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I understand that a stoichiometric mixture contains a balanced mixture of air and fuel and both fuel and air are completed consumed. For example, the ideal stoichiometric mixture for propane is approximately 4.01% of the fuel which is equivalent of 23.91:1 ARF by mass.

I also understand each fuel has its own explosive limit. As an example of Propane, the lower explosive limit is 2.1% (lean mixture) and upper explosive limit is 9.5% (rich mixture).

The explosive pressure (or energy) will be larger with 4.01% of fuel mixture compared with 2.1% of fuel mixture assuming the mixture is in a defined container with no fuel saturation. Following this pattern, if the gas mixture reaches to its rich state, fuel level at 9.5%, would the explosive pressure (or energy) be greater than the stoichiometric mixture? Can this be calculated?

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  • $\begingroup$ Do you have a way of knowing/measuring %propane before reaction happens and %propane after reaction happens? $\endgroup$
    – Sam202
    Nov 1, 2022 at 16:43
  • $\begingroup$ @Sam202 I can calculate the %propane based on O2 analyzer before reaction because it's fuel mixing with air. After combustion, it might be difficult because of the product of CO, CO2, H2O and etc. $\endgroup$ Nov 1, 2022 at 17:37
  • $\begingroup$ Do you know the initial mass or amounts (moles) of each gas (oxygen, nitrogen, propane) before reaction? $\endgroup$
    – Sam202
    Nov 1, 2022 at 19:22
  • $\begingroup$ @Sam202 Yes, I can calculate or estimate nitrogen and propane based on the oxygen level measured. $\endgroup$ Nov 1, 2022 at 22:19

3 Answers 3

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Yes, the stoichiometric mixture is the most energetic and creates the highest pressure.

If the mix is not stoichiometric, then some of the energy of the explosion goes into heating up and expanding the unreacted component, whether that is fuel or air.

To use your example, the combustion of propane can be represented as

$$\ce{C3H8 + 5O2 -> 3CO2 + 4H2O}$$

This creates 7 moles of gas (three of $\ce{CO2}$ and four of $\ce{H2O}$) for each 6 moles of gas that react (five of $\ce{O2}$ and one of propane). Propane has a standard heat of combustion of -2220 kJ/mol. So if a stoichiometric mix is present, each mole of propane that combusts will release 2220 kJ of energy into the $\ce{CO2}$ and the $\ce{H2O}$ that is formed.

If a non-stoichiometric mix is used, say the propane is double (or the oxygen is half) of the stoichiometric mix, then the 2220 kJ released from burning one mole of propane will be released to (a) the 3 moles of $\ce{CO2}$ and the 4 moles $\ce{H2O}$ that is formed, but also to (b) the unreacted 1 mole of propane. When the energy of combustion must be "shared" by more moles of gas, the resulting temperature and thus pressure increases from combustion/explosion are less intense.

If the mixture were lean, i.e. with excess oxygen instead of propane, the unreacted oxygen would absorb some of the energy, with a similar decrease in obtained temperature and pressure.

