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Using EES (or other) software, write a general program to determine the heat transfer during the complete combustion of a hydrocarbon fuel \(\left(\mathrm{C}_{n} \mathrm{H}_{m}\right)\) at \(25^{\circ} \mathrm{C}\) in a steady-flow combustion chamber when the percent of excess air and the temperatures of air and the products are specified. As a sample case, determine the heat transfer per unit mass of fuel as liquid propane \(\left(\mathrm{C}_{3} \mathrm{H}_{8}\right)\) is burned steadily with 50 percent excess air at \(25^{\circ} \mathrm{C}\) and the combustion products leave the combustion chamber at \(1800\) \(\mathrm{K}\).

Short Answer

Expert verified
Question: Calculate the heat transfer per unit mass of fuel during the combustion of a hydrocarbon fuel, given a fuel sample of liquid propane with 50% excess air at 25°C, and the products leave the combustion chamber at 1800K. Answer: To calculate the heat transfer per unit mass of fuel during combustion, follow these steps: 1. Balance the combustion equation for the hydrocarbon fuel. 2. Adjust the balanced equation for excess air. 3. Determine specific enthalpies at input and output temperatures using EES (or other) software. 4. Calculate heat transfer per unit mass of fuel using the specific enthalpies and balanced equation. For the given sample case, use the balanced equation for liquid propane combustion and specific enthalpies calculated from EES software to find the heat transfer per unit mass of fuel.

Step by step solution

01

Write the balanced combustion equation

First, we need to balance the complete combustion equation for the hydrocarbon fuel \(\mathrm{C}_{n} \mathrm{H}_{m}\). The balanced equation should include oxygen, carbon dioxide, and water in addition to the fuel: $$\mathrm{C}_{n}\mathrm{H}_{m} + (\frac{n}{2} + \frac{m}{4})\mathrm{O}_{2} \rightarrow n\mathrm{CO}_{2} + \frac{m}{2}\mathrm{H}_{2}\mathrm{O}$$ For the given sample case of liquid propane, the balanced equation is: $$\mathrm{C}_{3}\mathrm{H}_{8} + 5\mathrm{O}_{2} \rightarrow 3\mathrm{CO}_{2} + 4\mathrm{H}_{2}\mathrm{O}$$
02

Account for excess air

Since we have 50% excess air, we need to multiply the amount of \(\mathrm{O}_{2}\) in the balanced equation by 1.5: $$\mathrm{C}_{3}\mathrm{H}_{8} + 7.5\mathrm{O}_{2} \rightarrow 3\mathrm{CO}_{2} + 4\mathrm{H}_{2}\mathrm{O} + 2.5\mathrm{O}_{2}$$
03

Calculate heat transfer

To calculate the heat transfer per unit mass of fuel during combustion, we need the specific enthalpies of each component at the specified temperatures. The heat transfer \(q\) can then be found by: $$q = (h_{CO2,out} - h_{CO2,in})n + (h_{H2O,out} - h_{H2O,in})\frac{m}{2} - (h_{O2,out} - h_{O2,in})x$$ where \(h_{in}\) and \(h_{out}\) are the specific enthalpies of each component at the input and output temperatures, and \(x\) is the excess oxygen ratio. The specific enthalpies can be determined using EES (or other) software with the given temperatures of 25°C and 1800K for input and output temperatures.
04

Organize the general program

Based on the previous steps, we can organize a general program for determining heat transfer during combustion as follows: 1. Input the hydrocarbon fuel, percent excess air, temperature of air and products. 2. Determine and balance the combustion equation. 3. Adjust the balanced equation for excess air. 4. Determine specific enthalpies at input and output temperatures using EES (or other) software. 5. Calculate heat transfer per unit mass of fuel according to the previously obtained information and equations. 6. Output the heat transfer per unit mass of fuel. For the given sample case, use the EES (or other) software to calculate specific enthalpies and eventually get the heat transfer per unit mass of fuel.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding the Steady-Flow Combustion Chamber
A steady-flow combustion chamber is an integral part of many engineering systems, such as internal combustion engines and gas turbines. In such devices, fuel combustion occurs in a controlled continuous process. The design of a steady-flow combustion chamber ensures that a consistent mixture of fuel and air enters, burns, and then exhausts as combustion products at steady rates, allowing for constant power output and efficient energy transfer.

When analyzing steady-flow combustion chambers, engineers are particularly interested in energy conservation. Applying the First Law of Thermodynamics to these systems guarantees that all heat and work interactions are accurately accounted for. Additionally, evaluating performance necessitates reliable calculations of the enthalpy or thermal content of the substances before and after combustion, which underscores the significance of specific enthalpy calculations in these systems.

Overall, understanding the principles of a steady-flow combustion chamber is essential for students in order to tackle problems related to energy conversion and efficiency in various thermodynamic applications.
Hydrocarbon Fuel Combustion Basics
The combustion of hydrocarbon fuels, substances primarily composed of carbon (C) and hydrogen (H), is a complex chemical process that releases energy. The general reaction for complete combustion involves a hydrocarbon reacting with oxygen (O2) to form carbon dioxide (CO2) and water (H2O), which can be represented by a balanced chemical equation.

For example, the complete combustion of propane \(\mathrm{C}_{3}\mathrm{H}_{8}\) can be represented as:\[\mathrm{C}_{3}\mathrm{H}_{8} + 5\mathrm{O}_{2} \rightarrow 3\mathrm{CO}_{2} + 4\mathrm{H}_{2}\mathrm{O}\]Understanding the stoichiometry of the reaction is crucial for calculating the amount of reactants needed and the energy released. The reaction balances carbon and hydrogen atoms on both sides to conform with the law of conservation of mass. Furthermore, mastering hydrocarbon fuel combustion analysis enables efficient design and operation of combustion systems, such as engines and furnaces, leading to optimized fuel use and reduced emissions.
The Role of Excess Air in Combustion
In the process of combustion, excess air is the additional air supplied over the theoretical air requirement for complete combustion of the fuel. This is often necessary to ensure complete burning as actual combustion is not perfectly efficient. Excess air is defined as the percentage of extra air supplied relative to the stoichiometric amount.

