Why Does Calculating Ethanol's AFR Yield Different Results from Standard Values?

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Discussion Overview

The discussion revolves around the calculation of the air-fuel ratio (AFR) for pure ethanol, exploring discrepancies between theoretical calculations and standard values. Participants examine the implications of assuming specific gas compositions and the importance of units in these calculations.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant calculates an AFR of approximately 9.9 for pure ethanol based on a balanced combustion equation, questioning the difference from the commonly accepted value of 9.
  • Another participant points out the assumption that oxygen constitutes 21% of a gas with a uniform molecular mass of 32 amu, suggesting that nitrogen's lighter mass must be considered.
  • A subsequent reply adjusts the calculation by incorporating nitrogen's presence, resulting in an AFR of 8.95, which is closer to the standard value.
  • Further consideration of argon leads to a refined AFR of 9.01, indicating a more accurate representation of air composition.
  • One participant emphasizes the importance of specifying whether percentages refer to weight or volume to avoid confusion in calculations.

Areas of Agreement / Disagreement

Participants express differing views on the assumptions made regarding gas composition and the implications for AFR calculations. There is no consensus on a single correct approach, as various methods yield different results.

Contextual Notes

Participants note limitations in their calculations, particularly regarding the assumptions about gas composition and the necessity of clarifying units of measurement.

D19A99G
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TL;DR
Calculation of the theoretical value of AFR of ethanol
First I have seen a few forums dotted about covering this subject already, however I couldn't find exactly what I was looking for so I apologise if its somewhere on here already.

I am trying to calculate the AFR value for pure ethanol.
I am only including oxygen in the category of "air" due to other elements being inert.

The balanced equation is C2H5OH + 3O2 = 3H2O + 2CO2.

When I then calculate the AFR while also considering that oxygen only makes up 21% of air, the value I get is approximately 9.9. Which is different to the widely used value of 9. Is there something I'm doing wrong or has the value of 9 only being achieved through practical tests which cannot be replicated via theory?

Ethanol mass = 46, Oxygen mass = 96
Air mass = Oxygen*4.76 = 456.96
AFR = 456.96/46 = 9.93
When; C=12amu, H=1amu, O=16amu
 
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You are assuming that oxygen makes up 21% of a gas of uniform molecular mass 32 amu. You have to account for the fact that nitrogen is lighter.
 
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Yeah ok, I can see what you mean. If I neglect the other gases which constitute the make up of "air" and assume oxygen = 21% and nitrogen = 79%. Then 3*(32+(3.76*28)) = 411.84. This would give me a AFR of 8.95 which is much closer.
Then considering Argon as well would give me a value of 9.01.

That makes sense thanks
 
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Sadly, you can't generally assume whether a generic 'percentage' is volume, mass, or whatever. Units matter. I use '%wt' and %vol' to prevent confusion (and collisions with Mars).
 
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