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?