When we have excess air, for instance, 50% as seen in the exercise, it means that we supply 1.5 times the stoichiometric required oxygen amount. This is crucial to avoid incomplete combustion, which can lead to carbon monoxide formation and unburnt fuel, adversely affecting efficiency and increasing pollutants. The presence of excess air can be accounted for in the balanced combustion equation as shown in the example of propane combustion:\[\mathrm{C}_{3}\mathrm{H}_{8} + 7.5\mathrm{O}_{2} \rightarrow 3\mathrm{CO}_{2} + 4\mathrm{H}_{2}\mathrm{O} + 2.5\mathrm{O}_{2}\]Students need to understand how to adjust combustion calculations for excess air in order to ensure their combustion efficiency and emissions calculations are realistic and applicable to real-world scenarios.
Specific Enthalpy Calculation in Combustion
Specific enthalpy, a thermodynamic property, quantifies the total heat content of a substance per unit mass, which includes internal energy and the work done by the substance to push against atmospheric pressure. In combustion processes, calculating the specific enthalpy changes of reactants and products is vital for determining the heat transfer involved.

In our exercise, the calculation of heat transfer follows the equation:\[q = (h_{CO2,out} - h_{CO2,in})n + (h_{H2O,out} - h_{H2O,in})\frac{m}{2} - (h_{O2,out} - h_{O2,in})x\]Here, \(h_{in}\) and \(h_{out}\) represent the specific enthalpies at the inlet and outlet temperatures respectively, \(n\) and \(m\) denote the number of moles, and \(x\) stands for the excess oxygen ratio. To find these values, one might use lookup tables, steam charts, or software like EES (Engineering Equation Solver).

Comprehension of specific enthalpy changes during combustion allows students to apply this knowledge to a range of problems not only in heat transfer but also in fields such as power generation, engine performance, and environmental engineering, thereby fostering a more comprehensive understanding of energy systems.

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Most popular questions from this chapter

An adiabatic constant-volume tank contains a mixture of \(1 \mathrm{kmol}\) of hydrogen \(\left(\mathrm{H}_{2}\right)\) gas and the stoichiometric amount of air at \(25^{\circ} \mathrm{C}\) and 1 atm. The contents of the tank are now ignited. Assuming complete combustion, determine the final temperature in the tank.

Acetylene gas \(\left(\mathrm{C}_{2} \mathrm{H}_{2}\right)\) at \(25^{\circ} \mathrm{C}\) is burned during a steady-flow combustion process with 30 percent excess air at \(27^{\circ} \mathrm{C} .\) It is observed that \(75,000 \mathrm{kJ}\) of heat is being lost from the combustion chamber to the surroundings per kmol of acetylene. Assuming combustion is complete, determine the exit temperature of the product gases.

An equimolar mixture of carbon dioxide and water vapor at 1 atm and \(60^{\circ} \mathrm{C}\) enter a dehumidifying section where the entire water vapor is condensed and removed from the mixture, and the carbon dioxide leaves at 1 atm and \(60^{\circ} \mathrm{C}\). The entropy change of carbon dioxide in the dehumidifying section is \((a)-2.8 \mathrm{kJ} / \mathrm{kg} \cdot \mathrm{K}\) \((b)-0.13 \mathrm{kJ} / \mathrm{kg} \cdot \mathrm{K}\) \((c) 0\) \((d) 0.13 \mathrm{kJ} / \mathrm{kg} \cdot \mathrm{K}\) \((e) 2.8 \mathrm{kJ} / \mathrm{kg} \cdot \mathrm{K}\)

What are the causes of incomplete combustion?

The furnace of a particular power plant can be considered to consist of two chambers: an adiabatic combustion chamber where the fuel is burned completely and adiabatically, and a heat exchanger where heat is transferred to a Carnot heat engine isothermally. The combustion gases in the heat exchanger are well mixed so that the heat exchanger is at uniform temperature at all times that is equal to the temperature of the exiting product gases, \(T_{p} .\) The work output of the Carnot heat engine can be expressed as $$ w=Q \eta_{c}=Q\left(1-\frac{T_{0}}{T_{p}}\right)$$ where \(Q\) is the magnitude of the heat transfer to the heat engine and \(T_{0}\) is the temperature of the environment. The work output of the Carnot engine will be zero either when \(T_{p}=T_{\mathrm{af}}\) (which means the product gases will enter and exit the heat exchanger at the adiabatic flame temperature \(T_{\mathrm{af}}\), and thus \(Q=0\) ) or when \(T_{p}=\) \(T_{0}\) (which means the temperature of the product gases in the heat exchanger will be \(T_{0}\), and thus \(\eta_{c}=0\) ), and will reach a maximum somewhere in between. Treating the combustion products as ideal gases with constant specific heats and assuming no change in their composition in the heat exchanger, show that the work output of the Carnot heat engine will be maximum when $$T_{p}=\sqrt{T_{\mathrm{af}} T_{0}}$$ Also, show that the maximum work output of the Carnot engine in this case becomes $$W_{\max }=C T_{\mathrm{af}}\left(1-\sqrt{\frac{T_{0}}{T_{\mathrm{af}}}}\right)^{2}$$ where \(C\) is a constant whose value depends on the composition of the product gases and their specific heats.

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