Thermodynamics 101

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Autodidact

Well-Known Member
First, lets not devolve into nit-picking over technical terms. The question I have is:

Disregarding other factors (such as timing and compression ratio), is the resulting gas pressure (or expansion, since holding volume constant and adding heat causes increased pressure, while holding pressure constant and adding heat causes expansion) of air/fuel combustion caused solely by the heat imparted to the air by the burning of the fuel? IOW, if I just heated up the same mass of air to the same temperature (magically or any other method), would I get the same increased pressure (or volume) as if I ignited a certain mixture of fuel/air?

Vigilant1

Well-Known Member
IOW, if I just heated up the same mass of air to the same temperature (magically or any other method), would I get the same increased pressure (or volume) as if I ignited a certain mixture of fuel/air?
No, because in addition to the heated air, you also have the increased pressure caused by a larger number of gas molecules after combustion. Before combustion, the fuel was almost all liquid, but after combustion the fuel had formed new molecules, and they were all gasses.

Here's some info I borrowed from this site, it gives the combustion equations for the primary constituents of gasoline. Notice that all the molecules on the right are gaseous, while all of the molecules on the left that were in the fuel were liquid droplets when they enter the cylinder with the cold fuel charge. For example, C8H18 is "Octane", and it boils at 125 deg C. C9H20 is nonane, it boils at 150 deg C.
Since, the pressure of a gas in a confined space depends almost entirely on the number of molecules and their temperature (what the molecules are and their weight doesn't matter), you can get some idea of the magnitude of the importance of the additional gas molecules just by counting the gaseous molecules on the left and the right. Of the 8 reactions, 7 result in a larger number of gaseous molecules than at the start of the reaction (the exception is the 4C8H7 + 39O2 -> 32CO2 + 7H20, but even that is fairly close, and it only accounts for 10% of the fuel). Of course, only the oxygen in the air is doing any reacting, so be sure to account for the "dead" N2 that is in the cylinder both before and after combustion.
In a sense, we can think of the cylinder acting like a steam engine does--liquid is introduced and it turns to gas, producing pressure on a piston. If we burn 5 gal/hour, that's 5 gallons of fuel that was turned into "hydrocarbon steam" to push on the piston.
In don't know how much of the increased pressure after combustion is due to the higher temps and how much is due to the increased number of gas molecules. I'd guess that the higher temps are quite a bit more important than the increased number of molecules.
pv=nRt

**************************
Gasoline is composed of 20% C8H18, 20% C9H20, 15% C10H22, 15% C11H24, 8% C6H6, 10% C7H8, 10% C8H7 and 2% C8H10 by mass. The Specific Gravity of Gasoline is .74

2C8H18 + 25O2 -> 16CO2 + 18H20
C9H20 + 14O2 -> 9CO2 + 10H20
2C10H22 + 31O2 -> 20CO2 + 22H20
C11H24 + 17O2 -> 11CO2 + 12H20
2C6H6 + 15O2 -> 12CO2 + 6H20
C7H8 + 9O2 -> 7CO2 + 4H20
4C8H7 + 39O2 -> 32CO2 + 7H20
2C8H10 + 21O2 -> 16CO2 + 10H2O
************************************

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Autodidact

Well-Known Member
No, because in addition to the heated air, you also have the increased pressure caused by the vaporization of the liquid fuel (or, really, the additional gaseous compounds that the liquid fuel turned into after combustion with the air).
Is the pressure from the fuel derived gaseous compounds small (trivial or negligible) or large?

Autodidact

Well-Known Member
Thank you! That helps.

pictsidhe

Well-Known Member
If we assume that the fuel is CnH2n-

CnH2n + 6N2 + 1 1/2 O2 = CO2 + H2O + 6N2

Then if the fuel was liquid, we have a small increase in gas molecules after combustion. If the fuel was evaporated, it's slightly less gas molecules after combustion. Reality is usually somewhere in the middle. Number of gas molecules before and after is much the same. Almost all of the expansion or pressure increase is due to the heat of combustion.

