Since the amount of mass LEAVING the combustion chamber is less than the mass going IN (The rest is converted to heat) it is possible to have too bog an exhaust valve.

Forced induction motors do well with a higher percentage of flow, but even there it CAN be too big, mostly because of shrouding and velocity.

I mean, it's not QUITE as simple as I laid it out, but it's close.

And, our heads flow remarkably well. Much better than the 1.94/1.50 Gen I heads.

I believe the LT4 exhaust valve fits, although I can't swear to it.
If it does, then it makes for a very nice valve, partially due to it's larger size, without being too large and partially due to the sodium cooling.

 

Last time I checked, mass in = mass out. Unless its nuclear powered .

 

 

What about overlap and blowby though?

 

With all due respect...
And I mean that, BTW. I have a tremendous amount of respect for a few people around here and you're one of them.

But you're wrong on this.

Air and fuel enter the engine and are burned, producing heat.

Whatever amount of A/F you put INTO the engine, you cannot BOTH produce heat AND get the same amount of A/F back out.

That would violate the laws of physics.

Conservation of energy and all that.

 

It doesn't matter if it's expressed as mass or volume, it still works the same.


Think of it like this:

You have a cord of firewood.
Then you make a fire.
Now, you have LESS than a cord of firewood.

 

I was just trying to get back on topic.

Eric, are you saying the reason the exhaust valve is normally smaller than the intake is because there is less flow through the exhaust?

 

Air and fuel enter and are burned - the heat energy produced is from the breaking of chemical bonds.

You get the same mass equivalent of air and fuel out, just the form of it is different - e.g., for methane

CH4 + 202 -> C02 + 2H20

Notice you no longer have "air and fuel" coming out, but the number and type of atoms are still the same - hence the mass will be the same (as the chemical bonds that are broken are massless, as they are actually just charge field interactions).



Actually to get a different mass out on a non-nuclear reaction would violate the law of conservation of mass, as well as the law of conservation of energy (since we don't have enough heat being produced to account for any loss of mass, that would mean energy was being destroyed if mass was actually lost but wasn't accounted for).

The energy, as pointed out above, comes from the breaking of chemical bonds. Here eneregy is being converted from bond form to heat energy - hence the law of conservation of energy is observed.




Chris

 

Trust me Eric... mass in = mass out. Chris B. explained where the energy comes from. These are called exothermic reactions. Combustion calculations are all based on mass balance.... what goes inta = what comes outta.... .

Your "cord of firewood" example seems to depend on what you can see... there is far less mass of ash laying in the pile then there was mass of firewood when you started. Problem is you are overlooking what you can't see, which is the huge mass of material that went up the chimney in the form of CO, CO2, H2O, etc.

 

With all due respect to you...you really might wanna go back to your introduction to physics text book before you make a statement like that.

Also, the reference to the firewood...here's how it actually works.

You have a cord of firewood.
Then you make a fire.
Now, you have something completely different. You have a VERY large volume of smoke (carbon) that you have released into the enviromnent and a lump of (mostly) carbon where the firewood was. If you mass all that smoke and carbon together...it will come out to the same mass you started with before the fire...all this despite releasing a lot of heat as well...the wood you started with had chemical bonds which held that energy...the process of burning is actually breaking down those bonds and releasing their energy as heat and light.

 

To the ones discussing the mass and energy issue -


Perhaps I am wrong.

If so, would someone be kind enough to explain to me how you can have, say; 100cc of air/fuel mixture, convert a sizeable percentage of it into heat by burning it and yet still end up with 100cc of combustion residue and unburned fuel/air?

 

Eric,

Go back and read Chris B, Injuneer and Mike454SS, they do a pretty good explaination of this.

It's not that the 100c of A/F is burned and then left as combustion residue and unburned fuel, alot of it is converted to other compounds and shot out of the tail pipe.

The actuall chemistry of the combustion process can be better explained by the Doc if he chimes in here. There can be increase in the number of molecules after the combustion process but the mass before and after has to be the same.

To quote Chris B

That's a pretty good explaination of it to me. I didn't even do well in College Physics.

The exhaust pressure in the port is more likely around 28-30psi where the intake is much less. In a engine with a good intake pulse tuing you are looking at up to 21psi, but that is for a very small amount of time. The average exhaust pressure vs the average intake pressure is a much wider margin than the peaks mostly due to the pressure drop when the piston going down the cylinder creating a low pressure in the cylinder. Most of the time it's about 10-12psi in the intake port, and hopefully it's higher than atmospheric pressure when the piston is going into dwell and heading up the bore. That right there is the major reason for the difference in the valve sizing.

Bret

 

I disagree.

Energy and mass are directly related through the formula E=MC^2.

You can convert mass into energy through chemical processes such as burning, but you lose mass in the process. Not as much mass as you lose in fission or fusion but you still lose mass.


So, if you have the exact same volume/mass of gasses leaving the combustion chamber as went into it, where did the heat come from?

 

 

You will not end up with 100cc - it is mass equivalents, not volume equivalents - but to explain how the mass remains equivalent, I would again use the burning of methane as an example

CH4 + 202 -> C02 + 2H20


In the above equation heat is produced, and mass is conserved - so where is the mass that has been converted into energy?




Yes, but this process only occurs in a nuclear conversion (by definition). Step back for a moment - let's say we convert just 1 MILLIONTH of a gram of fuel/air to energy (1ug) in a second of time - that is going to give us 89.98 MEGAWATTS of energy, or about 120,500 horsepower. And that is only 1 millionth of a gram. Hardly anything.

Realize if any mass -> energy conversion was taking place we would have orders of magnitude more energy released.




Sorry, the law of conservation of mass states explicitly that in all chemical processes the mass of reactants and products is the same before and after the reaction. Do a simple google for "law of conservation of mass" and you will find this readily.





You can't use "volume/mass" as interchangeable terms because they are not. The volume in a chemical reaction can easily change - e.g. add baking soda to vinegar and the resultant gas greatly increases the total volume. In the burning of gasoline you will end up with more moles of gas than you began with, so volume will increase even though mass is constant; additionally, the heat released from the reaction will cause the gasses to expand, greatly increasing the volume.




Chris