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    $\begingroup$ Thanks for your detailed explanation. I understand the chemistry and math side of your explanation. In order word, it is not a linear relation if I plot the graph of explosive pressure against gas mixture from LEL to UEL. I believe it is more of a bell curve where the maximum explosive pressure is measured at the peak of the bell curve which is close to stoichiometric mixture. $\endgroup$ Nov 1, 2022 at 18:54
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    $\begingroup$ @WilliamChao, yes, exactly. For similar reasons, the temperature and pressure released during combustion with air is much lower than when using a stoichiometric mix of fuel and pure oxygen. With air, a lot of the energy has to go into heating up and pressurizing the 79% of air that is nitrogen gas. Here is a somewhat related and definitely entertaining demonstration of a similar concept. $\endgroup$
    – Curt F.
    Nov 1, 2022 at 21:12
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    $\begingroup$ Great video of demonstrating the explosive energy between pure nitrogen and 2/3 nitrogen with 1/3 oxygen adding into the system. Thanks for finding that video. Is there any other literature you recommend that I can read more about? $\endgroup$ Nov 1, 2022 at 22:29
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    $\begingroup$ This is perfectly true in theory, assuming perfect mixing and complete reactions to equilibrium (so that even the very last propane molecule happens to encounter the last few oxygen molecules in time). In practice, the observed highest-energy mixture can be slightly non-stoichiometric if that helps avoid early terminations and incomplete combustion. Basically, it's a balance of two factors: excess unreacted molecules soaking up energy and local lack of reactants causing incomplete combustion. $\endgroup$
    – TooTea
    Nov 2, 2022 at 10:55
  • $\begingroup$ We were using two different methods for mixing the mixture: natural convection and forced convection with a fan. Since propane is heavier than air, it saturates on the bottom of the container which is where the sparker is located. This means there could be three different levels of mixture: rich mixture on the bottom, balanced mixture in the middle, and lean mixture on the top. It will reach to the point where there is no combustion since the mixture is too rich. Could be interesting to find out the mixture where the sparker is located. $\endgroup$ Nov 2, 2022 at 16:33
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No, slightly rich mixtures of propane and air have the highest explosive pressure as reported in DEFLAGRATION PARAMETERS OF PROPANE–AIR MIXTURES IN A CLOSED CYLINDRICAL VESSEL U.P.B. Sci. Bull., Series B, Vol. 73, Issue 3, pages 17-26 (2011).

Stoichiometric is 4.02% by volume propane as you say, while 4.70% is found to give the highest pressure.

This reference considers many different reaction products, not just the main reaction products. Solid graphite, and gaseous propane, CO2, CO, H2O, O2, N2, CH4, C2H2, C2H4, H2, NO, H, OH and O are all considered.

See also Laminar burning velocity and explosion index of LPG–air and propane–air mixtures Fuel 87 (2008) 39–57 which explains:

the maximum rate of flame propagation occurs on the rich side since additional fuel is needed to compensate for the effect of dissociation at higher temperature

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Assuming you do not want incomplete combustion of propane leading to the formation of poisonous $\ce{CO}$ gas, then propane would be the limiting reagent outside of stoichiometric conditions, according to the reaction:

$$\ce{C3H8 +5O2 -> 3CO2 +4H2O }$$

If you want to calculate how much energy (heat) is released as a function of the initial molar fraction of propane, we will consider the following:

Let:

$A$ represent propane.

$y_{Ao}$ = initial molar fraction of propane.

$n_{Ao}$ = initial moles of propane. [mol]

$n_{o}$ = initial total moles of gas. [mol]

$n$ = final total moles of gas. [mol]

$Q$ = heat released by combustion reaction. [kJ]

$\Delta\overline{H}^{\;°}$ = standard molar enthalpy of propane. [kJ/mol]

$X_A$ = conversion of propane.

The heat released by combustion of propane can be calculated by:

$$Q=n_{Ao}\;X_A\;\Delta\overline{H}^{\;o}$$

Assuming combustion is complete since oxygen is in excess, $X_A=1$, so we have:

$$Q=n_{Ao}\;\Delta\overline{H}^{\;o}$$

Since we're dealing with a reaction in gas phase, and the amount of gas molecules on the products side is higher than on the reactant side, the total final moles $n$ will be higher than the total initial moles $n_o$:

$$n=n_o(1+\epsilon)$$

We can calculate $\epsilon$ with:

$$\epsilon=\frac{\Delta n}{a}\;y_{Ao}=\frac{1}{1}\;y_{Ao}=y_{Ao}$$

Dividing both sides of the second equation by $n$, we get:

$$\frac{Q}{n}=\frac{n_{Ao}}{n_o(1+y_{Ao})}\;\Delta\overline{H}^{\;°}$$

Or equivalently:

$$\frac{Q}{n}=\frac{y_{Ao}}{1+y_{Ao}}\;\Delta\overline{H}^{\;°}$$

Finally, for propane: $\Delta\overline{H}^{\;°}=\;$-2043 kJ/mol

If we recognize that heat will be released to the surroundings , and we're only interested in the amount of heat produced per total moles of final mixture, we can ignore the negative sign, and we get:

$$\frac{Q}{n}=2043\frac{y_{Ao}}{1+y_{Ao}}$$

However, it's important to note that this equation is only valid for the following interval (since propane is only flammable within a certain composition range, and we assumed it was the limiting reagent):

$$y_{AL}\leq y_{Ao}\leq y_{AS}$$

Where $L$ represents the lower flammability limit composition of propane, and $S$ represents the stoichiometric composition of propane.

