Question

A fuel mixture of 60 percent by mass methane \(\left(\mathrm{CH}_{4}\right)\) and 40 percent by mass ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O}\right),\) is burned completely with theoretical air. If the total flow rate of the fuel is \(10 \mathrm{kg} / \mathrm{s}\), determine the required flow rate of air.

Short answer

Answer: The required flow rate of air for the complete combustion of the given fuel mixture is 139.08 kg/s.

Step-by-step solution

Determine the mass flow rates of methane and ethanol

Since the fuel mixture is 60% methane and 40% ethanol by mass, we can find the mass flow rates of each component by multiplying the total flow rate by the respective mass percentages: Mass flow rate of methane = 0.6 * 10 kg/s = 6 kg/s Mass flow rate of ethanol = 0.4 * 10 kg/s = 4 kg/s

Determine the molecular weights of methane, ethanol, and air

We need to determine the stoichiometric air-fuel ratios for the combustion of methane and ethanol. To do this, we will first need the molecular weights of methane, ethanol, and air. Using the periodic table, we find the following molecular weights: Molecular weight of methane (CH4) = 12 + 4 = 16 g/mol Molecular weight of ethanol (C2H6O) = 24 + 6 + 16 = 46 g/mol Molecular weight of air (assuming 79% nitrogen N2 and 21% oxygen O2) = 0.79 * (28) + 0.21 * (32) = 28.8 g/mol

Determine the stoichiometric air-fuel ratios for methane and ethanol

For complete combustion of methane and ethanol, we write the balanced chemical equations: Methane: CH4 + 2(O2 + 3.76N2) -> CO2 + 2H2O + 7.52N2 Ethanol: C2H6O + 3(O2 + 3.76N2) -> 2CO2 + 3H2O + 11.28N2 From the chemical equations, we can derive the stoichiometric air-fuel ratios by dividing the moles of air by the moles of fuel for each reaction: Air-fuel ratio for methane = 2 * (1 + 3.76) * (molecular weight of air) / (molecular weight of methane) = 2 * 4.76 * 28.8 / 16 = 17.2 kg_air/kg_CH4 Air-fuel ratio for ethanol = 3 * (1 + 3.76) * (molecular weight of air) / (molecular weight of ethanol) = 3 * 4.76 * 28.8 / 46 = 8.97 kg_air/kg_C2H6O

Calculate the required flow rate of air for complete combustion

Now that we have the stoichiometric air-fuel ratios for methane and ethanol, we can determine the required flow rate of air for the complete combustion of the fuel mixture by multiplying the mass flow rates of methane and ethanol with their respective air-fuel ratios: Required flow rate of air = (mass flow rate of methane * air-fuel ratio for methane) + (mass flow rate of ethanol * air-fuel ratio for ethanol) Required flow rate of air = (6 kg/s * 17.2 kg_air/kg_CH4) + (4 kg/s * 8.97 kg_air/kg_C2H6O) = 103.2 kg_air/s + 35.88 kg_air/s = 139.08 kg_air/s Thus, the required flow rate of air for the complete combustion of the given fuel mixture is 139.08 kg/s.