(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

A fuel gas consists of 75% butane (C_{4}H_{10}), 10% propane (C_{3}H_{8}) and 15% butene (C_{4}H_{8}) 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

C_{4}H_{10}+ 6½O_{2}⇒ 4CO_{2}+ 5H_{2}O

C_{3}H8 + 5O_{2}⇒ 3CO_{2}+ 4H_{2}O

C_{4}H_{8}+ 6O_{2}⇒ 4CO_{2}+ 4H_{2}O

n = PV / RT

C_{4}H_{10}:

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 O_{2}

10% excess of O_{2}= 0.197 x 1.1 = 0.217 kmol

Therefore N_{2}= 0.217 x (79/21) = 0.816 kmol

Total kmol of air: 0.217 + 0.816 = 1.033 kmol

Similar calculations for C_{3}H_{8}& C_{4}H_{8}give:

C_{3}H8: 0.00404 Kmol & Air: 0.1057 kmol

C_{4}H_{8}: 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)

C_{4}H_{10}: [0.0303 x (1.3287 / 1.3691)] x 111.7 = 3.28 MJ m^{-3}

C_{3}H_{8}: [0.00404 x (1.3287 / 1.3691)] x 105.2 = 0.41 MJ m^{-3}

C_{4}H_{8}: [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.

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

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