Loss of gaseous volume because of fuel combustion

[Zardo]

TL;DR Summary

How to determine the loss of gazeous volume when comparing volumes before and after combustion (e.g. automotive internal combustion engine)

Hi,

1st and general case:
2 H2 and 1 O2 take up 1 liter of volume at, for example, 20°C, before combustion. After combustion and cooling down to 20°C, how much volume do the exhaust gases take up?

Since the stoichiometric relationsship is 2 mols of H2 for every mole of O2, so those three moles need some volume. After combustion, 2 mols of H2O are generated for every 2 moles of H2 and 1 mole of O2, so the volume after combustion, once the gaz has cooled down to the initial temperature, will be 2/3 of the initial volume. Therefore, after combustion, there is a loss of volume (once the exhaust gases have cooled down).

2nd case, gasoline carbureted engine:
now the problem is, the displacement of the "ideal" cylinder is assumed to be 1 litre, and it is somehow able to draw 1 liter of mixed stoichiometric gases in (which does actually not happen in real engines, except if they are turbocharged). But the air flowing to the carburator has only 22% of oxygen content (let's just accept that). Then, inside the carburator, some of that air is displaced by fuel vapor which mixes with the air. So the O2 content of the mixture is actually lower than 22% (displaced by fuel vapor). The stoichiometric relationship is 14.7 : 1 (air to fuel as measured in grams air to grams fuel). But what is the relationsship as expressed in moles?

Let's simplify and say, gasoline = octane C8H18

Moles of air = (stoichiometric ratio of air in grams) * (moles of gasoline) * (molar mass of gasoline) / (molar mass of air)
Moles of air = (14.7) * (1 mol) * (114.23 g/mol) / (28.84 g/mol)
Moles of air ≈ 58.11 mol
Therefore, for every 1 mole of gasoline, approximately 58.11 moles of air are needed.

If the octane only reacts with O2 and no other byproducts are created:

C8H18 + 12.5 O2 -> 8 CO2 + 9 H2O


Absolutely, assuming complete combustion with only the O2 in the air reacting with 1 mole of octane, here's what you'll get:

Combustion reaction:

Octane (C8H18) reacts with oxygen (O2) to produce carbon dioxide (CO2) and water vapor (H2O). The balanced chemical equation for complete combustion is:

C8H18 + 12.5O2 -> 8CO2 + 9H2O

Moles of oxygen available:

Moles of air = 58.11 mol
We assume air is 22% oxygen (O2) by volume.
Moles of O2 in air = 0.22 * moles of air = 0.22 * 58.11 mol ≈ 12.78 mol
Limiting factor and moles of products:

According to the balanced equation, 1 mole of octane requires 12.5 moles of O2 for complete combustion. In this case, we have 1 mole of octane and 12.78 moles of O2.

Octane is the limiting factor because it will be consumed first.

Based on the limiting factor (octane) and the balanced equation, we can calculate the moles of products:

Moles of CO2 produced = (8 moles CO2 / 1 mole C8H18) * (1 mole C8H18) = 8 moles
Moles of H2O produced = (9 moles H2O / 1 mole C8H18) * (1 mole C8H18) = 9 moles
Remaining gases:

O2 (remaining): Since octane is the limiting factor, all the octane will react, but there might be some O2 left unreacted. We can calculate this by subtracting the O2 consumed from the total available O2.
Moles of O2 consumed = (12.5 moles O2 / 1 mole C8H18) * (1 mole C8H18) = 12.5 moles
Moles of O2 remaining = Moles of O2 in air - Moles of O2 consumed = 12.78 mol - 12.5 mol ≈ 0.28 mol
N2 (remaining): We assumed air is only composed of O2 (22%) and N2 (78%). Since we only consider O2 reacting, all the N2 from the air will remain unreacted.
Moles of N2 = 0.78 * total moles of air = 0.78 * 58.11 mol ≈ 45.33 mol
Summary of moles after combustion:

CO2: 8 mol
H2O: 9 mol
O2 (remaining): 0.28 mol
N2 (remaining): 45.33 mol

So it seems there is actually an increase in moles and therefore of 62.61/59.11 = 1.06 in volume in the case octane is used, since 8 + 9 + 0.28 +45.33 mol = 62.61 mols, but the ready-to-combust mixture was only 1 mol of octane plus 58.11 moles of air (total 59.11 moles)?

