Calculate the actual fuel:air ratio by volume and by mass

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Discussion Overview

The discussion revolves around calculating the fuel-to-air ratio by both volume and mass for a mixture of butane, propane, and butene. Participants explore the methodology for these calculations, addressing both the theoretical and practical aspects of the ratios involved.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant presents a calculation for the fuel-to-air ratio by volume, detailing the moles of oxygen required for combustion of the fuel gases.
  • Another participant questions the validity of the by-mass analysis, suggesting that it should directly convert moles from the by-volume analysis to mass without involving CO2.
  • Subsequent posts attempt to correct and refine the mass calculations based on the feedback received, with varying results presented by different participants.
  • Participants express confusion and seek clarification on the calculations, indicating a lack of consensus on the correct approach.
  • Some participants report their calculations for total mass of the fuel and the resulting fuel-to-air ratios, with differing values presented.
  • Final posts indicate that some participants believe their calculations are correct, while others express uncertainty and seek confirmation.

Areas of Agreement / Disagreement

There is no clear consensus on the correct method for calculating the fuel-to-air ratio by mass, as participants present differing calculations and express confusion over the process. Some participants agree on the validity of certain calculations, while others challenge them.

Contextual Notes

Participants' calculations depend on the accuracy of molecular weights and the assumptions made during the conversion from moles to mass. There are unresolved discrepancies in the calculations presented, particularly regarding the mass of the fuel components.

Who May Find This Useful

This discussion may be useful for individuals interested in combustion chemistry, fuel analysis, and those seeking to understand the complexities of calculating fuel-to-air ratios in practical applications.

PCal
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Homework Statement
A fuel gas consists of 75% butane (C4H10), 10% propane (C3H8) and 15% butene (C4H8) by volume. It is to be fed to the combustion chamber in 10% excess air at 25ºC, where it is completely burnt to carbon dioxide and water. The flue gases produced are to be used to generate 5 bar steam from water at 90ºC. Data: Net calorific value (MJ m–3) at 25ºC of: Butane (C4H10) = 111.7 MJ m–3 Butene (C4H8) = 105.2 MJ m–3 Propane (C3H8) = 85.8 MJ m–3 Air is 21% oxygen, 79% nitrogen by volume and 23.3% oxygen and 76.7% nitrogen by mass. Atomic mass of C = 12, O = 16, N=14 and H = 1.
Calculate actual fuel:air ratio by volume and by mass
I've done this a few times and got different answers every time, I think I might be correct with this one but I'm not sure so any help would be appreciated!
Relevant Equations
Balanced reactions
Butane C4H10 + 6.5O2 = 4CO2 + 5H20
Propane C3H8 + 5O2 = 3C02 + 4H2O
Butene C4H8 + 6O2 = 4CO2 + 4H20
By Volume:
Butane 0.75x6.5= 4.875 moles of O2
Propane 0.1x5= 0.5 moles of O2
Butene 0.15x6= 0.9 moles of O2
Total O2 4.875+0.5+9= 6.275 moles
Air required 6.275/0.21=29.88 moles/m^3
With 10% excess air 29.88x1.1= 32.868moles
1 mole fuel:32.87 moles of air

By Mass:
CO2 Moles (0.75x4)+(0.1x3)+(0.15x4)= 3.9
H2 Moles (0.75x10)+(0.1x8)+(0.15x8)= 9.5
O2 Moles (0.75x13)+(0.1x10)+(0.15x12)= 12.55
Nitrogen (12.55/21)x79= 47.21

Moles x Molecular mass
CO2 3.9x12= 46.8
H2 9.5x2= 19
O2 12.55x32= 401.6
Nitrogen= (401.6/23.3)x76.7=1322

Total Fuel= 46.8+19=65.8
Total Air= 401.6+1322=1723.6 +10% excess= 1895.96

1895.96/65.8= 1 mole fuel:28.81 mole of air
 
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Sorry I'm confused, I'm struggling to see anything in that thread that helps with calculating the ratios? It looks like that's for the next part of my question.
 
Your by-volume analysis looks OK. You took as a basis one mole of the fuel gas, and then determined the molar amounts of oxygen and nitrogen required. This was the correct thing to do.

But the by-mass analysis does not look OK. All you need to do is convert the moles from the by-volume analysis to mass for the by-mass analysis. This does not involve the CO2. Just multiply the numbers of moles of each species in part 1 by their molecular weights.
 
Oh amazing so...
Butane 4.875 x 58= 282.75g
Propane 0.5 moles x 44 = 22g
Butene 0.9 x 56 = 50.4g
Total mass= 355.15g

1895.96/355.15 = 5.33
= 1 : 5.33
Does this look better please?
 
You have 0.75 moles of butane having a mass of 43.5 grams.
 
Aaa I think I understand. So...
Butane 0.75 x 58 = 43.52g
Propane 0.1 x 44 = 4.4g
Butene 0.15 x 8.4g
Total mass= 56.32g
1895.96/56.32 = 33.664
= 1 : 33.67
 
PCal said:
Aaa I think I understand. So...
Butane 0.75 x 58 = 43.52g
Propane 0.1 x 44 = 4.4g
Butene 0.15 x 8.4g
Total mass= 56.32g
1895.96/56.32 = 33.664
= 1 : 33.67
32.97 moles of air has a mass of about 950 grams.
 
Butane 0.75 x 58 = 43.5g
Propane 0.1 x 44 = 4.4g
Butene 0.15 x 8.4g
Total mass= 56.3g

0.75 x 208 = 156g
0.1 x 160 = 16g
0.15 x 192 = 28.8g
Total O2 =200.8

(200.8/23.3) x 76.7 = 661

(200.8 + 661) x 1.1 = 947.97
947.97/ 56.3 = 16.83
1: 16.83
Please say this is correct?
 
  • #10
OK. This is correct.
 
  • #11
Chestermiller said:
OK. This is correct.
You haven’t said that because I asked you to?? Haha I’m currently doing a happy dance
 

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