Question

Ethanol gas \(\left(\mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O}\right)\) 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 CO. Determine (a) the theoretical kmols of \(\mathrm{O}_{2}\) required for complete combustion of one kmol of ethanol, (b) the balanced combustion equation for the incomplete combustion process, and \((c)\) the rate of heat transfer from the combustion process, in \(\mathrm{kW},\) when \(3.5 \mathrm{kg} / \mathrm{h}\) of fuel are burned when the reactants and products are at \(25^{\circ} \mathrm{C}\) with the water in the products remaining a gas.

Short answer

Answer: The theoretical kmols of O2 required for the complete combustion of Ethanol gas is 3 kmols. The rate of heat transfer during the combustion process is approximately 28.4 kW.

Step-by-step solution

Finding the balanced combustion equation for complete combustion of Ethanol gas

For complete combustion of Ethanol gas C2H6O, the balanced combustion equation is as follows: C2H6O + xO2 -> yCO2 + zH2O To balance the equation, we can set up the following relationships based on the coefficients: Carbon: 2y = 2 Hydrogen: 2z = 6 Oxygen: 2y + z = x Solving for y, z, and x, we obtain: y = 1 (for Carbon) z = 3 (for Hydrogen) x = 3 (for Oxygen) Therefore, the balanced combustion equation for complete combustion is: C2H6O + 3O2 -> 2CO2 + 3H2O

Find the theoretical kmols of O2 required for complete combustion

Using the balanced combustion equation from Step 1, we can see that one kmol of Ethanol gas requires 3 kmols of O2 for complete combustion. Thus, the theoretical kmols of O2 required for complete combustion is 3 kmols.

Find the balanced combustion equation for the incomplete combustion process

According to the given conditions, 90% of carbon is converted to CO2, and 10% is converted to CO. Therefore, we need to modify the complete combustion balanced equation accordingly. Let's replace y with 0.9 and introduce w as the coefficient for CO: C2H6O + xO2 -> 0.9(2)CO2 + wCO + 3H2O We need to balance the equation taking into consideration the incomplete combustion: Carbon: 0.9(2) + w = 2 Oxygen: 2(0.9(2)) + w = x Solving for w and x, we obtain: w = 0.2 (for Carbon) x = 3.4 (for Oxygen) Therefore, the balanced combustion equation for the incomplete combustion is: C2H6O + 3.4O2 -> 1.8CO2 + 0.2CO + 3H2O

Calculating the rate of heat transfer from the combustion process

Given, 3.5 kg/h of fuel is burned. First, we need to find the kmol flow rate of fuel. The molar mass of Ethanol gas (C2H6O) is: (2x12.01) + (6x1.008) + 16 = 46.068 kg/kmol Kmols of fuel burned per hour = (3.5 kg/h) / (46.068 kg/kmol) = 0.0759 kmol/h The Higher Heating Value (HHV) of Ethanol gas is approximately 1367 kJ/mol. Therefore, the heat release rate when 0.0759 kmol/h of fuel is burned will be: Heat release rate = (0.0759 kmol/h) * (1367 kJ/kmol) * (1 h / 3600 s) = 28.408 kW Consider that the reactants and products are at 25°C with the water in the products remaining a gas. The actual heat transfer rate might be slightly less than the heat release rate, but for this exercise, we can assume that the heat transfer rate is approximately the same as the heat release rate. Therefore, the rate of heat transfer from the combustion process is approximately 28.4 kW.