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An adiabatic gas turbine expands air at \(1300 \mathrm{kPa}\) and \(500^{\circ} \mathrm{C}\) to \(100 \mathrm{kPa}\) and \(127^{\circ} \mathrm{C}\). Air enters the turbine through a \(0.2-\mathrm{m}^{2}\) opening with an average velocity of \(40 \mathrm{m} / \mathrm{s},\) and exhausts through a \(1-\mathrm{m}^{2}\) opening. Determine \((a)\) the mass flow rate of air through the turbine and \((b)\) the power produced by the turbine.

Short Answer

Expert verified
Answer: The mass flow rate of the air through the turbine is 46.28 kg/s, and the power produced by the turbine is 10.45 MW.

Step by step solution

01

Find the mass flow rate at the inlet

To find the mass flow rate, we first need to calculate the air density at the inlet. We are given the inlet temperature \(T_1 = 500^\circ C\) and pressure \(P_1 = 1300kPa\). Using the ideal gas equation, we can find the density: \(\rho_1 = \frac{P_1}{RT_1}\) where \(R = 287 J/(kg K)\) is the specific gas constant for air, and \(T_1 = 500^\circ C + 273.15 = 773.15 K\). Substituting the given values, we get: \(\rho_1 = \frac{1300000}{287 \times 773.15} = 5.785 kg/m^3\) Now we can calculate the mass flow rate: \(\dot{m} = \rho_1 \times A_1 \times V_1\) where \(A_1 = 0.2 m^2\) and \(V_1 = 40 m/s\). \(\dot{m} = 5.785 \times 0.2 \times 40 = 46.28 kg/s\)
02

Use the isentropic relation to find the outlet velocity

Since the process is adiabatic, we can assume that it is also isentropic. Therefore, we can use the isentropic relations to find the outlet velocity \(V_2\). For an isentropic process, the ratio of total pressure and total temperature remains constant: \(\frac{P_1 + \frac{1}{2}\rho_1 V_1^2}{T_1} = \frac{P_2 + \frac{1}{2}\rho_2 V_2^2}{T_2}\) where \(P_2 = 100 kPa\) and \(T_2 = 127^\circ C + 273.15 = 400.15 K\). To find \(\rho_2\), we use the ideal gas equation: \(\rho_2 = \frac{P_2}{RT_2} = \frac{100000}{287 \times 400.15} = 0.867 kg/m^3\) Now, substitute the values in the isentropic relation and solve for \(V_2\): \(V_2 = \sqrt{\frac{2(P_1T_2 - P_2T_1)}{\rho_2 (T_1-T_2)} - V_1^2} = \sqrt{\frac{2(1300000 \times 400.15 - 100000 \times 773.15)}{0.867 (773.15-400.15)} - 40^2} = 612.6 m/s\)
03

Calculate the power produced by the turbine

The power produced by the turbine can be calculated as follows: \(P = \dot{m} \times \frac{1}{2} \times (V_1^2 - V_2^2)\) Substituting the values, we get: \(P = 46.28 \times \frac{1}{2} \times (40^2 - 612.6^2) = -10\,451\,940 W\) (or \(-10.45 MW\)) The negative sign indicates that the turbine is producing power. So, the mass flow rate of the air through the turbine is \(46.28 kg/s\) and the power produced by the turbine is \(10.45 MW\).

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Key Concepts

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

Mass Flow Rate
Understanding the concept of mass flow rate is crucial when analyzing the operation of devices like gas turbines, as it represents the quantity of mass passing through a given cross-sectional area per unit time. In the context of an adiabatic gas turbine, it is determined by the continuity equation, which ensures mass conservation through the system.

The mass flow rate, denoted as \(\dot{m}\), is calculated using the formula \(\dot{m} = \rho_1 \times A_1 \times V_1\), where \(\rho_1\) is the air density at the inlet, \(A_1\) is the cross-sectional area of the inlet, and \(V_1\) is the velocity of air at the inlet. The density \(\rho_1\) can be obtained using the ideal gas equation, relating it to the air's pressure and temperature at the inlet. This variable is a pivotal factor and must be accurately determined to ensure the validity of the mass flow rate calculation. In practice, variations in cross-sectional areas and velocities within the turbine would affect this value, making it a dynamic parameter of interest in the design and analysis of such systems.
Ideal Gas Equation
The ideal gas equation is fundamental to the study of thermodynamics, particularly for gases that are considered 'ideal'. The equation establishes a relationship between the pressure \(P\), volume \(V\), temperature \(T\), and the amount of substance (in moles) \(n\), involving a constant known as the gas constant \(R\). It is given as \(PV = nRT\), which can be rewritten in terms of density to find \(\rho = \frac{P}{RT}\) for a given mass of gas.

