Question

Methane \(\left(\mathrm{CH}_{4}\right)\) is burned completely with 80 percent excess air during a steady-flow combustion process. If both the reactants and the products are maintained at \(25^{\circ} \mathrm{C}\) and 1 atm and the water in the products exists in the liquid form, the heat transfer from the combustion chamber per unit mass of methane is \((a) 890 \mathrm{MJ} / \mathrm{kg}\) (b) \(802 \mathrm{MJ} / \mathrm{kg}\) \((c) 75 \mathrm{MJ} / \mathrm{kg}\) \((d) 56 \mathrm{MJ} / \mathrm{kg}\) \((e) 50 \mathrm{MJ} / \mathrm{kg}\)

Short answer

Question: Calculate the heat transfer from the combustion chamber per unit mass of methane for a complete combustion process using 80% excess air, with both reactants and products maintained at 25°C and 1 atm, and the water existing in liquid form. Answer: (b) 802 MJ/kg

Step-by-step solution

Write the balanced combustion equation

The general combustion equation for methane with oxygen is: \(C H_{4}+x O_{2} \rightarrow y C O_{2}+z H_{2} O\) To balance the equation, we have: \(C H_{4} + 2O_{2} \rightarrow C O_{2} + 2H_{2} O\)

Determine the amount of required and excess air

For complete combustion, 1 mole of methane requires 2 moles of oxygen, which is equivalent to 2 x 32 g = 64 g. Since the air composition is approximately 21% oxygen (\(O_2\)) and 79% nitrogen (\(N_2\)), we need 64 g / 0.21 = 304.76 g of air for complete combustion. Given that the air supply is 80% excess, the total air required becomes 304.76 g + (80/100) x 304.76 g = 548.57 g. The mole fraction of nitrogen in the air is 0.79, so the number of moles of nitrogen involved in the combustion process is 0.79 x 548.57 g / 28 g = 15.47 moles (approximately).

Identify the enthalpy values for reactants and products

Since the reactants and products are maintained at 25°C, we can assume their enthalpies to be approximately equal. The enthalpy of formation for the following components at 25°C are: - Methane (CH4): -74.85 kJ/mol - Oxygen (O2): 0 kJ/mol (since O2 is an element in its standard state) - Carbon dioxide (CO2): -393.51 kJ/mol - Water (H2O): -285.83 kJ/mol (in liquid form)

Calculate the heat transfer using the first law of thermodynamics

Using the first law of thermodynamics, we can calculate the heat transfer (Q) as follows: \(Q = \Delta H_{products} - \Delta H_{reactants}\) For reactants: - 1 mole of methane: 1 x -74.85 kJ/mol = -74.85 kJ - 4 moles of oxygen: 4 x 0 kJ/mol = 0 kJ For products: - 1 mole of carbon dioxide: 1 x -393.51 kJ/mol = -393.51 kJ - 2 moles of water: 2 x -285.83 kJ/mol = -571.66 kJ Now, substitute these values into the equation: \(Q = (-393.51 - 571.66) - (-74.85) = 802\) To convert the heat transfer per mole of methane into the heat transfer per kilogram, divide the result by the molar mass of methane (16.04 g/mol): \(Q = 802 \mathrm{kJ/mol} \cdot (1000 \mathrm{g/kg} / 16.04 \mathrm{g/mol}) = 802 \mathrm{MJ/kg}\)

Choose the correct answer from the given options

From the calculations, we found that the heat transfer from the combustion chamber per unit mass of methane is 802 MJ/kg. Therefore, the correct answer is: (b) \(802 \mathrm{MJ} / \mathrm{kg}\)

Key concepts

Complete Combustion

Complete combustion occurs when a fuel, such as methane (CH₄), reacts fully with the correct amount of oxygen, producing a limited number of products, typically carbon dioxide (CO₂) and water (H₂O). To achieve this, a stoichiometric (or perfect) mixture, which has the exact amount of oxygen required for combustion, is needed. However, in practical applications, some excess air is typically provided to ensure the complete combustion of the fuel, thereby maximizing energy release and minimizing harmful byproducts like carbon monoxide (CO) or unburnt hydrocarbons.

If a fuel burns with excess air, as in our exercise where methane combusts with 80 percent excess air, the combustion is assured to be complete because there is more than enough oxygen to react with all the fuel. This is essential for calculating the energy transfer in a combustion process as it assures all the fuel contributes to energy generation without leaving any unreacted. The excess oxygen does not participate in the combustion reaction, but it is accounted for when calculating the air-fuel ratio.

First Law of Thermodynamics

The first law of thermodynamics, also known as the law of energy conservation, states that energy within a closed system remains constant over time. It can neither be created nor destroyed but can change forms, such as from chemical energy in a fuel to thermal energy (heat) during combustion.

In the context of a steady-flow combustion process, the first law of thermodynamics helps us understand how energy is transferred when methane combusts. Energy can be transferred to or from a system via heat or work. In a combustion reaction, the chemical potential energy stored in the fuel is released as heat. By balancing the energy content before and after the combustion using enthalpy values of the reactants and products, we can calculate the net heat transfer, which is the difference between the enthalpy of the products and the reactants. This analysis is crucial to understand the amount of heat available for work or other applications after combustion.

Enthalpy of Formation

The enthalpy of formation is the heat change associated with the formation of one mole of a substance from its elements in their standard states. Standard states refer to the forms and conditions (usually 25°C and 1 atmosphere of pressure) at which substances are most stable. For elements in their standard states, like oxygen (O₂), the enthalpy of formation is considered zero, because they are already in their most stable form

The enthalpy values provided for reactants and products in our exercise are essential to determine the heat transfer during combustion. The enthalpy of formation for methane, carbon dioxide, and water allows us to calculate the total heat content before and after combustion, which, when subtracted, provides the heat transfer to the surroundings. This concept is vital as it sets the theoretical foundation for predicting the energy outcomes of chemical reactions and can be used to optimize processes for energy efficiency.

Excess Air in Combustion

Excess air in combustion refers to the amount of air above the stoichiometric requirement for a given fuel to completely oxidize all combustible components. It's expressed as a percentage of the stoichiometric air required. Excess air is crucial for complete combustion as it ensures all fuel is burned, which reduces pollutants and helps in the full release of energy.

In our exercise, 80% excess air means that the amount of air supplied is 80% more than what is theoretically needed for complete combustion. While excess air ensures all the methane is combusted, it also means nitrogen from the air (which doesn't participate in the combustion reaction) is heated up, leading to energy losses. Therefore, while some amount of excess air is necessary for complete combustion and to provide a margin of safety, too much excess air can lead to inefficiency. Calculating the exact amount of required air is critical in practical applications to minimize energy waste and optimize the energy yield of a combustion process.