Question

13.43 For a natural gas with a molar analysis of $86.5\mathrm{\%}{\mathrm{CH}}_4,8\mathrm{\%}$ ${\mathrm{C}}_2{\mathrm{H}}_6,2\mathrm{\%}{\mathrm{C}}_3{\mathrm{H}}_8,3.5\mathrm{\%}{\mathrm{N}}_2$, determine the lower heating value, in $\mathrm{kJ}$ per $\mathrm{kmol}$ of fuel and in $\mathrm{kJ}$ per $\mathrm{kg}$ of fuel, at ${25}^{\circ }\mathrm{C},1\mathrm{atm}$.

Short answer

The LHV is 848040 \text{kJ/kmol} or 41718.8 \text{kJ/kg}.

Step-by-step solution

- Write the Lower Heating Value (LHV)

The lower heating value (LHV) of each component in the fuel mixture is required. The LHV is the amount of heat released when a compound undergoes complete combustion and all the products return to a temperature of 25°C. Here are the lower heating values provided for each component in \text{kJ/kmol}:$\text{Double subscripts: use braces to clarify}$$\text{Double subscripts: use braces to clarify}$$\text{Double subscripts: use braces to clarify}$$\text{Double subscripts: use braces to clarify}$ as nitrogen does not combust.

- Calculate the Contribution of Each Component

Multiply each component's lower heating value by its molar percentage (expressed as a fraction):${\text{{LHV}}}_{\text{{mixture}}}=(0.865\times 802300)+(0.08\times 1427000)+(0.02\times 2044000)+(0.035\times 0)$

- Sum the Contributions

Calculate the total lower heating value for the fuel mixture by summing the contributions:${\text{{LHV}}}_{\text{{mixture}}}=693000\text{kJ/kmol}+114160\text{kJ/kmol}+40880\text{kJ/kmol}=848040\text{kJ/kmol}$

- Calculate Molar Mass of the Fuel Mixture

Compute the molar mass of the mixture by multiplying the molar masses of each component by their respective molar fractions: $\begin{array}{rlrlr}{\text{{Molar mass}}}_{\text{{mixture}}}=& (0.865\times 16.04\text{g/mol})+(0.08\times 30.07\text{g/mol})+(0.02\times 44.1\text{g/mol})+(0.035\times 28.02\text{g/mol})=& 16.0557\text{{g/mol}}+2.4056\text{{g/mol}}+0.882\text{{g/mol}}+0.9807\text{{g/mol}}=& 20.324\text{{g/mol}}=& 0.020324\text{{kg/mol}}\end{array}$

- Convert to \text{kJ/kg}

Finally, convert the lower heating value from \text{kJ/kmol} to \text{kJ/kg} by dividing by the molar mass.${\text{{LHV}}}_{\text{{mixture}}}=\frac{848040\text{kJ/kmol}}{20.324\text{kg/kmol}}=41718.8\text{kJ/kg}$

Key concepts

Molar Mass

Molar mass is the mass of one mole of a given substance. It is expressed in grams per mole (g/mol). To find the molar mass of a mixture, we need to compute the weighted average molar mass based on the molar fraction of each component. For instance, the given natural gas mixture contains methane (CH₄), ethane (C₂H₆), propane (C₃H₈), and nitrogen (N₂). Each component has its specific molar mass:

• Methane: 16.04 g/mol
• Ethane: 30.07 g/mol
• Propane: 44.1 g/mol
• Nitrogen: 28.02 g/mol

To determine the molar mass of the entire fuel mixture, we multiply the molar mass of each component by its molar fraction and then sum these values.

Combustion

Combustion is a chemical process where a fuel reacts with an oxidant, typically oxygen, to produce heat and combustion products like carbon dioxide (CO₂) and water (H₂O). The energy released during combustion is crucial for many applications, such as generating power or heat. In our exercise, we analyze the ideal combustion of a natural gas mixture, which primarily contains small hydrocarbon molecules. By understanding the proportions of each combustible component in the mixture, we can better predict the heat released and optimize fuel usage. Complete combustion implies that all carbon in the fuel is converted to CO₂ and all hydrogen to H₂O, without any remaining hydrocarbons or carbon monoxide (CO).

Fuel Mixture Analysis

Fuel mixture analysis involves understanding the composition of a fuel and its impact on various properties, such as its heating value. The given natural gas mixture composition is:

• 86.5% Methane (CH₄)
• 8% Ethane (C₂H₆)
• 2% Propane (C₃H₈)
• 3.5% Nitrogen (N₂)

By analyzing these percentages, we can determine how much energy each component contributes to the total heating value. For instance, methane, being the largest component, contributes the most to the overall energy content. Nitrogen does not combust and thus does not contribute to the heating value, but it might influence the combustion efficiency and emissions.

Enthalpy of Combustion

Enthalpy of combustion refers to the heat released when one mole of a substance burns completely in oxygen. This quantity is usually provided at standard conditions (25°C and 1 atm). In this exercise, we focus on the lower heating value (LHV), which neglects the latent heat of vaporization for any water produced. This means the water in the combustion products remains as vapor. The enthalpy of combustion for each fuel component in our mixture is:

• Methane (CH₄): 802.3 kJ/mol
• Ethane (C₂H₆): 1427 kJ/mol
• Propane (C₃H₈): 2044 kJ/mol
• Nitrogen (N₂): 0 kJ/mol (since nitrogen does not combust)

To find the LHV of the entire mixture, we calculate the weighted average based on these values and the molar percentage of each component.