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

[MCTachyon]

Homework Statement

[/B]
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?

 

[Chestermiller]

This exact same problem has been addressed on Physics Forums previously. Please use the Search to find it.

 

[MCTachyon]

Evening,

I have just read though this thread:

https://www.physicsforums.com/threads/net-calorific-value.888755/#post-5590572

But all that is answered there is "the net calorific value (CV) per m^3 of the fuel/air mix at 25°C"

I have already worked this out for myself. I can't see a solution to work out the net CV per kmol.

 

[MCTachyon]

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?

 

[Rogue]

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?

 

[MCTachyon]

DM me and I'll give you my findings.