Question

A gaseous fuel with 80 percent \(\mathrm{CH}_{4}, 15\) percent \(\mathrm{N}_{2},\) and 5 percent \(\mathrm{O}_{2}(\) on a mole basis) is burned to completion with 120 percent theoretical air that enters the combustion chamber at \(30^{\circ} \mathrm{C}, 100 \mathrm{kPa},\) and 60 percent relative humidity. Determine \((a)\) the air-fuel ratio and (b) the volume flow rate of air required to burn fuel at a rate of \(2 \mathrm{kg} / \mathrm{min}\).

Short answer

The main purpose of calculating the stoichiometric air-fuel ratio in this exercise is to determine the correct amount of air required for the complete combustion of the gaseous fuel mixture, which helps in calculating the actual air-fuel ratio and the mass and volume flow rates of air required to burn the fuel at a given rate.

Step-by-step solution

Calculate the stoichiometric air-fuel ratio

For complete combustion to occur, we need stoichiometric air, which can be determined by a balanced reaction. The gaseous fuel mixture is 80% CH4, 15% N2, and 5% O2 by the moles, and we write its combustion equation: $$0.8 \, \mathrm{CH}_{4} + 0.05 \, \mathrm{O}_{2} + 0.15 \, \mathrm{N}_{2} + x \, (\mathrm{O}_{2} + 3.76 \, \mathrm{N}_{2}) \rightarrow a \, \mathrm{CO}_{2} + b \, \mathrm{H}_{2}\mathrm{O} + c \, \mathrm{N}_{2}$$ For complete combustion, we balance the equation, which gives us: $$0.8 \, \mathrm{CH}_{4} + 2 \, \mathrm{O}_{2} + 0.15 \, \mathrm{N}_{2} + x \, (1+\, 3.76) \, \mathrm{O}_{2} = a \, \mathrm{CO}_{2} + 2 \, \mathrm{H}_{2}\mathrm{O} + c \, \mathrm{N}_{2}$$ Noting that in the equation, a = 0.8 because one mole of CH4 produces one mole of CO2, and b = 2 because it is a product of the combustion of CH4. Also, c = 0.15 because N2 does not react during combustion, so using the balanced equation, we can determine the stoichiometric air-fuel ratio. After determining the stoichiometric air, x, we can calculate the molar air-fuel ratio (A/F)_stoich.

Calculate the actual air-fuel ratio

Since the problem states that we have 120% of the theoretical (stoichiometric) air, we can calculate the actual air-fuel ratio as (A/F)_actual = 1.2 * (A/F)_stoich.

Determine the mass flow rate of air

Using the actual air-fuel ratio and the given fuel mass flow rate (2 kg/min), we can calculate the mass flow rate of air, m_air: $$m_\mathrm{air} = \mathrm{(A / F)_{actual} \times m_\mathrm{fuel}}$$ Replace (A/F)_actual and m_fuel with their corresponding values and calculate the mass flow rate of air m_air.

Calculate the volume flow rate of air

To calculate the volume flow rate of air (V_air) given temperature, pressure, and relative humidity, we can use the ideal gas law: $$PV = nRT$$ We can convert the mass flow rate of air, m_air, to a molar flow rate using the molar mass of air (M_air). Then, we can calculate the volume flow rate of air, V_air: $$V_\mathrm{air} = \frac{n_\mathrm{air} \cdot R \cdot T}{P}$$ Replace n_air, R, T and P with their corresponding values (include the effect of relative humidity) and calculate the volume flow rate of air required to burn the fuel at a rate of 2 kg/min.