I have read so many times that a lean mix buns hotter than a stockimetic one. Why is this, there is more fuel in the stocki and its not rich?

It just does.

http://forums.hypography.com/chemistry/ ... n-hot.html

You crack me up.
Here is my theory which could be wrong. I believe the peak temperature to be hotter in a lean engine but not the total heat. You see a lean mixture can burn the fuel very quickly because of the large amount of oxygen surrounding the fuel molecules. So the temperature can get very hot as the rate of burning is very fast. The stoichiometric mixture will burn slower but release more heat as there is more fuel burning. Slower burn means less temperature, more fuel buring = more BMEP or power

yer yer yer ... no need for incorrect scientific explanations (soz for the flame)... and it goes something like this

think of a deoderant can .. spray it for long enough it gets cold ..

ie decompression through a small hole causes heat loss

this occurs when fuel is pressurised into the engine through the injectors.

the engine temp is also lowered by the fuel absorbing heat along with your oil/water cooling systems ...

aint thermodynamics and physics grand

i hate both of em but was forced to study then for enginnering studies last yaer

rate of burning doesnt change the amount of energy which is released

How much flaming can you take? You're too cool to do it too Gladier so I'll let some others lay into you. Lots of mistakes... Peace bro.

Dug my head into books and looks like I'm wrong. Lean mixtures burn more slowly than stoichiometric, just as rich ones do. The book claims that in both these cases the temperature of expansion and exhaust is hotter than stochiometric because of the slower flame propagation.(then again I have a graph from a paper showing clearly that the exhaust temp drops from the AF ratio peak at 15.6 which is lean?). Infact every part of the combustion chamber is hotter when lean.

In any case, it is unlikely that the evaporation of the extra fuel in the stochiometric case as Gladier pointer out cools the mixture because unlike a rich mixture, which has unburned fuel, this fuel is burned, providing more heat. Also its well documented that the fuel spray mostly evaporates in the cylinder, not out of the nozzle (not that it makes any difference as it still absorbs heat of vaporisation).
About the heat Gladier, you are abolutely correct, but only for an adiabatic process, which although the simplified engine calcs assume it is, in reality there is heat loss, and a quicker ignition can provide hotter temp, but same heat.

I can understand why a richer mixture is cooler than stoichiometric, but the lean one has me stumped. Come on Gladier and Tekkie, get your brains working.

plus the lean mixture has less energy(heat) to put into the system compared to a rich mix so that makes a really good point... will have to sleep on it

Latent heat of evaporation....every bit of fuel that is put into a proper firing engine will evaporate. The injector, or carburator will disperse the fuel into fine mist, but not vapor. It takes a considerable amount of heat to vaporize the fuel. Some of the low teperature evaporating components of the fuel will vaporize in the manifold, and on the backside of the hot intake valve. Some will vaporize in the cylinder from the hot port walls...all of this depends on the availible heat, and heat transfer. After compression begins, the pressure slows down evaporation, even though heat is produced from compression. Not all of the fuel is yet a vapor, but the light ends (easily evaporated components) will be, and will be lit by the plug to start the combustion process. Now....every bit of fuel not yet vaporized will be vaporized by the heat of combustion....it is this latent heat used to vaporize excess fuel that makes rich mixtures burn cooler than lean mixtures.
Darcy

Yes, this all makes sense for rich mixtures, but not stoichiometric. I'm starting to wonder if not all stoichiometric fuel manages to get burned.

Even when running stiochy...the fuel not yet vaporized before combustion starts, will take some of the heat of combustion to vaporize it. Leaner= less unvaporized fuel to take heat away from the combustion process.
I'm sure we could pick at other factors effecting the heat,but it is my understanding that the latent heat for evaporation is the main controlling factor.

Still makes no sense as heat generated by extra fuel is way more that what it took away by evaporating. Only if it was excess fuel would your explanation be understandable Thanks for your input Mr. Pipe, hope you can solve this for us.

We can put as much fuel into an engine as we want...usually with just a keystroke on out laptop..., but air is what limits the power we can make. If we have a limit on our availible oxygen, then we also have a limit on how much fuel we can burn. Now don't forget that the mixture in the cylinder is not homogenous, so some areas of the chamber will fall outside the optimum range the completely combust...but we still use some of the heat of combustion to vaporize all of the fuel, even is there is not suitable conditions there to make it burn. Every drop of extra fuel that goes in will take heat away to vaporize it. Whether the conditions are present to burn it are a different matter.
....for some reason i think i'm doing a real bad job of explaining this...

Sounds ok to me mate.

sounds good .. but still doesnt help the lean is hotter ....

mk1owner .. care to explain your theory on quicker burning makes more heat?

Yes, but remember that I was wrong about lean burning faster. Also quicker burning does not produce more heat, but I believe a higher temperature. Quite simply, if the same amout of heat is released very quickly, there is less time for it to dissipate into head, block etc, and the combustion gas can obtain a higher peak temp. On the other hand , slow combustion allows more time for the heat to dissipate into environment lowering the peak temp.

Here is an example, try heating a cup of to boil water with a candle. It takes so long that all the heat dissipates and the water never boils. Now try with a bunsen burner and gas. Same amount of gas in calorific value as the wax. Its likely that you will be able to boil the water.

If the process was adiabatic, then temperatures would not be any different.
Anyway thats my thinking but I could be wrong....

Here is a graph clearly showing F/A ratio vs temperatures. It also shows affect of water injection but this can be ignored. Just look at the legend to pick out the graphs with fuel and air only. A/F is 1 divided by F/A.

If anyone is still interested, have a look here

http://mk1owner.angelfire.com/AFratio/index.html