If 1 % of the exhaust gases are typically blow-by gases in piston engines (meaning they pass the rings and end up in the crankcase), that would mean a gasoline engine would produce more blowby than an engine running on hydrogen for each combustion stroke of that cylinder, since the volume of the exhaust gases are not the same for a given volume of air drawn into the engine, right?

 

[jack action]

A mole is a unit of measurement for the amount of substance, not volume.

One mole contains exactly 6.02214076×10²³ elementary entities

You will have to consider the molar volume of each specific substance for your analysis:

the molar volume of a substance is the ratio of the volume occupied by a substance to the amount of substance, usually given at a given temperature and pressure. It is equal to the molar mass divided by the mass density.

Fortunately, for any ideal gas, the molar volume $V_m$ is:
$$V_m = \frac{RT}{p}$$
Where $R$ is a constant, $T$ the temperature and $p$ the pressure. So, your equation converted to volume - assuming the same pressure and temperature for the gasses before the reaction - would be:
$$2\frac{RT_{in}}{p_{in}} H_2 + 1\frac{RT_{in}}{p_{in}} O_2 = 2\frac{RT_{out}}{p_{out}} H_2O$$
Or:
$$p_{out} = \frac{2}{3}\frac{T_{out}}{T_{in}} p_{in}$$
In this case, the temperature will go up because converting hydrogen and oxygen to water is an exothermic reaction and since the volume is fixed by the combustion chamber, this means the pressure will change.

Repeating the exercise for the gasoline engine - assuming the fuel is in gaseous form like you did - will give:
$$p_{out} = \frac{62.61}{59.11}\frac{T_{out}}{T_{in}} p_{in}$$
Since, in a combustion chamber, the limiting factor is generally the highest temperature you can achieve for $T_{out}$, it will be the same in both cases. This means that $p_{out}$ will always be greater in a gasoline engine than in a hydrogen engine. Yes, this will make less blow-by in the hydrogen engine, but it will also make less force pushing on the piston in the same proportion, therefore less power for the same volume displacement.

 

[Zardo]

Thanks.

I know that it's an exothermic reaction. However, the blow-by gases rapidly cool down as they pass the rings and brush the cylinder wall and the crankcase. So they cool down to about 100°C (if the cylinder heads of the engine are liquid-cooled and the liquid typically has about 90°C).

In motorcycles, at least in Europe, blow-by gases leave the crankcase and cool further down to ambient temperature.

There's also a 2 litres aircraft engine (4 cylinder piston) using AVGAS or simple gasoline (about 95 octane rating) where the blow-by gases are actually used to push the oil from the dry sump back to the oil tank (no oil pump for pumping oil back to the tank, but there's an oil pump which sucks oil out of the tank in order to lubricate the engine).

I find it interesting to calculate the actual volume of the blow-by gases, assuming 1% of the exhaust gases are blow-by gases and they cool down to about 100°C which is higher than ambient air temperature (for example 15°C). However, in order to calculate the volume of blow-by gases per minute which pass the crankcase and blow oil back into the oil tank, then leave the oil tank via a small hole, it's necessary to know the amount of aspirated mixture per minute and then calculate the amount of total exhaust gazes per minute, of which 1% are assumed to be blow-by gases (in a healthy engine). For example, the engine has a 2 l displacement (4 cylinders of half a liter, each) and its RPM are 5000/min. So assuming the cylinder is entirely filled, this engine would suck in 1 litre of 15°C cool air 5000 times per minute (since the crankshaft needs two revolutions in order to suck in 2 litres of displacement) therefore there are 5000 l /min of admission mixture, then there is this 1.05 increase of volume because of the increases of moles and there is also an increase of volume as the blow-by gazes in the crankcase are about 100°C hot whereas the temperature of the air was about 15°C. Throttly fully open, of course.

So this means there are roughly 70 litres per minute of hot blow-by gases which pass the crankcase and the small tube blowing oil back to the oil tank : (273,15° K+100°K)/(273,15° K+15°K)*62.61/59.11 * 0.01 * 1 l * 5000 RPM = 68.58 litres of 100°C hot blow-by gases every minute which push the hot oil which has accumulated in the dry sump back into the oil tank.