Ethanol gas (C_2 H_6 O) is burned with 110 percent theoretical air. During the c

Question

Ethanol gas (C_2 H_6 O) is burned with 110 percent theoretical air. During the combustion process, 90 percent ot the carbon in the fuel is converted to CO_2 and 10 percent is converted to CO. Determine the theoretical kmols of O_2 required for complete combustion of one kmol of ethanol. the balanced combustion equation for the incomplete combustion process, and the rate of heat transfer from the combustion process, in kW, when 3.5 kg/h of fuel are burned when the reactants and products are at 25 degree C with the water in the products remaining a gas.

Explanation / Answer

Answer:

a)

Given Data;

Ethanol gas (C2H6O) is burned with 110 % excess air.

CO2 conversion = 90%

CO conversion = 10%

Theroetical kmols of O2 required = ?

Let us first write down the reaction equation,

C2H6O + 2.5 Air (O2 + 3.76 N2) gives CO2 + CO + 3 H2O + 9.4 N2

Therefore, 2.5 kmols of theoretical O2 required to burn 1 kmol of ethanol.

110 % Excess theoretical air, so equaiton will be

C2H6O + (1.1 * 2.5) Air (O2 + 3.76 N2) gives CO2 + CO + 3 H2O + (0.1 * 2.5) O2 + (1.1 * 2.75 * 3.76) N2

C2H6O + 2.75 Air (O2 + 3.76 N2) gives CO2 + CO + 3 H2O + (0.25) O2 + 11.374 N2

Therefore, for 110% excess air, 2.75 kmol of air is required for 1 kmol of ethanol.

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