Question

A gaseous fuel mixture that is 40 percent propane \(\left(\mathrm{C}_{3} \mathrm{H}_{8}\right)\) and 60 percent methane \(\left(\mathrm{CH}_{4}\right)\) by volume is mixed with the theoretical amount of dry air and burned in a steady-flow, constant pressure process at 100 kPa. Both the fuel and air enter the combustion chamber at \(298 \mathrm{K}\) and undergo a complete combustion process. The products leave the combustion chamber at 423 K. Determine (a) the balanced combustion equation, (b) the amount of water vapor condensed from the products, and \((c)\) the required air flow rate, in \(\mathrm{kg} / \mathrm{h},\) when the combustion process produces a heat transfer output of \(140,000 \mathrm{kJ} / \mathrm{h}\).

Short answer

a) The balanced combustion equation for the given fuel mixture is: $$0.4C_3H_8 + 0.6CH_4 + 3.2O_2 + 12.0N_2 \rightarrow 1.8CO_2 + 2.8H_2O + 12.0N_2$$ b) Since the combustion products leave at 423 K, above the saturation temperature at 100 kPa, there is no water vapor condensed from the products. c) The required air flow rate for the combustion process to produce a heat transfer output of 140,000 kJ/h is \(74,083.47\,\frac{\text{kg}}{\text{h}}\).

Step-by-step solution

Determine the Stoichiometric Combustion Equations

First, we need to find the stoichiometric combustion equations for propane and methane: 1) Propane (\(C_3H_8\)) combustion: $$C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O$$ 2) Methane (\(CH_4\)) combustion: $$CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O$$

Calculate the Mole Ratio for the Fuel Mixture

Given the fuel mixture contains 40% propane and 60% methane by volume, the amount of each fuel is as follows: 1) Propane: 0.4 moles of \(C_3H_8\) 2) Methane: 0.6 moles of \(CH_4\)

Calculate the Theoretical Amount of Dry Air

Using stoichiometry, calculate the moles of \(O_2\) required for complete combustion of the fuel mixture: 1) Propane \(O_2\): \(0.4 \times 5 = 2\,\text{moles}\) 2) Methane \(O_2\): \(0.6 \times 2 = 1.2\,\text{moles}\) Total moles of \(O_2\) required: \((2 + 1.2) = 3.2\,\text{moles}\) Assuming dry air contains 21% \(O_2\) and 79% \(N_2\) by volume, the moles of \(N_2\) in the air will be: \(\text{moles}\,N_2 = (3.2\,\text{moles}\,O_2) \times \frac{79}{21} = 12.0\,\text{moles}\,N_2\) Thus, the theoretical amount of dry air required is 3.2 moles of \(O_2\) and 12.0 moles of \(N_2\).

Write the Balanced Combustion Equation

The balanced combustion equation considering the fuel mixture proportions and the theoretical amount of dry air is as follows: $$0.4C_3H_8 + 0.6CH_4 + 3.2O_2 + 12.0N_2 \rightarrow 0.4 \times 3CO_2 + 0.6CO_2 + (0.4 \times 4 + 0.6 \times 2)H_2O + 12.0N_2$$ Simplifying the equation: $$0.4C_3H_8 + 0.6CH_4 + 3.2O_2 + 12.0N_2 \rightarrow 1.8CO_2 + 2.8H_2O + 12.0N_2$$ This is the balanced combustion equation for the given fuel mixture.

Calculate the Amount of Water Vapor Condensed from the Products

The products of combustion include 2.8 moles of water vapor. Since the products leave the combustion chamber at 423 K, and the saturation temperature at 100 kPa is 373 K (100°C), all the water vapor will remain as vapor and not condense. Therefore, there is no water vapor condensed from the products.

Calculate the Required Air Flow Rate

We are given that the heat transfer output of the combustion process is 140,000 kJ/h. Assuming complete combustion, the amount of heat released is equal to the heat of combustion of the fuel mixture: Heat of combustion for: 1) Propane: \(-2044 \frac{\text{kJ}}{\text{mol}}\) 2) Methane: \(-890 \frac{\text{kJ}}{\text{mol}}\) Total heat released by the fuel mixture: \((0.4 \times 2044) + (0.6 \times 890) = 1219.2\,\text{kJ/mol}\) To produce 140,000 kJ/h, we require: $$\frac{140,000 \,\text{kJ/h}}{1219.2\,\text{kJ/mol}} = 114.86\,\text{mol/h} \text{of fuel mixture}$$ As the total moles of 3.2 \(O_2\) and 12.0 \(N_2\) are required for complete combustion of the fuel mixture, the required air flow rate (in moles per hour) would be: $$114.86\,\text{mol/h} \times \left(3.2 + 12.0\right) = 1738.06\,\text{mol/h}$$ To convert moles per hour to \(\frac{\text{kg}}{\text{h}}\), we use the molecular masses of \(O_2\) (32 g/mol) and \(N_2\) (28 g/mol), and obtain: $$1738.06\,\text{mol/h} \times \left( 3.2 \times 32 + 12.0 \times 28 \right) \frac{\text{g}}{\text{mol}} \times \frac{1\,\text{kg}}{1000\,\text{g}} = 74,083.47 \frac{\text{kg}}{\text{h}}$$ The required air flow rate is \(74,083.47\,\frac{\text{kg}}{\text{h}}\).