Question

Benzene gas (C \(_{6} \mathrm{H}_{6}\) ) at 1 atm and \(77^{\circ} \mathrm{F}\) is burned during a steady-flow combustion process with 90 percent theoretical air that enters the combustion chamber at \(77^{\circ} \mathrm{F}\) and 1 atm. All the hydrogen in the fuel burns to \(\mathrm{H}_{2} \mathrm{O}\), but part of the carbon burns to CO. Heat is lost to the surroundings at \(77^{\circ} \mathrm{F}\), and the products leave the combustion chamber at 1 atm and 1900 R. Determine \((a)\) the heat transfer from the combustion chamber and ( \(b\) ) the exergy destruction.

Short answer

Question: Calculate the heat transfer and exergy destruction during the combustion of benzene gas (C6H6) when 90% theoretical air is used. The combustion process occurs at standard atmospheric pressure, and the specific enthalpies and exergies are obtained from standard enthalpy and exergy tables. Answer: To calculate the heat transfer (Q) and exergy destruction (Ed) during the combustion process, first write the balanced chemical equations for complete and partial combustion of benzene. Then, apply the energy equation and enthalpy balance to find Q, and calculate the work potential of reactants and products to determine Ed. Use standard enthalpy and exergy tables to find the required values.

Step-by-step solution

Write combustion equations for hydrocarbons

First, write the balanced chemical equations for the combustion of benzene. Since only 90% of theoretical air is used, there is an excess of fuel in the mixture, but part of the carbon burns to CO. Complete combustion equation: (1) C6H6 + (7.5 * 0.9)O2 + (7.5 * 0.9 * 3.76)N2 -> 6CO2 + 3H2O + (7.5 * 0.9 * 3.76)N2 Partial combustion equation: (2) C6H6 + (7.5 * 0.9)O2 + (7.5 * 0.9 * 3.76)N2 -> (6-x)CO2 + xCO + 3H2O + (7.5 * 0.9 * 3.76)N2 where x is the number of moles of CO formed during the combustion process.

Apply the energy equation and enthalpy balance to find the heat transfer

The energy equation for a steady-flow combustion process is: Q - (H_products - H_reactants) = 0 We need to find the heat transfer (Q) and enthalpy changes for reactants and products at the given conditions. We need to know the enthalpies of formation and absolute enthalpies at the specified temperatures using standard enthalpy tables. Let the enthalpy of formation at standard state be: hf_C6H6, hf_CO2, hf_CO and hf_H2O Let the enthalpies of reactants and products at their respective temperatures be: h_C6H6, h_O2, h_CO2, h_CO, h_H2O and h_N2 Now, calculate the enthalpies of reactants and products: H_reactants = h_C6H6 + 7.5 * 0.9 (h_O2 + 3.76 * h_N2) H_products = (6-x)h_CO2 + xh_CO + 3h_H2O + 7.5 * 0.9 * 3.76h_N2 Now, we can rewrite the energy equation as: Q = [(6-x)(h_CO2 + hf_CO2 - h_C6H6 - hf_C6H6) + x(h_CO + hf_CO - h_C6H6 - hf_C6H6) + 3(h_H2O + hf_H2O - h_O2 - hf_O2)] + 7.5 * 0.9 * 3.76(h_N2 - h_O2) After finding Q, we can proceed to find the exergy destruction.

Calculate exergy destruction

Exergy destruction (Ed) in the combustion process can be calculated by finding the work potential of the products (WP_products) and reactants (WP_reactants) and taking the difference: Ed = WP_reactants - WP_products Work potential (WP) can be calculated as: WP = sum(μ_i * n_i) where μ is the molar exergy and n_i is the number of moles. Using standard tables, find the exergy (μ) values of C6H6, O2, CO2, CO, H2O, and N2. Then, calculate the work potentials for reactants and products: WP_reactants = μ_C6H6 + 7.5 * 0.9(μ_O2 + 3.76 * μ_N2) WP_products = (6-x)μ_CO2 + xμ_CO + 3μ_H2O + 7.5 * 0.9 * 3.76μ_N2 Now, you can calculate the exergy destruction (Ed): Ed = WP_reactants - WP_products The final solutions are the heat transfer (Q) and exergy destruction (Ed).

Key concepts

Steady-Flow Combustion

Understanding steady-flow combustion is crucial for grasping how energy conversion systems, such as engines and power plants, operate efficiently. In simple terms, steady-flow combustion refers to a continuous and consistent chemical reaction process where fuel is burned to release heat energy. This process happens at a stable rate, meaning the mass flow rates of the reactants and products are constant over time.

In the context of the given exercise, benzene gas is burned in a combustion chamber where the flow of reactants (benzene and air) and products is steady. An important characteristic of this process is the law of conservation of mass, which states that mass can neither be created nor destroyed. This principle allows for the balancing of chemical equations related to the combustion, which is a critical step before any thermodynamic analysis can be conducted.

Enthalpy Balance

The concept of enthalpy balance is rooted in the first law of thermodynamics, which is all about the conservation of energy. Enthalpy, in simple terms, is the total heat content of a system. When dealing with a combustion process like the one in our exercise, we look at the enthalpy of the reactants (fuel and air) and compare it to the enthalpy of the products (combustion gases).

To establish an enthalpy balance, we determine the amount of heat absorbed or released during the process. This involves using enthalpy values from standard tables for each substance at the specified temperatures and states. By calculating the difference between the enthalpies of the products and reactants, we can find the heat transfer, labeled 'Q', which the exercise asks for. This step is the root of understanding how much energy is harnessed or lost in a combustion process.

Exergy Destruction

Exergy destruction is a thermodynamic concept that quantifies the irreversibility of a process—essentially the energy that cannot be used to do work. In thermodynamics, not all energy is created equal: some of it is bound to be 'destroyed' or degraded in quality due to factors like friction or heat loss.

This degradation is intrinsic to any real-world process, due to the second law of thermodynamics. The higher the exergy destruction, the less efficient the process is, because more useful energy is lost. The exercise requires the calculation of exergy destruction during the combustion of benzene, which serves as an indicator of the inefficiency present in the system. By evaluating the work potential of both reactants and products and taking their difference, one can assess the actual 'useful' energy lost, i.e., the exergy destruction.

Chemical Equations

Chemical equations are the language of chemical reactions, and understanding how to balance them is fundamental for anyone studying combustion processes. In these equations, the reactants are the substances that start the reaction and the products are what you get after the reaction.

The exercise involves balancing the chemical equation for the combustion of benzene, considering that there is not enough oxygen for complete combustion to occur—meaning part of the carbon burns to carbon monoxide (CO), not only to carbon dioxide (CO2). Writing and balancing the chemical equations allows us to visualize the stoichiometry—the quantitative relationship between reactants and products in a chemical reaction—and is vital for subsequent thermodynamic calculations.

Heat Transfer Thermodynamics

Heat transfer thermodynamics involves the study of how heat energy moves from one place to another. This can be due to temperature differences, with energy naturally flowing from a warmer object to a cooler one. In thermodynamic systems like combustion chambers, various forms of heat exchange—conduction, convection, and radiation—play roles in how energy is distributed and used.

The combustion process depicted in the exercise will inevitably lead to heat transfer to the surroundings. This loss of heat affects the performance and efficiency of the system. Knowing how to evaluate this heat transfer is crucial for optimizing fuel usage and minimizing energy waste, which translates directly to the operational cost and environmental impact.