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Diesel fuel \(\left(C_{12} H_{26}\right)\) at \(25^{\circ} C\) is burned in a steadyflow combustion chamber with 20 percent excess air that also enters at \(25^{\circ} \mathrm{C}\). The products leave the combustion chamber at 500 K. Assuming combustion is complete, determine the required mass flow rate of the diesel fuel to supply heat at a rate of \(2000 \mathrm{kJ} / \mathrm{s}\).

Short Answer

Expert verified
Answer: The mass flow rate of Diesel fuel required to supply heat at a rate of 2000 kJ/s is approximately 0.04545 kg/s.

Step by step solution

01

Balancing the combustion equation

The combustion equation for Diesel fuel is given by: C12H26 + x(O2 + 3.76N2) → 12CO2 + 13H2O + Nx. Now we balance the equation keeping in mind the 20% excess air. 1 C12H26 reacts with (12 + 13/2) O2 = 18.5 O2 (Stoichiometric air) With 20% excess air, the O2 available = 1.2 * 18.5 = 22.2 Hence, x = 22.2; N = 3.76 * 22.2 = 83.392 The final balanced equation is: C12H26 + 22.2(O2 + 3.76N2) → 12CO2 + 13H2O + 83.392N2 **Step 2: Determine the heat released per kg of Diesel fuel during combustion**
02

Heat released calculation

To determine the heat transfer, we must find the heat of combustion for Diesel fuel, which is 44,000 kJ/kg. This means that 1 kg of Diesel fuel releases 44,000 kJ of heat when burned. **Step 3: Calculate the mass flow rate of Diesel fuel**
03

Mass flow rate calculation

Now we can use the heat released per kg of Diesel fuel to determine the mass flow rate of Diesel required to provide 2000 kJ/s of heat. Let m be the mass flow rate of Diesel fuel in kg/s. We can write the energy balance equation as: Heat input = Heat output m * 44,000 kJ/kg = 2000 kJ/s Now, we solve for the mass flow rate of Diesel fuel: m = (2000 kJ/s) / (44,000 kJ/kg) = 0.04545 kg/s Therefore, the required mass flow rate of Diesel fuel to supply heat at a rate of 2000 kJ/s is approximately 0.04545 kg/s.

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Key Concepts

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

Combustion Equation Balancing
The process of balancing a combustion equation is essential in understanding the stoichiometry of a chemical reaction, which is particularly crucial when working with combustion reactions in engineering fields. Diesel fuel combustion, for example, involves the reaction between a hydrocarbon (\(C_{12}H_{26}\)) and oxygen (\(O_2\) from air) to produce carbon dioxide (\(CO_2\)), water (\(H_2O\)), and nitrogen, which does not react but is present due to excess air.

To balance the equation, first identify the amount of oxygen needed for stoichiometric combustion—that is, the exact amount required for complete combustion without any leftover fuel or oxygen. This typically involves a certain ratio of oxygen to fuel, based on the combustion chemistry of the hydrocarbon. Once you know the stoichiometric requirement, you can calculate the excess air supplied, which ensures complete combustion and is expressed as a percentage of the stoichiometric air. By multiplying the stoichiometric oxygen requirement by 1 + the excess air fraction (in this case, 20% or 0.2), you can obtain the actual amount of oxygen reacting.
Heat of Combustion
The heat of combustion is a fundamental property that quantifies the amount of energy released when a certain mass of fuel is burned. This value is critical for the design and analysis of combustion systems, such as engines and heaters. In our example with Diesel fuel (\(C_{12}H_{26}\)), a standard reference value for the heat of combustion, typically expressed in kilojoules per kilogram (\(kJ/kg\)), can be used to calculate the total energy released during the process.

Knowing the heat of combustion allows engineers to assess the efficiency of the fuel and the system using it. For instance, by comparing the energy content of the fuel to the energy output of a combustion chamber, we can determine the effectiveness of energy conversion and identify any losses that may occur in the form of unburned fuel, excess heat, or incomplete combustion.
Excess Air in Combustion
Excess air in combustion refers to the air supplied to the combustion chamber beyond what is theoretically needed for the complete oxidation of the fuel. Providing excess air ensures complete combustion, reducing pollutants like carbon monoxide and unburned hydrocarbons that can form from a deficiency of oxygen. However, too much excess air can lower the temperature of the combustion chamber, which can decrease the overall efficiency of the system.

In equations, the amount of excess air is represented by a multiplier of the stoichiometric air. For example, 20% excess air means that 1.2 times the stoichiometric amount is used. Adjusting the combustion equation to account for this excess is crucial in accurate calculations of the combustion process. Balancing the availability of excess air with the need for efficient combustion is a key consideration in the design and operation of combustion systems.
Energy Balance Equation
At the heart of understanding the relationship between fuel input and energy output in combustion processes lies the energy balance equation. This equation is a statement of the conservation of energy principle, which in the case of a combustion reaction, can be simplified to the equivalence of the heat produced by the combustion of fuel to the heat required or used by the system.

The energy balance for a steady-flow combustion process will take the form: Heat input by fuel combustion = Heat output demanded by the load. Using the heat of combustion of the fuel and the desired heat output, we can rearrange the equation to find the necessary mass flow rate of the fuel to meet a specified energy demand. This is crucial for the efficient design of energy systems, ensuring the demand is met without excess fuel usage, which could result in fuel wastage and additional cost.

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Most popular questions from this chapter

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.

One kilogram of butane \(\left(\mathrm{C}_{4} \mathrm{H}_{10}\right)\) is burned with \(25 \mathrm{kg}\) of air that is at \(30^{\circ} \mathrm{C}\) and \(90 \mathrm{kPa}\). Assuming that the combustion is complete and the pressure of the products is \(90 \mathrm{kPa},\) determine \((a)\) the percentage of theoretical air used and \((b)\) the dew-point temperature of the products.

Liquid ethyl alcohol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(\ell)\right)\) at \(25^{\circ} \mathrm{C}\) is burned in a steady- flow combustion chamber with 40 percent excess air that also enters at \(25^{\circ} \mathrm{C}\). The products leave the combustion chamber at \(600 \mathrm{K}\). Assuming combustion is complete, determine the required volume flow rate of the liquid ethyl alcohol, to supply heat at a rate of \(2000 \mathrm{kJ} / \mathrm{s}\). At \(25^{\circ} \mathrm{C}\) the density of liquid ethyl alcohol is \(790 \mathrm{kg} / \mathrm{m}^{3},\) the specific heat at a constant pressure is \(114.08 \mathrm{kJ} / \mathrm{kmol} \cdot \mathrm{K},\) and the enthalpy of vaporization is \(42,340 \mathrm{kJ} / \mathrm{kmol}\).

A fuel at 25 \(^{\circ} \mathrm{C}\) is burned in a well-insulated steady-flow combustion chamber with air that is also at \(25^{\circ} \mathrm{C}\). Under what conditions will the adiabatic flame temperature of the combustion process be a maximum?

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?

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