Gaseous E10 fuel is 10 percent ethanol (C2H6O) and 90

Chapter 15, Problem 64P

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Gaseous E10 fuel is 10 percent ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{8} \mathrm{O}\right)\) and 90 percent octane \(\left(\mathrm{C}_{8} \mathrm{H}_{18}\right)\) on a kmol basis. This fuel is burned with 110 percent theoretical air. During the combustion process, 90 percent of the carbon in the fuel is converted to \(\mathrm{CO}_{2}\) and 10 percent is converted to \(\mathrm{CO}\). Determine (a) the balanced combustion equation, (b) the dew-point temperature of the products, in \({ }^{\circ} \mathrm{C}\), for a product pressure of \(100 \mathrm{kPa}\), (c) the heat transfer for the process, in \(\mathrm{kJ}\), after \(2.5 \mathrm{~kg}\) of fuel are burned and the reactants and products are at \(25^{\circ} \mathrm{C}\) with the water in the products remaining a gas, and (d) the relative humidity of atmospheric air for the case where the atmospheric air is at \(258 \mathrm{C}\) and \(100 \mathrm{kPa}\) and the products are found to contain \(9.57 \mathrm{kmol}\) of water vapor per kmol of fuel burned.

                                     

Equation Transcription:

Text Transcription:

(C_2H_6O)

(C_8H_18)

CO_2

CO

^circC

100 kPa

kJ

2.5 kg

25^circC

258^circC

100 kPa

9.57kmol