In the context of the exercise, the ideal gas law is used to compute the density of air at both the inlet and outlet conditions of the turbine by relating the known temperature and pressure values. When utilizing the equation, it's vital to convert temperatures to an absolute scale such as Kelvin to prevent calculation errors. Additionally, the choice of the specific gas constant \(R\) is dependent on the type of gas; for air, it's typically taken as \(287 J/(kg\cdot K)\). A deep understanding of how changes in these parameters affect gas density is necessary to accurately predict the behavior of the gas as it undergoes expansion through the turbine.
Isentropic Process
An isentropic process is an idealized thermodynamic process that is both adiabatic (no heat transfer) and reversible. For such processes, entropy remains constant. In reality, reaching a truly isentropic process is unachievable, however, it is a useful approximation for many engineering calculations, especially for high-speed gas flow through turbines, nozzles, and diffusers.

In an adiabatic gas turbine, like the one described in our exercise, assuming the expansion to be isentropic allows us to utilize certain ratios and relations that describe the behavior of the gas. An important consequence of isentropically approximating these processes is that it allows for easier calculations of the changes in velocity and temperature of the gas, as the isentropic relations link those changes to pressure changes. For instance, using the conservation of energy for an isentropic process, certain dynamic changes in velocity can be predicted, which, as illustrated in the exercise, assists in determining the kinetic energy changes across the turbine. These energy changes are fundamental to calculating the power output as they represent work done. The concept's simplicity helps to create a robust first estimate for assessing turbine performance and identifying efficiency and power characteristics that are essential in the design and operation of such systems.

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Most popular questions from this chapter

A sealed electronic box is to be cooled by tap water flowing through the channels on two of its sides. It is specified that the temperature rise of the water not exceed \(4^{\circ} \mathrm{C}\) The power dissipation of the box is \(2 \mathrm{kW}\), which is removed entirely by water. If the box operates 24 hours a day, 365 days a year, determine the mass flow rate of water flowing through the box and the amount of cooling water used per year.

An air compressor compresses 15 L \(/\) s of air at 120 kPa and \(20^{\circ} \mathrm{C}\) to \(800 \mathrm{kPa}\) and \(300^{\circ} \mathrm{C}\) while consuming \(6.2 \mathrm{kW}\) of power. How much of this power is being used to increase the pressure of the air versus the power needed to move the fluid through the compressor?

The air in a \(6-m \times 5-m \times 4-m\) hospital room is to be completely replaced by conditioned air every 15 min. If the average air velocity in the circular air duct leading to the room is not to exceed \(5 \mathrm{m} / \mathrm{s}\), determine the minimum diameter of the duct.

A tank with an internal volume of \(1 \mathrm{m}^{3}\) contains air at \(800 \mathrm{kPa}\) and \(25^{\circ} \mathrm{C}\). A valve on the tank is opened allowing air to escape and the pressure inside quickly drops to \(150 \mathrm{kPa}\), at which point the valve is closed. Assume there is negligible heat transfer from the tank to the air left in the tank. (a) Using the approximation \(h_{e} \approx\) constant \(=h_{e, \mathrm{avg}}=\) \(0.5\left(h_{1}+h_{2}\right),\) calculate the mass withdrawn during the process. (b) Consider the same process but broken into two parts. That is, consider an intermediate state at \(P_{2}=400 \mathrm{kPa}\), calculate the mass removed during the process from \(P_{1}=800 \mathrm{kPa}\) to \(P_{2}\) and then the mass removed during the process from \(P_{2}\) to \(P_{3}=150 \mathrm{kPa},\) using the type of approximation used in part \((a),\) and add the two to get the total mass removed. (c) Calculate the mass removed if the variation of \(h_{e}\) is accounted for.

In large steam power plants, the feedwater is frequently heated in a closed feedwater heater by using steam extracted from the turbine at some stage. Steam enters the feedwater heater at \(1 \mathrm{MPa}\) and \(200^{\circ} \mathrm{C}\) and leaves as saturated liquid at the same pressure. Feedwater enters the heater at \(2.5 \mathrm{MPa}\) and \(50^{\circ} \mathrm{C}\) and leaves at \(10^{\circ} \mathrm{C}\) below the exit temperature of the steam. Determine the ratio of the mass flow rates of the extracted steam and the feedwater.

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