Substituting the values:

$$0.021\leq y_{Ao}\leq 0.0401$$

Conversely, when oxygen is the limiting reagent, an incomplete combustion of propane takes place, forming $\ce{CO}$ in addition to $\ce{CO2}$:

$$\ce{C3H8 +4O2 -> 2CO + CO2 +4H2O }$$

Since reaction is incomplete, conversion of propane is less than 1:

$$X_A<<1$$

$\epsilon$ for this reaction is different and calculated by:

$$\epsilon=\frac{\Delta n}{a}\;y_{Ao}=\frac{2}{1}\;y_{Ao}=2\;y_{Ao}$$

So $\frac{Q}{n}$ for this reaction is:

$$\frac{Q}{n}=\frac{y_{Ao}}{1+2\;X_A\;y_{Ao}}\;X_A\;\Delta\overline{H}^{\;°}$$

For this reaction, $\Delta\overline{H}^{\;°}=\;$ -1587 kJ/mol

So we have:

$$\frac{Q}{n}=1587\;X_A\;\frac{y_{Ao}}{1+2\;X_A\;y_{Ao}}$$

Which is valid for the interval:

$$y_{AS}< y_{Ao}\leq y_{AU}$$

$$0.0401< y_{Ao}\leq 0.095$$

If we assume only half of propane is converted into products, then $X_A=0.5$, and:

$$\frac{Q}{n}=793.5\;\frac{y_{Ao}}{1+y_{Ao}}$$

If you plot $\frac{Q}{n}$ vs $y_{Ao}$ for both reactions, while respecting the domain of $y_{Ao}$ in each case, you will notice the amount of heat produced per total moles of gas reaches its maximum value when stoichiometric composition is used:

$$y_{Ao}=y_{AS}\implies\frac{Q}{n}=\left(\frac{Q}{n}\right)_{max}$$

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  • $\begingroup$ Thank you for your derivation on calculating the heat released. I did an spreadsheet calculating the heat released on in each case (before and after stoichiometric mixture). With mole fraction of 0.0218, the heat released is 43.63 kJ. With mole fraction of 0.0401, the heat released is 78.85 kJ. With mole fraction of 0.0947, the heat released is 68.67 kJ. Does this mean that the highest energy released does not necessary need to be at the stoichiometric mixture? I thought the the mole fraction after 0.0401 will decrease as mole fraction increases. $\endgroup$ Nov 3, 2022 at 18:33
  • $\begingroup$ You're welcome. Firstly, it's important to note that those heat values you calculated are in kJ/mol, not kJ. If you want kJ, you need to know how many total moles of gas $n_o$ you had before propane combustion. I did not understand your last sentence, since it seems contradictory. $\endgroup$
    – Sam202
    Nov 3, 2022 at 18:48
  • $\begingroup$ Thank you for for the correction and apology for the confusion. I have plotted the Heat Released per mole versus Initial mole fraction of propane for both reactions. Please see link here. link. I understand two linear lines on the graph since it's two different reactions (completed and incomplete combustion). I thought the graph will be more of a bell curve shape. $\endgroup$ Nov 3, 2022 at 20:32
  • $\begingroup$ The functions of $\frac{Q}{n}$ vs $y_{Ao}$ are not linear, they're inverse functions. The "true" theoretical curve would be similar to what you're describing (bell with a maximum heat release per mole at stoichiometric composition), but that would be harder to obtain since you'd have to consider that molar enthalpy $\Delta \overline{H}$ is also a function of $y_{Ao}$, as formation of CO gas will alter gas composition and lower the amount of heat released per mol of total gas. $\endgroup$
    – Sam202
    Nov 3, 2022 at 20:51
  • $\begingroup$ Thank you for the clarification. I thought it's a simple plotting two functions on a same graph. I am not familiar with molar enthalpy as a function of molar fraction and maybe that's a topic for another day. This will give me a general idea of the amount of heat released at stoichiometric composition. $\endgroup$ Nov 3, 2022 at 21:22

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