable 8:1 ratio to 11:1 for great power output but need gasoline that has higher octane rating to prevent detonation. Not so good on economical fuel as higher octane is more expensive.
You can pick up more power by burning it more effectively for anyone particular engine. Engine control modules do this to make maximum use of the fuel.
but it all comes down to PV=rT.

Lets compare the old ignition point distributor to a new CD ignition. One triggers a spark over 0.025" spark plug gap at a few degrees of crankshaft rotation. The new CD ignition sparks a huge 0.060" spark plug gap over many degrees of crankshaft rotation. Same mechanicals but more effective use of same quantity of fuel.

The only way to get more pressure is to add more fuel, with all other things being constant. per your post above, if you had a closed vessel of a specific volume, and heated it, the pressure would increase as you have stated. The PV=rT would be in effect but we are talking about the internal combustion engine. In the IC we change to volume of the combustion chamber from a reliable 8:1 ratio to 11:1 for great power output but need gasoline that has higher octane rating to prevent detonation. Not so good on economical fuel as higher octane is more expensive.
You can pick up more power by burning it more effectively for anyone particular engine. Engine control modules do this to make maximum use of the fuel.
but it all comes down to PV=rT.

Lets compare the old ignition point distributor to a new CD ignition. One triggers a spark over 0.025" spark plug gap at a few degrees of crankshaft rotation. The new CD ignition sparks a huge 0.060" spark plug gap over many degrees of crankshaft rotation. Same mechanicals but more effective use of same quantity of fuel.
Thank you.

I've forgotten much of my thermo from college physics, but I think (at least part of) the answer as to why a richer fuel mixture makes more pressure/power is because you're converting more liquid (the raw fuel) into a gas.

Hi everyone,
Can anyone point me to any audio books on Holley carburettor tuning - great to have something to listen to on long drives! Thanks in advance

How can the slightly richer Air fuel ratio produce higher power while the temperature it produces is lesser than slightly leaner(slightly leaner than stoichiometric for best economy)?
FYI:I understood that ,at the stoichiometric ratio(precisely slightly leaner due to fast operation of engine) there will be enough oxygen for all the fuel to be burnt completely.So it results in higher temperature.
In case of a richer mixture, lesser than required oxygen combined with the cooling effect of the fuel(enthalpy of vapourisation), the temperature is lower.
But my doubt is,how can a lower temperature produce the best power(maximum)?
I thought, higher temperature=higher pressure=higher power.
The lower temperature refers to a better cooling more than a lower maximum temperature. Not only does the vaporization of the fuel cools the mixture, but more importantly, there is more matter within the cylinder. One of primary cooling system of an engine is exhaust gas carrying heat of the engine as it exits the cylinder.

If an engine with 14.7:1 fuel ratio has 14.7 kg of air entering the cylinder, then it has 1 kg of fuel as well, for a total of 15.7 kg of matter. Modifying the AFR to 12:1 means it has now 1.225 kg of fuel in the cylinder ( = ##1\ kg \times \frac{14.7}{12}## ), for a total of 15.925 kg of matter in the cylinder. That is a 1.4% increase in mass that can carry out more heat out of the engine.

Ketch22
The lower temperature refers to a better cooling more than a lower maximum temperature.

I can't understand this statement.Please explain.

When it is stated that rich fuel mixtures lower the temperature of engine, it refers to the coolant or oil temperature, not necessarily the maximum temperature within the cylinder.

If the maximum temperature within the cylinder increases, it sounds logical that the coolant temperature increases too. But if you have more mass within the cylinder that exits through the exhaust system, it also carries out more heat with it. Thus it is possible that the increase in cooling due to the extra mass in the cylinder more than compensate for the increase in temperature, so the coolant temperature actually decreases.

Charlie Cheap, Mohankpvk and Ranger Mike
First, I am new here but a long-time engine builder. As a practical exercise in building street-driven custom cars, I found opening up the exhaust not only helped power but seemed to help engine temperature. Here in Texas hot days in traffic can cause serious problems with modified older cars running later model engines. I am not an engineer but have read lots of literature on various mechanical subjects dealing with the internal combustion engine. As a self-taught engine builder, I also learned I am not the best instructor either...so here I am. What I do know is timing, fuel quality, and length of spark make all things happen, good or bad. For me timing seems to be most critical for excessive heat problems. Heat being energy, it is far better to have it PUSHING the piston down the cylinder rather than CHASING it or working AGAINST it on the way up. This explanation helps me understand the technical reason a bigger bang can make more power AND cool the engine...which seems to go against accepted physics. A closer look proves the laws of physics are still in-play with mechanical help.

A richer mixture lowers the exhaust gas temperature. In other words, less heat.

To tack onto Jack Actions comment about more cooing as there is more matter present to carry away the heat when an engine operates at a higher stoichiometric ratio. One must also consider that more matter in the same volume increases the density. A Higher density medium can transmit more power when applied in short bursts.
You have asked a great question which moved quickly from introductory to Laboratory hypothesis. The dyno test results say this is what happens. The Geeks wonder how does that work. The Engineers are still in discussion as to how does the effect come to be. Can you suggest a way for one of us to measure a definitive answer? If not welcome to the club that does not quite know it all yet. lol

A richer mixture lowers the exhaust gas temperature. In other words, less heat.
And so does a leaner mixture. It depends on the fuel air ratio that you are starting with:

This figure is from an article by John Deakin, who took the figure from the operator's manual for a Continental IO-550 engine. The article is an excellent discussion of the relationship between fuel air ratio, power, economy, exhaust gas temperature, and cylinder head temperature (this is an air cooled engine). The article: https://www.avweb.com/news/pelican/182084-1.html.