Autodidact

Well-Known Member
If we assume that the fuel is CnH2n-

CnH2n + 6N2 + 1 1/2 O2 = CO2 + H2O + 6N2

Then if the fuel was liquid, we have a small increase in gas molecules after combustion. If the fuel was evaporated, it's slightly less gas molecules after combustion. Reality is usually somewhere in the middle. Number of gas molecules before and after is much the same. Almost all of the expansion or pressure increase is due to the heat of combustion.
Thank you very much!

Autodidact

Well-Known Member
I get a molecule increase of 15% according to Vigilant1's info; if I did the calculation based on heated air, then multiplied by 1.15, would I be in the ballpark? Also, does increased compression ratio cause a corresponding increase in the heat added?

Vigilant1

Well-Known Member
Is the pressure from the fuel derived gaseous compounds small (trivial or negligible) or large?
Probably negligible, I'd guess less than 5%. But you could do the math using the reaction equations to figure out the total increased number of gas molecules (O2 molecules on the left side compared to the total number of gas molecules of all kinds on the right). Don't forget to account for the ratio of the different fuel components in the original mix (20% octane, 20% nonane, etc). When doing the proportions, also account for the 79% of the air that isn't oxygen (and remains unchanged by the reaction). The pressure on the piston is described by the ideal gas law (pV=nRT), where N is the number of moles. So, if the number of gas molecules increased by 5% after the reactions, then the pressure increased by 5%, too.

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Vigilant1

Well-Known Member
I get a molecule increase of 15% according to Vigilant1's info; if I did the calculation based on heated air, then multiplied by 1.15, would I be in the ballpark? Also, does increased compression ratio cause a corresponding increase in the heat added?
When you did the math, did you:
1) Account for the various ratios of each constituent of the fuel (rather than assuming they were all present in the same amounts)
2) Account for the fact that all the equations only use the O2 in the air, the 79% of the air that is not O2 doesn't affect the reactions.

Sorry, I'm too lazy to do the math, but I suspect the change in total number of gas molecules is less than the 15% you got.

Autodidact

Well-Known Member
the 79% of the air that is not O2 doesn't affect the reactions.
Doh!:gig: I gotta get some sleep, I'll think about it some more tomorrow...

Autodidact

Well-Known Member
C8H18 > 0.251852

C9H2O > 0.25333

C10H22 > 0.19091

C11H24 > 0.19167

2C6H6 > 0.084706

C7H8 > 0.11

4C8H7 > 0.0907

C8H10 > 0.0226087

= 1.1957767

* (1-.79)

= 0.2511

+ 0.79

= 1.04

OK, I found the ratios of resultant molecules divided by beginning molecules and multiplied them times the percentages of the gasoline constituents, and then multiplied that times the 21% of air that is reactive to the fuel and got approximately a 4% increase in molecules.

I assume that at full throttle with a non-lean mixture, there are enough oxygen molecules to burn all of the fuel (but that never really quite happens?), so, I guess I need to find out what a stoichiometric mixture is, ie, I need to find out how much heat gasoline imparts to a certain volume of air...

Vigilant1

Well-Known Member
OK, I found the ratios of resultant molecules divided by beginning molecules and multiplied them times the percentages of the gasoline constituents, and then multiplied that times the 21% of air that is reactive to the fuel and got approximately a 4% increase in molecules.

I assume that at full throttle with a non-lean mixture, there are enough oxygen molecules to burn all of the fuel (but that never really quite happens?), so, I guess I need to find out what a stoichiometric mixture is, ie, I need to find out how much heat gasoline imparts to a certain volume of air...
It depends what you are trying to find out. You've already answered the question in the OP: Burning fuel produces 4% more pressure in the cylinder than would be obtained simply by heating the fuel-air mixture in some other way without combustion ("magic"). This is because the combustion, in addition to heat, increases the number of gas molecules by 4%.

