Net calorific value (CV) per kmol of an fuel/air mix

AI Thread Summary
The discussion focuses on calculating the net calorific value (CV) per kmol of a fuel gas mixture consisting of butane, propane, and butene, combusted with 10% excess air at 25°C. Initial calculations indicate the total kmol of fuel and air, leading to a preliminary net CV of 4.23 MJ m-3 per kmol. However, participants express uncertainty about the conversion to net CV per kmol and seek clarification on the calculations. References to similar problems on Physics Forums suggest that the approach may be straightforward, but further feedback is requested for confirmation. The thread highlights the complexity of combustion calculations and the need for precise conversions in thermodynamic assessments.
MCTachyon
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Homework Statement


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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 25°C, 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 90°C

Calculate:

The net calorific value (CV) per kmol of the fuel/air mix at 25°C?

Data:
Net calorific value (MJ m–3) at 25°C 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

Homework Equations



PV = nRT

The Attempt at a Solution



C4H10 + 6½O2 ⇒ 4CO2 + 5H2O
C3H8 + 5O2 ⇒ 3CO2 + 4H2O
C4H8 + 6O2 ⇒ 4CO2 + 4H2O

n = PV / RT

C4H10:

n = (100 x 0.75) / (8.314 x 298)

n = 0.0303 kmol

Amount of 10% excess air reacted with:

0.0303 x 6.5 = 0.197 kmol of O2

10% excess of O2 = 0.197 x 1.1 = 0.217 kmol

Therefore N2 = 0.217 x (79/21) = 0.816 kmol

Total kmol of air: 0.217 + 0.816 = 1.033 kmol

Similar calculations for C3H8 & C4H8 give:

C3H8: 0.00404 Kmol & Air: 0.1057 kmol

C4H8: 0.00605 Kmol & Air: 0.19 kmol

Total amount of kmol of fuel = 0.0404 kmol
Total amount of kmol of air = 1.3287 kmol

Total kmol fuel/air mix = 1.3691 kmol

Therefore the net calorific value (CV) per kmol of the fuel/air mix at 25°C:

This is where I now get stuck. (But I will attempt to finish the question)

C4H10: [0.0303 x (1.3287 / 1.3691)] x 111.7 = 3.28 MJ m-3

C3H8: [0.00404 x (1.3287 / 1.3691)] x 105.2 = 0.41 MJ m-3

C4H8: [0.00605 x (1.3287 / 1.3691)] x 85.8 = 0.54 MJ m-3

Therefore the net calorific value (CV) per kmol of the fuel/air mix at 25°C:

3.28 + 0.41 + 0.54 = 4.23 MJ m-3 per kmol.

Close?
 
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This exact same problem has been addressed on Physics Forums previously. Please use the Search to find it.
 
Actually there is a further post on page 2 I didn't see.

"so per kmol is 40.89 mols / 3.1804 = 0.013 Mj / Kmol"

Is it as simple as that?
 
MCTachyon said:
Actually there is a further post on page 2 I didn't see.

"so per kmol is 40.89 mols / 3.1804 = 0.013 Mj / Kmol"

Is it as simple as that?

Is there any further feedback on this?
 
DM me and I'll give you my findings.
 
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