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

In summary, Butane, Propane, and Butene all have different ratios of air to fuel, and the mass of each varies depending on the ratio.
  • #1
PCal
31
2
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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  • #3
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.
 
  • #4
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.
 
  • #5
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?
 
  • #6
You have 0.75 moles of butane having a mass of 43.5 grams.
 
  • #7
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
 
  • #8
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.
 
  • #9
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?
 
  • #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
 

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

1. How do you calculate the actual fuel:air ratio by volume?

The actual fuel:air ratio by volume is calculated by dividing the volume of fuel by the volume of air. This can be represented by the equation: Fuel:Air Ratio (volume) = Volume of Fuel / Volume of Air. This ratio is typically expressed in units of volume, such as liters or cubic feet.

2. What is the importance of calculating the actual fuel:air ratio by volume?

Calculating the actual fuel:air ratio by volume is important because it helps determine the efficiency of a combustion process. If the ratio is too high, there is excess fuel and not enough air, which can lead to incomplete combustion and wasted fuel. If the ratio is too low, there is excess air and not enough fuel, which can result in reduced power and increased emissions.

3. How is the actual fuel:air ratio by mass different from the ratio by volume?

The actual fuel:air ratio by mass is calculated by dividing the mass of fuel by the mass of air. This can be represented by the equation: Fuel:Air Ratio (mass) = Mass of Fuel / Mass of Air. Unlike the ratio by volume, this ratio takes into account the weight of the fuel and air, which can vary depending on the temperature and pressure of the surrounding environment.

4. What are the units of measurement for the actual fuel:air ratio by mass?

The actual fuel:air ratio by mass is typically expressed in units of mass, such as kilograms or pounds. This is because the ratio is based on the weight of the fuel and air, rather than their volume.

5. How can the actual fuel:air ratio by volume and by mass be used in practical applications?

The actual fuel:air ratio by volume and by mass can be used in various practical applications, such as in the design and optimization of combustion processes in engines and boilers. It can also be used in the development of fuel-efficient vehicles and in the analysis of emissions from combustion processes.

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