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

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1. Nov 7, 2017

### MCTachyon

1. The problem statement, all variables and given/known data

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

2. Relevant equations

PV = nRT

3. 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?

2. Nov 8, 2017

### Staff: Mentor

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

3. Nov 8, 2017

### MCTachyon

Evening,

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.

Best regards,

4. Nov 8, 2017

### 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?

5. Jan 3, 2018

### Rogue

Is there any further feedback on this?

6. Jan 3, 2018

### MCTachyon

DM me and I'll give you my findings.