The heat of combustion for gasoline is easy to find, but I'm not sure where you are going. Real-world losses from friction, unavailability of the heat contained in the vaporized water in the exhaust, etc are huge, and make it very hard to get valid conclusions about real-world engine performance beginning from a theoretical basis. In many cases, and depending on what you are trying to do, it is more fruitful to start with existing real-world engines and see what/how they do.

Autodidact

Well-Known Member
The book I'm reading (Design and Tuning of Competition engines) has very little math and a lot of qualitative discussion and I believe it said that at full throttle and medium rpm on a naturally aspirated engine the intake charge is approximately @ atmospheric pressure before the compression stroke. I'm trying to find a relatively easy way to calculate the maximum gas pressure on the piston. I assume that from the instant of ignition and/or greatest heat production until the exhaust valve opens, there is very little heat loss so that I should be able to use the volume @ best mechanical advantage on the crank pin to find the maximum instantaneous torque (taking into account rod angle). I know that there is bearing friction and friction of the piston on the cylinder wall as well as drag from non combustion strokes of some of the other cylinders, but these are relatively minor and the simple calculation would give a slightly higher torque value which for my purposes should be a conservative approximation.

Vigilant1

Well-Known Member
I'm trying to find a relatively easy way to calculate the maximum gas pressure on the piston.
I'm sure this has been measured directly in actual test engines.
But if you want to calculate the pressure in the cylinder, you would just use the ideal gas law/equation. pV=nRt. Solve for P. You know the V (volume of the cylinder at the point of interest), you know the temp (would typical EGTs of spark ignition engines be close enough?). R is a constant. "n" is the only ting you'd need to work to find, it is the number of mols of gas molecules contained in the after-combustion mix. If I were you, I'd disregard the 4% gain in molecules (just add 4% to the pressure when you get done), and figure out how many mols of air were in the initial charge and how many mols of fuel. Fuel mols could be figured by using the stoichiometric fuel/air mix for these engines (what, about 14:1?), or compute back starting with the GPH of fuel the engine burns, convert to mass, then divide by the number of ignition events per hour.
Gasoline has a molar mass of about 102 grams per mol.

Autodidact

Well-Known Member
Thank you, that's very helpful - I still have a lot of thinking and reading to do, but it looks possible for me at this point.

Highplains

Well-Known Member
Wouldn't the maximum gas pressure be somewhat related to torque?

rv6ejguy

Well-Known Member
I've got some tests showing pressure plots on actual aero engines. Peak cylinder pressure was around 550-1400 psi depending on whether atmo or supercharged.

Imep varies from about 200 to 700 psi for atmo and supercharged engines respectively.

Stoichiometric ratio for gasolines range from about 14.7 to around 15.3.

Autodidact

Well-Known Member
Wouldn't the maximum gas pressure be somewhat related to torque?
Exactly. That's what I am wanting to calculate, maximum instantaneous torque - simplistically and useably (ie, a conservative approximation) for a specific purpose.

The book I'm reading points out that while compression without combustion (isothermal compression) rises linearly with increasing compression ratio, compression caused by combustion (adiabatic compression) rises exponentially with increasing compression ratio. So there seems to be some kind of nonlinear connection between combustion pressure and compression ratio, and that implies complexity in calculation to me, but hopefully not too much.

The thing is, when you increase compression ratio you not only increase combustion force, you also increase negative force on the crank pin during the compression stroke so that it appears to me that there is a larger spread between max positive torque and max negative torque which would cause an increasing max/mean torque ratio with an increase in compression ratio, meaning that (possibly) the max/mean torque charts you see on the epi-eng website and in various books are only typical examples and cannot be used for design of, say, transmission components.

Stoichiometric ratio for gasolines range from about 14.7 to around 15.3.
Thanks, Ross.

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