Question

One lbmol of methane \(\left(\mathrm{CH}_{4}\right)\) undergoes complete combustion with stoichiometric amount of air in a rigid container. Initially, the air and methane are at 14.4 psia and \(77^{\circ} \mathrm{F}\). The products of combustion are at \(800^{\circ} \mathrm{F}\). How much heat is rejected from the container, in Btu/lbmol fuel?

Short answer

Question: Calculate the heat rejected from a rigid container during the combustion of one lbmol of methane with the stoichiometric amount of air, given the initial conditions of 14.4 psia and 77°F and final temperature of 800°F. Answer: The heat rejected from the container during the combustion of one lbmol of methane is 23151.25 Btu/lbmol.

Step-by-step solution

Write the combustion equation for methane and stoichiometric amount of air

The combustion reaction of one lbmol of methane with stoichiometric amount of air can be written as: \(\mathrm{CH}_{4} + 2(\mathrm{O}_{2} + 3.76\mathrm{N}_{2}) \rightarrow \mathrm{CO}_{2} + 2(\mathrm{H}_{2}\mathrm{O}) + 7.52\mathrm{N}_{2}\)

Calculate the initial state properties

The initial conditions are given as 14.4 psia and \(77^{\circ}\mathrm{F}\). We can use the ideal gas equation of state to calculate the initial specific volume and specific enthalpy. The initial specific volume \(v_1\) can be calculated using the ideal gas law as follows: \(v_{1} = \frac{RT_{1}}{P_{1}}\) Convert the temperature to Rankine: \(T_1 = 77 + 460 = 537 \, \mathrm{R}\), R = 1.986 (Ra specific gas constant for air) \(v_{1} = \frac{1.986 \times 537}{14.4 \times 144} = 0.4865 \, \mathrm{ft}^3/\mathrm{lbmol}\) Now, we determine \(h_1\) using air enthalpy data tables. For the given initial state, the specific enthalpy is \(h_1 = 150.75 \, \mathrm{Btu}/\mathrm{lbmol}\) (from air property data table)

Calculate final state properties

The products of combustion are at \(800^{\circ}\mathrm{F}\). Since the combustion is complete, the final mixture is composed of \(\mathrm{CO}_{2}, \mathrm{H}_{2}\mathrm{O}\), and \(\mathrm{N}_{2}\). We need to find the specific volume \(v_{2}\) and specific enthalpy \(h_{2}\) for the final state. For a rigid container, the specific volume remains constant, so \(v_{2} = v_{1} = 0.4865 \, \mathrm{ft}^3/\mathrm{lbmol}\). To find \(h_2\), we can use property tables for the combustion products: \(h_{2}(\mathrm{CO}_{2}) = 7214 \, \mathrm{Btu}/\mathrm{lbmol}\) \(h_{2}(\mathrm{H}_{2}\mathrm{O}) = 1791 \, \mathrm{Btu}/\mathrm{lbmol}\) \(h_{2}(\mathrm{N}_{2}) = 1607 \, \mathrm{Btu}/\mathrm{lbmol}\) Weighting these values by the stoichiometric coefficients, the final state enthalpy for the mixture is: \(h_{2} = 1 \times h_{2}(\mathrm{CO}_{2}) + 2 \times h_{2}(\mathrm{H}_{2}\mathrm{O}) + 7.52 \times h_{2}(\mathrm{N}_{2}) = 23302 \, \mathrm{Btu}/\mathrm{lbmol}\)

Apply the first law of thermodynamics

Now, we can use the first law of thermodynamics for the closed system to determine the heat rejected. The first law of thermodynamics for a closed system undergoing a process is given by: \(Q_{out} - W_{out} = m(\Delta h)\) Since the process happens in a rigid container, there is no work done (\(W_{out} = 0\)). Therefore, \(Q_{out} = m(\Delta h)\) Substituting the values calculated in step 3: \(Q_{out} = (1\, \mathrm{lbmol})(h_{2} - h_{1})\) \(Q_{out} = (1\, \mathrm{lbmol})(23302 - 150.75) \, \mathrm{Btu}/\mathrm{lbmol}\)

Calculate the heat rejected from the container

\(Q_{out} = 23151.25\, \mathrm{Btu}/\mathrm{lbmol}\) Therefore, the heat rejected from the container during the combustion of one lbmol of methane is 23151.25 Btu/lbmol.

Key concepts

Stoichiometric Combustion

Understanding stoichiometric combustion is essential for analyzing many chemical processes, including the combustion of fuels. Stoichiometry involves the calculation of reactants and products in chemical reactions. In stoichiometric combustion, the fuel is burned with the exact amount of oxygen required for complete combustion, meaning all the fuel is converted into combustion products like carbon dioxide and water, with no left-over reactants.

This process is vital for two reasons. Firstly, stoichiometric combustion ensures maximum efficiency as all fuel is utilized. Secondly, it reduces the creation of undesirable by-products such as carbon monoxide or unburnt hydrocarbons that can occur with incomplete combustion.

In the exercise provided, stoichiometric amounts of air (oxygen and nitrogen) react with methane to yield specific products. This precise calculation not only helps in understanding the reaction mechanism but also aids in determining the energy change associated with the process, which is crucial for energy balance calculations.

Ideal Gas Law

The ideal gas law is a fundamental equation in thermodynamics that relates the pressure, volume, temperature, and number of moles of a gas. The equation is given as \( PV = nRT \), where \( P \) is the pressure, \( V \) is the volume, \( n \) is the amount of substance in moles, \( R \) is the universal gas constant, and \( T \) is the temperature in Kelvin or Rankine.

This law assumes the gas behaves ideally, meaning the particles have negligible volume and do not exert forces on each other, except during collisions. While real gases do not perfectly adhere to these assumptions, the ideal gas law provides an excellent approximation for many gases under standard conditions.

In our exercise, the ideal gas law is used to determine the specific volume of the methane-air mixture before combustion. This is essential for tracking how the gas properties change throughout the process. Since the container is rigid, volume remains constant, which simplifies further calculations, showing how an understanding of gas behavior supports the analysis of thermodynamic systems.

First Law of Thermodynamics

The first law of thermodynamics, also known as the law of energy conservation, states that energy cannot be created or destroyed in an isolated system. The change in internal energy of a system is equal to the heat added to the system minus the work done by the system on the surroundings. Mathematically, it is expressed as \( \triangle U = Q - W \), where \( \triangle U \) is the change in internal energy, \( Q \) is the heat exchanged, and \( W \) is the work done.

This law is pivotal in understanding energy transfers during chemical reactions, such as combustion. When applying this principle to our exercise, we consider the combustion taking place in a closed, rigid container. This means there is no work done by or on the system (\( W = 0 \)), simplifying our formula to \( \triangle U = Q \), where \( Q \) represents the heat rejected from the system.

By using the specific enthalpies of the initial and final states, the first law facilitates the calculation of heat transfer, allowing students to connect the conservation of energy with tangible chemical processes.