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We have purchased an oxygen analyzer that is calibrated with pure nitrogen and pure oxygen. It measures 20.91% when it samples air in the atmosphere. The analyzer is reading in %volume.

We have built a container that allows gas to escape. It is initially filled with air. The container is about 6 cu.ft and we have about 4 inch diameter hole with a mechanism to close and open the hole. The oxygen analyzer is on the other side of the room which the gas mixture unit is isolated in the explosive room designed to withstand explosion. We then introduce propane into the system. As more propane we have introduced into the container, the oxygen decreases to 20.14%.

We want to use the measurement of oxygen to calculate the air-fuel ratio. The combustion reaction is:

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

so a stoichiometric air-fuel mixture should have 1 part propane and 5 parts pure oxygen, or 1 part propane and 25 parts air, approximately (by volume).

Does the nitrogen-to-oxygen ratio remain constant (79/21=3.76) when propane is introduced in the container and partially displaces the air?

Karsten
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William Chao
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1 Answers1

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If we assume that the Temperature and Pressure conditions under which this is done are such that all gases behave like ideal gases, then we have:

$$X_{air}+X_{fuel}=1$$

Solving for $X_{air}$:

$$X_{air}=1-X_{fuel}$$

Dividing by $X_{fuel}$:

$$\frac{X_{air}}{X_{fuel}}=\frac{1-X_{fuel}}{X_{fuel}}$$

We also have:

$$X_{O_2}+X_{N_2}+X_{fuel}=1$$

As you noted, the ratio of $N_2$ : $O_2$ in air is:

$$\frac{X_{N_2}}{X_{O_2}}=3.76$$

So we have:

$$X_{N_2}=3.76\;X_{O_2}$$

Substituting and solving for $X_{fuel}$:

$$X_{O_2}+3.76\;X_{O_2}+X_{fuel}=1$$

$$X_{fuel}=1-4.76\;X_{O_2}$$

Substituting into the air/fuel ratio equation:

$$\frac{X_{air}}{X_{fuel}}=\frac{1-1+4.76\;X_{O_2}}{1-4.76\;X_{O_2}}$$

Simplifying:

$$\frac{X_{air}}{X_{fuel}}=\frac{4.76\;X_{O_2}}{1-4.76\;X_{O_2}}$$

This is the Air-Fuel ratio in terms of molar fractions or volume fractions, since we're assuming all species are ideal gases.

The volume or molar fraction of oxygen needed to satisfy stoichiometric relationship with propane, while using a different $N_2 : O_2$ ratio $\alpha$ can be calculated by:

$$X_{O_2}+X_{N_2}+X_{fuel}=1$$

$$X_{O_2}+\alpha\;X_{O_2}+0.2X_{O_2}=1$$

Finally:

$$X_{O_2}=\frac{1}{1.2+\alpha}$$

$$\alpha=\frac{X_{N_2}}{X_{O_2}}$$

Sam202
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  • Thank you for your detail explanation. This is still based on the fact that N2/O2 ratio stays constant when fuel is introduced into the system. Thank you for confirming that. Since we want to determine the Stoichiometric Mixture of propane from the oxygen reading, I am using 78.08% as N2 and 20.95% as O2 for the calculation which gives the N2/O2 ratio of 3.73. The stoichiometric mixture for propane is 15.67. If using N2/O2 ratio of 3.76, the oxygen level at stoichiometric mixture will be approximately 19.75%. With N2/O2 ratio of 3.73, it will be approximately 19.89%. – William Chao Oct 18 '22 at 18:53
  • You're welcome. The stoichiometric oxygen percentages I calculated with 3.73 and 3.76 ratios are slightly different (20.28% and 20.16%, respectively), but they are close to yours. I edited my response above to include my calculation. – Sam202 Oct 18 '22 at 19:41
  • My apology. The stoichiometric mixture of propane-air that I have stated (15.67) was incorrect. That is ratio by mass. The Stoichiometric mixture for propane in volume should be 23.9. After recalculation, I have got 20.165% of O2 with the ratio of 3.76 and 20.306% of O2 with the ratio of 3.73. This leads the percentage of fuel to be 4.01% which is closer to Propane's LEL of 2.1%. I expect the percentage of propane's stoichiometric mixture tp be somewhat in the middle of 6.1%. – William Chao Oct 18 '22 at 20:43
  • Is your goal to determine %$O_2$ required for %fuel to be lower than propane's LEL, or what condition in terms of LEL are you trying to satisfy? – Sam202 Oct 18 '22 at 20:56
  • The goal is to determine the %O2 and % fuel at stoichiometric mixture since there is no easy way to measure the composition of air-fuel mixture. We thought it is easier to use an oxygen analyzer to monitor the % fuel in the container. – William Chao Oct 18 '22 at 21:11
  • In terms of molar percentages, the stoichiometric %fuel would be equal to the product of %$O_2$ you're measuring, times 0.2. This would be equal in terms of volume percentages if all gases behave like ideal gases. If they do not behave like ideal gases, a correction factor must be included. What Pressure and Temperature conditions are you using for this operation? (this is important to know since it will determine if the gases behave ideally or not). – Sam202 Oct 18 '22 at 21:25
  • The air is in standard atmosphere pressure and room temperature. The fuel (propane) has bee injected into a container with small leak, which is close to atmosphere pressure at 70°F. I have also added a link of the excel spreadsheet for calculating mixture of N2, O2, and propane. See [Link] (https://docs.google.com/spreadsheets/d/1muH6MUjvY-8YFfMFDvxZzJSWSw1vJTDF/edit?usp=sharing&ouid=105155522786863337336&rtpof=true&sd=true) – William Chao Oct 18 '22 at 21:56
  • Under those conditions, the three gases behave reasonably ideally. In that case, what i would do is: (1) Measure the %$O_2$ with the instrument you have. (2) calculate %fuel with: $X_{fuel}=1-X_{O_2}(1+\alpha)$ (3) Calculate stoichiometric %$O_2$ with: $X_{O_2}=5;X_{fuel}$ – Sam202 Oct 18 '22 at 22:11
  • When %$O_2$ measured is equal to 20.1613%, you will have reached the point of stoichiometric composition, assuming $\alpha$ = 3.76 – Sam202 Oct 18 '22 at 22:41
  • Thank you for confirming with that and the oxygen level of 20.1613% is what I found to be the point of stoichiometric composition. If my calculation is correct, this % fuel at the oxygen level of 20.1613% would be 4.03%. – William Chao Oct 19 '22 at 16:09
  • I am just surprised that the point of stoichiometric composition is not somewhere in the middle of UEL and LEL. I thought that the point of stoichiometric would be somewhat 6.1% of fuel. – William Chao Oct 19 '22 at 16:13